A good motivation why Chebyshev uses (1) can be found in this master thesis. Namely, if one defines ψ(x)=∑n≤xΛ(n) with Λ the von Mangoldt function, then T(x)=∞∑n=1ψ(xn) and thus ψ(x)=∞∑n=1μ(n) T(xn) See Murty's book for example for these formulas. Because T(x)≈xlogx−x, ψ(x) can be approximated as ψ(x)≈∑n≤xμ(n)(xnlogxn−xn)≈∑n≤xμ(n)n(xlogx−x)−(∑n≤xμ(n)nlogn)x The prime number theorem, which was not proved yet at the time of Chebyshev, states that ψ(x)∼x. It thus seems natural to use some kind of approximation to the Möbius function ˜μ≈μ so that ˜μ(n)=0 for large n and ∑˜μ(n)n=0 in order to remove the term xlogx. Chebyshev took ˜μ(1)=1, ˜μ(2)=−1, ˜μ(3)=−1, ˜μ(5)=−1, ˜μ(30)=1, and other ˜μ(n)=0. For this choice, it follows indeed that 1−12−13−15+130=0 To get close to the prime number theorem, one also would like that −∑˜μ(n)nlogn≈1. Chebyshev's choice gives −∑˜μ(n)nlogn≈0.92 which is quite close to 1. In the following, I will take instead ˜μ(1)=1, ˜μ(2)=−2, and other ˜μ(n)=0. For this simpler choice, ∑˜μ(n)n=1−22=0 and ∑˜μ(n)nlogn≈0.69 Because 0.69 is farther from 1 than Chebyshev's value 0.92, one thus expects this simpler choice for ˜μ to give weaker results than Chebyshev's results. I now follow Chebyshev's reasoning from his paper but on this simpler choice of ˜μ. Additionally, I use Landau's large O notation whereas Chebyshev's uses explicit inequalities. I thus study the function T(x)−2 T(x2) Because T(x)=∑∞n=1ψ(xn), one has T(x)−2 T(x2)=ψ(x)−ψ(x2)+ψ(x3)−ψ(x4)+⋯ This series is alternating and decreasing, thus ψ(x)−ψ(x2)≤T(x)−2 T(x2)≤ψ(x) Because T(x)=xlogx−x+O(logx), it follows that T(x)−2 T(x2)=xlogx−x+O(logx)−2(x2logx2−x2+O(logx2))=−x+x⋅log2+x+O(logx)=log2⋅x+O(logx) The terms involving xlogx have cancelled. This is not a coincidence, but a consequence of the choice of ˜μ. The function ψ(x) has thus the following lower bound log2⋅x+O(logx)≤ψ(x) To obtain an upper bound , one can proceed as follows (Chebyshev uses a different argument) ψ(x)−ψ(x2)≤log2⋅x+O(logx)ψ(x2)−ψ(x4)≤log2⋅x2+O(logx)ψ(x4)−ψ(x8)≤log2⋅x4+O(logx)⋯ Because ψ(x2m)=0 if m>logxlog2, adding the above inequalities gives the upper bound ψ(x)≤2⋅log2⋅x+O(log2x)
Comments
- From (3) and (4) it follows that log2≤lim infx→∞ψ(x)x≤lim supx→∞ψ(x)x≤2log2 This is weaker than the prime number theorem limx→∞ψ(x)x=1
- The inequalities (3) and (4) are not strong enough to prove Bertrand's postulate. Indeed, I can only deduce that ψ(x)−ψ(x2)≥log2⋅x−2log2⋅x2+O(log2x)≥O(log2x) because the coefficient of the factor x has cancelled. Ramanujan has modified the reasoning above so that Bertrand's postulate does follow, see Murty p 38.
- In the next post, I will repeat the argument using explicit inequalities instead of using the big O notation.
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