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Why is $e^x$ the only nontrivial function with a repeating derivative, i.e. is its own derivative?

It says so in the Wikipedia article about $e$. Is there a proof of this that I (a calculus AB student) could understand? Thanks!

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marked as duplicate by Ittay Weiss, Lord_Farin, Belgi, Julian Kuelshammer, Davide Giraudo Jun 3 '13 at 9:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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This will help:math.stackexchange.com/q/58097/35983 –  Gautam Shenoy Jun 3 '13 at 9:10

4 Answers 4

up vote 4 down vote accepted

Technically $y=0$ is also a function with a repeating derivative but in answer to your question, what you're essentially doing is solving the differential equation $\frac{\mathrm{d}y}{\mathrm{d}x}=y$ to which the only non-trivial solution is $y=A e^x$. (The differential equation is not hard to solve)

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Oh ok thanks I should've thought of that –  Ovi Jun 3 '13 at 9:11
    
no problem at all –  john Jun 3 '13 at 9:12

$y = A e^x$ is a solution of the linear differential equation $$ y' = \frac{dy}{dx} = y $$ with an initital condition $y(0)=A$. By the Picard-Lindelöf theorem there exists a unique solution to that equation with intial condition $y(0)=A$. For $A = 0$ you get the solution $y(x) = 0$ (the constant zero function) which trivially has a repeating derivative. For $A \ne 0$ you get the solution $y(x) = Ae^x = A \exp(x)$.

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Let $y = f(x)$ be a function that satisfies the differential equation: $$ \dfrac{dy}{dx} =y $$

We will solve this ODE by separation of variables. Observe that this is equivalent to: $$ \begin{align*} \dfrac{1}{y}dy &= dx \\ \int \dfrac{1}{y}dy &= \int dx \\ \ln|y|&=x+C \\ |y|&=e^{x+C} \\ |y|&=e^Ce^{x} \\ y&=\pm e^Ce^{x} \\ y&=Ke^{x} \\ \end{align*} $$ where $K=\pm e^C$ is some constant. This proves that the only types of functions that are equal to their own derivative are functions that are scalar multiples of $y=e^x$.

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Since taking the derivative is a linear operator, we have for any constant $c$: $$ (cf(x))' = cf'(x) $$ Thus for any solution $f(x)$ to your question, we would also have $cf(x)$.
Therefore without loss of generality (and assuming the $f(0) \ne 0 $) we can choose $$f(0) = 1$$ Now the function is determined by the fact that the derivative is known, and you can make a computer plot the function as tiny linear segments with that slope. What you get is the exponential function.

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