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Hello,

I've tried to find the solution to my problem, but none of my attempts was succesful.

I have a function which is one-to-one in a particular domain which I am interested in. I would like to get the inverse function of it only in this domain. Here is my function and plot for xp=0..10000:

x := xp-> (-1)*720.5668720*sinh(0.2043018094e-3*xp-0.8532729286)+84952.59423+4.014460003*10^5*arcsinh(0.1219272144e-1*sinh(0.2043018094e-3*xp-0.8532729286)-0.2032498888)

I would appreciate any help,

Iza

Hello,

I tried to solve below equation, but it gives me zeros result. Please help me to find their inverse laplace. 

 It will be clearer if was pasted on Maple:

 

restart

Ps := [P[0], P[1], P[2], P[3], P[4]]:

eqs := [P[0](s) =~ (P[1](s)*mu[1]+P[2](s)*mu[2]+P[3](s)*mu[3]+P[4](s)*mu[4])/(s+lambda[1]+lambda[2]+lambda[3]+lambda[4]), P[1](s) = lambda[1]*P[0](s)/(s+mu[1]), P[2](s) = lambda[2]*P[0](s)/(s+mu[2]), P[3](s) = lambda[3]*P[0](s)/(s+mu[3]), P[4](s) = lambda[4]*P[0](s)/(s+mu[4])];

Ls := solve(eqs, Ps(s))[];

P(t)=~inttrans[invlaplace]~(rhs~(Ls), s, t);

 

Thank you

 

 

Hello

I have a complex set of Markov Processes in reliability application. To make them simpler for me, as a beginner in Maplesoft, I solve them manually to reach a point where I need inverse Laplace for a set of equations. For illustration, I used a simple example below. If I get the concepts for below example, I can apply them on more complicated systems, as following:

P0(s) = 1/(s+λ)+υ*P1(s)/(s+λ)

P1(s)=γ*P0(s)/(s+υ)

Mannuly I find that:

P0(t)=υ/(s+λ)+λ*exp(-(λ+υ)t)/(υ+λ)

P1(t)=υ/(s+λ)-λ*exp(-(λ+υ)t)/(υ+λ)

Please help me step by step to understand how to solve such inverse Laplace. 

Thank you,

Hello please,

Can some one help me find inverse of this fucntion 

f(x) = binomial(x+r-1, x)*((d*x+1)/(1+d*x+(1/2)*b))^x*((1/2)*b/(1+d*x+(1/2)*b))^r/(d*x+1)

 

 

Thanks.

Hallo:

I am trying to find the inverse fourier transform given a vector (of real) as input. I get the following error message:

Error, (in DiscreteTransforms:-InverseFourierTransform) entries of input data must all evaluate to complex floats, got '[.7644453211]'

I am unable to understand this message. I would much appreciate help to resolve this issue.

Thanks

Hi,

I was trying to find the solution for two theta variables in a couple of simultaneous equations (infact this is an iverse kinematics problem for a two link system pendulum).
The following are the initial inputs/equations to be manipulated:


Then I use the folowing command to rearrange for the theta values which I am after:

which gives me the result:

This is all fine until I give in values for l1, l2, x and y:


results:

I have a RootOf in there with a _Z term poping up here and there. I know that this configuration of the two link mechanism in fact dows have a solution and that these numbers are reasonable. Thus I have three questions:

Why does this happen?
What does the "signum" mean here?
how do I go about getting the nummerical values?

Many thanks,
- pjf

Hello,

I have 20 by 20 matrix (with only two symbolic components, say a and b), other entries of the matrix are either populated with zeros or real numbers. I am trying to obtain the inverse.

 

All the command I have tried returned an error message. In particular M^(-1) yields:

Error, (in radnormal/ifactors) too many levels of recursion.

 

I would appreciate any suggestion that could solve this issue. Thank you

 

ps: I am essentially trying to solve Ax=b. The commands within LinearAlgebra all yielded similar error messages.

where A^T is the transpose of A and it's given that (A^T).A is not invertible.

I am stuck as to how to arrive at the solution for x in this case. I initially thought I could multiply both sides by the inverse of A^T reducing it to Ax=b but that was obviously wrong since A^T is itself not invertible(it is singular).

Dear All

 

I have a question:

How can I computet the laplace inverse of a function that is below!

My version of Maple is 13:

It should be noted that r>0, \rho>0.5, y\in R.

 

> restart;
> with(MTM);
print(??); # input placeholder
> ilaplace(r^(rho-.5)*abs(y)^(rho-1)*exp(-r*y)/GAMMA(rho-.5), s);

Dear all,

I have a question, why is the output of the inverse Laplace transformation if the signal is multiplied by itself not just convoluted in time domain:

restart:
with(inttrans):
u0(s):=laplace(u0(t),t,s):
ul(s):=laplace(ul(t),t,s):

invlaplace(u0(s)*ul(s),s,t);
invlaplace(u0(s)*u0(s),s,t);

 

Thanks!

 

a+c^2

5*a+b^5

2*a+b^2*c+3*c

assume system of polynomials are above, how to test whether inverse exist

and find inverse of them

 

Hi,

 

I am trying to find the inverse of 8x8 generic symbolic matrix. Everytime I evaluate the program I get the following error:

 

Error, (in expand/bigprod) Maple was unable to allocate enough memory to complete this computation. Please see ?alloc

 

Can anyone guide me how to increase the memory allocation? And would that solve the problem for me?

 

Any help is appreciated.

 

Thanks!!

Dear all,

It's very convenient to define a DE or PDE through Differential Operator D, for example,

((D[1, 1]+D[1, 2]+D[2, 2])(z))(x, y) = exp(x)*sin(y)

Is it possible to realize Inverse Operator Method of Operator D? How to solve the following equation if we rewrite the pde through inverse operator method?

(z)(x, y)=((D[1, 1]+D[1, 2]+D[2, 2])^(-1))exp(x)*sin(y)

 

Thanks a lot.

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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