For fun and for the challenge, I'm trying to solve a few of Arnold's problems. I did Problem 5 and I would like to check if my solution is correct.
Problem 5: Calculate the $100$th derivative of the function $$\frac{1}{x^2+3x+2}$$ at $x=0$ with $10\%$ relative error.
Spoiler alert: my solution below. If you want to attempt solving the trivium some day, don't read ahead.
Solution: We have $$\frac{1}{x^2+3x+2} = \frac{1}{(x+1)(x+2)} = \frac{1}{x+1}-\frac{1}{x+2}\ .$$ Therefore, $$\frac{d^{100}}{dx^{100}}\frac{1}{x^2+3x+2} = 100!\left(\frac{1}{(x+1)^{101}}-\frac{1}{(x+2)^{101}}\right)\ .$$ At $x=0$, this gives $$100!\left(1-\frac{1}{2^{101}}\right)\approx\sqrt{200\pi}\left(\frac{100}{e}\right)^{100}\approx 25\left(\frac{100}{e}\right)^{100}$$ where we used Stirling's approximation and the approximations $$1-\frac{1}{2^{100}}\approx 1\ ,\qquad \sqrt{200\pi}\approx 25\ .$$
I believe this to be correct. Is it enough though, or do we need a numerical value at the end (which would be pretty hard to obtain without a computer...)? Are there other ways to obtain a solution?
$$f (x)=\frac {1}{1+x}-\frac {1}{2+x}$$
$$=1-x+x^2-...+(-1)^nx^n-$$ $$1/2 (1-x/2+x^2/4-...+(-1)^nx^n/2^n+$$ $$x^n\epsilon (x) $$
$$\implies f^{(100)}(0)=100!(1-\frac {1}{2^{101}}) $$