RE: suite

[Hall, Benjamin]

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.

Octave's home on the web:  http://www.octave.org
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