An Infinite Radical Solved in Two Ways

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These problems are easy but fun to solve. They also have educational value imo

leonhardeuler

I just noticed that the second radical was equal to x and substituted it in, which quickly gives both solutions

pNsB

This is very easy and quite interesting. I solved it in my head.

matrix

Hey SyberMath, another great explanation! I figured it out what x is in my head. I only found 1 solution x=0. X=2 is another solution. Thanks a lot!

carloshuertas

Learnt something new today.But tomorrow is my sst exam !!! Wish me luck PLEASE ! I need it syber math !! Following you for an year . I feel you are my teacher!

cube

Exercise dur! Rezolvari super elegante! Cel putin a doua este traznet! Next? Next! Super! Superb!

sberacatalin

I have a question.
If you solve it using algebra you get the solutions x=0 or 2 while if you try to solve it using geometry you get x=0 or 1, 93....
Why?
Should not we get the same solutions in both ways?

Chrisoikmath_

The correct and rigorous method to investigate this is to take a sequence of functions f such that f[n + 1] = sqrt{x + f[n](x)} = g{x}{f[n](x)}, where g is a family of functions such that g{x} : [–x, ♾) —> [0, ♾), where g{x}(t) = sqrt(x + t). For every real x, g{x} is continuous, so if f converges, then f(x) converges to L(x), such that g{x}[L(x)] = L(x). Therefore, sqrt[x + L(x)] = L(x) = [x + L(x)] – x = sqrt[x + L(x)]^2 – x, hence sqrt[x + L(x)]^2 – sqrt[x + L(x)] – x = 0. Notice that sqrt[x + L(x)]^2 – sqrt[x + L(x)] – x = sqrt[x + L(x)]^2 + 2·(–1/2)·sqrt[x + L(x)] + 1/4 – (x + 1/4) = (sqrt[x + L(x)] – 1/2)^2 – sqrt(x + 1/4)^2, so |sqrt[x + L(x)] – 1/2| = sqrt(x + 1/4), which also demands that x > –1/4 or x = –1/4.

For sqrt[x + L(x)] < 1/2 or sqrt[x + L(x)] = 1/2, sqrt[x + L(x)] – 1/2 = –sqrt(x + 1/4), and for sqrt[x + L(x)] > 1/2 or sqrt[x + L(x)] = 1/2, sqrt[x + L(x)] – 1/2 = sqrt(x + 1/4). Also, since g{x}(t) > 0 or g{x}(t) = 0, it follows that L(x) = 0 or L(x) > 0. For sqrt[x + L(x)] < 1/2, it follows that 1/2 – sqrt(x + 1/4) < 1/2, so –sqrt(x + 1/4) < 0, which is trivial. So x + L(x) = 1/4 – sqrt(x + 1/4) + x + 1/4 = x + 1/2 – sqrt(x + 1/4), which is equivalent to L(x) = 1/2 – sqrt(x + 1/4). Notice, though, x + L(x) = [1/2 – sqrt(x + 1/4)]^2, so sqrt[x + L(x)] = |1/2 – sqrt(x + 1/4)| = 1/2 – sqrt(x + 1/4), implying 1/2 – sqrt(x + 1/4) = 0 or 1/2 – sqrt(x + 1/4) > 0, which means sqrt(x + 1/4) = 1/2 or sqrt(x + 1/4) < 1/2, so x = 0 or x < 0. This encompasses x being in the interval [–1/4, 0]. L(x) = 1/2 – sqrt(x + 1/4) for x in [–1/4, 0]. For sqrt[x + L(x)] > 1/2, sqrt[x + L(x)] = 1/2 + sqrt(x + 1/4), and this works for [–1/4, ♾). sqrt[x + L(x)] = 1/2 implies L(x) = 1/4 – x, and one has that 1/2 – sqrt(x + 1/4) = 1/2 + sqrt(x + 1/4) = 1/2, which requires x = –1/4, and which requires L(–1/4) = 1/2. g{–1/4}(1/2) = 1/2, so this works regardless of which L is chosen.

Using a fixed-point theorem, it is required that |g{x}'[L(x)]| < 1. g{x}'[L(x)] = [1 + L'(x)]/(2·g{x}'[L(x)]) = [1 + L'(x)]/[2·L(x)] = (1 +/– 1/[2·sqrt(x + 1/4)])/[1 +/– 2·sqrt(x + 1/4)] = (+/–)/[2·sqrt(x + 1/4)], so 1/[2·sqrt(x + 1/4)] < 1 is required. This gives that 1/2 < sqrt(x + 1/4), which means 1/4 < x + 1/4, which implies x > 0. So f converges to L defined by L(x) = 1/2 + sqrt(x + 1/4) for all x > 0, regardless of what f[0] is.

Now we must consider the case x = 0, which the fixed-point theorem does not determine. For this, it matters what f[0](0) is. g{0}(t) = sqrt(t), so f[n](0) = {f[0](0)}^(1/2^n). So L(0) = 0 if f[0](0) = 0, and L(0) = 1 if f[0](0) > 1. This means L(0) is actually not well-defined. So simply, f converges to L, where L : (0, ♾) —> R, L(x) = 1/2 + sqrt(x – 1/4).

It is now that we can solve the equation L(x) = x, which the video does, but I suggest a method that does not involve extraneous solutions. Namely, x = 1/2 + sqrt(x + 1/4) is equivalent to x + 1/4 = 3/4 + sqrt(x + 1/4) = sqrt(x + 1/4)^2, so sqrt(x + 1/4)^2 – sqrt(x + 1/4) – 3/4 = sqrt(x + 1/4)^2 – sqrt(x + 1/4) + 1/4 – 1 = [sqrt(x + 1/4) – 1/2]^2 – 1 = [sqrt(x + 1/4) – 3/2]·[sqrt(x + 1/4) + 1/2] = 0, which is equivalent to sqrt(x + 1/4) – 3/2 = 0, which is equivalent to sqrt(x + 1/4) = 3/2, which is equivalent to x + 1/4 = 9/4, which implies x = 2.

angelmendez-rivera

#suggestion
If
Where t=x^-2x+2
Find the sum of all real x that satisfy the system.

abhinavbhutadab

My solution:

Define sqrt(x+sqrt(x+...)) recursively with a_0 = x>=0, a_n+1 = sqrt(x+a_n). Assuming (a_n) is convergent with limit a it follows that a = sqrt(x+a) => a^2 - a - x = 0 => a_(+/-) = (1+/-sqrt(4x))/2. The rest is obvious.

It's left to prove that (a_n) is in convergent.

1. (a_n) is clearly monoton increasing as x>=0.

2. (a_n) is bounded by x =< a_n =< a.

Thus (a_n) convergent.

IsomerSoma

By the way, 0 is _not_ a solution to the equation. The infinitely nested radical is not even defined at 0, since it can be equal to 0 or 1, depending on the initial condition chosen.

angelmendez-rivera

شكرا جزيلا اعجبتني طريقه الثانيه بارك الله فيك

sabrenhatam

x=sqrt(x+sqrt(x+...)). It can be shown that the infinite radical is well-defined for nonnegative x.

Square both sides: x^2 = x+sqrt(x+sqrt(x+...)), so x^2 - x = sqrt(x+sqrt(x+...)) = x. Thus x^2-2x = 0, so x=0 or x=2.

EDIT: Also I'm going to nitpick something you said in method 1. The closed form of the function is (1+sqrt(1+4x))/2 for x>0, but for x=0 this doesn't work: it would give you f(0) = 1, which is wrong (obviously f(0) = 0). For x=0, the other solution to the quadratic is the correct one. I point this out because this is a good, nontrivial example of a sequence of continuous functions which converges to a discontinuous function.

seanfraser