Probability that planar Brownian motion doesn't "encircle" 0

Question: Suppose $B_t$ is a standard Brownian motion in $\mathbb{R}^2$ and $T = \text{inf}\{t : |B_t| = 1\}$. Let $E$ denote the event that $0$ is contained in the unbounded component of $\mathbb{R}^2 \setminus B[0, T]$. Do there exist $\lambda < \infty$ and $\alpha > 0$ such that for all $|x| < 1$, we have$$\mathbb{P}^x(E) \le \lambda |x|^\alpha?$$
Comment: Another related question: "[Twisted random walks](http://mathoverflow.net/q/138680/6094)."
Comment: I guess a preliminary question would be, how to see that $E$ is measurable?
Comment: I changed the title to something a little more self-contained.  Feel free to edit further.
Comment: Related: http://mathoverflow.net/questions/202944/area-enclosed-by-brownian-motion-without-winding-number
Comment: In some sense, the question is really asking for the probability that BM *does* encircle 0 (i.e. what is really needed is a lower bound for the probability that 0 gets encircled).
Comment: Does the Brownian motion start at $x$?
Comment: Clearly $\alpha$ can be at most $1/2$, because $\mathbb P^x(E) \geq \sqrt{x}$. View the random walk as taking place in the complex plane, with $x$ a real number. Consider the harmonic function $\operatorname{Re}\sqrt{B_t}$ taking the principal branch with a cut at the negative real axis. This is a martingale until $B_t$ touches the negative real axis, takes a value of $0$ on that axis, at most $1$ on the unit circle, and is $\sqrt{x}$ at the start, so the probability that $B_t$ touches the unit circle without ever touching the negative real axis is at least $\sqrt{x}$.
Answer: The probability is asymptotic to $\lambda |x|^{1/4}$.  This was proved by Schramm, Werner and myself using the Schramm-Loewner evolution. The exponent is called the disconnection exponent for Brownian motion.
Answer: Divide the interval $[|x|,1]$ into $-\log_2 |x|$ intervals $[r_i,r_{i+1}]$. Radially, the BM performs a (simple) random walk between the circles of these radii, and there will be at least $-\log_2 |x| $ steps to this random walk before it hits $1$. During each step  the
tangential component of the BM has probability $>\delta$ to encircle the origin, some
$\delta>0$ (the tangential component is a time changed BM on the circle with time change
bounded within a constant factor of $r_i^2$).
So
$$P(E)\leq (1-\delta)^{-\log_2 |x|}$$
This computation does not  give the correct exponent $\alpha$.
Answer: Take any $s$ and $r$ such that $0\le s\le r\le1$, and let $p(s,r)$ denote the probability that the Brownian motion in $\mathbb R^2$ starting at a point at distance $s$ from the origin $O$ will encircle $O$ before hitting the circle of radius $r$ centered at $O$. Then for any $u\in[s,r]$
$$(1)\qquad p(s,r)\ge p(s,u)+[1-p(s,u)]p(u,r).
$$
The other crucial observation is that $p(s,r)=f(s/r)$ for some function $f\colon[0,1]\to[0,1]$ and all $s$ and $r$ such that $0\le s\le r\le1$ and $r>0$. This observation follows because, for the Brownian motion, space rescaling is equivalent to appropriate time rescaling. So, with $a:=s/u$ and $b:=u/r$, (1) yields
$$f(ab)\ge f(a)+[1-f(a)]f(b)
$$
and then
$$g(ab)\le g(a)g(b),
$$
where $g:=1-f$ and $0<a,b\le1$. Note that $f(1)=0$ and hence $g(1)=1$.
Also, $g$ is nonnegative and nondecreasing, since $p(s,r)$ is obviously nondecreasing in $r$.
Using ``path corridors'' that are (say) unions of rectangles, one can show that $0<g(1/2)<1$.
Take now any $a\in(0,1/2]$ and let $k:=k_a:=\lfloor\log_2\frac1a\rfloor$, so that $a^{1/k}\le1/2$ and $k\ge1$, and hence $k\ge\frac12\,\log_2\frac1a$. Then
$$g(a)\le g(a^{1/k})^k\le g(1/2)^k\le g(1/2)^{\frac12\,\log_2\frac1a}=a^\alpha,
$$
where $\alpha:=-\frac12\,\log_2 g(1/2)\in(0,\infty)$. That is,
$$P^x(E)=1-p(|x|,1)=g(|x|)\le|x|^\alpha$$
for all $x$ with $|x|\le1/2$. If now $|x|\in[1/2,1]$, then $P^x(E)\le1\le(2|x|)^\alpha$. So,
$$P^x(E)\le2^\alpha |x|^\alpha$$
for all $x$ with $|x|\le1$, as desired.
Comment: I don't see why (1) is an equality.  Let $C_r$ be the circle of radius $r$ centered at $O$. It sounds like you are saying "If the BM is going to encircle $O$ before hitting $C_r$, and it hasn't done so by the time it hits $C_u$ (at some point $y$), then the path started at $y$ must encircle $O$ before hitting $C_r$."  But that's not true - what if the path from $x$ to $y$ wound halfway around the origin, and after $y$ it completed the other half?  Then the combined path encircles the origin, while neither of the pieces before or after $y$ does so.
Comment: @NateEldredge However, we do have $g(ab) \leq g(a) g(b)$ and $g(a) \leq 1$, so $g$ is monotonic. This means that for each $\alpha$, either $g(a)  \leq \lambda |x|^\alpha$ for some $\lambda$ or $g(a) > |x|^\alpha$. Given that we have upper and lower bounds, this means that there exists a precise asymptotic constant $\alpha$ such that $|x|^\alpha <g(a) < O( |x|^{\alpha+\epsilon})$.
Comment: I have corrected the answer, in view of the comments by **Nate Eldredge** and **Will Sawin**.
Answer: This is an extended comment to observe that $E$ is measurable.
Let $U$ be the set of $\omega \in C([0,1]; \mathbb{R}^2\}$ that have not encircled the origin by time 1; i.e. such that $0$ is in the unbounded component of $\mathbb{R}^2 \setminus \omega([0,1])$.  I claim $U$ is open (with respect to the uniform norm $\|\cdot\|_\infty$).
Suppose $\omega \in U$, so that $0$ is in the unbounded component.   Choose any $y \in \mathbb{R}^2$ with $|y| > \|\omega\|_{\infty} + 1$, so $y$ is also in the unbounded component.    Now the unbounded component  is a connected open set, so it is path connected; let $\gamma \in C([0,1]; \mathbb{R}^2)$ be a path joining $0$ to $y$ which does not intersect $\omega$.  Set $\epsilon = \inf\{|\omega(t) - \gamma(s)| : s,t \in [0,1]\}$ which is strictly positive by compactness.  If $\|\omega - \tilde{\omega}\| < 1 \wedge \epsilon$ then $\gamma$ does not intersect $\tilde{\omega}$ either,  so both $0$ and $y$ are in the same component of $\mathbb{R}^2 \setminus \tilde{\omega}([0,1])$.  Since $\|\tilde{\omega}\| \le \|\omega\| + \|\omega - \tilde{\omega}\| < \|\omega\|+1$, $y$ is also in the unbounded component  of $\mathbb{R}^2 \setminus \tilde{\omega}([0,1])$, hence so is 0.  So $\tilde{\omega} \in U$.
By the same argument, for any $t \ge 0$, the set $U_t$, consisting of all those $\omega \in C([0,\infty) ; \mathbb{R}^2)$ which have not encircled the origin by time $t$, is open.
So let $\tau(\omega) = \inf\{t : \omega \notin U_t\}$ be the first time at which $\omega$ has encircled the origin.  Since $U_s \supset U_t$ for $s < t$, we can take the infimum over the rationals instead, and see that $\tau : C([0,\infty); \mathbb{R}^\infty) \to [0,\infty]$ is Borel.  Indeed, $\tau$ is a stopping time.  And $E$ is simply the event $\{\tau > T\}$.
Likewise, the function $p(s,r)$ in  Iosif Pinelis's answer does make sense, since we simply have $p(s,r) = P^{x_0}(\tau < T_r)$ where $T_r = \inf\{t : |B_t| = r\}$ and $x_0$ is any point with $|x_0|=s$ (by symmetry it does not matter which such $x_0$ is chosen).
Comment: This nicely rounds up the solution of the problem. My only, very minor suggestion here is to remove the assumption $\epsilon<1$ (which may need some justification) and then replace the condition $\|\omega-\tilde{\omega}\|<\epsilon$ by $\|\omega-\tilde{\omega}\|<1\wedge\epsilon$.


Roots of the Hilbert polynomial of an invertible sheaf

Question: Let $X$ be a smooth, projective variety over an algebraically closed field of characteristic zero. Fix a polarisation on $X$. Let $\mathcal{L}$ be an invertible sheaf on $X$ with Hilbert polynomial, say $P(t)$. Denote by $P'(t)$ the Hilbert polynomial of the dual $\mathcal{L}^\vee$. Is there any relation between the roots of $P(t)$ and $P'(t)$? In particular, I am looking for examples when $P(t)=P'(t+b)$ for some integer $b$.
Comment: Are you measuring the Hilbert polynomial of $\mathcal{L}$ with respect to a specified ample invertible sheaf $\mathcal{A}$ on $X$, i.e., $P(d) = \chi(X,\mathcal{A}^{\otimes d}\otimes \mathcal{L})$ for every integer $d\geq 0$?  Or is $P(t)$ the Hilbert polynomial of $\mathcal{L}$ "with respect to itself", i.e., $P(d)=\chi(X,\mathcal{L}^{\otimes d})$ for every integer $d\geq 0$?
Comment: @JasonStarr with respect to a specified ample invertible sheaf.
Comment: In general there is no direct relationship between $P(t)$ and $P'(t)$. However, by Snapper's Lemma, the function $p(a,b)=\chi(X,\mathscr L^{\otimes a} \otimes \mathcal O(b))$ is a polynomial in $\mathbb Q[a,b]$; you are interested in comparing $p(1,b)$ with $p(-1,b)$. Of course it is easy to make examples where these are very different (already on $\mathbb P^n \times \mathbb P^m$, I believe).
Answer: For the sake of simplicity, let us assume that $\mathcal{L}$ is a polarization, and that we are computing the Hilbert polynomial of $\mathcal{L}$ with respect to itself, i.e. $P(t)=\chi(X, \, \mathcal{L}^{\otimes t})$.
If $\mathcal{L}$ is such that $K_X=(\mathcal{L^{\vee}})^{\otimes b}$, then by Serre duality we obtain  $$P'(t+b)=\chi(X, \, (\mathcal{L}^{\vee})^{\otimes t+b}) = \chi(X, K_X \otimes \mathcal{L}^{\otimes t+b})=\chi(X,  \, \mathcal{L}^{\otimes t})= P(t).$$
For instance, if $X=\mathbb{P}^{n-1}$ and $\mathcal{L}=\mathcal{O}_{\mathbb{P}^{n-1}}(1)$, we have $K_X = (\mathcal{L}^{\vee})^{\otimes n}$ and so $$P'(t+n)=P(t).$$


Does SharePoint Foundation 2010 limit the number of users somehow?

Question: I am evaluating SharePoint Foundation 2010 (free edition) for a prospect with 50 users. Does Foundation 2010 limit the number of users somehow? Thank you!
Answer: No it doesn't; SharePoint Foundation is Free to use. Only the SharePoint Server (Standard & Enterprise) uses the Client Access License (CAL) model which is limited by CAL licenses per user.
Comment: SharePoint limit the number of users to 2 million per site collection (or that's what is supported) [SharePoint Server 2010 capacity management: Software boundaries and limits](http://technet.microsoft.com/en-us/library/cc262787.aspx)
Comment: Yeah actually it is a maximum of 5 million total simultaneous user for the entire farm. It can hold more but I think thats when you hit a performance peak.
Answer: In addition, this CAL model does not technically limit your usage of SP. But, for legal purposes you must ensure you have the right amount of CAL for your scenario.


Installing dotnet-sdk-5.0 on ubuntu 20.10

Question: So I followed the instructions from microsoft but I can't install dotnet-sdk-5.0 on my ubuntu 20.10 server. I keep getting this:

Does anybody know what can I do?
Answer: Time passed quickly, so did packages and their installation methods on Ubuntu. It's end of year 2023, Microsoft has stopped support of .NET 5.0, so you cannot find this version of .NET in Microsoft's repository for Ubuntu 22.04 (like the apt-cache command in another answer) or use any instructions in official Installation webpages. At the moment, if you wanna install .NET 5.0, follow the commands below:
wget https://packages.microsoft.com/config/ubuntu/20.04/packages-microsoft-prod.deb
sudo dpkg -i packages-microsoft-prod.deb
sudo apt-get update
sudo apt-get install -y dotnet-runtime-5.0

In the second command, if you are prompted
==> Package distributor has shipped an updated version.
...
*** microsoft-prod.list (Y/I/N/O/D/Z) [default=N] ?

It's safe to just answer Y.
In the last command, you will probably be given:
The following packages have unmet dependencies:
 dotnet-runtime-5.0 : Depends: dotnet-runtime-deps-5.0 (>= 5.0.17) but it is not going to be installed
                      Depends: dotnet-hostfxr-5.0 (>= 5.0.17) but it is not going to be installed
E: Unmet dependencies. Try 'apt --fix-broken install' with no packages (or specify a solution).

Just run sudo apt --fix-broken install as suggested and this command will automatically install the missing dependent packages for you as well as dotnet-runtime-5.0.
I tested the above commands on Ubuntu 22.04.
Answer: Did you remember to run sudo apt update before trying to install? If so, let's confirm if the package is available in your Apt cache:

Open a Terminal (if one is not already open)

Check apt-cache for the dotnet-sdk packages:
$ sudo apt-cache dump | grep dotnet-sdk

You should get something that looks like this:
Package: dotnet-sdk-2.1
Package: dotnet-sdk-3.1
Package: dotnet-sdk-5.0

Install the package that you'd like:
$ sudo apt install dotnet-sdk-5.0

Which should then result in:
Reading package lists... Done
Building dependency tree
Reading state information... Done
The following additional packages will be installed:
  aspnetcore-runtime-5.0 aspnetcore-targeting-pack-5.0 dotnet-apphost-pack-5.0 dotnet-host dotnet-hostfxr-5.0 dotnet-runtime-5.0
  dotnet-runtime-deps-5.0 dotnet-targeting-pack-5.0 netstandard-targeting-pack-2.1
The following NEW packages will be installed
  aspnetcore-runtime-5.0 aspnetcore-targeting-pack-5.0 dotnet-apphost-pack-5.0 dotnet-host dotnet-hostfxr-5.0 dotnet-runtime-5.0
  dotnet-runtime-deps-5.0 dotnet-sdk-5.0 dotnet-targeting-pack-5.0 netstandard-targeting-pack-2.1
0 to upgrade, 10 to newly install, 0 to remove and 3 not to upgrade.
Need to get 91.9 MB of archives.
After this operation, 382 MB of additional disk space will be used.
Do you want to continue? [Y/n]

Hope this helps you identify why apt couldn't find the package on your system.


How to change the field precision step?

Question: I need to change the field precision input step (that one which is used when you press SHIFT and drag the field value) I want something like 0.001 instead of 0.1
Answer: I'm assuming that you mean existing fields as opposed to a Custom Property that you create (where you can control the step value)?
There appears to be nothing in "Edit :: Preferences :: Interface" that provides this ability.   If the number of your target fields is limited...
You could create such Custom Property (on a scene-common empty object or the object in question or ... here in World properties):

and then RMB:: Copy Field as New Driver,

and then RMB :: Paste Driver into a field you want to step at a finer rate.

But that's certainly of limited value, works with pure numeric fields (gets confusing for fields that want to deal in measurements, and is exhausting if you want such control over many, arbitrary fields.
example:

How to change the field precision step?


Conversion of House Sale Proceeds in INR to USD or CDN$ for transfer to Canada

Question: I am a Canadian Resident that just sold my house in India and I understand I can repatriate the proceeds in INR converting to US$ or CDN$ in my NRE account and then move it to my Canadian Bank.  Which currency should I convert INR to ? What are the pros and cons?
Comment: Are you moving the money to the US or to Canada?
Answer: A sales proceed in India cannot be credited to NRE Account. There are limitations to credit of funds in NRE, broadly the money has to come from outside India in designated currencies.
As the funds you have are due to sale of Capital Asset, the Rupee is not fully convertible on account of Capital. You need apply for relevant permissions with RBI under FEMA before you repatriate the funds.
A CA will be able to help you complete the formalities.
The CDN$ is not traded in India and you would be able to rate shop a bit better with USD. However if in the end you need another conversion from USD to CDN$, then it defeats the purpose. Converting into CDN$ in India would be more tedious.
Answer: The capital gains on the sale of the house are taxable income to you in India
and you cannot repatriate the proceeds without a certificate from the
Income Tax Department (a Chartered
Accountant (CA) can help you obtain this) that the tax has been paid/withheld.
In some circumstances, you can invest the proceeds in specified bonds issued by
quasi-governmental agencies (Rural Electrification Authority?) for a period
of three years (earning a lower rate of interest than is available on the
open market), in which case no capital gains tax is assessed or payable. The
interest on these bonds is taxable income to you, though. After three years,
you can repatriate the entire proceeds (subject to a limitation
of $100,000? per year) back to Canada.


Why does a Bermudan option have a higher implied volatility than its European counterpart?

Question: I get that the premium for an earlier exercise should be higher to compensate the seller but intuitively you would think that the spot has "less room to run" in a potentially shorter period of time (due to a potential earlier exercise compared to the European counterpart).... am i thinking about this correctly? What am I missing?
Comment: what is the meaning of implied volatility for Bermudans?
Comment: Are Bermudan options not typically valued using a binomial tree approach? I assume the implied volatility would (or could) be the volatility that when used as input in a binomial tree model produces the observed market price of the Bermudan.
Answer: Optionality always benefits the holder of the option (since the payoff will be the higher of exercising the option or not), so an option with more "optionality" (in this case, more opportunities to choose exercise) will always be worth more.
I don't see where your question has anything to do with "implied" volatility.


Visualizing Orthogonal Polynomials

Question: Recently I was introduced to the concept of Orthogonal Polynomials through the poly() function in the R programming language. These were introduced to me in the concept of polynomial transformations in order to do a linear regression. Bear in mind that I'm an economist and, as should be obvious, am not all that smart (choice of profession has an odd signaling characteristic). I'm really trying to wrap my head around what Orthogonal Polynomials are and how, if possible, to visualize them. Is there any way to visualize orthogonal polynomials vs. simple polynomials?
Answer: Helge presented the continuous case in his answer; for the purposes of data fitting in statistics, one usually deals with discrete orthogonal polynomials. Associated with a set of abscissas $x_i$, $i=1\dots n$ is the discrete inner product
$$\langle f,g\rangle=\sum_{i=1}^n w(x_i)f(x_i)g(x_i)$$
where $w(x)$ is a weight function, a function that associates a "weight" or "importance" to each abscissa. A frequently occurring case is one where the $x_i$ are equispaced, $x_{i+1}-x_i=h$ where $h$ is a constant, and the weight function is $w(x)=1$; for this special case, special polynomials called Gram polynomials are used as the basis set for polynomial fitting. (I won't be dealing with the nonequispaced case in the rest of this answer, but I'll add a few words on it if asked).
Let's compare a plot of the regular monomials $x^k$ to a plot of the Gram polynomials:

On the left, you have the regular monomials. The "bad" thing about using them in data fitting is that for $k$ high enough, $x^k$ and $x^{k+1}$ are nigh-indistinguishable, and this spells trouble for data-fitting methods since the matrix associated with the linear system describing the fit is dangerously close to becoming singular.
On the right, you have the Gram polynomials. Each member of the family does not resemble its predecessor or successor, and thus the underlying matrix used for fitting is a lot less likely to be close to singularity.
This is the reason why discrete orthogonal polynomials are of interest in data fitting.
Comment: Thanks for taking the time to explain this. The issue of multicollinearity  (singular matrix) is exactly the context where I was using the poly() function which creates orthogonal polynomials. Thank you for grabbing hold of my context and sharing an answer that really gets at my specific domain of interest.
Comment: A few small notes: 1. in the limit of infinitely many abscissas (the *discrete* becomes *continuous*), the Gram polynomial becomes the Legendre polynomial. 2. The orthogonal polynomials associated with nonequispaced abscissas are easily constructed through the so-called Stieltjes procedure. Different abscissas correspond to different basis sets. The reason the equispaced case is much nicer is that one can factor out the spacing $h$ from the appropriate equations. 3. Any good old book on the difference calculus should have some mention on the Gram polynomials.
Comment: jd:You're very much welcome. I have to say I wish someone had explained it like that to me when I was first learning about this stuff.
Answer: Start here
Comment: That was actually a really helpful resource. Thank you.
Answer: I am not sure of your math-background, so I am trying to keep it simple, without oversimplifying some ideas. First off polynomials are nice for various reasons, e.g. a polynomial of degree $n$ has at most $n$ zeros. However, there are still many polynomials and it makes sense to choose VERY nice ones: orthogonal polynomials.
To choose orthogonal polynomials, one has a problem at hand, which comes with a way to measure functions $f: \Bbb R \to\Bbb R$ by an expression of the form
$$
 \mathcal{E}(f) = \int_{-\infty}^{\infty} f(x)^2 w(x) dx,
$$
where $w(x) > 0$ is a weight that satisfy $\int w(x) dx = 1$. One should think of $\mathcal{E}$ as an energy.
Now the orthogonal polynomial of degree $n$ can be defined as the polynomial $P_n(x) = x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$, where $a_{n-1}, \dots, a_0$ are real numbers, that minimizes $\mathcal{E}(P_n)$. It is this minimization property that is responsible for some of the power of orthogonal polynomials.
At this point let me also say that it is through the weight $w$ that your problem enters the definition of orthogonal polynomials. And that one also has orthonormal polynomials, which satisfy $\mathcal{p_n} = 1$. These are given by $p_n = \frac{1}{\sqrt{\mathcal{E}(P_n)}} P_n$.
Comment: You must have meant $\int w(x)p_n(x)^2dx=1$ instead of $p_n=1$.
Answer: I think of the space of polynomials on R as a set of graphs arranged round the real line like the pages of a book round the axis. Polynomials which have almost the same graph are close to each other; then orthogonal polynomials are those which fall at right angles in the picture, and the linear combinations generate the space just as the basis vectors of $\mathbb{R}^n$ generate $\mathbb{R}^n$.
Comment: PS If you have had a decent linear algebra course you will be familiar with the concept of a vector space as a collection of abstract vectors satisfying certain axioms. You can see that the set of polynomials on $\mathbb{R}$ satisfies these axioms and so is a vector space. An orthogonal set of polynomials then generates the whole space in roughly the same way that an orthogonal basis for an ordinary vectors space does.

Visualizing Orthogonal Polynomials


Capacitor Voltage in RC Network Circuit

Question:
I am  understanding a rc network shown in the above image . I need to calculate the Value Vc after i give 24 volts step input for 16 μs. i used capacitor charging equations Vc=Vinput(1-e^-t/rc) equation for outer loop but how do i take inner loop into calculation. Laplace method  gives very  huge polynomial . Could you please tell me if i can use time domain equation to calculate the value of Vc ? i am stuck on this problem for last 2 days . Kindly Help me out folks .
Comment: It's not clear from your question that you realise you can't consider the outer loop in isolation, i.e. it's not simply a 24V step and a time constant, \$\small \tau=261k\Omega\: \times 470pF\$. This is because the inner 'loop' affects \$\small V_C\$
Comment: exactly @Chu . so there will be some effect by inner loop. How do i write its equation ? Can you give any hints about it
Comment: You've tried Laplace - that's one method. The other method is to derive the differential equations and solve simultaneously. The amount of effort is about the same by either method. So it's whichever method you're most comfortable with.
Comment: ok. but how do i write differential equation for  inner loop ? can you just give an example ? @Chu
Comment: Assume \$\small I_1\$ and \$\small I_2\$ flow through \$\small C_1\$ and \$\small C_2\$,respectively. Then use KCL/KVL to form two simultaneous equations, then solve. I don't have the time!
Answer: \$ F(s) = \frac{V_C}{V_1} = k \cdot \frac{1+a_1 \cdot s}{1+b_1 \cdot s+b_2 \cdot s^2}\$
\$k = \frac{R_2+R_3}{R_1+R_2+R_3}\$
\$a_1 = \frac{R_2 R_3 C_1}{R_2+R_3}\$
\$b_1 = \frac{C_1 R_3 (R_1+R_2) +C_2 R_1 (R_2+R_3)}{R_1+R_2+R_3}\$
\$b_2 = \frac{C_1 C_2 R_1 R_2 R_3}{R_1+R_2+R_3}\$
Inverse Laplace of the step response (\$F(s) \cdot \frac{24}{s}\$) at \$t=16 \cdot 10^{-6}\$ gives 2.26538 V.
Comment: I start with impedance Z1(s) which is defined as 1/(s*C1) parallel R3 plus series resistor R2. Next impedance is Z2(s), which is Z1(s) parallel 1/(s*C2). Resulting voltage is Vc=V1*Z2(s)/(Z2(s)+R1).
Comment: While this answer is useful for finding the answer, it could be more useful if you detailed to the user how you reached this finding.
Comment: Basically following this approach: https://en.wikipedia.org/wiki/Voltage_divider
Comment: @UweD I appreciate your Big help and I elaborated  the given steps on paper .

Capacitor Voltage in RC Network Circuit


Performance reduction activating trace-functionality in kernel

Question: Is there any performance reduction to be expected from just activating trace (ftrace) functionality in the Linux kernel, but not using it at run time?
Answer: Ftrace inserts nops into the text code, and these get turned into jumps to the tracing when enabled. There may be a slight impact on performance just by increasing the pressure on instruction cache, but all my benchmarks have kept it below the noise. There's been some cases that enabling ftrace actually improved performance. But that was probably just a fluke, where the added nops moved the hot paths into better cache localities.


Accessing an OLAP data present in Oracle Cubes using C# .NET?

Question: I am a newbie to ORACLE.  I would like to access an OLAP cube in Oracle using C#.  Could you please suggest efficient ways of doing the same?
Answer: Ok, so now i found a library provided by Oracle: Oracle Data Provider for .NET that may help access OLAP data from Oracle Cubes.
Any hints?
Answer: Oracle OLAP Option cubes can be accessed via SQL using the OLAP_TABLE function OLAP_TABLE Syntax
This requires detailed knowledge of the cube structure and also the OLAP DML language used to manage the OLAP cubes.  This system is basically the old IRI / Oracle Express  system and language embedded in the Oracle Database Server.
As it's standard SQL used to extract the data views can be defined to handle the OLAP_TABLE functionality on the server.
The dbms_aw PLSQL package provides functions and procedures for executing statements and programs withing the OLAP cube. It also provides methods for maintainting and building cubes


What is the definition of inertial mass?

Question: What is the definition of inertial mass? I can see two options, either it's the coefficient associated with the object being accelerated in Newton's 2nd Law, or it's the coefficient relating momentum to velocity. Or they are equivalent.
Answer: Well, if it is constant, then it's equivalent.
Inertial mass is the resistance that a body offers when a force tries to change its state of movement.
So it is the $m$ that appears here
$$p=m\dot{x}$$
and here
$$F=m\ddot{x}.$$
But if it's not constant you can see it as the magnitude that appears in
$$ p=m\dot{x} $$
because in the force expression, there is a new term due to the mass variation. But the definition is the same.


why call the output port of a element to "src pad" in gstreamer?

Question: I wonder the strange name policy of gstreamer. in a pipeline, the sink pad is actually a input port and the src pad is a output port, but the sink element is the  end element, src element is the start element? I think it is wired to understand, could anyone show me more details.
Answer: The "src pad" might be an output for that block but it can also be seen as a source of data for the next block.
Similarly the "sink pad" might be an input but from the other perspective it is also a place where data can be poured into from other blocks.
It is just a matter of changing your perspective on the flow of data to look at it from other units and not the unit you are currently looking at. Where do you as a unit push data to? It certainly wouldn't be a source, I want to push data to a sink somewhere and let them deal with it. I also don't want to pull my data from a sink, I want to get it from a source of data.


What is causing these brown holes that go through my young cherry tree's leaves?

Question: What is making these holes in my cherry tree's leaves? And if it's a threat to the health of the plant, how can I stop these holes from forming?
Answer: The holes and the missing part looks like attacks from a larvae. It could be damages made by beetles too.
It does not resembles attacks from snails.
Take a look a this list:

Prunus species are used as food plants by the caterpillars of a number
of Lepidoptera (butterflies and moths).

Sweet cherries are called P. avium
Use "mouse over" on the names in the list for photos (See e.g. Geometridae). Perhaps you can recognize one from your region.
To keep your tree healthy it is important that you in late summer/autumn remove all leaves and fruits from the ground.
Also be sure to attract insect eating birds.


Prove that $G=\langle g_1,...,g_n\rangle$ if $G$ abelian.

Question: Let $G$ a finite abelian group. Are there $g_1,...,g_m$ s.t. $$G=\langle g_1,...,g_m\rangle$$
where $g_i\neq g_j^m$ for all $m$ and all $i\neq j$ ? For cyclic group no problem, but I can't see such a group if $G$ is not cyclic, any example ?
Comment: If $g_i = g_j^m$ for some $i, j, m$, then just remove $g_i$ from the presentation. Eventually you won't be able to do this and then you're done.
Comment: So one of the $g_i$s is the identity?
Comment: Klein four-group is abelian but not cyclic, and can be generated by any two of its non-identity elements. Also every non-identity element has order 2, so this property holds for the generator.
Answer: Choose some arbitrary $g\in G$ for which there exists no $h\in G$ with $\langle{g}\rangle < \langle{h}\rangle$; such a $g$ must exist by the fact that $G$ is finite. (Otherwise, we'd have an infinite sequence of distinct subgroups of $G$.) Consider $G/\langle{g\rangle}$ and induct on $\#G$. (It may also be worth considering what fails in a divisible group, such as $\mathbb{Q}/\mathbb{Z}$.)


Linux Mint 16 - How to launch script from desktop without "Run in Terminal, Display, Cancel, Run" prompt

Question: Is it possible to remove the  "Run in Terminal, Display, Cancel, Run" prompt that I get each time I double-click to run a script from the desktop?
Answer: You can also do right-click/Create Launcher on the desktop to create a launcher for your script -- then when you double-click the launcher it will just run without giving you that prompt.
Answer: I managed to find an option within Nemo --> Edit --> Preferences --> Behaviour --> Executable Text Files --> "Run executable text files when they are opened."


How can angle of attack for a T-tail configuration be 21 degrees without stalling and also be flying at its trimmed condition?

Question: I was reading a paper that looks at deep stall for T-tail configurations, and figure 2 suggests that the first trim condition for the aircraft in question occurs at approximately 21 degrees. Surely this would be a bad condition to fly at as it would result in stall due to flow separation over the wing, and if this doesn't occur the nose would be pointed so far up the aircraft would constantly be in climb? Excuse me if I'm missing something fundamental here. I've linked the paper here: https://research-information.bris.ac.uk/ws/portalfiles/portal/312655220/2022_SciTech_deep_stall.pdf
Comment: I would not take that graph too seriously - it is only meant to show the general principle.
Comment: The curve in figure 2 typically shifts up and down more or less proportionally to the elevator deflection. Most probably that particular curve has been drawn for an elevator with a quite positive (donwward) deflection. Pity that the source of the picture is not given.
Comment: @sophit As Peter pointed out, the graph is for illustration only, therefore it is described as "**Typical** pitching moment". It is an example the author drew himself (therefore no source) and most likely quickly drew himself.
Comment: @U_flow: I perfectly understand that. Anyway that is a scientific paper and if you make something up you should clearly write it otherwise people starts to wonder about the content, as this question demonstrates.
Comment: Thanks all, that clarifies things for me a lot more! :)
Answer:
the first "trim" condition...occurs at 21 degrees

It's a shame they did not show the whole Cm vs AoA curve.
The aircraft is most likely already stalled here.
What they are showing is the overall aircraft Cm for various AoA if no control inputs are added, relative to a given AoA and trim setting where pitch rate = 0
One would expect positive Cm at stall for certain asymmetric airfoils$^1$, while many others remain negative$^4$.
The graph shows increasingly negative Cm (greater static stability) as the stall deepens and lift producing airflow over the wing collapses.  At this point the still unshadowed tail (and rear fuselage) is providing enough pitch down force (by way of basic longitudinal stability) to over come upward pitching forces of the forward fuselage and any portion of remaining wing lift ahead of the CG.
Deep stall occurs (Cm > 0) when turbulence from the wing greatly reduces the stabilator coefficient of lift.  The plane will no longer pitch down even though it's angle of attack is very high.
Cm remains positive until a second "stable" zone appears, but at a hopelessly high AoA.
It would behoof the engineers to provide enough stabiliator control authority to keep Cm negative post stall.  The model in this study may have been purposely created to be stall prone$^3$ in order to generate the graph showing the dangers of deep stall.
$^1$ even if the Cm of the wing remains negative, an increasing Cm, combined with the trim setting of the tail, can cause an aircraft to become staticly unstable as pitch increases.  Wing sweep with wing tip washout$^5$ can greatly improve pitch stability$^2$.
$^2$ Slats are also a great way to enhance the safety of swept wings in high AoA situations by increasing wingtip "downwash" (lowering local wingtip AoA).  Stalls on swept wings usually start at the wing tips, causing Cm to go positive, leaving the stabilator as the last line of defense.
$^3$ exceeding aft CG limits is an excellent way to do this
$^4$ a benefit of "olde tyme" undercamber, which can be created with a setting of slats and a little flap
$^5$ early Dunne D.8 aircraft criticized as being "too stable"
Comment: The value of the pitching moment coefficient only shows how far the airplane is away from a trimmed condition. To learn about stability you need to look at the **derivative over lift coefficient** or angle of attack. A negative derivative shows positive static stability.
Comment: @PeterKämpf I agree with that, and it highlights the need for (available) control authority beyond trim.  Thankfully, the graphs help me while I continue to learn the calculus.


iOS Core Data Persistance After restarting app

Question: With Core Data, I am able to save and fetch objects in memory; however, when I restart my app (in the simulator) and fetch the results, the object is fetched but all the attributes are nil.  I have included all protocol methods in my appdelegate and having initialized my managedObjectContext as such:
managedObjectContext = [self managedObjectContext];
in - (BOOL)application:(UIApplication *)application didFinishLaunchingWithOptions:(NSDictionary *)launchOptions
Here is what I am getting in the logs...
Correct:
Fetching results log (app not restarted) --> Name: new entry!!!!
Incorrect:
Fetching results log (app restarted) --> Name: (null)
     -But, object still exists in persistantStore
--
<code>//Save entry
Entry *entry = (Entry *)[NSEntityDescription insertNewObjectForEntityForName:@"Entries" inManagedObjectContext:managedObjectContext];
[entry setTitle:@"new entry!!!!"];
NSError *error;
if(![managedObjectContext save:&error]){
    NSLog(@"SAVE ERROR!!!");
}

//Fetch entry
NSError *error;
NSFetchRequest *fetchRequest = [[NSFetchRequest alloc] init];
NSEntityDescription *entity = [NSEntityDescription entityForName:@"Entries"  inManagedObjectContext:managedObjectContext];
[fetchRequest setEntity:entity];
NSArray *fetchedObjects = [managedObjectContext executeFetchRequest:fetchRequest     error:&error];
for (NSManagedObject *info in fetchedObjects) {
    NSLog(@"Name: %@", [info valueForKey:@"title"]);
}
</code>
Answer: Are you sure your Entry class realizes the title property using @dynamic and not @synthetize?


Characteristic Function and Random Variable Transformation

Question: Let $X$ be a random variable, and let $\phi_X(t)$ be its characteristic function. Let $Y = f(X)$ be a transformation of the random variable $X$ where $f$ is increasing and one-to-one. Is there a direct functional relationship between the characteristic function of the transformation $f(X)$ and the characteristic function $\phi_X(x)$? Meaning if $\phi_Y(t)$ is the characteristic function of the random variable $f(X)$, can we write $\phi_Y(t)$ in terms of $\phi_X(t)$?
Comment: I think if there was any relation, it'd be quite well known. fourier transforms don't have special behaviour under transformations.
Answer: If $Y=g(X)$ (let me use $g$, reserving $f$ for the densities) :
Then $$ \phi_Y(t)=E(e^{i Yt})=\int_{-\infty}^{\infty} f_Y(y) e^{i y t} dy= \int_{-\infty}^{\infty} f_X(x) e^{i g(x) t} dx=\\
=\frac{1}{2 \pi}  \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \phi_X(s) e^{-i s x } ds \, e^{i g(x) t}  dx
$$
Of course, this is rather trivial (  $ \phi_X(t)  \to f_X(x) \to f_Y(y) \to \phi_Y(t)$) and formal (the formula is not practical, it does not give a nice or simple relationship between $\phi_Y(t)$ and $\phi_X(t)$ but I don't see why you'd expect that it can be simplified.
Answer: Here's I think what would be a counterexample: if $X \sim R[0,1]$ and $Y \sim N(X, \sigma^2)$, then MGF of $X$ would be $\frac{e^t-1}{t}$ and MGF of $Y$ would be $e^{xt+\frac{\sigma^2 t^2}{2}}$ and I doubt there's a way to express the latter via the former.
Comment: This was meant as a counterexample. I didn't say it's impossible for some functions.
Comment: Thanks. But in your case $Y=f(X)$ is not increasing and one-to-one or am I missing something?
Comment: What about something simpler. Consider $X \sim N(\mu , \sigma)$ and consider $f(x)= x/2$. (Even you could make it simpler by taking $\mu=0$ and $\sigma=1$).
Answer: This is more of a comment than answer and is for someone stumbling into this post.
For linear transformations of the form $Y=aX+b$, we can say that $\phi_Y(t)=e^{itb}\phi_X(at)$. If we are dealing with random vectors, then for transformation $AX+B$, we have $\phi_Y(t)=e^{it^\top B}\phi_X(A^\top t)$.
Source: https://www.statlect.com/fundamentals-of-probability/joint-characteristic-function and references therein.
Comment: please use the comment function for comments.


Will a Canon dSLR retain settings if the battery is removed for an extended period of time?

Question: If I remove the rechargeable BP battery from my Canon EOS 50D for an extended period of time, will the internal lithium battery preserve all of my settings (including custom functions, date/time, etc.)?
Comment: We're talking like 5+ years.
Comment: You'd should qualify 'extended'. Months, no problem. Years, will be because the internal rechargeable battery eventually runs out (even if on some high-end models it is replaceable).
Answer: Yes.
All modern canon dslr's have a small lithium battery (similar to the one that is used to power your computers CMOS) that powers the internal memory that retains the settings.


cgpdftraster quit unexpectedly: Report sent to apple; just after installing Sierra ; canon 640 will not print from rear tray

Question: Canon 640 will not load & print from rear tray after installing Serria?
Answer: From Cannon KB : Canon PIXMA MP640

Please Note: There are currently no macOS 10.12 Sierra printer drivers available for PIXMA products. This issue is being investigated and we hope to have these drivers available soon.

I guess you'll either have to wait, or go back to El Capitan.
It's always wise to check compatibility for mission-critical apps & devices before jumping to a new OS.


How to become a Stand-by Block Producer

Question: Is there a document / wiki etc. out there that provides details on how to become a stand-by block producer for EOS?
Comment: Your aim should be to become a block producer.
Answer: Stand-by block producers are the ones that are between 21 and +-50 in rank of votes. Once you have moved into that area in the rank table you are a stand-by block producer.
See also https://www.reddit.com/r/eos/comments/8s2kpc/howto_become_a_block_producer/

How to become a Stand-by Block Producer


What is the Red White and Blue suit in the Iron Man 3 trailer?

Question:
What is the red white and blue suit shown in the Iron Man 3 Trailer?
Comment: It looks like War Machine's armor, repainted. Has the USAF insignia, the large gun on his back, and apparently has a military color guard with him. My guess is it's James Rhodes.
Comment: @SachinShekhar So link to the question?
Comment: @GabeWillard I think you're right. See my answer for a "zoom and enhance" on the image.
Answer: If you zoom in on that image you can see that it has "Lt. Col. James Rhodes" stenciled on the chest plate.

So, although that does look like the Iron Patriot armor from the comics, I think it's just War Machine with a new paint job.
Original image source: First 'Iron Man 3' Poster and New Images Reveal Cool Armor and the Film's Villain
Answer: It is a reference to the Iron Patriot Armor - <code>"full battlesuit patterned after the heroic images of Captain America and Iron Man"</code>.
As shown by Bill the Lizard above, it's Lt Col James Rhodes driving the armor.
Comment: With the US Airforce insignia on the left forearm and what looks like a Vulcan cannon on the shoulder (Uncle Gaspacho anyone?) my guess is it's Rhodey, but that's just a guess.


Will having reversed hot and cold damage my faucet?

Question: On our old kitchen tap handles, the cold is on the left side and hot on the right. The new taps are the exact opposite, cold being on the right and hot on the left side. Is it okay to just connect them as they are as long as we, obviously, remember what side the cold and hot are on? Can it ruin anything like the cold water hose/pipes, etc.?
Comment: Unless the faucet has a thermostatic temperature control device, it doesn't physically matter which side carries hot water, the same materials are used on both sides.
Answer: This is an easy fix. Your supply lines were simply installed to the wrong shutoff.
If you look under your sink there should be two shutoff valves. One hot, one cold. The supply lines (may be rubber or have a stainless sheath or plastic) go from these valves to your faucet.
Turn the shutoffs off. Open both sides of the faucet to relieve pressure. Unhook the lines at the shutoff and put them on the opposite shutoff. Turn the shutoffs back on. You should now have cold on the right and hot on the left.
This is such a simple repair there is no reason to "just live with it" and as @Steven states it could be a minor hazard for guests.
Answer: The biggest risk is someone being burned.  In North America, cold should always be on the right side and hot on the left.  If you are outside of this region you should check with local standards.
It sounds like you either bought a faucet from a different region, or maybe your original faucet was installed backwards.
With regards to the actual plumbing, it shouldn't make a difference which side you are hooked up to.
Comment: I'm not sure what would be more confusing: to have the faucets labeled wrong but be on the correct side conventionally, or to be labeled correctly but on the wrong side conventionally. If the left/right side is a regional convention then I would strongly suggest getting a locally-appropriate faucet. Depending on how hot your water is, burns from a faucet could be severe.
Comment: Not only is it usually simple to switch the supply if that's what's reversed, it's also often possible to easily reverse the labels if that's what's wrong. Single lever faucets may have replacement handles with regionally appropriate labels.
Comment: @HenryJackson Funny! I just lived 9 months with my kitchen faucet labeled backwards (it said that pulling the joystick to the left would be cold but was actually hot), and we just switched the hoses yesterday, and I'm now thinking it was better to have the incorrect label than to buck North American convention and habit. It feels weird to have "hot" on the right. Obviously, if I could figure out an easy enough way to correct the blue and red dots built into the joystick, I'd switch the hoses back. And I might switch the hoses back *anyway*.


Which is stronger, Kaio-Ken or Super Saiyan?

Question: Is there any evidence that points to which is stronger, either reaching max level of Kaio-Ken or the max level of Super Saiyan? When Goku reached the highest level of Kaio-Ken, he seemed like he could take out the strongest Super Saiyans.
Answer: Kaio-ken is clearly inferior to his Super Saiyan state. Once Goku is able to sustain his Super Saiyan abilities, he stops even using Kaio-ken after a time. He was unable to assume the powers of the Super Saiyan when he returned to life to battle Freiza, so he used the power extending abilities of the Kaio-Ken which increased his power by multiples. Goku was the only known practitioner of the Kaio-Ken technique.
The Kaio-ken was not a natural extension of the Saiyan powers, it was taught to Goku by King Kai, who invented the power but was not capable of using it. The greatest weakness of the Kaio-Ken was the terrible strain it placed on Goku's body. After its use Goku was completely depleted of energy and unable to utilize any physical abilities for quite some time.

When Goku first mastered the Kaio-ken, the furthest he could handle was the Kaio-ken x4, but through training at 100x gravity and then later with his dramatic strength increase after his fight with Captain Ginyu, he was able to withstand astounding multiplications of the Kaio-ken technique, such as Kaio-ken x10. The highest multiplication for the Kaio-ken ever seen was the Kaio-ken x20, which Goku used against Frieza. --Dragonball Wiki

The Super Saiyan state is a more natural extension of the Saiyan species and once mastered can be maintained much more effectively and without the side-effects of the Kaio-Ken. It however requires a Saiyan to achieve a significantly higher state of battle mastery and thus was unseen for quite some time among the Saiyans.
Given a confrontation between a Saiyan with Kaio-ken and a Saiyan without it, the Kaio-ken, especially if the practitioner is experienced, is superior. However, a confrontation between the Kaio-Ken and the Super Saiyan would almost always leave the Kaio-Ken user, defeated. Super Saiyan level 1 is far superior to the Kaio-Ken even at the times 20 level.

In Dragon Ball Z: The Return of Cooler, Goku again uses the Kaio-ken on Meta Cooler, but it again proves far too feeble, forcing Goku to go Super Saiyan soon after. --Dragonball Wiki

It has been shown (one time) that the Kaio-ken could be stacked on top of the Super Saiyan powers but the strain on Goku was so great it was believed it could even kill him. He used this power arrangement in a battle against Pikkon.
Comment: Important note: Super Saiyan + Kaio-ken probably would have killed Goku if he wasn't already dead at the time.
Comment: This needs to be edited.  Goku didn't return to life to battle Frieza, he returned to life to battle the Saiyans.  Goku never died in his fight with Frieza and never lost the Super Saiyan ability during that fight.  He also was not able to transform into a Super Saiyan during his initial battle with Frieza, hence the use of Kaio-Ken.
Comment: Super-kaio-ken would have done no harm to goku, it could have just exosted him a bit, because goku had a great power than pikon, but the only think he needed to master was the speed, he was unable to keep up with the speed of pikon, that is why it appeared as if he was loosing the fight but he then uses the instant transmission technique to win
Comment: Actually, in the manga he went to i think 30 in the frieza fight when tien was being a smart alek and said Goku has already won because he could go x20 but King Kai replied saying he was already using the Kaio Ken x30 (Might've been 40)
Comment: Any evidence of compounding the two power ups? SSJ4 + Kaioken x40 or something crazy. ;)
Comment: It's probably easier to think of this in terms of power levels.. Kaio-Ken would multiply your power level by *x*. Super Saiyan on the other hand increases it by some number eg. in Kid Buu saga, Goku's SSJ2 power level was roughly 3,000,000,000, however SSJ3 was roughly 8,400,000,000. This is what makes the combination so dangerously effective. (Numbers pulled from Dragonball wikia forum)
Answer: Super Saiyan is stronger. Goku could take Friza in his final form as a Super Saiyan but couldn't scratch him in Kiaoken x 20. Not only that, but the number of times Goku would have to increase the Kiaoken output it would put too much strain on his body, so once it wore off he would either not be able to fight anymore or he would die.
Answer: Technically, there is no "Max level of Kaio-Ken". You could go up to Kaio-Ken 2,000,000 if your body could withstand it.
The highest level of Super Saiyan is Super Saiyan 4 or Super Saiyan God super saiyen, going by canon.
Because your power is increased, your ability to withstand more of the Kaio-Ken should be increased as well. So I would say  the "Max level of Kaio-Ken" is stronger by far.
Comment: As mentioned in Thaddeus's answer, Goku _did_ use Kaio-ken on top of Super Saiyan once.
Answer: Sounds like kaioken could be used to acclamante the body to strain to train for SS. A higher level SS should also be able to use a higher level of kaioken. Furthermore, sounds like kaioken shoulda been something that krillin, tien, and yamcha jumped on during their time with king kai. Since there's no SS transformation for humans, that would be an awesome tool for those characters. Plus, by later in the show (say, cell saga) I think we can safely assume that those characters are at least about as strong as goku was after he fought Ginyu but before he went SS, meaning they should each be resilient enough for masterful execution of kaioken - x10 or even more.
Answer: Super Saiyan is superior because kaioken has the side affect of strain and as king Kai said "total obliteration" while in super saiyan it multiplies the saiyans strength several times and the strain is minimal to non existent.
In conclusion super saiyan is superior due to the fact that even if super Saiyan  has a limit and kaioken doesn't ,kaioken is only to be used in short bursts as king Kai has said that it's only limit is your own and no one not even goku at his peak can reach kaioken 2,000,000 and if anyone did they would die or it would wear off leaving the fighter drained and useless. Super saiyan is superior in every class except limits but the limits are the users limit deeming it inferior and I am sorry if this angers anyone but it is true.
-
King Kai said this when Goku completed his training on other world.
He was explaining how the Kaioken worked and that it was only to be used in short bursts
-
I did some research and realized that in the episode counting down king Kai explains that kaioken could lead to your own destruction meaning since super saiyan has a lack of that risk and multiplies all attributes x 50 for an unprecidented amount of time it means super Saiyan is superior and Kaiden can only be sustained for a short period of time. Supersaiyan is superior.
Answer: Super Saiyan is stronger than Kaio-Ken
Super Saiyan is a natural extension of the Saiyan race, which is accomplished by gradual training and experience.
Before the Cell's Tournament, when Goku and Gohan are training in the Hyperbolic Time Chamber, Goku recommends that Gohan should keep his Saiyan state, even when they weren't training, so the body would get used to it.
Kaio-Ken is essentially a power multiplier, boosting strength, speed, reflexes and all other fighting abilities.
This technique is used to temporarily overload the user's body with enhanced capacities for a limited time, depending on each body's resistance.
Not to mention that Kaio-Kun is very rough to the body:

The technique, however, has its substantial flaws. If the user multiplies their base ki too much, the uprise of ki could easily obliterate them, or severely damage their body, as it did Goku's.
  After using the Kaio-ken x4 to overpower Vegeta's Galick Gun, Goku's entire body was thrashed, and he could barely move without hurting himself.
Source: Dragon Ball Wikia

At the moment, there is a new reference to Kaio-Ken being used by Goku as SSGSS. Theoretically, it's not recommend that a Super Saiyan uses Kaio-Ken since this process would much likely kill the user.
As a SSGSS, it is now possible to surpass this restriction:

Goku obtained the Super Saiyan God Super Saiyan form - which possessed great ki control, utilizing the Kaio-ken technique alongside it became possible.
Source: Dragon Ball Wikia

Conclusion: Since Kaio-Ken is limited by the user's body fortitude, developing the Super Saiyan form allows someone's body to reach higher Kaio-Ken level multipliers.
This means that Super Saiyan form determines whether or not Kaio-Ken is safe to use.
Answer: Kaio-ken is technically the more powerful technique in terms of max raw power output possible. Someone could, in theory, push their body to become use to ever higher levels of it. The draw back is extreme damage and exhaustion to the body. It's like over clocking a CPU, the higher you go the more power you're going to get out of it, but you're damaging it, and it will quickly overheat and crash.
Super Saiyajin on the other hand is more like adding another core to the the CPU that you can activate and deactivate. You can add more power and access it longer, but the CPU takes more damage from normal wear and tear over the course of it's life. The more you use it, the more you lessen the life span of the CPU even though you won't crash. Also, so long as you don't over clock or over heat, you're going to be stuck at given power output for each core you add.
So which is "stronger" or better? Depends on context.
Do you have a short time to live? Or, conversely, a long time to live? Do you need a greater boost right this second? Do you already have transformations? Can your body hand the stressing? Can your body handle transformations? Based on these questions the answer changes... If you're at Super Saiya-jin 3 and you're seeing the constraints of the transformations you can get to already I'd think it's is probably wiser to invest my time into getting to higher levels of Kaio-ken. On the other hand if there is a known legendary transformation that is ultra powerful, I'd go with looking for that transformation. Likewise. if I have a long time to live and train I'm going for training Kaio-ken, not mastering Super Saiya-jin.
The answer is situational, but consider that Super Saiya-jin 3 is something like x200 or x400 and it takes 1 year to get to x20. Think about the results and possible results SSJ3 is, what, 10 years down the line and very problematic, leaving you in a state worse of than your max Kaio-ken does. If you progress at x20/year and we compare it to SSJ3, I'd much rather bet on Kaio-ken than SSJ3 and another 10 years down the line you'll have x400, and 10 years after that you'll have x600, where as if you just go with SSJ3, you'll still be stuck at x200 or x400.
So I'm gonna side with Kaio-ken overall. 3 years after getting Kaioken you can do x60, 3 years after getting SSJ you can do x50, 4 years it is x80 vs x50. so this becomes increasingly obvious which is "better" overall, but if you can get SSJ/SSJ2 there is no reason not to have it and use it instead of Kaio-ken x50 and x100? because SSJ/SSJ2 are decisively better provided you don't mind shortening your life or would prefer not to be physically exhausted/injured afterwards.
Answer: I would say that Kaioken and Super Saiyan are equal because there technically is no max level of Kaioken as mentioned ealier, and there is no specified amount in power increase from Kaioken as Goku only said Kaioken increases the physical abilities of the user and never gave how much Kaioken increases the user's power by and it is said that Super Saiyan increases the Saiyans power by 50x the normal power.
In conclusion, Kaioken has no specific amount of power increase and technically no max level but puts an insane amount of strain on the body depending on the level of Kaioken being used, therefore it can not be maintained for that long however once Super Saiyan is mastered it can be maintained longer and used more effectively than Kaioken because it doesn't put that much of a strain on the body as Kaioken however  Super Saiyan does have a max level so that is why in my opinion they are both equal.
Comment: This doesn't introduce any new information, and only acts as a specualtive counter to the accepted answer.


Quickened spells and concentration

Question: I understand that I can't cast two spells per round (PHB p.202) even when using the Quickened Spell metamagic feat. However, is it allowable to cast a concentration spell like Crown of Madness on one turn, then on the following turn to use my main action to maintain that spell, and use Quickened Spell to cast another (non-concentration) spell with my bonus action?
Comment: Welcome to RPG.se, I see you've already taken the tour :) - As a suggestion, I would rethink the title of the question. It's more about using an action to maintain a spell than concentration itself. For example, Invisibility is a completely different concentration spell that doesn't share the same answer (as casting a spell in your next turn makes the spell end and it doesn't need an action to maintain it). Also, there's no need to state the game system or edition in the title - it's already tagged as dnd-5e, so we know it's about D&D 5e.
Comment: (My comment was too large, sorry) Part 2: Also, it's not exactly about Quickened Spells - it's about casting spells as a Bonus Action (the method doesn't matter). Although Sorcerer doesn't have many spells that use Bonus Action as casting time, Expeditious Retreat is a simple example that applies.
A possible title would be something like "Can I use an action to maintain a spell and cast another spell as a bonus action?". I'm not editing it myself because I'm not completely sure I understood your intentions behind the question.
Answer: Yes, you can do it
Maintaining the spell is not casting it.

On your subsequent turns, you must use your action to maintain control over the target, or the spell ends. Also, the target can make a Wisdom saving throw at the end of each of its turns. On a success, the spell ends.

Note that it says "use your action to maintain control over the target" - that's not casting a spell. If it was worded like "you must use your action to cast this spell again, without using spell slots", for example, you wouldn't be able to cast another spell as a bonus action in the same turn.
One simple way to see that you are not casting the spell again is that you don't expend any additional spell slots, even though the spell you would be casting is not a cantrip and it doesn't state you don't spend a spell slot (i.e. there is no specific rule in the spell telling you to go against the general rule - which would be to spend a spell slot).
As a note, the fact that it is a concentration spell doesn't matter. The only thing this changes is that the spell you are casting in the next round/turn must be non-concentration, as mentioned by you.

Side note

I understand that I can't cast two spells per round (PHB p.202) [...]

This is wrong.
You are not allowed to cast a spell as bonus action and then another (non-cantrip) spell as an action. (For clarification, you can't do it the other way around either, i.e., you can't cast a (non-cantrip) spell as an action and then cast another as a bonus action. For curiosity, you can't cast a (non-cantrip) spell as an action and a cantrip as a bonus action either. The cantrip has to be the action.)
You can, however, use more than one spell in your turn. For example, you can cast one spell as an action and another as a reaction, or use Action Surge (either multicass into fighter or EK) and use two spells as actions in the same turn. (Trivially, you can combine both the things I mentioned and cast three spells in a turn)
Comment: @KamilDrakari added that clarification. Also added a random curiosity about that ruling as well. (I call it curiosity because there are few bonus action cantrips, and you usually don't want to cast a spell with them.)
Comment: I would clarify your side note to also indicate that you can't go in the other order either, casting a spell with your action and then with your bonus action as well.
Comment: @kamildrakari you actually can cast an action cantrip and then a bonus action spell, nothing disallows for this.
Comment: @Medix2 see, I knew just saying "a spell" instead of "a non-cantrip spell" would come back to bite me...
Answer: We need to break this down a bit.

I understand that I can't cast two spells per round (PHB p.202) even when using the Quickened Spell metamagic feature.

You can't consume two spell slots in a single round, but there's nothing stopping you from casting a spell as an action or bonus action, and then casting a cantrip as a, respectively, bonus action or action. So long as one is a cantrip, the other can be whatever.

cast a concentration spell like Crown of Madness on one turn, then on the following turn to use my main action to maintain that spell, and use Quickened Spell to cast another (non-concentration) spell with my bonus action?

Concentration spells don't require an action to maintain unless stated otherwise. They only require that you not cast another spell that requires concentration, and that you not encounter an effect which breaks concentration (usually damage). So if you cast another spell the next turn, regardless of whether it takes an action or a bonus action, the only stipulation is that it not require concentration. If it doesn't, then you're free to cast the spell and maintain concentration on your previous spell.
Spells which specifically require you to spend an action to maintain concentration have that precise condition: spend an action maintaining the spell, and the spell's duration is maintained. Performing a bonus action on top of that, so long as the bonus-action spell doesn't require concentration, won't break the original spell's concentration.
Comment: "You can't consume two spell slots in a single round" -> Wrong. You could cast one spell with your action, cast  a reaction spell like Shield if your movement provoked an opportunity attack and cast a third with Action Surge. That's three spell slots used in one turn.
Comment: might also want to add in action surge allows 2 spells in one round
Comment: Even if your Bonus Action spell is a cantrip, you can only cast a cantrip with your Action if you cast any spell with a Bonus Action.
Comment: What GreySage said. [The rule](https://www.dndbeyond.com/compendium/rules/basic-rules/spellcasting#BonusAction) doesn't talk about multiple leveled spells per turn; it says that if you cast a spell as a bonus action, "You can't cast another spell during the same turn, except for a cantrip with a casting time of 1 action." This means that if you cast anything as a bonus action, even a cantrip, you can't cast any new spell on the same turn except a cantrip. (Maintaining concentration on a previously cast spell doesn't violate this rule.)
Comment: @Vylix: Nope. If you cast a spell as a bonus action, you can not cast another non-cantrip spell that turn (either before or after the bonus-action spell).
Comment: Concentration spells don't require an action to maintain, but some spells (which are concentrations spells) do, such as Crown of Madness. I could have worded that better, but my question still stands.
Comment: @V2Blast I'm always a bit confused about that rule, can I use Action to cast spell then use Bonus Action to cast another spell?

Quickened spells and concentration


Wordpress Thesis: Tweetmeme button remains after we got rid of the plugin

Question: We are having problems with getting rid of a Tweetmeme button from our Wordpress thesis blog.
The button somehow remains after we removed the plugin.
First we installed the plugin to the blog, but since we had problems, we asked our developer to fix the issue (He did something to the code of the plugin). After that, we switched to the official Tweet button and removed the Tweetmeme plugin.
When we see the source of a blog post, there is a Tweetmeme code there.
How can we get rid of the button/code?
Answer: I'd bet that he just hard coded the button into your theme. Look for 'tweetmeme' in your theme's <code>.php</code> files.


Why is angular momentum of the Earth/Moon system conserved?

Question: Why is the angular momentum of the Earth/Moon system conserved, apparently unaffected by external forces such as force of the Sun?
Comment: Welcome to SE.Physics!  Since this question has attracted 3 answers already, I'm guessing that other folks are getting the gist of what's being asked, however I'm a bit unclear about it personally.  Are you basically asking how the conservation of angular momentum can apply to partial subsets of the universe?
Comment: But momentum is conserved about centre of rotation
Answer: The total gravitational force of the sun acting on the earth-moon system acts through the center of mass of the system: it therefore cannot apply a torque about the center of mass, and so angular momentum about the center of mass is conserved.
Answer: Angular momentum is conserved when the net external torque is zero (you're thinking about linear momentum) via $\tau = \frac{dL}{dt}$. Anyway, as long as we assume that the earth, moon, and planets are all orbiting around in a disk (which is a decent approximation), then the position vector to the Earth/Moon system from any other of the planets is in the same plane as their perturbing gravitational forces, and furthermore these other planets are sufficiently far away from the Earth/Moon system so that the position vector and force vector are essentially parallel, meaning that the torque, $\tau = \vec{r} \times \vec{F}_{Grav} $, is zero, and thus angular momentum is conserved.
Answer: Actually the angular momentum of the Earth/Moon system conserved is not conserved. Considering just the earth, moon and sun in the solar system, this is the infamous three body problem for which there is no formulaic solution. The angular momentum is "mostly" conserved however.


What's the best way to reduce odor in my well water?

Question: Updating 60+yr old house, which has a 300 ft well , it has bad smelling water. How to fix cheapest & best way. This house was built by my stepfather, he said they dug til he say white sand . I don't know where to start. Test water before pipes and after.  There's got to be bacteria, fungus , somewhere. It stops smelling if ran about 10 minutes or so. Its so strong tho, it makes me scared to drink much less using it period is iffy to me! Thinking new pipes were the reason, put in New & still smells just as bad. Bathroom Also has stone shower that wasn't sealed or anything to keep mold or mildew at bay. Do I need to replace it all or is there a way to save it? Its the newest of the updated things but could be 5- 10 years since its been exposed to the same water from well. Nothing major there tho I removed silicone around edge of tub & scrubbed majorily, but am asking before I tear it out to be safe. Also can the water have any harmful reactions to humans or animals if they've used this water for prolonged years? ~ Tonya
Comment: Sealing the stone in the shower won't do much (or anything) to prevent mold. You need to vent the humid air out of the space (open a window or install an extractor fan).
Answer: You said the water smells bad unless run for 10 minutes this points to air in your pressure tank. With a failed bladder in an old tank air contacts the water in the tank and it develops a bad taste/smell. If I wanted to sanitize a well that deep I would pour several gallons of bleach into the well and let it sit for at least several hours then pump the well until the bleach smell is gone. As far as the stone shower I would not be concerned as much, if you want to sanitize that I would use a 3% hydrogen peroxide and water, spray the stone work then rince. Since your smell disipears after several minutes check your pressure tank as it may be the source of the bad smell.
Comment: The house I grew up in developed smelly well water over a couple of years. It was a failed bladder in the tank, as you said. We cut the old tank open, and OMG!! I can't describe how foul it was! One of the worse things I have seen, given that we had been drinking the water...
Comment: Actually you dump bleach in the well while recirculating water down the well with a hose, and run until you smell beach at the hose. Then run water at every faucet until you smell bleach, THEN wait several hours or overnight, then flush it out. You have to disinfect the whole pump/piping system, not just the well itself. And a few days of regular use later, you pull a sample and send it to a certified testing laboratory. [Having killed any temporary infestations in the pipes, you need to see what's in the water itself.)
Comment: You go your route I will do it the way. my well testing guy showed me this method. the issue is in the well and pressure tank that is eliminated with flow. Putting a high concentration of bleach in the pressure tank will shorten the bladder life.
Comment: Replacing the tank fixed it, though. We never had to disinfect the well or the plumbing.
Answer: If the system does not already have purification in place, Installing a Point-of-Use Water Filter or a Whole-House-Filter is probably best.


Calculating eccentricity from semi-major axis, position and velocity of particles

Question: I'm trying to calculate the eccentricity of a binary system's body knowing the bodies' velocities, positions, masses and the semi-major axis. I am only trying to calculate the eccentricity for one of the bodies but can't seem to get the formulas combined right to get the eccentricity only using these values.  Any advice would be brilliant, thanks!
Answer: Hint 1:solve around the barycenter, maybe you can use the reduced mass
Hint 2:  Try doimg what I call 'overloading the system'. Start by assuming you know everything (eccentricity, both axes, position/velocity at any point). Now derive as many relations as possible.
Comment: I can see how to do it using the true anomaly but calculating that for an arbitrary 3D system knowing only these seems awkward. Is that what's necessary or is there an easier method?
Comment: @Mithra: `knowing only these` see hint 2. Pretend to know everything, then find relations and reduce your variables. And btw, this is only a 2D system. I don't know about any easier methods.
Answer: First calculate $ h = || \vec x \times \vec v ||$.
Then the eccentricity is:
$ e= \sqrt{ 1 - \frac{h^2}{G(m+n) a} } $
Where $a$ is the semi-major axis.


Transformers: Short-circuit test

Question: In a short-circuit test where the secondary winding is short circuited, how come the primary voltage isn't zero (in real life) when it should ideally be zero?
Answer: The voltage on the primary side shouldn't ideally be zero as you claim.
If you short the secondary winding of a transformer then all the power you put into the primary winding will be dissipated in the internal resistance of the secondary winding.
Lets take the following example:
We have a 1:1 transformer with an internal resistance in the secondary winding of 1 Ohm. If we apply 1V AC to the primary winding it will cause 1V to be applied across the 1 Ohm resistance of the secondary winding (because it is shorted) and hence 1A will run in the secondary winding.
Since the transformer is 1:1 the same current (1A) will also run in the primary winding.
However as Rohat correctly states in his reply: If the transformer was ideal (ie. no internal resistances etc.) then the current in the primary winding would TEND TOWARDS infinity. And it follows that if the output impedance of your source is not zero ohm, then the voltage will drop to zero, but not because of your transformer, your transformer just wants to draw infinite current, it is your source which is not able to deliver it.
Answer: In an ideal transformer, input power equals to output power: \$v_{pri} \cdot i_{pri} = v_{sec} \cdot i_{sec}\$
If you short the secondary then the secondary voltage will be zero, thus the secondary current will be infinite: \$i_{sec} = v_{pri} \cdot i_{pri} / 0 = \infty \$.
Since the primary is driven from a source with zero (or nearly-zero) output impedance, the primary voltage cannot be zero. Instead, the primary current will reach infinity.
When it comes to practice, the non-zero impedances show themselves: the source's output impedance, the windings' DC resistances, finite magnetizing inductances, etc. They limit the output current according to Ohm's law.

Transformers: Short-circuit test


Have they used the LHC to see if any subparticles fly off a particle before collision?

Question: Have they used the LHC to see if a particle gives off any subparticles precollision?
Answer: I suppose the question is from an interested student.

Have they used the LHC to see if a particle gives off any subparticles precollision?

What is the LHC? It is a cirular collider, where protons are formed into two opposing beams of very high energy and with manipulation with magnetic fields, the beams made to collide head on. Then instruments are all around the head on collision region to measure the results of the collision.
Protons are stable, they do not decay. The beam of protons going around is made so that protons move parallel to each other in each beam before brought into a collision. Since they are charged and in a magnetic field, they do give off photons, which are the expected synchroton radiation, compensated by the magnets in the tunnel, described in the link, so that they do not lose momentum and come with a fixed known momentum at the collision point of the beams.
If  protons would shed other particles the calculations for the beams would not work, because they are made for synchroton radiation only , so this is an experimental proof that no other then photons are shed before collision.
So the answer is that the successful manipulation of the proton beams shows that only electromagnetic radiation, photons, is shed , which is compensated with calculations to keep the beam momentum fixed.
This answers for the LHC. For other accelerators, the beam could consist of particles that decay, as Kaons, and then a decay could happen , which means that the beam would have one less Kaon, as the decay parts would fly apart out of the beam line.
Answer: Yes. But it isn't to find out if particles decay all by themselves. Many do. This can be useful.
This video is about DUNE, an experiment at Fermilab. Fermilab is one of the premier subatomic crash test facilities. They study many things, but neutrinos is one of their specialties. The video explains why. What is the DUNE experiment?
These videos explain more about neutrinos. Neutrinos: Nature's Ghosts? Subatomic Stories: The amazing neutrino
This video tells how to make a neutrino beam. If you take a beam of protons and smash it into a stationary target, you get lots of pion particles. These are unstable and decay into a muon and a neutrino. You smash these into a second stationary target to remove the muons. The neutrinos keep going. Now you have a netrino beam. How do you make a neutrino beam?


I've tilled wet soil before doing proper research. What can i do to minimize the damage?

Question: I'm new to gardening, and I decided that I am going to create a new garden bed in my yard. The problem is, I began tilling without knowing of the damage that can be done by tilling very wet soil. Now I've got a muddy mess in my yard and I'm worried I've ruined the area. If I let the area dry for a few days then till again, would that help remedy the situation? Any advice is greatly appreciated!
Comment: What kind of soil do you have? I really, really hope that it's not clay.
Answer: Well, going at this without much information, I'm pretty sure that you've accidentally destroyed your soil's tilth, and there's no quick fix for that. More tilling won't help at all and can even make things worse.
You can try this test to see how bad things are (maybe the weather has already done this for you) - pick a spot and water it very well, perhaps an inch/2.5cm in an hour or so. If after that hour you have ponding on the soil, then the soil's structure has been destroyed. You could try adding compost, etc. and LIGHTLY tilling, but it won't change anything until it decays, and even then it won't do much because you'd have to add yards and yards of compost to have much improvement. Covering the soil with woodchips from an arborist also helps, but we're talking YEARS of chips to get tilth in maybe the top 1-2"/2-5cm.
One caveat on the test I mentioned above - do not do that test if you're uphill from a neighbor. There was someone in my neighbor who hired "landscapers" (i.e. "two guys with a skidsteer and no experience or education") to revamp their landscape. These guys ran their skidsteer all over the yard after 3 inches of rain, then tilled it and seeded a lawn. Two days later it rained heavily again, and the homeowners received a visit from the police because the rain (and grass seed) ran off their impervious landscape downhill into a neighbor's basement. Much angst and money-changing-hands ensued.


What does this mean in Ruby language?

Question: Run the following code,
a = [1, 2, 3, 4, 5]
head, *tail = a
p head
p tail

You will get the result
1
[2, 3, 4, 5]

Who can help me to explain the statement head,*tail = a, Thanks!
Answer: head, *tail = a means to assign the first element of the array a to head, and assign the rest of the elements to tail.
*, sometimes called the "splat operator," does a number of things with arrays. When it's on the left side of an assignment operator (=), as in your example, it just means "take everything left over."
If you omitted the splat in that code, it would do this instead:
head, tail = [1, 2, 3, 4, 5]
p head # => 1
p tail # => 2

But when you add the splat to tail it means "Everything that didn't get assigned to the previous variables (head), assign to tail."
Comment: nice answer! :) but you really should change your profile pic, you look like an overly groomed poodle ;) hehe
Comment: Haha. I haven't got that one before. I'll take it under advisement. :)
Comment: @banister: His current pic (maybe not the same one used back then) makes him look like Justin Bieber!
Answer: First, it is a parallel assignment. In ruby you can write
a,b = 1,2

and a will be 1 and b will be 2. You can also use
a,b = b,a

to swap values (without the typical temp-variable needed in other languages).
The star * is the pack/unpack operator. Writing
a,b = [1,2,3]

would assign 1 to a and 2 to b. By using the star, the values 2,3 are packed into an array and assigned to b:
a,*b = [1,2,3]

Answer: I don't know Ruby at all, but my guess is that the statement is splitting the list a into a head (first element) and the rest (another list), assigning the new values to the variables head and tail.
This mechanism is usually referred (at least in Erlang) as pattern matching.

What does this mean in Ruby language?


Cannot See Validation Rules

Question: My boss has asked that we modify all validation rules in our org. We're creating a custom setting that will allow admins to bypass all validation rules (validation rules will not apply to admins in our org). In order to do this, I believe you have to manually add the custom setting to your validation rule. I've queried in Workbench and found the names of every validation rule in our org. When I go to settings>object manager I cannot see the validation rule. My boss found the same when querying Tooling API. Any ideas?
Answer: Please note that you need to navigate to the object the rules are on, at which point you will see Validation Rules as one of your view options, along with Details, Fields & Relationships, etc.

Second important note, a Custom Setting is not best practice for this requirement. You should instead use a Custom Permission which you can add to any Profile or Permission Set you wish to be exempt.

Cannot See Validation Rules


How to fix a partially uprooted tree? Please help

Question: A few days back due to strong winds a peepal tree(i.e. Ficus religiosa or sacred fig) was partially uprooted near my house. It is approximately at a 45° angle. It is 7-8 meters tall. Almost half the roots are out of the surface. How can i fix it?
Answer: Trees are remarkably tough, there have been cases where trees have been partially uprooted and then grown well once stood back up. I would try to stand it back up and stake it well to support it whilst the roots grow and support the tree back in place.
Answer: When we travel in hurricane-prone regions such as the low lying islands of the Bahamas it is a fairly common sight to see fruit-laden trees supported by sturdy forked sticks at a precarious angle and surviving quite well. The problem is that when the roots are pulled from the ground they are subjected to the drying winds and desiccate rapidly making it difficult for them to re-establish themselves if reintroduced to the ground. And even if they do form new roots, the tree is weakened on that side making it vulnerable to the next prevailing wind from that direction.
Massive root loss often triggers a distress crop of fruit; the weight of the fruit only contributes to the difficulty, pulling the tree even further over on the leaning side. So for fruiting crops there is a vested interest in keeping that tree going just to get the crop while you work on a replacement tree. But a replacement may take a while to obtain either by home grown or purchase, and may take some time to achieve the size of the existing, which is why a supply of heavy forked sticks and big dimension lumber comes in handy.
The fruit of the F. religiosa is not significant as food, so we can set that aside for a moment. However in some countries this Ficus is regarded as a religious symbol - there may be local sensitivities with respect to keeping specimens in good order. No doubt it will be easy for you to gauge from discussions with neighbours whether this is an issue.
Answer: I just want to advice based on my own experience.
I had a large plum fruit tree (even higher than your tree, 8-10 m high) that was uprooted one winter almost identically to your tree - 45 degrees. I should say I live in a really windy area. The tree even stayed that way for several months. With the help of a friend and some ropes we just "returned" the tree to its original position. No special care afterwards.
The tree continued living as if nothing happen. :)
I would just perhaps advice fixing the tree to the neighboring tree for a season or two, just in case. In my case, it wasn't needed.


How are static variables with the same name in different functions identified by the System?

Question: AFAIK, we can have two static variables with the same name in different functions? How are these managed by the compiler and symbol table? How are their identities managed seperately?
Answer: Compilers don't store static variables' names in the linking symbol table.  They are just some memory that is part of the module as far as the linker is concerned. (this may not be 100% true in all cases but it is effectively true)
The names of static variables are usually included within the debugging symbol table.
When you feed a .c file to the compiler it keeps up with the names of all known symbols so that it can recognize them for what they are when they come up in future code.  It also remembers them so that it can give useful error/warning messages, but it pretty much forgets about them when generating output files (unless debugging symbols are being generated).
Answer: They are likely mangled in the table, in a similar way to how overloaded functions are implemented.
See dumpbin /symbols foo.obj if you want to peek at the table, or use objdump on linux.
Answer: It depends on the compiler, but some embedded ones simply add a number to the end of each duplicate name. That way each variable has a unique name.


is it possible to keep the texture attached to my project?

Question: every time i remove a texture from a file it also vanish from my project and it get replaced with a pink texture instead
is there a possible way to keep the texture attached to my project even if i removed them from my PC ?
Comment: Textures and other images used in the project are not saved as part of the .blend file by default unless they are **packed**.
Comment: https://blender.stackexchange.com/questions/56683
Answer: To make sure that all the images you use are part of your blend file you can go into the File menu > External Data > Automatically Pack Into .Blend. If you don't want to make it automatic but just want to pack the current images, choose Pack All Into .Blend.

If you want to do it for just one or several images, you can go in the UV Editor or Image Editor, select the image you want, and Image menu > Pack.


Is there something wrong with my question?

Question: I recently reached out on Stack Overflow to get help with an issue. I provided detailed screenshots and the specific part of the code where I'm getting the answer, but it keeps getting downvoted and no one answers it. Is this a bad question?
Firebase Web Read from Real-Time Database
Comment: Keeps getting downvoted? It's got 1 downvote and 1 upvote, that's not most people's definition of something that keeps happening, since it happened only once.
Comment: I'm voting to reopen this question as the linked duplicate is not appropriate at all. The linked question on main included screenshots, but it also included the code. The guidance about not uploading images of code/data/errors is intended to be for questions that *only* show the code/data/errors, which doesn't apply here.
Comment: NOTE: The OP even offered a bounty... the question in focus has a negative score at this time.
Comment: Related (posted 20 minutes later): *[How do I get unbanned from asking questions?](https://meta.stackoverflow.com/questions/420646/)*
Answer: The issue here (which is fairly common on sites such as this) is that you're providing screenshots, and only provide your real code much further down in the post. You write

I'm trying to display the user's price that they entered in the database, but I'm getting "undefined' back instead of the value that was entered. I didn't get any errors in the console either. How can I fix this?

but don't even say what language you're using. I see that you use Java below, but this should be stated up front as well as likely a little more context about what the issue is.
I want to emphasize how unhelpful screenshots are—and note this isn't personal. Screenshots and images are pretty much never helpful in my history of giving help with technical answers. We're writing on computers—we need text to work with, more specifically, code that's presented in a clear manner. The reason for this is that it allows us to better reproduce your error, so we can get to the bottom of what the real issue is. When the question begins in a sort of vague manner, no context is given, and only later on do you get to the code, it makes it more difficult for people who want to help you. And the only way to do that is with code, not images.
I don't think the question is bad at all, since there's likely a very legitimate answer. Rather, I think that it is poorly posed, and can be improved by clear explanations and the general context of your query.
Comment: *"but don't even say what language you're using."* You don't need to tell us what language you are asking about, that's what tags are for. "I am using Java" is actually just noise, as if they are using Java then it should be tagged with Java, and this we already know what language is being asked about
Comment: [[tag:java]] *isn't* in the tags in the linked question, [[tag:javascript]] is, which isn't the same.
Comment: I don't agree that it's just noise. But you are right, java was in the tags of the OP (at least as of now), so it was stated to begin with.
Answer: Don't use screenshots to convey textual information
Please read Why should I not upload images of code/data/errors when asking a question?. The reasoning applies very broadly.
This is not a discussion forum
Questions should be clear, explicit and direct. Don't tell us things that are obvious. We can see the screenshot and the code. We can see that some of the code is HTML, some is JavaScript and some is CSS; and it's normal for these things to be combined.
Try to find problems yourself first
Please read How much research effort is expected of Stack Overflow users? and https://ericlippert.com/2014/03/05/how-to-debug-small-programs/. Study the behaviour of the code by yourself. Examine what happens, step by step. Make sure you have some idea of what you expect to happen at each step (in order for your logic to work as intended), and then see exactly where something different happens. When you find something specific, use a search engine to look for existing explanations - on or off Stack Overflow.
If you use console.log to check what the code is doing, make 100% sure you know which console.log call is producing what output. What does "I'm getting 'undefined' back" actually mean? What is giving you that result? What do you think the results should look like instead? Why?
If you are still stuck, create a Minimal Reproducible Example. If the problem is that the code does not successfully retrieve the price information, why do we need to see code that retrieves (or attempts to retrieve) several other pieces of information? If you remove that part of the logic, does the error still occur? If the problem is that reading the data does not work, why do we need to see code to set the data? Does the code seem to behave differently when getting data that was already in the database, vs. data that was just set by that method? Exactly what steps are required to cause the problem?
Comment: If you're using `console.log`, make sure you know about [browsers' lazy evaluation of what is being logged](https://stackoverflow.com/questions/4057440/is-chrome-s-javascript-console-lazy-about-evaluating-objects) so that you are not confused about asynchronous behavior.
Comment: Examples for the last part are [this](https://stackoverflow.com/questions/73863661/how-can-i-get-my-buttons-to-hide-display-properly-when-i-click-them) and [this](https://stackoverflow.com/questions/73841591/how-can-i-get-my-cards-to-display-and-function-properly) (about 400 lines of "code", and 700 lines of "code", respectively). References or similar could be incorporated in the answer.
Comment: [Here](https://stackoverflow.com/questions/73861244/how-can-i-get-my-vertical-tab-button-to-switch-tabs-properly#comment130421260_73861244) it was claimed nearly 1,700 lines of "code" was the minimum.


Why is my cactus losing its thorns?

Question: I bought this cactus four months ago and it had some fake flowers glued to it. I took them off by force and with them some of the thorns were off too. I was hoping that I wasn't going to harm the plant by doing it but since then I think the cactus has started losing some other thorns too.
It is placed where it gets plenty of light ( not direct sunlight ) and I water it once in two weeks because the pot does not have a drainage hole in it
Answer: This probably isn't fatal in itself, but it is telling you the plant isn't happy.
The biggest problem is a pot with no drainage holes. Fix that right now! No cacti like having their roots permanently wet. The pictures don't show how deep the pot is, but the roots may have grown go right to the bottom of the pot and are sitting in water all the time.
Another problem may be lack of light. Full sunlight is ideal. If that is impossible in winter, the best plan is to stop watering completely for a few months (say October to March) and let the plant go dormant until the light levels improve. When you start watering again, you are more likely to get flowers (real ones!) and the plant will start to grow quickly after a dormant period.
Answer: Yeah, like @alphazero said, get the pot situation resolved as quickly as possible. You can drill holes in that pot with a masonry drill bit or get a pot that has holes.
The problem with the sparse spines is probably a period of stress along with fake flowers being added and removed, followed now by new growth this looks normal. Throw in repotting somewhere along the way.
The places where the areoles are gone will never have spines again. They will become less noticeable as the plant grows. The area of short/missing spines (cactus have spines, roses have thorns, but call them whatever you want, it doesn't hurt anyone!) will always be shorter/more stunted than the rest of the plant. Some cactus have variable spine growth where they are long in the summer and shorter in the winter. They eventually even out (mostly) and you no longer notice it.
Once you get the pot resolved figure out a way to get it more direct sunshine, the more, the better. Slowly introduce it to sunshine so it gets an hour for a few days and add an hour until it's fully in the sun.
Fertilize it with half strength balanced fertilizer occasionally (don't do this if you can't find more sun for it.)
Water when it's mostly dry. You aren't going to hurt it by leaving it a few days too long or too short. Somewhere in between watering every day and watering once every three months.
All that said, I don't see any big issues other than the pot, that will kill it. It's a nice plant.
Good luck!

Why is my cactus losing its thorns?


Inductive wireless mobile charger circuit

Question: In wireless mobile charging circuit, is there any formula of spacing which varifies that by increasing diameter and turns, spacing (range) can be increased?
Answer: The transferred power between two coils depends on the coupling inductance between them, which depends on several things, such as axial spacing, number of turns and radius of the coils, according to this formula:
$$
L_{12} = N_1.N_2.\frac{\mu_0}{2}.r_1.r_2.\int_0^{2\pi}\dfrac{cos\alpha}{\sqrt{b^2+r_1^2+r_2^2-2.r_1.r_2.cos\alpha}}d\alpha
$$
Being \$b\$ the axial spacing, \$r_i\$ the radius of the coils and \$N_i\$ the number of turns.
You can see how increasing the number of turns and radius will allow you to increase the axial distance to obtain the same coupling inductance, therefore the same transfered power.
See reference paper (page 5) Limitation of inductive power transfer for consumer applications
Comment: This one http://ir.lib.uwo.ca/cgi/viewcontent.cgi?article=2482&context=etd it is usefull and free
Comment: You linked a document that requires purchase in order to read it.
Answer: The efficiency of the power transfer depends on the coupling between the inductors "k" and their quality factor "Q". ("Q" = the ratio of the inductance "L" to the resistance "R" of a coil) The quality factor Q is mainly dependent on the shape and size of the coil as well as the materials used, usually you may expect values around 100.The coupling "k" is determined by the distance between the inductors (we mark the distance by "z") and the ratio of two coils diameters D2 / D1. The coupling is further determined by the shape of the coils and the angle between them. The coupling factor is a value between 0 and 1. The value of 1 expresses perfect coupling, so all flux generated in the transmitter penetrates the receiver coil. By the other hand, a 0 value expresses a system where transmitter and receiver coils are independent of each other. Poor coupling can be linearly compensated by a better quality factor and vice versa. So the efficiency drops dramatically at larger distance (z/D > 1) or at a large size difference of the coil (D2/D1 < 0.3). A very high efficiency (>90%) can be achieved at close distance (the ratio z/D < 0.1) and for coils of similar size (D2/D1 = 0.5 ~ 1). For more details about this refer this link wireless charger