MAKING SUBJECT OF FOMULAR

HOW DO I MAKE X SUBJECT OF FOMULAR IN THIS EQUATION x = exp(x+y).

I have tried the following code

syms x y

solve('x = exp(x+y)',x)

and this give me *

[sym empty].*

How best can I do it.

[Chibuzo Chukwu]

I am also stoked with a situation of the same form. All out gigs should please look into this kind of problem and suggest a way out.

Answer 1

[Walter Roberson]

In older versions of MATLAB,

solve('x = exp(x+y)', 'x')

In more modern versions

syms x y
solve(x == exp(x+y), x)

However, MATLAB is not able to find a solution. A solution exists, and is

-lambertw(-exp(y))

but MATLAB is not strong on Lambert W processing.

[Walter Roberson]

Chibuzo Chukwu comments to me

Your contribution has been very useful

Answer 2

[Chibuzo Chukwu]

Thanks for your response. I have tried to read about the Lambert W function and it has been helpful but can you look at this problem?

There are series of values for x and a and it's required to have a linear plot of a against x in the equation connecting them

ln(x^alpha * a) = c- 2*x*m

To determine values alpha and m from the graph

Is there anyway this can be rearrange to achieve this?

[Walter Roberson]

I am not clear as to what the inputs are, and what is to be determined? Is the task to fit scalars c and m given inputs x, alpha, and a ?

[Chibuzo Chukwu]

I forgot to add that c is just a constant which I have litle or no interest on. However, its not known as well.

[John D'Errico]

Please stop adding answers. Moved from an answer to a comment.

"The inputs are x and a and the parameters to determine are m and alpha"

Anyway, this is a completely different question. Askit as a new question, rather than resurrecting an old question into a zombie question.

[Walter Roberson]

Are x and a restricted to positive ?

[Chibuzo Chukwu]

Please pardon my use of the forum. I so new here. x and a are all positive numbers

[Walter Roberson]

First approximations, which can be used in fitting:

alpha = ((-x2+x3)*ln(a1)+(x1-x3)*ln(a2)-ln(a3)*(x1-x2))/((x2-x3)*ln(x1)+(-x1+x3)*ln(x2)+ln(x3)*(x1-x2))
c = ((x2*ln(a3)-ln(a2)*x3)*ln(x1)+(-x1*ln(a3)+ln(a1)*x3)*ln(x2)+ln(x3)*(ln(a2)*x1-x2*ln(a1)))/((x2-x3)*ln(x1)+(-x1+x3)*ln(x2)+ln(x3)*(x1-x2))
m = ((-ln(a2)+ln(a3))*ln(x1)+(ln(a1)-ln(a3))*ln(x2)-ln(x3)*(ln(a1)-ln(a2)))/((2*x2-2*x3)*ln(x1)+(-2*x1+2*x3)*ln(x2)+2*ln(x3)*(x1-x2))

If you calculate sum of squared residues (log(x^alpha * a) -( c- 2*x*m))^2 for a number of x and a, then you can use standard calculus techniques: differentiate ssr for m, solve that for m, substitute into ssr, differentiate with respect to alpha, solve for alpha: then there appear to be patterns for m and alpha that you could probably write formulae for with a bit of study: it looks like there are probably accessible analytical optima. However, these will be in terms of c, and solving for c analytically appears to be too messy by the time you get to 5 terms.

Perhaps a hybrid approach then: construct the sum of squared residues, use calculus to find optimal m and alpha in terms of c, and then do a numeric minimization to find the optimal c, after which you can back-construct alpha and then m.

[Chibuzo Chukwu]

I wish to contact you directly. Probably, I will need to send you the data set which in some cases can be more than 60 rows for a and m.

[Walter Roberson]

60 rows can easily be attached here as a .mat or xlsx file.

[Chibuzo Chukwu]

I have attached the data. The goal is to get a single value of alpha and m

[Walter Roberson]

The column heading of the file implies that the second column is Ln(a) rather than a itself. I will assume that is true.

T = readtable('Spectrum.xls');   %xlsread 'basic' does not support Unicode
x = T{:,1};
Ln_a = T{:,2};
syms alpha m c real
ssr = sum( (alpha*log(x) + Ln_a - (c + 2 * x * m)).^2 );
sol_m = simplify(solve(diff(ssr, m), m));
ssr_m = simplify(subs(ssr, m, sol_m));
sol_alpha = simplify(solve(diff(ssr_m, alpha), alpha));
ssr_m_alpha = simplify(subs(ssr_m, alpha, sol_alpha));

best_c = simplify(solve(diff(ssr_m_alpha,c),c));
best_alpha = simplify( subs(sol_alpha, c, best_c) );
best_m = simplify( subs(subs(sol_m, c, best_c), alpha, best_alpha) );

Each of those best_* variables is a rational value. You will probably want to double() them to get floating point approximations.

>> double([best_alpha, best_c, best_m])
ans =
          1.03482316989617          6.11693587912047      0.000599853145856567

[Chibuzo Chukwu]

Thanks for your response and the wanderful piece of code to what I asked. To be honest, your code addressed my initial question but it seems its not exactly what I need.

My purpose is to find the value of alpha and m for which when Ln_a is plotted against x it gives a straight line (or linear graph). Looking at the equation you will observe that its a power equation but I'm interested on at the top left end of the graph of Ln_a Vs x gives a linear graph.

I only hope I have communicated my intentions well. Thanks alot so far

[Walter Roberson]

? Ln_a and x are inputs. If you plot Ln_a against x you will get whatever you get, regardless of alpha and m values.

[Chibuzo Chukwu]

Yes, your observations are correct. What I am saying is, say there are two points p1 and p2 of the line that meets this conditions (ie the linear part of the graph of the plot of Ln_a aginst x). Then can the code be designed to only perform this solution for this range (ie between p1 and p2) of data other than the entire data as you did before?

[Walter Roberson]

No. If part of your data has a linear fit then it makes no sense to fit a power model to it.

If you want a linear fit then set alpha to 0. That would get ln a on the left side and linear terms on the right leading to aa simple polynomial fit.

[Chibuzo Chukwu]

Thanks a lot. Your insight has been very useful and infact you've answered my question very satisfactory. I will get back to you if I find any other obstacle. Thanks a million.