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Wolfram Community forum discussion about Problem with simplifying expression containing Conjugate[]. Stay on top of important topics and build connections ...

(I am using Mathematica 12). I hope I am not fooling myself, but why does Mathematica not evaluate the following expression to zero: Conjugate[a] ...

Problem 1. It is fine to try. In[1]:= FullSimplify[x Conjugate[x]] Out[1]= Abs[x]^2. However it doesn't work with. In[2]:= FullSimplify[x y Conjugate[x y]]

Mathematica does not automatically assume that variables are real ... ing our own complex conjugation function as suggested by Zimmer-.

Conjugate does not always propagate into arguments: Differentiating Conjugate is not possible: The limit that defines the derivative is direction dependent and therefore does not exist:

Applications (5)Sample problems that can be solved with this function ; Define a scalar product for complex‐valued lists: ; Implement a · transformation: ; Write a ...

Conjugate and Simplify. Even if I explicitly specify that a and b are complex, the expression is not evaluated: Only when I give explicit numerical values, the simplification is done. Mariusz Iwaniuk, engineer,hobbyist. The Assumptions should not even be necessary here; FullSimplify [Conjugate [a] Conjugate [b] - Conjugate [a b]] by itself will ...

Conjugate and Simplify. Even if I explicitly specify that a and b are complex, the expression is not evaluated: Only when I give explicit numerical values, the simplification is done. Mariusz Iwaniuk, engineer,hobbyist. The Assumptions should not even be necessary here; FullSimplify [Conjugate [a] Conjugate [b] - Conjugate [a b]] by itself will ...

29.11.2021 · Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. ... Q&A for work. Connect and share ... $ is not conjugate to $\gamma(a)$ along $\gamma$.

This tells us that the conjugate of any matrix gives a new matrix which represents the same linear map (so the action on the group is of exactly the same type. Even though it will now map the same vector to a different one as the preconjugated matrix, if you write the original vector in terms of the new basis, the new matrix will map it to the same vector as before, represented in the …

Conjugate does not always propagate into arguments: Differentiating Conjugate is not possible: The limit that defines the derivative is direction dependent and therefore does not exist:

ALL it does is add some declarations to a list on the wall. Some functions in Mathematica look at that list, many do not. Simplify looks at the ...

If you want Conjugate in the answer use ComplexExpand[Re[PP], TargetFunctions -> {Conjugate}]. Whatever you do, you will get a very long and complicated expression. Using Re alone is never going to "work" (well, it actually does "work", but does not expand - a different matter) except when dealing with numbers - how could it? The answers depends on which symbols are real and which are not. $\endgroup$ –

But it still not in the shape I prefer. – xslittlegrass. Nov 12 '12 at 16:20. Add a comment ...

It's true that Conjugate can be annoying, but it's understandable that you're having difficulty removing it in an automated way. The culprit are the square roots, which are actually harmless but aren't recognized by Mathematica to be harmless.. So if you don't want to manually remove Conjugate but also don't want to waste time rewriting the expression to rearrange the square …

Conjugate and transpose the column matrix back into a row matrix: ConjugateTranspose [ vec ] conjugates the entries but does not change the shape of vec : ConjugateTranspose works for symbolic matrices:

Wolfram Community forum discussion about ComplexExpand, Conjugate, ... and it accually took me hours to find this problem in a very big calculation.

mathematica transpose does not work when using NumberForm and OutputForm. Ask Question Asked 6 years, 6 months ago. Active 6 years, 6 months ago.

This answer is useful. 4. This answer is not useful. Show activity on this post. The expression is in general complex even with all variables real since if the argument of any Sqrt is negative the Sqrt will be complex. expr = Conjugate [ 1/Sqrt [1 + (-2 + es + Cos [kx] + Cos [ky] + Sqrt [ (-2 + es + Cos [kx] + Cos [ky])^2 + Sin [kx]^2 + Sin [ky]^2])^2/ (Sin [kx]^2 + Sin [ky]^2)]];

Conjugate and transpose the column matrix back into a row matrix: ConjugateTranspose [ vec ] conjugates the entries but does not change the shape of vec : ConjugateTranspose works for symbolic matrices:

Mathematica is not associated with Mathematica Policy Research, Inc. ... All these functions work, in general, by doing a search, starting at some initial values and taking steps that decrease (or for FindMaximum, ... but it often does not converge quickly. In subsequent

Nov 17, 2012 · Show activity on this post. FullSimplify fails to recognize that: a*Conjugate [b] + b*Conjugate [a] = 2 Re [a*b] I have some very complex equations that could be simplified greatly if Mathematica could recognize this simple identity. (and that a*Conjugate [b] - b*Conjugate [a] = 2 Im [a*b]). See, Mathematica will not finish solving my equations ...

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(Note that in my actual work, such Conjugate expressions are embedded in a much larger expression and I do not know in advance whether the expression Conjugate ...

16.11.2012 · Show activity on this post. FullSimplify fails to recognize that: a*Conjugate [b] + b*Conjugate [a] = 2 Re [a*b] I have some very complex equations that could be simplified greatly if Mathematica could recognize this simple identity. (and that a*Conjugate [b] - b*Conjugate [a] = 2 Im [a*b]). See, Mathematica will not finish solving my equations ...

30.07.2014 · I am applying the algorithm found in Stewart's text on Numerical Solution of Markov Chains. All the literature I have found has asserted that the Conjugate Gradient method only works for symmetric positive definite matrices. My generator matrices are not SPD whatsoever, but the Conjugate Gradient method is still converging to the correct solution.