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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User:Michael Retriever/Binomial theorem generalization

From Wikipedia, the free encyclopedia

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