
Re: sum of exponentials
Posted:
Nov 23, 2012 3:46 PM


"dwi" wrote in message <k8o3vo$ia$1@newscl01ah.mathworks.com>... > A correction: > x8=(x7*e^(1))/e^(1) > ie without using x1,x2,x3 but only the immediate previous nonzero values          With the correction you made, here is my modified code:
% An example: x = [12,17,21,0,0,0,31,0,0]; e = exp(1); % < I assume this is what 'e' is
% The code: a = 0; b = 1; f = 0; for k = 1:length(x) if x(k) ~= 0 a = x(k) + a*f*e^(1); b = 1 + b*f*e^(1); f = 1; else x(k) = a/b; f = 0; end end
% The results:
[(x(3)+x(2)*e^(1)+x(1)*e^(2))/(1+e^(1)+e^(2)); x(4); x(5); x(6)] =
19.21081095724738 19.21081095724738 19.21081095724738 19.21081095724738
[x(7); x(8); x(9)] =
31 31 31
> Thank you for your answer but if I use this the index of the exponent never changes.I want it to decrease as k increases, and i don't know hot to calculate a sum like this.
In the earlier thread where you said, "but if I use this the index of the exponent never changes", I don't know what you meant, but that code was doing just what you had asked for at that time.
Roger Stafford

