How can I expand these expressions with Maclaurin series :

$\displaystyle sin(sinx)$

$\displaystyle sin(cosx)$

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- Dec 27th 2009, 12:15 PMgilyosMaclaurin series question
How can I expand these expressions with Maclaurin series :

$\displaystyle sin(sinx)$

$\displaystyle sin(cosx)$ - Dec 27th 2009, 03:45 PMAnonymitySquared
The same way you would expand anything else.

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

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

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

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

And then insert them into the basic formula:

$\displaystyle 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? - Dec 27th 2009, 11:39 PMDrexel28
This is not easy in the way that you'd think. Finding the exact Maclaurin series for this is not easy. But, note that $\displaystyle \sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^{2n+1}x^{2n+1}}{(2n+1)!}\implies\sin\left(\sin( x)\right)=\sum_{n=0}^{\infty}\frac{(-1)^{2n+1}\sin^{2n+1}(x)}{(2n+1)!}$

- Dec 28th 2009, 03:29 AMHallsofIvy
- Dec 28th 2009, 09:22 PMDrexel28