Solve function is unable to find an explicit solution

Question

Sandra Maria Cherian 2018-6-30

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https://ww2.mathworks.cn/matlabcentral/answers/408215-solve-function-is-unable-to-find-an-explicit-solution

评论： Sandra Maria Cherian 2018-7-7

This is my code. It is not there entirely.It is almost like this.

clc;
% Read image
I=imread('03022017_subset.tif');
sigma0_VH=I(:,:,1);
sigma0_VV=I(:,:,2);
LIA=I(:,:,3);  %Local Incidence angle
[m,n,p]= size(I);
er=sym('er',[m,n]);     % declaring dielectric constant as symbolic variable matrix
o=ones(m,n);
L=4;
pow=10;
x=0;
s=0.5;
w=((pi.*(L.^(2)))./pow).*(exp(-(((L.^(2)).*((((2.244.*sind(LIA).*cosd(x))-(1.121997.*(sind(LIA)))).^2)+((2.244.*sind(LIA).*sind(x)).^2))))./(4.*pow)));
SIvv=symsum(Ivv,z,1,Inf);
eqsigmavv=sigma0_VV-((1.258877268./4.*pi).*(exp(-2.*1.258877268.*(s.^2).*((cosd(LIA)).^2))).*((abs(SIvv)).^2).*(w./factorial(pow)))==0;
dcvv=solve(eqsigmavv,er);

solve function of 'dcvv' returning an answer of 1 by empty. But it is not what expected to get. The value of eqsigmavv is very lengthy. It is showing truncated in display. And some variable 'r' which is never declared in the programe is also included in it.

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Walter Roberson 2018-6-30

在 MATLAB Online 中打开

You should never use non-trivial floating-point constants with solve(). You have

(1.258877268./4.*pi)

as part of your expression. How is that intended to be interpreted? Is that intended to be 1258877268 / 1000000000 / 4, exactly, multiplied by the indefinitely precise transcendental value pi ? Is it intended to be "take the nearest representable binary floating point number to 1.258877268, divide that by 4, an operation which just happens to preserve the same number of digits in binary floating point, and do a binary floating point multiplication by the nearest representable binary floating point number to the transcendental value pi, letting the multiplication round-off fall however it happens to, and use that mathematically uncertain binary floating point value in the rest of the calculation?

Is it even divide by 4 and multiply by pi that you want, or do you want to divide by (4 times pi) ?

Remember that solve() is to try to find exact closed form solutions, indefinitely precise -- to try to find theoretical solutions. It is a mistake to ask for the theoretical solution to an expression whose value you cannot even predict the value of.

Sandra Maria Cherian 2018-7-3

编辑：Walter Roberson 2018-7-3

在 MATLAB Online 中打开

I=imread('03022017_subset.tif');
sigma0_VH=I(:,:,1);
sigma0_VV=I(:,:,2);
ter=sym('ter',[3,3]);     % declaring dielectric constant as symbolic variable matrix
to=ones(3,3);
LIAT=[37.760056,36.951263,36.990906;37.904781,37.156265,37.195721;38.194260,37.566406,37.605480];
% Equation for Ft
tRv0=(to-sqrt(ter))./(to+sqrt(ter));
tFt=8*(tRv0.^2) .*((sind(LIAT)).^2).*((cosd(LIAT)+(sqrt(ter-(sind(LIAT)).^2)))./(cosd(LIAT).*(sqrt(ter-(sind(LIAT).^2)))));
tSt0=(abs(to+((8.*tRv0)./(tFt.*cosd(LIAT))))).^(-2);
L=4;
pow=10;
x=0;
s=0.5;
tw=((pi.*(L.^(2)))./pow).*(exp(-(((L.^(2)).*((((2.244.*sind(LIAT).*cosd(x))-(1.121997.*(sind(LIAT)))).^2)+((2.244.*sind(LIAT).*sind(x)).^2))))./(4.*pow)));
% Equation for St
syms tz;
tStn=symsum(((((1.121997.*s.*cosd(LIAT)).^(2.*tz))./(factorial(tz))).*tw),tz,1,Inf);
tStd=symsum(((((1.121997.*s.*cosd(LIAT)).^(2.*tz))./(factorial(tz))).*(((abs((tFt)+(((2.^(pow+2)).*tRv0./((exp((1.121997.*s.*cosd(LIAT)).^(2))).*cosd(LIAT)))))).^(2)).*tw)),tz,1,Inf);
tSt=(((abs(tFt)).^2).*tStn)./tStd;
tRv=(cosd(LIAT)-sqrt(ter-(sind(LIAT)).^2))./(cosd(LIAT)+sqrt(ter-(sind(LIAT)).^2));
tRtv=tRv+((tRv0-tRv).*(to-(tSt./tSt0)));
tfvv=2.*tRtv./cosd(LIAT);
tTv=to+tRtv;
tTvm=to-tRtv;
tsq=sqrt(ter-(sind(LIAT)).^2);
tFvv=(((((sind(LIAT)).^2)./cosd(LIAT))-(tsq./ter)).*(tTv.^2))-(2.*((sind(LIAT)).^2).*((to./cosd(LIAT))+(to./tsq)).*tTv.*tTvm)+(((((sind(LIAT)).^2)./cosd(LIAT))+(ter.*(to+(sind(LIAT)).^2)./tsq)).*(tTvm.^2));
syms tz;
tIvv=((2.*1.121997.*s.*cosd(LIAT)).^tz).*tfvv.*(exp(-1.*1.258877268.*(s.^2).*((cosd(LIAT)).^2)))+((1.121997.*s.*cosd(LIAT)).^tz).*tFvv;
tSIvv=symsum(((abs(tIvv)).^2).*tw./factorial(tz),tz,1,10);
tsigma0_VV=[0.11262583,0.86308515,1.1438491;0.14032036,1.0475565,1.6186293;0.21424975,0.61734170,0.90188682];
teqsigmavv=tsigma0_VV-((1.258877268./(4.*3.14159)).*(exp(-2.*1.258877268.*(s.^2).*((cosd(LIAT)).^2))).*(tSIvv))==0;
tdcvv = solve(teqsigmavv,ter,'IgnoreAnalyticConstraints',1);

This is the entire code

Walter Roberson 2018-7-3

I think I will need the .tif for testing.

You did not answer my questions about how you intend your floating point constants to be interpreted: do you intend them to be interpreted as implicit rationals in base 10, or do you intend them as implicit rationals in base 2 (which is how floating point is stored internally) ?

If you had the floating point constant 0.333333 in your code, would you intend that to represent the rational 1 divided by 3, or would you intend it to represent the rational 333333 divided by 1000000, or would you intend it to represent the rational 3002396749180579 divided by 9007199254740992 ? These choices matter when you are using solve(), since solve() asks for the exact solution, and it is a mistake to ask for the exact solution for an expression with vague input values.

Perhaps you should be using vpasolve()

Sandra Maria Cherian 2018-7-4

I tried vpasolve. It is also giving 0 by 1 as solution.

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Answer 1

Sandra Maria Cherian 2018-7-4

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This is the image.I want to represent 1/3 as 0.3333

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Walter Roberson 2018-7-4

The tif did not get attached.

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Answer 2

Sandra Maria Cherian 2018-7-5

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/408215-solve-function-is-unable-to-find-an-explicit-solution#answer_327546

编辑：Sandra Maria Cherian 2018-7-5

在 MATLAB Online 中打开

Image is unable to attach. Can you please try it with following matrix.

ter=sym('ter',[3,3]); 
to=ones(3,3); 
LIAT=37.760056,36.951263,36.990906;37.904781,37.156265,37.195721;38.194260,37.566406,37.605480]; 
tRv0=(to-sqrt(ter))./(to+sqrt(ter)); tFt=8*(tRv0.^2) .*((sind(LIAT)).^2).*((cosd(LIAT)+(sqrt(ter-(sind(LIAT)).^2)))./(cosd(LIAT).*(sqrt(ter-(sind(LIAT).^2)))));
tSt0=(abs(to+((8.*tRv0)./(tFt.*cosd(LIAT))))).^(-2);
% surface roughness spectra W(n)
 L=4; pow=10; x=0; s=0.5; 
tw=((pi.*(L.^(2)))./pow).*(exp(-(((L.^(2)).*((((2.244.*sind(LIAT).*cosd(x))-(1.121997.*(sind(LIAT)))).^2)+((2.244.*sind(LIAT).*sind(x)).^2))))./(4.*pow)));
% Equation for St syms tz; tStn=symsum(((((1.121997.*s.*cosd(LIAT)).^(2.*tz))./(factorial(tz))).*tw),tz,1,Inf);
tStd=symsum(((((1.121997.*s.*cosd(LIAT)).^(2.*tz))./(factorial(tz))).*(((abs((tFt)+(((2.^(pow+2)).*tRv0./((exp((1.121997.*s.*cosd(LIAT)).^(2))).*cosd(LIAT)))))).^(2)).*tw)),tz,1,Inf); tSt=(((abs(tFt)).^2).*tStn)./tStd;
tRv=(cosd(LIAT)-sqrt(ter-(sind(LIAT)).^2))./(cosd(LIAT)+sqrt(ter-(sind(LIAT)).^2)); tRtv=tRv+((tRv0-tRv).*(to-(tSt./tSt0)));
tfvv=2.*tRtv./cosd(LIAT);
tTv=to+tRtv; tTvm=to-tRtv; tsq=sqrt(ter-(sind(LIAT)).^2); tFvv=(((((sind(LIAT)).^2)./cosd(LIAT))-(tsq./ter)).*(tTv.^2))-(2.*((sind(LIAT)).^2).*((to./cosd(LIAT))+(to./tsq)).*tTv.*tTvm)+(((((sind(LIAT)).^2)./cosd(LIAT))+(ter.*(to+(sind(LIAT)).^2)./tsq)).*(tTvm.^2));
syms tz; tIvv=((2.*1.121997.*s.*cosd(LIAT)).^tz).*tfvv.*(exp(-1.*1.258877268.*(s.^2).*((cosd(LIAT)).^2)))+((1.121997.*s.*cosd(LIAT)).^tz).*tFvv; tSIvv=symsum(((abs(tIvv)).^2).*tw./factorial(tz),tz,1,10);
tsigma0_VV=[0.11262583,0.86308515,1.1438491;0.14032036,1.0475565,1.6186293;0.21424975,0.61734170,0.90188682];
teqsigmavv=((tsigma0_VV-((1.258877268./(4.*3.14159)).*(exp(-2.*1.258877268.*(s.^2).*((cosd(LIAT)).^2))).*(tSIvv)))==0);
tic tdcvv = vpasolve(teqsigmavv,ter); 
toc

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Sandra Maria Cherian 2018-7-5

编辑：Walter Roberson 2018-7-5

在 MATLAB Online 中打开

1.121997 is a constant for this particular image. I applied directly in the equation. I think there is a problem in the line: tRtv=tRv+((tRv0-tRv).*(to-(tSt./tSt0))); solution given by this equation contains some unexpected variables (r and n)

This is a solution:

tRtv(1,1)

ans =

((ter1_1^(1/2) - 1)/(ter1_1^(1/2) + 1) - ((ter1_1 - 6755037812346739/18014398509481984)^(1/2) - 1780232650527177/2251799813685248)/((ter1_1 - 6755037812346739/18014398509481984)^(1/2) + 1780232650527177/2251799813685248))*((51375416058454120524490299147421638592250773504*abs((18014398509481984*(ter1_1^(1/2) + 1)*(ter1_1 - 6755037812346739/18014398509481984)^(1/2))/(6755037812346739*((ter1_1 - 6755037812346739/18014398509481984)^(1/2) + 1780232650527177/2251799813685248)*(ter1_1^(1/2) - 1)) - 1)^2*abs((ter1_1 - 6755037812346739/18014398509481984)^(1/2) + 1780232650527177/2251799813685248)^2*abs(ter1_1^(1/2) - 1)^4*((4685593773734083*exp(63834711222653326585719000350569/324518553658426726783156020576256))/1125899906842624 - 4685593773734083/1125899906842624))/(14849716343180055507990398757828129180178400307*abs(ter1_1 - 6755037812346739/18014398509481984)*abs((36893488147419103232*ter1_1^(1/2) - 36893488147419103232)/(8668920821841493*ter1_1^(1/2) + 8668920821841493) - ( *6755037812346739\\\r\n*((ter1_1 -* 6755037812346739/18014398509481984)^(1/2) + 1780232650527177/2251799813685248)*(ter1_1^(1/2) - 1)^2)/(1780232650527177*(ter1_1^(1/2) + 1)^2*(ter1_1 - 6755037812346739/18014398509481984)^(1/2)))^2*abs(ter1_1^(1/2) + 1)^4*(exp(63834711222653326585719000350569/324518553658426726783156020576256) - 1)) - 1) - ((ter1_1 - 6755037812346739/18014398509481984)^(1/2) - 1780232650527177/2251799813685248)/((ter1_1 - 6755037812346739/18014398509481984)^(1/2) + 1780232650527177/2251799813685248)

Also solutions are very lengthy. simplify() is not working.

Walter Roberson 2018-7-6

在 MATLAB Online 中打开

Slightly better (at least for my implementing function):

09709586957955585, 4.50416224972559576, 5.13972527850544925
23993701810814461, 4.92827672315921195, 6.09692340175298675
57220009290275931, 3.87915792241145541, 4.59424410673230277

I tested with the range 20 to 90 as bounds, and all of the entries except possibly the first two quickly pinned themselves against the 20 lower bound (it looked like one of the first two might perhaps have been marginally different than 20 for the minimum value under the circumstance that all of the other values were 20.)

With that code, there is no way that there is a solution with any of the components required to be as large as 20.

Sandra Maria Cherian 2018-7-7

I finally solved it. The symbolic matrix ter is converted to a single symbolic variable and for finding tdcvv a for loop is constructed.Still the values are not the expected.Almost same as what you got. Now I am working on it.

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