# Taylor Series

1. Jun 16, 2015

### Cpt Qwark

Expanding the series to the $$n^{th}$$ derivative isn't so hard, however I'm having trouble with the summation. Any tips for the summation?
e.g. taylor series for $$sinx$$ around x=0 in summation notation is $$\sum^\infty_{n=0} \frac{x^{4n}}{2n!}$$
Thanks.

2. Jun 16, 2015

### HallsofIvy

Staff Emeritus
No, it isn't. For one thing, sin(x) is an odd function while your series includes only even power of x. The Taylor's series for sin(x) about x= 0 is $$\sum_{n=o}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}$$. What you have appears to be the Taylor's series, about x= 0, for $cos(x^2)$, except that the denominator should be (2n)! rather than 2n!.

In any case, what do you mean "having trouble with the summation". What are you trying to do?

3. Jun 16, 2015

### MrAnchovy

No, the Taylor series sum around x=0 (i.e. the Maclaurin series sum) for $\sin x$ is $$\sum_{k=0}^{\infty} \frac{(-1)^k x^{(1+2 k)}}{(1+2 k)!}$$. How did you get to the expression you wrote?

Last edited: Jun 16, 2015
4. Jun 16, 2015

### Cpt Qwark

Yeah sorry turns out it was mistook for another expression.
Anywas, what I meant was I had trouble rewriting the taylor/maclaurin series with a summation notation (Σ). Are there supposed to be patterns that you're supposed to recognise (such as the negative sign for sine and cosine functions) or something?

5. Jun 16, 2015

### MrAnchovy

I'm not always in favour of Khan Academy but this might help.

6. Jun 16, 2015

### MrAnchovy

Or is it just getting from $x - \frac{x^3}{3!} +\frac{x^5}{5!} -\frac{x^7}{7!} +\frac{x^9}{9!} - ...$ to the summation formula that is giving you problems?

If so then yes, you need to practice recognising parts of terms like this:
• first note you can always write $x$ as $\frac{x^1}{1!}$
• now notice you have odd numbers 1, 3, 5, 7, 9...: you can generate these with $2k + 1$ - that gives you $\frac{x^{2k+1}}{(2k+1)!}$
• now you just need the alternating + and - signs: -1 to an even power is 1 and to an odd power is -1 so, making sure you start off with the right one (you want the 0th term to have $1 = (-1)^0$ not $-1 = (-1)^{0+1}$) you have $(-1)^k$
• put them all together, add the sum remembering to go from $k=0$ - full marks!