Dear Users!

Hoped everyone fine with everything! I want to define a square matrix P whose elements are

p[i,j]=<φ[i],ψψ[j]>

printlevel := 2; for i from 0 while i <= 2^k-1 do for j from 0 while j <= M-1 do varphi[i, j] := t^j end do end do;
printlevel := 2; for i from 0 while i <= 2^k-1 do for j from 0 while j <= M-1 do phi[M*i+j+1] := varphi[i, j] end do end do;
printlevel := 2; for i from 0 while i <= 2^k-1 do for j from 0 while j <= M-1 do psi[i, j] := 2^((1/2)*k)*sqrt(2*j+1)*(sum((-1)^(j+i1)*factorial(j+i1)*(2*t-i)^i1/(factorial(j-i1)*factorial(i1)^2), i1 = 0 .. j)) end do end do;
printlevel := 2; for i from 0 while i <= 2^k-1 do for j from 0 while j <= M-1 do `&psi;&psi;`[M*i+j+1] := psi[i, j] end do end do;
 

take k=2,M=3

Let

f(x)= 2x^2 -2, x >= 0

Find d/dx f^-1(x)|(subscript x=0.)

Note that f(1)=0.

Use d/dx f^-1(x)= 1/(f'[f^-1(x)])

Hello

I was trying to introduce vector r_vec that has 3 components in x,y,z

I've attached my file, I've 2 questions here

first, why isn't the vector shown in as r_vec = () ei + () ej + () ek instead appears as a column vector

second, why doesn't it accept differentating

thank you
 

Download tst1.mwtst1.mw

I have the following pertubation problem I want to use maple to expand for me.

We have epsilon := eps;

x(t,eps):= x_{-1}(t)/eps+x_0(t)+x_1(t)*eps

z(t,eps):=z_{-1}(t)/eps+z_0(t)+z_1(t)*eps

I want to expand a Taylor series of the following function upto some arbitray order of eps, i.e O(eps^3) or higher (depending on my mood :-)), around t=0, f(x(t,eps),z(t,eps),cos(t/eps),sin(t/eps)).

Anyone has any suggestion how to use maple 2017.3 to do this?

Thanks!

c)Graph y=5x² and the tangent line at the point (-1,5) in the same coordinate system.

Manually, I was able to find the answer for (a). But how do I key in the Maple command for question (b) and (c) ?

Please help and thank you in advance.

My code.mw

Footnote: Can we use variation of parameters method in this question?

Hello Guys, I hope you are all fine. I have been struggling with creating an animation of the points (x,y) in maple. I have tried this example 
L := [[1, 1], [3, 2], [3.4, 6], [5, 3, 7], [3, 7, 9, 1], [2, 6, 8, 4, 5]];
animate(PointPlot, [L[trunc(t)]], t = 1 .. 6, frames = 150)
but in my case it shows two points at different location means it takes x and y seperate value and showed it on 1 and 2 on x axis but i want to animate it as the location of point. Please help me. 
Thank you in anticipation.

I have several functional equations in equally many unknown functions of at least two variables, plus parameters.  ("collect" works just for single equations, right?)

I know that for certain parameter ranges, all the functions involved will be quadratic, and I know some coefficients are zero.  That gives me some  coefficients to determine.  I want to

1. specify the functional equations [done in a very primitive low-tech way in the attachment, using atomic variables rather than indices ... have I done correctly?!?] 
2. get Maple to collect coefficients (the K's and the L's in the attachment; the variables are (y,z))
3. get Maple to state an equation system these coefficients have to satisfy (these will unfortunately be coupled quadratics)
4. get Maple to solve that equation system if possible, and if not: to tell me when (= for what parameter values, parameters being the "remaining letters" in the attachment) I have specified enough coefficients
5. in case of a solution, get Maple to tell me which coefficients are real and positive (for those that are solution of quadratic eq's: whether a positive solution exists)

Phew. I am still a complete newbie. Edit: Attachment link: STcoeff2match.mw where the equations themselves are EQ0, EQ1 and EQ2 at the bottom. Copying and pasting them, they look like this (download STcoeff2pastedEQs.mw)

Hi guys, I am trying to solve a system of differential equations, I have done the hand written calculations and I know the answer however I need to put it in a maple code for a generic system which I will work on over time. Here is what I have so far, 

restart;

eqn[1]:=-1/8*D[4](a)(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(u(x, y, t), x, x), diff(u(x, y, t), y, x), diff(u(x, y, t), x, t), diff(u(x, y, t), y, t), diff(u(x, y, t), t, t))=0;

eqn[2]:=-1/8*D[5](a)(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(u(x, y, t), x, x), diff(u(x, y, t), y, x), diff(u(x, y, t), x, t), diff(u(x, y, t), y, t), diff(u(x, y, t), t, t))=0;

eqn[3]:=-1/8*D[6](a)(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(u(x, y, t), x, x), diff(u(x, y, t), y, x), diff(u(x, y, t), x, t), diff(u(x, y, t), y, t), diff(u(x, y, t), t, t))=0;

eqn[4]:=-1/8*D[7](a)(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(u(x, y, t), x, x), diff(u(x, y, t), y, x), diff(u(x, y, t), x, t), diff(u(x, y, t), y, t), diff(u(x, y, t), t, t))=0;

eqn[5]:=-1/8*D[8](a)(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(u(x, y, t), x, x), diff(u(x, y, t), y, x), diff(u(x, y, t), x, t), diff(u(x, y, t), y, t), diff(u(x, y, t), t, t))=0;

eqn[6]:=-1/8*D[9](a)(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(u(x, y, t), x, x), diff(u(x, y, t), y, x), diff(u(x, y, t), x, t), diff(u(x, y, t), y, t), diff(u(x, y, t), t, t))=0;

eqn[7]:=-1/8*D[10](a)(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(u(x, y, t), x, x), diff(u(x, y, t), y, x), diff(u(x, y, t), x, t), diff(u(x, y, t), y, t), diff(u(x, y, t), t, t))=0;

eqn[8]:=-1/8*D[11](a)(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(u(x, y, t), x, x), diff(u(x, y, t), y, x), diff(u(x, y, t), x, t), diff(u(x, y, t), y, t), diff(u(x, y, t), t, t))-1/2=0;

eqn[9]:=-1/8*D[12](a)(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(u(x, y, t), x, x), diff(u(x, y, t), y, x), diff(u(x, y, t), x, t), diff(u(x, y, t), y, t), diff(u(x, y, t), t, t))=0;

dsolve({seq(eqn[i],i=1..9)},a(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(u(x, y, t), x, x), diff(u(x, y, t), y, x), diff(u(x, y, t), x, t), diff(u(x, y, t), y, t), diff(u(x, y, t), t, t)));

Then I get an error return which says:

Error, (in dsolve) too many arguments; some or all of the following are wrong: [{u(x, y, t)}, a(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(diff(u(x, y, t), x), x), diff(diff(u(x, y, t), x), y), diff(diff(u(x, y, t), t), x), diff(diff(u(x, y, t), t), y), diff(diff(u(x, y, t), t), t))].

I know that if I replace u(x,y,t) with a dummy variable U, and its derivative with Ux,Uy,... and so on then it will work, but I need the function u(x,y,t) to be part of the solution.

I know the solution should give me:

a(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(diff(u(x, y, t), x), x), diff(diff(u(x, y, t), x), y), diff(diff(u(x, y, t), t), x), diff(diff(u(x, y, t), t), y), diff(diff(u(x, y, t), t), t)) = -4*diff(u(x,y,t),x,x) + F(x,y,t),

where F(x,y,t) is the constant function.

Please any help would be great!!
 

I am trying to define a periodic piecewise function where essentially for every positive integer I have an output of '1' and the output is '0' everywhere else. I want to use this function in defining a Partial Differential Equation as well.

This is how I have been going about it after reading through the answers on similar questions posted before:

f := piecewise(t::posint, 1, 0_otherwise):
p := 1:
fperiodic := eval(f, t = t-p*trunc(t/p)):

But when I plot it to just as a test, I get a dead plot:
plot(fperiodic, t = 0 .. 5)

                     
I am hoping for something like:

Please help!
TIA

The summation takes too long time. Please help me
 

Hello,

     I've been using the invhilbert procedure from the inttrans package, but I'm running into a small problem. I'm attempting to apply invhilbert to an unknown function, and then later evaluate that function. However, in one particular case (bad, below), it does not produce the expected output. Curiously, I noticed that if I did *two* substitutions (good, below), it produces the expected result.

with(inttrans):

pde := inttrans:-invhilbert(f(t,s),s,x):
def := g = ((t) -> exp(t)*sin(B)):

bad := f = ((t,x) -> (1 + exp(t)*sin(B))*sin(x+A)):
good := f = ((t,x) -> (1 + g(t))*sin(x+A)):

eval['recurse'](pde, [good,def]);
# -cos(A) exp(t) sin(B) cos(x) + sin(A) exp(t) sin(B) sin(x) - cos(A) cos(x) + sin(A) sin(x)
eval['recurse'](pde, [bad,def]);
# -exp(t) sin(B) cos(x) + sin(A) sin(x) - cos(A) cos(x)

As a side-note: this discrepancy was very delicate. Removing any of terms (for instance, A) causes both to give the same, correct answer.

For this particular problem, I was able to manually replace exp(t)*sin(b) with the function g(t) and get the correct result, but I was hoping for a more automated approach (I need to apply it to many equations). Is there any way to get the correct result from equation bad?

Thank you very much!

code_(2).mw

 Dear All, If I have a square with lenght b and width 2a

.The question is how can I make Maple plot this square as points

.I want to define a function f(x_[i],y_[i]) for all sides.

Thanks

restart;

with(VectorCalculus);

x := Vector([VectorCalculus:-`*`(VectorCalculus:-`+`(R, VectorCalculus:-`*`(r, cos(p))), cos(t)), VectorCalculus:-`*`(VectorCalculus:-`+`(R, VectorCalculus:-`*`(r, cos(p))), sin(t)), VectorCalculus:-`*`(r, sin(p))]);

s := [p, t];

g := Matrix(nops(s), nops(s));

for i to nops(s) do

for j to nops(s) do

g[i, j] := simplify(DotProduct(diff(x, s[i]), diff(x, s[j])))

end do

end do;

g;

with(Physics);

Setup(dimension = 2);

Coordinates(X = s);

Setup(metric = g);

g_[];

g_[mu, nu].Ricci[mu, nu]

I have one question already here: In the last step I calculate the Riemann-Scalar.

I was wondering if also this works:

SubstituteTensorIndices({rho = mu, sigma = nu}, Riemann[mu, nu, rho, sigma, Array])

but it is kinda weird. The LHS is the expression I want and the RHS keeps this Array format. What do I need to specify instead of Array in order for this to work properly...

Original Question:

Further I do the following...

Define(A[mu], F);

F[mu, nu] := D_[mu](A[nu](X));

g_[mu, nu].F[`~mu`, `~nu`];

SumOverRepeatedIndices(%);

SubstituteTensorIndices(nu = `~mu`, F[mu, nu]);

SumOverRepeatedIndices(%)

The first result using g_[mu,nu] gives me a contravariant D_ and a covariant A while the second approach (SubstituteTensorIndices) gives me the same result with D_ being covariant and A now being contravariant.

I tried to enforce the contravariance with `~mu` but it didn't work.

So is there a way to enforce this?

A last question:

If I replace D_ above by d_ the partial derivatives do not expand out.

What is the reason for this?

Similarly

Christoffel[mu, nu, nu].A[mu];

SumOverRepeatedIndices(%)

does not give me the real expression in terms of each component A[1] and A[2] the first time I use SumOverRepeatedIndices...I need to do it twice, why is that?

I'm also happy about general advices to improve the way I implemented everything.

Like is there any meaning for signature() if I specify the metric as I did above?

What is better to use here: Physics or Tensor Package?