Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

Is it possible to export data from Maple to Excel where data are generated by Maple using Optimization:-LSSolve or DirectSearch:-SolveEquations function?

I have attached a file, where n is the primary variable. For n=1, 2, 3, .......... 20, I have got solution individually. But it is more time consuming event. Is it possible to use array or a vector for the variable n and solve equations at a time for all value of n and stored a excel file?
Any advice for me? @Preben Alsholm , @tomleslie or anybody
system_equations_(10.12.2008).mw
 

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?

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.

Edit: The worksheet Extrapolant-ver2.mw presents a slightly different version of the previous result. Here the square's corners are treated symmetrically, leading to a more pleasing interpolating function.

PS: A challenge.  Extend the result to 3D, that is, construct a function u(x,y,z) on the cube [−1,1]×[−1,1]×[−1,1] which matches prescribed functions on the cube's six faces.

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

Hello,

Please , how can we plot two figures with different X-axis and the same Y-axis ?

Thank you

 

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