大学生数学竞赛，不是数学建模，分为数学组和非数学组，我是非数学组。
全国初赛只考高数，全国总决赛考高数和线性代数。当年我是我们学校唯一 一个进入全国总决赛的，非数学组就我一个，数学组全军覆没。
下面贴一下我当年的习题和笔记：
上半册：
36个习题的小结
极限题 1 2 3 4 5 12 17 18 21 23 35
微分中值定理 7 8 9 10 11 13 14 15 16 20.5
微分方程25 30
导数6 18.5 24 26 27 28 29 31
多解题5 11 19
易错题12
施笃兹定理1 2 19
三角换元4
达布定理 11 32
1，设 X 1 ∈ ( 0 , 1 ) , X n + 1 = X n ( 1 − X n ) X_{1} \in(0,1), \quad X_{n+1}=X_{n}\left(1-X_{n}\right) \quad X1∈(0,1),Xn+1=Xn(1−Xn) 证明 lim n → ∞ n X n = 1 \lim _{n \rightarrow \infty} n X_{n}=1 limn→∞nXn=1
证明:
若 0 < X n < 1 0<X_{n}<1 0<Xn<1，则 0 < X n + 1 = X n ( 1 − X n ) < 1 0<X_{n+1}=X_{n}\left(1-X_{n}\right)<1 0<Xn+1=Xn(1−Xn)<1
∴ ∀ n , 0 < X n < 1 , \therefore \forall{n}, \,0<X_{n}<1, \quad ∴∀n,0<Xn<1,且 { X n } \left\{X_{n}\right\} {
Xn} 递减
∴ lim n → ∞ X n = c \therefore \lim _{n \rightarrow \infty} X_{n}=c ∴limn→∞Xn=c
∴ c = c ( 1 − c ) \therefore c=c(1-c) ∴c=c(1−c) 即 c = 0 c=0 c=0
因 1 X n + 1 = 1 X n + 1 1 − X n \frac{1}{X_{n+1}}=\frac{1}{X_{n}}+\frac{1}{1-X_{n}} Xn+11=Xn1+1−Xn1
∴ lim n → ∞ 1 n X n = lim n → ∞ 1 X n + 1 − 1 X n n + 1 − n = lim n → ∞ 1 1 − X n = 1 ( \therefore \lim _{n \rightarrow \infty} \frac{1}{n X_{n}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{X_{n+1}}-\frac1{X_n}}{n+1-n}=\lim _{n \rightarrow \infty}\frac{1}{1-X_{n}}=1( ∴limn→∞nXn1=limn→∞n+1−nXn+11−Xn1=limn→∞1−Xn1=1(施笃兹定理 ) ) )
∴ lim n → ∞ n X n = 1 \therefore\lim_{n \rightarrow \infty}n X_{n}=1 ∴limn→∞nXn=1
2，设 a 0 = 0 , a 1 = 1 + sin ( − 1 ) , a n = 1 + sin ( a n − 1 − 1 ) a_{0}=0, \quad a_{1}=1+\sin (-1), \quad a_{n}=1+\sin \left(a_{n-1}-1\right) \quad a0=0,a1=1+sin(−1),an=1+sin(an−1−1) 求 n → ∞ lim 1 n ∑ k = 1 n a k _{n \rightarrow \infty}^{\lim } \frac{1}{n} \quad \sum_{k=1}^{n} a_{k} n→∞limn1∑k=1nak
解: 设 b n = a n − 1 b_{n}=a_{n}-1 bn=an−1
则 b n = sin b n − 1 b_{n}=\sin b_{n-1} bn=sinbn−1
∴ ∣ b n ∣ = ∣ sin b n − 1 ∣ ≤ ∣ b n − 1 ∣ \therefore\left|b_{n}\right|=\left|\sin b_{n-1}\right| \leq\left|b_{n-1}\right| ∴∣bn∣=∣sinbn−1∣≤∣bn−1∣
∴ lim n → ∞ b n = c \therefore \quad \lim _{n \rightarrow \infty} b_{n}=c ∴limn→∞bn=c
∴ c = sin c \therefore c=\sin c ∴c=sinc 即 c = 0 c=0 c=0
∴ lim n → ∞ a n = 1 \therefore \lim _{n \rightarrow \infty} a_{n}=1 ∴limn→∞an=1
∴ lim n → ∞ ∑ k = 1 n a k n = lim n → ∞ a n + 1 n + 1 − n = 1 \therefore \lim _{n \rightarrow \infty} \frac{\sum_{k=1}^{n} a_{k}}{n}=\lim _{n \rightarrow \infty} \frac{a_{n+1}}{n+1-n}=1 \quad ∴limn→∞n∑k=1nak=limn→∞n+1−nan+1=1 (施笃兹定理)
3， 设 b n = ∑ k = 0 n 1 c n k 求 n → ∞ l i m b n { 设b }_{n}=\sum_{k=0}^{n} \frac{1}{c_{n}^{k}} { 求 }_{n \rightarrow \infty}^{l i m} b_{n} 设bn=∑k=0ncnk1求n→∞limbn
解 : b n = ∑ k = 0 n k ! ( n − k ) ! n ! { 解:\,} b_{n}=\frac{\sum_{k=0}^{n} k !(n-k)!}{n !} 解:bn=n!∑k=0nk!(n−k)!
∴ ( n + 2 ) b n = ∑ k = 0 n k ! ( n − k ) ! ( n + 1 − k + k + 1 ) n ! \therefore(n+2) b_{n}=\frac{\sum_{k=0}^{n} k !(n-k) !(n+1-k+k+1)}{n !} ∴(n+2)bn=n!∑k=0nk!(n−k)!(n+1−k+k+1)
= ∑ k = 0 n k ! ( n + 1 − k ) ! + ∑ k = 0 n ( k + 1 ) ! ( n − k ) ! n ! =\frac{\sum_{k=0}^{n} k !(n+1-k) !+\sum_{k=0}^{n} (k+1) !(n-k) !}{n !} =n!∑k=0nk!(n+1−k)!+∑k=0n(k+1)!(n−k)!
= ∑ k = 0 n k ! ( n + 1 − k ) ! + ∑ k = 0 n + 1 k ! ( n + 1 − k ) ! − ( n + 1 ) ! n ! =\frac{\sum_{k=0}^{n} k !(n+1-k) !+\sum_{k=0}^{n+1} k !(n+1-k) !-(n+1) !}{n !} =n!∑k=0nk!(n+1−k)!+∑k=0n+1k!(n+1−k)!−(n+1)!
= 2 ∑ k = 0 n + 1 k ! ( n + 1 − k ) ! − 2 ( n + 1 ) ! n ! =\frac{2 \sum_{k=0}^{n+1} k !(n+1-k) !-2(n+1) !}{n !} =n!2∑k=0n+1k!(n+1−k)!−2(n+1)!
= 2 ( n + 1 ) b n + 1 − 2 ( n + 1 ) =2(n+1) b_{n+1}-2(n+1) =2(n+1)bn+1−2(n+1)
即 b n + 1 = ( n + 2 ) b n 2 ( n + 1 ) + 1 { 即 } b_{n+1}=\frac{(n+2) b_{n}}{2(n+1)}+1 即bn+1=2(n+1)(n+2)bn+1
若 2 < b n < 2 + 6 n , n > 2 { 若 } 2<b_{n}<2+\frac{6}{n}, \quad n>2 若2<bn<2+n6,n>2
则 b n + 1 < n + 2 2 ( n + 1 ) ( 2 + 6 n ) + 1 = 2 + 1 n + 1 + 3 ( n + 2 ) n ( n + 1 ) ≤ 2 + 1 n + 1 + 5 n + 1 = 2 + 6 n + 1 b n + 1 > n + 2 n + 1 + 1 > 2 则 2 < b n + 1 < 6 n + 1 因 2 < b 3 = 8 3 < 4 ， 故 2 < b n < 2 + 6 n （ n > 2 ） 成 立 ∴ lim n → ∞ b n = 2 \text { 则 } b_{n+1}<\frac{n+2}{2(n+1)}\left(2+\frac{6}{n}\right)+1=2+\frac{1}{n+1}+\frac{3(n+2)}{n(n+1)} \leq 2+\frac{1}{n+1}+\frac{5}{n+1}=2+\frac{6}{n+1} \\ b_{n+1}>\frac{n+2}{n+1}+1>2 \\ 则2<b_{n+1}<\frac{6}{n+1} \\ 因2<b_{3}=\frac{8}{3}<4 ，故2<b_{n}<2+\frac{6}{n} （n>2）成立 \\ \therefore \lim _{n \rightarrow \infty} b_{n}=2 则 bn+1<2(n+1)n+2(2+n6)+1=2+n+11+n(n+1)3(n+2)≤2+n+11+n+15=2+n+16bn+1>n+1n+2+1>2则2<bn+1<n+16因2<b3=38<4，故2<bn<2+n6（n>2）成立∴limn→∞bn=2
4，设 a 1 = 1 2 , a n = 1 + a n − 1 2 , 求 极 限 lim n → ∞ a 1 a 2 ⋯ a n a_{1}=\sqrt{\frac{1}{2}} \,,\,a_{n}=\sqrt{\frac{1+a_{n-1}}{2}} \,,\,求极限 \lim _{n \rightarrow \infty} a_{1} a_{2} \cdots a_{n} a1=21,an=21+an−1,求极限limn→∞a1a2⋯an
解: a n − 1 = 2 a n 2 − 1 a_{n-1}=2 a_{n}^{2}-1 an−1=2an2−1 类似 cos 2 θ = 2 cos 2 θ − 1 \cos 2 \theta=2 \cos^2 \theta-1 cos2θ=2cos2θ−1
a 1 = 2 2 = cos π 4 a_{1}=\frac{\sqrt{2}}{2}=\cos \frac{\pi}{4} a1=22=cos4π
若 a n − 1 = cos π 2 n a_{n-1}=\cos \frac{\pi}{2^{n}} an−1=cos2nπ 则 a n = cos π 2 n + 1 a_{n}=\cos \frac{\pi}{2^{n+1}} an=cos2n+1π
∴ ∀ n , 有 a n = cos π 2 n + 1 \therefore \forall{n}, 有a_{n}=\cos \frac{\pi}{2^{n+1}} ∴∀n,有an=cos2n+1π
∴ a 1 a 2 ⋯ a n = cos π 2 2 cos π 2 3 ⋯ cos π 2 n + 1 = 1 2 n sin π 2 n + 1 \therefore a_{1} a_{2} \cdots a_{n}=\cos \frac{\pi}{2^{2}} \cos \frac{\pi}{2^{3}} \cdots \cos \frac{\pi}{2^{n+1}}=\frac{1}{2^{n} \sin \frac{\pi}{2^{n+1}}} ∴a1a2⋯an=cos22πcos23π⋯cos2n+1π=2nsin2n+1π1
∴ lim n → ∞ a 1 a 2 ⋯ a n = 2 π \therefore \lim _{n \rightarrow \infty} a_{1} a_{2} \cdots a_{n}=\frac{2}{\pi} ∴limn→∞a1a2⋯an=π2
5， a 1 > 0 , b 1 > 0 , c 1 > 0 , a 1 + b 1 + c 1 = 1 , a n + 1 = a n 2 + 2 b n c n , b n + 1 = b n 2 + 2 a n c n , c n + 1 = c n 2 + 2 a n b n , 证明 lim n → ∞ a n 存在并求它 解: a n + 1 + b n + 1 + c n + 1 = ( a n + b n + c n ) 2 = 1 方法1 ( 我的方法) a n + 1 2 + b n + 1 2 + c n + 1 2 = ∑ a n 4 + 4 a n b n c n ∑ a n + 4 ∑ a n 2 b n 2 = ( a n 2 + b n 2 + c n ) 2 + 2 ( ∑ a n 2 b n 2 + 2 a n b n c n ∑ a n ) = ( a n 2 + b n 2 + c n 2 ) 2 + 2 ( a n b n + b n c n + c n a n ) 2 设 S n = a n 2 + b n 2 + c n 2 则 1 3 ≤ S n < a n + b n + c n = 1 ∴ S n + 1 = S n 2 + 2 ( 1 − S n 2 ) 2 = 3 2 S n 2 − S n + 1 2 = S n + 1 2 ( S n − 1 ) ( 3 S n − 1 ) < S n a_{1}>0, b_{1}>0, c_{1}>0, a_{1}+b_{1}+c_{1}=1, a_{n+1}=a_{n}^{2}+2 b_{n} c_{n}, b_{n+1}=b_{n}^{2}+2 a_{n} c_{n}, c_{n+1}=c_{n}^{2}+2 a_{n} b_{n}, \quad \text { 证明 } \lim _{n \rightarrow \infty} a_{n} \text { 存在并求它 }\\ \text { 解: } a_{n+1}+b_{n+1}+c_{n+1}=\left(a_{n}+b_{n}+c_{n}\right)^{2}=1\\ \text { 方法1 }\left(\right.\text { 我的方法) }\\ \quad a_{n+1}{ }^{2}+b_{n+1}{ }^{2}+c_{n+1}{ }^{2}=\sum a_{n}{ }^{4}+4 a_{n} b_{n} c_{n} \sum a_{n}+4 \sum a_{n}{ }^{2} b_{n}{ }^{2}\\ =\left(a_{n}^{2}+b_{n}^{2}+c_{n}\right)^{2}+2\left(\sum a_{n}^{2} b_{n}^{2}+2 a_{n} b_{n} c_{n} \sum a_{n}\right)\\ =\left(a_{n}^{2}+b_{n}^{2}+c_{n}^{2}\right)^{2}+2\left(a_{n} b_{n}+b_{n} c_{n}+c_{n} a_{n}\right)^{2}\\ \text { 设 } S_{n}=a_{n}^{2}+b_{n}^{2}+c_{n}^{2} \text { 则 } \frac{1}{3} \leq S_{n}<a_{n}+b_{n}+c_{n}=1\\ \therefore S_{n+1}=S_{n}^{2}+2\left(\frac{1-S_{n}}{2}\right)^{2}=\frac{3}{2} S_{n}^{2}-S_{n}+\frac{1}{2}=S_{n}+\frac{1}{2}\left(S_{n}-1\right)\left(3 S_{n}-1\right)<S_{n} a1>0,b1>0,c1>0,a1+b1+c1=1,an+1=an2+2bncn,bn+1=bn2+2ancn,cn+1=cn2+2anbn, 证明 limn→∞an 存在并求它 解: an+1+bn+1+cn+1=(an+bn+cn)2=1 方法1 ( 我的方法) an+12+bn+12+cn+12=∑an4+4anbncn∑an+4∑an2bn2=(an2+bn2+cn)2+2(∑an2bn2+2anbncn∑an)=(an2+bn2+cn2)2+2(anbn+bncn+cnan)2 设 Sn=an2+bn2+cn2 则 31≤Sn<an+bn+cn=1∴Sn+1=Sn2+2(21−Sn)2=23Sn2−Sn+21=Sn+21(Sn−1)(3Sn−1)<Sn
∴ lim n → ∞ S n 存在, 设为 c ( 1 3 ≤ c < 1 ) 则 c = 3 2 c 2 − c + 1 2 即 c = 1 3 ∴ lim n → ∞ a n 2 + b n 2 + c n 2 = 1 3 由于柯西不等式, lim n → ∞ a n = lim n → ∞ b n = lim n → ∞ c n = 1 3 方法 2 (书上的方法) (1)设 M n = max { a n , b n , c n } , m n = min { a n , b n , c n } 若 a n ≥ b n ≥ c n , 则 M n = a n , m n = c n \therefore \lim _{n \rightarrow \infty} S_{n} \text { 存在, 设为 } c\left(\frac{1}{3} \leq c<1\right)\\ \text { 则 } c=\frac{3}{2} c^{2}-c+\frac{1}{2} \text { 即 } c=\frac{1}{3}\\ \therefore \lim _{n \rightarrow \infty} a_{n}^{2}+b_{n}^{2}+c_{n}^{2}=\frac{1}{3}\\ \text { 由于柯西不等式, } \lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} b_{n}=\lim _{n \rightarrow \infty} c_{n}=\frac{1}{3}\\ \text { 方法 } 2 \text { (书上的方法) }\\ \text { (1)设 } M_{n}=\max \left\{a_{n}, b_{n}, c_{n}\right\}, m_{n}=\min \left\{a_{n}, b_{n}, c_{n}\right\}\\ \text { 若 } a_{n} \geq b_{n} \geq c_{n}, \text { 则 } M_{n}=a_{n}, \quad m_{n}=c_{n} ∴limn→∞Sn 存在, 设为 c(31≤c<1) 则 c=23c2−c+21 即 c=31∴limn→∞an2+bn2+cn2=31 由于柯西不等式, limn→∞an=limn→∞bn=limn→∞cn=31 方法 2 (书上的方法) (1)设 Mn=max{
an,bn,cn},mn=min{
an,bn,cn} 若 an≥bn≥cn, 则 Mn=an,mn=cn
a n + 1 = a n 2 + 2 b n c n < a n 2 + a n b n + a n c n = a n = M n b n + 1 = b n 2 + 2 a n c n < a n 2 + a n b n + a n c n = a n = M n c n + 1 = c n 2 + 2 a n c n < a n c n + a n b n + a n 2 = a n = M n ∴ M n + 1 = max { a n + 1 , b n + 1 , c n + 1 } < M n 同理 m n + 1 > m n ( 2 ) M n + 1 − m n + 1 ≤ ( M n − m n ) 2 (证明和上面差不多, 略) ∴ lim n → ∞ M n = lim n → ∞ m n 即 lim n → ∞ a n = lim n → ∞ b n = lim n → ∞ c n = 1 3 a_{n+1}=a_{n}^{2}+2 b_{n} c_{n}<a_{n}^{2}+a_{n} b_{n}+a_{n} c_{n}=a_{n}=M_{n}\\ b_{n+1}=b_{n}^{2}+2 a_{n} c_{n}<a_{n}^{2}+a_{n} b_{n}+a_{n} c_{n}=a_{n}=M_{n}\\ c_{n+1}=c_{n}^{2}+2 a_{n} c_{n}<a_{n} c_{n}+a_{n} b_{n}+a_{n}^{2}=a_{n}=M_{n}\\ \therefore M_{n+1}=\max \left\{a_{n+1}, b_{n+1}, c_{n+1}\right\}<M_{n}\\ \text { 同理 } m_{n+1}>m_{n}\\ (2)M_{n+1}-m_{n+1} \leq\left(M_{n}-m_{n}\right)^{2} \quad \text { (证明和上面差不多, 略) }\\ \therefore \lim _{n \rightarrow \infty} M_{n}=\lim _{n \rightarrow \infty} m_{n}\\ \text { 即 } \lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} b_{n}=\lim _{n \rightarrow \infty} c_{n}=\frac{1}{3} an+1=an2+2bncn<an2+anbn+ancn=an=Mnbn+1=bn2+2ancn<an2+anbn+ancn=an=Mncn+1=cn2+2ancn<ancn+anbn+an2=an=Mn∴Mn+1=max{
an+1,bn+1,cn+1}<Mn 同理 mn+1>mn(2)Mn+1−mn+1≤(Mn−mn)2 (证明和上面差不多, 略) ∴limn→∞Mn=limn→∞mn 即 limn→∞an=limn→∞bn=limn→∞cn=31
6，求 y = a r c t a n x y=arctan\,x y=arctanx 在 x = 0 x=0 x=0 处的 n n n 阶导数
解: ( 1 + x 2 ) y ′ = 1 \left(1+x^{2}\right) y'=1 (1+x2)y′=1
由莱布伦茨公式, n ( n − 1 ) y ( n − 1 ) + 2 n x y ( n ) + ( 1 + x 2 ) y ( n + 1 ) = 0 n(n-1) y^{(n-1)}+2n x y^{(n)}+\left(1+x^{2}\right) y^{(n+1)}=0 n(n−1)y(n−1)+2nxy(n)+(1+x2)y(n+1)=0
∴ \therefore ∴ 在 x = 0 x=0 x=0 处, y ( n + 1 ) = − n ( n − 1 ) y ( n − 1 ) y^{(n+1)}=-n(n-1) y^{(n-1)} y(n+1)=−n(n−1)y(n−1)
∴ \therefore ∴ 当 n n n 为偶数时, y ( n ) ( 0 ) = 0 \quad y^{(n)}{ }_{(0)}=0 y(n)(0)=0
当 n n n 为奇数时, y ( n ) ( 0 ) = ( − 1 ) n − 1 2 ⋅ ( n − 1 ) ! y^{(n)}{ }_{(0)}=(-1)^{\frac{n-1}{2}} \cdot(n-1) ! y(n)(0)=(−1)2n−1⋅(n−1)!
下册
知识点
4， ( sin x ) ( n ) = sin ( x + n ⋅ π 2 ) (\sin x)^{(n)}=\sin \left(x+n \cdot \frac{\pi}{2}\right) (sinx)(n)=sin(x+n⋅2π)
( cos x ) ( n ) = cos ( x + n ⋅ π 2 ) (\cos x)^{(n)}=\cos \left(x+n \cdot \frac{\pi}{2}\right) (cosx)(n)=cos(x+n⋅2π)
5， e x = 1 + x + x 2 2 ! + x 3 3 ! + e^{x}=1+x+\frac{x^{2}}{2 !}+\frac{x 3}{3 !}+ ex=1+x+2!x2+3!x3+
sin x = x − x 2 3 ! + x 3 5 ! − x 7 7 ! ⋯ \sin x=x-\frac{x^{2}}{3 !}+\frac{x^{3}}{5 !}-\frac{x^{7}}{7 !} \cdots sinx=x−3!x2+5!x3−7!x7⋯
cos x = 1 − x 3 2 ! + x 4 4 ! − x 6 6 ! \cos x=1-\frac{x^{3}}{2 !}+\frac{x^{4}}{4 !}-\frac{x^{6}}{6 !} cosx=1−2!x3+4!x4−6!x6
ln ( 1 + x ) = x − x 2 2 + x 3 3 + ⋯ \ln (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}+\cdots ln(1+x)=x−2x2+3x3+⋯
ln ( 1 − x ) = − ( x + x 2 2 + x 3 3 + ⋯ ) \ln (1-x)=-\left(x+\frac{x^{2}}{2}+\frac{x^{3}}{3}+\cdots\right) ln(1−x)=−(x+2x2+3x3+⋯)
6，柯西不等式:
设f(x), g(x)在区间[a,b]上连续, 则
( ∫ a b f 2 ( x ) d x ) ( ∫ a b g 2 ( x ) d x ) ⩾ ( ∫ a b f ( x ) g ( x ) d x ) 2 \left(\int_{a}^{b} f^{2}(x) d x\right)\left(\int_{a}^{b} g^{2}(x) d x\right) \geqslant\left(\int_{a}^{b} f(x) g(x) d x\right)^{2} (∫abf2(x)dx)(∫abg2(x)dx)⩾(∫abf(x)g(x)dx)2
7，积分中值定理:
设f(x)在区间[a,b]上可连续, g(x)在区间[a,b]上可积且不变号，则
∃ ε ∈ [ a , b ] , ∫ a b f ( x ) g ( x ) d x = f ( ε ) ∫ a b g ( x ) d x \exists \varepsilon \in[a, b], \int _a^b f(x) g(x) d x=f(\varepsilon) \int _a^b g(x)dx ∃ε∈[a,b],∫abf(x)g(x)dx=f(ε)∫abg(x)dx