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∫
0
∞
sin
t
t
dt
=
π
2
{\displaystyle \int _{0}^{\infty }{\frac {\sin {t}}{t}}{\text{ dt}}={\frac {\pi }{2}}}
∫
0
∞
sin
t
t
⋅
sin
(
t
101
)
(
t
101
)
dt
=
π
2
{\displaystyle \int _{0}^{\infty }{{\frac {\sin {t}}{t}}\cdot {\frac {\sin {\left({\frac {t}{101}}\right)}}{\left({\frac {t}{101}}\right)}}}{\text{ dt}}={\frac {\pi }{2}}}
∫
0
∞
sin
t
t
⋅
sin
(
t
101
)
(
t
101
)
⋅
sin
(
t
201
)
(
t
201
)
dt
=
π
2
{\displaystyle \int _{0}^{\infty }{{\frac {\sin {t}}{t}}\cdot {\frac {\sin {\left({\frac {t}{101}}\right)}}{\left({\frac {t}{101}}\right)}}\cdot {\frac {\sin {\left({\frac {t}{201}}\right)}}{\left({\frac {t}{201}}\right)}}}{\text{ dt}}={\frac {\pi }{2}}}
⋮
{\displaystyle \vdots \!\,}
∫
0
∞
sin
t
t
⋅
sin
(
t
101
)
(
t
101
)
⋅
sin
(
t
201
)
(
t
201
)
⋯
sin
(
t
100
n
+
1
)
(
t
100
n
+
1
)
dt
=
π
2
{\displaystyle \int _{0}^{\infty }{{\frac {\sin {t}}{t}}\cdot {\frac {\sin {\left({\frac {t}{101}}\right)}}{\left({\frac {t}{101}}\right)}}\cdot {\frac {\sin {\left({\frac {t}{201}}\right)}}{\left({\frac {t}{201}}\right)}}\cdots {\frac {\sin {\left({\frac {t}{100n+1}}\right)}}{\left({\frac {t}{100n+1}}\right)}}}{\text{ dt}}={\frac {\pi }{2}}}
True for every
n
<
15341178777673149429167740440969249338310889
,
{\displaystyle n<15341178777673149429167740440969249338310889,}
but false for all
n
{\displaystyle n}
∫
0
∞
sin
t
t
dt
=
π
2
{\displaystyle \int _{0}^{\infty }{\frac {\sin {t}}{t}}{\text{ dt}}={\frac {\pi }{2}}}
∫
0
∞
sin
t
t
⋅
sin
(
t
2
)
(
t
2
)
dt
=
π
2
{\displaystyle \int _{0}^{\infty }{{\frac {\sin {t}}{t}}\cdot {\frac {\sin {\left({\frac {t}{2}}\right)}}{\left({\frac {t}{2}}\right)}}}{\text{ dt}}={\frac {\pi }{2}}}
∫
0
∞
sin
t
t
⋅
sin
(
t
2
)
(
t
2
)
⋅
sin
(
t
4
)
(
t
4
)
dt
=
π
2
{\displaystyle \int _{0}^{\infty }{{\frac {\sin {t}}{t}}\cdot {\frac {\sin {\left({\frac {t}{2}}\right)}}{\left({\frac {t}{2}}\right)}}\cdot {\frac {\sin {\left({\frac {t}{4}}\right)}}{\left({\frac {t}{4}}\right)}}}{\text{ dt}}={\frac {\pi }{2}}}
⋮
{\displaystyle \vdots \!\,}
∫
0
∞
sin
t
t
⋅
sin
(
t
2
)
(
t
2
)
⋅
sin
(
t
4
)
(
t
4
)
⋯
sin
(
t
2
n
)
(
t
2
n
)
dt
=
π
2
{\displaystyle \int _{0}^{\infty }{{\frac {\sin {t}}{t}}\cdot {\frac {\sin {\left({\frac {t}{2}}\right)}}{\left({\frac {t}{2}}\right)}}\cdot {\frac {\sin {\left({\frac {t}{4}}\right)}}{\left({\frac {t}{4}}\right)}}\cdots {\frac {\sin {\left({\frac {t}{2^{n}}}\right)}}{\left({\frac {t}{2^{n}}}\right)}}}{\text{ dt}}={\frac {\pi }{2}}}
True for all
n
{\displaystyle n}
Behind this "magic" is the following: Let
(
a
n
)
{\displaystyle \left(a_{n}\right)}
a
0
=
1.
{\displaystyle a_{0}=1.}
∫
0
∞
∏
n
=
0
N
sin
a
n
t
a
n
t
dt
=
π
2
{\displaystyle \int _{0}^{\infty }{\prod _{n=0}^{N}{\frac {\sin {a_{n}t}}{a_{n}t}}}{\text{ dt}}={\frac {\pi }{2}}}
∑
1
N
a
n
≤
1.
{\displaystyle \sum _{1}^{N}a_{n}\leq 1.}
∑
n
=
1
N
1
100
n
+
1
{\displaystyle \sum _{n=1}^{N}{\frac {1}{100n+1}}}
1
{\displaystyle 1}
N
{\displaystyle N}
10
43
.
{\displaystyle 10^{43}.}
In contrast,
∑
n
=
1
N
1
2
n
{\displaystyle \sum _{n=1}^{N}{\frac {1}{2^{n}}}}
∑
n
=
1
N
1
2
n
=
1
−
1
2
N
<
1
{\displaystyle \sum _{n=1}^{N}{\frac {1}{2^{n}}}=1-{\frac {1}{2^{N}}}<1}
N
.
{\displaystyle N.}
An example that diverges even slower is
a
n
=
1
(
100
n
+
1
)
⋅
ln
(
100
n
+
1
)
.
{\displaystyle a_{n}={\frac {1}{\left(100n+1\right)\cdot \ln {\left(100n+1\right)}}}.}
N
≈
10
10
43
.
{\displaystyle N\approx 10^{10^{43}}.}
± 1± 1± 1± 1± 3+45 +270 +1200 +5000 900 +19000 500 +66000 000 000 +230000 000
The resonant angle
ϕ
{\displaystyle \phi }
ϕ
=
12
⋅
λ
−
7
⋅
λ
N
−
5
⋅
ϖ
−
1
⋅
Ω
{\displaystyle \phi ={\rm {12\cdot \lambda -{\rm {7\cdot \lambda _{\rm {N}}-{\rm {5\cdot \varpi -{\rm {1\cdot \Omega }}}}}}}}}
This angle is librating intermittently, hence why Buie does not classify Haumea as a resonant object.[1] Haumea's ascending node
Ω
{\displaystyle \Omega }
here for the results of my orbit simulation.
Haumea and the other objects in the Haumea family occupy a region of the Kuiper belt where multiple resonances (including the 3:5, 4:7, 7:12, 10:17 and 11:19 mean motion resonances) interact, leading to the orbital diffusion of that collision family.[2] While Haumea is in a weak 7:12 resonance, other objects in the Haumea family are known to temporarily occupy some of the other resonances. For instance, (19308) 1996 TO66
Mathematicians : Let's calculate 50 trillion digits of
π
{\displaystyle \pi }
Astronomers : Let's assume
π
≈
1
{\displaystyle \pi \approx 1}
π
2
≈
10
{\displaystyle \pi ^{2}\approx 10}
