The problem is:

I don't want to spoil it by telling you the answer. But here are a few ideas:

1) Given the form of the right hand side (i.e. the shape as a function of x) how many solutions do you think this equation has?

2) Substituting the rhs of the equation into itself, you get a new equation with more radicals. Keep going and you get an infinite series. This should allow you to find a couple of solutions. It may not be rigorous, but given (1) above and direct substitution to verify that they are indeed solutions you are on safe ground.

Andre

x²-4x+3=0 -> x₁=3, x₂=1
The GraphCalc 4.0 shows me, that there are no more roots, but I must prove this statement.
I think one way is the function analysis, or even better is the indirect proof.
Let

_____ Ö-3+4x

=x+a

, where a ¹ 0

From the equation:

x1=2-a+

____ Ö1-4a

,

x2=2-a-

____ Ö1-4a

, where a £ 1/4

We return to the original equation:

-3+4

_________ Ö(-3+4a)+4x

=x2

16(4a-3)+64x=(x²+3)², and I don't see the continuation.
Is the proof faulty? If yes, then where is the mistake?

Peter

Yatir,

I don't think that is true. The 2nd degree equations can also have 0, 1, 2 roots!

The analysis of GraphCalc 4.0 for my problem:


Peter

Well done Peter. Your solutions are correct and the only real ones. Judging by your formulae you did it by generating infinite nested radicals and deriving sqrt(-3+4x)=x, which is how I did it.

If you like this sort of problem, you may wish to look here as well.

As for a proof that these are all the real solutions:

A physicist would either plot the graph of the RHS, note 2 intersections with y=x and say 'it's obvious'; or he would diligently compute the second derivative of the RHS, note that it is negative (i.e. that the curve is like an upside down parabola) and say 'it is obvious'.

For example, if 1 < x < 3, prove that

æ  ú Ö

-3+4   æ Ö -3+4   _____ Ö-3+4x

is strictly greater than x. Thus there cannot be a solution for 1 < x < 3.

Let me know how you get on.

Andre

Now suppose we can prove that

g(x₁(1-D)+x₂D) > g(x₁)(1-D)+g(x₂)D

for 0 < D < 1 (and any x₁, x₂)

Substituting the particular case x₁=1, x₂=3, we have proven

g(x) > x

for 1 < x < 3 Obviously if g(x) > x then there cannot be a solution to g(x)=x!

You should then extend the argument to g(g(x)) etc.

Andre

Andre,

Many thanks for your help! I try solve the problem on base your ideas, ...but now I see on the Mathworld here this formula:
(1)

first you must prove that the RHS converges, then set:
t=RHS
x=t
(x² -a)/b=t=x
Solving the upper equation will give you (1)

Yatir

Yes! Thanks Yatir. The RHS converges, see mathworld.wolfram.com
Herschfeld's Convergence Theorem.

I think the proof of sum formula of Continued Fraction is:


Sources: The Museum of Infinite Nested Radicals and Continued Fraction

The finish of the proof has already been given by Yatir.

Peter

The way I solved it was as follows:

1. First 'guess' what the solutions are by substituting the RHS into itself repeatedly. So

   
   

   x=

   æ Ö

   -3+4   ______ Ö-3+...

   
   

   suggesting (ignore convergence) that

   x²-4x+3=0

i.e. x=1,3. (call these x₁, x₂)
2. Substitute x₁ and x₂ into your original equation to verify that they are solutions (but not necessarily the only ones)
3. Now we prove that these are the only solutions. We do this in 3 parts: look at possible solutions x < x₁, solutions x₁ < x < x₂, solutions x > x₂.

Are there any solutions for x > x₂? Write x=x₂+D[D > 0]

And let

g(x)=

_____ Ö-3+4x

We want to prove g(x) < x

i.e.

g(x₂+D) < x₂ + D

so

Ö

-3+4(3+D)

< 3+D

square:

-3+12 + 4D < 9 + 6D+ D²

which is true for all D > 0 This argument can be run backwards so we have shown that g(x) < x.

Now the RHS of the original equation is g(g(g(x)))=g(g(x-d)) < g(g(x)) [d > 0]

=g(x-d) < g(x) < x

So there is no way that there can be a solution for x=g(g(g(x))) since we have shown the RHS is less than x!

Let me know if you have any problems proving the other 2 cases.

Andre

How about repeated squaring and then solving?

Yatir

Yatir, did squaring and solving give you the solution? I didn't/haven't tried it since on squaring you get a polynomial of order 8. On dividing out the two solutions we know gives a polynomial of order 6. Does it turn out to be obvious as to why this resulting sextic equation has no real roots?

Andre

Actualy it does!

x^6 + 4·x^5 + 25·x^4 + 88·x^3 + 427·x^2 + 1444x +5179=0

It obviously doesn't have any more real roots!

Yatir

Why does it obviously have no more real roots? (or are you being sarcastic?)

Although I don't have a proof right now, when I look at the polynomial, the LHS seems to grow pretty fast (for negative as well).

Yatir

-I meant obviously as in not in a proof but by intuition, I may be wrong.

-3+4(1+D) > (1+D)²

-3+4+4D > 1+2D+D²

0 > D²-2D

0 > D(D-2)

Here 0 < D < 2, so the RHS is negative. (See the GraphCalc diagram in this topic.)

III. case: x < x₁

Let D > 0. (If D < 0, see I. and II. cases)

-3+4(1-D) < (1-D)²

-4D < -2D+ D²

0 < D² + 2D

0 < D(D+2)

Andre, thanks for the very nice proof!

Peter

2sin((1)p/4)=Ö2

2sin((1+1/2)p/4)=

Ö

2+Ö2

2sin((1+1/2+1/4)p/4)=

æ Ö

2+ Ö 2+Ö2

The pattern looks pretty clear, however note also that:

2sin((1+1/2+1/4-1/8)p/4)=

æ  ú Ö

2+   æ Ö 2+ Ö 2-Ö2

Note the minus signs! You may like to experiment a little more and see if you can come up with the general case.

Andre