Binomial expansion? A really difficult one ?

Question

Expand (1-3x)^1/2 ascending powers of x up to and including the term x^3 simplifying each term as far as possible


State the range of values of x for which the expansion is valid

Show that       root 10 = 5 root (2/5) 

Hence by using your expansion with a suitable value of x obtain an approximation to 5 significant figures for root 10 

Calculate to 3 significant figures the percentage error in the approximation obtained in part (d) 

*sorry for using the word root instead of the symbol as there is no symbol on my keyboard

Anonymous · Accepted Answer

(1 - 3x)^(1/2)
= 1 + (1/2)(-3x)/1! + (1/2)(-1/2)(-3x)^2/2! + (1/2)(-1/2)(-3/2)(-3x)^3/3!
= 1 - (3/2)x - (9/8)x^2 - (27/16)x^3

sqrt(10) = sqrt( 25(2/5) ) = 5 sqrt( 2/5 )

Taking x=(1/5), then 1-3x = (2/5), so

(2/5)^(1/2) = 1 - (3/2)(1/5) - (9/8)(1/5)^2 - (27/16)(1/5)^3
= 1283/2000 = 0.6415

Thus sqrt(10) ~ (5)(0.6415) = 3.2075

This differs from the correct value by

1 - (3.2075 / sqrt(10)) = 0.0143, or 1.43% relative error

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Binomial expansion? A really difficult one ?

1 Answer