LonelyWarrior

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

@Yoku10086 还没我的大🤣

喜欢看腹肌还是晨勃肉棒？ #腹肌 #肉棒 #鸡吧

@pickover One of the oldest videos

Snowball Fight (1897). None of these people are alive today. Whatever became of their hopes and dreams? Credit: Lumière brothers, public domain

@Bad_____bunny99 The grey hair and the mustache are almost everything about his look

Nolan without a gray hair and a mustache

@ROBOHead_3 @HoshiHiiragi 并非，背后男的女的都不知道

@HoshiHiiragi 此男子一天粉丝数翻了20倍我不行了…

lady gaga你最好只是点错了。

@chris_juravich @Diarytells The answer is -π arctan(2ln(2)/π). If possible, I’ll post a detailed solution under this thread later.

@Diarytells This one’s not as easy as it looks. 👀

@Diarytells x->0+: (1+e^x)/2=1+x/2+x^2/4+x^3/12+o(x^3) (1+e^x)/2 (1-x/2)=[1+x/2+x^2/4+x^3/12+o(x^3)]-[x/2+x^2/4+x^3/8+o(x^3)]=1-x^3/24+o(x^3) Numerator: =ln(1-x^3/24+o(x^3))=-x^3/24+o(x^3) Denominator: =x^3/6+o(x^3) limit=(-x^3/24+o(x^3))/(x^3/6+o(x^3))=-1/4

Evaluate the limit

People need to endure some hardships, make some foolish mistakes, and confront their own immaturity in order to grow.

Starting today, I will no longer share any photos of myself—whether selfies or pictures of any part of my body. Anything I’ve posted before will also be deleted. Don’t ask me why. All I want to say is that after going through certain experiences, a person’s views on some things can change drastically, and some harm cannot be erased.

3 sum[(2k+1)k^2(k+1)^2,{k,1,n}] =3 sum[((k+1)^2-k^2)k^2(k+1)^2,{k,1,n}] (Method 1) =3 sum[k^2(k+1)^4-(k-1)^2 k^4-k^4((k+1)^2-(k-1)^2),{k,1,n}] =3 (n^2(n+1)^4)-12sum[k^5,{k,1,n}] (Method 2) =3 sum[(k-1)^4k^2-k^4(k+1)^2+k^2(8k^3+8k),{k,1,n}] =3 (-n^4(n+1)^2)+24sum[k^5,{k,1,n}]+24sum[k^3,{k,1,n}] By comparing the two methods, an appropriate linear combination can eliminate the summation of higher-order terms, thereby reducing the computational workload: (Method 1)*(2/3)+(Method 2)*(1/3) =2 (n^2(n+1)^4)+ (-n^4(n+1)^2)+8 sum[k^3,{k,1,n}] =n^2(n+1)^2(n^2+4n+2)+8sum[k^3,{k,1,n}] To compute sum[k^3,{k,1,n}], we consider: (n+1)^4-1=sum[(k+1)^4-k^4,{k,1,n}]=sum[4k^3+6k^2+4k+1,{k,1,n}]=4sum[k^3,{k,1,n}]+6*n(n+1)(2n+1)/6+4*n(n+1)/2+n ->sum[k^3,{k,1,n}]=n^2(n+1)^2/4 Then, we arrive at the final answer: n^2(n+1)^2(n^2+4n+2)+2n^2(n+1)^2 =n^2(n+1)^2(n+2)^2.

【問題】工夫せよ. #mathworld4

@HhG2cL9UOLRSz4I sum[arctan(2/n^2),{n,1,inf}]=sum[arctan([(n+1)-(n-1)]/[1+(n+1)(n-1)]),{n,1,inf}]=sum[arctan(n+1)-arctan(n-1),{n,1,inf}]=lim_{k->inf} arctan(k+1)+arctan(k)-arctan(0)-arctan(1)=3π/4.

今日は無限級数の問題です 作為的な問題です

@10ios4 @Diarytells フェルマー：この問題については実に巧妙な解法を思いついたが、この余白はそれを書くにはあまりにも狭すぎる。

@Diarytells 非常に解けそうな感じがするが、私は凄く眠たいので寝ます。

@BibianaGre797 @nickback33 @Diarytells Prima seguimi

@nickback33 @Diarytells Apprezzo molto i tuoi post: sono sinceri e incoraggianti. Mi piacerebbe entrare in contatto con te e conoscerti meglio, ma volevo essere rispettosa e fare il primo passo. Se ti interessa, sentiti libero/a di seguirmi e ti seguirò a mia volta 🥰🥰🥰

Evaluate the integral

@7cecee When he dies

At what age, Men be honest !!

You didn't impose any restrictions on f — no conditions on domain, continuity, monotonicity, differentiability, or anything of the sort. In that case, there are infinitely many solutions. For example: f(x)=3^{3^(1+floor(log_3(log_3(x)))) [1+1/2 (log_3(x)/3^(floor(log_3(log_3(x))))-1)^2]}, x>1. It doesn't have to be f(x)=x^3.

#math problem 02-05-2026 Algebra problem. 🎖️

@sonukg4india (100!)^50<(100^100)^50=10^10000 50^(100!)>10^(100)! (100)!>10000 Done.

Compare the two numbers. Show full working

I should definitely take a close look at the geometric principles behind this if I have time. Analyze why this kind of dense packing and periodicity occurs. Write parametric equation for each curve.

@fly_in_X 43 I’m basically just talking to myself on this platform, and hardly anyone likes discussing math with me, so I think this score is already pretty good.

两个X帐号自检工具，一个查评分的，一个查健康度(是否被ban)的。 评分：tweethunter.io/tweepcred-calc… 健康度:hisubway.online/shadowban/ 蓝V新人可以收藏一下，经常要用到。

@grok @black_xkrcd @Riazi_Cafe_en By contrast, I’m only π cos(1) meters tall

@Ionely_warrior @black_xkrcd @Riazi_Cafe_en That's about 74.69 inches—roughly 6'3". Tall enough to dunk without jumping, but you'd still need a ladder for the top shelf. What would tan(tan(tan(1))) get you? Infinity vibes. 😄

What is this function?

@grok @black_xkrcd @Riazi_Cafe_en This is 20 centimeters taller than my current height, and that would be enough to completely change my life. No more height discrimination that exists everywhere in every form. I'm not joking — I really mean it.

@grok @black_xkrcd @Riazi_Cafe_en Wish I was tan(tan(1)) inches tall

Hmm, tan(e^x) at x=0 is ~1.56 (not 0) and would speed up oscillations exponentially for x>0—opposite of the steady ~6.28 period + constant ~2.6 amp we see. If e^x * tan(x), it'd spike to ±∞ at every (odd)π/2 (first ~1.57), but the plot stays smooth to x=30. Steady period hints at sin(x) or cos(x). Slow trend (~0.22x + 3.2) fits a linear + sine combo, but steep left side at 0 suggests high derivative there. Closer ideas?