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How to find inverse function of a multivariable function

for example

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

f := x^2 + y^3

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

La aplicacion de las matrices en su más claro ejemplo, dirigido explicitamente a la criptografia; solo hay que tener conocimiento de algebra de matrices y calculo de matriz inversa. (versión español).

The application of the matrices in its most obvious example, explicitly directed to cryptography; just have to have knowledge of matrix algebra and inverse matrix calculation. (version english).

 

Criptografia.mw

 

Lenin Araujo Castillo

Physics Pure

Computer Science

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I got a question about the inverse laplace transform of the classic second order system:

Step_Respon=1/(s*((s^2/omega_n^2)+2*zeta*s/omega_n+1))

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Hi,

I have an inverse kinematics model for a double inverted pendulum. There are multiple solutions for each joint angles due to symmetry and I have to choose the appropriate ones. I thought I can use the previous choices and compare them with the current multiple solutions, so I could choose the ones which are closer to the previous ones. However, since the double inverted pendulum moves in in both region at the equilibrium, this aproach does not work. Could you advise...

Hi guys, I'm trying to solve the following function:

 

>with(inttrans):

>invlaplace (15/(s3 + 6s2 + 15s + 15), s, t)

but I'm not getting a viable answer - can maple answer this?

 

I've also simplified the above function a bit to: H(s) = 15/[(s + 2.3222)(s2 + 3.6778s + 6.4594)] and then tried to find it's inverse laplace but still don't get an answer.

inverse of a matrix...

July 08 2012 golnaz 5

Hello every body

I wrote a program and i need to inverse a matrix but i get "error (in linear algebra:- matrix inverse)" singular matrix

can you help me to do that. the matrix is:

Matrix([[1.0000, 0.7056, 0.7965, 0.6928, 0.5941, 0.5169], [0.7056, 1.0000, 0.5653, 0.6449, 0.4601, 0.4376], [0.7965, 0.5653, 1.0000, 0.5521, 0.5526, 0.4153], [0.6928, 0.6449, 0.5521, 1.0000, 0.4271, 0.4242], [0.5941, 0.4601, 0.5526, 0.4271, 1.0000, 0.2999], [0.5169, 0.4376, 0.4153, 0.4242, 0.2999, 1.0000]])

Hi,

I need to get inverse kinematics model of a double inverted pendulum. I got a model but it gives error when I run. It seems to me that the inverse kinematics model is not quite right. If you are able to help me, I will be very pleased.DoublePendulumI.msim 

Thanks for your help....

 

 

 

 

 

 

 

Im asked on one of my problem sheets to write a procedure to inverse a square matrix using LU factorization and using partial pivoting. I can do all of these using simple pen and paper but am struggling towrite the procedure, especially to involve the partial pivoting and to then solve the linear system. Any ideas how I can do this question?

Hi,

I would like to calculate the inverse fourier transform from this: 

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And I get this:

Hello,

i am currently trying to get some equation of motions. For this i have to invert a matrix which looks like this:

A = GL * M * GL.'     (6x6)

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Hey,

 

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I have tried to get...

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