MaplePrimes Questions

A have som problems when i Will save a project. It Will take 2-5 Minutes where my other freinds never Experince this? Also i have some problems when i pauses writing, when i take notes in class, it Will give me the loading circle on mac( iam using a MacBook Air 2018) 

i hope you Can help:)

thx

Hi, im trying to optimize a function with a constraint. 

I've tried the Optimize package but I can't seem to make it work. 

I've attached  an image to my question, witht the function and the constraint. I want to find the optimal "t_x"

Function: rho*ln((-beta*tau + rho)/(1 + t__x)) + sigma*ln(sigma/(1 + t__y)) + beta*tau + B - rho - sigma - beta*tau*ln((-beta*tau + rho)/(1 + t__x))

Constaint: -t__x*(-beta*tau + rho)/(1 + t__x) + t__y*sigma/(1 + t__y) = R

Hope you can help 

Dear Colleagues,

I am trying to solve the following system of ode

odeSystem := {diff(y1(x), x) = -x*y2(x)-(1+x)*y3(x), diff(y2(x), x) = -x*y1(x)-(1+x)*y4(x), diff(y3(x), x) = -x*y1(x)-(1+x)*y4(x)-5*x*cos((1/2)*x^2), diff(y4(x), x) = -x*y2(x)-(1+x)*y3(x)+5*x*sin((1/2)*x^2), y1(0) = 5, y2(0) = 1, y3(0) = -1, y4(0) = 0};
systemSol := dsolve(odeSystem);

However, the result displayed is not explicit and contains integral sign. Please, I need help to obtain explicit result.

Thank you.

Please see the attached file; I'm attempting to do some calculations with the 'PDETools' package; notice the first term in equation (4), where sqrt(x2+y2) is not canceling in the fraction, despite using the 'simplify' command; why is this happening, and how can I achieve complete simplification?

Ques_Mapleprime.mw

with(PDEtools):

DepVars := [u(x, y, t), U(xi, eta)]; 1; alias(u = u(x, y, t))

[u(x, y, t), U(xi, eta)]

 

u

(1)

xi[1] := 1/2*(x^2+y^2); 1; xi[2] := t; 1; u := (h(t)+(x^2+y^2)*(1/2))*arccos(x/sqrt(x^2+y^2))/t+U(xi[1], xi[2])

(1/2)*x^2+(1/2)*y^2

 

t

 

(h(t)+(1/2)*x^2+(1/2)*y^2)*arccos(x/(x^2+y^2)^(1/2))/t+U((1/2)*x^2+(1/2)*y^2, t)

(2)

(diff(u, x))*(diff(u, y))

(x*arccos(x/(x^2+y^2)^(1/2))/t-(h(t)+(1/2)*x^2+(1/2)*y^2)*(1/(x^2+y^2)^(1/2)-x^2/(x^2+y^2)^(3/2))/((1-x^2/(x^2+y^2))^(1/2)*t)+(D[1](U))((1/2)*x^2+(1/2)*y^2, t)*x)*(y*arccos(x/(x^2+y^2)^(1/2))/t+(h(t)+(1/2)*x^2+(1/2)*y^2)*x*y/((x^2+y^2)^(3/2)*(1-x^2/(x^2+y^2))^(1/2)*t)+(D[1](U))((1/2)*x^2+(1/2)*y^2, t)*y)

(3)

collect(simplify(subs(1/2*(x^2+y^2) = xi, t = eta, (x*arccos(x/(x^2+y^2)^(1/2))/t-(h(t)+(1/2)*x^2+(1/2)*y^2)*(1/(x^2+y^2)^(1/2)-x^2/(x^2+y^2)^(3/2))/((1-x^2/(x^2+y^2))^(1/2)*t)+(D[1](U))((1/2)*x^2+(1/2)*y^2, t)*x)*(y*arccos(x/(x^2+y^2)^(1/2))/t+(h(t)+(1/2)*x^2+(1/2)*y^2)*x*y/((x^2+y^2)^(3/2)*(1-x^2/(x^2+y^2))^(1/2)*t)+(D[1](U))((1/2)*x^2+(1/2)*y^2, t)*y))), D, 'distributed')

(1/4)*(2*(y^2/(x^2+y^2))^(1/2)*(x^2+y^2)^(1/2)*eta*x^3+2*(y^2/(x^2+y^2))^(1/2)*(x^2+y^2)^(1/2)*eta*x*y^2)*(2*(y^2/(x^2+y^2))^(1/2)*(x^2+y^2)^(1/2)*eta*x^2+2*(y^2/(x^2+y^2))^(1/2)*(x^2+y^2)^(1/2)*eta*y^2)*(D[1](U))(xi, eta)^2/(y*(x^2+y^2)^2*eta^2)+(1/4)*((2*arccos(x/(x^2+y^2)^(1/2))*x^3*(x^2+y^2)^(1/2)*(y^2/(x^2+y^2))^(1/2)+2*arccos(x/(x^2+y^2)^(1/2))*x*(x^2+y^2)^(1/2)*(y^2/(x^2+y^2))^(1/2)*y^2-x^2*y^2-y^4-2*h(eta)*y^2)*(2*(y^2/(x^2+y^2))^(1/2)*(x^2+y^2)^(1/2)*eta*x^2+2*(y^2/(x^2+y^2))^(1/2)*(x^2+y^2)^(1/2)*eta*y^2)+(2*(y^2/(x^2+y^2))^(1/2)*(x^2+y^2)^(1/2)*eta*x^3+2*(y^2/(x^2+y^2))^(1/2)*(x^2+y^2)^(1/2)*eta*x*y^2)*(2*arccos(x/(x^2+y^2)^(1/2))*x^2*(x^2+y^2)^(1/2)*(y^2/(x^2+y^2))^(1/2)+2*arccos(x/(x^2+y^2)^(1/2))*(x^2+y^2)^(1/2)*(y^2/(x^2+y^2))^(1/2)*y^2+x^3+x*y^2+2*h(eta)*x))*(D[1](U))(xi, eta)/(y*(x^2+y^2)^2*eta^2)+(1/4)*(2*arccos(x/(x^2+y^2)^(1/2))*x^3*(x^2+y^2)^(1/2)*(y^2/(x^2+y^2))^(1/2)+2*arccos(x/(x^2+y^2)^(1/2))*x*(x^2+y^2)^(1/2)*(y^2/(x^2+y^2))^(1/2)*y^2-x^2*y^2-y^4-2*h(eta)*y^2)*(2*arccos(x/(x^2+y^2)^(1/2))*x^2*(x^2+y^2)^(1/2)*(y^2/(x^2+y^2))^(1/2)+2*arccos(x/(x^2+y^2)^(1/2))*(x^2+y^2)^(1/2)*(y^2/(x^2+y^2))^(1/2)*y^2+x^3+x*y^2+2*h(eta)*x)/(y*(x^2+y^2)^2*eta^2)

(4)

``

Download Ques_Mapleprime.mw

Hi, is it possible to convert mathematica files into maple. i was trying to upload mathematica file here but got failed, so uploading mathematica file into pdf. could anyone can help me to convert these file into maple.

Conversion.pdf

Could you help me to convert the following maple solution expressed by the hypergeom function to the LegendreP and Q function?

diff(T[3](t), t, t)+3*(diff(a(t), t))*(diff(T[3](t), t))/a(t)+(2*(diff(a(t), t, t))/a(t)+6*(diff(a(t), t))^2/a(t)^2+(-Omega^2+6)/a(t)^2)*T[3](t)

diff(diff(T[3](t), t), t)+3*(diff(a(t), t))*(diff(T[3](t), t))/a(t)+(2*(diff(diff(a(t), t), t))/a(t)+6*(diff(a(t), t))^2/a(t)^2+(-Omega^2+6)/a(t)^2)*T[3](t)

(1)

"a(t) :=Zeta*(1-(1-t/(Zeta^()))^(2))^(1/(2)) "

proc (t) options operator, arrow, function_assign; Zeta*(1-(1-t/Zeta)^2)^(1/2) end proc

(2)

ODE2 := diff(T[3](t), t, t)+3*(diff(a(t), t))*(diff(T[3](t), t))/a(t)+(2*(diff(a(t), t, t))/a(t)+6*(diff(a(t), t))^2/a(t)^2+(-Omega^2+6)/a(t)^2)*T[3](t)

diff(diff(T[3](t), t), t)+3*(1-t/Zeta)*(diff(T[3](t), t))/((1-(1-t/Zeta)^2)*Zeta)+(2*(-(1-t/Zeta)^2/((1-(1-t/Zeta)^2)^(3/2)*Zeta)-1/((1-(1-t/Zeta)^2)^(1/2)*Zeta))/(Zeta*(1-(1-t/Zeta)^2)^(1/2))+6*(1-t/Zeta)^2/((1-(1-t/Zeta)^2)^2*Zeta^2)+(-Omega^2+6)/(Zeta^2*(1-(1-t/Zeta)^2)))*T[3](t)

(3)

generalsol := dsolve(ODE2, T[3](t))

T[3](t) = _C1*hypergeom([1/2+(-Omega^2+1)^(1/2), 1/2-(-Omega^2+1)^(1/2)], [1-((1/2)*I)*15^(1/2)], (1/2)*t/Zeta)*t^(-((1/4)*I)*15^(1/2)-1/4)*(2*Zeta-t)^(((1/4)*I)*15^(1/2)-1/4)+_C2*(-(-2*Zeta+t)*t)^(((1/4)*I)*15^(1/2)-1/4)*hypergeom([((1/2)*I)*15^(1/2)+1/2+(-Omega^2+1)^(1/2), ((1/2)*I)*15^(1/2)+1/2-(-Omega^2+1)^(1/2)], [1+((1/2)*I)*15^(1/2)], (1/2)*t/Zeta)

(4)

convert(_C1*hypergeom([1/2+sqrt(-Omega^2+1), 1/2-sqrt(-Omega^2+1)], [1-I*sqrt(15)*(1/2)], t/(2*Zeta))*t^(-I*sqrt(15)*(1/4)-1/4)*(2*Zeta-t)^(I*sqrt(15)*(1/4)-1/4), LegendreP)

_C1*GAMMA(1-((1/2)*I)*15^(1/2))*(-t/Zeta)^(((1/4)*I)*15^(1/2))*LegendreP(-1/2+(-Omega^2+1)^(1/2), ((1/2)*I)*15^(1/2), 1-t/Zeta)*t^(-((1/4)*I)*15^(1/2)-1/4)*(2*Zeta-t)^(((1/4)*I)*15^(1/2)-1/4)/((2*Zeta-t)/Zeta)^(((1/4)*I)*15^(1/2))

(5)

convert(_C2*(-t*(t-2*Zeta))^(I*sqrt(15)*(1/4)-1/4)*hypergeom([I*sqrt(15)*(1/2)+1/2+sqrt(-Omega^2+1), I*sqrt(15)*(1/2)+1/2-sqrt(-Omega^2+1)], [1+I*sqrt(15)*(1/2)], t/(2*Zeta)), LegendreQ)

_C2*(-t*(-2*Zeta+t))^(((1/4)*I)*15^(1/2)-1/4)*hypergeom([((1/2)*I)*15^(1/2)+1/2+(-Omega^2+1)^(1/2), ((1/2)*I)*15^(1/2)+1/2-(-Omega^2+1)^(1/2)], [1+((1/2)*I)*15^(1/2)], (1/2)*t/Zeta)

(6)

NULL

Download solve3.mw

Like this

Thank you.

I was wondering if it is possible to write Maple code in 2D notation inside .mpl and have it converted automatically to standard Maple 1D math.

For example, suppose I have A.mpl file with the content


#--- this is inside A.mpl file

ode:=y''(x)+y'(x)=sin(x);
ic:=y(0)=1,y'(0)=0;
dsolve([ode,ic],y(x));

Right now, reading this file in worksheet or even in document mode gives error:

        read "A.mpl"

Error, on line 1, syntax error, unexpected single forward quote:
ode:=y''(x)+y'(x)=sin(x);

But in document mode, one can type this in, and Maple will accept it. 

Since I do not use document mode and most of my code is in mpl files, I was wondering if one can write y'(x) instead of diff(y(x),x) and have maple automatically convert the code as it reads it somehow?

I find it much easier and more natural to write   y''(0)=1  than (D@@2)(y)(0)=1  for example.

I do understand the technical problems parsing this as has special meaning in Maple. But Maple does handle this input OK if typed in by hand in document 2D mode. It just needs to extend this to reading it from mpl files.

May be a future release of Maple will support this (using some switches or option when reading the file?)

Hi all!

I was wondering if it was possible to execute different subsections in my sheet based on if statements? I'm calculating two different processes and by being able to switch between the two with if statemens it would save me a lot of variables. Lets say i have three subsections in my sheet:

  • Variables
  • Process 1 calculation
  • Process 2 calculation

So by using an if statement i can choose which calculation that should be used, and therefore i can use the same variables. Or at least that's what i hope :) 

Thanks!

The two points on the sphere has radius r=1/2. So, the distance is 1.
The three points on the sphere has radius r=1/sqrt(3). So, all distances are 1.
The four points on the sphere has radius r=sqrt(6)/4. So, all distances are 1.
What is the smallest sphere radius for 5 points and above numbers?  

Find the smallest sphere radius and 3D position of each point.

Answer of smaller radius is better.

Tokoro.


 

All distances of point to point on the sphere to be integer.
The two points on the sphere has radius r=1/2. So, the distance is 1.
The three points on the sphere has radius r=1/sqrt(3). So, all distances are 1.

The four points on the sphere has radius r=sqrt(6)/4. So, all distances are 1.
What is the smallest sphere radius for 5 points and above numbers?  Find the smallest sphere radius and 3D position of each point.

with(plots)with(GraphTheory); with(LinearAlgebra); with(SpecialGraphs)

with(plottools)

NULL

The answer of 4 points. r=sqrt(6)/4

   

plots[display]([c, nodes, plotlinesx], scaling = constrained, style = wireframe, axes = boxed)

 

NULL


 

Download 3D-4points.mw

The two points on the circle has radius r=1/2. So, the distance is 1.
The three points on the circle has radius r=1/sqrt(3). So, the all distances are 1.
What is the smallest circle radius for 6 points and above numbers?  


 

All distances of point to point on the circle to be integer.
The two points on the circle has radius r=1/2. So, the distance is 1.
The three points on the circle has radius r=1/sqrt(3). So, the all distances are 1.
What is the smallest circle radius for 4 points and above numbers?  

with(plots)with(plottools)

The answer of 5 points. r=16/sqrt(15)

   

plots[display](Cc5, plotlinesx5, nodes)

 

 

NULL


 

 

Hey sorry again, with a given triangle how can I inscribe it in a circle and how do I find the circumcenter?

Hello, I'm just beginning to appreciate and enjoy the use of math to solve practical problems I encounter while working on sound synthesis and music composition, my main occupation. I've had no formal training on math but take it as a hobby and try to tutor myself with the occasional aid of books, videos and a couple of really patient friends. As they're currently not around, I bring here the issue I'm now struggling with:

I'm trying to model a function which would output the time taken for a voltage controlled amplifier to raise or decrease its volume from one level point to another at a given scaled rate (which you can set in the device, but is not expressed in db/sec, only with an integer from from 1 to 99). When the beginning and end level points are the min. and max. that the device allows, the time (y) to rate scale (x) function plots a nice exponential curve (of the form y=b*e^-cx). However, when the min and max levels are respectivly augmented or reduced and the volume range the amplifier has to cross gets smaller, the time taken at a given rate changes not as steadily. So for this situation, for each of the scaled rates I want to consider I draw a curve: on the x axis I plot increments of the amount summed or substrated to the min or max levels, and continue marking time on y. And then I get an exponential of the same form as before but only for some values of x: the curve suddenly flattens and approaches a horizontal asymptote when approaching low values. Rather by chance I figured that something with the form y=1+e^-e^x might to the trick. My question is how to adjust that function to my plotted curve, given that I know the actual function that leads the simple exponential portion: 16.7*e^-0.01x for x=22 onwards, y=14.7 from x=0 to x=22.

Any help would be greatly appreciated

Consider matrices A and B below; how one can plot basis vectors of column space in 2d, and plane or line spanned by basis of row space in 3D?

with(LinearAlgebra):
A := Matrix([[2, 3, 5], [1, 2, 7]]);

ColumnSpace(A);
RowSpace(A);
 
B := Matrix([[6, 4, 2], [3, 2, 1]]);
 
ColumnSpace(B);
RowSpace(B);
 

How can I simplify the expression by applying equivalent infinitesimal  in symbolic computation?

For example,for small x,  limit(x/sin(x) + x, x = 0),  I want to get 1+x, rather than 1. In fact, how can I set an assumption that x is small and let the expression be simplified or replaced by other equivalent infinitesimal terms?

sol := {c[1, 1] = 18.00000000, c[1, 2] = -0., u[1, 1] = -.9000000000, u[1, 2] = 0., x[1, 1] = 3.600000000, x[1, 2] = -0.}
How we can assign each in solution1 individually as a Variable.

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