Taylor Series Expansion of f(x) at 0

[esrever10]

Homework Statement

Expand the function $f(x)=\sin x / (\cosh x + 2)$ in a Taylor series around the origin going up to $x^3$

Relevant Equations

$f(x)=f(a)+f^{(1)}(a)(x-a)+\frac{1}{2!}f(2)(a)(x-a)^2+...$

First I got $f(0)=0$,

Then I got $f'(x)(0)=\frac{\cos x(2+\cosh x)-\sin x\sinh x}{(2+\cosh x)^2}=1/3$

But when I tried to got $f''(x)$ and $f'''(x)$, I felt that's terrible, If there's some easy way to get the anwser?

 

[Orodruin]

Please show your work.

 

[Office_Shredder]

Please show your work.

I don't get this, they showed us how they computed the first term of the series.

Esrever, a neat trick is you can write out the numerator and denominator as series, then apply $\frac{1}{1-y}=1+y+y^2...$ to the denominator where $y$ needs to be all the stuff that goes to zero. Then a bit of algebra on what's left gets you the solution.

This is not that much easier, but probably feels more fulfilling than computing tedious derivatives.

 

[Orodruin]

I don't get this, they showed us how they computed the first term of the series.

If you look at the last edit time of the OP you will see that it is later than when I posted. The original OP did not include this.

 

[Office_Shredder]

If you look at the last edit time of the OP you will see that it is later than when I posted. The original OP did not include this.

Sorry, my mistake for missing that.

 

[FactChecker]

If you find the series for sin(x) and cosh(x) out to enough terms, then you can do regular multiplication and long division of polynomials.

 

[pasmith]

You already have f(0) and f'(0). You (should) know that f''(0) = 0 because f is odd, so all you need is f'''(0). What you don't need is the general expression for f'''(x) in terms of x; an expression for f'''(0) in terms of f(0), f'(0) and f''(0) will suffice.

Start with <br /> (2 + \cosh x)f(x) = \sin x and differentiate both sides three times using the product rule: <br /> \sum_{n=0}^3 \binom{3}{n} f^{(n)}(x) \frac{d^{3-n}}{dx^{3-n}}(2 + \cosh x) = -\cos x. Now substitute x = 0 and solve for f&#039;&#039;&#039;(0).

 

[Orodruin]

I'd probably go with the method of post #3. I do find it simpler than brute forcing by computing the derivatives.