Binomial Expansion in Two Methods
Expanding the Binomial by Two Methods
There are two ways to obtain the coefficients of expanding the binomial in the form (a ∙ x + b)^n:
1. Binomial Theorem:
(a ∙ x + b)^n = Σ(comb(n, k) ∙ (a ∙ x)^k ∙ b^(n -k), k=0, n)
where comb(n,k) = n! ÷ (k! × (n - k)!)
2. Maclaurin Series (Taylor series at point x=0):
f(x) = (a ∙ x + b)^n
f(x) = f(0) + f'(0) ∙ x + f''(0)÷2! ∙ x^2 + f'''(0)÷3! ∙ x^3 + ... + f^n(0)÷n! ∙ x^n
Let's illustrate this through several examples.
Example 1: (2∙x+3)^2
Binomial Theorem:
(2∙x+3)^2
= comb(2,0)∙(2∙x)^2 + comb(2,1)∙(2∙x)∙3 + comb(2,2)∙3^2
= 4∙x^2 + 12∙x + 3
Maclaurin Series:
f(x) = (2∙x+3)^2, f(0)=9
f'(x) = 4∙(2∙x+3), f'(0)=12
f''(x) = 8, f''(0)=8
(2∙x+3)^2 = 9 + 12÷1! ∙ x + 8÷2! ∙ x^2 = 9 + 12∙x + 4∙x^2
Example 2: (x + 5)^3
Binomial Theorem:
(x+5)^3
= comb(3,0)∙x^3 + comb(3,1)∙x^2∙5 + comb(3,2)∙x∙5^2 + comb(3,3)∙5^3
= x^3 + 15∙x^2 + 75∙x + 125
Maclaurin Series:
f(x) = (x+5)^3, f(0) = 125
f'(x) = 3∙(x+5)^2, f'(0) = 75
f''(x) = 6∙(x+5), f''(0) = 30
f'''(x) = 6, f'''(0) = 6
(x + 5)^3 = 125 + 75÷1! ∙ x + 30÷2! ∙x^2 + 6÷3! ∙ x^3
= 125 + 75∙x + 15∙x^2 + x^3
Example 3: A general binomial: (a∙x+b)^2
Binomial Theorem:
(a∙x+b)^2
= comb(2,0)∙(a∙x)^2 + comb(2,1)∙(a∙x)∙b + comb(2,2)∙b^2
= a^2∙x^2 + 2∙a∙b∙x + b^2
Maclaurin Series:
f(x) = (a∙x+b)^2, f(0) = b^2
f'(x) = 2∙(a∙x+b)∙a, f'(0) = 2∙a∙b
f''(x) = 2∙a^2, f''(0) = 2∙a^2
(a∙x+b)^2 = b^2 + (2∙a∙b)÷1! ∙x + (2∙a^2)÷2! ∙x^2 = b^2 + 2∙a∙b∙x + a^2∙x^2
Two ways of obtaining the expansion of the binomial.
Eddie
