三千大千世界什么意思

This is the second equality in a sequence going back to Euler (the first one is the familiar $2^4=4^2$).

Let

$x=\left(\dfrac{n+1}{n}\right)^{n}$

$y=\left(\dfrac{n+1}{n}\right)^{n+1}$

Then

$x^y=\left(\dfrac{n+1}{n}\right)^{n×\left(\dfrac{n+1}{n}\right)^{n+1}}$ $=\left(\dfrac{n+1}{n}\right)^{\left(\dfrac{(n+1)^{n+1}}{n^n}\right)}$

and

$y^x=\left(\dfrac{n+1}{n}\right)^{(n+1)×\left(\dfrac{n+1}{n}\right)^{n}}$ $=\left(\dfrac{n+1}{n}\right)^{\left(\dfrac{(n+1)^{n+1}}{n^n}\right)}$

Thereby $x^y=y^x$. Here

$x=2.25=9/4=(3/2)^2, y=3.375=27/8=(3/2)^3$

which matches the Euler form with $n=2$.

Euler also showed that the formula above contains all the positive rational solutions for $x^y=y^x$ apart from $y=x$.

The values of $ ^y=y^x$ with this formula decrease with increasing $n$. Thus $16=2^4=4^2$ is greater than the limiting value of $e^e(\approx 15.154)$, with $2.25^{3.375}$ and other values in the sequence falling in-between.

This is the second equality in a sequence going back to Euler (the first one is the familiar $2^4=4^2$).

Let

$x=\left(\dfrac{n+1}{n}\right)^{n}$

$y=\left(\dfrac{n+1}{n}\right)^{n+1}$

Then

$x^y=\left(\dfrac{n+1}{n}\right)^{n×\left(\dfrac{n+1}{n}\right)^{n+1}}$ $=\left(\dfrac{n+1}{n}\right)^{\left(\dfrac{(n+1)^{n+1}}{n^n}\right)}$

and

$y^x=\left(\dfrac{n+1}{n}\right)^{(n+1)×\left(\dfrac{n+1}{n}\right)^{n}}$ $=\left(\dfrac{n+1}{n}\right)^{\left(\dfrac{(n+1)^{n+1}}{n^n}\right)}$

Thereby $x^y=y^x$. Here

$x=2.25=9/4=(3/2)^2, y=3.375=27/8=(3/2)^3$

which matches the Euler form with $n=2$. The next member of the sequence, with $n=3$, would be

$(64/27)^{256/81}=(256/81)^{64/27}$

Euler also showed that the formula above contains all the positive rational solutions for $x^y=y^x$ apart from $y=x$.

The values of $ x^y=y^x$ with this formula decrease with increasing $n$. Thus $16=2^4=4^2$ is greater than the limiting value of $e^e(\approx 15.154)$, with $2.25^{3.375}$ and other values in the sequence falling in-between.

This is the second equality in a sequence going back to Euler (the first one is the familiar $2^4=4^2$).

Let

$x=\left(\dfrac{n+1}{n}\right)^{n}$

$y=\left(\dfrac{n+1}{n}\right)^{n+1}$

Then

$x^y=\left(\dfrac{n+1}{n}\right)^{n×\left(\dfrac{n+1}{n}\right)^{n+1}}$ $=\left(\dfrac{n+1}{n}\right)^{\left(\dfrac{n+1^{n+1}}{n^n}\right)}$

and

$y^x=\left(\dfrac{n+1}{n}\right)^{(n+1)×\left(\dfrac{n+1}{n}\right)^{n}}$ $=\left(\dfrac{n+1}{n}\right)^{\left(\dfrac{n+1^{n+1}}{n^n}\right)}$

Thereby $x^y=y^x$. Here

$x=2.25=9/4=(3/2)^2, y=3.375=27/8=(3/2)^3$

which matches the Euler form with $n=2$.

Euler also showed that the formula above contains all the positive rational solutions for $x^y=y^x$ apart from $y=x$.

The values of $ ^y=y^x$ with this formula decrease with increasing $n$. Thus $16=2^4=4^2$ is greater than the limiting value of $e^e(\approx 15.154)$, with $2.25^{3.375}$ and other values in the sequence falling in-between.

This is the second equality in a sequence going back to Euler (the first one is the familiar $2^4=4^2$).

Let

$x=\left(\dfrac{n+1}{n}\right)^{n}$

$y=\left(\dfrac{n+1}{n}\right)^{n+1}$

Then

$x^y=\left(\dfrac{n+1}{n}\right)^{n×\left(\dfrac{n+1}{n}\right)^{n+1}}$ $=\left(\dfrac{n+1}{n}\right)^{\left(\dfrac{(n+1)^{n+1}}{n^n}\right)}$

and

$y^x=\left(\dfrac{n+1}{n}\right)^{(n+1)×\left(\dfrac{n+1}{n}\right)^{n}}$ $=\left(\dfrac{n+1}{n}\right)^{\left(\dfrac{(n+1)^{n+1}}{n^n}\right)}$

Thereby $x^y=y^x$. Here

$x=2.25=9/4=(3/2)^2, y=3.375=27/8=(3/2)^3$

which matches the Euler form with $n=2$.

Euler also showed that the formula above contains all the positive rational solutions for $x^y=y^x$ apart from $y=x$.

The values of $ ^y=y^x$ with this formula decrease with increasing $n$. Thus $16=2^4=4^2$ is greater than the limiting value of $e^e(\approx 15.154)$, with $2.25^{3.375}$ and other values in the sequence falling in-between.

This is the second equality in a sequence going back to Euler (the first one is the familiar $2^4=4^2$).

Let

$x=\left(\dfrac{n+1}{n}\right)^{n}$

$y=\left(\dfrac{n+1}{n}\right)^{n+1}$

Then

$x^y=\left(\dfrac{n+1}{n}\right)^{n×\left(\dfrac{n+1}{n}\right)^{n+1}}$ $=\left(\dfrac{n+1}{n}\right)^{\left(\dfrac{n+1^{n+1}}{n^n}\right)}$

and

$y^x=\left(\dfrac{n+1}{n}\right)^{(n+1)×\left(\dfrac{n+1}{n}\right)^{n}}$ $=\left(\dfrac{n+1}{n}\right)^{\left(\dfrac{n+1^{n+1}}{n^n}\right)}$

Thereby $x^y=y^x$. Here

$x=2.25=9/4=(3/2)^2, y=3.375=27/8=(3/2)^3$

which matches the Euler form with $n=2$.

Euler also showed that the formula above contains all the positive rational solutions for $x^y=y^x$ apart from $y=x$.

This is the second equality in a sequence going back to Euler (the first one is the familiar $2^4=4^2$).

Let

$x=\left(\dfrac{n+1}{n}\right)^{n}$

$y=\left(\dfrac{n+1}{n}\right)^{n+1}$

Then

$x^y=\left(\dfrac{n+1}{n}\right)^{n×\left(\dfrac{n+1}{n}\right)^{n+1}}$ $=\left(\dfrac{n+1}{n}\right)^{\left(\dfrac{n+1^{n+1}}{n^n}\right)}$

and

$y^x=\left(\dfrac{n+1}{n}\right)^{(n+1)×\left(\dfrac{n+1}{n}\right)^{n}}$ $=\left(\dfrac{n+1}{n}\right)^{\left(\dfrac{n+1^{n+1}}{n^n}\right)}$

Thereby $x^y=y^x$. Here

$x=2.25=9/4=(3/2)^2, y=3.375=27/8=(3/2)^3$

which matches the Euler form with $n=2$.

Euler also showed that the formula above contains all the positive rational solutions for $x^y=y^x$ apart from $y=x$.

The values of $ ^y=y^x$ with this formula decrease with increasing $n$. Thus $16=2^4=4^2$ is greater than the limiting value of $e^e(\approx 15.154)$, with $2.25^{3.375}$ and other values in the sequence falling in-between.