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Re: Harmonic numbers
From: |
Jaroslav Hajek |
Subject: |
Re: Harmonic numbers |
Date: |
Fri, 26 Feb 2010 07:15:06 +0100 |
On Thu, Feb 25, 2010 at 11:43 PM, Iliopoulos Vasileios
<address@hidden> wrote:
> Hi All,
>
> I am beginner in GNU Octave. I would like to ask you how to compute
> in Octave the folloing expresssion
>
> \sum_{j=1}^{n}harmonic(j)
>
>
> Where harmonic(j):=\sum_{i=1}^{j}1/i
>
>
> Thanks
>
You can generate the first n harmonic numbers by
h = cumsum (1 ./ (1:n));
And sum them using sum:
x = sum (h);
Or you can figure out the formula that holds for harmonic numbers:
sum (h(1:n)) = (n+1)*h(n) - n,
and use that to compute more accurately.
--
RNDr. Jaroslav Hajek, PhD
computing expert & GNU Octave developer
Aeronautical Research and Test Institute (VZLU)
Prague, Czech Republic
url: www.highegg.matfyz.cz
- Harmonic numbers, Iliopoulos Vasileios, 2010/02/25
- Re: Harmonic numbers,
Jaroslav Hajek <=