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!1x2+3!1x3+...+n!1xn+o(xn)
sin
x
\sin x
sinx
sin
x
=
x
−
1
3
!
x
3
+
1
5
!
x
5
−
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.
.
+
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!1x3+5!1x5−...+(2n+1)!1x2n+1+o(x2n+2)
cos
x
\cos x
cosx
cos
x
=
1
−
1
2
!
x
2
+
1
4
!
x
4
−
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.
.
+
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!1x2+4!1x4−...+(2n)!1x2n+o(x2n+1)
ln
(
1
+
x
)
\ln (1 + x)
ln(1+x)
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−21x2+31x3−...+(−1)n−1n1xn+o(xn)
(
1
+
x
)
m
{(1 + x)^m}
(1+x)m
(
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 \over {1 + x}}
1+x1
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 \over {\sqrt {1 + x} }}
1+x1
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+x1=1−21x+2×41×3x2−2×4×61×3×5x3+...+(−1)n−1(2n)!!(2n−1)!!xn+o(xn)
稍复杂的函数泰勒展开求法
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+a1x+...+anxn+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+a1x+...+anxn+o(xn),g(x)=b1x+b2x2...+bnxn+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[b1x+...+bnxn+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[b1x+...+bnxn+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
