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MAKING SUBJECT OF FOMULAR

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lloyd mukunza 2014-9-20

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评论： Walter Roberson 2018-12-17

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.

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Chibuzo Chukwu 2018-12-13

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.

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回答（2 个）

Walter Roberson 2018-12-13

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在 MATLAB Online 中打开

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.

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Walter Roberson 2018-12-17

Chibuzo Chukwu comments to me

Your contribution has been very useful

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Chibuzo Chukwu 2018-12-13

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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?

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Walter Roberson 2018-12-13

在 MATLAB Online 中打开

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.

Walter Roberson 2018-12-17

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 2018-12-17

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.

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