numeric solve issue for an equation involving a logarithm

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?

[Azzi Abdelmalek]

what is b(i)?

[Mac Sampson]

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)');

end

[Azzi Abdelmalek]

 (2:1:((n/2)-2)/1)    % why /1 ?

[Mac Sampson]

thanks for catching that. that was a typo

Answer 1

[Azzi Abdelmalek]

use

 log10

not

 Log10

[Mac Sampson]

thanks for the tip.I changed it, but i am still getting an error message of:

??? 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)');

[Walter Roberson]

() is used for function calls in MuPAD. Use [] for indexing. Or better yet, use simple variables since you aren't importing any value from the workspace.

b(i) = solve('r=1/2*log10(b)+1/2*b','b');

Remember, when you use a string argument to solve(), values are not imported from the MATLAB workpace! If you want to import the value of r(i) then don't use a string argument to solve:

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

[Azzi Abdelmalek]

p=1;
n=1;
t=1;
n=10
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(['-' num2str(r(i)) '+.5*log10(b) + .5*b'],'b');
end

[Mac Sampson]

could you explain this a bit more?

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

[Azzi Abdelmalek]

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]

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

Answer 2

[Matt Tearle]

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

[Mac Sampson]

is b(i) = fzero(@(b) (log10(b) + b)/2 - r(i),b(i-1));

really the correct way to type it in? I am getting all kinds of syntax errors popping up

[Matt Tearle]

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.

[Mac Sampson]

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]

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 3

[Matt Tearle]

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);

[Mac Sampson]

here is the revised code, per your instruction:

clear all;

p=input('TauesNtm1 Value?');

n=input('Nt Value?');

t=input('Taues1 Value?');

(5)imax = (n/2)-2;

(6)r = zeros(imax,1);

(7)syms b

(9)for i = 2:imax;

 (10)     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));

(11)    bsolve(i)=solve((.5*log10(b) + b)/2 == r(i),b);

(12)

(14)bnum = double(bsolve);

here is the error message I am getting:

Warning: Size vector should be a row vector with integer elements. > In testloopwithgraph2 at 6 ??? Error using ==> char Conversion to char from logical is not possible.

Error in ==> solve>getEqns at 160 vc = char(v);

Error in ==> solve at 84 [eqns,vars] = getEqns(varargin{:});

Error in ==> testloopwithgraph2 at 11 bsolve(i)=solve((.5*log10(b) + b)/2 == r(i),b);

[Walter Roberson]

Either update your MATLAB version to R2012a or later, or use

bsolve(i)=solve((.5*log10(b) + b)/2 - r(i),b);

[Mac Sampson]

after the adjustment to subtracting r from both sides as suggested, this is working for giving me the a values, but still gives no b (tau) values except to say conj(b)

Warning: Size vector should be a row vector with integer elements. > In testloopwithgraph2 at 6 Q values Columns 1 through 7

         0   -0.0236   -0.0182   -0.0128   -0.0074   -0.0020    0.0035

Columns 8 through 14

    0.0089    0.0143    0.0197    0.0252    0.0306    0.0360    0.0414

Columns 15 through 21

    0.0468    0.0523    0.0577    0.0631    0.0685    0.0739    0.0794

Columns 22 through 23

    0.0848    0.0902

Tau Values conj(b)

[Walter Roberson]

By solving the general case in advance, you can avoid having to do the solve() each time. The general solution is

b(i) = (1/2)*10000^r(i)*exp(-4*r(i)*ln(10))*lambertw(2*ln(10)*exp(4*r(i)*ln(10)))/ln(10)

However, lambertw() is defined only by the Symbolic Math toolbox, which gives a fair bit of overhead if you need to do this many times. For MATLAB code for a numeric version of the principle branch of lambertw(), see http://www.mathworks.com/matlabcentral/newsreader/view_thread/32527

[Mac Sampson]

that was a great step forward. Thanks. Unfortunately, I tried to invert the function as well in mathematica and it spit out the wrong answers for b, just like it has now done with the new code. It works if i manually substitute a[[1]] into the original r(1)=0.5*log10(b)+0.5*b and gives me the right answer, but not if i try to invert that function.

[Walter Roberson]

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 ?

[Mac Sampson]

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]

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 4

[Mac Sampson]

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!

[Walter Roberson]

Vote early, vote often!