There's a classic paradox in math that shows you can build a bridge that balances on the end of a table and extends infinitely far.

The normal method is very slow.

But I'm going to show you a faster way to build it.

[MUSIC PLAYING] Say I give you 10 identical one centimeter long blocks.

You want to balance them off the edge of a table so they make a little bridge, like this.

How should you do it so the bridge extends as far from the table as possible?

Pause here if you want to think about it.

Let's start with one block.

You can push it exactly halfway off the table, but no more.

At that moment, half its weight is on the table, and half its weight is off the table, which means that its center of mass is exactly at the edge of the table.

Now you place a second block under the first so that it aligns with the end of the table.

You want to push the two blocks together as far off the table as you can without them falling, until their center of mass is exactly at the edge of the table.

The distance you'll push them until this happens is somewhere between 0 and one half centimeters.

So we'll just call it a variable x.

If you push them distance x off the table, then where is the center of mass?

We'll assume each block weighs one gram.

The first block has 1/2 plus x grams past the table and one half minus x grams over the table.

The second block has x grams past the table and 1 minus x grams over the table.

This means that in total, there is 1/2 plus 2x grams past the table and 3/2 minus 2x grams over the table.

To make the center of mass exactly at the edge of the table, the mass past the table should be equal to the mass over the table.

So one half plus 2x should equal 3/2 minus 2x.

With a little algebra, we can see that this happens when x equals one quarter.

Let's put a third block under, flush with the table.

How far can you push this whole thing?

Let the variable x represent the distance they'll move.

The first block has mass 3/4 plus x off the table and one quarter minus x on the table.

The second is one quarter plus x off and 3/4 minus x on.

And the third has mass x off the table and 1 minus x on.

In order for the center of mass to be directly over the edge of the table, the right column and left column must be equal.

So 1 plus 3x equals 2 minus 3x, which implies x equals 1/6.

When we put the fourth block under the first block, we'll actually move past the edge of the table, which complicates the method we've been using.

But let's look at what we've learned.

The first block extends one half centimeter out from the second.

The second extends one quarter centimeter out from the third.

And the third extends 1/6 centimeters.

We can rewrite that as 1/2 times 1/1 and 1/2 times 1/2 and 1/2 times 1/3.

That leaves us with a pretty good guess that the fourth block will extend 1/8 centimeters, which is 1/2 times 1/4.

And the nth block will extend 1/2 times 1 over n centimeters.

Here's the proof.

Let's assume we have n minus 1 blocks balanced so their center of mass is exactly over the edge of the table and one block underneath flush against the table.

We can replace these with point masses.

So the n minus 1 blocks become a point of weight n minus 1 at its center of mass directly above the edge of the table.

And the single block is a point of weight 1 at its center.

Now we slide these point masses together until their collective center of mass is directly over the edge of the table.

Again, we'll let x represent that distance.

Then there is one gram at distance one half minus x to the left and n minus 1 grams at distance x to the right.

To balance them, we set them equal.

After a little algebra, we see that x equals 1 over 2n.

So the nth block always slides over 1 over 2n centimeters.

Here's what's crazy about that.

After we put down n blocks, we add up all their individual extensions to discover that the length the bridge will extend over the table is equal to 1/2 times 1/1 plus 1/2 plus 1/3 all the way up to 1/n.

So what's the longest bridge you can build with 10 blocks?

It's this sum, which is roughly 1.46 centimeters.

What if you have 20 blocks?

Then your bridge extends roughly 1.8 centimeters.

Actually, for an arbitrary number n, the sum 1 plus 1/2 plus 1/3, all the way up to 1/n, is pretty well approximated by the natural log of n. The natural log is the inverse of exponentiation.

And the graph of the natural log function looks like this.

It grows really slowly.

So after we balance 100 blocks on the table, the bridge is roughly 1/2 times the natural log of 100, or 2.3 centimeters over the edge.

After 1,000 blocks, it's roughly 1/2 times the natural log of 1,000, or 3.5 centimeters over the edge.

As we add more and more blocks, the bridge is not growing very fast.

But what happens if we use infinitely many blocks?

Our formula says that the distance the bridge will extend off the table is equal to 1/2 times 1 plus 1/2 plus 1/3 plus 1/4 and so on.

The infinite sum, 1 plus 1/2 plus 1/3 plus 1/4 and so on, is known as the harmonic series.

By the way, it's an example of an infinite series.

Some infinite series are finite.

You can add together infinitely many numbers and get a finite one.

Like 1/2 plus 1/4 plus 1/8 plus 1/16 plus 1/32 equals 1.

And here's an awesome visual proof of that fact.

But the harmonic series does not equal a finite number.

Mathematicians are hesitant to say that something equals infinity.

So instead, we say the harmonic series diverges, which means that the sum is bigger than any finite number.

Basically, nontechnically, that means it equals infinity.

Of course, there are physical limitations that mean our arbitrarily long bridge wouldn't actually stand.

In fact, if you try it with cards or blocks or books, you'll find it's very hard to obtain the theoretical distances we've derived.

In the description, there's a link to a fun article exploring these limitations.

So with infinitely many blocks, we can build a bridge to infinity.

But it's a mighty slow process.

There's actually a pretty terrible math joke about this.

The harmonic series is known to diverge, but it's never been observed to do so.

Anyway, can you build a bridge to infinity faster?

Before, we restricted ourselves to stacking just one block on top of another.

With this restriction, the fastest growing bridge you can build is the one we just made.

But, somewhat counterintuitively, if you allow yourself to stack two blocks on top of one, you can build a bridge to infinity faster.

This isn't exactly a bridge, except for spiders and other things that can hang upside down.

But it actually grows off the table way faster than our other bridge.

What does that mean, more precisely?

Well, remember that after placing n blocks down on our previous bridge, it extended off the table roughly 1/2 times the natural log of n. After n blocks of this new bridge, it extends roughly 1/2 times n to the 1/3.

Actually, we don't have to pay much attention to the constants, since they don't matter much when n is very large.

The old bridge grows roughly like the natural log of n, whose graph looks like this.

And the new bridge grows roughly like n to the 1/3, whose graph looks like this.

The new bridge is growing exponentially faster, which is kind of amazing.

Mathematicians Mike Patterson and Uri Zwick created the super fast bridge in a 2009 article.

They called it a parabolic stack.

You can't actually build the parabolic stack brick by brick there are certain points where it would topple.

But you can build it in layers like this, and it will be stable after each layer.

Shortly after, they proved, along with some collaborators, that the fastest any bridge can grow is n to the 1/3.

The n to the 1/3 is multiplied by a constant number.

And in fact, you can make the constant bigger than the one for the parabolic stack above.

But we only consider the term n to the 1/3 asymptotically, that is, as n gets big.

Here's your challenge for the day, build the best-- prettiest, biggest, funniest, whatever you judge to be the best-- bridge you can, and send us a picture.

It can be the original harmonic construction, the parabolic stack version, or your own creation.

Link to it in the comments, tweet us @pbsinfinite or email it to us at pbsinfiniteseries@gmail.com.

See you next time on "Infinite Series."

Hey, all.

I just wanted to respond to some of the comments about our second episode on Shor's algorithm.

MultiMurmaider says, "I fail to see how the world can benefit from the destruction of all internet security.

Sounds more like world economic collapse."

OK, so first off, I thought the math of Shor's algorithm was cool, not the potential for world economic collapse.

But I don't think we actually have to worry about that.

Even if we develop quantum computers, and even if we implement Shor's algorithm, which would do away with our modern methods of cryptography, it's very likely that we'll develop new ones, and we have other ways to secure information that use quantum computers.

So we probably shouldn't worry too much about the total destruction of all security.

Marshall Webb said, "You say a few times that under the transformation a quantum computer would be very likely to give the right answer.

But how likely would it be?

And with computers making billions of calculations, wouldn't a small error rate become problematic?"

That's a really good question about quantum computers in general, not just Shor's algorithm.

Quantum computers work probabilistically.

So they have a chance of making an error.

But we don't really have to worry about that.

10xTinThOuGhT give a really solid answer as to why.

They pointed out that, first, you can usually check the answer a quantum computer gives you with a classical computer.

So in the case of Shor's algorithm, you can just use a classical computer to check whether the numbers it produced were actually the factors of n, or whether it was actually a period it gave you, or something like that.

They also pointed out that, too, you can run a computation several times and that minimizes the error.

If you have a small error each time, the odds that you have an error after many times are very, very small.

And so we can just run it several times.

It's no problem to do that with a quantum computer.

Quantum computers are really, really fast.

So it doesn't matter if we have to do it 10, 15 times, no problem.

All right.

See you next week.