Taylor series prob

[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

[cookiemonster]

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).

cookiemonster

[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}

[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}