Prove the following Power Series is monotonic
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Your request for help is too vague. Where are you stuck?
In other words, what have you thought about or tried thus far?
Please be specific.
Also, check out the post titled "Read Before Posting", for additional guidelines on what to post.
Thank you! :cool:
In other words, what have you thought about or tried thus far?
Please be specific.
Also, check out the post titled "Read Before Posting", for additional guidelines on what to post.
Thank you! :cool:
Well, first, how do I show that the series converges for \(\displaystyle -1 \le x \le 1\)?
And then, what specifically should I do to show that it is monotonic? What theorem do I use? What is the approach?
And finally, I believe that to show that f(x)=1.5 and f(x)=-0.5 has something to do with f taking its min and max values at the end of the interval. But again i'm not sure specifically what is the formal explanation and how to bind everything together.
I started with splitting f to \(\displaystyle f_{e}(x) = \sum_{n=1}^{\infty} \frac{x^{2n}}{(2n)^{\frac{3}{2}}}\) and \(\displaystyle f_{o}(x) = - \sum_{n=1}^{\infty} \frac{x^{2n-1}}{(2n-1)^{\frac{3}{2}}}\)
But i don't really know how to take it from here and what tools to use.
Would appreciate any help
And then, what specifically should I do to show that it is monotonic? What theorem do I use? What is the approach?
And finally, I believe that to show that f(x)=1.5 and f(x)=-0.5 has something to do with f taking its min and max values at the end of the interval. But again i'm not sure specifically what is the formal explanation and how to bind everything together.
I started with splitting f to \(\displaystyle f_{e}(x) = \sum_{n=1}^{\infty} \frac{x^{2n}}{(2n)^{\frac{3}{2}}}\) and \(\displaystyle f_{o}(x) = - \sum_{n=1}^{\infty} \frac{x^{2n-1}}{(2n-1)^{\frac{3}{2}}}\)
But i don't really know how to take it from here and what tools to use.
Would appreciate any help
The ratio test can be used to show that the radius of convergence is [-1,1].
Here is an interesting identity that shows if x = 1, then the series converges.
\(\displaystyle \displaystyle \sum_{k=1}^{\infty}\frac{(-1)^{n}}{n^{p}}=\frac{1-2^{p-1}}{2^{p-1}}\zeta(p)\)***
Which means the series \(\displaystyle \displaystyle\sum_{k=1}^{\infty}\frac{(-1)^{n}}{n^{\frac{3}{2}}}=\frac{1}{2}(\sqrt{2}-2)\zeta(3/2)\approx -.765...\)
If x = -1, then \(\displaystyle \displaystyle\sum_{k=1}^{\infty}\frac{1}{n^{\frac{3}{2}}}=\zeta(3/2)\approx 2.61...\)
When a series is strictly monotone, then it is increasing or decreasing.
\(\displaystyle \begin{array}{c|c|c}\hline a_{n+1}/a_{n}& >1 &\text{increasing}\\ \hline a_{n+1}/a_{n}& <1&\text{decreasing}\\ \hline a_{n+1}/a_{n}& \geq 1&\text{non decreasing}\\ \hline a_{n+1}/a_{n}& \leq 1&\text{non-increasing}\end{array}\)
***. The \(\displaystyle \zeta(p)\) symbol is the Riemann Zeta function. It is used extensively in the evaluation of various series.
Here is an interesting identity that shows if x = 1, then the series converges.
\(\displaystyle \displaystyle \sum_{k=1}^{\infty}\frac{(-1)^{n}}{n^{p}}=\frac{1-2^{p-1}}{2^{p-1}}\zeta(p)\)***
Which means the series \(\displaystyle \displaystyle\sum_{k=1}^{\infty}\frac{(-1)^{n}}{n^{\frac{3}{2}}}=\frac{1}{2}(\sqrt{2}-2)\zeta(3/2)\approx -.765...\)
If x = -1, then \(\displaystyle \displaystyle\sum_{k=1}^{\infty}\frac{1}{n^{\frac{3}{2}}}=\zeta(3/2)\approx 2.61...\)
When a series is strictly monotone, then it is increasing or decreasing.
\(\displaystyle \begin{array}{c|c|c}\hline a_{n+1}/a_{n}& >1 &\text{increasing}\\ \hline a_{n+1}/a_{n}& <1&\text{decreasing}\\ \hline a_{n+1}/a_{n}& \geq 1&\text{non decreasing}\\ \hline a_{n+1}/a_{n}& \leq 1&\text{non-increasing}\end{array}\)
***. The \(\displaystyle \zeta(p)\) symbol is the Riemann Zeta function. It is used extensively in the evaluation of various series.
I didn't think you had. You do not need to. I was just showing that for curiosity and showing what it converges to for x=-1 and 1.
As I said, the ratio test can be used to show the radius of convergence.
Then, if it is strictly monotonic, try the ratio of \(\displaystyle a_{n}\) and \(\displaystyle a_{n+1}\)
Which is the ratio test. \(\displaystyle \displaystyle \frac{(-1)^{n+1}x^{n+1}}{(n+1)^{\frac{3}{2}}}\cdot \frac{n^{\frac{3}{2}}}{(-1)^{n}x^{n}}\)
\(\displaystyle =\displaystyle \lim_{n\to \infty}\frac{n^{\frac{3}{2}}|x|}{(n+1)^{\frac{3}{2}}}=|x|\)
converges if |x|<1, diverges if |x|>1.
Radius of convergence is 1. Interval of convergence is [-1,1].
\(\displaystyle \displaystyle \frac{k^{\frac{3}{2}}|x|}{(k+1)^{\frac{3}{2}}}<1 \;\ for \;\ n\geq 1\)
so it is decreasing. The derivative could also be used to show it is strictly monotone.
As I said, the ratio test can be used to show the radius of convergence.
Then, if it is strictly monotonic, try the ratio of \(\displaystyle a_{n}\) and \(\displaystyle a_{n+1}\)
Which is the ratio test. \(\displaystyle \displaystyle \frac{(-1)^{n+1}x^{n+1}}{(n+1)^{\frac{3}{2}}}\cdot \frac{n^{\frac{3}{2}}}{(-1)^{n}x^{n}}\)
\(\displaystyle =\displaystyle \lim_{n\to \infty}\frac{n^{\frac{3}{2}}|x|}{(n+1)^{\frac{3}{2}}}=|x|\)
converges if |x|<1, diverges if |x|>1.
Radius of convergence is 1. Interval of convergence is [-1,1].
\(\displaystyle \displaystyle \frac{k^{\frac{3}{2}}|x|}{(k+1)^{\frac{3}{2}}}<1 \;\ for \;\ n\geq 1\)
so it is decreasing. The derivative could also be used to show it is strictly monotone.
Oh, thanks a lot for your help
Some follow up questions:
Why is the interval r is [-1,1] and not (-1,1)?
And I need to provide a little rigorous proof, so can you tell me on what do I base the fact that if a_n+1/a_n < 1 then the series is decreasing? On what theorem do you base this?
Some follow up questions:
Why is the interval r is [-1,1] and not (-1,1)?
And I need to provide a little rigorous proof, so can you tell me on what do I base the fact that if a_n+1/a_n < 1 then the series is decreasing? On what theorem do you base this?