Taylor Series of an arctangent function

Question

The question is what is the Taylor series expansion of $\tan^{-1}(\frac{2x}{1-x^2})$ at $x=0$? I know that $\tan^{-1}(x)=x-x^3/3+x^5/5+...$

This is complicated to replace $x$ with $2x/1-x^2$ is there a neater way to compute this?

Answer 1

Notice that after simplifications,

$$\left( \arctan \frac {2x} {1-x^2} \right) ' = \frac 1 {1 + \frac {4x^2} {(1-x^2)^2}} \frac {2(1-x^2) - 2x(-2x)} {(1-x^2)^2} = \frac 2 {1+x^2} = (2 \arctan x)' ,$$

whence it follows that

$$\arctan \frac {2x} {1-x^2} - 2 \arctan x = C$$

with $C \in \Bbb R$ a constant. To find out $C$ just evaluate the equality at $x=0$, to get

$$\arctan 0 - 2 \arctan 0 = C ,$$

whence $C=0$ and thus

$$\arctan \frac {2x} {1-x^2} = 2 \arctan x$$

and the required series is obvious now if you now the series of $\arctan$.

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Alternatively, use the formula

$$\tan (u+v) = \frac {\tan u + \tan v} {1 - \tan u \tan v}$$

with $u = v$ to get

$$\tan (2u) = \frac {2 \tan u} {1 - \tan^2 u} ,$$

whence

$$2u = \arctan \frac {2 \tan u} {1 - \tan^2 u}$$

whence, denoting $x = \tan u$, we get

$$2 \arctan x = \arctan \frac {2x} {1-x^2} .$$

These trigonometric computations are a bit formal, since I haven't payed attention to the domain of definition of the above functions, but they can be made perfectly rigorous with just a bit of supplementary work.

Answer 2

You could try using the Taylor series expansion of $\dfrac{2x}{1-x^2}$: $$\dfrac{2x}{1-x^2}=2x\left(1+x^2+x^4+\cdots\right)=2x+2x^3+2x^5+2x^7+\cdots$$ which is valid if $|x|<1$.

Answer 3

Let $f(x)=\arctan(\frac{2x}{1-x^2})$

then

$$f'(x)=\frac{1}{(1+(\frac{2x}{1-x^2})^2)}\frac{2-2x^2+4x^2}{(1-x^2)^2}$$

$$=\frac{2}{1+x^2}$$

thus

$$f(x)=f(0)+2\arctan(x)$$

so ...

Answer 4

As $$\tan(2\arctan x)=\frac{2\tan(\arctan x)}{1-\tan^2(\arctan x)},$$

you simply have

$$2\arctan x=\arctan\frac{2x}{1-x^2}=2\left(1-\frac{x^3}3+\frac{x^5}5-\cdots\right).$$