[Journal of mathematics at home University] 307th questions about Mathematics Analysis in 2006 of Huazhong Normal University

 

1. ($30 '$) Calculate the question.

(1) $ \ DPS {\ lim _ {x \ to 1} \ frac {\ sin ^ 2 (x-1) \ sin \ frac {1} {X-1 }}{ e ^ {X-1}-1 }}$.

(2) set $ Y = x ^ x + A ^ x $, evaluate $ y' $.

(3) $ \ DPS {\ int \ sex {\ ln x + \ frac {1 }{\ ln x }}\ RD x }$.

(4) set $ \ DPS {f (x, y) = x ^ y + (Y-1) ^ 2 \ arcsin \ SQRT {\ frac {x} {y }}$, evaluate $ f_x '(x, 1) $.

(5) $ \ DPS {\ iint_d (x + y) e ^ {x ^ 2 + y ^ 2} \ RD x \ RD y} $, $ d = \ sed {(x, y); x ^ 2 + y ^ 2 \ Leq 1} $.

(6) Evaluate $ \ DPS {I = \ int_l x \ sin Y \ RD X-\ cos y \ RD x} $, where $ L $ is the sine curve from $ O (0, 0) $ to $ A (\ Pi, 0) $ Y = \ SiN x $.

 

Answer:

(1) by $ \ Bex \ lim _ {x \ to1} \ frac {\ sin ^ 2 (x-1 )} {e ^ {X-1}-1} = \ lim _ {T \ to 0} \ frac {t ^ 2} {e ^ T-1} = \ lim _ {T \ to 0} \ frac {2 t} {e ^ t} = 0 \ EEx $ Zhi $ \ DPS {\ frac {\ sin ^ 2 (x-1 )} {e ^ {X-1}-1} = O (1) \ (x \ to 1)} $, it multiplied by a bounded amount $ \ DPS {\ sin \ frac {1} {X-1} $ is still an infinitely small amount, and the original limit $ = 0 $.

(2) $ \ Bex y' = (E ^ {x \ ln x}) '+ A ^ x \ ln a = x ^ X (\ ln x + 1) + A ^ x \ ln. \ EEx $

(3) know by segment credits $ \ Bex \ mbox {Original credits} = x \ ln x-\ int \ frac {1} {x \ ln x} \ cdot x \ RD X + \ int \ frac {1} {\ ln x} \ RD x = x \ ln x + C. \ EEx $

(4) $ \ Bex f_x '(x, 1) = \ frac {\ P (x ^ y) }{\ p x }|_{ (x, 1 )} = Yx ^ {Y-1} | _ {(x, 1)} = 1. \ EEx $

(5) from symmetry, original points $ = 0 $.

(6) by the Green formula, $ \ Bex I + \ int _ \ PI ^ 0-\ cos 0 \ RD x = \ iint _ {0 \ Leq x \ Leq \ pi \ atop 0 \ Leq Y \ leq \ SiN x} 0 \ RD x \ rd y \ rA I =-\ pi. \ EEx $

 

2. ($ 20' $) set $ f (x) $ on $ (A, + \ infty) $, and $ f' (x) $ on $ (A, X, + \ infty) $ bounded upper. proof:

(1) $ f (x) $ consistent continuity on $ (A, + \ infty) $;

(2) $ \ DPS {f (a ^ +) = \ lim _ {x \ To a ^ +} f (x)} $ exists, however, $ \ DPS {\ lim _ {x \ to + \ infty} f (x)} $ does not necessarily exist;

(3) If $ \ DPS {\ lim _ {x \ to + \ infty} f (x)} $ exists, and $ \ Bex \ lim _ {x \ To a ^ +} f (x) = \ lim _ {x \ To a ^ +} f (x ), \ EEx $ then $ f' (x) $ has at least one zero point on $ (A, + \ infty) $.

 

Proof:

(1) set $ | f' | \ Leq M $, then $ \ Bex \ forall \ ve> 0, \ exists \ Delta = \ frac {\ ve} {m}> 0, \ st | X-x' | <\ Delta \ rA | f (x) -F (x') | = | f' (\ xi) | \ cdot | X-x' | <\ ve. \ EEx $

(2) from $ \ Bex \ forall \ ve> 0, \ exists \ Delta = \ frac {\ ve} {m}> 0, \ st a <X, x' <A + \ Delta \ rA | f (x)-f (x ') | <\ ve \ EEx $ and the kernel convergence criterion of the Function Limit $ F (a ^ +) $. $ f (x) = \ SiN x $ knows $ F (+ \ infty) $ does not necessarily exist.

(3) Use the reverse verification method. if $ F' $ has no zero point, set $ F'> 0 $. while $ F $ strictly increments, $ \ Bex x <A + 1 <A + 2 <Y \ rA f (x) <f (a + 1) <f (a + 2) <f (y ). \ EEx $ order $ x \ To a ^ + $, $ Y \ to + \ infty $ we found $ \ Bex f (a ^ +) \ Leq f (a + 1) <f (a + 2) \ Leq F (+ \ infty ). \ EEx $ This is in conflict with the question. therefore, there is a conclusion.

 

3. ($ 20' $) set $ f (x) $ continuous on $ [] $, $ F (0) = F (1) $. proof:

(1) $ \ DPS {x_0 \ In \ SEZ {0, \ frac {1} {2 }}$ makes $ \ DPS {f (x_0) = f \ sex {x_0 + \ frac {1} {2 }}$;

(2) Try to speculate: For any positive integer $ N $, whether $ \ DPS {x_0 \ In \ SEZ {0, \ frac {n-1} {n }}$ make $ \ DPS {f (x_0) = F \ sex {x_0 + \ frac {1} {n }}$, and prove your conclusion.

 

Proof:

(1) set $ \ DPS {f (x) = f (x)-f \ sex {x + \ frac {1} {2 }}$, then $ \ Bex F (0) f \ sex {\ frac {1} {2 }=\ SEZ {f (0) -F \ sex {\ frac {1} {2 }}\ cdot \ SEZ {f \ sex {\ frac {1} {2}-F (1 )} =-\ SEZ {f (0)-f \ sex {\ frac {1} {2 }}^ 2 \ Leq 0. \ EEx $ is established by the Intermediate Value Theorem of the continuous function.

(2) Our asserted conclusion is true. in fact, set $ \ DPS {f (x) = f (x)-f \ sex {x + \ frac {1} {2 }}$, then $ \ bee \ label {307_3_sum} f (0) + f \ sex {\ frac {1} {n }}+ \ cdots + f \ sex {\ frac {n-1} {n }}= 0. \ EEE $ use reverse verification to prove the conclusion. if $ \ Bex f (x) \ NEQ 0, \ quad x \ In \ SEZ {0, \ frac {n-1} {n }}, \ EEx $ may be set to $ F> 0 $, which is in conflict with \ eqref {307_3_sum.

 

4. ($10 '$) set $ f (x) $ to be continuous on $ [0, ++ \ infty) $, and $ f (x)> 0 $, note $ \ Bex \ varphi (x) = \ frac {\ int_0 ^ x TF (t) \ RD t} {\ int_0 ^ x F (t) \ RD t }. \ EEx $

(1) Evaluate $ \ DPS {\ lim _ {x \ to 0 ^ ++} \ varphi (t)} $;

(2) proof: $ \ varphi (x) $ increases monotonically on $ (0, ++ \ infty) $.

 

Proof:

(1) $ \ Bex \ lim _ {x \ to 0 ^ ++} \ varphi (X) = \ lim _ {x \ to 0 ^ +} \ frac {XF (x)} {f (x)} = 0. \ EEx $

(2) $ \ beex \ Bea \ varphi '(x) & =\ frac {XF (x) \ int_0 ^ XF (t) \ rd t-f (x) \ int_0 ^ x TF (t) \ RD t} {\ SEZ {\ int_0 ^ x F (t) \ RD t} ^ 2 }\\\&=\ frac {f (x) \ int_0 ^ x (x-t) f (t) \ RD t} {\ SEZ {\ int_0 ^ x F (t) \ RD t} ^ 2} \ &> 0, \ quad x> 0. \ EEA \ eeex $

 

5. ($ 10' $) proof: If $ \ DPS {\ VSM {n} a_n} $ absolutely converges, then $ \ Bex \ VSM {n} a_n (a_1 + A_3 + \ cdots + A _ {2n-1}) \ EEx $ is also absolutely converged.

 

Proof: from $ \ DPS {\ VSM {n} a_n} $ absolute convergence knowledge $ \ DPS {\ VSM {n} | a_n |}$ convergence. according to the comparison principle, $ \ DPS {\ VSM {n} | A _ {2n-1} |}$ convergence, its part and $ \ Bex s_n = | a_1 | + | A_3 | + \ cdots + | A _ {2n-1} | \ EEx $ are bounded, set the ing to $ m> 0 $. then $ \ Bex | a_n (a_1 + A_3 + \ cdots + A _ {2n-1}) | \ Leq m \ cdot | a_n |. \ EEx $ conclusions on the reuse of comparison principles.

 

6. ($ 15' $) set $ f (x) $ to continuous on $ \ DPS {\ SEZ {0, \ frac {\ PI} {2 }}$. proof:

(1) $ \ sed {\ sin ^ NX} $ inconsistent convergence on $ \ DPS {\ SEZ {0, \ frac {\ PI} {2 }}$;

(2) $ \ sed {(\ sin ^ Nx) f (x) }$ in $ \ DPS {\ SEZ {0, \ frac {\ PI} {2 }}$ the condition for consistent convergence is $ \ DPS {f \ sex {\ frac {\ PI} {2 }}= 0} $.

 

Proof:

(1) Use the reverse verification method. if $ \ sed {\ sin ^ NX} $ is uniformly converged on $ \ DPS {\ SEZ {0, \ frac {\ PI} {2 }}$, then the Limit Function $ \ Bex g (x) =\ sedd {\ BA {ll} 0, & 0 \ Leq x <\ frac {\ PI} {2} \ 1, & X = \ frac {\ PI} {2} \ EA} \ EEx $ continuous. this is a conflict.

(2) $ \ rA $: by $ \ sed {(\ sin ^ Nx) f (x) }$ in $ \ DPS {\ SEZ {0, \ frac {\ PI} {2 }}$ consensus Upper Limit Function $ \ Bex h (x) = \ sedd {\ BA {ll} 0, & 0 \ Leq x <\ frac {\ PI} {2} \ f \ sex {\ frac {\ PI} {2 }}, & amp; X = \ frac {\ PI} {2} \ EA} \ EEx $ continuous, $ \ Bex 0 =\lim _ {x \ To \ frac {\ PI} {2} h (x) = f \ sex {\ frac {\ PI} {2 }}. \ EEx $ \ la $: by $ \ DPS {f \ sex {\ frac {\ PI} {2 }=0 }$ $ \ bee \ label {307_6_con} \ forall \ ve> 0, \ exists \ Delta> 0, \ ST \ frac {\ PI} {2}-\ Delta \ Leq x \ Leq \ frac {\ PI} {2} \ rA | (\ sin ^ Nx) f (x) | \ Leq | f (x) | =\ sev {f (x) -F \ sex {\ frac {\ PI} {2 }}< \ ve. \ EEE $ and for $ \ DPS {x \ In \ SEZ {0, \ frac {\ PI} {2}-\ Delta} $, $ \ Bex | (\ sin ^ Nx) f (x) | \ Leq \ sev {\ sin ^ n \ sex {\ frac {\ PI} {2}-\ Delta} \ cdot \ max | f | \ Leq \ sex {\ frac {\ PI} {2}-\ Delta} ^ n \ cdot \ max | f |. \ EEx $ then $ \ bee \ label {307_6_lim} \ exists \ n = N (\ delta) = N (\ ve)> 0, \ st n \ geq n \ Ra \ Max _ {x \ In \ SEZ {0, \ frac {\ PI} {2}-\ Delta} | (\ sin ^ Nx) f (x) | <\ ve. \ EEE $ conclusion: \ eqref {307_6_con} And \ eqref {307_6_lim.

 

7. ($10 '$) set $ f (x, y, z) $ is the $ N $ function on $ \ BBR ^ 3 $ (for any $ T> 0 $, $ F (TX, Ty, Tz) = t ^ NF (x, y, z) $) with a first-order continuous partial derivative, $ f_z (x, y, z) \ NEQ 0 $. if the equation $ f (x, y, z) = 0 $ determines the approximate implicit function $ z = g (x, y) $. proof: $ z = g (x, y) $ must be a homogeneous function.

 

Proof:

(1) Permit: Micro functions $ F (x_1, \ cdots, X_n) $ is $ N $ RMB homogeneous function only when $ \ Bex \ sum X_ I F _ {X_ I} '= NF. \ EEx $ \ rA $: For any fixed $ x_1, \ cdots, X_n $, for $ \ Bex F (tx_1, \ cdots, tx_n) = t ^ NF (x_1, \ cdots, X_n) \ EEx $ get the value of $ T $ at $ T = 1 $ after evaluate. $ \ la $: For any fixed $ x_1, \ cdots, X_n $, note $ \ Bex \ PHI (t) =\ frac {1} {t ^ n} f (tx_1, \ cdots, tx_n), \ EEx $ then $ \ PHI (1) = f (x_1, \ cdots, x_n) $, $ \ Bex \ Phi '(t) = \ frac {\ sum (tx_ I) F _ {X_ I}' (tx_1, \ cdots, tx_n) -n f (tx_1, \ cdots, tx_n)} {t ^ {n + 1 }}= 0. \ EEx $ so $ \ PHI (t) = \ PHI (1) $.

(2) Evidence-based questions: $ \ beex \ Bea xz_x + yz_y & = x \ cdot \ sex {-\ frac {f_x '} {f_y'} + Y \ cdot \ sex {-\ frac {f_y'} {f_z' }}\\& =-\ frac {xf_x '+ yf_y'} {f_z' }\\\& =\ frac {zf_z'} {f_z '} \ quad \ sex {xf_x '+ yf_y' + zf_z' = NF = 0 }\& = z. \ EEA \ eeex $

 

8. ($ 20' $) set $ f (x, y) $ to have a second-order continuous partial derivative on $ \ BBR ^ 2 $. proof:

(1) For $ \ BBR ^ 2 $ any smooth and simple closed curve $ L $, always $ \ Bex \ int_l \ frac {\ p f} {\ P \ VEC {n }}\ rd s = \ iint_d \ sex {\ frac {\ P ^ 2f} {\ P x ^ 2} + \ frac {\ P ^ 2 f} {\ P y ^ 2 }}\ RD x \ RD y, \ EEx $ where $ \ VEC {n} $ is the outer normal direction of $ L $, $ \ DPS {\ frac {\ p f }{\ p \ VEC {n }}$ $ f (x, y) $ Number of Wizard along $ \ VEC {n} $, $ d $ is a bounded closed area enclosed by $ L $;

(2) $ f (x, y) $ is the Harmonic Function on $ \ BBR ^ 2 $ (I .e. $ \ DPS {\ frac {\ P ^ 2f} {\ P x ^ 2} + \ frac {\ P ^) 2 f} {\ P y ^ 2} = 0} $) is a smooth and closed curve $ L $ for $ \ BBR ^ 2 $, always $ \ DPS {\ int_l \ frac {\ p f} {\ P \ VEC {n }}\ rd s = 0 }$.

 

Proof:

(1) $ \ beex \ Bea \ int_l \ frac {\ p f }{\ p \ VEC {n }}\ rd s & =\ int_l \ SEZ {f_x '\ cos (\ VEC {n }, \ VEC {x}) + f_y '\ cos (\ VEC {n}, \ VEC {y })} \ RD s \ & =\ int l f_x '\ RD y + f_y' (-\ rd x) \ & =\ iint_d (F _ {XX} ''+ F _ {YY}'') \ RD x \ RD y. \ EEA \ eeex $

(2) apparently.

 

9. ($15 '$) set $ N $ to a positive integer. The given equation $ x ^ N + x = 1 $.

(1) This equation only has a unique positive root $ x_n \ In () $.

(2) $ \ DPS {\ vlm {n} x_n = 1} $.

 

Proof:

(1) set $ f_n (x) = x ^ N + X-1 $, then $ \ Bex f_n (0) =-1 <0 <1 = f_n (1 ). \ EEx $ A root on $ (0, 1) $ f_n (x) = 0 $. \ Bex f_n' (x) = NX ^ {n-1} + 1> 0, \ quad x> 0, \ EEx $ and $ f_n (X) = 0 $ only this root $ x_n $ in $ (0, \ infty) $.

(2) by $ \ Bex f_n (x)> F _ {n + 1} (x ), \ quad 0 <x <1 \ EEx $ Zhi $ \ Bex 0 = f_n (x_n) = F _ {n + 1} (x_n ). \ EEx $ strictly incrementing by $ F _ {n + 1} $ X _ {n + 1}> x_n $. therefore, the increment of $ \ sed {x_n} $ has an upper bound. the limit $ L \ Leq 1 $ exists. if $ L <1 $, $ \ Bex 1 = x_n + x_n ^ n \ Leq x_n + L ^ n. \ EEx $ order $ n \ To \ infty $1 \ Leq L $. this is a conflict. therefore, $ L = 1 $.

[Journal of mathematics at home University] 307th questions about Mathematics Analysis in 2006 of Huazhong Normal University