\magnification=1200
%\font\small=cmbx10
\font\usl=cmsl10 \font\tt=cmr10 \font\ita=cmti10 \font\sita=cmti10
scaled 1200 \font\mmm=cmbx10 \font\pealk=cmbx10 scaled \magstep2
\font\glav=cmbx10 scaled \magstep2 \font\titl=cmr10 scaled
\magstep2 \font\kaks=cmr10
%\font\kaheksa=cmr8
\font\kuus=cmr6 \font\kaksk=cmr10 scaled \magstep3
\font\kuusk=cmr8 scaled \magstep3

\hsize=17,5truecm
\input amssym.def
\input amssym
\input cyracc.def
\font\bold=cmbx10 \font\tencyr=wncyr9 scaled \magstep1
\font\eu=eufm10 \font\tencyi=wncyi9 scaled \magstep1
\font\tencyb=wncyb9 scaled \magstep1 \font\toncyb=wncyb9 scaled
\magstep2
\def\cyr{\tencyr\cyracc}
\def\cyb{\tencyb\cyracc}
\def\cob{\toncyb\cyracc}
\def\cyi{\tencyi\cyracc}
\def\enn{{\Bbb N}}
\def\prl{\parallel}
\def\ds{\displaystyle}
\nopagenumbers \baselineskip=18pt \cyr \centerline{\cob
Resheniya.}\vskip 5mm

{\mmm 1.} Tak kak $$|a_n|={|na_n|\over n} \leq {1\over 2}
\Bigl(n^2a_n^2+{1\over n^2}\Bigr),$$ to s\cydot hodimost\cprime\
ryada $\ds\sum_k |a_k| $ sleudet iz s\cydot hodimosti ryadov
$\ds\sum_n n^2 a_n^2$ i $\ds \sum_n {1\over n^2}$.\vskip 5mm

{\mmm 2.} Tak kak podintegral\cprime nye funkcii v obeih chastyah
ravenstva $\ds \int_1^{f(x)} e^{u^2} du=\int_0^x {udu\over f(u)}$
nepreryvny i $f$ differenciruema, to my mozhem
prodifferencirovat\cprime\ \`eto ravenstvo. Imeem
$$f'(x)e^{f^2(x)}={x\over f(x)}\ \Longleftrightarrow\
e^{f^2(x)}f(x) f'(x)={\bigl(e^{f^2(x)}\bigr)'\over 2}=x\
\Longleftrightarrow\ e^{f^2(x)}=x^2+C.$$ Sledovate\cprime no
$f(x)=\sqrt{\ln(x^2+C)}$. Opredelim konstantu $C$. Sperva zametim,
chto $C>0$. Dalee, tak funkciya $f(x)=\sqrt{\ln(x^2+C)}$ prinimaet
minimal\cprime noe znachenie pri $x=0$, to $f(0)=\sqrt{\ln(C)}=1$.
Sledovatel\cprime no $C=e$ i $f(x)=\sqrt{\ln(x^2+e)}$.\vskip 5mm

{\mmm 3.} Rassmotrim funkciyu $\ds f(x):=\Bigl(x+{1\over
x}\Bigr)^a$ i vychislim e\"e vtoruyu proizvodnuyu:
$$\eqalign{f''(x)&= {\Bigl({ x^2+1\over
x}\Bigr)^aa(ax^4-2ax^2+a+4x^2+1-x^4)\over x^2(x^2+1)^2}\cr &=
{\Bigr({x^2+1\over
x}\Bigr)^aa\bigl(a(x^2-1)^2+(2+\sqrt{5}-x^2)(x^2-2+\sqrt{5})\bigr)\over
x^2(x^2+1)^2}>0\cr}$$ pri $x\in (0,1)$. Sledovatel\cprime no $f$
vypukla vniz pri $x\in (0,1)$, po\`etomu dlya lyubyh
$\{x_1,\ldots,x_n\}\subset (0,1)$ imeem $${1\over n}\sum_{k=1}^n
f(x_k)={1\over n} \sum_{k=1}^n\Bigl(x_k+{1\over x_k}\Bigr)^a\geq
f\Bigl({1\over n} \sum_{k=1}^n x_k\Bigr)=\Bigl((\sum_{k=1}^n
x_k)/n+{n\over \sum_{k=1}^n x_k}\Bigr)^a.$$ Tak kak po usloviyu
$x_1+x_2+\ldots +x_n=1$, to $${1\over
n}\sum_{k=1}^n\Bigl(x_k+{1\over x_k}\Bigr)^a\geq \Bigl({1\over
n}+n\Bigr)^a,$$ otkuda i sleduet trebuemoe neravenstvo.\vskip 5mm

{\mmm 4.} Zametim, chto $$a_n=\sum_{k=0\atop 2|k}^n
C^k_n(\sqrt{3})^k,\quad \sqrt{3}b_n=\sum_{k=0\atop 2\not|k}^n
C^k_n(\sqrt{3})^k.$$ S drugo\u i storony
$\ds(1-\sqrt{3})^n=\sum_{k=0}^n
(-1)^kC^k_n(\sqrt{3})^k=a_n-\sqrt{3}b_n$. Po\`etomu
$$\eqalign{a_n&={1\over
2}\bigl((1+\sqrt{3})^n+(1-\sqrt{3})^n\bigr),\cr b_n&={1\over
2\sqrt{3}}\bigl((1+\sqrt{3})^n-(1-\sqrt{3})^n\bigr).\cr}$$
Sledovatel\cprime no $$\lim_{n\to \infty} {a_n\over
b_n}=\sqrt{3}\lim_{n\to\infty}{(1+\sqrt{3})^n+(1-\sqrt{3})^n\over
(1+\sqrt{3})^n-(1-\sqrt{3})^n}=\sqrt{3}.$$\vskip 3mm


{\mmm 5.} Oboznachim $P:=3E-3A$ i $Q:=3A+2E$. Togda
$$\eqalign{X^{-1}&=Q^{-1}+{(P/3)^{-1}\over
3}=Q^{-1}+P^{-1}=P^{-1}(P+Q)Q^{-1}=P^{-1}(5E)Q^{-1}\cr &
=5(QP)^{-1}=5(6E+3(A-3A^2))^{-1}={5E\over 6}.\cr}$$
Sledovatel\cprime no $X=6E/5$.\vskip 5mm

{\mmm 6.} Dokazhem utverzhdenie indukcie\u i\ po $r$. Pust\cprime\
$r = 0$ i pokazhem, chto fukciya $u(n) = a^n$, gde $a \not = 0$
yavlyaet\cydot sya resheniem rekurrentnogo sootnosheniya (1) togda
i tol\cprime ko togda, kogda $a$ yavlyaet\cydot sya kornem $f(x).$

Podstavlyaya $u(n) = a^n$ v rekurrentnoe sootnoshenie $$u(n+k) =
a_0 u(n) + a_1 u(n+1)+ \ldots + a_{k-1} u(n+k-1),\eqno(1)$$
poluchim: $$\eqalign{a^{n+k} = a_0 a^n + a_1 a^{n+1} + \ldots +
a_{k-1}a^{n+k-1}\ &\Longleftrightarrow\ a^n ( a^k - a_{k-1}a^{k-1}
- \ldots - a_0) = 0\cr &\Longleftrightarrow\ a^n f(a) = 0.\cr}$$

Dopustim, chto $v(n) = n^{r-1}a^n$ yavlyaet\cydot sya resheniem
rekurrentnogo sootnosheniya (1) i $a$ -- koren\cprime\ $f(x)$
kratnosti ne menee $r$ i pokazhem $u(n) = n^{r}a^n$ yavlyaet\cydot
sya resheniem  (1) togda i tol\cprime ko togda, kogda $a$ --
koren\cprime\ $f(x)$ kratnosti ne menee $r+1$. Podstavlyaya
$n^{r}a^n$ v (1) poluchim: $$(n+k)^r a^{n+k} = a_0 n^r a^n + a_1
(n+1)^{r}a^{n+1} + \ldots + a_{k-1}(n+k-1)^r a^{n+k-1}.$$ Otkuda
$$(n+k)v(n+k) = a_0 n v(n) + a_1 (n+1) v(n+1) + \ldots +
a_{k-1}(n+k-1)v(n+k-1).$$ Tak kak $v(n)$ yavlyaet\cydot sya
resheniem (1), to nam neobhodimo pokazat\cprime , chto $v(n)$
udovletvoryaet ravenstvu $$ k v(n+k) = a_1 v(n+1) + \ldots +
a_{k-1}(k-1)v(n+k-1)\eqno(2)$$
 togda i tol\cprime ko togda, kogda $a$ --
koren\cprime\ $f(x)$ kratnosti ne menee $r+1$. Netrudno
zametit\cprime , chto sootnosheniyu (2) sootvet\cydot stvuet
mnogochlen $x f'(x)$ i po induktivnomu prepolozheniyu $v(n)$
yavlyaet\cydot sya resheniem (2) togda i tol\cprime ko togda,
kogda $a$ -- koren\cprime\ $f'(x)$ kratnosti ne menee $r$. Takim
obrazom $a$ dolzhno yavlyat\cprime sya kornem $f(x)$ kratnosti ne
menee $r+1$.
\end
