Thread: Maclaurin series for x²/(1- cos x)?

Okay, so I have a function defined like this:
f(x) = x^2 / (1 - cos x), when x is not 0
f(0) = 2

I'm supposed to find Maclaurin series for this, but I can't figure out how I should do it. All derivatives in the point x=0 have the value 0, so I'm only left with 2 + 0 + 0 + 0...
I really doubt that this is the answer I'm looking for.

Originally Posted by Koaske

Okay, so I have a function defined like this:
f(x) = x^2 / (1 - cos x), when x is not 0
f(0) = 2

I'm supposed to find Maclaurin series for this, but I can't figure out how I should do it. All derivatives in the point x=0 have the value 0, so I'm only left with 2 + 0 + 0 + 0...
I really doubt that this is the answer I'm looking for.

Hi Koaske.

My understanding is that to get f'(0), you can't use the formula for f'(x) obtained by using the quotient rule, etc., because this only applies when $\displaystyle x \neq 0$. Instead, you need to look at right- and left-derivatives, defined by

$\displaystyle lim_{h->0^+}(f(a+h)-f(a))/h$

and

$\displaystyle lim_{h->0^-}(f(a+h)-f(a))/h$

respectively, at the point x=a=0. If both exist and are equal, then that is the value of the derivative at that point. I worked some of it on paper and had to use L'Hopital's rule, then got lazy and switched to Mathematica, anyway I found that f'(0)=0, but f''(0)=1/3, and I didn't try to find out the higher order derivatives.

Hope that helps.

EDIT: It seems that f(x) is infinitely differentiable, which makes consideration of right- and left-derivatives unnecessary. So to get that f''(0)=1/3, it is enough to evaluate $\displaystyle lim_{x->0}f''(x)=1/3$.

EDIT 2: Did the problem require you to characterize all terms of the Maclaurin series, or just some of the first terms? Because even if I'm completely lazy and let Mathematica do all the work for me, I still don't see a pattern to characterize the terms...

$\displaystyle \{f^{(n)}\} = 0, \dfrac{1}{3}, 0, \dfrac{1}{5}, 0, \dfrac{5}{21}, 0, \dfrac{7}{15}, 0, \dfrac{15}{11}, 0, \dfrac{7601}{1365}, ... \ \text{for}\ n = 1, 2, 3, ...$

Let's write...

$\displaystyle \frac{x^{2}}{1-\cos x} = \sum_{n=0}^{\infty} a_{n}\cdot x^{n}$ (1)

... and then we search a way to find the $\displaystyle a_{n}$. Remembering that is...

$\displaystyle \cos x = \sum_{k=0}^{\infty} (-1)^{k}\cdot \frac{x^{2k}}{(2k)!}$ (2)

... with simple steps we obtain...

$\displaystyle \frac{1 - \cos x}{x^{2}} = \sum_{k=0}^{\infty} (-1)^{k}\cdot \frac{x^{2k}}{(2k+2)!} = \frac{1}{2} - \frac{x^{2}}{24} + \frac{x^{4}}{720} - \dots$ (3)

Now the $\displaystyle a_{n}$ are found imposing the conditions ...

$\displaystyle \sum_{n=0}^{\infty} a_{n}\cdot x^{n} \cdot \sum_{k=0}^{\infty} (-1)^{k}\cdot \frac{x^{2k}}{(2k+2)!} = 1$ (4)

... and solving for the $\displaystyle a_{n}$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$