questions on Binomial expansion?

Question

a) expand (1+x/2)^1/4   in ascending powers of x, up yo and including the term is x^2 simplifying your terms as far as possible.
b) by substituting x=1/8 into your expression , obtain an approximation to 4 decimal places of quad root of 17
c)expand (1-x/2)^-1/4 in ascending powers of x up to and including the terms in x^2, simplifying your terms as far as possible
d) using your answer to parts a) and c) , expand (2+x / 2-x) ^ 1/4 in ascending powers of x up to and including the terms in x^2 , simplifying your terms as far as possible 

e)use your answers from part (d) to find an estimate to 4 decimal places of 
0 $ integration symbol $ 0.2  (2+x / 2-x) ^ 1/4 dx

kb · Accepted Answer

a) (1 + x/2)^(1/4) = 1 + (1/4) * (x/2)^1 + [(1/4)(1/4 - 1)/2!] * (x/2)^2 + ..., via binomial series
.........................= 1 + x/8 - 3x^2/128 + ...

b) Let x = 1/8:
(1 + (1/8)/2)^(1/4) ≈ 1 + (1/8)/8 - 3(1/8)^2/128
==> (17/16)^(1/4) = (1/2) * 17^(1/4) ≈ 1 + (1/8)/8 - 3(1/8)^2/128
==> 17^(1/4) ≈ 2[1 + (1/8)/8 - 3(1/8)^2/128] = 2.0305...
--------
c) (1 - x/2)^(-1/4) = 1 + (-1/4) * (-x/2)^1 + [(-1/4)(-1/4 - 1)/2!] * (-x/2)^2 + ..., via binomial series
.........................= 1 + x/8 + 5x^2/128 + ...
 
d) ((2 + x)/(2 - x))^(1/4)
= ((1 + x/2)/(1 - x/2))^(1/4)
= (1 + x/2)^(1/4) (1 - x/2)^(-1/4)
= (1 + x/8 - 3x^2/128 + ...) (1 + x/8 + 5x^2/128 + ...), by (a),(c)
= 1 + (1/4)x + (5/128 + 1/64 - 3/128)x^2 + ... 
= 1 + (1/4)x + (1/32)x^2 + ... 

e) Using part d:
∫(x = 0 to 0.2) ((2 + x)/(2 - x))^(1/4) dx
= ∫(x = 0 to 0.2) (1 + (1/4)x + (1/32)x^2 + ...) dx
= (x + (1/8)x^2 + (1/96)x^3 + ...) {for x = 0 to 0.2}
≈ 0.2 + (1/8)(0.2)^2 + (1/96)(0.2)^3 
≈ 0.2051.

I hope this helps!

chenier · Answer

the 1st 3 phrases of a binomial enlargement (q+p)^n are nC0*q^(n-0)*p^0 , nC1*q^(n-one million)*p^one million and nC2*q^(n-2)*p^2 for the period of this query q = one million, p = px and n = 9 2d term is 9C1*one million^8*(px)^one million = 36x 9*one million*px = 36x 9px = 36x p = 36x/9x p = 4 0.33 term is 9C2*one million^7*(px)^2 = qx^2 36*one million*p^2x^2 = qx^2 36p^2x^2 = qx^2 36p^2 = q q = 36*4^2 = 36*sixteen = 576

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