Items tagged with inverse

Is there a command in Maple to produce a table of z values given F(z) where F is the CDF of the Standard Normal Distribution? I know of the command ProbabilityTable to generate a table of z, F(z) values.  What I would really like is F, Inv(F) table of values. I guess  I could write my own code to do this but was wondering if there is an easier way to do this.


I want to find the inverse of a 11x11 matrix which I imported from excel using the import data tool. When I try to find the the inverse it gives me this error:


K:=ExcelTools:-Import("C:\\Assignment 2.xlsx", "Q2", "V7:AF17");

K := Vector(4, {(1) = ` 1..11 x 1..11 `*Array, (2) = `Data Type: `*anything, (3) = `Storage: `*rectangular, (4) = `Order: `*Fortran_order})

kkk:=convert(K, matrix):

Error, (in MatrixInverse) MatrixInverse expects its 1st argument, M, to be of type {Matrix, [Matrix(square)], [Vector, Matrix(square)], [Matrix(square), Matrix(square), Matrix(square)]} but received kkk


Can someone please help me out??? Thank you


I was playing around with the example "Simple Inverse Kinematic Problem" and found somethings to be odd:

the angular motion seems to be calculated from between pi and negative pi and this has some effects when using position block to move a joint.

the original angular displacement is graph below

the angular displacement after ik calculations have been performed:

if you run the simulation it seems to copy and mirror the input pendulum, however if you disable one of the IK solutions you see that infact its motion isn't like the input.

this becomes more prevailant when you use a position block to force rotation on a joint instead of using the 'prescribed rotation' blocks that comes with the example.

My question then:
why does this happen?
how do I work around this?

the importance that the motion is follow precisely becomes more prevailent when we want to extract other values such as vel, accel, torque. they are incorrect and very jumpy. Also simply put the angular displacement is wrong, how do I fix it?

(side note: I'm thinking  it has to do with the way arctan is calculated in maple limits it to stay in range -pi to pi
"For real arguments x, y, the two-argument function arctan(y, x), computes the principal value of the argument of the complex number x+Iy, so −π < arctan(y,x) ≤ π." from


Dear Community,

Would someone have a good and easy to understand/implement description of the Den Iseger algorithm for the numerical inversion of Laplace transform? Even better if someone would have a Maple script to do it, that would be superb.

Tx in advance,

best regards


I would like to apply inverse Laplace transform to U(x,p), which is defined by

For simplicity with my calculations, I assumed p:=i*beta^2. That is why I have the following equation after applying Laplace transform

(beta=0 is not a pole, that is why I removed the last term in my calculations later. Because there is no contribution) where

Here p and beta are complex values, we can write Re(p)=-2*Re(beta)*Im(beta), Im(p)=(Re(beta))^2-(Im(beta))^2 due to p:=i*beta^2. I numerically compute the roots of h(beta), you can find the numerical values of beta (I assumed digits are 50 due to accuracy )

Finally, I would like to plot U(x,t) with the values t=0.8, lambda=1, L=10, k=1. For checking the figure give t=0 and observe that U(x,0)=0.

I am expecting the plot is more or less like the following figure

PS: I already tried to solve and plot the problem, but I could not find where I make a mistake. I  share the worksheet below. Thank you!



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


Perla D'andrea



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]


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


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

N=2 →

N := 2:
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):

N=3 →

N := 3:
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):

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.

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

Therefore det(A)= CN-1 * e[a1,a2,] * 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.

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(`$`(a[q] = b[q], q = 1 .. N), e)*ML:
for r to N do s[r] := sum(sum(s[r-1], a[r] = 1 .. N), b[r] = 1 .. N) end do:
A := Matrix(N, N, proc (i, j) options operator, arrow; evalf((3*i+2*j)/(i+2*j)) end proc):
for x to N do for y to N do
e := product(product(signum(a[j]-a[i]), j = i+1 .. N-1), i = 1 .. N-1):
ML := product(AA[a[k], b[k]], k = 1 .. N-1):
S[0, x, y] := e*subs(`$`(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);

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))

Hi and thanks in advance

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?


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,



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:



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




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)


Mannuly I find that:



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)





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.


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