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I'm trying to calculate the 100th derivative of $$f(x) = \dfrac{1}{1+x^2}$$ at $x=0$.

So far, I've had only found a way to do it, and that is rewriting it as $$f(x) = \dfrac{1}{2i} \bigg(\dfrac{1}{x-i} + \dfrac{1}{x+i}\bigg)$$ and using higher derivative formula for each term in the parentheses. However, my teacher didn't allow the use of complex numbers in calculating derivatives, as we only defined the derivative in the set of real number (with the definition of limit), although I knew that the final number would always be a real number.

I tried to set $x=\tan(t)$, but it also didn't work, because I don't know if there is a chain rule for higher order derivatives.

Is there any way to solve this problem without using complex numbers?

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  • $\begingroup$ Hint: Use power series expansion. $\endgroup$ Oct 1 '18 at 3:50
  • 1
    $\begingroup$ Did you learn about the taylor expansion? $\endgroup$
    – J1U
    Oct 1 '18 at 3:51
  • $\begingroup$ I've just learned about it this morning, so, yes $\endgroup$ Oct 1 '18 at 16:14
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Notice that since $\sum x^n = \frac{1}{1-x} $, then

$$ \frac{1}{1+x^2} = \sum_{n=0}^{\infty} (-1)^n x^{2n} $$

Therefore, $f^{(100)}(0) = 100! \cdot a_{100} = 100! $

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Direct computation is possible. Using the Leibniz rule for products,

$$(1+x^2)f(x)=1$$

$$2xf(x)+(1+x^2)f'(x)=0$$

$$2f(x)+2\,2xf'(x)+(1+x^2)f''(x)=0$$

$$3\,2f'(x)+3\,2xf''(x)+(1+x^2)f'''(x)=0$$

$$6\,2f''(x)+4\,2xf'''(x)+(1+x^2)f''''(x)=0$$

$$10\,2f'''(x)+5\,2xf''''(x)+(1+x^2)f'''''(x)=0$$

$$\cdots$$

Then

$$f(0)=1$$

$$f'(0)=0$$

$$f(0)+f''(0)=0$$

$$2f'(0)+f'''(0)=0$$

$$6\,2f''(0)+f''''(0)=0$$

$$10\,2f'''(0)+f'''''(0)=0$$

$$\cdots$$

From this, every even derivative is zero and

$$(2k+1)(2k-1)f^{(2k-1)}(0)+f^{(2k+1)}(0)=0.$$

This recurrence is easy to solve.

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