Question

What fun, useful, mathematical fact do you know that is not normally taught at school?

Answer 1

How to evaluate "towers of exponents", such as `2^(2^(2^2))`, and how to work out the last digit of `2^n,` `ninNN`.

Explanation:

In order to evaluate these "towers", we start at the top and work our way down.

So:

`2^(2^(2^2))=2^(2^4)=2^16=65,536`

On a similar, but slightly unrelated note, I also know how to work out the last digits of `2` raised to any natural exponent. The last digit of `2` raised to something always cycles between four values: `2,4,8,6`.

`2^1=2,` `2^2=4,` `2^3=8,` `2^4=16`

`2^5=32,` `2^6=64,` `2^7=128,` `2^8=256`

So if you want to find the last digit of `2^n`, find which place it is in the cycle, and you'll know its last digit.

Answer 2

If `n > 0` and `a` is an approximation to `sqrt(n)`, then:

`sqrt(n) = a+b/(2a+b/(2a+b/(2a+b/(2a+b/(2a+...)))))`

where `b = n-a^2`

Explanation:

Suppose we want to find the square root of some number `n > 0`.

Further we would like the result to be some kind of continued fraction that repeats at each step.

Try:

`sqrt(n) = a+b/(2a+b/(2a+b/(2a+b/(2a+b/(2a+...)))))`

`color(white)(sqrt(n)) = a+b/(a+a+b/(2a+b/(2a+b/(2a+b/(2a+...)))))`

`color(white)(sqrt(n)) = a+b/(a+sqrt(n))`

Subtract `a` from both ends to get:

`sqrt(n)-a=b/(a+sqrt(n))`

Multiply both sides by `sqrt(n)+a` to get:

`b = (sqrt(n)-a)(sqrt(n)+a) = n-a^2`

So if `a^2` is a little less than `n`, then `b` will be small and the continued fraction will converge quicker.

For example, if we have `n=28` and choose `a=5`, then we get:

`b = n-a^2 = 28-5^2 = 28-25=3`

So:

`sqrt(28) = 5+3/(10+3/(10+3/(10+3/(10+3/(10+...)))))`

which gives us approximations:

`sqrt(28) ~~ 5+3/10 = 5.3`

`sqrt(28) ~~ 5+3/(10+3/10) = 545/103 ~~ 5.29126`

`sqrt(28) ~~ 5+3/(10+3/(10+3/10)) = 5609/1060 ~~ 5.2915094`

A calculator tells me `sqrt(28) ~~ 5.291502622`

So this is not converging particularly quickly.

Alternatively, we might put `n=28` and `a=127/24` to find:

`b = n-a^2 = 28-127^2/24^2 = 28-16129/576 = (16128-16129)/576 = -1/576`

So:

`sqrt(28) = 127/24-(1/576)/(127/12-(1/576)/(127/12-(1/576)/(127/12-...)))`

giving us approximations:

`sqrt(28) ~~ 127/24 = 5.291bar(6)`

`sqrt(28) ~~ 127/24-(1/576)/(127/12) = 32257/6096 ~~ 5.29150262467`

That's converging a lot faster.

Answer 3

You can find approximations to square roots using a recursively defined sequence.

Explanation:

`color(white)()`
The method

Given a positive integer `n` which is not a perfect square:

• Let `p = floor(sqrt(n))` be the largest positive integer whose square does not exceed `n`.

• Let `q = n-p^2`

• Define a sequence of integers by:

  `{(a_1 = 1), (a_2 = 2p), (a_(i+2) = 2pa_(i+1)+qa_i" for " i >= 1) :}`

Then the ratio between successive terms of the sequence will tend towards `p+sqrt(n)`

`color(white)()`
Example

Let `n=7`.

Then `p = floor(sqrt(7)) = 2`, since `2^2=4 < 7` but `3^2 = 9 > 7`.

Then `q=n-p^2 = 7-2^2 = 3`

So our sequence starts:

`1, 4, 19, 88, 409, 1900, 8827, 41008,...`

In theory the ratio between consecutive terms should tend towards `2+sqrt(7)`

Let's see:

`4/1 = 4`

`19/4 = 4.75`

`88/19 ~~ 4.63`

`409/88 ~~ 4.6477`

`1900/409 ~~ 4.6455`

`8827/1900 ~~ 4.645789`

`41008/8827 ~~ 4.645746`

Note that `2+sqrt(7) ~~ 4.645751311`

`color(white)()`
How it works

Suppose we have a sequence defined by given values of `a_1, a_2` and a rule:

`a_(n+2) = 2p a_(n+1) + q a_n`

for some constants `p` and `q`.

Consider the equation:

`x^2-2px-q = 0`

The roots of this equation are:

`x_1 = p+sqrt(p^2+q)`

`x_2 = p-sqrt(p^2+q)`

Then any sequence with general term `Ax_1^n+Bx_2^n` will satisfy the recurrence rule we specified.

Next solve:

`{ (Ax_1+Bx_2 = a_1), (Ax_1^2+Bx_2^2 = a_2) :}`

for `A` and `B`.

We find:

`a_1x_2-a_2 = Ax_1(x_2-x_1)`

`a_1x_1-a_2 = Bx_2(x_1-x_2)`

and hence:

`A=(a_1x_2-a_2)/(x_1(x_2-x_1))`

`B=(a_1x_1-a_2)/(x_2(x_1-x_2))`

So with these values of `x_1, x_2, A, B` we have:

`a_n = Ax_1^n+Bx_2^n`

If `q < 3p^2` then `abs(x_2) < 1` and the ratio between successive terms will tend towards `x_1 = p+sqrt(p^2+q)`

Answer 4

Modular division

Explanation:

Modular division is just the same as division except the answer is the remainder instead of the actual value. Rather than the `-:` symbol, you use the `%` symbol.

For example, usually, if you were to solve `16-:5` you would get `3` remainder `1` or `3.2`. However, using modular division, `16%5=1`.

Answer 5

Evaluating squares with summations

Explanation:

Normally, you should know squares such as `5^2=25`. However, when numbers get bigger such as `25^2`, it gets harder to know off the top of your head.

I realized that after a while, squares are just sums of odd numbers.

What I mean is this:

`sum_(n=0)^k 2n+1` where `k` is the base value minus `1`

So `5^2` could be written as:

`sum_(n=0)^4 2n+1`

That will give you:

`1+3+5+7+9`

This, in fact, is `25`.

Since the numbers are always incrementing by `2`, I could then add the first and last number and then multiply by `k/2`.

So for `25^2`

`sum_(n=0)^24 2n+1=1+3+...+49`

So I can just do `(49+1)(25/2)` and get `25^2` which is `625`.

It's not really practical but it's interesting to know.

`color(white)()`
Bonus

Knowing that:

`n^2 = overbrace(1+3+5+...+(2n-1))^"n terms" = ((1+(2n-1))/2)^2`

allows us to solve some problems about differences of squares.

For example, what are all the solutions in positive integers `m, n` of `m^2-n^2 = 40` ?

This reduces to finding what sums of consecutive odd integers add up to `40`...

`40 = overbrace(19+21)^"average 20"`

`color(white)(40) = (1+3+...+21)-(1+3+...+17)`

`color(white)(40) = ((1+21)/2)^2+((1+17)/2)^2`

`color(white)(40) = 11^2-9^2`

`40 = overbrace(7+9+11+13)^"average 10"`

`color(white)(40) = (1+3+...+13)-(1+3+5)`

`color(white)(40) = ((1+13)/2)^2-((1+5)/2)^2`

`color(white)(40) = 7^2-3^2`