Remainder Term for a Taylor Series and Polynomials If $f(x)=e^x$ and $f_n(x)$ is the $n^{th}$ order Taylor Polynomial of this function, then we know that the Taylor Theorem states there exists $t\in [x_0, x]$ such that the remainder term of is given by,
$$r_n(x)=\frac{f^{(n+1)}(t)}{(n+1)!}(x-x_0)^{n+1}=\frac{e^t}{(n+1)!}(x-x_0)^{n+1}$$
as $f^{(n+1)}(t)=f(t)=e^t$. Now we also know,
$$f=f_n+r_n$$
$$\Rightarrow e^x = f_n(x) + \frac{e^t}{(n+1)!}(x-x_0)^{n+1}$$
On the right hand side, $f_n$ is a polynomial and so is the expression for the remainder term. However $e^x$ is not a polynomial. How is it possible to write a non polynomial function in terms of a polynomial? We can use a similar procedure and arrive at this contradiction for other non polynomial functions as well. So what is the gap in my understanding?
$t$ is just a value between $x_0$ and $x$ and $e^t$ is how to evaluate this constant.
It does depend on $x$, but nobody said that Taylor polynomial is expressing the function in full. It has an error in its approximation.
The term with $t$ is that error term, so it talks about an error in Taylor polynomial approximation. It is not part of polynomial and of course that it depends on $x$, because typically more you are away from $x_0$ worse the error.
Just because it has $(x-x_0)^{n+1}$ in it does not make it a part of the polynomial, there are other expressions for the error term that look less like an additional polynomial term.
So, yes, besides $(x-x_0)^{n+1}$ the error term may have one additional dependency on $x$, or it may not, we do not know. All we know is that $t$ is between the two. In general it is movable regardless if this move is expressible as a function on $x$ or not.
Typically this $t$ is used to say that max error cannot be worse than, in your case, the maximum value of $e^t$ where $t$ is between $x$ and $x_0$ together with other values in that term. Some functions will have this max value fixed for quite some possible range between $x$ and $x_0$, some will not. But anyway $t$ is not really useful beside deciding this max error, because we know nothing else about $t$ except that it is between $x$ and $x_0$.
If $t$ would be a fixed value then the error term would be in form of $k(x-x_0)^{n+1}$, meaning every function would be expressible as a polynomial of $n+1$ degree, and unless the function is such a polynomial that is not the case.
$t$ is just a value between $x_0$ and $x$ and $e^t$ is how to evaluate this constant.
It does depend on $x$, but nobody said that Taylor polynomial is expressing the function in full. It has an error in its approximation.
The term with $t$ is that error term, so it talks about an error in Taylor polynomial approximation. It is not part of polynomial and of course that it depends on $x$, because typically more you are away from $x_0$ worse the error.
Just because it has $(x-x_0)^{n+1}$ in it does not make it a part of the polynomial, there are other expressions for the error term that look less like an additional polynomial term.
So, yes, besides $(x-x_0)^{n+1}$ the error term may have one additional dependency on $x$, or it may not, we do not know. All we know is that $t$ is between the two. In general it is movable regardless if this move is expressible as a function on $x$ or not.
Typically this $t$ is used to say that max error cannot be worse than, in your case, the maximum value of $e^t$ where $t$ is between $x$ and $x_0$ together with other values in that term. Some functions will have this max value fixed for quite some possible range between $x$ and $x_0$, some will not. But anyway $t$ is not really useful beside deciding this max error, because we know nothing else about $t$ except that it is between $x$ and $x_0$.
If $t$ would be a fixed value then the error term would be in form of $k(x-x_0)^{n+1}$, meaning every function would be expressible as a polynomial of $n+1$ degree, and unless the function is such a polynomial that is not the case.