# Taylor series prob

1. Apr 17, 2004

### dbzgtjh

Help me out with this Taylor series problem:

The Taylor series for sin x about x = 0 is x-x^3/3!+x^5/5!-... If f is a function such that f '(x)=sin(x^2), then the coefficient of x^7 in the Taylor series for f(x) about x=0 is?

thanks

2. Apr 17, 2004

You can use term-by-term differentiation and integration for Taylor Series. So just integrate the x^6 term for the Taylor Series of f'(x).

Last edited: Apr 17, 2004
3. Apr 20, 2004

### NuPowerbook

Well, since you know that the Taylor series for $$\sin x=x-\frac{x^3}{3!} + \frac{x^5}{5!}-\frac{x^7}{7!}+\cdots$$ then you can just plug in x^2 for x in the Taylor expansion, so it would become:$$\sin x^2=x^2-\frac{x^6}{3!}$$. Now you can integrate $$f'(x)$$ as the taylor approximation, with: $$\int x^2-\frac{x^6}{3!}\,dx$$ which is equal to $$\frac{1}{3}x^3-\frac{1}{7}\cdot\frac{x^7}{3!} = \frac{x^3}{3}-\frac{x^7}{3!\cdot7}$$. So this would make the coefficient $$-\frac{1}{42}$$

4. Apr 20, 2004

### Ebolamonk3y

I got same. $$\frac{-1}{42}$$

I used the $$\sin x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}$$ then you plug in $$x^2$$... the same... and then differentiate... you get $$2x\sinx$$ and you divide both sides with $$2x$$... you get $$\frac{2x^3}{2!}-\frac{4x^6}{4!}+\frac{6x^10}{6!}$$ Integrate... look at $$x^7$$ $$\frac{-4x^7}{7*4!}$$ same as $$\frac{-1}{42}$$