(e^(i * pi)) + 1 = 0 | Hacker News Reader

(e^(i * pi)) + 1 = 0

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41 pointsshawndumas48 comments

I remember idly putting e^(i*pi) into my TI-8x in high school when I was bored, entirely at random, and assuming I'd broken my calculator when it returned -1.

If you like that kind of surprise you might wish to revisit cos^2 x + sin^2 x = 1. Try it with x a complex number. Why on earth does that work? It is not as though a complex number makes sense as an angle. More at http://lydianrain.wordpress.com/2009/01/06/summoning-cthulhu...

You could also look at Chebyshev polynomials

T_0(x) = 1

T_1(x) = x

T_2(x) = 2x^2 - 1

T_3(x) = 4x^3 -3x

It is easy enough to find recurrance relations and plots on the web. Notice that all the graphs go from -1 to 1 and illustrate that the polynomials are strangely well behaved in this range, oscillating up and down between -1 and 1. Why is that?

T_n(x) = cos (n arccos x)

There is this nice trignometric formula, but since cosine is always between -1 and 1 it makes no sense to try using its inverse function, arccos, outside this range. On the other hand the polynomials have no implied restriction on x, so if you have a calculator, such as a Common Lisp REPL, with full support for complex numbers you can use the trignometric formula outside the -1 to 1 range and observe it agreeing with the polynomials.

In high school, I remember learned this from reading James Gleick's book about Feynman. Feynman had written in one of his notebooks, as a fifteen year-old: THE MOST REMARKABLE FORMULA IN MATH: e^(i*pi) + 1 = 0

I marched in to my calculus teacher the next day and stammered about how unbelievable this was. A few weeks later he helped us work through why it's true.

(Edit: ha, typo. Thanks BoppreH.)

e^(i*pi) + 1 = 0

It's a plus there.

There's actually a reasonably intuitive reason this one works. (At least for me, the Taylor expansions just look like a crazy coincidence). If you try to think about complex powers using the normal definition of exponentiation, you'll just get a headache. Multiply e by itself i times?

However, there's another way to think about exponentiation. An exponential function's derivative with respect to the exponent is proportional to the value of the function, with the base determining the constant of proportionality. This fact can be intuited. Consider the following non-rigorous observation: x^(n+1) = x * (x^n), so x^(n+1) - (x^n) = (x-1) (x^n).

Now think about y = e^x, where d/dx (e^x) = (e^x). In words, the growth rate of y is the current value of y. Start with e^0=1, so the growth rate is 1. If we increase x by dx, we increase y by dx. If we increase x by idx, we increase y by idx. So, if we draw a vector from the origin to the position of y on the complex plane, the vector will be in the real direction, and y will be increasing in the complex direction, perpendicular to said vector. So the absolute value of y isn't increasing.

Now, in general, whatever the value of y, an increase of x by idx will increase y by iydx. Multiplication by i has the effect of a ninety-degree rotation to the left in the complex plane; try some examples if you don't accept this. So, whatever the current value of y, a small increase in x will move y perpendicular to the vector pointing to y. This results in a circular motion on the complex plane, as long as the i component of x is increased. When y gets halfway through this circle, it's back on the real axis, at -1. When does it get there? Well, since y has moved in a circle of radius 1, it has traveled a distance (not displacement) of magnitude pi to get through a semicircle. Since |dy/dx| started as one, and hasn't changed, that means the absolute value of x has increased by pi. Since we have only increased the imaginary component of x, then x must be i * pi. So y = -1 when x = i * pi. e^(i * pi)= -1; e^(i * pi) +1 = 0.

According to Howard Eves in Foundations and Fundamental Concepts of Mathematics (pages 176-7), this and much of Euler's work was achieved purely by formal manipulation. He had great intuition, although that sometimes led him to "absurdities" in e.g. limits.

I realize that most of you probably knew this already but I didn't and it is staggering...

Also cool is that Google 'knows':

http://www.google.com/search?aq=f&q=(e^(i+*+pi))+%2B+1&#...

Why would it not? The Google calculator supports complex numbers in general.

http://www.google.com/search?hl=en&source=hp&q=%28e%...

Dude, I just learned what i was a month ago. I am a product of public school. I am trying; be nice.

Haha, but don't take it badly. I was just mentioning that you can use it for other equations, since you mentioned it like a pre-programmed special case.

I think he was referring to the (unnecessary) "why would it not?".

Ah, nostalgia-hit. I still remember sitting in class and watching the teacher arrive at that equation step by step, and when it was there it was just about the most mind-blowing thing I'd ever seen.

"You mean all that messy junk we've been learning actually combines to form something... simpler?"

Interestingly - this equation contains the 5 most important constant in maths.

And the three most common operators.

I'd say the four most common operators: Addition, multiplication, exponentiation, and test-for-equality. :-)

Those three operators are the first three hyper operators. 1st (Addition), 2nd (Multiplication) and 3rd (Exponentiation).

I wonder if throwing in the 4th hyper operator, tetration, and the value phi produces anything.

Thanks for introducing me to tetration; I had never heard of that before.

http://en.wikipedia.org/wiki/Tetration

If pi == 6.28 [1] then we could have a more elegant version

e^(i*pi) == 1

[1]: http://www.math.utah.edu/~palais/pi.pdf, http://www.math.utah.edu/~palais/pi.html

Debatable, I suppose.

I think the inclusion of 1 is pretty significant, in terms of the contrast between 0 and 1. Maybe that sounded strange -- oh well, I'm not a mathematician by any means. :)

Part of the beauty in the original equation is that it contains every object thought to be elemental in arithmetic and moreover, it contains every one of them exactly once.

It contains 0, 1, +, , ^, which are sufficient to express all of elementary natural arithmetics (unless I'm mistaken) -- all natural numbers can be expressed by adding enough ones to a zero (0, 0+1, 0+1+1, ...), and if I recall correctly, + and aren't enough to state all the truths in natural arithmetics so ^ is needed.

There is a very interesting paper on this called "The saga of the High School Identities" by S. Burris (http://www.math.uwaterloo.ca/~snburris/htdocs/MYWORKS/PREPRI...)

Moreover, it contains three very attractive (if not the most attractive) mathematical constants we know. There is so much we don't know about their relationship. We have no idea if any of these numbers is transcendental or not: pi + e, pi − e, pi*e, pi/e, pi^pi, e^e, pi^e. We don't even know if half of them are even irrational or not (therefore, we don't know if pi is just another facet of e, or maybe an entirely different object!)

So to see all this (and more; one could say so much more) in one equation so terse is so surreal a feeling.

> 0, 1, +, , ^, which are sufficient to express all of elementary natural arithmetics (unless I'm mistaken)

Well, personally I don't favor zeros in equation. What's the point of 1+0=1 ?

Leaving 0 on the right side looks so unbalanced.

The reason I subconsciously did it was that natural numbers have an inherent structure to them. You can describe all natural numbers using only the objects 0 and succ, where succ = (+ 1).

An interesting fact is that this is sufficient to express + and :

a + 0 = a a + (succ b) = succ (a + b)

a 1 = a a * (succ b) = a + a * b

One way to look at structures (like, say, natural numbers) is too look at what really defines them, as in, what is absolutely necessary and sufficient to define them? In this case, 0 and succ is.

Note that I'm not saying you're wrong, everyone is entitled to their own definition of beauty. Isn't that the beauty of it?

> I subconsciously did it was that natural numbers have an inherent structure to them.

I understand and concur you :)

But I don't think Euler's identity is meant to serve this purpose. Maybe some other formula.

It's hard to un-like something once you already liked it very much. First impression matters. How about e^(i*pi)-1==0 instead?

I think your aesthetic sense in this case may have been affected by years of pro-pi propaganda. The 'beautiful' equation requires a rearrangement (i.e., moving the -1 to the other side), which is a bit kludgy, to say the least. Moreover, the superior (IMHO) version in the parent comment can be put into words in a rather compelling way: the exponential of the imaginary unit times the circle constant is unity. (If you're still hung up on the appearance of 0, just add it to the RHS. ;-) The same can't be said for the standard version.

P.S. Please email me if you're interested in pursuing this further; I have a project (launching June 28) that I think might convince you of the error of your ways. :-)

I have some secret plans in this direction. Please email me if you'd like to be part of it (my address is in my profile). Hint: It's launching on June 28. :-)

Your version contains less information, since it's a result of taking a square of both sides of e^(i3.14...) = -1 (which is equivalent to the original), and taking a square is not injective. In other words, your version would be true also if e^(i3.14...) was equal to 1. So it's arguably less interesting then the original.

its got the five most beautiful constants in one formula e, pi , i , 1 and 0 and hence i chose it for my handle :)

I would put phi before a lot of those.

Probably my favorite equation, and the proofs for the formula are quite nice too.

Perhaps interestingly, I have a friend who accepted this, but after proof upon proof could not agree that 0.999... is exactly equal to 1. He eventually conceded, but it was a pretty long period of time to that point and it wasn't through proof.

You should always ask if they agree that 1/3 = 0.33333... then you just need to add them up :)

I find this easy to accept:

  lim_{n->inf} 1 - 1/10^n = 1

But I'm not quite sure if I agree that 0.999... = lim 0.999...

edit: according to Wikipedia, they're today defined to be equal. While convenient, it's not obvious to me.

edit2: ah-ha! I just remembered the proof that had once upon a time satisfied me: there is a number between any two distinct numbers. there isn't one between 0.999... and 1, so they aren't distinct.

Bonus points for realizing it yourself and editing your post accordingly!

Makes sense. The fact that e^(ipi) = -1 seems crazy at first, but no one has a preconceived notion of what it should equal, so when you show a simple proof, of course they'll accept it.

But people's whole mathematical careers they've worked with decimals and developed firm ideas of how they work. So they idea that 0.999... = 1 seems to defy their years of decimal practice, which makes it hard to accept.

Fun anecdote: In my calc BC class, we did an infinite series question that proved that .999... = 1 (it was an infinite series of 9/(10^n), so it was adding 9/10 + 9/100 + 9/1000 + ...), which blew the minds of half of our class.

During the argument that ensued, our teacher said "Well, it doesn't actually equal 1. It'll never quite get there. It just gets infinitely close."

Needless to say, what little respect for her I did have at that point in the year was gone after that.

Hi everyone.I came across the equetion e^(i)pi)1=0 in the strangest place.It was wrien in binary. Thats not the wird bit. It was in a crop circle at a whoping 200ft diameter.http://www.cropcircleconnector.com/2010/wilton/wilton2010a.h... Their must be some realy bord intelligent pranksters out their. Or are we getting pointed in a direction to help us advance. Is their more to this equetion than we actually know?

In ASCII. This equation is famous, and crafting proper crop circles requires some familiarity with computers for the GPS and line art. This crop circle, while clever, is not particularly astounding, and is in fact proof against extraterrestrial crop circles.

I remember a while back they polled google engineers and asked them what they thought was the most beautiful equation in math. I believe this one came in the winner.

Once you get involved with differential equations, especially applied differential equations this is seen quite often.

All trigonometric identities can be easily derived from the Euler's formula. Instead of learning the whole course called "trigonometry" at school we'd have been just taught this formula!

e^ix = cos x + i sin x

It really has nothing to do with the value of i, it's just a convention. It really just says that cos pi = -1 which is nice but that's it.

By that reasoning, a lot of math is a "convention".

The link between exponential and sinusoidal functions is of fundamental importance in many fields--obviously anything dealing with complex numbers, which is not just math but physics, electric engineering, computer graphics, and so on... and the complex definition of e^x is really the only way to define it such that it retains all its important properties in a natural way, revealing the link between trig and exponentiation. Euler's identity is fundamental, not conventional.

It is actually far from just a convention. It has everything to do with the Taylor series of the functions e^x, sin x and cos x.

e^x = 1 + x/1 + x^2/(2!) + x^3/(3!) + ...

If you substitute ix for x, you get:

e^{ix} = 1 + ix/1 + (ix)^2/(2!) + (ix)^3/(3!) + ... = = 1 + ix + (-x^2)/(2!) + (-i)x^3/(3!) + ...

You can also derive the Taylor series for sin x and cos x, which are:

sin x = x - x^3/(3!) + x^5/(5!) + ...

cos x = 1 - x^2/(3!) + x^4/(4!) + ...

Therefore you can see that e^{ix} = cos x + i(sin x). Of course, I haven't shown how to derive the Taylor series (http://en.wikipedia.org/wiki/Taylor_series) for these functions, but I have to stop somewhere.

I hope that I have shown you that this isn't just an arbitrary convention, in fact far from it. The formula is so beautiful also because of the many intricate relationships between all these elementary mathematical concepts, including complex analysis, trigonometric functions, series, etc.

It should also be noted that this is just one of the many (in fact, aleph zero many) different ways to prove this equivalence.

Edit: Some formatting corrections, sorry, new here.

I can't resist not showing this clever derivation of the e Maclaurin series.

I referenced the fact that

e^x = 1 + x/1 + x^2/(2!) + x^3/(3!) + ...

Now, one of the many equivalent definitions of e is that it is the only real number for which this holds:

d/dt (e^t) = e^t

This means that e^t is a real function which is so smooth, that no matter how many times you differentiate it, you always get the same function (it turns out that it is the only such real function)

Now, we know that e^0 = 1 (since x^0 = 1, where x != 0). Therefore, in its Maclaurin series, the only term not depending on x should therefore be 1 (otherwise e^0 wouldn't be 1).

So know we know that e^x looks something like this:

e^x = 1 + (something)

Now we can ask ourselves which this question: since d/dx e^x = e^x, what must also be in the e^x series, if 1 belongs to it? Well, whatever differentiates to 1, so now we know that

e^x = 1 + x + (something)

(because d/dx (1 + x + something) = 1 + x + d/dx something)

Now we can again ask this question for our current form; what must we differentiate in order to obtain 1 + x? And thus we get

e^x = 1 + x + x^2/2 + something

This way, if we write the two (equivalent) series this way:

e^x = 1 + x + x^2/2 + d/dx e^x = 1 + x + x^2/2 + x^3/(2*3) +

And we can complete it with the infinite Maclaurin series.

Now this is less formal than it should be and it would probably make the formalists cringe, but I hope you get the idea. You can actually apply the same principle for sin x and cos x, except in their case, they're actually mutually derived from each other. I'll leave that as an exercise for the reader (oh how fun it is to say that after reading this phrase countless times)

Great. Now I'm going to spend the rest of the morning trying to figure out whether you can really do it that way.

[Edit: Ah yes. The formalism goes like this: Consider the above series. By the above argument, it differentiates to itself. QED. ]

Ah, but how do you know that e^x is the only thing that differentiates to itself? :)

You don't need to. The uniqueness part of the definition is never used in that argument. (In fact, a*e^x also differentiates to itself, for any a; but that's a trivial case.)

Uniqueness almost follows from that argument. It's now easy to see that exp is the only analytic function satisfying exp' = exp and exp(0) = 1: if you have another one, by the same argument, it has the same Maclaurin expansion, hence is the same function.

However, I don't know how to prove uniqueness over all functions, not just analytic ones.

I wouldn't say it's "just" a convention. That implies it's arbitrary. It might look arbitrary, but it's not: for example, d/dx ((cos x + i sin x)/e^ix) = 0, as required. This doesn't hold if you pick (say) e^ix = cos x - i sin x.

this is the crux of a lot of electrical engineering.

old, lol

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