Find the least integer n such that f (x) is O(xn) for each of these functions. a) f (x) = 2×2 + x3 log x b) f (x) = 3×5 + (log x)4 c) f (x)

Question

Find the least integer n such that f (x) is O(xn) for each of these functions. a) f (x) = 2×2 + x3 log x b) f (x) = 3×5 + (log x)4 c) f (x) = (x4 + x2 + 1)/(x4 + 1) d) f (x) = (x3 + 5 log x)/(x4 + 1)

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Charlie 2 weeks 2021-11-15T08:27:19+00:00 1 Answer 0 views 0

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    2021-11-15T08:28:28+00:00

    Answer:

    a)4

    b)5

    c)0

    d)0

    Step-by-step explanation:

    a) For x>0 we have that logx<=x, so f(x)<=2x^2+x^4

    The largest power of x is a smallest n for which f(x) is O(x^n). Beacuse of f(x)<=2x^2+x^4, n is 4.

    Now we have to find C and k for O(x^2).

    We know that for x>0 it is x^2>x>2, so we have:

    k=2 and

    |f(x)|<= | 2x^2+x^4|<= |2x^2|+|x^4|=2x^2+x^4 <x^2*x^2+x^4=2x^4

    it follows that C=2.

    For a different k we will have other C.

    b) The largest power of x in f(x) is smallest n of O(x^n), so n=5.

    For x>0 we have logx<=x and for x>1 we have x^5>x^4.

    SO for k=1, we have x>1, and :

    |f(x)|<=|3x^5|+|log^4x|=3x^5+(logx)^4<=3x^5+x^4<3x^5+x^5=4x^5

    It fallows that C=4.

    c) f(x)=1+\frac{x^2}(x^4+1}, so n=0.

    For k=0, we have x>0, and

    |f(x)|<=1+\frac{x^2}(x^4+1}<=1+1=2x^0

    It fallows that C=2.

    d) Using x>0 we have |f(x)|<=\frac{x^4+5x}{x^4+1}.

    When x>3 then  x^4+5x<x^4+1, so we have for k=3, n=0

    |f(x)|<=1=x^0.

    It fallows that C=1.

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