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Dear all,

I woul line integrate e gradient moltiply to a vector. I will try to explain better.

I have define this function:

phi(xi,eta)=(2*xi-1)*(xi-1)*(2*eta-1)*(eta-1)

and I have a vector a

v=(0,1)

I would like to applu a dot product between the gradient of phi an the vector and integrate the results. 

I have already try in many way but without succeed. Someone could please help me?

Thanks 

I have updated  Maple from 18.01 to 18.02, but there is something strange happened to me. I can not use int anymore. Here is my codes:

restart;
int(sin(x),x);
Error, (in int) wrong number (or type) of arguments: invalid option value passed to indefinite integration: {}

kernelopts(version);
print(`output redirected...`); # input placeholder
    Maple 18.00, X86 64 LINUX, Feb 10 2014, Build ID 922027

 

Here is the screenshot:

Hello,

I have a question: Why gives Maple no result for:

expr2 := sin(k*(t-tau))*sin(k*tau);
Int(simplify(expr2, symbolic), tau = 0 .. t);

It is possible with Wolfram Alpha: http://www.wolframalpha.com/input/?i=Int%28sin%28sqrt%28k%2Fm%29*%28t-tau%29%29*sin%28sqrt%28k%2Fm%29*tau%29%2C+tau+%3D+0+..+t%29

 

Thanks!

int(int(y, 0 <= y, x^2+y^2+z^2 <= 1));

Error, (in int) integration range or variable must be specified in the second argument, got 0 <= y

int(int(y, y = 0, x^2+y^2+z^2 = 1));

Error, (in int) integration range or variable must be specified in the second argument, got y = 0

Hi all

kx,ky is the wavenumber, how can I get the 4 cases of piecewise function according to kx=0,kx≠0 and ky=0,ky≠0. Thanks

J := `assuming`([4*(int(int(JJ*exp(-I*(kx*x+ky*y))*sin(2*l*pi*x/a)*sin(2*k*pi*y/b), x = 0 .. a, AllSolutions), y = 0 .. b, AllSolutions))/(a*b)], [k::posint, l::posint, a > 0, b > 0, JJ > 0])

Hi all

Assume that we have following vectors:

> V1 := [1/9, -5/9, 7/9, 1/9, -5/9, 7/9, 1/9, -5/9, 7/9];

>  V2:=t*V1;

and we want to compute the integral of V2, namely:

>  Int(seq(V2[i],i=1..9),t=0..1);
>

            [       5 t  7 t         5 t  7 t         5 t  7 t]
      V2 := [t/9, - ---, ---, t/9, - ---, ---, t/9, - ---, ---]
            [        9    9           9    9           9    9 ]

Error, (in print/Int) invalid input: IntegrationTools:-GetOptions expects its 1st argument, v, to be of type Integral, but received Int(1/9*t,-5/9*t,7/9*t,1/9*t,-5/9*t,7/9*t,1/9*t,-5/9*t,7/9*t,t = 0 .. 1)

 how we integrate from V2? why answer is wrong in my code?

thanks a lot

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

I have a great problem with this integral and Maple gives two answers completely different:

 

int(x^-5/3*cos((x-1)*h), x = 0..infinity)

so I get two different results :

 

-(27/8)*h^2+3/2+(27/8)*h^(7/6)*LommelS2(11/6, 1/2, h)

 

or this:

 

-(27/8)*h^2+3/2+(27/8)*h^(7/6)*LommelS1(11/6, 1/2, h)

In the first integral A get Lommels2 and If I get the Integral by using Taylor of cos((x-1)*h) and after that I resum I get Lommels1.

 

Thank you.

 

 

my int is in below.

my_int.mw

my int has hyperbolic and trigonal parts thus when they are multiplied , Maple is not able to integrate!

 

Now, How can I solve this integral?

thank you for help

my int is:

 

``

restart

eq1 := m*(diff(diff(w(x, t), t), t))+diff(diff(EIy*(diff(diff(w(x, t), x), x))+(EIz-EIy)*(diff(diff(w(x, t), x), x))*(theta(x)+phi(x, t))^2, x), x)-Pz = 0

l := 16; m := .75; EIy := 0.2e5; EIz := 0.400e7; GJ := 0.1e5; mj := .1; Pz := 5000; Mz := 0; theta := proc (x) options operator, arrow; 0 end proc

w := proc (x, t) options operator, arrow; q[1](t)*(cosh(1.8751*x/l)-cos(1.8751*x/l)+(-1)*.734096*(sinh(1.8751*x/l)-sin(1.8751*x/l)))+q[2](t)*(cosh(4.6941*x/l)-cos(4.6941*x/l)+(-1)*1.01847*(sinh(4.6941*x/l)-sin(4.6941*x/l)))+q[3](t)*(cosh(7.8548*x/l)-cos(7.8548*x/l)+(-1)*.999224*(sinh(7.8548*x/l)-sin(7.8548*x/l))) end proc

phi := proc (x, t) options operator, arrow; q[4](t)*sqrt(2)*sin(1.5708*x/l)+q[5](t)*sqrt(2)*sin(4.7124*x/l)+q[6](t)*sqrt(2)*sin(7.8540*x/l) end proc

NULL

f[1] := int(lhs(eq1)*(cosh(1.8751*x/l)-cos(1.8751*x/l)-.734096*(sinh(1.8751*x/l)-sin(1.8751*x/l))), x = 0 .. l)

``

Hi all,

As we know that the indefinite integral in Maple is defined up to a piecewise constant.

For example,

 

Due to the indeterminate range of variable n, int returns the piecewise function.

But in the other similar case, we get a generic solution rather than a piecewise function,

 

Why not to returns the result like this,

 

Thanks for any help.

I have an expression like this:

Since it is linear I want Maple to rewrite it into this:

(with the benefit that Maple then can solve it at least up to a point). i have tried to conceive a rule to do that but got stuck relatively quickly. Does anybody have a way to do this (in some genrality)?

Thanks,

Mac Dude.

 

limit(t*(int(exp(-t*tan(x)), x = 0 .. (1/2)*Pi)), t = infinity)?

Is it possible to find it in Maple? The command

MultiSeries:-limit(t*(int(exp(-t*tan(x)), x = 0 .. (1/2)*Pi)), t = infinity);
outputs
.

 

restart:with(plots):

h3:=((1+lambda*m*x/a0+phi*((4/Pi*sum((-1)^(n+1)/(2*n-1)*cos(2*Pi*(2*n-1)*x),n=1..infinity)))));

q:=Q-1:

f:=sin(alpha)/E;

DP3:=Int(f-((q*(k+1)*(k+2))/((1-tau)^(k+1)*((h3^(k+2)))*(k+1+tau)))^(1/k),x=0..1);

E:=0.2:phi:=0.2:alpha:=0.1:k:=1:lambda:=0.1:a0:=0.5:m:=0.1:tau:=0.1:

plot((DP3),Q=0..1,axes=box,linestyle=1,color=[red]);

I am unable to plot DP3 vs Q, not only it take very long time but didn't give any output.

Please have a look.

 

Cheers!

 


firstly apologies in advance for stuff in this question such as "triangle symbol",  my computer is pretty old. 


ok so i was confused a bit here, what i'm trying to do is write a maple procedure that computes Af for a given f contained in V . except we only need to correct the bug in the script below. This script demonstrates such a procedure in the case that omega is a square. The domain is given here as the negative set of a function F contained in V .  I have left in notes where/what i think we need to do but i dunno how to...

N:=10 ; # Global Var
F:=(x,y)->sgn(abs(x-N/2)+abs(y-N/2)-N/4);
Average := proc(F, f0) local f, i, j;
f := f0; # !!!!!!!!!!!!!! something is bad here...
for i to N do for j to N do
if F(i, j) < 0 then
f[i, j] := (f0[i - 1, j] + f0[i + 1, j] + f0[i, j + 1] + f0[i, j - 1])/4 ;
end if;
end do;end do;
return f;
end proc;
f0:=Matrix(N,F); # just to have something to test the procedure
Average(F,f0); # does not return the expected average, modifies f0

 

the necessary information we were given to produce this so far was..

Let N be a positive integer and [N] = {i contained in N | 1<= i <=N }  Let "Omega" C {(i,j) contained in [N] x [N] | 2<=i,j<=N-1} be a subset. Let V = R^([N]x[N]) be the vector space of real valued functions [N]x[N] -> R
and A, "triangle symbol":V->V (average) and "triangle symbole" (Laplacian) be the linear maps such that
[Af](i; j) = f(i; j)      if (i; j) not contained in "Omega"   OR

                             [f(i, j + 1) + f(i, j - 1) + f(i + 1, j) + f(i - 1, j)]/4 if (i,j) is contained in "Omega"

["traingle symbol"f](i,j) =  0 if (i,j) isnt contained in "Omega"   OR

                            ( f(i,j) - [f(i, j + 1) + f(i, j - 1) + f(i + 1, j) + f(i - 1, j)]/4 )    if (i,j) is contained in "Omega"

 Please and thank you for any help in advance <3

                           

Hi all

I dont know why some Z1 appears on the screen and the code does not converge.
please help me
thanks alooooot


restart;
n:=3;
nn:=3;
m:=1;
BB:=1;
BINF:=5:
pr:=7;
digits:=10;
>
eq1:=diff(f(tau),tau$3)+((3/5)*f(tau)*diff(f(tau),tau$2))-(1/5)*(diff(f(tau),tau$1))^2+((2/5)*tau*diff(h(tau),tau$1))-((2/5)*h(tau))-BB*diff(f(tau),tau$1)=0;
eq11:=(1/pr)*diff(h(tau),tau$3)+(3/5)*f(tau)*diff(h(tau),tau$2)=0;

h(tau):=sum(p^i*h[i](tau),i=0..nn);
f(tau):=sum(p^i*f[i](tau),i=0..n);

 

H1:= p*(diff(f(tau),tau$3)+((3/5)*f(tau)*diff(f(tau),tau$2))-(1/5)*(diff(f(tau),tau$1))^2+((2/5)*tau*diff(h(tau),tau$1))-((2/5)*h(tau))-BB*diff(f(tau),tau$1))+(1-p)*(diff(f(tau),tau$3)):
H11:= p*((1/pr)*diff(h(tau),tau$3)+(3/5)*f(tau)*diff(h(tau),tau$2))+(1-p)*(diff(h(tau),tau$3)):
>
eq2:=simplify(H1):
eq22:=simplify(H11):
eq3:=collect(expand(eq2),p):
eq33:=collect(expand(eq22),p):
eq4:=
convert(series(collect(expand(eq2), p), p, n+1), 'polynom');
eq44:=
convert(series(collect(expand(eq22), p), p, n+1), 'polynom');
for i to n do
s[i] := coeff(eq4, p^i) ;
print (i);
end do:
for i to nn do
ss[i] := coeff(eq44, p^i) ;
print (i);
end do:
s[0]:=diff(f[0](tau), tau$3);
ss[0]:=diff(h[0](tau), tau$3);
ics[0]:=f[0](0)=0, D(f[0])(0)=0, D(f[0])(BINF)=0;
icss[0]:=h[0](BINF)=0, D(h[0])(0)=1, D(h[0])(BINF)=0;

dsolve({s[0], ics[0]});
f[0](tau):= rhs(%);
#dsolve({ss[0], icss[0]});
h[0](tau):= -exp(-tau); #;rhs(%);

>
>
for i to n do
f[ii-1](tau):=convert(series(f[ii-1](tau), tau, nn+1), 'polynom');
h[ii-1](tau):=convert(series(h[ii-1](tau), tau, nn+1), 'polynom');
s[i]:=simplify((s[i]));
ics[i]:=f[i](0)=0, D(f[i])(0)=0, D(f[i])(BINF)=0;
dsolve({s[i], ics[i]});
f[i](tau):=rhs(%);
ss[i]:=(ss[i]);
icss[i]:=h[i](BINF)=0, D(h[i])(0)=0, D(h[i])(BINF)=0;
dsolve({ss[i], icss[i]});
h[i](tau):=rhs(%);

end do;

f(tau):=sum((f[j])(tau),j=0..n);
with(numapprox):

 

plot(diff(f(tau),tau),tau=0..5,color=blue,style=point,symbol=circle,symbolsize=7,labels=["tau","velocity"]);
plot(pade(diff(f(tau),tau), tau, [7, 7]),tau=0..5,color=blue,style=point,symbol=circle,symbolsize=7,labels=["tau","velocity"]);

 

 

int(BesselJ(2, r*k)*BesselJ(1, 1500*k), k = 0 .. infinity);

(of course, calculated numerically) against r=0..40 ?

PS. It is possible in Mathematica 9.0.1.0 .

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