Disregard #12, I\'m just a bit confused with all of #13. Will award points to pe

Question

Disregard #12, I'm just a bit confused with all of #13. Will award points to person who does this precise.

ln 1 + x/1 - x - tanx at x = 0.0015. f(x) = 1/ 1 + 2x2 - 1 + x2 at x = 5.00 times 10-3. f(R, h) = R - R2 + h2 for R = 1.374 times 103 kn and h = 1.00 m. F(x) = 1 - 1/ 1 - x for x = 2.5 times 10-13 Determine the order, O(xp), of the following functions. You may need to use series expansions in powers of x when x rightarrow 0, or series expansions in powers of 1/x when x rightarrow infinity . x(1 - x) as x rightarrow infinity. x5/4/1 - cosx as x rightarrow 0. x/x2 - 1 as x rightarrow infinity . x2 + x - x as x rightarrow infinity .

Explanation / Answer

Assuming you know the basis expansion (at order 1 and 2) at x=0 :

sqrt(1-x) = 1+O(x) or 1-1/2x+O(x^2)
cos(x) = 1+O(x) or 1-x^2/2 + O(x^3)
1/(1-x) = 1+O(x) or 1+x+O(x^2)

Remember also that f(x) = O(x^p) near 0 means that |f(x)| <= M|x^p| in some neighboorhood of zero : |x| < delta
Which is also equivalent to the fact that lim | f(x)/x^p | is finite (easy to check)


a) sqrt(x(1-x)) = sqrt(x) * sqrt(1-x) = sqrt(x) * (1+O(x)) = sqrt(x)+O(x^(3/2)) = O(x^(1/2))

(Indeed sqrt(x(1-x))/x^(1/2) has limit 1 as x->0, so finite)

b) x^(5/4)/(1-cos(x)) = x^(5/4)/(1-(1-x^2/2+O(x^2)) = x^(5/4)/(x^2/2+O(x^3)) = O(1/x^(3/4))

(Indeed x^(5/4)/(1-cos(x))/(1/x^(3/4)) has limit 0 as x->0, so finite)


c) x/(x^2-1) = (1/x) / (1-1/x^2) = 1/x * (1+O(1/x)) = 1/x + O(1/x^2) = O(1/x)

(Indeed (x/(x^2-1))/(1/x) = x^2/(x^2-1) has limit 1 as x->+inf, so finite)

d) sqrt(x^2+x)-x = x*(sqrt(1+1/x)-1) = x*(1+1/(2x)+O(1/x^2)-1) = 1/2 + O(1/x) = O(1) = O(x^0)

Indeed (sqrt(x^2+x)-x)/1 has a finite limit (=1/2)

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