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Topic: sum of exponentials
Replies: 9   Last Post: Nov 24, 2012 7:16 PM

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 Roger Stafford Posts: 5,929 Registered: 12/7/04
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 non-zero 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

Date Subject Author
11/22/12 dwi
11/22/12 Roger Stafford
11/23/12 dwi
11/23/12 dwi
11/23/12 dwi
11/23/12 dwi
11/23/12 Roger Stafford
11/24/12 dwi
11/24/12 Roger Stafford
11/24/12 dwi