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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