I want solve PDE equation with imaginary unit ,but Maple give me an Error:

Download PDE_with_Imaginary_Unit.mw

If I delete Imaginary unit from equation Maple give me a solution.

Thanks .

Hi, I am trying to add some useful inverse Laplace transforms to Maple's integral tables. I am frustrated that these transforms will only be recognized  if they are entered in exactly the same format. For example, if I type:

restart;

with(inttrans):

addtable(invlaplace, 1/(s^y*sqrt(a + s^x)), Sum(a^n*sqrt(Pi)*t^(-1 + (1/2 + n)*x + y)/(GAMMA(1/2 - n)*GAMMA(1 + n)*GAMMA(1/2*x + n*x + y)), n = 0 .. infinity), s, t, {a, x, y}, y::Range(-infinity, infinity), x::Range(-infinity, infinity), a::Range(-infinity, infinity));

savetable(invlaplace, `InvlaplaceTable.m`):

Then:

restart;

with(inttrans):

read `InvlaplaceTable.m`:

invlaplace(1/(s^y*sqrt(a + s^x)), s, t);#Ok works for general

invlaplace(1/(s^3*sqrt(a + s^x)), s, t);# Dosen't WORK ?# For y = 3.

invlaplace(1/(s^3*sqrt(s^3 + a)), s, t);# Dosen't WORK ?# For y = 3. and x = 3

value(eval(invlaplace(1/(s^y*sqrt(a + s^x)), s, t), [y = 3, x = 3]))#Workaround !!!

Thanks.

Hi!

How to calculate  a value in MAPLE:

My code:

evalf(eval(diff(n*Zeta(n, 3), n), n = 3)); give me:

#-0.3740436824 + 3.*eval(diff(Zeta(n, 3), n), {n = 3}) ,it should be only:-0.3740436824

OR:

fdiff(n*Zeta(n, 3), [n], n = 3);

#fdiff(n -> n*Zeta(n, 3), [1], [3]) ???

It's a Bug  or (As Designed / Not a Bug) ?

Thanks in advance.

Mathematica code:

D[n*Derivative[n][Zeta][3], n] /. n -> 3 // N;

(* -0.374044*)

Hi, fairly simple question,

I solve a simple equation:

solve(x^x = 4, allsolutions, explicit);

#(2*(I*Pi*_Z7+ln(2)))/LambertW(_Z9, 2*ln(2)+(2*I)*Pi*_Z7)

Maple  gave me solution with prefix _Z for integer values, but why  _Z9 must be exactly equal  Zero  to be correct ?.

If _Z9 is integer it can also take other values than zero ?

Thanks in advance.

More information see attached file:

Download Allsolution.mw

Hello

The problem is translate Mathematica code to Maple to find  numerical solution using int(numeric).

I have a more complicated example and here I gives a very simplified version.

I use  successive approximations to solve integral-equation with symbolic int it's easy to do,but with numeric int  I'm failed

MMA code:

func[x_, 0] := x
ifunc[0][x_] := x
func[x_?NumericQ, n_Integer] := x + NIntegrate[(x - y)*ifunc[n - 1][y], {y, 0, x}]
ifunc[j_Integer] := ifunc[j] = Interpolation[Table[{x, func[x, j]}, {x, -3, 3, 0.2}]]

Plot[{Sinh[x], ifunc[n][x] /. n -> 4}, {x, -3, 3}]

My first attempt to translate:

ifunc := proc (n, x) options operator, arrow; x end proc;

ifunc(0, x) := x;

func(x, 0) := x;

func := proc (x, n) x+int((x-t)*ifunc(n-1, t), t = 0 .. x, numeric) end proc;

T := proc (j) option remember;

Interpolation:-Interpolate([seq(x, x = -2 .. 2, .1)], [seq(func(x, j), x = -2 .. 2, .1)], method = cubic)

end proc;

plot([sinh(x), (T(4))(x)], x = -2 .. 2);

See attached file for more info.

Thanks.

test_numeric_volterra.mw

EDITED :----------------------------------------------------------

Third attempt:

func(x, 0) := x;

(ifunc(0))(x) := x:

func := proc (x, n) option remember; x+int((x-t)*(ifunc(n-1))(t), t = 0 .. x, numeric) end proc;

ifunc := proc (j) local f; option remember;

ifunc(0) := proc (t) options operator, arrow; t end proc;

f := proc (t) options operator, arrow;

CurveFitting:-Spline([seq(x, x = -3 .. 3, .1)], [seq(func(x, j), x = -3 .. 3, .1)], x, degree = 1) end proc end proc;

n := 4; plot([sinh(x), (ifunc(n))(x)], x = -3 .. 3)# for n=4 diverge !!!