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Matrix and inverse matrix...

August 13 2016
1 6

Hello all

I try to make matrix of size 21*21 but maple doen't run , I don't Know why ?

Also I want to comput the inverse but I get the same problem.

Do you have any idea .

Thank you

expression with inverse trig funtion...

July 03 2016
1 4

Hello

I am having issue in finding the explanation on how to solve inverse trig funtion and expression with inverse trig funtions. I do not understand my school book and I was hoping that the software would have given me an extra help in understanding my school problems.

Thank you very much

Regards,

Perla D'andrea

June 30 2016
1 1

HOW DO I GET A STEP BY STEP SOLUTION FOR INVERSE Z TRANSFORM HELPPP!!!!!!

Levi-Civita symbol and inverse of matrix...

May 14 2016
0 1

Hi

We know determinant of a square matrix A[ij] (i,j ∈ {1,2,3}) is equal to the following expression

det(A) = 1/6 * e[ijk] * e[pqr] * A[ip] * A[jq] * A[kr]

in which e[ijk] is a third order Tensor (Permutation notation or Levi-Civita symbol) and has a simple form as follows:

e[mnr] = 1/2 * (m-n) * (n-r) * (r-m).

The (i,j) minor of A, denoted Mij, is the determinant of the (n − 1)×(n − 1) matrix that results from deleting row i and column j of A. The cofactor matrix of A is the n×n matrix C whose (i, j) entry is the (i, j) cofactor of A,

C[ij]= -1 i+j * M[ij]

A-1=CT/det(A)

The general form of Levi-Civita symbol is as bellow:

I want to write a program for finding inverse of (NxN) matrix:

N=2 →

restart;
N := 2:
with(LinearAlgebra):
f := (1/2)*(sum(sum(sum(sum((m-n)*(p-q)*A[m, p]*A[n, q], q = 1 .. 2), p = 1 .. 2), n = 1 .. 2), m = 1 .. 2)):
A := Matrix(N, N, proc (i, j) options operator, arrow; evalf((37*i^2+j^3)/(2*i+4*j)) end proc):
f/Determinant(A);

N=3 →

restart;
N := 3:
with(LinearAlgebra):
f := (1/24)*(sum(sum(sum(sum(sum(sum((m-n)*(n-r)*(r-m)*(p-q)*(q-z)*(z-p)*A[m, p]*A[n, q]*A[r, z], m = 1 .. N), n = 1 .. N), r = 1 .. N), p = 1 .. N), q = 1 .. N), z = 1 .. N)):
A := Matrix(N, N, proc (i, j) options operator, arrow; 10*i^2/(20*i+j) end proc):
f/Determinant(A);

The results of above programs are equal to 1 and validation of method is observed.

If we can write the general form of determinant then we can find the inverse of any square non-singular matrices.

Now I try to write the mentioned program.

restart;
with(linalg):
N := 7:
Digits := 40:
e := product(product(signum(a[j]-a[i]), j = i+1 .. N), i = 1 .. N):
ML := product(A[a[k], b[k]], k = 1 .. N):
s[0] := e*subs($(a[q] = b[q], q = 1 .. N), e)*ML: for i to N do s[i] := sum(sum(s[i-1], a[i] = 1 .. N), b[i] = 1 .. N) end do: A := Matrix(N, N, proc (i, j) options operator, arrow; evalf((3*i+j)/(i+2*j)) end proc): # arbitrary matrix CN:=simplify(s[N]/det(A)); Therefore det(A)= CN-1 * e[a1,a2,..an] * e [b1,b2,.., bn] * A[a1,b1] * A[a2,b2] * ... * A[an,bn]. The correction coefficient is CN(for N)/CN(for N-1) = N!/(N-1)! =N. restart: with(linalg): N := 4: Digits := 20: e := product(product(signum(a[j]-a[i]), j = i+1 .. N), i = 1 .. N): ML := product(A[a[k], b[k]], k = 1 .. N): s[0] := e*subs($(a[q] = b[q], q = 1 .. N), e)*ML:
for r to N do s[r] := sum(sum(s[r-1], a[r] = 1 .. N), b[r] = 1 .. N) end do:
A := Matrix(N, N, proc (i, j) options operator, arrow; evalf((3*i+2*j)/(i+2*j)) end proc):
DET:=S[N]:
for x to N do for y to N do
e := product(product(signum(a[j]-a[i]), j = i+1 .. N-1), i = 1 .. N-1):
ML := product(AA[a[k], b[k]], k = 1 .. N-1):
S[0, x, y] := e*subs($(a[q] = b[q], q = 1 .. N-1), e)*ML: for r to N-1 do S[r, x, y] := sum(sum(S[r-1, x, y], a[r] = 1 .. N-1), b[r] = 1 .. N-1) end do: f[y, x] := (-1)^(x+y)*subs(seq(seq(AA[t, u] = delrows(delcols(A, y .. y), x .. x)[t, u], t = 1 .. N-1), u = 1 .. N-1), S[N-1, x, y]) end do: end do: Matrix(N, N, f)/(DET)*(24/6); A^(-1); CN for N=4 and N=3 is 24 and 6 respectively i.e. CN(4)/CN(3)=24/6. When I use bellow procedure the error "(in S) bad index into Matrix" is occurred. Please help me to write this algorithm by using procedure. Thank you restart; with(linalg): Digits := 40: n := 7: S := proc (N) local e, ML, s, i: e := product(product(signum(a[j]-a[i]), j = i+1 .. N), i = 1 .. N): ML := product(A[a[k], b[k]], k = 1 .. N): s[0] := e*subs($(a[q] = b[q], q = 1 .. N), e)*ML:
for i to N do s[i] := sum(sum(s[i-1], a[i] = 1 .. N), b[i] = 1 .. N) end do
end proc:
A := Matrix(n, n, proc (i, j) options operator, arrow; evalf((3*i+j)/(i+2*j)) end proc): # arbitrary matrix
CN := simplify(S(n)/det(A))

How do I plot a function...?...

April 23 2016
1 12

I have a function like

w=f(r) on the other hand

z=g(r) wher r is common between these two.

how i could plot w in term of z?

Inverse function in restricted domain...

December 30 2015
1 2

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

Can I solve this Inverse Laplace?...

November 10 2015
0 4

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

How to find invlaplace?...

November 08 2015
1 1

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)

Thank you,

Inverse functions...

October 27 2015
1 44

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.

Unable to understand the error in inverse FT...

May 11 2015
1 6

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

solving simultaneous equations, gettin _Z in resul...

March 29 2015
1 2

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

Inverting a Large Matrix...

March 12 2015
1 6

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.

find solutions to the normal equations (A^T).A.x=(...

February 09 2015
0 1

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).

Laplace inverse of a function...

January 30 2015
0 2

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);

Inverse Laplace Transformation...

November 19 2014
0 4

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!

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