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I want the taylor series expansion around some value $a$ of the function $f(x)=\frac{1}{1+x^2}$. I used the general formula \begin{eqnarray} f(x) = f(a) + \sum_{n=1}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n \end{eqnarray} But unfortunately, I cannot compute any general formula for $f^{(n)}(a)$. The first derivative is $$ f^{(1)}(x)= -\frac{2x}{(1+x^2)^2}.$$The second derivative is $$ f^{(2)}(x)= \frac{6x^2-2}{(1+x^2)^3}.$$The third derivative is $$ f^{(3)}(x)= \frac{24x(x^2-1)}{(1+x^2)^4}.$$The fourth derivative is $$ f^{(4)}(x)= -\frac{24(5x^4-10x^2+12)}{(1+x^2)^5}$$. The fifth derivative is $$ f^{(5)}(x)= \frac{240x(3x^4-10x^2+3)}{(1+x^2)^5}$$.

What is the $n$-th derivative of the function for working with the above taylor series which I want to use to prove something?

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    $\begingroup$ your second derivative and your fourth derivative are equal, are you sure that's the case? $\endgroup$ – AmateurMathPirate Jul 8 '18 at 10:44
  • $\begingroup$ @AmateurMathPirate, Thank you. I correct it. $\endgroup$ – user85361 Jul 8 '18 at 11:02
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$$2f(x)=\frac1{1+ix}+\frac1{1-ix}.$$ Therefore $$2f^{(n)}(x)=\frac{(-i)^nn!}{(1+ix)^{n+1}}+\frac{i^nn!}{(1-ix)^{n+1}}.$$

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  • $\begingroup$ Thank you but I want real values and also around a value $a$. $\endgroup$ – user85361 Jul 8 '18 at 8:39
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    $\begingroup$ That's fine; whenever $x$ is real, then so is $f^{(n)}(x)$. @user85361 $\endgroup$ – Lord Shark the Unknown Jul 8 '18 at 8:41
  • $\begingroup$ There isn't any convergence requirement like $\vert x\vert <1$? $\endgroup$ – user85361 Jul 8 '18 at 8:43
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    $\begingroup$ @user85361 As you can see, this formula is valid for all real $x$. $\endgroup$ – Lord Shark the Unknown Jul 8 '18 at 8:44
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Another form of the $n$-th derivative: $$f^{(n)}(x)=(-1)^n n!\frac{\sin((n+1)\cot^{-1} x)}{(1+x^2)^{(n+1)/2}}.$$ It's easy to prove by induction.

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  • $\begingroup$ Could we find a tight upper bound for this function or lower bound when the term is negative? $\endgroup$ – user85361 Jul 8 '18 at 11:04
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    $\begingroup$ could you please provide a reference for this excellent equation? $\endgroup$ – user85361 Jul 8 '18 at 12:09

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