Find the Maclaurin Series for the function $f(x) = x^2e^{x^2}$ and determine the interval on which it is valid.
We can build up this series from one we alread know, namely the series for $e^x$.
We can plug $x^2$ into $\displaystyle \sum_{n=0}^\infty \frac{x^n}{n!}$ and then multiply by $x^2$ to get the series we want.
For the interval where it is valid, apply the ratio test to this series. What happens in the limit?
Since the limit goes to zero, this series converges everywhere. We also know that the series for $e^x$ converges everywhere.
We can write the series as $\displaystyle \sum_{n=0}^\infty \frac{1}{n!} x^{2n+2}$ and this is valid for all $x$.
Find the Taylor series of $f(x) = e^{3x}$ centered at $c = -1$ and determine the interval on which it is valid.
We can't use our known series for $e^x$ here, because we aren't centered at zero. So we need to work this out using derivatives.
Take a few derivatives of $f(x) = e^{3x}$ and see if you can find a pattern for the derivatives.
The pattern is that $\displaystyle f^{(k)}(x) = 3^k e^{3x}$ so that $\displaystyle f^{(k)}(-1) = 3^k e^{-3}$. Then we can write the series.
Use the ratio test to determine where this series is valid. It should look like the series for $e^x$
The Taylor Series is $\sum_{n=0}^\infty \frac{3^n}{n!}e^{-3}(x+1)^n$ and it converges for all x.
Find the Taylor Series of $\displaystyle f(x) = \frac{1}{(x - 8)^2}$ centered at $c=4$ and determine the interval where it is valid.
You could work this out directly using derivatives, or you could work with the series and relate it to something we already know about. (I would recommend the latter.)
Consider the function $g(x) = \frac{1}{8-x}$. What is the derivative of this function? How can I write this function as a power series centered at 4? To do this you'll want to get it into the form $\frac{1}{1 - a(x-4)}$ for some constant $a$.
Since $g(x) = \frac{1}{4} \frac{1}{1 - \frac{1}{4}(x-4)}$, we can write $\displaystyle g(x) = \frac{1}{4} \sum_{n=0}^\infty \frac{1}{4^n}(x-4)^n$.
Then, we can differentiate this series term by term to get the series for $f(x)$. For the interval where it is valid, we know where the series for $g(x)$ is valid.
The Taylor Series for this $f$ is $\displaystyle \frac{1}{4} \sum_{n=1}^\infty \frac{n}{4^{n}} (x-4)^{n-1}$ and it is valid on $(0,8)$.
Express the definite integral $\displaystyle \int_0^1 \tan^{-1}(x)\ dx$ as an infinite series. This will be a series of numbers because the answer should be a number.
To do this, we need to write $\tan^{-1}(x)$ as a power series, and then integrate it. How do we get to inverse tangent as a power series?
We can find the power series for $\displaystyle \frac{1}{1+x^2}$ using the tricks from 10.6, and then integrate this to get to inverse tangent.
Since $\displaystyle \frac{1}{1+x^2} = \sum_{n=0}^\infty (-1)^n x^{2n}$, we know that $\tan^{-1}(x) = \sum_{n=0}^\infty \frac{(-1)^n}{2n}x^{2n+1}$. We then need to integrate this one more time for our desired series.
The series we get is $\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{2n(2n+1)}x^{2n+2}$. Then we can plug in 1 to evaluate this series.
This integral is $\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{2n(2n+1)}$.
Use Taylor Series to find the value of $f^{(9)}(0)$ for the function $f(x) = xe^{-x^2}$.
These derivatives show up as the coefficients of the Taylor Series, so if we find the Taylor Series for this function, we can get the values of the derivatives at zero.
The Taylor Series for $f(x)$ is $\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{n!}x^{2n+1}$.
To get the coefficient of the $x^9$ term, we need $n=4$. The coefficient of this term is $\frac{(-1)^4}{4!}$.
The coefficient of this term is $\frac{f^{(9)}(0)}{9!}$, and we can then solve for $f^{(9)}(0)$.
$\displaystyle f^{(9)}(0) = \frac{9!}{4!} = 9\cdot 8 \cdot 7 \cdot 6\cdot 5 = 15120$
Find the Maclaurin Series for the function $f(x) = (x^2 + 1) \cos(2x)$.
Find the Taylor Series for $f(x) = \sin{x}$ centered at $c=\pi$.