I have
f1[x_, y_] := x^2 - 10 x + y^2 + 8;
f2[x_, y_] := x*y^2 + x - 10 y + 8;
f[x_, y_] := {f1[x, y], f2[x, y]} ;
x0 = {0, 0};
I want to evaluate f[x_, y_] in x0, so f[0, 0]
I am doing this but does not work, what is the correct way?
MatrixForm[f[{x0}]]
I get f[{{0, 0}}]
but want {8, 8} instead
In[61]:= f ## x0
Out[61]= {8, 8}
What went wrong? When you evaluate f[{x0}] this equals f[{{0,0}}], which doesn't match the defined pattern for f. f##x0, which is shorthand for Apply[f,x0], replaces the head of x0 (which internally equals List[0,0], hence its head is List), with f. You then get f[0,0] which matches the argument pattern of f. You then get the correct result.
Related
I am evaluating a partial derivative
fx = D[m[x, y], x]
The output gives me an output in terms of x and y. I'm trying to evaluate the function using
fx1 = fx/.{x-> 1.0, y-> 2.0}
but keeps giving me an answer like
0.471328[1.0, 2.0]
But I only want the 0.471328
I can't even begin to guess what code you have written, and not shown, to get where you are, but
In[3]:= Head[0.471328[1.0, 2.0]]
Out[3]= 0.471328
will give you what you are asking for.
Let me show one correct way to do this:
m[x_, y_] := x^2 + y^2 + x y
fx[x_, y_] := D[m[x, y], x]
fx[x, y] /. {x -> 1.0, y -> 2.0}
(*
=> 4.
*)
I am trying to evaluate the derivative of a function at a point (3,5,1) in Mathematica. So, thats my input:
In[120]:= D[Sqrt[(z + x)/(y - 1)] - z^2, x]
Out[121]= 1/(2 (-1 + y) Sqrt[(x + z)/(-1 + y)])
In[122]:= f[x_, y_, z_] := %
In[123]:= x = 3
y = 5
z = 1
f[x, y, z]
Out[124]= (1/8)[3, 5, 1]
As you can see I am getting some weird output. Any hints on evaluating that derivative at (3,5,1) please?
The result you get for Out[124] leads me to believe that f was not cleared of a previous definition. In particular, it appears to have what is known as an OwnValue which is set by an expression of the form
f = 1/8
(Note the lack of a colon.) You can verify this by executing
g = 5;
OwnValues[g]
which returns
{HoldPattern[g] :> 5}
Unfortunately, OwnValues supersede any other definition, like a function definition (known as a DownValue or, its variant, an UpValue). So, defining
g[x_] := x^2
would cause g[5] to evaluate to 5[5]; clearly not what you want. So, Clear any symbols you intend to use as functions prior to their definition. That said, your definition of f will still run into problems.
At issue, is your use of SetDelayed (:=) when defining f. This prevents the right hand side of the assignment from taking on a value until f is executed later. For example,
D[x^2 + x y, x]
f[x_, y_] := %
x = 5
y = 6
f[x, y]
returns 6, instead. This occurs because 6 was last result generated, and f is effectively a synonym of %. There are two ways around this, either use Set (=)
Clear[f, x, y]
D[x^2 + x y, x];
f[x_, y_] = %
f[5, 6]
which returns 16, as expected, or ensure that % is replaced by its value before SetDelayed gets its hands on it,
Clear[f, x, y]
D[x^2 + x y, x];
f[x_, y_] := Evaluate[%]
I want to write a
Module Arg[f_,n_]
that takes a function f (having <=n arguments) and a natural number n and outputs the n-th argument of the function f.
As an example, suppose that f is defined by
f[a_,b_]=a^2+b^2.
Then,
Arg[f[s,t],1]
should be s;
while
Arg[f[u,v],2]
should be v.
My question is whether this is possible. If so, what should I write in the place of "???" below?
Arg[f_,n_] := Module[{}, ??? ]
Note that I don't want to specify a_ and b_ in the definition of Arg like
Arg[f_,a_,b_,n_]
EDIT: "Arg" is just my name for the module not the internal function Arg of Mathematica.
Perhaps
SetAttributes[arg, HoldFirst];
arg[f_[x___], n_] := {x}[[n]]
f[a_, b_] := a^2 + b^2.
arg[f[arg[f[s, t], 1], t], 1]
arg[f[s, t], 2]
(*
-> s
-> t
*)
arg[ArcTan[f[Cos#Sin#x, x], t], 1]
(*
-> x^2. + Cos[Sin[x]]^2
*)
Assuming your second example should give u, this should do the job:
ClearAll[arg];
SetAttributes[arg, HoldFirst];
arg[g_, n_] := Module[
{tmp, ret},
Unprotect[Part];
tmp = Attributes[Part];
SetAttributes[Part, HoldFirst];
ret = Part[g, n];
ClearAttributes[Part, HoldFirst];
SetAttributes[Part, tmp];
Protect[Part];
ret
]
so that
f[a_, b_] = a^2 + b^2.;
arg[f[s, t], 1]
gives s.
This is very heavy-handed though, so I expect someone will find something better soon enough.
This is a bit better (doesn't redefine built-in functions even temporarily):
ClearAll[arg2];
SetAttributes[arg2, HoldFirst];
arg2[g_, n_] := Hold[g][[1, n]]
I need to find fixed points of iterative map x[n] == 1/2 x[n-1]^2 - Mu.
My approach:
Subscript[g, n_ ][Mu_, x_] := Nest[0.5 * x^2 - Mu, x, n]
fixedPoints[n_] := Solve[Subscript[g, n][Mu, x] == x, x]
Plot[
Evaluate[{x,
Table[Subscript[g, 1][Mu, x], {Mu, 0.5, 4, 0.5}]}
], {x, 0, 0.5}, Frame -> True]
I'll change notation slightly (mostly so I myself can understand it). You might want something like this.
y[n_, mu_, x_] := Nest[#^2/2 - mu &, x, n]
fixedPoints[n_] := Solve[y[n, mu, x] == x, x]
The salient feature is that the "function" being nested now really is a function, in correct format.
Example:
fixedPoints[2]
Out[18]= {{x -> -1 - Sqrt[-3 + 2*mu]},
{x -> -1 + Sqrt[-3 + 2*mu]},
{x -> 1 - Sqrt[ 1 + 2*mu]},
{x -> 1 + Sqrt[ 1 + 2*mu]}}
Daniel Lichtblau
First of all, there is an error in your approach. Nest takes a pure function. Also I would use exact input, i.e. 1/2 instead of 0.5 since Solve is a symbolic rather than numeric solver.
Subscript[g, n_Integer][Mu_, x_] := Nest[Function[z, 1/2 z^2 - Mu], x, n]
Then
In[17]:= fixedPoints[1]
Out[17]= {{x -> 1 - Sqrt[1 + 2 Mu]}, {x -> 1 + Sqrt[1 + 2 Mu]}}
A side note:
Look what happens when you start very near to a fixed point (weird :) :
f[z_, Mu_, n_] := Abs[N#Nest[1/2 #^2 - Mu &, z, n] - z]
g[mu_] := f[1 + Sqrt[1 + 2*mu] - mu 10^-8, mu, 10^4]
Plot[g[mu], {mu, 0, 3}, PlotRange -> {0, 7}]
Edit
In fact, it seems you have an autosimilar structure there:
I have a function, let's say for example,
D[x^2*Exp[x^2], {x, 6}] /. x -> 0
And I want to replace 6 by a general integer n,
Or cases like the following:
Limit[Limit[D[D[x /((-1 + x) (1 - y) (-1 + x + x y)), {x, 3}], {y, 5}], {x -> 0}], {y -> 0}]
And I want to replace 3 and 5 by a general integer m and n respectively.
How to solve these two kinds of problems in general in mma?
Many thanks.
Can use SeriesCoefficient, sometimes.
InputForm[n! * SeriesCoefficient[x^2*Exp[x^2], {x,0,n}]]
Out[21]//InputForm=
n!*Piecewise[{{Gamma[n/2]^(-1), Mod[n, 2] == 0 && n >= 2}}, 0]
InputForm[mncoeff = m!*n! *
SeriesCoefficient[x/((-1+x)*(1-y)*(-1+x+x*y)), {x,0,m}, {y,0,n}]]
Out[22]//InputForm=
m!*n!*Piecewise[{{-1 + Binomial[m, 1 + n]*Hypergeometric2F1[1, -1 - n, m - n,
-1], m >= 1 && n > -1}}, 0]
Good luck extracting limits for m, n integer, in this second case.
Daniel Lichtblau
Wolfram Research
No sure if this is what you want, but you may try:
D[x^2*Exp[x^2], {x, n}] /. n -> 4 /. x -> 0
Another way:
f[x0_, n_] := n! SeriesCoefficient[x^2*Exp[x^2], {x, x0, n}]
f[0,4]
24
And of course, in the same line, for your other question:
f[m_, n_] :=
Limit[Limit[
D[D[x/((-1 + x) (1 - y) (-1 + x + x y)), {x, m}], {y, n}], {x ->
0}], {y -> 0}]
These answers don't give you an explicit form for the derivatives, though.