Homework Help: Power series (part 2)

1. Jul 21, 2014

MissP.25_5

Hi.
I have another question about power series. I am having problem with the summarizing of the sum (writing in $\sum_{}^{}$ form).

Here is the question:
Let α be a real number that is not 0.
Let $$f(z)=e^{{\alpha}Ln(z+1)}$$
For integer n>0, find $$f^n(0).$$

My partial solution:
$$f(z)=e^{{\alpha}Ln(z+1)}=(z+1)^{\alpha}$$

$$f^1(z)=α(z+1)^{α-1}$$

$$f^2(z)=α(\alpha-1)(z+1)^{α-2}$$

$$f^3(z)=α(α-1)(α-2)(z+1)^{α-3}$$

The answer for $$f^n(z)$$ is

$$f^n(z)=\alpha(\alpha-1)...(\alpha-n+1)(z+1)^{\alpha-n}.$$

I am confused because when n=1,
$$f^1(z)=\alpha(z+1)^{\alpha-1}$$
but in the answer, there is the coefficient (α-1), this is where I am confused.

Last edited: Jul 21, 2014
2. Jul 21, 2014

CAF123

Did you make a typo in writing down the answer? There does not seem to be any $n$ index on the right hand side.

3. Jul 21, 2014

MissP.25_5

You're right, there is a typo error. Give me a few minutes to correct it.

4. Jul 21, 2014

MissP.25_5

$$f^n(z)=\alpha(\alpha-1)...(\alpha-n+1)(z+1)^{\alpha-n}$$

5. Jul 21, 2014

CAF123

Yes, that's better. I think maybe it is just the notation that is confusing you. The quantity $\alpha (\alpha - 1) \dots (\alpha - (n-1))$ just means you write out all terms starting from $\alpha$ and ending with the term corresponding to that particular of $n$. Putting a factor of $\alpha - 1$ in the general formula just serves to tell you that by increasing $n$ from 1 to 2, you should put in another factor of $\alpha$ lowered by 1. So, by extrapolating this knowledge to the general case, you see that $f^n(z)$ corresponds to the case where you have decreased $\alpha$ by 1 $n-1$ times and multiplied together all the consecutive terms.

6. Jul 21, 2014

MissP.25_5

Ah, now I get why it's written that way, it's because it is corresponding to n. Thank you for explaining that. But I don't get this part, though. What do you mean by this:

So, by extrapolating this knowledge to the general case, you see that $f^n(z)$ corresponds to the case where you have decreased $\alpha$ by 1 $n-1$ times and multiplied together all the consecutive terms.

7. Jul 21, 2014

CAF123

By computing the $n^{\text{th}}$ derivative of $f(z)$, you have decreased the value of $\alpha$ by $n-1$ times, right? E.g for n=2, you have factors $\alpha$ and $(\alpha - 1)$. When I say 'multiply together all consecutive terms', I mean the general form for the $n^{\text{th}}$ derivative includes an overall term with all those factors multiplied together, e.g for n=2, you have a term $\alpha(\alpha-1)$.

8. Jul 21, 2014

MissP.25_5

Got it. So, the final answer is just plug in 0 into the function, right?

$∴f^n(0)= α(α-1)...(α-n+1)$

9. Jul 21, 2014

CAF123

Right.

10. Jul 21, 2014

MissP.25_5

Thanks. Problem solved.

11. Jul 21, 2014

MissP.25_5

Hey, just one last question.

In the question, $f(z)=e^{{\alpha}Ln(z+1)}$ and I have written it as $f(z)=(z+1)^{\alpha}$.
This is correct, right? Because someone said to me that it's not. He said:
If z is not a real number, $e^{\alpha \, \mathrm{Ln}\,\left( z + 1 \right) } \neq \left( z + 1 \right) ^{\alpha }$ as the logarithm laws in general do not carry from the real numbers to the complex numbers...

12. Jul 21, 2014

Ray Vickson

It is OK for the "principle branch" of the function $\ln$. What you did is valid as long as $z+1$ stays away from the negative real axis; certainly it is OK if $\text{Re}(z+1) > 0$, or $\text{Re}(z) > -1$.

13. Jul 21, 2014

HallsofIvy

You understand that $e^{\alpha ln(z+ 1)}= (z+ 1)^\alpha$, right?

14. Jul 21, 2014

MissP.25_5

Yes i do.

15. Jul 22, 2014

MissP.25_5

There's actually a continuation to this question. I have to expland f(z) into maclaurin series and then find the radius of convergence. To expand it, we have to use the exponential function method, right? Which is:
$$e^z=\sum_{n=0}^{\infty}\frac{z^n}{n!}=1+\frac{z^2}{2!}+\frac{z^3}{3!}+...$$

In this problem $f(z) = e^{αLn(z+1)}$, how can I expand it into maclaurin series? And, how do I find the radius of convergence?

Last edited: Jul 22, 2014