User talk:Calogero

Statement of Theorem
Let $\N$ be the natural numbers.

Then: $\forall n, r \in \N_{> 0}$

$n^r$ can be expressed as the sum of $n$ consecutive odd numbers, the first of which is $n^{r-1} - n + 1$

Proof
$n^r =n*n^{r-1}$

$=\sum_{i=0}^{n-1} n^{r-1}$

$=\sum_{i=0}^{n-1} n^{r-1} - n - 1 + n + 1$

$=\sum_{i=0}^{n-1} n^{r-1} - n + 1 + n - 1$

$=[\sum_{i=0}^{n-1} n^{r-1} - n + 1] + \sum_{i=0}^{n-1} n-1$

$=[\sum_{i=0}^{n-1} n^{r-1} - n + 1] + n(n-1)$

$=[\sum_{i=0}^{n-1} n^{r-1} - n + 1] + 2\frac{n(n-1)}{2}$

$=[\sum_{i=0}^{n-1} n^{r-1} - n + 1] + 2\sum_{i=0}^{n-1} i$

$=[\sum_{i=0}^{n-1} n^{r-1} - n + 1] + \sum_{i=0}^{n-1} 2i$

$=\sum_{i=0}^{n-1} n^{r-1} - n + 1 + 2i$

Proof that the first term is always odd
n is either even or odd

if n is even, $n = 2m$, then $(2m)^{r-1}$ for any power $r$ is also even

since the difference of two even numbers is even, so $(2m)^{r-1} - 2m$ is even

hence, $(2m)^{r-1} - 2m + 1$ must be odd

alternately, if n is odd, $n = 2m+1$, then $(2m+1)^{r-1}$ is also odd

since the difference of two odd numbers is even, $(2m+1)^{r-1} - (2m+1)$ is even

and again, $(2m+1)^{r-1} - (2m+1) + 1$ must be odd

Background
For squares, $r=2$, the formula reduces to the sum of the first $n$ odd numbers, a fact which I had known since my youth.

But then I was reading a history of Mathematics, in which I learned that Nichomachus of Gerasa observed that cubical numbers are the sum of successive odd numbers. The example given in the book was $2^3 = 8 = 3+5$

$3^3 = 27 = 7 + 9 + 11$

$4^3 = 64 = 13 + 15 + 17 + 19$

Noticing that the value for $n^3$ was the sum of $n$ consecutive odd numbers, I was curious to see if this worked for $n^4$. It was in playing with fourth powers that I noticed the pattern, generalized it, and then derived the proof.

Statement of Theorem
With Continuous scoring, at each step, the benefit of achieving the next question and the cost of failing it are greater the lower one's score is and lesser the higher one's score is. These two tendencies meet at 50%, thus promoting mediocrity. Finally, the impact of each step decreases with the square of the number of questions to that point.

Preamble
A continuous score is the score achieved by someone attempting a process that awards a point for each question achieved. It is called continuous, because the score is displayed at each step of the process.

Mediocrity is defined as a 50% score.

We define the following variables:

$q$ the number of questions attempted to the given point in the process

$a$ the number of questions achieved to that point

$f$ the number of questions failed to that point

$s$ the continuous score to that point

Hence $q=a+f$ and $s=\frac{a}{q}$

and so $f=q-a$

Proof
We first consider the benefit of achieving the next question.

In that case, the score increases from $\frac{a}{q}$ to $\frac{a+1}{q+1}$

with a resulting benefit of $\frac{a+1}{q+1} - \frac{a}{q}$

$=\frac{q(a+1)}{q(q+1)} - \frac{(q+1)a}{q(q+1)}$

$=\frac{q(a+1)-(q+1)a}{q(q+1)}$

$=\frac{qa+q -(qa+a)}{q(q+1)}$

$=\frac{qa+q-qa-a}{q(q+1)}$

$=\frac{q-a}{q(q+1)}$

$=\frac{f}{q(q+1)}$

We next consider the cost of failing the next question.

In that case, the score decreases from $\frac{a}{q}$ to $\frac{a}{q+1}$

with a resulting cost of $\frac{a}{q+1} - \frac{a}{q}$

$=\frac{qa}{q(q+1)} - \frac{(q+1)a}{q(q+1)}$

$=\frac{qa -(q+1)a}{q(q+1)}$

$=\frac{qa -(qa+a)}{q(q+1)}$

$=\frac{qa-qa-a}{q(q+1)}$

$=\frac{-a}{q(q+1)}$

In summary, the benefit of achieving the next question is $=\frac{f}{q(q+1)}$

and the cost of failing it is $=\frac{-a}{q(q+1)}$

Now the greater one's score is the lower is their value of $f$ and hence the lesser is the benefit in achieving the next question. As well, the higher is their value of $a$, and thus the greater is the cost of their failing the next question.

Conversely, the lesser one's score is the greater is their value of $f$ and hence the greater is the benefit in achieving the next question. As well, the lower is their value of $a$, and thus the lesser is the cost of their failing the next question.

These two tendencies meet when $a=f$ or when the score is $50\%$.

Finally, we note that the denominator of both the benefit and cost at each step is more the square of the number of questions, viz. $q(q+1) = q^2+q$. Thus at each step, there is a diminishing impact on the score with more than the square of the number of questions to that point.

The General, Weighted Scoring Case
It might be argued that there exists a weighted scoring scheme that would overcome these difficulties, but such is not the case.

Proof of General, Weighted Scoring Case
To deal with the more general case, we define the following:

$w_i$ the points assigned at the $i^{th}$ question

so, $\sum_{i=1}^{q} w_i$ is the total points up to and including the $i^{th}$ question

$a(i) \in \{0,1\}$ whether or not the player achieved the $i^{th}$ question

$f(i) \in \{0,1\}$ whether or not the player failed the $i^{th}$ question

note that $f(i) = 1 - a(i)$

so now the score at question q becomes

$s=\frac{\sum_{i=1}^{q} w_ia(i)}{\sum_{i=1}^{q} w_i}$

As before, we first consider the benefit of achieving the next question.

In that case, the score increases from

$\frac{\sum_{i=1}^{q} w_ia(i)}{\sum_{i=1}^{q} w_i}$

to

$\frac{\sum_{i=1}^{q} w_ia(i) + w_{q+1}}{\sum_{i=1}^{q+1} w_i}$

with a resulting benefit of

$\frac{\sum_{i=1}^{q} w_ia(i) + w_{q+1}}{\sum_{i=1}^{q+1} w_i} - \frac{\sum_{i=1}^{q} w_ia(i)}{\sum_{i=1}^{q} w_i}$

$=\frac{[\sum_{i=1}^{q} w_ia(i) + w_{q+1}] * \sum_{i=1}^{q} w_i -\sum_{i=1}^{q} w_ia(i) * \sum_{i=1}^{q+1} w_i}{\sum_{i=1}^{q+1} w_i * \sum_{i=1}^{q} w_i}$

$=\frac{\sum_{i=1}^{q} w_ia(i)* \sum_{i=1}^{q} w_i + w_{q+1} * \sum_{i=1}^{q} w_i -\sum_{i=1}^{q} w_ia(i) * [\sum_{i=1}^{q} w_i + w_{q+1}]}{\sum_{i=1}^{q+1} w_i * \sum_{i=1}^{q} w_i}$

$=\frac{\sum_{i=1}^{q} w_ia(i)* \sum_{i=1}^{q} w_i + w_{q+1} * \sum_{i=1}^{q} w_i -\sum_{i=1}^{q} w_ia(i) * \sum_{i=1}^{q} w_i - \sum_{i=1}^{q} w_ia(i) * w_{q+1}}{\sum_{i=1}^{q+1} w_i * \sum_{i=1}^{q} w_i}$

$=\frac{w_{q+1} * \sum_{i=1}^{q} w_i - \sum_{i=1}^{q} w_ia(i) * w_{q+1}}{\sum_{i=1}^{q+1} w_i * \sum_{i=1}^{q} w_i}$

$=\frac{w_{q+1} * [\sum_{i=1}^{q} w_i - \sum_{i=1}^{q} w_ia(i) ]}{\sum_{i=1}^{q+1} w_i * \sum_{i=1}^{q} w_i}$

$=\frac{w_{q+1} * \sum_{i=1}^{q} [w_i - w_ia(i)] }{\sum_{i=1}^{q+1} w_i * \sum_{i=1}^{q} w_i}$

$=\frac{w_{q+1} * \sum_{i=1}^{q} w_i(1 - a(i)) }{\sum_{i=1}^{q+1} w_i * \sum_{i=1}^{q} w_i}$

$=\frac{w_{q+1} * \sum_{i=1}^{q} w_if(i) }{\sum_{i=1}^{q+1} w_i * \sum_{i=1}^{q} w_i}$

We next consider the cost of failing the next question.

In that case, the score decreases from

$\frac{\sum_{i=1}^{q} w_ia(i)}{\sum_{i=1}^{q} w_i}$

to

$\frac{\sum_{i=1}^{q} w_ia(i)}{\sum_{i=1}^{q+1} w_i}$

with a resulting cost of

$\frac{\sum_{i=1}^{q} w_ia(i)}{\sum_{i=1}^{q+1} w_i} - \frac{\sum_{i=1}^{q} w_ia(i)}{\sum_{i=1}^{q} w_i}$

$=\frac{\sum_{i=1}^{q} w_ia(i) * \sum_{i=1}^{q} w_i -\sum_{i=1}^{q} w_ia(i) * \sum_{i=1}^{q+1} w_i}{\sum_{i=1}^{q+1} w_i * \sum_{i=1}^{q} w_i}$

$=\frac{\sum_{i=1}^{q} w_ia(i) * \sum_{i=1}^{q} w_i -\sum_{i=1}^{q} w_ia(i) * [\sum_{i=1}^{q} w_i+ w_{q+1}]}{\sum_{i=1}^{q+1} w_i * \sum_{i=1}^{q} w_i}$

$=\frac{\sum_{i=1}^{q} w_ia(i) * \sum_{i=1}^{q} w_i -\sum_{i=1}^{q} w_ia(i) * \sum_{i=1}^{q} w_i - \sum_{i=1}^{q} w_ia(i) *w_{q+1}]}{\sum_{i=1}^{q+1} w_i * \sum_{i=1}^{q} w_i}$

$=\frac{-\sum_{i=1}^{q} w_ia(i) *w_{q+1}}{\sum_{i=1}^{q+1} w_i * \sum_{i=1}^{q} w_i}$

$=\frac{-w_{q+1}*\sum_{i=1}^{q} w_ia(i) }{\sum_{i=1}^{q+1} w_i * \sum_{i=1}^{q} w_i}$

In summary, the benefit of achieving the next question is

$=\frac{w_{q+1} * \sum_{i=1}^{q} w_if(i) }{\sum_{i=1}^{q+1} w_i * \sum_{i=1}^{q} w_i}$

and the cost of failing it is

$=\frac{-w_{q+1}*\sum_{i=1}^{q} w_ia(i) }{\sum_{i=1}^{q+1} w_i * \sum_{i=1}^{q} w_i}$

And, as in the special case where $w_i = 1, \forall i \in \{1,...,q\}$, for whatever values of these $w_i$, the benefit is a function of $f(i)$ and the cost a function of $a(i)$. Thus the same considerations hold in this general case.

Finally, we note that the denominator of both the benefit and cost at each step is the square of the total number points up to that question, viz.

$\sum_{i=1}^{q+1} w_i * \sum_{i=1}^{q} w_i$

$=[\sum_{i=1}^{q} w_i + w_{q+1}] * \sum_{i=1}^{q} w_i$

$=[\sum_{i=1}^{q} w_i]^2 + w_{q+1} * \sum_{i=1}^{q} w_i$

And again, at each step, there is a diminishing impact on the score with more than the square of the total number of points to that questions.

Statement of Theorem
$\forall n, m \in \N_{> 0}, n | \prod_{i=0}^{n-1} m+i$

Proof
There are three cases of products of consecutive integers:

1. The product consists of only positive integers

2. The product consists of only negative integers

3. The product includes 0

In Case 1, we have a set of n consecutive numbers being multiplied.

That is, $m+i for i=0,...,n-1$

We define $R_n(k)$ as the remainder when dividing $k$ by $n$.

$\forall k > 0 \in \N, R_n(k) \in {0,...,n-1}$

Since $R_n(k+1) = R_n(k) + 1$

then among all the product factors, one of them must have a remainder of 0, and hence be divisible by $n$.

Since this is a factor of the product, the product itself is divisible by $n$.

Case 2 reduces to Case 1,

with only a question of whether the product is positive or negative,

which has no bearing on the product being divisible by n.

In Case 3, the product is always 0, and any number divides 0

Consequent Formulae
1. $4 | n(n^3-2n^2 -n+2)$

$ (n-2)(n-1)n(n+1)$

$= n(n-1)(n+1)(n-2)$

$= n(n^2-1)(n-2)$

$= n(n^3-2n^2 -1n+2)$

2. $5 | n(n^4-5n^2+4)$

$ (n-2)(n-1)n(n+1)(n+2)$

$ = n(n-2)(n+2)(n-1)(n+1)$

$ = n(n^2-4)(n^2-1)$

$ = n(n^4-n^2-4n^2+4)$