What fun, useful, mathematical fact do you know that is not normally taught at school?
5 Answers
How to evaluate "towers of exponents", such as
Explanation:
In order to evaluate these "towers", we start at the top and work our way down.
So:
On a similar, but slightly unrelated note, I also know how to work out the last digits of
So if you want to find the last digit of
If
#sqrt(n) = a+b/(2a+b/(2a+b/(2a+b/(2a+b/(2a+...)))))#
where
Explanation:
Suppose we want to find the square root of some number
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
#sqrt(n)-a=b/(a+sqrt(n))#
Multiply both sides by
#b = (sqrt(n)-a)(sqrt(n)+a) = n-a^2#
So if
For example, if we have
#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
So this is not converging particularly quickly.
Alternatively, we might put
#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.
You can find approximations to square roots using a recursively defined sequence.
Explanation:
The method
Given a positive integer
-
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
Example
Let
Then
Then
So our sequence starts:
#1, 4, 19, 88, 409, 1900, 8827, 41008,...#
In theory the ratio between consecutive terms should tend towards
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
How it works
Suppose we have a sequence defined by given values of
#a_(n+2) = 2p a_(n+1) + q a_n#
for some constants
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
Next solve:
#{ (Ax_1+Bx_2 = a_1), (Ax_1^2+Bx_2^2 = a_2) :}#
for
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
#a_n = Ax_1^n+Bx_2^n#
If
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
For example, usually, if you were to solve
Evaluating squares with summations
Explanation:
Normally, you should know squares such as
I realized that after a while, squares are just sums of odd numbers.
What I mean is this:
So
That will give you:
This, in fact, is
Since the numbers are always incrementing by
So for
So I can just do
It's not really practical but it's interesting to know.
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
This reduces to finding what sums of consecutive odd integers add up to
#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#