关于圆周率
作者:互联网
关于圆周率
圆周率为什么是一个定值
我们都知道,圆周率是一个圆的周长与直径的比值,我们在这里介绍一种利用数列极限来证明圆周率收敛于 π \pi π。
设单位圆内接正 n n n 边形的周长为 L n L_n Ln ,则 L n = n s i n 18 0 ∘ n L_n = nsin\frac{180^\circ}{n} Ln=nsinn180∘。数列 L n L_n Ln 应该收敛与单位元的半周长,及圆周率 π \pi π 。现在我们来严格证明这一点。
证明:令
t
=
18
0
∘
n
(
n
+
1
)
t = \frac{180^\circ}{n(n+1)}
t=n(n+1)180∘,则当
n
≥
3
n \geq 3
n≥3 的时候,
n
t
≤
4
5
∘
nt \leq 45^\circ
nt≤45∘,且
tan
n
t
=
tan
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(
n
−
1
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t
+
t
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=
tan
(
n
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1
)
t
+
tan
t
1
−
tan
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n
−
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t
tan
t
≥
tan
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t
+
tan
t
\tan nt = \tan \left((n-1)t + t\right) = \frac{\tan (n-1)t + \tan t}{1-\tan (n-1)t \tan t} \geq \tan(n-1)t + \tan t
tannt=tan((n−1)t+t)=1−tan(n−1)ttanttan(n−1)t+tant≥tan(n−1)t+tant
同理:
tan
(
n
−
1
)
t
≥
tan
(
n
−
2
)
t
+
tan
t
,
tan
(
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)
t
≥
tan
(
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−
3
)
t
+
tan
t
,
⋯
\tan (n-1)t \geq \tan(n-2)t + \tan t,\tan(n-2)t \geq \tan(n-3)t + \tan t,\cdots
tan(n−1)t≥tan(n−2)t+tant,tan(n−2)t≥tan(n−3)t+tant,⋯
所以:
tan
n
t
≥
tan
(
n
−
1
)
t
+
tan
t
≥
tan
(
n
−
2
)
t
+
2
tan
t
≥
⋯
≥
n
tan
t
\tan nt \geq \tan (n-1)t + \tan t \geq \tan (n-2)t + 2\tan t \geq \cdots \geq n\tan t
tannt≥tan(n−1)t+tant≥tan(n−2)t+2tant≥⋯≥ntant
故
sin
(
n
+
1
)
t
=
sin
n
t
cos
t
+
cos
n
t
sin
t
=
sin
n
t
cos
t
×
(
cos
n
t
sin
t
sin
n
t
cos
t
)
=
sin
n
t
cos
t
×
(
1
+
tan
t
tan
n
t
)
∵
tan
n
t
≥
n
tan
t
→
t
a
n
t
tan
n
t
≤
tan
t
n
tan
t
=
1
n
∴
s
i
n
(
n
+
1
)
t
≤
sin
n
t
cos
t
×
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1
+
1
n
)
=
sin
n
t
cos
t
×
n
+
1
n
≤
n
+
1
n
sin
n
t
∴
n
sin
(
n
+
1
)
t
≤
(
n
+
1
)
sin
n
t
\begin{aligned} \sin (n+1)t = & \sin nt \cos t + \cos nt \sin t \\ = & \sin nt \cos t \times \left( \frac{\cos nt \sin t}{\sin nt \cos t} \right) = \sin nt \cos t\times\left(1 + \frac{\tan t}{\tan nt} \right)\\ \\ \because \tan nt \geq n \tan &t \rightarrow \frac{tant}{\tan nt} \leq \frac{\tan t}{n\tan t} = \frac{1}{n} \\ \\ \therefore sin(n+1)t &\leq \sin nt \cos t \times (1 + \frac{1}{n}) = \sin nt \cos t \times \frac{n+1}{n} \leq \frac{n+1}{n}\sin nt \\ \\ \therefore n \sin (n+1)t& \leq (n+1)\sin nt \end{aligned}
sin(n+1)t==∵tannt≥ntan∴sin(n+1)t∴nsin(n+1)tsinntcost+cosntsintsinntcost×(sinntcostcosntsint)=sinntcost×(1+tannttant)t→tannttant≤ntanttant=n1≤sinntcost×(1+n1)=sinntcost×nn+1≤nn+1sinnt≤(n+1)sinnt
因此,当
n
≥
3
n \geq 3
n≥3 的时候:
L
n
=
n
sin
18
0
∘
n
≤
(
n
+
1
)
sin
18
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∘
n
+
1
=
L
n
+
1
L_n = n \sin \frac{180^\circ}{n} \leq (n+1)\sin\frac{180^\circ}{n+1} = L_{n+1}
Ln=nsinn180∘≤(n+1)sinn+1180∘=Ln+1
又因为,单位圆内接正 n n n 边形的面积:
S n = n sin 18 0 ∘ n cos 18 0 ∘ n < 4 ∴ n sin 18 0 ∘ n < 4 cos 18 0 ∘ n \begin{aligned} S_n = n\sin\frac{180^\circ}{n} \cos\frac{180^\circ}{n} < 4 \\ \\ \therefore n\sin \frac{180^\circ}{n} < \frac{4}{\cos\frac{180^\circ}{n}} \end{aligned} Sn=nsinn180∘cosn180∘<4∴nsinn180∘<cosn180∘4
所以当 n ≥ 3 n \geq 3 n≥3 的时候
L n = n sin 18 0 ∘ n < 4 cos 18 0 ∘ n ≤ 4 cos 6 0 ∘ = 8 L_n = n\sin \frac{180^\circ}{n} < \frac{4}{\cos \frac{180^\circ}{n}} \leq \frac{4}{\cos 60^\circ} = 8 Ln=nsinn180∘<cosn180∘4≤cos60∘4=8
综上所述,数列
{
n
sin
18
0
∘
n
}
\lbrace n \sin \frac{180^\circ}{n} \rbrace
{nsinn180∘} 单调增且有上界,所以它是收敛的,这个极限就是圆周率
π
\pi
π,即:
lim
n
→
∞
n
sin
18
0
∘
n
=
π
\lim_{n \to \infty}n \sin \frac{180^\circ}{n} = \pi
n→∞limnsinn180∘=π
圆周率的一种无穷级数展开
tan π 4 = 1 → π = 4 arctan ( 1 ) \begin{aligned} & \tan \frac \pi 4 = 1 \rightarrow \pi = 4 \arctan(1) \\ \\ \end{aligned} tan4π=1→π=4arctan(1)
令 f ( x ) = arctan ( x ) f(x) = \arctan(x) f(x)=arctan(x),则 f ′ ( x ) = 1 1 + x 2 f'(x) = \frac 1 {1+x^2} f′(x)=1+x21。令 g ( t ) = 1 1 + t g(t) = \frac 1 {1+t} g(t)=1+t1,则:
g
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=
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⋅
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∴
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(
0
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=
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n
⋅
n
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\begin{aligned} g^{(n)}(t) = (-1)^n \cdot n! (t + 1)^{-n-1} \quad \therefore g^{(n)}(0) = (-1)^n \cdot n! \end{aligned}
g(n)(t)=(−1)n⋅n!(t+1)−n−1∴g(n)(0)=(−1)n⋅n!
把
g
(
t
)
g(t)
g(t) 在零点泰勒展开:
g ( t ) = ∑ n = 0 ∞ ( − 1 ) n ⋅ t n g(t) = \sum_{n = 0}^{\infty} (-1)^n \cdot t^n g(t)=n=0∑∞(−1)n⋅tn
所以:
f
′
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=
1
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x
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=
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n
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f'(x) = \frac 1 {1 + x^2} = g(x^2) = \sum_{n=0}^{\infty} (-1)^n \cdot x ^{2n}
f′(x)=1+x21=g(x2)=n=0∑∞(−1)n⋅x2n
对这个式子求定积分得到原函数:
f
(
x
)
−
f
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0
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=
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∫
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⋅
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∫
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d
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∞
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⋅
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2
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1
2
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∴
f
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∞
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\begin{aligned} f(x) - f(0) = &f(x) = \int_0^x \sum_{n=0}^{\infty}(-1)^n \cdot z^{2n}dz \\ = &\sum_{n=0}^{\infty} (-1)^n \int_0^x z^{2n}dz = \sum_{n=0}^\infty (-1)^n \cdot \frac{x^{2n+1}}{2n+1} \\ \\ \therefore f(1) = \sum_{n=0}^\infty (-1)^n &\cdot \frac{1}{2n+1} \end{aligned}
f(x)−f(0)==∴f(1)=n=0∑∞(−1)nf(x)=∫0xn=0∑∞(−1)n⋅z2ndzn=0∑∞(−1)n∫0xz2ndz=n=0∑∞(−1)n⋅2n+1x2n+1⋅2n+11
综上所述:
π
=
4
×
(
1
−
1
3
+
1
5
−
1
7
+
1
9
−
⋯
)
\pi = 4 \times (1 - \frac 1 3 + \frac 1 5 - \frac 1 7 + \frac 1 9 - \cdots)
π=4×(1−31+51−71+91−⋯)
标签:cos,frac,圆周率,geq,关于,tan,nt,sin 来源: https://blog.csdn.net/ID246783/article/details/121085904