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## RE: suite

**From**: |
Hall, Benjamin |

**Subject**: |
RE: suite |

**Date**: |
Sat, 11 Dec 2004 15:37:30 -0500 |

If you write out some of the terms you get
U_0 = x
U_1 = a * x + b
U_2 = a ( a*x + b ) + b
U_3 = a ( a ( a*x + b ) + b ) + b = a^3*x + a^2b + ab + b
U_n = a^(n-1)*x + a^(n-2)*b + a^(n-3)*b + ... ab + b
so you could calculate U_n as follows
U_n = a^(n-1)*U_1 + sum( cumprod( repmat(a, [1 n-2] ) ) * b )
It's harder to read than Mike's suggestion and you only get U evaulated for
a single case plus its is only marginally faster than the for-loop below on
my machine:
0.66 seconds for n=200 repeated 20 times for "for loop"
0.55 seconds for formula above with n=200 and repeated 20 times
So much for the fancy formula...
Ben
-----Original Message-----
From: Mike Miller [mailto:address@hidden
Sent: Saturday, December 11, 2004 1:12 PM
To: adel.essafi
Cc: help
Subject: Re: suite
On Sat, 11 Dec 2004, adel.essafi wrote:
>* hi list*
>* please, how can I define a suite in octave*
>* U(n+1)=3*U(n)+2 (for example)*
>* and calculate u(1000)*
A simple method:
n=1000;
U=zeros(1,n);
U0=4; #or whatever you want
U(1)=3*U0+2;
for i=2:n,
U(i)=3*U(i-1)+2;
end
That would work, but there are probably much nicer ways to do this and
other people will have more sophisticated ideas than mine!
Mike
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Octave is freely available under the terms of the GNU GPL.
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Subscription information: http://www.octave.org/archive.html
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**suite**, *adel\.essafi*, `2004/12/11`
**RE: suite**,
*Hall, Benjamin* **<=**