# Taylor Expansion

• Dec 8th 2009, 07:56 PM
jzellt
Taylor Expansion
What is the Taylor Expansion of:

sin(sinx) at x=0 up to terms of order 3 ?

I know that the Taylor of sin(x) is: x - x^3/3 + ...

But then what is the Taylor of sin(x - x^3/3 + ...) ?

Thanks.
• Dec 8th 2009, 08:16 PM
Prove It
Quote:

Originally Posted by jzellt
What is the Taylor Expansion of:

sin(sinx) at x=0 up to terms of order 3 ?

I know that the Taylor of sin(x) is: x - x^3/3 + ...

But then what is the Taylor of sin(x - x^3/3 + ...) ?

Thanks.

$\displaystyle \sin{X} = X - \frac{X^3}{3!} + \frac{X^5}{5!} - \frac{X^7}{7!} + \dots - \dots$.

In your case, $\displaystyle X = \sin{x}$.

So $\displaystyle \sin{(\sin{x})}$

$\displaystyle = \left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots - \dots\right)$

$\displaystyle - \frac{\left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots - \dots\right)^3}{3!}$

$\displaystyle + \frac{\left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots - \dots\right)^5}{5!}$

$\displaystyle - \frac{\left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots - \dots\right)^7}{7!} + \dots - \dots$

Very long I know, and some simplifying needs doing as well...
• Dec 9th 2009, 05:44 AM
HallsofIvy
The fact that you only need "to order 3" simplifies a lot!