# Computing Taylor series for $\arctan(x^2 -1)$ by performing substitution on an existing Taylor series for $\arctan x$

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?

• Welcome to Mathematics SE. Take a tour. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. May 19 at 6:45
• 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 $-1$. May 19 at 7:15
• Hint Differentiate, use the binomial theorem, integrate and don(t forget the integration constant. May 19 at 9:32

$$\arctan(x^2-1)=f(x^2)$$ 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 $$\arctan(z-1)=-\frac\pi4+\sum_{n=1}^\infty\left(2^{-n/2}\sin\frac{n\pi}4\right)\frac{z^n}n.$$