This is a homework. I'd appreciate if you didn't give away answer straightaway but instead pointed me to the right direction.

From huge majority of sources the definition of $\mathcal{O}(n)$ is:

$f \in \mathcal{O}(g) \Leftrightarrow \exists c \> \exists{k} \> \forall x > k : f(x) \leq c\cdot g(x)$

but in Papadimitriou's book there is a different definition I haven't seen anywhere before:

The order of $f$, denoted $\mathcal{O}(f)$, is the set of all functions $g:N\rightarrow N$ with the property that there are positive natural numbers $c>0$ and $d>0$ such that, for all $n\in N$, $g(n) \leq c \cdot f(n)+d.$

I assume the first definition evaluates the functions as continuous functions whereas the second definition is concerned only about the points on $n \in N$. Therefore undefined values that asymptotically goes to $\pm\infty$ are not an issue anymore.

For example if we take $f(n)=\frac{1}{n}$ and $g(n)=1$. We know that $f(n)\in g(n)$ because $\lim_{x\to\infty}\frac{\frac{1}{x}}{1}=0$. We find the value for $f(1)$ and raise the $g(n)$ above this point such as: $f(n) \leq 6\cdot f(n)$ or $f(n) \leq f(n)+6$

Is my assumption correct? If it is how can I prove this equivalence?

This is a homework. I'd appreciate if you didn't give away answer straightaway but instead pointed me to the right direction.

From huge majority of sources the definition of $\mathcal{O}(n)$ is:

$f, g : N \to R^+$

$f \in \mathcal{O}(g) \Leftrightarrow \exists{c,n_0} \in N \> \forall n > n_0 : f(n) \leq c\cdot g(n)$

but in Papadimitriou's book there is a different definition I haven't seen anywhere before:

The order of $f$, denoted $\mathcal{O}(f)$, is the set of all functions $g:N\rightarrow N$ with the property that there are positive natural numbers $c>0$ and $d>0$ such that, for all $n\in N$, $g(n) \leq c \cdot f(n)+d.$

I assume the first definition evaluates the functions as continuous functions whereas the second definition is concerned only about the points on $n \in N$. Therefore undefined values that asymptotically goes to $\pm\infty$ are not an issue anymore.

For example if we take $f(n)=\frac{1}{n}$ and $g(n)=1$. We know that $f(n)\in g(n)$ because $\lim_{x\to\infty}\frac{\frac{1}{x}}{1}=0$. We find the value for $f(1)$ and raise the $g(n)$ above this point such as: $f(n) \leq 6\cdot f(n)$ or $f(n) \leq f(n)+6$

Is my assumption correct? If it is how can I prove this equivalence?

This is a homework. I'd appreciate if you didn't give away answer straightaway but instead pointed me to the right direction.

From huge majority of sources the definition of $\mathcal{O}(n)$ is:

$f \in \mathcal{O}(g) \Leftrightarrow \exists c \> \exists{k} \> \forall x > k : f(x) \leq c\cdot g(x)$

but in Papadimitriou's book there is a different definition I haven't seen anywhere before:

The order of $f$, denoted $\mathcal{O}(f)$, is the set of all functions $g:N\rightarrow N$ with the property that there are positive natural numbers $c>0$ and $d>0$ such that, for all $n\in N$, $g(n) \leq c \cdot f(n)+d.$

I assume the first definition evaluates the functions as continuous functions whereas the second definition is concerned only about the points on $n \in N$. Therefore undefined values that asymptotically goes to $\pm\infty$ are not an issue anymore.

For example if we take $f(n)=\frac{1}{n}$ and $g(n)=1$. We know that $f(n)\in g(n)$ because $\lim_{x\to\infty}\frac{\frac{1}{x}}{1}=0$. We find the value for $f(1)$ and raise the $g(n)$ above this point such as: $f(n) \leq 6\cdot f(n)$ or $f(n) \leq f(n)+6$

Is my assumption correct? If it is how can I prove this equivalence?

Thank you.

This is a homework. I'd appreciate if you didn't give away answer straightaway but instead pointed me to the right direction.

From huge majority of sources the definition of $\mathcal{O}(n)$ is:

$f \in \mathcal{O}(g) \Leftrightarrow \exists c \> \exists{k} \> \forall x > k : f(x) \leq c\cdot g(x)$

but in Papadimitriou's book there is a different definition I haven't seen anywhere before:

The order of $f$, denoted $\mathcal{O}(f)$, is the set of all functions $g:N\rightarrow N$ with the property that there are positive natural numbers $c>0$ and $d>0$ such that, for all $n\in N$, $g(n) \leq c \cdot f(n)+d.$

I assume the first definition evaluates the functions as continuous functions whereas the second definition is concerned only about the points on $n \in N$. Therefore undefined values that asymptotically goes to $\pm\infty$ are not an issue anymore.

For example if we take $f(n)=\frac{1}{n}$ and $g(n)=1$. We know that $f(n)\in g(n)$ because $\lim_{x\to\infty}\frac{\frac{1}{x}}{1}=0$. We find the value for $f(1)$ and raise the $g(n)$ above this point such as: $f(n) \leq 6\cdot f(n)$ or $f(n) \leq f(n)+6$

Is my assumption correct? If it is how can I prove this equivalence?