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Anonymous05/11/26, 16:16No.16974219Euler's number
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?
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Anon05/11/26, 16:21No.16974221
Take it up with Lawvere.
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Anon05/11/26, 17:07No.16974274
It's an arbitrarily defined number for algebraic tricks.
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Anon05/11/26, 17:15No.16974286
Because [math] \frac{\mathrm{d}}{\mathrm{d}x} e^x = e^x[/math]
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Anon05/11/26, 19:56No.16974391
The 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.
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Anon05/11/26, 20:11No.16974404
The most important mathematical function was discover by bussiness major of rennaissance *mic drops*
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Anon05/12/26, 11:00No.16974765
[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]
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Anon05/14/26, 08:52No.16975883
To anybody who did derivatives in their lives, you will eventually notice that certain functions, when derived, tend to be more and more similar to the original function. In particular, [eqn] f(x) = a^x [/eqn],
You can check this yourself in fucking desmos even. It just so happens, that there's one point where a derivative of a function is equal to its function, and that's when [math]a = e[/math].
You can do that yourself as pic related.Now the reason you see [math]e[/math] everywhere, is because any quantity whose change is linearly proportional to itself ie. [math] \frac {dN(t)} {dt} \propto N(t) [/math] will have a differential equation in the form [math] {dN(t)} \frac{1}{N(t)} = \lambda {dt} [/math] will have a solution such that [math] \ln{N(t)} = \lambda t + C [/math] and therefore [math] N(t) = e^{\lambda t+C} [/math]. Now notice that I can continuously keep derivating that: [math] \frac{d}{dt} N(t) = \lambda e^{\lambda t+C} [/math] and [math]e[/math] will never disappear. The first case was simply a special case when [math] \lambda t + C = 0 [/math].It's simple: change of many things in nature are proportional to that quantity itself, therefore exponential function models it through basic mathematical and logical rules. If this is not understandable, then you might as well argue multiplication is wrong or something.
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Anon05/14/26, 17:41No.16976124
The natural logarithm is defined as the signed area under one the simplest hyperbolas. What is the [math]x>0[/math] such that
[eqn]\int_{1}^{x}{\frac{1}{t}}{dt} = 1 [/eqn]The answer is Euler's constant
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Anon05/15/26, 07:27No.16976389
Since (1+x/n) gives a first order change, let's say you want to rotate the complex number a + ib. Try and rotate it counterclockwise infinitesimally by an infinitely small part [math] (1+x/n)(a+ib) = (a - \tfrac{\Delta}{n} b) + i(b + \tfrac{\Delta}{n} a) [/math] . This means x must be equal to [math] i\Delta[/math] That's a first order change. To get infinite order, the object that rotates a+ib is [math] e^{i\Delta} [/math] , or just theta.
Instead of complex number a + ib, you could do vector [a,b]^T, and you'd find that (1+x/n) is such that 1 is the 2x2 identity matrix I, and x is a [math] \theta [/math] times the 2x2 matrix i that squares to -I.If independently sampling from any distribution, the larger the random sample, the sampling distribution of the mean becomes more and more gaussian (look up central limit theorem). The end of the proof relies on the doing [math] (1 - \tfrac{ t^2/2}{n} )^n [/math] , which goes to the gaussian [math] e^{-t^2/2} [/math] . In some way, gaussian functions maximize entropy.If you want to solve the differential equation [math] \tfrac{d}{dx}^n f(x) = f(x),\ f(0) = 1 [/math], you could imagine that the 'n' basic solutions are of the form [math] \tfrac{d}{dx}^n f(x) = a_i^n f(x),\ a_i^n = 1 [/math], meaning you'd like to find the various n-roots of unity. The solution of course is [math] e^{a_i x} [/math] where i is from 0 to n-1. Lots of differential equations will often involve e^x in some way then. Solutions to the heat/diffusion equation, solutions to wave equations, etc.When n=2, we get sin and cos, meaning these two can be written in terms of e^x in some way. Fouriers Theorem tell us that period functions can be written in terms of sines and cosines, meaning they can also be written in terms of e^x. Extend the period to infinite size and you can get the Fourier Transform. Instead of sticking to only i in the Fourier Transform, you can try any complex number s, and now you got the Laplace Transform.