Evaluate [expr] causes expr to be evaluated even if it appears as the argument of a function whose attributes specify that it should be held unevaluated.

By using Evaluate, you can get any argument of a function evaluated immediately, even if the argument would usually be evaluated later under the control of the function. An exception to this is when the function has the HoldAllComplete attribute; in this case, the contents of the function are not modified by the evaluator.

The simplest examples of evaluation involve using definitions such as f [x_] =x^2 which transform one expression directly into another. But evaluation is also the process used to execute programs written in the Wolfram Language.

Apr 26, 2012 · You can use functions like Apply and Map to evaluate a function in a list of points, for example. f[x_, y_] := -4 x + x^2 - 4 y - y^2 pts = {{1, -2}, {2, -3}, {3, -2}, {2, -1}}; Apply[f, pts, {1}] (* out: {1, -1, 1, -1} *) or using @@@ as a short hand for Apply[ ...., {1}] f @@@ pts

Jun 5, 2013 · In Maple, you right click and can evaluate an expression (whatever it is) at a specific point say: Expr.= abcddafjosjfoj, then right click and simply evaluate at c=-34 say. But how does one do the correspoinding operation in Mathematica?

Oct 20, 2013 · Evaluate is then used to evaluate the symbolic expression first (nullifying the HoldAll-Attribute) and afterwards evaluating the obtained expression at certain numeric values for the plotting-variable x.

5 days ago · Function evaluation in Mathematica is indicated by square brackets. That is, while in mathematical notation, we write \( f(x), \) in Mathematica the correct syntax is f[x]. Ordinary parentheses are used exclusively for algebraic grouping.

Sep 23, 2020 · I need to evaluate a function in 2D but it has three unknowns, the two arguments (x, y) and alpha. I need to evaluate x and y between {0,100}, while alpha between {0,1}. The function is: f(x, y) = ...

How about function? The file states: A small test. test[exp_] := Function[x, exp] test[x^2] (*Function[x$, x^2]*) So we see that the formal parameters of function have been renamed to x$ and no longer match the x you have put in. Here are …

sol = DSolve[f'[x] == 1/x , f[x], x] ; f[x] /. sol. (*{C[1] + Log[x]}*) It would be easy to substitute a value for x now, but you still have the constant $C [1]$ because you have not provided initial conditions yet. This code fixes that: to get a numerical solution at $x=4$. Be respectful.