Items tagged with max max Tagged Items Feed

maximum and minimum functions...

November 14 2013 adel-00 25

I tried to get the maximum and minimum values of the following function. From the plot I get them but its not accurate. Please advise me to get them accurate.

 

F:=0.85:B:=0.5:

K:=N->(N*(1+F*N/(N^2+B^2-F*N)));

 

implicitplot(((N^2+B^2-F*N)*K=N*(N^2+B^2-F*N+F*N),K=0..10,N=0..10,view=[0..5,0..4],numpoints=90000,axes=boxed,thickness=2,color=black,font=[1,1,20],tickmarks=[3, 3],linestyle=1));

 

Hi,

1-Triying to plot a function divided by its maximum value,sometimes it works with some parameters that means, the max.value of the plot is 1.

But when i change the data the max. value in the plot in graeter than 1 which is wrong!! should be 1.

dont know why??

2- Changing different data in the parameters, the programme takes long long time then i stop it?

 

please help me with these two problems.


restart:
>
------------------------- Defining the nature of the variables used ----------------------
assume(T,real):Digits:=25:n:=1:tau:=Pi:
theta:=0:phi:=0:
lambda:=n;Omega:=1:Gamma:=0.01:
--------------------- Input---------------------------------
1

J1

term1:=(exp((Gamma+I*d)*tau)-1)/(2*(Gamma+I*d)):
Ak1:=d->(exp((Gamma+I*(d+0.5*Omega-2*lambda*k/Gamma))*tau)-1)/(Gamma+I*(d+0.5*Omega-2*lambda*k/Gamma))+(exp((Gamma+I*(d-0.5*Omega+2*lambda*k/Gamma))*tau)-1)/(Gamma+I*(d-0.5*Omega+2*lambda*k/Gamma)):
Ak2:=d->(exp((Gamma+I*(d-0.5*Omega-2*lambda*k/Gamma))*tau)-1)/(Gamma+I*(d-0.5*Omega-2*lambda*k/Gamma))+(exp((Gamma+I*(d+0.5*Omega+2*lambda*k/Gamma))*tau)-1)/(Gamma+I*(d+0.5*Omega+2*lambda*k/Gamma)):
term2:=(evalf(-0.25*sum(BesselJ(k,Omega*Gamma/(4*n))*Ak1(d)+BesselJ(k,-(Omega*Gamma)/(4*n))*Ak2(d),k=0..50))):
J1:=(term1+term2):
J1mod:=(Re(J1))^2+(Im(J1))^2:
###### J2#########################
Ak1:=d->(exp((Gamma+I*(d+0.5*Omega-2*lambda*k/Gamma))*tau)-1)/(Gamma+I*(d+0.5*Omega-2*lambda*k/Gamma))-(exp((Gamma+I*(d-0.5*Omega+2*lambda*k/Gamma))*tau)-1)/(Gamma+I*(d-0.5*Omega+2*lambda*k/Gamma)):
Ak2:=d->(exp((Gamma+I*(d-0.5*Omega-2*lambda*k/Gamma))*tau)-1)/(Gamma+I*(d-0.5*Omega-2*lambda*k/Gamma))-(exp((Gamma+I*(d+0.5*Omega+2*lambda*k/Gamma))*tau)-1)/(Gamma+I*(d+0.5*Omega+2*lambda*k/Gamma)):

J2:=(evalf(-0.25*sum(BesselJ(k,Omega*Gamma/(4*n))*Ak1(d)+BesselJ(k,-Omega*Gamma/(4*n))*Ak2(d),k=0..100))):
######################

J2mod:=(Re(J2))^2+(Im(J2))^2:
J3 same as J1differ in sign
term1:=(exp((Gamma+I*d)*tau)-1)/(2*(Gamma+I*d)):
Ak1:=d->(exp((Gamma+I*(d+0.5*Omega-2*lambda*k/Gamma))*tau)-1)/(Gamma+I*(d+0.5*Omega-2*lambda*k/Gamma))+(exp((Gamma+I*(d-0.5*Omega+2*lambda*k/Gamma))*tau)-1)/(Gamma+I*(d-0.5*Omega+2*lambda*k/Gamma)):
Ak2:=d->(exp((Gamma+I*(d-0.5*Omega-2*lambda*k/Gamma))*tau)-1)/(Gamma+I*(d-0.5*Omega-2*lambda*k/Gamma))+(exp((Gamma+I*(d+0.5*Omega+2*lambda*k/Gamma))*tau)-1)/(Gamma+I*(d+0.5*Omega+2*lambda*k/Gamma)):
term2:=(evalf(0.25*sum(BesselJ(k,Omega*Gamma/(4*n))*Ak1(d)+BesselJ(k,-Omega*Gamma/(4*n))*Ak2(d),k=0..100))):
J3:=term1+term2:
J3mod:=(Re(J3))^2+(Im(J3))^2:
J4 same as J2 but -0.25-->2


J4:=-2*J2:
######################

J4mod:=(Re(J4))^2+(Im(J4))^2:

calculate the spectrum

 

Spec:=d->(exp(-2*Gamma*tau)*(J1mod*cos(theta/2)^2+J2mod+J3mod*sin(theta/2)^2-0.5*Re(J3*J4*sin(theta)*exp(I*phi))+0.5*Re(J1*J4*sin(theta)*exp(-I*phi)))):

with(plots):

tit:=sprintf("l=%g,W=%g,G=%g",lambda,Omega,Gamma):
Smax1:=max(seq(evalf(Spec(d)),d=-100..100)):
plot(evalf(Spec(d)/Smax1),d=-15..15,axes=boxed,title=tit,color=black,font=[2,3,18],thickness=2,tickmarks=[3,3],titlefont=[SYMBOL,14],font=[1,1,18],linestyle=1);

 

 

 

 

Hi,

I'm plotting this function of (x) which is very complicated expression anyway I plotted it..Now I want to plot the f(x)/max value of f(x)???

I appriciated for any advice...

would like to normalize it, how to find x such as 5.6 in this example

 

l := [20,8,22,38,49,36,59,41,32,21,23];

sl := gfun[listtoseries](l, x, egf);

g := convert(sl, ratpoly); 

g := g - subs(x=0, g);

g := g/subs(x=5.6, g); 

Hello

I am trying to plot solution of ode0 together with the maximum and minimum values but I am having difficulty since the first plot is a solution and second is values. I should have a plot with two line one represent the solution of ode0 and second (the max an min). Any advise or suggestion?

This is the code:

> restart;
with(DEtools); with(plots); Nsols := 5; Ntstep := 10;
 k := 0; A := 0.37e-1; B := 0.2e-6;
ode0 := diff(U(t), t) = -(A+B*U(t))*U(t);

Dear Mapleprimes,

 

I am trying to solve an Optimization problem analytically. I set up a lagrangian as follows:

 

L:= Objective function + mu_1 (constraint 1) + mu_2 (constraint 2)

 

I maximise with respect to two variables, call them x1 and x2

 

I want constraint one to bind but not constraint 2. Hence, I set mu_2=0. The I do:

 

diff(L,x1); diff(L,x2), diff(L,mu1), diff(L,mu2)

I'm using Linearalgebra where I have a 10x1 matrix. I want to find the closest element of the matrix to origo, and the furthest away from origo. Can anyone help me to a understandable solution? :)

Find the derivative of f(x)=|(x^3)-8*(x^2)+5*x+4|-0.5*x;x in [-1,7]

Find critical points of f(x) and dertimine the local maxima and local minima.

Output: Two lists of points (x,y), a list of local minima and a list of local maxima.

Hint: you may use Maple package Student[Calculus1]]

     use first derivative test to avoid 'kink' point i.e. undifferentiable point

     set delta=0.0001, test derivative around critical point x+delta and x-delta...

I need this quite often, but never found a nice way of getting the index of minimal or maximal value of a Vector (or a list or whatever). Is there an easier way than the following?

N := 15:
L := LinearAlgebra[RandomVector](N):
m := min(L):
for i to N do if L[i] = m then print(i); break end if end do;

I'm having trouble maximizing this funciton:

 

S=  (w1*E1+w2*E2+w3*E3)/(CoVar[1,2]), subject to the constraint w1+w2+w3=1

 

I need it to choose w1, w2, and w3 to maximize S

Hey,

I've got a nondifferentiable bounded multivariable function and I'd like to find a good approxiation of it's minumum and maximum.
Basicly, I'd like to automaticly be able to return the maximum and minimum of the height of the 3D-plot i have. That would just be perfect.
However, I can't find anything that does anything like this.....

Hello. I need to solve a problem where:

  A is defined over a real inteval [c...d]   and   B is defined over another real interval [d....e]. 

I want maple  to take max(A,B) and tell me which interval is larger( asume no intersection)

For example

A=[0....1] and B=[2...3]

Answer should be max(A,B)=B

Thank you

I have 2 arrays of data that I have calculated ( X1 and Y1). I plot them and can see the min and the max points. I need these points. I use the command:

P:=ExtremePoints(X1, Y1);

And it gives me:

 

P:=ExtremePoints(X1, [4, 2,3,1,5,0.3,4]);

I expected that P will be an array and when I used the command:

N:=ArrayNumElems(P);

It will give me the total number of the min and max, but it gives me a mistake.

I will...

As title

I need help in finding the maximum point of each curve (please see maple file link attached), or if possible display each maximum point (under different "ga") on the 3Dplot

Your help is appreciated

 

Correction.mw

"Find the local maxima and minima in some noisy signal"

I have a data array which I want to determine the peaks in the measurements.
For this task I tried to translate the code found  from MatLab to Maple.
As you can see I get the values(peaks) on the x axis displaced.
By using GlobalSearch get the same peaks with the coordinates correctly.
As much as I seek and analyze command lines can not find the error or errors.

1 2 Page 1 of 2