hi

please help me for solve this equations?

thanksazizi.mw
 

Download azizi.mw

I am trying to achieve an intuitive (non-mathematical) understanding of parallel transport on the unit sphere.
A vector parallel transported along a unit sphere's geodesic clearly maintains as constant its initial orientation vis-a-vis the geodesic i.e. with regard to its moving bases vectors.

Dr. Lopez's application "Visualizing a Parallel Field in a Curved Manifold" displays animation of the parallel transport of a vector along a latitude, which is not a geodesic. As seen by an observer moving with it, the vector rotates clockwise with regard to its moving bases vectors. The rate of rotation increases with higher latitudes.

Then to what is the transported vector maintaining parallelism?

I have uploaded a Maple 2016 worksheet Parallel_transport_on_the_unit_sphere.mw which mostly copies Dr. Lopez's application and, for additional clarity, adds the unit sphere, the moving bases vectors and the moving tangent plane to his display.

> restart;
> with(plots);
> pr := .72; p := 0; n := [2, 3, 4, 5]; s := 1; a := .2; b := 1;
> R1 := 2*n/(n+1);
                               2 [2, 3, 4, 5]
                              ----------------
                              [2, 3, 4, 5] + 1
> R2 := 2*p/(n+1);
                                      0
>
>
> for j to nops(n) do R1 := 2*n[j]/(1+n[j]); R2 := 2*p/(1+n[j]); 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)+pr*s^f(eta)*(diff(theta(eta), eta))+R2*pr*s*(diff(f(eta), eta))*theta(eta)+2*(a*(diff(f(eta), eta))+b*theta(eta))/(n[j]-1) = 0, f(0) = 0, (D(f))(0) = 1+b*((D@@2)(f))(0), (D(f))(5) = 0, theta(0) = 1+s*(D(theta))(0), theta(5) = 0], numeric, method = bvp); fplt[j] := plots[odeplot](sol1, [eta, diff(diff(f(eta), eta), eta)], axes = boxed); tplt[j] := plots[odeplot](sol1, [[eta, theta(eta)]], axes = boxed) end do;
>
> plots:-display([seq(fplt[j], j = 1 .. nops(n))]);

> plots:-display([seq(tplt[j], j = 1 .. nops(n))]);

Dear sir

In the above problem graph, i am getting all the lines are in same color then how to identify the lines of different values like n=2,3,4,5,6(or can we set different color for different values of n for each line)

For the ODE system with boundary conditions, I was able to obtain solutions for n=0, but not for n>0. I obtained the error, Initial newton iteration is not converging. Anyone knows the solution for this? I am open to all suggestions and any help would be greatly appreciated:)

ODE_solution.mw

How do I convert an equation into a vectorial form in Maple?

Say you hav f_1(v)=-1/2*C*p*v*V

where V=[vx, vy] and v=|V|.

v=46m/s^2, A=18m^2, p=1,1

Important Question
Anyone figured out how to upload a worksheet since the revamp? Big green up-arrow appears no longer functional

Less important question
Mapleprimes input window can no longer be stretched horizontally - is this deliberate?

Trying to write a generat purpose procedure insteat of having 3 or 4 different named procceduers.

Need to determine the input types of say A,B,C

Proc(A,B,C)

1) A,B,C could be for the form 3/2 , 5x-7 , x^2-sqrt(x+3)  i.e numbers or formulas

2) A,B,C could be 2 elenent lists [2,3]  , [5x-6y,8x+2] ,[2,sqrt(3)]  i.e 2D points

3) A,B,C could be 3 element lists as above    i.e 3D points

4) A,B,C could be 2D vectors

5) A,B,C could be 3D vectors

A, B, C will not be of mixed types but would be good to check A, B, C for consistency and if not retrun an error

Looking to calculate output based on

If type 1) ......

elif type 2).....

.

.

elif type 5)...

end if

return.....

end proc

> restart;
> with(plots);
> pr := .72; p := 0; n := 1; s := 1; a := [-0.5,0.0,0.5]; b := 1;
> R1 := 2*n/(n+1);
                                      1
> R2 := 2*p/(n+1);
                                      0
>
>
> for j to nops(a) do R1 := 2*n/(n+1); R2 := 2*p/(n+1); 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)+pr*s^f(eta)*(diff(theta(eta), eta))+R2*pr*s*(diff(f(eta), eta))*theta(eta)+2*(a[j]*(diff(f(eta), eta))+b*theta(eta))/(n+1) = 0, f(-.5) = 0, (D(f))(0) = 1+b*((D@@2)(f))(0), (D(f))(5) = 0, theta(-.5) = 1+s*(D(theta))(0), theta(5) = 0], numeric, method = bvp); fplt[j] := plots[odeplot](sol1, [eta, diff(diff(f(eta), eta), eta)], color = ["blue", "black", "orange"]); tplt[j] := plots[odeplot](sol1, [eta, theta(eta)], color = setcolors(["red", "Coral"])) end do;
Error, (in dsolve/numeric/process_input) boundary conditions specified at too many points: {0, 5, -1/2}, can only solve two-point boundary value problems
>
> plots:-display([seq(fplt[j], j = 1 .. nops(a))], color = [green, red]);

> plots:-display([seq(tplt[j], j = 1 .. nops(a))], color = [green, red]);

Dear sir,

In this program i m not getting the solution for decimal values and i do not have idea about the how to set different color for multiple lines(i tried for different set of colors but it shows that only for first color )

Hi,

I want to find the inverse of a 11x11 matrix which I imported from excel using the import data tool. When I try to find the the inverse it gives me this error:

-----------------------------------------------------------------------

K:=ExcelTools:-Import("C:\\Assignment 2.xlsx", "Q2", "V7:AF17");

K := Vector(4, {(1) = ` 1..11 x 1..11 `*Array, (2) = `Data Type: `*anything, (3) = `Storage: `*rectangular, (4) = `Order: `*Fortran_order})

with(LinearAlgebra):
kkk:=convert(K, matrix):
KK:=MatrixInverse(kkk);

Error, (in MatrixInverse) MatrixInverse expects its 1st argument, M, to be of type {Matrix, [Matrix(square)], [Vector, Matrix(square)], [Matrix(square), Matrix(square), Matrix(square)]} but received kkk

Can someone please help me out??? Thank you

Hi there,

I have a big polynomial expression involving powers of x and y, that comes from expanding a function in powers of x and y in polynomial form (I use series(convert(series(a,x=0,10),polynom),y=0,10) ). I want to multiply each of the terms by the factorial of the power of x and y it has. How can I do this?
I tried using Physics[Coefficient](a,x) but I get the error: it cannot compute the degree of the expression.
I tried using a double for with a double coeff to get each of the coefficients and the maybe be able to multiply them but I get the error "unable to compute coeff".

Is it because as expanding the series I have the term +O(y^11) that it cannot compute it?


[Edit]
I managed to substitute the x terms using subs(x^3=3!*x^3,x^5=5!*x^5,a). Obviously this is not very efficient since I need to write the substitution for each term, and since the ploynom is grouped in powers of y, this does not work for y (neither does algusbs).
 

[Edit 2]:

an example of it would be:
 

with result 

And I want the coefficients for each x and y power to be multiplied by the factorial of those powers.

Thank you!

Hi,

Does anyone no how to normalize wavefunctions such that the plot on the right looks like the one on the left.

Here's my code:-

schro := {diff(psi(x), x, x)-(alpha*x^4+x^2-energy)*psi(x) = 0};

ic := {psi(3) = 0, (D(psi))(3) = 1};

Ic := [{psi(3) = 0, (D(psi))(3) = -1}, %$2];

E := [1.06538, 3.306, 5.74796];

schro1 := [seq(subs(energy = e, alpha = .1, schro), e = E)];

soln1 := [seq(dsolve(schro1[i] union Ic[i], {psi(x)}, type = numeric), i = 1 .. nops(E))]; with(plots);

display(seq(odeplot(soln1[i], [x, psi(x)], -3 .. 3, color = [red, blue, green][i]), i = 1 .. nops(E)))

Thanks in advance

Dear All

For six parameters, I have corresponding list of their values and there are eight values for every parameter. I need to put these values in a formula to obtain a list of output values. There are two formulas one for 'P' and next is for 'RL'. I have used value of 'P' to calculate value of 'RL'. There are some complex number too, for which I have used modulus and final value is calculated by using 'evalf', but this command is not returning proper values for list as required. But this command works fine when I use single value from every list to calculate RL.

The Maple sheet attached herewith.

List.mw

Regards

Hi,

I was playing around with the example "Simple Inverse Kinematic Problem" and found somethings to be odd:

the angular motion seems to be calculated from between pi and negative pi and this has some effects when using position block to move a joint.

the original angular displacement is graph below

the angular displacement after ik calculations have been performed:

if you run the simulation it seems to copy and mirror the input pendulum, however if you disable one of the IK solutions you see that infact its motion isn't like the input.

this becomes more prevailant when you use a position block to force rotation on a joint instead of using the 'prescribed rotation' blocks that comes with the example.

My question then:
why does this happen?
how do I work around this?

the importance that the motion is follow precisely becomes more prevailent when we want to extract other values such as vel, accel, torque. they are incorrect and very jumpy. Also simply put the angular displacement is wrong, how do I fix it?

(side note: I'm thinking  it has to do with the way arctan is calculated in maple limits it to stay in range -pi to pi
"For real arguments x, y, the two-argument function arctan(y, x), computes the principal value of the argument of the complex number x+Iy, so −π < arctan(y,x) ≤ π." from https://www.maplesoft.com/support/help/Maple/view.aspx?path=invtrig)

pjf

Hello people in mapleprimes,

I want to simplify the next expression which has 1/k as its exponent,

especially, I want to collect for T.　I hope you will teach me how to do it.

(F__X*(Omega+1)/(F__I*(beta-1)*T))^(1/k)*(T/phi)^(beta/k)

If I do as

simplify(%)assuming(symbolic);

the output is

F__X^(1/k)*(Omega+1)^(1/k)*F__I^(-1/k)*T^((beta-1)/k)*(1/(beta-1))^(1/k)*phi^(-beta/k)

But, as all variables has 1/k as its exponent, I want to collect it to (...)^(1/k).

Is this possible?

taro