Is there an intuitive reason why the constant $e$ to the power of $x$ has a derivative that equals the value of the function? I know that this is the result of differentiating, and I've seen several proofs of how you work out the derivative, I was just wondering why this is and why it is the case for a number estimated from constant growth?
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1If you compute the derivative of $b^x$, you find that the answer is $b^x$ multiplied by an annoying constant. So we then ask, is there a special value of $b$ that makes that annoying constant equal to $1$. The answer is $b = e$. In my mind this is how we discover the constant $e$. – littleO Jan 16 '20 at 11:11
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Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. – N. F. Taussig Jan 16 '20 at 11:24
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1https://www.youtube.com/watch?v=m2MIpDrF7Es for a very nice video about what @littleO was saying – user1936752 Jan 16 '20 at 11:46
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How do you define $e^x$? The answer is different depending on the definition. For example, if you define $\exp$ as "the function whose derivative is equal to itself, with value $1$ at $x=0$", then it's trivial. – Patrick Stevens Jan 16 '20 at 11:48
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I'm not sure I'm just not getting the obvious here, I was just wondering, is there a reason why b is equal to the constant we get when we estimate what value we get from constant growth? Is the fact that e makes the constant equal to 1 related to the fact that it is derived from estimating constant growth? – Martin Jan 16 '20 at 13:01
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1@user1936752 This is exactly the video I would have recommended! – Andrew Chin Jan 16 '20 at 14:17
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Maybe a nice way to see it is the power series expansion: $$\forall x\in \mathbb{R} \colon e^x =1 + x + \frac {x^2} {2!} + \frac {x^3} {3!} + \cdots $$
S. Maths
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By definition,$$\frac{d}{dx}a^x=\lim_{h\to0}\frac{a^{x+h}-a^x}{h}.$$Each exponential function has the nice property $a^{x+h}=a^xa^h$, so we can take out an $h$-independent $e^x$ factor, making the above limit the original function $a^x$ times just a number, namely $\lim_{h\to0}\frac{a^h-1}{h}$. This limit is in turn, by definition, the derivative of $a^x$ at $x=0$. Now if we gradually increase $a$ from just above $0$ to not quite $\infty$, $a^x$ will get steeper and steeper at $x=0$. And $e$ is just the choice of $a$ for which the slope is $1$, so that $e^x$ is its own derivative.
J.G.
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Something that might lend some intuition, is that $$ e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n=\lim_{n\to\infty}\left(1+\frac xn\right)^{n-1} $$ and $$ \frac{\mathrm{d}}{\mathrm{d}x}\left(1+\frac xn\right)^n=\left(1+\frac xn\right)^{n-1} $$ It takes a little bit to make this rigorous, but intuition and rigor often follow different paths.
robjohn
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I really like this answer, but I wonder how one would intuitively draw the connection between the statements $$e=\lim\limits_{n\to\infty}\left(1+\frac1n\right)^n\quad\text{and}\quad e^x=\lim\limits_{n\to\infty}\left(1+\frac{x}{n}\right)^n.$$ – Andrew Chin Jan 16 '20 at 14:21
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Perhaps $$ \begin{align} \lim_{n\to\infty}\left(1+\frac xn\right)^n &=\lim_{n\to\infty}\left(\left(1+\frac xn\right)^{n/x}\right)^x\ &=\left(\lim_{n\to\infty}\left(1+\frac xn\right)^{n/x}\right)^x \end{align} $$ – robjohn Jan 16 '20 at 14:37
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I understand this line of thinking, but this assumes $x\neq0$, does it not? – Andrew Chin Jan 16 '20 at 14:40
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The right side looks to be in the form $(1^\infty)^0=1^{\infty\cdot0}$. Should this be a problem? – Andrew Chin Jan 16 '20 at 14:47
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What I meant was $ \lim\limits_{n\to\infty}\left(1+\frac0n\right)^n=1$ and $e^0=1$. – robjohn Jan 16 '20 at 14:54
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If $x$ changes by $1$, then the function changes from $e^x$ to $e^{x+1} = e\cdot e^x.$
If $x$ changes by $2$, then the function changes from $e^x$ to $e^{x+2} = e^2 e^x.$
If $x$ changes by $h$, then the function changes from $e^x$ to $e^{x+2} = e^h e^x.$ In each case the change in the function is $e^x(e^h-1)$, which is $e^x$ times some number. This is true for any base, but for $e$, the "some number" over $h$ happens to tend to $1$.
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The slope of the tangent line at point $x$ is always equal to the value of $f(x) = e^x$. This may be the intuition you're looking for. And this is true for every point $x$. And it is true for every point $x$ only for functions of the form $e^x + const$. That makes $f(x) = e^x$ quite "special" if you think about it from geometrical perspective.
How to see this? Draw in one plane: 1) the graph of $e^x$, 2) the unit circle, 3) the line S with the equation $x=1$ (on which values of $\tan$ are measured).
Take any point $x_0$ and draw the tangent line $L_0$ to the graph (of $e^x$) at the point $(x_0, e^{x_0})$. Now draw a line $L_1$ passing through $(0,0)$ and parallel to the tangent line $L_0$. Note the height at which $L_1$ intersects the line $S$. The special ting here is that this height is always equal to $e^{x_0}$ i.e. to the value (at point $x_0$) of the function itself. In a way that property is quite remarkable especially since it holds for every value of $x$ ($x_0$ was taken arbitrary here).
Why remarkable? Because as we know no other function (apart from the functions $g(x) = e^x + const$) has this property (no other function has it for every $x$, I mean).
peter.petrov
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I don't think OP asked for intuition for the meaning of the statement, rather for why it is true. – Wojowu Jan 16 '20 at 11:27
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@Wojowu Hm... I think the title and the question clearly designate the opposite. The title itself talks about $intuition$. – peter.petrov Jan 16 '20 at 11:28
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Let's see what the OP thinks... I think I understood their ask correctly. – peter.petrov Jan 16 '20 at 11:31
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@Wojowu Btw, that is exactly why it is true. Because the meaning of the derivative is just that and it happens to be equal to the value of the function... and that happens to be true for every $x$. That's exactly what intuition, insight, etc. is about. – peter.petrov Jan 16 '20 at 11:37
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Thank you for this explanation, I think I might've worded my question incorrectly. I was thinking more along the lines of, the intuition to why a constant defined in a way that estimates the value of constant growth would give us a number that when taken to the power x would give us a derivative that is equal to the value of the function. Sorry for any misunderstanding. – Martin Jan 16 '20 at 12:49