Hello everyone, here is some of the links I was talking about: math.stackexchange.com/questions/2191147/a-infinite-and-alternating-square-root-of-2 math.stackexchange.com/questions/4049946/the-infinitely-nested-radicals-problem-and-ramanujans-wondrous-formula www.dgp.toronto.edu/~mjmcguff/math/nestedRadicals.pdf
Alright, my time to shine and present a solution that doesn't involve quartics! We're going to define two numbers, namely: a = sqrt(2 + sqrt(2 - sqrt(2 + sqrt(2 - ...)))) b = sqrt(2 - sqrt(2 + sqrt(2 - sqrt(2 + ...)))) It's easy to see that a = sqrt(2 + b) and b = sqrt(2 - a). Square both equations to get a^2 = 2 + b and b^2 = 2 - a, then subtract both to get a^2 - b^2 = a + b Setting everything on the left side gets us to a^2 - bˆ2 - (a + b) = 0, which can be simplified to (a + b)(a - b) - (a + b) = 0 and then (a + b)(a - b - 1) = 0 We now have two cases, the first of which being a + b = 0. Both a and b are defined as square roots, which implies that they cannot be negative numbers, thus the only way they could sum to 0 is if both are 0; however, this breaks the equations a = sqrt(2 + b) and b = sqrt(2 - a). The other case takes us to a - b - 1 = 0, which gives us b = a - 1, and therefore aˆ2 = 2 + (a - 1) which leads us to a^2 - a - 1 = 0, whose only positive root is a = phi, which is indeed the correct answer.
x=1 does work *if* you allow for sqrt(1) = -1, which technically is a solution. In fact, the other two solutions also work if you let yourself take a different combination of negative values of the square root function
I was chatting to some people and this integral came up, Integral of (e^x * sec^2(x)(sin(2x) + 2)dx its not super hard but its pretty interesting. Thought you might enjoy it.
Let x be the solution, since it is an iteration, √(2 + √( 2 - x )) = x ( xx - 2 )^2 = 2 - x x^4 - 4xx + x + 2 = 0 ( x - 1 )( x + 2 )( xx - x - 1 ) = 0 x = 1 or -2 or ( 1 + √5 )/2 or ( 1 - √5 )/2 The two negative values are rejected, square root is non-negative, Checking shows ( 1 + √5 )/2 is a solution but 1 is not, Note that this is the golden ratio 🤗
The (-2) value seems to work right when one plugs it in for the value of x in the equation unlike 1....doesnt that contradict of what you say....i know negetive values can't be accepted but why and why exactly is - 2 working when plugged in...
2 - x = x^4 - 4x^2 + 4 ... (x-1)(x+2)(x^2 - x - 1) = 0 x can not be negative so x != -2 and x != -1/φ x definitely greater than 1 so x != 1 at the rest we have x = φ - golden ratio
x = sqrt(2+sqrt(2-sqrt(...))) = sqrt(2+sqrt(2-x)) x^2 - 2 = sqrt(2-x) The condition : 1. x >= 0 2. x > sqrt(2) 3. 2-x => 0 -> x =< 2 4. x^2 - 2 >= 0 => x =< -sqrt(2) (doesnt satisfy c1) or x => sqrt(2) The condition for this equation is going to be sqrt(2) < x =< 2 (between sqrt(2) and 2 or equal 2) x^4 - 4x^2 + 4 = 2-x x^4 - 4x^2 + x + 2 = 0 x^2 * (x^2 - 4) + (x+2) = 0 (x+2)(x^2 * (x-2) + 1) = 0 Divide by x+2 because x = -2 is not a solution x^3 - 2x^2 + 1 = 0 x^3 = 2x^2 - 1 Now im going to try use integer factorization in hope that one of the solution is an integer x^3 = (sqrt(2)x + 1)(sqrt(2)x - 1) This equation on the right has 2 factorization term meanwhile on the left you have 3 factorization term, let's try to invite 1 as the 3rd factor on the right hand side x^3 = (sqrt(2)x + 1)(sqrt(2)x - 1)*1 So there are 2 possibilities : (sqrt(2)x+1 = 1 AND sqrt(2)x - 1 = 1) OR x^3 = 1 (sqrt(2)x = 0 AND sqrt(2)x = 2) OR x = 1 (x = 0 AND x = sqrt(2)) [either one doesnt satisfy) OR x = 1 x = 1 x^3 - 2x^2 + 1 = (x-1)(x^2 - x - 1) x^2 - x - 1 = 0 as we all know the above solution is either x = -1/phi or x = phi x = -1/phi = -0.618 doesnt satisfy and so does x = 1 So the only solution is x = phi = 1.618 which make sense because phi^2 = phi+1 = 2.618 and that number is bigger than the 2 in the sqrt function in the very beginning
Okay... and now solve this for an exact value of x without using approximation... √(2 + √(2 - √( 2 + x ))) = x (...if you've finished it you can also try this: √(2 - √(2 + √( 2 - x ))) = x ;) )
Why on earth would you make this so complicated? Let the whole thing equal x. If you square it and subtract 2, you get x again. In other words, x^2 - x - 2 = 0. This is equal to (x-2)*(x+1).
Ramanujan summation is just a way of assigning a value to an expression. The actual Cauchy sum diverges so u can't say inf =-1/12. The real value is "inf" or more rigorously, the sum doesn't have a value
Hello everyone, here is some of the links I was talking about:
math.stackexchange.com/questions/2191147/a-infinite-and-alternating-square-root-of-2
math.stackexchange.com/questions/4049946/the-infinitely-nested-radicals-problem-and-ramanujans-wondrous-formula
www.dgp.toronto.edu/~mjmcguff/math/nestedRadicals.pdf
Alright, my time to shine and present a solution that doesn't involve quartics!
We're going to define two numbers, namely:
a = sqrt(2 + sqrt(2 - sqrt(2 + sqrt(2 - ...))))
b = sqrt(2 - sqrt(2 + sqrt(2 - sqrt(2 + ...))))
It's easy to see that a = sqrt(2 + b) and b = sqrt(2 - a). Square both equations to get a^2 = 2 + b and b^2 = 2 - a, then subtract both to get a^2 - b^2 = a + b
Setting everything on the left side gets us to a^2 - bˆ2 - (a + b) = 0, which can be simplified to (a + b)(a - b) - (a + b) = 0 and then (a + b)(a - b - 1) = 0
We now have two cases, the first of which being a + b = 0. Both a and b are defined as square roots, which implies that they cannot be negative numbers, thus the only way they could sum to 0 is if both are 0; however, this breaks the equations a = sqrt(2 + b) and b = sqrt(2 - a).
The other case takes us to a - b - 1 = 0, which gives us b = a - 1, and therefore aˆ2 = 2 + (a - 1) which leads us to a^2 - a - 1 = 0, whose only positive root is a = phi, which is indeed the correct answer.
Damn thats an awesome approach
Wow! This is nice!
@@SyberMath thanks! I'm pretty sure I saw this technique used on another problem you solved a while ago, can't remember which one though.
@@AlexandreRibeiroXRV7 No problem. Me neither! 😁
That's really good
Very nice. If you change 2 with 3 you get a similar equation: √(3 + √( 3 - x )) = x; and the solution is x = 2
Another great explanation, SyberMath! Thanks a lot!
Sure. Thank you!
x=1 does work *if* you allow for sqrt(1) = -1, which technically is a solution. In fact, the other two solutions also work if you let yourself take a different combination of negative values of the square root function
But sqrt function is strictly limited to a non negative range
Only if it is defined as the principle value. It can also be defined as a multi-valued function.
@@FaerieDragonZook a real function can not be multivalued by definition of a function. It would just be a relation between 2 sets otherwise
Another fantastic video! I an currently in calculus, but I learn so many algebra tricks from you!
I was chatting to some people and this integral came up, Integral of (e^x * sec^2(x)(sin(2x) + 2)dx its not super hard but its pretty interesting. Thought you might enjoy it.
Glad to hear that! 💖
Let x be the solution, since it is an iteration,
√(2 + √( 2 - x )) = x
( xx - 2 )^2 = 2 - x
x^4 - 4xx + x + 2 = 0
( x - 1 )( x + 2 )( xx - x - 1 ) = 0
x = 1 or -2 or ( 1 + √5 )/2 or ( 1 - √5 )/2
The two negative values are rejected, square root is non-negative,
Checking shows ( 1 + √5 )/2 is a solution but 1 is not,
Note that this is the golden ratio 🤗
Use formula (where a = 2 in this case) (sqrt(4a-3)+1)(1/2). If the first sign is a minus sign instead of a plus sign, it is (sqrt(4a-3)-1)(1/2).
How are you find this Formula, random indian guys?
First time but passed great time to your live🤗 love it allot💖
Thank you!
Thank you for not naming the video something like "a very GOLDEN problem"
The (-2) value seems to work right when one plugs it in for the value of x in the equation unlike 1....doesnt that contradict of what you say....i know negetive values can't be accepted but why and why exactly is - 2 working when plugged in...
Absolutely superb! Magnific! Mai vrem!
Mulțumesc! Vor veni mai multe...
I love how many of these pop out the golden ratio quadratic equation
Me too! 🤩
I think we have to demonstrate that we have convergence before calculate the limit, but I don't know how to do it.
I knew this had some golden touch to it ✨
really nice golden sotution!!
How did you prove the convergence of sqrt(2+sqrt(2-sqrt(2+sqrt(2-...))))?
This was a golden explanation(solution).
Thank you!
2 - x = x^4 - 4x^2 + 4
...
(x-1)(x+2)(x^2 - x - 1) = 0
x can not be negative so x != -2 and x != -1/φ
x definitely greater than 1 so x != 1
at the rest we have x = φ - golden ratio
Really interesting !
Thanks!
very nice question
x = sqrt(2+sqrt(2-sqrt(...))) = sqrt(2+sqrt(2-x))
x^2 - 2 = sqrt(2-x)
The condition :
1. x >= 0
2. x > sqrt(2)
3. 2-x => 0 -> x =< 2
4. x^2 - 2 >= 0 => x =< -sqrt(2) (doesnt satisfy c1) or x => sqrt(2)
The condition for this equation is going to be
sqrt(2) < x =< 2 (between sqrt(2) and 2 or equal 2)
x^4 - 4x^2 + 4 = 2-x
x^4 - 4x^2 + x + 2 = 0
x^2 * (x^2 - 4) + (x+2) = 0
(x+2)(x^2 * (x-2) + 1) = 0
Divide by x+2 because x = -2 is not a solution
x^3 - 2x^2 + 1 = 0
x^3 = 2x^2 - 1
Now im going to try use integer factorization in hope that one of the solution is an integer
x^3 = (sqrt(2)x + 1)(sqrt(2)x - 1)
This equation on the right has 2 factorization term meanwhile on the left you have 3 factorization term, let's try to invite 1 as the 3rd factor on the right hand side
x^3 = (sqrt(2)x + 1)(sqrt(2)x - 1)*1
So there are 2 possibilities :
(sqrt(2)x+1 = 1 AND sqrt(2)x - 1 = 1) OR x^3 = 1
(sqrt(2)x = 0 AND sqrt(2)x = 2) OR x = 1
(x = 0 AND x = sqrt(2)) [either one doesnt satisfy) OR x = 1
x = 1
x^3 - 2x^2 + 1 = (x-1)(x^2 - x - 1)
x^2 - x - 1 = 0
as we all know the above solution is either x = -1/phi or x = phi
x = -1/phi = -0.618 doesnt satisfy and so does x = 1
So the only solution is x = phi = 1.618 which make sense because phi^2 = phi+1 = 2.618 and that number is bigger than the 2 in the sqrt function in the very beginning
*Such a cool experience* 😎😌
Glad to hear that!
It was also a "golden" experience!
Cool problem!!!
Thanks!
I used synthetic division at last
Okay... and now solve this for an exact value of x without using approximation... √(2 + √(2 - √( 2 + x ))) = x (...if you've finished it you can also try this: √(2 - √(2 + √( 2 - x ))) = x ;) )
Similar to that of a Putnam problem only the number there was 7
Really? I did not know that!
Why on earth would you make this so complicated? Let the whole thing equal x. If you square it and subtract 2, you get x again. In other words, x^2 - x - 2 = 0. This is equal to (x-2)*(x+1).
That’s not right!
@@SyberMath Oh, I missed the minus sign. So you have to end up with a quartic. Never mind. I'll be leaving now.
@@paulkolodner2445 No worries! it would be nice if that was the case. ☺
Nice and easy😀
I got 1,-1 and -2 as solution but 1 is the answer.
Try Ramanujan Summation series.
♾ = -1/12
That is wrong
Ramanujan summation is just a way of assigning a value to an expression. The actual Cauchy sum diverges so u can't say inf =-1/12. The real value is "inf" or more rigorously, the sum doesn't have a value
( 1 + √5 )/2
504//3.03.22. Решение не понятно.
I don’t like this type of sequence…not a well defined series, surely?
It is well defined I think
Let a1=√2 and a_n = √(2-a_(n-1)) if n is even and a_n = √(2+a_(n-1)) if n is odd. Now take the odd indexed terms of this series and you are done
Why not?