若已知展开式sinx/x=1-x^2/3!+x^4/5!-x^6/7!+···对x∈R,x≠0成立,则由于sinx/x=0有无穷多个根:±π,±2π,···,±nπ,····,于是1-x^2/3!+x^4/5!-x^6/7!+···=(1-x^2/π^2)(1-x^2/2^2·π^2)···(1-x^2/n^2·π^2)···,利用
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/03 15:39:55
![若已知展开式sinx/x=1-x^2/3!+x^4/5!-x^6/7!+···对x∈R,x≠0成立,则由于sinx/x=0有无穷多个根:±π,±2π,···,±nπ,····,于是1-x^2/3!+x^4/5!-x^6/7!+···=(1-x^2/π^2)(1-x^2/2^2·π^2)···(1-x^2/n^2·π^2)···,利用](/uploads/image/z/12060852-60-2.jpg?t=%E8%8B%A5%E5%B7%B2%E7%9F%A5%E5%B1%95%E5%BC%80%E5%BC%8Fsinx%2Fx%3D1-x%5E2%2F3%21%2Bx%5E4%2F5%21-x%5E6%2F7%21%2B%C2%B7%C2%B7%C2%B7%E5%AF%B9x%E2%88%88R%2Cx%E2%89%A00%E6%88%90%E7%AB%8B%2C%E5%88%99%E7%94%B1%E4%BA%8Esinx%2Fx%3D0%E6%9C%89%E6%97%A0%E7%A9%B7%E5%A4%9A%E4%B8%AA%E6%A0%B9%EF%BC%9A%C2%B1%CF%80%2C%C2%B12%CF%80%2C%C2%B7%C2%B7%C2%B7%2C%C2%B1n%CF%80%2C%C2%B7%C2%B7%C2%B7%C2%B7%2C%E4%BA%8E%E6%98%AF1-x%5E2%2F3%21%2Bx%5E4%2F5%21-x%5E6%2F7%21%2B%C2%B7%C2%B7%C2%B7%3D%EF%BC%881-x%5E2%2F%CF%80%5E2%29%281-x%5E2%2F2%5E2%C2%B7%CF%80%5E2%29%C2%B7%C2%B7%C2%B7%281-x%5E2%2Fn%5E2%C2%B7%CF%80%5E2%29%C2%B7%C2%B7%C2%B7%2C%E5%88%A9%E7%94%A8)
若已知展开式sinx/x=1-x^2/3!+x^4/5!-x^6/7!+···对x∈R,x≠0成立,则由于sinx/x=0有无穷多个根:±π,±2π,···,±nπ,····,于是1-x^2/3!+x^4/5!-x^6/7!+···=(1-x^2/π^2)(1-x^2/2^2·π^2)···(1-x^2/n^2·π^2)···,利用
若已知展开式sinx/x=1-x^2/3!+x^4/5!-x^6/7!+···对x∈R,x≠0成立,则由于sinx/x=0有无穷多个根:±π,±2π,···,±nπ,····,于是1-x^2/3!+x^4/5!-x^6/7!+···=(1-x^2/π^2)(1-x^2/2^2·π^2)···(1-x^2/n^2·π^2)···,利用上述结论可得1+1/2^2+1/3^2+···+1/n^2+···=
若已知展开式sinx/x=1-x^2/3!+x^4/5!-x^6/7!+···对x∈R,x≠0成立,则由于sinx/x=0有无穷多个根:±π,±2π,···,±nπ,····,于是1-x^2/3!+x^4/5!-x^6/7!+···=(1-x^2/π^2)(1-x^2/2^2·π^2)···(1-x^2/n^2·π^2)···,利用
把(1-x^2/π^2)(1-x^2/2^2·π^2)···(1-x^2/n^2·π^2)···,展开可以得到1+a1*x^2+a2*x^4+a3*x^6.,由于两边的等式关系式可以得到对应项的系数相等即 a1=-1/3!;由于展后后 :a1=-1/π^2-1/(2^2·π^2)-1/(3^2·π^2)-..-/n^2·π^2
两边乘以-1*π^2得到;
-a1*π^2=1/1+1/(2^2)+1/(3^2)+..+/n^2;
把系数a1=-1/3!带入的到;
1/1+1/(2^2)+1/(3^2)+..+/n^2=π^2/3!=π^2/6;
证明:1=0.99999.....