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If $f(x)=\frac{1}{x^2+x+1}$, find $f^{(36)} (0)$.
So far I have tried letting $a=x^2+x+1$ and then finding the first several derivatives to see if some terms would disappear because the third derivative of $a$ is $0$, but the derivatives keep getting longer and longer. Am I on the right track? Thanks!
We can write:
$$1+ x + x^2 = \frac{1-x^3}{1-x}$$
Therefore:
$$f(x) = \frac{1-x}{1-x^3} $$
We can then expand this in powers of $x$:
$$f(x) = (1-x)\sum_{k=0}^{\infty}x^{3 k}$$
which is valid for $\left|x\right|<1$. The coefficient of $x^{36}$ is thus equal to $1$, so the 36th derivative at $x = 0$ is $36!$ .
Let $\omega$ be a complex cube root of $1$. Then Partial Fraction representation of $f(x)$ is given by
$f(x) = \dfrac{1}{x^2+x+1} = \dfrac{1}{(x-\omega)(x-\omega^2)} = \dfrac{1}{\omega - \omega^2}\Big(\dfrac{1}{x-\omega} - \dfrac{1}{x - \omega^2}\Big)$.
Find successive derivatives to show that
$f^{(36)}(x) = \dfrac{1}{\omega - \omega^2}(36! (x-\omega)^{-37} - 36! (x - \omega^2)^{-37})$.
Let $x = 0$ and use $\omega^3 = 1$.
Use $$x^2+x+1=\left(x+\frac{1}{2}+\frac{\sqrt3}{2}i\right)\left(x+\frac{1}{2}-\frac{\sqrt3}{2}i\right)$$
By this hint we obtain: $$\left(\frac{1}{x^2+x+1}\right)^{(36)}_{x=0}=\left(\frac{1}{\left(x+\frac{1}{2}+\frac{\sqrt3}{2}i\right)\left(x+\frac{1}{2}-\frac{\sqrt3}{2}i\right)}\right)^{(36)}_{x=0}=$$ $$=\left(\frac{1}{\sqrt3i}\left(\frac{1}{x+\frac{1}{2}-\frac{\sqrt3}{2}i}-\frac{1}{x+\frac{1}{2}+\frac{\sqrt3}{2}i}\right)\right)^{(36)}_{x=0}=$$ $$=\frac{1}{\sqrt3i}\left(\frac{36!}{\left(x+\frac{1}{2}-\frac{\sqrt3}{2}i\right)^{37}}-\frac{36!}{\left(x+\frac{1}{2}+\frac{\sqrt3}{2}i\right)^{37}}\right)_{x=0}=$$ $$=\frac{1}{\sqrt3i}\left(\frac{36!}{\left(\frac{1}{2}-\frac{\sqrt3}{2}i\right)^{37}}-\frac{36!}{\left(\frac{1}{2}+\frac{\sqrt3}{2}i\right)^{37}}\right)=\frac{1}{\sqrt3i}\left(\frac{36!}{\frac{1}{2}-\frac{\sqrt3}{2}i}-\frac{36!}{\frac{1}{2}+\frac{\sqrt3}{2}i}\right)=36!$$
Hint 1: Can you write your function as a sum of powers of $x$? If so, that's a Maclaurin series, and finding the 36th derivative (at $0$) from that should be pretty easy.
Hint 2: What's a series expression for $\frac{1}{1+y}$ (at least when $y$ is small)?
Hint 3: Have you heard of "completing the square"?
Do you know a smooth function can be expressed uniquely in a Taylor series around a point? Therefore $f(x)$ can be expressed uniquely around zero as $$f(x)=\sum_{n=0}^{36}\frac1{n!}f^{(n)}(0)x^n+o(x^{36}).$$
We also note that when $x$ is small, we have that $$\frac1{1+(x+x^2)}=\sum_{m=0}^{36}(-1)^m(x+x^2)^m+o(x^{36})$$ according to geometric series expansion.
Then we have to pick out all the $x^{36}$ terms and sum their coefficients up. Still a tedious task because we have to look from $m=18$ all the way to $m=36$. But the complexity might be smaller than directly computing the derivatives.
Edit: not particularly tedious. Since for each $m$ in question $(x+x^2)^m$ has only one $x^{36}$ term. For $m=18$ it is $(-1)^{18}=1$, obviously. For $m=19$ it is $(-1)^{19}{19\choose 2}=-{19\choose 2}$. For $m=20$ it is $(-1)^{20}{20\choose 4}$ and so on. There is clearly a pattern in it.
A non power/taylor series method using Leibniz' generalize product rule:
$$ (fg)^{n}(x) = \sum_{r = 0}^{n} \binom{n}{r} f^{n-r}(x) g^{r}(x)$$
Let $f(x) = \dfrac{1}{x^2 + x + 1}$ and $g(x) = x^2 + x +1 $
Then $f(x) g(x) = 1$. We apply the rule to $fg$.
If $a_k = f^{k}(0)$ the we get the recurrence relation
$$a_k + ka_{k-1} + k(k-1)a_{k-2} = 0$$
With $a_{0} = 1$ and $a_1 = -1$
Now let $b_k = \dfrac{a_k}{k!}$, then we get
$$b_k + b_{k-1} + b_{k-2} = 0$$
with $b_{0} = 1$ and $b_{1} = -1$. Thus we get that the $b_i$ are
$$1, -1, 0, 1, -1, 0, 1, \dots$$
We see that $b_{3k} = 1$, $b_{3k+1} = -1$ and $b_{3k+2} = 0$.
Thus $$a_{36} = b_{36}\cdot 36! = 36!$$
