January 20, 2009
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Follow Up, numbers, Topology and geometry

In response to the post “Stacking Cannonballs” Trevor H. writes:

I was very intrigued by the recursive sequence you mentioned in the past two episodes–the sequence that begins with 1 and each successive term is the average of all the previous terms times some constant. I have always been fascinated by Pascal’s triangle and all of its surprise appearances in mathematics. Also, my fist encounter with doing mathematics for fun out of my own curiosity was to find a formula for triangle numbers. Like Kyle, I was inspired by bowling pin arrangements. The experience was very rewarding and I have been in love with mathematics ever since.

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January 6, 2009
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answers, Follow Up, numbers

Given a difference table, as we considered back in EV. What’s the Difference , how do we come up with a polynomial that gives the values on the top row?

For example, suppose we have

**-1** -1 3 35 143 399 899 . . . . .
**0** 4 32 108 256 500 . . . . .
**4** 28 76 148 244 . . . . .
**24** 48 72 96 . . . . .
**24** 24 24 . . . . .

What is the polynomial P(n), of degree four, that gives

P(0) = -1 P(1) = -1 P(2) = 3 P(3) = 35 P(4) = 143 , etc.

Can this be expressed simply in terms of the leading values on the left of the table: -1, 0, 4, 24, 24?

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August 16, 2008
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answers, Follow Up, infinity, logic, paradoxes

That the worm falls off the end of the rope depends on the fact that the incredible

harmonic series

1 + 1/2 + 1/3 + 1/4 + . . .

diverges to infinity, growing as large as you please!

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July 5, 2008
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Follow Up

** Why can there be no computable bound to the Busy Beaver Function?**

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April 17, 2008
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Follow Up, guests, paradoxes

We present a recording of Raymond Smullyan’s lecture at the Gathering for Gardner, March 30, 2008; Newcomb’s paradox really is a stumper.

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January 9, 2008
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Follow Up, math puzzles, numbers

We also asked, on this week’s segment how to label the faces of some ordinary dice, with twelve different numbers (we did say different didn’t we?) so that every roll produces a prime number. This puzzle is from the fascinating site www.primepuzzles.net. Don’t peek!

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October 9, 2007
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answers, calculusey stuff, Follow Up, The Mathcast

We can say a bit more about the Princess’s escape.

Amazingly, an optimal path for the Princess is to swim in a half circle of radius 1/8 that of the lake, then dash out to the edge.

We’ll give an analytic proof, but we could give a totally synthetic (geometric) proof as well.

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August 16, 2007
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Follow Up, numbers, The Mathcast

What numbers can 1,2,4,8,16,… etc “form”? Well, *every* number can be “formed” by summing various powers of 2. For example, 13 = 1 + 4 + 8.

In this way, we could say that a power of 2, say 64, is “full of divisors” since it has enough divisors to form any number up to 64. Its divisors are of course 1, 2, 4, 8, 16 and 32, and we can form any number from 1 to 63 by summing up these divisors as needed.

But what other numbers of “full of divisors”?

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June 8, 2007
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Follow Up, The Mathcast

A correspondent writes:

Greetings,

I think that in the long run both strategies are equivalent. This game doesn’t favor any player.

Demonstration

Chaim Expected Gains = 3 * 1/4 + 1 * 1/4 = 1

Kyle Expected Gains = 1/4 * 2 + 1/4 * 2 = 1

This is so if both of us pick H half of the time, and pick T half of the time.

But!

If I know Kyle is going to pick H half of the time and T half of the time, I should adjust my strategy. I can do better by always picking H; the payout would then be

C: 3*1/2 = 3/2

K: 2*1/2 = 1

Net 1/2 in my favor!!

Conversely, if I am picking H half of the time and T half of the time, Kyle should adjust his strategy and choose T all of the time; this comes out to

C: 1*1/2 = 1/2

K: 2*1/2 = 1

Net 1/2 in Kyle’s favor– rats!

John von Neumann’s celebrated result is that both players have an * optimal * strategy, one that *cannot * be exploited by the other player. If we both play optimally, is the game balanced?

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May 6, 2007
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answers, Follow Up, infinity, logic, paradoxes, The Mathcast

Amusingly, this problem has exactly the same solution as the proof that there are as many rational numbers as there are counting numbers. And the proof generalizes: one stork can catch three frogs, or ten or fifty.

Here are some bonus problems:

- The stork can catch the frog even if it can start at any
*rational* number and hop any fixed * rational* distance each step.
- However, if the frog can start at
* any real number* or hop any real distance, the stork has no strategy that guarantees a catch. This is, in effect, the same as proving that the real numbers are not countable.

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