[Journal of mathematics at home University] 250th questions about the Postgraduate Entrance Exam of mathematics analysis at the University of Chinese Emy of Sciences in 2013

1 ($25 '$) calculation:

(1) ($10 '$) $ \ DPS {\ lim _ {n \ To \ infty} \ sin ^ 2 \ sex {\ pi \ SQRT {n ^ 2 + N }}$.

Answer: $ \ beex \ Bea \ mbox {Original limit} & =\ LiM _ {n \ To \ infty} \ sin ^ 2 \ sex {\ pi \ SQRT {n ^ 2 + n}-\ pi n }\\\&=\ LiM _ {n \ To \ infty} \ sin ^ 2 \ frac {\ pi n }{\ SQRT {n ^ 2 + n} + N }\\\&=\ sin ^ 2 \ sex {\ lim _ {n \ To \ infty} \ frac {\ PI} {\ SQRT {1 + \ frac {1} {n }}+ 1 }\\\\=\ sin ^ 2 \ frac {\ PI} {2 }\\& = 1. \ EEA \ eeex $

 

(2) ($ 15' $) $ \ DPS {\ lim _ {n \ To \ infty} a_n }$, where $ \ DPS {A_1 = 1, A _ {n + 1} = 1 + \ frac {1} {a_n} \ (n \ geq 1)} $.

Answer: Order $ A = (1 + \ SQRT {5})/2 $, then $ \ beex \ Bea | A _ {n + 1}-A | & =\ sev {\ frac {1} {A}-\ frac {1} {a_n} \ frac {| a_n-a |} {a_na} \ Leq \ frac {1} {A} | a_n-a | \ & \ Leq \ cdots \\ & \ Leq \ frac {1} {A ^ n} | a_1-a | \ & \ to 0 \ quad \ sex {n \ To \ infty }. \ EEA \ eeex $

 

2 ($ 15' $) set $ f (x) $ continuous, $ \ DPS {g (x) = \ int_0 ^ x F (x-T) \ sin t \ RD t} $, trial certificate: $ \ Bex g'' (x) + g (x) = f (x), \ quad g (0) = G' (0) = 0. \ EEx $

Proof: by $ \ Bex g (x) = \ int_0 ^ x F (x-T) \ sin t \ rd t = \ int_0 ^ x F (s) \ sin (X-S) \ RD s \ EEx $ Zhi $ \ Bex G' (x) = \ int_0 ^ x F (s) \ cos (X-S) \ rd s, \ quad g'' (x) = f (x)-\ int_0 ^ x F (s) \ sin (X-S) \ rd s. \ EEx $ then $ \ Bex g'' + G = F, \ quad g (0) = G' (0) = 0. \ EEx $

 

3 ($ 15' $) calculates the maximum curvature of the curve $ Y = e ^ x $.

Answer: curvature $ \ Bex \ Kappa (x) = \ frac {y''} {(1 + y' ^ 2) ^ {3/2 }}=\ frac {e ^ x} {(1 + e ^ {2x}) ^ {3/2 }}. \ EEx $ by $ \ Bex \ Kappa '(x) = \ frac {e ^ X (1-2e ^ {2x})} {(1 + e ^ {2x }) ^ {5/2 }}\ EEx $ $ \ Bex \ MAX \ Kappa = \ Kappa \ sex {-\ frac {\ ln 2} {2 }=\ frac {2 \ SQRT {3 }}{ 9 }. \ EEx $

 

4 ($2 \ times 15' = 30' $) Calculate points

(1) $ \ DPS {I =\int _ {1/4} ^ {1/2} \ RD Y \ int _ {1/2} ^ {\ SQRT {y} e ^ {Y/x} \ RD x + \ int _ {1/2} ^ 1 \ rd y \ int_y ^ {\ SQRT {y} e ^ {Y/x} \ RD x} $.

Answer: $ \ beex \ Bea I =\int _ {1/2} ^ 1 \ RD x \ int _ {x ^ 2} ^ x e ^ {Y/x} \ RD y = \ int _ {1/2} ^ 1 x (e-e ^ X) \ RD x = \ frac {3e-4 \ SQRT {e }}{ 8 }. \ EEA \ eeex $

 

(2) $ \ DPS {J = \ iint _ \ Omega | x ^ 2 + y ^ 2-1 | \ RD x \ RD y }, $ \ Bex \ Omega = \ sed {(x, y); \ 0 \ Leq x \ Leq 1, \ 0 \ Leq Y \ Leq 1 }. \ EEx $

Answer: $ \ beex \ Bea J & =\ int_0 ^ 1 \ RD x \ int_0 ^ {\ SQRT {^ 2} (^ 2-y ^ 2) \ RD y + \ int_0 ^ 1 \ RD x \ int _ {\ SQRT {3o^ 2 }}^ 1 (x ^ 2 + y ^ 2-1) \ RD y \\\\=\ frac {\ PI} {8} + \ sex {\ frac {\ PI} {8}-\ frac {1} {3 }}\ \ & =\ frac {\ PI} {4}-\ frac {1} {3 }. \ EEA \ eeex $

 

5 ($ 15' $) discusses the series $ \ DPS {\ sum _ {n = 1} ^ \ infty \ frac {n ^ 2} {(x + 1/n) ^ n }}$ convergence and consistent convergence (including the consistent convergence of inner and closed ).

Answer: by $ \ beex \ Bea \ lim _ {n \ To \ infty} \ sev {\ frac {(n + 1) ^ 2} {(x + 1/(n + 1) ^ {n + 1 }}{\ frac {n ^ 2} {(x + 1/n) ^ n }}& =\ LiM _ {n \ To \ infty} \ sev {\ sex {\ frac {x + 1/n} {x + 1/(n + 1 )}} ^ n \ frac {1} {x + 1/(n + 1 )}} \\<=\ frac {1 }{| x |}\ LiM _ {n \ To \ infty} \ frac {\ sex {1 + 1/(nx )} ^ {NX \ cdot 1/x} {\ SEZ {1 + 1/(n + 1) x)} ^ {(n + 1) X \ cdot N/(n + 1) x )}} \\&=\ frac {1 }{| x |}\ EEA \ eeex $ that is, when $ | x |> 1 $, the original series converges; original Series divergence when $ | x | <1 $; when $ x = 1 $, the original series is $ \ DPS {\ sum \ frac {n ^ 2} {(1 + 1/n) ^ n }}$ divergence (general items do not tend to $0 $ ); when $ x =-1 $, the original series is $ \ DPS {\ sum \ frac {n ^ 2} {(-1 + 1/n) ^ n} = \ sum \ frac {(-1) ^ NN ^ 2} {(1 + 1/(-N )) ^ {-n \ cdot (-1) }}$ divergence (the wildcard does not tend to be $0 $ ). therefore, the convergence field of the original series is $ \ sed {X; | x |> 1} $. by $ \ Bex \ frac {n ^ 2} {[(1 + 1/n) + 1/N] ^ n }=\ frac {n ^ 2} {(1 + 2/N) ^ {n/2 \ cdot 2 }}\ geq \ frac {n ^ 2} {e ^ 2} \ EEx $ know that the original series is at $ \ sed {X; | x |> 1} $ inconsistent convergence. end with $ \ beex \ Bea \ sev {\ frac {n ^ 2} {(x + 1/n) ^ n }}& \ Leq \ frac {n ^ 2} {(1 + \ ve-1/N) ^ n} \ quad \ sex {| x | \ geq 1 + \ ve }\\& \ Leq \ frac {n ^ 2} {(1 + \ ve/2) ^ n} \ quad \ sex {n \ geq 2/\ ve} \ & \ Leq \ sex {\ frac {n} {\ SQRT [3] {1 + \ ve /2 }}^ 2 \ sex {\ frac {1} {\ SQRT [3] {1 + \ ve/2 }}^ n \ & \ Leq \ sex {\ frac {2} {2 + \ ve} ^ {n/3} \ quad \ sex {n \ Gg 1} \ EEA \ eeex $ know the original series at $ \ sed {X; | x | \ geq 1 + \ ve} \ (\ forall \ ve> 0) $ consistent convergence.

 

6 ($2 \ times 10' = 20' $)

(1) proof: When $0 <x <\ PI/2 $, $2/\ pI <\ SiN x/x <1 $.

Proof: $ \ Bex \ SiN x = \ int_0 ^ x \ cos t \ rd t <\ int_0 ^ x \ rd t = x; \ EEx $ by $ (\ SiN x) ''=-\ SiN x <0 $ \ SiN x $ at $ (0, \ PI/2) $ is a concave function, and $ \ beex \ Bea \ SiN x & =\ sin (1-2x/\ PI) \ cdot 0 + 2x/\ pi \ cdot \ PI/2) \ & \ geq (1-2x/\ PI) \ cdot \ sin 0 + 2x/\ pi \ cdot \ sin \ PI/2 = 2x/\ pi. \ EEA \ eeex $

 

(2) set the function $ f (x) $ in the closed range $ [a, B] $ the secondary function can be micro, and $ f'' (x) <0 $, then $ \ Bex \ frac {f (a) + F (B )} {2} \ Leq \ frac {1} {B-a} \ int_a ^ B f (t) \ RD T. \ EEx $

Proof: by $ f'' (x) <0 $ $ F $ is a concave function, while $ \ beex \ Bea \ int_a ^ B f (x) \ rd x & =\ int_a ^ B f \ sex {\ frac {B-x} {B-a} \ cdot A + \ frac {X-A} {B-} \ cdot B} \ RD x \ & \ geq \ int_a ^ B \ SEZ {\ frac {B-x} {B-a} \ cdot F () + \ frac {X-A} {B-a} \ cdot F (B )} \ RD x \\\\=\ frac {B-a} {2} [F (a) + F (B)]. \ EEA \ eeex $

 

7 ($15 '$) evaluate the function $ \ DPS {f (x, y) = x ^ 2 + y ^ 2 + \ frac {3} {2} x + 1} $ in the set $ \ DPS {G = \ sed {(x, y ); 4x ^ 2 + y ^ 2-1 = 0 }}$.

Answer: use the resource name as the resource Resource Name (*, Y; \ lambda) = x ^ 2 + y ^ 2 + \ frac {3} {2} x + 1 + \ Lambda (4x ^ 2 + y ^ 2-1 ), \ EEx $ is composed of $ \ beex \ Bea l_x & = 2x + \ frac {3} {2} + 8 \ Lambda x = 0, \ l_y & = 2y + 2 \ Lambda y = 0, \ L _ \ Lambda & = 4x ^ 2 + y ^ 2-1 = 0 \ EEA \ eeex $ $ \ BEX (x, y) = \ sex {\ frac {1} {4 }, \ frac {\ SQRT {3 }}{ 2 }}\ mbox {or} \ sex {\ PM \ frac {1} {2}, 0 }. \ EEx $ again by $ \ Bex f \ sex {\ frac {1} {4 }, \ frac {\ SQRT {3 }}{ 2 }=\ frac {35} {16}, \ quad f \ sex {\ frac {1} {2 }, 0} = 2, \ quad f \ sex {-\ frac {1} {2 }, 0 }=\ frac {1} {2} \ EEx $ \ Bex \ max_gf =\ frac {35} {16 }, \ quad \ min_gf =\frac {1} {2 }. \ EEx $

 

8 ($ 15' $) set the infinite real sequence $ \ sed {a_n }, \ sed {B _n} $ meet $ \ Bex a _ {n + 1} = B _n-\ frac {na_n} {2n + 1}, \ Quad (n = 1, 2, \ cdots ). \ EEx $ test certificate:

(1) If $ \ sed {B _n} $ is bounded, $ \ sed {a_n} $ is also bounded;

(2) If $ \ sed {B _n} $ converges, $ \ sed {a_n} $ also converges.

Proof:

(1) set $ \ Bex M = \ MAX \ sed {2 \ sup_n | B _n |, | a_1 | }. \ EEx $ we use mathematical induction to learn easily $ | a_n | \ Leq M $. in fact, the recurrence step is $ \ Bex | A _ {n + 1} | \ Leq | B _n | + \ sev {\ frac {na_n} {2n + 1} \ Leq \ Frac. {m} {2} + \ frac {m} {2} = m. \ EEx $

(2) set $ B _n \ to B $ at $ \ Bex a _ {n + 1} = B _n-\ frac {na_n} {2n + 1 }, \ Quad (n = 1, 2, \ cdots ). \ EEx $ the upper limit and lower limit are $ \ Bex \ varlimsup _ {n \ To \ infty} a_n \ Leq B-\ frac {1} {2} \ varliminf _{ n \ To \ infty} a_n, \ varliminf _ {n \ To \ infty} a_n \ geq B-\ frac {1} {2} \ varlimsup _ {n \ To \ infty} a_n. \ EEx $ then $ \ beex \ Bea \ varlimsup _ {n \ To \ infty} a_n \ Leq B-\ frac {1} {2} \ SEZ {B -\ frac {1} {2} \ varlimsup _ {n \ To \ infty} a_n} \ Ra \ varlimsup _ {n \ To \ infty} a_n \ Leq \ frac {2B }{ 3 }, \\\ varliminf _ {n \ To \ infty} a_n \ geq B-\ frac {1} {2} \ SEZ {B-\ frac {1} {2} \ varliminf _ {n \ To \ infty} a_n} \ Ra \ varliminf _ {n \ To \ infty} a_n \ geq \ frac {2B} {3 }. \ EEA \ eeex $ therefore $ a_n \ to 2B/3 $.