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

5 Answers
Apr 25, 2017

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.

Apr 25, 2017

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.

Apr 29, 2017

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)

Apr 29, 2017

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.

Apr 29, 2017

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