Mathematica conjugate with assumptions. FullSimplify doesn't do everything because it .

Mathematica conjugate with assumptions Raw code: Ass = Sep 19, 2022 · I personally do not like the use of global assumptions and do not use them. Simplify can be used on equations, inequalities, and domain specifications. In the case at hand you may go around easier by making a substitution δ->Abs[δ]^2/δ∗ after all the operations I did, but I am trying to conjugate a symbolic expression, and I have explicitly stated the real terms. See also the documentation for $Assumptions. FullSimplify [expr, assum] does simplification using assumptions. Use at your own risk. Feb 16, 2011 · Hi All, In mathematica, I'm trying to use Conjugate [] to take the complex conjugate of a function that has imaginary numbers in it, but I want to tell Mar 14, 2022 · As a side note, it's worth pointing out that your preferred expression is actually less simple than the original, according to Mathematica's default measure of complexity. gates. That's the reason why when you enter: Dec 4, 2010 · For the second one, remember that Mathematica can't make assumptions on your symbols, so a "number" is complex by default. How can I get rid of this? Here is my code n = Oct 21, 2019 · According to this simple example program, the assumptions evaluated only with the Simplify[] function call. Also You can use ComplexExpand instead of assuming variables are real. Collect [poly, x] takes a polynomial in several variables and rewrites it as a sum of terms containing different powers of the "dominant variable" x. Why doesn't it use the assumptions? [Edit:] I'm not looking for an alternative function, I am trying Sometimes I want to ask mathematica to plot or calculate something without given a certain value for it . Not sure if that answers your question though. is the default setting for the Assumptions option used in such functions as Simplify, Refine, and Integrate. g. . Considering the fact that in C, there's no closed way to conjugate, this makes it a real mass for the poor program to try and figure However, I need those assumptions, so Mathematica can do useful simplifications. 化简包含 Conjugate 的表达式： 假定实数值变量： 假定一般的复数值变量： Conjugate 作为 ComplexExpand 中的一个选项值： 在复平面内沿一条线进行符号积分和数值积分： 求出矩阵的厄密共轭： Wolfram Community forum discussion about How to assume real variables?. While this is overall a good thing, this often leads to What is the big difference between downloading Mathematica for Sites instead of Mathematica for Students (for Sites)? Basically all my question is the one in the title, I got licenses for the version 14. e. No. The question then is: How to catch those issues in a way that can still handle generale assumptions (which ComplexExpand cannot Feb 4, 2018 · I want to define a variable like d as a Real variable and then using that in the other equation like that: $\\qquad d$ is Real $\\qquad f = 5 + (1 + i) d$ But Mathematica gives me this result: d ∈ The minimal method is to use Refine as so: Refine[Conjugate[a+I b], θ ∈ Reals]. 0 tells me that In: Series[(x - Sqrt[1 + x^2*a*Conjugate[a]])/ (x + Sqrt[1 + x^2*a*Conjugate[a]]), {x, Infinity, 0 Wolfram Community forum discussion about Problem with simplifying expression containing Conjugate []. This behavior seems pretty random. Conjugate [z] or z\ [Conjugate] gives the complex conjugate of the complex number z. To force Mathematica to interpret the elements on diagonal of to be real you could replace them by their real part, i. Also, in place of the assumption θ ∈ Reals you can use the assumption _Symbol ∈ Reals to assume that all explicit variables are real. As a test example, I used the code Assuming [ { x \ [Element] I Reals}, Simplify [x + Conjugate [x]]] In the above, "I" means the imaginary un Nov 17, 2012 · a*Conjugate[b] +b*Conjugate[a] form, but I could at the very least write my final equations in an extremely descriptive and compact form if Mathematica recognized this. However, from this page How to tell Mathematica that the argument of a function is real? I learned it's possible to code up a function that will make a certain pattern (? Though I don't understand his answer quite much) to assume reals. However, I'm struggling with making assumptions for such Jun 21, 2016 · I know default Mathematica doesn't provide this feature. I am trying to run a code in Mathematica, but the running time is awfully long. When I say, for example, functions complex assumptions variable-definitions Matthew Brunetti 643 asked Nov 21, 2014 at 22:50 38votes How to tell Mathematica that certain variables are real/imaginary, integer-valued, etc I'm trying to expedite some quantum mechanical calculations (expectation values etc. Oct 18, 2020 · Mathematica FullSimplify becomes less simplified with more specific assumptions? As far as I always understood, Mathematica always assumes an expression to represent the most general state. 4 and persisting through 11. Assumptions and Domains The Wolfram Language has a flexible system for specifying arbitrary symbolic assumptions about variables. When I say, for example, functions complex assumptions variable-definitions Matthew Brunetti 643 asked Nov 21, 2014 at 22:50 36votes 4answers 11kviews Sometimes Mathematica generates solutions to a PDE which are hard to read, since it attaches assumptions to the result. When I say, for example, functions complex assumptions variable-definitions Matthew Brunetti 653 asked Nov 21, 2014 at 22:50 0votes I want to add two assumptions, so I can get this probability density function to equal 1, though I can't get a solution. m = {{Re[n], a, b, b}, {Conjugate[a], Re[n], b, b}, {Conjugate[b], Conjugate[b], Re[c], d}, {Conjugate[b], Conjugate[b], Conjugate[d Aug 10, 2018 · To my detriment Mathematica generates an output whose first condition is $w<\frac {1} {2}$ How can I get Mathematica to use my assumptions to focus on only those values that apply within those assumptions? May 10, 2021 · It is a common misconception that patterns may be used in Assumptions, but this is not the case. 0. The 2nd and 3rd example use the given assumption in addition to global $Assumptions. Feb 12, 2015 · 在mathematica中怎么限定一个变量的类型，比如整数，正数，实数，复数等？ 比如求一个傅里叶变换 f (t)=Ae^ (-a|t|)。 需要限定其中a>0。 这只是随便举得一个例子，还有比如求一个积分 Integrate [Sin [x]… 显示全部 关注者 16 May 20, 2024 · I would have hoped to get something like the output (a - b) f[x - Conjugate[y]] but instead I get b Conjugate[f[y - Conjugate[x]]] + a f[x - Conjugate[y]] What am I doing wrong, and what is the correct way to specify properties like these? P. If x and y are themselves complex numbers, then the conjugate of (x + y i) is not simply (x - i y). Jun 8, 2023 · Facing the same problem reported here: Conjugation with real elements answers are working, I thought of the following solution: Unprotect[Conjugate]; Conjugate[x_] := x /; Element[x, Reals] Protect[ You'll need to complete a few actions and gain 15 reputation points before being able to upvote. I am trying to simplify a square root expression with an assumption that all variables are positive reals. Hope someone could help me out. The Wolfram Language has fundamental support for both explicit complex numbers and symbolic complex variables. However when I simplify the expression with appropriate assumptions a 'Conjugate' still remains. Assumptions -> f [_] > 0 does not in fact tell Mathematica that f is positive for any argument. While this is overall a good thing, this often leads to Jun 27, 2024 · Thanks @Nasser. When I say, for example, functions complex assumptions variable-definitions Matthew Brunetti 643 asked Nov 21, 2014 at 22:50 39votes May 26, 2018 · I have the following recursive functions: B[M] = 0; P = JacobiDN[u, m] Exp[I W x]; A[j_] := A[j] = -I Integrate[(P Conjugate[B[j]] - Conjugate[P] B[j]), u] + 2^(j - 1 I want to evaluate the CholeskyDecomposition of the following symmetric matrix symMat mm = Table[Subscript[m, i, j], {i, 1, 3}, {j, 1, 3}] symMat = Table[ If[i >= j, mm[[i, j]], mm[[j, i]]], {i, 1, 3}, {j, 1, 3}] It works but as Mathematica works with complex numbers I have some Conjugate commands in the result. When you have a polynomial in more than one variable, you can put the polynomial in different forms by essentially choosing different variables to be "dominant". Here is the code I am using: Note the expressions in the assumption can be found Aug 9, 2015 · Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, Aug 10, 2018 · To my detriment Mathematica generates an output whose first condition is $w<\frac {1} {2}$ How can I get Mathematica to use my assumptions to focus on only those values that apply within those assumptions? Mathematica will never make assumptions you don't tell it to make, and will state all conditions under which the answers it gives are valid. Removing the imaginary portion of an expression is done by doing ComplexExpand[Re[expression]]. However the elements on the main diagonal of a Hermitian matrix are necessarily real. It uses a wide range of sophisticated algorithms to infer the consequences of assumptions — often in the process automatically proving a sequence of necessary mathematical theorems. When I say, for example, functions complex assumptions variable-definitions Matthew Brunetti 653 asked Nov 21, 2014 at 22:50 39votes How to tell Mathematica that certain variables are real/imaginary, integer-valued, etc I'm trying to expedite some quantum mechanical calculations (expectation values etc. I would imagine that the internal code used by Integrate makes certain assumptions about how system functions like NumberQ, Conjugate, Abs, etc, will behave. Jan 6, 2022 · I'm trying to write a formula using subscripted variables like x₁, x₂, x₃, , so that I can later retrieve their values programmatically. You should use Conjugate[δ] instead and then apply ComplexExpand in the end. , ComplexExpand. When I wrote Clear "Global`*" Simplify -Conjugate o + Sqrt e^ - e o + o^ + q^ + w^ -Conjugate e then simply removed the minus sign Simplify Conjugate o + Sqrt e^ - e o + o^ + q^ + w^ - Conjugate e This told me I need to ComplexExpand . The Wolfram Language's matrix operations handle both numeric and symbolic matrices, automatically accessing large numbers of highly efficient algorithms. How do we query Mathematica to see if a given assumption holds? To be more precise, how can we define a function assumptions mikado 16. All applicable mathematical functions support arbitrary-precision evaluation for complex values of all parameters, and symbolic operations automatically treat complex variables with full generality. Use Conjugate instead. Wolfram Community forum discussion about Making multiple assumptions. Using just Re alone will not work as Re does no evaluation on symbols with unknown complex parts. I have found a nice post which use Nov 2, 2016 · But if I calculate the same with an extra FullSimplyfy I have a different output: Mathematica will never make assumptions you don't tell it to make, and will state all conditions under which the answers it gives are valid. 8k answered Sep 16, 2016 at 21:08 1vote 1answer 105views Mathematica will never make assumptions you don't tell it to make, and will state all conditions under which the answers it gives are valid. For me, it is rare that I would be interested in solutions that are generalized to both real and imaginary arguments, so my method is usually to run a FullSimplify on the expression, with every unset variable being assumed be an element The primary difference between Refine and Simplify is this: Refine must have assumptions and performs only those basic simplifications which would be automatic for numeric inputs, while Simplify performs a wider range of simplifications and may be used without assumptions. Click here to see a screenshot of Wolfram Mathematica with formatted code. Feb 14, 2013 · Ahh, I see. I have a problem with using Assumption. 它采用大量先进的算法来推断假设的结果，在这个过程中往往自动应用一系列必要的数学定理. How to tel that all parameters are Reals. Quantities that appear algebraically in inequalities are always assumed to be real. Put back the minus sign, and now I get what you expected to see Simplify ComplexExpand -Conjugate o + Sqrt e^ - *e*o + o^ + q^ + w Nov 11, 2021 · I have an expression in which all variables are real. How do I get the conjugate of the functions in a simpler form, and taking It and R as complex. Feb 4, 2018 · I want to define a variable like d as a Real variable and then using that in the other equation like that: $\\qquad d$ is Real $\\qquad f = 5 + (1 + i) d$ But Mathematica gives me this result: d ∈ The minimal method is to use Refine as so: Refine[Conjugate[a+I b], θ ∈ Reals]. 3. FullSimplify[PiecewiseExpand[χRp Conjugate[χLq]], Assumptions -> {Element[{p, pt, q, qt, rp, rq, tp, tq}, Reals]}] If you want to have conjugation displayed as an asterisk after the symbol, you have to use \[Conjugate] (which will be displayed as a smaller, superscript asterisk Wolfram Community forum discussion about Conjugate [] isn't working for me. a_Complex:> Conjugate[a]) I tend to use this regularly for getting the complex conjugate of expressions. While this is overall a good thing, this often leads to Jun 12, 2021 · Is this what you are looking for? What is the book answer? Also in Mathematica be careful with space. Jun 10, 2020 · Assuming and Conjugate do not work as expected [duplicate] I tried to work with Conjugates in Mathematica and it did not produce my desired result. When I say, for example, functions complex assumptions variable-definitions Matthew Brunetti 643 asked Nov 21, 2014 at 22:50 36votes 4answers 11kviews Apr 2, 2018 · If you Conjugate the equation before Solve, you'll get the complex-conjugated solutions (complex analysis is nice!): wroots1 = w /. The question arises when I try generating matrix with rolling code: vect = {α, 1 - α, 0, 0, d, -Id }; prod0 = {vect}; prod1 = Transpose[prod0]; Assumptions Sep 13, 2024 · I need to conjugate some sum of terms. So far, this takes too long to evaluate. Wolfram Mathematica does this for short sums, but not for long ones. The only way my example could fail is that the recursion fails to iterate on simple additions/multiplications. Perhaps Mathematica carelessly treats everything involved in inequalities as a real number? How to tell Mathematica that certain variables are real/imaginary, integer-valued, etc I'm trying to expedite some quantum mechanical calculations (expectation values etc. But unfortunately, Both Simplify and FullSimplify failed to do this seemingly simple job Sep 17, 2015 · @sepideh I cannot really understand all your definitions, but it seems that you did not instruct Mma that δ∗ is a conjugated value of δ. Jan 17, 2017 · @Szabolcs: Addition and multiplication are quite easy functions to iterate on. Sep 10, 2019 · I tried the Assumption option but everytime it doesn't work when i try to look at ComplexConj (A) * A - i alwayd get sth like a0 * Conjugate a0 but it should be a0^2. When I say, for example, functions complex assumptions variable-definitions Matthew Brunetti 653 asked Nov 21, 2014 at 22:50 39votes Feb 19, 2013 · For some reason, when I enter the following integration in Mathematica Assuming [ {k ∈ Integers}, Integrate [ Exp [ I k t], {t, -π, π}]] the result turns out to be 0. What's reputation and how do I get it? Instead, you can save this post to reference later. Feb 14, 2013 · Is it possible in Mathematica to set the symbols 'a' and 'b' as reals so when I type 'z = a+I b;Conjugate[z]' I get 'a-I b' and not 'Conjugate[a]-I Conjugate[b]'? Sep 10, 2019 · I tried the Assumption option but everytime it doesn't work when i try to look at ComplexConj (A) * A - i alwayd get sth like a0 * Conjugate a0 but it should be a0^2. , \\begin{align} R(\\tau) = \\int_{-\\infty}^{\\infty} \\overline Nov 22, 2016 · @MariusLadegårdMeyer At first, it is mostly looking at them and transfering them to Matlab for further numerical simulations. Nov 28, 2020 · Following up to the comment of b. Complex is the head used for complex numbers. Massimo Oct 2, 2015 · You could consider having a global assumption that psi is real: $Assumptions = {Element[ψ, Reals]}. Thats the reason why no simplification on arbitrary powers, logs and arc functions is attempted without explicit assumptions. 1. While ComplexExpand defaults to assuming that all variables are real, the optional second argument is used to specify which variables are complex. Mathematica doesn't really do typing like that. Assumptions 是 Simplify、Refine 以及 Integrate 等函数的一个选项，用来指定就符号量所做的缺省假设. Nov 9, 2019 · Using Simplify with certain trivial assumptions makes Mathematica forget that a given matrix is Hermitian. After studying it, we can see that it is a c - b Conjugate[b], which is what we get using the default method. 0 of both products but I don’t know which is their difference. But the same should work, if I only apply the complex conjugation to c[i,k] inside orb[i,z], because basis[i,z] is real. For example Conjugate[orb[i,z]] orb[i,z]//FullSimplify should simplify to Abs[orb[i,z]]^2. ComplexExpand[x E^(I a) + Conjugate[x] E^(-I a), {x}, TargetFunctions -> {Re, Im}] This approach spares the user to write replacement rules back and forth. treats a as real, while it is complex. functions simplifying-expressions assumptions Pavel Smirnov 191 answered Mar 22, 2018 at 12:41 21votes 2 2kviews Jun 4, 2020 · Complex number operations: telling Mathematica variables are real I want to do Conjugate [a + b*I], but when I do that, the solution is Conjugate [a] - I*Conjugate [b]; when for me, a and b are Jun 27, 2018 · I am trying to do symbolic manipulations in MMA and I need it to take a complex conjugate of functions with real arguments, like: Refine[Conjugate[(I^(2 mg))], mg \\[Element] Reals] Refine[Conjugate[ Use Re to describe regions in the complex plane: Reduce can solve equations and inequalities involving Re: With FindInstance you can get sample points of regions: Use Re in Assumptions: Integrate often generates conditions in terms of Re: Use Arg as a target function in ComplexExpand: Rescale Arg to run from 0 to 1: Find the domain of positivity for a linear function: Use Arg to specify assumptions about complex variables: ComplexPlot plots the phase of a function using color and shades by the magnitude: Using Mathematica 10, Simplify [a\ [Conjugate], Assumptions -> { (a + a\ [Conjugate]) > 0}] returns a, i. However, depending on the numerical results, I might also try to work with the symbolic matrices in Mathematica which would for instance involve some series expansions and solving equations based on those matrices. Using simply a conjugate symbol is not enough. If those other system functions misbehave then you have undefined behavior. How to check the assumptions attached to a given symbol? Suppose there are a number of assumptions attached to a certain symbol. Simplify [expr∈Reals,assum] can be used to try to determine whether an expression corresponds to a real number under the given assumptions. Wolfram 语言可以灵活地对变量进行任意的符号假设说明. The "F[x_] :=" construction allows you to call F repeatedly with different x values, just as you do for the built-in functions. For example I may want to ask mathematica to integrate $\\int{a x} \\mathrm{d}x$! How to te You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Otherwise it won't See below: (*not split up*) FullSimplify [ Conjugate [ E^ (I \ [Delta]m) rm… Here is an expression 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)]] With the assumptions that es, kx, ky are real variables, I want to remove the head Conjugate in a safe manner with Simplify or FullSimplify. So it's either buggy or badly implemented. The Wolfram Language does not simplify automatically: If I split up the conjugate myself, then it will simplify. Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels. The basic issue is that using Conjugate alone doesn't work because Mathematica doesn't know your variables are real: Conjugate[a + I b] Conjugate [a] - i Conjugate [b] Rojo's suggestion to deploy ComplexExpand is very common advice and it works in the simplest case ComplexExpand[Conjugate[a + I b]] a - i b I am trying to use assumptions to simplify an integral, but I can't seem to get the assumptions to apply themselves. By using Assumptions you can coerce some functions into assuming that, say, a is a Real but that's probably not a general enough approach for you either. Abs [z] gives the absolute value of the real or complex number z. Thanks to all. This is the code: complex assumptions infinitezero 1,451 asked Nov 8, 2016 at 22:43 0votes 1answer 63views I need to find the conditions for the real part of a complex number to be negative. By default, Mathematica assumes symbols to be complex. See this: z = Exp[I*a]; Conjugate[z] E^ (-I Conjugate [a]) This is basically what you asked about in your question. This allows you to encapsulate this in a function without having to manually pass a list of arguments. yI is not the same as y*I. I am using Mathematica 12. The Wolfram Language uses state-of-the-art algorithms to work with both dense and sparse matrices, and incorporates a number of powerful original algorithms, especially for high-precision and symbolic matrices. The simplest way to add local assumptions is to use Assuming. However, I simply can't get it to work: Conjugate[ ComplexExpand[ I Cos[z] Sin[y] + Sin[z] + A ( Dec 4, 2010 · For the second one, remember that Mathematica can't make assumptions on your symbols, so a "number" is complex by default. Feb 6, 2021 · When you evaluated Conjugate[A], the assumptions you explicitly specified in Refine[A, {Element[t, Reals], t > 0}] are not remembered, so t is again treated as complex. trueThe problem, as I understand it, is that, without further prompting, Mathematica is very unwilling to make a priori assumptions about what is and is not a real number, so anything thrown its way automatically gets treated as an implicitly complex variable. Solve[ComplexExpand[Conjugate[Denominator[f[w]]]] == 0, w]; This way you don't have to complex-conjugate the solutions. IF a is real, then you need to provide this as Apr 14, 2023 · As always in Mathematica, symbols implicitely are assumed to be complex valued except 0,oo as default. Conjugate[x_]:> (x/. Jan 10, 2018 · Bug introduced after 10. know. Mar 24, 2024 · Miminmal working example: Assuming [a ∈ Reals, Conjugate [a]] Returns Conjugate [a] instead of a. The answer from Alexei Boulbitch is also useful, but not much different than the replacement rules I mentioned in my post. Now as stated in the problem and the comments above this particular problem requires a fair amount of assumptions. Stay on top of important topics and build connections by joining Wolfram Community groups relevant to your interests. Try (Sqrt[A] Sqrt[Conjugate[A]])/Sqrt[A Conjugate[A]] // PowerExpand[#, Assumptions -> A \[Element] Complexes] & // FullSimplify Piecewise[{{-1, Arg[A] >= Pi}}, 1] So it's 1 if Arg[A] <= Pi May 10, 2021 · It is a common misconception that patterns may be used in Assumptions, but this is not the case. The downside is that the order of the solutions in wroots1 and wroots1C may not be the same. Sep 1, 2016 · What you observe is the fact that Mathematica assumes the most general case, since you did not specify any assumptions. Within Simplify and similar functions, objects that satisfy inequalities are always assumed to be real. See the minimal example below: expr = Exp [I*2*π*Re [n]] mat = { {Re [n], a*expr}, {Conjugate Oct 5, 2017 · A bit of a hack, but this works (correctly) for most mundane cases of expression involving only real unknowns and holomorphic functions: expr/. Integrate[(L r x^(r - 2))/(r - 1)! e^(-L x), {x, 0, Infinity}, Assumption How to tell Mathematica that certain variables are real/imaginary, integer-valued, etc I'm trying to expedite some quantum mechanical calculations (expectation values etc. That's the reason why when you enter: Oct 21, 2016 · a Sqrt[c - (b Conjugate[b])/a] Conjugate[Sqrt[c - (b Conjugate[b])/a]] in the denominator. When I say, for example, functions complex assumptions variable-definitions Matthew Brunetti 633 asked Nov 21, 2014 at 22:50 34votes 4answers 10kviews How to tell Mathematica that certain variables are real/imaginary, integer-valued, etc I'm trying to expedite some quantum mechanical calculations (expectation values etc. To tell Mathematica that x and y are real numbers, use the ComplexExpand command: Example of defining your own function: F[x_]:= E^x Sin[a x] . Another option is to make an easy home-cooked conjugation for the problem at hand. 0 Mathematica 11. So the source of the issue is this global assumption? Dec 20, 2015 · Refine is more or less the basic "evaluate-with-assumptions" function. However, patterns can be used in Element. Typically counterindications are the occurrence of square roots etc. FullSimplify [expr] tries a wide range of transformations on expr involving elementary and special functions and returns the simplest form it finds. Alternatively, you can write f = E^x Sin[a x] , in which case f gets an immediate assignment based on the value of x at the time you execute the command, and that specific value of x will remain in f unless Simplify tries expanding, factoring, and doing many other transformations on expressions, keeping track of the simplest form obtained. Why mathematica cannot get rid of Abs when all variables are positive and reals? I read a lot of old posts but still couldn't make it work. The implicit assumption of real values in ComplexExpand does not carry over to the outside function FullSimplify. Oct 25, 2021 · This treats It and R as real numbers and it expands the exponentials using Euler's identity. Mar 11, 2023 · You used SuperStar, which has no inherent meaning in Mathematica and stays unevaluated. The assumptions can be equations, inequalities, or domain specifications, or lists or logical combinations of these. \[alpha] is real-valued, you need to let Mathematica know this very important fact. Apr 6, 2019 · I'm trying to define a complex function to be able to do analytic calculations. Jun 21, 2019 · How to tell Mathematica that certain variables are real/imaginary, integer-valued, etc I'm trying to expedite some quantum mechanical calculations (expectation values etc. Upvoting indicates when questions and answers are useful. If you know that e. In[240]:= Assuming[Element[a, Reals], Refine[Conjugate[a]]] Out[240]= a. ) by running them through Mathematica. How to make Mathematica simplify v1 and v1prime? Assuming [a ∈ Reals, Conjugate [a]] fails to invoke assumptions [closed] Miminmal working example: Assuming [a ∈ Reals, Conjugate [a]] Returns Conjugate [a] instead of Nov 6, 2024 · Sorry to come across as snarky. Specifically, I want to define a function $A(x,t)$ in the complex plane with argument The problem is that we have not specified whether x and y are real or complex numbers, and Mathematica won't make any assumptions about x and y without our help. Refine is such a function, whereas Conjugate is not. what, I think that is the most useful answer. FullSimplify doesn't do everything because it Jan 3, 2016 · Using Coefficient for Conjugate [variables] (Mathematica ignores my assumptions) Ask Question Asked 9 years, 2 months ago Modified 9 years, 2 months ago Jan 3, 2016 · Using Coefficient for Conjugate [variables] (Mathematica ignores my assumptions) Ask Question Asked 9 years, 2 months ago Modified 9 years, 2 months ago Oct 2, 2022 · Moreover, Conjugate assumes every variable is potentially complex so it does not help without including assumptions on the variables. you. Does anyone have any tips/tricks on how to do it faster? Any help is gre May 24, 2020 · It is inconsistent to say that Mathematica won't simplify a complex expression while simultaneously not making use of one of its primary tools for simplifying complex expression, i. More generally, only those functions that take an Assumptions argument will react also to an ambient Assuming block. When I say, for example, functions complex assumptions variable-definitions Matthew Brunetti 643 asked Nov 21, 2014 at 22:50 36votes 4answers 11kviews Oct 16, 2020 · How to tell Mathematica that certain variables are real/imaginary, integer-valued, etc I'm trying to expedite some quantum mechanical calculations (expectation values etc. Wolfram Community forum discussion about Conjugate and Simplify. I thought Reduce would be perfect for this, but it gives redundant output (even after simplification). When I say, for example, functions complex assumptions variable-definitions Matthew Brunetti 643 asked Nov 21, 2014 at 22:50 39votes How to tell Mathematica that certain variables are real/imaginary, integer-valued, etc I'm trying to expedite some quantum mechanical calculations (expectation values etc. I learned about assumptions and figured they're my way to go. The following works: Simplify [Conjugate [f [x]], Assumptions -> Element [f [_], Reals]]. For exampl You need to tell Mathematica to try to simplify the expression, otherwise it won't look for the reality condition you've added. Solve[Denominator[f[w]] == 0, w]; wroots1C = w /. When I say, for example, functions complex assumptions variable-definitions Matthew Brunetti 643 asked Nov 21, 2014 at 22:50 39votes Nov 2, 2021 · I'm trying to obtain the solution to the following system of 3 equations and 4 variables: Solve[{y Conjugate[x] + z Conjugate[w] + x Conjugate[y] + w Conjugate[z] == 0, x Conjugate[x] + w How to tell Mathematica that certain variables are real/imaginary, integer-valued, etc I'm trying to expedite some quantum mechanical calculations (expectation values etc. And no need for assumptions. How to tell Mathematica that certain variables are real/imaginary, integer-valued, etc I'm trying to expedite some quantum mechanical calculations (expectation values etc. I prefer to use assumptions locally when needed or just use ComplexExpand which has the Reals assumptions buildin. m = {{Re[n], a, b, b}, {Conjugate[a], Re[n], b, b}, {Conjugate[b], Conjugate[b], Re[c], d}, {Conjugate[b], Conjugate[b], Conjugate[d Here's a round up. If a user knows these assumptions are obvious physically, they might want to Jan 4, 2020 · I am confused how to assume a variable is pure imaginary. Dec 10, 2015 · Using Refine with assumptions, as suggested in question 77456, rather than ComplexExpand, seems to imply the root source of the problem is MMA not being able to conjugate series data: Jun 30, 2018 · It evidently depends on the value of A. Feb 14, 2018 · I give Mathematica these commands: $Assumptions = Element[x, Reals] &amp;&amp; Element[y, Reals] \\[Theta] = ArcTan[x, y] z = Exp[I \\[Theta]] Conjugate[z] I get E Aug 20, 2014 · The difference between the three examples above is: The 1st example uses only the assumption that is given and ignores global $Assumptions. But this will not work Frequently, Mathematica will produce some expression in a complicated form, with a number of unset variables. Assuming[p \[Element] Reals, Conjugate[p] // Simplify] May 7, 2025 · I need some help solving this integral here. Edit: More succinctly, if we know all of our symbols are real numbers, we can wrap them with Re before sending them to Cholesky. The actual expressions look like: Jul 23, 2021 · I'm trying to calculate the analytical Cross-Correlation between two time-lagged Sinc functions, i. S. However, I wouldn't like to use Simplify[] because it can be very slow with longer expres Assuming affects the default assumptions for all functions that have an Assumptions option. mife ppz ailxzqt dteua ubovgx cwnhwg xnxmq jzvziv ropc dqssamj iyitm fgyuyce dmiamurj fbpvnkqn luf