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- Dictionaryfunction/ˈfʌŋ(k)ʃn/
noun
- 1. an activity that is natural to or the purpose of a person or thing: "bridges perform the function of providing access across water" Similar
- 2. a relation or expression involving one or more variables: "the function (bx + c)"
verb
- 1. work or operate in a proper or particular way: "her liver is functioning normally" Similar Opposite
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Souler. 11 1 1 3. 2. use := instead of = and read about Set and SetDelayed. – eldo. Oct 18, 2015 at 17:53. 3. Note that if you do not include a space between x and y in xy, the latter will be considered as an independent variable of name xy. This being said, defining f[x_, y_] = (y/x) + x y and asking for f[1, 2] works fine for me and returns 4.
Mar 1, 2015 · 1. This works: f[u_, x_] := D[u, x] + a[x] u. By way of explanation, everything is an expression, and there is nothing particularly special about functions. You and I know that this definition doesn't have lot of meaning for objects "u" that aren't functions, but Mathematica doesn't need to know that u is a function.
Each function should document any side effects it makes. Things like creating a file, or updating a file, and such are considered a side-effect and can't be prevented by the language. But should be documented in the function header, so someone reading this function later on, can see right away what are its side-effects. $\endgroup$ –
Sep 12, 2016 · First argument of f function is first element of sequence, second argument is standard sequence specification, so {n} will give n-th element of sequence, {min, max} will give list containing elements from min to max, max will give elements from 1 to max.
Sorted by: You cannot make definitions with patterns on the left-hand side in the first argument of a scoping construct (such as Module). You need do that in the body of the Module. You should also use a different symbol for the internal function parameter. norm[x_] :=. Module[{fun1, fun2}, fun1[p_] := p^2 + p - 1;
I'd like to define a function by way of the output of a definite integral with symbolic bounds. For instance, F[m_,k_] := Integrate[x,{x,m,k}] would define F[m_,k_] := (1/2)(k^2 - m^2) This seems to work fine for simple examples, but I've run into the cases where the evaluation takes much longer than the indefinite integral. For instance running.
Usually, we define a function as . F[x_] := x + 10 where the right-hand side is an expression. Now, I would like define a function where the right-hand side is the solution of a set of equations. Like . optV[V_, r_, λ_, μ_, η_, σ_] := V /.
Mar 15, 2018 · I think it's easier just to define this straight up, rather than compute something procedurally. f[1, 0] = 77; f[0, 1] = 66; f[_, _] = 0; Mathematica is fundamentally an expression rewriting system, so telling it how to rewrite expressions directly like this is usually clearer, faster, and easier to debug.
$\begingroup$ Yes the function MakeRegion does not have a return value but it does define the function region which is the function that I use for the numerical calculations. Here is a simple proof of concept example: Clear[example] example[a_] := ( Clear[fun]; fun[x_, y_] := Evaluate[x Total[RandomVariate[NormalDistribution[], a]] + y ]; ); $\endgroup$
So again, yes you can define a function that has: any number of arguments, any number of local variables, routines that set local variables, and; return any, all, or any combination of the local variables. Note one can also define local functions inside a function then use them just as you would any other. A simple if kind of silly example follows:
