Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

I've made calculation as array := function(array1,array2). Then summ of all array[i] all in one cycle. If I look in summ variable I see fu(0+fu() +.. agebraical representation. I tried evalf(summ). still algebraical. How to derive numerically?

how can the graphic 3D as .EPS

f(x,y)=sin(x)*cos(x)

I want to calculate the following summation for a,b;

 

when it is odd, 

vanishes.

MY TRY:

restart:
J:=proc(a,b,c) local u,v,p,w;
p:=(a+b+c):
if p=2*k then
u:=b+c-a: 
v:=c+a-b: 
w:=a+b-c:
return (-1)^p*sqrt(binomial(2*u,u)*binomial(2*v,v)*binomial(2*w,w)/((2*p+1)*binomial(2*p,p)))
else 
return 0
end if
end proc:
aa:=2:bb:=4:
sum(J(aa,bb,cc)*J(aa,bb,cc)*sqrt(2*(2*aa+1)*(2*bb+1)*(2*cc+1))*psi[1,cc],cc=abs(bb-aa)..(aa+bb));

 

Hi
I am confused by solve and SolveEqutions and their options!
Solving same equation results different solutions.
Which  solution is correct and why?
How i can be sure one of the solutions is correct about the other equations?
I attached my code.

Thanks.

Test2_MaplePrimes.mw

Its difficult for me to explain what I'm looking for so I will try with an example:

Example 1

Is there anyway any function that will say for example the sum of squares of integers for a given n say n=4, 1^2 + 2^2 + 3^2 + 4^2 = 30 that will give the equation ((n^3)/3) + ((n^2)/2) + (n/6) that gives the sum for a given n?

It would be determined from the series 1^2 + 2^2 + 3^2 + 4^2 + ... or the equation for the elements 1^2, 2^2, 3^2, 4^2 the equation that gives the sum; in this case the well known ((n^3)/3) + ((n^2)/2) + (n/6) .

 

So far I've got FindSequenceFunction[{1, 4, 9, 16}, n] but this only finds the individual elements not an equation for the sum of them. Of course I already know the equation for producing the individual elements.

How can plot  f := proc (x, y) options operator, arrow; sin(x)*cos(y) end proc  as surface and contour (contour is project on surface)

Hi

I want to solve quadratic eqution involving more than 2 parameters...want to analize unique soltuions and real roots also want to plot the real and unique region on graphs ....thanx

the equation is -delta*(Q*S*alpha*b-a*alpha^2+M*c+b*delta) where alpha is the varible and rest are paramters 

 

Hei!

In my Maple project I want to put a simple sin graph on top of an image. But how do I create a background image with a graph on top?


 

restart

PI := proc (p, e, q) options operator, arrow; p*S(p, e, q)+v*(q-S(p, e, q))-w*q-g(e)+(w-c)*q end proc

proc (p, e, q) options operator, arrow; p*S(p, e, q)+v*(q-S(p, e, q))-w*q-g(e)+(w-c)*q end proc

(1)

S := proc (p, e, q) options operator, arrow; q-(int(F(x), x = 0 .. q)) end proc

proc (p, e, q) options operator, arrow; q-(int(F(x), x = 0 .. q)) end proc

(2)

NULL

PI(p, e, q)

p*(q-(int(F(x), x = 0 .. q)))+v*(int(F(x), x = 0 .. q))-w*q-g(e)+(w-c)*q

(3)

g := proc (e) options operator, arrow; (1/2)*mu*e^2 end proc

proc (e) options operator, arrow; (1/2)*mu*e^2 end proc

(4)

``

F := proc (x) options operator, arrow; int(f(x), x = 0 .. q) end proc

proc (x) options operator, arrow; int(f(x), x = 0 .. q) end proc

(5)

P := proc (p, e, q) options operator, arrow; simplify(eval(PI(p, e, q), [alpha = 50, w = 10, mu = 10, c = 5, v = 1, f(x) = 1/2])) end proc

proc (p, e, q) options operator, arrow; simplify(eval(PI(p, e, q), [alpha = 50, w = 10, mu = 10, c = 5, v = 1, f(x) = 1/2])) end proc

(6)

Diff_p := diff(P(p, e, q), p)

q-(1/2)*q^2

(7)

Diff_e := diff(P(p, e, q), e)

-10*e

(8)

Diff_z := diff(P(p, e, q), q)

-p*q+p+q-5

(9)

``

``


 

Download Profit_code.mw

 

 

 

why i can not evaluate 29 polynomial. maple try to evaluate last 7hr, how many time required too solve it?

 

How to change the numbers shown in the diagram shapes from y=0.5- to rigth form y=-0.5a.mw

Download a.mw


 

 

 

 

 

 

The attached worksheet develops a procedure for extrapolating boundary data from a square to its interior.  Specifically, let's consider the square [−1,1]×[−1,1] and the continuous functions u1, u2, u3, u4 defined on its edges. The procedure constructs a continuous function u(x,y) in the interior of the square which matches the boundary data.

The function u(x,y) is necessarily discontinuous at a corner if the boundary data of the edges meeting at that corner are inconsistent.  However, if the boundary data are consistent at all corners, then u(x,y) is continuous everywhere including the boundary.

Here is what the extrapolating function looks like:

proc (x, y) options operator, arrow; (1/2)*(1-y)*(u__1(x)-(1/2)*(1-x)*u__1(-1))+(1/2)*(x+1)*(u__2(y)-(1/2)*(1-y)*u__2(-1))+(1/2)*(y+1)*(u__3(x)-(1/2)*(x+1)*u__3(1))+(1/2)*(1-x)*(u__4(y)-(1/2)*(y+1)*u__4(1))+(1/4)*(u__1(-1)-u__4(-1))*(-x^2+1)*(1-y)/((x+1)^2+(y+1)^2)^(1/2)+(1/4)*(u__2(-1)-u__1(1))*(-y^2+1)*(x+1)/((y+1)^2+(1-x)^2)^(1/2)+(1/4)*(u__3(1)-u__2(1))*(-x^2+1)*(y+1)/((1-x)^2+(1-y)^2)^(1/2)+(1/4)*(u__4(1)-u__3(-1))*(-y^2+1)*(1-x)/((1-y)^2+(x+1)^2)^(1/2) end proc

 

The worksheet Extrapolant.mw contains the details of the derivation, and an example.

Top part of worksheet gives me a and b; lower part does not

Please tell me where I go wrong. Thanks

Test1.mw


 

restart

p := (alpha+c*(beta-k^2/mu)+z-Lambda(z))/(2*beta-k^2/mu)

(alpha+c*(beta-k^2/mu)+z-Lambda(z))/(2*beta-k^2/mu)

(1)

x := simplify((p-c)/(p-v))

mu*(beta*c+Lambda(z)-alpha-z)/(Lambda(z)*mu+((-c+2*v)*beta-z-alpha)*mu+k^2*(c-v))

(2)

``

xx := p-c-(p-v)*x

(alpha+c*(beta-k^2/mu)+z-Lambda(z))/(2*beta-k^2/mu)-c-((alpha+c*(beta-k^2/mu)+z-Lambda(z))/(2*beta-k^2/mu)-v)*mu*(beta*c+Lambda(z)-alpha-z)/(Lambda(z)*mu+((-c+2*v)*beta-z-alpha)*mu+k^2*(c-v))

(3)

Lambda := proc (z) options operator, arrow; int((z-u)*phi(u), u = 0 .. z) end proc

proc (z) options operator, arrow; int((z-u)*phi(u), u = 0 .. z) end proc

(4)

``yy := eval(xx, [alpha = 50, mu = 10, c = 5, v = 1, beta = 2, k = 2, phi(u) = 1/2])

100/9+(5/18)*z-(5/72)*z^2-10*(136/9+(5/18)*z-(5/72)*z^2)*(-40+(1/4)*z^2-z)/((5/2)*z^2-544-10*z)

(5)

``

zvalue := solve(yy, z)

z

(6)

yy

100/9+(5/18)*z-(5/72)*z^2-10*(136/9+(5/18)*z-(5/72)*z^2)*(-40+(1/4)*z^2-z)/((5/2)*z^2-544-10*z)

(7)

``


 

Download Dummy_Code.mw

Preliminaries: 

and 

I wrote a code for above. You can Download the code for Above:  

 

---------------------------------------------------------------------------------------------------------------------------------

Now, we can calculate the Matrix  as follows  for k=2, M=3,

---------------------------------------------------------------------------------------------------------------------------------


Question:  How can we find the Matrix  for ?

(If it is not possible to find the Matrix for any k and M,

can we find it for k=3, M=4? )

Before finding Matrix,   we must find to 

in terms of any of Psi_i,j    like

and others.

 

But how? :)
Update:

I found elements of Matrix 
NOW, How to create  Matrix A:=  by the elements in the_code.mw

J:=proc(j1,j2,j3,m1,m2,m3)
local i,f:
i:=max(0,j2-j3-m1,j1-j3+m2):
f:=min(j1+j2-j3,j1-m1,j2+m2):
if m1+m2+m3<>0 then:
0
elif j3>j1+j2 then:
# printf("Error. Does not satisfy the triangle relation");
0
elif j3<abs(j1-j2) then:
# printf("Error. Does not satisfy the triangle relation");
0
elif abs(m1)>j1 or abs(m2)>j2 or abs(m3)>j3 then:
0
else:
simplify(((-1)^(j1-j2-m3))*((((j1+j2-j3)!*(j1-j2+j3)!*(-j1+j2+j3)!*(j1+m1)!*(j1-m1)!*(j2+m2)!*(j2-m2)!*(j3+m3)!*(j3-m3)!)/(j1+j2+j3+1)!))^(1/2)*sum(((-1)^t)/((j1+j2-j3-t)!*(j1-m1-t)!*(j2+m2-t)!*(j3-j2+m1+t)!*(j3-j1-m2+t)!*t!),t=i..f));
end if;
end proc:

###### That is matrix which we want to find ###############
#psi_1aa.psi_1bb:=add(J(aa,bb,c,0,0,0)*J(aa,bb,c,0,0,0)*sqrt(2*(2*aa+1)*(2*bb+1)*(2*c+1))*psi(1,c),c=abs(aa-bb)..(aa+bb));
#psi_2aa.psi_2bb:=add(J(aa,bb,c,0,0,0)*J(aa,bb,c,0,0,0)*sqrt(2*(2*aa+1)*(2*bb+1)*(2*c+1))*psi(2,c),c=abs(aa-bb)..(aa+bb));
#Let A:= Psi.Transpose(Psi). So, it is matrix which we want to find.


aa:=0:bb:=0: 
A(1,1):=add(J(aa,bb,c,0,0,0)*J(aa,bb,c,0,0,0)*sqrt(2*(2*aa+1)*(2*bb+1)*(2*c+1))*psi(1,c),c=abs(aa-bb)..(aa+bb)); 
# it is element of A in first row and column or psi_10.psi_10
                               
aa:=1:bb:=0: 
A(2,1):=add(J(aa,bb,c,0,0,0)*J(aa,bb,c,0,0,0)*sqrt(2*(2*aa+1)*(2*bb+1)*(2*c+1))*psi(c),c=abs(aa-bb)..(aa+bb)); 
# it is element of A in second row and column or psi_11.psi_10 
#We can proceed so so for aa=0..M-1 and bb=0..M-1

#Similarly; Let's find psi_20.psi_20
A(M+1,1):=add(J(aa,bb,c,0,0,0)*J(aa,bb,c,0,0,0)*sqrt(2*(2*aa+1)*(2*bb+1)*(2*c+1))*psi(2,c),c=abs(aa-bb)..(aa+bb));
 # it is element of A in M+1 row and column or psi_20.psi_10

 

Best regards.

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