a formula for π 2 (a definition)

There is no easy way for definition of $\pi$ .
One of the most simplest definition of $\pi$ is using the distance.

You must know $\pi$ is equivalent to the length of a semicircular circumference.
That is, $\pi$ is double the length of the circumference of the quadrant.

The graph $C$ is drawn by the function $y=f(x)$ on the interval $x\in[a,b]$ .
$f(x)$ is differentiable and $f'(x)$ is continuous on $[a,b]$ .
Then, the length $l$ of $C$ is
\[ l=\int_a^b \sqrt{1+f'(x)^2}dx .  \]

The function of a circular whose center is origin and radius is 1 is
\[ x^2+y^2=1 .  \]
Thus, the length of quadrant circumference is
\[ l=\int_0^1\frac{1}{\sqrt{1-x^2}}dx  \]
because
\[ y=\sqrt{1-x^2} ,  \]
\[ y'=-\frac{x}{\sqrt{1-x^2}} ,  \]
\[ (y')^2=\frac{x^2}{1-x^2} ,   \]
therefore,
\[ l=\int_0^1\sqrt{1+\frac{x^2}{1-x^2}} dx  . \]

We will define
\[ 2l= 2\int_0^1\frac{dx}{\sqrt{1-x^2}}=\pi  . \]

a formula for π

There are many formulas for π .

One of most known formulas is
\[ \int_{-\infty}^{\infty}\frac{dx}{1+x^2}=\pi, \qquad(-\infty\lt x\lt \infty) .  \]

Please put $x=f(t)=\tan t=\frac{\sin t}{\cos t},\quad (t\in (-\pi/2,\pi/2))$ .


As we already have known
\[ \cos't=-\sin t,\quad \sin' t=\cos t ,    \]
we will obtain 
\[ \frac{dx}{dt}=\frac{df(t)}{dt}=f'(t)=\frac{\cos^2 t+\sin^2 t}{\cos^2 t}=\frac{1}{\cos^2 t}  . \]

Then,
\[ f'(t)=1+f^2(t), \quad\mbox{and}\quad dx=f'(t)dt  \]
$f$ is a monotonically increasing mapping of $(-\pi/2,\pi/2)$ onto $(-\infty,\infty)$.
Thus,
\[ \int_{-\infty}^{\infty}\frac{dx}{1+x^2}=\int_{-\pi/2}^{\pi/2}\frac{f'(t)dt}{1+f^2(t)}=\int_{-\pi/2}^{\pi/2}dt=\pi  \]

Napier's constant

The function $e^x$ is one of most important function,
where $e$ is called Napier's constant.

\[ e=\lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^n=2.71828182845\cdots    \]
or
\[ \frac{1}{e}=\lim_{n\rightarrow \infty}\left(1-\frac{1}{n}\right)^n=0.367879\cdots \]

The function $e^x$ is defined as follow.
\[ e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}  \]
$e^x$ has many important properties. one is
\[ (e^x)'=e^x, \quad or\quad \frac{d}{dx}e^x=e^x  \]
[proof]
By definitions,
$(e^x)'=\frac{d}{dx}e^x=\lim_{h\rightarrow 0}\frac{e^{x+h}-e^x}{h}$
It is, as
$\lim_{h\rightarrow 0}\frac{e^{x+h}-e^x}{h}=e^x\lim_{h\rightarrow 0}\frac{e^{h}-1}{h}=e^x$ , then
 \[ \lim_{h\rightarrow 0}\frac{e^{h}-1}{h}=1 \]
 is required. At first, if $n=\frac{1}{h}$, then
\[ e=\lim_{h\rightarrow 0}\left(1+h\right)^{1/h}    \]
Thus,
\[ \log e=\log \lim_{h\rightarrow 0}\left(1+h\right)^{1/h} =\lim_{h\rightarrow 0}\log \left(1+h\right)^{1/h}    \]
\[ \lim_{h\rightarrow 0}\frac{\log (1+h)}{h}=1   \]
So, if we put $1+h=e^t$ , then $h=e^t-1$ . As $h\rightarrow 0$ , $t\rightarrow 0$ .
\[ \lim_{t\rightarrow 0}\frac{\log e^t}{e^t-1}=1  \]
\[ \lim_{t\rightarrow 0}\frac{t\log e}{e^t-1}=1  \]
\[ \lim_{t\rightarrow 0}\frac{t}{e^t-1}=1  \]
\[ \lim_{t\rightarrow 0}\frac{1}{\left(\frac{e^t-1}{t}\right)}=1  \]
Then, changing $t$ to $h$, we will get
\[ \lim_{h\rightarrow 0}\frac{e^{h}-1}{h}=1  \]

coffee break 22 (my PC 3)

Over almost two months,  I verified the migration of about 20 games. 
 
About thirty games complete the synchronization process only. 

Although the migration of my own data is completed,
I can hardly log in with many types of accounts. 

The PC of win10 and Ryzen 7 is running smoothly.
However,  the memory's clock is low,  the CPU's heat is concerned and 
even if I delete the files in usb memory, the display of files does not disappear immediately.

I want to adjust gradually. 

coffee break 21 (my PC 2)

In ryzen7, as I heard that memory is important in drawing better performance,
I changed the built-in memory.

CORSAIR→GSkill F4-3200C14D-16GFX

As mentioned previously, I assembled a planned PC.
Since I used parts of existing PC, it took two days to complete the whole assembly.

Data migration has not ended yet, and there are games that will not start on win10. I am doing it little by little every day.　

coffee break 20 (my PC)

I am troubled by personal computer malfunction and irregular actions.
Sometimes it crashes and it takes a restart suddenly.
Looking at the error log of Windows it is KP 41.

This trouble has once been encountered.
The problem was solved by replacing the power supply unit.

I have been using PC for about 5 years while exchanging parts,
so I'm thinking about using a all new tone .　

The specifications of the PC considered are as follows.
cpu: ryzen7 2700x,
mb: asrock x470 taichi, 
mem: corsair ddr4-2666mhz 8×2gb,
os: win10 home,
base storage: crucial ssd 500gb, 
gpu: asus gtx1070,
psu: corsair 750w
and so on.

I am planning to make it using summer vacation.

coffee break 19 (I see it)

“Je le vois, mais je ne le crois pas!”
      (“I see it, but I don’t believe it!”)   R. Dedekind

coffee break 18 (cantor's paradise)

"No one shall drive us from the paradise which Cantor has created for us."  D.Hilbert

This paradise means axiomatic set theory.

field

A field has two operations, the addition and the multiplication.

It is an abelian group under addition, with 0 as additive identity.

The nonzero elements form an abelian group under multiplication,
and the multiplication is distributive over addition.

In order to avoid existential quantifiers, fields can be defined by two binary operations
(addition and multiplication),
two unary operations (yielding the additive and multiplicative inverses, respectively),
and two nullary operations (the constants 0 and 1).

ring

On a set $A$ (not empty), additional operation and
multiplicative operation are defined.

If $a,b\in A$ ,

(1)$\rho_1:(a,b)\rightarrow a\rho_1b\in A$
(2)$\rho_2:(a,b)\rightarrow a\rho_2b\in A$

You may prefer the expression $a+b$ to $a\rho_1b$ and $a\times b$ to $a\rho_2b$ .

In addition, the following conditions are satisfied.

(1)for the additional operation $\rho_1$ , $(A,\rho_1)$ is the abelian group.
(2)the multiplicative operation $\rho_2$ is associative.
 $(a\rho_2 b)\rho_2 c=a\rho_2(b\rho_2 c)\qquad (a,b,c\in A)$
(3)the multiplicative operation $\rho_2$ is distributive
   for any addtional operations from both sides.
 $a\rho_2(b\rho_1 c)=(a\rho_2 b)\rho_1(a\rho_2 c)$ ,
 $(b\rho_1c)\rho_2 a=(b\rho_2 a)\rho_1(c\rho_2 a)$ .
(4)there exists an element $e$ such that $e\rho_2 a=a\rho_2 e=a$ for any $a\in A$ .

Then, $(A,\rho_1,\rho_2)$ is called the ring.

The difference between a ring and a group is that a ring has two operations
and a group has one operation.

If the multiplivative operation is commutative $a\rho_2 b=b\rho_2 a$ ,
then $(A,\rho_1,\rho_2)$ is the abelian ring.

The integer $(\mathbb{Z},+,\times)$ is an abelian ring.

Let $A$ be a family of sets (not empty, but has the empty set).

(1)if $A_i\in A$ ,then $A_i^c\in A$ .
(in other words, $A_i\cup A_j\in A$ and $A_i/A_j\in A$ ($A_i,A_j\in A, i\ne j$)
(2)$\cup_{n=1}^{\infty}A_n\in A$ .

This $A$ is called a $\sigma-$ring .

$\sigma-$ring is one of bases of the lebesgue theory.

group

After we choose two elements from a set $G$ ,
let us make a new set
\[ G\times G=\left\{(a,b) ; a,b\in G \right\} .  \]

This set $G\times G$  is called a direct product or a cartesian product of the set $G$ .
In general, the cartesian product of $G$ and $H$ is a set of ordered pairs of two different sets.
\[  \left\{ \lt a,b\gt ; a\in G, b\in H  \right\}  \]

There is a mapping or function $\rho$ from the set $G\times G$ to the set $G$.
\[ \rho : G\times G \rightarrow G  \]
Thus,
\[ \rho((a,b))=x\in G  \]
or as the mapping $\rho$ is a binary operation,
\[  a\rho b=x  \]

If the operation $\rho$ from $G\times G$ to $G$ satisfies the below condition,

(1)$a\rho(b\rho c)=(a\rho b)\rho c\qquad (a,b,c\in G)$
(2)for any element $a$ in $G$ , there exists a element $e\in G$ such that $a\rho e=a$ and $e\rho a=a$ .
(3)for any element $a$ in $G$ , there exists a element $y\in G$ such that $a\rho y=y\rho a=e$ .

then $(G,\rho)$ is called the group.

Additionally, if $a\rho b=b\rho a$ (commutativity) is satisfied at all times, the groups is called the abelian group.

If the operation $\rho$ may be $+$ , $(\mathbb{Z},+)$ is one of abelian groups,
and the element $e$ is $0$ ,and the element $y$ is $-a$.

When the set $\mathbb{R}^*$means real numbers except $0$ , $(\mathbb{R}^*, \times)$ becomes the abelian group.
(you must know $e=1$ and $y=\frac{1}{a}$.)

The collection of N-square matrix becomes a group (not an abelian group).

A group is an algebraic structure.

next steps

Explanation of the foundation of mathematics has been completed.

We have gotten numbers and basic rules of calculation and logics on sets.

However, these are not enough, because every sets do not satisfied.
Especially rules of calculation depend on sets.

Numbers can exist in every sets, as whose elements are empty set. 

Real numbers satisfies all calculation rules.
Integers satisfies rules of addition, subtraction, and multiplication.

From next posts, we are going to look for the properties of sets
of which calculation rules are limited.