[Journal of mathematics at home University] Question of 242nd Zhejiang University Mathematics Analysis Postgraduate Entrance Exam

Source: Internet
Author: User

1 ($4 \ times 10' = 40' $)

(1) $ \ DPS {\ lim _ {x \ to 0} \ frac {\ SiN x-\ arctan x }{\ Tan x-\ arcsin x }}$.

Answer: $ \ beex \ Bea \ lim _ {x \ to 0} \ frac {\ SiN x-\ arctan x} {\ Tan x-\ arcsin x} & = \ lim _ {x \ to 0} \ frac {\ SEZ {X-\ frac {x ^ 3} {6} + O (x ^ 3 )} -\ SEZ {X-\ frac {x ^ 3} {3} + O (x ^ 3 )}} {\ SEZ {x + \ frac {x ^ 3} {3} + O (x ^ 3 )} -\ SEZ {x + \ frac {x ^ 3} {6} + O (x ^ 3 )}} \ & =\ frac {-\ frac {1} {6} + \ frac {1} {3 }{\ frac {1} {3}-\ frac {1} {6 }\\& = 1. \ EEA \ eeex $

(2) $ \ DPS {\ int_0 ^ \ pi \ frac {\ cos 4 \ Theta} {1 + \ cos ^ 2 \ Theta} \ RD \ Theta} $.

Answer: $ \ beex \ Bea \ mbox {original points} & =\ int_0 ^ \ pi \ frac {\ cos4 \ Theta} {1 + \ frac {1 + \ cos2 \ Theta} {2 }}\ RD \ Theta \ & = 2 \ int_0 ^ \ pi \ frac {\ cos 4 \ Theta} {3 + \ cos2 \ Theta} \ RD \ Theta \ \ & =\ int_0 ^ {2 \ PI} \ frac {\ cos 2 t} {3 + \ cos t} \ RD t \ quad \ sex {T = 2 \ Theta} \ & =\ int_0 ^ {2 \ PI} \ frac {2 \ cos ^ 2t-1} {\ cos T + 3} \ RD t \ & =\ int_0 ^ {2 \ PI} \ frac {2 (\ cos T + 3) ^ 2-12 (\ cos T + 3) + 17} {\ cos T + 3} \ RD t \\\&=\ int_0 ^ {2 \ PI} 2 (\ cos T + 3) -12 + \ frac {17} {\ cos T + 3} \ RD t \ & =-6 \ cdot 2 \ PI + 17 \ int_0 ^ \ pi \ int _{ -\ PI} ^ \ pi \ frac {1} {3-\ cos s} \ RD s \ Quad (S = \ pi-T) \ & =-12 \ PI + 34 \ int_0 ^ \ pi \ frac {1} {3-\ cos s} \ RD s \ & =-12 \ PI + 34 \ int_0 ^ \ pi \ frac {1} {2 + 2 \ sin ^ 2 \ frac {s} {2 }}\ rd s \ & =-12 \ PI + 34 \ int_0 ^ \ pi \ frac {\ cos ^ 2 \ frac {s} {2} + \ sin ^ 2 \ frac {s} {2 }}{ 2 \ cos ^ 2 \ frac {s} {2} + 4 \ sin ^ 2 \ frac {s} {2 }}\ rd s \ & =-12 \ PI + 34 \ int_0 ^ \ pi \ frac {2 \ RD \ tan \ frac {s} {2 }}{ 2 + 4 \ tan ^ 2 \ frac {s} {2 }}\\& =- 12 \ PI + 34 \ int_0 ^ \ infty \ frac {\ RD t} {1 + 2t ^ 2} \ & =-12 \ PI + 34 \ cdot \ frac {1} {\ SQRT {2 }}\ int_0 ^ \ infty \ frac {\ RD (\ SQRT {2} t )} {1 + (\ SQRT {2} t) ^ 2 }\\\&=\ sex {\ frac {17 }{\ SQRT {2 }}- 12} \ pi. \ EEA \ eeex $ \ bzj students who have learned the complex variable function can also use the number of records for calculation or verification: $ \ beex \ Bea \ mbox {Original credits} & =\ int_0 ^ {2 \ PI} \ frac {2 \ cos ^ 2t-1} {3 + \ cos t} \ RD T \ & =\ int _ {| z | = 1} \ frac {2 \ sex {\ frac {z + Z ^ {-1 }}{ 2 }}^ 2 -1} {3 + \ frac {z + Z ^ {-1 }}{ 2 }}\ frac {\ RD z} {iz }\\\\=\ frac {1} {I} \ int _ {| z | = 1} \ frac {z ^ 4 + 1} {z ^ 2 (6z + Z ^ 2 + 1 )} \ rd z \ & =\ frac {1} {I} \ cdot 2 \ pi I \ SEZ {\ underset {z = 0} {\ res} \ frac {z ^ 4 + 1} {z ^ 2 (6z + Z ^ 2 + 1 )} + \ underset {z =-3 + 2 \ SQRT {2 }}{\ res} \ frac {z ^ 4 + 1} {z ^ 2 (6z + Z ^ 2 + 1 )}} \ & = 2 \ pi \ SEZ {\ sex {\ frac {z ^ 4 + 1} {z ^ 2 (6z + Z ^ 2 + 1 )}} '| _ {z = 0} + \ frac {z ^ 4 + 1} {z ^ 2 [Z-(-3-2 \ SQRT {2})]} | _ {z =-3 + 2 \ SQRT {2 }}\\\& = 2 \ pi \ sed {-6 + \ frac {1} {4 \ SQRT {2 }}\ frac {z ^ 4 + 1} {z ^ 2} | _ {z =-3 + 2 \ SQRT {2 }}\\& = 2 \ pi \ sed {-6 + \ frac {1} {4 \ SQRT {2 }}\ SEZ {(17-12 \ SQRT {2 }) + (17 + 12 \ SQRT {2 })}} \\& = 2 \ pi \ SEZ {-6 + \ frac {17} {2 \ SQRT {2 }}\\\&=\ sex {\ frac {17 }{ \ SQRT {2 }}- 12} \ pi. \ EEA \ eeex $ \ ezj

(3) set $ D =\sed {(x, y); x ^ 2 + y ^ 2 \ Leq \ SQRT {3}, X \ geq 0, Y \ geq 0 }$, $[1 + x ^ 2 + y ^ 2] $ indicates the maximum integer not exceeding $1 + x ^ 2 + y ^ 2 $. calculate dual points $ \ Bex \ iint_dxy [1 + x ^ 2 + y ^ 2] \ RD x \ RD y. \ EEx $

Answer: $ \ beex \ Bea \ mbox {Original credits} & =\ iint _ {0 \ Leq x ^ 2 + y ^ 2 <1 \ atop x \ geq 0, Y \ geq 0} XY \ RD x \ RD y + 2 \ iint _ {1 \ Leq x ^ 2 + y ^ 2 <\ SQRT {3} \ atop x \ geq 0, Y \ geq 0} XY \ RD x \ rd y \ & =\ int_0 ^ 1 \ RD r \ int_0 ^ \ frac {\ PI} {2} r \ cos \ Theta \ cdot r \ sin \ Theta \ cdot r \ RD \ Theta + 2 \ int_1 ^ {\ SQRT [4] {3 }}\ RD r \ int_0 ^ \ frac {\ pi} {2} r \ cos \ Theta \ cdot r \ sin \ Theta \ cdot r \ RD \ Theta \ & =\ frac {1} {8} + \ frac {1} {4 }\\&=\ frac {3} {8 }. \ EEA \ eeex $

(4) set $ \ DPS {s_n = \ frac {1} {\ SQRT {n }}\ sex {1 + \ frac {1} {\ SQRT {2 }}+ \ cdots + \ frac {1 }{\ SQRT {n }}}$, evaluate $ \ DPS {\ lim _ {n \ To \ infty} s_n} $.

Answer: The sotlz formula, $ \ Bex \ mbox {Original limit }=\ LiM _ {n \ To \ infty} \ frac {\ frac {1 }{\ SQRT {n }}{\ SQRT {n}-\ SQRT {n-1 }}=\ LiM _ {n \ To \ infty} \ frac {\ SQRT {n} + \ SQRT {n-1 }}{\ SQRT {N }}= 2. \ EEx $

 

2 ($ 10' $) demonstrate whether a continuous function defined on $ \ BBR $ exists so that $ F (f (x) = E ^ {-x} $.

Proof: Use the reverse verification method. If such $ F $ exists

(1) $ F $ is a single shot: $ \ Bex f (x) = f (y) \ rA e ^ {-x} = f (x )) = f (y) = E ^ {-y} \ rA x = y. \ EEx $

(2) $ F $ monotonous. the reverse verification method is also used. if $ F $ is not monotonous, $ \ Bex \ exists \ A <B <C, \ st F (a) \ Leq F (B) \ geq F (c) \ mbox {or} f (a) \ geq F (B) \ Leq F (c ). \ EEx $ \ Bex F (a) \ Leq F (B) \ geq F (C), \ quad F (a) \ geq F (c ), \ EEx $ is known by the mediated Value Theorem of a continuous function $ \ Bex \ exists \ D \ In (B, c), \ st F (d) = f (). \ EEx $ this conflicts with $ F $.

(3) Since $ F $ is monotonous, we know that $ f \ circ f $ increments, which is in conflict with $ e ^ {-x} $. Therefore, we have a conclusion.

 

3. Discussion of function level $ \ DPS {\ sum _ {n = 1} ^ \ infty \ frac {\ SQRT {n + 1}-\ SQRT {n} {n ^ convergence and consistent convergence of X }}$.

Answer: by $ \ Bex \ sum _ {n = 1} ^ \ infty \ frac {\ SQRT {n + 1}-\ SQRT {n }}{ n ^ x} = \ sum _ {n = 1} ^ \ infty \ frac {1} {n ^ X (\ SQRT {n + 1} + \ SQRT {n })} \ EEx $ knows the original function when $ \ Bex x + \ frac {1} {2}> 1 \ rA x> \ frac {1} {2} \ EEx $ level convergence. when $ \ DPS {A> \ frac {1} {2} $, $ \ Bex \ frac {1} {n ^ X (\ SQRT {n + 1} + \ SQRT {n })} \ Leq \ frac {1} {n ^ {x + \ frac {1} {2 }}\ Leq \ frac {1} {n ^ {A + \ frac {1} {2 }}\ Quad (x \ geq ), \ EEx $ we know that the level of the original function is uniformly converged on $ [A, \ infty) $. finally, by $ \ DPS {\ sum _ {n = 1} ^ \ infty \ frac {1} {(n + 1) ^ {1 + \ frac {1} {n} $ divergent knowledge $ \ Bex \ exists \ ve_0> 0, \ forall \ n, \ exists \ P, \ ST \ sum _ {k = n + 1} ^ {n + p} \ frac {1} {(n + 1) ^ {1 + \ frac {1} {n }}\ geq 2 \ ve_0, \ EEx $ and $ \ Bex \ sum _ {k = n + 1} ^ {n + p} \ frac {\ SQRT {k + 1}-\ SQRT {k }}{ K ^ {\ frac {1} {2} + \ frac {1} {k }}=\ sum _ {k = n + 1} ^ {n + p} \ frac {1} {k ^ {\ frac {1} {2} + \ frac {1} {k }}\ sex {\ SQRT {k + 1} + \ SQRT {k }}\ geq \ frac {1} {2} \ sum _ {k = n + 1} ^ {n + p} \ frac {1 }{( n + 1) ^ {1 + \ frac {1} {n }}\ geq \ ve_0. \ EEx $ therefore, the level of the original function is inconsistent on $ \ DPS {\ sex {\ frac {1} {2}, \ infty} $.

 

4 ($ 15' $) set $ f (x), g (x), \ varphi (x) $ to continuous functions on $ [a, B] $, and $ g (x) $ is monotonically increasing, $ \ varphi (x) \ geq 0 $, and can satisfy any $ x \ in [a, B] $, $ \ Bex f (x) \ Leq g (x) + \ int_a ^ x \ varphi (t) f (t) \ rd t. \ EEx $ proof: For any $ x \ in [a, B] $, all $ \ Bex f (x) \ Leq g (x) e ^ {\ int_a ^ x \ varphi (s) \ RD s }. \ EEx $

Proof: Set $ \ Bex f (x) = \ int_a ^ x \ varphi (t) f (t) \ RD t, \ EEx $ then $ \ beex \ Bea f' (x) & =\ varphi (x) f (x) \ Leq \ varphi (x) g (x) + \ varphi (x) f (x), \ f' (x)-\ varphi (x) f (x) & \ Leq \ varphi (X) g (x), \\\ SEZ {f (x) e ^ {-\ int_a ^ x \ varphi (t) \ RD t} '& \ Leq \ varphi (x) g (x) e ^ {-\ int_a ^ x \ varphi (t) \ RD t} \ f (x) e ^ {-\ int_a ^ x \ varphi (t) \ RD t} & \ Leq \ int_a ^ x \ varphi (t) g (t) e ^ {-\ int_a ^ t \ varphi (s) \ RD s} \ rd t, \ f (x) & \ Leq \ int_a ^ x \ varphi (t) g (t) e ^ {\ int_t ^ x \ varphi (s) \ RD s} \ RD t, \ EEA \ eeex $ \ beex \ Bea f (x) & \ Leq g (x) + f (x) \ & \ Leq g (x) + \ int_a ^ x \ varphi (t) g (t) e ^ {\ int_t ^ x \ varphi (s) \ RD s} \ RD t \ & \ Leq g (t) + g (x) \ int_a ^ x \ varphi (t) e ^ {\ int_t ^ x \ varphi (s) \ RD s} \ RD t \ quad \ sex {g \ mbox {increment }\\& = g (x) + g (x) \ SEZ {-e ^ {\ int_t ^ x \ varphi (s) \ RD s }}_{ T = A} ^ {T = x }\\& = g (x) + g (x) \ SEZ {-1 + e ^ {\ int_a ^ x \ varphi (s) \ RD s }\\& = g (x) e ^ {\ int_a ^ x \ varphi (s) \ RD s }. \ EEA \ eeex $

 

5

(1) $ \ DPS {\ lim _ {n \ To \ infty} \ int_0 ^ \ frac {\ PI} {2} \ sin ^ NX \ RD x = 0} $;

(2) $ \ DPS {\ lim _ {n \ To \ infty} \ int_0 ^ \ frac {\ PI} {2} \ SiN x ^ n \ RD x = 0} $.

Proof:

(1) for any $ \ DPS {\ Delta \ In \ sex {0, \ frac {\ PI} {2 }}$, $ \ beex \ Bea \ int_0 ^ \ frac {\ PI} {2} \ sin ^ NX \ rd x & =\ int_0 ^ {\ frac {\ PI} {2 }-\ Delta} \ sin ^ N x \ RD x + \ int _ {\ frac {\ PI} {2}-\ Delta} ^ \ frac {\ PI} {2} \ sin ^ NX \ RD x \ & \ Leq \ sex {\ frac {\ PI} {2}-\ Delta} \ sin ^ n \ sex {\ frac {\ pi} {2}-\ Delta} + \ Delta \ & \ equiv I _1 + I _2. \ EEA \ eeex $ to $ \ forall \ ve> 0 $, $ \ DPS {0 <\ Delta <\ frac {\ ve} {2 }$, and $ \ DPS {I _2 <\ frac {\ ve} {2 }$; then the $ \ Delta $, $ \ Bex \ exists \ n, \ forall \ n> N, \ mbox {has} I _1 <\ frac {\ ve} {2 }. \ EEx $ therefore, $ \ DPS {\ lim _ {n \ To \ infty} \ int_0 ^ \ frac {\ PI} {2} \ sin ^ NX \ RD x = 0} $.

(2) by (the second point of the integral value theorem:

(A) If $ f \ In \ mathscr {r} [a, B] $, $ g \ searrow $, $ g \ geq 0 $, then $ \ DPS {\ exists \ Xi \ in [a, B], \ ST \ int_a ^ BF (x) g (x) \ RD x = g () \ int_a ^ \ Xi f (x) \ rd x;} $

(B) If $ f \ In \ mathscr {r} [a, B] $, $ g \ nearrow $, $ g \ geq 0 $, then $ \ DPS {\ exists \ ETA \ in [a, B], \ ST \ int_a ^ BF (x) g (x) \ RD x = g (B) \ int _ \ ETA ^ B f (x) \ RD X;} $

(C) If $ f \ In \ mathscr {r} [a, B] $, $ G $ is monotonous, then $ \ DPS {\ exists \ Xi \ in [, b], \ ST \ int_a ^ BF (x) g (x) \ RD x = g (a) \ int_a ^ \ Xi f (x) \ RD x + g (B) \ int _ \ Xi ^ BF (x) \ RD X .} $) $ \ beex \ Bea \ int_0 ^ \ frac {\ PI} {2} \ SiN x ^ n \ rd x & =\ int_0 ^ 1 \ SiN x ^ n \ RD x + \ int_1 ^ \ frac {\ PI} {2} \ SiN x ^ n \ RD x \ & \ Leq \ int_0 ^ 1 x ^ n \ RD x + \ int_1 ^ \ frac {\ PI} {2} \ frac {1} {NX ^ {n-1 }}\ cdot NX ^ {n-1} \ SiN x ^ n \ RD x \ & = \ frac {1} {n + 1} + \ frac {1} {n} \ int_1 ^ \ Xi \ RD (-\ Cos x ^ N) \ quad \ sex {\ mbox {point second medium value theorem }\\\& \ Leq \ frac {1} {n + 1} + \ frac {2} {n} \ EEA \ eeex $ conclusion.

Note: students who have learned real-time variable functions can also use Lebesgue to control the convergence theorem and draw a conclusion immediately.

 

6 (1) ($5 '$) construct a closed range $ [-] $ a function that can be traced everywhere, so that its derivative is unbounded on $ [-] $.

(2) ($15 '$) set the function $ f (x) $ in $ (a, B) $, which proves that $ (\ Alpha, \ beta) exists) \ subset (a, B) $ to make $ f' (x) $ bounded within $ (\ Alpha, \ beta) $.

Proof:

(1) function $ \ DPS {f (x) on $ [-] $) = \ left \ {\ BA {ll} x ^ 2 \ sin \ frac {1} {x ^ 2}, & X \ NEQ 0, \ 0, & amp; X = 0, \ EA \ right .} $ then $ \ DPS {f' (X) = \ left \ {\ BA {ll} 2x \ sin \ frac {1} {x ^ 2}-\ frac {2} {x} \ cos \ frac {1 }{ x ^ 2 }, & X \ NEQ 0, \ 0, & x = 0 \ EA \ right .} $ unbounded.

(2) get $ C <D $ to make $ [c, d] \ subset (a, B) $, which is set by $ \ Bex [C, d] = \ cup _ {n = 1} ^ \ infty e_n, \ quad e_n = \ sed {x \ in [c, d]; \ | f' (X) | \ Leq n} \ EEx $ and baire theorem $ E _ {n_0} $ has an internal point. that is, evidence.

 

7. Set the two mixed partial derivatives of $ f (x, y) $ F _ {XY} (x, y) $, $ F _ {Yx} (x, y) $ exists near $ (0, 0) $ and $ F _ {XY} (x, y) $ is continuous at $ (0, 0) $. proof: $ F _ {XY} (0, 0) = F _ {Yx} (0, 0) $.

Proof: $ \ beex \ Bea F _ {Yx} (0, 0) & =\ LiM _ {x \ to 0} \ frac {f_y (x, 0) -f_y (0, 0 )} {x} \ & =\ LiM _ {x \ to 0} \ frac {1} {x} \ lim _ {Y \ to 0} \ SEZ {\ frac {f (X, y)-f (x, 0)} {y}-\ frac {f (0, Y)-f (0, 0 )} {Y }}\\\=\ LiM _ {x \ to 0} \ frac {1} {x} \ lim _ {Y \ to 0} \ frac {[F (x, y)-f (0, y)]-[F (x, 0)-f (0, 0)]} {y} \ & =\ LiM _ {x \ to 0} \ frac {1} {x} \ lim _ {Y \ to 0} \ frac {1} {Y} \ SEZ {\ frac {f (x, y)-f (0, Y)} {f (x, 0)-f (0, 0)}-1} \ SEZ {f (x, 0) -F (0, 0 )} \ & =\ LiM _ {x \ to 0} \ frac {1} {x} \ lim _ {Y \ to 0} \ frac {1} {y} \ Sez {\ frac {f_x (sx, y)} {f_x (sx, 0)}-1} [F (x, 0)-f (0, 0)] \ & =\ LiM _ {x \ to 0} \ frac {1} {x} \ lim _ {Y \ to 0} \ frac {f_x (sx, Y) -f_x (sx, 0)} {y} \ cdot \ frac {f (x, 0)-f (0, 0)} {f_x (sx, 0 )} \ & =\ LiM _ {x \ to 0} \ frac {1} {x} \ lim _ {Y \ to 0} f _ {XY} (sx, Ty) \ frac {f_x (\ Theta X, 0) x} {f_x (sx, 0 )} \ & =\ LiM _ {x \ to 0} \ lim _ {Y \ to 0} f _ {XY} (sx, Ty) \ frac {f_x (\ Theta X, 0)} {f_x (sx, 0 )} \ & =\ LiM _ {x \ to 0} \ frac {1} {x} \ lim _ {Y \ to 0} f _ {XY} (0, 0 ), \ quad \ sex {F _ {XY} \ mbox {continuous at origin }}. \ EEA \ eeex $

 

8 ($ 20' $) known pairs real number $ n \ geq 2 $, formula $ \ Bex \ sum _ {P \ Leq n} \ frac {\ ln p} {p} = \ ln n + O (1 ), \ EEx $ the sum is the sum of all prime numbers not greater than $ N $ p $. verification: $ \ Bex \ sum _ {P \ Leq n} \ frac {1} {p} = C + \ ln n + O \ sex {\ frac {1 }{ \ ln n }}, \ EEx $ the sum is also the sum of all prime numbers not greater than $ N $, $ C $ is a constant irrelevant to $ N $.

Proof: Set $ \ Mu $ to counting measure on $ \ BBN $, so that $ \ DPS {\ Mu (n) =\left \{\ BA {ll} 1, & n \ mbox {is a prime number}, \ 0, & n \ mbox {is a combination number }. \ EA \ right .} $ is easy to draw conclusions by using the simple Riemann-Stieltjes points technique. for more information, see link.

[Journal of mathematics at home University] Question of 242nd Zhejiang University Mathematics Analysis Postgraduate Entrance Exam

Contact Us

The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email.

If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days.

A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

  • Sales Support

    1 on 1 presale consultation

  • After-Sales Support

    24/7 Technical Support 6 Free Tickets per Quarter Faster Response

  • Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.