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.
*)
Related
Here's the problem statement:
Two non-linear inter-dependent, initial value first order differential equations were solved using NDSolve to yield an analytical solution. The solution was used to calculate another parameter, as a function of the same x value.
Let's say we have the ODEs as:
X'[t]=a*S[t]*X[t]/(b+S[t]
S'[t]=-a*S[t]*X[t]/(c(b+S[t])) where a,b,c are also known constants
X[0]=constant
S[0]=constant
soln = NDSolve[{X'[t]=a*S[t]*X[t]/(b+S[t],S'[t]=-a*S[t]*X[t]/(c(b+S[t])),X[0]=constant,S[0]=constant},{X,S},{t,0,50}]
The solution is of the form
X-> InterpolatingFunction[{{0.0,50}},<>],S->InterpolationFunction[{{0.0,50}},<>}}
Now the new parameter is: Yvalue=(S[t]/.soln)+(X[t]/.soln)
I'm trying to figure out the correct syntax to calculate Yvalue for an entered t value.
Ex- One needs to calculate Yvalue at t=0.1,0.56, 2.3 etc
Thank you for your time.
Regards,
Ankur
NDSolve demands that all parameters be given specific numeric values. If you assign values to a,b,c,X[0],S[0] and carefully match up all your parens and carefully use == versus = correctly, then this can work
In[1]:= a = 2; b = 3; c = 5;
soln = NDSolve[{X'[t] == a*S[t]*X[t]/(b + S[t]),
S'[t] == -a*S[t]*X[t]/(c(b+S[t])), X[0]==7, S[0]==11}, {X,S}, {t,0,50}][[1]]
Out[2]= {X -> InterpolatingFunction[{{0.,50.}}, <>],
S -> InterpolatingFunction[{{0.,50.}}, <>]}
In[3]:= Yvalue = S[t] + X[t] /. soln /. t -> 0.1
Out[3]= 18.9506
In[4]:= Yvalue = S[t] + X[t] /. soln /. t -> 0.56
Out[4]= 25.6919
In[5]:= Yvalue = S[t] + X[t] /. soln /. t -> 2.3
Out[5]= 61.9823
and even
In[6]:= Plot[S[t] + X[t] /. soln, {t, 0, 50}, PlotRange -> {0, 70}]
Out[6]= ...PlotSnipped...
I want to use the solution of Maximization, defined as a function, in another function. Here's an example:
f1[y_] := x /. Last[Maximize[{Sin[x y], Abs[x] <= y}, x]] (* or any other function *)
This definition is fine, for example if I give f1[4], I get answer -((3 \[Pi])/8).
The problem is that when I want to use it in another function I get error. For example:
FindRoot[f1[y] == Pi/4, {y, 1}]
Gives me the following error:
ReplaceAll::reps: {x} is neither a list of replacement rules nor a valid dispatch table, and so cannot be used for replacing. >>
FindRoot::nlnum: The function value {-0.785398+(x/.x)} is not a list of numbers with dimensions {1} at {y} = {1.}. >>
I've been struggling with this for several days now! Any comment, idea, help, ... is deeply appreciated! Thank you very much!
When y is not a number, your Maximize cannot be resolved, in which case the Last element of it is x, which is why you get that odd error message. You can resolve this by clearing the bad definition of f1 and making a new one that ensures only numeric arguments are evaluated:
ClearAll[f1]
f1[y_?NumericQ] := x /. Last[Maximize[{Sin[x y], Abs[x] <= y}, x]]
FindRoot[f1[y] == \[Pi]/4, {y, 1}]
(* {y -> 0.785398} *)
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 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.
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: