problem x =√(a - √(a + √(a - √(a + ... The leading radical pattern including sign change repeats starting at the third radical sign in from the left. Since this pattern is then repeated to infinity and equals x, we have x =√(a - √(a + √(a - √(a + ... = √[a - √(a + x)] This is a closed form expression that is highly dependent upon the value of a. The equation is: x = √[a - √(a + x)] x² = a - √(a + x) √(a + x) = a - x² Squaring both sides (which may introduce extraneous solutions): a + x = ( a - x² )² = a² - 2 a x² + x⁴ x⁴- 2 a x²- x + ( a² - a ) = 0 This is a quartic with no cubic term. We can attempt to factor this into the product of 2 quadratics. We must keep in mind we are looking for positive x values. x⁴- 2 a x²- x + ( a² - a ) = 0 This factors into (-x² + b x + a )[-x²- b x + (a-1)] = 0 for b an unknown coefficient. Solving for b: x⁴- [(a-1)+b²-a ]x² +[b(a-1)-a b]x + a²-a = 0 -2 a = b²-1 Solution for b is b = 1 (-x² + x + a )[-x²- x + (a-1)] = 0 Use the 0 product property to find x solutions. 2 solution sets: 1) from factor -x² + x + a = 0 x = 1/2 - √(4a+1) / 2 x = 1/2 + √(4a+1) / 2 For the radicand to be positive we need 4a+1 ≥ 0 4a ≥ -1 a ≥ -1/4 The correct positive branch is x = 1/2 + √(4a+1) / 2 This includes where a = 0 since 0 ≥ -1/4 2) from factor -x² - x + a -1 = 0 x = 1/2 - √(4a-3) / 2 x = 1/2 + √(4a-3) / 2 Here we need 4a-3 ≥ 0 4a ≥ 3 a ≥ 3/4 The correct positive branch is x = 1/2 + √(4a-3) / 2 This doesn't work when a = 0. answer 1/2 + √(4a+1) / 2
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? ☺
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
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?
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_
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
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.
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.
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* --------------------------------------- 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
The solution is the second one (and with minus), because it's the only one compatible with a=0.
problem
x =√(a - √(a + √(a - √(a + ...
The leading radical pattern including sign change repeats starting at the third radical sign in from the left. Since this pattern is then repeated to infinity and equals x, we have
x =√(a - √(a + √(a - √(a + ...
= √[a - √(a + x)]
This is a closed form expression that is highly dependent upon the value of a.
The equation is:
x = √[a - √(a + x)]
x² = a - √(a + x)
√(a + x) = a - x²
Squaring both sides (which may introduce extraneous solutions):
a + x = ( a - x² )²
= a² - 2 a x² + x⁴
x⁴- 2 a x²- x + ( a² - a ) = 0
This is a quartic with no cubic term. We can attempt to factor this into the product of 2 quadratics. We must keep in mind we are looking for positive x values.
x⁴- 2 a x²- x + ( a² - a ) = 0
This factors into
(-x² + b x + a )[-x²- b x + (a-1)] = 0
for b an unknown coefficient. Solving for b:
x⁴- [(a-1)+b²-a ]x² +[b(a-1)-a b]x + a²-a = 0
-2 a = b²-1
Solution for b is
b = 1
(-x² + x + a )[-x²- x + (a-1)] = 0
Use the 0 product property to find x solutions.
2 solution sets:
1) from factor -x² + x + a = 0
x = 1/2 - √(4a+1) / 2
x = 1/2 + √(4a+1) / 2
For the radicand to be positive we need
4a+1 ≥ 0
4a ≥ -1
a ≥ -1/4
The correct positive branch is
x = 1/2 + √(4a+1) / 2
This includes where a = 0 since 0 ≥ -1/4
2) from factor -x² - x + a -1 = 0
x = 1/2 - √(4a-3) / 2
x = 1/2 + √(4a-3) / 2
Here we need
4a-3 ≥ 0
4a ≥ 3
a ≥ 3/4
The correct positive branch is
x = 1/2 + √(4a-3) / 2
This doesn't work when a = 0.
answer
1/2 + √(4a+1) / 2
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? ☺
It's a great observation! 😄
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
Very nice!
The most important part is missing. The technique is educative.
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?
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_
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
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.
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.
Of course x is non negative.
a >= 1 and a = 0 (x = 0 then).
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*
---------------------------------------
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