Conversation started Sep 9, 2012 at 13:20.
N3buchadnezzar
N3buchadnezzar
Sep 9, 2012 13:20
I am trying to show that the function $$f(a) = \lim_{h \to 0} \frac{a^h - 1}{h} $$
equals $\log a$. Is it enough to show that the function satisfies $f(ab)=f(a)+f(b)$ and $f(1)=0$ ?
Jonas Teuwen
Jonas Teuwen
@JayeshBadwaik I know, that is what I mean.
@N3buchadnezzar Does that uniquely determine the logarithm for you?
Jayesh Badwaik
Jayesh Badwaik
@JonasTeuwen Okies :-)
Jonas Teuwen
Jonas Teuwen
@N3buchadnezzar As in: how do you define the logarithm?
N3buchadnezzar
N3buchadnezzar
@JonasTeuwen I know there are various ways to define the logarithm, but the simplest I guess is to use the integral definition
Jonas Teuwen
Jonas Teuwen
@N3buchadnezzar That is important.
@N3buchadnezzar But then you must show that the solution to your functional thingie is unique.
N3buchadnezzar
N3buchadnezzar
Sep 9, 2012 13:31
$$\log a = \int_1^a \frac{\mathrm{d}x}{x}$$
@JonasTeuwen Yes, and I was unsure about the $f(1)=0$, I do not think the restrictions are strong enough.. =(
Bleh
Jonas Teuwen
Jonas Teuwen
@N3buchadnezzar So, what about trying to compute $f(a) - \log a$?
And bound it.
Maybe it requires rephrasing the definition.
N3buchadnezzar
N3buchadnezzar
My motivation is to either express e as a sum or as a limit
Asumption: There exists a function $f(x) = a^x$, where $a$ is some real number, such that $f'(x)=f(x)$
Now I want to use the definition of the derivative to show this, it leads me to
Jonas Teuwen
Jonas Teuwen
Hmm.
N3buchadnezzar
N3buchadnezzar
$$ \lim_{h \to 0} \frac{a^{x+h}-a^x}{h} = a^x $$
Jonas Teuwen
Jonas Teuwen
You use the existence theorem?
And then you call that $e$.
Or rather, after uniqueness.
N3buchadnezzar
N3buchadnezzar
Sep 9, 2012 13:36
$$ \lim_{h \to 0} \frac{a^h - 1}{h} = 1$$
Jonas Teuwen
Jonas Teuwen
So you want to show that for small $h$ that $$a^h - 1 \sim h \int_1^a \frac{\text{d}x}{x}?$$
N3buchadnezzar
N3buchadnezzar
So I need to find a number "a", that satisfies that equation. And I do not want to plug in some formula for e and show that it fits. I want to derive either a closed familiar sum or limit.
Jonas Teuwen
Jonas Teuwen
First you have to show it exists bro.
Otherwise your manipulations are only formally true.
N3buchadnezzar
N3buchadnezzar
@JonasTeuwen Yeah!
Jonas Teuwen
Jonas Teuwen
Let us do what I always do when I don't know what to do: split.
$\int_1^a = \sum \int_1^{1 + a 2^{-m}}$.
Mm.. Boring.
That $a$ is retard, let us make it $2$.
So, $$2^h - 1 \sim h \int_1^2 \frac{\text{d}x}{x}.$$
And $$\int_1^2 = \sum_m \int_{1 + 2^{- m - 1}}^{1 + 2^{-m}}.$$
N3buchadnezzar
N3buchadnezzar
Sep 9, 2012 13:41
Hmm
Jonas Teuwen
Jonas Teuwen
So now let us take given $\epsilon > 0$ an $M$ such that $2^{-M} < \epsilon$.
Well, you catch my drift. I need to get some beer.
 
Conversation ended Sep 9, 2012 at 13:41.