answer from llm qwq32b is: let i=k; i₃=3^(2^i); => sum termᵢ=2ⁱ(i₃+1-2)/(i₃+1)=2ⁱ-(2ⁱ⁺¹)/(i₃+1) split into two sums: first one is geometric series, second 2ⁱ⁺¹/(i₃+1) = aᵢ-aᵢ₊₁, where aᵢ= 2ⁱ⁺¹/(i₃-1) __ 2ⁱ⁺¹(1/(i₃-1) - 2/(i₃*i₃ - 1)) = 2ⁱ⁺¹(i₃+1-2)/((i₃+1)*(i₃-1)) = 2ⁱ⁺¹/(i₃+1) so sum telescopes leaving only a₀-a_n = 1-2^(n+1)/((3^(2^n)) - 1) ??? qwq32b insist that valid answer is s_n=2^n-2+2^n/((3^(2^(n-1))) - 1), and it is wrong valid answer s_n=2^n-2+2^(n+1)/((3^(2^(n))) - 1) i did ask it write python program and check numerically which one is correct, my version was correct :) (??? is point where i spotted mistake in qwq32b math, it did get lost in constant -1 +1 +1 ... and made -1 for some reason in final index) p.s. it does gets many integrals correctly including exponentially fascinating one.
@@maths_505Bro it's me from another account 😅 I was trying to not focus on math by switching into that account to watch cinema or random content Guess what , youtube saw through that and showed your video and I started watching again from another account 😂 also.
@@maxvangulik1988 Hmm… not sure but I suggest you copy some syntax online. Things like Σₙ xⁿ/n! is way more easier to read than things like Š[n](x^n/n!) which is still better than raw latex like \sum_{n}\frac{x^{n}}{n!}
Your rapper name is M.C. Gamma.
Another cracking case! Happy new year man! All the best for 2025!
Happy new year homie
Hi,
"ok, cool" : 0:16 , 1:15 , 3:34 , 7:55 ,
"terribly sorry about that" : 6:19 , 7:45 .
6:35 it should be gamma prime of one I believe...
0:20 What is that dead whistle?
Thank you for your effort.
Just two "terribly sorry about that* today :(
@@zubii2017 terribly sorry about that
@@maths_505I could see your reply coming from a mile away
@owenvaughn3191 🤣🤣🤣
This integral is already solved by Maths 505 before one year 😂😂
@ 6:19 Should be Gamma(1), not Gamma(2).
isn’t the sum at 7:15 supposed to be negative?
the negative cancelled after differentiation
Nice video , can you solve this question please : sum from 0 to n-1 of 2^k (3^(2^k)-1)/(3^(2^k)+1) .
answer from llm qwq32b is:
let i=k; i₃=3^(2^i); => sum termᵢ=2ⁱ(i₃+1-2)/(i₃+1)=2ⁱ-(2ⁱ⁺¹)/(i₃+1)
split into two sums: first one is geometric series,
second 2ⁱ⁺¹/(i₃+1) = aᵢ-aᵢ₊₁, where aᵢ= 2ⁱ⁺¹/(i₃-1)
__ 2ⁱ⁺¹(1/(i₃-1) - 2/(i₃*i₃ - 1)) = 2ⁱ⁺¹(i₃+1-2)/((i₃+1)*(i₃-1)) = 2ⁱ⁺¹/(i₃+1)
so sum telescopes leaving only a₀-a_n = 1-2^(n+1)/((3^(2^n)) - 1) ???
qwq32b insist that valid answer is s_n=2^n-2+2^n/((3^(2^(n-1))) - 1),
and it is wrong
valid answer s_n=2^n-2+2^(n+1)/((3^(2^(n))) - 1)
i did ask it write python program and check numerically which one is correct, my version was correct :) (??? is point where i spotted mistake in qwq32b math, it did get lost in constant -1 +1 +1 ... and made -1 for some reason in final index)
p.s. it does gets many integrals correctly including exponentially fascinating one.
Fantastic
A me sembra semplicemente Γ"(2)...derivò due volte la funzione gamma .I=(ω(2))^2+ψ1(2)=0,4225^2-(π^2/6-1)=0,823693..
hi bro
Yo
@@maths_505Bro it's me from another account 😅
I was trying to not focus on math by switching into that account to watch cinema or random content
Guess what , youtube saw through that and showed your video and I started watching again from another account 😂 also.
Noice
t=e^x
dt=e^x•dx
I=int[0,♾️](t•ln^2(t)e^-t)dt
I=Ř''(2)
Ř'(x)=¥(x)Ř(x)
Ř''(x)=Ř(x)(¥,(x)+¥^2(x))
Ř(2)=1
¥(2)=1-ř
¥,(2)=pi^2/6-1
I=Ř''(2)=pi^2/6-2ř+ř^2
Why don’t you write ∫{0→∞}, Γ, ψ, γ, π, instead of those uncommon symbols?
@ I'm working with an iphone 12 keyboard
@@maxvangulik1988 Hmm… not sure but I suggest you copy some syntax online. Things like Σₙ xⁿ/n! is way more easier to read than things like Š[n](x^n/n!) which is still better than raw latex like \sum_{n}\frac{x^{n}}{n!}
@ i would write that as "sum[n=1,♾️](x^n/n!)"
I always specify both bounds