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

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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

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