Question

I need to prove this mathematically and I'm not sure what to do.

3. (a) Find the best possible relationship using one of the notations: 0, 2, O, o, w, for the following pairs of functions: n3 + 6n1.5 + 3100 and nig8 – 10n1.6 – 9000; nlgn and n1.01; 3n and (3.01)”; 7" and n!. Justify each answer. (b) Function f(n) = 21 +5000n - 60000 when n < 75 and f(n) = n2 – 100n for 150 > n > 75 and f(n) = nlogn – 50n for n > 150. Write f(n) in asymptotic notation in the simplest possible form.

Answer 1

a) Since $\lg 8=3$, both $n^3&plus;6n^{1.5}&plus;3100$ and $n^{\lg 8}-10n^{1.6}-9000$ are in $\Theta(n^3)$. Hence,

$n^3&plus;6n^{1.5}&plus;3100\in\Theta(n^{\lg 8}-10n^{1.6}-9000)$

For $n\lg n, n^{1.01}$ , we note that

${\frac{n\lg n}{n^{1.01}}}={\frac{\lg n}{n^{0.01}}}$

But L'Hospital's rule applied to

${\frac{\lg x}{x^{0.01}}}$

shows that

$\lim_{n\rightarrow\infty}{\frac{n\lg n}{n^{1.01}}}=\lim_{n\rightarrow\infty}{\frac{\lg n}{n^{0.01}}}=0$

Therefore, $n\lg n=o(n^{1.01})$ .

For $3^n,(3.01)^n$, we see that

${\frac{(3.01)^n}{3^n}}=(1&plus;{\frac 1{300}})^n$

Therefore,

$\lim_{n\rightarrow\infty}{\frac{(3.01)^n}{3^n}}=\lim_{n\rightarrow\infty}(1&plus;{\frac 1{300}})^n=\infty~\Rightarrow~\lim_{n\rightarrow\infty}{\frac{3^n}{(3.01)^n}}=0$

which implies

$3^n\in o((3.01)^n)$

For $7^n,n!$ we note that by Stirling approximation we have

${\frac{n\ln 7}{\ln (n!)}}={\frac{n\ln 7}{n\ln n-n&plus;O(\ln n)}}$

which implies

$\lim_{n\rightarrow\infty}{\frac{n\ln 7}{\ln (n!)}}=\lim_{n\rightarrow\infty}{\frac{n\ln 7}{n\ln (n/e)&plus;O(\ln n)}}=\lim_{n\rightarrow\infty}{\frac{\ln 7}{\ln (n/e)&plus;O(n^{-1}\ln n)}}=0$

Therefore,

$7^n\in o(n!)$

b) Let

$C_1=\max_{0\leq n\leq 75}|2^n&plus;500n-60000|,~~~C_2=\max_{75<n< 150}|n^2-100n|$

Note that

$\begin{align*}\lim_{n\rightarrow\infty}{\frac{n\lg n}{f(n)}}&=\lim_{n\rightarrow\infty}{\frac{n\lg n}{n\lg n-50n}}\\ &=\lim_{n\rightarrow\infty}{\frac 1{1-{\frac{50}{\lg n}}}}\\ &={\frac 1{1-\lim_{n\rightarrow\infty}{\frac{50}{\lg n}}}}\\ &={\frac 1{1-0}}\\ &=1\end{align*}$

Thus, we have

$\begin{align*}f(n)\in \Theta(n\lg n)\end{align*}$