The bigO notation

Question

I am studying the use of Big O notation online ( use in Taylor Series Etc.) but I am not understanding it right from its definition itself. It says, f is said to be in O(x^(n)) if |f(x)|0 and x->infinity. After that I have been given following examples, Cases as x->0 1. 5x+3x^(2) is in O(x) but not in O(x^(2)) 2. Sin(x) is in O(x) but not in O(x^(2)). 3. ln(1+x)-x is in O(x^(2)) but not in O(x^(3)). 4. 1-cos(x^(2)) is in O(x^(4)) but not in O(x^(5)). 5. sqrt(x) is not in O(x^(n)) for any n>=1. 6. e^(-1/x^(2)) is in O(x^(n)) for all n. Cases as x-> infinity 1. arctan(x) is in O(1) as well as O(x^(n)) for any n>=0. 2. x*sqrt(1+x^(2)) is in O(x^(2) but not in O(x^(3/2)). and many more. I am not able to apply the definition to all above examples to check. Can you please help. I am familier with the relative growth of functions, Loptials rule etc. Thanks and regards, Rahul.

Answer 1

Limit as x→0, f(x)=

1. 5x+3x²: g(x)=x: x=0⁻, |f(x)|<6x; x=0⁺, |f(x)|<6x. This means that we can find a constant c (in this case the example is 6), such that |5x+3x²|≤cx. x(5+3x) approaches 5x as x gets smaller. We could have picked 8, because even when x=1, 5x+3x²=8. When x=0.01 (for example), f(0.01)=0.0503 and f(-0.01)=0.0497, compared with cg(x)=cx. So an arbitrary value of c that is greater than the coefficient of x will suffice to show that f(x)∈O(g(x)). Now we turn to g(x)=x². Since x is the predominant part of f(x) as x→0, no positive multiple of x² is going to satisfy |5x+3x²|≤cx², hence f(x)∉O(x²).
2. sin(x): g(x)=x: 0≤|sin(x)|≤1 so c≥1 enables |sin(x)|≤cx. As x→0 sin(x)→x, but is always less than x as can be seen by graphing f(x) and g(x) or using the Taylor expansion of sin(x). When g(x)=x², a graph of these functions shows that for 0≤x≤0.88 (approx), f(x)≥g(x), and no value of c can change this—confirmed by the Taylor series for sin(x) which commences with x. When x→0, x²<x, so f(x)∉O(x²).
3. ln(1+x)-x: g(x)=x²: ln(1+x)=x-x²/2+x³/3-x⁴/4+... (Taylor expansion) so subtracting x from this expansion we get: f(x)=-x²/2+x³/3-x⁴/4+...; when x→0⁻, the first term is always negative, and the second, smaller term is also negative. Therefore |f(x)| consists of a series of positive quantities, the first term being x²/2 which is smaller than x². The second term becomes insignificant as x approaches zero, so |f(x)|-x²=-x²/2. When x→0⁺, the first term of |f(x)| stays the same and the second term is negative, becoming insignificant as x approaches zero, so the situation is similar to x→0⁻, and f(x)∈O(x²). Now we try g(x)=x³. |f(x)|-x³=x²/2±x³/3+...-x³. In this case the x² term predominates and |f(x)|-x³>0, so f(x)∉O(x³).
4. 1-cos(x²): g(x)=x⁴: cos(x⁴)=1-(1-x⁴/2!+x⁸/4!-...). When x→0 we can approximate this to x⁴/2!. x⁴ is always positive, so |f(x)|-g(x)=x⁴/2-x⁴=-x⁴/2. Therefore |f(x)|<g(x), and f(x)∈O(x⁴). When g(x)=x⁵, x⁴/2 predominates over x⁵, so we have x⁴/2-x⁵=x⁴(½-x). Since x→0, ½-x→½ and |f(x)|-g(x)>0, therefore f(x)∉O(x⁵). No constant c can change the inequality for all x→0.
5. √x: g(x)=xⁿ for n≥1: we can assume positive square root. When x=1,  f(x)=g(x) for all n; but we need to compare f and g when x→0. √x>x for 0<x<1, because, squaring the inequality: x>x²⇒1>x, that is, x<1 and x cannot be negative since the square root of a negative number is imaginary. So |f(x)|-x>0 (n=1). So we have at least one value of n where |f(x)|>g(x) and therefore f(x)∉O(x⁵).
6. e^(-1/x²): g(x)=xⁿ: when -1<x<1, 1/x² is large, approaching ∞ as x→0, and e^-∞→0; consider ln(-1/x²) and ln(xⁿ)⇒-1/x² and nln(x). For example, x=0.1: we get -100 and -2.3n, so n>50 to match f(x). When x=0.01, we have -10⁴ and -4.6n, so n>2000 to match f(x). It’s clear that the difference between these two quantities gets wider as x→0, and so as exponents for e, f(x) approaches zero from the positive side much faster than g(x) can. There is no value of c such that cg(x)>f(x) the closer x gets to zero. Therefore, f(x)∈O(xⁿ).

More to follow (if space permits)...

CLUE: Look at expansions for trig/exp/log functions and use similar logic to Q3. For example, expansion of cos(x) uses even powers of x, so higher odd powers in g(x) may become insignificant when comparing |f(x)| with g(x) and its positive multiples.

Limit as x→∞ f(x)=

1. arctan(x): g(x)=1: arctan is an angle, and if c=π/2, |arctan(x)|<π/2, and cg(x)=π/2. This angle is in the first quadrant. So f(x)∈O(1). x⁰=1 is a special case of xⁿ for n≥0. The asymptote for |f(x)| is π/2. So cg(x)=cxⁿ, and when c=π/2, |f(x)|≤πxⁿ/2 ∀n≥0 as x→∞. Therefore, f(x)∈O(xⁿ).
2.