Finding f^n(0) for the Power Series e^(αLn(z+1))

[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.

 

[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.

 

[MissP.25_5]

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

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

 

[MissP.25_5]

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

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

 

[CAF123]

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

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.

 

[MissP.25_5]

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.

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.

 

[CAF123]

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.

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 $\alpha(\alpha-1)$.

 

[MissP.25_5]

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 $\alpha(\alpha-1)$.

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

$∴f^n(0)= α(α-1)...(α-n+1)$

 

[CAF123]

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

$∴f^n(0)= α(α-1)...(α-n+1)$

Right.

 

[MissP.25_5]

Right.

Thanks. Problem solved.

 

[MissP.25_5]

Right.

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...

 

[Ray Vickson]

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...

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 $\text{Re}(z) > -1$.

 

[HallsofIvy]

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

 

[MissP.25_5]

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

Yes i do.

 

[MissP.25_5]

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 $\text{Re}(z) > -1$.

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?