MATLAB Answers

Mac Sampson

numeric solve issue for an equation involving a logarithm

Asked by Mac Sampson
on 21 Sep 2012

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/ 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
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?


b(i) is the new variable i want to solve for. here is the complete loop:

p=input('TauesNtm1 Value?');
n=input('Nt Value?');
t=input('Taues1 Value?');
for i=(2:1:((n/2)-2)/1);
      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));
      b(i)=solve('r(i)=.5*log10(b(i)) + .5*b(i)');
 (2:1:((n/2)-2)/1)    % why /1 ?

thanks for catching that. that was a typo



No products are associated with this question.

4 Answers

Answer by Azzi Abdelmalek
on 21 Sep 2012
Edited by Azzi Abdelmalek
on 21 Sep 2012






could you explain this a bit more?

syms B
r(i) = [...]
b(i) = solve(r(i) = 1/2*log10(B)+1/2*B, B);

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 

so this code would go within the loop and without noting r and b as r(i) and b(i)?

Answer by 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));


Matt Tearle
on 21 Sep 2012

Yes, assuming you want to use the previous value for b as your initial guess. What errors do you get? It's entirely possible that fzero is failing to find the root.

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
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...

Answer by Matt Tearle
on 21 Sep 2012
Edited by 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);
bnum = double(bsolve);


I'm not sure what you mean about inverting the function? Do you mean that given a particular b value, you want to find the r value ?

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.

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.

Answer by 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!

  1 Comment

Vote early, vote often!

Join the 15-year community celebration.

Play games and win prizes!

Learn more
Discover MakerZone

MATLAB and Simulink resources for Arduino, LEGO, and Raspberry Pi

Learn more

Discover what MATLAB® can do for your career.

Opportunities for recent engineering grads.

Apply Today

MATLAB Academy

New to MATLAB?

Learn MATLAB today!