What is the physical interpretation of this formula?

[Maurice]

If the following formula is expanded for an odd value of k, the formula is, arithmetically, obviously true, but not so obviously true when k is even.

for all positive integers k, and kCj represents or C(k, j)

If kCj means the number of arrangements of k distinguishable objects taken j at a time, when order does not matter, then physically, what does this formula actually mean, and why physically should it have a value of 1?

 

[HallsofIvy]

First, I think "physical" is the wrong word- that is not quite what you mean. A mathematical formula does NOT HAVE a single "physical interpretation" and there is NO "physical" reason a formula should or should not be true.

Now, this may be what you are thinking of: If we take the nth power of a+ b, then each term will have "i" copies of a and "j" copies of b where i+ j= n- that is, \(\displaystyle a^ib^j\). The coefficien of \(\displaystyle a^ib^j\) will be the number of different ways we can order "i" copies of a and "j" copies of b. For example \(\displaystyle (a+ b)^2= a(a+ b)+ b(a+ b)= aa+ ab+ ba+ bb= a^2+ 2ab+ b^2\) and \(\displaystyle (a+ b)^3= (a+ b)(aa+ ab+ba+ bb)= a(aa+ ab+ ba+ bb)+ b(aa+ ab+ ba+ bb)= aaa+ aab+ aba+ abb+ baa+ bab+ bba+ bbb= aaa+ (aab+aba+ baa)+ (abb+ bab+ bba)+ bbb= a^3+ 3a^2b+ 3ab^2+ b^3\). That is as close to a "physical" reason for the coefficients as you will have.

So we have \(\displaystyle (a+ b)^n= \sum_{j=0}^n \begin{pmatrix} n \\ i \end{pmatrix} a^jb^{n-j}\) (the "binomial formula") and, in particular, taking a= -1, b= 1, \(\displaystyle (-1+ 1)^n= \sum_{j=0}^n \begin{pmatrix} n \\ j \end{pmatrix} (-1)^j(1)^{n-j}= \sum_{j=0}^n \begin{pmatrix} n \\ j\end{pmatrix}(-1)^j\).

But \(\displaystyle (-1+ 1)^n= 0^n= 0\) so \(\displaystyle \sum_{j=0}^n \begin{pmatrix} n \\ j\end{pmatrix}(-1)^j= 0\).

Your sum is not from 0 to n, it is from 1 to n so is missing the "0th" term which is \(\displaystyle (-1)^n\). What that is depends upon whether n is even or odd.

For n odd, \(\displaystyle (-1)^n= -1\) so \(\displaystyle \sum_{j=0}^n \begin{pmatrix} n \\ j\end{pmatrix}(-1)^j= -1+ \sum_{j=1}^n \begin{pmatrix} n \\ j\end{pmatrix}(-1)^j= 0\). Adding 1 to both sides, \(\displaystyle \sum_{j=1}^n \begin{pmatrix} n \\ j\end{pmatrix}(-1)^j= 1\).

For n even, \(\displaystyle (-1)^n= 1\) so \(\displaystyle \sum_{j=0}^n \begin{pmatrix} n \\ j\end{pmatrix}(-1)^j= 1+ \sum_{j=1}^n \begin{pmatrix} n \\ j\end{pmatrix}(-1)^j= 0\). Subtracting 1 from both sides, \(\displaystyle \sum_{j=1}^n \begin{pmatrix} n \\ j\end{pmatrix}(-1)^j= -1\).

 

[Maurice]

Thank you very much, for your reply. I am totally amazed by what you have done in such a short time.

I think I get your message. I was hoping for an insight in terms of arrangements, but perhaps I am a little naive. I am not a mathematician, I come from a physics background and like to try to get a physical feel for equations, but you are correct in that this is rarely possible.

I am a little confused by your calculation though, has the power of -1 as ( j+1), and I cannot see values for a and b that will generate a power of ( j+1). In the above equation, the value is independent of whether k is odd or even. Could you please explain what am I missing?

Many thanks.

 

[Dr.Peterson]

I am a little confused by your calculation though, View attachment 22416has the power of -1 as ( j+1), and I cannot see values for a and b that will generate a power of ( j+1). In the above equation, the value is independent of whether k is odd or even. Could you please explain what am I missing?

Just multiply both sides of Halls' final equation by -1, which will distribute over all the terms.