MaplePrimes Questions

How to find inverse function of a multivariable function

for example

f := x^2 + y + z^3

f := x^2 + y^3

Gentlepeople.

Given 6 teams [A,B,C,D,E,F] I require that each team plays each other once.

 I can generate the pairs:

lst:=combinat:-choose({A,B,C,D,E,F}, 2);  

{{A, B}, {A, C}, {A, D}, {A, E}, {A, F}, {B, C}, {B, D}, {B, E}, {B, F},  {C, D}, {C, E}, {C, F}, {D, E}, {D, F}, {E, F}}

On a competion night three pairs will compete and it will take 5 nights for all teams to play each other.

now I need some eredite richard who can show me how to produce the five night output:

(instead of comma a 'v' needs to be mapped)

AvF

BvD

CvF

.....

 

I am trying to invert a function and, during the course, wind up with something like,

 

a_1*V^P*V + a_2*V^P*V^2 + a_3*V^P*V^3 + a_4*V^P*V^3 + a_5*V^P*V^3 + a_6*V^P*V^4 + \cdots

 

is there a way to get the poly into a form 

 

(a_1+a_2+...)*V^P + (b_1+b_2+...)*V^(P+1) + (c_1+c_2+...)*V^(P+2) + .....

 

?

Thanks. Attached is my Maple 16 sheet.

restart

with(LargeExpressions)

``

main := V^P*(V-1)^Q = c^(P-Q)*X^Q*(V-c^2)^Q

V^P*(V-1)^Q = c^(P-Q)*X^Q*(V-c^2)^Q

(1)

``

assume(Q, real); additionally(Q > 0); additionally(Q, integer)

assume(P, real); additionally(P > 0); additionally(P, integer)``

NULL

``

you := proc (s, n) V[1] := s; for i to n do V[i+1] := (`@`(simplify, expand))(c^((P-Q)/P)*X^(Q/P)*(-c^2+V[i])^(Q/P)/(V[i]-1)^(Q/P), symbolic) end do end proc

proc (s, n) local V, i; V[1] := s; for i to n do V[i+1] := (`@`(simplify, expand))(c^((P-Q)/P)*X^(Q/P)*(V[i]-c^2)^(Q/P)/(V[i]-1)^(Q/P), symbolic) end do end proc

(2)

``

``

START*HERE

START*HERE

(3)

``

Rearranging "main" leads to the following.

``

main2 := X^Q = V^P*(V-1)^Q/(c^(P-Q)*(V-c^2)^Q)

X^Q = V^P*(V-1)^Q/(c^(P-Q)*(V-c^2)^Q)

(4)

``

It can't take a series due to the series about zero, so I split it into

``

split := X^Q = V^P*(a[1]+a[2]*V+a[3]*V^2+O(V^3))

X^Q = V^P*(a[1]+a[2]*V+a[3]*V^2+O(V^3))

(5)

````

Then I'll get a polynomial expression, switch places of the two variables' leading terms, and substitute into the "original" polynomial (split).

``

````

  one := (V-1)^Q/(c^(P-Q)*(V-c^2)^Q)

(V-1)^Q/(c^(P-Q)*(V-c^2)^Q)

(6)

``

two := `assuming`([simplify(`assuming`([series(one, V, 3)], [V > c^2, c > 1]))], [real])

series(c^(-Q-P)-(c^(-Q-P-2)*Q*(c^2-1))*V+((1/2)*c^(-Q-P-4)*Q*(-c^4+c^4*Q+1+Q-2*Q*c^2))*V^2+O(V^3),V,3)

(7)

``

three := convert(two, polynom)

c^(-Q-P)-c^(-Q-P-2)*Q*(c^2-1)*V+(1/2)*c^(-Q-P-4)*Q*(-c^4+c^4*Q+1+Q-2*Q*c^2)*V^2

(8)

``

four := simplify(expand(V^P*three))

V^P*c^(-Q-P)-V^(P+1)*c^(-Q-P)*Q+V^(P+1)*c^(-Q-P-2)*Q-(1/2)*V^(P+2)*c^(-Q-P)*Q+(1/2)*V^(P+2)*c^(-Q-P)*Q^2+(1/2)*V^(P+2)*c^(-Q-P-4)*Q+(1/2)*V^(P+2)*c^(-Q-P-4)*Q^2-V^(P+2)*c^(-Q-P-2)*Q^2

(9)

``

five := collect(four, V, factor)

V^P*c^(-Q-P)-V^(P+1)*c^(-Q-P)*Q+V^(P+1)*c^(-Q-P-2)*Q-(1/2)*V^(P+2)*c^(-Q-P)*Q+(1/2)*V^(P+2)*c^(-Q-P)*Q^2+(1/2)*V^(P+2)*c^(-Q-P-4)*Q+(1/2)*V^(P+2)*c^(-Q-P-4)*Q^2-V^(P+2)*c^(-Q-P-2)*Q^2

(10)

collect(five, V)

V^P*c^(-Q-P)-V^(P+1)*c^(-Q-P)*Q+V^(P+1)*c^(-Q-P-2)*Q-(1/2)*V^(P+2)*c^(-Q-P)*Q+(1/2)*V^(P+2)*c^(-Q-P)*Q^2+(1/2)*V^(P+2)*c^(-Q-P-4)*Q+(1/2)*V^(P+2)*c^(-Q-P-4)*Q^2-V^(P+2)*c^(-Q-P-2)*Q^2

(11)

factor(five, V)

Error, (in factor) 2nd argument, V, is not a valid algebraic extension

 

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Download invert2.mw

Hi Maple Prime-ers!

I have a question about efficiency.  I have a set of algebraic equations with some polynomials, that I would like to solve at different points.  I've tried using a for-loop and a map-loop.  Here is a example:

 

n:=10000;  #Number of solving points
eq1:={b = ''a^2'', c = b^3/2, d = c^(1/2)*4 + b^2}; #Equation to solve

a := convert([seq(i,i=1..n)],Vector);  #timesteps

ans := Vector[column](n)

## Try solving in a for-next loop
t1 := time():
for q from 1 to n do
ans(q):=solve(subs({'a' = a(q)},eq1)):
od:
t2 := time() - t1;

## try solving in a map loop
t1s := time():
ans_s := map(q->solve(subs({'a' = a(q)},eq1)),a);
t2s := time() - t1s;

On my computer (2.2Ghz, 2 cores), these both take 115s to solve.  Using Map over For-Next did not speed up computational speed.  

The problem I wish to tackle has 12 equations, invovles 5th order polynomials, and n ~= 300000.  Solving this set of equations takes 2-3 hours.

Anyone know a more efficient method?  Thanks for reading :D

 

Please look at the following set.

restart; Z := {x1*(h1+h3) = 0, b1*x1+b4*x3 = 0, f2*x2+f3*x1 = 0, g2*x2+g4*x3 = 0, d1*x1+d2*x2+d4*x3 = 0}; for i to numelems(Z) do print(Z[i]) end do; "++++++++++++++++"; h3 := -h1; x3 := x2; g2 := -g4; Z; print("The following loop gives wrong result! The last element d1*x1+d2*x2+d4*x2=0 of set Z have disappeared!"); for i to numelems(Z) do print(Z[i]) end do

{x1*(h1+h3) = 0, b1*x1+b4*x3 = 0, f2*x2+f3*x1 = 0, g2*x2+g4*x3 = 0, d1*x1+d2*x2+d4*x3 = 0}

 

x1*(h1+h3) = 0

 

b1*x1+b4*x3 = 0

 

f2*x2+f3*x1 = 0

 

g2*x2+g4*x3 = 0

 

d1*x1+d2*x2+d4*x3 = 0

 

"++++++++++++++++"

 

{0 = 0, b1*x1+b4*x2 = 0, f2*x2+f3*x1 = 0, d1*x1+d2*x2+d4*x2 = 0}

 

"The following loop gives wrong result! The last element d1*x1+d2*x2+d4*x2=0 of set Z have disappeared!"

 

0 = 0

 

b1*x1+b4*x2 = 0

 

f2*x2+f3*x1 = 0

 

0 = 0

(1)

``

``

Why?

Download Set.mw

Notepad displays alt+1, alt+2 and alt+3 as ☺☻♥

Just wondering why Maple displays them as square boxes?  Shouldn't it have the font capability to display them as well?

 what does index mean in maple? for example, if i had RootOf(20+60Z+335Z2+825Z3+1629Z4+2520Z5, index=3) ?

 

 

I'd expect the following to give the result "c+2".

> c := a+b

> simplify(a+b+2)

How can I let Maple know that I'd prefer it to write "c" in place of "a+b" when possible?

Hi! I'm new to Maple. Is there a better way to represent matrices? This is what I've done so far, but the indices (C11, C12, n1, etc) look a little weird and superfluous. Am I doing it the "Maple way"?

 

Examples:

C=Matrix([[C11,C12],[C12,C22]])

 

The result of the following lines in Maple is V := [5., 3.,5. 11., 6.] that I think should be V := [5., 3.,5. 0., 6.], is there somthing wrong?

 

v1:= Vector([-28., -63., -17., -55., 17.], datatype= float[8]):

V:= LinearAlgebra:-Modular:-Mod(11, v1, float[8]);

bug.mw



I have three vectors a, b, c. Knowing that the scalar of product a . b = 1, a. c = 1, b.c=1 and modul of a equal to 1, modul of b equal to 4 and modul of c equal to 4. Now I consider two vectors u = -6*a + b + c and v = 3*a -4*b + c. How to find the angle of two vectors u and v?

Hello, Please how do I compute cdf of student t distribution in maple Tξ+1(.). I have a function that i nvolves student t distribution but finding it difficult to compute student t in the funcion. I am new to maple.

Hi all,

I tried to create the*.exe file by C and call the kernel of maple for calculation.

To understand the OpenMaple, I ran the OpenMaple C code sample, “simple.c” in “<Maple>\ samples\OpenMaple\simple” through Microsoft Visual C++ 2008 but got the incorrect result:

 

It’s seems that the error is happen due to the wrong maple directory. But according to the description of "kernelopts" in help, the value of mapledir cannot be set.

 

Refer to the description of  "OpenMaple,Examples" in help , both header file path and library file path had been set:

O/S: Windows (32-bit)

Header file Directories: "C:\Program Files\Maple 17\extern\include"

Library Directories: "C:\Program Files\Maple 17\bin.win"

Environment Variable: "C:\program files\Maple 17\bin.win"

 

I'd appreciate any help on this topic. Thank a lot.

Dear Maple experts:

I have a very simple question:

I have a polynomial p(x) with integer coefficients and a prime p.  I want to reduce the exponents of this polynomial mod p.  So for example, if p(x) = 1 + x^3 + x^17 + x^19 and p is 17. Then I want Maple to output 1+x^3+1+x^2.

I know how to reduce the coefficients mod p, but not the exponents. Can someone suggest how I might go about doing this.  Thank you very much for your help.

Best wishes

Sunil

if ((a.d-b.c)=0) then print([0,0,0,0])
fi;
MARK:=(a,b,c,d)->[d/(a.d-b.c),-b/(a.d-b.c),-c/(a.d-b.c),a/(a.d-b.c)];

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