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/sci/ - Science & Math


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10765275 No.10765275 [Reply] [Original]

If you can't prove this you are a brainlet and you don't belong on /sci/

>> No.10765279

>Look at me guys I passed Calculus!

>> No.10765288
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10765288

prove integral of e^x using Riemann sums

>> No.10765290

>>10765279
>Look at me guys i just memorized all the diferentiation rules and didn't even bother to think about them!

>> No.10765315

woah bro just use series expansion and differentiate

>> No.10765382

>>10765315
If you did this without proving the series converges uniformly, you dont pass

>> No.10765401

>>10765315
its easier to use the fact the the anti derivative of e^x is e^x, the derivative follows naturally

>> No.10765509

e = lim n -> inf (1 + 1/n)^n
e^x = lim n -> inf ((1 + 1/n)^n)^x
1/n = dx
d/dx e^x = lim n -> inf (((1 + 1/n)^n)^x+1/n - ((1 + 1/n)^n)^x/(1/n)
= lim n -> inf (((1 + 1/n)^n)^x)*((1 + 1/n)^n)^1/n) - ((1 + 1/n)^n)/(1/n)
= lim n -> inf (1 + 1/n)^nx
= e^x

>> No.10765510

>>10765288
shit dude I have not even tried this before

>> No.10765563
File: 1.78 MB, 2847x3544, 3E269B53-213D-47D6-A7AB-9E66D1C0630A.jpg [View same] [iqdb] [saucenao] [google]
10765563

>>10765288
How do I do this last limit without Le Hospital’s rule?

>> No.10765592

Help me out here niggers, I hate L'hospitals it feels disingenuous

>> No.10765617

>>10765563
Taylor expansion

>> No.10765632

>>10765617
huh...guess that works thanks anon. Series are fine too.

>> No.10767173

>>10765563
cute handwriting !

>> No.10767206
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10767206

>>10765275
x =1

d / d * 1 = e^1 / e^1
d / d = 1

>> No.10767306
File: 65 KB, 1605x345, TookerLimit.jpg [View same] [iqdb] [saucenao] [google]
10767306

>>10765275

(e^x)' = lim h -> inf (e^(x+h)-e^x)/h
= lim h -> inf e^x*(e^h-e^0)/h=e^x.

>> No.10767315

>>10765509
this is the best answer because it uses the actual definition of e. patrician anons would know this from baby rudin

>> No.10767318

>>10767315
rudin is a meme

>> No.10767319

>>10767173
its hideous

>> No.10767322

>>10765617
Taylor expansion also presupposes the derivative, so it'd be a circular argument. L'Hopital's rule is really just a special case of Taylor Series if you think about it.

>>10765509
is the closest to actual proof so far

>> No.10767367
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10767367

>>10765563
ok, now prove it for x^n

>> No.10767375

>>10767318
no it’s not. that was my textbook for analysis 1 at columbia university. it’s a serious textbook and only brainless who refuse to (or can’t) use their brains to follow it and work the exercises shill the “baby rudin is a meme” meme

>> No.10767389

>>10767375
it was also my book for analysis 1. then i got a better book for analysis 2

>> No.10767476

>>10767367
Haha. I’ve actually done this, but only for natural numbers. There was a question in spivak’s calculus chapter 2 about the sum of 1^k + ... + n^k and I used that to find the limit in this Riemann sum. Still... not done it for rational/irrational numbers because I’d have to use some series.

>> No.10767477

>>10767173
Thanks
>>10767319
Shut the fuck up nigger. Post handwriting

>> No.10767500

>>10767477
you write like a female child

>> No.10767563

>>10767500
Then I take it back anon. I’m a preschool math teacher, maybe the little girls’ handwritings have influenced my own, I think that’s unfortunate.

>> No.10767608

>>10765275
lim w -> 0 of ln(w+1)/w = 1 (this is obvious from graphing 1/w, and taking the area under the curve).
So lim v -> 1 of (v-1)/ln(v) = 1.
So lim h -> 0 of (e^h - 1)/h = 1.
So lim h -> 0 of (e^(x+h) - e^x)/(x+h - x) = e^x.
So d/dx of e^x = e^x.

>> No.10767639

>using L'Hopital to prove derivative of a function
>need the derivative to use lhopital
>ISHYGDDT

>> No.10767648

>>10765275
btw
d/dx e^x = e^x lim_(dx-> 0) ((e^dx -1)/dx)
= e^x lim_(dx-> 0) (1 + sum_(n=0)^(infty) ((dx)^n/(n+1)!)
(hand wave that the series of functions is well-behaved enough to exhange the limit with the sum)
= e^x(1+0)

>> No.10767659

The d's cancel, so you're left with 1/x, which is basically 0 so it has no effect on the rest of the equation and you're left with e^x = e^x

>> No.10767686

>>10765275
Because the exponent of x is 1. You can prove with the power rule

>> No.10767902

>>10765290
That's what your suppose to do.

>> No.10768147
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10768147

By exponential-power analogy theorem, [math]\forall x\;\in\; \mathbf R,\; \mathrm e^x \;=\; \exp\,x[/math].
By definition, [math]\exp' \;=\; \exp[/math].
Therefore, [math] \forall x \;\in\; \mathbf R,\; \frac{\mathrm d}{\mathrm dx} \mathrm e^x \;=\; \exp'\,x \;=\; \exp\,x \;=\; \mathrm e^x [/math].
Whoa! Calc 1 is so hard!

>> No.10768158

>>10765275
When I try to solve this I i get e=(1+dx)^(1/dx) where dx approaches 0. So I test my solution by plugging really small values for dx and I get something that increasingly approaches e.

For example:
(1+0.00000000000001)^(1/0.00000000000001) = 2.71611003409

If I add an additional 0 I get

(1 + 0.000000000000001)^(1 / 0.000000000000001) =
3.03503520655

One more 0 and I get 1. Is this a calculator problem or what?

>> No.10768443

>>10768158
100% the calculator can't handle it, it rounded down to 1^1, not before giving you the true value of the euler constant of course, known by only the smartest of engineers; the only ones capable of doing such a precise aproximation.

>> No.10768501

>>10767206
I mean yeah screw series and integrals

>> No.10768504

>>10765275
stop spamming this shit faggot. you already had a thread about it and you learned there that you are a fag.

>> No.10768507

>>10767319
no it is not, edgelord.

>> No.10769187

>>10765275
d/dx (e^x) = e^x lne

>> No.10769468

>>10765275
By definition.

>> No.10769503

>>10767500
the word you're looking for is girl, E.T

>> No.10769561
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10769561

>>10769468
>By defenition

>> No.10769773

>>10767206
God damnit, I fucking spit my drink out laughing