常见泰勒展开公式及复杂泰勒展开求法

https://blog.csdn.net/weixin_45792450/article/details/104404432

e x = 1 + x + 1 2 ! x 2 + 1 3 ! x 3 + . . . + 1 n ! x n + o ( x n ) {e^x} = 1 + x + {1 \over {2!}}{x^2} + {1 \over {3!}}{x^3} + ... + {1 \over {n!}}{x^n} + o({x^n}) ex=1+x+2!1​x2+3!1​x3+...+n!1​xn+o(xn)

sin ⁡ x = x − 1 3 ! x 3 + 1 5 ! x 5 − . . . + 1 ( 2 n + 1 ) ! x 2 n + 1 + o ( x 2 n + 2 ) \sin x = x - {1 \over {3!}}{x^3} + {1 \over {5!}}{x^5} - ... + {1 \over {(2n + 1)!}}{x^{2n + 1}} + o({x^{2n + 2}}) sinx=x−3!1​x3+5!1​x5−...+(2n+1)!1​x2n+1+o(x2n+2)

cos ⁡ x = 1 − 1 2 ! x 2 + 1 4 ! x 4 − . . . + 1 ( 2 n ) ! x 2 n + o ( x 2 n + 1 ) \cos x = 1 - {1 \over {2!}}{x^2} + {1 \over {4!}}{x^4} - ... + {1 \over {(2n)!}}{x^{2n}} + o({x^{2n + 1}}) cosx=1−2!1​x2+4!1​x4−...+(2n)!1​x2n+o(x2n+1)

ln ⁡ ( 1 + x ) = x − 1 2 x 2 + 1 3 x 3 − . . . + ( − 1 ) n − 1 1 n x n + o ( x n ) \ln (1 + x) = x - {1 \over 2}{x^2} + {1 \over 3}{x^3} - ... + {( - 1)^{n - 1}}{1 \over n}{x^n} + o({x^n}) ln(1+x)=x−21​x2+31​x3−...+(−1)n−1n1​xn+o(xn)

( 1 + x ) m = 1 + m x + m ( m − 1 ) 2 x 2 + . . . + m ( m − 1 ) . . . ( m − n + 1 ) n ! x n + o ( x n ) {(1 + x)^m} = 1 + mx + {{m(m - 1)} \over 2}{x^2} + ... + {{m(m - 1)...(m - n + 1)} \over {n!}}{x^n} + o({x^n}) (1+x)m=1+mx+2m(m−1)​x2+...+n!m(m−1)...(m−n+1)​xn+o(xn)

1 1 + x = 1 − x + x 2 − x 3 + . . . + ( − 1 ) n − 1 x n + o ( x n ) {1 \over {1 + x}} = 1 - x + {x^2} - {x^3} + ... + {( - 1)^{n - 1}}{x^n} + o({x^n}) 1+x1​=1−x+x2−x3+...+(−1)n−1xn+o(xn)

1 1 + x = 1 − 1 2 x + 1 × 3 2 × 4 x 2 − 1 × 3 × 5 2 × 4 × 6 x 3 + . . . + ( − 1 ) n − 1 ( 2 n − 1 ) ! ! ( 2 n ) ! ! x n + o ( x n ) {1 \over {\sqrt {1 + x} }} = 1 - {1 \over 2}x + {{1 \times 3} \over {2 \times 4}}{x^2} - {{1 \times 3 \times 5} \over {2 \times 4 \times 6}}{x^3} + ... + {( - 1)^{n - 1}}{{(2n - 1)!!} \over {(2n)!!}}{x^n} + o({x^n}) 1+x ​1​=1−21​x+2×41×3​x2−2×4×61×3×5​x3+...+(−1)n−1(2n)!!(2n−1)!!​xn+o(xn)

分解函数法 f ( x ) = f 1 ( x ) + f 2 ( x ) + . . . + f n ( x ) f(x) = {f_1}(x) + {f_2}(x) + ... + {f_n}(x) f(x)=f1​(x)+f2​(x)+...+fn​(x)，对各函数分别展开，然后按多项式次数汇总即可

求导积分法 f ′ ( x ) = a 0 + a 1 x + . . . + a n x n + o ( x n ) f'(x) = {a_0} + {a_1}x + ... + {a_n}{x^n} + o({x^n}) f′(x)=a0​+a1​x+...+an​xn+o(xn)，左右两边积分即得原函数的泰勒展开 对两边求导可得导函数泰勒展开，对两边积分可得原函数泰勒展开

复合函数法 若 f ( x ) = a 0 + a 1 x + . . . + a n x n + o ( x n ) , g ( x ) = b 1 x + b 2 x 2 . . . + b n x n + o ( x n ) 若f(x) = {a_0} + {a_1}x + ... + {a_n}{x^n} + o({x^n}),g(x) = {b_1}x + {b_2}{x^2}... + {b_n}{x^n} + o({x^n}) 若f(x)=a0​+a1​x+...+an​xn+o(xn),g(x)=b1​x+b2​x2...+bn​xn+o(xn)

则 f ( g ( x ) ) = a 0 + a 1 [ b 1 x + . . . + b n x n + o ( x n ) ] + . . . 则f(g(x)) = {a_0} + {a_1}[{b_1}x + ... + {b_n}{x^n} + o({x^n})] + ... 则f(g(x))=a0​+a1​[b1​x+...+bn​xn+o(xn)]+...

+ a n [ b 1 x + . . . + b n x n + o ( x n ) ] n + o ( x n ) + {a_n}{[{b_1}x + ... + {b_n}{x^n} + o({x^n})]^n} + o({x^n}) +an​[b1​x+...+bn​xn+o(xn)]n+o(xn)

由上述方法加上初等函数泰勒展开，很容易算出诸如 arctan ⁡ x , 1 1 + x 2 , 1 + x \arctan x,{1 \over {1 + {x^2}}},\sqrt {1 + x} arctanx,1+x21​,1+x ​之类的泰勒展开