# Power Series Representation of Functions, Taylor Series

I'm having some issues trying to understand Taylor Series, Power Series as a representation of a Function, integrate of a Power Series and how it all works (or should work) together. I have a question example where I can't understand how to proceed. Here it is:

Consider the function $f(x) = \sum_{k=1}^\infty \frac{(-1)^k(k+1)}{5^kk!}(x-3)^k$ to any value of x.
1. Find the Taylor Serie of $\int f(x)dx$ at $x = 3$

-- It is given a power series, how am I supposed to integrate it and evaluate at x = 3? Don't you need the function this series is represanting to evaluate it and find the taylor series? How do I even integrate this series?

2. Write $I = \int_0^3f(x)dx$ as a sum of a numerical serie.

-- This one I almost have no idea what it is referring to. Is it the same integral as the first part but evaluated inside 0 and 3?

Thanks to all.

• Perhaps your textbook has a result about "integrate term-by-term" for a power series. – GEdgar Jan 10 '18 at 1:38
• I strongly suspect that your text, probably in the same section as this problem, has the result that $\int \sum a_nx^n dx= \sum \int a_n x^n dx= \sum\frac{1}{n+1}a_nx^{n+1}$. – user247327 Jan 10 '18 at 1:42

Someone else can give a more detailed answer. Here's a quick explanation

1. The Taylor series around the point $x=3$ does not mean evaluating the function at $x=3$, but rather a local approximation for $f(x)$ for values of $x$ that are close to $3$. This is a function of $x$, not a single number. What the question is asking, in simpler terms:

If $f(x)$ has an anti-derivative so that $F'(x) = f(x)$, what is the Taylor series expansion of $F(x)$ for values of $x$ near $3$?

Since $f(x)$ already has a Taylor expansion around $x=3$ (every term is a power of $(x-3)$), all you need to do is integrate this series representation term by term.

1. This question asks you to evaluate the series at specific points. Once you have a series expression for $F(x)$ computed in part 1, you'll need to compute $F(3) - F(0)$. You can do this by just plugging in the appropriate values of $x$.

There are more intricacies to this, such as the notion of the radius of convergence of the Taylor series. While the Taylor series is generally very good local approximation of a function (say for values close to $x=3$, like $x=3.1$ or $x=2.9$), we want to know how far we can stray from the central point before it becomes a bad approximation (is it still accurate at $x=0$? How about at $x=10$?). Some series will converge everywhere (if taken enough terms), while some only for a limited range (for example $0 < x < 6$). This will depend on the specific function.

If the Taylor series of a function is convergent everywhere, we can treat it as an equivalent form (we can interchange $f(x)$ with its Taylor series in computations). This is what the exercise seems to be doing, and what might be the source of your confusion.

• I thank you so much for this answer. My calculus book does have integrate term-by-term and all of that, but I wasn't able to get some points you explained. – João Ghignatti Jan 10 '18 at 2:07
• You're welcome. Glad it helped! – Dylan Jan 10 '18 at 2:28

Integrate the power series $$f(x) = \sum_{k=1}^\infty \frac{(-1)^k(k+1)}{5^kk!}(x-3)^k$$ term by term. Since $$\int (x-3)^kdx= \frac {1}{k+1} (x-3)^{k+1}+c$$ we get $$\int f(x) dx= \sum_{k=1}^\infty \frac{(-1)^k(k+1)}{5^k(k+1)!}(x-3)^{k+1}+C$$ Where C is a constant.

Evaluating at x=3 implies that the indefinite integral $$\int f(x) dx= C$$ For the definte integral we evaluate $$\int_0^3 (x-3)^kdx= \frac {1}{k+1} (-3)^{k}$$ Thus $$\int_0^3 f(x) dx=\sum_{k=1}^\infty \frac{(-1)^k(k+1)}{5^k(k+1)!}(-3)^{k}$$