In my head, if you denest one it's the same expression so say it equals x. We square both sides and minus one to get the same expression, which is x, equals x^2-1. Just move it over and famous phi formula
I think you made a mistake, you cant suppose inductively that Xn(1+√5)/2 and then prove that X(n+1)>(1+√5)/2 by your own same steps shown from the time stamp 6:20
it's okay, more precisely my above reply meant "i relooked at that section of the video and didn't find a mistake" To state what we have done more rigorously, we first proved "x1 < golden ratio". This is our base case "P(1)". Then we proved "for all n in Z+, if xn < golden ratio then xn+1 < golden ratio" This is our induction step "for all n in Z+, if P(n) then P(n+1)". Proving both the base case and induction step means we have proven "for all n in Z+, P(n)". That is, we have proven "for all n in Z+, xn < golden ratio". I did not check to see if the following statement is true, but i think this was the point you were trying to make: "for all n in Z+, if xn > golden ratio then xn+1 > golden ratio". Even if this statement was true, i don't see how this would imply "we cannot suppose inductively xn < golden ratio". In the induction step for the proof in the video, we really just let n be an arbitrary positive integer and assumed xn < golden ratio. Let me know if there is anything that is unclear
@@iliekmathphysics If we inductively assume that, "For every n in Z+ Xn>(1+√5)/2 then it implies that X(n+1)>(1+√5)/2" [you can check if it is true or not] which is a contradiction to your proof that "For every n in Z+ X(n+1)(1+√5)/2 and why not Xn
This statement: "For all n in Z+, if xn > golden ratio then xn+1 > golden ratio." is true, but i don't see how this means there is a problem with the proof.
Pfft! It's as easy as solving a quadratic equation; just need to form a recursive relationship. Let, √(1+√(1+√(1+...))) = x. So, x^2 = 1 + √(1+√(1+√(1+...))). Thus, x^2 = 1 + x [∵ √(1+√(1+√(1+...))) = x] Solving this quadratic equation leads us to the Golden ratio. ∴ x = (1±√5)/2 = ϕ QED.
![If a sequence converges to x, then all of its subsequences converge to x (Proof) [ILIEKMATHPHYSICS]](http://i.ytimg.com/vi/vNI2eIUQWQg/mqdefault.jpg)
![If a sequence converges to x, then all of its subsequences converge to x (Proof) [ILIEKMATHPHYSICS]](/img/tr.png)
Keep it up bro, this is very good. The way you explain is just different from every math's related youtubers
To prove the fact that it's increasing, use proof by induction...
Base case: x1 = 1, x2 = sqrt(2), so x2 > x1. Easy base case.
Inductive case: Assume x(n) > x(n-1). Then 1 + x(n) > 1 + x(n-1). Therefore, sqrt(1 + x(n)) > sqrt(1 + x(n-1)). Therefore x(n+1) > x(n). Inductive case done.
I just love these! every singe step clearly explained! :))
🎉🎉🎉🎉🎉🎉 thank you very much for your beautiful work!!!!!!
I love your channel
In my head, if you denest one it's the same expression so say it equals x. We square both sides and minus one to get the same expression, which is x, equals x^2-1. Just move it over and famous phi formula
Nice
The continued fraction of phi is 1+1/1+1/1+.... then there is
a relegion of these two formulas .
nice.
bro this is in the as level maths textbook as well 💀
x = sqrt(1+sqrt(1+sqrt(1+...)))
x = sqrt(1+x)
x^2 - x - 1 = 0
so x = phi :)
technically +- phi from the quadratic; but the inf series of roots and ones in the original problem gives only positive
you need to prove convergence first
@@marcosmaldonado7890 ah.
@@yante7 its not to be annoying or anything sometimes doing tricks like that do not work on every case
WAIT THIS IS TRUE
I think you made a mistake,
you cant suppose inductively that Xn(1+√5)/2 and then prove that X(n+1)>(1+√5)/2 by your own same steps shown from the time stamp 6:20
there is no mistake there
@@iliekmathphysics Sorry then, but could you please explain how it is not a mistake?
it's okay, more precisely my above reply meant "i relooked at that section of the video and didn't find a mistake"
To state what we have done more rigorously, we first proved "x1 < golden ratio". This is our base case "P(1)".
Then we proved "for all n in Z+, if xn < golden ratio then xn+1 < golden ratio" This is our induction step "for all n in Z+, if P(n) then P(n+1)".
Proving both the base case and induction step means we have proven "for all n in Z+, P(n)". That is, we have proven "for all n in Z+, xn < golden ratio".
I did not check to see if the following statement is true, but i think this was the point you were trying to make: "for all n in Z+, if xn > golden ratio then xn+1 > golden ratio". Even if this statement was true, i don't see how this would imply "we cannot suppose inductively xn < golden ratio". In the induction step for the proof in the video, we really just let n be an arbitrary positive integer and assumed xn < golden ratio.
Let me know if there is anything that is unclear
@@iliekmathphysics If we inductively assume that, "For every n in Z+ Xn>(1+√5)/2 then it implies that X(n+1)>(1+√5)/2" [you can check if it is true or not] which is a contradiction to your proof that "For every n in Z+ X(n+1)(1+√5)/2 and why not Xn
This statement: "For all n in Z+, if xn > golden ratio then xn+1 > golden ratio." is true, but i don't see how this means there is a problem with the proof.
Pfft! It's as easy as solving a quadratic equation; just need to form a recursive relationship.
Let, √(1+√(1+√(1+...))) = x.
So, x^2 = 1 + √(1+√(1+√(1+...))).
Thus, x^2 = 1 + x [∵ √(1+√(1+√(1+...))) = x]
Solving this quadratic equation leads us to the Golden ratio.
∴ x = (1±√5)/2 = ϕ
QED.
Or just solve x = sqrt(1+x)
You need to prove it converge first before solve this equation.