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How to represent $sumlimits_n=0^infty a_knz^kn$ in terms of $f(z) = sumlimits_n=0^infty a_nz^n$

3

$begingroup$

Given a power series $f(z) = sumlimits_n=0^infty a_nz^n$ where $zin mathbbC$ and radius of convergence $R.$ Then my goal is to find,
$$sum_n=0^infty a_knz^kn$$
for $|z|<R$ and $kin mathbbN.$

So I tried two examples and I think there is a connection with roots of unity.
beginalign*
sum_n=0^infty a_2nz^2n &=frac12(f(z)+f(i^2z))\
sum_n=0^infty a_3nz^3n &=frac12(f(z)+f(iz) + f(i^2z))
endalign*
My guess is that
$$sum_n=0^infty a_knz^kn =frac12(f(z)+f(iz) + f(i^2z)+ cdots + f(i^k-1z))$$
if $k$ is odd. For $k$ even I am not sure. Any generalizations of this fact or proof ideas will be much appreciated.

complex-analysis power-series

share|cite|improve this question

edited Apr 1 at 7:47

Martin R

31k33561

asked Apr 1 at 7:15

model_checkermodel_checker

4,45521931

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is $n$? An integer? A number ranging from $-infty$ up to $infty$? Or just starting by $0$ up to $infty$?
  $endgroup$
  – mrtaurho
  Apr 1 at 7:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Starting from $0$ and upto infinity.
  $endgroup$
  – model_checker
  Apr 1 at 7:18
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your formula for $sum a_3n z^3n$ is not correct.
  $endgroup$
  – Kavi Rama Murthy
  Apr 1 at 7:24
  
  

  
  
  
  

add a comment | 

3

$begingroup$

Given a power series $f(z) = sumlimits_n=0^infty a_nz^n$ where $zin mathbbC$ and radius of convergence $R.$ Then my goal is to find,
$$sum_n=0^infty a_knz^kn$$
for $|z|<R$ and $kin mathbbN.$

So I tried two examples and I think there is a connection with roots of unity.
beginalign*
sum_n=0^infty a_2nz^2n &=frac12(f(z)+f(i^2z))\
sum_n=0^infty a_3nz^3n &=frac12(f(z)+f(iz) + f(i^2z))
endalign*
My guess is that
$$sum_n=0^infty a_knz^kn =frac12(f(z)+f(iz) + f(i^2z)+ cdots + f(i^k-1z))$$
if $k$ is odd. For $k$ even I am not sure. Any generalizations of this fact or proof ideas will be much appreciated.

complex-analysis power-series

share|cite|improve this question

edited Apr 1 at 7:47

Martin R

31k33561

asked Apr 1 at 7:15

model_checkermodel_checker

4,45521931

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is $n$? An integer? A number ranging from $-infty$ up to $infty$? Or just starting by $0$ up to $infty$?
  $endgroup$
  – mrtaurho
  Apr 1 at 7:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Starting from $0$ and upto infinity.
  $endgroup$
  – model_checker
  Apr 1 at 7:18
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your formula for $sum a_3n z^3n$ is not correct.
  $endgroup$
  – Kavi Rama Murthy
  Apr 1 at 7:24
  
  

  
  
  
  

add a comment | 

$begingroup$

Given a power series $f(z) = sumlimits_n=0^infty a_nz^n$ where $zin mathbbC$ and radius of convergence $R.$ Then my goal is to find,
$$sum_n=0^infty a_knz^kn$$
for $|z|<R$ and $kin mathbbN.$

So I tried two examples and I think there is a connection with roots of unity.
beginalign*
sum_n=0^infty a_2nz^2n &=frac12(f(z)+f(i^2z))\
sum_n=0^infty a_3nz^3n &=frac12(f(z)+f(iz) + f(i^2z))
endalign*
My guess is that
$$sum_n=0^infty a_knz^kn =frac12(f(z)+f(iz) + f(i^2z)+ cdots + f(i^k-1z))$$
if $k$ is odd. For $k$ even I am not sure. Any generalizations of this fact or proof ideas will be much appreciated.

complex-analysis power-series

share|cite|improve this question

edited Apr 1 at 7:47

Martin R

31k33561

asked Apr 1 at 7:15

model_checkermodel_checker

4,45521931

$endgroup$

share|cite|improve this question

edited Apr 1 at 7:47

Martin R

31k33561

asked Apr 1 at 7:15

model_checkermodel_checker

4,45521931

share|cite|improve this question

edited Apr 1 at 7:47

Martin R

31k33561

asked Apr 1 at 7:15

model_checkermodel_checker

4,45521931

edited Apr 1 at 7:47

Martin R

31k33561

edited Apr 1 at 7:47

Martin R

31k33561

asked Apr 1 at 7:15

model_checkermodel_checker

4,45521931

asked Apr 1 at 7:15

model_checkermodel_checker

4,45521931

1 Answer
1

active

oldest

votes

3

$begingroup$

Let $omega ne 1$ be a $k$-th root of unity: $omega^k = 1$. Then
$$
sum_j=0^k-1 omega^nj = begincases
k & text if $n$ is a multiple of $k$, \
0 & text otherwise.
endcases
$$
Consequently,
$$
frac 1k sum_j=0^k-1 f(omega^j z ) = sum_n=0^infty a_knz^kn
$$

For $k=2$ this gives your result
$$
sum_n=0^infty a_2nz^2n =frac12(f(z)+f(-z))\
$$
For $k=3$ the correct result is
$$
sum_n=0^infty a_3nz^3n =frac13(f(z)+f(omega z) + f(omega^2 z))
$$
with $omega = - frac 12 pm fracsqrt 32$.

share|cite|improve this answer

edited Apr 1 at 7:33

answered Apr 1 at 7:28

Martin RMartin R

31k33561

$endgroup$

add a comment | 

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3

$begingroup$

Let $omega ne 1$ be a $k$-th root of unity: $omega^k = 1$. Then
$$
sum_j=0^k-1 omega^nj = begincases
k & text if $n$ is a multiple of $k$, \
0 & text otherwise.
endcases
$$
Consequently,
$$
frac 1k sum_j=0^k-1 f(omega^j z ) = sum_n=0^infty a_knz^kn
$$

For $k=2$ this gives your result
$$
sum_n=0^infty a_2nz^2n =frac12(f(z)+f(-z))\
$$
For $k=3$ the correct result is
$$
sum_n=0^infty a_3nz^3n =frac13(f(z)+f(omega z) + f(omega^2 z))
$$
with $omega = - frac 12 pm fracsqrt 32$.

share|cite|improve this answer

edited Apr 1 at 7:33

answered Apr 1 at 7:28

Martin RMartin R

31k33561

$endgroup$

add a comment | 

3

$begingroup$

Let $omega ne 1$ be a $k$-th root of unity: $omega^k = 1$. Then
$$
sum_j=0^k-1 omega^nj = begincases
k & text if $n$ is a multiple of $k$, \
0 & text otherwise.
endcases
$$
Consequently,
$$
frac 1k sum_j=0^k-1 f(omega^j z ) = sum_n=0^infty a_knz^kn
$$

For $k=2$ this gives your result
$$
sum_n=0^infty a_2nz^2n =frac12(f(z)+f(-z))\
$$
For $k=3$ the correct result is
$$
sum_n=0^infty a_3nz^3n =frac13(f(z)+f(omega z) + f(omega^2 z))
$$
with $omega = - frac 12 pm fracsqrt 32$.

share|cite|improve this answer

edited Apr 1 at 7:33

answered Apr 1 at 7:28

Martin RMartin R

31k33561

$endgroup$

add a comment | 

$begingroup$

Let $omega ne 1$ be a $k$-th root of unity: $omega^k = 1$. Then
$$
sum_j=0^k-1 omega^nj = begincases
k & text if $n$ is a multiple of $k$, \
0 & text otherwise.
endcases
$$
Consequently,
$$
frac 1k sum_j=0^k-1 f(omega^j z ) = sum_n=0^infty a_knz^kn
$$

For $k=2$ this gives your result
$$
sum_n=0^infty a_2nz^2n =frac12(f(z)+f(-z))\
$$
For $k=3$ the correct result is
$$
sum_n=0^infty a_3nz^3n =frac13(f(z)+f(omega z) + f(omega^2 z))
$$
with $omega = - frac 12 pm fracsqrt 32$.

share|cite|improve this answer

edited Apr 1 at 7:33

answered Apr 1 at 7:28

Martin RMartin R

31k33561

$endgroup$

share|cite|improve this answer

edited Apr 1 at 7:33

answered Apr 1 at 7:28

Martin RMartin R

31k33561

answered Apr 1 at 7:28

Martin RMartin R

31k33561

answered Apr 1 at 7:28

Martin RMartin R

31k33561