0
$\begingroup$

Consider the power series:

$$ \sum_{n=0}^{\infty} a_n (x - c)^n $$

Now consider its derivative:

$$ \sum_{n=1}^{\infty} n a_n (x - c)^{n-1} $$

We can say at first that the Radius of Convergence for the original power series is

$$ R = \lim_{n \to \infty} |a_{n+1} / a_{n}| $$

(via the Ratio Test).

On the other hand, can we not also say that the radius of convergence for the derivative of the power series is

$$ \lim_{n \to \infty} \left|\frac{(n+1) a_{n+1}}{n a_{n}} \right| = |a_{n+1} / a_{n}| = R? $$

via the same argument? Is my reasoning correct? That is, is the argument that the Radius of Convergence the same for both a power series and its derivative really this simple? :)

$\endgroup$
  • $\begingroup$ I changed the title. $\endgroup$ – user1770201 Sep 4 '16 at 16:32
  • $\begingroup$ The radius of convergence is not always given by the ratio test. When the ratio test does give it, however, your argument works. $\endgroup$ – zhw. Sep 4 '16 at 17:05
  • $\begingroup$ @zhw: Do you mean that the ratio test only gives the radius of convergence AFTER you have shown that the power series does indeed converge? $\endgroup$ – user1770201 Sep 4 '16 at 17:06
  • 1
    $\begingroup$ No, I'm saying what I wrote. For example $x + x^2/2^2 + x^3 + x^4/2^4 + x^5 + x^6/2^6 + \cdots $ has radius of convergence 1, but the ratio test fails miserably here. $\endgroup$ – zhw. Sep 4 '16 at 17:10
2
$\begingroup$

Observe that

$$\lim\sup_{n\to\infty}\sqrt[n]{|a_n|}=\lim\sup_{n\to\infty}\sqrt[n]{|na_n|}$$

since $\;\sqrt[n]n\xrightarrow[n\to\infty]{}1\;$ , so both power series convergence radius are the same.

$\endgroup$
  • $\begingroup$ What if this limsup equals $1$? By the root test, this may be inconclusive. $\endgroup$ – user1770201 Sep 4 '16 at 17:35
  • $\begingroup$ @user1770201 You seem to have missed the point. Google "Cauchy-Hadamard formula" or something like that: it gives you the radius of convergence $\;R\;$ by means of the formula $$R=\lim\sup_{n\to\infty}\frac1{\sqrt[n]{|a_n|}}$$ and thus what my answer proves is that both a power series and its derivative power series have the same radius of convergnece. $\endgroup$ – DonAntonio Sep 4 '16 at 17:46
  • $\begingroup$ @DonAntonio I was looking for a proof of the fact that $\limsup_{n \rightarrow \infty} |a_n|^\frac{1}{n} = \limsup_{n \rightarrow \infty} |na_n|^\frac{1}{n} $. What is the standard way of doing it? $\endgroup$ – Anu Feb 4 '18 at 6:39
  • $\begingroup$ @Anu you could use, I guess, that $$\limsup_{n\to\infty}\sqrt[n]n=\lim_{n\to\infty}\sqrt[n]n=1\ldots$$ $\endgroup$ – DonAntonio Feb 4 '18 at 8:46
0
$\begingroup$

Suppose the radius of convergence of $\sum a_nx^n$ is $R.$ Then $\sum a_nx^n$ converges absolutely for $x\in (-R,R).$ Now fix an $x_0 \in (-R,R),$ and choose $y\in (-R,R)$ with $|y| > |x_0|.$ Then $n|a_nx_0^n| = |a_ny^n|n|x_0/y|^n.$ Because $|x_0/y| < 1,$ $n|x_0/y|^n \to 0.$ Since $\sum |a_ny^n| < \infty,$ $\sum |na_nx_0^n| < \infty.$ It follows that the radius of convergence of $\sum na_nx^n$ is at least $R.$

$\endgroup$
  • $\begingroup$ This is wrong. The coefficients of x^n is not na_n $\endgroup$ – John Mitchell Feb 24 at 14:46
  • $\begingroup$ @JohnMitchell I don't understand your comment. I showed that for each $x_0\in (-R,R),$ the series $\sum na_nx_0^n$ converges absolutely. This implies the ROC of $\sum na_nx^n$ is at least $R.$ $\endgroup$ – zhw. Feb 24 at 17:06
  • $\begingroup$ No what you've shown is that $\sum na_nx^n$ converges. We've to show $\sum (n+1)a_(n+1)x^n$ converges $\endgroup$ – John Mitchell Mar 4 at 18:32
  • $\begingroup$ @JohnMitchell That follows trivially from what I showed. $\endgroup$ – zhw. Mar 4 at 19:10

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.