The Most Beautiful formula

May 25, 2016
Summer School
I A C S
112/28/2018

A three fold challenge
Yet in another School ?
Moreover, in Summer ?
More-moreover in a
Post-Lunch Session ?
212/28/2018

What is
COMMON
among the
images?
412/28/2018

They are Beautiful,
the Most Beautiful …
1312/28/2018

‘all mathematics is beautiful, yet
some is more beautiful than the
other’
paraphrasing George Orwell
1412/28/2018

 What is the most beautiful formula?
 How and Why is it Most Beautiful?
 How and why something is Beautiful?
 A Few Remarks on Mathematics, Mathematicians,
and Formula…
1512/28/2018

 The moving power of mathematical invention is
not reasoning but imagination.
Augustus de Morgan
 Imagination is more important than knowledge.
Albert Einstein
How can it be that mathematics, being after all a
product of human thought independent of experience,
is so admirably adapted to the objects of reality?
Albert Einstein 1612/28/2018

 There is no branch of mathematics,
however abstract, which may not some
day be applied to phenomena of the real
world.
Nikolai Lobatchevsky.
 A mathematician is a blind man in a
dark room looking for a black cat which
isn’t there.
 Charles Darwin.
1712/28/2018

Lt y →xU = 1 ; Buddhahood (enlightened)
y → x is “Human Revolution”,
“Casting off the transient and Revealing the True”.
LIFE
A philosophical Version of the Mathematical Identity
U = x/y y = Bodhisattva, x = true identity
“Yui Butsu, Yo Butsu”
1812/28/2018

 Does an infinite series necessarily have a finite Answer?
 Convergence
 S = ?
S = 1+x +x2 +x3+ . . .+xn-1 + xn
S.x= x+ x2 +x3 +. . . + xn +xn+1
S.(1-x) = 1 - xn+1
S = (1 – xn+1)/(1-x)
S => 1/(1-x) [since xn+1 =0 for large n]
Check it using calculator and see the convergence
2012/28/2018

The hare and the tortoise
decide to race
2312/28/2018

Since I run twice as fast as you do, I
will give you a half mile head start.
Thanks!
2412/28/2018

½ ¼ 8
1
16
1
The hare quickly reaches the turtle’s
starting point – but in that same time
The turtle moves ¼ mile ahead.
2612/28/2018

½ ¼ 8
1
16
1
By the time the rabbit reaches the
turtle’s new position, the turtle
has had time to move ahead.
2712/28/2018

½ ¼ 8
1
16
1
No matter how quickly the hare
covers the distance between himself
and the turtle, the turtle uses that
time to move ahead.
2812/28/2018

½ ¼ 8
1
16
1
Can the hare ever catch the
turtle???
2912/28/2018

1+1/2 +1/4 +1/8 + . . . ?
3012/28/2018

1 mile 2 miles
1+1/2 +1/4 +1/8 + . . . = 2 ???
3312/28/2018

let x = ½
S = 1+1/2 +1/4 +1/8 + . . .
S = 1/(1-x) = 1/(1- ½) = 1/(1/2) = 2
3412/28/2018

 A zeroth order reaction [At = A0 - k t ]
 t1/2 = f(? ? ?) [ half-life time]
= A0 /(2k) [k = rate constant]
 tcomp = A0 / k [completion time]
 A0 → A0 /2 → A0 /4 → A0 /8 → → → zero
 tcomp = t1 + t2 + t3 + …
 A0 / k = A0 /2k + A0 /4k + A0 /8k + …
 A0 / k = (A0 /k) (1/2 + 1 /4 +1 /8 + … )
 1 = (1/2 + 1 /4 +1 /8 + … )
 2 = (1+ 1/2 + 1 /4 +1 /8 + … )
3712/28/2018

 A zeroth order reaction [At = A0 - k t ]
 tƟ = f(? ? ?) [ half-life time]
= A0 /(2k) [k = rate constant]
 tcomp = A0 / k [completion time]
 A0 → A0 /2 → A0 /4 → A0 /8 → → → zero
 tcomp = t1 + t2 + t3 + …
 A0 / k = A0 /2k + A0 /4k + A0 /8k + …
 A0 / k = (A0 /k) (1/2 + 1 /4 +1 /8 + … )
 1 = (1/2 + 1 /4 +1 /8 + … )
 2 = (1+ 1/2 + 1 /4 +1 /8 + … )
3812/28/2018

1/8
1/4
1/8
1
1/2
4212/28/2018

1/16
1/4
1/8
1
1/2
4312/28/2018

1/16 1/32
1/32
1/4
1/8
1
1/2
4412/28/2018

Physicist Richard Feynman wrote in his
notebook in bold scrawl, when he was
fifteen.
4512/28/2018

 Kanser and J. Newman, in Mathematics and the
Imagination.
4612/28/2018

 A discovery of the
remarkable connection
between the
exponential and the
trigonometric functions,
by all time great
Mathematician,
Euler.
 Euler was a great
experimental
mathematician.
 He played with
formulas like a
child playing
with toys.
4712/28/2018

Can I expand an arbitrary function f(x) around x=a
as an infinite series?
f(x) =c0 + c1(x-a) +c2(x-a)2 + c3(x-a)3+ . . .+cn (x-a)n ;
But how to get all those {ci}?
4912/28/2018

at x=a,
f(a) = c0
f’ (x) = c1 + 2c2 (x-a) + 3c3 (x-a)2 + . . . + ncn (x-a)n-1 ;
fˊ (a) = c1
f’’ (x) = 2c2 + 3.2c3 (x-a) + . . .+ n(n-1)cn (x-a)n-2 ;
fˊˊ(a) = 2c2 
fˊˊ(a)/2 = c2
fˊˊˊ(a) = 3.2.c3  c3 = fˊˊˊ(a) / 3.2 or
fˊˊˊ(a) /3! = c3
5012/28/2018

at x=a,
f(a) = c0
fˊ (a) = c1
fˊˊ(a)/2 = c2
fˊˊˊ(a) /3! = c3
Similarly …
f(n)(a) /n! = cn
5112/28/2018

Thus we get the Taylor’s series
f(x) =c0 + c1(x-a) +c2(x-a)2 + c3(x-a)3+ . . .+ cn (x-a)n ;
with Cn = f(n)(a) /n!
f(x) = f(a) + (x-a) fˊ (a) + {(x-a)2/2!} fˊˊ(a)
+ {(x-a)3/3!} fˊˊˊ(a) + . . .;
take a=0, to get the Maclaurin Series:
f(x) = f(0) + x fˊ (0) + (x2/2!) fˊˊ(0) + (x3/3!) fˊˊˊ(0) + . . .
5212/28/2018

Example 2
f(x) = f(0) + x fˊ (0) + (x2/2!) fˊˊ(0) + (x3/3!) fˊˊˊ(0) + . . .
f(x)= sin(x) = x - x3/3! + x5/5! – x7/7! + x9/9! - …
5412/28/2018

 Replacing x by iy is like playing with meaningless
symbols, but Euler had enough faith in his formula to
make it meaningful. Thus we get,
eiy = 1 + iy - y2/2! - iy3/3! + y4/4! + iy5/5! - y6/6! – iy7/7! + …
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5612/28/2018
eiy = 1 + iy - y2/2! - iy3/3! + y4/4! + iy5/5! - y6/6! – iy7/7! + …
eiy = (1 - y2/2! + y4/4! - y6/6! + …) +
i (y - y3/3! + y5/5! - y7/7! + …)

 eiy = 1 + iy - y2/2! - iy3/3! + y4/4! + iy5/5! - y6/6! – iy7/7! +
eiy = (1 - y2/2! + y4/4! - y6/6! + …) +
i (y - y3/3! + y5/5! - y7/7! + …)
 sin(x) = x - x3/3! + x5/5! – x7/7! + x9/9! - …
 cos(x) = 1 - x2/2! + x4/4! – x6/6! + x8/8! - …
eiy = cos (y) + i sin(y), similarly e-iy = cos (y) - i sin(y)
Take y = π,
eiπ = cos (π) + i sin(π) = -1 + 0
eiπ = -1 5712/28/2018

 It involves three most important mathematical operations,
namely
 addition,
 multiplication and
 exponentiation.
 It connects the five most important constants in
mathematics:
e, π, i, 0 and 1.
5812/28/2018

 It includes four major branches of classical
mathematics:
 arithmetic (through 0 and 1 ),
 algebra (by i),
 geometry (by π), and
 analysis (by e).
5912/28/2018

A Home Task
π/4 = 1 – 1/3 +1/5 -1/7 +1/9 - …
6012/28/2018

1. Use eiπ = -1 = i2 to prove ln i = iπ/2;
4. evaluate ln(1+ i) – ln(1- i) = 2 i (1 – 1/3 +1/5 -1/7 + 1/9 - …)
5. manipulate to have ln {(1+i)/(1-i)} = ln i
6. equate (4) and (5) to get ln i = 2 i (1 – 1/3 +1/5 -1/7 + 1/9 - …)
7. equate (1) and (7) to find
π/4 = 1 – 1/3 +1/5 -1/7 +1/9 - …
Enough Hints
6112/28/2018

eiπ = -1 = i2 Solution
ln eiπ = ln i2
iπ = 2 ln i
ln i = iπ/2;
ln(1+ i) – ln(1- i) = 2 i (1 – 1/3 +1/5 -1/7 + 1/9 - …)
ln{(1+i)/(1-i)} = ln{(1+i)(1+i)/(1-i)(1+i)}
= ln(1+i2+ 2i)/(1- i2)=ln(2i/2) = ln i
So, ln i = 2 i (1 – 1/3 +1/5 -1/7 + 1/9 - …)
Or, i π/2 = 2 i (1 – 1/3 +1/5 -1/7 +1/9 - …)
π/4 = 1 – 1/3 +1/5 -1/7 +1/9 - …
6212/28/2018

Often we say, Simple is beautiful. If
it’s so, then the formula must be
beautiful.
6312/28/2018

if you like to relate
beauty with
usefulness and significance,
then also we must say:
The formula is beautiful.
6512/28/2018

 The most beautiful is
something lying in
repaying the debt of Gratitude.
6612/28/2018

The Most Beautiful formula

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