ok c is the answer

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Identify the FALSE statement:

$A)$ $f(n)=\theta(f(\frac{n}{2})$ $implies$ $f(n^{2})=\theta(f(\frac{n^{2}}{2}))$

$B)$ $f(n)=O(g(n))$ $implies$ $log(f(n))=O(log(g(n)))$ where $n\geq2$

$C)$ $f(n)=O(g(n))$ $implies$ $2^{f(n)}=O(2^{g(n)})$

$D)$ $f(n)+g(n)=\theta(Max(f(n),g(n)))$

$A)$ $f(n)=\theta(f(\frac{n}{2})$ $implies$ $f(n^{2})=\theta(f(\frac{n^{2}}{2}))$

$B)$ $f(n)=O(g(n))$ $implies$ $log(f(n))=O(log(g(n)))$ where $n\geq2$

$C)$ $f(n)=O(g(n))$ $implies$ $2^{f(n)}=O(2^{g(n)})$

$D)$ $f(n)+g(n)=\theta(Max(f(n),g(n)))$

0

suppose F(n)=2logn and G(n)=logn here f(n)=O(g(n))

but

2^{f(n)}=2^{2logn}

2^{g(n)}=2^{logn}

now assume 2^{logn}=x

then 2^{f(n)}= x^{2}

and

2^{g(n)}=x

so 2^{f(n)}=O(2^{g(n)}) is not hold

this was expalnation ??

0

f(n)=2logn and g(n)=logn here f(n)=O(g(n))??

i think f(n)=logn and g(n)=2logn here f(n)=O(g(n))

how you write this?

i think f(n)=logn and g(n)=2logn here f(n)=O(g(n))

how you write this?

0

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