User:Michael Retriever/Binomial theorem generalization

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This is an explanation of how to generalize the binomial theorem and binomial coefficients to any real non-natural number.


The binomial theorem where n is a natural number is


(x+y)^n=\sum_{k=0}^n{n \choose k}x^{(n-k)}y^k\,


where the definition of the binomial coefficient used is


{n \choose k} = \frac{n!}{k!(n-k)!}\, as long as k ≤ n


First, let's observe the results of the binomial coefficients for n=5 and k={0,1,2,3,4,5}


{5 \choose 0} = \frac{1\cdot 2\cdot 3\cdot 4\cdot 5}{(1)(2\cdot 3\cdot 4\cdot 5)} = \frac{1}{1} = \frac{1}{0!}\,

{5 \choose 1} = \frac{1\cdot 2\cdot 3\cdot 4\cdot 5}{(1)(1\cdot 2\cdot 3\cdot 4)} = \frac{1}{1}\,(5) = \frac{1}{1!} (5)\,

{5 \choose 2} = \frac{1\cdot 2\cdot 3\cdot 4\cdot 5}{(1\cdot 2)(1\cdot 2\cdot 3)} = \frac{1}{1\cdot 2}\,(5\cdot 4) = \frac{1}{2!} (5\cdot 4)\,

{5 \choose 3} = \frac{1\cdot 2\cdot 3\cdot 4\cdot 5}{(1\cdot 2\cdot 3)(1\cdot 2)} = \frac{1}{1\cdot 2\cdot 3}\,(5\cdot 4\cdot 3) = \frac{1}{3!} (5\cdot 4\cdot 3)\,

{5 \choose 4} = \frac{1\cdot 2\cdot 3\cdot 4\cdot 5}{(1\cdot 2\cdot 3\cdot 4)(1)} = \frac{1}{1\cdot 2\cdot 3\cdot 4}\,(5\cdot 4\cdot 3\cdot 2) = \frac{1}{4!} (5\cdot 4\cdot 3\cdot 2)\,

{5 \choose 5} = \frac{1\cdot 2\cdot 3\cdot 4\cdot 5}{(2\cdot 3\cdot 4\cdot 5)(1)} = \frac{1}{1\cdot 2\cdot 3\cdot 4\cdot 5}\,(5\cdot 4\cdot 3\cdot 2\cdot 1) = \frac{1}{5!} (5\cdot 4\cdot 3\cdot 2\cdot 1)\,


We can see that the result of the binomial coefficients can also be expressed as


{n \choose k} = \frac{1}{k!} \prod_{m=0}^{k-1}{(n - m)}\,


The advantatge of this expression is that it doesn't operate n as a factorial, and therefore n can be either a natural number or a real non-natural number. Since n can be any real number in our expression, we will write the expression from now on as


{r \choose k} = \frac{1}{k!} \prod_{m=0}^{k-1}{(r - m)}\,


This has a direct implication within the binomial theorem, as it makes the formula work with any real number in the exponent.


(x+y)^r=\sum_{k=0}^r{r \choose k}x^{(r-k)}y^k\,


The spicey bit here is that, in the summatory, if r isn't a natural number, k will never reach the value r. That is why the true form of the above expression is


(x+y)^r=\sum_{k=0}^\infty{r \choose k}x^{(r-k)}y^k\,


As an example, we can test this formula with (5+4)^{1/2}\,


(5+4)^{1/2} = \sqrt{9} = 3\,


(5+4)^{1/2} = \sum_{k=0}^\infty{0.5 \choose k}\,5^{(0.5-k)}\,4^k = \,


= {0.5 \choose 0}\,5^{(0.5)} + {0.5 \choose 1}\,5^{(-0.5)}\,4 + {0.5 \choose 2}\,5^{(-1.5)}\,4^2 + {0.5 \choose 3}\,5^{(-2.5)}\,4^3\ + \dots =\,


= 2.2360679774998\,+\,0.8944271909999\,-\,0.1788854382000\,+\,0.0715541752800\,-\,\cdots = 3


The binomial coefficient where r is a real non-natural number is very useful in infinite expansions and formulas, such as the ones used for the perimeter of an ellipse.

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