What is the newton's generalised binomial theorem?

Answer 1

It's better to think about the ordinary binomial theorem first.

Consider a binomial (x + y), and raising it to a power, say squaring it.

(x + y)^2 = (x + y)(x + y) = x^2 + 2xy + y^2

Now try cubing it.

(x + y)^3 = (x + y)(x + y)(x + y) = x^3 + 3x^2 y + 3xy^2 + y^3

It becomes very tedious to do this. The binomial theorem allows us to expand binomial expressions to a power very quickly.

The generalised binomial theorem is, as it says, 'generalised' - the 'original' binomial theorem only allows us to expand binomial expressions to a power which is a whole number (0, 1, 2, 3 ... etc) but not numbers such as 1/2, 1/3 or -1.

Newton's generalised binomial theorem allows us to expand binomial expressions for any _rational_ power. (that is any number which can be expressed as a ratio of two integers - not something horrible like the cube root of three)

So now we can expand things like (x + y)^0.5, (1 - x)^-1 and all that malarky - this has some fairly deep significances, such as allowing numerical approximations of surds and bears relevance to some power series.

For example, take (1 - x)^-4, using Newton's generalised binomial theorem it can be seen that

(1 - x)^-4 = 1 + 4x + 10x^2 + 20x^3 ...

Each expansion for a rational exponent of the binomial expressions creates an infinite series.

The actual calculations are best left to a site which can show you the mathematical notation, but if you can do the normal binomial theorem - the nuances of this one will be easy to grap.

Answer 2

Newton's generalized binomial theorem states that for any real number x and any real number r, the binomial expansion of (1 + x)^r converges if |x| < 1. The formula for the expansion is given by (1 + x)^r = 1 + rx + r(r-1)x^2/2! + r(r-1)(r-2)x^3/3! + ... + r(r-1)(r-2)...(r-n+1)x^n/n! for non-negative integer n.