How can I prove that a function that is its own derivative exists? And how can I prove that this function is of the form $a(b^x)$?
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There are two ways you could show it. The harder route would be to prove the existence and uniqueness theorem for ordinary differential equations, thus showing there exists solutions to $y'=y$. The more direct way would be to just construct the function $e^x$ and show that it's its own derivative. You would start by defining $$\ln(x) = \int_1^x \frac{1}{t}\, dt$$ and prove that it's a strictly increasing function on $(0,\infty)$ with range $(-\infty, \infty)$. It follows that $\ln(x)$ has an inverse, which we should dub $e^x$. As for finding the derivative of this new and mysterious function: $$y=e^x$$ $$\ln(y)=x$$ Taking the $x$ derivative of both sides, $$\frac{y'}{y} = 1$$ $$\implies y'=y$$ And do show that every function which is its own derivative is a constant multiple of $e^x$, suppose that $f'=f$. Then, noting that $e^x$ is nowhere zero, $$\frac{d}{dx} \frac{f(x)}{e^x} = \frac{f'(x)e^x-f(x)e^x}{(e^x)^2} = \frac{f(x)e^x-f(x)e^x}{(e^x)^2} = 0$$ Therefore, $$\frac{f(x)}{e^x}$$ is constant since it has a connected domain, and so $f(x) = ce^x$ for some $c$. |
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$f(x) = 0$ is trivially its own derivative, and is of the form $a(b^x)$ for $a=0$ and any positive $b$. That's all we need to solve the problem posed. |
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If you postulate a solution to $y=y'$ of the form $g(x)=\sum_{k=0}^\infty a_kx^k$, by the equality of the power series on both sides of the equality one gets $$ a_{k+1}=(k+1)a_k,$$and then one deduces that $$a_k=\frac{a_0}{k!},\ \ \ \ k=1,2,\ldots.$$So $$y(x)=a_0\,\sum_{k=0}^\infty\frac{x^k}{k!}.$$One can then focus on the case where $a_0=1$, say $g(x)=\sum_{k=0}^\infty\frac{x^k}{k!}$. Define $e=g(1)$. Using the series one can show that $$ g(x+y)=g(x)g(y).$$ It follows that $$\tag{1}g(x)=e^x$$ for $x$ rational. As $g$ is continuous (infinitely differentiable, even), it has to be $e^x=g(x)$ for irrational $x$, too. Thus $$ y(x)=a_0\,e^x. $$ |
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An intuitive answer: For smooth functions and "small" $h$, we have $$f'(x)\approx\frac{f(x+h)-f(x)}h.$$ Then $f'(x)=f(x)$ yields $$f(x+h)\approx(1+h)f(x),$$ and $$f(x+2h)\approx(1+h)f(x+h)\approx(1+h)^2f(x),$$ $$\cdots$$ $$f(x+nh)\approx(1+h)^nf(x).$$ Now with $nh=1$, $$f(x+1)\approx \left(1+\frac1n\right)^n f(x),$$ which should ring a bell. |
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