LonelyWarrior

Life is tough, but this kind of toughness is exactly what I need.

[LonelyWarrior @Ionely_warrior]

No one bothers to read my posts anyway, I’ll say whatever I feel like from now on

[LonelyWarrior @Ionely_warrior]

I have my reasons for hating Turkish netizens.

Today the algorithm actually put a complete, living reason right in front of my eyes.

[Marko @Gudura23FB]

1.75 boyunda herifin her tarafı kas olsa ne olur?

vücut çalışan adam en az 1.85 ama mümkünse 1.90 olmalı ki vücut bir şeye benzesin.

1.75 adamın vücuduyla artistlik yapmaya çalışması komik.

[LonelyWarrior @Ionely_warrior]

If we assign a score of 0 to the reference image on the left and 100 to the one on the right, with the score varying linearly based on the amount of leg hair, and ask two AIs to rate my legs on this scale, ChatGPT gave me a 23, while Grok gave me a 30.

Overall, I have a solid physical foundation, but there is still plenty of room for improvement to transform my body into the shape I desire.

[LonelyWarrior @Ionely_warrior]

@Pravin_00S @grok The lower level is the fitness zone, the upper level is the protein & carb refueling zone 😄

[LonelyWarrior @Ionely_warrior]

@JoKaMath @MaureenNeesp6k So (3) might be correct, but I don't yet know how to prove it.

[LonelyWarrior @Ionely_warrior]

@JoKaMath @MaureenNeesp6k For m=6 we could choose k=100557819

[LonelyWarrior @Ionely_warrior]

@MaureenNeesp6k Strictly speaking, this problem should ask for the limit as h approaches 0 from the positive direction.
The answer is (f(0)+f(1)/2)π. The primary contributions to the integral come from the intervals [-δ,δ] and [1-δ,1].

[LonelyWarrior @Ionely_warrior]

@TheFigen_ I don't think hamburgers are unhealthy.

[LonelyWarrior @Ionely_warrior]

@Rainmaker1973 Actually, the key is that the person’s mass is negligible compared to the mass of the hemisphere, so the combined center of gravity of the person and the hemisphere always lies inside the hemisphere, and thus it does not tip over.

[LonelyWarrior @Ionely_warrior]

Due to the pairwise differences in orbital periods given by Kepler’s third law, rotational symmetry makes the angle α between the planets’ position vectors relative to the star uniformly distributed over [0,2π). Given orbital radii r_i and r_j, the distance between the two planets is given by the law of cosines as

d_ij=sqrt(r_i^2+r_j^2−2r_i r_j cos(α)).

Consequently, the “average” distance between the planets, namely the expected value of their distance, is

E[d_ij]=1/(2π) int[sqrt(r_i^2+r_j^2−2r_i r_j cos(α)),{0,2π}].

These are all elliptic integrals and cannot be expressed in elementary closed form. For comparison, numerical integration gives

E[d_12]=1/(2π) int[sqrt(10−6cos(α)),{0,2π}]≈3.08,
E[d_13]=1/(2π) int[sqrt(26−10cos(α)),{0,2π}]≈5.05,
E[d_14]=1/(2π) int[sqrt(50−14cos(α)),{0,2π}]≈7.04,
E[d_23]=1/(2π) int[sqrt(34−30cos(α)),{0,2π}]≈5.46,
E[d_24]=1/(2π) int[sqrt(58−42cos(α)),{0,2π}]≈7.33,
E[d_34]=1/(2π) int[sqrt(74−70cos(α)),{0,2π}]≈7.93.

Therefore, the answer to the problem is:
On average, the planets closest to planets 1, 2, 3, and 4 are planets 2, 1, 1, and 1, respectively.

(Of course, if the goal is merely to compare their magnitudes, I could try to provide an analytical method for determining the ordering of the corresponding elliptic integrals.)

[LonelyWarrior @Ionely_warrior]

@NoContextHumans Being able to support your entire body in mid-air with just your hands is already really impressive. Right now, even doing push-ups with my feet still supported feels tiring for me.

May I keep getting stronger.

[LonelyWarrior @Ionely_warrior]

Let ω be a cube root of unity. Then
ω^10+ω^8+1=ω+ω^2+1=0.
Thus, ω is a root of x^10+x^8+1. Similarly, ω^2 is also a root.
Therefore, (x−ω)(x−ω^2)=x^2+x+1 is a factor of x^10+x^8+1.

Now let ξ=exp(i π/3). Then ξ^2=ω, and
ξ^10+ξ^8+1=ω^5+ω^4+1=ω^2+ω+1=0.
Thus, ξ is a root of x^10+x^8+1. Since the polynomial has real coefficients, conj(ξ) is also a root.
Therefore, (x−ξ)(x−conj(ξ))=x^2−x+1 is another factor.

We have now found two quadratic factors. Dividing the original polynomial by their product gives the remaining factor. Finally,
x^10+x^8+1=(x^2+x+1)(x^2−x+1)(x^6−x^2+1).

[LonelyWarrior @Ionely_warrior]

For 0≤x<1, the equation is clearly satisfied.

Now suppose
floor(sqrt[20](x))=floor(sqrt[26](x))=n≥1.
Then
n^20≤x<(n+1)^20, n^26≤x<(n+1)^26.
These two intervals overlap if and only if n^26<(n+1)^20, and when they do, their intersection is n^26≤x<(n+1)^20.

For n=1, this condition is obviously satisfied, since 1<2^20.

For n≥2, define g(n)=(n+1)^20/n^26.
Then
0<g(n)/g(n−1)=(1−1/n²)^20 (1−1/n)^6<1,
so g(n) is strictly decreasing.

Thus, we only need to find an integer N≥2 such that
g(N)≥1≥g(N+1).
The possible values of n are then 2,3,…,N.

Now,
g(2)=3^20/2^26=9^10/2^26>8^10/2^26=2^4>1,
g(3)=4^20/3^26=4^20/9^13=4(8/9)^13.
By the binomial theorem,
(9/8)^13=(1+1/8)^13
>1+13/8+78/64+286/512=1127/256>4,
so g(3)<1.

Therefore N=2, and the only possible values of n are 1 and 2.

The solution set is 0≤x<2^20 or 2^26≤x<3^20.

[LonelyWarrior @Ionely_warrior]

My initial idea was quite simple: first estimate the range of d_12(x), and then narrow it down step by step.

If x has N digits in base 12, then clearly
1<=d_12(x)<=11N.
We have N=floor(log_12(x))+1.

Since floor(20262026 log_12(20262026))+1
<20262026*7+1<12^8,
1<=d_12(20262026^20262026)<11*12^8<12^9.
1<=d_12(d_12(20262026^20262026))<=11*9=99.

Since 99=8*12+3, among the positive integers from 1 to 99, the one with the largest d_12 value is 7A=95, and d_12(95)=18.
1<=d_12(d_12(d_12(20262026^20262026)))<=18.

Similarly, among the positive integers from 1 to 18, the one with the largest d_12 value is A=11, and d_12(11)=11.
1<=d_12(d_12(d_12(d_12(20262026^20262026))))<=11.

However, this is still not enough to determine its value uniquely. At this point, I originally thought that I would have to compute the exact value at every stage; otherwise, there would be no way to solve the problem.

Then I noticed another important but easily overlooked property:
x≡d_12(x) mod 11.

Indeed, write x as
x=sum[a_n 12^n,{n,0,N}], 0<=a_n<=11.
Then
d_12(x)=sum[a_n,{n,0,N}],
so
x-d_12(x)
=sum[a_n(12^n-1),{n,0,N}]
=11 sum[a_n(1+12+…+12^(n-1)),{n,1,N}],
which is a multiple of 11.

Therefore, modulo 11,
d_12(d_12(d_12(d_12(20262026^20262026))))
≡20262026^20262026
≡2026^20262026 10001^20262026
≡(11184+2)^20262026 (11*909+2)^20262026
≡2^40524052
≡2^mod(40524052,10) (by Fermat’s little theorem)
≡2^2
≡4.

Since we already know that
1<=d_12(d_12(d_12(d_12(20262026^20262026))))<=11, its value can only be 4.
That solves the problem.

[LonelyWarrior @Ionely_warrior]

Let me see if reposting increases the views👀

[LonelyWarrior @Ionely_warrior]

By the AM-GM inequality,
the original expression
>=2sqrt(2)sin(π/2 cos(x))sin(π/2 sin(x))/(sin(x)cos(x)).
Define the latter appropriately at its removable discontinuities to obtain the continuous function
2sqrt(2)sinc(π/2 cos(x))sinc(π/2 sin(x)).
Its minimum is sqrt(2)π. We now prove this.

Let u=π/2 cos(x), v=π/2 sin(x). Then the problem reduces to finding the minimum of
sinc(u)sinc(v) subject to u^2+v^2=π^2/4.
Since sinc is an even function, it suffices to consider 0<=x<π/2, i.e. u,v>0 or (u,v)=(π/2,0).

The Lagrangian is
sinc(u)sinc(v)-λ(u^2+v^2-π^2/4).
The stationary-point conditions are
sinc’(u)sinc(v)=2λu, sinc(u)sinc’(v)=2λv.
When u,v>0, eliminating λ and simplifying gives
(u cot(u)-1)/u^2=(v cot(v)-1)/v^2.
Since f(t)=(t cot(t)-1)/t^2 is monotonic on (0,π/2), we obtain
u=v=π/(2sqrt(2)).
When (u,v)=(π/2,0), since sinc’(0)=0, it is easy to verify that the stationary-point conditions are satisfied.
Therefore, all the stationary points are
(π/(2sqrt(2)),π/(2sqrt(2))) and (π/2,0).

Since sinc(u)sinc(v) is globally continuous and smooth, its extrema must be attained at stationary points. Its values at these two stationary points are, respectively,
sinc(π/(2sqrt(2)))^2 and 2/π.
Moreover, g(t)=ln(sinc(sqrt(t))) is strictly concave on [0,π^2/4], so
g(π^2/8)>1/2(g(0)+g(π^2/4)),
which, after simplification, gives
sinc(π/(2sqrt(2)))^2>2/π.

Therefore, the minimum of
2sqrt(2)sinc(π/2 cos(x))sinc(π/2 sin(x))
is
2sqrt(2)·π^2/4·2/π=sqrt(2)π.

[LonelyWarrior @Ionely_warrior]

@lEGczaaPvb75343 @sadfatmind70699 @doritenholm Oh really😂

[LonelyWarrior @Ionely_warrior]

By the AM-GM inequality,
the original expression
>=2sqrt(2)sin(π/2 cos(x))sin(π/2 sin(x))/(sin(x)cos(x)).
Define the latter appropriately at its removable discontinuities to obtain the continuous function
2sqrt(2)sinc(π/2 cos(x))sinc(π/2 sin(x)).
Its minimum is sqrt(2)π. We now prove this.

Let u=π/2 cos(x), v=π/2 sin(x). Then the problem reduces to finding the minimum of
sinc(u)sinc(v) subject to u^2+v^2=π^2/4.
Since sinc is an even function, it suffices to consider 0<=x<π/2, i.e. u,v>0 or (u,v)=(π/2,0).

The Lagrangian is
sinc(u)sinc(v)-λ(u^2+v^2-π^2/4).
The stationary-point conditions are
sinc’(u)sinc(v)=2λu, sinc(u)sinc’(v)=2λv.
When u,v>0, eliminating λ and simplifying gives
(u cot(u)-1)/u^2=(v cot(v)-1)/v^2.
Since f(t)=(t cot(t)-1)/t^2 is monotonic on (0,π/2), we obtain
u=v=π/(2sqrt(2)).
When (u,v)=(π/2,0), since sinc’(0)=0, it is easy to verify that the stationary-point conditions are satisfied.
Therefore, all the stationary points are
(π/(2sqrt(2)),π/(2sqrt(2))) and (π/2,0).

Since sinc(u)sinc(v) is globally continuous and smooth, its extrema must be attained at stationary points. Its values at these two stationary points are, respectively,
sinc(π/(2sqrt(2)))^2 and 2/π.
Moreover, g(t)=ln(sinc(sqrt(t))) is strictly concave on [0,π^2/4], so
g(π^2/8)>1/2(g(0)+g(π^2/4)),
which, after simplification, gives
sinc(π/(2sqrt(2)))^2>2/π.

Therefore, the minimum of
2sqrt(2)sinc(π/2 cos(x))sinc(π/2 sin(x))
is
2sqrt(2)·π^2/4·2/π=sqrt(2)π.

[LonelyWarrior @Ionely_warrior]

If n is allowed to be real, the problem becomes trivial. In that case, one could ask some more interesting questions—for example, what is the sum of the reciprocal squares of all such real values of n? This sum converges.

Incidentally, we generally don’t use the letter n for a real-valued variable, but that’s not a big deal; it’s mostly a matter of convention.

[LonelyWarrior @Ionely_warrior]

@sadfatmind70699 @doritenholm This passage is actually about two people: Dante, the medieval Italian poet, and Ding Zhen, the Tibetan internet celebrity who went viral on the Chinese internet a few years ago.

[LonelyWarrior @Ionely_warrior]

@pichuaizi96776 @doritenholm I guess the girlfriend was just trying to make a joke, but when the boyfriend responded in the same way, she felt that he wasn’t being sincere.

By the way, do you guys always speak like that?