Power Series: The Binomial Series

hbghlyj

math.mcgill.ca/~labute/courses/255w03/L22.pdf

The Taylor series for the function $f(x)=(1+x)^{\alpha}$ about $x=0$ is
$$
\sum_{n=0}^{\infty} \frac{\alpha(\alpha-1) \cdots(\alpha-n+1)}{n !} x^{n}=1+\alpha+\frac{\alpha(\alpha-1)}{2 !} x+\cdots+\frac{\alpha(\alpha-1) \cdots(\alpha-n+1)}{n !} x^{n}+\cdots
$$
This series is called the binomial series. We will determine the interval of convergence of this series and when it represents $f(x)$. If $\alpha$ is a natural number, the binomial coefficient
$$
\left(\begin{array}{l}
\alpha \\
n
\end{array}\right)=\frac{\alpha(\alpha-1) \cdots(\alpha-n+1)}{n !}
$$
is zero for $\alpha>n$ so that the binomial series is a polynomial of degree $\alpha$ which, by the binomial theorem, is equal to $(1+x)^{\alpha}$. In what follows we assume that $\alpha$ is not a natural number.
If $a_{n}$ is the $n$-th term of the binomial series, we have
$$
\frac{a_{n+1}}{a_{n}}=\frac{\alpha-n}{n+1} x \rightarrow-x \text { as } n \rightarrow \infty
$$
so that the radius of convergence of the binomial series is 1.
When $x=-1$, we have
$$
\frac{a_{n+1}}{a_{n}}=\frac{n-\alpha}{n+1} \quad \text { and } \quad \lim _{n \rightarrow \infty} n\left(1-\frac{a_{n+1}}{a_{n}}\right)=\alpha+1 .
$$
Since $a_{n}$ has constant sign for $n>\alpha$, Raabe's test applies to give convergence for $\alpha>0$ and divergence for $\alpha<0$.

If $x=1$, the series becomes alternating for $n>\alpha$. By Raabe's test the series converges absolutely if $\alpha>0$. If $\alpha \leq-1$ then $\left|a_{n+1}\right| \geq\left|a_{n}\right|$ so that the series diverges. The remaining case is $-1<\alpha<0$. In this case $\left|a_{n}\right|>\left|a_{n+1}\right|$ by Alternating series test we only have to show that $a_n \to 0$. Setting $u=1+\alpha$, we have
$$
\left|a_{n}\right|=\prod_{k=1}^{n}\left(1-\frac{u}k\right) \Longrightarrow \log \left|a_n\right|=\sum_{k=1}^{n} \log \left(1-\frac{u}k\right)<-u \sum_{k=1}^{n} \frac{1}{k} \rightarrow-\infty
$$
which implies that $a_{n} \to 0$.


Theorem (Binomial Theorem). The interval of convergence $I$ of the binomial series is
$$
[-1,1] \text { if } \alpha>0, \quad(-1,1] \text { if }-1<\alpha<0, \quad(-1,1) \text { if } \alpha \leq-1 \text {. }
$$
The convergence at the endpoints is absolute $\Longleftrightarrow \alpha>0$. On $I$ we have
$$
(1+x)^{\alpha}=\sum_{n=0}^{\infty} \frac{\alpha(\alpha-1) \cdots(\alpha-n+1)}{n !} x^{n}
$$
Proof. We only have to prove the last statement. By Taylor's theorem, we have
$$
(1+x)^{\alpha}=\sum_{k=0}^{n-1} \frac{\alpha(\alpha-1) \cdots(\alpha-k+1)}{k !} x^k+R_{n}(x)
$$
where
$$
R_{n}(x)=\frac{1}{(n-1) !} \int_{0}^{x} f^{(n)}(t)(x-t)^{n-1} d t=\frac{1}{(n-1) !} \int_{0}^{x} \alpha(\alpha-1) \cdots(\alpha-n+1)(1+t)^{\alpha-n}(x-t)^{n-1} d t
$$
Using the first mean value theorem for integrals, we obtain
$$
R_{n}(x)=\frac{\alpha(\alpha-1) \cdots(\alpha-n+1)}{(n-1) !}(1+\theta x)^{\alpha-n}(x-\theta x)^{n-1} \int_{0}^{x} d t
$$
where $0<\theta<1$. Simplifying, we get
$$
R_{n}(x)=c_{n}(x t) \alpha x(1+\theta x)^{\alpha-1} \quad \text { with } \quad c_{n}(s)=\frac{(\alpha-1) \cdots(\alpha-n+1)}{(n-1) !} s^{n-1}
$$
and $t=(1-\theta) /(1+\theta x)$. Then $(1+s)^{\alpha-1}=\sum_{n=1}^{\infty} c_{n}(s)$. Now let $x \in I$. Since $0<t<1$ if $x>-1$, we have $|x t|<1$ and so the series $\sum_{n=1}^{\infty} c_{n}(x t)$ converges if $x>-1$. So its $n$-th term $c_{n}(x t)$ converges to zero. If $x=-1 \in I$, we have $t=1$. Since the series for $(1+x)^{\alpha}$ converges for $x=-1$ we have $\alpha>0$ and hence $\alpha-1>-1$. Since the series $\sum_{n=1}^{\infty} c_{n}(s)$ converges at $s=1$ if $\alpha>-1$, we have $c_{n}(-1) \rightarrow 0$ since $\left|c_{n}(-1)\right|=\left|c_{n}(1)\right| \rightarrow 0$.
QED

Example 1. For $|x|<1$ we have,
$$
\sin ^{-1}(x)=\int_{0}^{x} \frac{d t}{\sqrt{1-t^{2}}}
$$
By the binomial theorem, we have
$$
\left(1-x^{2}\right)^{-1 / 2}=1+\frac{1}{2} x^{2}+\frac{1 \cdot 3}{2 \cdot 4} x^{4}+\cdots+\frac{1 \cdot 3 \cdots(2 n-1)}{2 \cdot 4 \cdots 2 n} x^{2 n}+\cdots
$$
for $|x|<1$. Integrating, we get
$$
\sin ^{-1}(x)=x+\frac{1}{2} \frac{x^{3}}{3}+\frac{1 \cdot 3}{2 \cdot 4} \frac{x^{5}}{5}+\cdots+\frac{1 \cdot 3 \cdots(2 n-1)}{2 \cdot 4 \cdots 2 n} \frac{x^{2 n+1}}{2 n+1}+\cdots
$$
The series converges when $x=1$ by Raabe's test since
$$
n\left(1-\frac{a_{n+1}}{a_{n}}\right)=\frac{6 n^{2}+5 n}{4 n^{2}+10 n+6} \rightarrow \frac{3}{2}>1
$$
Since the series for $x=-1$ is the negative of the above series, $[-1,1]$ is the interval of convergence of the power series. Since the series is continuous on its interval of convergence and $\sin ^{-1}(x)$ is continuous there as well, we see that the power series expansion is valid on $[-1,1]$. It follows that
$$
\frac{\pi}{2}=1+\frac{1}{2} \cdot \frac{1}{3}+\frac{1 \cdot 3}{2 \cdot 4} \cdot \frac{1}{5}+\cdots+\frac{1 \cdot 3 \cdot(2 n-1)}{2 \cdot 4 \cdots 2 n} \cdot \frac{1}{2 n+1}+\cdots
$$
We leave to the reader the task of proving that the remainder after $n$ terms is less than $2 / \sqrt{n+2}$ for $n \geq 10$. This would give an estimate of $n=4000000$ to get $\pi$ correct to 2 decimal places.
Since $\sin ^{-1}(1 / 2)=\pi / 6$, we also have
$$
\frac{\pi}{6}=\frac{1}{2}+\frac{1}{2} \cdot \frac{1}{3}\left(\frac{1}{2}\right)^3+\frac{1 \cdot 3}{2 \cdot 4} \cdot \frac{1}{5}\left(\frac{1}{2}\right)^5+\cdots+\frac{1 \cdot 3 \cdots(2 n-1)}{2 \cdot 4 \cdots 2 n} \cdot \frac{1}{2 n+1}\left(\frac{1}{2}\right)^{2 n+1}+\cdots
$$
which converges more rapidly than the previous series. In fact, to compute $\pi$ to 2 decimal places, 3 terms suffice. (3.1390625)

Example 2. The substitution $x=\sin \theta$ reduces the improper integral
$$
K=\int_{0}^{1} \frac{d x}{\sqrt{\left(1-x^{2}\right)\left(1-k^{2} x^{2}\right)}}\quad\left(k^{2}<1\right)
$$
to the integral
$$
K=\int_{0}^{\pi / 2} \frac{d \theta}{\sqrt{1-k^{2} \sin ^{2} \theta}}
$$
But
$$
\left(1-k^{2} \sin ^{2} \theta\right)^{-1 / 2}=1+\frac{1}{2} k^{2} \sin ^{2} \theta+\frac{1 \cdot 3}{2 \cdot 4} k^{4} \sin ^{4} \theta+\cdots \frac{1 \cdot 3 \cdots(2 n-1)}{2 \cdot 4 \cdots 2 n} k^{2 n} \sin ^{2 n} \theta+\cdots
$$
with $\left|k^{2} \sin ^{2} \theta\right|<k^{2}<1$ so that we can integrate term by term to get
$$
\begin{aligned}
K &=1 \frac{\pi}{2}+\frac{1}{2} k^{2} \int_{0}^{\pi / 2} \sin ^{2} \theta d \theta+\frac{1 \cdot 3}{2 \cdot 4} k^{4} \int_{0}^{\pi / 2} \sin ^{4} \theta d \theta+\cdots \frac{1 \cdot 3 \cdots(2 n-1)}{2 \cdot 4 \cdots 2 n} k^{2 n} \int_{0}^{\pi / 2} \sin ^{2 n} \theta d \theta+\cdots \\
&=\frac{\pi}{2}\left(1+\left(\frac{1}{2}\right)^{2} k^{2}+\left(\frac{1 \cdot 3}{2 \cdot 4}\right)^{2} k^{4}+\cdots\left(\frac{1 \cdot 3 \cdots(2 n-1)}{2 \cdot 4 \cdots 2 n}\right)^{2} k^{2 n}+\cdots\right)
\end{aligned}
$$
since
$$
\int_{0}^{\pi / 2} \sin ^{2 n} \theta d \theta=\frac{1 \cdot 3 \cdots(2 n-1)}{2 \cdot 4 \cdots 2 n} \frac{\pi}{2}
$$

hbghlyj

We leave to the reader the task of proving that the remainder after $n$ terms is less than $2 / \sqrt{n+2}$ for $n \geq 10$.

如何证明呢
\[\frac{1}{n!} \int_0^1\arcsin^{(n+1)}(t)(1-t)^n\mathrm d t<\frac2{\sqrt{n+2}}\]

1. Table[Pi/2.-Normal[Series[ArcSin[t],{t,0,n}]]/.t->1,{n,10,15}]

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{0.254105, 0.231733, 0.231733, 0.21438, 0.21438, 0.200415}

1. Table[1/n! NIntegrate[(1-t)^n(D[ArcSin[x],{x,n+1}]/.x->t),{t,0,1}],{n,10,15}]

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{0.254105, 0.231733, 0.231733, 0.21438, 0.21438, 0.200415}

1. N/@Table[Pi/2-(Normal[Series[ArcSin[t],{t,0,n}]]/.t->1)-2/Sqrt[n+2],{n,10,15}]

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{-0.323245, -0.322967, -0.30279, -0.302018, -0.28562, -0.284656}

1. ListLinePlot[{Table[Pi/2-(Normal[Series[ArcSin[t],{t,0,n}]]/.t->1),{n,10,100}],Table[2/Sqrt[n+2],{n,10,100}]}]

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