Polynomials, dice and pensions!!

OK a weird title for a blog. I'm typing this with a few loose thoughts in my head. OK to business, the journey begins...

let

p+q=1

This is quite simple, but I'll further restrain our space with both p & q are positive but less than one. Now as seen before in the previous blog, if there is a game or situation where someone has n attempts, if they see a certain situation they stop, if not they try again (persistence). For example you have 4 tries to throw a six, if you get it on the first go, then you stop, if not try again for up to 4 attempts. Looking at this situation q, represents the probability that you "go again", p the probability you "stop". All the permutations for n attempts are (as previously shown)

p + pq + pq² + pq³ + .. + pqⁱ +.. + pq^(n-1) + qⁿ == 1

or

p(1 + q + q² + q³ + .. + qⁱ +.. + q^(n-1) )+ qⁿ == 1

Well the above is fine, except that most people would not recognise it, rather gravitate to the expansion of (p+q)ⁿ == 1

On the face of it, there is absolutely nothing wrong with either formulae. So lets run with this, recognising that qⁿ will cancel from both formulae.

p(1 + q + q² + q³ + .. + qⁱ +.. + q^(n-1) ) + qⁿ == 1

pⁿ +np^(n-1)q .. ⁿC_i pⁱq^(n-i) .. + npq^(n-1) + qⁿ == 1

Well the qⁿ cancels, and we can divide through by p to get

1 + q + q² + q³ + .. + qⁱ +.. + q^(n-1) ==

p^n-1 +np^(n-2)q .. ⁿC_i p^(i-1)q^(n-i) .. + nq^(n-1)

The top formulae has a very well known simplification, namely (1-qⁿ)/(1-q) or (1-qⁿ)/p (as p+q=1). The formulae is thus

(1-qⁿ)/p = sum(i=0..n-1) ⁿC_i p^(i-1)q^(n-i)

Looking carefully at the above this is merely a re-write of (p+q)ⁿ .

This can also be broken down to look like the following..

(p+q)ⁿ + p(p+q)⁽ⁿ⁺¹⁾ + p²(p+q)^(n-2) + .. pⁱ(p+q)^(n-i) + .. pⁿ

So we are now back to our original formulae as p+q=1!!

However, lest assume p+q!=1. let us assume p+q=K instead, and use a similar method (that I stumbled upon by chance) to see where it leads. so

(p+q)ⁿ = kⁿ

and thus

sum(i=0..n) ⁿC_i pⁱq^(n-i) = kⁿ

rearranging

sum(i=0..n) ⁿC_i p^(i-1)q^(n-i) = (kⁿ-qⁿ)/p

and so we can transform the above into

(p+q)^n-1 + p(p+q)^(n-2) + p²(p+q)^(n-3) + .. pⁱ(p+q)^(n-i-1) + .. p^n-1 = (kⁿ-pⁿ)/(k-p)

finally

k^(n-1) + pk^(n-2) + p²k^(n-3) + .. p^(n-2)k + p^(n-1) = (kⁿ-pⁿ)/(k-p)

An alternate method..

So lets do something similar, let

S = aⁿ + a^(n-1)b + a^(n-2)b² + .. a^(n-i)bⁱ .. + bⁿ

Sa = aⁿ⁺¹ + aⁿb + a^(n-1)b² + .. a^(n-i+1)bⁱ .. + abⁿ

Sb = aⁿb+ a^(n-1)b² + a^(n-2)b³ + .. a^(n-i)bⁱ⁺¹ .. + bⁿ⁺¹

Sa - Sb = aⁿ⁺¹ - bⁿ⁺¹

and so

S = (aⁿ⁺¹ - bⁿ⁺¹)/(a-b)

This is identical to above formula (and far quicker to run through!!!)