Let A and B two real closed intervals.I define b(x) as B+x for any real x ; more precisely, if B=[B1, B2], b(x) = [B1+x, B2+x]I want to build a function f(x) such that :

For example : if A=[0, 2] and B=[-2,-1], then

I coded a few variants using piecewise or Heaviside functions. In some sense I have already answered my own question ... but no one is neither elegant nor concise.

I wonder if there exist a Maple function that returns the measure of the intersection of two real intervals (when they overlap) and 0 otherwise ?

Hi all!

as shown below, how can get a result without 'R':

p_com(z,t):=Re(exp(I*omega*t-I*k*(lambda[r]+I*lambda[i])*z)) assuming omega::real,t::real,k::real,lambda[r]::real,lambda[i]::real,z::real

Thanks very much!

why the the software can't plot the function like x^(4/3)*sin(1/x) or x^(1/3)

it could only plot where x>0,but the value is does exist where x<0.

Thanks in advance for your help.

Hi all,

I tried to find the real solution of the unlinear integral equation:

exp(-h^2/T)*(Int(exp(-x^2/T)*BesselI(0, h*x/T)*x, x = 0 .. 1))/T

but I got the warning and an complex solution:

solve(subs(T = 1, eq)-.99 = 0, h)

Warning, solutions may have been lost

-1.232350242*10^(-32)-1.130417828*I

Can anyone help me to find a real solution for this issue (if possible)...?

I would like to thank you in advance.

In the following problem though b and c are same (except the way denominator 2 is hanfled), command ' a-b ' readily answers zero, but a-c not so. Why? Only on condition of assumption real it gives zero!

Why the answer is not given as zero?

Download what_is_the_difference_between_b_and_c.mw

What difference therms b and c make for Maple? Are they not same?

Ramakrishnan V

rukmini_ramki@hotmail.com

hi all

i have a set of complex numerics, so:

1- i want the numeric with least valence(potency) in imagin particle,

2- i want print the real particle of this numeric.

for example:

A:= .5464691235-.4473247264*I, -.4563184747+1.*10^(-14)*I, .5464691235+.4473247264*I

i want print: -.4563184747

plz help

If for example I type :

It gives out this :

I want dsolve to know I'm using real numbers so it gives out something like :

I tried some assumptions and stuff like assume(a,real), but I didn't manage to figure it out.

If that matters, I'm using Maple 18 student.

EDIT : I know dsolve is not necessary for this particular example, but I want to know if it's possible with dsolve or maybe an other tool that can handle ODE.

Thank you in advance !

I have this matrix

and I want a matrix which components are de real parte of the components of A. I have tried this:

but it doesn't work.

What is wrong?

Thank you so much.

Hello,

I would like to assume a matrix to have only real components.

I have seen that the function assume has some features to assume properties for matrix. But, I didn't find the one that I want : "assume a matrix to have only real components".

This assumation should allow me to suppress this kind of choice in my code :

if A::{complexcons, undefined} then evalf(1/A); elif A::rtable and ArrayTools:-NumElems(A) = 1 then Vector([evalf(1/A(1))]); else evalf(LinearAlgebra:-MatrixInverse(A)); end if;

Do you have some ideas ?

P.S: The matrix should be also a square matrix. So, the first code line will probably be : assume(a, 'SquareMatrix')

Thanks a lot for your help.

consider quadratic equation ax^2+bx+c =0 :the coefficients vary between -1 and +1 . just like this :-1<a<+1 , -1<b<+1 , -1<c<+1 ;how can some one proove that this equation should have real answers ?! can anybody help ? thanks in advance.

Can I use Maple to solve equation like for a,b,c?

a,b,c are real numbers and I need to solve it in real domain.

I have the following characteristic equation by use of maple. How do I find a condition on x, that will return real eigenvalues and complex eigenvalues?

So here is the issue: I have a 50 by 50 tridiagonal matrix. The entries in the first row, first column are -i*x and the last row last column is -i*x; these are along the main diagonal, where i is complex and x is a variable. Everything in between these two entries is 0. Above and below the main diagonal the entries are -1. My issue is that I have to find a conditon on x that makes the eigenvalues real. I am completely new to maple and have no programming experience.. Can someone show me how to this?

Hi,

I got the Real and Imaginary of an expression J1

assume(d,real):

Gamma:=0.04:tau:=10*Pi:j:=0:

J1:=(exp((1-I*d)*Gamma*tau)-1)/((1-I*d));

J1mod:=simplify((Re(J1))^2+(Im(J1))^2): (I works here this amont is real)

################

but when I change the expression for J1 to be

J1:=((2*e^(-2^(-j-1)*(1-I*d))-e^(-2^(-j)*(1-I*d))-1)*exp((1-I*d)*Gamma*tau)-1)/((1-I*d)):

J1mod:=simplify((Re(J1))^2+(Im(J1))^2):

J1mod here is complex(I dont know why? it doesnt separate the real and the im )

Any comments will help

Thanks

I need you help to understand this problem.

Let alpha=0.1.;

When I do : (-0.2)^alpha I get a complex number.

I must find a real number.

What's the problem

Thanks for your idea.

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