Alex0099

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1 years, 92 days

MaplePrimes Activity


These are replies submitted by Alex0099

@Mariusz Iwaniuk 

Yes, of course. After one minutes I got the final result in the another form, and warning "computation more out try again", and so, it is each iteration

@Mariusz Iwaniuk 

I have a full Mathematica version, but for double summation step-by-step solution not available!

@Mariusz Iwaniuk 

I understand code, but if i evaluate sum manually I don't get this result, and I don't understand how it works. It is a pity that Mathematica not be able to display step-by-step solutions...

In any case, it is clear that this sum converges

@vv 

I found a few articles, and this sum looks like sum, evaluating with "Mock Theta Function" methods, but this sum not trivial, and If I use Mathematica and try to get step-by-step solution, this sum is not converge! So, it's prove, that this sum converge very fast...

Theorem should be useful  - thanks for advice!

@vv 

Let me clarify, why did the order of summation change? And another question, but in Maple there is a tool for more detailed calculations, for example, such as Tutors -> One Variables -> Integrals...

So, if the double series convereges very fast, I probably, should prove it (for example using limits).

@mmcdara

 You are right, tmp := RHS(t)*pm(i,1,t) is RHS(t)...

Exuse me, for my question, I would also like to clarify: if I pass a negative number as j (because, for example, my wavelet or scaling function can start at -2, where my function begins), then it turns out that such a cycle will not work, right?

It is a pity, of course, that the final answer cannot be represented as a linear combination with the coefficients C, D, but simply written as a final function. But this seems to be just the implementation is not suitable for this task ...

And thank you so much for the fixes!

@Carl Love 

Apparently, only people with a very strong mathematical background are able to solve such a problem .. Thank you for the post, in any case)

@Carl Love 

Thanks for advise, I fix it 5 hours ago, but this fact it's very terrible, so I get the same error 
 

Error, (in LinearAlgebra:-LinearSolve) inconsistent system
Error, (in SumTools:-DefiniteSum:-ClosedForm) summand is singular in the interval of summation

 

I try to implement full code, but I get a same errors:

 

restart; with(LinearAlgebra)

h1 := proc (x) options operator, arrow; piecewise(0 < x and x < 1/2, 1, 1/2 < x and x < 1, -1, 0) end proc

proc (x) options operator, arrow; piecewise(0 < x and x < 1/2, 1, 1/2 < x and x < 1, -1, 0) end proc

(1)

hi := proc (j, k, t) local a, b, c, m; m := 2^j; a := k/m; b := (k+1/2)/m; c := (k+1)/m; return piecewise(a <= t and t < b, 1, b <= t and t < c, -1) end proc

J := 2; N := 2^J; hd := Vector(N); H := Matrix(N, N); T := Vector(N); hd[1] := h1(t); for i to N do T[i] := (i-1/2)/N end do; for j from 0 to J-1 do m := 2^j; for k from 0 to m-1 do i := m+k+1; hd[i] := hi(j, k, t) end do end do

NULL

for i to N do for j to N do H[i, j] := eval(hd[i], t = T[j]) end do end do; pn := proc (i, n, t) if n = 1 then return int(hd[i], t) end if; return int(pn(i, n-1, t)) end proc

NULL

RHS := proc (x) options operator, arrow; piecewise(-2 <= x and x < -1, 1, -1 <= x and x < 0, -1, 0 <= x and x < 1/2, 2, 1/2 <= x and x < 1, -2, 0) end proc

proc (x) options operator, arrow; piecewise(-2 <= x and x < -1, 1, -1 <= x and x < 0, -1, 0 <= x and x < 1/2, 2, 1/2 <= x and x < 1, -2, 0) end proc

(2)

R := Vector(N); TMP := Matrix(N, N); A := Matrix(N, N)

Vector(4, {(1) = 0, (2) = 0, (3) = 0, (4) = 0})

 

Matrix(4, 4, {(1, 1) = 0, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (2, 1) = 0, (2, 2) = 0, (2, 3) = 0, (2, 4) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 0, (3, 4) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = 0})

 

Matrix(%id = 36893490055698982180)

(3)

"for i from 1 to N do  R[i] := evalf(RHS(T[i])):  tmp := RHS(x)*pn(i,1,t)  for j from 1 to N do  TMP[i,j]:=eval(tmp, t = T[j]):  od:  od:"

Error, unterminated loop

"for i from 1 to N do  R[i] := evalf(RHS(T[i])):  tmp := RHS(x)*pn(i,1,t)  for j from 1 to N do  TMP[i,j]:=eval(tmp, t = T[j]):  od:  od:"

 

A := Transpose(LinearSolve(Transpose(H+TMP), R)); sol := sum('A[m0]*pn(m0, 2, t)', m0 = 1 .. N); y := unapply(sol, t)

Error, (in SumTools:-DefiniteSum:-ClosedForm) summand is singular in the interval of summation

 

NULL

NULL


 

Download wqw1.mw

For my post I need a help for wavelt decomposition, and I don't know how to implement and decide earlier noticed problems
 

@tomleslie@mmcdara 2796 

Thank so much for your invaluable and much-needed help! 

@tomleslie 

Excuse me, I mean, how to find these terms without using Maple (so I'm trying to understand how to get terms purely mathematical, the theoretical idea for evaluating it)...

@tomleslie

It's clear, but I'm interested what how to analyze this psi(x) without maple, maybe you know some works of literature, references, i.e., you yourself should know how to get these coeffs? 

Do I understand correctly that this 4*l^2*sin(n*Pi/2) and  l^2((-1^n)-1) the same thing?

@tomleslie 

Sorry, but it is by no means my desire, to give the solution in such detail!

@tomleslie 


Thank you for the much-needed help! Just let me ask you one more question! How is it all analytic to make a change of variable x and get an interval (-l/2;l/2) in order to examine the function for even and odd parity, somehow the maple calculates these coefficients using the commands you gave?

@tomleslie 

Thanks for this worksheet file and solution to the equation! I just need to get the result step by step, so I separate the variables, solve two different ODEs, but at the moment of calculating the expansion coefficients psi (x), there was a problem! Could you tell me, maybe how I can still consider such a case in maple and implement it: perhaps I should replace x so that the interval becomes (-l / 2; l / 2), and think about when the integrand will be even / odd ...

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