I am am mathmatica notebook to find an analytical solution to the follow constrained optimization problem:
Max y^(1-b)(x^b(1-a(x/(x+1)))) s.t. M = Px+qy
x,y
I have tried the following code:
Maximize[{y^(1-b)(x^b(1-a(x/(x+1)))), M==Px+qy}, {x,y}]
and in returns the same function as an output. In the function a, b, M, P, and q are all parameters. I have also tried assigning the parameters arbitrary values to test to see if mathmatica is not sure how to deal with the parameters. I used to following code:
Maximize[{y^(1-0.5)(x^0.5(1-0.75(x/(x+1)))), 1000=5x+5y},{x,y}]
and it returns the same function. However, if I remove the constraint it will solve the optimization problem.
Maximize[{y^(1-0.5)(x^0.5(1-0.75(x/(x+1))))},{x,y}]
{7.2912*^59,{x->2.89727*^60,y->2.93582*^60}}
I am not sure what to do. After reading about constrained optimization problem the syntax appears to be correct. Sorry, it this question is really basic I am very new to mathmatica, also since I am using a notebook I could not past the output from the first two lines in.
The constraint is incorrectly specified, it should be 1000 == 5 x + 5 y. Maximize works better with exact numbers.
Maximize[{Rationalize[y^(1 - 0.5) (x^0.5 (1 - 0.75 (x/(x + 1))))],
1000 == 5 x + 5 y}, {x, y}] // N
(* {25.7537, {x -> 96.97, y -> 103.03}} *)
Related
I'm puzzled by what I think is a mistake in a partial derivative I'm having Mathematica do for me.
Specifically, this is what I have:
Derivative I'd like to take
I'm trying to take the partial derivative of the following w.r.t. the variable θ (apologies for the formatting):
f=(1/4)(-4e((1+θ)/2)ψ+eN((1+θ)/2)ψ+eN((1+θ)/2-θd)ψ)-s
But the solution Mathematica produces seems very different from the one I get when I take the derivative myself. While Mathematica says the partial derivative of f w.r.t. θ is:
(1/4)eψ(N-2)
By hand, I get and am quite confident the correct answer is instead:
(1/4)eψ(N(1-d)-2)
That is, Mathematica is producing something that drops the variable d when it is differentiating. I've explored different functions that take a derivative in Mathematica, and the possibility that maybe some of the variables I'm using (such as d) might be protected or otherwise special, but I can't say that I know why the answer's so off. This is the first time in the notebook that d appears, so it is not set to 0. For context, I'm trying to confirm that the derivative of the function is positive for values of the variables in certain ranges, and we have d>0 and d<(1/2). Doing this all by hand works but I'm trying to confirm with Mathematica as I will be dealing with more complicated functions and need to make sure I'm having Mathematica produce the right derivatives.
Your didn't add spaces in eN and θd, so it thinks they're some other 2-character variables.
Adding spaces between them gives your expected result:
f[θ,e,N,ψ,d,s] = (1/4) (-4 e ((1+θ)/2) ψ + e N ((1+θ)/2) ψ + e N ((1+θ)/2 - θ d) ψ) - s;
D[f[θ, e, N, ψ, d, s], θ] // FullSimplify
(* 1/4 e (-2 + N - d N) ψ *)
I HAVE EQUATION
y - 7(e^x/x)dydx=0
How to find analytic solution in Mathematica?
My work :
I simplify the equations become y'=yx/7e^x
I run in Mathematica,
DSolve[y'[x] == (y[x] x)/(7 e^x), y[x], x]
I get result:
{{y[x] -> E^(1/7 E^-x (-1 - x)) C[1]}}.
Questions:
Are my simplifications correct?
Do I type correct code in step 2?
How to find exact value of C in the result, because I want to use to find y' for given x value
Thank you for the answer
This is an algorithm question that I've been struggling with. I figured I could get some insight here. I need to make the following function in Haskell:
Declare the type and define a function that takes two numbers as input and finds their product by addition. That is, add the first number, as many times as second number, to itself.
My problem is that this is basically just multiplying two numbers together, but it says that I need to do it with addition. Does anyone have any clue on how to do this?
This is all I can come up with (it's not right): (x + x) * y
Thank you
if a is the first number and b the second
sum $ take a $ cycle [b]
should do ot
mult (x, y):
sum = 0
for 1 to y:
sum = sum + x
return sum
This is just the algorithm. I do not know Haskell. So the lambda expression in the other answer may be more appropriate. Also, I use an intermediate variable.
PS: forget the previous embarrassing recursive algorithm
Work it out by induction.
We know the answer to one simple (the simplest) problem: multiplying anything by 0 yields 0. So we write:
mul x 0 = 0
Now, the inductive step: we can build a solution to a bigger problem, if we know a solution to the smaller problem; that way we can always reduce any big problem to the smallest problem, for which we know the solution. So, for any y, the solution for y+1 can be found by adding x to the solution for y: mul x (y+1) = x + (mul x y). In Haskell we can't write (y+1) on the left hand side, so we write equivalently:
mul x y = x + (mul x (y-1))
This function will keep adding x until y is zero.
Try this also
multiply::(Num a,Eq a) => a -> a -> a
multiply a 0 = 0
multiply a b = a + multiply a (b - 1)
main = print $ multiply 5 7
I have a basic problem in Mathematica which has puzzled me for a while. I want to take the m'th derivative of x*Exp[t*x], then evaluate this at x=0. But the following does not work correct. Please share your thoughts.
D[x*Exp[t*x], {x, m}] /. x -> 0
Also what does the error mean
General::ivar: 0 is not a valid variable.
Edit: my previous example (D[Exp[t*x], {x, m}] /. x -> 0) was trivial. So I made it harder. :)
My question is: how to force it to do the derivative evaluation first, then do substitution.
As pointed out by others, (in general) Mathematica does not know how to take the derivative an arbitrary number of times, even if you specify that number is a positive integer.
This means that the D[expr,{x,m}] command remains unevaluated and then when you set x->0, it's now trying to take the derivative with respect to a constant, which yields the error message.
In general, what you want is the m'th derivative of the function evaluated at zero.
This can be written as
Derivative[m][Function[x,x Exp[t x]]][0]
or
Derivative[m][# Exp[t #]&][0]
You then get the table of coefficients
In[2]:= Table[%, {m, 1, 10}]
Out[2]= {1, 2 t, 3 t^2, 4 t^3, 5 t^4, 6 t^5, 7 t^6, 8 t^7, 9 t^8, 10 t^9}
But a little more thought shows that you really just want the m'th term in the series, so SeriesCoefficient does what you want:
In[3]:= SeriesCoefficient[x*Exp[t*x], {x, 0, m}]
Out[3]= Piecewise[{{t^(-1 + m)/(-1 + m)!, m >= 1}}, 0]
The final output is the general form of the m'th derivative. The PieceWise is not really necessary, since the expression actually holds for all non-negative integers.
Thanks to your update, it's clear what's happening here. Mathematica doesn't actually calculate the derivative; you then replace x with 0, and it ends up looking at this:
D[Exp[t*0],{0,m}]
which obviously is going to run into problems, since 0 isn't a variable.
I'll assume that you want the mth partial derivative of that function w.r.t. x. The t variable suggests that it might be a second independent variable.
It's easy enough to do without Mathematica: D[Exp[t*x], {x, m}] = t^m Exp[t*x]
And if you evaluate the limit as x approaches zero, you get t^m, since lim(Exp[t*x]) = 1. Right?
Update: Let's try it for x*exp(t*x)
the mth partial derivative w.r.t. x is easily had from Wolfram Alpha:
t^(m-1)*exp(t*x)(t*x + m)
So if x = 0 you get m*t^(m-1).
Q.E.D.
Let's see what is happening with a little more detail:
When you write:
D[Sin[x], {x, 1}]
you get an expression in with x in it
Cos[x]
That is because the x in the {x,1} part matches the x in the Sin[x] part, and so Mma understands that you want to make the derivative for that symbol.
But this x, does NOT act as a Block variable for that statement, isolating its meaning from any other x you have in your program, so it enables the chain rule. For example:
In[85]:= z=x^2;
D[Sin[z],{x,1}]
Out[86]= 2 x Cos[x^2]
See? That's perfect! But there is a price.
The price is that the symbols inside the derivative get evaluated as the derivative is taken, and that is spoiling your code.
Of course there are a lot of tricks to get around this. Some have already been mentioned. From my point of view, one clear way to undertand what is happening is:
f[x_] := x*Exp[t*x];
g[y_, m_] := D[f[x], {x, m}] /. x -> y;
{g[p, 2], g[0, 1]}
Out:
{2 E^(p t) t + E^(p t) p t^2, 1}
HTH!
Currently I use a single equation with different combination of known/unknown parameters. As I don't have any fancy calculator it would be much easier to define the equation in Mathematica and passing known parameters to calculate unknown values.
I would be very thankful if anyone of you could give an example solution (if possible using given equation).
Let's say we have an equation of satellite speed at given point in the elliptical orbit:
v = sqrt(u(2/r - 1/a))
where
v = speed
u = constant 3.986 * 10^14 m^3/s^2
r = radius (distance from the center of the Earth)
a = semi major axis of the ellipse
This equation can be used to calculate the speed or for example we know what is the speed needed for a manoeuvre to move the cargo to other orbit and have to model the orbit (a) at given radius (r)
Thanks!
You can define equations in Mathematica using the ":=" operator. To define the example equation:
v[u_, r_, a_] := Sqrt[u*(2/r-1/a)]
I'm not sure how to generalize it to solve for any unknown...If I figure it out I'll get back to you.
You may want to try something like:
Solve[v[1, r, 7]==15, r]
that will solve for r assuming you know v, u, and a... you can then change each of the paramaters for the unknown...
A little bit late :) ... but Reduce[] does what you want. We define a function:
solveForMe[rules_] := Reduce[( v == Sqrt[3.986*10^14 *(2/r - 1/a)]) /. rules];
and invoke it with any valid combination for the assignments. For example:
In[72]:= Off[Reduce::ratnz];
solveForMe[{a -> 7 10^6, r -> 7 10^6}]
solveForMe[{v -> 10, r -> 7 10^6}]
solveForMe[{v -> 10, a -> 7 10^6}]
The output is :
Out[73]= v == 7546.05
Out[74]= a == 3.5*10^6
Out[75]= r == 1.4*10^7
HTH! ...