Maple Questions and Posts

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hi

how i can draw this equation in maple in 3D?

(x-y)^2+(x-z)^2+(z-x)^2=3

thanks

PLOT3D.mw


 

I'm modeling the interaction between rotation and vibration in water molecule. I've come up with the solution functions, but I wanted to make some illustrative animations to fully understand what is going on. I made an animation (in the very end of the attached file) showing vibration of molecule and I want the trajectory of each (of two) particles to be traced after it, how could it be done? Also I want to have the movement of the angular momentum vector (vector with components [Jx=J*cos(varphi)*sin(theta), Jy=J*sin(varphi)*sin(theta), Jz=J*cos(theta)]) be animated simultaneously with the vibration of molecule (on the same plot). Is it hard to achieve?

File: ClassicalTrajectoriesH2X_morse.mw

 

Hello all,

 

Is here anyone who could help me with this Integral.

I want to determine P(y=2) in which the Nu is a constant and also is unknown.

In deed I want to derive the P which only has one unknown parameter(Nu). But I cannot solve this indefinite integral although the \alpha_1 , \alpha_2 and B are known constants. The Nu parameter mustn't be valued preferely.Indefinite_INtegral.mw

> restart;
> with(plots);
> setoptions(title = `Family Plot`, axes = boxed);
> pr := .71; n := 1; p := 0; q := 0; b := 0; l := 0; s := 0; m := 0;
>
>
> R1 := 2.*n/(1+n);
                                 1.000000000
> R2 := 2.*p/(1+n);
                                     0.
> sol1 := dsolve([diff(diff(diff(f(eta), eta), eta), eta)+f(eta)*(diff(diff(f(eta), eta), eta))+R1*(1-(diff(f(eta), eta))^2) = 0, diff(diff(theta(eta), eta), eta)+.71*f(eta)*(diff(theta(eta), eta))-.71*(diff(f(eta), eta))*theta(eta)*R2 = 0, f(0) = 0, (D(f))(0) = 1.8+b*((D@@2)(f))(0), (D(f))(18) = 0, theta(0) = 1+s*(D(theta))(0), theta(18) = 0], numeric, method = bvp[midrich]); plots[odeplot](sol1, [eta, theta(eta)], color = red, axes = boxed);

Dear sir/madam

In my program i want to plot multiple lines for diffrent values of n, but i do not have such type of idea please can you help me.

Hello there. I have to solve a simple line integral (3*y-x)dL from A(2;1) to B(3;-1). A and B are points in the line described by function y=5-2x. I found a function LineInt. Using paramether Line I wrote LineInt(3*y-x, Line(<2, 1>, <3, -1>)), it doesn't work with error below:

Error, (in Line:-ModuleCopy) invalid input: invalid specification of line

It seems that the problem is with first argument, that is the equation under the integral sign. Or no... How to write it properly? Thanks.

 


Here is my code. I am deriving soil compression line for oedometric testing with porosity, below is a part of the derivation.

I would like to differentiate the (Sst/Vst)/(Spt/Vpt) by dsm first then integrate it with dsm ranges from 0 to dmax to get my final answer...

Hope someone can help me on writing the code... Thank you very much!!!

restart

``

Sst := `&beta;ss`*[dsx(dsm)^(2-Ds)-dsm^(2-Ds)]/(2-Ds);

`&beta;ss`*[dsx(dsm)^(2-Ds)-dsm^(2-Ds)]/(2-Ds)

(1)

Spt := `&beta;ps`*[dpx(dsm)^(2-Dp)-dpm(dsm)^(2-Dp)]/(2-Dp);

`&beta;ps`*[dpx(dsm)^(2-Dp)-dpm(dsm)^(2-Dp)]/(2-Dp)

(2)

Vst := `&beta;sv`*[dsx(dsm)^(3-Ds)-dsm^(3-Ds)]/(3-Ds);

`&beta;sv`*[dsx(dsm)^(3-Ds)-dsm^(3-Ds)]/(3-Ds)

(3)

Vpt := `&beta;pv`*[dpx(dsm)^(3-Dp)-dpm(dsm)^(3-Dp)]/(3-Dp);

`&beta;pv`*[dpx(dsm)^(3-Dp)-dpm(dsm)^(3-Dp)]/(3-Dp)

(4)

``

(dpm(dsm)/dpx(dsm))^(2-Dp) = (dsm/dsx(dsm))^(2-Ds);

(dpm(dsm)/dpx(dsm))^(2-Dp) = (dsm/dsx(dsm))^(2-Ds)

(5)

dpx := proc (dsm) options operator, arrow; C1*dsx(dsm)*(dsm/dsx(dsm))^((3-Ds)*Ds) end proc;

proc (dsm) options operator, arrow; C1*dsx(dsm)*(dsm/dsx(dsm))^((3-Ds)*Ds) end proc

(6)

dpm := proc (dsm) options operator, arrow; C1*dsx(dsm)*(dsm/dsx(dsm))^((3-Ds)*Ds+Dp-Ds) end proc;

proc (dsm) options operator, arrow; C1*dsx(dsm)*(dsm/dsx(dsm))^((3-Ds)*Ds+Dp-Ds) end proc

(7)

eval(Sst*Vpt/(Vst*Spt));

`&beta;ss`*[dsx(dsm)^(2-Ds)-dsm^(2-Ds)]*(3-Ds)*(2-Dp)*`&beta;pv`*[(C1*dsx(dsm)*(dsm/dsx(dsm))^((3-Ds)*Ds))^(3-Dp)-(C1*dsx(dsm)*(dsm/dsx(dsm))^((3-Ds)*Ds+Dp-Ds))^(3-Dp)]/((2-Ds)*`&beta;sv`*[dsx(dsm)^(3-Ds)-dsm^(3-Ds)]*`&beta;ps`*[(C1*dsx(dsm)*(dsm/dsx(dsm))^((3-Ds)*Ds))^(2-Dp)-(C1*dsx(dsm)*(dsm/dsx(dsm))^((3-Ds)*Ds+Dp-Ds))^(2-Dp)]*(3-Dp))

(8)

``

``

``

``


Download GEO_Assignment_3.mw

n := 5:
z1 := exp(2*3.14*I*k1/n)*cosh(z)^(2/n);
z2 := exp(2*3.14*I*k2/n)*sinh(z)^(2/n);
xx := Re(z1);
yy := Re(z2);
uu := cos(alpha)*Im(z1) + sin(alpha)*Im(z2);

i find that the 3d graph has many intersection points to itself

how to find these intersection points of calabi yau ?

 

 

Hello. I want to replicate the results of this table in Maple.

My "attempt" uses the inbuilt dictionary, but it will only find the words of length n, not the frequency of letters in each ....

word_length.mw

I have a problem integrating a solution from fsolve.   I read in another post on this forum that the solution was to use unapply.   This works if I then set up the integration as suggested (i.e., without giving the argument to the function), but not if you do it in a way that seems logical to me (i.e.,the first version of the int command marked ‘fails’ below.   if you can plot a function why can’t you integrate it ?).  

Anyway the real problem I have is if I want to use the solution found using fsolve as the argument of another function (h below) and then integrate that.  I assume the final line fails because of the same reason the initial attempt to integrate g(x) fails. However, I can’t figure out what the equivalent notation would be if I wanted to omit the ‘x’ variable.   I tried using unapply again, and also putting in quotes, but nothing works.

> restart;

> g:=unapply('fsolve(a*y^2-sin(y),y=2)',a);

> plot('g(x)',x=1..2);

> evalf(Int('g(x)',x=1..2));#this fails

> evalf(Int(g,1..2));#this works fine

> h:=x->x*sin(x);

> h(g(1.0));

> h(g(2.0));

> evalf(Int(h(g(x)),x=1..2));# this fails

Hello,

With the Explore function, the use of sliders is very convenient to test the sensibility of a result with regard to a parameter.

However, It is also very convenient to specify a accurate value to a parameter.

Consequently, i would like to combine the use of slider (usually defined in the default mode) with the use of the option "controller=textarea".

Do you have ideas to combine the use of slider and the use of textarea for the definition of the parameters in the Explore function ?


Here you can find a example of the theta4 function depending of 8 parameters (xp3,xp4,zp3,zp4,phi3,phi4, gamma3, gamma4).

TestExplore_4.mw

I manage to use either the sliders or the textarea option but not both.

Thank you in advance for your help and ideas.

I mean the root of the equation

GAMMA(n-1/n)*GAMMA(1/n)/(n*GAMMA(n)) = 1

belonging to RealRange(Open(1),4). It should be noticed there are solutions outside this interval. Here is my try.

 

``

solve({n > 1, GAMMA(n-1/n)*GAMMA(1/n)/(n*GAMMA(n)) = 1, n < 4}, [n])``

[]

(1)

`in`(which*is*wrong, view*of)

simplify(eval(GAMMA(n-1/n)*GAMMA(1/n)/(n*GAMMA(n)), n = (1/2)*sqrt(5)+1/2))

1

(2)

Also

Student[Calculus1]:-Roots(A = 1, n = 1 .. 4)

[1.618033989]

(3)

There is a substitute

fsolve(GAMMA(n-1/n)*GAMMA(1/n)/(n*GAMMA(n)) = 1, n = 1 .. 4)

1.618033989

(4)

NULL

identify(%)

(1/2)*5^(1/2)+1/2

(5)

``

There is a shade of hope that GAMMA(n-1/n)*GAMMA(1/n)/(n*GAMMA(n))  can be simplified.

Download solution.mw

 PS. An SCR was submitted by me.

The following product

 

(product(mu^x[i]/factorial(x[i]), i = 1 .. n))

 

does not simplify to the most obvious form whatever I try

 

mu^(sum(x[i], i = 1 .. n))/(product(factorial(x[i]), i = 1 .. n))

 

What can it be?

 

 

I have recently acquired Maple 2016 and wanted to see how its numerical integration compared to previous version (in this instance, 2015 and 18). This integration is a tougher problem than the usual "textbook" case using a well behaved function. The integrand presented in the worksheet below is a small example but it can get much larger.

I am calculating a triple integral numerically from a function read in from a file which contains Laguerre polynomials. Some simplifications are done first and then that is fed into the integration. In the example script below the input has been put into the program to make it simpler.

So far it appears Maple 18 is faster than 2015 (in this case anyway) and 2016 does not appear to like the syntax I am using even though it runs fine on 18 and 2015 (it does not like the simplify(expr1,LaguerreL) or sqrt parts).

Looking at the stats of the calculation runs:

Maple 18:

memory used=0.52MiB, alloc change=0 bytes, cpu time=20.33s, real time=20.49s, gc time=0ns

answer = 0.160262735437965


Maple 2015:

memory used=350.84KiB, alloc change=0 bytes, cpu time=28.77s, real time=29.24s, gc time=0ns

answer = 0.160262735437309

What is interesting is that Maple 18 is allocating more memory in order to solve the problem compared to 2015. Does anyone have any ideas why this is occuring? Also has there been a syntax change from 2015 -> 2016 which I have not been aware of. Is there a different way to write the script to run in 2016?

Here is the worksheet:

Maple_numeric_speed.mw

Thank you in advance

- Yeti

Dear all,

 

I am trying to plot in the same graph two sets of data points given by vectors. Please see the example below. Any help will be highly appreciated. Thanks!

 

question.mw

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