Maple 12 Questions and Posts

These are Posts and Questions associated with the product, Maple 12

In the link below I attempt to solve 2 trig series which are essentially equivalent as indicated by the numerical output of eq (5).  The series  represented by S13 & S14 has arguments of the trig functions that realizes that only the odd terms for k yield non-zero results.  The case represented S11 & S12 by makes no such presumption; nonetheless, all cases agree within reason numerically.  Now to find min/max values taking the derivative is needed which is simply done by removing the integral as indicated by Q1 through Q6.

Now resolving the roots works OK for Q6 because beta = 2*pi *t/T conveniently collapsed the numerator into factorable expressions.  Resolving the roots for Q3 did not work so well because what I think is that the expression in red has multiple roots so it only spits out t as the solution?  I expressed the angle alpha in terms of beta & probably need to resolve kappa to somehow get the expression in red to collapse into a factored expression, but I am not sure how to execute this.  When I solve for kappa I get ZERO.

Does anyone have suggestions?  Remember I demonstrated that both series are practically idendical numerically; hence, there derivatives should be as well as long as both series are well behaved functions.  So the solutions must be the same as well.

trig_series_solns.mw

with(plots):
xy := MatrixMatrixMultiply(Matrix([[cos(t),sin(t)],[sin(t),cos(t)]]),Matrix([[x],[y]]));
animate3d([t, xy[1][1], xy[2][1]], x = 1 .. 10, y = 1 .. 10, t = -Pi .. Pi,coords=spherical);
animate3d([x, xy[1][1], xy[2][1]], x = 1 .. 10, y = 1 .. 10, t = -Pi .. Pi,coords=spherical);
animate3d([t, xy[1][1], xy[2][1]], x = -10 .. 10, y = -10 .. 10, t = -Pi .. Pi,coords=spherical);
 
would like to see how it rotate a ball
 

See below, why the difference + what is the distinction between the 2?  Notice that they yield different results with the is command.

restart; P := a[0]+sum(4*(int(cos(2*Pi*(2*k-1)*x/T), x = 0 .. t))/T, k = 1 .. n); Q := a[0]+4*(int(sum(cos(2*Pi*(2*k-1)*x/T), k = 1 .. n), x = 0 .. t))/T; is(expand(P = Q)), is(combine(P = Q)); is(value(eval(P = Q, n = 10)))

false

(1)

restart; P := a[0]+Sum(4*(Int(cos(2*Pi*(2*k-1)*x/T), x = 0 .. t))/T, k = 1 .. m); Q := a[0]+Int(Sum(4*cos(2*Pi*(2*k-1)*x/T)/T, k = 1 .. m), x = 0 .. t); is(expand(P = Q)), is(combine(P = Q)); is(value(eval(P = Q, m = 10)))

true

(2)

sum(4*(int(cos(2*Pi*(2*k-1)*x/T), x = 0 .. t))/T, k = 1 .. n)

((1/2)*I)*(-ln(-(1+exp((2*I)*Pi*t/T))/(exp((2*I)*Pi*t/T)-1))+exp((2*I)*Pi*(1+2*n)*t/T)*LerchPhi(exp((4*I)*Pi*t/T), 1, 1/2+n)+ln((1+exp((2*I)*Pi*t/T))/(exp((2*I)*Pi*t/T)-1))-LerchPhi(exp(-(4*I)*Pi*t/T), 1, 1/2+n)*exp(-(2*I)*Pi*(1+2*n)*t/T))/Pi

(3)

a[0]+Sum(4*(Int(cos(2*Pi*(2*k-1)*x/T), x = 0 .. t))/T, k = 1 .. m)

a[0]+Sum(4*(Int(cos(2*Pi*(2*k-1)*x/T), x = 0 .. t))/T, k = 1 .. m)

(4)

``


 

Download sum_versus_Sum.mw

 

Below is the output from test relation that seems contradictory.  All I did was swap the order of operation from summing an integral to taking the integral of the sum.  The summation and integration variables are independent so I would think the statement for S4 would yield a TRUE result.  Also, I got a FALSE return on a well known trig identity.

Am I missing something subtle or even obvious?

equivalence_discrepancy.mw

I am still working on the worksheet linked below.  But I have run into trouble solving the equation symbolicly within for the variable t.  I think this is due to the fact there are multiple solutions and the commands I employed through the GUI interface is not capable of handling this issue?  For example, the solutions for sin(pi*t/T) would be N*T.  MAPLE is simply stating t=0.  So I think this is why my solutions are failing to produce results.

What other commands should I be employing?

solving_transcendental.mw

with(LinearAlgebra):
test1 := Matrix([[1,-1],[-1,1]]);
for i from 1 to 20 do
print(Eigenvectors(test1)[2]);
od:

i run the same command 20 times, but sometimes left has negative 1 , sometimes right has negative 1

why position is like random , is my maple has virus?

the result are not consistent, i am learning quantum computation, will this influence the quantum computation and result?

i am doing up and down and change to a differential equation, Pauli equation

It seems only one Pauli equation

i find 3*3 matrix 

has 3 different kinds of matrix

one is row 1 and row 2

second is row 1 and row 3

Third is row 2 and 3

column 1 is constant 

Is there any one has the same experience to predict these 3 kind of matrix, and know what I mentioned?

 

I executed finding the roots of the derivative of a series expansion containing 500 terms.  I did it 2 ways.  The 1st using fsolve & the 2nd using RootFinding.  The fsolve took over 20 minutes to find a single root within a specified range while the RootFinding took less than 60 seconds to find all roots within a larger range.  I do not know of the inner mechanisms of either command, but why is this the case?  Why would the algorithms differ?  My results are in the link below.

fsolve_vs_RootFinding.mw

Matrix([[xx3[1,2],xx3[1,3]],[xx3[2,2],xx3[2,3]]])
Matrix(2, 2, {(1, 1) = (1/6)*sqrt(3)+(1/2)*I, (1, 2) = (1/6)*sqrt(3)-(1/2)*I, (2, 1) = (1/6)*sqrt(3)-(1/2)*I, (2, 2) = (1/6)*sqrt(3)+(1/2)*I})
expect output be
but these example are wrong
((1/6)*sqrt(3)+(1/2)*I)*Matrix([[1,-I].[-I,1]])
but these example are wrong
MatrixMatrixMultiply(Matrix([[(1/6)*sqrt(3),(1/2)],[(1/6)*sqrt(3),(1/2)]]),Matrix([[1,-I],[1,I]]));
concept like this output

Below is MAPLE code to simplify a series.  MAPLE expresses the result in terms of functions which many people are not familiar with.  Is there a way to express the answer in terms of more conventional functions expecially if N is a positive integer?


 

Cn := ((-I)*(1/2))*(2*(I*Pi*n*tau-(2*I)*Pi*n)*cos(Pi*n*tau/T)-T*(2*I)*sin(Pi*n*tau/T)+(4*I)*Pi*n)/(Pi^2*n^2); S4 := a[0]+sum(Cn*sin(2*Pi*n*x/T), n = 1 .. k); a[0] := 0; T := 4; tau := 2; Cn; S5 := unapply(S4, k, x); T := simplify(S5(N, x))

convert(T, StandardFunctions);

(-polylog(2, exp(-((1/2)*I)*Pi*(x-1)))*N^2-exp(-((1/2)*I)*Pi*N*(x+1))*LerchPhi(exp(-((1/2)*I)*(x+1)*Pi), 2, N)*N^2+polylog(2, exp(((1/2)*I)*(x+1)*Pi))*N^2+exp(-((1/2)*I)*Pi*N*(x-1))*LerchPhi(exp(-((1/2)*I)*Pi*(x-1)), 2, N)*N^2+polylog(2, exp(-((1/2)*I)*(x+1)*Pi))*N^2-exp(((1/2)*I)*Pi*N*(x+1))*LerchPhi(exp(((1/2)*I)*(x+1)*Pi), 2, N)*N^2-polylog(2, exp(((1/2)*I)*Pi*(x-1)))*N^2+exp(((1/2)*I)*Pi*N*(x-1))*LerchPhi(exp(((1/2)*I)*Pi*(x-1)), 2, N)*N^2+exp(((1/2)*I)*Pi*N*(x+1))-exp(((1/2)*I)*Pi*N*(x-1))+exp(-((1/2)*I)*Pi*N*(x+1))-exp(-((1/2)*I)*Pi*N*(x-1))-I*exp(-((1/2)*I)*x*Pi*N)*LerchPhi(exp(-((1/2)*I)*x*Pi), 1, N)*N^2*Pi-I*ln(1-exp(-((1/2)*I)*x*Pi))*N^2*Pi+I*exp(((1/2)*I)*x*Pi*N)*LerchPhi(exp(((1/2)*I)*x*Pi), 1, N)*N^2*Pi+I*ln(1-exp(((1/2)*I)*x*Pi))*N^2*Pi-I*exp(((1/2)*I)*x*Pi*N)*N*Pi+I*exp(-((1/2)*I)*x*Pi*N)*N*Pi)/(N^2*Pi^2)

(1)

``


 

Download simplify.mw

When attempting to numerically solve for a function using fsolv it is possible that the function has multiple roots.  So to focus on a particular region you specify a range such as:

xmax := fsolve(S, x = 0 .. 1/2)

Is it possible fsolve may not resolve the solution due to the fact that delta x is not small enough or does fsolve autonomously adjust delta x in order to find the solution?  If not, how do you manually dictate the delta x for the interval specified?

Below is a link to my worksheet.  I am attempting to evaluate a Fourier series for a particular number of terms & at a specific location xmax.  As far as I can tell xmax is assessed correctly.  Howver, when I go to evaluate the Fourier series for x = xmax using either the 'value' or 'evalf' command they do on seem to recognize that x = xmax.

So I am guessing I must have some syntax problem.  Can anyone show me what I have wrong?

command_syntax.mw

interface(prettyprint=0):
interface(screenwidth=500):
with(LinearAlgebra):

expect 

Matrix([[a1,a2,3],[5,6,7],[9,10,12]])

but

it print datatype = anything,storage = rectangular,order = Fortran_order,shape  and (2,1) etc

Matrix(3,3,{(2, 1) = 1, (3, 1) = 1, (3, 2) = 1},datatype = anything,storage = rectangular,order = Fortran_order,shape = []), 

The MAPLE worksheet associated with the link below attempts to generate a sequence of signals to be symmetric about t=0.  It worked for the 1st plot, but not for the 3rd plot.  I believe the problem resides with the Heaviside function.  I need to somehow create a function that is the mirror image of the Heaviside function.  Does anyone know how to do that?

untitled4.mw

The link below has my code for generating 2 matrices.  The 1st one does not generate flfoating point numerical data; whereas, the 2nd one does.  What is wrong with the 1st case?  I am attempting to single out one harmonic which works in the 2nd case.  Also, is there a way I can generate a spectrum of S2(k= 1 to 100, t= 0 to 1)?

?untitled6.mw

 

1.
tanh(1-x) = sum(p(ii)*x^q(ii), ii=0..infinity) or product(p*x^q(ii), ii=0..infinity) ?
2.
tanh(1-x)*1/(1-x) = sum(p(ii)*x^q(ii), ii=0..infinity) or product(p*x^q(ii), ii=0..infinity) ?
3.
tanh(x) = sum(p(ii)*x^q(ii), ii=0..infinity) or product(p*x^q(ii), ii=0..infinity) ?

Remark: it may not be possible to use diff to find p(ii)

update

series(tanh(1-x), x=0);
with(OrthogonalSeries):
Coefficients(series(1/(1-x), x=0));
coeffs(series(tanh(1-x), x=0));
coeffs(series(tanh(1-x), x=0),x);
Error, invalid arguments to coeffs;
 
and is it possible to find q(ii) only if assume p(ii) all are one?
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