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

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.

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.

#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)#

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#.

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#