## Saturday, January 29, 2022

### Binomial Expansion in Two Methods

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

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