Sorry, not A.

Is D correct?

Is D correct?

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+2 votes

Best answer

$A. f(n)$= Θ($2^{n+3}$)

$f(n) = 2^{n-4} =\Large \frac{2^n}{2^4}$

let $g(n) = 2^{n+3} = 2^{n}.2^3$

$\large \frac{f(n)}{g(n)} = \frac{1}{2^3.2^4}$

this tells us $f(n)$= Θ($2^{n+3}$) is true

$B.$ $f(n)$=Ω($n^{1000}$)

$f(n)$ is exponential and $n^{1000}$ is polynomial so $f(n)$=Ω($n^{1000}$) is true

C. $f(n)$=Ο($2^{n-10}$) is also true just like A

So **none of them are false**

0 votes

For the first one, we have to show that $2^{n-4} \leq c \times 2^{n-10}$ for all $n \geq n_0$.

This is true for any $c \geq 2^6$ as we get by solving the inequality.

For the second one, we have to prove that $f(n) \geq c \times g(n)$ for all $n \geq n_0$

$\implies 2^{n-4} \geq c \times n^{1000}$.

This is true for all $n_0$ when $n - 1000 \log n > 4$. You can get this inequality by doing using the limit definition while comparing terms.

For the third, we have to show $c_1 \times 2^{n-3} \leq 2^{n-4} \leq c_2 \times 2^{n+3}$.

You can find $c_1, c_2$ according to this condition.

Hence, all three are true.

This is true for any $c \geq 2^6$ as we get by solving the inequality.

For the second one, we have to prove that $f(n) \geq c \times g(n)$ for all $n \geq n_0$

$\implies 2^{n-4} \geq c \times n^{1000}$.

This is true for all $n_0$ when $n - 1000 \log n > 4$. You can get this inequality by doing using the limit definition while comparing terms.

For the third, we have to show $c_1 \times 2^{n-3} \leq 2^{n-4} \leq c_2 \times 2^{n+3}$.

You can find $c_1, c_2$ according to this condition.

Hence, all three are true.

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