# Thread: representation of a arccos as taylor expansio

1. ## representation of a arccos as taylor expansio

I'm having trouble expressing arccos(x) as a Maclaurin series. I realized that once you differentiate it, the function will just expand rapidly, so is there a easy way approaching this problem.

2. ## Maybe this is what you are talking about

Originally Posted by lllll
I'm having trouble expressing arccos(x) as a Maclaurin series. I realized that once you differentiate it, the function will just expand rapidly, so is there a easy way approaching this problem.
but did you consider using a similar method to your last one by saying $arcos(x)=\int{\frac{-1}{\sqrt{1-x^2}}}$?

3. Originally Posted by lllll
so is there a easy way approaching this problem.
Yes.

Have you considered the Binomial series - Wikipedia, the free encyclopedia?

4. ## Just wondering

Originally Posted by Krizalid
This is what I was implying in my post...do waht I said then say $(1-x^2)^{\alpha}$ with $\alpha=\frac{-1}{2}$ in this case...so does that work or is the binomial expansion limited to x not $x^{y},y>1$?

5. Originally Posted by lllll
I'm having trouble expressing arccos(x) as a Maclaurin series. I realized that once you differentiate it, the function will just expand rapidly, so is there a easy way approaching this problem.
Yes

$f(x)=\cos^{-1}(x)$ then

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

expand this in a power series using the binomial series then integrate

$\binom{k}{n}=\frac{k(k-1)(k-2) \cdot \cdot \cdot (k-n+1)}{n!}$

$(1-x^2)^{-1/2}=1-\frac{1}{2}x^2+\frac{\frac{-1}{2}\frac{-3}{2}}{2!}x^4- \cdot \cdot \cdot =$

$-f'(x)=\sum_{n=0}^{\infty} \binom{k}{n}(-1)^nx^{2n} \iff f'(x)=\sum_{n=0}^{\infty} \binom{k}{n}(-1)^{n+1}x^{2n}$

Integrating both sides we get

$\cos^{-1}(x)=\int f'(x)dx=\int \sum_{n=0}^{\infty} \binom{k}{n}(-1)^{n+1}x^{2n}$

Finally

$f(x)=\sum_{n=0}^{\infty} \binom{k}{n} \frac{(-1)^{n+1}x^{2n+1}}{2n+1}$

6. Originally Posted by TheEmptySet
Yes

$(1-x^2)^{-1/2}=1-\frac{1}{2}x^2+\frac{\frac{-1}{2}\frac{-3}{2}}{2!}x^4- \cdot \cdot \cdot =$
how did you get this?

7. Originally Posted by lllll
how did you get this?
This is newtons extention of the binomial theorem to fractional and negative values

$
\binom{k}{n}=\frac{k(k-1)(k-2) \cdot \cdot \cdot (k-n+1)}{n!}$

lets expand a few

$\binom{-\frac{1}{2}}{0} =1$ by definition

$\binom{-\frac{1}{2}}{1} =\frac{\frac{-1}{2}}{1!}$

$\binom{-\frac{1}{2}}{2} =\frac{\frac{-1}{2}\frac{-3}{2}}{2!}$

$\binom{-\frac{1}{2}}{3} =\frac{\frac{-1}{2}\frac{-3}{2}\frac{-5}{2}}{3!}$

and so on

8. Sorry I think this is what you need

$(1+x)^k=\sum_{n=0}^{\infty}\binom{k}{n}x^n$

so

$(1+(-x^2))^k=\sum_{n=0}^{\infty}\binom{k}{n}(-x^2)^n=\sum_{n=0}^{\infty}\binom{k}{n}(-1)^nx^{2n}$