Answer :
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 :
[tex]|f(x)|<=|3x^5|+|log^4x|=3x^5+(logx)^4<=3x^5+x^4<3x^5+x^5=4x^5[/tex]
It fallows that C=4.
c) [tex]f(x)=1+\frac{x^2}(x^4+1}[/tex], so n=0.
For k=0, we have x>0, and
[tex]|f(x)|<=1+\frac{x^2}(x^4+1}<=1+1=2x^0[/tex]
It fallows that C=2.
d) Using x>0 we have [tex]|f(x)|<=\frac{x^4+5x}{x^4+1}[/tex].
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.
In each scenario, we must identify the smallest integer (n) such that [tex]f(x)\ \ is\ \ O (x^{n})[/tex] implying there must be constants C and k such that[tex]|f(x)| \geq C|x^n| \ \ for \ \ x > k[/tex].
- Since [tex]x^3 \log x[/tex] is not [tex]O (x^3)[/tex] (since the [tex]\log x[/tex] factor expands indefinitely as x grows), [tex]n = 3[/tex] is too small. In the other hand, certainly [tex]\log x[/tex] grows more slowly than x, so [tex]2x^2 + x^3 \log x \leq 2x^4 +x^4 = 3x^4[/tex]. Since, [tex]n = 4[/tex] is the answer, with [tex]C = 3[/tex] and [tex]k = 1[/tex].
- The [tex](\log \ x)^4[/tex] is insignificant compared to the term, so the answer is [tex]n = 5[/tex]. Formally we can take [tex]C = 4 \ \ and\ \ k = 1[/tex]
- For large x this fraction is close to [tex]1[/tex] (It is demonstrated by dividing the numerator and denominator by [tex]x^4[/tex]). Therefore the answer is [tex]n=0[/tex], in other words this function is [tex]O(x^{0}) = O(1)[/tex]. Formally we can write[tex]f(x) \leq \frac{3x^4}{x^4} = 3[/tex] for all [tex]x > 1[/tex], so we can take [tex]C = 3[/tex] and [tex]k = 1[/tex].
- Here the answer is [tex]n =-1[/tex] since for large [tex]x , f(x)[/tex] is close to [tex]\frac{1}{x}[/tex]. Formally we can write [tex]f(x) < \frac{6x^3}{x^4} = \frac{6}{x}[/tex] for all [tex]x > 1[/tex] , so we can take [tex]C = 6[/tex] and [tex]k = 1[/tex].
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