Heartiest congratulations to Mr Joseph Schooling, who has achieved the legendary accomplishment of a Gold medal at the Olympics! Good job!
Here is some analysis of factors contributing to Joseph’s Schooling’s success:
Height
Joseph’s height is 1.84 m (close to the ideal height of 1.90 m). In swimming, a tall height is desirable, as that would lead to an instant lead the moment you jump into the water. But this is not basketball, too tall (e.g. over 2 m) is probably not good as that will lead to increased drag and bad aquadynamics. So judging by height of former Olympic swimmers, roughly 1.90 m is the ideal height.
In fact, in 2006, “Schooling’s parents, Colin and May, send him for a bone test – which calculates growth potential – to see how far he can go in the sport. The test reveals that he will hit 1.90 metres, which is an optimum height to excel in the sport at the highest levels.” (http://www.todayonline.com/sports/schooling-story)
Motivation / Training Hard
This is probably the most important factor. Even when just eight years old, Joseph Schooling woke up his dad at 4.30 am to request to go for swimming training.
Schooling also took training to the highest level by going overseas (to Texas) for his training.
Joseph Schooling also has a motivational poster (see below) to remind himself daily of the times he need to achieve for a podium finish.
Supportive Parents (and relatives)
Joseph Schooling’s grand-uncle Lloyd Valberg, a former high jumper, was Singapore’s first ever Olympian at the 1948 London Games. He was the one who inspired Joseph to aim for the Olympics, when Joseph was just 6 years old.
“Colin (Joseph’s father) and his wife, May, decided to groom their only son to the best of their ability, going so far as to ensure he had access to swimming facilities to train during every vacation they took during his childhood. This would continue into his formative years, when Joseph was sent first to the famed Bolles School in United States to train under renowned coach Sergio Lopez and now the University of Texas to continue his development at the best possible environment.” (from https://sg.sports.yahoo.com/news/colin-schooling–the-world-has-taken-notice-of-joseph-and-singapore-150934197.html)
Bounceback from Adversity
In 2012 Olympics, Joseph met with extreme bad luck in the form of his goggles and swim suit not being approved, leading to a last minute change which would affect his timings.
Despite that, Joseph continued to persevere in his training, which led to his fantastic results in 2016.
Excellent Coach
Joseph Schooling’s coach is Sergio Lopez, “who is a former international top swimmer from Spain, who won the bronze medal in the 200 meters breaststroke at the 1988 Summer Olympics in Seoul” (Wikipedia).
Coach is very important, Michael Phelps credits his success to his legendary coach Bob Bowman, who has written a book The Golden Rules: 10 Steps to World-Class Excellence in Your Life and Work
Government Support
The Singapore government has humanely allowed a deferment from National Service (2 year mandatory conscription into the army) for Joseph Schooling. This is a crucial factor as for Olympic athletes, uninterrupted training is of utmost importance.
Joseph Schooling T-shirts
Finally, here are some cool Joseph Schooling Shirts suitable for his ardent fans. (Note that Red/White are Singapore’s national colors.)
be a continuous real-valued function defined on a closed interval ![[a,b]](https://s0.wp.com/latex.php?latex=%5Ba%2Cb%5D&bg=ffffff&fg=1a1a1a&s=0&c=20201002)
. Let 
be the function defined, for all 
in 
, and 
for all 
, 
be a differentiable curve given by the vector function 
, 
.
variables whose gradient vector 
is continuous on 
such that 
for all 
is constant.
by Stewart) or too advanced. Many analysis books skip multivariable calculus altogether and just focus on measure and integration.
be a 
times differentiable function at the point 
. Then there exists 
such that 
and 
.
, 
)
.
where 
.
, 
be both closed (or both open) subsets of a topological space 
such that 
, and let 
also be a topological space. If both 
and 
are continuous, then 
is continuous.
be a closed subset of 
is closed in 
, which is continuous. Hence 
for some set 
is closed (in 
is also closed (in 
, each of the following statements is equivalent to the statement that ![F[x]](https://s0.wp.com/latex.php?latex=F%5Bx%5D&bg=ffffff&fg=1a1a1a&s=0&c=20201002)
with at least one root in 
splits over ![|\text{Aut}(E/F)|=[E:F]](https://s0.wp.com/latex.php?latex=%7C%5Ctext%7BAut%7D%28E%2FF%29%7C%3D%5BE%3AF%5D&bg=ffffff&fg=1a1a1a&s=0&c=20201002)
, that is, the number of automorphisms equals the degree of the extension.
.
and 
is the corresponding fixed field (the set of those elements in 
).
is an intermediate field of 
is the set of those automorphisms in 
which fix every element of 
, the trivial subgroup of 
.
be a PID. Let 
be a nonzero prime ideal. Suppose 
. Then 
for some 
.
implies 
or 
. Since 
, we have 
for some 
. Then, 
which implies 
, thus 
is a unit. Hence 
.
“. He is a very good author of Christian books.
is a finite group, 
is the center of 
are all the conjugacy classes in 
be an element in 
for each 
. Then we have: ![\displaystyle |G|=|Z(G)|+\sum_{i=1}^r[G:C_G(g_i)].](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%7CG%7C%3D%7CZ%28G%29%7C%2B%5Csum_%7Bi%3D1%7D%5Er%5BG%3AC_G%28g_i%29%5D.&bg=ffffff&fg=1a1a1a&s=0&c=20201002)
.
. Then 
for all 
. Hence 
consists of a single element 
. Then
![\displaystyle |\text{Orb}(g_i)|=[G:\text{Stab}(g_i)]=[G:C_G(g_i)].](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%7C%5Ctext%7BOrb%7D%28g_i%29%7C%3D%5BG%3A%5Ctext%7BStab%7D%28g_i%29%5D%3D%5BG%3AC_G%28g_i%29%5D.&bg=ffffff&fg=1a1a1a&s=0&c=20201002)
and 
be real or complex sequences.
converges to 
converges to 
.
has radius of convergence 
, and another power series 
has radius of convergence 
, then their Cauchy product converges to 
and has radius of convergence at least the minimum of 
.
is a continuous map from a closed subset 
, then there exists a continuous map 
with 
for all 
in 
, i.e., if 
, there is a continuous function 
on 
and a closed set 
for which 
. Let 
be a sequence of measurable functions on 
is a function that is holomorphic (analytic) and injective, then 
for all 
in 
for some 
. Using Taylor series, 
for all 
, with 
, 
and 
where 
, we write 
where 
.
on a small circle centered at 
has at least two zeroes there.
but sufficiently close to 
, 
be the two roots of 
, 
. If 
, then 
for some analytic function 
. This means 
, which implies that 
. There are some patterns that can be detected, maybe Big Data can help.
holomorphic and fixes zero.
can be remembered as “Range of f” subset of “Domain”.
can be remembered as some sort of “Contraction Mapping”.
, or 
, then 
where 
. Remember it as “
be the open unit disk in the complex plane 
centered at the origin and let 
.
for all 
and 
for some 
with 
on 
, and 
. Note that 
on 
denote the closed disk of radius 
centered at the origin. The Maximum Modulus Principle implies that, for 
, given any 
, there exists 
on the boundary of 
such that 
we get 
, thus 
at some point of 
where ![\displaystyle |\text{Orb}(x)|=[G:\text{Stab}(x)]=\frac{|G|}{|\text{Stab}(x)|}.](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%7C%5Ctext%7BOrb%7D%28x%29%7C%3D%5BG%3A%5Ctext%7BStab%7D%28x%29%5D%3D%5Cfrac%7B%7CG%7C%7D%7B%7C%5Ctext%7BStab%7D%28x%29%7C%7D.&bg=ffffff&fg=1a1a1a&s=0&c=20201002)
by 
is a subgroup of 
, then 
. Thus 
, which implies 
, thus 
is well-defined.
, then 
. Thus 
.
does not divide 
.
, 
, 
, 
.
or 
and 
or 
. Since 
, 
, 
be the Sylow 
. Thus 
. Since 
there is a non-identity element in 
. Its order has to be 
.
so 
is a subgroup of 
thus 
. 
is cyclic, thus it has a unique subgroup 
of order 
.
be generators for 
, where 
. (We may conclude this since the action of 
, then 
.
, then 
Choosing a different 
amounts to choosing a different generator 
on 
have the same number of zeroes inside 
in the disk 
. Let 
, 
, then
.
such that for every finite disjoint collection 
of open intervals in 
then 
almost everywhere, the derivative is Lebesgue integrable, and 
for all 
for all 
is abelian iff 
, 
is trivial.
, 
, we have
, thus 
. Then for any 
, 
. Multiplying by 
on the left gives 
for any 
for all 
so 
is trivial.
. Then we have 
, so 
)
. Then each of the following four assertions is equivalent to the measurability of 
Sets)
.
.
Sets)
.
.
. By the definition of outer measure, there is a countable collection of open intervals 
which covers 
. Then 
. Since 
-finite, 
of measurable sets, each of which has finite outer measure.
, there is an open set 
containing 
for which 
. The set 
is open, it contains 
that contains 
. Define 
. Then 
is measurable):
is measurable. 
is measurable.
(iii)):
is measurable iff 
such that 
.
which is closed. Note that 
.
(iv)):
.
