# Thread: Obtain a MacLaurin Series for a function with given conditions for x

1. ## Obtain a MacLaurin Series for a function with given conditions for x

Hi,

The question asks to find a Maclaurin series for the function

f(x) = (x-sinx)/x^3 when x does not = 0

and

f(x) = 1/6 when x = 0

Since sinx = x - x^3/3! + x^5/5! - ...

x-sinx = x^3/3! - x^5/5! + x^7/7! - ...

and (x-sinx)/x^3 = 1/3! - x^2/5! + x^4/7! - ... = 1/6 - x^2/5! + x^4/7! - ...

I noticed that I can make a series out of the second term onward, and in doing so would make f(x) = 1/6 when x = 0. This gives me 1/6 - series(n=0,infinity) ((-1)^(x+1) * x^(2n+2))/(2n+3)!

Would that be my answer? Just curious, because I've never done a problem like this before.

2. ## Re: Obtain a MacLaurin Series for a function with given conditions for x

Your series would just be \displaystyle \begin{align*} \frac{1}{3!} - \frac{x^2}{5!} + \frac{x^4}{7!} - \dots + \dots \end{align*}. I don't know why you're subtracting it from \displaystyle \begin{align*} \frac{1}{6}\end{align*}.

3. ## Re: Obtain a MacLaurin Series for a function with given conditions for x

Originally Posted by Prove It
Your series would just be \displaystyle \begin{align*} \frac{1}{3!} - \frac{x^2}{5!} + \frac{x^4}{7!} - \dots + \dots \end{align*}. I don't know why you're subtracting it from \displaystyle \begin{align*} \frac{1}{6}\end{align*}.
But how would I write that as a series in sigma notation without it being equal to 0 when x = 0? I was assuming the answer had to be in sigma notation.

4. ## Re: Obtain a MacLaurin Series for a function with given conditions for x

It wouldn't be equal to 0 when x = 0, the constant term is $\displaystyle \frac{1}{3!} = \frac{1}{6}$, like we require.

Anyway, in sigma notation, we could write this series as $\displaystyle \sum_{n = 0}^{\infty}\frac{(-1)^{n}x^{2n}}{(2n + 3)!}$.

5. ## Re: Obtain a MacLaurin Series for a function with given conditions for x

Originally Posted by Prove It
It wouldn't be equal to 0 when x = 0, the constant term is $\displaystyle \frac{1}{3!} = \frac{1}{6}$, like we require.

Anyway, in sigma notation, we could write this series as $\displaystyle \sum_{n = 0}^{\infty}\frac{(-1)^{n}x^{2n}}{(2n + 3)!}$.
Oh thanks, I was thinking 0^0 = 0 or undefined; I couldn't really find a definite answer online, does 0^0 = 1? That's why I was confused.

0^0 - Wolfram|Alpha

6. ## Re: Obtain a MacLaurin Series for a function with given conditions for x

No, $\displaystyle 0^0$ is undefined.

7. ## Re: Obtain a MacLaurin Series for a function with given conditions for x

Would that be my answerReplica Handbags,Replica Wallets For Sale
? Just curious, because I've never www.louisvuittonend.com
done a problem like this before.