Compute the first four non-zero terms in the Taylor series for the function $\arctan(x^2 -1)$ centred about the point $a = 0$.

I think there is easy way to solve this, starting from $$\arctan x \approx x - \frac 1 3 x^3 + \frac 1 5 x^5 + \dots $$ instead of differentiating step by step, which is a lot of work.

But when I substitute $x^2 -1$ into the Taylor series for $\arctan x$, the answer is different from the correct answer and I'm not sure why. Can't we find a Taylor series by making use of an existing Taylor series?

  • 2
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    $\begingroup$ I don't see how plugging in is much easier, you have to evaluate an infinite series for each term since the coefficient of $x^{2n}$ will have contributions from each $(x^2-1)^k$ for $k \ge n$. Second, and maybe the bigger issue is that if $x$ is near $0$ then $x^2 - 1$ is near $-1$, while the series for $\arctan$ you're using is around $0$ and (I think) has radius of convergence $1$, so is not convergent on any neighborhood of $-1$. $\endgroup$
    – ronno
    May 19 at 7:15
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    $\begingroup$ Hint Differentiate, use the binomial theorem, integrate and don(t forget the integration constant. $\endgroup$ May 19 at 9:32

1 Answer 1


$\arctan(x^2-1)=f(x^2)$ where $f(z)=\arctan(z-1)$ and $$f'(z)=\frac1{1+(1-z)^2}.$$ To find the power series for $f'(z)$ and hence for $f(z)$, use partial fractions $$f'(z)=\frac1{2i}\left(\frac1{1-i-z}-\frac1{1+i-z}\right)$$ and the geometric series, or a ready-to-use formula (obtained the same way): $$\frac1{1-2r\cos\phi+r^2}=\sum_{n=1}^\infty\frac{\sin n\phi}{\sin\phi}r^{n-1}\qquad(|r|<1)$$ at $r=z/\sqrt2$ and $\phi=\pi/4$. Integrating the resulting series, we get $$\arctan(z-1)=-\frac\pi4+\sum_{n=1}^\infty\left(2^{-n/2}\sin\frac{n\pi}4\right)\frac{z^n}n.$$

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    $\begingroup$ (As explained in the comments to the OP, the substitution is a bad idea.) $\endgroup$
    – metamorphy
    May 19 at 10:26

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