Summing Taylor Series: Tips & Tricks

[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.

 

[HallsofIvy]

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.

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?

 

[pbuk]

(edit: didn't notice HallsOfIvy had already answered)

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?

 

[Cpt Qwark]

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?

(edit: didn't notice HallsOfIvy had already answered)

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?

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?

 

[pbuk]

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

 

[pbuk]

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!