- 284 名前：１３２人目の素数さん [2022/12/31(土) 07:31:25.08 ID:RFpv/yWr.net]
- >>43
kが奇数で0<t<πのとき sin(kπ-t)=sint>0 |sin(kπ-t)|=sin(kπ-t)=sint
kが偶数のとき sin(kπ-t)=-sint<0 |sin(kπ-t)|=-sin(kπ-t)=sint
a[k]=1/(1+(kπ/n)^2)と置く kπ/n=tanθ dk/dθ=n/π/(cosθ)^2
Σ[k=1,2n]a[k]>∫[1,2n]a[k]dk=n/π∫[arctan(π/n),arctan(2π)]dθ
Σ[k=1,2n]a[k-1]<1+Σ[k=1,2n]a[k]<1+∫[0,2n]a[k]dk=1+n/π*arctan(2π)
2Σ[k=1,2n]a[k-1]/n→arctan(2π)/π 2Σ[k=1,2n]a[k]/n→arctan(2π)/π
b[n]=∫[0,2π] |sin(nx)|/(1+x^2) dx=1/n∫[0,2nπ] |sin(t)|/(1+(t/n)^2) dx
=1/nΣ[k=1,2k]∫[(k-1)π,kπ] |sin(t)|/(1+(t/n)^2) dx
=1/nΣ[k=1,2k]∫[0,π] |sin(kπ-t)|/(1+((kπ-t)/n)^2) dx
=1/nΣ[k=1,2k]∫[0,π] sint/(1+((kπ-t)/n)^2) dx
1/nΣ[k=1,2k]a[k]∫[0,π]sintdx<b[n]<1/nΣ[k=1,2k]a[k-1]∫[0,π]sintdx
2Σ[k=1,2k]a[k]/n<b[n]<2Σ[k=1,2k]a[k-1]/n 与式=arctan(2π)/π
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