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I am really sorry if this question sounds stupid or too obvious. Even though I know how to apply all of these when calculating limits or sums of series, I still have questions that I need to answer.

1) Taylor's theorem says that a k-times differentiable function can be approximated by a k-th order Taylor's polynomial in the neighborhood of some given point.

Okay, this seems absolutely clear. We have a function and we want to linearize this function near some $a$.

2) Taylor's series is a representation of a function that is infinitely differentiable at a real or complex number a.

Should we say this representation is true in the neighborhood of $a$?

Let's say we want to express $e^x$ as a power series (at $a=0$). Do we need to say that this power series holds in the neighborhood of zero? Or just at $0$?

3) If I understand correctly, a power series is technically the same as a Taylor series, but most of the time we are interested in using Taylor's series. Is that correct?

Can you please clarify the questions above for me?

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  • $\begingroup$ Different series have different radius of convergence. $\endgroup$ – Karl Aug 22 '18 at 17:31
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Given any sequence $n\mapsto a_n\in{\mathbb C}$ $(n\geq0)$ of complex numbers you can put on the blackboard the formal power series $$\sum_{n=0}^\infty a_n z^n\ .\tag{1}$$ Such an object makes sense, e.g., in combinatorics, even if this series does not converge for any $z\ne0$. If there is a $\rho>0$ such that $(1)$ converges for $|z|<\rho$ then $$f(z):=\sum_{k=0}^\infty a_n z^n\qquad(|z|<\rho)$$ is a well defined function on the open disc of radius $\rho$. If you are just interested in the real environment and all $a_n\in{\mathbb R}$ then the real-valued function $f$ is well defined on the interval $\>]{-\rho},\rho[\>$. It is then proven in Calculus 102 that this $f$ is in fact infinitely differentiable, and that $$a_n={f^{(n)}(0)\over n!}\qquad(n\geq0)\ .$$ This is saying that the given series is in fact the Taylor series at $0$ of the function it defines: $$f(x)=\sum_{n=0}^\infty{f^{(n)}(0)\over n!} x^n\ .$$ Conversely things are a little different: Given a function $f$ which is $N\geq0$ times differentiable at $0$ we can set up its Taylor polynomial $j_0^Nf$ of degree $N$ at $0$: $$j_0^N f(x):=\sum_{n=0}^N{f^{(n)}(0)\over n!} x^n\ .$$ In Calculus 102 it is shown that this is the unique polynomial $p_N$ of degree $\leq N$ satisfying $$f(x)-p_N(x)=o(x^N)\qquad(x\to0)\ .\tag{1}$$ Since these $j_0^Nf$ approximate $f$ ever better in the neighborhood of $x=0$ when $N\to\infty$ we are led to the question whether in fact $f$ is "represented by its Taylor series", i.e., whether $$f(x)=j_0^\infty f(x):=\sum_{n=0}^\infty{f^{(n)}(0)\over n!} x^n\tag{2}$$ holds in a neighborhood $|x|<\rho$ of $x=0$. In order to answer this question one has to analyze the remainder term $o(x^N)$ in $(1)$ for the particular $f$ under consideration.

An example: If $f(x):=\log(1+x)$ then one needs "Taylor's theorem with remainder" and estimates for $f^{(n)}(x)$ $\bigl(|x|\leq{1\over2}\bigr)$ to prove that $(2)$ is true for this $f$ and $|x|\leq{1\over2}$. One then obtains that indeed $$\log(1+x)=x-{x^2\over2}+{x^3\over3}-\ldots\qquad\bigl(|x|<{1\over2}\bigr)\ .$$ But there are examples of $C^\infty$ functions for which $(2)$ does not hold, e.g., $f(x):=e^{-1/x}$ $(x>0)$ and $:=0$ $(x\leq0)$. I won't go into this.

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Should we say this representation is true in the neighborhood of $a$?

One property of any power series is its "radius of convergence." The Taylor series will converge to the value of the function within the radius of convergence (at least, this will be true for most "normal" functions you'd encounter in an elementary calculus class). Different series will have different radii of convergence.

Let's say we want to express $e^x$ as a power series (at $a=0$). Do we need to say that this power series holds in the neighborhood of zero? Or just at $0$?

In the case of $e^x$, you can show that the associated Taylor series converges on the entire real line (i.e., the radius of convergence is infinite), so the Taylor series "holds" (i.e., converges to the exact value $e^x$) everywhere. (Although convergence will be faster near $0$, which is why you'll want to use the Taylor polynomial approximations only near $0$.)

If I understand correctly, a power series is technically the same as a Taylor series, but most of the time we are interested in using Taylor's series. Is that correct?

A Taylor series is a type of power series. A power series is just some series where the $n^{th}$ term has an $x^n$ involved. A Taylor series is a series of that type, with the coefficients set by the derivatives of some function.

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    $\begingroup$ A Taylor series in general does not necessarily converge to the original function even on its radius of convergence. $\endgroup$ – Ian Aug 22 '18 at 17:35
  • $\begingroup$ @Ian I guess I should probably throw a disclaimer in there about that... $\endgroup$ – BallBoy Aug 22 '18 at 17:37
  • $\begingroup$ Thank you for answering. When calculating limits, when $x->0$, we can use the McLaurin formula where the remainder $R_{n}{x}$ is written in the Peano form. The Taylor theorem says that a function can be represented (conditions)... when $x->a$, i.e in the neighborhood of $a$. I should have provided a different example (other than $e^{x}$). Let's say there's some $f(x)$ for which we can write the corresponding McLaurin series. Even though the McLaurin series is a special case of the Taylor series (when $a=0$), we no longer say that our approximation "works" in the neighborhood of $0$. $\endgroup$ – Don Draper Aug 22 '18 at 17:53
  • $\begingroup$ @TimurSharapov I'm afraid I don't understand your comment. Perhaps you can provide a concrete example that illustrates what you mean? $\endgroup$ – BallBoy Aug 22 '18 at 17:56
  • $\begingroup$ @Y.Forman, for example, when calculating limits where $x->0$, we can write that $\tg{x}=x+\frac{x^{3}}{3}+o(x^{3})$. This is Taylor polynomial for $tgx$ at $a=0$, which "holds" in the neighborhood of $0$. We couldn't use this representation if $x$ approached $1$ instead of $0$. Is that correct? $\endgroup$ – Don Draper Aug 22 '18 at 18:09

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