numeric solve issue for an equation involving a logarithm
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I am trying to solve an equation involving a common logorithm within a loop. I first solve an equation to get r(i). then I take that r(i) value and plug it into an equation to solve for b(i).The second equation is:
b(i)=solve('r(i)=.5*Log10(b(i)) + .5*b(i)')
It is having a lot of trouble solving this. This is the error message:
Warning: Could not find an exact (case-sensitive) match for 'Log10'.
/Applications/MATLAB_R2009aSV.app/toolbox/matlab/elfun/log10.m is a
case-insensitive match and will be used instead.
You can improve the performance of your code by using exact
name matches and we therefore recommend that you update your
usage accordingly. Alternatively, you can disable this warning using
warning('off','MATLAB:dispatcher:InexactCaseMatch').
This warning will become an error in future releases.
> In testloopwithgraph at 6
??? Error using ==> mupadengine.mupadengine>mupadengine.feval at 162
Error: no indeterminate(s) [numeric::solve]
Error in ==> solve>mupadSolve at 232
list = feval(symengine,'mlfsolve',eqns,vars);
Error in ==> solve at 93
[R,symvars,order] = mupadSolve(eqns,vars);
Error in ==> testloopwithgraph at 7
b(i)=solve('r(i)=.5*Log10(b(i)) + .5*b(i)');
I can numericaly solve this equation in mathematica with just a warning about how the output might not cover all values, but it does give me values. In my matlab code it won't give me values at all. Is there a way around this?
4 Comments
Answers (4)
Azzi Abdelmalek
on 21 Sep 2012
Edited: Azzi Abdelmalek
on 21 Sep 2012
use
log10
not
Log10
6 Comments
Azzi Abdelmalek
on 21 Sep 2012
Edited: Azzi Abdelmalek
on 21 Sep 2012
do you want to use symbolic toolbox. In your case, implicit solution could'nt be found, unless you use just
syms b
%with r known
syms b
sol=solve(-r(1)+0.5*log10(b)+0.5*b,b)
Matt Tearle
on 21 Sep 2012
If I understand your intent correctly, you're trying to solve the equation
r_i = (Log10(b) + b)/2
for b, given a (numeric?) value of r_i. Then you want to store that b value as b_i. (Repeat for i+1)
If so, then (1) you don't need to do this symbolically and (2) you are getting the error because solve can't figure out what the variable is.
I'd suggest using fzero:
for i = 2:n
% get r(i)
% solve for b(i) using b(i-1) as an initial guess
b(i) = fzero(@(b) (log10(b) + b)/2 - r(i),b(i-1));
end
4 Comments
Matt Tearle
on 21 Sep 2012
Yes, I understand that, but fzero is a numerical solver, so it requires an initial guess for the solution. When doing stuff like this in a loop, it's not uncommon to use the previous solution as the starting point for the next solution. If r(i) varies somewhat slowly and smoothly with i, that would make sense. But it looks like maybe that's not the case here.
So, see my new answer...
Matt Tearle
on 21 Sep 2012
Edited: Matt Tearle
on 21 Sep 2012
Didn't see your comment with your code. This works:
imax = (n/2)-2;
r = zeros(imax,1);
syms b
for i = 2:imax;
r(i)=.5*log10((2*p)^((1 - 2*(i - 2))/(n - 4))*(t/2)^((2*(i - 2))/(n - 4)))+(i - 2)*(((n - 4)*p-2*p-.5*t)/(n - 4));
bsolve(i)=solve((log10(b) + b)/2 == r(i),b);
end
bnum = double(bsolve);
8 Comments
Walter Roberson
on 22 Sep 2012
b(i) in terms of r(i) is the formula I show above, involving LambertW .
I tested in Maple and the results appear to be correct. However, in cases where r(i) is complex, there can be multiple solutions.
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