Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

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

 Messages: [ Previous | Next ]
 dwi Posts: 17 Registered: 10/18/12
Re: sum of exponentials
Posted: Nov 24, 2012 7:16 PM

"Roger Stafford" wrote in message <k8r036\$nn7\$1@newscl01ah.mathworks.com>...
> "dwi" wrote in message <k8qlsv\$ktj\$1@newscl01ah.mathworks.com>...
> > Ok, I understand now how this works. But still, you said the result will be
> > x(3)+x(2)*e^(-1)+x(1)*e^(-2))/(1+e^(-1)+e^(-2));
> > while I want
> > (x(3)*e^(-1)+x(2)*e^(-2)+x(1)*e^(-3))/(e^(-1)+e^(-2)+e^(-3));
> > Also, how would your code change if I had e^(-1/20), e^(-2/20), e^(-3/20) etc?

> - - - - - - - -
> The two expressions
>
> (x(3)*e^(-1)+x(2)*e^(-2)+x(1)*e^(-3))/(e^(-1)+e^(-2)+e^(-3))
>
> and
>
> (x(3)+x(2)*e^(-1)+x(1)*e^(-2))/(1+e^(-1)+e^(-2))
>
>
> are identically equal. Just divide the numerator and denominator of the first expression by e^(-1) to get the second expression. What you want and what this code produces are the same thing.
>
> As to your second question, just the two lines
>
> a = x(k) + a*f*e^(-1);
> b = 1 + b*f*e^(-1);
>
> would need to be changed to:
>
> a = x(k) + a*f*e^(-1/20);
> b = 1 + b*f*e^(-1/20);
>
> Roger Stafford

Ok, thank you!

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