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How does one prove that the exponential and logarithmic functions are inverse using the definitions:

$$e^x= \sum_{i=0}^{\infty} \frac{x^i}{i!}$$ and $$\log(x)=\int_{1}^{x}\frac{1}{t}dt$$

My naive approach (sort of ignoring issues of convergence) is to just apply the definitions straightforwardly, so in one direction I get:

\begin{align}\log(e^x)&=\int_{1}^{e^x}\frac{1}{t}dt\\ &=\int_{0}^{e^x-1}\frac{1}{1+t}dt\\ &=\int_0^{e^x-1}\sum_{j=0}^\infty (-1)^jt^jdt \\ &=\sum_{j=0}^\infty (-1)^j \int_0^{e^x-1} t^jdt\\ &=\sum_{j=0}^\infty \frac{(-1)^j}{j+1}(e^x-1)^{j+1}\\ &=\sum_{j=0}^\infty \frac{(-1)^j}{j+1} \sum_{k=0}^{j+1} \frac{n!}{k!(n-k)!}e^{x(n-k)}(-1)^k\\ &=\sum_{j=0}^\infty \sum_{k=0}^{j+1} \frac{(-1)^{j+k}n!}{(j+1)k!(n-k)!} e^{x(n-k)}\\ &=\sum_{j=0}^\infty \sum_{k=0}^{j+1} \frac{(-1)^{j+k}n!}{(j+1)k!(n-k)!} \sum_{\ell=0}^\infty \frac{(-1)^\ell}{\ell !}(n-k)^\ell x^\ell\\ &=\sum_{j=0}^\infty \sum_{k=0}^{j+1} \sum_{\ell=0}^\infty \frac{(-1)^{j+k+\ell}n!(n-k)^\ell x^\ell}{(j+1)k!\ell!(n-k)!} \end{align}

and I cant see at all that this is equal to $x$. My guess is I'm going about this all wrong.

user140776
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  • Any reason why you don't want to show it the easy way---Let one be $f$ and the other $g$; then show $f(g(x)) = g(f(x)) = x$? (When you say $log$, I am assuming you mean the natural log.) – DDS Jul 13 '19 at 03:40
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    thats what Im trying to do – user140776 Jul 13 '19 at 03:42
  • @user140776 Your sum $\sum_{k=0}^n$ should be $\sum_{k=0}^{j+1}$. Trying working with that and then, as you have done here, swap the order of summation. Also, one minor point: you should assume $x \in [0,\log(2))$ to be able to use the geometric series for $\frac{1}{1+t}$. – Dzoooks Jul 13 '19 at 04:00
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    @Dzoooks Ah, yes, good points. I'll edit that for corectness. – user140776 Jul 13 '19 at 04:24
  • You could apply the fundamental theorem of calculus to do it. It would simplify what you're doing greatly. – Cameron Williams Jul 13 '19 at 05:33
  • And one might notice a fun fact from this question, that $$\sum_{i=0}^{\infty} \frac{(\ln x)^i}{i!}=x$$ !!! – Fareed Abi Farraj Jul 13 '19 at 05:42

2 Answers2

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$\log(e^x)=\int_1^{e^x}\frac{1}{t}dt=\int_0^x\frac{1}{e^u}e^u du=x$ and since it's easy to prove that $e^x$ is bijective then $\log$ is its inverse.

Julien D
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  • It suffices to show $\exp(x) > 0$ for all $x$, as $\exp' = \exp$ (a fact you've used already). I'm not certain how best to show this, without using the Picard existence-uniqueness theorem for differential equations. – Theo Bendit Jul 13 '19 at 04:13
  • @DustanLevenstein Right, I was more talking about $\exp(x) > 0$. – Theo Bendit Jul 13 '19 at 04:15
  • Ah, understood, my bad. – Dustan Levenstein Jul 13 '19 at 04:16
  • @TheoBendit this can be shown by showing directly that $e^xe^{-x}$ has derivative zero, and is therefore a constant, equal to $1$. Since $e^{x}$ is easily seen to be positive for $x \ge 0$, it follows that $e^x$ is positive for all $x$. It should also be noted that it is imperative to show this for the calculation Julien demonstrated to make sense - you can't integrate $\frac{1}{t}$ across $0$. – Dustan Levenstein Jul 13 '19 at 04:33
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    Am I missing something? You have shown that $\log(\exp(x))=x$, but how do you know that $\exp(\log(x))=x$? – user1551 Jul 13 '19 at 04:44
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    Because of the uniqueness of the inverse. $exp$ is one-to-one then it admits an inverse $g$, or $log\circ exp=id$ then $log=log\circ exp\circ g=id\circ g=g$. – Julien D Jul 13 '19 at 05:06
  • By a "one-to-one function", do you mean it's bijective? (Because this term usually refers to an injective function.) – user1551 Jul 13 '19 at 05:25
  • @user1551. One-to-one is, as you said, a synonym for injective .But the $\log$ function is bijective from $\Bbb R^+$ to $\Bbb R$. – DanielWainfleet Jul 13 '19 at 06:42
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    Oh yes sorry, bijective! – Julien D Jul 13 '19 at 06:53
  • @DanielWainfleet That of course is true, but it isn't what the answer says. – user1551 Jul 13 '19 at 06:55
  • @JulienD I see. Thanks for the clarification. – user1551 Jul 13 '19 at 06:55
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We use the fact that $g(x)=\exp(x)$ is the unique function $g:\mathbb{R} \rightarrow (0,\infty)$ such that $g'(x)=g(x)$ and $g(0)=1.$

Since $f(x)=\log(x)$ is evidently bijective, it must have the inverse $f^{-1}:\mathbb{R} \rightarrow (0,\infty)$. By the well-known theorem on the derivative of the inverse function, $$(f^{-1}){'}(x)=\frac{1}{f'(f^{-1}(x))}=\frac{1}{1/(f^{-1}(x))}=f^{-1}(x)$$ for every $x>0$. Furthermore, since $f(1)=\log(1)=0,$ $f^{-1}(0)= 1$.

It follows that $f^{-1}(x)$ must be $\exp(x).$

Fractal
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  • Good idea. I fixed some errors and added some elaboration. – Dustan Levenstein Jul 13 '19 at 13:26
  • I believe you made an error when you edited 1/(1/(f^(-1)(x))) to 1/f(f^(-1)(x)). Note that I defined f(x) to be log(x), not e^x. I re-edited the solution. – Fractal Jul 13 '19 at 17:27
  • Oh, I see. I think you have to use uniqueness of solutions to differential equations to conclude that $f^{-1} = \exp$ though. What I had written for $f = \exp$ only uses the constant of integration. – Dustan Levenstein Jul 13 '19 at 20:49
  • You can use the harmonic series to show that $\log(x)$ is bijective. I wouldn't call that "evident" without explanation though. – Dustan Levenstein Jul 13 '19 at 20:50
  • The definition of log(x) using the integral shows that log(x) is continuous and strictly increasing on its domain, so we can indeed quickly conclude that it is a bijection from (0, inf) onto its range. I suppose proving that its range is R may take a bit more time, but such verification is not difficult. – Fractal Jul 14 '19 at 04:21