重要极限的证明
1. 求证:sin(π/(2n+1))sin(2π/(2n+1))sin(3π/(2n+1))……sin(nπ/(2n+1))=√(2n +1)/2^n, Sol:复数方法:
复数方程 z^(2n+1) =1的根是 a1,a2,a3,... ,a(2n),1。
其中,ak =cos(2kπ/(2n+1))+i sin(2kπ/(2n+1)),k =1,2,... ,2n 。 所以,ak =(a1)^k
所以,z^(2n+1) -1=(z-a1)(z-a2)...(z-a(2n))(z-1) ,即
(z-a1)(z-a2)...(z-a(2n))=(z^(2n+1) -1)/(z-1) =z^(2n)+z^(2n-1) +... +z +1。
两边令z =1,并取模,则:
|1-a1|×|1-a2|×...... ×|1-a2n|=2n +1......... (*)
因为,|1-ak|=√|(cos(2kπ/(2n+1))-1)) +i sin(2kπ/(2n+1))|=2×sin(kπ/(2n+1)),所以由(*)式得:
2^n×sin(π/(2n+1))sin(2π/(2n+1))sin(3π/(2n+1))……sin(nπ/(2n+1))=2n +1。
所以,sin(π/(2n+1))sin(2π/(2n+1))sin(3π/(2n+1))……sin(nπ/(2n+1))=√(2n+1)/2^n
2. 三角函数
求证:sin(π/(2n+1))sin(2π/(2n+1))sin(3π/(2n+1))……sin(nπ
/(2n+1))=√(2n +1)/2^n.
证:sin(π/(2n+1))sin(2π/(2n+1))sin(3π/(2n+1))........sin(nπ/(2n+1))=√(2n +1)/2^n
设Z=cos2π/(2n+1)+ isin2π/(2n+1)
则x^(2n+1)=1的根为1,z,...z^2n
得x^2n+...+x+1=(x-z)(x-z^2)...(x-z^2n)
2n+1=|(1-z)||(1-z^2)|... |(1-z^2n)|...(1)
又|(1-z^k)|=2sinkπ/(2n+1)...(2)
|1-z^k| = |1-(cos(2kπ/(2n+1)) +sin(2kπ/(2n+1)) )|
=|1-cos(2kπ/(2n+1))) -sin(2kπ/(2n+1)) )|
=√((1-2cos(2kπ/(2n+1)) +cos^2 (2kπ/(2n+1))) + sin^2 (2kπ/(2n+1))) =√(2-2cos(2kπ/(2n+1)) )
=√(4sin^2(kπ/(2n+1))
=2sin(kπ/(2n+1)
故
2n+1 =( n(π/(2n+1)). n(2π/(2n+1)) n(3π/(2n+1))........ n(2nπ/(2n+1)) 两边开方, 得
sin(π/(2n+1))sin(2π/(2n+1))sin(3π/(2n+1))........sin(nπ/(2n+1)) =√(2n+1) / 2^n
另外那个类似, 可以尝试自己证一下.
3. 为什么sin π/n+sin2π/n......+sin(n-1)π/n=cotπ/2n?
解:2 sin [π/(2n)]·sin(π/n)= cos [π/n -π/(2n)]- cos [π/n +π/(2n)]= cos [π/(2n)]- cos [3π/(2n)]2 sin [π/(2n)]·sin(2π/n) = cos [2π/n -π/(2n)]- cos [2π/n+π/(2n)]= cos [3π/(2n)]- cos
[5π/(2n)]2 sin [π/(2n)]·sin(3π/n)= cos [3π/n -π/(2n)]- cos [3π/n +π/(2n)]= cos [5π/(2n)]- cos [7π/(2n)]……2 sin [π/(2n)]·sin[(n-1)π/n]= cos [(n-1)π/n -π/(2n)]- cos [(n-1)π/n +π/(2n)]= cos [(2n-3)π/(2n)]- cos [(2n-1)π/(2n)]
故:2 sin [π/(2n)] ·{sin(π/n)+sin(2π/n)+......+sin[(n-1)π/n]}= cos [π/(2n)]- cos [(2n-1)π/(2n)]= cos [π/(2n)]- cos [π-π/(2n)]=2 cos [π/(2n)]
故:sin(π/n)+sin(2π/n)+......+sin[(n-1)π/n]= cos[π/(2n)]/ sin
[π/(2n)]= cot [π/(2n)]
4. 级数sin n/(n+1)收敛还是发散, 如果收敛, 是绝对收敛还是条件收敛, 为什么? Sol:收敛,Dirichlet 判别法. 这是最典型的一个用Dirichlet 判别法判别收敛的例子.sinn 的部分和=[sin1/2(sin1+sin2+...+sinn)]/sin1/2(积化和差公式)=[cos1/2-cos(2n+1)/2)]/sin1/2,于是有界,1/(n+1)单调递减趋于0, 收敛. 不绝对收敛.|sinn/(n+1)|>=sin^2n/(n+1)=[1-cos(2n)]/2(n+1).类似用Dirichl et 判别法知道级数cos2n/(n+1)收敛, 但级数1/(n+1)发散, 于是易知不绝对收敛. 建议记住这个典型例子.
o n ln c n +ln c 1
n +... +ln c n 求lim =I . 2x →∞n
2n
n ln o n ln c n +ln c 1+... +ln c =n ln 2-ln n =ln 2-1ln n n n sol :≤ n 2n 2n n
I =ln2
5. 求sin π/n*sin2π/n*…*sin(n-1)π/n的值, 用复数思想
6. 三角函数连乘(正弦) 求证:sin[π/(2n+1)]*sin[2π/(2n+1)]*sin[3π/(2n+1)]*……*sin[nπ/(2n+1)]=(根号下2n-1)/2^n
Sol: 7. 证一般项级数∑sin √(n^2+1)π条件收敛
Sol:∵sin √(n²+1)π =[(-1)^n]sin[√(n²+1)π-n π]
=[(-1)^n]sin[√(n²+1)-n]π
=[(-1)^n]sin{1/[√(n²+1)+n]}π
lim(n→∞)[sin{1/[√(n²+1)+n]}π]/(1/n) =lim(n→∞)n π/[√(n²+1)+n]
=π/2
∴∑sin{1/[√(n²+1)+n]}与∑1/n有相同的敛散性, 即∑sin{1/[√(n²+1)+n]}π发散 lim(n→∞)sin{1/[√(n²+1)+n]}π=0,且sin{1/[√[(n+1)²+1]+(n+1)]}π≤sin{1/[√(n²+1)+n]}π
由莱布尼兹判别法知lim[(-1)^n]sin{1/[√(n²+1)+n]}π收敛
∴原级数条件收敛
其他回答:sin √(n^2+1)π=(-1)^n sin(√(n^2+1)π+nπ)
再利用分子有理化可得:(-1)^n sin(π/[根号(n^2+1)+n])
利用 Dirichlet判别法可知级数收敛。
而它的绝对值级数可以等价为:sin(π/[根号(n^2+1)+n])~π/[根号(n^2+1)+n]~1/n即发散。
9.Sin(π/n) ×sin(2π/n) ×sin(3π/n) ×…×sin[(n-1)π/n]=n×2^(1-n) 这等式怎么证? 大概要从哪个方面入手? sin(π/n) ×sin(2π/n) ×sin(3π/n) ×…×sin[(n-1)π/n]=n×2^(1-n) 用复数
w=cos(2π/n)+isin(2π/n)
w'=cos(2π/n)-isin(2π/n)
z^n=1
(z-1)(z^(n-1)+z^(n-2)+……+z+1)=0
z^(n-1)+z^(n-2)+……+z+1=(z-w)(z-w^2)(z-w^3)……(z-w^(n-1)) 令
z=1
n =(1-w)(1-w^2)(1-w^3)…(1-w^(n-1))
1-w^k=2sinkπ/n(sinkπ/n+icoskπ/n)
|1-w^k|=|2sinkπ/n(sinkπ/n+icoskπ/n)|=|2sinkπ/n||(sinkπ/n+icoskπ/n)|=|2sinkπ/n|=2sin(kπ/n)
取模
|n|=|(1-w)(1-w^2)(1-w^3)…(1-w^(n-1))|
|n|=|(1-w)||(1-w^2)||(1-w^3)|…|(1-w^(n-1))|
n=2^(n-1)sin(π/n)sin(2π/n)……sin[(n-1)π/n]
得证