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2020.04.28
カテゴリ： カテゴリ未分類
朝２時間、山の中で楽しい遊びをしてきた。
山椒とタケノコを採った。 鹿が近くでないていた。 鶯は今、庭近くで啼いている。
0^0 is two values 1 and 0; 1 is the principal. We can enjoy many applications:
\subsection{$0^0= 1,0$}\index{$0^0= 1,0$}
By the standard definition, we will consider
$$
0^0 = \exp (0 \log 0) =\exp 0 = 1,0.
$$
The value $1$ is famous which was derived by\index{Abel, N.} N. Abel, meanwhile, \index{Michiwaki, H.}H. Michiwaki had directly derived it as 0 from the result of the division by zero. However, we now know that $0^0= 1,0$ is the natural result.
We will see its reality.
\medskip
{\bf For $0^0= 1$:}
\medskip
\index{ $0^0= 1$}
In general, for $ z \ne 0$, from $z^0 = e^{0 \log z}$, $z^0 = 1$, and so, we will consider that $0^0= 1$ in a natural way.
For example, in the elementary expansion
$$
( 1 + z)^n = \sum_{k=0}^n {}_nC_k z^k
$$
the formula $0^0= 1$ will be convenient for $k=0$ and $z=0$.
$$
\exp z = \sum_{k=0}^{\infty} \frac{1}{k!} z^k
$$
in order to have a sense of the expansion at $z=0$ and $k=0$, we have to accept the formula $0^0= 1$.
In the differential formula
\frac{d^n}{dx^n} x^n = n x^{n-1},$$
in the case $n=1$ and $x=0$, the formula $0^0= 1$ is convenient and natural.
In the Laurent expansion, if $0^0=1$, it may be written simply as
$$
f(z) = \sum_{n=-\infty}^{\infty} C_n (z - a)^n,
$$
for $f(a) = C_0$.
\bigskip
{\bf For $0^0= 0$:}
\medskip
\index{$0^0= 0$}
For any positive integer $n$, since $z^n =0$ for $z=0$, we wish to consider that $0^0= 0$ for $n=0$.
For the expansion
$$
\frac{t}{\exp t -1} =\sum_{n=0}^\infty \frac{B_n}{n!} t^n,
$$
with the Bernoulli's constants $B_n$, the usual value of the function at $t=0$ is $1$ and this meets the value $0^0=1$. Meanwhile, by the division by zero, we have the value $0$ by the method
$$
\frac{t}{\exp t -1}|_{t=0} = \frac{0}{\exp 0 -1} = \frac{0}{0}=0
$$
and this meets with $0^0=0$.
Note that by the division by zero calculus, we have the value $0$
(V. V. Puha: 2018.7.3.6:01).
\medskip
{\bf Philip Lloyed's question} (2019.1.18): \index{Lloyed's question} {\it What is the value of the equation \index{Lloyed, P.}
$$
x^x =x
$$
?
}
By the equation
$$
x(x^{x-1} - 1) = 0,
$$
we have $x=0$ and $x=1$ therefore, we have also $0^0=0$.
P. Lloyed discovered also the solution $-1$, as we see the result directly and interestingly.
\medskip
Khandakar Kawkabum Munir Saad asked the question for the equation $2^x=0$ in Quora: 2019.7.4.17:00. We can give the solution $x=0$. Therefore, for the very interesting equation $x^2 = 2^x$ we have the trivial solutions $2$ and $4$ and furthermore, the solution $0$.
\medskip
山椒とタケノコを採った。 鹿が近くでないていた。 鶯は今、庭近くで啼いている。
0^0 is two values 1 and 0; 1 is the principal. We can enjoy many applications:
\subsection{$0^0= 1,0$}\index{$0^0= 1,0$}
By the standard definition, we will consider
$$
0^0 = \exp (0 \log 0) =\exp 0 = 1,0.
$$
The value $1$ is famous which was derived by\index{Abel, N.} N. Abel, meanwhile, \index{Michiwaki, H.}H. Michiwaki had directly derived it as 0 from the result of the division by zero. However, we now know that $0^0= 1,0$ is the natural result.
We will see its reality.
\medskip
{\bf For $0^0= 1$:}
\medskip
\index{ $0^0= 1$}
In general, for $ z \ne 0$, from $z^0 = e^{0 \log z}$, $z^0 = 1$, and so, we will consider that $0^0= 1$ in a natural way.
For example, in the elementary expansion
$$
( 1 + z)^n = \sum_{k=0}^n {}_nC_k z^k
$$
the formula $0^0= 1$ will be convenient for $k=0$ and $z=0$.
$$
\exp z = \sum_{k=0}^{\infty} \frac{1}{k!} z^k
$$
in order to have a sense of the expansion at $z=0$ and $k=0$, we have to accept the formula $0^0= 1$.
In the differential formula
\frac{d^n}{dx^n} x^n = n x^{n-1},$$
in the case $n=1$ and $x=0$, the formula $0^0= 1$ is convenient and natural.
In the Laurent expansion, if $0^0=1$, it may be written simply as
$$
f(z) = \sum_{n=-\infty}^{\infty} C_n (z - a)^n,
$$
for $f(a) = C_0$.
\bigskip
{\bf For $0^0= 0$:}
\medskip
\index{$0^0= 0$}
For any positive integer $n$, since $z^n =0$ for $z=0$, we wish to consider that $0^0= 0$ for $n=0$.
For the expansion
$$
\frac{t}{\exp t -1} =\sum_{n=0}^\infty \frac{B_n}{n!} t^n,
$$
with the Bernoulli's constants $B_n$, the usual value of the function at $t=0$ is $1$ and this meets the value $0^0=1$. Meanwhile, by the division by zero, we have the value $0$ by the method
$$
\frac{t}{\exp t -1}|_{t=0} = \frac{0}{\exp 0 -1} = \frac{0}{0}=0
$$
and this meets with $0^0=0$.
Note that by the division by zero calculus, we have the value $0$
(V. V. Puha: 2018.7.3.6:01).
\medskip
{\bf Philip Lloyed's question} (2019.1.18): \index{Lloyed's question} {\it What is the value of the equation \index{Lloyed, P.}
$$
x^x =x
$$
?
}
By the equation
$$
x(x^{x-1} - 1) = 0,
$$
we have $x=0$ and $x=1$ therefore, we have also $0^0=0$.
P. Lloyed discovered also the solution $-1$, as we see the result directly and interestingly.
\medskip
Khandakar Kawkabum Munir Saad asked the question for the equation $2^x=0$ in Quora: 2019.7.4.17:00. We can give the solution $x=0$. Therefore, for the very interesting equation $x^2 = 2^x$ we have the trivial solutions $2$ and $4$ and furthermore, the solution $0$.
\medskip
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