Answers to questions about 2012 mathematics analysis postgraduate entrance exams of Huazhong Normal University

From xulu1352

 

I.

(1) proof: by $ {x }_{ 1 }}=\ frac {1} {2 }, {x }_{ 2 }}=\ frac {3} {8}, {x }_{ 3 }}=\ frac {55} {128 }, \ cdots $. I guess $ \{{ x }_{ 2n + 1 }}\}$ monotonically decreases, $ \{{ x }_{ 2n }}\} $ monotonically incrementing

Use the number method to verify first $ \ SQRT {2}-1 \ le {x }_{ 2n + 1 }}\ Le \ frac {1} {2 }, \ frac {3} {8} \ le {x }_{ 2n + 1 }}\ Le \ SQRT {2}-1 $

(1) When $ n = 1 $, $ \ SQRT {2}-1 {x }_{ 1 }}=\ frac {1} {2} $ hengchengli

(2) When $ n = 2k-1 $ is set, $ \ SQRT {2}-1 \ le {x }_{ 2k-1 }}\ Le \ frac {1} {2} $, then

\ [{X }_{ 2 k + 1 }=\ frac {1} {2} (416_{ 2 k} ^ {2 }) =\ frac {1} {2} [1-\ frac {1} {4} {(_ {2k-1} ^ {2 })} ^ {2}] \ in [\ SQRT {2}-1, \ frac {1} {2}) \]

That is, when $ n = 2 k + 1 $

From (1) (2: $ \ SQRT {2}-1 \ le {x }_{ 2n + 1 }}\ Le \ frac {1} {2} $

Likewise, it can be proved that $ \ frac {3} {8} \ le {x }_{ 2n + 1 }}\ Le \ SQRT {2}-1 $

Then, we prove that $ \{{ x }_{ 2n + 1 }\}$ monotonically decreases, and $ \{{{ x }_{ 2n }}\}$ monotonically increases, mathematical induction

(1) When $ n = 1 $, $ {x }_{ 1 }}=\ frac {1} {2} \ frac {55} {128 }={{ x }_{ 3 }}$ hengchengli

(2) When $ n = 2k-1 $ is set, $ {x }_{ 2k-1 }}\ Ge {x }_{ 2 k + 1 }}$, then

$ {X }_{ 2 K + 3 }}- {x }_{ 2 k + 1 }}=\ frac {1} {8} [{() _ {2k-1} ^ {2 })} ^ {2 }}- {(416_{ 2 k + 1} ^ {2 })} ^ {2 }}] =\ frac {1} {8} (X _ {2 k + 1} ^ {2}-X _ {2k-1} ^ {2 }) (2-x _ {2 k + 1} ^ {2}-X _ {2k-1} ^ {2}) \ le 0 $

That is, when $ n = 2 k + 1 $

From (1) (2), we can see that: $ \{{ x }_{ 2n + 1 }}\}$ monotonically decreasing

Likewise, it can be proved that $ \{{ x }_{ 2n }}\}$ increases monotonically.

Therefore, $ \{{ x }_{ 2n + 1 }\}$ monotonically decreases, $ \ SQRT {2}-1 \ le {x }_{ 2n + 1 }}\ Le \ frac {1} {2 }$, which is known by the monotonic bounded principle: $ \{{ x }_{ 2n + 1 }\}$ convergence

Set $ \ underset {n \ to + \ infty} {\ mathop {\ lim }}\, {x }_{ 2n + 1 }}= L \ in [\ SQRT {2}-1, \ frac {1} {2}] $, by $ {x }_{ 2 K + 3 }=\ frac {1} {2} [1-\ frac {1} {4} {(416 _{ 2 k + 1} ^ {2 })} ^ {2}] $, pairs on both sides $ n \ to + \ infty $

Then $ L = \ frac {1} {2} [1-\ frac {1} {4 }{{ (1-{L} ^ {2 }})} ^ {2}] $, obtain $ L = \ SQRT {2}-1 $

That is, $ \ underset {x \ To \ infty} {\ mathop {\ lim }}\, {x }_{ 2n + 1 }}=\ SQRT {2}-1 = A $

Likewise, it can be proved that $ \ underset {x \ To \ infty} {\ mathop {\ lim }}\, {x }_{ 2n }}=\ SQRT {2}-1 = A $

So $ \ underset {n \ to + \ infty} {\ mathop {\ lim }}\, {x }_{ n }}=\ underset {n \ to + \ infty} {\ mathop {\ lim }}\, {x }_{ 2n }}=\ underset {n \ to + \ infty} {\ mathop {\ lim }}\, {x }_{ 2n + 1 }}= a $

(3) proof: set $ {A }_{ n }}={ {x }_{ n }}- A = {x }_{ n }}- (\ SQRT {2 }-1) $

Therefore

$ \ Underset {n \ to + \ infty} {\ mathop {\ lim }}\, \ left | \ frac {A }_{ n + 1 }}{{ A }_{ n }}\ right | =\ underset {n \ to + \ infty }{\ mathop {\ Lim }}\, \ left | \ frac {x }_{ n + 1 }}- A }{{ x }_{ n }}- A }\ right | =\ underset {n \ to + \ infty} {\ mathop {\ lim }}\, \ left | \ frac {1} {2} (416_{ n} ^ {2 }) -A }{{ x }_{ n }}- a} \ right | =\ underset {n \ to + \ infty }{\ mathop {\ Lim }}\, \ frac {1} {2} \ left | \ frac {416_{ n} ^ {2}-(2 \ SQRT {2}-2 )} {x }_{ n }}- (\ SQRT {2}-1 )} \ right | =\ underset {n \ to + \ infty} {\ mathop {\ lim }}\, \ frac {1} {2} \ left | \ frac {X _ {n} ^ {2}-{(\ SQRT {2}-1 )} ^ {2 }}{{ x }_{ n }}- (\ SQRT {2}-1)} \ right | = A1 $

So $ \ sum \ limits _ {n = 1} ^ {+ \ infty} {({x }_{ n }}}-a) $ absolute convergence

II,

(1) The reason is as follows:

To prove that $ {\ sin} ^ {3 }}\ left | f (x) \ right | $ is consistent and continuous on $ I $, you only need to prove that $ \ left | f (x) \ right | $ is consistent and continuous on $ I $.

Because $ f (x) $ is consistent and continuous on $ I $, for any $ \ varepsilon 0 $, any $ {x }_{ 1 }}, {x }_{ 2 }}\ in I $, $ \ Delta 0 $ exists. When

$ \ Left | {x }_{ 1 }}- {x }_{ 2 }}\ right | \ Delta $, $ \ left | f ({x }_{ 1})-f ({x }_{ 2}) \ right | \ varepsilon $

Therefore, for any $ \ varepsilon 0 $, any $ {x }_{ 1 }}, {x }_{ 2 }}\ in I $, $ \ Delta 0 $ exists. When

$ \ Left | {x }_{ 1 }}- {x }_{ 2 }}\ right | \ Delta $, $ \ left | f ({x }_{ 1}) \ right |-\ left | f ({x }_{ 2 }}) \ right | \ Le \ left | f ({x }_{ 1})-f ({x }_{ 2 }}) \ right | \ varepsilon $

That is, $ \ left | f (x) \ right | $ is consistent and continuous on $ I $.

For any $ \ varepsilon 0 $, any $ {x }_{ 1 }}, {x }_{ 2 }}\ in I $, $ \ Delta 0 $ exists. When

$ \ Left | {x }_{ 1 }}- {x }_{ 2 }}\ right | \ Delta $

$ \ Left | {\ sin} ^ {3 }}\ left | f ({x }_{ 1 }}) \ right |-{\ sin} ^ {3 }}\ left | f ({x }_{ 1 }}) \ right | = \ left | (\ sin \ left | f ({x }_{ 1 }}) \ right |-\ sin \ left | f ({x }_{ 1}) \ right |) ({\ sin} ^ {2 }}\ left | f ({x }_{ 1 }}) \ right | + \ sin \ left | f ({x }_{ 1 }}) \ right | \ sin \ left | f ({x }_{ 2 }}) \ right | + {\ sin} ^ {2 }}\ left | f ({x }_{ 2 }}) \ right | $ \ le 3 \ left | \ sin \ left | f ({x }_{ 1 }}) \ right |-\ sin \ left | f ({x }_{ 1 }}) \ right | \ le 3 \ left | \ sin F ({x }_{ 1 }}) -\ sin F ({x }_{ 2}) \ right | \ le 3 \ left | f ({x }_{ 1 }}) -F ({x }_{ 2}) \ right | $

Therefore, we can see from the consistent convergence of the composite functions: $ {\ sin} ^ {3 }}\ left | f (x) \ right | $ consistent continuity on $ I $

(4) The reason is as follows:

To prove that $ {\ sin} ^ {3} f (x) $ is consistent and continuous on $ I $, you only need to prove $ f (x) $ consistent continuity on $ I $

Because $ \ left | f (x) \ right | $ is consistent and continuous on $ I $, any $ {x }_{ 1 }}, {x }_{ 2 }}\ in I $, $ \ Delta 0 $ exists. When

$ \ Left | {x }_{ 1 }}- {x }_{ 2 }}\ right | \ Delta $, by $ \ left | f ({x }_{ 1}) \ right |-\ left | f ({x }_{ 2 }}) \ right | \ frac {\ varepsilon} {2} $

1: If $ F ({x }_{ 1}), F ({x }_{ 2}) $ is the same number, $ \ left | f ({x }_{ 1})-f ({x }_{ 2 }}) \ right | \ frac {\ varepsilon} {2} \ varepsilon $

2: If $ F ({x }_{ 1}) and F ({x }_{ 2}) $ are of the same number) $ continuous, $ y $ between $ {x }_{ 1 }}, {x }_{ 2 }}$ makes $ f (y) = 0 $

So $ \ left | f ({x }_{ 1})-f ({x }_{ 2 }}) \ right | \ Le \ left | f ({x }_{ 1}) \ right | + \ left | f ({x }_{ 2 }}) \ right | = \ left | f ({x }_{ 1}) \ right |-\ left | f (y) \ right | + \ left | f ({x }_{ 2}) \ right |-\ left | f (y) \ right | \ varepsilon $

From 1, 2, we can see that $ f (x) $ is consistent and continuous on $ I $.

Based on the consistent continuity of the composite function, we can see that $ {\ sin} ^ {3} f (x) $ is consistent and continuous on $ I $.

Iii. proof:

Adequacy: counterproof method: Assume that $ f (x) \ Le \ frac {F (B)-f (a)} is available for any $ x \ In (a, B) $ )} {B-a} $

Make $ g (x) = f (x)-\ frac {F (B)-f (a)} {B-a} (X-a) $

When $ x \ In (a, B) $, $ g (x) = f (x)-\ frac {F (B)-f ()} {B-a} \ le 0 $

Therefore, $ g (x) $ decreases monotonically on $ [a, B] $.

While $ G (A) = g (B) = f (a) $

So when $ x \ in [a, B] $, $ g (x) = f (a) $

So $ f (x) = \ frac {F (B)-f (a)} {B-a} (X-A) + f (a) $ conflict

Therefore, $ \ Xi \ In (a, B) $ must exist, so that $ F (\ xi) \ frac {F (B)-f ()} {B-a} $

Necessity:

Make $ g (x) = f (x)-\ frac {F (B)-f (a)} {B-a} (X-a) $

Then $ G (A) = g (B) = f (a) $

Proof Method:

1: If $ f (x) $ is a constant function, $ g (x) = f (x)-\ frac {F (B)-f ()} {B-a }=- \ frac {F (B)-f (a)} {B-a}, X \ in [a, B] $

This corresponds to the presence of $ \ Xi \ In (a, B) $, making $ F (\ xi) \ frac {F (B)-f ()} {B-a} $ conflict

2: If $ f (x) $ is a linear function, we can see that $ g (x) = f (x)-\ frac {F (B)-f ()} {B-a} $ is a constant.

Because $ \ Xi \ In (a, B) $ exists, $ F (\ xi) \ frac {F (B)-f ()} {B-a} $ Yes, $ g (x) 0 $

Therefore, $ g (x) $ increases monotonically at $ [a, B] $.

While $ G (A) = g (B) = f (a) $ conflict

As shown in the preceding figure, $ f (x) $ is not a regular or linear function.

IV,

(1) proof: Set $ f \ left (X, Y \ right) ={{ x} ^ {2 }}+ y-\ cos \ left (xy \ right) $,

Obviously, $ f \ left (0, 1 \ right) = 0 $,

$ {F }_{ y }}\ left (X, Y \ right) = 1 + x \ sin \ left (xy \ right) $,

$ {F }_{ y }}\ left (0, 1 \ right) = 1 \ Ne 0 $. The implicit function has a theorem,

$ \ Delta 0 $ exists, and a continuous and micro function on $ \ left [-\ delta, \ Delta \ right] $ Y = Y \ left (x \ right) $ exists, $ Y \ left (0 \ right) = 1 $,

$ F \ left (X, Y \ left (x \ right) \ equiv 0 $, $ x \ In U (0) $

(3) proof: (1:

$ {F }_{ x }}\ left (X, Y \ right) = 2x + Y \ sin \ left (xy \ right) $,

$ {Y} \ left (x \ right) =-\ frac {f }_{ x }}\ left (X, Y \ right )} {f }_{ y }}\ left (X, Y \ right) }=- \ frac {2x + Y \ sin \ left (xy \ right )} {1 + x \ sin \ left (xy \ right) }$,

When $ 0x \ Delta $, ($ \ Delta 0 $ is small enough), $ {y} \ left (x \ right) 0 $, $ Y \ left (x \ right) $ strictly monotonically decreases on $ \ left [0, \ Delta \ right] $;

$-\ Delta x0 $, $ {y} \ left (x \ right) 0 $, $ Y \ left (x \ right) $ strictly monotonically incrementing on $ \ left [-\ delta, 0 \ right] $,

V,

(1) solution: (1) definition of partial derivative:

$ {F }_{ x }}( 0, 0) = \ underset {\ delta x \ to 0 }{\ mathop {\ lim }}\, \ frac {f (\ delta X, 0)-f (0, 0 )} {\ delta x }=\ underset {\ delta x \ to 0 }{\ mathop {\ lim }}\, \ delta x \ cos \ frac {1 }{\ left | \ delta x \ right |}= 0 $

$ {F }_{ y }}( 0, 0) = \ underset {\ Delta Y \ to 0 }{\ mathop {\ lim }}\, \ frac {f (0, \ Delta y)-f (0, 0 )} {\ Delta y }=\ underset {\ Delta Y \ to 0 }{\ mathop {\ lim }}\, \ Delta Y \ cos \ frac {1 }{\ left | \ Delta Y \ right |}= 0 \ $

(2) When $ (x, y) \ NE (0, 0) $,

$ {F }_{ x} (x, y) = 2x \ cos \ frac {1} {\ SQRT {x} ^ {2 }}+ {y} ^ {2 }}}+ \ frac {x} {\ SQRT {x} ^ {2} + {y} ^ {2 }}}\ sin \ frac {1} {\ SQRT {x} ^ {2 }}+ {y} ^ {2 }}}$

$ {F }_{ y} (x, y) = 2y \ cos \ frac {1} {\ SQRT {x} ^ {2 }}+ {y} ^ {2 }}+ \ frac {y} {\ SQRT {x} ^ {2} + {y} ^ {2 }}}\ sin \ frac {1} {\ SQRT {x} ^ {2 }}+ {y} ^ {2 }}}$

So $ {f_x} (x, y) =\left \{\ begin {array} {ll} 2x \ cos \ frac {1 }{{{ \ SQRT {{ x ^ 2 }+ {y ^ 2 }}}} + \ frac {x} {\ SQRT {x ^ 2} + {y ^ 2 }}}\ sin \ frac {1 }{{ \ SQRT {x ^ 2 }+ {y ^ 2 }}}}, \ hbox {$ {x ^ 2} + {y ^ 2} \ Ne 0 $} \ 0, \ hbox {$ {x ^ 2} + {y ^ 2} = 0 $ .} \ end {array} \ right. $
$ {F_y} (x, y) =\left \{\ begin {array} {ll} 2y \ cos \ frac {1 }{{{ \ SQRT {{ x ^ 2 }+ {y ^ 2 }}}} + \ frac {y} {\ SQRT {x ^ 2} + {y ^ 2 }}}\ sin \ frac {1 }{{ \ SQRT {x ^ 2 }+ {y ^ 2 }}}},, \ hbox {$ {x ^ 2} + {y ^ 2} \ Ne 0 $} \ 0, \ hbox {$ {x ^ 2} + {y ^ 2} = 0 $ .} \ end {array} \ right. $

(2) set $ Y = kx $, so $ \ underset {x \ to 0} {\ mathop {\ lim }}\, \ frac {x} {\ SQRT {x} ^ {2 }}+ {y} ^ {2 }}=\ frac {1} {\ SQRT {1 + {k} ^ {2 }}$ related to $ K $

Therefore, $ \ underset {x \ to 0 }{\ mathop {\ lim }}\, {f }_{ x }}( X, 0) $ does not exist, therefore, $ {f }_{ x} (x, y) $ is not continuous at (0, 0.

Similarly, $ {f }_{ y} (x, y) $ is not continuous at (0, 0.

(3) If $ u = f (x, y) $ is set

$ \ Deltau-du = [F (\ delta X, \ Delta y)-f ()]-[{f }_{ x) \ delta x + {f }_{ y} (0, 0) \ deltay] = (\ Delta {x} ^ {2 }}+ \ Delta {y} ^ {2 }}) \ cos \ frac {1} {\ SQRT {\ Delta {x} ^ {2 }}+ \ Delta {y} ^ {2 }}$

Because $ \ underset {\ delta x \ to 0, \ Delta Y \ to0} {\ mathop {\ lim }}\, \ frac {\ Delta U-du} {\ SQRT {\ Delta {x} ^ {2 }}+ \ Delta {y} ^ {2 }}= \ underset {\ delta x \ to 0, \ Delta Y \ to 0 }{\ mathop {\ Lim }}\, \ SQRT {\ Delta {x} ^ {2 }}+ \ Delta {y} ^ {2 }}\ cos \ frac {1} {\ SQRT {\ Delta {x} ^ {2 }}+ \ Delta {y} ^ {2 }}= 0 $

Therefore, $ f (x, y) $ can be smaller than (0, 0.

6. Solution: Because $ Y = x + y, y (0) = 1 $

Obtain $ y =-+ 2 {e} ^ {x} $ by Constant Variation

Note $ {A }_{ n }}= y (\ frac {1} {n }) -1-\ frac {1} {n} = 2 ({e} ^ {\ frac {1} {n}-1-\ frac {1} {n }) $

By $ \ underset {n \ to + \ infty} {\ mathop {\ lim }}\, \ SQRT [N] {A }_{ n }}={ e} ^ {\ underset {n \ to + \ infty} {\ mathop {\ lim} }\, \ frac {\ ln {A }_{ n }}{ n }}={{ e} ^ {\ underset {n \ To ++ \ infty} {\ mathop {\ lim }}\, \ frac {\ ln2 ({e} ^ {\ frac {1} {n }}- 1-\ frac {1} {n })} {n }}\ overset {n = \ frac {1} {x }}{\ mathop {= }}\, {e} ^ {\ underset {x \ to + {0} ^ {+ }}{\ mathop {\ lim }}\, \ frac {\ ln 2 ({e} ^ {x}-32a )} {\ frac {1} {x }}={ {e} ^ {-\ underset {x \ to + {0} ^ {+ }}{\ mathop {\ lim }}\, \ frac {x} ^ {2} ({e} ^ {x}-1 )} {e} ^ {x}-3o}}={ {e} ^ {-\ underset {x \ to + {0} ^ {+ }}} {\ mathop {\ lim }}\, \ frac {x} ^ {2} [x + O (x)]} {\ frac {x} ^ {2 }}{ 2} + O ({x} ^ {2}) }}= 1 $

So the convergence radius of the Power Series $ r = 1 $

Because $ x = 1 $, $ Y (\ frac {1} {n }) -1-\ frac {1} {n} = 2 ({e} ^ {\ frac {1} {n}-1-\ frac {1} {n }) = \ sum \ limits _ {n = 1} ^ {+ \ infty} {\ frac {2} {n} ^ {2 }}+} \ sum \ limits _ {n = 1} ^ {+ \ infty} {o (\ frac {2 }{{ n} ^ {2 }}}}) $ convergence

When $ x =-1 $, we can see that the number of series converges according to the lebreitz discriminant method.

So the Convergence Domain of This idempotent series is $ [-] $

VII,

(1) proof: make $ U ={{ t} ^ {2 }}\ rightarrowt =\sqrt {u }, dt = \ frac {1} {2 \ SQRT {u} du $, so $ \ int _ {x} ^ {x + c} {\ sin {t} ^ {2 }}dt = \ frac {1} {2} \ int _ {x} ^ {2 }}^{{ (x + C )} ^ {2 }}{\ frac {\ Sinu }{\ SQRT {u }}du $

Functions $ \ frac {1} {\ SQRT {u }}$ in $ [{x} ^ {2}, {(x + C )} ^ {2}] $ decrease on and $ \ frac {1} {\ SQRT {u }}\ GE 0 $, which is known by the value theorem in the second part of the integral:

$ \ Xi \ in [{x} ^ {2}, {(x + C)} ^ {2}] $ exists, making

$ \ Int _ {x} ^ {x + c} {\ sin {t} ^ {2 }}dt = \ frac {1} {2} \ int _{ {x} ^ {2 }}^{{ (x + C )} ^ {2 }}{\ frac {\ Sinu} {\ SQRT {u }}du =\ frac {1} {2x} \ int _ {{ x} ^ {2 }}^ {\ Xi} {\ sin UDU = \ frac {1} {2x} (\ cos {x} ^ {2}-\ cos \ XI )} $

Therefore, $ \ left | \ int _ {x} ^ {x + c} {\ sin {t} ^ {2} DT \ right | \ Le \ frac {1} {x} $

(2) Yes, for the following reasons:

Proof:

Make $ f (x) = \ int _ {x} ^ {x + c} {\ sin {t} ^ {2 }}dt = \ frac {1} {2} \ int _{ {x} ^ {2 }}^{{ (x + C )} ^ {2 }}{\ frac {\ Sinu} {\ SQRT {u }}du =-\ frac {1} {2} \ frac {\ cosu }{\ SQRT {u }}|_{{ x} ^ {2 }}{{ (x + C )} ^ {2 }}- \ frac {1} {4} \ int _ {x} ^ {2 }}{{ (x + C )} ^ {2 }}{\ frac {\ Sinu }{{ u} ^ {\ frac {3} {2 }}} du $

$ =\ Frac {\ cos {x} ^ {2 }}{ 2x}-\ frac {\ cos {(x + C )} ^ {2 }}{ 2 (x + C )} -\ frac {1} {4} \ int _ {x} ^ {2 }}^{{ (x + C )} ^ {2 }}{\ frac {\ Sinu }{{ u} ^ {\ frac {3} {2 }}} du $

$ {U }_{ 0 }}\ in [{x} ^ {2 },{ {(x + C) }^{ 2}] $, make $ \ left | \ frac {\ cos {u }_{ 0 }}{ U _ {0} ^ {\ frac {3} {2 }}\ right | \ frac {1 }{{ U }^{\ frac {3 }{ 2 }}}$, therefore, when $ x0 $

$ \ Left | f (x) \ right | \ Le \ left | \ frac {\ cos {x} ^ {2 }}{ 2x} \ right | + \ left | \ frac {\ cos {{ (x + C )} ^ {2 }}{ 2 (x + C )} \ right | + \ frac {1} {4} \ int _ {x} ^ {2 }}{{ {(x + C )} ^ {2 }}{\ left | \ frac {\ cos U }{{ u} ^ {\ frac {3} {2 }}\ right |} du \ frac {1} {2x} + \ frac {1} {2 (x + C )} + \ frac {1} {4} \ int _ {x} ^ {2 }}^{{ (x + C )} ^ {2 }}{\ frac {1 }{{ u} ^ {\ frac {3} {2 }}} du $

$ =\ Frac {1} {2x} + \ frac {1} {2 (x + C )} + \ frac {1} {4} (-2 {u} ^ {-\ frac {1} {2 }}}) | _ {x} ^ {2 }}^ {(x + C)} ^ {2 }}=\ frac {1} {x} $

8. Proof: For any point on $ {r} ^ {2} $ ({x }_{ 0 }}, {y }_{ 0}) $, make $ {L }_{ 1 }}=\{ x |{{ (X-{x }_{ 0 }})} ^ {2 }={{ R} ^ {2 }}\}$. The direction is counter-clockwise.

According to the Green formula:

$ \ Int _ {L} {PDX + qdy =}\ int _ {L + {L }_{ 1 }}{ PDX + qdy = \ iint _ {d} {[\ frac {\ partialq} {\ partial x }}- \ frac {\ partial p} {\ partial y}] dxdy = [\ frac {\ partial Q }{ \ partialx}-\ frac {\ partial p} {\ partial y}] {{|}_ {M }}\ PI {{ R }^ {2 }}= 0 $

$ D $ is a graph surrounded by $ L + {L }_{ 1 }}$, $ m \ In d $

On the other hand, we can see from the mean point theorem:

$ \ Int _ {L} {PDX + qdy =}-\ int _ {L }_{ 1 }}{ p (x, {y }_{ 0 }}) dx =-P ({x }_{ 1 }},{ y }_{ 0 }}) \ cdot2r $

$ ({X }_{ 1 }},{ y }_{ 0 }}) \ in {L }_{ 1 }}$

Compare the two statements:

$ [\ Frac {\ partialq} {\ partial x}-\ frac {\ partial p} {\ partial y}] {{|}_{ M }}\ PI {{ r} ^ {2 }}=-P ({x }_{ 1 }}, {y }_{ 0 }}) \ cdot 2R \ rightarrow [\ frac {\ partialq }{\ partial x}-\ frac {\ partial P }{\ partial y}] {{|}_{ m }} \ frac {\ PIR} {2} =-P ({x }_{ 1 }}, {y }_{ 0}) $

$ R \ to 0 $: $ P ({x }_{ 0 },{ y }_{ 0 }}) = 0 $, from the arbitrary nature of $ ({x }_{ 0 }}, {y }_{ 0}) $, $ p (x, y) = 0 $

As a result, $ [\ frac {\ partial Q} {\ partialx}-\ frac {\ partial p} {\ partial y}] {{|}_{ m} $, make $ r \ to 0 $ know $ [\ frac {\ partial Q }{\ partial x}-\ frac {\ partiphosphatase }{\ partial y}] {{| }_ {({x }_{ 0 }}, {y }_{ 0 }}) }}= 0 $, from $ ({x }_{ 0 }}, {y }_{ 0 }}) $ arbitrary, $ \ frac {\ partial Q} {\ partial x} = 0 $

 

Answers to questions about 2012 mathematics analysis postgraduate entrance exams of Huazhong Normal University