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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!

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    $\begingroup$ Hint: Complete the square - some stuff cleans up. $\endgroup$ Commented Jul 2, 2017 at 3:20
  • $\begingroup$ You can try other way round think in an infinite series, and its nth derivative and its setting it to zero gives you coefficient in series, so this just become a power series question.. $\endgroup$
    – Ash Pd
    Commented Jul 2, 2017 at 3:22
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    $\begingroup$ @ Sean Roberson How would I "complete the square"? $\endgroup$
    – chrismc
    Commented Jul 2, 2017 at 3:28
  • $\begingroup$ May I ask why anyone would care about anything past the fifth derivative, or past the second or third if it's not physics-related? $\endgroup$
    – DonielF
    Commented Jul 3, 2017 at 2:50
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    $\begingroup$ @DonielF: you can ask why, but that's the difference between scientists/engineers and mathematicians... $\endgroup$
    – smci
    Commented Jul 3, 2017 at 5:19

6 Answers 6

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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 $36!$ .

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    $\begingroup$ How did you get from $f(x) = \frac{1-x}{1-x^3}$ to $f(x) = (1-x)\sum_{k=0}^{\infty}x^{3 k}$? $\endgroup$
    – chrismc
    Commented Jul 2, 2017 at 3:52
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    $\begingroup$ @chrismc You apply the formula for the geometric series to $$\frac{1}{1-x^3}$$ where instead of $x$ you have $x^3$, that's why you get $\left(x^3\right)^k = x^{3k}$ in the sumand. $\endgroup$ Commented Jul 2, 2017 at 4:17
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    $\begingroup$ What if $|x| \ge 1$? $\endgroup$
    – AHB
    Commented Jul 2, 2017 at 11:53
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    $\begingroup$ @AHB This is beside the point, since we only care about small $x$. The reason we can do so is the uniqueness of Maclaurin (or Taylor) expansion around a particular point. $\endgroup$
    – Vim
    Commented Jul 2, 2017 at 12:46
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    $\begingroup$ @AHB - it's the radius of convergence if you're interested in further reading. This particular Maclaurin series will not converge by adding more terms if $|x| > 1$ because $x$ represents the ratio in the geometric series. So if you have an infinite series with terms $a_n=x^n$ and the ratio, $x$, is 2, starting at $n=0$, for example, you'd be adding up the infinite sum 1 + 2 + 4 + 8 + 16 + 32, etc... which diverges to infinity. For any value of $|x|≥1$. $\endgroup$
    – rb612
    Commented Jul 4, 2017 at 18:45
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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 $\omega^3 = 1$.

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    $\begingroup$ What? Is this jokingly supposed to be hard to understand, like an obfuscated C contest but for math? $\endgroup$ Commented Jul 2, 2017 at 19:00
  • $\begingroup$ Been a while since I've done partial fractions, but ${(x-\omega)(x-\omega^2)} = {x^2+x+1}$? Would've though the middle terms signs don't work, as they'd be negative $\endgroup$ Commented Jul 2, 2017 at 22:50
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    $\begingroup$ @JeopardyTempest $(x-\omega)(x - \omega^2) = x^2 - (\omega + \omega^2) x + \omega^3 = x^2 - (-1) x + 1 = x^2 + x + 1$. $\endgroup$
    – jgsmath
    Commented Jul 2, 2017 at 23:56
  • $\begingroup$ Indeed, yep, got it, my mistake... I thought you'd later said omega = 1, not omega^3 = 1. Whoops $\endgroup$ Commented Jul 3, 2017 at 6:13
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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: $$\left(\frac{1}{x^2+x+1}\right)^{(36)}_{x=0}=\left(\frac{1}{\left(x+\frac{1}{2}+\frac{\sqrt3}{2}i\right)\left(x+\frac{1}{2}-\frac{\sqrt3}{2}i\right)}\right)^{(36)}_{x=0}=$$ $$=\left(\frac{1}{\sqrt3i}\left(\frac{1}{x+\frac{1}{2}-\frac{\sqrt3}{2}i}-\frac{1}{x+\frac{1}{2}+\frac{\sqrt3}{2}i}\right)\right)^{(36)}_{x=0}=$$ $$=\frac{1}{\sqrt3i}\left(\frac{36!}{\left(x+\frac{1}{2}-\frac{\sqrt3}{2}i\right)^{37}}-\frac{36!}{\left(x+\frac{1}{2}+\frac{\sqrt3}{2}i\right)^{37}}\right)_{x=0}=$$ $$=\frac{1}{\sqrt3i}\left(\frac{36!}{\left(\frac{1}{2}-\frac{\sqrt3}{2}i\right)^{37}}-\frac{36!}{\left(\frac{1}{2}+\frac{\sqrt3}{2}i\right)^{37}}\right)=\frac{1}{\sqrt3i}\left(\frac{36!}{\frac{1}{2}-\frac{\sqrt3}{2}i}-\frac{36!}{\frac{1}{2}+\frac{\sqrt3}{2}i}\right)=36!$$

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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"?

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  • $\begingroup$ Hint 2 basically leads to my answer. Which doesn't look particularly clean, though. $\endgroup$
    – Vim
    Commented Jul 2, 2017 at 3:35
  • $\begingroup$ Yeah...that's why I put in hint 3, which makes it much nicer. $\endgroup$ Commented Jul 2, 2017 at 3:36
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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.

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  • $\begingroup$ Shouldn't the upper limit of the summation be infinity since your are writing a taylor series for the function? $\endgroup$
    – chrismc
    Commented Jul 2, 2017 at 3:36
  • $\begingroup$ @chrismc alright, maybe a more proper terminology is Maclaurin series. $\endgroup$
    – Vim
    Commented Jul 2, 2017 at 3:37
  • $\begingroup$ So would the upper limit of the summation be or not be infinity? I'm kind of confused. $\endgroup$
    – chrismc
    Commented Jul 2, 2017 at 3:38
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    $\begingroup$ Oh ok sorry. What does the o mean? $\endgroup$
    – chrismc
    Commented Jul 2, 2017 at 3:44
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    $\begingroup$ @chrismc please google "little-o notation". $\endgroup$
    – Vim
    Commented Jul 2, 2017 at 3:45
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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 $$a_{36} = b_{36}\cdot 36! = 36!$$

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