# Taylor Series

• Nov 3rd 2010, 09:38 AM
blitze105
Taylor Series
Hello all,
Unfortunately I was absent during one of my classes this week when the professor covered Taylor Series. i have gone through the book and finished all of the first section of the homework. The second section, however, is not in my book... and I can't find it online.

If any one could help me find examples on these types of problems or simply point me in the right direction I would greatly appreciate it. I am NOT asking for answers.

1) Find an expression for the general term of each of the series below. Use n as your index, and pick your general term so that the sum giving the series starts with n=0.

2)find the sum of each convergent series.

Attachment 19569

Sorry for the Image, I just think it looks cleaner and I dont know how to do the text formatting for the problems.
~Blitze
• Nov 3rd 2010, 01:12 PM
Opalg
Quote:

Originally Posted by blitze105
Hello all,
Unfortunately I was absent during one of my classes this week when the professor covered Taylor Series. i have gone through the book and finished all of the first section of the homework. The second section, however, is not in my book... and I can't find it online.

If any one could help me find examples on these types of problems or simply point me in the right direction I would greatly appreciate it. I am NOT asking for answers.

1) Find an expression for the general term of each of the series below. Use n as your index, and pick your general term so that the sum giving the series starts with n=0.

$x^4\sin x^2 = x^6 - \dfrac{x^{10}}{3!} + \dfrac{x^{14}}{5!} - \dfrac{x^{18}}{7!} +\ldots$

2)find the sum of each convergent series.

$1- \dfrac{2^2}{2!} +\dfrac{2^4}{4!} - \dfrac{2^6}{6!} + \ldots + \dfrac{(-1)^n2^{2n}}{(2n)!} + \ldots$

For 1), you are supposed to find an expression for the n'th term in the first series (in a similar style to the expression for the n'th term in the second series above). You may find it helpful to write down the first few terms of the series with their index numbers below them, like this:

$\begin{array}{rcrrrr}x^4\sin x^2 &= &x^6 &- \frac{x^{10}}{3!} &+ \frac{x^{14}}{5!} &- \frac{x^{18}}{7!} \\ (n&=&0&1&2&3)\end{array}$

First thing to notice is that the terms alternate in sign, so the n'th term will have to have a factor $(-1)^n$ in it. Then it has a power of x, starting at $x^6$ in the n=0 term, and going up by 4 at a time. So the n'th term will need a factor $x^{4n+6}$. Finally, there is a factorial in the denominator, starting with 1! in the n=0 term (although that doesn't explicitly appear there), and going up by 2 at a time. So the denominator in the n'th term should be $(2n+1)!$. Having done all that, you should be able to write down the formula for the n'th term (and check that it gives the right answers for n=1,2 and 3).

For problem 2), you are supposed to spot that this is the value of the series $1- \frac{x^2}{2!} +\frac{x^4}{4!} - \frac{x^6}{6!} + \ldots + \frac{(-1)^nx^{2n}}{(2n)!} + \ldots$ when x=2. You are also supposed to recognise this as the Taylor series for a familiar (trigonometric) function. That should enable you to write down the sum of the series.