Binomial theorem

The binomial theorem, also known as the binomial expansion, describes a method for expanding brackets containing two terms raised to a power as a polynomial. More specifically, it describes how to express a bracket of the form (x+a)ⁿ as a polynomial with terms b_kx^ka^n-k, for various coefficients b_k. When n is natural number, the binomial theorem produces a finite length polynomial, but can also be extended to non-integer n. For n not being a natural number, the binomial expansion of (x+a)ⁿ is:

$(x+a)^{n}=\sum _{k=0}^{n}{n \choose k}x^{k}a^{n-k}$

Here, $n \choose k$ is the binomial coefficient and pronounced "n choose k". It is derived from combinatorics and equal to:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {n \choose k} = \frac{n!}{(n-k)!k!} }

These coefficients for constant n form the rows of Pascal's triangle. The name comes from the Latin, bi-nomin, meaning two names (terms).

Proof

The binomial theorem for integer n can easily be proved using proof by induction. Supposing the theorem is true for n=N. Then:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x+a)^{N+1} = (x+a) \sum^N_{k=0} {N \choose k} x^k a^{N-k} }

Multiplying out the bracket, and substituting j=k+1 gives:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x+a)^{N+1} = \sum^N_{k=0} {N \choose k} x^k a^{N+1-k} + \sum^{N+1}_{j=1} {N \choose j-1} x^j a^{N+1-k} }

Separating the k=0 and j=N+1 terms, the sums can be combined as:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x+a)^{N+1} = a^{(N+1)} + \sum^N_{k=1} \left( {N \choose k}+{N \choose k-1} \right) x^k y^{(N+1)-k} + {N+1 \choose N+1}x^{N+1} }

Using a combinations identity, that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {N \choose k}+{N \choose k-1}={N+1 \choose k}} , this can be rewritten as:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x+a)^{N+1} = \sum^{N+1}_{k=0} {N+1 \choose k} x^k a^{N-k} }

Hence if the theorem holds for n=N, it must hold for n=N+1. As it is true for n=0, it is true for all natural numbers.

Generalisation to Non-Integers

When the binomial theorem is extended to non-integer n or n<0, the series it creates is infinite:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x+a)^n = \sum^{\infty}_{k=0} {n \choose k} x^k a^{n-k} = a^n + nxa^{n-1} + \frac{n(n-1)}{2!} x^2 a^{n-2} + \frac{n(n-1)(n-2)}{3!} x^3 a^{n-3}... }

As it is an infinite series, it is often referred to as the "binomial series". The binomial coefficients are generalised so that n need not be a natural number as:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {n \choose k} = \frac{1}{k!} \prod^{k}_{i=1}(n-k+1) = \frac{n(n-1)(n-2)...(n-k+1)}{k!} }

for k>0 and 1 for k=0. As this series contains an infinite number of terms, one must worry whether it converges or not. In the case of the binomial expansion, it always converges if |x/a|<1 or if n a natural number.

Other generalisations exist such as the multinomial expansion for brackets with more than two terms and the multi-binomial expansion where several binomial brackets are multiplied together.

Example

As an example, consider expanding (x+y)⁴. It would take a while doing it by hand. Using the binomial theorem, we quickly derive:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x+y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 +y^4 }

These coefficients form the 5th row of Pascal's triangle. As an example with negative n, consider (x+1)^-1:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x+1)^{-1} = 1 - x + \frac{1}{2} x^2 - \frac{1}{6} x^3 + \frac{1}{24} x^4 -... }

which will converge for |x|<1.

See also

• Polynomial
• Combinatorics
• Pascal's triangle