Maple does not simplify the expression A*[sin(x)]^2+A*[cos(x)]^2=A, where A is any expression.

How can I make it do so? 

Thanks,

Maple 17 64-bit on Windows 8

Hi!

Let F(z) (with z complex) a given function. I want to compute F^n(z0), i.e. the composition of F with itself n-times, where z0 is a given point (complex).

Is correct the following procedure to compute F^k(z0)?

App := proc (k, z0) local z1, z2, j; z1 := z0; z2 := NULL; for j to k do z2 := F(z1); z1 := z2 end do; return z2 end proc

Many thanks in advance for your comments.

Hello!! Please help me,I need to solve a system of linear algebraic equations by running, and I solved the built-in command solve

restart;
with(plots):
f:=unapply(-x^2+1,x);
mu[1]:=unapply(1/(t^2+1),t);
mu[2]:=unapply(1/(t-5),t); 
g:=unapply(t^3-7*x,[t,x]);
l:=2; T:=3;
n:=10: m:=n: 
h:=evalf(l/n); 
tau:=evalf(T/m);
for k from 0 to n do 
x[k]:=h*k:
end do: 
for j from 0 to m do 
t[j]:=tau*j: 
end do:
ss:=evalf({seq(seq((y[k,j+1]-y[k,j])/tau=(y[k-1,j]-2*y[k,j]+y[k+1,j])/h^2+g(t[j],x[k]),k=1..n-1),j=0..m-1),seq(y[0,j]=mu[1](t[j]),j=1..m),seq(y[k,0]=f(x[k]),k=0..n),seq(y[n,j]=mu[2](t[j]),j=1..m)});
#s:=evalf(solve(ss,{seq(seq(y[k,j],k=0..n),j=0..m)}));

Good evening!!! I have a task to implement the task of Cauchy by the method of Milne, wrote the code, but did not understand it until the end, help to understand? what's wrong?
First calculate four "initial" values by the method of Runge-Kutta methods, then use the method of Milne, the Fact that two times running, perhaps extra?

restart;
with(plots):
a:=0; b:=1; eps:=evalf(10^(-3)):
f:=unapply(2*x*(x^2+y),x,y);
G:=simplify(dsolve({diff(y(x),x)=f(x,y(x)),y(a)=1}));                    
N:=15: h:=(b-a)/N:
for i from 0 to N do 
x[i]:=a+i*h: 
end do:
y[0]:=1;
s[0]:=1;
for i from 0 to 2 do 
t[1]:=evalf(h*f(x[i],y[i])):
t[2]:=evalf(h*f(x[i]+h/2,y[i]+t[1]/2)): 
t[3]:=evalf(h*f(x[i]+h/2,y[i]+t[2]/2)):
t[4]:=evalf(h*f(x[i]+h,y[i]+t[3])):
y[i+1]:=evalf(y[i]+(t[1]+2*t[2]+2*t[3]+t[4])/6):
q[1]:=evalf(h*f(x[i],s[i])):
q[2]:=evalf(h*f(x[i]+h/2,s[i]+q[1]/2)): 
q[3]:=evalf(h*f(x[i]+h/2,s[i]+q[2]/2)):
q[4]:=evalf(h*f(x[i]+h,s[i]+q[3])):
s[i+1]:=evalf(s[i]+(q[1]+2*q[2]+2*q[3]+q[4])/6):
end do;
for i from 3 to N-1 do 
y[i+1]:=evalf(y[i-3]+((4*h)/3)*(2*f(x[i],y[i])-f(x[i-1],y[i-1])+2*f(x[i-2],y[i-2]))):
s[i+1]:=evalf(s[i-1]+(h/3)*(f(x[i+1],y[i+1])+4*f(x[i],s[i])+f(x[i-1],s[i-1]))):
d[i+1]:=abs(y[i+1]-s[i+1])/29:
if abs(d[i+1]) < eps then y[i]:=y[i]:
else y[i]:=s[i];
end if: end do;
s1:=plot(rhs(G),x=a..b,color=yellow):
s2:=pointplot({seq([x[k],y[k]],k=0..N)}): 
display(s1,s2);

I want to write a procedure called Resistance which calculates and displays the equivalent resistance to three resistors R1, R2 and R3

If the resistances are connected in series then Rser = R1 + R2 + R3

If the resistances are connected in parallel then (R1 * R2 * R3) / (R1 * R2 + R1 * R3 + R2 * R3) and after that, I must write an algorithm which I will test this resistance but it does not work please help me
 

Download RESISTANCE.mw

Good day.

I have a problem with the numerical approximation of incomplete elliptic integrals of the first kind in the case when the argument and the modulus are complex numbers. 

Let's consider an incomplete elliptic integral of the first kind EllipticF((a+I*b)*x, c+I*d), where a,b,c,d are arbitrary real numbers and x is very large number. Here is the code:

restart;
R0 := EllipticF((a+I*b)*x, c+I*d);
with(MultiSeries, series):
R1 := series(R0, x = infinity, 2);
a := .5; b := 1.75; c := 10; d := 12.5; x := 10^6;
evalf(R0); evalf(R1);

Below you can see that answers are different

.1680611942+.2058775337*I
-0.4563467782e-1-.3592094173*I+O(1/1000000000000000000)

While the same code for an incomplete elliptic integral of the second kind gives the correct approximation:

restart;
R00 := EllipticE((a+I*b)*x, c+I*d);
R11 := series(R00, x = infinity, 2);
a := .5; b := 1.75; c := 10; d := 12.5; x := 10^6;
evalf(R00); evalf(R11);
-1.687498740*10^7+2.374999011*10^7*I
-1.687501260*10^7+2.375000989*10^7*I+O(1/1000000)

What is the difference?
 

I want to visualize data provided by the Maximize command. The output I get is of the following form (example):

 [.358700275060090779, [p[0] = .192413186080606, p[1] = 0.906594292940704e-1, p[2] = 0.677108912885780e-1, p[3] = 0.609551830556988e-1, p[4] = 0.589744573790909e-1, p[5] = 0.585737058072817e-1, p[6] = 0.589744573787748e-1, p[7] = 0.609551830550955e-1, p[8] = 0.677108912877626e-1, p[9] = 0.906594292931833e-1, p[10] = .192413186079858]]
I want to plot the sequence p[0], ..., p[10] with the command LineChart (for example).

My question is: How to automatically "clean up" output as above so I can feed it to LineChart?

Hello,

I've used maple to solve a bunch of differential equations symbolically and the equations are pretty massive and so I don't particularly want to type by hand again on MS Word so if someone can tell me how to copy to word so it appears in an editable equation format I'll be very grateful.

Kind Regards

Ben

When I try to calculate the derivative of a covariant metric with respect to the corresponding contravariant metric, or vice versa, the result is correct up to the sign, which is wrong:

diff(g_[nu, tau], g_[~mu, ~eta]);

Maple's result: g_[eta, nu] g_[mu, tau]

Correct result: -g_[eta, nu] g_[mu, tau]

diff(g_[~nu, ~tau], g_[mu, eta]);

Maple's result: g_[~eta, ~nu] g_[~mu, ~tau]

Correct result: -g_[~eta, ~nu] g_[~mu, ~tau]

I've loaded DifferentialGeometry, Tensor, Physics. Is this my fault, or Maple's?

hi

i got a problem to solve 2 nonlinear  pde equation that are not coupled but there is 8 coupled boundary condition. i can use pdsolve in maple.anyone can help me in numerical solving of these pdes?

Can Maple perform huge calculations via parallel-pooling (i.e working on multiple or all CPUs) similar to Matlab?

Hi!

I want to plot the approximation of a surface by polynomials. The surface is given by (x,y,f(x,y)) where f(x,y) is given by the following expression

proc (x) options operator, arrow; (sum(i*cos((i+1)*(-2+4*x[1])+i), i = 1 .. 5))*(sum(i*cos((i+1)*(-2+4*x[2])+i), i = 1 .. 5)) end proc

with both variables varying in the interval [0,1]. Then, by using the Bernstein polynomials of two variables (see, for instance, this paper for details  https://www.sciencedirect.com/science/article/pii/0021904589900956), the graph of the resulting (plot3d) surface (x,y,p(x,y))  it is not even like to the original surfaces.

Please, see this PDF of what I have done:  plots.pdf

Some idea or suggestion?

Thanks!

I am trying to create an array whose elements are from another matrix. Suppose I have a matrix of n cross m dimension my array will have a dimension of 1 cross n*m. When I create  array I am getting error. The array contains only last element of matrix. The expected answer is given in attachment.

sigma.mw

im trying to input this variable with value structure

> xxx := x[2, 0], x[2, 1], x[2, 2], x[1, 0], x[1, 1]

check the type

> whattype(xxx);
exprseq

but when i try to use Get tools from maplets package, its give me error

> xxx:=Get`('xxx1'::exprseq)
invalid argument(s): xxx1::exprseq

is it possible to passing exprseq from maplet input into procedure?
or is there another way to input it (xxx) ?

You are please to check maple file:    robust_rev1.mw

case close, many thanks @tomleslie and @vv

Hello.

Can you please tell how the guess vector is defined in Newton's method in "fsolve" if not set initial interval for unknowns? Maybe someone knows what "norm of errors", "new norm" and "incr" in "infolevel[fsolve]" are?

Thanks!