“Unable to prove literally” error when trying to create a symbolic function based on an existing function











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2
down vote

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I need to find inverse of virtual value function of lognormal random variables. This is what I tried to do:



syms flogn(x,p1,p2)
% assume(p2<=0); %Adding this doesn't change the error
flogn(x,p1,p2) = x - logncdf(x,p1,p2,'upper')/lognpdf(x,p1,p2);
glogn = finverse(flogn,x);


But I got the error:



Error using symengine
Unable to prove 'p2 <= 0' literally. Use 'isAlways' to test the statement mathematically.

Error in sym/subsindex (line 810)
X = find(mupadmex('symobj::logical',A.s,9)) - 1;

Error in sym/privsubsasgn (line 1085)
L_tilde2 = builtin('subsasgn',L_tilde,struct('type','()','subs',{varargin}),R_tilde);

Error in sym/subsasgn (line 922)
C = privsubsasgn(L,R,inds{:});

Error in logncdf>locallogncdf (line 73)
sigma(sigma <= 0) = NaN;

Error in logncdf (line 47)
[varargout{1:max(1,nargout)}] = locallogncdf(uflag,x,varargin{:});

Error in Untitled5 (line 3)
flogn(x,p1,p2) = x - logncdf(x,p1,p2,'upper')/lognpdf(x,p1,p2);


I also tried with beta distribution, and got a similar error. How can I use logncdf with symbolic variables?










share|improve this question
























  • This issue is similar to symbolic indexing.
    – Bebs
    Nov 21 at 10:06










  • This issue is very similar to that one.
    – Bebs
    Nov 21 at 11:12

















up vote
2
down vote

favorite












I need to find inverse of virtual value function of lognormal random variables. This is what I tried to do:



syms flogn(x,p1,p2)
% assume(p2<=0); %Adding this doesn't change the error
flogn(x,p1,p2) = x - logncdf(x,p1,p2,'upper')/lognpdf(x,p1,p2);
glogn = finverse(flogn,x);


But I got the error:



Error using symengine
Unable to prove 'p2 <= 0' literally. Use 'isAlways' to test the statement mathematically.

Error in sym/subsindex (line 810)
X = find(mupadmex('symobj::logical',A.s,9)) - 1;

Error in sym/privsubsasgn (line 1085)
L_tilde2 = builtin('subsasgn',L_tilde,struct('type','()','subs',{varargin}),R_tilde);

Error in sym/subsasgn (line 922)
C = privsubsasgn(L,R,inds{:});

Error in logncdf>locallogncdf (line 73)
sigma(sigma <= 0) = NaN;

Error in logncdf (line 47)
[varargout{1:max(1,nargout)}] = locallogncdf(uflag,x,varargin{:});

Error in Untitled5 (line 3)
flogn(x,p1,p2) = x - logncdf(x,p1,p2,'upper')/lognpdf(x,p1,p2);


I also tried with beta distribution, and got a similar error. How can I use logncdf with symbolic variables?










share|improve this question
























  • This issue is similar to symbolic indexing.
    – Bebs
    Nov 21 at 10:06










  • This issue is very similar to that one.
    – Bebs
    Nov 21 at 11:12















up vote
2
down vote

favorite









up vote
2
down vote

favorite











I need to find inverse of virtual value function of lognormal random variables. This is what I tried to do:



syms flogn(x,p1,p2)
% assume(p2<=0); %Adding this doesn't change the error
flogn(x,p1,p2) = x - logncdf(x,p1,p2,'upper')/lognpdf(x,p1,p2);
glogn = finverse(flogn,x);


But I got the error:



Error using symengine
Unable to prove 'p2 <= 0' literally. Use 'isAlways' to test the statement mathematically.

Error in sym/subsindex (line 810)
X = find(mupadmex('symobj::logical',A.s,9)) - 1;

Error in sym/privsubsasgn (line 1085)
L_tilde2 = builtin('subsasgn',L_tilde,struct('type','()','subs',{varargin}),R_tilde);

Error in sym/subsasgn (line 922)
C = privsubsasgn(L,R,inds{:});

Error in logncdf>locallogncdf (line 73)
sigma(sigma <= 0) = NaN;

Error in logncdf (line 47)
[varargout{1:max(1,nargout)}] = locallogncdf(uflag,x,varargin{:});

Error in Untitled5 (line 3)
flogn(x,p1,p2) = x - logncdf(x,p1,p2,'upper')/lognpdf(x,p1,p2);


I also tried with beta distribution, and got a similar error. How can I use logncdf with symbolic variables?










share|improve this question















I need to find inverse of virtual value function of lognormal random variables. This is what I tried to do:



syms flogn(x,p1,p2)
% assume(p2<=0); %Adding this doesn't change the error
flogn(x,p1,p2) = x - logncdf(x,p1,p2,'upper')/lognpdf(x,p1,p2);
glogn = finverse(flogn,x);


But I got the error:



Error using symengine
Unable to prove 'p2 <= 0' literally. Use 'isAlways' to test the statement mathematically.

Error in sym/subsindex (line 810)
X = find(mupadmex('symobj::logical',A.s,9)) - 1;

Error in sym/privsubsasgn (line 1085)
L_tilde2 = builtin('subsasgn',L_tilde,struct('type','()','subs',{varargin}),R_tilde);

Error in sym/subsasgn (line 922)
C = privsubsasgn(L,R,inds{:});

Error in logncdf>locallogncdf (line 73)
sigma(sigma <= 0) = NaN;

Error in logncdf (line 47)
[varargout{1:max(1,nargout)}] = locallogncdf(uflag,x,varargin{:});

Error in Untitled5 (line 3)
flogn(x,p1,p2) = x - logncdf(x,p1,p2,'upper')/lognpdf(x,p1,p2);


I also tried with beta distribution, and got a similar error. How can I use logncdf with symbolic variables?







matlab symbolic-math inverse






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edited Nov 21 at 10:43









Adriaan

12.4k63160




12.4k63160










asked Nov 12 at 2:13









cauthon14

69211




69211












  • This issue is similar to symbolic indexing.
    – Bebs
    Nov 21 at 10:06










  • This issue is very similar to that one.
    – Bebs
    Nov 21 at 11:12




















  • This issue is similar to symbolic indexing.
    – Bebs
    Nov 21 at 10:06










  • This issue is very similar to that one.
    – Bebs
    Nov 21 at 11:12


















This issue is similar to symbolic indexing.
– Bebs
Nov 21 at 10:06




This issue is similar to symbolic indexing.
– Bebs
Nov 21 at 10:06












This issue is very similar to that one.
– Bebs
Nov 21 at 11:12






This issue is very similar to that one.
– Bebs
Nov 21 at 11:12














1 Answer
1






active

oldest

votes

















up vote
2
down vote



accepted










Note that you could have minimalize your sample code even more, like this:



syms flogn(x,p1,p2)
flogn(x,p1,p2) = logncdf(x,p1,p2);


It's shorter, produces the same error and helps us to focus on the source of the error message.





So, the error doesn't come from trying to inverse a function, but from trying to use an existing function for numerical calculations with symbolic variables.



The error comes from the fact that you want to create a symbolic function flogn based on existing function logncdf, and logncdf has a multiple comparisons.



With the command edit logncdf, you can read the source code of the function and see comparison at lines 73 and 76.



% Return NaN for out of range parameters.
sigma(sigma <= 0) = NaN;

% Negative data would create complex values, which erfc cannot handle.
x(x < 0) = 0;


Matlab cannot compare symbols so it throws errors.



Depending on what you really need, you can have different solutions.




  • Do you really need to symbolize the function flogn? Couldn't you just write it as a function then calculate the inverse of it (if it can be inversed...)?


  • If you really want to keep the symbolization, you can also rewrite your own function logncdf (with another name) so it does not have the comparisons. But it's still not guaranteed that you will find an inverse.







share|improve this answer























  • Thanks for the minimization suggestion. Will keep in mind the next time.
    – cauthon14
    Nov 21 at 21:00






  • 1




    I don't need to symbolize the function. But how do I compute the inverse without it? For now, I have circumvented the problem, but I would still like to know your thoughts on this. Is creating an array X of numbers and then computing y = fzero(f(x)-x) on all x in X the only way?
    – cauthon14
    Nov 21 at 21:13










  • @cauthon14 Hi! I would recommend using fzero just like you said. I suggest reading this or a lot of questions about similar topics have been ask here, in SO. If you still don't find a solution in existing questions in SO, fell free to ask a new question. Regards.
    – Bebs
    Nov 22 at 7:47











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










Note that you could have minimalize your sample code even more, like this:



syms flogn(x,p1,p2)
flogn(x,p1,p2) = logncdf(x,p1,p2);


It's shorter, produces the same error and helps us to focus on the source of the error message.





So, the error doesn't come from trying to inverse a function, but from trying to use an existing function for numerical calculations with symbolic variables.



The error comes from the fact that you want to create a symbolic function flogn based on existing function logncdf, and logncdf has a multiple comparisons.



With the command edit logncdf, you can read the source code of the function and see comparison at lines 73 and 76.



% Return NaN for out of range parameters.
sigma(sigma <= 0) = NaN;

% Negative data would create complex values, which erfc cannot handle.
x(x < 0) = 0;


Matlab cannot compare symbols so it throws errors.



Depending on what you really need, you can have different solutions.




  • Do you really need to symbolize the function flogn? Couldn't you just write it as a function then calculate the inverse of it (if it can be inversed...)?


  • If you really want to keep the symbolization, you can also rewrite your own function logncdf (with another name) so it does not have the comparisons. But it's still not guaranteed that you will find an inverse.







share|improve this answer























  • Thanks for the minimization suggestion. Will keep in mind the next time.
    – cauthon14
    Nov 21 at 21:00






  • 1




    I don't need to symbolize the function. But how do I compute the inverse without it? For now, I have circumvented the problem, but I would still like to know your thoughts on this. Is creating an array X of numbers and then computing y = fzero(f(x)-x) on all x in X the only way?
    – cauthon14
    Nov 21 at 21:13










  • @cauthon14 Hi! I would recommend using fzero just like you said. I suggest reading this or a lot of questions about similar topics have been ask here, in SO. If you still don't find a solution in existing questions in SO, fell free to ask a new question. Regards.
    – Bebs
    Nov 22 at 7:47















up vote
2
down vote



accepted










Note that you could have minimalize your sample code even more, like this:



syms flogn(x,p1,p2)
flogn(x,p1,p2) = logncdf(x,p1,p2);


It's shorter, produces the same error and helps us to focus on the source of the error message.





So, the error doesn't come from trying to inverse a function, but from trying to use an existing function for numerical calculations with symbolic variables.



The error comes from the fact that you want to create a symbolic function flogn based on existing function logncdf, and logncdf has a multiple comparisons.



With the command edit logncdf, you can read the source code of the function and see comparison at lines 73 and 76.



% Return NaN for out of range parameters.
sigma(sigma <= 0) = NaN;

% Negative data would create complex values, which erfc cannot handle.
x(x < 0) = 0;


Matlab cannot compare symbols so it throws errors.



Depending on what you really need, you can have different solutions.




  • Do you really need to symbolize the function flogn? Couldn't you just write it as a function then calculate the inverse of it (if it can be inversed...)?


  • If you really want to keep the symbolization, you can also rewrite your own function logncdf (with another name) so it does not have the comparisons. But it's still not guaranteed that you will find an inverse.







share|improve this answer























  • Thanks for the minimization suggestion. Will keep in mind the next time.
    – cauthon14
    Nov 21 at 21:00






  • 1




    I don't need to symbolize the function. But how do I compute the inverse without it? For now, I have circumvented the problem, but I would still like to know your thoughts on this. Is creating an array X of numbers and then computing y = fzero(f(x)-x) on all x in X the only way?
    – cauthon14
    Nov 21 at 21:13










  • @cauthon14 Hi! I would recommend using fzero just like you said. I suggest reading this or a lot of questions about similar topics have been ask here, in SO. If you still don't find a solution in existing questions in SO, fell free to ask a new question. Regards.
    – Bebs
    Nov 22 at 7:47













up vote
2
down vote



accepted







up vote
2
down vote



accepted






Note that you could have minimalize your sample code even more, like this:



syms flogn(x,p1,p2)
flogn(x,p1,p2) = logncdf(x,p1,p2);


It's shorter, produces the same error and helps us to focus on the source of the error message.





So, the error doesn't come from trying to inverse a function, but from trying to use an existing function for numerical calculations with symbolic variables.



The error comes from the fact that you want to create a symbolic function flogn based on existing function logncdf, and logncdf has a multiple comparisons.



With the command edit logncdf, you can read the source code of the function and see comparison at lines 73 and 76.



% Return NaN for out of range parameters.
sigma(sigma <= 0) = NaN;

% Negative data would create complex values, which erfc cannot handle.
x(x < 0) = 0;


Matlab cannot compare symbols so it throws errors.



Depending on what you really need, you can have different solutions.




  • Do you really need to symbolize the function flogn? Couldn't you just write it as a function then calculate the inverse of it (if it can be inversed...)?


  • If you really want to keep the symbolization, you can also rewrite your own function logncdf (with another name) so it does not have the comparisons. But it's still not guaranteed that you will find an inverse.







share|improve this answer














Note that you could have minimalize your sample code even more, like this:



syms flogn(x,p1,p2)
flogn(x,p1,p2) = logncdf(x,p1,p2);


It's shorter, produces the same error and helps us to focus on the source of the error message.





So, the error doesn't come from trying to inverse a function, but from trying to use an existing function for numerical calculations with symbolic variables.



The error comes from the fact that you want to create a symbolic function flogn based on existing function logncdf, and logncdf has a multiple comparisons.



With the command edit logncdf, you can read the source code of the function and see comparison at lines 73 and 76.



% Return NaN for out of range parameters.
sigma(sigma <= 0) = NaN;

% Negative data would create complex values, which erfc cannot handle.
x(x < 0) = 0;


Matlab cannot compare symbols so it throws errors.



Depending on what you really need, you can have different solutions.




  • Do you really need to symbolize the function flogn? Couldn't you just write it as a function then calculate the inverse of it (if it can be inversed...)?


  • If you really want to keep the symbolization, you can also rewrite your own function logncdf (with another name) so it does not have the comparisons. But it's still not guaranteed that you will find an inverse.








share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 21 at 10:16

























answered Nov 20 at 9:25









Bebs

8083923




8083923












  • Thanks for the minimization suggestion. Will keep in mind the next time.
    – cauthon14
    Nov 21 at 21:00






  • 1




    I don't need to symbolize the function. But how do I compute the inverse without it? For now, I have circumvented the problem, but I would still like to know your thoughts on this. Is creating an array X of numbers and then computing y = fzero(f(x)-x) on all x in X the only way?
    – cauthon14
    Nov 21 at 21:13










  • @cauthon14 Hi! I would recommend using fzero just like you said. I suggest reading this or a lot of questions about similar topics have been ask here, in SO. If you still don't find a solution in existing questions in SO, fell free to ask a new question. Regards.
    – Bebs
    Nov 22 at 7:47


















  • Thanks for the minimization suggestion. Will keep in mind the next time.
    – cauthon14
    Nov 21 at 21:00






  • 1




    I don't need to symbolize the function. But how do I compute the inverse without it? For now, I have circumvented the problem, but I would still like to know your thoughts on this. Is creating an array X of numbers and then computing y = fzero(f(x)-x) on all x in X the only way?
    – cauthon14
    Nov 21 at 21:13










  • @cauthon14 Hi! I would recommend using fzero just like you said. I suggest reading this or a lot of questions about similar topics have been ask here, in SO. If you still don't find a solution in existing questions in SO, fell free to ask a new question. Regards.
    – Bebs
    Nov 22 at 7:47
















Thanks for the minimization suggestion. Will keep in mind the next time.
– cauthon14
Nov 21 at 21:00




Thanks for the minimization suggestion. Will keep in mind the next time.
– cauthon14
Nov 21 at 21:00




1




1




I don't need to symbolize the function. But how do I compute the inverse without it? For now, I have circumvented the problem, but I would still like to know your thoughts on this. Is creating an array X of numbers and then computing y = fzero(f(x)-x) on all x in X the only way?
– cauthon14
Nov 21 at 21:13




I don't need to symbolize the function. But how do I compute the inverse without it? For now, I have circumvented the problem, but I would still like to know your thoughts on this. Is creating an array X of numbers and then computing y = fzero(f(x)-x) on all x in X the only way?
– cauthon14
Nov 21 at 21:13












@cauthon14 Hi! I would recommend using fzero just like you said. I suggest reading this or a lot of questions about similar topics have been ask here, in SO. If you still don't find a solution in existing questions in SO, fell free to ask a new question. Regards.
– Bebs
Nov 22 at 7:47




@cauthon14 Hi! I would recommend using fzero just like you said. I suggest reading this or a lot of questions about similar topics have been ask here, in SO. If you still don't find a solution in existing questions in SO, fell free to ask a new question. Regards.
– Bebs
Nov 22 at 7:47


















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