Either f(x)=x[SUP]3[/SUP]+x-1; To calculate:

[larry91]

Either f(x)=x³+x-1; To calculate: \frac{x_2}{x_1}+\frac{x_3}{x_2}+\frac{x_1}{x_3}.

I tried a lot of approaches but I didn't get the solution! May you help me please?

 

[Mark44]

Fixed your LaTeX. Tip: Don't use HTML tags [ sup] and [ sub] inside LaTeX tags. Instead, use ^{n} or _{n}, respectively.

Either f=x³+x-1; To calculate: \frac{x_2}{x_1}+\frac{x_3}{x_2}+\frac{x_1}{x_3}.

I tried a lot of approaches but I didn't get the solution! May you help me please?

I have no idea what you're asking. Either f = x³ + x - 1 OR what?

And how are you supposed to calculate the expression you give when you don't show values for x₁, x₂, and x₃?

Please provide the complete problem statement.

 

[larry91]

From Viete's relations => x_1+x_2+x_3=0,x_1x_2+x_1x_3+x_2x_3=1,x_1x_2x_3=1;
From these relations you should to calculate S;
x_1, x_2, x_3 are the solution of the F ecuation;

 

[Mark44]

What does this mean?

Either f=x³+x-1

Why did you write "either"?
Is f a number or did you mean f(x) = x³+x-1?
Are x₁, x₃, and x₃ solutions of the equation f(x) = 0? I.e., solutions of the equation x³+x-1 = 0?

 

[larry91]

Yes! Sorry! f(x1) = f(x2) = f(x3) = 0;

 

[Mark44]

What about "either"? What did you mean by that?

 

[larry91]

Sorry once again for mistake!

f(x) = x³+x-1; where x₁, x₂, x₃ are the solutions of the equation f(x) = x³+x-1;
To calculate : \frac{x_2}{x_1}+\frac{x_3}{x_2}+\frac{x_1}{x_3}.

f(x) is a polynomial function;

 

[Mark44]

Have you combined the three fractions to make a single fraction?

What does it mean to say that f(x₁) = 0? f(x₂) = 0? f(x₃) = 0?

 

[larry91]

So, I'll give you an example!
f(x) = x³+x-1;
To calculate: \frac{1}{x_1+x_2}+\frac{1}{x_1+x_3}+\frac{1}{x_2+x_3}

From Viete's relations => x₁+x₂+x₃ = 0 =>
=> x₁+x₂ = -x₃
x₁x₂+x₁x₃+x₂x₃=1
x₁x₂x₃=1

So, we have to calculate : -\frac{1}{x_3}-\frac{1}{x_2}-\frac{1}{x_1} = -(\frac{x_1x_2+x_1x_3+x_2x_3}{x_1x_2x_3}) = 1/1 = 1;

:)

 

[Mark44]

I don't need an example of a different problem. I just want to know what work you have done on this problem. Have you combined the three fractions, yet?

What does it mean to say that f(x1) = 0? f(x2) = 0? f(x3) = 0?

 

[larry91]

Ok!

I try this: \frac{x_2}{x_1}+\frac{x_3}{x_2}+\frac{x_1}{x_3} = \frac{x_2^2x_3}{x_1x_2x_3}+\frac{x_3^2x_1}{x_1x_2x_3}+\frac{x_1^2x_2}{x_1x_2x_3}=
x₂²x₃+x₃²x₁ + x₁²x₂;
And from here I tried a lot of different approaches but I didn't succeeded to solve it!

 

[Mark44]

OK, that's what I get when I combine the three fractions.

Next, what does it mean that f(x₁) = 0? And the same for the other two values.

 

[Mark44]

The reasons that I'm asking all these questions are:
1) To find out what you have done. In your first post you didn't show any of the work that you have done. Before anyone can help you here, you need to show some evidence of trying to work the problem.
2) To see how much you understand this problem. I didn't get a complete explanation of what this problem was about until post #7. If you don't have a good understanding of just what the problem says, it's much harder to solve it.

 

[Dick]

You are being a little harsh here, Mark44. I think larry91 has a pretty good idea of what the Viete relations mean and has a legitimate problem. Post 9 was an attempt to prove that. I keep getting bumped off the net and loosing stuff so I'm going to do this one in several posts. Bear with me.

 

[Dick]

Ok, so the Viete relations allow you to solve for completely symmetric functions of the roots x1, x2 and x3. The immediate problem here is that F(x1,x2,x3)=x2/x1+x3/x2+x1/x3 is not a completely symmetric function of the roots. There are two different permutations of the roots that give you different answers. Check F(1,2,3) and F(3,2,1). They are different. Hence, it's not a pure Viete problem. The final hope is that maybe some special property of x^3+x-1 makes them equal. Stay tuned for part 3.

 

[Dick]

Ok, let's check the special property guess. Let's drag out the heavy artillery. I'm using the computer algebra system Maxima to actually approximate the roots. I find x1=.6823278, x2=1.1615414*i - .3411639 and x3 is the conjugate of x2, of course.
Check them. They work ok. Now compare F(x1,x2,x3) with F(x3,x2,x1). They are pretty different. Bad news. You can't compute the value of x2/x1+x3/x2+x1/x3 using Viete. There are two possible values. And neither looks very pleasant.

 

[larry91]

I tried this:
We can assume that F(x1,x2,x3) is a symetric ecuation; => F(x₁,x₂,x₃) = F(x_s(1),x_s(2),x_s(3)), where s is a permutation of the {1, 2, 3} set;

F(x₁,x₂,x₃) = F(x₂,x₁,x₃) => \frac{{x_2 }}{{x_1 }} + \frac{{x_3 }}{{x_2 }} + \frac{{x_1 }}{{x_3 }} = \frac{{x_3 }}{{x_1 }} + \frac{{x_1 }}{{x_2 }} + \frac{{x_2 }}{{x_3 }} \Rightarrow \frac{{x_2 }}{{x_1 }} + \frac{{x_3 }}{{x_2 }} + \frac{{x_1 }}{{x_3 }} - \frac{{x_3 }}{{x_1 }} - \frac{{x_1 }}{{x_2 }} - \frac{{x_2 }}{{x_3 }} = 0
\frac{{x_2 - x_3 }}{{x_1 }} + \frac{{x_3 - x_1 }}{{x_2 }} + \frac{{x_1 - x_2 }}{{x_3 }} = 0
\begin{array}{l}<br /> x_1 + x_2 + x_3 = 0 \Rightarrow - x_3 = x_1 + x_2 ; - x_2 = ... \\ <br /> \Rightarrow \frac{{x_2 - x_3 }}{{x_1 }} + \frac{{x_3 - x_1 }}{{x_2 }} + \frac{{x_1 - x_2 }}{{x_3 }} = \frac{{2x_2 + x_1 }}{{x_1 }} + \frac{{2x_3 + x_2 }}{{x_2 }} + \frac{{2x_1 + x_3 }}{{x_3 }} = 0 \\ <br /> \Rightarrow 2(\frac{{x_2 }}{{x_1 }} + \frac{{x_3 }}{{x_2 }} + \frac{{x_1 }}{{x_3 }}) + 3 = 0 \Rightarrow 2S + 3 = 0 \Rightarrow S = - \frac{3}{2} \\ <br /> \end{array}

 

[Dick]

I tried this:
We can assume that F(x1,x2,x3) is a symetric ecuation; => F(x₁,x₂,x₃) = F(x_s(1),x_s(2),x_s(3)), where s is a permutation of the {1, 2, 3} set;

F(x₁,x₂,x₃) = F(x₂,x₁,x₃) => \frac{{x_2 }}{{x_1 }} + \frac{{x_3 }}{{x_2 }} + \frac{{x_1 }}{{x_3 }} = \frac{{x_3 }}{{x_1 }} + \frac{{x_1 }}{{x_2 }} + \frac{{x_2 }}{{x_3 }} \Rightarrow \frac{{x_2 }}{{x_1 }} + \frac{{x_3 }}{{x_2 }} + \frac{{x_1 }}{{x_3 }} - \frac{{x_3 }}{{x_1 }} - \frac{{x_1 }}{{x_2 }} - \frac{{x_2 }}{{x_3 }} = 0
\frac{{x_2 - x_3 }}{{x_1 }} + \frac{{x_3 - x_1 }}{{x_2 }} + \frac{{x_1 - x_2 }}{{x_3 }} = 0
\begin{array}{l}<br /> x_1 + x_2 + x_3 = 0 \Rightarrow - x_3 = x_1 + x_2 ; - x_2 = ... \\ <br /> \Rightarrow \frac{{x_2 - x_3 }}{{x_1 }} + \frac{{x_3 - x_1 }}{{x_2 }} + \frac{{x_1 - x_2 }}{{x_3 }} = \frac{{2x_2 + x_1 }}{{x_1 }} + \frac{{2x_3 + x_2 }}{{x_2 }} + \frac{{2x_1 + x_3 }}{{x_3 }} = 0 \\ <br /> \Rightarrow 2(\frac{{x_2 }}{{x_1 }} + \frac{{x_3 }}{{x_2 }} + \frac{{x_1 }}{{x_3 }}) + 3 = 0 \Rightarrow 2S + 3 = 0 \Rightarrow S = - \frac{3}{2} \\ <br /> \end{array}

The whole point I'm trying to make is that F(x1,x2,x3) is NOT a symmetric equation. Don't you believe me? You are getting the real part right, it's -3/2. But if you actually compute the three roots and plug them in you'll find F(x1,x2,x3) and F(x2,x1,x3) are complex conjugates. They aren't equal. G(x1,x2,x3)=F(x1,x2,x3)+F(x2,x1,x3) would be symmetric and you CAN compute that.

 

[larry91]

Ok! Thanks!

 

[epenguin]

You need to read in your algebra textbook the chapter on 'symmetric polynomials' (or maybe 'symmetric functions of roots' or something) where it is explained that problems like this can always be solved and all such symmetric polynomials like your x₂²x3+... expressed in terms of the standard ones (x₁ + x₂ +...), (x₁x₂ + x₁x₃ + ..). ...

OK, just to get you into the idea this is not as hard as it looks, try expand

(x₁ + x₂ + x₃)(x₁x₂ + x₁x₃ + x₂x₃)

Each bracket is something you've met already, and their product was going to give you the expression you are asked about plus something else, so it's not like it all came from the blue.

 

[epenguin]

Sorry, in above comment I had not looked carefully enough, I just saw you seemed not to have solved it, but it seems instead you had got as far as I had.

Could you check if you have reported the question correctly? Because I mean

x₂/x₁ + x₃/x₂ + x₁/x₃

doesn't really mean anything. I mean x₁, x₂, x₃ are just names for three different roots. So even if you know the three roots you are free to name them x₁, x₂, x₃ in any order. Therefore only a symmetric function of them, which is independent of this choice means anything unless you are given more information that pins an identity on each one, as Dick I think is saying. I suspect the question was find Σx₁²x₂ (six terms) which you can solve as I indicated.

 

[Dick]

Or maybe it's find x1/x2+x2/x1+x2/x3+x3/x2+x1/x3+x3/x1. I was guessing it was more like that. Yes, the problem is that the problem as stated isn't totally symmetric. And that is what I was saying.

 

[epenguin]

Yes I really meant Σ(x₁/x₂) (six terms) which = (Σx₁²x₂)/x₁x₂x₃