Proof:Taylor expansion of inverse trigonometric functions

I find it quite difficult to remember the Taylor expansion of inverse trigonometric functions.Actually in school we have been just taught the series (for finding limits in calculus without teaching us the proof).

Can someone provide a short proof of these expansions (mainly for inverse trigonometric functions) so that I can derive them even if I forget.

• Even a link will suffice.
– user220382
Jun 26 '15 at 11:18

You can remember the generalized binomial formula

$$(1+x)^p=1+px+\frac{p(p-1)x^2}2+\frac{p(p-1)(p-2)x^3}{3!}+\frac{p(p-1)(p-2)(p-3)x^4}{4!}\cdots,$$

and apply it for $p=-1$ (fairly easy)

$$\frac1{1+x}=1-x+x^2-x^3+x^4\cdots,$$

or $p=-1/2$,

$$\frac1{\sqrt{1+x}}=1- \frac{x}2+ \frac{3\,x^2}{2^2\cdot 2}- \frac{3\cdot5\,x^3}{2^3\cdot3!}+ \frac{3\cdot5\cdot7\,x^4}{2^4\cdot4!}\cdots.$$

Then for the $\arctan$, substitute $x^2$ for $x$ and integrate.

$$\int\frac{dx}{1+x^2}=x-\frac{x^3}3+\frac{x^5}5-\frac{x^7}7+\frac{x^9}9\cdots,$$ and for the $\arcsin$ substitute $-x^2$ for $x$ and integrate $$\int\frac{dx}{\sqrt{1-x^2}}=x+ \frac{x^3}{2\cdot3}+ \frac{3\,x^5}{2^2\cdot 2\cdot5}+ \frac{3\cdot5\,x^7}{2^3\cdot3!\cdot7}+ \frac{3\cdot5\cdot7\,x^9}{2^4\cdot4!\cdot9}\cdots.$$

For the same "price", you also get $\ln$, the natural logarithm,

$$\int\frac{dx}{1+x}=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\frac{x^5}5\cdots.$$

Remember the derivatives of the inverse trigonometric functions, whose Taylor expansions are easier to remember: $$(\arctan x)'=\frac{1}{1+x^2}=\sum_{n=0}^\infty(-1)^nx^{2n}.$$ $$(\arcsin x)'=\frac{1}{\sqrt{1-x^2}}=(1-x^2)^{-1/2}=\sum_{n=0}^\infty(-1)^n\binom{-1/2}{n}x^{2n}.$$ Integrate to obtain the desired expansions.

The Taylor series is one of the most fundamental series and are very useful in finding different limits.

Hint: Take the polynomial $$\sum^{\infty}_{i=0} a_i(x-a)^{i}.$$ Look at the derivatives for this polynomial(first derivative, second derivative and so on..)

Look at the first derivative of this polynomial. That is,

$f'(x)=a_1+2a_2(x-a)+3a_3(x-a)^2+4a_4(x-a)^3+.....$ Now let $x=0$, you get that $f'(0)=a_1$.(You can prove this by finding out values of ( $a_1,a_2$...)

Now keep on taking derivatives and check what would happen when $a=x$.

This series for $a=0$ is called the Maclaurin series and is given by $$f(x)=\sum_{n=0}^{\infty}\frac{f^{n} 0}{n!} {x^n}$$

Now when $a\neq 0$ you get the Taylor series

• I could understand a thing :-(....please say a bit more...
– user220382
Jun 26 '15 at 11:22
• Okay let me expand it a bit...
– user210387
Jun 26 '15 at 11:22