## view, display and plot question?...

I have been able implement this procedure here.

AreaBetween := overload([proc(fnc::{list}, a::anything, b::anything, col::list) option overload; plottools:-transform(unapply([x, y + fnc[2]], x, y))(plot([fnc[1]] + [-fnc[2]], x = a .. b, col = [col[2], col[3]], gridlines, size = [600, 600], thickness = 3, filled = [color = col[1]])); end proc])

1) How do get to construct a larger canvas? Meaning if I want a xy axis, x = -10..10, y = -10..10 however only x = a..b is suppose to be shaded. Can I do that with my procedure?

2) Only the lower graph is colored currently. How do I get it to color the upper graph?

## fourier transform of erf(x) ?...

Hello;

Maple 2019.2.1 on windows 10

Is there a trick to make inttrans:-fourier(erf(x),x,k) return the Fourier transform of the error function? Now, using Maple 2019.2.1 it returns unevaluated. But direct application of Fourier transform integral does return the correct result. So why inttrans does not work?

 > inttrans:-fourier(erf(x),x,k)

 > 1/sqrt(2*Pi)*int(erf(x)*exp(-I*x*k),x=-infinity..infinity)

 >

## How to select functions from a list...

Hi,

I have a set of functions of the form P :=  { a(t), b(t), f(g(t)), u(v(w(t)), A(a(t), b(t)), ... } and I want to obtain the set Q := { a(t), b(t), g(t), w(t) } (that is the set of the innermost functions)
Since more than an hour I'm fighting without success with select and parmatch and I'm desperatly askink for your help.

PS : the list P only contains functions of t, never of t and of another variables.

## plotting error ...

Any one help me  to remove the error.

I want to plot the curve for different values of alpha.

here is my codes.

restart:
with(linalg):with(plots):
ge[1]:=diff(u[1](x,t),t)=alpha*diff((u[2](x,t)-1)*diff(u[1](x,t),x),x)+(16*x*t-2*t-16*(u[2](x,t)-1))*(u[1](x,t)-1)+10*x*exp(-4*x):
ge[2]:=diff(u[2](x,t),t)=diff(u[2](x,t),x\$2)+alpha*diff(u[1](x,t),x)+4*(u[1](x,t)-1)+x^2-2*t-10*t*exp(-4*x):
bc1[1]:=u[1](x,t)-1:
bc1[2]:=u[2](x,t)-1:
bc2[1]:=3*u[1](x,t)+diff(u[1](x,t),x)-3:
bc2[2]:=5*diff(u[2](x,t),x)-evalf(exp(4))*(u[1](x,t)-1):
IC[1]:=u[1](x,0)=1:
IC[2]:=u[2](x,0)=1:
NN:=2:
N:=2:
L:=1:
for i to NN do
dydxf[i]:=1/2*(-u[2,i](t)-3*u[0,i](t)+4*u[1,i](t))/h:
dydxb[i]:=1/2*(u[N-1,i](t)+3*u[N+1,i](t)-4*u[N,i](t))/h:
dydx[i]:=1/2/h*(u[m+1,i](t)-u[m-1,i](t));
d2ydx2[i]:=1/h^2*(u[m-1,i](t)-2*u[m,i](t)+u[m+1,i](t)):od:
for i to NN do bc1[i]:=subs(diff(u[1](x,t),x)=dydxf[1],
diff(u[2](x,t),x)=dydxf[2],u[1](x,t)
=u[0,1](t),u[2](x,t)=u[0,2](t),x=0,bc1[i]):od:
for i to NN do bc2[i]:=subs(diff(u[1](x,t),x)=dydxb[1],
diff(u[2](x,t),x)=dydxb[2],u[1](x,t)
=u[N+1,1](t),u[2](x,t)=u[N+1,2](t),x=L,bc2[i]):od:
for i to NN do eq[0,i]:=bc1[i];eq[N+1,i]:=bc2[i]:od:
for i from 1 to N do eq[i,1]:=diff(u[i,1](t),t)= subs(diff(u[1](x,t),x\$2) =
subs(m=i,d2ydx2[1]),
diff(u[2](x,t),x\$2) = subs(m=i,d2ydx2[2]),diff(u[1](x,t),x) =
subs(m=i,dydx[1]),diff(u[2](x,t),x) = subs(m=i,dydx[2]),u[1](x,t)=u[i,1](t),
u[2](x,t)=u[i,2](t),x=i*h,rhs(ge[1])):od:

for i from 1 to N do eq[i,2]:=diff(u[i,2](t),t)= subs(diff(u[1](x,t),x\$2) =
subs(m=i,d2ydx2[1]),
diff(u[2](x,t),x\$2) = subs(m=i,d2ydx2[2]),diff(u[1](x,t),x) =
subs(m=i,dydx[1]),diff(u[2](x,t),x) = subs(m=i,dydx[2]),u[1](x,t)=u[i,1](t),
u[2](x,t)=u[i,2](t),x=i*h,rhs(ge[2])):od:

for i to NN do u[0,i](t):=(solve(eq[0,i],u[0,i](t))):od:

for i to NN do u[N+1,i](t):=(solve(eq[N+1,i],u[N+1,i](t))):od:

h:=L/(N+1):

for i from 1 to N do eq[i,1]:=eval(eq[i,1]):od:
for i from 1 to N do eq[i,2]:=eval(eq[i,2]):od:

eqs:=seq(seq((eq[i,j]),i=1..N),j=1..NN):
Y:=seq(seq(u[i,j](t),i=1..N),j=1..NN):

ICs:=seq(u[i,1](0)=rhs(IC[1]),i=1..N),seq(u[i,2](0)=rhs(IC[2]),i=1..N):

sol:=dsolve({eqs,ICs},{Y},type=numeric,stiff=true,maxfun=1000000,abserr=1e-6,relerr=1e-5,output=listprocedure):

Warning, The use of global variables in numerical ODE problems is deprecated, and will be removed in a future release. Use the 'parameters' argument instead (see ?dsolve,numeric,parameters)
for j to NN do for i to N do U[i,j]:=subs(sol,u[i,j](t)):od:od:

for i to NN do U[0,i]:=subs(u[1,1](t)=U[1,1],u[1,2](t)=U[1,2],
u[2,1](t)=U[2,1],u[2,2](t)=U[2,2],u[0,i](t)):od:
for i to NN do U[N+1,i]:=eval(subs(u[N,1](t)=U[N,1],u[N,2](t)=U[N,2],
u[N-1,1](t)=U[N-1,1],u[N-1,2](t)=U[N-1,2],u[N+1,i](t))):od:
tf:=1.:
M:=30:
T1:=[seq(tf*i/M,i=0..M)]:
PP:=matrix(N+2,M+1):
for i from 1 to N+2 do PP[i,1]:=evalf(subs(x=(i-1)*h,rhs(IC[1]))):od:
for i from 1 to N+2 do for j from 2 to M+1 do
PP[i,j]:=evalf(subs(t=T1[j],U[i-1,1](t))):od:od:

G1:=[seq([ seq([(i-1)*h,T1[j],PP[i,j]], i=1..N+2)], j=1..M+1)]:
t=0.02:
pars:=[0.1,0.5,1,2,5];
clr:=[black,red,green,gold,blue];
for m from 1 to 5 do
G1[m]:=plot([seq(subs(alpha=pars[m],G1[i])],i=0..N+1)],thickness=3,color=clr[j]):od:

display({seq(G1[i],i=1..5)},title="Figure ",axes=boxed,labels=[x,u]);
restart:

Eksamensopgaver_samlet.mw

Hey

Can anyone help me with this file?

I don't know how it happend, but the file somehow got corrupted. Is there anything i can do?

Thanks

## s0lve diff(U, x, t) = U by admion decomposition ...

s0lve  diff(U, x, t) = U by admion decomposition method

## How to set a equation like Diff or diff as symbol?...

if there is a equation subs(...)

how to make a similar symbol like Diff(f(t),t) and diff(f(t),t) use for a new solver procedure use?

## Code in Proc Command ...

Hello Everyone;
I want to make this code complete in proc envirement (Proc command in maple). Kindly help me. Thanks in advance.

## a math question...

There are 10 questions. each question has one negetive point and three positive points. what is the number of students so that there will not be any repeatitive grades so all the students have distinct grades. Thanks in advance

## Code conversion From Python...

Hello Everyone,

I am new to Maple and I have designed an algorithm to compute multivariate differential dimension Grobner bases. Now I want to add this Maple as a package but I do not how to do that.

Thank you so much for your help.

My code can be found at https://github.com/Gsparsh/Grobner-bases

## How to generate a vector by calculating the entrie...

Hi,

I am a newbie and therefore have a rather simple question for the community:

By reading in my measurement data (topographic data) I have generated column vectors:
for i from 1 to k do
B[i] := Vector(A[1 .. k, i]);
end do:
where k is a predefined rank of the original square matrix. The arithmetic mean was then calculated for each individual column vector:
for i from 1 to k do
Am[i] := add(B[i][n], n = 1 .. k)/k;
end do:

Now I try to create new column vectors by subtracting the corresponding mean value from each entry of the original vectors, something like that:
for i from 1 to k do
C[i]:=B[i][n]-Am[i],n=1..k

But I just can't get it right. Can anyone help me? Thanks in advance!

Cheers
Christian

## How to make entries in the last row to be the sam...

For example, the column dimension is 6 and I ran the following command,

A := Mod(2,Matrix(3,6,(i,j)->if i<2 and j < 3 then 1  elif i = 2 and j > 2 and j <5 then 1 else 0 end if),float[8])

the result is the following matrix with the last row containing all 0s.

A := Matrix(3, 6, [[1., 1., 0., 0., 0., 0.], [0., 0., 1., 1., 0., 0.], [0., 0., 0., 0., 0., 0.]])
Actually, I want the last row containing all 1s but I do not know how.
Any help?

## i want to write a expression without any quotes...

i want to write a expression without any quotes . for example in this case i want to wrate E*1    E*2

## Error, (in dsolve/numeric/process_input) system mu...

I can remove this error.?