A fractional part limit.

We look at a nice limit related to the fractional part function.

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[mrmanning]

Find yourself someone who loves you the way Michael Penn loves the floor function

[gnarlybonesful]

Wow very counterintuitive - the fractional part is converging to 1 even though 1 cannot be a fractional part of any number. Neat problem.

[Czeckie]

gotta admit this was unexpected! the solution method was pretty cute, but I am extremely surprised the limit exists. I would guess the sequence oscillates.

[f-th]

Two things I wanna say:

1) Let A = 2+sqrt2, and B = 2-sqrt2. That A^n + B^n is an integer follows from the fact that A is an algebraic integer, so its powers and its conjugate B are also algebraic integers (and so B^n as well). The conjugate of A^n is also B^n, so they are both the roots of the same polynomial with integer coefficients (since they are algebraic integers). So if you take (x-A^n)(x-B^n), the coefficients are integers, in particular if you expand this product, -(A^n+B^n) is an integer, and therefore A^n+B^n is an integer.

2) The conclusion of the problem is that higher and higher powers of A get closer to being integers. Aside from A being an algebraic integer, this hinges on the fact that B was between 0 and 1. You can generalize this by allowing A to be any algebraic integer, and all its other conjugates are complex numbers with absolute value less than 1. For example, if you take A = the golden ratio phi, and its conjugate is B = -1/phi with 0 < |B| < 1, you'll see that A^n + B^n is an integer, and A^n gets closer and closer to being an integer.

[Adam-rtir]

3:20-6:00 alternatively consider automorphism f(a+sqrt(2)b) = a-sqrt(2)b of Z[sqrt(2)], and notice the expression is invariant under it, so it must be an integer.

[ivanklimov]

looking at a limit of a fractional part is very interesting, usually in limits we completely disregard fractional parts since they have such little impact on a limiting expression that goes to infinity. great video overall.

[lexyeevee]

fun fact: in many programming languages, you can compute the fractional part using the modulo operator, as `x % 1`, since x is congruent to {x} mod 1 if you extend modular arithmetic to the reals

[SakiKnin]

Man, u are posting vids every day, what a great channel to be part of. Thanks!

[experiencedimposter]

For the limit you get that {(2+√2)ⁿ}=1-(2-√2)ⁿ. This is true for all natural n, but not for all real numbers approaching infinity. The series does not converge unless only considering integers, so I think it needs to be mentioned.

[criskity]

Trying a few examples on my calculator: seems it converges to 1. Not rigorous, but hey. And that's a good place to stop.

[tomasstride]

Fascinating. The limit of the fractional part is not a fraction, which of course is entirely consistent . It shows the mysterious nature of limits.

[ritam]

Solved this myself. This technique is taught in coaching (cram school) under "Binomial Theorem" in my country.

[annevanderbijl]

Working with fractional parts, that's kinda like working in modulo 1, right?

[ChefSalad]

This reminds me of my favorite approximation for π: 3+(√2)/10. It's kind of like 3.1 but where you replace the tenths digit (the 1 in 3.1) with √2. It's unreasonably good as an approximation as well. It's short by only ~½‱ (~0.005%).

For those unfamiliar with the permyriad symbol (‱), it means "for every ten thousand", similar to the percent (% - "for every hundred") and permille (‰ - "for every thousand") symbols.

[TomMarAlem]

Totally intuitive. The square root of two has gotta be between one and two so the numerator of the sum has gotta be 3.
I find it striking how fast it must be converging.

Is there a gauge to measure how fast a function converges?
Not having revised maths in quite a while I would presume it to be the derivative.

[aa-lrjk]

Would'n be easier just point out that {(1+a)^n}+{(1-a)^n} is always integer, where a = 2^0.5, and since {(1-a)^n} tends to 0 as n tends to infinity, we have lim{(1+a)^n}=1?

[henry]

Another way to show that (2+sqrt(2))^n+(2-sqrt(2))^n is an integer:
Let f(n) = (2+sqrt(2))^n+(2-sqrt(2))^n, then:
f(0) = 2
f(1) = 4
f(n+2) = 4f(n+1) - 2f(n) (because 2+sqrt(2) and 2-sqrt(2) are the roots of x^2 = 4x -2)
So f(n) is an integer by induction (where n is a positive integer).

[elasmarsaadallah]

!!!

Do not confuse, in general we have :
lim {Kₙ} ≠ { lim Kₙ } and in some case the first one exist but the seconde one don't exist :
For example for Kₙ = (-1)ⁿ :
lim {Kₙ} = 0 and lim Kₙ ∄ ⇒ { lim Kₙ } ∄

To see that lim {Kₙ} ≠ { lim Kₙ }, we can take the simple example Kₙ = 1-1/n :

We have { Kₙ } = Kₙ so :

lim { Kₙ } = lim Kₙ = 1

{ lim Kₙ } = {1} = 0

[kangjaeho]

Such a cool intuitive problem! I had fun solving it.

[jfcrow]

Also {(1+sqrt(2))^n} limit equals zero