CMS: Smoothed Complex Series
Han de Bruijn
Consider the following function f defined by the complex power series:
f(z) = Sum(n=0,oo) c_n z^n ; c_n and z complex, n = natural >= 0
Suppose this series has a (real) radius of convergence = r > 0.
Consider the function F defined by the slightly modified power series:
F(z) = Sum(n=0,oo) C_n z^n
Where the C_n are defined by: C_n = s^(n^2).c_n , for real s < 1 .
F(z) could be called a smoothed complex function (: take s close to 1)
Theorem (HdB). F(z) is an _entier_ function.
Proof(?). Define R = r/s^n . Then R approaches infinity for n -> oo
and |C_n.R^n| = s^(n^2).|c_n|.(r/s^n)^n = |c_n|.r^n = bounded.
Interpretation. Let s = exp(-(2.pi/N)^2/2) = exp(-(d.2.pi/L)^2/2) .
Then the series, when specified for z = exp(i.t) = the unit circle,
corresponds with a Fourier series expansion for an arbitrary Gaussian
smoothed closed curve with thickness d and length L. And it has almost
converged already for n = N . As argued in:
http://hdebruijn.soo.dto.tudelft.nl/jaar2004/Fransen3.pdf
Right or wrong? Any pointers to existing theory are quite welcome.
Han de Bruijn
Han de Bruijn
> CMS = "Complex Made Simple", the book by David C. Ullrich.
>
> Consider the following function f defined by the complex power series:
>
> f(z) = Sum(n=0,oo) c_n z^n ; c_n and z complex, n = natural >= 0
>
> Suppose this series has a (real) radius of convergence = r > 0.
> Consider the function F defined by the slightly modified power series:
>
> F(z) = Sum(n=0,oo) C_n z^n
>
> Where the C_n are defined by: C_n = s^(n^2).c_n , for real s < 1 .
Oops, 0 < s < 1 .
David C. Ullrich
<Han.de...@DTO.TUDelft.NL> wrote:
>CMS = "Complex Made Simple", the book by David C. Ullrich.
>
>Consider the following function f defined by the complex power series:
>
> f(z) = Sum(n=0,oo) c_n z^n ; c_n and z complex, n = natural >= 0
>
>Suppose this series has a (real) radius of convergence = r > 0.
>Consider the function F defined by the slightly modified power series:
>
> F(z) = Sum(n=0,oo) C_n z^n
>
>Where the C_n are defined by: C_n = s^(n^2).c_n , for real s < 1 .
>
>F(z) could be called a smoothed complex function (: take s close to 1)
>
>Theorem (HdB). F(z) is an _entier_ function.
Yes (the English spelling would be "entire").
>Proof(?). Define R = r/s^n . Then R approaches infinity for n -> oo
> and |C_n.R^n| = s^(n^2).|c_n|.(r/s^n)^n = |c_n|.r^n = bounded.
More or less. One slight error is that |c_n|r^n need not be bounded.
(Look again at the stuff about the radius of convergence of a power
series: Say S is the set of rho > 0 such that |c_n| rho^n is bounded.
Then r is the sup of S; you're assuming that r is actually an element
of S, which need not be so.)
Also you need to show that C_n R^n is bounded for every fixed
R > 0; the fact that your R is actually R_n makes the way you
wrote things a little confusing (although since R_n -> infinity
showing C_n (R_n)^n is bounded is actually more than enough).
You meant this:
Fix rho with 0 < rho < r; then |c_n| rho^n is bounded. Now
suppose R > 0. If n is large enough, say n > N, we have
R < rho/s^n. So for n > N we have
|C_n.R^n| <= s^(n^2).|c_n|.(rho/s^n)^n = |c_n|.rho^n
Hence C_n R^n is bounded.
>Interpretation. Let s = exp(-(2.pi/N)^2/2) = exp(-(d.2.pi/L)^2/2) .
>Then the series, when specified for z = exp(i.t) = the unit circle,
>corresponds with a Fourier series expansion for an arbitrary Gaussian
>smoothed closed curve with thickness d and length L. And it has almost
>converged already for n = N . As argued in:
>
>http://hdebruijn.soo.dto.tudelft.nl/jaar2004/Fransen3.pdf
>
>Right or wrong?
Well yes, one could regard your F as the result of convolving
f with a gaussian (hence smoothing it). I have no idea what
that bit about "almost converged already for n = N" means...
>Any pointers to existing theory are quite welcome.
>Han de Bruijn
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
Han de Bruijn
> On Mon, 25 May 2009 15:31:51 +0200, Han de Bruijn
> <Han.de...@DTO.TUDelft.NL> wrote:
>
>>CMS = "Complex Made Simple", the book by David C. Ullrich.
>>
>>Consider the following function f defined by the complex power series:
>>
>> f(z) = Sum(n=0,oo) c_n z^n ; c_n and z complex, n = natural >= 0
>>
>>Suppose this series has a (real) radius of convergence = r > 0.
>>Consider the function F defined by the slightly modified power series:
>>
>> F(z) = Sum(n=0,oo) C_n z^n
>>
>>Where the C_n are defined by: C_n = s^(n^2).c_n , for real s < 1 .
>>
>>F(z) could be called a smoothed complex function (: take s close to 1)
>>
>>Theorem (HdB). F(z) is an _entier_ function.
>
> Yes (the English spelling would be "entire").
Right.
>>Proof(?). Define R = r/s^n . Then R approaches infinity for n -> oo
>> and |C_n.R^n| = s^(n^2).|c_n|.(r/s^n)^n = |c_n|.r^n = bounded.
>
> More or less. One slight error is that |c_n|r^n need not be bounded.
> (Look again at the stuff about the radius of convergence of a power
> series: Say S is the set of rho > 0 such that |c_n| rho^n is bounded.
> Then r is the sup of S; you're assuming that r is actually an element
> of S, which need not be so.)
Quite right.
> Also you need to show that C_n R^n is bounded for every fixed
> R > 0; the fact that your R is actually R_n makes the way you
> wrote things a little confusing (although since R_n -> infinity
> showing C_n (R_n)^n is bounded is actually more than enough).
>
> You meant this:
>
> Fix rho with 0 < rho < r; then |c_n| rho^n is bounded. Now
> suppose R > 0. If n is large enough, say n > N, we have
> R < rho/s^n. So for n > N we have
>
> |C_n.R^n| <= s^(n^2).|c_n|.(rho/s^n)^n = |c_n|.rho^n
>
> Hence C_n R^n is bounded.
Right ! with an exclamation mark: the Theorem is true.
>>Interpretation. Let s = exp(-(2.pi/N)^2/2) = exp(-(d.2.pi/L)^2/2) .
>>Then the series, when specified for z = exp(i.t) = the unit circle,
>>corresponds with a Fourier series expansion for an arbitrary Gaussian
>>smoothed closed curve with thickness d and length L. And it has almost
>>converged already for n = N . As argued in:
>>
>>http://hdebruijn.soo.dto.tudelft.nl/jaar2004/Fransen3.pdf
>>
>>Right or wrong?
>
> Well yes, one could regard your F as the result of convolving
> f with a gaussian (hence smoothing it). I have no idea what
> that bit about "almost converged already for n = N" means...
It means that the "relative error" is expected to be = exp(-(2.pi)^2/2)
which is approximately 10^(-9) and for practical purposes small enough.
(Well, sort of, please don't shoot me for this) That's great! So now we
can make entire functions out of arbitrary complex power series without
being concerned about any radius of convergence. And we can introduce a
sensible (computationally advantageous) cut-off at n = N , effectively
turning these entire functions into polynomials.
>>Any pointers to existing theory are quite welcome.
Encouraging that I've been not so far off in guessing where my purported
proof could receive your criticism. Thank you very much !
Han de Bruijn
David C. Ullrich
<Han.de...@DTO.TUDelft.NL> wrote:
If the goal is just to turn entire functions into polynomials it's
much simpler to just let the polynomials be the partial sums
of the power series.
>>>Any pointers to existing theory are quite welcome.
>
>Encouraging that I've been not so far off in guessing where my purported
>proof could receive your criticism. Thank you very much !
>
>Han de Bruijn
David C. Ullrich