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.

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

  • $\begingroup$ 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. $\endgroup$ – João Ghignatti Jan 10 '18 at 2:07
  • $\begingroup$ You're welcome. Glad it helped! $\endgroup$ – 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 $$ \int_0^3 f(x) dx=\sum_{k=1}^\infty \frac{(-1)^k(k+1)}{5^k(k+1)!}(-3)^{k}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.