Exercise 8.4.4

Question

Define $e = E(1)$. Show $E(n) = e^n$ and $E(m/n) = \left(\sqrt[n]{e}\right)^m$ for all $m,n \in \mathbf{Z}$.

Ulisse Mini · Accepted Answer

We have for $n \geq 1$ that
$$E(n) = E\left(\sum^n_{i=1} 1\right) = \prod^n_{i=1} E(1) = e^n $$
We also have for $n=0$, $E(0) = 1 = e^0$, by the standard definition of $a^n$ for any $a \in \mathbf{R}$. Finally for $n > 0$, $e^{-n} = 1/e^n = 1/E(n) = E(-n)$, so $e^n = E(n)$ for all $n \in \mathbf{Z}$.

By definition $\sqrt[n]{e}$ is the unique positive number which satisfies $(\sqrt[n]{e})^n = e$. $E(1/n)$ satisfies this equality, since
$$E(1) = E\left(\sum^n_{i=1} \frac{1}{n}\right)= \prod^n_{i=1} E(1/n) $$
so $\sqrt[n]{e} = E(1/n)$. Finally
$$E(m/n) = E\left(\sum^m_{i=1} \frac{1}{n}\right) = \prod^m_{i=1}E(1/n) = \left(\sqrt[n]{e}\right)^m$$

Exercise 8.4.4

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Exercise 8.4.4

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