Items tagged with real

Hi there,

            Recently, I encountered a problem. I have a function( omega as its variable)  (18)


I tried to find a point where its first derivative equals 0. In this case, Maple returned four solutions. In my

question, both beta, gamma, R4 and C2 >0, I want it to return a real positive solution, the first term

in (19) (i.e. 1/(sqrt(beta *gamma) *1/R4 C2).


I know it is easy to find out the positive real roots in this case. This question seems to make no sense.

However, sometime I came across an expression complicated enough that I cannot tell whether it is real


Is there a approach to find a real positive solution of an symbolic eqution?

Thanks in advance!


                                                                     A University student in BeiHang University, Beijing

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 :

  1. if  A and b(x) do not overlap then f(x) = 0
  2. otherwise f(x) is some expression of the covering length

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

  1. f(x) = 0 if  -1+x < 0 or -2+x > 2
  2. otherwise f(x) = L   where L is the measure of the intersection of A and b(x)

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



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!

a := (1/2)*(kappa*omega^2+omega^3)*(Y+(1/2)*(-sqrt(N)*omega^(3/2)*sin(theta[2])*cos(varphi[2])*lambda__b+sqrt(N)*omega^(3/2)*sin(theta[1])*cos(varphi[1])*lambda__a)/(kappa*omega^2+omega^3))^2/omega:

b := (1/2)*(kappa*omega^2+omega^3)*(Y+(1/2)*(-sqrt(N)*omega^(3/2)*sin(theta[2])*cos(varphi[2])*lambda__b+sqrt(N)*omega^(3/2)*sin(theta[1])*cos(varphi[1])*lambda__a)/(kappa*omega^2+omega^3))^2/omega:




c := (1/2)*(kappa*omega^2+omega^3)*(Y+(-sqrt(N)*omega^(3/2)*sin(theta[2])*cos(varphi[2])*lambda__b+sqrt(N)*omega^(3/2)*sin(theta[1])*cos(varphi[1])*lambda__a)/(2*(kappa*omega^2+omega^3)))^2/omega:








Why the answer is not given as zero?





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

Ramakrishnan V

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 :

`assuming`([dsolve({-7 = a(x)^2*b(x), a(x) = 3*b(x)}, {a(x), b(x)})], [(b(x))::real])

It gives out this :

`assuming`([dsolve({-7 = a(x)^2*b(x), a(x) = 3*b(x)}, {a(x), b(x)})], [(b(x))::real])

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

[{a(x) = -21^(1/3)}, {b(x) = (-21^(1/3))*(1/3)}]

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

A := Matrix([[x+I*y, z+I*w], [-z+I*w, x-I*y]])

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.


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
elif A::rtable and ArrayTools:-NumElems(A) = 1 then
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 |a-b|+\sqrt{2b+c}+c^2-c+1/4=0 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?


I got the Real and Imaginary of an expression J1 




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



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

Any comments will help


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