Maple 2015 Questions and Posts

These are Posts and Questions associated with the product, Maple 2015

Dear friends! I am facing problem to solve the below system of ODEs numerically please find the mistake and correct it.

alpha := -1; R := 2; m := 2; Pr := 7; Le := 1.25; Nt := .2; Nb := .2; g := .5; K1 := .1; Q := .5

Eq1 := eta^3*(diff(F(eta), eta, eta, eta, eta))+alpha*(eta^4*(diff(F(eta), eta, eta, eta))+eta^3*(diff(F(eta), eta, eta))-eta^2*F(eta))-2*eta^2*(diff(F(eta), eta, eta, eta))+3*eta*(diff(F(eta), eta, eta))-3*(diff(F(eta), eta))+eta*R*(diff(F(eta), eta))^2-3*eta*R*F(eta)*(diff(F(eta), eta, eta))+3*R*F(eta)*(diff(F(eta), eta))+3*eta^2*R*F(eta)*(diff(F(eta), eta, eta, eta))-eta^2*(diff(F(eta), eta))*(diff(F(eta), eta, eta))-M^2*(eta^3*(diff(F(eta), eta, eta))-eta^2*(diff(F(eta), eta))); Eq2 := eta*(diff(G(eta), eta, eta))+alpha*Pr*eta^2*(diff(G(eta), eta))+R*Pr*F(eta)*(diff(G(eta), eta))+Nb*eta*(diff(G(eta), eta))*(diff(H(eta), eta))+Nt*eta*(diff(G(eta), eta))^2+diff(G(eta), eta)+Q*Pr*eta*G(eta) = 0; Eq3 := eta*(diff(H(eta), eta, eta))+alpha*Le*Pr*eta^2*(diff(H(eta), eta))+R*Le*Pr*F(eta)*(diff(H(eta), eta))+Nt*eta*(diff(G(eta), eta, eta))/Nb+Nt*(diff(G(eta), eta))/Nb+diff(H(eta), eta)-g*Le*Pr*eta*H(eta)-Le*Pr*K1*eta = 0;

IC1 := F(0) = 0, F(1) = 1, (D(F))(0) = 0, (D(F))(1) = 0, (D(G))(0) = 0, G(1) = 1, (D(H))(0) = 0, H(1) = lambda; dsys1 := {Eq1, Eq2, Eq3, IC1}; dsol1 := dsolve(dsys1, numeric, continuation = lambda, range = 0 .. 1);

dsol1x := subs(dsol1, F(eta));

dsol1y := subs(dsol1, G(eta)); dsol1z := subs(dsol1, H(eta));
 

With my best regards and sincerely.

Muhammad Usman

School of Mathematical Sciences 
Peking University, Beijing, China

Hi everybody, 

The sequence 
    writeto(MyFile)
    showstat(Myproc)

keeps printing the result on the screen

How is it possible to redirected the output of showstat(Myproc) into a file ?

Thanks in advance

how to convert a nested for loop to iterative version with stack

#my testing for wildcard to one
#after testing, it loop a very long time and not stop
ppp := [[0,0,0,x],[0,0,1,0],[0,1,0,0],[1,0,0,0]]:
check1 := [seq(0,ii=1..nops(ppp))];
ttt1 := [[0,0,0,1],[0,0,1,0],[0,1,0,0],[1,0,0,0]]:
mmmeaght1 := [seq(0,ii=1..nops(ppp[1]))]:
bbb1 := [seq(0,ii=1..nops(ppp[1]))]:
emap := [(xx) -> if [xx < 0 assuming x > 0] then 0 else 1 end if, (xx) -> evalf(1/(1+exp(xx)))]:
#trace(perceptronrule1);
MM(ppp, ttt1, mmmeaght1, bbb1, check1, emap);
 

when test wildcard variable for input, would like to assume x > 0 then

i try assuming x > 0 , got error

 

I want to calculate the intersection between three circles.
I know that in this case i can calculate intersection of only the first and second equation, but I need this for a interactive component.

The command "intersection"[GEOMETRY] work only with 2 circles.

I did this but it doesn't work.

Thanks.

I used the command line betwen two poin, and i saw the graphic.

The line passes for my two point, but i I would like it started in the first point and finished in the second.

Thanks.

when test most simple case one to one, and many to one these two reasonable cases, it run a very long time without exit the program.

when i test with the example in book Neural Network Design, it can output correctly but only for book example.

restart:
with(ExcelTools):
with(ListTools):
with(plots):
with(LinearAlgebra):
zipplus := proc(mm, pp)
return zip((x,y) -> x+y, mm, pp)
end proc:
zipminus := proc(mm, pp)
return zip((x,y) -> x-y, mm, pp)
end proc:
zipstar := proc(mm, pp)
return zip((x,y) -> x*y, mm, pp)
end proc:

metara := proc(pp,meaght,bb,mapp,deep)
if deep > 0 then
 pp2 := metara([seq(pp[i], i = 1 .. nops(pp))],meaght,bb,mapp,deep-1):
 mp := zip((x,y) -> x*y,pp2,meaght):
 mpsam := sum(mp[m],m=1..nops(pp2)):
 mpb := [seq(0, i = 1 .. nops(pp2))]:
 for ii from 1 to nops(bb) do
  mpb[ii] := evalf(mpsam + bb[ii]);
 od:
 pa := [seq(0, i = 1 .. nops(pp2))]:
 for ii from 1 to nops(bb) do
  pa[ii] := evalf(mapp[deep](mpb[ii])):
 od:
 return pa:
else
 return pp:
end if:
end proc:

perceptronrule1 := proc(p, t1, meaght1, b1, checksum, mapp)
 meaght3 := meaght1:
 b3 := b1:
 checksum2 := checksum;
 print(p[1]):
 while sum(checksum2[jj], jj=1..nops(p)) <> nops(p) do
  #print("sum(checksum2[jj], jj=1..nops(p))");
  #print(sum(checksum2[jj], jj=1..nops(p)));
  for ii from 1 to nops(p) do
   #print("metara(p[ii],meaght3,b3,mapp,1)");
   #print("p[ii]");
   #print(p[ii]);
   #print("b3");
   #print(b3);
   #print("meaght3");
   #print(meaght3);
   #print(metara(p[ii],meaght3,b3,mapp,1));
   #print("t1[ii]");
   #print(t1[ii]);
   e := zipminus(t1[ii], metara(p[ii],meaght3,b3,mapp,1));
   #print("e");
   #print(e);
   meaght2 := meaght3 + zipstar(e,p[ii]);
   #print("meaght2");
   #print(meaght2);
   #print("meaght3");
   #print(meaght3);
   #print("b3");
   #print(b3);
   b2 := b3 + e;
   #print("b2");
   #print(b2);
   #print("b3");
   #print(b3);
   #print("checksum2");
   #print(checksum2);
   diff1 := zipminus(meaght2, meaght3):
   diff2 := zipminus(b2, b3):
   if sum(diff1[m],m=1..nops(diff1)) = 0 and sum(diff2[m],m=1..nops(diff2)) = 0 then
    checksum2[ii] := 1:
   else
    checksum2[ii] := 0:
    b3 := b2:
    meaght3 := meaght2:
   end if:
  od:
 od:
 return [meaght3, b3, checksum];
end proc:

#Example from book
ppp := [[2,2],[1,-2],[-2,2],[-1,1]]:
check1 := [seq(0,ii=1..nops(ppp))];
ttt1 := [[0,0],[1,1],[0,0],[1,1]]:
mmmeaght1 := [seq(0,ii=1..nops(ppp[1]))]:
bbb1 := [seq(0,ii=1..nops(ppp[1]))]:
emap := [(x) -> if x < 0 then 0 else 1 end if, (x) -> evalf(1/(1+exp(x)))]:
#trace(perceptronrule1);
perceptronrule1(ppp,ttt1,mmmeaght1,bbb1,check1,emap);

#my testing for one to one
#after testing, it loop a very long time and not stop
ppp := [[0,0,0,1],[0,0,1,0],[0,1,0,0],[1,0,0,0]]:
check1 := [seq(0,ii=1..nops(ppp))];
ttt1 := [[0,0,0,1],[0,0,1,0],[0,1,0,0],[1,0,0,0]]:
mmmeaght1 := [0,0,0,0]:
bbb1 := [0,0,0,0]:
emap := [(x) -> if x < 0 then 0 else 1 end if, (x) -> evalf(1/(1+exp(x)))]:
#trace(perceptronrule1);
perceptronrule1(ppp,ttt1,mmmeaght1,bbb1,check1,emap);

#my testing for many to one
#after testing, it loop a very long time and not stop
ppp := [[1,1,0,0],[0,1,1,0],[0,1,0,1]]:
check1 := [seq(0,ii=1..nops(ppp))];
ttt1 := [[1,0,0,0],[0,1,0,0],[0,0,0,1]]:
mmmeaght1 := [0,0,0,0]:
bbb1 := [0,0,0,0]:
emap := [(x) -> if x < 0 then 0 else 1 end if, (x) -> evalf(1/(1+exp(x)))]:
#trace(perceptronrule1);
perceptronrule1(ppp,ttt1,mmmeaght1,bbb1,check1,emap);

#my testing for one to many
#after testing, it loop a very long time and not stop
ppp := [[1,0,0,0],[0,1,0,0],[0,0,0,1]]:
check1 := [seq(0,ii=1..nops(ppp))];
ttt1 := [[1,1,0,0],[0,1,1,0],[0,1,0,1]]:
mmmeaght1 := [0,0,0,0]:
bbb1 := [0,0,0,0]:
emap := [(x) -> if x < 0 then 0 else 1 end if, (x) -> evalf(1/(1+exp(x)))]:
#trace(perceptronrule1);
perceptronrule1(ppp,ttt1,mmmeaght1,bbb1,check1,emap);

Hello,

i need to represent a circle using the command [geometry] circle because i have a problem and i have to calculate an 

intersection between two circles whit command intersection() which is ruled by Package geometry.

For example, with the first circle  i did like the help page says

 But it's Error:

centre of circle(3300;2200)

radius=110

I don't understand the eror. Can you help me?
Thanks in advance.

Hello.

I've just installed Maple 2015 and when I started it, this happened

It seemed that something was wrong with the graphics or java. But I really don't know what to do next.

Please help me, thank you.

Say I define the following variables.

These are all nineth roots of unity. An equivalent definition would be:

 

In fact, the following code shows that aa[i] /a[i] =1 for all i, so one would concluse aa[i]=a[i]:

for i to 9 do 
simplify(a[i]/aa[i])
end do

But when I try to check via "Equal":

I get as output

                              true
                             false
                             false
                             false
                             false
                             false
                              true
                             false
                             false
The problem goes even further since one representation is accepted as a solution of a linear equation system while the other is not.

 

Another curiosity:

gives just the same expression, whereas simplifying the same expression to the third power gives 0.

I am writting a program that needs to rename variables by increasing the second index of a variable, all the variables will be named y[something,number].

e.g.

y[a,2]->y[a,3]

If I was doing this outside maple I can see how I could use regular expressions, but I can't see how to do it in maple

http://www.maplesoft.com/applications/view.aspx?SID=4229

from book example, it seems assumed that input size of data such as list size or matrix size is the same as

trained data set size, but this need to hard code infinite number of types of size

What is the method to programming neural network when input size is smaller or changing and not equal to size of trained data set?

Hello! Hope everyone would be fine. I want to solve the following system of ODEs please help to find the numerical solution

N := .6; alpha := .4; beta := .1; Nt := .2; Pr := .5; Nb := .1; s := .2; lambda[1] := 1; delta := .5; gm := 1; Sc := 1:L:=1:

Eq1 := (alpha*s+1)*(diff(F(eta), eta, eta, eta))-(F(eta)+(1/2)*s*eta)*(diff(F(eta), eta, eta))+((1/2)*(diff(F(eta), eta))-s)*(diff(F(eta), eta))-2*(G(eta)^2-(1-gm)^2)-2*lambda[1]*(H(eta)+N*Y(eta))-(alpha+beta-(1/4)*delta*(diff(F(eta), eta, eta, eta)))*(diff(F(eta), eta, eta))^2-(alpha-2*beta)*(diff(F(eta), eta))*(diff(F(eta), eta, eta, eta))-(2*(alpha-beta-(1/4)*delta*(diff(F(eta), eta, eta, eta))))*(diff(G(eta), eta))^2-(2*(alpha-(1/4)*delta*(diff(F(eta), eta, eta))))*G(eta)*(diff(G(eta), eta, eta)) = 0; Eq2 := (alpha*s+1)*(diff(G(eta), eta, eta))-F(eta)*(diff(G(eta), eta))+G(eta)*(diff(F(eta), eta))+s*(1-gm-G(eta)-(1/2)*eta*(diff(G(eta), eta)))-(1/2)*alpha*s*eta*(diff(G(eta), eta, eta, eta))+((3/2)*alpha+beta)*G(eta)*(diff(F(eta), eta, eta, eta))-((1/2)*alpha+beta)*(diff(F(eta), eta))*(diff(G(eta), eta, eta))-delta*((diff(F(eta), eta, eta))^2+6*(diff(G(eta), eta))^2)*(diff(G(eta), eta, eta)) = 0; Eq3 := (diff(H(eta), eta, eta))/Pr-F(eta)*(diff(H(eta), eta))+(1/2)*H(eta)*(diff(F(eta), eta))-s*(2*H(eta)+(1/2)*eta*(diff(H(eta), eta)))+Nb*(diff(H(eta), eta))*(diff(Y(eta), eta))+Nt*(diff(H(eta), eta))^2 = 0; Eq4 := (diff(Y(eta), eta, eta))/Sc-F(eta)*(diff(Y(eta), eta))+(1/2)*Y(eta)*(diff(F(eta), eta))-s*(2*Y(eta)+(1/2)*eta*(diff(Y(eta), eta)))+Nt*(diff(H(eta), eta, eta))/Nb = 0;

IC1 := F(0) = 0, (D(F))(0) = 0, G(0) = gm, H(0) = 1, Y(0) = 1; IC2 := (D(F))(L) = 0, G(L) = 1-gm, (D(G))(L) = 0, H(L) = 0, Y(L) = 0; dsys1 := {Eq1, Eq2, Eq3, Eq4, IC1, IC2}; dsol1 := dsolve(dsys1, numeric, output = listprocedure, range = 0 .. L);

dsol1f := subs(dsol1, F(eta));

dsol1g := subs(dsol1, G(eta)); dsol1h := subs(dsol1, H(eta)); dsol1y := subs(dsol1, Y(eta));

With my best regards and sincerely.

Hi, there

How can I find the recurrence relation  for second derivative of sequence of functions  f-{n}(x)=\frac{(1-x^2)^n}{n!} in  maple 15?

please specify the commands.

we know the solution f"_{n}(x)=2(1-2n)f_{n-1}(x)+4f_{n-2}(x)

Regards

M.R. Yegan

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