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Borel Transform

Nomenclature in complex analysisconvergence of complex power series - infinite convergence radiusradius of convergence of half iterate of sinh(z)?Extending an analytic function to an entire functionRelation between simple pole and radius of convergenceAn analogue for Fourier transform on the relation between Fourier and Laurent seriesDoes a holomorphic function converges and equal to its Taylor series?Functions of exponential type and associated functions in the sense of Borel.Fourier transform of meromorphic functionConvergence of a power series: general question

0

$begingroup$

Suppose a complex series with complex coefficients $$a=sum_n=0^inftyc_nz^-n-1$$
Then its Borel transform is defined by $$Ba(x)=sum_n=0^inftyfracc_nn!x^n$$ ($x$ is a complex number).

The claim is that if the series $a$ has a non-zero radius of convergence, then the Borel transform $Ba(x)$ is an entire function(that is, holomorphic everywhere in the complex plane) of exponential type(its growth is bounded by an exponential function).

If the radius of convergence of the Borel tranform is finite, then that of the original series is zero.

How does one prove these claims?

sequences-and-series complex-analysis summation

share|cite|improve this question

edited Mar 12 at 15:11

Andrews

1,2691421

asked Mar 12 at 10:45

Mani JhaMani Jha

94

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  en.wikipedia.org/wiki/Cauchy%E2%80%93Hadamard_theorem
  $endgroup$
  – Giuseppe Negro
  Mar 12 at 10:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Done. Thanks a lot.
  $endgroup$
  – Mani Jha
  Mar 12 at 11:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, but I still can't prove the exponential bound
  $endgroup$
  – Mani Jha
  Mar 12 at 11:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The point is, I guess, that $|c_n|^1/n$ has a finite limsup, because of the condition on the radius of convergence of $a$. Thus, $$frac^1/n(n!)^1/n to 0.$$From this it should arise an exponential bound, I think...
  $endgroup$
  – Giuseppe Negro
  Mar 12 at 11:21
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Suppose a complex series with complex coefficients $$a=sum_n=0^inftyc_nz^-n-1$$
Then its Borel transform is defined by $$Ba(x)=sum_n=0^inftyfracc_nn!x^n$$ ($x$ is a complex number).

The claim is that if the series $a$ has a non-zero radius of convergence, then the Borel transform $Ba(x)$ is an entire function(that is, holomorphic everywhere in the complex plane) of exponential type(its growth is bounded by an exponential function).

If the radius of convergence of the Borel tranform is finite, then that of the original series is zero.

How does one prove these claims?

sequences-and-series complex-analysis summation

share|cite|improve this question

edited Mar 12 at 15:11

Andrews

1,2691421

asked Mar 12 at 10:45

Mani JhaMani Jha

94

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  en.wikipedia.org/wiki/Cauchy%E2%80%93Hadamard_theorem
  $endgroup$
  – Giuseppe Negro
  Mar 12 at 10:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Done. Thanks a lot.
  $endgroup$
  – Mani Jha
  Mar 12 at 11:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, but I still can't prove the exponential bound
  $endgroup$
  – Mani Jha
  Mar 12 at 11:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The point is, I guess, that $|c_n|^1/n$ has a finite limsup, because of the condition on the radius of convergence of $a$. Thus, $$frac^1/n(n!)^1/n to 0.$$From this it should arise an exponential bound, I think...
  $endgroup$
  – Giuseppe Negro
  Mar 12 at 11:21
  
  

  
  
  
  

add a comment | 

$begingroup$

Suppose a complex series with complex coefficients $$a=sum_n=0^inftyc_nz^-n-1$$
Then its Borel transform is defined by $$Ba(x)=sum_n=0^inftyfracc_nn!x^n$$ ($x$ is a complex number).

The claim is that if the series $a$ has a non-zero radius of convergence, then the Borel transform $Ba(x)$ is an entire function(that is, holomorphic everywhere in the complex plane) of exponential type(its growth is bounded by an exponential function).

If the radius of convergence of the Borel tranform is finite, then that of the original series is zero.

How does one prove these claims?

sequences-and-series complex-analysis summation

share|cite|improve this question

edited Mar 12 at 15:11

Andrews

1,2691421

asked Mar 12 at 10:45

Mani JhaMani Jha

94

$endgroup$

share|cite|improve this question

edited Mar 12 at 15:11

Andrews

1,2691421

asked Mar 12 at 10:45

Mani JhaMani Jha

94

share|cite|improve this question

edited Mar 12 at 15:11

Andrews

1,2691421

asked Mar 12 at 10:45

Mani JhaMani Jha

94

edited Mar 12 at 15:11

Andrews

1,2691421

edited Mar 12 at 15:11

Andrews

1,2691421

asked Mar 12 at 10:45

Mani JhaMani Jha

94

asked Mar 12 at 10:45

Mani JhaMani Jha

94