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Thread starter 14 years ago
#1
f(x)=4/√(1 + 2/3 x)
-3/2 <x< 3/2

1.)
a.) Show that f(1/10) = √(15).

b.) Expand f(x) in ascending powers of x up to and including the term in x².

c.) Use your expansion to obtain an approximation for √(15), giving answer as an exact fraction.

d.) Show that 244/63 is a more accurate approximation for √(15).

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The series expansion of (1+5x)^(3/5), in ascending powers of x up to and including the x³ term is;

1+3x+px²+qx³, |x|<(1/5).

2.)
a.) Find the values of P & Q.

b.) Use the expansion with a suitable value of x to find an approximate value for (1.1)^3/5.

c.) Obtain the value of (1.1)^3/5 from your calculator & hence find the percentage error in the answer to part b. 0
Thread starter 14 years ago
#2
Anyone at all? 0
14 years ago
#3
Question 1:
f(x)=4/√(1 + 2/3 x)
-3/2 <x< 3/2

a) Substituting x = 1/10 in to the formula, we have
f(1/10) = 4/√(1 + 2/3*1/10) = 4/√(16/15) = √15

b) Rewrite the function again as well as formula: (1 + X)^n
f(x) = 4*( 1 + 2/3x)^ (-1/2)
Using binomial series, its function can be expressed:
f(x) = 4*{1 + (-1/2)*(2/3)*x + [(-1/2)*(-1/2 - 1)*(2/3x)^2]/2 + ...}
4 - 4/3x + 4/3x^2 + ...

c) Substituting x = 1/10 into the series,
f(1/10) = 4 - 4/3*1/10 + 4/3*(1/10)^2 = 97/25

d) Using the answer above we get the approximate value of √15 is 3.88
Otherwise, 244/63 is 3.8730 is close to the real value of √15 is 3.87298

Question 2:
a) p = [(3/5)*(3/5 - 1)]*25 /2 = - 3
q = [ (3/5)*(3/5 - 1)*(3/5 - 2)]*125/6 = 7

b) we have f(x) = 1 + 3x - 3x² + 7x³
(1.1)^3/5 = (1 + 5*1/50)^3/5
Hence, substituting with x = 1/50, the value is
(1.1)^3/5 = f(1/50) = 1 + 3*1/50 - 3*(1/50)^2 + 7(1/50)^3 = 1.058856

c) By calculator : (1.1)^3/5 = 1.058852853
Percentage of error is : (1.058856 - 1.058852853)/1.058852853*100%
= 2.97*10^(-4) %
It's very small.
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