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

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Answers (4)

Azzi Abdelmalek
Azzi Abdelmalek on 21 Sep 2012
Edited: Azzi Abdelmalek on 21 Sep 2012
use
log10
not
Log10
  6 Comments
Azzi Abdelmalek
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)
Mac Sampson
Mac Sampson on 21 Sep 2012
so this code would go within the loop and without noting r and b as r(i) and b(i)?

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Matt Tearle
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
Mac Sampson
Mac Sampson on 21 Sep 2012
i don't want to use previous values of b. the first equation gives me a value for r(i). I then want to plug in r(i) into the next equation to solve for b. this gives me one value for r(i) at that iteration and one value for b(i) at that iteration. I then want it to return to the beginning of the loop and do it all over again with i+1, giving me two new values for r(i+1) and b(i+10
Matt Tearle
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...

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Matt Tearle
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
Mac Sampson
Mac Sampson on 22 Sep 2012
no. r(i)=0.5*log10(b)+0.5*b is the equation i am using. I know r(i) from the previous equation in the loop. I want to solve for b. When I try to invert the function and get b(i) in terms of r(i), it spits out the wrong values.
Walter Roberson
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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Mac Sampson
Mac Sampson on 22 Sep 2012
Thanks so much to all of you. I believe with your help I have found what is wrong and will embark on tracking down and fixing the problem. Thanks again for all your help!

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