关于arctanx的麦克劳林展开式推导:
先把结论写上:
arctanx=x−13x3+15x5−⋯+(−1)nx2n+12n+1+⋯(−1⩽x⩽1)=∑n=0∞(−1)nx2n+12n+1(−1⩽x⩽1)\begin{aligned}
arctanx=&x-\frac{1}{3}x^3+\frac{1}{5}x^5-\cdots+(-1)^{n}\frac{x^{2n+1}}{2n+1}+\cdots
&(-1 \leqslant x \leqslant 1)\\
=&\sum_{n=0}^\infty(-1)^{n}\frac{x^{2n+1}}{2n+1}&(-1 \leqslant x \leqslant 1)\\
\end{aligned}arctanx==x−31x3+51x5−⋯+(−1)n2n+1x2n+1+⋯n=0∑∞(−1)n2n+1x2n+1(−1⩽x⩽1)(−1⩽x⩽1)
关于这个式子,最简洁的证明用到了级数的一些知识;第二种是我自己瞎jb推的,比较繁琐,也不严谨,但是学完了泰勒展开就能推
方法一
思想:先求导然后展开然后积分
摘自教科书!!!!!
求导,再由等比级数展开:
(arctanx)′=11+x2=1−x2+x4−⋯+(−x2)n+⋯ (−1<x<1)=∑n=0∞(−x2)n(−1<x<1)\begin{aligned}
(arctanx)'=&\frac{1}{1+x^2}\\
=&1-x^2+x^4-\cdots+(-x^2)^n+\cdots\ \ \ &(-1<x<1)\\
=&\sum_{n=0}^{\infty}(-x^2)^n&(-1<x<1)
\end{aligned}(arctanx)′===1+x211−x2+x4−⋯+(−x2)n+⋯ n=0∑∞(−x2)n(−1<x<1)(−1<x<1)
利用幂级数的逐项可积性可得(左右两边积分):
arctanx=∫0x11+x2dx=∫0x[1−x2+x4−⋯+(−x2)n+⋯ ]dx (−1<x<1)=x−13x3+15x5−⋯+(−1)nx2n+12n+1+⋯(−1<x<1)\begin{aligned}
arctanx=&\int_0^x\frac{1}{1+x^2}dx\\
=&\int_0^x[1-x^2+x^4-\cdots+(-x^2)^n+\cdots]dx\ \ \ \ &(-1<x<1)\\
=&x-\frac{1}{3}x^3+\frac{1}{5}x^5-\cdots+(-1)^{n}\frac{x^{2n+1}}{2n+1}+\cdots&(-1<x<1)\\
\end{aligned}arctanx===∫0x1+x21dx∫0x[1−x2+x4−⋯+(−x2)n+⋯]dx x−31x3+51x5−⋯+(−1)n2n+1x2n+1+⋯(−1<x<1)(−1<x<1)
或者也可以简便的写成这样:
arctanx=∫0x11+x2dx=∫0x[∑n=0∞(−x2)n]dx(−1<x<1)=∑n=0∞[∫0x(−x2)ndx](−1<x<1)=∑n=0∞(−1)nx2n+12n+1(−1<x<1)\begin{aligned}
arctanx=&\int_0^x\frac{1}{1+x^2}dx\\
=&\int_0^x{\bigg[}\sum_{n=0}^{\infty}(-x^2)^n{\bigg]}dx&(-1<x<1)\\
=&\sum_{n=0}^\infty{\bigg[}\int_0^x(-x^2)^ndx{\bigg]}&(-1<x<1)\\
=&\sum_{n=0}^\infty(-1)^{n}\frac{x^{2n+1}}{2n+1}&(-1<x<1)\\
\end{aligned}arctanx====∫0x1+x21dx∫0x[n=0∑∞(−x2)n]dxn=0∑∞[∫0x(−x2)ndx]n=0∑∞(−1)n2n+1x2n+1(−1<x<1)(−1<x<1)(−1<x<1)
由于x=±1x=\pm1x=±1时,级数±∑n=0∞(−1)n12n+1\pm\sum_{n=0}^\infty(-1)^{n}\frac{1}{2n+1}±∑n=0∞(−1)n2n+11为交错级数,由Leibniz判别法易知其收敛。再根据幂级数的连续性定理得到
arctanx=∑n=0∞(−1)nx2n+12n+1(−1⩽x⩽1)\begin{aligned}
arctanx=\sum_{n=0}^\infty(-1)^{n}\frac{x^{2n+1}}{2n+1}(-1\leqslant x \leqslant 1 )
\end{aligned}arctanx=n=0∑∞(−1)n2n+1x2n+1(−1⩽x⩽1)
arctanxarctanxarctanx的展开式可以用来求π\piπ,不过收敛速度很慢,
取x=1x=1x=1得到:
π4=1−13+15−17+⋯\begin{aligned}
\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots
\end{aligned}4π=1−31+51−71+⋯
注解:没学过等比级数的可以这样理解
11+x2=1−x2+x4−⋯+(−x2)n+⋯(−1<x<1)\begin{aligned}
\frac{1}{1+x^2}=1-x^2+x^4-\cdots+(-x^2)^n+\cdots(-1<x<1)
\end{aligned}1+x21=1−x2+x4−⋯+(−x2)n+⋯(−1<x<1)
由等比数列公式推出:
q≠1q\ne1q=1时,
1+q+q2+⋯+qn−1=1∗(1−qn)1−q\begin{aligned}
1+q+q^2+\cdots+q^{n-1}=\frac{1*(1-q^n)}{1-q}
\end{aligned}1+q+q2+⋯+qn−1=1−q1∗(1−qn)
当∣q∣<1\vert{q}\vert<1∣q∣<1时,
limn→+∞qn=0\begin{aligned}
\lim_{n\rightarrow+\infty}q^n=0
\end{aligned}n→+∞limqn=0
因此等式两边变为:
1+q+q2+⋯+qn−1+⋯=11−q (∣q∣<1)\begin{aligned}
1+q+q^2+\cdots+q^{n-1}+\cdots=\frac{1}{1-q}\ \ \ (\vert{q}\vert<1)
\end{aligned}1+q+q2+⋯+qn−1+⋯=1−q1 (∣q∣<1)
然后把q=(−x2)q=(-x^2)q=(−x2)代入即可。
方法二
上一个方法很精辟了,但是我还是写下我初学时的脑洞吧……
思路在于,麦克劳林展开式需要求在0这点的n阶导数,那我们就求呗。注意先算n阶导数的表达式然后把0代入是不行的,不像ex,sinx,cosxe^x,sinx,cosxex,sinx,cosx那样有规律,所以采用莱布尼茨公式(就是那个和二项式定理长得很像的)找递推关系:
y=arctanxy′=11+x2(1+x2)y′=1\begin{aligned}
y=arctanx\\
y'=\frac{1}{1+x^2}\\
(1+x^2)y'=1
\end{aligned}y=arctanxy′=1+x21(1+x2)y′=1
两边用莱布尼茨公式求n-1阶导数
注意到y′(n−1)=y(n)y'^{(n-1)}=y^{(n)}y′(n−1)=y(n)
Cn−10(1+x2)y(n)+Cn−11(2x)y(n−1)+Cn−12(2)y(n−2)=0\begin{aligned}
C_{n-1}^0(1+x^2)y^{(n)}+C_{n-1}^1(2x)y^{(n-1)}+C_{n-1}^2(2)y^{(n-2)}=0
\end{aligned}Cn−10(1+x2)y(n)+Cn−11(2x)y(n−1)+Cn−12(2)y(n−2)=0
x=0x=0x=0代入
y(n)∣x=0+(n−1)(n−2)y(n−2)∣x=0=0\begin{aligned}
y^{(n)}\vert_{x=0}+(n-1)(n-2)y^{(n-2)}\vert_{x=0}=0
\end{aligned}y(n)∣x=0+(n−1)(n−2)y(n−2)∣x=0=0
即隔一项有递推关系
因为y(0)∣x=0=arctan0=0y^{(0)}\vert_{x=0}=arctan0=0y(0)∣x=0=arctan0=0,得所有偶数阶导数为0,考虑奇数n=2k−1,k∈N∗n=2k-1,k\in\Bbb N^*n=2k−1,k∈N∗
y(2k−1)∣x=0=−(2k−2)(2k−3)y(2k−3)∣x=0=(−1)2(2k−2)(2k−3)(2k−4)(2k−5)y(2k−5)∣x=0=⋯=(−1)k−1(2k−2)(2k−3)⋯2⋅1⋅y(1)∣x=0=(−1)k−1(2k−2)!\begin{aligned}
y^{(2k-1)}\vert_{x=0}=&-(2k-2)(2k-3)y^{(2k-3)}\vert_{x=0}\\
=&(-1)^2(2k-2)(2k-3)(2k-4)(2k-5)y^{(2k-5)}\vert_{x=0}\\
=&\cdots\\
=&(-1)^{k-1}(2k-2)(2k-3)\cdots2\cdot1\cdot y^{(1)}\vert_{x=0}\\
=&(-1)^{k-1}(2k-2)!\\
\end{aligned}y(2k−1)∣x=0=====−(2k−2)(2k−3)y(2k−3)∣x=0(−1)2(2k−2)(2k−3)(2k−4)(2k−5)y(2k−5)∣x=0⋯(−1)k−1(2k−2)(2k−3)⋯2⋅1⋅y(1)∣x=0(−1)k−1(2k−2)!
然后代入麦克劳林展开式:
y=∑n=0∞y(n)∣x=0n!xn=∑k=1∞y(2k−1)∣x=0(2k−1)!x2k−1=∑k=1∞(−1)k−12k−1x2k−1=x−x33+x55−x77+⋯+(−1)n−1x2n−12n−1+⋯\begin{aligned}
y=&\sum_{n=0}^\infty\frac{y^{(n)}\vert_{x=0}}{n!}x^n\\
=&\sum_{k=1}^\infty\frac{y^{(2k-1)}\vert_{x=0}}{(2k-1)!}x^{2k-1}\\
=&\sum_{k=1}^\infty\frac{(-1)^{k-1}}{2k-1}x^{2k-1}\\
=&x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots+(-1)^{n-1}\frac{x^{2n-1}}{2n-1}+\cdots
\end{aligned}y====n=0∑∞n!y(n)∣x=0xnk=1∑∞(2k−1)!y(2k−1)∣x=0x2k−1k=1∑∞2k−1(−1)k−1x2k−1x−3x3+5x5−7x7+⋯+(−1)n−12n−1x2n−1+⋯