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[SOLVED] Interval of Convergence

MacLaddy

Member
Jan 29, 2012
52
Hello all,

Again I find myself at odds with my online class. Somehow, and with two problems in a row, I am finding the reciprocal answer to what Math Lab is telling me.

I would be very appreciative is someone could check my work.

Find the limit of convergence, and the radius.

\(\displaystyle \sum \frac{k^2x^{2k}}{k!}\)

Using the ratio test

\(\displaystyle \lim_{k\rightarrow \infty}\frac{(k+1)^2x^{2k+2}}{(k+1)k!}*\frac{k!}{k^2x^{2k}}\)

That should be an absolute value, but I don't know how to input that...

This should simplify down to,

\(\displaystyle x^2\lim_{k\rightarrow \infty}\frac{k^2+2k+1}{k^3+k^2}\) (with absolute values inputed)

Which should give an interval of convergence of,

\(\displaystyle 0<x^2<0\), R=0, [0,0]

My online class is showing R=\(\displaystyle \infty\) (\(\displaystyle -\infty,\infty\))

The last question gave an R=4, and I was showing R=1/4. I am reversing this somehow. Any help is appreciated.

Thanks,
Mac
 
Last edited:

MacLaddy

Member
Jan 29, 2012
52
Now this is getting weird. I'm definitely doing something funny.

This is the very next question

\(\displaystyle \sum \frac{x^{2k-1}}{10^{k-1}}\)

I am getting an interval of convergence of

\(\displaystyle (-\sqrt{\frac{1}{10}},\sqrt{\frac{1}{10}})\)

And the answer it is showing is

\(\displaystyle (-\sqrt{10},\sqrt{10})\)
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,040
Now this is getting weird. I'm definitely doing something funny.

This is the very next question

\(\displaystyle \sum \frac{x^{2k-1}}{10^{k-1}}\)

I am getting an interval of convergence of

\(\displaystyle (-\sqrt{\frac{1}{10}},\sqrt{\frac{1}{10}})\)

And the answer it is showing is

\(\displaystyle (-\sqrt{10},\sqrt{10})\)
Hi there. (Wave)

Ok, so let's try the Ratio Test.

\(\displaystyle \left( \frac{x^{2(k+1)-1}}{10^{(k+1)-1}} \right) \left( \frac{10^{k-1}}{x^{2k-1}} \right) = \left( \frac{x^{2k}x^1}{10^{k}} \right) \left( \frac{10^{k}10^{-1}}{x^{2k}x^{-1}} \right) \)

After a bunch of cancelling you get \(\displaystyle \frac{x10^{-1}}{x^{-1}}=\frac{x^2}{10}\)

Now you need to find when \(\displaystyle \left| \frac{x^2}{10} \right| < 1 \)

That's just \(\displaystyle |x^2| < 10\) which gives the answer your book has. I checked my work a couple times for computational errors and don't believe I have any, but it's possible.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Hello all,

Again I find myself at odds with my online class. Somehow, and with two problems in a row, I am finding the reciprocal answer to what Math Lab is telling me.

I would be very appreciative is someone could check my work.

Find the limit of convergence, and the radius.

\(\displaystyle \sum \frac{k^2x^{2k}}{k!}\)

Using the ratio test

\(\displaystyle \lim_{k\rightarrow \infty}\frac{(k+1)^2x^{2k+2}}{(k+1)k!}*\frac{k!}{k^2x^{2k}}\)

That should be an absolute value, but I don't know how to input that...

This should simplify down to,

\(\displaystyle x^2\lim_{k\rightarrow \infty}\left| \frac{k^2+2k+1}{k^3+k^2} \right|\) ... (CB1)

Which should give an interval of convergence of,

\(\displaystyle 0<x^2<0\), R=0, [0,0]

My online class is showing R=\(\displaystyle \infty\) (\(\displaystyle -\infty,\infty\))
You can show that your series is convergent for all \(x\) by comparing it to a suitably modified version of the exponential series which does have an infinite radius of convergence.

You will also get this result if you use (CB1) and do the algebra correctly

CB
 
Last edited:

MacLaddy

Member
Jan 29, 2012
52
Hi there. (Wave)

Ok, so let's try the Ratio Test.

\(\displaystyle \left( \frac{x^{2(k+1)-1}}{10^{(k+1)-1}} \right) \left( \frac{10^{k-1}}{x^{2k-1}} \right) = \left( \frac{x^{2k}x^1}{10^{k}} \right) \left( \frac{10^{k}10^{-1}}{x^{2k}x^{-1}} \right) \)

After a bunch of cancelling you get \(\displaystyle \frac{x10^{-1}}{x^{-1}}=\frac{x^2}{10}\)

Now you need to find when \(\displaystyle \left| \frac{x^2}{10} \right| < 1 \)

That's just \(\displaystyle |x^2| < 10\) which gives the answer your book has. I checked my work a couple times for computational errors and don't believe I have any, but it's possible.
Ahh, that's a bit of a palm to forehead moment.

For some reason I was separating the $x^2$ out, so that I was working with \(\displaystyle -\frac{1}{10}<x^2<\frac{1}{10}\). That makes a lot more sense now. I can't explain why I was thinking that.

You can show that your series is convergent for all \(x\) by comparing it to a suitably modified version of the exponential series which does have an infinite radius of convergence.

You will also get this result if you use (CB1) and do the algebra correctly

CB
Still not sure I understand this one. If I take the limit of what's on the right, it becomes zero. At least I thought it did. So if I multiply $x^2$ by 0 I will just end up with \(\displaystyle -1<0<1\). I'm definitely missing a step here. I see my mistake from the other two problems, but I can't quite rectify it with this one. I'll probably figure it out soon, hopefully, since I saw my mistake from the other ones.

Thanks again, it is much appreciated.
 
Last edited:

Jameson

Administrator
Staff member
Jan 26, 2012
4,040
Still not sure I understand this one. If I take the limit of what's on the right, it becomes zero. At least I thought it did. So if I multiply $x^2$ by 0 I will just end up with \(\displaystyle -1<0<1\). I'm definitely missing a step here. I see my mistake from the other two problems, but I can't quite rectify it with this one. I'll probably figure it out soon, hopefully, since I saw my mistake from the other ones.

Thanks again, it is much appreciated.
When you take the limit and get 0 regardless of x then the radius of convergence is infinity because it just doesn't matter what the value of x is, the limit will always be less than 1 so it's always true.
 

MacLaddy

Member
Jan 29, 2012
52
I may have seen one way to work this, but I probably am not understanding it in the correct- or proper- terms.

If I were to take the limit shown, and leave $x^2$ in the numerator instead of pulling it out front, then improperly I could say \(\displaystyle -1<\frac{x^2}{\infty}<1 = -\infty<x^2<\infty\)

Is that "sort-of" the right thinking? It at least turns out correctly in this scenario, but I know that would be improper.
 

MacLaddy

Member
Jan 29, 2012
52
When you take the limit and get 0 regardless of x then the radius of convergence is infinity because it just doesn't matter what the value of x is, the limit will always be less than 1 so it's always true.
Well, when you put it simply like that then it makes perfect sense. (Smile)

Thanks again, both of you.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Still not sure I understand this one. If I take the limit of what's on the right, it becomes zero. At least I thought it did. So if I multiply $x^2$ by 0 I will just end up with \(\displaystyle -1<0<1\). I'm definitely missing a step here. I see my mistake from the other two problems, but I can't quite rectify it with this one. I'll probably figure it out soon, hopefully, since I saw my mistake from the other ones.

Thanks again, it is much appreciated.
\(\displaystyle |x^2|\lim_{k\rightarrow \infty}\left| \frac{k^2+2k+1}{k^3+k^2} \right|\) ... (CB1)
\[|x^2|\left| \frac{k^2+2k+1}{k^3+k^2} \right|=\left|\frac{x^2(k+1)}{k^2}\right|\]

Now for fixed \(x\) this goes to zeros as \(k \to \infty\) which is less than 1 so the series converges for all \(x\)

CB