What is the Radius of Convergence for zsin(z^2)?

[aruwin]

Hello.
I need explanation on why the answer for this problem is $R=\infty$.

Here's the question and the solution.

Expand the function into maclaurin series and find the radius of convergence.
$zsin(z^2)$

Solution:
$$zsin(z^2)=z\sum_{n=0}^{\infty}(-1)^n\frac{z^{2(2n+1)}}{(2n+1)!}$$

Divide both sides by z,

$$sin(z^2)=\sum_{n=0}^{\infty}(-1)^n\frac{z^{2(2n+1)}}{(2n+1)!}$$

So here's the calculation but I don't know how to get the radius of convergence. Answer is $\infty$.

 

[Prove It]

Hello.
I need explanation on why the answer for this problem is $R=\infty$.

Here's the question and the solution.

Expand the function into maclaurin series and find the radius of convergence.
$zsin(z^2)$

Solution:
$$zsin(z^2)=z\sum_{n=0}^{\infty}(-1)^n\frac{z^{2(2n+1)}}{(2n+1)!}$$

Yes...

Divide both sides by z,

$$sin(z^2)=\sum_{n=0}^{\infty}(-1)^n\frac{z^{2(2n+1)}}{(2n+1)!}$$

WHY?! You were asked to find the MacLaurin Series for $\displaystyle \begin{align*} z\sin{ \left( z^2 \right) } \end{align*}$ WHICH YOU HAVE!

I don't know how to get the radius of convergence. Answer is $\infty$.

If the MacLaurin Series for $\displaystyle \begin{align*} \sin{(z)} \end{align*}$ is convergent for all z (it is - check with the ratio test), then so is any composition such as $\displaystyle \begin{align*} \sin{ \left( z^2 \right) } \end{align*}$ and multiplying by $\displaystyle \begin{align*} z \end{align*}$ is just multiplying a number by another number, still giving a number...

 

[aruwin]

Yes...
WHY?! You were asked to find the MacLaurin Series for $\displaystyle \begin{align*} z\sin{ \left( z^2 \right) } \end{align*}$ WHICH YOU HAVE!
If the MacLaurin Series for $\displaystyle \begin{align*} \sin{(z)} \end{align*}$ is convergent for all z (it is - check with the ratio test), then so is any composition such as $\displaystyle \begin{align*} \sin{ \left( z^2 \right) } \end{align*}$ and multiplying by $\displaystyle \begin{align*} z \end{align*}$ is just multiplying a number by another number, still giving a number...[/size]

Oh, I divided that part with z to use it to find the radius by ratio test.
I got R= 4. What is my mistake here?

 

[Nono713]

How does $\lvert -4n^2 + 10n + 6 \rvert$ go to four as $n$ tends to infinity? There's a mistake in the before-last step, I think you might have been thinking of something else and weren't paying attention :p

 

[aruwin]

How does $\lvert -4n^2 + 10n + 6 \rvert$ go to four as $n$ tends to infinity? There's a mistake in the before-last step, I think you might have been thinking of something else and weren't paying attention :p

Well, because the other 2 terms are getting closer to 0 while 4 doesn't? As you can see in my work, I divided all terms by $n^2$ and the only significant term left is 4. I am not sure, though.

 

[Nono713]

Well, because the other 2 terms are getting closer to 0 while 4 doesn't? As you can see in my work, I divided all terms by $n^2$ and the only significant term left is 4. I am not sure, though.

It is true that $10n + 6$ grows much slower in comparison to $4n^2$, but that's not what the problem is asking - you're trying to find the limit of that polynomial as $n$ goes to infinity, and that limit does go to infinity. An easy way to show that it does, is to show that the expression (in absolute values) eventually increases by at least $1$ every successive $n$ (but in general, any polynomial goes to either positive or negative infinity as its variable tends to infinity).

 

[aruwin]

It is true that $10n + 6$ grows much slower in comparison to $4n^2$, but that's not what the problem is asking - you're trying to find the limit of that polynomial as $n$ goes to infinity, and that limit does go to infinity. An easy way to show that it does, is to show that the expression (in absolute values) eventually increases by at least $1$ every successive $n$ (but in general, any polynomial goes to either positive or negative infinity as its variable tends to infinity).

I'm sorry that I am still not quite clear with this. But how does it differ from this limit? In this one, the limit is 1/4 instead of infinity and the working is almost the same as this one.
I posted the question before on this thread:
http://mathhelpboards.com/calculus-10/how-calculate-complicated-factorial-11367.html#post53249

Is it because the simplification gives a fraction while this one has a whole number?

 

[aruwin]

It is true that $10n + 6$ grows much slower in comparison to $4n^2$, but that's not what the problem is asking - you're trying to find the limit of that polynomial as $n$ goes to infinity, and that limit does go to infinity. An easy way to show that it does, is to show that the expression (in absolute values) eventually increases by at least $1$ every successive $n$ (but in general, any polynomial goes to either positive or negative infinity as its variable tends to infinity).

Oh, I get it now! I just googled about limits and learned it again. I think I understand now why the lmit is infinity. Thanks!