Why does e^x differentiate to e^x?

How would you go about doing that?

The reasoning isn't even correct, let alone a proof


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How you answer your question depends on how you define $e^x$. We can define the number $e$ as the unique number $e \in \mathbb{R}$ such that $\dfrac{d}{dx}[e^x] = e^x$. Or, we can look at it from first principles as in your example, but that requires a certain knowledge of limits which I don't want to go into if you haven't covered it yet.

Alternatively we can define the function $\exp x$ by its power series
$\exp x = \displaystyle \sum_{n=0}^{\infty} \dfrac{x^n}{n!}$

Since we're allowed to differentiate power series term-by-term, we find that...
$\dfrac{d}{dx} (\exp x) = \displaystyle \sum_{n=0}^{\infty} \dfrac{d}{dx} \left( \dfrac{x^n}{n!} \right)$

$= \displaystyle \sum_{n=0}^{\infty} \dfrac{nx^{n-1}}{n!}$

$= \displaystyle \sum_{n=1}^{\infty} \dfrac{x^{n-1}}{(n-1)!}$

$= \displaystyle \sum_{n=0}^{\infty} \dfrac{x^n}{n!}$

$= \exp x$

The result that $\exp x = e^x$ is another proof entirely, but for the purposes of your question you can assume it's true.

The proof that $\dfrac{a^h-1}{h} \to \ln a$ as $h \to 0$ is quite complicated; if you've just done A-levels then there's no point in you seeing it because it'll just look like garble... just accept it as truth So given that $\displaystyle \lim_{h \to 0} \dfrac{a^h-1}{h} = \ln a$, we get that $\displaystyle \lim_{h \to 0} \dfrac{e^h-1}{h} = \ln e = 1$.


Nor do I (and nor did he by the looks of it), I think it's nonsense

Wow, a load of people replied while I was typnig my first response, sorry to keep people waiting.
I'll post this reply to Nuodai, then work on answering the other posts.



I really hate to say this, but I don't understand (I've been saying that far too often lately). I've never heard of power series before, and I'm trying to understand this, but I just can't make sense of it - a search on google for an explanation of power series gave results that I still wasn't able to follow.




I didn;t know it approached natural log a, the book just said that for e it approached 1.

But I've only done AS levels, so it sounds like the question of how this works is one I won't encounter for some time. Which is horribly furstrating, since it means not understanding how these rules work, instead just applying them.

Sorry about the normal text, I still can't type very much Latex, Hope it's followable. There is all the possibility that I have made a mistake or a bad assumption as I haven't had any formal education on limits. So valid proof or algebraic parlour trick?

why not? assuming a continuous number line etc.

I really hate to say this, but I don't understand (I've been saying that far too often lately). I've never heard of power series before, and I'm trying to understand this, but I just can't make sense of it - a search on google for an explanation of power series gave results that I still wasn't able to follow.

I didn;t know it approached natural log a, the book just said that for e it approached 1.

But I've only done AS levels, so it sounds like the question of how this works is one I won't encounter for some time. Which is horribly furstrating, since it means not understanding how these rules work, instead just applying them.

It's not defined at 0. You don't need it to be defined at 0 for the limit to exist. (The limit considers what happens as we get close to 0, not what happens at 0).

It's pretty complicated to explain and depends on how you define $e^x$
Mathematically, the definition of a limit says that if we want the limit to be the value 'y' then we have to be able to get $\frac{e^h - 1}{h}$ to be close to y for h that are sufficiently close to 0. Infact, if you give me a positive number, I have to show that for small enough |h|, $\frac{e^h - 1}{h}$ is at most that positive number away from y (in this case y=1).

Basically, don't think of it as a coincidence that e to the x differentiates to itself. Think of it as the definition of e. The coincidence is that this number = 2.71...

I'd say don't worry too much about it right now, rely on maybe a graph plotter like wolfram alpha. If you want to you can come back to this when you've done all of C4 and you'll see how it all beautifully links up, especially if you're doing FP2 where you'll learn about power series.

http://www.wolframalpha.com/input/?i=%28e^h-1%29%2Fh

Look at what's happening when h is going closer and closer to 0 in the graph i linked.

Sadly it's not valid at all "Equals" and "tends to" mean very different things, and you can't treat "tends to" like you do "equals". You've got them confused along the way. Saying "$e^x=(1+\frac{1}{x})^x$ as $x \to \infty$" doesn't make any sense, for example.

Stuff

Sorry about the normal text, I still can't type very much Latex, Hope it's followable. There is all the possibility that I have made a mistake or a bad assumption as I haven't had any formal education on limits. So valid proof or algebraic parlour trick?

You didn't justify your claim at all. What you claimed is that for some $y \in \mathbb{R}$, it's true that for all $x \in \mathbb{R}$
$\dfrac{dy^x}{dx} < y^x$
and that there exists another number such that the < is a >.

Assuming that this is true is as big an assumption as just assuming that there exists $y \in \mathbb{R}$ such that $\dfrac{dy^x}{dx} = y^x$. You could have just said "assume the result is true; therefore the result is true".
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Oh, I was being lazy around that part and the brackets got too much for me, basically throughout the whole thing i'm working inside the initial limit.

Sadly it's not valid at all "Equals" and "tends to" mean very different things, and you can't treat "tends to" like you do "equals". You've got them confused along the way. Saying "$e^x=(1+\frac{1}{x})^x$ as $x \to \infty$" doesn't make any sense, for example.


You didn't justify your claim at all. What you claimed is that for some $y \in \mathbb{R}$, it's true that for all $x \in \mathbb{R}$
$\dfrac{dy^x}{dx} < y^x$
and that there exists another number such that the < is a >.

Assuming that this is true is as big an assumption as just assuming that there exists $y \in \mathbb{R}$ such that $\dfrac{dy^x}{dx} = y^x$. You could have just said "assume the result is true; therefore the result is true".


Don't worry too much about it... the problem with being taught Maths is that, in order to make it understandable, things are often presented in the wrong order. The idea is that we can define a function $\exp x$ as follows:
$\exp x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots$

Differentiating this we see fairly clearly that $\dfrac{d}{dx} \exp x = \exp x$.


Don't worry about it; I didn't know these things properly until half-way through my first year of uni. You're better understanding things in less depth, but actually understanding them, than you are attempting to go into the real deep nitty-gritty of everything and just being confused.

Clearly: the function $\mathbb{F}(y) = \dfrac{dy^x}{dx} - y^x$ is either going to be positive, negative or zero. That's not a big assumption to make.

The jump is in assuming that there must be a continuous progression from +ve to -ve, going through 0 on the way. The assumption is that $\mathbb{F}(y) = \dfrac{dy^x}{dx} - y^x$ is a smooth function.

I thought that with differentiation, the differentiated function was the function you got when h was 0? After deriving a form that you can put h=0 into without dividing or multiplying everything by zero.
Because you want the gradient of the tangent to your equation.

Clearly: the function $\mathbb{F}(y) = \dfrac{dy^x}{dx} - y^x$ is either going to be positive, negative or zero. That's not a big assumption to make.

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I see. Looks like I will have to work extra hard in analysis

Out of curiosity is there a valid way of manipulating that limit to complete the proof ?

You can't ever divide by 0. The quotient $\frac{f(x+h)-f(x)}{h}$ does not exist when $h=0$. You're only considering what happens when h gets closer and closer to 0.

For example, suppose we want to find the derivative of the function $f(x)=x$
We calculate that $\frac{f(x+h)-f(x)}{h}$ is 1 whenever h is non zero, so as h gets closer and closer to 0, this thing is still 1, so the limit is 1. I don't care what happens at 0 to calculate that limit- I care what happens when we get close to 0 (I also don't care what happens at h=50 for example).

But given the value of f(x), eg f(x)=x^2, isn't the goal to find the new equation by removing h, by setting h to 0?
So, you end up with 2x.

I'm getting a feeling of deja vu here

It's a tautological result that $\displaystyle \lim_{h \to 0} h = 0$; i.e. that $h \to 0$ as $h \to 0$. We don't actually set $h=0$, we just look at it's behaviour as $h \to 0$. So, like in your last thread, when we had
$\displaystyle \lim_{h \to 0} [2x+h]$
what we actually did was take out the 2x (because it doesn't depend on h), so we're left with
$2x + \displaystyle \lim_{h \to 0} h$
but we know that $\displaystyle \lim_{h \to 0} h = 0$, so this is just equal to $2x$.

In the case $\displaystyle \lim_{h \to 0} h$ it looks like there's no difference between looking at the behaviour of h as it tends to zero and just setting h=0; but there is a difference, and that's what we use here.

Yeah, me too.

I think it's this difference between setting h to 0 and using lim h = 0 that I don't understand.
Do you think you can explain the difference? In the previous thread I though that you actually set h to 0, but you just did it after rearranging so that you got rational results.