1+2+3+... = -1/12 ... huh?

Recently a video did its rounds on social media, showing how the sum of the natural numbers was $\frac{-1}{12}$. Naturally there was some skepticism from some of my friends, so I decided to write up a short note to clarify the situation.

You can find the video here, and for the more mathematically inclined, a blog post by Terry Tao.

What I'm going to show is that there is a slightly more rigorous way of interpreting the manipulations performed in the YouTube video, while pointing out the parts which are still not fully kosher.

(Could I start by saying that the sum doesn't work "because of physics"? Thank you.)

Infinite Sums

In case your response upon watching that video was "Wait a minute! That sum is clearly infinity" - you're not wrong. The typical way to define an infinite sum is as the limit of its partial sums i.e. 

$$S = a_1 + a_2 + a_3 + \cdots = \sum _{n=1} ^ \infty a_n $$

is defined to be 

$$ lim _{N \rightarrow \infty} \sum_{n=1} ^N a_n$$

and according to this definition, the sum $1+2+3+\cdots$ is certainly $\infty$.

And it gets worse - there are sums whose partial sums don't even have a limit. Here's an example which is actually from the video; the sum

$$S_1=1-1+1-1+1-1+ \cdots$$

has partial sums alternating between 1 and 0. In the video, he says that since the partial sums alternate in this way, the sum must be the "average" of these two values i.e. 1/2. 

I'm going to show you another way of seeing this. We won't actually use this sum, but it's a good way to start.

Start with the series expansion

$$\frac{1}{1-x} = \sum _{n >= 0} x^n$$

Let's first see why the fraction $\frac{1}{1-x}$ has that series expansion (you've probably seen this before):

$$(1-x) (1+x+x^2+x^3\cdots) = (1+x+x^2+x^3\cdots) - x (1+x+x^2+x^3\cdots)$$

which becomes

$$ = (1+x+x^2+x^3\cdots) - (x+x^2+x^3\cdots) $$
just by multiplying by $x$. Now all the powers of $x$ cancel, leaving only

$$ (1-x) (1+x+x^2+x^3\cdots) = 1$$

and therefore

$$(1+x+x^2+x^3\cdots) = \frac{1}{1-x}$$

Now this is fine as far as algebra goes i.e. symbolic manipulation. But what happens when we want to plug in values for $x$, say $x=500$?

Well, you can't. The problem is that the infinite sum will just stop making sense. The values of $x$ for which the sum does make sense is called its domain of convergence. For the sum $\sum x^n$, the domain of convergence is just the interval $-1<x<1$ (take my word for it - if we get into that level of detail I'll never get done here).

Okay, so 

1. When $x$ is between -1 and 1, the value of the fraction $\frac{1}{1-x}$ matches with the limit of partial sums i.e. the infinite sum of $\sum x^n$.
2. When $x=1$, both the fraction and the sum clearly blow up - the fraction gets a zero in the denominator and the sum just becomes $1+1+1+1+\cdots$. 
3. But what happens at $x=-1$?

Let's compute - I'm going to plug in some values of $x$, starting at $-0.1$ and going down to $-0.9$. You can imagine $x$ moving to the left on the number line as you go down the rows of this table: 

For each row, I've listed the first few terms of the series along with their signs so that it looks like a sum (as if I've plugged values of $x$ into $1+x+x^2+\cdots$). Then in the sixth column I have the infinite sum itself (well, not quite infinite, I did the computation with $1+x+\cdots + x^{1000}$). The final column has the value of the fraction for this value of $x$.

Here's what you're supposed to notice about this table:

1.  The sum $\sum x^n$ matches the fraction $\frac{1}{1-x}$. Of course if I had shown more digits after the decimal points they would eventually have differed, because the sum only used 1000 terms.
2. Both of these get closer to $1/2$ as $x$ gets closer to $-1$ (the bottom rows)
3. The second column is getting closer to -1, the third column to +1, the fourth to -1 and the fifth to +1.

I've attempted to draw a graph which to show (3). The different colors correspond to the rows of the table. Take a look at each block of colored bars.

1. The first one is $x^0$, which is just 1, no matter what $x$ is
2. The second one is $x^1$, so you can see the dark blue bar is for $x=-0.5$ and the red one is for $-1$
3. The third one is for $x^2$, etc...

Every time you move from blue to red in a colored block, it's like going down the rows of the table.

Now what I'm trying to show you here is how the individual terms of the series $1+x+x^2+\cdots$ get closer to $1-1+1-1+\cdots$ as $x \rightarrow -1$. So when you look at the series, this is a way you can picture it changing as $x \rightarrow -1$.

I would have been nice if I had also indicated the sums of these terms, but my graphing skills only go so far... anyway from the table you can see that they approach $1/2$. 

(And of course if you plug $x=-1$ into the fraction $\frac{1}{1-x}$, you get $1/2$, no magic there). 

So what have you just seen? What I just showed you was that $1+x+x^2+\cdots \rightarrow 1/2$ as $x \rightarrow -1$, i.e 

$$ 1-1+1-1+1- \cdots = 1/2$$
To summarize, I'm going to state all this in a very particular way:

1. There is a function $f(x) = \frac{1}{1-x}$,
2. with a series expansion $1+x+x^2+x^3\cdots$
3. and there is also a point $x=-1$ 
4. which allows us to view the sum $1-1+1-\cdots$ as $f(-1)$.

LET ME REITERATE: The sum $1-1+1-\cdots$ has no meaning according to the definitions of infinite sums. We are specifically interpreting it as the specialization of a series at a point on the boundary of the domain of convergence.

The sum $S_2$

The next sum in the video is 

$$ S_2 = 1 - 2 + 3 - 4 + \cdots $$

which they claim equals $1/4$.

If you believe what I said in the last section, this one is easy. Just differentiate

$$ \frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots $$

with respect to $x$; you get

$$ \frac{1}{(1-x)^2} = 1 + 2x + 3x^2 + 4x^3 + \cdots$$

and plug in $x=-1$ to get

$$ \frac{1}{4} = 1 - 2 + 3 - 4 + \cdots$$

A different kind of series

We just interpreted $1 - 2 + 3 - 4 + \cdots$ as the specialization of $$1 + 2x + 3x^2 + 4x^3 + \cdots$$ at $x=-1$. Now we're going to look at it differently, as the specialization at $x=-1$ of

$$\frac{1}{1^x} - \frac{1}{2^x} + \frac{1}{3^x} - \frac{1}{4^x} \cdots$$

Here comes the non-kosher part: I'm going to ask you to believe that since
$$1 - 2 + 3 - 4 + \cdots = 1/4$$
using the series
$$1 + 2x + 3x^2 + 4x^3 + \cdots,$$
it continues to be $1/4$ even when we use this new series

$$\frac{1}{1^x} - \frac{1}{2^x} + \frac{1}{3^x} - \frac{1}{4^x} \cdots$$

Accepting this for the moment, the final series we need is

$$ g(x) = \frac{1}{1^x} + \frac{1}{2^x} + \frac{1}{3^x} + \frac{1}{4^x} \cdots$$

Very similar, but now we have a + sign on every term.

Step 1. Make a modification of $g(x)$ which only uses even numbers.

This is easy - we just divide it by $2^x$; watch:

$$\frac{g(x)}{2^x} = \frac{1}{2^x} \left ( \frac{1}{1^x} + \frac{1}{2^x} + \frac{1}{3^x} + \frac{1}{4^x} \cdots \right )$$

$$ = \frac{1}{2^x} + \frac{1}{4^x} + \frac{1}{6^x} + \cdots$$

Cool.

Step 2. Now we subtract it from $g(x)$, first once:

$$ g(x) - \frac{g(x)}{2^x} =  \frac{1}{1^x} +  \frac{1}{3^x} +  \frac{1}{5^x} + \cdots $$

and then again:

$$ g(x) - \frac{g(x)}{2^x}- \frac{g(x)}{2^x} = \frac{1}{1^x} - \frac{1}{2^x} + \frac{1}{3^x} - \frac{1}{4^x} + \frac{1}{5^x}+ \cdots $$

(Looking familiar!) Let's simplify the left-hand side by grouping the $g(x)$'s

$$g(x) (1 - \frac{1}{2^x}- \frac{1}{2^x} ) = \frac{1}{1^x} - \frac{1}{2^x} + \frac{1}{3^x} - \frac{1}{4^x} + \frac{1}{5^x}+ \cdots$$

$$g(x) \left (1 - \frac{2}{2^x} \right ) = \frac{1}{1^x} - \frac{1}{2^x} + \frac{1}{3^x} - \frac{1}{4^x}  +  \frac{1}{5^x}+ \cdots $$

If we plug $x=-1$ into the left-hand side, we get $g(-1) (1-4)$, and if we plug it into the right-hand side we get $1 - 2 + 3 - 4 + \cdots =1/4$. 

Diving by $1-4$ gives us $$g(-1) = -1/12$$

And... what is $g(-1)$? Go ahead, look at the definition, you will get

$$1+2+3+4+\cdots = -1/12$$

The Non-Kosher Step

Why am I uncomfortable with saying that 

$$\frac{1}{1^x} - \frac{1}{2^x} + \frac{1}{3^x} - \frac{1}{4^x} \cdots$$
approaches $1/4$ as $x \rightarrow -1$?

Remember that whole bit above where I drew the graphs of those series and said "well, the series approaches 1/4 as $x$ approaches -1"? I showed you a table and hopefully convinced you that what I was doing wasn't complete nonsense. (Okay, I drew the graphs for $S_1$, but it works for $S_2$ as well. Try it in Excel or something).

We can't do that here.

Not only can we not approach $x = -1$, we can't even cross $x=+1$! The series converges when $x>1$, and blows up at $x=1$, but even if you try to evaluate it at $x=0.99$ it won't work. It just blows up.

So how did I jump from $x=1$, across 0 all the way to $x=-1$?

Well, that's why this is the non-kosher step. And I don't have an explanation which doesn't involve more machinery.

But it's not "because of the physics".

Playing with roots of unity

This post doesn't have a goal, I'm just recording some behavior I came across. Sometimes people approach me at parties and tell me how interested their children are into math; now I can point them here.

 Let $p$ be prime, and let $\zeta$ be a $p$-th root of unity. This means that $\zeta^p=1$ (but $\zeta \neq 1$). For every integer $k$, defined $\epsilon_k$ to be $\zeta^k + \zeta^{-k}$.

 Note that $\epsilon_0 = 2$, and $\epsilon_k = \epsilon _{-k}$. Let's see how these multiply:
$$ \begin{align} \epsilon_k \epsilon_l &= (\zeta^k + \zeta^{-k})(\zeta^l + \zeta^{-l}) \\ &= (\zeta ^ {k+l} + \zeta ^ {k-l} + \zeta ^ {-k+l} + \zeta ^ {-k-l}) \\ &= \epsilon_{k+l} + \epsilon _ {k-l} \end{align} $$
Now I will show that all the $\epsilon_k$'s can actually be expressed as polynomials in $\epsilon_1$. First let's expand powers of $\epsilon_1$ using the binomial expansion: $$ \begin{align} \epsilon_1^n &= (\zeta + \zeta^{-1})^n \\ &= \sum_{l=0}^n \binom{n}{l} \zeta^l \zeta^{-(n-l)} \\ &= \sum_{l=0}^n \binom{n}{l} \zeta^{2l-n} \end{align} $$

Let's add the terms in pairs, starting with the first and last: $$ \binom{n}{0} \zeta^{-n} + \binom {n}{n} \zeta^n. $$ This is just $\epsilon_n$.

Now add the second and second-to-last terms: $$ \begin{align} \binom{n}{1} \zeta^{2-n} &+ \binom {n}{n-1} \zeta^{2(n-1)-n} \\ \binom{n}{1} \zeta^{2-n} &+ \binom {n}{n-1} \zeta^{n-2} \end{align} $$ which is $n \epsilon_{n-2}$.

In this way, every term pairs up with another term, giving an integer times an $\epsilon_k$ (remember that binomial coefficients are always integers!). When $n$ is even, there is also the middle term $\binom{2l}{l}$, coming from $l=n/2$. This term is actually even making it an integral multiple of $\epsilon_0$.

To summarize: every $\epsilon_1 ^n$ is an integral sum of $\epsilon_k$'s for $k$ <= $n$. Even better, it is $\epsilon_n$ + an integral sum of $\epsilon_k$'s for $k$ < $n$! Which means we can turn it around to say that $\epsilon_n$ is $\epsilon_1^n$ minus an integral sum of $\epsilon_k$'s for $k$ < $n$.

Why is this good? Because continuing in this way, we can replace each $\epsilon_k$ with an $\epsilon_1^k$ plus/minus some other stuff, which we can then replace in the next step. And each time our coefficients stay integers.

So we get that $\epsilon_n$ is an integral polynomial in $\epsilon_1$. Here are the first few for $n$ <= $4$, I'm just going to write $\epsilon$ for $\epsilon_1$ now: $$ \begin{align} \epsilon_1 &= \epsilon \\ \epsilon_2 &= \epsilon^2 - 2 \\ \epsilon_3 &= \epsilon^3 - 3 \epsilon \\ \epsilon_4 &= \epsilon^4 - 4 \epsilon^2 + 2 \end{align} $$

As I find time I will study these polynomials and record my results here. Feel free to make contributions in the comments!

A Calculation involving Circles

This is a computation I did for a friend last month. We wanted to describe the set of points the ratio of whose distances from two other points remained fixed.

Let me say that better: I have two points in the plane, $S$ and $T$, and am given a positive fixed quantity $\alpha$. I want to describe the set of points $P$ for which

$$ \frac{\text{distance} (P,S)} {\text{distance}(P,T)} = \sqrt {\alpha}$$
The reason I put in a square-root is because I'm actually going to work with the square of this formula.

Notice that if $\alpha = 1$, then I just get a line. So I'm only going to consider the cases in which $\alpha \neq 1$.

I'll start by naming the coordinates of these points:
$$S= (a_1, b_1), T = (a_2, b_2), P= (x, y)$$

Then the ratio of distances reads as

$$\frac{(x-a_1)^2+(y-b_1)^2}{(x-a_2)^2+(y-b_2)^2} = \alpha$$

Now I'm going to do some algebra.

$$(x-a_1)^2+(y-b_1)^2 = \alpha((x-a_2)^2+(y-b_2)^2)$$

$$x^2-2a_1x+a_1^2 + y^2-2yb_1+b_1^2 = \alpha(x^2-2a_2x+a_2^2 + y^2-2yb_2+b_2^2)$$

$$x^2-2a_1x+a_1^2 + y^2-2yb_1+b_1^2 = \alpha x^2-2a_2\alpha x+\alpha a_2^2 + \alpha y^2-2y\alpha b_2+\alpha b_2^2$$

$$(1-\alpha)x^2 + (- 2a_1 + 2\alpha a_2 )x +(a_1^2-\alpha a_2^2) + (1-\alpha) y^2 + (- 2 b_1 + 2 \alpha b_2 )y + (b_1^2 - \alpha b_2^2) = 0$$

Now since $\alpha \neq 1$, I can divide by $(1-\alpha)$:

$$\left [x^2 + \frac{- 2a_1 + 2\alpha a_2 }{1-\alpha} x + \frac{a_1^2-\alpha a_2^2}{1-\alpha} \right ] + \left [ y^2 + \frac{- 2 b_1 + 2 \alpha b_2 }{1-\alpha}y + \frac{b_1^2 - \alpha b_2^2}{1-\alpha} \right ] = 0$$

Now I'm going to use a clever little transformation called "completing the square", which is basically a convenient rewriting of

$$X^2+2AX+B$$
as
$$ (X+A)^2+(B-A^2)$$

It's the same, you can check. I do this to both the $x$ component and the $y$ component (separately!), and get

$$\left [ \left (x + \frac{- a_1 + \alpha a_2 }{1-\alpha} \right )^2 + \left ( \frac{a_1^2-\alpha a_2^2}{1-\alpha} - \left ( \frac{-a_1 + \alpha a_2 }{1-\alpha} \right ) ^2 \right ) \right ]
+ \\
\left [ \left (y + \frac{- b_1 + \alpha b_2 }{1-\alpha} \right )^2 + \left ( \frac{b_1^2-\alpha b_2^2}{1-\alpha} - \left ( \frac{- b_1 + \alpha b_2 }{1-\alpha} \right ) ^2 \right ) \right ] = 0
$$

Now I move the constants (the parts not involving $x$ or $y$) to the right-hand-side of the $=$ sign. Watch those signs!

$$\left (x - \frac{a_1 - \alpha a_2}{1-\alpha} \right )^2
+ \\
\left (y - \frac{b_1 - \alpha b_2}{1-\alpha} \right )^2
=
\left ( \left ( \frac{- a_1 + \alpha a_2 }{1-\alpha} \right ) ^2 - \frac{a_1^2-\alpha a_2^2}{1-\alpha} \right ) + \left ( \left ( \frac{ - b_1 + \alpha b_2}{1-\alpha} \right ) ^2 - \frac{b_1^2-\alpha b_2^2}{1-\alpha} \right )
$$

And because $(-1)^2=-1$,

$$\left (x - \frac{a_1 - \alpha a_2}{1-\alpha} \right )^2
+ \\
\left (y - \frac{b_1 - \alpha b_2}{1-\alpha} \right )^2
=
\left ( \left ( \frac{a_1 - \alpha a_2}{1-\alpha} \right ) ^2 - \frac{a_1^2-\alpha a_2^2}{1-\alpha} \right ) + \left ( \left ( \frac{b_1 - \alpha b_2}{1-\alpha} \right ) ^2 - \frac{b_1^2-\alpha b_2^2}{1-\alpha} \right )
$$

END OF LONG CALCULATION.

What did this give us? Well, that final relation describes a circle of $(x,y)$ points, whose center is

$$\left ( \frac{a_1 - \alpha a_2}{1-\alpha}, \frac{b_1 - \alpha b_2}{1-\alpha} \right )$$

which we can write as
$$\frac{S-\alpha T}{1-\alpha}$$

The square of the radius of this circle is

$$ { \left ( \left ( \frac{a_1 - \alpha a_2}{1-\alpha} \right ) ^2 - \frac{a_1^2-\alpha a_2^2}{1-\alpha} \right ) + \left ( \left ( \frac{b_1 - \alpha b_2}{1-\alpha} \right ) ^2 - \frac{b_1^2-\alpha b_2^2}{1-\alpha} \right ) }
$$

$$=\frac{{(a_1-\alpha a_2)^2-(a_1^2 - \alpha a_2^2)(1-\alpha)+(b_1-\alpha b_2)^2-(b_1^2 - \alpha b_2^2)(1-\alpha)}}{(1-\alpha)^2}
$$

Let's simplify the numerator. We start by expanding it out:

$$(a_1^2-2\alpha a_1 a_2+\alpha^2a_2^2)-(a_1^2 - \alpha a_2^2 - a_1^2\alpha + \alpha^2 a_2^2) + (b_1^2-2\alpha b_1 b_2+\alpha^2b_2^2)-(b_1^2 - \alpha b_2^2 - b_1^2\alpha + \alpha^2 b_2^2)
$$
Lots of these terms cancel one another, and in fact we are left with
$$\alpha(a_1^2-2a_1a_2+a_2^2) + \alpha(b_1^2-2b_1b_2+b_2^2)
$$

$$=\alpha ((a_1-a_2)^2 + (b_1-b_2)^2)$$

So the radius of the circle is (the positive value of)

$$\frac{\sqrt{\alpha ((a_1-a_2)^2 + (b_1-b_2)^2)}}{1-\alpha}
$$

which is $\sqrt\alpha/(1-\alpha)$ times the distance between $R_1$ and $R_2$.

In summary

The set of points which have a fixed ratio of distances from two other fixed points is either

a. A line, if that ratio is 1
b. Another circle, whose center is $\frac{R_1-\alpha R_2}{1-\alpha}$ and radius is $\sqrt\alpha/|1-\alpha| |R_1 - R_2|$.