[Journal of mathematics at home University] 303rd questions about Mathematics Analysis in 2004 of Huazhong Normal University

 

 

1. ($50 '= 10' + 10' + 15' + 15' + 15' $) Calculate the following limits:

(1) $ \ DPS {\ lim _ {x \ to 0} (\ Cos x) ^ \ frac {1} {\ sin ^ 2x }}$.

(2) $ \ DPS {\ vlm {n} \ SQRT [N] {1 + \ frac {1} {2} + \ frac {1} {3} + \ cdots + \ frac {1} {n }}$.

(3) $ \ DPS {\ lim _ {x \ to + \ infty} x ^ \ frac {7} {4} \ sex {\ SQRT [4] {x + 1} + \ SQRT [4] {X-1}-2 \ SQRT [4] {x }}$.

(4) $ \ DPS {\ vlm {n} \ sin \ frac {\ PI} {2n} \ sum _ {k = 1} ^ n \ sin \ frac {k \ PI }{ N }}$.

 

Answer:

(1) $ \ beex \ Bea \ mbox {Original limit} & = \ lim _ {x \ to 0} \ SEZ {1 + (\ cos x-1 )} ^ {\ frac {1} {\ cos X-1} \ cdot \ frac {\ cos X-1} {\ sin ^ 2x }}\\ & = e ^ {\ lim _ {x \ to 0} \ frac {-\ SiN x} {2 \ SiN x \ Cos x }}\\& = e ^ {-\ frac {1} {2 }}. \ EEA \ eeex $

(2) by $ \ Bex 1 <\ SQRT [N] {1 + \ frac {1} {2} + \ cdots + \ frac {1} {n }}< \ SQRT [n] {n} \ EEx $ and $ \ Bex \ SQRT [N] {n }=\ exp \ sex {\ frac {1} {n} \ ln n} \ to 1 \ quad \ sex {n \ To \ infty} \ EEx $ knows the original limit $ = 1 $.

(3) $ \ beex \ Bea \ mbox {Original limit} & =\ LiM _ {x \ to + \ infty} x ^ 2 \ sex {\ SQRT [4] {1 + \ frac {1} {x }}+ \ SQRT [4] {1-\ frac {1} {x}-2 }\\\\=\ LiM _ {T \ 0} \ frac {\ SQRT [4] {1 + t} + \ SQRT [4] {1-T}-2} {t ^ 2} \ & = \ lim _{ T \ to 0} \ frac {\ SEZ {1 + \ frac {1} {4} X-\ frac {3} {32} x ^ 2 + O (x ^ 2 )} + \ SEZ {1-\ frac {1} {4} X-\ frac {3} {32} x ^ 2}-2} {t ^ 2} \ & = -\ frac {3} {16 }. \ EEA \ eeex $

(4) $ \ beex \ Bea \ mbox {Original limit} & =\ vlm {n} \ frac {\ sin \ frac {\ PI} {2n }{\ frac {\ pi} {2n }}\ cdot \ frac {\ PI} {2} \ cdot \ sum _ {k = 1} ^ n \ frac {1} {n} \ sin \ frac {k \ PI} {n} \ & =\ frac {\ PI} {2} \ int_0 ^ 1 \ sin \ pi x \ RD x \ & = 1. \ EEA \ eeex $

 

2. ($15 '$) set $ f (x) $, $ g (x) $ continuous on $ [a, B] $, on $ (A, B) $ inner can be imported. If $ X_1 and X_2 $ are two zeros of $ f (x) $ in the range $ [a, B] $, it is proved: $ \ Xi \ In (a, B) $ exists, so that $ \ Bex f' (\ xi) g (\ xi) + f (\ XI) g' (\ xi) = 0. \ EEx $ (original title: $ f' (\ xi) + f (\ xi) G' (\ xi) = 0 $)

 

Proof: Take $ f (x) = f (x) g (x) $, then $ F (X_1) = f (X_2) = 0 $. by the Rolle Theorem, $ \ Xi \ In (x_1, x_2) \ subset (a, B) $ causes $ \ Bex 0 = f' (\ XI) = f' (\ xi) g (\ xi) + f (\ xi) G' (\ XI ). \ EEx $

 

3. ($15 '$) set $ f (x) $ continuous on $ [a, B] \ (B> A> 0) $, on $ (A, B) $ can be imported, proving that $ \ Xi, \ ETA $ exists in $ (a, B) $ to make $ \ Bex f' (\ XI) = \ frac {\ ETA ^ 2f '(\ ETA)} {AB }. \ EEx $

 

Proof: $ \ beex \ Bea \ frac {f' (\ ETA )} {-\ frac {1} {\ ETA ^ 2 }}&=\ frac {F (B)-f ()} {\ frac {1} {B}-\ frac {1} {A }}\\\&=\ frac {F (B)-f ()} {B-a} (-AB) \\& = f' (\ xi) (-AB ). \ EEA \ eeex $

 

4. ($ 15' $) set $ f (x) $ Riann product on $ [a, B] $, proof: $ e ^ {f (x )} $ Riann can accumulate on $ [a, B] $.

 

Proof: it is bounded by $ F $ Riann product knowledge $ F $, set to $ M $, and $ \ beex \ Bea | E ^ {f (x )} -E ^ {f (y)} | & = e ^ \ Xi | f (x)-f (y) | \ quad \ sex {\ Xi \ mbox {In} f (x), f (y) \ mbox {{}\\\& \ Leq e ^ m | f (x)-f (y) |, \ EEA \ eeex $ the determination theorem of the amplitude property of the Riann product is known.

 

5. ($15 '$) set $ f_n' (x) \ (n =, \ cdots) $ continuous on $ [a, B] $, $ g (x) $ is also continuous on $ [a, B] $, and any $ X_1 in $ [a, B] $, x_2 $ and positive integer $ N $ \ bee \ label {303_5_eq} | f_n (X_1)-f_n (X_2) | \ Leq \ frac {m} {n} | x_1-x_2 | \ quad \ sex {M> 0 }. \ EEE $ proof: $ \ Bex \ vlm {n} \ int_a ^ B g (x) f_n' (x) \ RD x = 0. \ EEx $

 

Proof: \ eqref {303_5_eq} knows $ \ Bex | f_n' (x) | \ Leq \ frac {m} {n }, \ EEx $ and $ \ Bex \ sev {\ int_a ^ B g (x) f_n' (X) \ RD x} \ Leq \ frac {m} {n} \ int_a ^ B | G (x) | \ RD x \ to 0 \ quad \ sex {n \ To \ infty }. \ EEx $

 

6. ($ 15' $) set $ f_n (x) \ (n = 1, 2, \ cdots) $ to be continuous on $ [a, B] $, and $ \ sed {f_n (x) }$ uniformly converges on $ [a, B] $ to $ f (x) $. proof:

(1) $ m> 0 $ exists, so that $ N $ has $ | f_n (x) | \ Leq M $, $ | f (x) | \ Leq M $;

(2) If $ f (x) $ is a continuous function on $ (-\ infty, + \ infty) $, then $ F (f_n (x )) $ uniformly converge to $ F (f (x) $.

 

Proof:

(1) The Limit Function of the Continuous Function Column with consistent convergence knows $ F $ continuity, while the boundary $ M_1> 0 $. and by $ \ Bex \ Max _ {x \ in [a, B]} | f_n (x)-f (x) | \ to 0 \ quad \ sex {n \ To \ infty} \ EEx $ \ Bex \ exists \ n, \ st n \ geq n \ Ra \ Max _ {x \ in [a, B]} | f_n (x)-f (x) | \ Leq 1 \ Ra \ Max _ {x \ in [a, B]} | f_n (x) | \ Leq M_1 + 1. \ EEx $ so, $ \ sed {f_n (x) }$, $ f (x) $ bounded $ \ Bex \ Max _ {x \ in [, b]} | F_1 (x) | + \ cdots + \ Max _ {x \ in [a, B]} | f _ {N-1} (X) | + M_1 + 1 \ equiv M. \ EEx $

(2) The $ F $ continuous knowledge $ F $ is consistent and continuous on $ [-M, m] $, while $ \ Bex \ forall \ ve> 0, \ exists \ Delta> 0, \ st | y_1 | \ Leq M, \ | Y_2 | \ Leq m, \ | y_1-y_2 | <\ Delta \ rA | f (y_1)-f (Y_2) | <\ ve. \ EEx $ and the $ f_n $ uniformly converges to $ F $ Zhi $ \ beex \ Bea \ exists \ n, \ st n \ geq N, \ x \ in [a, B] & \ rA | f_n (x)-f (x) | <\ Delta \ & \ rA | f (f_n (x )) -F (f (x) | <\ ve. \ EEA \ eeex $

 

8. ($ 15' $) set the function $ f (x, y) $ in a neighboring area of $ (x_0, y_0) $ there is a continuous second-order partial derivative, and $ \ Bex F (x_0, y_0) = 0, \ quad f_x '(x_0, y_0) = 0, \ quad f_y' (x_0, y_0)> 0, \ quad F _ {XX} ''(x_0, y_0) <0. \ EEx $ proof: the implicit function identified by the equation $ f (x, y) = 0 $ Y = f (x) $ gets the minimum value at $ x_0 $.

 

Proof: by $ \ Bex f (x, y) = 0, \ quad y = f (x) \ EEx $ $ \ Bex f_x '+ f_y 'F' = 0 \ Ra \ mbox {at} x_0 \ mbox ,} f' =-\ frac {f_x '} {f_y'} = 0; \ EEx $ seek again, with $ \ Bex F _ {XX} ''+ F _ {XY} ''f' + F _ {Yx} ''f' + F _ {YY} ''(F ') ^ 2 + f_y 'f'' = 0 \ Ra \ mbox {at} x_0 \ mbox ,} f'' =-\ frac {f_x ''} {f_y '}> 0. \ EEx $ so $ Y = f (x) $ get the minimum value at $ x_0 $.

 

1. ($ 15' $) set $ A_1, A_2, \ cdots, a_n $ to $ N $ different numbers on the number field $ \ BBP $, solving Linear Equations $ \ Bex \ BA {rrrrrrrrl} X_1 & + & X_2 & + & \ cdots & + & x_n & = & 1 \ a_1x_1 & + & a_2x_2 & + & \ cdots & + & a_nx_n & = & a_n \ A_1 ^ 2x_1 & + & A_2 ^ 2x_2 & + & \ cdots & + & a_n ^ 2x_n & = & a_n ^ 2 \\ \ cdots & \ A_1 ^ {n-1} X_1 & + & A_2 ^ {n-1} x_2 & + & \ cdots & + & a_n ^ {n-1} x_n & = & a_n ^ {n-1} \ EA \ EEx $

 

A: Because of the coefficient matrix's determinant $ \ NEQ 0 $, the original linear equations have only one solution. obviously, $ (0, \ cdots,) ^ t $ is a solution, which is the only solution of the original linear equations.

 

2. ($ 15' $) set $ \ BBP $ to a number field, $ A \ In \ BBP ^ {n \ times N }$, $ M (X) = x ^ 3 + 2x + 1 $ is the smallest polynomial of $ A $. Evaluate $ A ^ {-1} $.

 

Proof: $ \ Bex a ^ 3 + 2a + E = 0 \ rA A (-a ^ 2-2e) = E \ rA a ^ {-1} =-a ^ 2-2e. \ EEx $

 

3. ($ 30' $) set $ \ BBP $ to a number field, $ A = (A _ {IJ}) = (\ al_1, \ cdots, \ al_n) \ In \ BBP ^ {n \ times N }$, $ A _ {NN} $ algebraic remainder formula $ A _ {NN} \ NEQ 0 $.

(1) proof: $ \ al_1, \ cdots, \ Al _ {n-1} $ linear independence;

(2) When $ | A | = 0 $, evaluate the basic solution of the linear equations $ A ^ * x = 0 $, $ A ^ * $ is the adjoint matrix of $ A $.

 

Proof:

(1) the column vector of the matrix specified by $ A _ {NN} \ NEQ 0 $ Zhi $ A _ {NN} $ \ Bex \ sex {\ BA {L} A _ {11 }\\ vdots \ A _ {n-1, 1} \ EA}, \ quad \ cdots, \ quad \ sex {\ BA {L} A _ {1, n-1} \ vdots \ A _ {n-1, n-1} \ EA} \ EEx $ linear independence, while the $ \ al_1, \ cdots, \ Al _ {n-1} $ after adding a component is also linear independence.

(2) by $ | A | = 0 $ Zhi $ \ rank (a) \ Leq n-1 $. if $ \ rank (A) = n-2 $, $ A ^ * = 0 $, which conflicts with $ A _ {NN} \ NEQ 0 $. therefore, $ \ rank (A) = n-1 $, $ \ rank (a ^ *) = 1 $. $ \ Bex a ^ * A = | A | E = 0 \ rA a ^ * (\ al_1, \ cdots, \ Al _ {n-1}) = 0, \ EEx $ we know that $ \ al_1, \ cdots, \ Al _ {n-1} $ is the basic solution of $ A ^ * x = 0 $.

 

4. ($ 30' $) set $ \ BBP $ to a number field, $ \ Bex v =\sed {A \ In \ BBP ^ {n \ times n }; A ^ t = A}, \ quad V_2 = \ sed {B \ In \ BBP ^ {n \ times n}; B \ mbox {upper Triangle Matrix }}. \ EEx $

(1) prove that $ v_1 and V_2 $ are all linear subspaces of $ \ BBP ^ {n \ times n} $;

(2) prove $ \ Bex \ BBP ^ {n \ times n} = v_1 + V_2, \ quad \ BBP ^ {n \ times n} \ NEQ v_1 \ oplus V_2. \ EEx $

 

Proof:

(1) apparently true.

(2) For any $ C \ In \ BBP ^ {n \ times N }$, $ \ beex \ Bea C & =\ sex {\ BA {CCCC} C _ {11} & C _ {12} & \ cdots & C _ {1N} \ c _ {21} & C _ {22} & \ cdots & C _ {2n} \ vdots & \ vdots \ C _ {N1 }& C _ {N2} & \ cdots & C _ {NN} \ EA} \ & = \ sex {\ BA {CCCC} C _ {11} & C _ {21} & \ cdots & C _ {N1} \ C _ {21} & C _ {22} & \ cdots & C _ {N2} \ vdots &\ vdots & \ vdots \ C _ {N1} & C _ {N2} & \ cdots & C _ {NN} \ EA} + \ sex {\ BA {CCCC} 0 & C _ {12}-C _ {21} & \ cdots & C _ {1N}-C _ {N1} \ 0 & 0 & \ cdots & C _ {2n} -C _ {N2 }\\\ vdots & \ vdots \ 0 & 0 & \ cdots & 0 \ EA }\\& \ In v_1 + V_2. \ EEA \ eeex $ but $ \ Bex 0 \ NEQ \ sex {\ BA {CCC} 1 & \\\& \ ddots \\& & 1 \ EA }\ in v_1 \ cap V_2, \ EEx $ and $ \ BBP ^ {n \ times n} \ NEQ v_1 \ oplus V_2. $

 

5. ($30 '$) set $ p (x) $ to an irrevocable polynomial on the number field $ \ BBP $, $ \ Al \ NEQ 0 $ to $ p (x) $.

(1) prove that the constant term $ p (x) $ is not equal to zero;

(2) prove to any positive integer $ M $, $ (p (x), x ^ m) = 1 $;

(3) set $ p (x) = x ^ 3-2x + 2 $, evaluate $ \ DPS {\ frac {1 }{\ Al ^ 5 }}$.

 

Proof:

(1) Use the reverse verification method. if the constant term $ p (x) $ is $0 $, then $ \ Bex p (x) = c_nx ^ N + \ cdots + c_1x = XQ (x ), \ EEx $ you $ \ Bex P (\ Al) = \ Al Q (\ Al) \ rA Q (\ Al) = 0 \ Ra \ deg Q (x) \ geq 1. \ EEx $ therefore, $ p (x) $ is decomposed into the product of polynomials where the number of times is less than $ \ deg p (x) $. this is in conflict with $ p (x) $. therefore, there is a conclusion.

(2) set $ (p (x), x ^ m) = d (x) $, then $ d (x) \ mid p (x) $. by $ p (x) $ unknown $ \ Bex d (x) = 1 \ mbox {or} d (x) = \ frac {1} {C ^ n} p (x ). \ EEx $ if $ \ DPS {d (x) =\frac {1} {C ^ n} p (x) }$, then $ \ beex \ Bea d (x) \ mid x ^ m & \ rA p (x) \ mid x ^ m \ & \ rA x ^ m = R (x) p (x) \ & \ rA 0 \ NEQ \ Al ^ m = Q (\ Al) P (\ Al) = 0. \ EEA \ eeex $ this is a conflict. therefore, $ d (x) = 1 $.

(3) by $ \ Bex x ^ 5 = (x ^ 2 + 2) p (x) + R_1 (x), \ quad R_1 (X) =-2x ^ 2 + 4x-4, \ EEx $ \ Bex p (x) = \ sex {-\ frac {x} {2}-1} R_1 (X) -2 \ EEx $ Zhi $ \ beex \ Bea 2 = \ sex {-\ frac {x} {2}-1} R_1 (x)-P (x) \\&=\ sex {-\ frac {x} {2}-1} \ SEZ {x ^ 5-(x ^ 2 + 2) p (x )} -P (x ). \ EEA \ eeex $ order $ x = \ Al $ \ Bex 2 = \ sex {-\ frac {\ Al} {2}-1} \ Al ^ 5 \ Ra \ frac {1} {\ Al ^ 5} =-\ frac {\ Al + 2} {4 }. \ EEx $

 

6. ($ 20' $) set $ N $ real quadratic form $ F (x_1, \ cdots, X_n) = x ^ tax $ after Orthogonal Linear replacement $ x = QY $ ($ q $ is an orthogonal array) convert to $ \ Bex Y_1 ^ 2 + 2y_2 ^ 2 + \ cdots + ny_n ^ 2. \ EEx $ proof:

(1) The feature value of $ A $ is $1, 2, \ cdots, N $;

(2) orthogonal arrays $ B $ make $ A = B ^ 2 $.

 

Proof:

(1) by $ \ Bex f (x) = x ^ tax = y ^ TQ ^ taqy = \ sum_ I iy_ I ^ 2 \ EEx $ Q \ Bex Q ^ Taq = \ diag (1, 2, \ cdots, n ). \ EEx $ the similarity matrix has the same feature value.

(2) get $ \ Bex B = Q \ diag (1, \ SQRT {2}, \ cdots, \ SQRT {n }) Q ^ t \ EEx $ has a conclusion.

 

7. ($ 20' $) set $ \ SCRA $ to linear transformation of the number field $ \ BBP $ top $ N $ Dimension Linear Space $ V $, $ \ Al \ In V $, $ \ SCRA ^ {n-1} (\ Al) \ NEQ 0 $, $ \ SCRA ^ N (\ Al) = 0 $. proof:

(1) $ \ Al, \ SCRA (\ Al), \ cdots, \ SCRA ^ {n-1} (\ Al) $ is a group of bases of $ V $;

(2) set $ W $ to $ \ SCRA $-invariant subspaces, $ A_1, A_2, \ cdots, a_n \ In \ BBP, \ A_1 \ NEQ 0 $, and the vector $ \ Bex \ Beta = A_1 \ Al + A_2 \ SCRA (\ Al) + \ cdots + a_n \ SCRA ^ {n-1} (\ Al) \ in W, \ EEx $ then $ W = V $.

 

Proof:

(1) set $ \ Bex K_0 \ Al + k_1 \ SCRA (\ Al) + \ cdots + K _ {n-1} \ SCRA ^ {n-1} (\ Al) = 0, \ EEx $ with $ \ SCRA $ Action $ N-2 $ Times found $ K_0 \ SCRA ^ {n-1} (\ Al) = 0 $. since $ \ SCRA ^ {n-1} (\ Al) = 0 $, we have $ K_0 = 0 $. in the preceding formula, $ \ Bex K_1 \ SCRA (\ Al) + \ cdots + K _ {n-1} \ SCRA ^ {n-1} (\ Al) = 0. \ EEx $ use $ \ SCRA $ Action $ N-2 $ same discovery $ k_1 = 0 $. and so on. we get all $ K_ I = 0 $.

(2) from $ W $ is $ \ SCRA $-constant sub-space knowledge $ \ beex \ Bea \ SCRA (\ beta) & = A_1 \ SCRA (\ Al) + A_2 \ SCRA ^ 2 (\ Al) + \ cdots + A _ {n-1} \ SCRA ^ {n-1} (\ Al) \ in W, \\\ SCRA ^ 2 (\ beta) & = A_1 \ SCRA ^ 2 (\ Al) + A_2 \ SCRA ^ 3 (\ Al) + \ cdots + A _ {N-2} \ SCRA ^ {n-1} (\ Al) \ in W, \ cdots & = \ cdots, \ SCRA ^ {N-2} (\ beta) & = A_1 \ SCRA ^ {N-2} (\ Al) + A_2 \ SCRA ^ {n-1} (\ Al) \ in W, \\\ SCRA ^ {n-1} (\ beta) & = A_1 \ SCRA ^ {n-1} (\ Al) \ in W. \ EEA \ eeex $ by $ A_1 \ NEQ 0 $ and the last formula $ \ SCRA ^ {n-1} (\ Al) \ in W $. into the penultimate formula: $ \ SCRA ^ {N-2} (\ Al) \ in W $. and so on. we get $ \ Bex \ Al, \ SCRA (\ Al), \ cdots, \ SCRA ^ {n-1} (\ Al) \ in W. \ EEx $ so $ W = V $.

[Journal of mathematics at home University] 303rd questions about Mathematics Analysis in 2004 of Huazhong Normal University