[Home Squat University Mathematics magazine] The No. 435 issue of the seventh Chinese College Students Mathematics Contest Preliminary Examination Questions (mathematics, October 2015) Reference solutions

I. ($ $) Set $L _1$ and $L _2$ is a two-plane line in space. In the standard Cartesian coordinate system, the line $L _1$ the point of the $a $, in the unit vector $v $ for the straight line direction, and the line $L _2$ the point with the $b $, in the unit vector $w $   To a straight line direction.   (1). Proof: There is a unique point $P \in l_1$ and $Q \in l_2$ makes two-point connection $PQ $ while perpendicular to $L _1$ and $L _2$. (2). $P $ point and $Q $ point coordinates (expressed in $a, b,v,w$).

It is proved that the equations of test instructions, $L _1,l_2$ are $$\bex x=a+sv\ (S\IN\BBR) and \quad y=b+tw\ (T\IN\BBR) respectively. \eex$$ set $P \in l_1,\ q\in l_2$ meet $$\bex pq\perp l_1,\quad pq\perp l_2, \eex$$ if $P, q$ coordinates are $p, q$, $$\bex p=a+sv,\qu Ad Q=b+tw,\quad \sef{q-p,v}=0,\quad \sef{q-p,w}=0.  \eex$$ the above equation can be solved only, and the proof is completed. In fact, $$\bex Q-P=B-A+TW-SV, \eex$$ $$\bex \sedd{\ba{ll} \sef{b-a,v}+\sef{tw-sv,v}=0\\ \sef{b-a,w}+\sef{tw-sv,w}=0 \ea}, \eex$$ $$\bex \sedd{\ba{ll}-s+\sef{v,w}t=\sef{a-b,v}\\-\sef{v,w}s+t=\sef{a-b,w} \ea}\quad\sex{|v|=|w|=1}, \eex$$ $$ \bex S=\frac{\sef{a-b,v}-\sef{a-b,w}\sef{v,w}}{-1+\sef{v,w}^2},\quad t=\frac{-\sef{a-b,w}+\sef{a-b,v}\sef{v,w}}{ -1+\SEF{V,W}^2}. \eex$$ so $$\bex p=a-\sef{a-b,\frac{v-\sef{v,w}w}{1-\sef{v,w}^2}}v,\quad q=b-\sef{a-b,\frac{-w+\sef{v,w}v}{1-\sef{ V,w}^2}}w. \eex$$

Two. ($ $ $) $A $ for $4$-order compound array, which satisfies the relation of traces: $\tr a^i=i$, $i =1,2,3,4$. The determinant of $A $.

Answer: Set $A $ with a eigenvalues of $\lm_j\ (1\leq J\leq 4) $, then the eigenvalues of the $A ^i$ are $\lm_j^i\ (1\leq J\leq 4) $. Press test instructions, $$\bex s_i\equiv \sum_{j=1}^4 \lm_j^i=i,\quad 1\leq i\leq 4. \eex$$ is set $A $ of the characteristic polynomial $$\bex f (\lm) =\prod_{j=1}^4 (\lm-\lm_j) =\lm^4+a_1\lm^3+a_2\lm^2+a_3\lm+a_4, \eex$$ by Newton formula (see PKU Higher garbage Fourth edition 49th), $$\beex \bea s_1+a_1&=0,\\ s_2+s_1a_1+2a_2&=0,\\ s_3+s_2a_1+s_1a_2+3a_3&=0,\\ s_4+s_3a_1+s_ 2a_2+s_1a_3+4a_4&=0. \eea \eeex$$ will $s _i=i$ in order to find $$\bex A_1=-1,\quad a_2=-\frac{1}{2},\quad A_3=-\frac{1}{6},\quad}. \eex$$ So, $$\bex | A|=A_4=\FRAC{1}{24}. \eex$$

Three. ($ $) Set $A $ for $n $ order solid Phalanx, its $n $ value is an even number. Try to prove that there are only 0 solutions to the matrix equation $$\bex x+ax-xa^2=0 \eex$$ about $X $.

Proof: (1). To prove the conclusion: set $A, the characteristic polynomial of b$ is $f (\LM), G (\LM) $, if $ (f (\LM), G (\LM)) =1$, then $g (A) $ reversible. In fact, $$\beex \bea &\quad 1=u (\LM) F (\lm) +v (\LM) g (\LM) \ &\ra E=u (a) f (a) +v (a) g (a) \ &\ra e=v (a) g (a) \quad\sex {\mbox{hamilton-caylay Theorem}}. \eea \eeex$$ (2).  Again the conclusion: if $A, the eigenvalues of b$ are different, then there are only 0 solutions $AX =xb$. In fact, $$\beex \bea ax=xb&\ra a^2x=aax=axb=xbb=xb^2\\ &\ra \cdots\\ &\ra F (A) x=xf (b) \ &\ra 0=Xf (b) \ \ & \ra 0=x\quad\sex{\mbox{by (1) and} (f (\LM), G (\LM)) =1}. \eea \eeex$$ (3). Finally prove the topic. The original matrix equation can be converted to $$\bex (e+a) x=xa^2. \eex$$ Note that the eigenvalues of the $E +a$ are odd, and the eigenvalues of the $A ^2$ are even numbers. From (2) stand $X =0$.

Four. ($ $) sequence $\sed{a_n}$ meet relational $$\bex a_{n+1}=a_n+\frac{n}{a_n},\ Quad A_1>0. \eex$$ Verification: $\dps{\vlm{n}n (a_n-n)}$ exist.   

Proof: Write $$\bee\label{435:4} \bea a_{n+1}-(n+1) &=a_n-n+\frac{n}{a_n}-1     =a_n-n-\frac{a_n-n }{a_n}\\     &=\sex{1-\frac{1}{a_n}} (A_n-n). \eea \eee$$ used mathematical induction to prove $$\bee\label{435:4:bounds} n\leq a_n\leq n+b\quad\sex{n\geq 2,\ b=a_2-2}. \eee$$ in fact, $$\beex \bea a_2&=a_1+\frac{1}{a_1}\geq 2,\\ a_n\geq n\ra a_{n+1}&=a_n+\frac{n}{a_n}\\ &=n+\sex {a_n-n+\frac{n}{a_n}}\\ &\geq n+\sex{n-n+\frac{n}{n}}\quad\sex{f (x) =x-n+\frac{n}{x}\mbox{in}[\sqrt{n},\infty) \mbox{on increment}}\\ &=n+1;\\ a_2&=2+b,\\ a_n\leq n+b\ra A_{n+1}&=a_n+\frac{n}{a_n}     \ Leq n+b+\frac{n}{n}     = (n+1) +b. \eea \eeex$$   by (1)-\eqref{435:4:bounds} known $$\beex \bea a_{n +1}-(n+1) &\geq\sex{1-\frac{1}{n}} (a_n-n), \ \ n\sez{a_{n+1}-(n+1)}&\geq (n-1) (a_n-n) \      &\geq \cdots\\     &\geq a_2-2=b,\\ a_{n+1}-(n+1) &\leq \sex{1-\frac{1}{n+b}} ( a_n-n), \ \ (n+b) \sez{a_{n+1}-(n+1)}&\leq \sez{(n-1) +b} (a_n-n) \ &\leq \cdots\\ &\leq (1+b) (a_2-2) = (1+b) b. \eea \eeex$$ thereupon $$\bex (n-1) (A_ N-n) \leq n\sez{a_{n+1}-(n+1)} \leq (N+b) \sez{a_{n+1}-(n+1)} \leq (1+b) B. \eex$$ This indicates that the $\sed{(n-1) (a_n-n)}$ has an upper bound, while the limit exists. Also by $$\bex N (a_n-n) =\frac{n}{n-1}\cdot (n-1) (a_n-n) \eex$$ The conclusion of the evidence is established.   

Five. ($ $) Set $f (x) $ is $[0,+\infty) $ bounded continuous function, $h (x) $ is $[0,+\infty) $ on continuous function, and $\dps{\int_0^{+\infty}h (x) \rd x=a<1}$. The constructor columns are as follows: $$\bex g_0 (x) =h (x), \quad g_n (x) =f (x) +\int_0^x H (t) g_{n-1} (t) \rd T\quad (n=1,2,\cdots). \eex$$ Verification: $\sed{g_n (x)}$ converges to a continuous function, and the limit function is obtained.

Proof: Make $$\bex A_n=\max_{0\leq x<\infty} |g_n (x)-g_{n-1} (x) |, \eex$$ then $$\beex \bea a_n&\leq \int_0^\infty |h (t) |\cdo T |g_{n-1} (t)-g_{n-2} (t) |\rd t     \leq \int_0^\infty |h (t) \rd T\cdot a_{n-1}\\ &\leq a a_{n-1}\ Leq \cdots\\ &\leq a^{n-1}a_1. \eea \eeex$$ then $$\bex \vsm{n}a_n\quad\sex{\leq a_1\vsm{n}a^{n-1} \eex$$ converge. Also by $$\bex G_n (x) =g_0 (x) +\sum_{k=1}^n [G_k (x)-g_{k-1} (x)],\quad 0\leq x<\infty \eex$$ and Superior series Discriminant method $g _n$ uniformly convergent. Set the Limit function to $g (x) $, then $g $ continuous and satisfy $$\bex g (x) =f (x) +\int_0^x H (t) g (t) \rd T. \eex$$ Order $$\bex g (x) =\int_0^x H (t) g (t) \rd T, \eex$$ Then $$\beex \bea g ' (x) &=h (x) g (x) =h (x) [F (x) +g (x)],\\ G ' (x)-H (x) g (x) &=h (x) f (x), \ \sez{e^{-\int_0^x h (t) \rd t}g (x )} ' &=e^{-\int_0^x h (t) \rd t}h (x) f (x), \ e^{-\int_0^x h (t) \rd T} G (x) &=\int_0^x e^{-\int_0^s H (t) \rd T}h (s) f (s) \ Rd s,\\ G (x) &=\int_0^x e^{\int_s^x H (t) \rd T}h (s) f (s) \rd s,\\ G (x) &=f (x) +\int_0^x e^{\int_s^x H (t) \rd T}h (s) f (s) \rd S. \eea \eeex$$   

Six. ($ $) Set $f (x) $ is a continuous function with a lower bound or upper bound on the $\bbr$ and has a positive number $a $ makes $$\bex f (x) +a\int_{x-1}^x F (t) \rd T \eex$$ constant. Verification: $f (x) $ is constant.

proof: May be set $f $ has a lower bound, so $\dps{m=\inf_{x\in [0,\infty]}f (x)}$, $$\bex g (x) =f (x)-M\GEQ 0. \eex$$ according to test instructions, $\exists\ c\in\bbr,\st$ $$\bee\label{435:6:eq} \bea f (x) +a\int_{x-1}^x F (t) \rd t&=c,\\ g (x) +a\int_{x-1} ^x g (t) \rd t&=f (x)-m+a\int_{x-1}^x [F (t)-m]\rd t=c-(a+1) m\equiv M\geq 0,\\ g (x) &=m-a\int_{x-1}^x g (t) \rd T. \eea \eee$$ the right end can be guided, and both ends can be derivative simultaneously, $$\beex \bea G ' (x) &=-a[g (x)-G (x-1)],\\ [E^{ax}g (x)] ' &=ag (x-1) \geq 0\\ &\ra m=g (x) +a\ Int_{x-1}^x g (t) \rd T     =g (x) +a\int_{x-1}^x e^{at}g (t) \cdot e^{-at}\rd t\\ &\quad\quad\ \ & Nbsp;       \leq g (x) +ae^{ax}g (x) \int_{x-1}^x e^{-at}\rd t\\ &\quad\quad\ \         =g (x) e^a,\\ &\ra g (x) \geq me^{-a}\\ &\ra 0=\inf_{x\in[0,\ Infty)}g (x) \geq me^{-a}\\ &\ra 0=m=g (x) +a\int_{x-1}^x g (t) \rd t\quad\sex{g (t) \geq 0}\\ &\ra g\equiv 0. \eea \eeex$$ 

[Home Squat University Mathematics Magazine] NO. 435 session Seventh Chinese College Students Mathematics Contest Preliminary Examination Question (Mathematics class, October 2015) reference answer