Well Veeky Forums?

When you take a derivative, you get rid of a constant, whatever it may be. So when you take the antiderivative, you add a constant, but you don't know what it is, so we label it as a c we add to to equation, hence, + C.

Even if the constant is 0?

[eqn]
\begin{aligned}
\frac{d}{dx} e^x &= \lim_{\Delta x \to 0} \frac{e^{x + \Delta x} - e^x}{\Delta x} \\
&= \lim_{\Delta x \to 0} \frac{e^x (e^{\Delta x} - 1)}{\Delta x} \\
&= e^x \left( \lim_{\Delta x \to 0} \frac{e^{\Delta x} - 1}{\Delta x} \right) \\
&= e^x \left( 1 \right) \\
&= e^x \\
\int f'(&x) \; dx = f(x) + C \\
\ldots \\
\int e^x \; &dx = e^x + C
\end{aligned}
[/eqn] If you want anything rigourous, you're looking in the wrong place.

The constant could be 0, but you can't possibly know if all you're given is an indefinite integral

inb4 any of you fags try anything funny with the limit...

Well, Veeky Forums?

*brain explodes* xD

yeah, as the other user mentioned, it could be zero, but we just can't know with only that information. We musn't assume it is zero, or any other number, so we have it represented by C

That's gay as fuck lmao.. Fuck a nigga names Isaac ole gay boi fuc boi

A cool historical fact is that because of Calculus we got into the Industrial Revolution. But you may be onto something, fuck that faggot.

>because of Calculus we got into the industrial Revolution
Elaborate.

wtf reverse image searching that got me here warosu.org/sci/thread/7128104
The fuck is this shit user
The fuck.

It gave us the ability to find the immediate rate of change at a point. Before calculus, we could only find the slope of a line (the rate of change) from two points, and it would therefore not be the immediate rate of change when, say, x=3. (y2-y1) / (x2-x1) With the concept of the limit, we were able to accurately measure for example the velocity at a specific point, acceleration, pretty much a lot if not all the formulas you use in the mechanical side of hs physics were derived using calculus.

So basically Calculus helped the white devil's exploit the poor black man.

Oh my god guys... this... this is us.. this is Veeky Forums literally but renamed. warosu.org/sci/
Why does this exist?

that's an archive

>never heard of warosu
am i being baited

(You)

I assume if you take the power series expansion and plug it into the FTC, you can massage some algebra into it and it falls right out.

That's usually how proofs go.

give a proof as to why anyone should give a proof

Use the definition of e^x as the inverse of lnx then find its derivative. After, use the FTC.

Integration is the inverse of differentiation. Show that the derivative of e^x +C is e^x and you're golden.

Why is this not valid for n^x?

>circular logic
Lmao ok

>Gaize this is us omg
>Why does this exist

I'm sorry, but your mother never should have kept drinking with you in her stomach.

Lrn2valid

What was circular, he used the FTC and a derivative.

Also, any proof for this integral will appear circular. This is because it is it's own result.

I would also like to remind you that the complex exponential is literally a circle.

>the complex exponential
"[math]1^1/k[/math]"?

Ugh.

>[math]1^{1/k}[/math]?

xD

Thanks for (You)

I need it before this math final. My entire life is riding on this test. Needless to say I'm in edge.

np my man

:/

I no longer understand

The proof is trivial and is left as an exercise to the reader.

All I did was post the limit-defintion of the derivative and applied to to [math]e^x[/math] then showed the fundamental theorem of calculus.

Other proofs still wind up being "circular" in their logic. For instance...
[eqn]
\begin{aligned}
x = &\; \ln{y} \quad \Rightarrow \quad y = e^x \\
\ldots \\
\frac{dy}{dx} &= \frac{1}{\frac{1}{y}} \\
&= \left( \frac{y}{1} \right) 1 \\
&= y \\
&= e^x \\ \\
\int &e^x \; dx = e^x + C
\end{aligned}
[/eqn] There most definite exists a few other crazy proofs but google exists and this one is the only other one I can come up off the top of my head. Thanks for wasting a few minutes of my morning.

Correct

It's circular because in order to take the limit as x-->0 of (e^x-1)/x, you ordinarily have to invoke L'hopital's rule, which will use the fact that d/dx e^x = e^x.

i.e. circular

[math]e^x[/math] is different to [math]a^x[/math] in that [math]e = \sum_{k=0}^{\infty} \frac{1}{n!} = 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \ldots + \frac{1}{(n - 1)!} + \frac{1}{n!}[/math].

[eqn]
\begin{aligned}
a^x &= e^{x \ln{a}} \\
\frac{d}{dx} a^x &= \frac{d}{dx} e^{x \ln{a}} \quad \quad u = x \ln{a} \\
&= \frac{d}{du} e^u \; \frac{d}{dx} x \ln{a} \\
&= e^{x \ln{a}} \cdot \ln{a} \\
&= a^x \ln{a}
\end{aligned}
[/eqn]

You're implying that I solved it algebraically via L'Hôpital's rule, which I didn't [because I'm a lazy fuck]. Refer to

The post No.8037627 does not seem like a Science & Math question. The integral of e^x is easily computable by the use of limits and derivatives. It is a simple deduction. It's simple. So I do think that it is an off-forum and inappropriate question. Recently, in contemporary education there has been use of television and computer programs to learn Mathematics and Science. The particular picture document "advanced mathematics.gif" (4 KB, 400x300) is like such pattern, with reference to the recent trend in public education, so it is a homework problem. Homework problems are not appropriate for Veeky Forums because they are not conducive to actual learning. Specifically, Veeky Forums is more for learning after homework or experimental thoughts with pictures.

Then how did you solve it? If you can solve it without using the fact that e^x is its own derivative, then your proof is non-circular. Otherwise, you have no proof.

And don't tell me that you "plugged it into a computer" because again, you will have not proved anything.

>don't tell me that you "plugged it into a computer"
I already announced that hours ago. Too late.

If you want an actual proof and not a bullshitted that can be written by me and everybody else in a few minutes on LaTeX, remind yourself that you're on Veeky Forums.

Another method of "proof" that most would suggest is a series expansion of [math]e^x[/math] but that also results in circular logic because that can only be done assuming [math]\frac{d}{dx} e^x = e^x[/math].

:^)

which definition of e^x are we using?
taylor series, differential equation, inverse of log, or something obscure like lim_n (1+x/n)^n, (sum 1/n!)^x

LOL SEX

exp(x) is literally defined to be the function that is its own derivative. Therefore it is also its own antiderivative.

Define [math] A : f(x) \to \int_0^xf(t)dt [/math]

Now we're looking for f, s.t.
[math]
Af = f-1
[/math]

[math]
\Leftrightarrow (id-A)f =1
[/math]

[math]
\Leftrightarrow f = (id-A)^{-1}1
[/math]

[math]
\Leftrightarrow f = \sum_{n=1}^\infty A^n(1)
[/math]
And because
[math]
A^n(1)=\frac{x^n}{n!},
[/math]
we get
[math]
f(x) = \sum_{n=1}^\infty \frac{x^n}{n!}
[/math]

of course the sum should be from n=0

Can you elaborate on how inv (id - A) is A^n?

geometric series
proving that ||A||

.... can you do that with linear transformations? You'd have to define an inner product... seems a bit fuzzy to me

her you go
en.wikipedia.org/wiki/Neumann_series

Ah, okay. So I guess proving the operator norm is sorta trivial in this case, since it's obvious from the definition of f that c

i actually ommited it, because it's nontrivial.
you'd first have to define the norm on the space, A acts upon (should be [math] C^{\infty} (\mathbb{R}) [/math] to include every function with a taylor series.)
Now you try to find a norm on that space, that induces an operator norm such that ||A||

Would anybody mind filling me in on this (to me) odd notation?

Am I right in assuming that [math]dx[/math] is to be understood as [math]\Delta x[/math], with the added property that the change is infinitesimal?

Am I right in assuming that [math]\frac {dy}{dx}[/math] is to be understood as an infinitesimal change in y with respect to an infinitesimal change in x?

If the answers to the previous two paragraphs are yes, how in the name of fuck should I interpret [math]\frac {d}{dx}[/math]? An infinitesimal change (in what) with respect to an infinitesimal change in x?

just don't.
It's notation and shouldn't be treated as a number
>inb4 some autist starts the nonstandard analysis meme again

Here are some notation examples you might want to comment on.

[math]\int e^x dx[/math]
[math]\int e^x \frac {dy}{dx}[/math]
[math]\int e^x \frac {d}{dx}[/math]

I'm asking because I don't understand what the notation means, I thought I made that clear, but maybe I was too ambiguous.

[math]\frac{d}{dx}[/math] is to be interpreted as an operator that may be applied to a function, resulting in the infinitesimal change in that function with respect to an infinitesimal change in x.

>Explain this notation to me
>Hurr it's not a number

well okay then
d/dx is notation for the differential operator with respect to x meaning
[math]
\left[\frac{\mathrm{d} }{\mathrm{d} x}f\right](x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}
[/math]
dx is just integral notation, clarifying with respect to which variable you are integrating.
Historically it meant something like an infinitesimal quantity, but it doesn't quite fit in with the way people nowadays do analysis. Limits were introduced to avoid having to deal with infinitesimals.
There is a branch of analysis, that tries to give things like infinitesimals a meaning and do some work with it, but nobody actually bothers to do research in this area anymore, because it doesn't yield results like limit based analysis.

e^(x+1)/(x+1)

It's called the "most general integral" because it represents every possible expression of that form regardless of the constant term.

Reminder that the concept of "indefinite integrals" as taught in high school calculus are cancer and this term isn't even used in higher level math classes.

>Limits were introduced to avoid having to deal with infinitesimals
I don't really have much to do with mathematics, but why are limits not infinitesimals?
lim x-> 0 is fundamentally no different from 0.

>Another method of "proof" that most would suggest is a series expansion of exex but that also results in circular logic because that can only be done assuming ddxex=exddxex=ex.
>:^)
You can define e^x in terms of power series, then it isn't circular.

You can also define e to be the number that makes that limit 1. You have to proof such a number exists, but otherwise the proof isn't circular.

the epsilon delta definition of a limit avoids having to work with infinitesimals.
limits are just real numbers (as opposed to hypperreals, which contain infinitesimals), since we defined them in a way to be the closure of Q under the limit operation.
your limit is literally the number 0, not "something bigger than zero, but smaller than every other real number", or in other words infinitesimal

>closure of Q under the limit operation.

"Indefinite integrals are never used in my higher math classes, I'm so smart!"

If you are using definite integrals with any bounds other than zero or infinity, you're an engineer.

>If you use orthogonal projection, you're an engineer
>If you use an inner product, you're an engineer

Topology.

Youre dumb

I should have foreseen that this triggers the wildburger

on an unrelated sidenote, here's some cool vid of him defining e^x in the historical way as the inverse of the antiderivative of 1/x
youtu.be/HRD9X-2Bmdw?t=36m21s
His mathhistory videos are all pretty nice

I'm not sure but I guess it's ζ (1)

I think I get what you're saying, but why does it matter?

Take your pedophile cartoons back to .

Fucking degenerate

Sakurako is qt and you can't deny.

easier to work with / yielding more results
according to my analysis prof - never done any nonstandard analysis myself

Integral of e^x is e^x + C. Make it a power series and build from that definition if you want to prove it.

Why the fuck are people taking this seriously.

These threads are still far better than the threads involving tinfoilery and pseudoscience.

But then you have to prove that it is ok to integrate it term by term.

In that case let's build from a definition of the natural numbers. :^)

bump