It tends to a constant at infinity, ok big deal, many expressions do that. Why do I have to deal with this bastard everywhere from now on?
>>16974219Take it up with Lawvere.
>>16974219It's an arbitrarily defined number for algebraic tricks.
>>16974219Because [math] \frac{\mathrm{d}}{\mathrm{d}x} e^x = e^x[/math]
>>16974286Doubt.
>>16974219The number e, per se, isn't directly of most importance, but rather it's the function [math] \displaystyle \lim_{n\rightarrow\infty} (1+ \tfrac{x}{n})^n = \lim_{n\rightarrow\infty} (1+ \tfrac{1}{n})^{nx} = e^x = \text{exp}(x)[/math]because then what happens if you replace "x" with the linear operator [math] \Delta x \tfrac{d}{dx} [/math]?For an infinitely differentiable funciton, f(x), then [math] \displaystyle e^{\Delta x \tfrac{d}{dx} } \cdot f(x) = \lim_{n\rightarrow\infty} (1+ \tfrac{\Delta x}{n} \tfrac{d}{dx})^n \cdot f(x) = (1 + \Delta x \tfrac{d}{dx} + \tfrac{\Delta x}{2!} \tfrac{d}{dx}^2 + \dots) f(x) = f'(x + \Delta x)[/math],and of course you can get that [math] \tfrac{d}{dx} e^x = \text{exp}'(x) = \text{exp}(x) = e^x [/math]e^x is inherently associated with derivatives and thus calculus. The form (1+x/n) is associated with the first order change, (1+x/n)^2 is second order, etc.
>>16974391Shit, I did not mean to write f'(x + dx), it's supposed to be f(x + dx), my b
>>16974391>e^x is inherently associated with derivatives and thus calculus*e^x is inherently associated with CHANGE and thus calculus
The most important mathematical function was discover by bussiness major of rennaissance *mic drops*
>>16974397No.
>>16974391[math] \ displaystyle \ lim_{n\ rightarrow\ infty} (1+ \ tfrac{x}{n})^n = \ lim_{n\ rightarrow\ infty} (1+ \ tfrac{1}{n})^{nx} = e^x = \ text{exp}(x)[/math]because then what happens if you replace "x" with the linear operator [math] \ Delta x \ tfrac{d}{dx} [/math]?For an infinitely differentiable funciton, f(x), then[math] \ displaystyle e^{\ Delta x \ tfrac{d}{dx} } \ cdot f(x) = \ lim_{n\ rightarrow\ infty} (1+ \ tfrac{\ Delta x}{n} \ tfrac{d}{dx})^n \ cdot f(x) = (1 + \ Delta x \ tfrac{d}{dx} + \ tfrac{\ Delta x}{2!} \ tfrac{d}{dx}^2 + \ dots) f(x) = f'(x + \ Delta x)[/math],and of course you can get that[math] \ tfrac{d}{dx} e^x = \ text{exp}'(x) = \ text{exp}(x) = e^x [/math]
>>16974391[math] \displaystyle \lim_{n \rightarrow \infty} (1+ \tfrac{x}{n})^n = \lim_{n \rightarrow \infty} (1+ \tfrac{1}{n})^{nx} = e^x = \text{exp}(x)[/math]because then what happens if you replace "x" with the linear operator [math] \Delta x \tfrac{d}{dx} [/math]?For an infinitely differentiable funciton, f(x), then[math] \displaystyle e^{ \Delta x \tfrac{d}{dx} } \cdot f(x) = \lim_{n \rightarrow \infty} (1+ \tfrac{ \Delta x}{n} \tfrac{d}{dx})^n \cdot f(x) = (1 + \Delta x \tfrac{d}{dx} + \tfrac{ \Delta x}{2!} \tfrac{d}{dx}^2 + \dots) f(x) = f'(x + \Delta x)[/math],and of course you can get that[math] \tfrac{d}{dx} e^x = \text{exp}'(x) = \text{exp}(x) = e^x [/math]
>>16974391[math] \displaystyle \lim_{n \rightarrow \infty} (1+ \tfrac{x}{n})^n = \lim_{n \rightarrow \infty} (1+ \tfrac{1}{n})^{nx} = e^x = \text{exp}(x) [/math] [math] \Delta x \tfrac{d}{dx} [/math][math] \displaystyle e^{ \Delta x \tfrac{d}{dx} } \cdot f(x) = \lim_{n \rightarrow \infty} (1+ \tfrac{ \Delta x}{n} \tfrac{d}{dx})^n \cdot f(x) = (1 + \Delta x \tfrac{d}{dx} + \tfrac{ \Delta x}{2!} \tfrac{d}{dx}^2 + \dots) f(x) = f'(x + \Delta x) [/math][math] \tfrac{d}{dx} e^x = \text{exp}'(x) = \text{exp}(x) = e^x [/math]
>>16974391>>16974680>>16974683>>169746884chan latex broken
>>16974379It's true by definition
[math]{\rm pexp}_n(z) := \left(1 + \dfrac {z} {n} \right)^n[/math][math]= \sum_{k=0}^n \Big(\prod_{j=1}^{k-1}\left(1-\dfrac{k-j}{n}\right)\Big)\dfrac {1} {k!} z^k[/math][math]\frac{\mathrm d}{\mathrm d z}{\rm pexp}_n(z) = \dfrac{1}{1+z/n}{\rm pexp}_n(z)[/math]
>>16974762I define that you're braindead. It's true by definition.
>>16974767You can define exp(x) as the function that is it's own derivate and all properties including e and that it an exponential function just pop out
>>16974773I can define you're braindead. A natural consequence is anything you say lacks merit. It's entirely self consistent!
>>16974379It's true by construction
>>16974776>I can define you're braindead. A natural consequence is anything you say lacks merit. It's entirely self consistent!That's not a very good analogy. A better one would be (e.g.) to define 'Retard' as "the poster whose opinions are their own derivatives", but then I'd need to do some work to show this object I've defined is actually (You). I'm leaving that as an exercise for the reader.
>>16975572Kekd
