Initial Value problem and showing that $ x(t_n) - x_n = 1/2ht_ne^{tn} + O(h^2)$ I have a question which I can't for the life of me figure out.  The questions starts by giving $$x'(t) = x(t), x(0) = 1, x(t) = e^t$$
So they give you the solution and basically ask you to apply the Euler method with step size $h$ to show $$x_n = (1+h)^{t_n/h} , t_n = nh , ~~ n = 0,1....$$
Now I have gotten this out quite simply however, the next part asks use the formula $$log(1+h) = h - 1/2h^2 + O(h^3)$$ to show that $$ x(t_n) - x_n = 1/2ht_ne^{tn} + O(h^2)$$
Now I have tried a couple of different options and i feel it should be easier than it is. The furthest I can sort of get is $$x_n =e^{t_n(1 - 1/2h+O(h^2))}$$ (but I have so many different types of solutions and maybe working with Taylor series) any help would be greatly appreciated

Now you split
$$
e^{t_n(1−1/2h+O(h^2))}=e^{t_n}·e^{ht_n(−1/2+O(h))}
$$
and apply the exponential series to the second factor
$$
e^{ht_n(−1/2+O(h))}=1-\frac12 t_nh + O(h^2)
$$
and insert this to get the desired error formula. One could go one step further and get hard bounds using
$$
|e^x-1|\le |x|·\frac2{2-|x|}\quad\text{or}\quad
|e^x-1-x|\le\frac{x^2}{2}·\frac{3}{3-|x|}
$$

It is ways faster to go the logarithm-exponential way to find the term $e^{t_n}$ together with the exact form of the error. The direct way to extract the exponential series with error terms from the binomial formula is much more messy.
From general limits one knows of course that
$$
\lim_{n\to\infty}\left(1+\frac tn\right)^n=e^t.
$$
but to directly derive the error terms fills more paper than the way over the logarithm.
