/sci/ - Science & Math

>>5598879
was that first one meant to be a capital 'E'?

anyway, e^x is a function, whereas 2 and 7 are constants, so it's a totally different ballgame, brah

>>5598884
probably not
and i ment 2^x and 7^x, sorry i'm tired and procrastinating from actually doing math right now

>>5598884
Constants are functions too, you fucktard.

>>5598895
no they arn't, fag

>>5598884
it's because Y^x differentiates to Y^xlogY
so e^x differentiates to e^xlog(e)
and we all know that logs and 'e's cancel each other out, dont we?
..dont we?

yeh, anyway, they do

so that cancels to one,
and leaves just e^x
so e^x differentiates to e^x
usee?

simple, rite?

>>5598906
e^(xlog(e)) is different from e^(log(e)^(x))

retard

the perfunctory answer is that (a^x)' = ln a . a^x

so when does ln a = 1?

or, for what number n does (logb a)/(logb n) = 1? (for any b greater than 1)

the answer is e

>>5598906
>>5598912
Thanks,
but why does (a^x)' =ln a . a^x?
or is that just something i need to accept and move along

>>5598906
>no they arn't, fag

Yes They Are, Pleb. Take You're Meds.

>>5598911
w/e, aspie cunt

>>5598919
i don't know how to prove it's true, but you can check that it isn't false for any particular case

>>5598922
no they fucking arnt! a function is something with inputs and outputs, a constant, aint!
so YOU'RE the fucking pleb!

>>5598948
My sides. What meds are you taking?

>>5598919
http://tutorial.math.lamar.edu/Classes/CalcI/DiffExpLogFcns.aspx

If you look at the graphs of 1^x, 2^x, 3^x, 4^x and so on, and look into the derivatives in between 2^x and 3^x you'll see that the graph of the gradient function is very similar to the actual function for numbers getting closer to 2.7, (e). e, the value you are talking about, is the value where e^x is its own gradient function. It is how it is almost its definition.

>>5598919
>but why does (a^x)' =ln a . a^x?
that's just how differentiating works
i just accept it and use it
there is no why

>>5598948
input: any number
output: 6

>>5598952
none of your fucking business! thats what!

>>5598968
well i dont need your fucking help!

>>5598965
EK pls. I want to help you.

>>5598895
>Constants are functions too, you fucktard.
what does this even mean? Is 7 a function? A function is a relationship. f(x)=7 is a function, but 7 is not a function.

>>5598971
Come on, you know you need help. How's Harriet btw?

>>5598973
>f(x)=7 is a function

no, f is a function.

f(x) is a value

f(x) = 7 is an equation

you could easily give a function the name 7

Here's a good start.

>>5598979
>>5598973

<span class="math">7: \mathbb{Z} \mapsto \mathbb{Z},[/spoiler]
<span class="math">7(x) = x+1[/spoiler]

>>5598989
BAM!

>>5598976
i dont need any fucking help, im fine.
and shes fine

>>5598998
You know you're lying. You're not happy. All of us can see it.

>>5598988
OOOOOOOOOOOOOH it's back to parametric differentiation
that makes sense now, thanks branon

>>5599002
Who's branon? And you're welcome.

>>5599028
bronon?
2edgy5u

bro anon

<span class="math">\frac{d}{dx}e^x \Leftrightarrow \lim_{h \rightarrow 0} \frac{e^(x+h)-e^x}{h} \Leftrightarrow \lim{h \rightarrow 0} e^x \frac{e^h-1}{h} \Leftrightarrow e^x[/spoiler]

>>5598989
Well, if we're working in the integers, <span class="math">7_{\mathbb Z}[/spoiler] is the set of ordered pairs of naturals (a,b) such that <span class="math">a=b+7_{\mathbb N}[/spoiler]. By definition, a set of ordered pairs is a relation, and it is trivial to show that <span class="math">7_{\mathbb Z}[/spoiler] is itself a function.

>>5599040
That assumes that the derivative of e^x at x=0 is 1.

>>5598906
If constants are functions, why does one often consider the sheaf of locally constant functions modulo the constant ones in topology?

>>5599115
Where can I learn about sheaf theory? Do you have a good book?

>>5598963
>typical EK post
why are you even here?

>>5599088
Well duh, e^0=1. The trick is, that only for number e~2.7... lim h->0 (e^h-1)/h =1

>>5599088
No it doesn’t.
<span class="math">
\lim_{h \rightarrow 0} \frac{e^h-1}{h} = \lim_{h \rightarrow 0} \sum_{n=1}^\infty \frac{h^{n-1}}{n!} = \lim_{h \rightarrow 0} \sum_{n=0}^\infty \frac{h^n}{(n+1)!}
[/spoiler]
The radius of convergence of <span class="math">\sum_n \frac{h^n}{(n+1)!} [/spoiler] is 1, so the limit <span class="math">\sum_{n=0}^\infty \frac{h^n}{(n+1)!} [/spoiler] is continuous for every |h|<1. So
<span class="math">
\lim_{h \rightarrow 0} \sum_{n=0}^\infty \frac{h^n}{(n+1)!} = \sum_{n=0}^\infty \lim_{h \rightarrow 0} \frac{h^n}{(n+1)!} = 1.
[/spoiler]

>>5598919
<span class="math"> \frac{d}{dx} a^x = \frac{d}{dx} e^{\ln(a) x} = \ln(a) e^{\ln(a) x} = \ln(a) a^x. [/spoiler]

>>5598988
Nice answer. This uses the fact that you can define ln(a) as the integral of 1/x from 1 to a + Fundamental Theorem of Calculus.

>>5599115
What you just said implies that constants are functions. They're boring functions, but functions nonetheless.

>>5598879
because thare's no way to capitalize 2 and 7.

>>5598919
<span class="math">
\frac{\partial }{\partial x}a^x=\frac{\partial }{\partial x}exp(ln(a)\cdot x)=\frac{\partial }{\partial x}(ln(a)\cdot x)\cdot exp(ln(a)\cdot x)
[/spoiler]
<span class="math">
=ln(a)\cdot exp(ln(a)\cdot x)=ln(a) \cdot a^x
[/spoiler]

>>5599205
If you're using the taylor series definition, the derivative is easier to find because you just differentiate termwise. The only reason you would use the limit definition is if you are starting with a different definition of e^x, and haven't derived the series expansion yet. Also, the series expansion has radius of convergence infinity, not 1. You know, since e^z is entire and all.

>>5599199
Obviously e^0 = 1, but you would have to prove that <span class="math">e'(0) = 1[/spoiler] which is a different statement.

Let's play a contructive game that isn't done often enough for noobies (for evidence, please see all of the idiots on the board who think they're explaining it when they're actually just stating the answer)

From the definition of the derivative:

<span class="math"> \frac{d}{dx} log_n(x) = lim_{h -> 0} \frac{log_n(x+h) - log_n(x)}{h} [/spoiler]
<span class="math"> = log_n((1 + \frac{h}{x})^{\frac{1}{h}}) [/spoiler]
<span class="math"> = log_n((1 + \frac{h}{x})^{\frac{1}{h}*\frac{x}{x}}) [/spoiler]
<span class="math"> = \frac{1}{x}log_n((1 + \frac{h}{x})^{\frac{x}{h}}) [/spoiler]

And then we sub u = h/x still goes to zero and say <span class="math"> lim (1 + u)^{\frac{1}{u}} = e [/spoiler]

Thus <span class="math"> \frac{d}{dx} log_n(x) = \frac{log_n(e)}{x} [/spoiler]
More importantly, the derivative of ln|x| is 1/x.

Now lets use this to differentiate n^x implicitly. n^x = y implies xln|n| = ln|y| implies ln|n| = y'/y implies y' = yln|n| which is y when n = e

Consider your minds expanded.

>>5599220
You're right about the radius of convergence; I have no idea why I thought it would be 1. Differentiating the series termwise also works, but in most cases you don't know that yet when you're trying to solve this problem.

really guys?
explanations relying on lnx isn't really valid to explain e^x related concepts, i hope you're seeing the circularity of this
anyway
op, try to think of a function that is equal to its derivative
we could try this infinite sum perhaps:

f(x) = 1 + x + x^2/2 + x^3/3! + ... = f ' (x)
because it goes on forever
then let's just substitute 1 for x: it returns the euler's number, e
ok, let's try x=2, strangely f(2)=e^2 and f(3)=e^3...
you see the pattern?
e^x = f(x), where only e satisfies the power representation of the infinite sum
or taylor series if you will