[Jia Liwei university mathematics magazine] 058th Lanzhou University 2011 mathematical analysis postgraduate exam reference

$ \ BF abstract $: This article provides a reference for the 2011 mathematical analysis postgraduate exam of Lanzhou University.

 

$ \ BF keywords $: Lanzhou University; mathematical analysis; postgraduate exam

 

1 calculation question ($6 \ times 10' = 60' $ ).

(1) limit $ \ Bex \ lim _ {x \ to 0} \ frac {e ^ {X-\ SiN x}-e ^ \ frac {x ^ 3} {6}} {x ^ 5} \ EEx $

Answer: $ \ Bex \ mbox {Original limit} & =& \ lim _ {x \ to 0} \ frac {e ^ \ Xi \ sex {X-\ SiN x-\ frac {x ^ 3} {6 }}{ x ^ 5} \ quad \ sex {\ Xi \ mbox {at} 0, between x \ mbox {}\\\&\lim _ {x \ to 0 }\frac {-x ^ 5/120 + O (x ^ 7 )} {x ^ 5 }\\&=&-\ frac {1} {120 }. \ EEx $

 

(2) set $ x \ geq 0 $, $ \ DPS {f (x) = \ int_0 ^ X (u-u ^ 2) \ sin U \ RD u} $, calculates the maximum value of $ f (x) $.

A: from $ \ Bex f' (x) = (X-x ^ 2) \ sin ^ 2x \ left \ {\ BA {ll}> 0, & 0 <x <1 \ <0, & x> 0 \ EA \ right. \ EEx $ $ F $ the maximum value on $ x \ geq 0 $ is $ \ Bex F (1) = \ int_0 ^ 1 (u-u ^ 2) \ sin ^ 2u \ RD u =-2 \ cos 1-\ SIN 1. \ EEx $

 

(3) set $ \ Omega = [0, 1] ^ 3 \ subset \ BBR ^ 3 $, $ p $ to a real number, determine generalized points $ \ Bex \ iint _ \ Omega \ frac {\ RD x \ rd y \ RD z} {\ sex {x ^ 2 + y ^ 2 + Z ^ 2} ^ \ frac {p} {2 }}\ EEx $.

Answer: you have noticed that $ \ B _1 \ subset \ Omega \ subset B _ {\ SQRT {3 }}\ quad \ sex {B _r = \ sed {(x, y, z ); \ x ^ 2 + y ^ 2 + Z ^ 3 \ Leq R ^ 2 }}\ EEx $ we know the convergence and divergence of the original generalized points and $ \ Bex \ iint _ {B _1} \ frac {\ RD x \ RD Y \ RD z} {\ sex {x ^ 2 + y ^ 2 + Z ^ 2} ^ \ frac {p} {2 }}=\ int_0 ^ 1 \ frac {4 \ Pi R ^ 2} {R ^ p} \ RD r = 4 \ pi \ int_0 ^ 1 \ frac {1} {r ^ {P-2 }}\ RD r \ EEx $ same. therefore, when $ P <3 $, the original integral converges. When $ p \ geq 3 $, the original integral divergence.

 

(4) evaluate the power series $ \ DPS {\ sum _ {n = 0} ^ \ infty \ sex {\ frac {1} {3 ^ {n + 1}-(-2) the Convergence domains and functions of ^ {n + 1} x ^ n} $.

Answer: $ \ Bex \ mbox {Original Series} & =& \ frac {1} {3} \ sum _ {n = 0} ^ \ infty \ sex {\ frac {x} {3 }}^ N-(-2) \ sum _ {n = 0} ^ \ infty (-2x) ^ n \ quad \ sex {\ sev {x} <3 \ mbox {And} \ sev {2x} <1 }\\\\\=& \ frac {1} {3} \ frac {1} {1-\ frac {x} {3} + 2 \ frac {1} {1 + 2x} \ quad \ sex {\ sev {x} <\ frac {1} {2 }}\\\&\frac {1} {3-x} + \ frac {2} {1 + 2x} \ quad \ sex {\ sev {x} <\ frac {1} {2 }}. \ EEx $

 

(5) calculate $ \ DPS {\ int _ {()} ^ )} \ frac {x \ rd x + Y \ RD y }{\ SQRT {x ^ 2 + y ^ 2 }}$, path along the curve $ \ DPS {y ^ 4 = 2x ^ 2-2 (x> 0, Y> 0)} $.

Answer: $ \ Bex \ mbox {original second curve points }=\ int _ {()} ^ )} \ RD \ SQRT {x ^ 2 + y ^ 2 }=\ SQRT {x ^ 2 + y ^ 2} | ^ )} = \ SQRT {14}-1. \ EEx $

 

(6) calculate $ \ DPS {\ iint _ {\ varsigma} Z \ RD x \ RD y + x \ rd y \ rd z + Y \ RD x \ RD z} $, $ \ varsigma $ is the cylindrical surface $ x ^ 2 + y ^ 2 = 1 $ is flat $ z = 0 $ and $ z = 3 $ is the outer side of the cut part.

Answer: $ \ Bex \ mbox {original second-Type Curved Surface points} & =& \ iint _ {x ^ 2 + y ^ 2 \ Leq 1 \ atop 0 \ Leq Z \ Leq 3} \ SEZ {\ frac {\ P z} + \ frac {\ P x} + \ frac {\ P (-y )} {\ p y }}\ RD x \ rd y \ rd z \\& &-\ iint _ {x ^ 2 + y ^ 2 \ Leq 1 \ atop z = 3} Z \ RD x \ rd y \ & = & 3 \ pi-3 \ pi \ & = & 0. \ EEx $

 

2 answer questions ($3 \ times 6' = 18' $ ).

(1) If $ f (x) $ is defined near $ x = x_0 $ and $ \ Delta> 0 $ and constant $ \ Alpha, L $, set $ \ Bex \ sev {f (x)-f (x_0)} for any $ x \ In (x_0-\ delta, x_0 + \ delta) $ )} \ Leq L \ sev {x-x_0} ^ \ Alpha, \ EEx $ then $ f (x) $ can be exported at $ x_0 $.

Answer: Error! For example, function $ f (x) =\ SQRT {x} $ where $ x_0 = 0 $ is suitable for the condition $ \ DPS {\ alpha = \ frac {1} {2 }$, however, it cannot be exported at the origin.

 

(2) If the binary function $ f (x, y) $ is defined in a neighborhood of $ (x_0, y_0) $ and $ \ DPS {\ lim _ {(X, y) \ To (x_0, y_0)} f (x, y)} $ exists, then $ \ DPS {\ lim _ {Y \ To y_0} \ lim _ {x \ To x_0} f (x, y)} $ must exist.

Answer: Error! For example, if the function $ \ DPS {f (x, y) = Y \ sin \ frac {1} {x }}$ is at $ (0, 0) $, the limit is fixed, but the limit does not exist for multiple times (first $ x $ and then $ y $.

 

(3) binary functions $ f (x, y) $ in $ (x_0, y_0) $ micro, then the partial derivative of $ f (x, y) $ is in $ (x_0, y_0) $ points are continuous.

Answer: Error! For example, the function $ \ Bex f (x, y) =\left \{\ BA {ll} x ^ 2 \ sin \ frac {1} {x }, & X \ NEQ 0 \ 0, & x = 0 \ EA \ right. \ EEx $ is the inverse example.

3 ($ 15' $) known $ x \ in [-\ Pi, \ Pi] $, proof: $ \ DPS {\ sev {\ sin \ frac {x} {2 }}\ Leq \ frac {\ sev {x }}{\ PI }}$.

Proof: $ \ Bex \ sev {\ sin \ frac {x} {2 }=\ sin \ frac {\ sev {x }}{ 2} \ geq \ frac {2} {\ PI} \ cdot \ frac {\ sev {x }}{ 2 }=\ frac {\ sev {x }}{\ PI }. \ EEx $

 

4 ($ 15' $) proof: Symbolic Functions $ \ Bex \ mathrm {SGN} (x) =\left \ {\ BA {lll}-1, & x <0 \ 0, & x = 0 \ 1, & x> 0 \ EA \ right. \ EEx $ on $ [-] $ is the product of Riann, but the original function does not exist on $ [-] $.

Proof: $ \ mathrm {SGN} (\ cdot) $ only has two skip intervals, and Riann can accumulate. Because of this, it does not have the original function, after all, the function allows a maximum of the second type of interruptions (refer to the 020th Lanzhou University 2005 mathematical analysis postgraduate exam reference ).

 

5 ($ 15' $) set $ f (x) $ to the real function on $ \ BBR $, $ \ DPS {f (x, y) =\ frac {f (y-x)} {2x }$ and $ \ DPS {f (1, Y) = \ frac {1} {2} y ^ 2-y + 5} $. take $ x_0> 0 $ and define $ \ Bex X _ {n + 1} = f (X_n, 2x_n) \ quad \ sex {n \ geq 0 }. \ EEx $ proof: $ \ DPS {\ lim _ {n \ To \ infty} X_n} $ exists and its value is obtained.

Proof: By question, $ \ Bex F (1, y) = \ frac {f (Y-1 )} {2 }=\ frac {1} {2} y ^ 2-y + 5, \ EEx $ \ Bex F (Y-1) = y ^ 2-2y + 10 = (Y-1) ^ 2 + 9, \ EEx $ \ Bex f (x) = x ^ 2 + 9, \ EEx $ and $ \ bee \ label {lz115: EQ} X _ {n + 1} = f (X_n, 2x_n) = \ frac {f (x_n )} {2x_n }=\ frac {x_n ^ 2 + 9} {2x_n }. \ EEE $

(1) $ X _ {n + 1} \ geq 3 $ (mean inequality, $ n \ geq 0 $ );

(2) $ \ DPS {\ frac {X _ {n + 1 }}{ x_n }=\ frac {1} {2} + \ frac {9} {2x_n ^ 2} \ Leq 1} $ ($ n \ geq 1 $) zhi \ footnote {monotonic bounded theorem .} $ \ DPS {\ lim _ {n \ To \ infty} x_n = A} $ exists. on \ eqref {lz115: in EQ}, $ n \ To \ infty $ \ Bex a =\frac {A ^ 2 + 9} {2a} \ rA A = 3. \ EEx $

 

6 ($ 15' $) set $ f (x, y) $ in the region $ d \ subset \ BBR ^ 2 $, each independent variable $ x $ and $ y $ are continuous, and when $ x $ is fixed, $ y $ is monotonous. proof: $ F $ is a binary continuous function on area $ d $.

Proof: Refer to 018th Lanzhou University 2007 mathematical analysis postgraduate exam reference answers.

 

7 ($ 12' $) sets the Function Column $ \ sed {f_n (x) }_{ n = 1} ^ \ infty $ to be continuous on $ [] $, it can be imported within $ () $ and $ \ sed {f_n (x) }_{ n = 1} ^ \ infty $ is uniformly bounded on $ [] $, $ \ DPS {\ sed {f_n' (x) }_{ n = 1} ^ \ infty} $ is uniformly bounded on $ (0, 1) $. proof: The Function Column $ \ sed {f_n (x) }_{ n = 1} ^ \ infty $ has a child column with consistent convergence.

Proof: only after \ footnote {certificate \ eqref {lz117: eq} is verified, $ n_l \ To \ infty $ can be obtained .} $ \ exists \ sed {n_k} \ subset \ sed {n} $ to make $ \ bee \ label {lz117: eq} \ forall \ ve> 0, \ exists \ n> 0, \ s. t. \ left. \ BA {RR} n_k, n_k \ geq n \ x \ in [0, 1] \ EA \ right \} \ Ra \ sev {F _ {n_k} (X) -F _ {n_l} (x)} \ Leq \ ve. \ EEE $ this can be completed in three steps.

(1) first \ footnote {use $ \ Bex \ sev {f_n (x)-f_n (x')} \ Leq \ sup _ {\ Xi \ In (0, 1 )} \ sev {f_n' (\ xi)} \ cdot \ sev {X-x '}. \ EEx $} description $ \ sed {f_n (x)} $, that is, $ \ Bex \ forall \ ve> 0, \ exists \ Delta> 0, \ left. \ BA {RR} \ sev {X-x'} \ Leq \ Delta \ n \ In \ BBN \ EA \ right \} \ Ra \ sev {f_n (X) -f_n (x')} \ Leq \ frac {\ ve} {3 }. \ EEx $

(2) followed by \ footnote {multiple applications of Weierstrass clustering theorem .} description: $ \ exists \ sed {n_k} \ subset \ sed {n} $ enable $ \ sed {F _ {n_k} (x_j )} _ {k = 1} ^ \ infty $ convergence. here, $ x_j = J \ Delta \ in [0, 1] $ is a finite number of $ [0, 1] $. therefore, for any of the above $ \ ve> 0 $, $ \ Bex \ exists \ n> 0, \ s. t. \ left. \ BA {RR} n_k, n_l \ geq n \ forall \ x_j \ EA \ right \} \ Ra \ sev {F _ {n_k} (x_j) -F _ {n_l} (x_j)} \ Leq \ frac {\ ve} {3 }. \ EEx $

(3) final description \ eqref {lz117: eq }:$ $ \ Bex x \ in [0, 1] \ Ra \ exists \ X_j, \ s. t. \ sev {x-x_j} \ Leq \ delta, \ EEx $ and $ \ Bex \ sev {F _ {n_k} (X) -F _ {n_l} (x)} & \ Leq & \ sev {F _ {n_k} (x)-f _ {n_k} (x_j )} + \ sev {F _ {n_k} (x_j)-f _ {n_l} (x_j) }\\& & + \ sev {F _ {n_l} (x_j) -F _ {n_l} (x) }\\& \ Leq & \ ve. \ EEx $