1

Prove that $\exp(x) = e^x$ for every real number $x$. (Hint: first prove the claim when $x$ is a natural number. Then prove it when $x$ is an integer. Then prove it when $x$ is a rational number. Then use the fact that real numbers are the limits of rational numbers to prove it for all real numbers. You may find the exponent laws to be useful.)

The author defines $e: = \exp(1) = \sum_{n=0}^\infty 1/n!$. I can show that $\exp(x) = e^x$ when $x$ is an integer.

When $x$ is a rational number, let $x = \frac{n}{m}$ for $n, m \in \mathbb{N}$ (if $x$ is negative, I can use the result that $\exp(-x) = 1/\exp(x)$). Then, $\exp(n/m) = \exp(1/m + ... + 1/m) = \exp(1/m) \cdot\cdot\cdot \exp(1/m)$. Thus, the problem is reduced to showing $\exp(1/m) = e^{1/m}$. How can I show this?

shk910
  • 3,599
  • What is the definition of $e^x$? Using the limit definition? If so, https://math.stackexchange.com/questions/69806/prove-the-definitions-of-e-to-be-equivalent seems to answer? – Gareth Ma Mar 28 '20 at 13:03
  • Also $\exp(z)=e^z$ for every complex number. – Dietrich Burde Mar 28 '20 at 13:06
  • 1
    From context it seems that $\exp(x)$ is defined as $\sum_{n=0}^\infty \frac{x^n}{n!}$ and $e = \exp(1)$. – Umberto P. Mar 28 '20 at 13:09
  • Using whatever definition of $\exp$ to prove that it satisfies $\exp(a+b) =\exp(a) \exp(b) $ and this gives your identity $\exp(x) =e^x$ for all rational $x$. This is all algebra can do. Beyond that you need a definition of irrational powers. Do that and your problem is solved with a little more effort. – Paramanand Singh Mar 29 '20 at 16:06

1 Answers1

1

You could just take $n = m$. Then $$e = \exp(1) = \exp(m/m) = \exp(1/m)^m$$ and raise each side to the $1/m$ power.

Umberto P.
  • 52,165