eaitmpr

[eaitmpr @eaitmpr]

@microfusio91945 (1+x)^(3/x)
=e(3ln(1+x)/x)
=e(3(x-x²/2+o(x²))/x)
=e(3-3x/2+o(x))
=e³.e(-3x/2+o(x))
=e³.(1-3x/2+o(x))

( (1+x)^(3/x)-e³)/x=-3e³/2+o(1)

[eaitmpr @eaitmpr]

@Rashid5890 Σ_1..n i=n(n+1)/2
Σ_n..p i =(p-n+1)(p+n)/2

S_k=Σ_k..n+k-1 i =(n+k-1-k+1)(k+n+k-1)/2=n(2k+n-1)/2

Σ_1..n S_k
= n.Σk +Σn(n-1)/2
=n²(n+1+n-1)/2=n³

n³ mod n²-1 = an+b:
n²-1=(n-1)(n+1) =>
1=a+b
-1=-a+b
b=0, a=1
n³ mod n²-1=n

Answer=2026

[eaitmpr @eaitmpr]

@Diarytells 15.
∫_π..π x²⁰²⁶dx/(1+sinx+√(1+sin²x))
∫_-π..π x²⁰²⁶(1+sinx-√(1+sin²x))dx/2sinx
0 (odd) +2∫_0..π x²⁰²⁶sinxdx/2sinx (even)
∫_0..π x²⁰²⁶dx=π²⁰²⁷/2027

[eaitmpr @eaitmpr]

@Diarytells ...=ln2-2ln(1+ln2)

14.
∫_0..1 sinπxdx/(sinπxΓ²(x)+π)
∫_0..1 πdx/Γ(x)Γ(1-x)(πΓ(x)/Γ(1-x)+π)
∫_0..1 dx/Γ(x)(Γ(x)+Γ(1-x))
∫_0..1 dx/Γ(1-x)(Γ(1-x)+Γ(x)) (King’s)
1/2.∫_0..1 dx/Γ(x)Γ(1-x)
1/2π.∫_0..1 sinπxdx=1/2π.[-cosπx/π]=1/π²

[eaitmpr @eaitmpr]

@microfusio91945 x^(x+1)/(1+x)^x
=x/(1+1/x)^x
=x.e[-xln(1+1/x)]
=x.e[-x(1/x-1/2x²+o(1/x²))]
=x.e(-1+1/2x+o(1/x))
=x.e-1.(1+1/2x+o(1/x))
=x/e+1/2e+o(1)

Hence lim=1/2e

[eaitmpr @eaitmpr]

@microfusio91945 Then
I(a,b)=I(0,b) +∫_0..a dI(u,b)/db.du
=0+∫ 2π(ln(u+b)-lnu)du
=2π[(u+b)ln(u+b)-u-(ulnu-u)]
=2π((a+b)ln(a+b)-alna-blnb)

[eaitmpr @eaitmpr]

@microfusio91945 I(a,b)=?

dI/da=∫ 2a/(x²+a²).ln(1+b²/x²)dx
d²I/dadb=∫ 4abdx/(x²+a²)(x²+b²)
=4ab/(b²-a²).∫ (1/(x²+a²)-1/(x²+b²))dx
=4ab/(b²-a²)(π/2a-π/2b)
=2π/(a+b)

dI/da(a,b)=dI/da(a,0) +∫_0..b d²I/dadb(a,v).dv
=0+∫ 2πdv/(a+v) =2π(ln(a+b)-lna)
...

[eaitmpr @eaitmpr]

@Diarytells (sinx.(sinx.(sinx.(..)^1/5)^1/4)^1/3)^1/2
=sinx^(1/2+1/2.3+1/2.3.4+...)
=sinx^(Σ1/n! -2)

?
=∫ sinx^(e-2)dx
=Β((e-1)/2,1/2)/2
=Γ((e-1)/2)Γ(1/2)/2Γ((e-1)/2+1/2)
=√πΓ((e-1)/2)/2Γ(e/2)

[eaitmpr @eaitmpr]

@microfusio91945 1<k<n: 1/(n+1)<1/(n+1/k)<1/(n+1/n)
Σ sin(πk/n)/(n+1) < Σ < Σ sin(πk/n)n/(n²+1)

Σsin(πk/n)=Im{Σe(iπk/n)}=Im{2/(1-e(iπ/n))}=1/tan(π/2n)

Σ ~ n/tan(π/2n) ~2/π

[eaitmpr @eaitmpr]

@microfusio91945 Σ 1/(n+k) = 1/n.Σ 1/(1+k/n) ~ ∫_0..1 dt/(1+t) =[ln(1+t)] =ln2

1/√(n²+t) decreasing :
∫_0..n dt/√(n²+t) > Σ 1/√(n²+k) > ∫_1..n+1 dt/√(n²+t)
2[√(n²+n)-√n²] > Σ > 2[√(n²+n+1)-√(n²+1)]
2/(√(1+1/n)+1) > Σ > 2/(√(1+1/n+1/n²)+√(1+1/n²))
Lim Σ =2/(1+1)=1

...

[eaitmpr @eaitmpr]

@Diarytells Let u=y/x then
x²(u'x+u)=2u²x²+6x²u+3x²
u'/(2u²+5u+3)=1/x
u'/(u+1)(2u+3)=u'/(u+1)-2u'/(2u+3)

ln(u+1)-ln(2u+3)=lnx+c
(u+1)/(2u+3)=cx
u+1=cx(2u+3)=2cxu+3cx
u=(3cx-1)/(1-2cx)=y/x

u(1)=1/1=(3c-1)/(1-2c), c=2/5
y=x(6x-5)/(5-4x)

Lim (y/x)=-3/2

[eaitmpr @eaitmpr]

@Diarytells sin²(6π[π²x])=0 for all x
[x]=0 on ]0;1[ and =1 on ]1;π/2[
so
I=∫_0..1 x³dx +∫_1..π/2 x³dx/2
=1/4-0 +((π/2)⁴/4-1/4)/2
=1/8+π⁴/(8.2⁴)
=(2⁴+π⁴)/128

αβ=2π

[eaitmpr @eaitmpr]

@microfusio91945 Let a_1=1/2², a_n=[Π_2..n (1-1/k²)]/(n+1)²
a_1=1/4<1/2
Suppose a_n<1/2n for some n, then
a_n+1 =(n+1)²a_n.(1-1/(n+1)²)/(n+2)²
=a_n.((n+1)²-1)/(n+2)²
=a_n.n/(n+2)
<1/2(n+2)<1/2(n+1)
Hence a_n<1/2n for all n
Hence Σ_1..n a_k <n.1/2n<1/2

[eaitmpr @eaitmpr]

@ZeldaAnecdotes Faux, celui de TOTK arrive à escalader des falaises en portant 20 épées, 6 boucliers, 4 arcs, 900 flèches, 80 fagots de bois, 90 bombes, 20 melons glagla, 50 pierres soneau, 15 tenues, etc.

[eaitmpr @eaitmpr]

@Diarytells ln Π [(π/2)/arctan(n+k)]
=Σ [2/π(n+k)+4/π²(n+k)²+o(1/n⁴)]
=1/n.Σ 2/π(1+k/n) +1/n².Σ 4/π²(1+k/n)²+Σo(1/n⁴)
=∫_0..1 2dt/π(1+t) +O(1/n)
=2ln2/π

Lim=2^(2/π)

[eaitmpr @eaitmpr]

@Diarytells arctan(n+k)=π/2-arctan(1/(n+k))
=π/2-1/(n+k)+1/3(n+k)³+o(1/n³)

ln[(π/2)/arctan(n+k)]=-ln(1-2/π(n+k)+2/π(n+k)³+o(1/n³))=2/π(n+k)+4/π²(n+k)²+o(1/n⁴)

...

[eaitmpr @eaitmpr]

@chris_juravich @CPierre67 @microfusio91945 I'm too lazy to type on a small pad \frac{1}{n+1} in an editor instead of 1/(n+1), and I know tweets can be difficult to read sometimes, sorry for that 😓
Is there a known app that converts "generic quick writing" into correct TeX command ready to copypaste in an editor ?

[eaitmpr @eaitmpr]

@chris_juravich @CPierre67 @microfusio91945 Oh it's ok thx!
I've written in LateX in the past, just not on a mobile phone yet 😇

[eaitmpr @eaitmpr]

@Diarytells t=x/2018:
∫_0..∞ x^(α-1).e(-x/2018)dx
=2018^α.Γ(α)=2018^α.(α-1)!

Σ_1..∞ (...)
= Σ_1..∞ 2018^(α-1)/(α-1)!
= Σ_0..∞ 2018ⁿ/n!
=e(2018)

ln(...)=2018