Something of a trinity
(Wink from Diety)
Why do we measure angles in radians? Because then d/dx (sin x) = 1 at x = 0, and sin x ≈ x for small x.
In my opinion drilling down too much on conventions misses the point of math.
e pops up quite often when taking limits on a surprising number of varied phenomena. It is much more than a mere convention, unless you subscribe to the nihilistic, anti-epistemological notion that all of mathematics is merely convention. It seems to be the center of the conceptual space particularly around questions of relative and absolute scale.
It's true you can use any base for computing things but some are more natural than others in that specific parametrizations have natural interpretations especially when it comes to physics (timescales, information-theoretic optimality, etc.).
It does!
e^z, defined as the series \sum_{n = 0}^\infty z^n/n!, can only be a function of a dimensionless number z.
sin(z) and cos(z), defined as power series, technically also work this way. And that's OK, because angles are dimensionless: a radian is just C/(2πr), where C is the circumference of a circle of radius r. But it is sometimes convenient to pick your favorite number of radians, like π/180 of them, and call that a degree, and then to say that sin(x degrees) is the same as sin(xπ/180 radians).
With this convention, where the left-hand side of e^(ix) = sin(x) + icos(x) is a function of a dimensionless variable, and the right-hand side can be viewed as a function of a dimensioned argument only in the sense written above, it really is the case that the equation written is true, but the equation e^(ix) = sin(x degrees) + icos(x degrees) is false.
(On the other hand, you could make the case that e^(ix) is really a function of an angle, where its value is the complex number that lies on the unit circle at that angle. Then you do recover a "dimensioned" version of e^(ix) = sin(x) + i*cos(x) that's valid even if you measure angles in degrees.)
(iπx/180)
e = cosd x + i sind x
ix
π/180
-> e = cosd x + i sind x
π/180
-> let f = e
ix
f = cosd x + i sind x
Probem is, f doesn't have nice properties like: d x x
- f /= f
dx
There is something uniquely special about the unit circle, and about using the unscaled distance around the unit circle as the measure of the angle. |[cos(x+δ) + i sin(x+δ)] - [cos(x) + i sin(x)]| = δ
This is also related to how we can estimate sin(x) = x for small values next to zero, if using radians.In radians, the derivative sin'(x) is cos(x), and cos'(x) is -sin(x). Derviation just shifts the waveform left by ninety degrees. In units other than radians, we get wacky constant terms that change at each step.
That's related to how e^x is its own derivative.
A more practical way to measure angles would be in rotations. 0° = 0, 360° = 1, 90° = 0.25, etc. It would remove a ton of 2π and 4π² factors from a lot of equations in physics.
For example, try to work out the Taylor series for sin(x) using degrees (or rotations). It's awful.
Fourier transform would have 4π² instead of 2π under the exponent, no big deal.
The Euler's formula gets a factor of 2π under the exponent though. Given its wide application, it adds plenty of noise, of course.
I agree with you that this is completely fine though. I also find it more natural to think of “how many percent of a turn” an angle is than how many “degrees” or “radians” something is, since we use base-10 everywhere else. My workaround is to mostly write everything in terms of sin(2πτ), cos(2πτ), and exp(2πiτ) when I can, where τ measures turns.
That’s not true. If the unit is degrees, d/dx sin(x) = pi/180 * cos(x).
Implicit in that statement is the use of the series definition of sine, which is a "meaningful" or "natural" definition insofar as it represents some function we care about. I'll address at the end what happens if we assume that definition regardless of its independent plausibility, but first consider:
The semantic meaning of sine, at least from its historical roots and how you might independently uncover it from earlier fields like geometry instead of later fields like differential equations, is that given an angle (in some units, we'll touch on that in a moment) we'd like to know the ratio two sides of a particular triangle associated with that angle inscribed in a circle. Given a choice of units for the angle, the triangle is fixed, and so the result (that ratio of side lengths) is also fixed.
Suppose you want to know how that ratio varies with respect to the angle. You can imagine a change of coordinates `y = cx` and consider the derivative of `sin(x)` vs `sin(y/c)`. The latter will have a numeric value `1/c` times less than the former. E.g., imagine a whole circle represented `1` angle instead of `2pi`. Then converting from our normal radians baseline to that new unit you have `y = (1/2pi)x`, and the derivative of the semantic ratio we're considering with respect to the new measure of angle is multiplicatively `2pi` greater than the original.
Going back to your series definition, suppose we pick that series as the definition of sine, independent of units. The problem that arises is that particular uses of sine do have units, and converting from the problem you care about to your particular from-on-high chosen definition of sine will run into the exact sort of problem our `y = cx` paragraph above touched on. The derivative with respect to the quantity of interest still has an extra `1/c` factor, and the fact that our God-blessed choice of sine is independent of units didn't actually solve anything in the composite problem.
What is "the point of math?"
So I'm curious to know about this other... non-conventional point.
90 degrees is one fourth of a circle, it would be so much more intuitive if we'd use "1/4th of something" rather than "1/2 radians" to express this
This is perhaps the most unnatural equation (well identity) in maths. It doesn't fall out anywhere, you would never write it down and solve for e, it's a special case of a more general result you would get first, and it's symbol soup for precisely the reason that the identity itself confers no understanding.
exp/log are natural because you almost can't help but discover them as they appear in so many different seemingly unrelated places.
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.
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.
(Off-by-one errors, they're not just for programmers!)
Relationship to pi; Euler's formula:
ix
e = cos x + isin x
derivative is itself: d x x
-- e = e
dx
Ah, that explains why you didn't see this obvious one coming.
Multiplication of complex number in the x + iy form treats them exactly like any other binomial factors, using the "FOIL" rule:
F O I L
(a + bi)(c + di) = ac + adi + bci + bdi^2
= ac + (ad + bc)i - bd
= ac - bd + (ad + bc)i
I don't know about self-evident; you pretty much have to do it that way. If you hand (a + bi)(c + di) to someone who knows basic algebra, but has no idea what i is, they will come up with the first FOIL expression above; after that, we have to know that i^2 is -1.> nor in the geometric interpretation.
That isn't self-evident, but when you gemoetrically work out what the multiplication is doing, it's just "add the arguments (angles), multiply the moduli (distances from origin)".
Very simple!
You have to do it that way because it's the only way to get a field out of R^2, but I don't think it's at all self-evident. The "we have to know that i^2 is -1" is doing some heavy lifting in your explanation. In the context of "complex numbers are just a 2d plane" there's no a priory reason why you'd think that (0,1)*(0,1) has any particular meaning, let alone that it's equal to (-1,0).
If we interpret vectors in R^n as dislocation (or as forces, or...), then vector addition behaves exactly as one would expect. I think with enough motivation a bright school kid could come up with the formula for vector addition. And it works for every dimension.
Vector multiplication only works for 2d (ignoring the less well behaved cross product that also only works in certain dimensions) and I don't think you'd come up with it naturally.
Also, historically, the 2d plane isn't how complex numbers were discovered. They first appeared as weird quantities that made solving (real) cubic equations simpler but were regarded as just temporary artifacts. The realisation that they can be identified with R^2 came much later.
* 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, which amounts to $2.44
* Compounded n times in a year, you get 100/n percent interest every 1/n year, which amounts to (1+1/n)^n dollars
* So continuous compound interest is the limit as n approaches infinity, which amounts to $2.71828 at the end of the year
(This is a great problem to give to pre-calc students to see if they can figure out the calculation for themselves.)
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 is e. In general, when you evaluate this function at any real value of x, the result is e^x.
But the Euler equation e^iπ = -1 has nothing to do with exponentiating e, it's just a notational convention that is defined to be the series above. When you evaluate that series at x=iπ, the result is -1. In general, the value of the series for any x=iy is cos(y) + i*sin(y).
It's that simple.
Can't the same thing be said about using fractions on the exponent? Exponentiation is actually just repeated multiplication (a^n=a*a*...*a, repeated n times), but you can't do that when n is a fraction or irrational anymore than you can do it when it's imaginary.
We have to define what it means for an exponent to be non-integer: for fractions we might define a^(b/c) as the root of the equation x^c=a^b, and to allow irrationals I think you need some real analysis (it's been a while, but I think the usual way is to first define exp and log, and then say that a^b=exp(b*log(a)), which is kind of cheating because we have to define exp first!).
There's a very intuitive way to "see" that e^ix=cos(x)+i*sin(x): all you have to do is to treat complex numbers like you would any other number, and "believe" the derivative rule for complex numbers (so (e^(ix))'=ie^(ix)). Then you can just graph f(x)=e^(ix) for real x by starting at x=0 (when clearly f(x)=1) and from there take small steps in the x axis and use the derivative to find the value of the next step with the usual formula f(x+dx)=f(x)+f'(x)*dx.
Doing that you realize the image of e^(ix) just traces a circle in the complex plane because every small step in the x direction makes e^(ix0) walk a small step perpendicular to the line going from 0 to e^(ix0), simply because multiplying by i means rotating 90 degrees.
a^b, for positive a and irrational b can also be defined as lim (x -> b, x € Q) a^x - which is possible because Q is dense in R. This is a pretty natural way of extending a function to the reals.
The way we extend exponentiation to complex exponents is IMHO much less straightforward.
> The way we extend exponentiation to complex exponents is IMHO much less straightforward.
I think it depends on how much you're used to dealing with complex numbers. In college, I was always taught to prove that Euler formula by replacing ix for x in that series, and then noting that the alternating signs and presence/absence of i in the terms allowed you to separate it into two series for cosine and sine. That always felt awkward, like there was no way anyone could just come up with that naturally.
Many years later I found that construction with graphing e^(ix) by taking small steps using f(x+dx)=f(x)+f'(x)*dx, and everything clicked: how exponentials work in the real axis is pretty different from the imaginary axis, but both are completely intuitive and unavoidable once you understand that.
Even later I "discovered" the connection with group theory[1]; that one still blows my mind.
[1] This is a really nice explanation: https://www.youtube.com/watch?v=mvmuCPvRoWQ
Sure you can. You know what 2^n is and you want that 2^(1/3) 2^(1/3) 2^(1/3)= 2^1=1. That uniquely defines the exponential function on the rationals. For the real numbers you need some amount of continuity or measurability, but then it is also uniquely determined.
> but I think the usual way is to first define exp and log, and then say that a^b=exp(b*log(a)), which is kind of cheating because we have to define exp first!).
No, you don't just "say" that. You prove it. Big difference.
Whether a^b=exp(b*log(a)) is a definition or a proof really depends on how exactly you define certain terms (e.g. exp). What's certainly a theorem that requires a proof is that the definition of a^b (for irrational b) via limits of rational exponents and the one via exp are equivalent.
That depends on your point of view. You can also view exponentiation as "really" being e's infinite series, and it so happens that that matches what you get in the case of repeated multiplication. The advantage of that is now you can start exponentiating a lot more than just numbers. Here's 3blue1brown on raising e to the power of a matrix: https://www.youtube.com/watch?v=O85OWBJ2ayo
In general there is fruit in viewing infinite series as the fundamental building block of a lot of math and non-infinite series as the special case. I won't claim which is "real" or "correct", though, just point out that there is value in viewing "repeated multiplication" as the special case rather than the "real" thing. Of course you can always view exponentiation as the generalization too.
Yes, it can. In fact, it is really useful to think of the "usual" definition of exponentiation as repeated multiplication as nothing more than a special case of a much more general concept, which is evaluating a function whose defining property is that it is its own derivative.
I've seen (1+1/n)^n before, but never seen an explanation of why I might ever want to use something of that form. I've used the e^ix notation extensively, but again I've never really cared, because to me it was just a compact representation of sin and cos together. Likewise, all the proofs in that article are still a bit "so at this point on this carefully chosen graph, the gradient is e" and I think "Who cares? You carefully chose the graph to prove a point, I'll never see it in the real world."
And then the example with compounding interest - immediately I can see the application. It's definitely a good way of explaining it, although maybe it'd be even more grounded if it had n=12 and n=365 as examples. When you notice that the actual values seem to be converging, then you can try plugging in ever bigger and bigger numbers. This way you can discover the value of e for yourself and that process of discovery leads to a better understanding than rote learning of an abstract thing you haven't mentally visualised yet. All the other explanations are useful later, and they allow you to see it in different situations, but having at least one "this is why it's tangibly useful" hook at the start is definitely a massive help in understanding something.
https://betterexplained.com/articles/an-intuitive-guide-to-e...
I'd argue it's a bit more, it's a natural way to extend domain of exponentiation, as we do before this with naturals -> integers -> reals.
Yes, but it's not a priori clear that the series actually even converges for all x, and the fact that power series are differentiable term-by-term within their radius of convergence also requires proof.
- 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 these very different motivations lead to the same thing.
The "has nothing to do with exponentiating e" I would strongly disagree with. It has everything to do with exponentiating and is exactly the only way exponentiation can work. So afterwards you can pretend you didn't know that and define exponentiation by using e. Same for matrix exponentials, semigroups etc.
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."
Or, to put it another way, you will blunder into this somewhere around month two of calculus 1, unavoidably.
Of course, that doesn't show how it will show up in all sorts of other places; "the one and only function that is its own derivative" strikes me as more likely to be something we encounter everywhere.
and its derivative then is (by the chain rule)
ln(b) exp( ln(b) * x ) = ln(b) * b^x.
That's true, but it's begging the question because you can't define ln without already knowing about e.
You have to go back to first principles:
d(b^x) := lim(∂->0):(b^(x+∂) - b^x)/∂ = ((b^x)(b^∂) - b^x)/∂ = (b^x)(b^∂-1)/∂
So d(b^x) is itself multiplied by lim(∂->0):(b^∂-1)/∂. But now what? How do you evaluate that limit? How do you show that e is the magic value of b that makes that limit turn out to be 1? And in particular, how do you show that to someone whose only background knowledge is how to differentiate polynomials?
IMHO it's a lot easier to see that e is the value to which the polynomial series that is its own derivative converges at x=1.
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.
It has, that's the beauty of it. You can define as usual the function x -> e^x on the real line. Now, complex analysis tells us that if this function can be extended to a holomorphic function on the whole complex plane, then the extension is unique. And, in fact, this function does admit such an extension, so you can compute e^z for any complex number z, and in this way one gets e^iπ = -1.
e^(iπ) is (e^i)^π.
There is a concrete complex number e^i:
[1]> (exp 1)
2.7182817
[2]> (expt (exp 1) #c(0 1))
#C(0.54030234 0.84147096)
See, it's around 0.54 + 0.84i. It's on the unit circle.When you raise this number to pi, you get -1.
[3]> (expt * pi)
#C(-1.0 1.2776314E-7)
This means it's the pi-th root of -1; let's try it: [4]> (expt -1 (/ pi))
#C(0.5403023058681397174L0 0.84147098480789650666L0)
and e must be the i-th root of this: [5]> (expt * (/ #c(0 1)))
#C(2.7182818284590452354L0 -4.6847612413106414363L-20)
Yes; it is all literally exponentiation which we can approximate with concrete floating-point numbers that know nothing about the formula we are exploring.Yes, that is true, of course. But consulting a Lisp REPL merely demonstrates that it is true. It does not explain why it is true.
You'd get further, pedagogically speaking, by pointing out that (e^i)^i = e^(i*i) = e^(-1) = 1/e, and so e^i is a number that is in some sense "half way" between e and 1/e when you are exponentiating. As an analogy, consider:
e^(1/2) * e^(1/2) = e^(1/2 + 1/2) = e^1 = e
as a demonstration that e^(1/2) is a number that is in some sense "half-way" between 1 and e when you are multiplying, a.k.a. the square root of e. But that still leaves unanswered the question of what "half-way" means when exponentiating rather than multiplying.
Exponents give us expoential decay/growth in the real number line, but periodicity in the imaginary domain, which is strange.
In physics, this lets us analyze decaying or amplifying oscillations in a unified way. (Laplace transform and all that.)
The other constants fundamental to science, like the gravitational constant or the speed of light, can only be measured, not discovered from nothing. We aren't even sure how constant they actually are, there might be extremely tiny variations in either time or space that our instruments just can't measure yet. In theory, other universes could exist where these constants are "set" a little bit differently; whether we could live in such universes is another matter entirely.
E, on the other hand, comes from pure mathematics. As long as fractions, addition and exponentiation work the same way in another hypothetical universe, this strange E number is going to have the same strange value.
One of my favorites is https://en.wikipedia.org/wiki/Feigenbaum_constants . Pops up in many situations, and as I understand it, it is not even well-understood why it pops up in as many places as it does. Intuitively one would expect that many of the places it pops up ought to have their own local constant of some sort, but instead this one keeps popping up. 7-year-old Numberphile video on it: https://www.youtube.com/watch?v=ETrYE4MdoLQ
Neither of those things are sort of empirically measured, and in any formula where they show up, you could theoretically absorb them into other constants -- and in fact the Einstein gravitational constant does exactly that -- it's defined as (8*pi*G)/c^4, absorbing pi into newton's gravitational constant (for historical reasons, when using plank units, they set G and c to 1, so it ends up just being 8pi -- _reduced_ planck units set the whole constant to 1). It's just frequently easier to separate out e and pi for the purposes of actually working out the math.
A few months later I talked to a super advanced math genius kid who had a signature that said e^(i*pi)+1=0 and I asked him if that was Euler's number. He was a super quiet skittish guy that rarely talked. His eyes lit up and he spent the next 2 hours teaching me about Taylor series and showed me how to prove it.
It remains the most fascinating math equation I have ever seen.
He had a lot of issues and dropped out because he couldn't pass a history class. I found that guy on Facebook 20 years later and thanked him. He didn't remember that but he was so happy that he made such a big impact on me.
The number is what it is cause it's "increment of increment" is the same as it's "increment" when you are using exponentiation.
1 - x <= e^{-x} so e^x <= 1/(1 - x) for x < 1
(1 + x/n)^n <= e^x <= (1 - x/n)^{-n} for x < 1
Letting n go to infinity gives e^x = \sum_{n=0}^infy x^n/n! using Newton's binomial formula.
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.