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Problem: Using the Chebyshev inequality related to the prime number theorem, show that for each integer $n \geq 2$, that there is at least one prime number $p$ with $n<p\leq50n$.

Attempt: So I know that: $0.23 \frac {n} {ln (n)} \leq \pi(n) \leq 5.6 \frac {n} {ln (n)}$, and

$0.23 \frac {50n} {ln (50n)} \leq \pi(50n) \leq 5.6 \frac {50n} {ln (50n)}$

What I thought of doing at first is showing that $f(x) = 0.23 \frac {50x} {ln (50x)}-5.6 \frac {x} {ln (x)}$ is positive and strictly increasing on the interval $(2, \infty)$, but this function is only positive starting at about $n=41$, and is larger than $1$ at about $n=45$, so this doesn't seem like a very clever approach (I would have to test all values of $n$ smaller than $45$). The goal would have been to show that $|\pi(50n)-\pi(n)| \geq 0.23 \frac {50x} {ln (50x)}-5.6 \frac {x} {ln (x)} > 1$.

I'm also looking for an argument for why $f(x) = 0.23 \frac {50x} {ln (50x)}-5.6 \frac {x} {ln (x)}$ is strictly increasing on $ (e, \infty)$ without trying to find the zeroes of the derivative, which is a messy equation.

Any insights on how to approach this problem efficiently appreciated.

IntegrateThis
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  • @ChristianBlatter oops typo. – IntegrateThis Mar 26 '19 at 19:14
  • Huh?? Let $p=2$ and that will be true for all $n$. If you want some other prime $p$ then for any $n$ there is always a prime $p < n$ and if $p < n$ then $p <n < 50n$. Are you sure you wrote this right? .. oh, and it's not true from $n = 2$ there is no prime $p$ so that $p < 2 < 100$. – fleablood Mar 26 '19 at 19:14
  • @fleablood, sorry typo in the post, should have checked first, my bad. – IntegrateThis Mar 26 '19 at 19:16
  • @IntegrateThis Where did you get the constant factors of $0.23$ and $5.6$? I did a search for Chebyshev inequality, but can't find them anywhere. The closest thing I found is, in the Encylopedia of Mathematics Chebyshev theorems on prime numbers, #6 which says "For $x$ larger than some $x_0$ the inequality $0.9212\ldots \lt \frac{\pi(x)\ln(x)}{x} \lt 1.1055\ldots$". Your current values are so relatively far apart that, as you've seen, you can't directly use your stated inequality for your proof. – John Omielan Mar 26 '19 at 20:27
  • @IntegrateThis Regardless of the factors used on the left & right, a small enhancement for checking is that, as $\pi(n)$ is always an integer, you can use the ceiling function (i.e., smallest integer $\ge$ value) for the LHS and the floor function (i.e., largest integer $\le$ value) for the RHS. – John Omielan Mar 26 '19 at 20:30
  • @JohnOmielan the constants come from my professor's notes. So you mean that $0<\pi(n)<6 n/ln(n) $ and $11 n/ln(50n) < \pi(50n) < 280 n /ln(50n)$ – IntegrateThis Mar 26 '19 at 20:41
  • @IntegrateThis No, I mean using the ceiling & floor functions around the entire values, i.e., $\left\lceil 0.23\frac{n}{\ln(n)}\right\rceil$ and $\left\lfloor 5.6\frac{n}{\ln(n)}\right\rfloor$. Thus, for $n = 100$, $0.23\frac{n}{\ln(n)} = 4.9943\ldots$, so $\left\lceil 0.23\frac{n}{\ln(n)}\right\rceil = 5$, and $5.6\frac{n}{\ln(n)} = 121.6024\ldots$, so $\left\lfloor 5.6\frac{n}{\ln(n)}\right\rfloor = 121$. You can use the ceiling function on the LHS as the value of $\pi(n)$ can't be integer part of the LHS or anything less, & similarly you can use the floor function on the RHS. – John Omielan Mar 26 '19 at 20:50

2 Answers2

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I would have to test all values of $n$ smaller than $45$

$47$ is prime. It is between $n$ and $50n$ for all $n\in [1,45]$. Test done.

Now, estimating $f(x)=a\frac{cx}{\ln(cx)}-b\frac{x}{\ln x}$ without getting too grubby: $$f(x)=\frac{acx}{\ln x+\ln c}-\frac{bx}{\ln x} = \frac{(ac-b)\cdot x\ln x -b\ln c\cdot x}{\ln x(\ln x+\ln c)}$$ If $x \ge e$, $x\ln x \ge x > 0$. If also $ac-b > b\ln c > 0$, then $(ac-b)\cdot x\ln x > b\ln c\cdot x$ and $f(x)$ is positive. So, then, estimating for our specific values $a\approx 0.23$, $b\approx 5.6$, $c = 50$, $\ln c \approx 3.9 < 4$, $$ac-b \approx 5.9,\quad b\ln c < 22.4$$ Yeah, not good enough for $f$ to be positive. On the other hand, we can effectively improve our constant by pushing out further. If we take $x\ge 50$ instead, then $x\ln x \ge x\ln 50$, and the numerator estimate becomes $$(ac-b)\cdot x\ln x - b\ln c\cdot x \ge \ln(50)[(ac-b)x - bx] \approx \ln(50)\cdot 0.3x > 0$$ We have now shown that for sufficiently large $x$, $\pi(50x) > \pi(x)$. We don't need to show that $\pi(50x) \ge \pi(x) +1$; after all, the prime-counting function $\pi$ only changes when we cross a prime, so we can only have $\pi(50x) > \pi(x)$ if there's at least one prime in $(x,50x]$.

So then, this inequality works for all $x\ge 50$. For $2\le x< 50$, we need something else - like an explicit prime in $(x,50x]$. Since $50 < 53 < 2\cdot 50$, the fact that $53$ is prime suffices to cover all of these cases.

jmerry
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You do not have to investigate any functions. Since $\pi(50n)\ge 0.23\,\frac{50n}{\ln(50n)}$ and $\pi(n)\le 5.6\,\frac n{\ln n}$, it suffices to show that $$ 0.23\,\frac{50n}{\ln(50n)} > 5.6\,\frac n{\ln n}, $$ which simplifies to $$ \frac{11.5}{\ln(50n)} > \frac{5.6}{\ln n}. $$ In view of $11.5>2\cdot 5.6$ and $\ln(50n)=\ln n+\ln 50$, this will follow from $$ 2\ln n \ge \ln n + \ln 50; $$ equivalently, from $\ln n\ge \ln 50$.

The last inequality is obviously true if $n\ge 50$, and to complete the proof just notice that in the remaining case $2\le n\le 49$, one has $n<53<50n$ (so that you can take $p=53$).

W-t-P
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  • @IntegrateThis: Just curious: what can be the reason to prefer a longer and more complicated solution (function investigation, parameters etc) to a shorter and simpler one (which in addition was the first to appear)? – W-t-P Mar 29 '19 at 08:23
  • Ya I felt odd choosing between the two answers. I guess I felt like the first answer had more work involved to write it, really I have no idea maybe you are right and I should have chose yours. – IntegrateThis Mar 29 '19 at 20:42