I have four spindle tori governed by the following equations:

I could plot the surfaces using implicitplot3d and I can imagine the volume common to these surfaces but I could not visualize it. So, am looking for a way to plot the volume covered by the surfaces such that f1<0, f2<0, f3<0 and f4<0. I know that it's easy in case of a 2D filled plot but is there any way this could be done for the 3D case? Any mathematical advice as to how to characterize or calculate this volume would also be great.

Why Maple does not simplify to 1

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

I'm trying to use the continued fractions expansion to simplify a fraction, and I need to cut the expansion off for large values (for example when a denominator is greater than 20). I have a list of the outputs but I'm struggling to find the correct commands to use in order to do this.

p := ContinuedFraction((q-1)/2^t);
print(p);
cf := Term(p, 0..100);
print(cf);
R := Convergent( p, 100);
r := denom(R);
printf("The order is %d"\n, r);
 

This is the section of code, any help would be appreciated, thanks.

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

Download hilbert.mw

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

To demonstrate Maple 2018’s new Python connectivity, we wanted to integrate a large Python library. The result is the DeepLearning package - this offers an interface to a subset of the Tensorflow framework for machine learning.

I thought I’d share an application that demonstrates how the DeepLearning package can be used to recognize the numbers in images of handwritten digits.

The application employs a very small subset of the MNIST database of handwritten digits. Here’s a sample image for the digit 0.

This image can be represented as a matrix of pixel intensities.        

The application generates weights for each digit by training a two-layer neural network using multinomial logistic regression. When visualized, the weights for each digit might look like this.

Let’s say that we’re comparing an image of a handwritten digit to the weights for the digit 0. If a pixel with a high intensity lands in

• an intensely red area, the evidence is high that the number in the image is 0
• an intensely blue area, the evidence is low that the number in the image is 0

While this explanation is technically simplistic, the application offers more detail.

Get the application here

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!

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!