An Interesting Infinite Radical

[@pje188yt]

Its interesting that when a = 2, the solution is phi - 1, and if you start the alternating signs with +, the solution is phi.

[@jorgechavesfilho]

The solution is the second one (and with minus), because it's the only one compatible with a=0.

[@MarieAnne.]

The solution x = (1−√(4a+1))/2 only works with a = 0. For all other values a > 0, this gives x < 0, which is clearly not true from original equation. For a ≥ 1, the solution x = (√(4a−3)−1)/2 seems to be the only one that works. Example: a = 7 √(7−√(7+√(7−√(7+√(7−√(7+√(7−√(7+√(7−√(7)))))))))) ≈ 2.0000002609 (√(4a−3)−1)/2 = (√(4*7−3)−1)/2 = (√(25)−1)/2 = (5−1)/2 = 2

[@onegreengoat9779]

Using your notation, we are assuming that the sequential terms converge if an x exists. We know to satisfy the domain and the range of the square root function both a>=0 and x>=0 are real numbers. If a=0, then x=0 is the only possible solution, which is the (1-sqrt(1+4a))/2 solution. If 0

[@alipourzand6499]

Great video! There is a connection between repeating nested radicals and repeating continued fractions. The equivalent alternating continued fraction will lead to the same result. Now the million dollar question is how square root and multiplicative inverse are related? ☺

[@SyberMath]

It's a great observation! 😄

[@scottleung9587]

Very nice!

[@Kaurenjalalifarahani]

let gamma = that expression , then gamma squared = a - gamma

[@aesthetics_ai]

as a physics student i have watched a lot of this kind of video , speically with towet power and immediatly i knew it was from the start the idea of sqrt ( a+ sqrt (a-x) = x

[@chaosredefined3834]

Problem: x = sqrt(a - sqrt(a + sqrt(a - sqrt(a + ........ Now, as syber points out, we can just sub in x, giving us x = sqrt(a - sqrt(a + x)). But, we can take it a step further, by introducing y = sqrt(a + x). This means that x = sqrt(a - y). So, y^2 = a + x, and x^2 = a - y. Subtracting these, we get y^2 - x^2 = (a + x) - (a - y), or y^2 - x^2 = x + y. By difference of two squares, we get (x + y)(y - x) = (x + y). Case 1: x + y = 0. If this is the case, that equation holds, so we have y = -x, so x = sqrt(a + x), or x^2 = x + a. This means that x^2 - x = a, so 4x^2 - 4x = 4a, and 4x^2 - 4x + 1 = 4a + 1. Therefore (2x - 1)^2 = sqrt(4a + 1)^2, so 2x - 1 = +/- sqrt(4a + 1), and thus x = (1 +/- sqrt(4a + 1))/2. Case 2: x + y =/= 0. If this is the case, we can divide by x + y, giving us y - x = 1. So, y = 1 + x, and so x = sqrt(a - 1 - x). This gives us x^2 = a - 1 - x => x^2 + x = a - 1 => 4x^2 + 4x = 4a - 4 => 4x^2 + 4x + 1 => 4a - 3 => 2x + 1 = +/- sqrt(4a - 3) => x = (-1 +/- sqrt(4a - 3))/2. So, answers are from the set: (1 +/- sqrt(4a + 1))/2 and (-1 +/- sqrt(4a - 3))/2

[@aarona3144]

Wolfram Alpha doesn't use AI to solve these problems. Before AI was even a thing WA was able to solve all of these problems using mathematical algorithms. Most likely it doesn't understand your query because of the ellipsis in your query. What might be nice though is if it had the ability to interpret that in the way that a user would intend it to. Maybe that's an update that they should add to their query parser algorithm?

[@vladimirkaplun5774]

The most important part is missing. The technique is educative.

[@yurenchu]

_Answer_ : √(a - √(a + √(a - √(a + √(...))))) = = 0 for a=0 = -½ + √(a - ¾) for a≥1 _Calculation_ : Let x = √(a - √(a + √(a - √(a + √(...))))) y = √(a + √(a - √(a + √(a - √(...))))) . x = √(a - y) y = √(a + x) x² = (a - y) y² = (a + x) Subtract the last two equations from each other: x² - y² = (a - y) - (a + x) x² - y² = -y - x x² + x = y² - y x² + x + 1/4 = y² - y + 1/4 (x + 1/2)² = (y - 1/2)² (x + 1/2) = (y - 1/2) OR (x + 1/2) = -(y - 1/2) x = y - 1 OR (x + 1/2) = -y + 1/2 x = (y - 1) OR x = -y Combine either solution with y² = (a + x) : Case #1: x = (y - 1) y² = a + (y-1) y² - y = a - 1 y² - y + 1/4 = a - 3/4 (y - 1/2)² = (4a - 3)/4 y - 1/2 = ± (1/2)√(4a - 3) y = 1/2 ± (1/2)√(4a - 3) ... x = (y-1) ... y = [1 + √(4a - 3)]/2 & x = [-1 + √(4a - 3)]/2 [case #1.1] OR y = [1 - √(4a - 3)]/2 & x = [-1 - √(4a - 3)]/2 [case #1.2] Check that both x≥0 and y≥0 : Case #1.1 : x = [-1 + √(4a - 3)]/2 ≥ 0 ==> √(4a - 3) ≥ 1 ==> (4a - 3) ≥ 1 ==> 4a ≥ 4 ==> a ≥ 1 y = [1 + √(4a - 3)]/2 ≥ 0 ==> √(4a - 3) ≥ -1 ==> (4a - 3) ≥ 0 ==> 4a ≥ 3 ==> a ≥ 3/4 ==> Solution for a ≥ 1 , x = [-1 + √(4a - 3)]/2 Case #1.2 : y = [1 - √(4a - 3)]/2 ≥ 0 ==> -√(4a - 3) ≥ -1 ==> √(4a - 3) ≤ 1 ==> (4a - 3) ≤ 1 ==> 4a ≤ 4 ==> a ≤ 1 x = [-1 - √(4a - 3)]/2 ≥ 0 ==> -√(4a - 3) ≥ 1 ==> not possible, as LHS is non-positive and RHS is positive. ==> No values of a with solutions. Case #2: x = -y y² = a + (-y) y² + y = a y² + y + 1/4 = a + 1/4 (y + 1/2)² = (4a + 1)/4 y + 1/2 = ±(1/2)√(4a + 1) y = -1/2 ± (1/2)√(4a + 1) ... x = -y ... y = [ -1 + √(4a + 1)]/2 & x = [1 - √(4a + 1)]/2 [case #2.1] OR y = [ -1 - √(4a + 1)]/2 & x = [1 + √(4a + 1)]/2 [case #2.2] Check that both x≥0 and y≥0 : Case #2.1 : y = [-1 + √(4a+1)]/2 ≥ 0 ==> √(4a+1) ≥ 1 ==> (4a+1) ≥ 1 ==> 4a ≥ 0 ==> a ≥ 0 x = [1 - √(4a+1)]/2 ≥ 0 ==> -√(4a+1) ≥ -1 ==> √(4a+1) ≤ 1 ==> 4a+1 ≤ 1 ==> 4a ≤ 0 ==> a ≤ 0 ==> Only solution for a = 0 : x = 0 and y = 0 . Case #2.2 : x = [1 + √(4a+1)]/2 ≥ 0 ==> √(4a+1) ≥ -1 ==> 4a+1 ≥ 0 ==> 4a ≥ -1 ==> 4a ≥ -1/4 y = [-1 - √(4a+1)]/2 ≥ 0 ==> -√(4a+1) ≥ 1 ==> not possible, as LHS is non-positive and RHS is positive. ==> No values of a with solutions. Conclusion: for a = 0 , x = 0 for a ≥ 1 , x = (-1+√(4a - 3))/2 = -½ + √(a - ¾) Check into original expression: a = 0 : √(a - √(a + √(a - √(a + √(...))))) = √(0 - √(0 + √(0 - √(0 + √(...))))) = 0 ==> works out! a ≥ 1 : √(a - √(a + √(a - √(a + √(...))))) = √(a - √(a - ½ + √(a - ¾) )) = = √(a - √(a - ¾ + ¼ + √(a - ¾) )) = √(a - √( (a - ¾) + ¼ + 2*(½)*√(a - ¾) )) = √(a - √( (√(a-¾) + ½)² )) = √(a - (√(a-¾) + ½) ) = √(a - √(a-¾) - ½ ) = √(a-¾ - √(a-¾) - ½ + ¾) = √( (a-¾) - 2*½*√(a-¾) + ¼) = √( (√(a-¾) - ½)² ) = (√(a-¾) - ½) = √(a - ¾) - ½ = -½ + √(a - ¾) ==> works out!

[@s1ng23m4n]

root(a - root(a + root(...))) = x root(a + root(a - root(...))) = y x^2 = a - y y^2 = a + x y^2 - x^2 = x + y (y - x)(y + x) = x + y (y + x)(y - x - 1) = 0 y = -x only if x = y = a = 0 y - x - 1 = 0 y = x + 1 x^2 = a - y = a - x - 1 x^2 + x + 1 - a = 0 x = (root(4a - 3) - 1) / 2

[@icfj77]

I have observed the four solutions without calculating it again, and only the one you say works. We need to take into account that a is greater than one, and x must be greater than 0.

[@stephenshefsky5201]

Assuming x is real, the original equation x = sqrt(a - sqrt(a + ...)) implies x >= 0, which implies from the solution x = (sqrt(4a - 3) - 1)/2 that a >= 1.

[@s1ng23m4n]

Of course x is non negative. a >= 1 and a = 0 (x = 0 then).

[@RashmiRay-c1y]

Let the given expression be E. The, we see that E^2=a-√(a+E) which gives a^2-a(2E^2+1) +(E^4-E)=0 whence, a=E^2+E+1, a=E^2-E. The first does not yield a real E for a=0. Recalling that E is positive, we get, from E^2_E-a=0, E=1/2[√(4a+1) +1]

[@guyhoghton399]

Nice bit of thinking outside the box. I think this is a bit simpler though: *_x = √( a - √( a + {√( a - √(a + ...) )} ) )_* ∴ *_x = √( a - √(a + x) )_* ... ① Let *_y = √(a + x)_* ∴ _y² = a + x_ ... ② _x = √(a - y)_ from ① ⇒ _x² = a - y_ Subtracting this from ②: _y² - x² = y + x_ ⇒ _(y + x)(y - x) = y + x_ ⇒ *_(y + x)(y - x - 1) = 0_* Case (i) *_y + x = 0_* ⇒ _x = -√(a + x)_ ... ③ ⇒ _x² = a + x_ ⇒ *_x² - x - a = 0_* However ... if _a_ is real then from its definition _x = √(a - √(a + ...))_ must be real and non-negative, which is contradicted by ③ unless _x = a = 0._ So *_x = 0_* is the only valid solution in this case, with _a = 0._ Case (ii) *_y - x - 1 = 0_* ⇒ _√(a + x) = x + 1_ ⇒ _a + x = x² + 2x + 1_ ⇒ *_x² + x - (a - 1) = 0_* ⇒ *_x = ½( -1 + √(4a - 3) )_* ( _x ≥ 0_ ) This solution is OK as long as _a ≥ 1_ E.g. _a = 7 ⇒ x = 2_

[@MarieAnne.]

x = √(a − √(a + √(a − √(a + ...)))) → a ≥ 0 Note 1: √(a − √(a + √(a − √(a + ...)))) ≥ 0 → x ≥ 0 Note 2: − √(a + √(a − √(a + ...))) ≤ 0 → √(a − √(a + √(a − √(a + ...)))) ≤ √a → x ≤ √a x = √(a − √(a + x)) x² = a − √(a + x) x² − a = −√(a + x) x^4 − 2ax² + a² = a + x a² − (2x²+1)a + (x^4−x) = 0 This is a quartic equation in x, but a quadratic equation in a. So we solve for a: a = [(2x²+1) ± √((2x²+1)² − 4(x^4−x))] / 2 a = [(2x²+1) ± √(4x^4 + 4x² + 1 − 4x^4 + 4x)] / 2 a = [(2x²+1) ± √(4x² + 4x + 1)] / 2 a = [(2x²+1) ± (2x+1)] / 2 a = x² + x + 1, a = x² − x We now have two quadratic equations in x: x² + x + 1 − a = 0, x² − x − a = 0 x = (−1±√(4a−3))/2, x = (1±√(4a+1))/2 When a = 0, then x = 0. The only solution where this works is x = (1−√(4a+1))/2 However, for all other values of a, this gives x < 0. The solution x = (−1−√(4a−3))/2 also gives x < 0. Since x ≥ 0, we can eliminate these two solutions for a > 0. From original equation we got x ≤ √a But the solution x = (1+√(4a+1))/2 = 1/2 + √(a+1/4) > √a So we can eliminate this solution, leaving us with *x = (√(4a−3)−1)/2 ≥ 0 → a ≥ 1* Solutions: For a = 0 → *√(a − √(a + √(a − √(a + ...)))) = 0* For a ≥ 1 → *√(a − √(a + √(a − √(a + ...)))) = (√(4a−3)−1)/2* --------------------------------------- Check using a = 7 √(7) ≈ 2.64575131106 √(7-√(7)) ≈ 2.08668365809 √(7−√(7+√(7))) ≈ 1.97338263207 √(7−√(7+√(7−√(7)))) ≈ 1.99639358455 √(7−√(7+√(7−√(7+√(7−√(7+√(7−√(7+√(7−√(7)))))))))) ≈ 2.0000002609 Value seems to be converging to 2 (√(4a−3)−1)/2 = (√(4*7−3)−1)/2 = (√(25)−1)/2 = (5−1)/2 = 2