# Maclaurin series question

• December 27th 2009, 12:15 PM
gilyos
Maclaurin series question
How can I expand these expressions with Maclaurin series :

$sin(sinx)$

$sin(cosx)$
• December 27th 2009, 03:45 PM
AnonymitySquared
Quote:

Originally Posted by gilyos
How can I expand these expressions with Maclaurin series :

$sin(sinx)$

$sin(cosx)$

The same way you would expand anything else.

Find $f'(x), f''(x), f'''(x),...,f^n(x)$ using the chain rule.

For example, $\frac{d}{dx} \sin(\cos(x)) = \cos(\cos(x)) \times (-\sin(x)) = - \sin(x) \cos(\cos(x))$

Then from these, find $f'(0), f''(0), f'''(0),...,f^n(0)$

For example, $f'(0) = -\sin(0) \cos((\cos(0)) = 0$

And then insert them into the basic formula:

$f(x) = f(0) + f'(0) x + \frac{f''(0)}{2}x^2 + \frac{f'''(0)}{6} x^3... + \frac{f^n (0)}{n!} x^n$

Which part of this process are you having trouble with?
• December 27th 2009, 11:39 PM
Drexel28
Quote:

Originally Posted by gilyos
How can I expand these expressions with Maclaurin series :

$sin(sinx)$

$sin(cosx)$

This is not easy in the way that you'd think. Finding the exact Maclaurin series for this is not easy. But, note that $\sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^{2n+1}x^{2n+1}}{(2n+1)!}\implies\sin\left(\sin( x)\right)=\sum_{n=0}^{\infty}\frac{(-1)^{2n+1}\sin^{2n+1}(x)}{(2n+1)!}$
• December 28th 2009, 03:29 AM
HallsofIvy
Quote:

Originally Posted by Drexel28
This is not easy in the way that you'd think. Finding the exact Maclaurin series for this is not easy. But, note that $\sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^{2n+1}x^{2n+1}}{(2n+1)!}\implies\sin\left(\sin( x)\right)=\sum_{n=0}^{\infty}\frac{(-1)^{2n+1}\sin^{2n+1}(x)}{(2n+1)!}$

But that last is not a Maclaurin series. What is your point?
• December 28th 2009, 09:22 PM
Drexel28
Quote:

Originally Posted by HallsofIvy
But that last is not a Maclaurin series. What is your point?

I never said it was? I was merely pointing out another possible root of attack.