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Hello,

• Is there a simple way to find the domain for the real solutions of f(x)?

• And is there a way to let maple get the part of f(x) with the sqrt?
   (not by typing it by hand as I dit below)

• Is there a way to write the summary of the found domains in one line?

Thanks for your help. 





restart:
# How to find the Domain for real solutions for x?
f(x):=(x-1+sqrt(x^2-3*x+2))/(x-1);
discont_for_x=discont(f(x),x);
# x<>+1 (because the de denom=0 is not allowed)
denom(f(x))=0;
x={solve(denom(f(x))=0,x)};
# x<=1 union  2<=x (because the part under the sqrt must be >=0 to give Real solutions)
sqrt(x^2-3*x+2);
0<=x^2-3*x+2;
x=solve(0<=x^2-3*x+2,x);




Hi, the title isn't great as I didn't know how to describe this really. I need to solve the following equation for b:

y = (1-exp(-x*b))/(1-exp(-50*b))

When I put a value for y in, this is fine and fsolve gives me a numeric real solution. However, even when using RealDomain, it does not give me a real solution if I leave y as it is, and instead gives a 'RootOf' solution, which I don't really understand. This is the same whether using solve or isolate:

b=-(1/50)RootOf(_Zx-50ln(-y+ye^(_Z)+1))

I have the values of x and y for multiple data points and can put them in an nx1 matrix. Is there a way to replace x and y with matrices (with real numbers in) and solve for each set of points for b (ie there would be n values of b)? Obviously I could go through and put in each value of x and y but this would take ages, so was just wondering if there's a quick way to do this.

I have tried by simply putting matrices instead of the letter but get the error:

Error, invalid input: exp expects its 1st argument, x, to be of type algebraic, but received Vector(50, {(1) = -50*b, (2) = -49*b,...

Thanks for your time

James

Real roots only...

January 17 2014 Syeda 25

I want to find real roots only.  Cannot we find a simplified formula for x in this case which gives only real roots? 

 

 

``

eq1 := a^2*x^3+Typesetting:-delayDotProduct(2*a*b-Typesetting:-delayDotProduct(a^2, e), x^2)+(-2*a*b*c^2-a*c+b^2)*x-c*b-d-b^2*e = 0:

``

# Formula

eq2 := A*x^3+B*x^2+C*x+E = 0:

``

NULL

a := .7438:

b := 15.12*z[1]+10.85*z[1]^2:

c := 18.92-17.76*z[2]:

d := -.9224:

e := 2.106-5.317*z[2]+2.87*z[2]^2:NULL

NULL

A := a^2:

B := -a^2*e+2*a*b:

C := -2*a*b*e^2-a*c+b^2:

E := -b^2*e-b*c-d:

``

eq2

.55323844*x^3+(-1.165120155+2.941568785*z[2]-1.587794323*z[2]^2+22.492512*z[1]+16.140460*z[1]^2)*x^2+(-1.4876*(15.12*z[1]+10.85*z[1]^2)*(2.106-5.317*z[2]+2.87*z[2]^2)^2-14.072696+13.209888*z[2]+(15.12*z[1]+10.85*z[1]^2)^2)*x-(15.12*z[1]+10.85*z[1]^2)^2*(2.106-5.317*z[2]+2.87*z[2]^2)-(15.12*z[1]+10.85*z[1]^2)*(18.92-17.76*z[2])+.9224 = 0

(1)

``

``# Putting z1 and z2 value

"(->)"

.55323844*x^3+14.11629660*x^2+83.26002702*x-3.52866181 = 0

(2)

 

"(->)"

[[x = 0.4208050385e-1], [x = -9.354079555], [x = -16.20375615]]

(3)

``

``

 

Download cubic.mw

My attempt:

RealDomain[solve]({x^2+y^2+z^2 = 3, x+y+z = 3}, {x,y,z});

             {x = -RootOf(_Z^2+(z-3)*_Z+z^2-3*z+3)-z+3, y = RootOf(_Z^2+(z-3)*_Z+z^2-3*z+3), z = z}

 

In fact, the system in the real domain has a unique solution x = 1, y = 1, z = 1. It is easy to find by hand, noting that the plane  x + y + z = 3  is tangent to the sphere  

Hi, was wondering if any of you could help me, when I try and find the real part of a function to plot, I get a float(undefined) error, however by just using evalf if gives gives me the real and comlex parts.

zetaroots.mw

The function i want to find realy parts for is f(x).

 

Thanks,

Matt

I propose a different proof of this remarkable identity (see  http://www.mapleprimes.com/posts/144499-Stunningly-Beautiful-Identity-Proved ) in which  directly constructed a polynomial, whose root is the value of LHS, and this is expressed in radicals.

For the proof, we need three simple identities with cubic roots (a, b, c -any real numbers):

Hi,

I'm using Maple to carry out some calculations in Tropical algebra, which requires taking minima of real numbers and infinity.

I'm currently using symbols rather than real numbers, which is causing a problem, I have (for example) the following lines of Maple code:

> assume(0 < a)
> min(a+infinity, 2*a+infinity)
               min(a~ + infinity, 2a~ + infinity)

I am trying to determine if a particular system of 15 polynomial equations in 9 variables has a real solution using Maple's RegularChains library.  I am using the IsEmpty command which returns true if there are no solutions and false otherwise.  In Maple 16, this can be done using the command:

IsEmpty( sys, R ) ;

where "sys" is a list of equations and "R" is a polynomial ring, both of which I define in the worksheet.  But this syntax only works...

I'm very new to Maple and I'm just curious as to how Maple computes its summations with the MatrixPower command.

if A = a real square matrix

c = some real constant

x = c*A

then Why is it, that when I try to use...

sum(MatrixPower(x,k), k = 0..3);

I get a non-real / ridiculous result, but when I type it out...

MatrixPower(x,0) + MatrixPower(x,1) + MatrixPower(x,2) + MatrixPower(x,3);

x^0 + x^1 + x^2 + x^3;

Dear members,

I would like to generate a tree (the simplest "stage tree"that I have in mind is in the attached file) ad to work with it. For that I need to  "replicate" [the stage tree which starts in O (with some value R)] after the nodes x2, y2, z2. The nodes x1, y1, z1 are terminal (with values f(R), g(R), h(R) ).

I found soemthing on the posts on mapleprimes, but it is only bor "binarytrees".

How may I proceed?.

 

I really...

I use Maple 15 and I want to take the real and the imaginary part of a simple expression e.g. 5*Dirac(x)+3*I, having already assumed x a real value. However, Maple seems to have problem with the Dirac function. The output looks like this:

> assume(x, 'real');
> Re(5*Dirac(x)+3*I);
5 Re(Dirac(x))
> Im(5*Dirac(x)+3*I);
3 + 5 Im(Dirac(x))
 
Any thoughts on how to overcome this are welcomed.

Dear All,

I am trying to solve the following equation in Maple but could not find the actual result:

P:=1/2 * Real (E x H*)

How to find the real value of (E x H) with H*= H conjugate.   And also which command to solve for the value P.

Your help will really be appreciated

Thanks in advance

A.Q

Soton

 

Hi,

I am trying to simplify the expression s as given below. (I am not sure why it comes up with all the vector caclulus notation in it but it should display okay when you enter it)

Because of the presence of the exponential imaginary fucntions I thought evalc might be useful but when I use it I get a huge expression with csgn appearing in it. To my knowledge csgn appears when assumptions are not correctly specified - is this so? I can't see any assumption...

Hi,

 

I'm trying to model a magnetic field with maple.

In order to do that, I compute a (huge) sum and integrals of real and posive numbers.

The problem is that for some points, when I compute the norm of this field I get an imaginary number with a very small imaginary part (egs -2, 27293.1844462+1.42305652280*10^(-84)*I ).

 

How can I get rid of it?

thx in advance 

Dear Sir / Madam,

Would you recommend me how to isolate the real and imaginary part and simplify the equation A? Maple cannot evaluate this equation to numeric value...

A := evalf(sqrt((Re(B))^2 + (Im(B))^2));
A := evalf(abs(B));

are not working.

I have attached my equation function.txt

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