北大本科06数分期中试题(李伟固命题)
前些年在百度文库(http://wenku.baidu.com/link?url=Y_HDeeYMcyEGuUJWt3fNqC7N08AEqMVfleNVGcv7hC2t9EVO0-MFFHWuqnLYyiDJ4H7ATgg1fOQGN1Lta2RW4Z4pOrm6aZ468WkrTqNqZzG)找到一份李伟固命制的06年北大期中考试题,当时感觉难度稍大,正好结识了Veer大神,便把这份题发给他。事实证明,V神秒得还是很顺利!
1.给定实数λi(1≤i≤n),满足n∑i=1λki>0(k=1,2,3,⋯).令f(x)=n∏i=111−λix.证明: f(k)(0)>0,k=1,2,3,⋯.
证明:令g(x)=lnf(x)=−ln(1−λ1x)−ln(1−λ2x)−⋯−ln(1−λnx)(x∈U(0;δ)使得对∀λi,有1−λix>0),则
g(x)=(n∑i=1λi)x+(12n∑i=1λ2i)x2+⋯+(1kn∑i=1λki)xk+⋯由函数幂级数展开的唯一性可知g(k)(0)=1kn∑i=1λki>0(x∈U(0;δ)).
另一方面f(x)=eg(x)(x∈U(0;δ)).首先注意到对任意可导函数F(x),有(eF(x))′=F′(x)eF(x).其次注意到对可导函数组F1,F2,⋯,Fs,有(F1F2F3⋯Fs)′=F′1F2F3⋯Fs+F1F2′F3⋯Fs+⋯+F1F2F3⋯Fs′,从而归纳可证
f(k)(x)=(eg(x))(k)=(∑j∈N+,ki∈N+g(k1)(x)g(k2)(x)⋯g(kj)(x))eg(x).由g(k)(0)>0,k=1,2,3,⋯且g(0)=0,所以f(k)(0)>0,k=1,2,3,⋯.
补充:可知f(x)=n∏i=111−λix=(1+λ1x+λ21x2+⋯)(1+λ2x+λ22x2+⋯)⋯(1+λnx+λ2nx2+⋯).由幂级数的乘积公式归纳可得
f(k)(0)=∑k1+k2+⋯+kn=k(k1,k2,⋯,kn)λk11λk22⋯λknn(其中ki∈N).若能通过n∑i=1λki>0(k=1,2,3,⋯)得出∑k1+k2+⋯+kn=k(k1,k2,⋯,kn)λk11λk22⋯λknn>0即可得到证明.
2.令D={u=(x,y)∈R2|‖. f(u)=f(x,y)是全平面上的连续可微函数满足\left\| {\nabla f\left( {0,0} \right)} \right\| = 1,\left\| {\nabla f\left( u \right) - \nabla f\left( v \right)} \right\| \le \left\| {u - v} \right\|.那么对于任意的u,v\in D,证明函数f|_D在D中唯一点处达到其最大值.
证明:对u\in D,有\left\| {\nabla f\left( u \right) - \nabla f\left( {0,0} \right)} \right\| \le \left\| {u - \left( {0,0} \right)} \right\|,即
1 - \left\| {\nabla f\left( u \right)} \right\| = \left\| {\nabla f\left( {0,0} \right)} \right\| - \left\| {\nabla f\left( u \right)} \right\| \le \left\| {\nabla f\left( u \right) - \nabla f\left( {0,0} \right)} \right\| \le \left\| u \right\|,亦即1 - \left\| u \right\| \le \left\| {\nabla f\left( u \right)} \right\|,则\nabla f\left( u \right) \ne \left( {0,0} \right),所以f的最大值只可能在边界上取得.
易知f在其边界上的函数可设为g\left( t \right) = f\left( {\frac{1}{2}\cos \theta ,\frac{1}{2}\sin \theta } \right),g'\left( \theta \right) = \frac{1}{2}\left( { - \sin \theta {f_x} + \cos \theta {f_y}} \right),\theta \in \left[ {0,2\pi } \right).现假设f在其边界上有两点取得最大值,不妨设为\theta=\theta_1和\theta=\theta_2,记{u_1} = \left( {\frac{1}{2}\cos {\theta _1},\frac{1}{2}\sin {\theta _1}} \right),{u_2} = \left( {\frac{1}{2}\cos {\theta _2},\frac{1}{2}\sin {\theta _2}} \right),则由g'\left( {{\theta _1}} \right) = g'\left( {{\theta _2}} \right) = 0可得- \sin {\theta _1}{f_x}\left( {{u_1}} \right) + \cos {\theta _1}{f_y}\left( {{u_1}} \right) = 0,即\nabla f\left( {{u_1}} \right)与\left( { - \sin {\theta _1},\cos {\theta _1}} \right)垂直即可.设\nabla f\left( {{u_1}} \right) = {a_1}\left( {\cos {\theta _1},\sin {\theta _1}} \right),由于f在u_1处取得最大值,则f在u_1点沿方向(\cos\theta_1,\sin \theta_1)的方向导数需大于等于0,所以a_1\geq 0.
另一方面由\left\| {\nabla f\left( {0,0} \right)} \right\| - \left\| {\nabla f\left( {{u_1}} \right)} \right\| \le \left\| {\nabla f\left( {0,0} \right) - \nabla f\left( {{u_1}} \right)} \right\|当且仅当{\nabla f\left( {{0,0}} \right)}与{\nabla f\left( {{u_1}} \right)}异向取等可知
\left\| {\nabla f\left( {0,0} \right)} \right\| - \left\| {{u_1}} \right\| \le \left\| {\nabla f\left( {{u_1}} \right)} \right\| = \left| {{a_1}} \right|,
即a_1\geq \frac12.同理\nabla f\left( {{u_2}} \right) = {a_2}\left( {\cos {\theta _2},\sin {\theta _2}} \right),a_2\geq \frac12.由于不等式取等需要与{\nabla f\left( {0,0} \right)}异向,故a_1\geq\frac12,a_2\geq\frac12中有一个是严格的.不妨设为a_1>\frac12,再设\overrightarrow {{r_1}} = \left( {\cos {\theta _1},\sin {\theta _1}} \right),\overrightarrow {{r_2}} = \left( {\cos {\theta _2},\sin {\theta _2}} \right),则
\begin{align*}&{\left\| {\nabla f\left( {{u_1}} \right) - \nabla f\left( {{u_2}} \right)} \right\|^2} - {\left\| {{u_1} - {u_2}} \right\|^2}\\= &{\left\| {{a_1}\overrightarrow {{r_1}} - {a_2}\overrightarrow {{r_2}} } \right\|^2} - {\left\| {\frac{1}{2}\overrightarrow {{r_1}} - \frac{1}{2}\overrightarrow {{r_2}} } \right\|^2} = a_1^2 + a_2^2 - 2{a_1}{a_2}\overrightarrow {{r_1}} \overrightarrow {{r_2}} - \left( {\frac{1}{4} - \frac{1}{2}\overrightarrow {{r_1}} \overrightarrow {{r_2}} + \frac{1}{4}} \right)\\= &a_1^2 + a_2^2 - \frac{1}{2} - \left( {2{a_1}{a_2} - \frac{1}{2}} \right)\overrightarrow {{r_1}} \overrightarrow {{r_2}} \left( \text{由于}{\overrightarrow {{r_1}}\text{与} \overrightarrow {{r_2}} \text{不同向},\text{所以}\overrightarrow {{r_1}} \overrightarrow {{r_2}} < 1,\text{且}2{a_1}{a_2} - \frac{1}{2} > 0} \right)\\> &a_1^2 + a_2^2 - \frac{1}{2} - \left( {2{a_1}{a_2} - \frac{1}{2}} \right) = {\left( {{a_1} - {a_2}} \right)^2} \ge 0,\end{align*}
因此{\left\| {\nabla f\left( {{u_1}} \right) - \nabla f\left( {{u_2}} \right)} \right\|^2} - {\left\| {{u_1} - {u_2}} \right\|^2} > 0,即\left\| {\nabla f\left( {{u_1}} \right) - \nabla f\left( {{u_2}} \right)} \right\| > \left\| {{u_1} - {u_2}} \right\|,矛盾,从而f在边界上只有一点取得最大值,即函数f|_D在D中唯一点处达到其最大值.
3.讨论级数\sum\limits_{n = 1}^{ + \infty } {{{\left( { - 1} \right)}^{n - 1}}\frac{{\sin \left( {\ln n} \right)}}{{{n^\alpha }}}},\alpha\in \mathbb{R}的收敛性.
解:(1)当\alpha\leq0时,我们可知\mathop {\lim }\limits_{n \to \infty } {\left( { - 1} \right)^{n - 1}}\frac{{\sin \left( {\ln n} \right)}}{{{n^\alpha }}}发散,从而级数\sum\limits_{n = 1}^{ + \infty } {{{\left( { - 1} \right)}^{n - 1}}\frac{{\sin \left( {\ln n} \right)}}{{{n^\alpha }}}}发散.
(2)当\alpha>0时,由于\mathop {\lim }\limits_{n \to \infty } {\left( { - 1} \right)^{n - 1}}\frac{{\sin \left( {\ln n} \right)}}{{{n^\alpha }}} = 0,我们讨论其前2n项和数列的收敛性即可,也就是级数\sum\limits_{n = 1}^{ + \infty } {\left[ {\frac{{\sin \left( {\ln \left( {2k - 1} \right)} \right)}}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{{\sin \left( {\ln \left( {2k} \right)} \right)}}{{{{\left( {2k} \right)}^\alpha }}}} \right]}的收敛性.
而
\begin{align*}&\left| {\frac{{\sin \left( {\ln \left( {2k - 1} \right)} \right)}}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{{\sin \left( {\ln \left( {2k} \right)} \right)}}{{{{\left( {2k} \right)}^\alpha }}}} \right|\\= &\left| {\sin \left( {\ln \left( {2k - 1} \right)} \right)\left( {\frac{1}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{1}{{{{\left( {2k} \right)}^\alpha }}}} \right) + \frac{{\sin \left( {\ln \left( {2k - 1} \right)} \right) - \sin \left( {\ln \left( {2k} \right)} \right)}}{{{{\left( {2k} \right)}^\alpha }}}} \right|\\\le &\left| {\sin \left( {\ln \left( {2k - 1} \right)} \right)} \right|\left[ {\frac{1}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{1}{{{{\left( {2k} \right)}^\alpha }}}} \right] + \left| {\frac{{2\cos \left[ {\frac{{\ln \left( {2k - 1} \right) + \ln \left( {2k} \right)}}{2}} \right]\sin \left[ {\frac{{\ln \left( {2k - 1} \right) - \ln \left( {2k} \right)}}{2}} \right]}}{{{{\left( {2k} \right)}^\alpha }}}} \right|\\\le &\left[ {\frac{1}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{1}{{{{\left( {2k} \right)}^\alpha }}}} \right] + \left| {\ln \left( {1 - \frac{1}{{2k}}} \right)} \right|\frac{1}{{{{\left( {2k} \right)}^\alpha}}}.\end{align*}
已知\left| {\ln \left( {1 - \frac{1}{{2k}}} \right)} \right|\frac{1}{{{{\left( {2k} \right)}^\alpha }}} \sim \frac{1}{{{{\left( {2k} \right)}^{\alpha + 1}}}},从而\sum\limits_{k = 1}^{ + \infty } {\left| {\ln \left( {1 - \frac{1}{{2k}}} \right)} \right|\frac{1}{{{{\left( {2k} \right)}^\alpha }}}}收敛.又\sum\limits_{k = 1}^{ + \infty } {\left[ {\frac{1}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{1}{{{{\left( {2k} \right)}^\alpha }}}} \right]}是收敛的,故\sum\limits_{n = 1}^{ + \infty } {\left| {\frac{{\sin \left( {\ln \left( {2k - 1} \right)} \right)}}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{{\sin \left( {\ln \left( {2k} \right)} \right)}}{{{{\left( {2k} \right)}^\alpha }}}} \right|}收敛,因此\sum\limits_{n = 1}^{ + \infty } {\left[ {\frac{{\sin \left( {\ln \left( {2k - 1} \right)} \right)}}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{{\sin \left( {\ln \left( {2k} \right)} \right)}}{{{{\left( {2k} \right)}^\alpha }}}} \right]}亦收敛.
综上所述, 当\alpha\leq 0时,级数发散;当\alpha>0时级数收敛.
3.求\mathop {\inf }\limits_{n \ge 1} \left\{ {\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \sum\limits_{k = 1}^n {\frac{{\cos kx}}{k}} } \right\}.
解:首先, \mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \sum\limits_{k = 1}^1 {\frac{{\cos kx}}{k}} = \mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \cos x = 0,\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \sum\limits_{k = 1}^2 {\frac{{\cos kx}}{k}} = \mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \left( {\cos x + \frac{{\cos 2x}}{2}} \right) = - \frac{1}{2}.
现记{f_n}\left( x \right) = \sum\limits_{k = 1}^n {\frac{{\cos kx}}{k}},则\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} {f_1}\left( x \right) = 0 \ge - \frac{1}{2},\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} {f_1}\left( x \right) = - \frac{1}{2} \ge - \frac{1}{2}.
现假设n=k-1(k\geq2,k\in\mathbb{N}_+)时,有\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} {f_n}\left( x \right) = \mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} {f_{k - 1}}\left( x \right) \ge - \frac{1}{2}成立;
当n=k时,注意到{f_n}^\prime \left( x \right) = \sum\limits_{k = 1}^n {\left( { - \sin kx} \right)} = \sum\limits_{i = 1}^k {\left( { - \sin ix} \right)} = \sum\limits_{i = 1}^k {\frac{{\cos \left( {i + \frac{1}{2}} \right)x - \cos \left( {i - \frac{1}{2}} \right)x}}{{2\sin \frac{x}{2}}}} ,
则{f_k}^\prime \left( x \right) = \frac{{\cos \left( {k + \frac{1}{2}} \right)x - \cos \frac{x}{2}}}{{2\sin \frac{x}{2}}} = - \frac{{\sin \frac{{k + 1}}{2}x\sin \frac{k}{2}x}}{{\sin \frac{x}{2}}},x \in \left[ {0,\frac{\pi }{2}} \right].
由f_k(x)的连续性可知\min f_k(x)的点只可能在端点或稳定点,从而令f'_k(x)=0,则x = \frac{{2j\pi }}{{k + 1}},\frac{{2j\pi }}{k},j \in \mathbb{Z}且\frac{{2j\pi }}{{k + 1}},\frac{{2j\pi }}{k} \in \left( {0,\frac{\pi }{2}} \right),而
{f_k}\left( {\frac{{2j\pi }}{{k + 1}}} \right) = \sum\limits_{i = 1}^n {\frac{{\cos i\left( {\frac{{2j\pi }}{{k + 1}}} \right)}}{i}} = {f_{k - 1}}\left( {\frac{{2j\pi }}{{k + 1}}} \right) + \frac{{\cos \frac{{2kj\pi }}{{k + 1}}}}{k} = {f_{k - 1}}\left( {\frac{{2j\pi }}{{k + 1}}} \right) + \frac{{\cos \frac{{2j\pi }}{{k + 1}}}}{k}.
由于\frac{{2j\pi }}{{k + 1}} \in \left[ {0,\frac{\pi }{2}} \right],所以\frac{{\cos \frac{{2j\pi }}{{k + 1}}}}{k} > 0,则{f_k}\left( {\frac{{2j\pi }}{{k + 1}}} \right) = {f_{k - 1}}\left( {\frac{{2j\pi }}{{k + 1}}} \right) + \frac{{\cos \frac{{2j\pi }}{{k + 1}}}}{k} \ge - \frac{1}{2}.
而{f_k}\left( {\frac{{2j\pi }}{k}} \right) = \sum\limits_{i = 1}^n {\frac{{\cos i\left( {\frac{{2j\pi }}{k}} \right)}}{i}} = {f_{k - 1}}\left( {\frac{{2j\pi }}{k}} \right) + \frac{1}{k} \ge - \frac{1}{2},又
{f_k}\left( 0 \right) = \sum\limits_{i = 1}^k {\frac{1}{i}} > - \frac{1}{2},{f_k}\left( {\frac{\pi }{2}} \right) = \sum\limits_{i = 1}^k {\frac{{\cos i\frac{\pi }{2}}}{i}} = \left\{ \begin{array}{l}\frac{1}{2}\left[ { - 1 + \left({\frac{1}{2} - \frac{1}{3}} \right) + \cdots + \left( {\frac{1}{{s - 1}} - \frac{1}{s}} \right)} \right],k = 2s + 1\\\frac{1}{2}\left[ { - 1 + \left( {\frac{1}{2} - \frac{1}{3}} \right) + \cdots + \left( {\frac{1}{{s - 1}} - \frac{1}{s}} \right)} \right],k = 2s\end{array} \right.,从而{f_k}\left( {\frac{\pi }{2}} \right) \ge - \frac{1}{2}.
综上归纳可知\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} {f_n}\left( x \right) = \mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \sum\limits_{k = 1}^n {\frac{{\cos kx}}{k}} \ge - \frac{1}{2},又\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \sum\limits_{k = 1}^2 {\frac{{\cos kx}}{k}} = - \frac{1}{2},因此
\mathop {\inf }\limits_{n \ge 1} \left\{ {\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \sum\limits_{k = 1}^n {\frac{{\cos kx}}{k}} } \right\} = - \frac{1}{2}.
4.函数f(x)在[0,1]上二次可导, f(0)=2,f'(0)=-2,f(1)=1.证明存在c\in (0,1),使得f(c)f'(c)+f''(c)=0.
证明:令F\left( x \right) = \frac{1}{2}{f^2}\left( x \right) + f'\left( x \right),则F(0)=\frac12\times 2^2-2=0.现假设不存在\xi\in (0,1]使得F(\xi)=0,则F在(0,1]上不变号.倘若\forall x\in (0,1]都有F(x)<0,即\frac{1}{2}{f^2}\left( x \right) + f'\left( x \right) < 0,则f'\left( x \right) < - \frac{1}{2}{f^2}\left( x \right) \le 0,则f在[0,1]上单调递减,由f(0)=2,f(1)=1可知1\leq f(x)\leq 2,f(x)\neq 0,从而\frac{{f'\left( x \right)}}{{{f^2}\left( x \right)}} < - \frac{1}{2}.由f,f'的连续性可知\int_0^1 {\frac{{f'\left( x \right)}}{{{f^2}\left( x \right)}}dx} < \int_0^1 {\left( { - \frac{1}{2}} \right)dx},即\left. { - \frac{1}{{f\left( x \right)}}} \right|_0^1 < - \frac{1}{2},得到-\frac12<-frac12,矛盾.
倘若\forall x\in (0,1]都有F(x)>0,即\frac{1}{2}{f^2}\left( x \right) + f'\left( x \right) > 0,由于f(1)=1\neq 0,则\forall x\in U^-(1,\delta),有\int_x^1 {\frac{{f'\left( t \right)}}{{{f^2}\left( t \right)1}}dt} > \int_x^1 {\left( { - \frac{1}{2}} \right)dt},即\left. { - \frac{1}{{f\left( t \right)}}} \right|_x^1 > \frac{{x - 1}}{2},从而\frac{1}{{f\left( x \right)}} > \frac{{x + 1}}{2}.
当x\in U^-(1,\delta)时,有\frac{1}{{f\left( x \right)}} > \frac{{x + 1}}{2} > 0\left( {\delta \le 1} \right),故f(x)在[0,1]上不为0,则\frac{1}{{f\left( 0 \right)}} > \frac{{0 + 1}}{2},则\frac12>\frac12,矛盾.
因此存在\xi\in (0,1]使得F(\xi)=0,从而F(0)=F(\xi)=0,由罗尔定理可知\exists c\in (0,1),使得
F'\left( c \right) = f\left( c \right)f'\left( c \right) + f''\left( c \right) = 0.
5.A和B是自然数\mathbb{N}的两个无穷子集,满足A\cap B=\text{空集},A\cup B=\mathbb{N},对于任意的自然数c>0,是否存在两个递增的数列\{a_n\},\{b_n\},\{a_n\}\in A,\{b_n\}\in B,使得\mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}} = c.
证明:对\forall l\geq s\in \mathbb{N}_+.记s\sim l=\{s,s+1,\cdots,l\},则若对\forall N\in \mathbb{N}_+, \exists n_2\geq n_1\geq N\geq c^2有x_{n_1}\sim x_{n_2}\subset B.令x_{n_1}=v_1,x_{n_2}=u_1,则v_1\sim u_1\subset B.若cv_1\sim cu_1+c[\sqrt{u_1}+1]\not\subset B,则\exists a\in A,b\in B,使得|a-cb|\leq c或|a-cb|\leq \sqrt{b}+1.
若cv_1\sim cu_1+c[\sqrt{u_1}+1]\subset B,令u_2=u_1+[\sqrt{u_1}+1],则cv_1\sim cu_2\subset B.以此类推,若始终成立c^kv_1 \sim c^k u_{k+1}\subset B,易知u_{k+1}=u_k+[\sqrt{u_k}+1]\geq u_k+\sqrt{u_1},即u_{k+1}-u_k\geq \sqrt{u_1},所以u_k有趋于无穷大的趋势,所以存在某个k_0\in \mathbb{N}_+,使得\frac{u_{k_0}}{v_1}>c.而c^{k_0}v_1\sim c^{k_0} u_{k_0+1}\subset B.令x_{n_{k_0}}\in B满足c^{k_0}v_1\sim c^{k_0}u_{k_0+1}\subset c^{k_0}v_1\sim x_{n_{k_0}}\subset B,x_{n_{k_0}}+1\in A,令a_1=x_{n_{k_0}}+1,则c^{k_0}v_1<\frac{a_1}{c}\leq x_{n_{k_0}+1},则\exists b_1\in B,使得\left| {\frac{{{a_1}}}{c} - {b_1}} \right| \le 1,即\left| {{a_1} - c{b_1}} \right| \le c.
若不始终成立c^kv_1\sim c^ku_{k+1},则\exists b_1\in B,a_1\in A有|a_1-cb_1|\leq c或|a_1-cb_1|\leq \sqrt{b_1}+1,即|a_1-cb_1|\leq \sqrt{a_1}+1.接着再取n_1\geq n_2\geq N,有x_{n_2}\geq x_{n_1}>a_1,b_1且x_{n_1}\sim x_{n_2}\subset B,则\exists a_2\in A,b_2\in B有|a_2-cb_2|\leq \sqrt{b_2}+1,\cdots,故存在递增数列\{a_n\}\subset A,\{b_n\}\subset B,有|a_n-cb_n|\leq \sqrt{b_n}+1,则\left| {\frac{{{a_n}}}{{{b_n}}} - c} \right| \le \frac{1}{{\sqrt {{b_n}} }} + \frac{1}{{{b_n}}},所以\mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}} = c.