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Advanced Mathematics (Seventh Edition) Tongji University exercises 1-10 personal solutions

2022-06-27 04:07:00 【Navigator_ Z】

Advanced mathematics （ The seventh edition ） Tongji University exercises 1-10

 

1.   false set up Letter Count f ( x ) stay close District between [ 0 ,   1 ] On even To continue , and And Yes [ 0 ,   1 ] On ren One spot x Yes 0 ≤ f ( x ) ≤ 1 .      try Prove bright [ 0 ,   1 ] in have to save stay One spot c , send have to f ( c ) = c （ c call by Letter Count f ( x ) Of No dynamic spot ） . \begin{aligned}&1. \ Hypothesis function f(x) In the closed zone [0, \ 1] Continuous on , And right [0, \ 1] A little x Yes 0 \le f(x) \le 1.\\\\&\ \ \ \ Try to prove [0, \ 1] There must be a point in c, bring f(c)=c（c Called a function f(x) The fixed point ）.&\end{aligned} ​1.  false set up Letter Count f(x) stay close District between [0, 1] On even To continue , and And Yes [0, 1] On ren One spot x Yes 0≤f(x)≤1.     try Prove bright [0, 1] in have to save stay One spot c, send have to f(c)=c（c call by Letter Count f(x) Of No dynamic spot ）.​​

Explain ：

   set up F ( x ) = f ( x ) − x , be F ( 0 ) = f ( 0 ) − 0 ≥ 0 , F ( 1 ) = f ( 1 ) − 1 ≤ 0    Such as fruit F ( 0 ) = 0 or F ( 1 ) = 0 , be 0 or 1 namely by f ( x ) Of No dynamic spot , Such as fruit F ( 0 ) > 0 And F ( 1 ) < 0 , be from zero spot set The reason is ,    have to Out have to save stay c ∈ ( 0 ,   1 ) , send F ( c ) = f ( c ) − c = 0 , namely f ( c ) = c , c by f ( x ) Of No dynamic spot . \begin{aligned} &\ \ set up F(x)=f(x)-x, be F(0)=f(0)-0 \ge 0,F(1)=f(1)-1 \le 0\\\\ &\ \ If F(0)=0 or F(1)=0, be 0 or 1 That is to say f(x) The fixed point , If F(0) \gt 0 And F(1) \lt 0, By the zero point theorem ,\\\\ &\ \ It must exist c \in (0, \ 1), send F(c)=f(c)-c=0, namely f(c)=c,c by f(x) The fixed point . & \end{aligned} ​   set up F(x)=f(x)−x, be F(0)=f(0)−0≥0,F(1)=f(1)−1≤0   Such as fruit F(0)=0 or F(1)=0, be 0 or 1 namely by f(x) Of No dynamic spot , Such as fruit F(0)>0 And F(1)<0, be from zero spot set The reason is ,   have to Out have to save stay c∈(0, 1), send F(c)=f(c)−c=0, namely f(c)=c,c by f(x) Of No dynamic spot .​​

2.   Prove bright Fang cheng x 5 − 3 x = 1 to Less Yes One individual root Medium On 1 and 2 And between . \begin{aligned}&2. \ Prove the equation x^5-3x=1 At least one root is between 1 and 2 Between .&\end{aligned} ​2.  Prove bright Fang cheng x5−3x=1 to Less Yes One individual root Medium On 1 and 2 And between .​​

Explain ：

   set up f ( x ) = x 5 − 3 x − 1 , be f ( x ) stay close District between [ 1 ,   2 ] On even To continue , And f ( 1 ) = − 3 < 0 , f ( 2 ) = 25 > 0 .    from zero spot set The reason is , have to Out ∃   ξ ∈ ( 1 ,   2 ) , send f ( ξ ) = 0 , ξ yes Medium On 1 and 2 And between Of Fang cheng Of root . \begin{aligned} &\ \ set up f(x)=x^5-3x-1, be f(x) In the closed zone [1, \ 2] Continuous on , And f(1)=-3 \lt 0,f(2)=25 \gt 0.\\\\ &\ \ By the zero point theorem , obtain \exists\ \xi \in (1, \ 2), send f(\xi)=0,\xi Is between 1 and 2 Between the roots of the equation . & \end{aligned} ​   set up f(x)=x5−3x−1, be f(x) stay close District between [1, 2] On even To continue , And f(1)=−3<0,f(2)=25>0.   from zero spot set The reason is , have to Out ∃ ξ∈(1, 2), send f(ξ)=0,ξ yes Medium On 1 and 2 And between Of Fang cheng Of root .​​

3.   Prove bright Fang cheng x = a   s i n   x + b , Its in a > 0 , b > 0 , to Less Yes One individual just root , and And it No super too a + b . \begin{aligned}&3. \ Prove the equation x=a\ sin\ x+b, among a \gt 0,b \gt 0, At least one positive root , And it does not exceed a+b.&\end{aligned} ​3.  Prove bright Fang cheng x=a sin x+b, Its in a>0,b>0, to Less Yes One individual just root , and And it No super too a+b.​​

Explain ：

   set up f ( x ) = x − a   s i n   x − b , be f ( x ) stay close District between [ 0 ,   a + b ] On even To continue , And f ( 0 ) = − b < 0 , f ( a + b ) = a [ 1 − s i n   ( a + b ) ] ,    When s i n   ( a + b ) < 1 when , f ( a + b ) > 0 , from zero spot set The reason is , have to Out ∃   ξ ∈ ( 0 ,   a + b ) , send f ( ξ ) = 0 , namely ξ by Fang cheng Of root ,    yes just root and And No super too a + b , When s i n   ( a + b ) = 1 when , f ( a + b ) = 0 , a + b yes full foot strip Pieces of Of just root . \begin{aligned} &\ \ set up f(x)=x-a\ sin\ x -b, be f(x) In the closed zone [0, \ a+b] Continuous on , And f(0)=-b \lt 0,f(a+b)=a[1-sin\ (a+b)],\\\\ &\ \ When sin\ (a+b) \lt 1 when ,f(a+b) \gt 0, By the zero point theorem , obtain \exists\ \xi \in (0, \ a+b), send f(\xi)=0, namely \xi Is the root of the equation ,\\\\ &\ \ Is a positive root and does not exceed a+b, When sin\ (a+b)=1 when ,f(a+b)=0,a+b Is the positive root that satisfies the condition . & \end{aligned} ​   set up f(x)=x−a sin x−b, be f(x) stay close District between [0, a+b] On even To continue , And f(0)=−b<0,f(a+b)=a[1−sin (a+b)],   When sin (a+b)<1 when ,f(a+b)>0, from zero spot set The reason is , have to Out ∃ ξ∈(0, a+b), send f(ξ)=0, namely ξ by Fang cheng Of root ,   yes just root and And No super too a+b, When sin (a+b)=1 when ,f(a+b)=0,a+b yes full foot strip Pieces of Of just root .​​

4.   Prove bright ren One most high Time power Of finger Count by p. Count Of generation Count Fang cheng a 0 x 2 n + 1 + a 1 x 2 n + ⋅ ⋅ ⋅ + a 2 n x + a 2 n + 1 = 0 to Less Yes One real root ,      Its in a 0 , a 1 , ⋅ ⋅ ⋅ , a 2 n + 1 all by often Count , n ∈ N . \begin{aligned}&4. \ It is proved that the exponent of any highest power is odd a_0x^{2n+1}+a_1x^{2n}+\cdot\cdot\cdot+a_{2n}x+a_{2n+1}=0 At least one real root ,\\\\&\ \ \ \ among a_0,a_1,\cdot\cdot\cdot,a_{2n+1} All are constants ,n \in N.&\end{aligned} ​4.  Prove bright ren One most high Time power Of finger Count by p. Count Of generation Count Fang cheng a0​x2n+1+a1​x2n+⋅⋅⋅+a2n​x+a2n+1​=0 to Less Yes One real root ,     Its in a0​,a1​,⋅⋅⋅,a2n+1​ all by often Count ,n∈N.​​

Explain ：

   When x Of most Yes value charge branch Big when , f ( x ) = a 0 x 2 n + 1 + a 1 x 2 n + ⋅ ⋅ ⋅ + a 2 n x + a 2 n + 1 Of operator Number take " On a 0 Of operator Number , When x by just when And a 0 Same as Number ,    When x by negative when And a 0 different Number , a 0 ≠ 0 . because by f ( x ) yes even To continue Of , it stay charge branch Big Of District between two End different Number , from zero spot set The reason is have to know it stay District between Inside    to Less Yes One spot It's about by zero , the With f ( x ) = 0 to Less Yes One individual real root . \begin{aligned} &\ \ When x When the absolute value of is sufficiently large ,f(x)=a_0x^{2n+1}+a_1x^{2n}+\cdot\cdot\cdot+a_{2n}x+a_{2n+1} The symbol of depends on a_0 The symbol of , When x Is timing and a_0 Same number ,\\\\ &\ \ When x When it is negative and a_0 Different sign ,a_0 \neq 0. because f(x) Is a continuous , It has different signs at both ends of a sufficiently large interval , It is known from the zero point theorem that it is in the interval \\\\ &\ \ Zero at least at one point , therefore f(x)=0 At least one real root . & \end{aligned} ​   When x Of most Yes value charge branch Big when ,f(x)=a0​x2n+1+a1​x2n+⋅⋅⋅+a2n​x+a2n+1​ Of operator Number take " On a0​ Of operator Number , When x by just when And a0​ Same as Number ,   When x by negative when And a0​ different Number ,a0​​=0. because by f(x) yes even To continue Of , it stay charge branch Big Of District between two End different Number , from zero spot set The reason is have to know it stay District between Inside    to Less Yes One spot It's about by zero , the With f(x)=0 to Less Yes One individual real root .​​

5.   if f ( x ) stay [ a ,   b ] On even To continue , a < x 1 < x 2 < ⋅ ⋅ ⋅ < x n < b （ n ≥ 3 ） , be stay ( x 1 ,   x n ) Inside to Less Yes One spot ξ ,      send f ( ξ ) = f ( x 1 ) + f ( x 2 ) + ⋅ ⋅ ⋅ + f ( x n ) n . \begin{aligned}&5. \ if f(x) stay [a, \ b] Continuous on ,a \lt x_1 \lt x_2 \lt \cdot\cdot\cdot \lt x_n \lt b（n \ge 3）, It's in (x_1, \ x_n) There's at least a little \xi,\\\\&\ \ \ \ send f(\xi)=\frac{f(x_1)+f(x_2)+\cdot\cdot\cdot+f(x_n)}{n}.&\end{aligned} ​5.  if f(x) stay [a, b] On even To continue ,a<x1​<x2​<⋅⋅⋅<xn​<b（n≥3）, be stay (x1​, xn​) Inside to Less Yes One spot ξ,     send f(ξ)=nf(x1​)+f(x2​)+⋅⋅⋅+f(xn​)​.​​

Explain ：

   because by f ( x ) stay District between [ a ,   b ] On even To continue , also because by [ x 1 ,   x n ] ⊂ [ a ,   b ] , the With f ( x ) stay [ x 1 ,   x n ] On even To continue .    set up M = m a x { f ( x )   ∣   x 1 ≤ x ≤ x n } , m = m i n { f ( x )   ∣   x 1 ≤ x ≤ x n } , be m ≤ f ( x 1 ) + f ( x 2 ) + ⋅ ⋅ ⋅ + f ( x n ) n ≤ M .    Such as fruit On Statement No etc. type by No etc. Number , from Medium value set The reason is have to know , ∃   ξ ∈ ( x 1 ,   x n ) , send f ( ξ ) = f ( x 1 ) + f ( x 2 ) + ⋅ ⋅ ⋅ + f ( x n ) n ,    Such as fruit On Statement No etc. type by etc. Number , m = f ( x 1 ) + f ( x 2 ) + ⋅ ⋅ ⋅ + f ( x n ) n , be Yes f ( x 1 ) = f ( x 2 ) = ⋅ ⋅ ⋅ = f ( x n ) = m ,    ren take x 2 , ⋅ ⋅ ⋅ , x n − 1 in One spot do by ξ , Yes ξ ∈ ( x 1 ,   x n ) , send f ( ξ ) = f ( x 1 ) + f ( x 2 ) + ⋅ ⋅ ⋅ + f ( x n ) n    Same as The reason is can Prove , f ( x 1 ) + f ( x 2 ) + ⋅ ⋅ ⋅ + f ( x n ) n = M . \begin{aligned} &\ \ because f(x) In the interval [a, \ b] Continuous on , Again because [x_1, \ x_n] \subset [a, \ b], therefore f(x) stay [x_1, \ x_n] Continuous on .\\\\ &\ \ set up M=max\{f(x)\ | \ x_1 \le x \le x_n\},m=min\{f(x)\ | \ x_1 \le x \le x_n\}, be m \le \frac{f(x_1)+f(x_2)+\cdot\cdot\cdot+f(x_n)}{n} \le M.\\\\ &\ \ If the above inequality is not equal , It is known from the intermediate value theorem that ,\exists\ \xi \in (x_1, \ x_n), send f(\xi)=\frac{f(x_1)+f(x_2)+\cdot\cdot\cdot+f(x_n)}{n},\\\\ &\ \ If the above inequality is equal ,m=\frac{f(x_1)+f(x_2)+\cdot\cdot\cdot+f(x_n)}{n}, Then there are f(x_1)=f(x_2)=\cdot\cdot\cdot=f(x_n)=m,\\\\ &\ \ Take whatever you like x_2,\cdot\cdot\cdot,x_{n-1} A little bit of it as \xi, Yes \xi \in (x_1, \ x_n), send f(\xi)=\frac{f(x_1)+f(x_2)+\cdot\cdot\cdot+f(x_n)}{n}\\\\ &\ \ The same is true ,\frac{f(x_1)+f(x_2)+\cdot\cdot\cdot+f(x_n)}{n}=M. & \end{aligned} ​   because by f(x) stay District between [a, b] On even To continue , also because by [x1​, xn​]⊂[a, b], the With f(x) stay [x1​, xn​] On even To continue .   set up M=max{ f(x) ∣ x1​≤x≤xn​},m=min{ f(x) ∣ x1​≤x≤xn​}, be m≤nf(x1​)+f(x2​)+⋅⋅⋅+f(xn​)​≤M.   Such as fruit On Statement No etc. type by No etc. Number , from Medium value set The reason is have to know ,∃ ξ∈(x1​, xn​), send f(ξ)=nf(x1​)+f(x2​)+⋅⋅⋅+f(xn​)​,   Such as fruit On Statement No etc. type by etc. Number ,m=nf(x1​)+f(x2​)+⋅⋅⋅+f(xn​)​, be Yes f(x1​)=f(x2​)=⋅⋅⋅=f(xn​)=m,   ren take x2​,⋅⋅⋅,xn−1​ in One spot do by ξ, Yes ξ∈(x1​, xn​), send f(ξ)=nf(x1​)+f(x2​)+⋅⋅⋅+f(xn​)​   Same as The reason is can Prove ,nf(x1​)+f(x2​)+⋅⋅⋅+f(xn​)​=M.​​

6.   set up Letter Count f ( x ) Yes On close District between [ a ,   b ] On Of ren It means two spot x 、 y , constant Yes ∣ f ( x ) − f ( y ) ∣ ≤ L ∣ x − y ∣ ,      Its in L by just often Count , And f ( a ) ⋅ f ( b ) < 0 . Prove bright ： to Less Yes One spot ξ ∈ ( a ,   b ) , send have to f ( ξ ) = 0 . \begin{aligned}&6. \ Let's set the function f(x) For a closed interval [a, \ b] Any two points on x、y, There is always |f(x)-f(y)| \le L|x-y|,\\\\&\ \ \ \ among L It's a normal number , And f(a)\cdot f(b) \lt 0. prove ： At least a little \xi \in (a, \ b), bring f(\xi)=0.&\end{aligned} ​6.  set up Letter Count f(x) Yes On close District between [a, b] On Of ren It means two spot x、y, constant Yes ∣f(x)−f(y)∣≤L∣x−y∣,     Its in L by just often Count , And f(a)⋅f(b)<0. Prove bright ： to Less Yes One spot ξ∈(a, b), send have to f(ξ)=0.​​

Explain ：

   ren take x 0 ∈ ( a ,   b ) , ∀   ε > 0 , take δ = m i n { ε L , x 0 − a , b − x 0 } , When ∣ x − x 0 ∣ < δ when ,    from false set up ∣ f ( x ) − f ( x 0 ) ∣ ≤ L ∣ x − x 0 ∣ < L δ ≤ ε , the With f ( x ) stay x 0 It's about even To continue . because by yes ren take x 0 ∈ ( a ,   b ) ,    the With f ( x ) stay ( a ,   b ) Inside even To continue . \begin{aligned} &\ \ Take whatever you like x_0 \in (a, \ b),\forall\ \varepsilon \gt 0, take \delta=min\left\{\frac{\varepsilon}{L},x_0-a,b-x_0\right\}, When |x-x_0| \lt \delta when ,\\\\ &\ \ By hypothesis |f(x)-f(x_0)| \le L|x-x_0| \lt L\delta \le \varepsilon, therefore f(x) stay x_0 Continuous . Because it is optional x_0 \in (a, \ b),\\\\ &\ \ therefore f(x) stay (a, \ b) Internal continuity . & \end{aligned} ​   ren take x0​∈(a, b),∀ ε>0, take δ=min{ Lε​,x0​−a,b−x0​}, When ∣x−x0​∣<δ when ,   from false set up ∣f(x)−f(x0​)∣≤L∣x−x0​∣<Lδ≤ε, the With f(x) stay x0​ It's about even To continue . because by yes ren take x0​∈(a, b),   the With f(x) stay (a, b) Inside even To continue .​​

7.   Prove bright ： if f ( x ) stay ( − ∞ ,   + ∞ ) Inside even To continue , And lim ⁡ x → ∞ f ( x ) save stay , be f ( x ) have to stay ( − ∞ ,   + ∞ ) Inside Yes world . \begin{aligned}&7. \ prove ： if f(x) stay (-\infty, \ +\infty) Internal continuity , And \lim_{x \rightarrow \infty}f(x) There is , be f(x) Must be in (-\infty, \ +\infty) There is a boundary inside .&\end{aligned} ​7.  Prove bright ： if f(x) stay (−∞, +∞) Inside even To continue , And x→∞lim​f(x) save stay , be f(x) have to stay (−∞, +∞) Inside Yes world .​​

Explain ：

   set up lim ⁡ x → ∞ f ( x ) = A , ∀   ε > 0 , take ε = 1 , ∃   X > 0 , When ∣ x ∣ > X when ,    Yes ∣ f ( x ) − A ∣ < 1 ⇒ ∣ f ( x ) ∣ ≤ ∣ f ( x ) − A ∣ + ∣ A ∣ < ∣ A ∣ + 1 .    because by f ( x ) stay [ − X ,   X ] On even To continue , from Yes world sex set The reason is have to Out , ∃   M > 0 , ∀   x ∈ [ − X ,   X ] , Yes ∣ f ( x ) ∣ ≤ M .    take M ′ = m a x { M ,   ∣ A ∣ + 1 } , namely Yes ∣ f ( x ) ∣ ≤ M ′ , ∀   x ∈ ( − ∞ ,   + ∞ ) . \begin{aligned} &\ \ set up \lim_{x \rightarrow \infty}f(x)=A,\forall\ \varepsilon \gt 0, take \varepsilon=1,\exists\ X \gt 0, When |x| \gt X when ,\\\\ &\ \ Yes |f(x)-A| \lt 1 \Rightarrow |f(x)| \le |f(x)-A|+|A| \lt |A|+1.\\\\ &\ \ because f(x) stay [-X, \ X] Continuous on , From the boundedness Theorem ,\exists\ M \gt 0,\forall\ x \in [-X, \ X], Yes |f(x)| \le M.\\\\ &\ \ take M'=max\{M, \ |A|+1\}, That is to say |f(x)| \le M',\forall\ x \in (-\infty, \ +\infty). & \end{aligned} ​   set up x→∞lim​f(x)=A,∀ ε>0, take ε=1,∃ X>0, When ∣x∣>X when ,   Yes ∣f(x)−A∣<1⇒∣f(x)∣≤∣f(x)−A∣+∣A∣<∣A∣+1.   because by f(x) stay [−X, X] On even To continue , from Yes world sex set The reason is have to Out ,∃ M>0,∀ x∈[−X, X], Yes ∣f(x)∣≤M.   take M′=max{ M, ∣A∣+1}, namely Yes ∣f(x)∣≤M′,∀ x∈(−∞, +∞).​​

8.   stay What Well strip Pieces of Next , ( a ,   b ) Inside Of even To continue Letter Count f ( x ) by One Cause even To continue ？ \begin{aligned}&8. \ Under what conditions ,(a, \ b) Continuous functions in f(x) Is uniformly continuous ？&\end{aligned} ​8.  stay What Well strip Pieces of Next ,(a, b) Inside Of even To continue Letter Count f(x) by One Cause even To continue ？​​

Explain ：

   Such as fruit f ( a + ) , f ( b − ) all save stay , set up F ( x ) = { f ( a + ) , x = a , f ( x ) , x ∈ ( a ,   b ) , f ( b − ) , x = b . , Prove have to F ( x ) stay District between [ a ,   b ] On even To continue ,    from and F ( x ) stay District between [ a ,   b ] On One Cause even To continue , also Just Yes F ( x ) stay District between ( a ,   b ) Inside One Cause even To continue , f ( x ) stay District between ( a ,   b ) Inside One Cause even To continue . \begin{aligned} &\ \ If f(a^+),f(b^-) All exist , set up F(x)=\begin{cases}f(a^+),x=a,\\\\f(x),x \in (a, \ b),\\\\f(b^-),x=b.\end{cases}, Prove F(x) In the interval [a, \ b] Continuous on ,\\\\ &\ \ thus F(x) In the interval [a, \ b] Uniformly continuous on , There are F(x) In the interval (a, \ b) Internal consistency and continuity ,f(x) In the interval (a, \ b) Internal consistency and continuity . & \end{aligned} ​   Such as fruit f(a+),f(b−) all save stay , set up F(x)=⎩⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎧​f(a+),x=a,f(x),x∈(a, b),f(b−),x=b.​, Prove have to F(x) stay District between [a, b] On even To continue ,   from and F(x) stay District between [a, b] On One Cause even To continue , also Just Yes F(x) stay District between (a, b) Inside One Cause even To continue ,f(x) stay District between (a, b) Inside One Cause even To continue .​​

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