Finding a Real Solution for a Complex Equation

[Nusc]

Homework Statement

I need to find a positive REAL value x such that

<br /> \frac{2-(4+x^2-x\sqrt (4+x^2) \sqrt (\frac{2+x^2+x\sqrt 4+x^2 }{8+2x^2})}{4\sqrt (4+x^2) } == 0<br />

x cannot be complex or zero! So far, I don't even think its possible.

Any suggestions?

Homework Equations

The Attempt at a Solution

 

[Dick]

The formatting on that is not very clear. But why do you need that? Is this the solution to some more basic problem?

 

[Nusc]

Is that not good enough?

 

[Dick]

Is that not good enough?

It is now, sort of. Wasn't before. I'm still curious why. If I'm reading it right, then x=0 doesn't work anyway, correct?

 

[Nusc]

You're right in this case.

 

[Defennder]

Well I could reduce it to x^3 -3x^2 -4=0

Since I don't know how to solve for exact solutions, I used http://www.1728.com/cubic.htm
and it appears that x= 3.3553013976081196 would be an approximate answer.

I noticed there is a double equal sign ('==') in the expression. Does that mean anything?

 

[epenguin]

Really checked through your transcription? - I mean they usually don't give questions with messy answers. In real life yes, but not so often in textbooks.

Is
\frac{2-(4+x^2-x\sqrt (4+x^2) \sqrt (\frac{2+x^2+x\sqrt (4+x^2) }{8+2x^2})}{4\sqrt (4+x^2) } == 0

a bit nearer by any chance?

 

[Nusc]

Well I could reduce it to x^3 -3x^2 -4=0

Since I don't know how to solve for exact solutions, I used http://www.1728.com/cubic.htm
and it appears that x= 3.3553013976081196 would be an approximate answer.

I noticed there is a double equal sign ('==') in the expression. Does that mean anything?

That's strange, I got mine to cancel out which is what I expected.

So nevermind.

 

[Nusc]

Well I could reduce it to x^3 -3x^2 -4=0

Since I don't know how to solve for exact solutions, I used http://www.1728.com/cubic.htm
and it appears that x= 3.3553013976081196 would be an approximate answer.

I noticed there is a double equal sign ('==') in the expression. Does that mean anything?

Actually I was wrong, how did you simplify it to a cubic?

 

[Nusc]

If you plot that function in mathematica, from say 0-5 you can find values of x that would yield an approximate answer to 0.

Does anyone know of a way to get the best possible result?

 

[DavidWhitbeck]

The cubic has an exact solution, you can look it up or you can just use Newton's method to find an approximate solution.

 

[Nusc]

The function I'm trying to approximate is not a cubic, I'm just wondering how the other guy got it. The function of interest is at the top. What's better in this case Bisection or Newton?

 

[DavidWhitbeck]

Have you tried solving for x? You haven't even simplified your expression! Throw the denominator away, and notice that there are two square root factors on one of the terms that cancel each other (up to a factor of 1/2). I could tell you what they are, but I would prefer to figure out what I'm talking about.

Once you do that it's easy to isolate the square root term and then square both sides of the equation. You will have either a cubic or a quartic.

 

[Defennder]

I'm not entirely sure I interpreted your expression correctly. For one thing, is it supposed to be \sqrt{4+x^2} or \sqrt{4} + x^2? It appears it should be the former. In that case, it's best written as

\frac{2-(4+x^2-x\sqrt{4+x^2} \sqrt{\frac{2+x^2+x\sqrt{4+x^2} }{2(4+x^2)}}}{4\sqrt{4+x^2}} = 0

Again, what does the double equal sign '==' mean?

Assuming my interpretation of your expression is correct, this time it reduces to 2x^4+5x^2+4=0.

That gives x^2=-\frac{5}{4} \pm \frac{\sqrt{7}}{4}i

Finding the square roots of these would hence give you 4 solutions. But then again you mentioned that you thought that x cannot be complex. Is that something stated by the question?

 

[rootX]

Use some numerical method:
Newton doesn't look good here
Bisection ..
or some good one

 

[Vid]

From a programming prospective a single equal sign is an assignment operator that gives the value of the right side to the left, and a double equal sign is a comparison operator that returns a value of true if the two sides are indeed equal.

 

[jimvoit]

Ballance the parentheses in the numerator so that the expression is unambiguous.

 

[Nusc]

Below is my MAPLE procedure but it gives me values that are very large...

restart;
bisection := proc (f1, c, d, TOL, N0)
local f, a, b, p, i, j, k;
f := f1;
p := evalf((1/2)*c+(1/2)*d);
a := c;
b := d;
i := 1;
j := 0;
k := 1;
while j = 0 and k <= N0 do
if evalf(f(a)*f(p)) < 0 then b := evalf(p)
else a := evalf(p) end if;
print(k);
print("Value of P:");
print(p);
if abs(f(p)) < TOL or N0 <= i then printf("Procedure completed successfully. Approximate value of P:"), P;
j := 1 end if;
k := k+1;
i := i+1;
p := evalf((1/2)*a+(1/2)*b)
end do
end proc

What can be done to improve the code?

 

[jimvoit]

The double equal sign is used in Mathematica to indicate an equation. I plotted the left side of the equation and found it asymtotic to the x axis, which indicates that there is no solution. That's no help with the algebra I know, and I may have transcribed the equation incorrectly..But that's that for what it's worth.

 

[Nusc]

It's not asymptotic to the x-axis... Try -1000..1000

 

[rootX]

you have to write one?!
Why don't simply use built in procedures (obviously they are superior)?

matlab:
>> syms x;
>> f = (2-(4+x^2-x*sqrt(4-x^2)*sqrt((2+x^2+2*x+x^2)/(8+2*x^2))))/(4*sqrt(4+x^2));
>> solve(f,x)

ans =

1.0042365002824003967092316690056-1.0497367979853078503097252196741*i
1.0042365002824003967092316690056+1.0497367979853078503097252196741*i

 

[Nusc]

Oh that's cool. I should acquaint myself with MATLAB... But it appears that function is irrelevant now as I realized something.
For example, a function say
<br /> Sin^2[\frac{1+\sqrt{5}t}{2}]<br />
is not periodic. It will never reach 1 or 0 after t.

<br /> \frac{1}{4(4+x^2)^{3/2}}(2*\sqrt{4+x^2}*Cosh[t x](2+(2+x^2)Cosh[t\sqrt{4+x^2}])-2((2+x^2)\sqrt{4+x^2}+2\sqrt{4+x^2}Cosh[t\sqrt{4+x^2}+x(4+x^2)Sinh[t x]Sinh[t \sqrt{4+x^2}]<br />
How do you make complicated functions periodic?

 

[jimvoit]

The original post was looking for a real solution

 

[jimvoit]

Also...the parentheses in the initial expression have not yet been ballanced. An assumption must be made as to where the missing ")" goes. In these situations one wonders if the problem has been transcribed incorrectly from a text and the authors solution is to a different problem.

 

[Nusc]

-Log[Exp[i*t*x]]==0
Exp[-Log[Exp[i*t*x]]]==0
Exp[-i*t*x]==Exp[0]
Exp[-i*t*x]==1
Log[Exp[-i*t*x]]==Log[1]
Log[Exp[-i*t*x]]==0

Is this correct, I need to find a solution to the first expression.