Riemann Hypothesis for Dummies?
First, i'm sorry if the title is a little bit offensive. I just wonder if there's such a thing as "Riemann Hypothesis For Dummies", while the original statement of the conjecture itself would require at least advance undergraduate complex analysis, and (no math undergraduate is dummy right?). I'm going to write down some analysis preliminary that will lead us to understand the statement of Riemann Hypothesis.
Recall that the Initial Riemann zeta function is defined via the Dirichlet series by
\[\zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^s}\qquad\Re(s)>1\]
Note that the function is clearly analytic on $\Re(s)>1$.
The statement of Riemann Hypothesis is (it is not the most simple form, perhaps the most simple form i know is the thing related to Liouville's Lambda Function )
[Riemann Hypothesis] All nontrivial zeros of the Riemann Zeta Function lie on the critical line $\Re(s)=\frac{1}{2}$.
The question immediately arises here, isn't the series diverges for $\Re(s) \leq 1$? Then why does it
mention $\Re(s)=\frac{1}{2}$? Also what is the nontrivial zero?
First we resolve the problem on defining the Riemann Zeta Function (without initial), we will use Riemann's Argument, since so far this function only defined for $\Re(s)>1$. How could we reach (read:define) it on $\Re(s)\leq 1$, that is to define it for all complex plane? Moreover, since the function is already analytic on $\Re(s) >1$, we want the function to be also analytic on the other region $\Re(s) \leq 1$.
One tool we can use here is the analytic continuation, for which we can fill the the value of the function outside its analytic domain resulting the other analytic function with a larger domain, this can be done if we can find another analytic function $F(s)$ such that the restriction of $F$ on $Re(s)>1$ is $\zeta(s)$ itself. Unfortunately, there is no hope for $s=1$, we will see this later, now we focus on finding an analytic function $F$ such that $F(s)=\zeta(s)$, for $\Re(s)>1$, this function must be analytic on the domain larger than just $\Re(s)>1$.
To find the extended function, we start with what we only have, for $\Re(s)>1$ we have
\[\begin{align*}\zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^s}&=\sum_{n=0}^{\infty}\frac{1}{(n+1)^s} \\ &=\sum_{n=1}^{\infty} \frac{1}{(n+1)^s}+1\\ &=\sum_{n=1}^{\infty}\frac{1}{(n+1)^s} + \sum_{n=1}^{\infty} \left(\frac{1}{n^{s-1}} - \frac{1}{(n+1)^{s-1}}\right)\\ &=\sum_{n=1}^{\infty}\left(\frac{1}{(n+1)^s}-\frac{n+1}{(n+1)^s}+\frac{n}{n^{s}}\right)\\ &=\sum_{n=1}^{\infty}n\left(\frac{1}{n^s}-\frac{1}{(n+1)^s}\right)\\ &=\sum_{n=1}^{\infty} n s\int_{n}^{n+1} x^{-s-1}\,dx\end{align*}\]
Since $\lfloor x \rfloor = n$ when $x \in [n,n+1)$, we have $\int_{n}^{n+1}\lfloor x \rfloor x^{-s-1}\,dx=n\int_{n}^{n+1} x^{-s-1}\,dx$. We have
\[\begin{align*}\zeta(s)&=s\sum_{n=1}^{\infty} n \int_{n}^{n+1} x^{-s-1}\,dx\\ &=s\sum_{n=1}^{\infty}\int_{n}^{n+1}\lfloor x \rfloor x^{-s-1}\,dx\\ &=s\int_1^{\infty}\lfloor x \rfloor x^{-s-1}\,dx\end{align*}\]
Write $\lfloor x \rfloor = x - \{x\}$ where $\{x\}$ is fractional part of $x$. We have
\[\begin{align*}\zeta(s)&=s \int_1^{\infty} x^{-s-1}\lfloor x \rfloor \,dx\\ &=s \int_1^{\infty}(x-\{x\})x^{-s-1}\,dx\\ &=s \int_1^{\infty} x^{-s}\, dx - s\int_1^{\infty} x^{-s-1}\{x\}\, dx\end{align*}\]
Since $s \int_1^{\infty} x^{-s} \,dx = s \frac{-x^{-s+1}}{-s+1}\mid_1^{\infty}=\frac{s}{s-1}$, we have
\[\zeta(s)=\frac{s}{s-1} - \int_1^{\infty}\{x\}x^{-s-1}\,dx\]
Via this representation of $\zeta(s)$, we see that $\zeta(s)$ has simple pole for $s=1$, and also since $\{x\} \in [0,1)$ the integral $\int_1^{\infty}\{x\}x^{-s-1}\,dx$ converges for $\Re(s)>0$, by comparison test with $\int_1^{\infty} x^{-\Re(s)-1} \, dx$. Thus, on this representation, $\zeta(s)$ is defined for $\Re(s)>0$. We actually have extend the domain of $\zeta(s)$ (applying analytic continuation) from $\Re(s)>1$ to $\Re(s)>0$, and resulting the simple pole in $s=1$ with residue 1. Still, our works to define $\zeta(s)$ for the whole complex plane, not finish yet.
To continue, we borrow a function from Advanced Calculus Course (in Universitas Indonesia we call it Calculus IV), Gamma Function.
Gamma function, initially defined by $\Gamma(s)=\int_0^{\infty}t^{s-1}e^{-t}\,dt$ for $s$ is real positive, since the integral also converges whenever $s$ is a complex number with $\Re(s)>0$, therefore the Gamma function is defined as a complex function, this function is analytic on where it is defined that is $\Re(s)>0$. However using this definition, we can't go anywhere other than $\Re(s)>0$, but luckily, using also analytic continuation, the Gamma Function can also extendable to an analytic function in the whole complex plane except for $s=0, -1,-2, \cdots, -n, \cdots$ (see the link above)
Using it then, Gamma Function is extendable to $\mathbb{C}/\{0,-1,-2,\cdots\}$. In the relation between $\zeta$ and $\Gamma$, consider
\[\Gamma\left(\frac{s}{2}\right)=\int_0^{\infty}t^{s/2-1}e^{-t}\, dt\]
Substitue $t=n^2\pi u$ we have
\[\Gamma\left(\frac{s}{2}\right)=\int_0^{\infty}\pi^{\frac{s-2}{2}} n^{s-2} u^{\frac{s-2}{2}} e^{-n^2 \pi u} (n^2 \pi)\,du=n^s \pi^{\frac{s}{/2}}\int_0^{\infty} u^{\frac{s}{2}-1} e^{-n^2 \pi u} \, du\]
Equivalent to
\[\pi^{-\frac{s}{2}}\Gamma\left(\frac{s}{2}\right)\left(\frac{1}{n^s}\right)=\int_0^{\infty} u^{\frac{s}{2}-1} e^{-n^2 \pi u} \, du\]
Taking the summation from $n=1$ to $\infty$ and since the integral on the left converges uniformly, we can exchange the integral and summation thus resulting
\[\pi^{-\frac{s}{2}}\Gamma\left(\frac{s}{2}\right) \zeta(s) = \int_0^{\infty} u^{\frac{s}{2}-1} \left(\sum_{n=1}^{\infty}e^{-n^2 \pi u}\right) \, du\]
We borrow again a function from advanced calculus, Jacobi-Theta Function, defined by $\theta(t)=\sum_{n=-\infty}^{\infty} e^{n^2 i\pi t}$. Therefore $\theta(it)=\sum_{n=-\infty}^{\infty} e^{-n^2 \pi t}$. Since "$\sum_{n=-\infty}^{0} e^{-n^2 \pi t} = \sum_{n=0}^{\infty} e^{-n^2\pi t}$", using this we have
\[\theta(it)=\sum_{n=0}^{\infty} e^{-n^2 \pi t}+ \sum_{n=1}^{\infty}e^{-n^2 \pi t}=2 \sum_{n=1}^{\infty} e^{-n^2 \pi t}+1.\]
Thus $\sum_{n=1}^{\infty}e^{-n^2 \pi t} = \frac{\theta(it)-1}{2}$. Therefore the zeta function is represented as
\[\zeta(s)=\frac{\pi^{\frac{s}{2}}}{\Gamma\left(\frac{s}{2}\right)}\int_0^{\infty} u^{\frac{s}{2}-1} \left(\frac{\theta(ui)-1}{2}\right)\, du\]
From this point we shall use the functional equation for $\theta$, that is $(-ix)^{\frac{1}{2}}\theta(x)=\theta\left(-\frac{1}{x}\right)$, which is quite hard to prove, this functional equation can be obtained from modular transform of theta function. Usually given as Jacobi Identities for theta function (See the Link above, but no proof there). Substituing $x \mapsto ix$ to the functional equation we have $x^{\frac{1}{2}} \theta(ix)=\theta\left(\frac{-1}{ix}\right)=\theta\left(\frac{i}{x}\right)$.
\[\begin{align*}&\int_0^{\infty}x^{s/2-1}\left(\frac{\theta(xi)-1}{2}\right)\,dx=\\ &\int_0^{1}x^{s/2-1}\left(\frac{\theta(xi)-1}{2}\right)\,dx+\int_1^{\infty}x^{s/2-1}\left(\frac{\theta(xi)-1}{2}\right)\,dx\end{align*}\]
Using substitution $x \mapsto \frac{1}{x}$ on the first integral, we have
\[\begin{align*}\int_0^1x^{s/2-1}\left(\frac{\theta(xi)-1}{2}\right)\, dx&= \int_{\infty}^{1}x^{-s/2+1}\left(\frac{\theta\left(\frac{i}{x}\right)-1}{2}\right)\left(\frac{-1}{x^2}\right)\, dx \\ &= \int_1^{\infty} x^{-s/2-1}\left(\frac{x^{1/2}\theta(ix)-1}{2}\right)\\ &= \int_1^{\infty} \frac{x^{-s/2-1/2}\theta(ix)-x^{-s/2-1/2}}{2}\\ &+ \int_1^{\infty} \frac{x^{-s/2-1/2}-x^{-s/2-1}}{2} \, dx\\ &=\int_1^{\infty} \frac{x^{-s/2-1/2}\theta(ix)-x^{-s/2-1/2}}{2}+\frac{1}{s(s-1)}\end{align*}\]
Thus we have
\[\int_0^{\infty}x^{s/2-1}\left(\frac{\theta(xi)-1}{2}\right)\,dx\]
\[=\frac{1}{s(s-1)}+\int_1^{\infty}(x^{-s/2-1/2}+x^{s/2-1})\left(\frac{\theta(xi)-1}{2}\right)\,dx\]
Hence the Riemann Zeta Function would takes form
\[\zeta(s)=\frac{\pi^{\frac{s}{2}}}{\Gamma\left(\frac{s}{2}\right)}\left(\frac{1}{s(s-1)}+\int_1^{\infty}(x^{-s/2-1/2}+x^{s/2-1})\left(\frac{\theta(xi)-1}{2}\right)\,dx\right)\]
Note that Gamma function has simple pole on $s=0$, thus $\frac{1}{\Gamma\left(\frac{s}{2}\right)}$ cancel the simple pole also at $s=0$ obtained from $\frac{1}{s(s-1)}$.
Since $\theta(xi)$ has exponential order, the above integral converges for every $s\in \mathbb{C}$, therefore the expression on the right performs an analytic function for all $s \in \mathbb{C}$ except for $s=1$ (in which the denominator is zero) and $s=-2,-4,-6, \cdots$ (in which $\Gamma\left(\frac{s}{2}\right)$ is not analytic) thus we have extend the zeta function to an analytic function defined on $\mathbb{C}/\{1,-2,-4,\cdots,-2k,\cdots\}$.
We actually have any other question to resolve, that is a nontrivial zeros of $\zeta(s)$, to make it short, we call $s$ the zero is trivial zero iff $s$ is a pole from the $\Gamma\left(\frac{s}{2}\right)$, that is $s=0,-2,-4,\cdots$. The other zeros is known to be located on $0\leq \Re(z) \leq 1$, are called nontrivial zeros. Therefore we have got the idea now, that the Riemann Hypothesis asks wether these nontrivial zeros always located on the $\Re(s)=\frac{1}{2}$.
It is known that more than billion of the first nontrivial zeros are all agree with the hypothesis, which seems likely the hypothesis to be true. But, as the problem is complicated enough to describe, the proof challenges many greatest minds in the world(s), as the most difficult problem in Number Theory.
Writing this post remind me with the quote from Albert Einstein, which is put on the wall of the seminar room in our university :D , "If you can't explain it simply, you don't understand it well enough" , well maybe I don't really understand this very well, because even to describe it, the miscellaneous properties of the other special functions are being used.
Recall that the Initial Riemann zeta function is defined via the Dirichlet series by
\[\zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^s}\qquad\Re(s)>1\]
Note that the function is clearly analytic on $\Re(s)>1$.
The statement of Riemann Hypothesis is (it is not the most simple form, perhaps the most simple form i know is the thing related to Liouville's Lambda Function )
[Riemann Hypothesis] All nontrivial zeros of the Riemann Zeta Function lie on the critical line $\Re(s)=\frac{1}{2}$.
The question immediately arises here, isn't the series diverges for $\Re(s) \leq 1$? Then why does it
mention $\Re(s)=\frac{1}{2}$? Also what is the nontrivial zero?
First we resolve the problem on defining the Riemann Zeta Function (without initial), we will use Riemann's Argument, since so far this function only defined for $\Re(s)>1$. How could we reach (read:define) it on $\Re(s)\leq 1$, that is to define it for all complex plane? Moreover, since the function is already analytic on $\Re(s) >1$, we want the function to be also analytic on the other region $\Re(s) \leq 1$.
One tool we can use here is the analytic continuation, for which we can fill the the value of the function outside its analytic domain resulting the other analytic function with a larger domain, this can be done if we can find another analytic function $F(s)$ such that the restriction of $F$ on $Re(s)>1$ is $\zeta(s)$ itself. Unfortunately, there is no hope for $s=1$, we will see this later, now we focus on finding an analytic function $F$ such that $F(s)=\zeta(s)$, for $\Re(s)>1$, this function must be analytic on the domain larger than just $\Re(s)>1$.
To find the extended function, we start with what we only have, for $\Re(s)>1$ we have
\[\begin{align*}\zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^s}&=\sum_{n=0}^{\infty}\frac{1}{(n+1)^s} \\ &=\sum_{n=1}^{\infty} \frac{1}{(n+1)^s}+1\\ &=\sum_{n=1}^{\infty}\frac{1}{(n+1)^s} + \sum_{n=1}^{\infty} \left(\frac{1}{n^{s-1}} - \frac{1}{(n+1)^{s-1}}\right)\\ &=\sum_{n=1}^{\infty}\left(\frac{1}{(n+1)^s}-\frac{n+1}{(n+1)^s}+\frac{n}{n^{s}}\right)\\ &=\sum_{n=1}^{\infty}n\left(\frac{1}{n^s}-\frac{1}{(n+1)^s}\right)\\ &=\sum_{n=1}^{\infty} n s\int_{n}^{n+1} x^{-s-1}\,dx\end{align*}\]
Since $\lfloor x \rfloor = n$ when $x \in [n,n+1)$, we have $\int_{n}^{n+1}\lfloor x \rfloor x^{-s-1}\,dx=n\int_{n}^{n+1} x^{-s-1}\,dx$. We have
\[\begin{align*}\zeta(s)&=s\sum_{n=1}^{\infty} n \int_{n}^{n+1} x^{-s-1}\,dx\\ &=s\sum_{n=1}^{\infty}\int_{n}^{n+1}\lfloor x \rfloor x^{-s-1}\,dx\\ &=s\int_1^{\infty}\lfloor x \rfloor x^{-s-1}\,dx\end{align*}\]
Write $\lfloor x \rfloor = x - \{x\}$ where $\{x\}$ is fractional part of $x$. We have
\[\begin{align*}\zeta(s)&=s \int_1^{\infty} x^{-s-1}\lfloor x \rfloor \,dx\\ &=s \int_1^{\infty}(x-\{x\})x^{-s-1}\,dx\\ &=s \int_1^{\infty} x^{-s}\, dx - s\int_1^{\infty} x^{-s-1}\{x\}\, dx\end{align*}\]
Since $s \int_1^{\infty} x^{-s} \,dx = s \frac{-x^{-s+1}}{-s+1}\mid_1^{\infty}=\frac{s}{s-1}$, we have
\[\zeta(s)=\frac{s}{s-1} - \int_1^{\infty}\{x\}x^{-s-1}\,dx\]
Via this representation of $\zeta(s)$, we see that $\zeta(s)$ has simple pole for $s=1$, and also since $\{x\} \in [0,1)$ the integral $\int_1^{\infty}\{x\}x^{-s-1}\,dx$ converges for $\Re(s)>0$, by comparison test with $\int_1^{\infty} x^{-\Re(s)-1} \, dx$. Thus, on this representation, $\zeta(s)$ is defined for $\Re(s)>0$. We actually have extend the domain of $\zeta(s)$ (applying analytic continuation) from $\Re(s)>1$ to $\Re(s)>0$, and resulting the simple pole in $s=1$ with residue 1. Still, our works to define $\zeta(s)$ for the whole complex plane, not finish yet.
To continue, we borrow a function from Advanced Calculus Course (in Universitas Indonesia we call it Calculus IV), Gamma Function.
Gamma function, initially defined by $\Gamma(s)=\int_0^{\infty}t^{s-1}e^{-t}\,dt$ for $s$ is real positive, since the integral also converges whenever $s$ is a complex number with $\Re(s)>0$, therefore the Gamma function is defined as a complex function, this function is analytic on where it is defined that is $\Re(s)>0$. However using this definition, we can't go anywhere other than $\Re(s)>0$, but luckily, using also analytic continuation, the Gamma Function can also extendable to an analytic function in the whole complex plane except for $s=0, -1,-2, \cdots, -n, \cdots$ (see the link above)
Using it then, Gamma Function is extendable to $\mathbb{C}/\{0,-1,-2,\cdots\}$. In the relation between $\zeta$ and $\Gamma$, consider
\[\Gamma\left(\frac{s}{2}\right)=\int_0^{\infty}t^{s/2-1}e^{-t}\, dt\]
Substitue $t=n^2\pi u$ we have
\[\Gamma\left(\frac{s}{2}\right)=\int_0^{\infty}\pi^{\frac{s-2}{2}} n^{s-2} u^{\frac{s-2}{2}} e^{-n^2 \pi u} (n^2 \pi)\,du=n^s \pi^{\frac{s}{/2}}\int_0^{\infty} u^{\frac{s}{2}-1} e^{-n^2 \pi u} \, du\]
Equivalent to
\[\pi^{-\frac{s}{2}}\Gamma\left(\frac{s}{2}\right)\left(\frac{1}{n^s}\right)=\int_0^{\infty} u^{\frac{s}{2}-1} e^{-n^2 \pi u} \, du\]
Taking the summation from $n=1$ to $\infty$ and since the integral on the left converges uniformly, we can exchange the integral and summation thus resulting
\[\pi^{-\frac{s}{2}}\Gamma\left(\frac{s}{2}\right) \zeta(s) = \int_0^{\infty} u^{\frac{s}{2}-1} \left(\sum_{n=1}^{\infty}e^{-n^2 \pi u}\right) \, du\]
We borrow again a function from advanced calculus, Jacobi-Theta Function, defined by $\theta(t)=\sum_{n=-\infty}^{\infty} e^{n^2 i\pi t}$. Therefore $\theta(it)=\sum_{n=-\infty}^{\infty} e^{-n^2 \pi t}$. Since "$\sum_{n=-\infty}^{0} e^{-n^2 \pi t} = \sum_{n=0}^{\infty} e^{-n^2\pi t}$", using this we have
\[\theta(it)=\sum_{n=0}^{\infty} e^{-n^2 \pi t}+ \sum_{n=1}^{\infty}e^{-n^2 \pi t}=2 \sum_{n=1}^{\infty} e^{-n^2 \pi t}+1.\]
Thus $\sum_{n=1}^{\infty}e^{-n^2 \pi t} = \frac{\theta(it)-1}{2}$. Therefore the zeta function is represented as
\[\zeta(s)=\frac{\pi^{\frac{s}{2}}}{\Gamma\left(\frac{s}{2}\right)}\int_0^{\infty} u^{\frac{s}{2}-1} \left(\frac{\theta(ui)-1}{2}\right)\, du\]
From this point we shall use the functional equation for $\theta$, that is $(-ix)^{\frac{1}{2}}\theta(x)=\theta\left(-\frac{1}{x}\right)$, which is quite hard to prove, this functional equation can be obtained from modular transform of theta function. Usually given as Jacobi Identities for theta function (See the Link above, but no proof there). Substituing $x \mapsto ix$ to the functional equation we have $x^{\frac{1}{2}} \theta(ix)=\theta\left(\frac{-1}{ix}\right)=\theta\left(\frac{i}{x}\right)$.
\[\begin{align*}&\int_0^{\infty}x^{s/2-1}\left(\frac{\theta(xi)-1}{2}\right)\,dx=\\ &\int_0^{1}x^{s/2-1}\left(\frac{\theta(xi)-1}{2}\right)\,dx+\int_1^{\infty}x^{s/2-1}\left(\frac{\theta(xi)-1}{2}\right)\,dx\end{align*}\]
Using substitution $x \mapsto \frac{1}{x}$ on the first integral, we have
\[\begin{align*}\int_0^1x^{s/2-1}\left(\frac{\theta(xi)-1}{2}\right)\, dx&= \int_{\infty}^{1}x^{-s/2+1}\left(\frac{\theta\left(\frac{i}{x}\right)-1}{2}\right)\left(\frac{-1}{x^2}\right)\, dx \\ &= \int_1^{\infty} x^{-s/2-1}\left(\frac{x^{1/2}\theta(ix)-1}{2}\right)\\ &= \int_1^{\infty} \frac{x^{-s/2-1/2}\theta(ix)-x^{-s/2-1/2}}{2}\\ &+ \int_1^{\infty} \frac{x^{-s/2-1/2}-x^{-s/2-1}}{2} \, dx\\ &=\int_1^{\infty} \frac{x^{-s/2-1/2}\theta(ix)-x^{-s/2-1/2}}{2}+\frac{1}{s(s-1)}\end{align*}\]
Thus we have
\[\int_0^{\infty}x^{s/2-1}\left(\frac{\theta(xi)-1}{2}\right)\,dx\]
\[=\frac{1}{s(s-1)}+\int_1^{\infty}(x^{-s/2-1/2}+x^{s/2-1})\left(\frac{\theta(xi)-1}{2}\right)\,dx\]
Hence the Riemann Zeta Function would takes form
\[\zeta(s)=\frac{\pi^{\frac{s}{2}}}{\Gamma\left(\frac{s}{2}\right)}\left(\frac{1}{s(s-1)}+\int_1^{\infty}(x^{-s/2-1/2}+x^{s/2-1})\left(\frac{\theta(xi)-1}{2}\right)\,dx\right)\]
Note that Gamma function has simple pole on $s=0$, thus $\frac{1}{\Gamma\left(\frac{s}{2}\right)}$ cancel the simple pole also at $s=0$ obtained from $\frac{1}{s(s-1)}$.
Since $\theta(xi)$ has exponential order, the above integral converges for every $s\in \mathbb{C}$, therefore the expression on the right performs an analytic function for all $s \in \mathbb{C}$ except for $s=1$ (in which the denominator is zero) and $s=-2,-4,-6, \cdots$ (in which $\Gamma\left(\frac{s}{2}\right)$ is not analytic) thus we have extend the zeta function to an analytic function defined on $\mathbb{C}/\{1,-2,-4,\cdots,-2k,\cdots\}$.
We actually have any other question to resolve, that is a nontrivial zeros of $\zeta(s)$, to make it short, we call $s$ the zero is trivial zero iff $s$ is a pole from the $\Gamma\left(\frac{s}{2}\right)$, that is $s=0,-2,-4,\cdots$. The other zeros is known to be located on $0\leq \Re(z) \leq 1$, are called nontrivial zeros. Therefore we have got the idea now, that the Riemann Hypothesis asks wether these nontrivial zeros always located on the $\Re(s)=\frac{1}{2}$.
It is known that more than billion of the first nontrivial zeros are all agree with the hypothesis, which seems likely the hypothesis to be true. But, as the problem is complicated enough to describe, the proof challenges many greatest minds in the world(s), as the most difficult problem in Number Theory.
Writing this post remind me with the quote from Albert Einstein, which is put on the wall of the seminar room in our university :D , "If you can't explain it simply, you don't understand it well enough" , well maybe I don't really understand this very well, because even to describe it, the miscellaneous properties of the other special functions are being used.
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