Finding the Coefficient in a Power Series Sum

[aruwin]

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 $\alpha$ 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}$

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

$$f^2(z)=\alpha(\alpha-1)(z+1)^{\alpha-2}$$

$$f^3(z)=\alpha(\alpha-1)(\alpha-2)(z+1)^{\alpha-3}$$

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

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

 

[Prove It]

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 $\alpha$ 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}$

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

$$f^2(z)=\alpha(\alpha-1)(z+1)^{\alpha-2}$$

$$f^3(z)=\alpha(\alpha-1)(\alpha-2)(z+1)^{\alpha-3}$$

The answer for $f^n(z)$ is
$$\alpha(\alpha-1)...(\alpha-n+1)(z+1)^{\alpha-n}$$.

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

If z is not a real number, $\displaystyle \begin{align*} \mathrm{e}^{\alpha \, \mathrm{Ln}\,\left( z + 1 \right) } \neq \left( z + 1 \right) ^{\alpha } \end{align*}$ as the logarithm laws in general do not carry from the real numbers to the complex numbers...

 

[aruwin]

If z is not a real number, $\displaystyle \begin{align*} \mathrm{e}^{\alpha \, \mathrm{Ln}\,\left( z + 1 \right) } \neq \left( z + 1 \right) ^{\alpha } \end{align*}$ as the logarithm laws in general do not carry from the real numbers to the complex numbers...

but Lnz=ln|z| + iArgz

 

[Prove It]

but Lnz=ln|z| + iArgz

And that should be enough to show that $\displaystyle \begin{align*} \alpha \, \textrm{Ln}\, \left( z + 1 \right) \neq \textrm{Ln}\,\left[ \left( z + 1 \right) ^{\alpha } \right] \end{align*}$, rather...

$\displaystyle \begin{align*} \alpha\,\textrm{Ln}\,\left( z + 1 \right) &= \alpha \left[ \ln{ \left| z +1 \right| } + \mathrm{i}\,\mathrm{Arg}\,\left( z + 1 \right) \right] \\ &= \alpha \ln{ \left| z + 1 \right| } + \mathrm{i}\,\alpha\,\textrm{Arg}\,\left( z + 1 \right) \\ &= \ln{ \left( \left| z + 1 \right| ^{\alpha} \right) } + \mathrm{i}\,\alpha\,\textrm{Arg}\,\left( z + 1 \right) \end{align*}$

which is of course, NOTHING like $\displaystyle \begin{align*} \textrm{Ln}\,{ \left[ \left( z + 1 \right) ^{\alpha} \right] } \end{align*}$...

 

[aruwin]

And that should be enough to show that $\displaystyle \begin{align*} \alpha \, \textrm{Ln}\, \left( z + 1 \right) \neq \textrm{Ln}\,\left[ \left( z + 1 \right) ^{\alpha } \right] \end{align*}$, rather...

$\displaystyle \begin{align*} \alpha\,\textrm{Ln}\,\left( z + 1 \right) &= \alpha \left[ \ln{ \left| z +1 \right| } + \mathrm{i}\,\mathrm{Arg}\,\left( z + 1 \right) \right] \\ &= \alpha \ln{ \left| z + 1 \right| } + \mathrm{i}\,\alpha\,\textrm{Arg}\,\left( z + 1 \right) \\ &= \ln{ \left( \left| z + 1 \right| ^{\alpha} \right) } + \mathrm{i}\,\alpha\,\textrm{Arg}\,\left( z + 1 \right) \end{align*}$

which is of course, NOTHING like $\displaystyle \begin{align*} \textrm{Ln}\,{ \left[ \left( z + 1 \right) ^{\alpha} \right] } \end{align*}$...

Then how should I start?

 

[aruwin]

And that should be enough to show that $\displaystyle \begin{align*} \alpha \, \textrm{Ln}\, \left( z + 1 \right) \neq \textrm{Ln}\,\left[ \left( z + 1 \right) ^{\alpha } \right] \end{align*}$, rather...

$\displaystyle \begin{align*} \alpha\,\textrm{Ln}\,\left( z + 1 \right) &= \alpha \left[ \ln{ \left| z +1 \right| } + \mathrm{i}\,\mathrm{Arg}\,\left( z + 1 \right) \right] \\ &= \alpha \ln{ \left| z + 1 \right| } + \mathrm{i}\,\alpha\,\textrm{Arg}\,\left( z + 1 \right) \\ &= \ln{ \left( \left| z + 1 \right| ^{\alpha} \right) } + \mathrm{i}\,\alpha\,\textrm{Arg}\,\left( z + 1 \right) \end{align*}$

which is of course, NOTHING like $\displaystyle \begin{align*} \textrm{Ln}\,{ \left[ \left( z + 1 \right) ^{\alpha} \right] } \end{align*}$...

I checked this on wolfram:
is e'^''('a ln'('x'+'yi')'')' '=' '('x'+'yi')''^'a - Wolfram|Alpha