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This is a frustrating article because it never explains why e is the natural logarithm base. To me, the easiest way to understand it is via continuous compound interest: * If you invest $1 at 100% interest for 1 year, you get $2 at the end * Compounded 2 times in a year, you get 100/2 = 50% interest every 1/2 year, which amounts to $2.25 * Compounded 4 times in a year, you get 100/4 = 25% interest every 1/4 year, whi…
e is natural because it's the smallest positive root of the equation e^(i pi) + 1 = 0 .
We want a^(i pi) + 1 = 0. Now,
a^(i pi) = e^(ln(a) i pi) = e^(i ln(a) pi) = cos( ln(a) pi) + i sin( ln(a) pi),
so we want cos( ln(a) pi) = -1, sin( ln(a) pi) = 0,
so ln(a) = 1, 3, 5, so a = e, e^3, e^5, ...
Thus indeed e^(i pi) + 1 = 0.
But also ln(a) = -1, -3, -5 work, so for example for
a = 1/e = 0.3678794412... > 0, we have
a^(i pi) + 1 = 0;
and of course for a = e^-99 = 1.0112214926104485... × 10^-43 etc.
Earlier quoted context omitted.
Nonsense, this equation doesn't even make any sense without a well defined notion of the exponential function, and then a well defined extension of said function into the complex numbers. You will already have e by the time you reach this equation because defining what exp(z) even means requires you already know the properties of e^x over the reals you wish to preserve. And Euler's formula comes from finding such a f…
> And Euler's formula comes from finding such a function and then defining it to be exp(z). That's certainly one way, but you can also define exp via its power series (which is easily proven to be convergent everywhere). Then, all the properties of exp, as well as Euler's formula, are actual theorems, not just definitions.
e is the unique real number satisfying 1 + x 1 - x (1 + x/n)^n Letting n go to infinity gives e^x = \sum_{n=0}^infy x^n/n! using Newton's binomial formula.
> e is the unique real number satisfying 1 + x I think if you replace e by 2e, this still holds. Therefore your definition of e is not unique.
(2e)^(-0.1) = 0.844243...
The former is not <= the latter.
Earlier quoted context omitted.
That may be how it was arrived at historically, but it is not the best way to explain it. e arises when you ask the question: is there a function that is its own derivative? And it turns out the answer is yes. It is this infinite series: 1 + x + x^2/2 + x^3/6 + ... x^n/n! ... which you can easily verify is its own derivative simply by differentiating it term-by-term. When you evaluate this function at x=1, the result…
Note that C + Cx + ... would have also worked. So why the 1? Just simplicity?
As I remember it e^x is the only function which is its own derivative, which I guess makes e “natural” in some sense.
Earlier quoted context omitted.
That may be how it was arrived at historically, but it is not the best way to explain it. e arises when you ask the question: is there a function that is its own derivative? And it turns out the answer is yes. It is this infinite series: 1 + x + x^2/2 + x^3/6 + ... x^n/n! ... which you can easily verify is its own derivative simply by differentiating it term-by-term. When you evaluate this function at x=1, the result…
Have you happened to have read Visual Complex Analysis ? If not I think you might like it. And a tangent from your username: I quite like how complex numbers and functions like CIS made it into Common Lisp.
Earlier quoted context omitted.
That may be how it was arrived at historically, but it is not the best way to explain it. e arises when you ask the question: is there a function that is its own derivative? And it turns out the answer is yes. It is this infinite series: 1 + x + x^2/2 + x^3/6 + ... x^n/n! ... which you can easily verify is its own derivative simply by differentiating it term-by-term. When you evaluate this function at x=1, the result…
For me these concept seem intuitively linked, simply because derivation measures the rate of change, and that compounding definition aims to find the compounding function with constant growth. The factorial definition only seems to make for an easier proof of the derivation of e^x.
Earlier quoted context omitted.
That may be how it was arrived at historically, but it is not the best way to explain it. e arises when you ask the question: is there a function that is its own derivative? And it turns out the answer is yes. It is this infinite series: 1 + x + x^2/2 + x^3/6 + ... x^n/n! ... which you can easily verify is its own derivative simply by differentiating it term-by-term. When you evaluate this function at x=1, the result…
The exponential function arises from multiple sources: - the solution of the ode you just stated. - compound interest. - The defining property of exponential functions is f(x+y)=f(x)f(y) with some normalization. - Moving on the unit circle is given by an exponential function because rotation is a group, i.e. a^(i(x+y)). Now choose the basis a such that you move with unit speed. - ... The nice thing is that all of the…
That was a bad way of phrasing it. I should have said something more along the lines of "... has nothing to do with multiplying e by itself iπ times. Multiplying something by itself n times is actually a special case of the more general concept of finding a function that is its own derivative."