# Thread:

1. ## Confused with this Taylor series problem.

The problem says, "Write down the Taylor series of $Cos(x)$ about $x=0$ and sum the series $\sum_{0}^{inf} (-\pi ^2)^n / (2n)! 9^n$"

First question is, when do you stop writing terms of the series if it doesn't tell you?

Also, what does the second series have to do with the first one? This seems like two totally different problems crammed into one. Am I wrong?

2. Taylor for cosx

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

our series

$\sum_{n=0}^{\infty} \frac{ (-\pi^2)^n}{(2n)! 9^n}$

$\sum_{n=0}^{\infty} \frac{ (-1)^n (\pi^2)^n}{9^n} \frac{1}{(2n)!}$

$\sum_{n=0}^{\infty} \frac{(-1)^n }{(2n)!}\left(\frac{\pi}{3}\right)^{2n}$

ahaa what do you think now

3. I am still not totally sure what they have to do with eachother.

4. the difference is just in second summation $\frac{\pi}{3}$ and in first $x$

so $\sum_{n=0}^{\infty} \frac{(-1)^n }{(2n)!}\left(\frac{\pi}{3}\right)^{2n}= \cos \frac{\pi}{3}= \frac{1}{2}$

5. Originally Posted by Mattpd
The problem says, "Write down the Taylor series of $Cos(x)$ about $x=0$ and sum the series $\sum_{0}^{inf} (-\pi ^2)^n / (2n)! 9^n$"

First question is, when do you stop writing terms of the series if it doesn't tell you?
You don't. A Taylor polynomial stops after a finite number of terms. A Taylor series is an infinite series.

Also, what does the second series have to do with the first one? This seems like two totally different problems crammed into one. Am I wrong?
The Taylor series for cos(x) is $\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}$

The sum you are given is either $\sum_{n=0}^\infty \frac{(-\pi^2)^n}{n!}9^n$ or $\sum_{n=0}^\infty \frac{(-\pi^2)^n9^n}{n!|9^n$ since, in what you wrote, it is not clear whether the $9^n$ is supposed to be in the numerator or denominator.

In either case, notice that we have n! in the denominator, as in the cos(x) series. Now, because of the " $x^{2n}$" in the cos(x) series, try to try powers as "2n". We can write $(-\pi^2)^n= (-1)^n \pi^{2n}$ and, of course, $9^n= 3^{2n}$

That is, depending upon whether the " $9^n$ is in the numerator or denominator we get

$\sum_{n=0}^\infty \frac{(-\pi^2)^n}{n!}9^n= \sum_{n=0}^\infty \frac{(-1)^n}{n!}\left(3\pi\right)^{2n}= cos(3\pi)= -1$
or

$\sum_{n=0}^\infty \frac{(-\pi^2)^n{n! 9^n}= \sum_{n=0}^\infty \frac{(-1)^n}{n!}\left(\frac{\pi}{3}\right)^n= cos(\frac{\pi}{3})= \frac{1}{2}$.
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