05th sets of simulated exam for Mathematics Competition Training of Gannan Normal University

1. (1) set $ f (x) $ to be bounded on $ [0, 1] $ and continuous at $ x = 1 $, test the limit $ \ DPS {\ vlm {n} n \ int_0 ^ 1 x ^ {n-1} f (x) \ RD x} $.

(2) calculate the following formula $ \ Bex \ int_0 ^ 1 \ cfrac {x ^ {n-1 }}{ 1 + x} \ RD x =\cfrac {A} {n} + \ cfrac {B} {n ^ 2} + O \ sex {\ cfrac {1} {n ^ 2 }}\ Quad (n \ To \ infty) the undetermined constants in \ EEx $ A, B $.

Answer: (1) from $ F $ continuous knowledge at $ x = 1 $ \ Bex \ forall \ ve> 0, \ exists \ Delta \ In (0, 1), \ st 1-\ Delta \ Leq x \ Leq 1 \ rA | f (x)-f (1) | <\ ve. \ EEx $ set $ | f (x) | \ Leq M, \ forall \ x \ in [0, 1] $, then $ \ beex \ Bea \ sev {n \ int_0 ^ 1 x ^ {n-1} f (x) \ rd x-F (1 )} & =\ sev {\ int_0 ^ 1 NX ^ {n-1} [F (x)-f (1)] \ RD x }\\\& \ Leq \ int_0 ^ {1-\ Delta} NX ^ {n-1} | f (x)-f (0) | \ RD x + \ int _ {1-\ Delta} ^ 1 NX ^ {n-1} | f (x)-f (0) | \ RD x \ & \ leq2m (1-\ delta) ^ N + \ ve [1-(1-\ delta) ^ N] \\& <2 m (1-\ delta) ^ N + \ ve. \ EEA \ eeex $ make $ n \ To \ infty $ \ Bex \ VLS {n} \ sev {n \ int_0 ^ 1 x ^ {n-1} f (x) \ rd x-F (1)} \ Leq \ ve. \ EEx $ then make $ \ ve \ to 0 ^ + $ a conclusion. (2) $ \ Bex a = \ vlm {n} n \ int_0 ^ 1 \ cfrac {x ^ {n-1} {1 + x} \ RD x = \ cfrac {1 }{ 2 }, \ EEx $ \ beex \ Bea B & =\ vlm {n} \ SEZ {n ^ 2 \ int_0 ^ 1 \ cfrac {x ^ {n-1} {1 + x} \ rd x-\ cfrac {n} {2 }\\\\&=\ vlm {n} n \ int_0 ^ 1 NX ^ {n-1} \ sex {\ cfrac {1} {1 + x}-\ cfrac {1} {2 }}\ RD x \ & =\ vlm {n} n \ int_0 ^ 1 \ sex {\ cfrac {1} {1 + x}-\ cfrac {1} {2 }}\ RD x ^ n \\\&=\ vlm {n} n \ int_0 ^ 1 \ cfrac {x ^ n} {(1 + x) ^ 2} \ RD x \\\&=\ vlm {n} \ cfrac {n} {n + 1} \ int_0 ^ 1 \ cfrac {(n + 1) x ^ n} {(1 + x) ^ 2} \ RD x \\\\\cfrac {1} {4 }. \ EEA \ eeex $

 

 

2. set $ f \ in C [0, + \ infty) $, $ A $ as a real number, the limited limit $ \ Bex \ vlm {x} \ SEZ {f (x) + A \ int_0 ^ x F (t) \ RD t} exists }. \ EEx $ proof: $ \ DPS {\ vlm {x} f (x)} $ exists and is zero.

Proof: note $ \ Bex f (x) = E ^ {ax} \ int_0 ^ x F (t) \ RD t, \ EEx $ \ Bex f '(x) = E ^ {ax} \ SEZ {f (x) + A \ int_0 ^ x F (t) \ RD t}, \ EEx $ \ Bex \ vlm {x} \ cfrac {f' (x )} {AE ^ {ax }}=\ cfrac {1} {A} \ vlm {x} \ SEZ {f (x) + A \ int_0 ^ x F (t) \ RD t} \ EEx $ exists. by the l' hosprac law, $ \ Bex \ vlm {x} \ int_0 ^ x F (t) \ rd t = \ vlm {x} \ cfrac {f (x )} {e ^ {ax }}=\ vlm {x} \ cfrac {f' (x)} {AE ^ {ax} \ EEx $. therefore, $ \ Bex \ vlm {x} f (x) = \ vlm {x} \ SEZ {f (x) + A \ int_0 ^ x F (t) \ RD t}-A \ vlm {x} \ int_0 ^ x F (t) \ RD t \ EEx $ exists. it is known that $ \ Bex \ vlm {x} \ int_0 ^ x F (t) \ RD t \ EEx $ F (+ \ infty) = 0 $ (otherwise, $ F (+ \ infty) = A \ NEQ 0 $. set $ a> 0 $, and $ \ Bex \ exists \ x> 0, \ st x \ geq x \ rA f (x) \ geq \ cfrac {A} {2}, \ EEx $ \ beex \ Bea \ int_0 ^ x F (t) \ rd t & = \ int_0 ^ XF (t) \ rd t + \ int_x ^ x F (t) \ RD t \ Quad (x \ geq X) \\& \ geq \ int_0 ^ XF (t) \ rd t + \ cfrac {A} {2} (X-x) \ To \ infty \ Quad (x \ To \ infty ). \ EEA \ eeex $ this is a conflict ).

 

 

3. (1) set $ n \ In \ BBN ^ + $, calculate points $ \ DPS {\ int_0 ^ {\ PI/2} \ cfrac {\ sin NX} {\ SiN x} \ RD x }. $

(2) calculate $ \ DPS {\ int_0 ^ {\ frac {\ PI} {2 }}\ cfrac {x ^ 2} {\ sin ^ 2x} \ RD x} $.

Answer: (1) by $ \ beex \ Bea 2 \ SiN x \ cdot \ cfrac {1} {2} & = \ SiN X, \ 2 \ SiN x \ cdot \ cos 2x & =\ sin 3x-\ SiN x, \ 2 \ SiN x \ cdot \ cos 4x & =\ sin 5x-\ sin 3x, \\\ cdots & =\ cdots, \ 2 \ SiN x \ cdot \ cos 2nx & =\ sin (2n + 1) x-\ sin (2n-1) X \ EEA \ eeex $ Zhi $ \ Bex 2 \ SiN x \ sex {\ cfrac {1} {2} + \ sum _ {k = 1} ^ n \ cos 2kx }=\ sin (2n + 1) x. \ EEx $ then $ \ Bex \ int_0 ^ {\ PI/2} \ cfrac {\ sin (2n + 1) x} {\ SiN x} \ RD x = \ int_0 ^ {\ PI/2} \ sex {1 + 2 \ sum _ {k = 1} ^ n \ cos 2kx} \ RD x = \ cfrac {\ PI} {2 }. \ EEx $ and then $ \ beex \ Bea 2 \ SiN x \ Cos x & = \ sin 2x, \ 2 \ SiN x \ cos 3x & = \ sin 4x-\ sin 2x, \ 2 \ SiN x \ cos 5x & = \ sin 6x-\ sin 4x, \\\ cdots <=\ cdots, \\ 2 \ SiN x \ cos (2n-1) x <=\ sin 2nx-\ sin (2N-2) X \ EEA \ eeex $ Zhi $ \ Bex 2 \ SiN x \ sum _ {k = 1} ^ n \ cos (2k-1) x = \ sin 2nx. \ EEx $ \ Bex \ int_0 ^ {\ PI/2} \ cfrac {\ sin 2nx} {\ SiN x} \ RD x = 2 \ int_0 ^ {\ PI/2} \ sum _ {k = 1} ^ n \ cos (2k-1) X \ RD x = 2 \ sum _ {k = 1} ^ n \ cfrac {(-1) ^ {k-1} {2k-1 }. \ EEx $ (2) $ \ beex \ Bea \ int_0 ^ {\ frac {\ PI} {2 }}\ frac {x ^ 2} {\ sin ^ 2x} \ rd x & = -\ int_0 ^ {\ frac {\ PI} {2} x ^ 2 \ RD \ cot x \ & = 2 \ int_0 ^ {\ frac {\ PI} {2 }} X \ cot x \ RD x \ & = 2 \ int_0 ^ {\ frac {\ PI} {2} X \ RD \ ln \ SiN x \ & =-2 \ int_0 ^ {\ frac {\ PI} {2 }}\ ln \ SiN x \ rd x. \ EEA \ eeex $ find $ \ beex \ Bea \ int_0 ^ {\ frac {\ PI} {2 }}\ ln \ SiN x \ rd x & =\ int_0 ^ {\ frac {\ PI} {2 }}\ ln \ Cos x \ RD x \ quad \ sex {\ frac {\ PI} {2}-x \ leftrightsquigarrow x} \ & =\ frac {1} {2} \ int_0 ^ {\ frac {\ PI} {2 }}\ ln \ SiN x + \ ln \ Cos x \ RD x \ & =\ frac {1} {2} \ int_0 ^ {\ frac {\ PI} {2 }}\ ln \ sin 2x \ rd x-\ frac {\ pi} {4} \ ln 2 \ & =\ frac {1} {4} \ int_0 ^ {\ frac {\ PI} {2 }}\ ln \ sin 2x \ RD x-\ frac {\ PI} {4} \ ln 2 \ quad \ sex {2x \ leftrightsquigarrow x }\\\\=\ frac {1} {4} \ SEZ {\ int_0 ^ {\ frac {\ PI} {2 }}\ ln \ SiN x \ RD x + \ int_0 ^ {\ frac {\ PI} {2 }}\ ln \ cos X \ RD x}-\ frac {\ PI} {4} \ ln 2 \ quad \ sex {X-\ frac {\ PI} {2} \ leftrightsquigarrow x }\\ & =\ frac {1} {2} \ int_0 ^ {\ frac {\ PI} {2 }}\ ln \ SiN x-\ frac {\ PI} {4 }\ ln 2. \ EEA \ eeex $ then $ \ Bex \ int_0 ^ {\ frac {\ PI} {2 }}\ ln \ SiN x \ RD x =-\ frac {\ pi} {2} \ ln 2, \ quad \ int_0 ^ {\ frac {\ PI} {2 }}\ frac {x ^ 2} {\ sin ^ 2x} \ RD x = \ pi \ ln 2. \ EEx $

 

 

4. set $ f (x) $ to the real-value function on $ \ BBR $. If $ \ Bex f (x) \ Leq \ liminf _ {Y \ to x} f (y), \ EEx $ is called $ f (x) $ in the lower half of $ x $; if $ f (x) $ is semi-consecutive at any $ x \ In \ BBR $, $ f (x) $ is semi-consecutive at $ \ BBR $; if $-f (x) $ is semi-continuous, it is called $ f (x) $ semi-continuous. test certificate:

(1) $ f (x) $ in the lower half of $ x $, if and only if $ \ Bex \ forall \ ve> 0, \ exists \ Delta> 0, \ st Y \ In U (x, \ delta) \ rA f (y) <f (x) + \ ve. \ EEx $

(2) $ F $ continuous at $ x $ when and only when $ f (x) $ is in the upper half of $ x $ consecutive and lower half consecutive.

(3) If $ F $ has lower bound and lower half continuity, define $ \ Bex f_k (x) =\inf _ {Y \ In \ BBR} \ sed {f (y) + k | x-y |}, \ quad k \ In \ BBN, \ EEx $ then $ f_k (x) $ is a continuous function, $ F_1 (X) \ Leq F_2 (x) \ Leq \ cdots \ Leq f_k (x) \ Leq \ cdots $, and $ \ Bex \ lim _ {k \ To \ infty} f_k (x) = f (x), \ quad \ forall \ x \ In \ BBR. \ EEx $ [This indicates that bounded semi-continuous functions can be approached by continuous function increments]

Proof: (1) $ \ rA $: by $ \ Bex f (x) = \ liminf _ {Y \ to x} f (y) = \ sup _ {\ Delta> 0} \ INF _ {Y \ in u (x, \ delta)} f (y) \ EEx $ and upper-validation definition $ \ Bex \ forall \ ve> 0, \ exists \ Delta> 0, \ ST \ INF _ {Y \ in u (x, \ delta)} f (y)> F (x)-\ ve. \ EEx $ \ la $: If $ \ Bex \ forall \ ve> 0, \ exists \ Delta> 0, \ INF _ {Y \ in u (x, \ delta)} f (y)> F (x)-\ ve, \ EEx $ \ Bex \ liminf _ {Y \ to x} f (y) = \ sup _ {\ Delta> 0} \ INF _ {Y \ in u (x, \ delta)} f (y) & \ geq f (x)-\ ve. \ EEx $ order $ \ ve \ to 0 ^ + $ \ DPS {\ liminf _ {Y \ to x} f (y) \ geq f (x )} $. in addition, $ \ Bex \ INF _ {Y \ in u (x, \ delta)} f (y) \ Leq f (x) \ Ra \ liminf _ {Y \ to x} f (y) = \ sup _ {\ Delta> 0} \ INF _ {Y \ in u (x, \ delta )} f (y) \ Leq f (x ). \ EEx $ (2) known by (1) $ f (x) $ in the upper half of $ x $ \ LRA $ \ Bex \ forall \ ve> 0, \ exists \ Delta> 0, \ st Y \ In U (x, \ delta) \ rA f (y) <f (x) + \ ve. \ EEx $ has a conclusion. (3) at $ \ Bex f (y) + k | x-y | \ Leq f (y) + k | X-z | + k | Z-y | \ EEx $ for the lower bounds of $ Y \ In \ BBR $, $ \ Bex f_k (X) \ Leq f_k (z) + k | X-z |. \ EEx $ change $ X, Z $. We found $ \ Bex f_k (z) \ Leq f_k (x) + k | Z-x |. \ EEx $ \ Bex | f_k (x)-f_k (z) | \ Leq k | X-z |. \ EEx $ so, $ f_k (x) $ continuous. in addition, it is clear that $ \ Bex F_1 (x) \ Leq F_2 (x) \ Leq \ cdots \ Leq F _ (x) \ Leq \ cdots, \ quad \ forall \ x \ In \ BBR. \ EEx $ forward certificate $ \ DPS {\ vlm {k} f_k (x) = f (x)} $. on the one hand, by $ \ beex \ Bea f_k (x) & =\ INF _ {Y \ In \ BBR} \ sed {f (y) + k | x-y | }\\& \ Leq f (x) \ quad \ sex {\ mbox {Get} y = x} \ EEA \ eeex $ zhi$ \ DPS {\ vlm {k} f_k (X) \ Leq f (x)} $. on the other hand, from $ F $ semi-continuous, $ \ Bex \ forall \ ve> 0, \ exists \ Delta> 0, \ st Y \ In U (X, \ delta) \ rA f (y)> F (x)-\ ve. \ EEx $ set $ f (x) \ geq M, \ forall \ x $, then $ \ beex \ Bea Y \ In U (x, \ delta) & \ rA f (y) + k | x-y | \ geq f (y)> F (x)-\ ve; \ Y \ not \ In U (X, \ delta) & \ rA f (y) + k | x-y | \ geq m + k \ Delta> F (x) -\ ve \ quad \ sex {\ mbox {as long as} k> \ cfrac {-M + f (x)-\ ve} {\ Delta }}. \ EEA \ eeex $ when $ k> \ cfrac {-M + f (x)-\ ve} {\ Delta} $, $ \ beex \ Bea f_k (x) & =\ INF _ {Y \ In \ BBR} \ sed {f (y) + k | x-y |} \ geq f (x)-\ ve; \ vlm {k} f_k (x) & \ geq f (x)-\ ve. \ EEA \ eeex $ make $ \ ve \ to 0 ^ + $ \ DPS {\ vlm {k} f_k (x) \ geq f (x)} $.

 

 

5. set $ A_1 = B, A_2 = C $, at $ n \ geq 3 $, defined by $ \ DPS {a_n = \ frac {A _ {n-1} + A _ {N-2} {2} $. test Certificate: $ \ DPS {\ lim _ {n \ To \ infty} a_n} $ exists and asks for it.

Answer: $ \ beex \ Bea a _ {n + 1}-a_n & =-\ frac {1} {2} (a_n-a _ {n-1 }) \ & =\ cdots \ & =\ sex {-\ frac {1} {2 }}^ {n-1} (a_2-a_1) \\&=\ sex {-\ frac {1} {2 }}^ {n-1} (C-B ), \ a_n & = \ sum _ {k = 1} ^ {n-1} (A _ {k + 1}-A_k) + a_1 \ & =\ sum _ {k = 1} ^ {n-1} \ sex {-\ frac {1} {2} ^ {k-1} (C-B) + B \ & = (C-B) \ frac {1-\ sex {-\ frac {1} {2 }}^ {n-1 }}{ 1-\ sex {-\ frac {1} {2 }}} + B \\\\\ cfrac {2 (C-B )} {3} + B =\cfrac {2C + B} {3}, \ Quad (n \ To \ infty ). \ EEA \ eeex $

 

 

6. set $ A_1> B _1> 0 $, $ \ DPS {A _ {n + 1 }=\ frac {2a_nb_n} {a_n + B _n }, \ B _ {n + 1 }=\ SQRT {A _ {n + 1} B _n}, \ n \ In \ BBN _ +} $. proof: $ \ sed {a_n} $ and $ \ sed {B _n} $ converge to the same limit.

Proof: by $ \ Bex \ frac {A _ {n + 1 }}{ a_n }=\ frac {2b_n} {a_n + B _n }, \ quad \ frac {A _ {n + 1 }}{ B _n }=\ frac {2a_n} {a_n + B _n }, \ quad \ frac {A _ {n + 1 }}{ B _ {n + 1 }}=\ SQRT {\ frac {A _ {n + 1 }}{ B _n} }, \ quad \ frac {B _ {n + 1 }}{ B _n }=\ SQRT {\ frac {A _ {n + 1 }}{ B _n }}, \ quad \ EEx $ mathematical induction knowledge $ \ Bex B _1 \ Leq B _2 \ Leq \ cdots \ Leq B _n \ Leq \ cdots \ Leq a_n \ Leq \ cdots \ Leq A_2 \ leq A_1. \ EEx $ this indicates that $ \ DPS {\ lim _ {n \ To \ infty} a_n, \ lim _ {n \ To \ infty} B _n} $ both exist. in $ B _ {n + 1 }=\ SQRT {A _ {n + 1} B _n} $, the limit between $ n \ To \ infty $ is equal.