2
$\begingroup$

Is there any intuitive argument or visual "proof" that $\exp' = \exp$? Suppose you have defined the Euler number $\mathrm{e}$ as limit of the sequence $(a_n)$ where $a_n = \left (1 + \frac{1}{n} \right)^n \quad \forall n > 0$, and that the $\exp(x)$ is introduced as $\mathrm{e}^x$. The use of the power series of $\exp$ should be avoided.

$\endgroup$
  • 4
    $\begingroup$ question: how do you introduce real powers without knowing the exp function? $\endgroup$ – mookid Oct 15 '14 at 13:53
  • $\begingroup$ If you can use that $\log'(x)=1/x$ and that $\log$ and $\exp$ are inverses, then it follows from the chain rule, as in math.stackexchange.com/a/31392/589. $\endgroup$ – lhf Oct 15 '14 at 14:05
  • $\begingroup$ See also math.stackexchange.com/questions/381397/definition-of-expx. $\endgroup$ – lhf Oct 15 '14 at 14:17
  • $\begingroup$ I agree with @mookid , there is a structural issue regarding properties you can take for granted $\endgroup$ – mvggz Oct 15 '14 at 14:21
  • $\begingroup$ You can prove that $e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n$ (hint: let $m=\frac nx$). Differentiate that. (P.S. As a bonus, the binomial theorem gives you the Taylor series!) $\endgroup$ – Akiva Weinberger Oct 15 '14 at 14:35
4
$\begingroup$

$$ \exp(x) = \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n $$

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Do you intend to commute $\lim$ with $d/dx$ ? $\endgroup$ – lhf Oct 15 '14 at 14:12
  • 1
    $\begingroup$ @lhf: Yeah. Rigorously justifying it, but the OP is only asking for an intuitive argument. $\endgroup$ – user14972 Oct 15 '14 at 14:15
  • 1
    $\begingroup$ I disagree. The author's definition of $e^x$ would be rather $\lim_{n\to\infty}\left(1+\frac1n\right)^{nx}$, and this definition itself requires to say what the non-integer powers are. $\endgroup$ – Start wearing purple Oct 15 '14 at 14:22
  • $\begingroup$ @O.L.: I opted to leave transforming between the two versions to the reader. $\endgroup$ – user14972 Oct 15 '14 at 14:35
2
$\begingroup$

Once you have defined $exp(x)=e^x$ we can work our way trhough the definition of derivative (which is quite intuitive) for demostrating that $exp'(x)=exp(x)$. The formal definition of derivative is:

$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$

When $h$ tends to $0$.

So, substituting $f(x)$ by $e^x$ we get that:

$$f'(x)=\lim_{h\to0}\frac{exp(x+h)-exp(x)}{h}$$

Working with the expression a bit we can rearrange it to:

$$f'(x)=\lim_{h\to0}\frac{e^x(e^h-1)}{h}$$

Now notice that as $h$ approaches $0$ the denominator of the fraction $h$ and the term $e^h-1$ also approach $0$. Furthermore, when being really close to $0$ they approach it at 'the same speed' and we say they are 'equivalent infinitesimals' which means that we can approximate one to the other. We can substitute then $e^h-1$ with $h$.

enter image description here

With this in mind we end up with:

$$f'(x)=\lim_{h\to0}\frac{e^x·h}{h}$$

We can cancel the $h$ and get to the final result:

$$exp'(x)=exp(x)$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ What would be a explanation that $\lim_{h\to0}a^h-1\not=h$ for $a\not=e$ $\endgroup$ – kingW3 Oct 18 '14 at 12:06
  • $\begingroup$ @kingW3 Note that $\lim_{h\to0}a^h-1$ can also be expressed as $\lim_{h\to0}e^{ln(a^h)}-1=\lim_{h\to0}e^{h·ln(a)}-1$which is the same case as stated above but with $h·ln(a)$ replacing $h$, which means we can aproximate $a^h-1$ to $h·ln(a)$ $\endgroup$ – Ioannes Oct 18 '14 at 17:30
2
$\begingroup$

Provided that you have defined general exponentiation in the first place, you can write

$$\frac{d}{dx} a^x = \lim_{h \to 0} \frac{a^{x+h} - a^x}{h} = a^x \lim_{h \to 0} \frac{a^h - 1}{h}.$$

You now need to intuitively argue that this last limit, as a function of $a$, ranges from $0$ to $\infty$ as $a$ ranges from $1$ to $\infty$. It is also clearly continuous. Therefore by the intermediate value theorem there is some $a$ such that the limit is $1$. We call this number $e$.

Defining general exponentiation in the first place is a huge mess. Indeed, I think that from the perspective of real analysis, it is easier to define $\exp$ by the ODE it satisfies, then prove it has an inverse $\ln$, then prove that $\exp(q \ln x) = x^q$ whenever $q$ is rational. (Provided you can independently prove the power rule for rational powers $q$, this last argument is actually quite simple.)

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Since $$ \frac{e^{x+h}-e^x}{h}=\frac{e^xe^h-e^x}{h}=e^x\frac{e^h-1}{h} $$ you only need to prove that $\exp'(0)=1$.

Perhaps this helps.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ But how to show that $\exp'(0) = 1$? $\endgroup$ – Julia Oct 15 '14 at 16:00
0
$\begingroup$

There is an elegant way to define e which is to start from

$$\lim_{x\to\infty} (1+1/x)^x = e$$

This definition allows the normal intuition associated do e (euler number), that makes the concept easy to explain

It starts from a hypothetical situation where has 100% interest in one year over some money. In this simple case, one double his money. 100 turns in 200. After, if one consider 2 periods ($x=2$), the factor that multiply the money is $(1 + 1/2 )^2 \approx 2.25$, because the interest sum to the principal at half period. In this case, $\$100$ turns $\$225$.

Now supposes that one uses $x=4$ periods. The total is $(1+1/4)^4 \approx 2.44$, i.e., 100 turns 244. If one uses $N=10$, the value is $(1+1/10)^10 \approx 2.59$ ($\$100$ converts in $\$259$)

Now there are 10 periods where the interest join with the money and serves as a basis for calculating subsequent interest, increasing the gain. For $n=100$, one reachs $2,70$ ($\$100$ changes to $\$270$). The money growth rate is lowering...

So one does the number of periods to increase towards infinity. In that situation, it's easy to notice that the mysterious number approaches $2.71828...$. $\$100$ transforms in almost $\$271.83$.

So

$$\lim_{x\to\infty} (1/x)^x = e$$

Now, replace $x$ for $1/h$. If $x\to\infty$, now one has $h\to 0$

$$\lim_{h\to 0} (1+h)^{1/h} = e$$

Is the starting point for the next steps:

$$ \lim_{h\to 0} (1+h) = \lim_{h\to 0}e^h $$ $$ \lim_{h\to 0} (h) = \lim_{h\to 0}(e^h - 1)$$

How $h\to 0$ but it's not 0, one can divide both sides for $h$ and invert the sides:

$$\lim_{h\to 0}(e^h-1) / h=1$$

Now we can calculate (e^x)' using the derivative definition:

$$\lim_{h\to 0}\left(\frac{e^{x+h}-e^x}{h}\right)=\lim_{h\to 0}\left(\frac{e^xe^h-e^x}{h}\right)= e^x\lim_{h\to 0}\left(\frac{e^h-1}{h}\right)$$

Well, this limit was calculated above and is equal to 1, so

$$(e^x)' = e^x $$

There is logic in this result since we are in a exponential function: the $e^x$ slope is increasing forever in increasing rate, that increases in increasing rate etc. It never stops.

If one has $a^x$ compared with $x^b$, where $a$ is a constant $> 1$ and $b$ is another constant, at some point the exponential exceeds the power, i.e., $a^x > x^b$ (for large value of $x$)

If one uses $log_e$ in both sides, this feature becomes clearly visible:

$$x\; \ln a = b \; \ln x$$

If $a>1$, the distance between $b$ and $\ln(a)$ can be huge but they are constants. However at some point $x$ will prevail over $\ln(x)$ in such a proportion that the exponential will leave the power function well behind.

In formal terms, it is said that:

$$\lim_{x\to \infty} a^x/x^b = \infty $$

PS: It can be easily proved by L'Hôpital's rule after apply $log_e$ in the numerator and denominator, but it's very intuitive.

Speed is variation of distance, acceleration is variation of speed etc. So a second degree function stops varying after second derivative, e.g, the acceleration is constant.

If one supposes $b=100$ exponent in a power function, if one differentiate 100 times gets a constant, i.e, it stops the rate of variation of variation of variation... The exponential function never stops. For that reason the derivative needs to be a exponential function, because the reasoning is the same. It's a recursive argument.

Any constant in exponent, can be converted to other constant in the base.

So

$$(a^x)' = (e^{x \ln a})' = \ln a \;e^{x \ln a} = \ln a\; a^x$$

it keeps completely the logic of exponential function.

One can imagine hyper exponential function, with an even more radical profile of increase, but that's it for now

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.