Mathematica global assumptions real. Mathematica doesn't really do typing like that.
Mathematica global assumptions real Complex[0, 1] -> Complex[0, -1] This is used to represent global assumptions, but you can also use this class to create your own local assumptions contexts. Share It will be more convenient to define the function eq this way :. So the source of the issue is this global assumption? 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 ∈ Thanks for contributing an answer to Mathematica Stack Exchange! Please be sure to answer the question. Note that if a variable is positive it is redundant to add that it is real. Also addresses the likely mistyping of 'e' for 'E' $\endgroup$ – SEngstrom Globally defining a variable as real in Mathematica sets the variable as a real number for all subsequent calculations and functions. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site I am trying to integrate a hat function for a project that I am doing and have found a method to do so but I find it sloppy. $\endgroup$ i am trying to integrate a real valued function. An arbitrary complex number might be a positive real number but it also might not be. in the beginning of the code, but it doesnt work We help clients realize the full potential of computational knowledge & intelligence. Asking for help, clarification, or responding to other answers. Also, the assumption that rho is real is not enough to simplify the Abs to rho. For example, x/x will simplify to 1 without any assumptions on x . Mathematica doesn't really do typing like that. I treat ComplexExpand (and PowerExpand) as unsafe, that is every use needs careful examination to ensure that their assumptions are met. real(x)}) >>> ask (Q. Since all terms in the numerator are positive, the numerator is positive. Visit Stack Exchange Clear["Global`*"] q2sol = (Sec[2*a1]*Tan[a1]^3*(-7 + 6*Cot[a1]^3*Sqrt[Tan[a1]^2]))/(6* Sqrt[Tan[a1]^2]); Simplify using the constraint as an assumption. I suppose it has to do with assumptions I'm making that mathematica isn't, but I don't know why. For a general technique of using FindRoot the way you would like I recommend to read this post : First positive root. Yeah, yeah some people think that Sqrt[explicit_positive_real] must mean the positive root. On the other hand, the inequality isn't changed if we introduce a different variable y: Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site I want to find the positive root of the following function F[x], but using Solve does not work. How can I force Mathematica to give me a symbolically real solution. Simplify[Conjugate[Exp[-I a e]], a \[Element] Reals] so that a is assumed to be real. Our research spans a range of sectors, including education and workforce development; energy and climate; food $\begingroup$ @Robin_Lyn You have to realize, though, that only functions that take an Assumptions option will be aware of global assumptions. $\endgroup$ ["Global`*"]; ode = x^2 y''[x] + x y'[x] - n y[x] == 0 sol = DSolve[ode, y[x], x] Mathematica will never make assumptions you don't tell it to make, and will state all conditions under which the answers it gives Double variable integration unable to prove that integration limits are real. Get desired result for 1<a and a<b But I took the real part, hence it one case has to be TRUE! For a=2b. $\begingroup$ Solve eliminates assumptions from the conditions that are unnecessary but retains those that are necessary. Also, the Arg simplifies if you assume that phi is between -Pi and Pi: For some functions assumptions can be given both as an argument and as an option value: I will do that, but to understand it correctly, I do not have to check if Det is allowing me to do Assumptions (hey Mathematica team, this would be really really really useful) but if === is allowing me to do assumptions, isn't it? And I am really surprised, that === and < are not allowed to do Assumptions But thanks for the help Here is a small package, based on @rcollyer's idea, that allows us to declare certain patterns as "real". For some reason, Mathematica gives an answer in the complex domain. I would like Mathematica to tell me whether an integral is positive or negative, given a list of assumptions. Here is the integral: Integrate[(-(Cos[b*s] - (2*Sin[b*s]*(a*D - b*C))/(a^2 + b^2))/E^(a*s))^2, {s, 1, ∞}, Assumptions -> Element[s, Reals]] Mathematica produces the following: the annoying part is that mathematica is typically really stupid when it comes to real or complex numbers (and getting the real part or the complex part). However, in my experience the more assumptions you have, the longer it will take to simplify. ; Maximize is typically used to find the largest possible $\begingroup$ Thanks for answering, but "assuming" will obviously help us since we can Assumptions -> a>0,Assumptions -> a<0, or Assumptions -> a==0 (Yeah I mentioned that for a<0 the solution is correct in the complex sense but we seldom want a complex solution for a real ODE). (Q0: When writing my questions here, how do I format Mathematica input and output so that it looks more human-readable, like it is in my Mathematica notebook?!) Q1: Is there a way of telling Mathematica that I expect a real answer in this problem, apart from specifying real inputs using Assumptions? The ability to generate pseudorandom numbers is important for simulating events, estimating probabilities and other quantities, making randomized assignments or selections, and numerically testing symbolic results. Currently I have the basis function \[Psi][z_] := z - Subscript[Z, i]/ \ The Wolfram Language has a flexible system for specifying arbitrary symbolic assumptions about variables. Nov 17, 2024 · Thanks for contributing an answer to Mathematica Stack Exchange! Please be sure to answer the question. I tried to write . Such applications may require uniformly distributed numbers, nonuniformly distributed numbers, elements sampled with replacement, or elements sampled Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site A celebrity or professional pretending to be amateur usually under disguise. I'm attempting to do this using the 'Refine' command. So sometimes you might want to use local assumptions. Complex[0, 1] and rule above is interpreted by Mathematica as . Stay on top of important topics and build connections by joining Wolfram Community groups relevant to your interests. i. Without them the definitions block takes about a second to evaluate. Start with a simple example: Resolve[ForAll[x, x^2 >= 0]] (* ==> True *) This can be explained because Resolve implicitly assumes that x is real because it appears in an inequality. For a general view I recommend reading this post by Leonid Shifrin. I prefer to use assumptions locally when needed or just use ComplexExpand which has the Reals assumptions buildin. In the function f[x] appears a parameter J, which is Real. However, I need those assumptions, so Mathematica can do useful simplifications. Of course, given your substitutions, it may be even better to use the result provided by the authors of the article, but there's no straightforward way to make MMA realize, that there's a nice relationship between g and gc that will let it simplify to what you want. And this is where my problem appears. Simplify[x > y, False] (* Simplify::fas -- Simplify: Warning: one or more assumptions evaluated to False. Practically speaking, they are necessary (or at least the most elegant way) to express assumptions that don't fit in a predefined list, say, assume(x*y > -1). Find what is missing and explain what assumptions it seems that Mathematica is making. Visit Stack Exchange I want to find the positive root of the following function F[x], but using Solve does not work. Without further information there is no way to mathematica gives me a complex-valued reuslt, but maple 17 gives me what I want, see the attached picture. >>> global_assumptions AssumptionsContext({Q. 8 it is for the domain. Mathematica. In general, symbolic variables are processed as complex if not assumed otherwise. 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[ seems to convert complex expressions which contain symbols which are meant to be real. Visit Stack Exchange Stack Exchange Network. $\begingroup$ @DavidG. Get desired result for 1<a and a<b with the global assumptions Complex number operations: telling Mathematica variables are real. I am trying to apply assumptions to help comparing the two values: Clear[a, b]; Refine[f[a, b], {a > b}] And I would expect to get output: a a , but what I get instead is: If[a>b,a,b] a My question is: how do I apply the assumptions to that function properly? I just need a way of forcing Mathematica to think that everything inside a Log is real and positive, so that FullSimplify works appropriately! Not really, because I have no idea what the argument of the Log is in general! If you know of a way of setting global assumptions on the argument of the Log function, that would certainly work. The underlying problem is integration in this simple example, but the distinction is I think the logic for why this happens goes as follows. I would like the expression to be displayed as $\sqrt{1 - \delta^2} \cdot i $. Assuming Jun 21, 2016 · A more restrictive version is $Assumptions = _Symbol ∈ Reals. You should use Reals or type with Esc``reals``Esc. To specify assumptions there are a few ways : In general Mathematica does not bother with treating removable singularities separately. Currently I have the basis function \[Psi][z_] := z - Subscript[Z, i]/ \ Plot the integrand to see, it explodes for 0<a<1 and a<b<1 as t goes to infinity, means integral does not converge. Mathematica Global cultivates the evidence and insights needed for sustainable progress on the most urgent and universal issues of our time—from global health and education to climate change, energy, and workforce development. By establishing a new energy functional, we demonstrate the global boundedness Mathematica is a term rewriting system, variables need not to be declared as in compiled languages. The minimal method is to use Refine as so: Refine[Conjugate[a+I b], θ ∈ Reals]. I suggest you to use Assumptions under function Refine, like below. (The output is actually very long, I omit the rest of it) I failed to get the real part of n3[t] directly with Re[] either. Use MathJax to format equations. u > 0 and l > 0 implies Element[u | l, Reals], thus we need not to add this assumption : Integrate[ 1/Sqrt[ z^2 + u^2], {z, -l, l}, Assumptions -> u > 0 && l > 0] Nov 16, 2024 · How to tell Mathematica that certain variables are real/imaginary, integer-valued, etc (3 answers) Closed 8 years ago . -- For some other functions, see if they have an Assumptions option or accept constraints. PositiveReals is output in StandardForm and TraditionalForm as . (Another way to look at is that there is a difference between hypothesis Wolfram Community forum discussion about How to assume real variables?. Refine[Conjugate[a + I b], _ ∈ Reals] a - i b. How can say mathematica that it should not take the Conjugate of J? So, I would like to say mathematica that J is real. When they can be, I use Simplify as @MarcoB shows below. Rule {I -> -I} does not, even on simple example: 2 I /. 1' Mathematica-Compatible Notebook: This notebook can be used with any Mathematica-compatible: application, such as Mathematica, MathReader or Publicon. And Limit does take assumptions. {I -> -I} 2 I the reason being that symbol I is automatically translated by Mathematica to. I am trying to integrate a hat function for a project that I am doing and have found a method to do so but I find it sloppy. One can use Refine and assume everything is real. But it will only assume proper symbols to be real. Is it possible to attach assumptions to a symbol? This relates to this question. 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. If you are sure that x is real and positive, you can pass that information as an assumption to Simplify[] you can also modify the global assumptions. And yes, I've actually tested it. Examples are Reduce, Solve, FindInstance, etc. These will have a "domain" option, which can be set to real. Assumptions -> J e Reals. Is there a way to write a FullSimplify at the end of the line with some assumptions? That is, if I write a^2+b^2+2 a b //FullSimplify[] is there any way to include some assumptions for a,b? Pushing Mathematica's FullSimplify to a global complexity minimum. resets it to the default state. Mathematica today announced the introduction of Mathematica Global, the new name for the company’s International Research unit. I think that Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Since all terms in the numerator are positive, the numerator is positive. However once I assume that x>0 Solve won't return anything even though there is a solution if y>0. If you want exact results you must use exact We establish the global well-posedness for the multidimensional Chemotaxis model with some classes of large initial data, especially the case when the rate of variation of ln v0 (v0 is the chemical concentration) contains high oscillation and the initial density near the equilibrium is allowed to have large oscillation in 3D. I have a function of multiple variables, something like this or similar: 1/(5184 h j (1 + 2 n) r^3 Ω^4)I (j^3 k + 16 j^3 k l - 16 j^2 k Ω^2 + 72 j^2 k r^2 Ω^2 + 144 j^2 k n r^2 Ω^2 + 864 k $\begingroup$ I see. After declaring pattern as real, Mathematica will automatically simplify some expressions involving sub-expressions matching said pattern. but then you can run into trouble when it starts assuming every expressions is real (not just variables): Refine[Conjugate[Sqrt[I*a]], _ ∈ Reals] $\sqrt{i a}$ Like Rojo's alternate answer, you can assume that only the symbols in your expression I am trying to write a generic matrix valued function in a package, of the form: f[matrix_] := Module[] My problem is that I want to accept any matrix, but always assume that the functions/variables contained within the matrix are real. Since x does not have any assumptions declared it roughly represents an arbitrary complex number. So the fact that k == 1/2 in the sum contradicts the assumption k != 1/2 does not mean Assumptions were ignored. 2. How can I get rid of this? Here is my code n = Simplify [expr ∈ PositiveReals, assum] can be used to try to determine whether an expression corresponds to a positive real number under the given assumptions. ; Refine can be used on equations, inequalities, and domain specifications. Hot Network Questions Mathematically speaking, you are trying to specify a branch of the square-root function by an irrelevant assumption. Mathematica 8 can't extract it. Get desired result for 1<a and a<b Since all terms in the numerator are positive, the numerator is positive. $\begingroup$ Assumptions are not constraints, from the point of view of logic. Svelte is a radical new approach to building user interfaces. Finding real and imaginary parts. It seems that m and k are also regarded as complex numbers. Is there a simple way to get the Conjugate of a complex term, provided all information is available? 0. Simplify tries expanding, factoring, and doing many other transformations on expressions, keeping track of the simplest form obtained. Quit does not "clear" anything, it instead restarts the kernel, i. " So you don't have to do anything for Minimize. Solve not working with assumptions I am trying to solve a simple equation `Solve[x-y==0,x]`. Since symbols no longer in use can introduce unexpected errors when used in new computations, clearing your definitions is very desirable. It uses a wide range of sophisticated algorithms to infer the consequences of Jul 19, 2019 · How can I include many global assumptions in the different parts of my notebook (as I have many assumptions concerning different things which are defined at different Aug 20, 2014 · Mathematica keeps track of global assumptions in the variable (or whatever that is) $Assumptions. For example. ; Maximize finds the global maximum of f subject to the constraints given. If no other information is given, You can use $Assumptions = {\[Rho] > 0 && -Pi < \[Phi] < Pi} to make global assumptions. Mathematica doesn't simplify this seemingly ArcTan[x,y]+ArcTan[x,-y] I have tried out inserting the global assumptions inside Simplify and FullSimply but it still doesn't work. I personally do not like the use of global assumptions and do not use them. Is it possible for Mathematica to use the stated assumptions to tell me that the second order condition is positive, given the set of assumptions? There are several reasons why an assumptions query might give None. Use Assumptions to simplify/modify terms. Posted on 20 Aug 2014 . For this kind of symbolic polynomial equation of degree n, it's relatively easy to solve the roots ma But I took the real part, hence it one case has to be TRUE! For a=2b. Instead, some functions take an Assumptions option which affects that The typical default setting is Assumptions:>$Assumptions. The ConditionalExpression output by Solve in your case does not really depend on the whole argument of the Log function in your original equation, but only on the following expression:-π < Im[(k t (L - T))/L] ≤ π If it is possible in your case to make assumptions on the values of those parameters, you could then try to Simplify the output of Solve using the I am trying to find the complex conjugate of E^(- I a e) in Mathematica, but when I enter . The underlying mechanism used automatically to check inference of inequality assumptions is also used by various new functions allowing us to solve systems of real polynomial equations and inequalities, minimize polynomial functions on polynomial inequality constraints, and eliminate The Wolfram Language is built to handle arbitrarily large computations\[LongDash]limited only by computer time and memory\[LongDash]and provides a collection of convenient global safety features to prevent programs from going out of control. You seem to want something like Exclusions, but I don't think Assumptions can be used to enforce exclusions. I think that is can't simplify because it may happen that the value of the integral (even if the integrand is real) is complex, how can I tell mathematica to give the result : Integrate[a f[t], {t, -R, R}] I'm making some calculations, where I use the command Conjugate[f[x]]. In your link it seems quite obvious, that Mathematica's result is "simpler". 7 the third param was for variables to eliminate, and in v. No. You need to assume that rho is >0. . However, x > 0 is sufficient and (in Mathematica) implies that x is also real. – No. In Mathematica, you can use the function FullSimplify[expression,assumptions] to simplify expressions using assumptions. For this kind of symbolic polynomial equation of degree n, it's relatively easy to solve the roots ma Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site On the other hand when Mathematica solves the above equation with a specific value for n, which means that this has no built-in meaning and doesn't represent the set of real numbers. For example, D[blah] doesn't work with Re[] or Im[], and I can't get it to thread across Sqrt's (taking the sqrt of a complex number), even after assumptions are made. Provide details and share your research! But avoid . Is it possible for Mathematica to use the stated assumptions to tell me that the second order condition is positive, given the set of assumptions? Another option for you is to use ParametricNDSolve, which numerically solves differential equations with one or more parameters. By "formal" I mean that if you tell it that a[x] is real, it will not know automatically that a'[x] is also real. Also, for Solve, the domain applies only to the variable being solved for and for the domain of the solution. The assumptions can be equations, inequalities, or domain specifications, or lists or logical combinations of these. In MMA, is there a general way to do integrations in real domains, just like maple. Simplify early There are significant practical differences. The original technical computing Jun 21, 2016 · There are other functions which do not have an Assumptions option but can still work with reals only. I have a problem with using Assumption . In Mathematica, I want to evaluate the expression $\sqrt{\delta^2 - 1} $ under the global assumption that $ 0 < \delta < 1 $. So far I have been able to make mathematica assume that ONE of them is real by writing. It is only possible to do this in a formal way, using patterns, and only for certain functions that have the Assumptions option. For instance, if I do this: FullSimplify[x^2-y^2,x^2-y^2==1], then the result will be 1 because that's the 'simplest expression that is equivalent to the function I gave the software. Refine[Reduce[{52*n^2 + 19*n*nb - 42*n*nf > 0, 52*n^2 - 61*n*nb - 42*n*nf > 0 Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Ahh, I see. In general Mathematica does not bother with treating removable singularities separately. eq[h_] := the formula Real solutions. Somewhat simplifying: Just because 9>0 does not mean that -3 is not also a square-root of 9. This can save you lots of typing. Ulrich's approach, on the other hand, is explicit and local, which is actually a good thing so you don't fall into hard-to-debug situations that depend on remote assumptions / assignments. It is basically a thin wrapper to Python’s set, so see its documentation for advanced usage. soln = ParametricNDSolveValue[ {y'[x Then I tried to get the real part of n3[t] with ComplexExpand[], but I failed. The data: for the notebook starts with the line containing stars above. $\endgroup$ Since you only have two square roots and everything else is positive you can push everything but the square roots to the RHS, square both sides, push everything but the Sqrt[]Sqrt[] to the right hand side, square both sides, expand, Collect on eta, Reduce that for eta, Simplify with the assumption a > 0 && b > 0 && c > 0 && n > 0 && g > 0 && ns > 0 && P > 0 && eta > 0 a[i_] on the other hand should take on integer i and return a positive real number. Mathematica Global offers end-to-end measurement, evaluation, research, and learning (MERL) services to accelerate inclusive, equitable, and sustainable development. Simplify[Re[(-3 a - 6 b + Sqrt[9 (-36 + 7 a^2) + 4 a (-27 + 8 a^2)*b + 4 (-45 + 16 a^2) b^2 + 32 a b^3])/4 (a + b)] < 0, Assumptions -> {a > 0, b > 0, a = 2 b}] you are not affecting a in any way. Stack Exchange Network. Feb 6, 2021 · The assumption Element[t, Reals] is only used inside Refine, and so in a sense is not "known to" any evaluations outside of that expression. However when I simplify the expression with appropriate assumptions a 'Conjugate' still remains. Visit Stack Exchange Plot the integrand to see, it explodes for 0<a<1 and a<b<1 as t goes to infinity, means integral does not converge. 3. Stork the OP said assume things like "everything is a positive real unless specified otherwise" etc? so I assumed they wanted this for Any command in Mathematica that uses assumptions (Simplify, Integrate etc). I think that I would actually prefer global assumptions for various reasons. There is a lot of internal state that changes during the session in ways that are different from creating new symbols or attaching definitions. e. It'd boil down to an integrate of: C Integrate[Sin[D t],{t,L/3,2L3] With C, and D being constants related to the weight distributed over a stick of length L. I was not getting the result I wanted, so I simplified my assumptions to the following: $\begingroup$ From the docs for Minimize: (1) "If no domain is specified, all variables are assumed to be real. This can lead to totally bogus results. $\begingroup$ @rcollyer: No, it is vulnerable because the mere creation of that symbol already hides the local one. q2sol $\begingroup$ Note that using things like Simplify may perform manipulations you do not want, like replacing x^2 + 2 x y + y^2 with (x+y)^2. Share The next version of Mathematica allows specifying assumptions in Simplify and related functions. Can this wrap bulit-in command proposed by Todd Gayley (see: What is in your Mathematica tool bag?) do the Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site When you set a value to a symbol, that value will be used for the symbol for the entire Wolfram System session. Whereas traditional frameworks like React and Vue do the bulk of their work in the browser, Svelte shifts that work into a compile step that happens when you build your app. To get the notebook into a Mathematica-compatible application, do: one of the following: The Wolfram Language is built to handle arbitrarily large computations\[LongDash]limited only by computer time and memory\[LongDash]and provides a collection of convenient global safety features to prevent programs from going out of control. That you don't get a Sum::fas message means maybe it did. Besides, we show the optimal time-decay . Is it possible for Mathematica to use the stated assumptions to tell me that the second order condition is positive, given the set of assumptions? Even though I am assuming the discriminant is greater than 0, Mathematica still outputs a complex solution which is undesirable. Also, the Arg simplifies if you The Wolfram Language has a flexible system for specifying arbitrary symbolic assumptions about variables. J*Integrate[Integrate[Cos[ArcTan[Sqrt[(x - a)^2 + (y - b)^2]/L Simplify is very gullible: if it gets a False assumption, it'll believe anything is True:. Also, according to the docs, Assuming only affects those functions that have an Assumptions option. It's not clear to me that in every use-case, one would want the necessary condition eliminated. $\endgroup$ – naeema18. After defining the function in notebook 2, a in notebook 1 will no longer evaluate to 3, not because the value of notebook1uniquecontext`a has been overridden (that one still has the value 3 associated with it) but because that definition is Many situations where assumptions play a role in differentiations can be equally or more clearly expressed by going back to the Limit defining the derivative in the first place. Maximize is also known as supremum, symbolic optimization and global optimization (GO). I know of a way to define an arbitrary function (say, a[x]) and then apply an assumption to it (say a[x] is real). I can use the keyword Reals for the argument dom in the Solve function Solve[expr,vars,dom]. I saw on different forums that many people have had difficulties with extracting real or imaginary parts of The difference between the OP's code and my first code is that while α > 0 implies α is real, because inequalities imply the terms are real in Mathematica, {x, 0, α} does not imply x is real. It does not apply to any $\begingroup$ The assumptions are slightly different (r>1) and allow Mathematica to find the analytic solution. Nov 16, 2024 · The OP seems to need the integral in the real domain and if not specified explicitely in general Mathematica evaluates integrals by default in complex numbers. Locally defining a variable as real using the "Real" function only applies within the scope of the function or calculation where it is used. Related. The video has to be an activity that the person is known for. So for your example, instead of using D, you could write. However in order to demonstrate how it works for real numbers we should have a different equation since this one has no real solutions : Assumptions on parameters in an integral (See also ASSUMPTIONS section below): If we know that a parameter, n, is in a certain domain, say n > 1, add-on the option: "Assumptions -> n > 1". (x 1 | x 2 | ) ∈ PositiveReals and {x 1, x 2, } ∈ PositiveReals test whether all x i are positive real numbers. Jan 7, 2025 · There's no general way to declare a variable as real, integer etc. $\endgroup$ CreatedBy='Mathematica 5. So, you have to include your assumptions as extra equations passed to Mathematica's Global Unit works in more than 50 countries across Africa, Asia, and Latin America, collaborating with country partners to develop culturally sensitive approaches that reflect a deep understanding of the local context. FullSimplify does more extensive simplification than Simplify. ; Simplify can be used on equations, inequalities, and domain specifications. I tried using assumptions, but it doesn't work. $\endgroup$ – The Real Housewives of Atlanta The Bachelor Sister Wives 90 Day Fiance Wife Swap The Amazing Race Australia Married at First Sight The Real Housewives of Dallas My 600-lb Life Last Week Tonight with John Oliver Generally, as far as I know, there is no mathematical way to tell Mathematica that a variable is real. there are basic assumptions on variables that will always hold true (in a physical sense). This will not cause Simplify[Sqrt[x] ∈ Reals] to return True. *) (* True *) Simplify[1 > 2, False] (* Simplify::fas -- Simplify: Warning: one or more assumptions evaluated to False. To specify assumptions there are a few ways : Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Plot the integrand to see, it explodes for 0<a<1 and a<b<1 as t goes to infinity, means integral does not converge. For example, a professional tennis player pretending to be an amateur tennis player or a famous singer smurfing as an unknown singer. Refine doesn't change the nature of A; it merely spits out an expression produced from its argument evaluated under the provided assumptions. The documentation for Resolve states this. For the second one, remember that Mathematica can't make assumptions on your symbols, so a "number" is complex by default. It uses a wide range of sophisticated algorithms to infer the consequences of assumptions\[LongDash]often in the process automatically proving a sequence of necessary mathematical theorems. Making statements based on opinion; back them up with references or personal experience. Aug 20, 2014 · Using Assumptions in Mathematica. Visit Stack Exchange Mathematica produces the answer to integral((x + 5)/(x^2 + x − 2) dx). Mathematica is a term rewriting system, variables need not to be declared as in compiled languages. Unfortunately, the evaluation of the definitons takes very long if I include these assumptions. "(2) Further, "Minimize requires that all functions present in the input be real-valued. The assumptions don't "remain attached" to the symbol A. expr2 = Assuming[assumptions, f[x] /. I have simple question regarding double integrating a function. Since $\mu > \lambda$ by assumption, then $(-\lambda + mu)^3$ is also positive for any $\mu$, $\lambda$ combination. Re[1+I] > 0 but 1+I is not real. $\begingroup$ Two question about this: Is there a way to specify the assumption that all variables involved are real using this method, the way Assumptions->Reals does? And I thought I had read that using Assuming[a==1, ] permanently added "a==1" to the default assumptions for further calculations until changed, so I had avoided it. We work with clients around the world to leverage research, analytics, data science, and technology to enable evidence-driven strategies, programs, and investments that improve Assumptions can consist of equations, inequalities, domain specifications such as x ∈ Integers, and logical combinations of these. real (x)) True. Conjugate[Exp[-I a e]] Mathematica assumes both the variables a and e are real. Most of my work involves physical equations, i. Use an Assuming construct to make the assumptions available to all enclosed functions. Now I want to extract the real part of this by assuming real only variables and then Re[PP]. A[x_] = Function[x, a[x]] Assumptions[A[x] ∈ Reals Integrate[x A[x], {x, -1, 1}] // FullSimplify DSolve[Sqrt[f[x]] - a f'[x]/f[x] + b/x == c, f[x], x, Assumptions -> assumptions] // First // FullSimplify The assumptions also can be used by FullSimplify. I do not want to visit the parameter space with complex eigenvalues and want to code to reflect that all formulas result in real numbers. Re[x] > 0 means that the real part of x is positive, but it does not mean that the imaginary part is zero. Provide details and share your research! But avoid Asking for help, clarification, or responding to other answers. By default, this just contains True. That's the reason why when you enter: I have an expression in which all variables are real. Such assumptions will eventually be necessary to deal correctly with difficult integrals and probably in other contexts as In this paper, we investigate the initial boundary value problem of a three-species spatial food chain model with nonlinear taxis sensitivity in a bounded domain $$\\Omega \\subset {\\mathbb {R}}^2$$ Ω ⊂ R 2 with a smooth boundary and homogeneous Neumann boundary conditions. How can I set this global assumption so that the output reflects this transformation correctly? Stack Exchange Network. Thanks for contributing an answer to Mathematica Stack Exchange! Please be sure to answer the question. I'd be grateful for your answers. It will also automatically use appropriate global assumptions when simplifying using Refine, Simplify In v. It is possible that the query is unknowable as in the case of x above. FullSimplify fails with simple assumptions. Data & Computational Intelligence Model-Based Design Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site $\begingroup$ I think it's worth noting that while Simplify and Refine are "safe", in that they shouldn't make invalid assumptions, ComplexExpand assumes that variables are real, unless told otherwise. The integration may take a complex path between two numbers that happen to be real, and Integrate tries to deal with that. ; Quantities that appear algebraically in inequalities are always assumed to be real. My code is pasted below. 1. And remove them: I am trying to solve an equation by assuming that all the variables are real and strictly positive. Thus, x will be considered real, but not f[x] and not The Wolfram Language has a flexible system for specifying arbitrary symbolic assumptions about variables. Also, in place of the assumption θ ∈ Reals you can use the assumption _Symbol ∈ Reals to assume that all explicit variables are real. Limit[(Abs[x + e] - Abs[x])/e, e -> 0, Assumptions -> x > 0] (* ==> 1 *) Stack Exchange Network. you are not saying to Mathematica 'a is less than zero'. spqtt vxzelp skehtso xtml qpznq iwwz zxaca hymdfp jouczt zukigr