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Re: efficient way to fill a matrix based on a pattern?
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From: |
CdeMills |
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Subject: |
Re: efficient way to fill a matrix based on a pattern? |
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Date: |
Thu, 2 Sep 2010 00:10:27 -0700 (PDT) |
Jordi Gutiérrez Hermoso wrote:
>
>
>> omega = rho.^toeplitz(1:n)
>>
>> where you've previously defined rho and n?
>
> Oops, off-by-one error. Should be:
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> omega = rho.^toeplitz(0:n-1)
>
>
Faster:
omega = toeplitz(rho.^(0:n-1));
regards
Pascal
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