Yes, $2^{2^{2^n}}$ is a good example for $p(n)$ and $q(n)$ could be $n!$.

[mathmari]

Hey!

Find an order $f_1, f_2, \dots f_{30}$ of the functions that satisfies the relations $f_1=\Omega(f_2), f_2=\Omega(f_3), \dots, f_{29}=\Omega(f_{30})$$$\frac{n}{\lg n} , \ \ n^{\lg n} ,\ \ (\sqrt{2})^{\lg n}, \ \ n^2, \ \ n!, \ \ (\lg n)! ,\ \ \left( \frac{3}{2} \right)^n ,\ \ n^3 ,\ \ \lg^2 n ,\ \ \lg(n!) ,\ \ 2^{2^n}, \ \ n^{\frac{1}{\lg n}}, \ \ \ln{\ln n}, \ \ e^{\log_{10} n }, \ \ n \cdot 2^n, \ \ n^{\lg{\lg n}}, \ \ \ln n, \ \ 1 ,\ \ 2^{\lg n}, \ \ (\lg n)^{\lg n}, \ \ e^n ,\ \ 4^{\lg n}, \ \ (n+1)!, \ \ \sqrt{\lg n}, \ \ \lg{\lg{\lg n}}, \ \ 2^{\sqrt{2 \lg n}}, \ \ n, \ \ 2^n ,\ \ n \lg n, \ \ 2^{2^{n+1}}$$
How can I find the order of the functions?? (Wondering) Do I have to find the limits of all the functions pairwise? Or is there an easier and faster way?? (Wondering)

I have done the following: The functions are:

$$\frac{n}{\lg n} , \ \ n^{\lg n} ,\ \ (\sqrt{2})^{\lg n}=\sqrt{n}, \ \ n^2, \ \ n!, \ \ (\lg n)! ,\ \ \left( \frac{3}{2} \right)^n ,\ \ n^3 ,\ \ \lg^2 n ,\ \ \lg(n!) ,\ \ 2^{2^n}, \ \ n^{\frac{1}{\lg n}}=2, \ \ \ln{\ln n}, \ \ e^{\log_{10} n }, \ \ n \cdot 2^n, \ \ n^{\lg{\lg n}}, \ \ \ln n, \ \ 1 ,\ \ 2^{\lg n}, \ \ (\lg n)^{\lg n}, \ \ e^n ,\ \ 4^{\lg n}=n^2, \ \ (n+1)!, \ \ \sqrt{\lg n}, \ \ \lg{\lg{\lg n}}, \ \ 2^{\sqrt{2 \lg n}}, \ \ n, \ \ 2^n ,\ \ n \lg n, \ \ 2^{2^{n+1}}$$We know that $$\log^a n < n^b < c^n, \forall a, b>0, \forall c>1$$

Is the following classification correct?? (Wondering)

Constants: $1, n^{\frac{1}{\lg n}}$

Logarithms: $\lg{\lg{\lg n}}, \lg^2 n, \lg(n!), \sqrt{\lg n}, (\lg n)^{\lg n}, \ln{\ln n}, \ln n$

Polynomials: $(\sqrt{2})^{\lg n}, n^{\lg n}, n^2, n^3, n,e^{\log_{10} n } ,4^{\lg n}, \frac{n}{\lg n}$

Exponentials: $n!, (\lg n)!, \left( \frac{3}{2} \right)^n, 2^{2^n}, n \cdot 2^n ,2^{\lg n}, e^n, (n+1)!, 2^{\lg n}, 2^n, n \lg n, 2^{\sqrt{2 \lg n}}, 2^{2^{n+1}}$

 

[I like Serena]

Hey! ;)

Do I have to find the limits of all the functions pairwise? Or is there an easier and faster way?? (Wondering)

I think so, although I'd skip the obvious ones.

For $f_1(n)=\Omega(f_2(n))$, we need that $f_1(n) \ge k\cdot f_2(n)$ for some constant $k$ and any sufficiently large $n$.
That is under the assumption that all functions are positive which they are.
It is true if
$$\lim_{n\to \infty} \frac{f_2(n)}{f_1(n)}$$
exists. (Nerd)

So in case of doubt we need to verify if the limit exists for 2 successive functions. (Thinking)

Is the following classification correct?? (Wondering)

Suppose we take $f_1=n^2$ and $f_2=n^3$, then we need that:
$$n^2=\Omega(n^3)$$
That is, $n^2$ is bounded from below by $n^3$.
Is that true? (Wondering)

 

[mathmari]

Suppose we take $f_1=n^2$ and $f_2=n^3$, then we need that:
$$n^2=\Omega(n^3)$$
That is, $n^2$ is bounded from below by $n^3$.
Is that true? (Wondering)

No, it is not true... It is $$n^3=\Omega(n^2)$$

I haven't find the right order yet...

Since it stands that $$\log^a n <n^b <c^n, \forall a, b>0 \text{ and } \forall c>1 \tag 1$$

I want to know which of the functions are the logarithmic functions , which are the polynomial functions and which are the exponential functions so that using the inequality $(1)$ I will not have to find the limits of all the functions pairwise.

Or can we not say for example that $n^{\lg \lg n}$ belongs to the polynomials, because it is a composition of a polynomial and a logarithmic function?? (Wondering)

 

[I like Serena]

No, it is not true... It is $$n^3=\Omega(n^2)$$

I haven't find the right order yet...

Ah, okay. (Blush)

Since it stands that $$\log^a n <n^b <c^n, \forall a, b>0 \text{ and } \forall c>1 \tag 1$$

I want to know which of the functions are the logarithmic functions , which are the polynomial functions and which are the exponential functions so that using the inequality $(1)$ I will not have to find the limits of all the functions pairwise.

Or can we not say for example that $n^{\lg \lg n}$ belongs to the polynomials, because it is a composition of a polynomial and a logarithmic function?? (Wondering)

I'm afraid that the functions you've classified as polynomials, are not in fact polynomials. That is simply because thay are not of the form $a_0+a_1 n +a_2 n^2 +...$.

However, we can say for instance: $n^{\lg \lg n} > n^k$ for any constant $k>0$ and any sufficiently large $n$.
That is, $n^{\lg \lg n}$ is asymptotically larger than any polynomial.
So we can classify it beyond the polynomials.
We still have to figure out where it is in relation to the exponentials though. (Wasntme)

 

[mathmari]

Can we formulate it as followed?? (Wondering)

$1$ and $n^{\frac{1}{\lg n}}=2$ are both constants so they are at the same equivalence class, therefore $1 \sim n^{\frac{1}{\lg n}}$

It stands that:

- $n^3 > n^2 \sim 4^{\lg n}=n^2 > n \sim 2^{\lg n}=n > (\sqrt{2})^{\lg n}=\sqrt{n}=n^{\frac{1}{2}} > e^{\log_{10} n }=n^{\frac{1}{\ln 10}} \Rightarrow n^3 > n^2 \sim 4^{\lg n} > n \sim 2^{\lg n} > (\sqrt{2})^{\lg n} > e^{\log_{10} n }$

- $e^n > 2^n > \left( \frac{3}{2} \right)^n$

- $2^{2^{n+1}}=2^{2^n 2}=4^{2^n} > 2^{2^n} \Rightarrow 2^{2^{n+1}} > 2^{2^n}$

- $\lg^2 n > \ln n > \sqrt{\lg n}=\lg^{\frac{1}{2}} n \Rightarrow \lg^2 n > \ln n > \sqrt{\lg n}$

$\lim_{n \rightarrow +\infty} \frac{2^{2^n}}{e^n}= \dots = \infty \Rightarrow 2^{2^n} > e^n$

Since it stands that $$c^n > n^b> \log^a n, \forall a, b >0 \text{ and } c>1$$ we have that:

$$2^{2^{n+1}} > 2^{2^n}> e^n > 2^n > \left( \frac{3}{2} \right)^n> n^3 > n^2 \sim 4^{\lg n} > n \sim 2^{\lg n} > (\sqrt{2})^{\lg n} > e^{\log_{10} n }> \lg^2 n > \ln n > \sqrt{\lg n} > 1 \sim n^{\frac{1}{\lg n}}$$

$\lim_{n \rightarrow +\infty} \frac{n}{\frac{n}{\lg n}}=\lim_{n \rightarrow +\infty} \lg n=\infty \Rightarrow n > \frac{n}{\lg n}$

$\lim_{n \rightarrow +\infty} \frac{\frac{n}{\lg n}}{(\sqrt{2})^{\lg n}}= \dots = \infty \Rightarrow \frac{n}{\lg n} > (\sqrt{2})^{\lg n}$

So, $$2^{2^{n+1}} > 2^{2^n}> e^n > 2^n > \left( \frac{3}{2} \right)^n> n^3 > n^2 \sim 4^{\lg n} > n \sim 2^{\lg n} > \frac{n}{\lg n} > (\sqrt{2})^{\lg n} > e^{\log_{10} n }> \lg^2 n > \ln n > \sqrt{\lg n} > 1 \sim n^{\frac{1}{\lg n}}$$

$\lim_{n \rightarrow +\infty} \frac{n^{\lg n} }{}=\lim_{n \rightarrow +\infty} n^{\lg n -3}=\infty \Rightarrow n^{\lg n} > n^3$

$\lim_{n \rightarrow +\infty} \frac{n^{\lg n}}{\left( \frac{3}{2} \right)^n}= \dots =0 \Rightarrow \left( \frac{3}{2} \right)^n > n^{\lg n}$

So, $$2^{2^{n+1}} > 2^{2^n}> e^n > 2^n > \left( \frac{3}{2} \right)^n> n^{\lg n}> n^3 > n^2 \sim 4^{\lg n} > n \sim 2^{\lg n} > \frac{n}{\lg n} > (\sqrt{2})^{\lg n} > e^{\log_{10} n }> \lg^2 n > \ln n > \sqrt{\lg n} > 1 \sim n^{\frac{1}{\lg n}}$$

$\lim_{n \rightarrow +\infty} \frac{n!}{2^{2^{n+1}}} = \dots =0 \Rightarrow 2^{2^{n+1}} > n!$

$\lim_{n \rightarrow +\infty} \frac{n!}{2^{2^n}}= \dots =0 \Rightarrow 2^{2^n} > n!$

$\lim_{n \rightarrow +\infty} \frac{n!}{e^n}= \dots =\infty \Rightarrow n!> e^n$

So, $$2^{2^{n+1}} > 2^{2^n}> n! > e^n > 2^n > \left( \frac{3}{2} \right)^n> n^{\lg n}> n^3 > n^2 \sim 4^{\lg n} > n \sim 2^{\lg n} > \frac{n}{\lg n} > (\sqrt{2})^{\lg n} > e^{\log_{10} n }> \lg^2 n > \ln n > \sqrt{\lg n} > 1 \sim n^{\frac{1}{\lg n}}$$

$\lim_{n \rightarrow +\infty} \frac{(\lg n)!}{n!}= \dots =0 \Rightarrow n! > (\lg n)!$

$\lim_{n \rightarrow +\infty} \frac{(\lg n)!}{e^n}= \dots =0 \Rightarrow e^n > (\lg n)!$

$\lim_{n \rightarrow +\infty} \frac{(\lg n)!}{2^n}= \dots =0 \Rightarrow 2^n > (\lg n)!$

$\lim_{n \rightarrow +\infty} \frac{(\lg n)!}{\left( \frac{3}{2} \right)^n}= \dots =0 \Rightarrow \left( \frac{3}{2} \right)^n > (\lg n)!$

$\lim_{n \rightarrow +\infty} \frac{(\lg n)!}{n^{\lg n}}= \dots =0 \Rightarrow n^{\lg n} > (\lg n)!$

$\lim_{n \rightarrow +\infty} \frac{(\lg n)!}{n^3}= \dots =\infty \Rightarrow (\lg n)! > n^3$

So, $$2^{2^{n+1}} > 2^{2^n}> n! > e^n > 2^n > \left( \frac{3}{2} \right)^n> n^{\lg n}> (\lg n)! > n^3 > n^2 \sim 4^{\lg n} > n \sim 2^{\lg n} > \frac{n}{\lg n} > (\sqrt{2})^{\lg n} > e^{\log_{10} n }> \lg^2 n > \ln n > \sqrt{\lg n} > 1 \sim n^{\frac{1}{\lg n}}$$

$\lim_{n \rightarrow +\infty} \frac{\lg(n!)}{\lg^2 n}= \dots =\infty \Rightarrow \lg(n!) > \lg^2 n$

$\lim_{n \rightarrow +\infty} \frac{\lg(n!)}{e^{\log_{10} n }}= \dots = \infty \Rightarrow \lg(n!) > e^{\log_{10} n }$

$\lim_{n \rightarrow +\infty} \frac{\lg(n!)}{(\sqrt{2})^{\lg n}}= \dots =\infty \Rightarrow \lg(n!) > (\sqrt{2})^{\lg n}$

$\lim_{n \rightarrow +\infty} \frac{\lg(n!)}{\frac{n}{\lg n} }= \dots =\infty \Rightarrow \lg(n!) > \frac{n}{\lg n} $

$\lim_{n \rightarrow +\infty} \frac{\lg(n!)}{n}= \dots =\infty \Rightarrow \lg(n!) > n$

$\lim_{n \rightarrow +\infty} \frac{\lg(n!)}{n^2 }= \dots =0 \Rightarrow n^2 > \lg(n!)$

So, $$2^{2^{n+1}} > 2^{2^n}> n! > e^n > 2^n > \left( \frac{3}{2} \right)^n> n^{\lg n}> (\lg n)! > n^3 > n^2 \sim 4^{\lg n} > \lg(n!) > n \sim 2^{\lg n} > \frac{n}{\lg n} > (\sqrt{2})^{\lg n} > e^{\log_{10} n }> \lg^2 n > \ln n > \sqrt{\lg n} > 1 \sim n^{\frac{1}{\lg n}}$$

$\lim_{n \rightarrow +\infty} \frac{\ln{\ln n}}{\ln n}= \dots =0 \Rightarrow \ln n > \ln{\ln n}$

$\lim_{ n \rightarrow +\infty} \frac{\ln{\ln n}}{\sqrt{\lg n}}= \dots =0 \Rightarrow \sqrt{\lg n} > \ln{\ln n}$

So, $$2^{2^{n+1}} > 2^{2^n}> n! > e^n > 2^n > \left( \frac{3}{2} \right)^n> n^{\lg n}> (\lg n)! > n^3 > n^2 \sim 4^{\lg n} > \lg(n!) > n \sim 2^{\lg n} > \frac{n}{\lg n} > (\sqrt{2})^{\lg n} > e^{\log_{10} n }> \lg^2 n > \ln n > \sqrt{\lg n} > \ln{\ln n} > 1 \sim n^{\frac{1}{\lg n}}$$

$\lim_{n \rightarrow +\infty} \frac{n \cdot 2^n}{2^n} \lim_{n \rightarrow +\infty} n=\infty \Rightarrow n \cdot 2^n > 2^n$

$\lim_{ n \rightarrow +\infty} \frac{n \cdot 2^n}{e^n}= \dots =0 \Rightarrow e^n > n \cdot 2^n$

So, $$2^{2^{n+1}} > 2^{2^n}> n! > e^n > n \cdot 2^n > 2^n > \left( \frac{3}{2} \right)^n> n^{\lg n}> (\lg n)! > n^3 > n^2 \sim 4^{\lg n} > \lg(n!) > n \sim 2^{\lg n} > \frac{n}{\lg n} > (\sqrt{2})^{\lg n} > e^{\log_{10} n }> \lg^2 n > \ln n > \sqrt{\lg n} > \ln{\ln n} > 1 \sim n^{\frac{1}{\lg n}}$$

$\lim_{ n \rightarrow +\infty} \frac{n^{\lg{\lg n}}}{n^{\lg n}}= \dots =0 \Rightarrow n^{\lg n} > n^{\lg{\lg n}}$

$\lim_{n \rightarrow +\infty} \frac{n^{\lg{\lg n}}}{(\lg n)!}= \dots =\infty \Rightarrow n^{\lg{\lg n}} > (\lg n)!$

So, $$2^{2^{n+1}} > 2^{2^n}> n! > e^n > n \cdot 2^n > 2^n > \left( \frac{3}{2} \right)^n> n^{\lg n}> n^{\lg{\lg n}} > (\lg n)! > n^3 > n^2 \sim 4^{\lg n} > \lg(n!) > n \sim 2^{\lg n} > \frac{n}{\lg n} > (\sqrt{2})^{\lg n} > e^{\log_{10} n }> \lg^2 n > \ln n > \sqrt{\lg n} > \ln{\ln n} > 1 \sim n^{\frac{1}{\lg n}}$$

$\lim_{n \rightarrow +\infty} \frac{(\lg n)^{\lg n}}{(\lg n)!}= \dots =\infty \Rightarrow (\lg n)^{\lg n} > (\lg n)!$

$\lim_{n \rightarrow +\infty} \frac{(\lg n)^{\lg n}}{n^{\lg{\lg n}}}= \dots =1 \Rightarrow (\lg n)^{\lg n} \sim n^{\lg{\lg n}}$

So, $$2^{2^{n+1}} > 2^{2^n}> n! > e^n > n \cdot 2^n > 2^n > \left( \frac{3}{2} \right)^n> n^{\lg n}> n^{\lg{\lg n}} \sim (\lg n)^{\lg n}> (\lg n)! > n^3 > n^2 \sim 4^{\lg n} > \lg(n!) > n \sim 2^{\lg n} > \frac{n}{\lg n} > (\sqrt{2})^{\lg n} > e^{\log_{10} n }> \lg^2 n > \ln n > \sqrt{\lg n} > \ln{\ln n} > 1 \sim n^{\frac{1}{\lg n}}$$

$\lim_{n \rightarrow +\infty} \frac{(n+1)!}{n!}=\lim_{n \rightarrow +\infty} \frac{(n!(n+1)}{n!}=\lim_{n \rightarrow +\infty} (n+1)=\infty \Rightarrow (n+1)! > n!$

$\lim_{n \rightarrow +\infty} \frac{(n+1)!}{2^{2^n}}= \dots =0 \Rightarrow 2^{2^n} > (n+1)!$

So, $$2^{2^{n+1}} > 2^{2^n}> (n+1)! > n! > e^n > n \cdot 2^n > 2^n > \left( \frac{3}{2} \right)^n> n^{\lg n}> n^{\lg{\lg n}} \sim (\lg n)^{\lg n}> (\lg n)! > n^3 > n^2 \sim 4^{\lg n} > \lg(n!) > n \sim 2^{\lg n} > \frac{n}{\lg n} > (\sqrt{2})^{\lg n} > e^{\log_{10} n }> \lg^2 n > \ln n > \sqrt{\lg n} > \ln{\ln n} > 1 \sim n^{\frac{1}{\lg n}}$$

$\lim_{n \rightarrow +\infty} \frac{\lg{\lg{\lg n}}}{\ln{\ln n} }= \dots =0 \Rightarrow \ln{\ln n} > \lg{\lg{\lg n}}$

So, $$2^{2^{n+1}} > 2^{2^n}> (n+1)! > n! > e^n > n \cdot 2^n > 2^n > \left( \frac{3}{2} \right)^n> n^{\lg n}> n^{\lg{\lg n}} \sim (\lg n)^{\lg n}> (\lg n)! > n^3 > n^2 \sim 4^{\lg n} > \lg(n!) > n \sim 2^{\lg n} > \frac{n}{\lg n} > (\sqrt{2})^{\lg n} > e^{\log_{10} n }> \lg^2 n > \ln n > \sqrt{\lg n} > \ln{\ln n} > \lg{\lg{\lg n}} > 1 \sim n^{\frac{1}{\lg n}}$$

$\lim_{n \rightarrow +\infty} \frac{2^{\sqrt{2 \lg n}}}{(\sqrt{2})^{\lg n}}= \dots =0 \Rightarrow (\sqrt{2})^{\lg n} > 2^{\sqrt{2 \lg n}}$

$\lim_{n \rightarrow +\infty} \frac{2^{\sqrt{2 \lg n}}}{e^{\log_{10} n }}= \dots =0 \Rightarrow e^{\log_{10} n } > 2^{\sqrt{2 \lg n}}$

$\lim_{n \rightarrow +\infty} \frac{2^{\sqrt{2 \lg n}}}{\lg^2 n}= \dots =\infty \Rightarrow 2^{\sqrt{2 \lg n}} > \lg^2 n$

So, $$2^{2^{n+1}} > 2^{2^n}> (n+1)! > n! > e^n > n \cdot 2^n > 2^n > \left( \frac{3}{2} \right)^n> n^{\lg n}> n^{\lg{\lg n}} \sim (\lg n)^{\lg n}> (\lg n)! > n^3 > n^2 \sim 4^{\lg n} > \lg(n!) > n \sim 2^{\lg n} > \frac{n}{\lg n} > (\sqrt{2})^{\lg n} > e^{\log_{10} n }> 2^{\sqrt{2 \lg n}} > \lg^2 n > \ln n > \sqrt{\lg n} > \ln{\ln n} > \lg{\lg{\lg n}} > 1 \sim n^{\frac{1}{\lg n}}$$

$\lim_{n \rightarrow +\infty} \frac{n \lg n}{\lg(n!)}= \dots =1 \Rightarrow n \lg n \sim \lg(n!)$

So, $$2^{2^{n+1}} > 2^{2^n}> (n+1)! > n! > e^n > n \cdot 2^n > 2^n > \left( \frac{3}{2} \right)^n> n^{\lg n}> n^{\lg{\lg n}} \sim (\lg n)^{\lg n}> (\lg n)! > n^3 > n^2 \sim 4^{\lg n} > \lg(n!) \sim n \lg n > n \sim 2^{\lg n} > \frac{n}{\lg n} > (\sqrt{2})^{\lg n} > e^{\log_{10} n }> 2^{\sqrt{2 \lg n}} > \lg^2 n > \ln n > \sqrt{\lg n} > \ln{\ln n} > \lg{\lg{\lg n}} > 1 \sim n^{\frac{1}{\lg n}}$$

Is it correct?? (Wondering) Could I improve something at the way I justified it?? (Wondering)

 

[mathmari]

Also, I have to find a non-negative function $g(n)$ such that for all the above function $f_i(n)$, $g(n)$ is nor $O(f_i(n))$ neither $\Omega (f_i(n))$.

Could you give me some hints how I could find such a function?? (Wondering)

 

[I like Serena]

So, $$2^{2^{n+1}} > 2^{2^n}> (n+1)! > n! > e^n > n \cdot 2^n > 2^n > \left( \frac{3}{2} \right)^n> n^{\lg n}> n^{\lg{\lg n}} \sim (\lg n)^{\lg n}> (\lg n)! > n^3 > n^2 \sim 4^{\lg n} > \lg(n!) \sim n \lg n > n \sim 2^{\lg n} > \frac{n}{\lg n} > (\sqrt{2})^{\lg n} > e^{\log_{10} n }> 2^{\sqrt{2 \lg n}} > \lg^2 n > \ln n > \sqrt{\lg n} > \ln{\ln n} > \lg{\lg{\lg n}} > 1 \sim n^{\frac{1}{\lg n}}$$

Is it correct?? (Wondering) Could I improve something at the way I justified it?? (Wondering)

Looks good to me. (Nod)

I selected the first 6 and put them in this plot.
It confirms that those are in the correct order. (Happy)

For a justification, I think you only need to give the limits that prove that your order is the correct one. No need to do so many more. It would be nice if some of the actual calculation of the non-trivial ones were shown. (Nerd)

Also, I have to find a non-negative function $g(n)$ such that for all the above function $f_i(n)$, $g(n)$ is nor $O(f_i(n))$ neither $\Omega (f_i(n))$.

Could you give me some hints how I could find such a function?? (Wondering)

So g is not bounded above by any of the functions, nor is it bounded from below. (Thinking)

That means that g keeps going up above the highest function, and keeps going below the lowest function.

Which function would be above all others?
Which function would be below all others?
Which function goes up and down? (Wondering)

 

[mathmari]

Looks good to me. (Nod)

I selected the first 6 and put them in this plot.
It confirms that those are in the correct order. (Happy)

For a justification, I think you only need to give the limits that prove that your order is the correct one. No need to do so many more. It would be nice if some of the actual calculation of the non-trivial ones were shown. (Nerd)

Ok... Nice! (Sun)

So g is not bounded above by any of the functions, nor is it bounded from below. (Thinking)

That means that g keeps going up above the highest function, and keeps going below the lowest function.

Which function would be above all others?
Which function would be below all others?
Which function goes up and down? (Wondering)

How can we find such a function?? (Wondering) I don't have any idea...

 

[I like Serena]

How can we find such a function?? (Wondering) I don't have any idea...

I think $2^{2^{2^n}}$ will go above all others, won't it? (Thinking)

 

[mathmari]

I think $2^{2^{2^n}}$ will go above all others, won't it? (Thinking)

Yes, but this function is bounded below, isn't it?? (Wondering)
We are looking for a function that isn't bounded below, don't we?? (Wondering)

 

[I like Serena]

Yes, but this function is bounded below, isn't it?? (Wondering)
We are looking for a function that isn't bounded below, don't we?? (Wondering)

That's why we have to tweak it to go up and down.
How could we do that?
Which functions do you know that go up and down? (Wondering)

 

[mathmari]

That's why we have to tweak it to go up and down.
How could we do that?
Which functions do you know that go up and down? (Wondering)

Do we have to multiply it with a trigonometric function?? (Wondering)

 

[I like Serena]

Do we have to multiply it with a trigonometric function?? (Wondering)

That is indeed one way to do it (sine wave).
Alternatively we can use a piece wise function (square wave).

Suppose we have functions $p(n)$ and $q(n)$ that are respectively above and below all functions.
Then $(1+\sin 2\pi n)p(n) + (1-\sin 2\pi n)q(n)$ should do the job.

Alternatively, we can pick:
\begin{cases}p(n) & \text{ if $n$ odd} \\ q(n) & \text{ if $n$ even}\end{cases}
(Nerd)

 

[mathmari]

Alternatively, we can pick:
\begin{cases}p(n) & \text{ if $n$ odd} \\ q(n) & \text{ if $n$ even}\end{cases}
(Nerd)

What functions could we take for example?? (Wondering)

The one of them $2^{2^{2^n}}$?? And the other one?? (Wondering)