power series
- Thread starter kezman
- Start date
skeeter
Elite Member
- Joined
- Dec 15, 2005
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- 3,216
well ... if what pka sez is true (and he's very reliable), then
\(\displaystyle \L\Sigma 2^n sin(\frac{x}{3^n})\) converges for all values of x that are of the form \(\displaystyle k \pi\), where k is an integer.
but ... I think this series converges for all x.
\(\displaystyle lim_{n\rightarrow \infty} \frac{sin(\frac{x}{3^n})}{\frac{x}{3^n}} \, = \, 1\)
so wouldn't this series have the same interval of convergence as
\(\displaystyle \L\Sigma \frac{2^n}{3^n} x \, = \, \Sigma (\frac{2}{3})^n x\) ???
\(\displaystyle \L\Sigma 2^n sin(\frac{x}{3^n})\) converges for all values of x that are of the form \(\displaystyle k \pi\), where k is an integer.
but ... I think this series converges for all x.
\(\displaystyle lim_{n\rightarrow \infty} \frac{sin(\frac{x}{3^n})}{\frac{x}{3^n}} \, = \, 1\)
so wouldn't this series have the same interval of convergence as
\(\displaystyle \L\Sigma \frac{2^n}{3^n} x \, = \, \Sigma (\frac{2}{3})^n x\) ???
If I start from the point of the geometric series.
For the two methods I have these results...
\(\displaystyle \begin{array}{l}
\sum\limits_{n = 0}^\infty {\sqrt[n]{{\left| {\left( {\frac{2}{3}} \right)^n } \right|\left| x \right|}}} < 1 \Rightarrow \sqrt[n]{{\left| x \right|}} < \frac{3}{2} \\
\sum\limits_{n = 0}^\infty {\frac{{\left| {\left( {\frac{2}{3}} \right)^n } \right|\frac{2}{3}\left| x \right|}}{{\left| {\left( {\frac{2}{3}} \right)^n } \right|\left| x \right|}}} < 1 \Rightarrow \frac{2}{3} < 1 \\
\end{array}\)
For the two methods I have these results...
\(\displaystyle \begin{array}{l}
\sum\limits_{n = 0}^\infty {\sqrt[n]{{\left| {\left( {\frac{2}{3}} \right)^n } \right|\left| x \right|}}} < 1 \Rightarrow \sqrt[n]{{\left| x \right|}} < \frac{3}{2} \\
\sum\limits_{n = 0}^\infty {\frac{{\left| {\left( {\frac{2}{3}} \right)^n } \right|\frac{2}{3}\left| x \right|}}{{\left| {\left( {\frac{2}{3}} \right)^n } \right|\left| x \right|}}} < 1 \Rightarrow \frac{2}{3} < 1 \\
\end{array}\)
pka
Elite Member
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- Jan 29, 2005
- Messages
- 11,976
NO! NOT AT ALL THAT. Here is how it works.
If \(\displaystyle \sum\limits_n {\left| {a_n } \right|}\) converges then \(\displaystyle \sum\limits_n {a_n }\) converges. (ABSOLUTE CONVERGENCE)
If \(\displaystyle \forall n,\;0 \le a_n \le b_n\) and \(\displaystyle \sum\limits_n {b_n }\) converges then \(\displaystyle \sum\limits_n {a_n }\) converges. (BASIC COMPARSION)
Recall that \(\displaystyle \left| {\sin (x)} \right| \le \left| x \right|\)
so that \(\displaystyle \sum\limits_n {\left| {2^n \sin \left( {\frac{x}{{3^n }}} \right)} \right|} \le \sum\limits_n {\left| {2^n \left( {\frac{x}{{3^n }}} \right)} \right| \le|x|\sum\limits_n {\left| {\left( {\frac{2}{3}} \right)^n } \right|} }\) the last series converges for any x.
If \(\displaystyle \sum\limits_n {\left| {a_n } \right|}\) converges then \(\displaystyle \sum\limits_n {a_n }\) converges. (ABSOLUTE CONVERGENCE)
If \(\displaystyle \forall n,\;0 \le a_n \le b_n\) and \(\displaystyle \sum\limits_n {b_n }\) converges then \(\displaystyle \sum\limits_n {a_n }\) converges. (BASIC COMPARSION)
Recall that \(\displaystyle \left| {\sin (x)} \right| \le \left| x \right|\)
so that \(\displaystyle \sum\limits_n {\left| {2^n \sin \left( {\frac{x}{{3^n }}} \right)} \right|} \le \sum\limits_n {\left| {2^n \left( {\frac{x}{{3^n }}} \right)} \right| \le|x|\sum\limits_n {\left| {\left( {\frac{2}{3}} \right)^n } \right|} }\) the last series converges for any x.