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December 9th, 2017, 04:56 PM   #1
jks
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Math Focus: Electrical Engineering Applications
Two Properties of Pascal's Triangle?

I was wondering about two properties of Pascal's Triangle (PT) that appear to hold true (as far as I have checked), and if they are indeed known properties. The properties involve power sums to 0 and power sums to the factorial of a row number, using the values of PT as coefficients. Unfortunately, Googling with "Pascal's Triangle", "sum", and "factorial" leads to a lot of noise to try to filter through, due to the nature of the terms of PT.

So I thought I would ping the forum to check for any existing knowledge.

The reason that I am asking is related to this thread as I will explain in another thread that I will start involving the solution to these types of problems. (I will also post another possible way of obtaining the solution in the referenced thread).

For reference, below is a portion of the left justified PT:

$\displaystyle \large {\begin{array}{c c}
& k & 1 & 2 & 3 & 4 & 5 & 6 \\
r \\
0 & & 1 \\
1 & & 1 & 1 \\
2 & & 1 & 2 & 1 \\
3 & & 1 & 3 & 3 & 1 \\
4 & & 1 & 4 & 6 & 4 & 1 \\
5 & & 1 & 5 & 10 & 10 & 5 & 1
\end{array}}$


The first property is:

For a given row $r$ and for each $k^{th}$ term in $r$ as given above, for any number $n$ (although I am only interested in $n \geq r$):

$\displaystyle \large PT_{r,1} \cdot n^r - PT_{r,2} \cdot (n-1)^r + PT_{r,3} \cdot (n-2)^r ... PT_{r,r+1} \cdot (n-r)^r = r!$ (alternating signs for each term).

For example, with $n=5$ and $r=4$:

$\displaystyle \large 1 \cdot 5^4 - 4 \cdot 4^4 + 6 \cdot 3^4 - 4 \cdot 2^4 + 1 \cdot 1^4 = 4! = 24 $

and with $n=6$ and $r=4$:

$\displaystyle \large 1 \cdot 6^4 - 4 \cdot 5^4 + 6 \cdot 4^4 - 4 \cdot 3^4 + 1 \cdot 2^4 = 4! = 24 $

so:

$\displaystyle \large 1 \cdot n^4 - 4 \cdot (n-1)^4 + 6 \cdot (n-2)^4 - 4 \cdot (n-3)^4 + 1 \cdot (n-4)^4 = 4! = 24 $

which can be shown to be true after expanding and simplifying.

This property is true for all of the values that I have checked numerically so far (up to $r=100$ for $n=r+1$ and for various other values for $r,n$). I can see why it might be true since PT is instrumental in calculating $(n-a)^r$, but I really don't have the skills to prove it analytically.


The second property is:

For a given row $r \gt 0$ and for each $k^{th}$ term in $r$ as given above, for any number $n$, although I am only interested in $n \geq r$ (and $r-q \geq 2$):

$\displaystyle \large PT_{r,1} \cdot n^{(r-q)} - PT_{r,2} \cdot (n-1)^{(r-q)} + PT_{r,3} \cdot (n-2)^{(r-q)} ... PT_{r,r+1} \cdot (n-r)^{(r-q)} = 0, \quad 1 \leq q \leq r$ (alternating signs for each term).

For example, with $n=5$, $r=4$, $q=1$:

$\displaystyle \large 1 \cdot 5^3 - 4 \cdot 4^3 + 6 \cdot 3^3 - 4 \cdot 2^3 + 1 \cdot 1^3 = 0$

and with $n=5$, $r=4$, $q=2$:

$\displaystyle \large 1 \cdot 5^2 - 4 \cdot 4^2 + 6 \cdot 3^2 - 4 \cdot 2^2 + 1 \cdot 1^2 = 0$

and with $n=7$, $r=5$, $q=3$:

$\displaystyle \large 1 \cdot 7^2 - 5 \cdot 6^2 + 10 \cdot 5^2 - 10 \cdot 4^2 + 5 \cdot 3^2 - 1 \cdot 2^2 = 0$

Again, I do not possess the skills to prove/disprove for all $n, r, q$.

Obviously, comments are welcome.
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December 9th, 2017, 09:09 PM   #2
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I am rather tired and am having some difficulty understanding your notation.

Using standard notation, the kth term of the nth row is

$\dbinom{n}{k} \equiv \dfrac{n!}{k! * (n - k)!} \text {, where } 0! = 1 \text { and } m! = m * (m - 1)! \text { if } m \in \mathbb Z \text { and } m > 0.$
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December 9th, 2017, 09:43 PM   #3
jks
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Math Focus: Electrical Engineering Applications
Oops, I was so caught up in my application with terms starting at 1 that I did not follow convention by starting $k$ at $0$. Plus, I used $r$ for $n$ (but at least it starts at 0). I will have to ask readers to bear with my mistakes.

My application is hopefully explained here.

Last edited by jks; December 9th, 2017 at 09:58 PM. Reason: Add reference to the problem I am trying to solve.
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June 17th, 2018, 12:28 PM   #4
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Both of these appear to be true due to Euler's Finite Difference Theorem as can be seen here in Equation 10.1 on page 45.

As stated in the reference:

$\displaystyle
\sum_{k=0}^n (-1)^k \binom{n}{k} k^j =
\begin{cases}
0, & 0 \leq j < n \\
(-1)^n n!, & j=n
\end{cases}
$

Of course, we can make the factorials all positive (largest $k$ is positive) by multiplying each side by $(-1)^n$:

$\displaystyle
\sum_{k=0}^n (-1)^{n+k} \binom{n}{k} k^j =
\begin{cases}
0, & 0 \leq j < n \\
(-1)^{2n} n!, & j=n
\end{cases}
$

Since $(-1)^{2n}=1$, we get:

$\displaystyle
\sum_{k=0}^n (-1)^{n+k} \binom{n}{k} k^j =
\begin{cases}
0, & 0 \leq j < n \\
n!, & j=n
\end{cases}
$


and multiplying the left side by $(-1)^{-2k}=1$, for each $k$ in turn, we get:

$\displaystyle
\sum_{k=0}^n (-1)^{n-k} \binom{n}{k} k^j =
\begin{cases}
0, & 0 \leq j < n \\
n!, & j=n
\end{cases}
$

which is the form that I have recently been working with.
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