Taylor Series Problem Solved: Coefficient of x^7

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

 

[Ebolamonk3y]

I got same. [tex]\frac{-1}{42}[/tex]

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