Does MATLAB have a feature to show step-by-step solution?

Question

Janice 2022-6-28

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1749685-does-matlab-have-a-feature-to-show-step-by-step-solution

评论： Janice 2022-6-28

Hello everyone.

I have a pretty complicated equation where I need to find the solution for B. I substituted the values of variables and separated the equation into 4 big "parts" to make it easier to read.

100-B==
100*exp(-(0.05)*m*(0.15))
    *normcdf(-((log(B)+(0.01)*m*(0.15)-log(100)+((0.05)-(0.03)-(0.2)^2/2)*m*(0.15))/((0.2)*sqrt(m*(0.15)))))
-exp(-((0.03)-(0.01))*m*(0.15))*B
    *normcdf(-((log(B)+(0.01)*m*(0.15)-log(100)+((0.05)-(0.03)+(0.2)^2/2)*m*(0.15))/((0.2)*sqrt(m*(0.15)))))
+(0.15)*(0.05)*100*symsum(exp(-(0.05)*(m-j)*(0.15)) ...
    *normcdf(-((log(B)+(0.01)*m*(0.15)-(log(B)+(0.01)*(m-j)*(0.15))+((0.05)-(0.03)-(0.2)^2/2)*(m-j)*(0.15)) ...
    /((0.2)*sqrt((m-j)*(0.15))))),j,0,m-1)
-(0.15)*((0.03)-(0.01))*exp((0.01)*m*(0.15))*B*symsum(exp(-((0.03)+(0.01))*(m-j)*(0.15))* ...
    normcdf(-((log(B)+(0.01)*m*(0.15)-(log(B)+(0.01)*(m-j)*(0.15))+((0.05)-(0.03)+(0.2)^2/2)*(m-j)*(0.15)) ...
    /((0.2)*sqrt((m-j)*(0.15))))),j,0,m-1)

where m denotes time and B denotes "price". I used vpasolve(<equation>,B) to solve the equation.

When I used m=4, it returns a real value for the solution, but starting from m=5, an imaginary component showed up. B is not supposed to have imaginary values at any point of time m.

I tried decomposing the equation and evaluating the value at each "part" at m=5 to figure out where the imaginary component originated from, but I cannot seem to find anything that results in a negative value inside the square root. Does MATLAB have a feature to show step-by-step solution?

Thank you very much!

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Christine Tobler 2022-6-28

If it's already defined as a symbol, j won't have any effect like that. I was just wondering as an outside possibility - if you had forgotten to define j, it might have been that its default value of sqrt(-1) was used instead.

Janice 2022-6-28

Oh I see, thank you very much!

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Answer 1

Torsten 2022-6-28

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1749685-does-matlab-have-a-feature-to-show-step-by-step-solution#answer_995555

在 MATLAB Online 中打开

m = 5;
B0 = 1;
B = fsolve(@(B)fun(B,m),B0)
B = 1
B = 1.0000
B = 2
B = 2.0000
B = 4.5000
B = 4.5000
B = 10.7500
B = 10.7500
B = 26.3750
B = 26.3750
B = 65.4375
B = 65.4375
B = 126.9775
B = 80.8225
B = 80.8225
B = 81.8887
B = 81.8887
B = 81.8346
B = 81.8346
B = 81.8344
B = 81.8344
Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
B = 81.8344
fun(B,m)
B = 81.8344
ans = 1.9267e-10
function res = fun(B,m)
B
sum1 = 0.0;
sum2 = 0.0;
for j=0:m-1
    sum1 = sum1 + exp(-(0.05)*(m-j)*(0.15)) ...
        *normcdf(-((log(B)+(0.01)*m*(0.15)-(log(B)+(0.01)*(m-j)*(0.15))+((0.05)-(0.03)-(0.2)^2/2)*(m-j)*(0.15)) ...
        /((0.2)*sqrt((m-j)*(0.15)))));
    sum2 = sum2 + exp(-((0.03)+(0.01))*(m-j)*(0.15))* ...
        normcdf(-((log(B)+(0.01)*m*(0.15)-(log(B)+(0.01)*(m-j)*(0.15))+((0.05)-(0.03)+(0.2)^2/2)*(m-j)*(0.15)) ...
        /((0.2)*sqrt((m-j)*(0.15)))));
end
res = 100*exp(-(0.05)*m*(0.15))...
    *normcdf(-((log(B)+(0.01)*m*(0.15)-log(100)+((0.05)-(0.03)-(0.2)^2/2)*m*(0.15))/((0.2)*sqrt(m*(0.15)))))...
    -exp(-((0.03)-(0.01))*m*(0.15))*B...
    *normcdf(-((log(B)+(0.01)*m*(0.15)-log(100)+((0.05)-(0.03)+(0.2)^2/2)*m*(0.15))/((0.2)*sqrt(m*(0.15)))))...
    +(0.15)*(0.05)*100*sum1 ...
    -(0.15)*((0.03)-(0.01))*exp((0.01)*m*(0.15))*B*sum2 - (100-B);
end

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Janice 2022-6-28

在 MATLAB Online 中打开

Thank you very much!

I'm really sorry I am fairly new to coding, but I am having trouble with defining the function.

I tried running the code

function res = fun(B,m)
B
sum1 = 0.0;
sum2 = 0.0;
for j=0:m-1
    sum1 = sum1 + exp(-(0.05)*(m-j)*(0.15)) ...
        *normcdf(-((log(B)+(0.01)*m*(0.15)-(log(B)+(0.01)*(m-j)*(0.15))+((0.05)-(0.03)-(0.2)^2/2)*(m-j)*(0.15)) ...
        /((0.2)*sqrt((m-j)*(0.15)))));
    sum2 = sum2 + exp(-((0.03)+(0.01))*(m-j)*(0.15))* ...
        normcdf(-((log(B)+(0.01)*m*(0.15)-(log(B)+(0.01)*(m-j)*(0.15))+((0.05)-(0.03)+(0.2)^2/2)*(m-j)*(0.15)) ...
        /((0.2)*sqrt((m-j)*(0.15)))));
end
res = 100*exp(-(0.05)*m*(0.15))...
    *normcdf(-((log(B)+(0.01)*m*(0.15)-log(100)+((0.05)-(0.03)-(0.2)^2/2)*m*(0.15))/((0.2)*sqrt(m*(0.15)))))...
    -exp(-((0.03)-(0.01))*m*(0.15))*B...
    *normcdf(-((log(B)+(0.01)*m*(0.15)-log(100)+((0.05)-(0.03)+(0.2)^2/2)*m*(0.15))/((0.2)*sqrt(m*(0.15)))))...
    +(0.15)*(0.05)*100*sum1 ...
    -(0.15)*((0.03)-(0.01))*exp((0.01)*m*(0.15))*B*sum2 - (100-B);
end

but I get an error message saying

"Error: Function definition are not supported in this context. Functions can only be created as local or nested functions in code files."

I am not quite sure where I did wrong..

Torsten 2022-6-28

编辑：Torsten 2022-6-28

在 MATLAB Online 中打开

Load all this in the editor and run it.

By "run" I mean click on the green RUN arrow to the right at the task bar.

m = 1:100;
B0 = 1;
B = arrayfun(@(m)fsolve(@(B)fun(B,m),B0),m)
%fun(B,m)
plot(m,B)
function res = fun(B,m)
sum1 = 0.0;
sum2 = 0.0;
for j=0:m-1
    sum1 = sum1 + exp(-(0.05)*(m-j)*(0.15)) ...
        *normcdf(-((log(B)+(0.01)*m*(0.15)-(log(B)+(0.01)*(m-j)*(0.15))+((0.05)-(0.03)-(0.2)^2/2)*(m-j)*(0.15)) ...
        /((0.2)*sqrt((m-j)*(0.15)))));
    sum2 = sum2 + exp(-((0.03)+(0.01))*(m-j)*(0.15))* ...
        normcdf(-((log(B)+(0.01)*m*(0.15)-(log(B)+(0.01)*(m-j)*(0.15))+((0.05)-(0.03)+(0.2)^2/2)*(m-j)*(0.15)) ...
        /((0.2)*sqrt((m-j)*(0.15)))));
end
res = 100*exp(-(0.05)*m*(0.15))...
    *normcdf(-((log(B)+(0.01)*m*(0.15)-log(100)+((0.05)-(0.03)-(0.2)^2/2)*m*(0.15))/((0.2)*sqrt(m*(0.15)))))...
    -exp(-((0.03)-(0.01))*m*(0.15))*B...
    *normcdf(-((log(B)+(0.01)*m*(0.15)-log(100)+((0.05)-(0.03)+(0.2)^2/2)*m*(0.15))/((0.2)*sqrt(m*(0.15)))))...
    +(0.15)*(0.05)*100*sum1 ...
    -(0.15)*((0.03)-(0.01))*exp((0.01)*m*(0.15))*B*sum2 - (100-B);
end