Group Abstract

Message Boards

WOLFRAM COMMUNITY

5.3K Views

18 Replies

21 Total Likes

View groups...

Follow this post

Share this post:

GROUPS:

Mathematics Calculus Equation Solving Wolfram Language

Finding roots of multivariable function

Byron Alexander

Posted 1 year ago

POSTED BY: Byron Alexander

18 Replies

Sort By:

Daniel Lichtblau

Daniel Lichtblau, Wolfram Research

Posted 1 year ago

We take a derivative after that. Will the logs be off by something other than a constant in the variable `phi`?

POSTED BY: Daniel Lichtblau

Byron Alexander

Posted 1 year ago

Hi, just a note for completeness of this post. I think there is a small problem in using PowerExpand to take the Log in this way. Consider that the Sine and Cosine functions take negative values hence you cannot use the Log property of taking the respective exponents of 4 and 2 and moving them to factors in front of the Log terms. If you take the Log of L[\phi, m1, m2, m3], I don't think the log-function which you obtained is correct for this reason. Consider the attached code, where I use FullSimplify (as Ivanov noted):

POSTED BY: Byron Alexander

Byron Alexander

Posted 1 year ago

Hi Daniel I am very interested in finding out why the different approaches in Mathematica yield different outcomes. If you have a chance could you please check out this post.

POSTED BY: Byron Alexander

Daniel Lichtblau

Daniel Lichtblau, Wolfram Research

Posted 1 year ago

Good question. I think one way is to find discontinuity points. Skipping some of the setup, here is hoq that could be done. ee = -m1 Log[2] + 4 m0 Log[Cos[\[Phi]/2]] + 4 m2 Log[Sin[\[Phi]/2]] + 2 m1 Log[Sin[\[Phi]]]; Solve[FunctionDiscontinuities[{ee, 0 <= \[Phi] <= 2Pi}, \[Phi]] && 0 <= \[Phi] <= 2Pi, \[Phi]] (* During evaluation of In[18]:= FunctionDiscontinuities::unkds: Warning: The set of discontinuities may be incomplete due to missing domain and discontinuity information for some of the functions involved. Out[19]= {{\[Phi] -> 0}, {\[Phi] -> 0}, {\[Phi] -> \[Pi]}, {\[Phi] -> \[Pi]}, {\[Phi] -> 2 \[Pi]}, {\[Phi] -> 2 \[Pi]}} *)

Good question. I think one way is to find discontinuity points. Skipping some of the setup, here is hoq that could be done.

ee = -m1  Log[2] + 4  m0  Log[Cos[\[Phi]/2]] + 
   4  m2  Log[Sin[\[Phi]/2]] + 2  m1  Log[Sin[\[Phi]]];
Solve[FunctionDiscontinuities[{ee, 0 <= \[Phi] <= 2*Pi}, \[Phi]] && 
  0 <= \[Phi] <= 2*Pi, \[Phi]]

(* During evaluation of In[18]:= FunctionDiscontinuities::unkds: Warning: The set of discontinuities may be incomplete due to missing domain and discontinuity information for some of the functions involved.

Out[19]= {{\[Phi] -> 0}, {\[Phi] -> 
   0}, {\[Phi] -> \[Pi]}, {\[Phi] -> \[Pi]}, {\[Phi] -> 
   2 \[Pi]}, {\[Phi] -> 2 \[Pi]}} *)

POSTED BY: Daniel Lichtblau

Byron Alexander

Posted 1 year ago

Thanks.

POSTED BY: Byron Alexander

Byron Alexander

Posted 1 year ago

One more query, is there a systematic way in Mathematica of identifying these indeterminate points from the log function itself. I don't think Mathematica has built-in functions to find these indeterminate critical points?

POSTED BY: Byron Alexander

Daniel Lichtblau

Daniel Lichtblau, Wolfram Research

Posted 1 year ago

Absolutely right, my method should also check for when the log is `-Infinity` (a boundary case, with this transformation). And, as you note, this happens at multiples of `Pi`. I should have caught that. Thanks for pointing out the missing solutions.

POSTED BY: Daniel Lichtblau

Denis Ivanov

Posted 1 year ago

POSTED BY: Denis Ivanov

Daniel Lichtblau

Daniel Lichtblau, Wolfram Research

Posted 1 year ago

(1) This should link to the mathematica.stackexchange crosspost. (2) The notebook should use consistent notation for `{P1,P2,P3}`. (3) You might try the suggestion from a comment (mine) in MSE. Specifically, convert exponentials to trigs, and optimize the log of the expression of interest. P1[\[Phi]_] := 2^-4ExpToTrig[ ((E^(I(\[Phi]/2)) + E^(-I(\[Phi]/2))) (E^(I(-(\[Phi]/2))) + E^(I(\[Phi]/2))))]^2; P2[\[Phi]_] := 2^-3 ExpToTrig[ (E^(I(\[Phi])) - E^(-I(\[Phi]))) (E^(I(-\[Phi])) - E^(I(\[Phi])))]; P3[\[Phi]_] := 2^-4 ExpToTrig[ ((E^(I(\[Phi]/2)) - E^(-I(\[Phi]/2))) (E^(I(-(\[Phi]/2))) - E^(I(\[Phi]/2))))]^2; L[\[Phi]_, m0_, m1_, m2_] := ((P1[\[Phi]])^m0)((P2[\[Phi]])^m1)((P3[\[Phi]])^m2) In[24]:= llog = PowerExpand[Log[L[\[Phi], m0, m1, m2]]] ( Out[24]= -m1 Log[2] + 4 m0 Log[Cos[\[Phi]/2]] + 4 m2 Log[Sin[\[Phi]/2]] + 2 m1 Log[Sin[\[Phi]]] ) In[26]:= Solve[D[llog, \[Phi]] == 0, \[Phi]] ( Out[26]= {{\[Phi] -> ConditionalExpression[ 2 (-ArcTan[Sqrt[m1 + 2 m2]/Sqrt[ 2 m0 + m1]] + \[Pi] ConditionalExpression[ 1, \[Placeholder]]), ConditionalExpression[1, \[Placeholder]] \[Element] Integers]}, {\[Phi] -> ConditionalExpression[ 2 (ArcTan[Sqrt[m1 + 2 m2]/Sqrt[ 2 m0 + m1]] + \[Pi] ConditionalExpression[1, \[Placeholder]]), ConditionalExpression[1, \[Placeholder]] \[Element] Integers]}} *)

(1) This should link to the mathematica.stackexchange crosspost.

(2) The notebook should use consistent notation for {P1,P2,P3}.

(3) You might try the suggestion from a comment (mine) in MSE. Specifically, convert exponentials to trigs, and optimize the log of the expression of interest.

P1[\[Phi]_] := 
  2^-4*ExpToTrig[ ((E^(I*(\[Phi]/2)) + 
         E^(-I*(\[Phi]/2))) (E^(I*(-(\[Phi]/2))) + 
         E^(I*(\[Phi]/2))))]^2;
P2[\[Phi]_] := 
  2^-3 *ExpToTrig[ (E^(I*(\[Phi])) - 
       E^(-I*(\[Phi]))) (E^(I*(-\[Phi])) - E^(I*(\[Phi])))];
P3[\[Phi]_] := 
  2^-4 *ExpToTrig[ ((E^(I*(\[Phi]/2)) - 
         E^(-I*(\[Phi]/2))) (E^(I*(-(\[Phi]/2))) - 
         E^(I*(\[Phi]/2))))]^2;
L[\[Phi]_, m0_, m1_, 
  m2_] := ((P1[\[Phi]])^m0)*((P2[\[Phi]])^m1)*((P3[\[Phi]])^m2)

In[24]:= llog = PowerExpand[Log[L[\[Phi], m0, m1, m2]]]

(* Out[24]= -m1 Log[2] + 4 m0 Log[Cos[\[Phi]/2]] + 
 4 m2 Log[Sin[\[Phi]/2]] + 2 m1 Log[Sin[\[Phi]]] *)

In[26]:= Solve[D[llog, \[Phi]] == 0, \[Phi]]

(* Out[26]= {{\[Phi] -> 
   ConditionalExpression[
    2 (-ArcTan[Sqrt[m1 + 2 m2]/Sqrt[
         2 m0 + m1]] + \[Pi] ConditionalExpression[
         1, \[Placeholder]]), 
    ConditionalExpression[1, \[Placeholder]] \[Element] 
     Integers]}, {\[Phi] -> 
   ConditionalExpression[
    2 (ArcTan[Sqrt[m1 + 2 m2]/Sqrt[
        2 m0 + m1]] + \[Pi] ConditionalExpression[1, \[Placeholder]]),
     ConditionalExpression[1, \[Placeholder]] \[Element] Integers]}} *)

POSTED BY: Daniel Lichtblau

Denis Ivanov

Posted 1 year ago

POSTED BY: Denis Ivanov

Byron Alexander

Posted 1 year ago

Many thanks for your response. I attach the modified code where I attempt to explicitly show the 4th and 5th solutions (the solutions other than $\phi = {0, \pi ,2\pi}$). Do you agree with this code and the computation of the fourth and fifth roots: Lastly, at the end of your response you stated Limit[DL[[Phi], m0, m1, m2], [Phi] -> #, Assumptions -> Element[{m0, m1, m2}, PositiveIntegers]] & /@ {0, [Pi], 2 [Pi]} {0, 0, 0} so all five solutions are valid. But it seems that you only tested the solutions $\phi = 0, \pi ,2\pi$ yielding derivatives of {0, 0, 0}, yet you state 'so all five solutions are valid'. Could you advise on this please.

POSTED BY: Byron Alexander

Denis Ivanov

Posted 1 year ago

Yes, you are wright, looks like I mess something ( We can get firstly: `FullSimplify /@ {P1[\[Phi]] , P2[\[Phi]], P3[\[Phi]]}` $\left\{\cos ^4\left(\frac{\phi }{2}\right),\frac{\sin ^2(\phi )}{2},\sin ^4\left(\frac{\phi }{2}\right)\right\}$ Cause $\sin^2\left(\alpha/2\right)=\frac{1-\cos\left(\alpha\right)}{2}$ we can check result directly and your last post looks like fully correct!

POSTED BY: Denis Ivanov

Byron Alexander

Posted 1 year ago

Many thanks for your response. Firstly, yes P02 = P1; P11 = P2; P20 = P3. Why is it that you seem to get a different function when you FullSimplify at the start, they don't seem to be equivalent. Please have a look at my adapted code using your what you suggest in your post: