Example

Diff(x1,t) = x2^2,
Diff(x2,t) = u,
y = x1

result in

Diff(y,t$2)-4*u^2*Diff(y,t) = 0
2*u*x2 - Diff(y,t^2) = 0
x1 - y = 0

is it possible a general and generic method to apply some steps on system below

Diff(u1, t) = u1*u2-u1*u3,
Diff(u2, t) = -u1*u2+u2*u3-(1/2)*(conjugate(u2-u3)*(u1*u2-u1*u3)+conjugate(u1)*(-u1*u2+u2*u3)-conjugate(u1)*(u1*u3-u2*u3))/conjugate(u1),
Diff(u3, t) = u1*u3-u2*u3+(1/2)*(conjugate(u2-u3)*(u1*u2-u1*u3)+conjugate(u1)*(-u1*u2+u2*u3)-conjugate(u1)*(u1*u3-u2*u3))/conjugate(u1),
y = u1

Here is the question:Prior to this question I was given f(z)=z^2+1, N(z)=(z^2-1)/(2z), T(z)=z-I/z+I such that T(N^k (z))=(T(z))^2^k. And L is a set of number on the real axis. Now the question is that given we have two regions of the complext plane as follow:

R₊ = {z : N^k{z) -> i as k -> ∞}; R_- = {z : N^k(z) -> -i as k -> ∞}.

Draw a diagram to illustrate these regions, the line L and the roots i and -i. We call R₊ the basin of attractionfor the root +i, and similarly R_-is the basin of attraction for the root -i.

 Show that if z is on the set L (the common boundary of the two regions R₊ and R_{_}, then N^k(z) stays on L for all values of k. (This is easy once you identify what L is.) So in this case iteration does not produce a root at all.

So basically my problem is that the fact I'm not very familar with the commands to draw such diagram, and I don't know much about Newton's method to compute complex roots. It would be appreciated if anyone can help me how to get start with the question. Thanks.

The dot product of vector v and u= ||v|| ||u||fosters also vector u*v=XvXu+Yv=A, v=<2,5>, u=<4,9>.

Dear users

A friend of of mine has a problem with an integral and since it's for his thesis, it's pretty important. 

That's why I ask it here cause I don't know where to ask it elsewhere, so if it's wrong posted, completely my bad.

l:=(y*o)/(v);

R:=(Phi*o)/v;

A:=5*(a*ln(R)+b);

P:=sqrt(1+4*k^2*l^2*(1-exp^(-l/a)));

M:=int((2)/(1+P),o);

With other words, I want to integrate the wole thing to the variable o, who appears in the variables l and R.

Somehow, when I put this in Maple, it won't solve it. Probably it's just a stupid fault or i just forget something, but i don't find it. Does anyone knows how to solve it?

Already a lot of thanks!

[> restart;
[> OPTIMAL:=proc(f,Condition::set)
local vars,L,S,SS,k,m,K,Fmax,Fmin,sol_min,sol_MIN,sol_max,sol_MAX;
vars:=indets(f,name) union indets(Condition,name);
L:=map(convert,Condition,equality);
K:=combinat[choose](L);
S:=NULL:
for k to nops(k) do if solve(K[k])<>NULL then
extrema(f,K[k],vars,'s||k'):S:=S,a||values(s||k):fi od:
SS:={}
for k to nops([s]) do
if type(s[k],set) then SS:=SS union evalc(s[k]) fi od;
SS:=remove(has,SS,l);
S:=NULL:
for k to nops(ss) do if type(SS[k],set(equation)) and
map(evalb,evalf(simplify(subs(SS[k],Condition))))={true}
then S:=S,vatue(ss[k]) fi od;
SS =[s];
Fmax:=-infinity:
for k to nops(SS) do
if evalf(simplify(subs(ss[k],f)))>evalf(Fmax)
then Fmax:=simplify(value(subs(SS[k],f)));sol_max:=SS[k]; fi;od;
sol_MAX:=sol_max:
for k to nops(SS) do
if Fmax=simplify(value(subs(SS[k],f)))
and (SS[k] minus sol_max)<>{} then
sol_MAX:=sol_MAX,Ss[k] fi od;
‘Fmax'=simplify(expand(Fmax)),sol_MAX:
Fmin:=infinity:
for k to nops(SS) do
if evalf(simplify((subs(SS[k],f))))<evalf(Fmin)
then Fmin:=simplify((subs(SS[k],f)));sol_min:=SS[k]; fi;od;
sol_MIN:=sol_min:
for k to nops(SS) do
if Fmin=simplify(value(subs(SS[k],f)))
and (SS[k] minus sol_min)<>{} then
sol_MIN:=sol_MlN,SS[k] fi od;
RETURN('F[min]'=simplify(expand(Fmin)),sol_MlN,
'F[max]'=simplify(expand(Fmax)),sol_max);
end: ↵

" Error, unable to match delimiters"
  Please, help me.
I don't understand. Thank you, thank you.

Does someone know how to ask Maple to compute intersection and union of ranges (for that matter Realrange)?

if I find fourier intergra solution if 1/pi∫ （form infintiy to 0）(a*cos(u*x)+u*sin(u*x))/a^2+u^2 du. can you polt the function to the integral converges? and can you plot the fourier intergral on a=1 and -1<=x<=3 with 0<=u<=20?

Hi

Is there any way to program a button component to stop current calculations and restart maple server?
Thank you.

If I type the following:

a := 1

a2 := 1.5

a*a2

When I press enter

I want it to display the expression without evaluating and then the result:

(1)*(1.5)

By default the program displays  1.5.  Is it possible to display it this way?

Please have a visit of this qestion `http://math.stackexchange.com/q/574843/8581`. I did a very elementary attempt there for example for a function:

 > with(plots):
      h := x->piecewise(x < -2, x+3, x <= 2, 5-x^2, 3-x):
      t:=x->h(-x):
      a:= plot(h(x), x = 0 .. 10, color = red, thickness = 3):
      b:= plot(t(x), x = -10 .. 0, color = green, thickness = 3):
      display(a,b);

But  I am eager to know  the formal codes if they exists. Thanks for your time.

I was successful in getting the Maple add-in to load into Excel 2013 but the help file accessed in the Add-In Ribbon is not found.  Does anyone know where in Excel that file can be loaded?

Hi, I have encountered a difficult question.

My answer is A=151,B = 47.

Could anyone tell me whether this answer is correct?

The question is as follow:

=∑((120n²+An+B)/(16ⁿ((512n⁴ + 1024n³ + 712n² + 194ⁿ + 15))  (n starts from 0 to infinity)

Thanks in advance.

Good morning sir.

I request your kind support to the above cited question.

With thanks & Regards

M.Anand

Assistant Professor in Mathematics

SR International Institute of Technology,

Hyderabad, Andhra Pradesh, INDIA.

Hi there..

I have a question on how to do a pointplot.

before plotting, I need to know the value of lambda[j] and  all the values of lambda already have.

so now I need to plot a graph with the values of lambda with different range and different colour,

Let say I have

> for j from 17 to 32 do k[j] := j+1;

x[j] := add(P[j, 1], j = j-1 .. j+2);

X[j] := add(P[j, 1]^2, j = j-1 .. j+2);

y[j] := add(P[j, 2], j = j-1 .. j+2);

Y[j] := add(P[j, 2]^2, j = j-1 .. j+2);

xy[j] := add(P[j, 1]*P[j, 2], j = j-1 .. j+2);

cx[j] := evalf(x[j]/k[j]);

cy[j] := evalf(y[j]/k[j]);

c11[j] := evalf(X[j]/k[j]-cx[j]^2);

c22[j] := evalf(Y[j]/k[j]-cy[j]^2);

c12[j] := evalf(xy[j]/k[j]-cx[j]*cy[j]);

C[j] := evalf(Matrix(2, 2, [[c11[j], c12[j]], [c12[j], c22[j]]]));

E[j] := simplify(fnormal(LinearAlgebra[Eigenvalues](C[j])));

if E[j][1] > E[j][2] then lambda[j] := E[j][2]/(E[j][1]+E[j][2]) else lambda[j] := E[j][1]/(E[j][1]+E[j][2])  end if;

lambda[j];

 end do;

the range of lambda [j] are as follows:

 0.02< lambda [j]<0.06

 0.06< lambda [j]<0.12

 0.12< lambda [j]<0.18

for i from 17 to 32, do if   0.02< lambda [j]<0.06 then green[i]:=P[i,j]; j:=i+1 elif

 0.06< lambda [j]<0.12 then red[i]:=P[i,j];j:=i+1 ; elif 0.12< lambda [j]<0.18 then blue[i]:=P[i,j];j:=i+1 end if;end do

how to do a point plot with the above situation so that in my plotting all the information are on the same graph.

All help is greatly appreciated.

Thanks

Im trying to plot the different values of shanks as points in this ] loop.

pp:=0.0:sumn:=0:sumnm1:=0:sumnp1:=0:

for i from 1 to 60 do

if(n>2) then sumnm1:=sumn fi;

if (n>1) then sumn:=sumnp1 fi; p:=1/evalf(2*beta*BesselJ(2,zeros[i])+epsilon*zeros[i]*BesselJ(0,zeros[i])+zeros[i]*BesselJ(1,zeros[i])); pp:=pp+p;

sumnp1:=pp;

if (n>2) then shanks:=(sumnp1*sumnm1-sumn*sumn)/(sumnp1-2*sumn+sumnm1) fi;

shanks;

od;

sumnp1: -2*sumn+sumnm1:

I have tried plotting it but it only plots the last value. 

Many thanks James