Solve non-linear equation with varying value

I have a equation to solve

exp(-10*z)-9*exp(-2*z)+(8-4*x)=0

here

x=linspace(0,1,101)

Since the equation itself would change to a new equation for each x, i am unable to figure out how to solve it. I need to solve for z. Can somebody suggest an approach to this.

Answer 1

[madhan ravi]

x=linspace(0,1,101);
Result=cell(1,numel(x)); % preallocate
for i = 1:numel(x)
Result{i}=fzero(@(z) exp(-10*z)-9*exp(-2*z)+(8-4*x(i)),[0 5]);
                                                    %   ^^^--initial guess interval
end
celldisp(Result)
[Result{:}] % double array

Gives:

>> [Result{:}]

ans =

  Columns 1 through 7

         0    0.0049    0.0096    0.0142    0.0187    0.0231    0.0274

  Columns 8 through 14

    0.0316    0.0357    0.0398    0.0438    0.0477    0.0517    0.0555

  Columns 15 through 21

    0.0594    0.0632    0.0670    0.0707    0.0744    0.0782    0.0818

  Columns 22 through 28

    0.0855    0.0892    0.0928    0.0964    0.1001    0.1037    0.1073

  Columns 29 through 35

    0.1109    0.1145    0.1181    0.1217    0.1252    0.1288    0.1324

  Columns 36 through 42

    0.1360    0.1396    0.1432    0.1468    0.1504    0.1540    0.1576

  Columns 43 through 49

    0.1612    0.1648    0.1685    0.1721    0.1758    0.1794    0.1831

  Columns 50 through 56

    0.1868    0.1905    0.1942    0.1979    0.2016    0.2054    0.2091

  Columns 57 through 63

    0.2129    0.2167    0.2205    0.2244    0.2282    0.2321    0.2359

  Columns 64 through 70

    0.2398    0.2438    0.2477    0.2517    0.2556    0.2596    0.2637

  Columns 71 through 77

    0.2677    0.2718    0.2759    0.2800    0.2842    0.2883    0.2925

  Columns 78 through 84

    0.2968    0.3010    0.3053    0.3096    0.3140    0.3183    0.3227

  Columns 85 through 91

    0.3272    0.3317    0.3362    0.3407    0.3453    0.3499    0.3545

  Columns 92 through 98

    0.3592    0.3640    0.3687    0.3735    0.3784    0.3833    0.3882

  Columns 99 through 101

    0.3932    0.3982    0.4033

>> 

[P K]

Hello Madan,Thanks for your response. But the z value should not be negative.

How you have decided that x0 =0 ? If i change x0, my answers would change. Can you explain a bit .

[madhan ravi]

Ravi Singh, x0 is the initial guess , why z shouldn't be negative ? , plot the graph to verify?

[P K]

Because I have a plot which based on the above equation. The maximum value of Z=0.4(approx) at x=1.Thats why i am saying.

[madhan ravi]

Perhaps try this with fsolve() :

x=linspace(0,1,101);
Result=cell(1,numel(x));
for i = 1:numel(x)
Result{i}=fsolve(@(z) exp(-10*z)-9*exp(-2*z)+(8-4*x(i)),[-5 5],optimoptions('fsolve','Display','off'));
                                                      %  ^^---- change it to 0 if you only want the positive roots
end
celldisp(Result)

Note: When plotting the graph two roots were sighted , fsolve is able to find those two roots.

fplot(@(z) exp(-10*z)-9*exp(-2*z)+(8-4*1)) 
                                    %  ^--- from 0 to 1 with 101 points to verify the roots
xlim([-1 1])  % to see the roots clearly
ylim([-1 1])
grid on

See the below graph the two roots can be sighted :

>> Result{end}

ans =

   -0.2339    0.4033 % as you said approximately 4 (the positive root)

[madhan ravi]

@Ravi Singh, See edited answer , if you got the answer to your question make sure to accept the answer else let know of your requirements which part is not clear.

[madhan ravi]

Anytime :)