來源: 經濟數學 Alpha C. Chian
科目: 經濟數學 Ch12 Optmization with equality Constrains
問題:
題目中說的12.22式 f(v)≧f(u) => { Σfj(u)(vj-uj) }
{ Σfj(v)(vj-uj) } ≧0
{ }
The function z=f(X1,X2)= X1X2 (X1,X2 ≧0） is quasiconcave .
(半凹的嗎?)
We shall now check it. Let u= (u1,u2) and v=(v1,v2)
be any two points in the domain. Then f(u)=u1u2 and f(v)=V1V2
assume that: f(v)≧f(u) or v1v2≧u1u2 (v1 v2 u1 u2≧0) (12.27)
since the partial derivitives of are f1=x2 and f2=x1 12.22式 amounts
to the condition that (請問這邊f的偏導數根下面推出的方程式有何關係?)
f1(u)(v1-u1)+f2(u)(v2-u2) = u2(v1-u1)-u1(v2-u2) ≧0
(為何透過導數可以判定需要都以 u去測度?
以及f1(u)和f2(u)為何等於 u2和u1 ?? 下面12.28的式子怎麼來的)
or upon rearangement ,
u2(v1-u1)≧u1(u2-v2) (12.28)
we need to consider four possiblities regarding the value of u1 and u2
First, if u1=u2=0 then 12.28 is trivially satisfied. Second ,
if u1=0 but u2>0 then 12.28 reduce to the condition u2v1≧0. which is again
(違反的意思嗎?? 那為何可推出使u2 v1不為負成立)
satisfied since u2 and v1 are both nonnegtive.
Third , if u1>0 and u2=0 then (12.28) reduce to the condition 0≧-u1v2,
(也同樣是違反0≧-u1v2嗎 那又為何等是成立?)
which is still satisfied.
Last, suppose that u1 and u2 are both possitive
so that u1 and u2 are both possitive , so that v1 and v2 are also positive.
(這邊比較沒問題)
Subtracting v2u1 from both sides of (12.27), we obtain
v2(v1-u1)≧u1(u2-v2) (12.29)
(這邊還看的懂 只是不知道同時減掉 v2u1的用意??)
Three subpossibilities now present themselves.
(不懂這個單字?? 推測??)
1.If u2=v2 then v1≧u1 .In fact , we should have v1>u1 since (v1,v2) are
distinct points. (不懂?? u2=v2不是會使等號右邊變成0? 那怎麼又推得
v1≧u1 以及distinct point 的 v1>u1? 不是也應該是 v1=u1??
The fact that u2=v2 and v1>u1 implies that condition(12.28 is satisfied)
2.If u2>v2 , then we must also have v1>u1 by(12.29) .Multiplying both sides
of (12.29) by u2/v2 , we get
u2(v1-u1)≧ u2/v2[u1(u2-v2)]>u1(u2-v2) [since u2/v2 >1 ] (12.30)
Thus 12.28 is again satisfied (這邊我看的懂 我比較沒意見)
3. The final subpossiblities is that u2<v2, implying that u2/v2 is a positive
fraction. In this case . the first line of (12.30) still holds.
The second line also holds , but now for a different reason:
(以上我還能夠理解 下面兩句就看不懂 )
a fraction (u2/v2) of a negative number (u2-v2)
is greater than the latter number itself.
請問 整個題目的中心主旨 就是在映證 原本
透過12.27的假設而來的 12.28恆等式嗎 u2(v1-u1)≧u1(u2-v2)
題目太複雜了 又有用到一些特別的英文 煩請高手 感激不盡
--
Anything, without blind insistence, is right.
--
科目: 經濟數學 Ch12 Optmization with equality Constrains
問題:
題目中說的12.22式 f(v)≧f(u) => { Σfj(u)(vj-uj) }
{ Σfj(v)(vj-uj) } ≧0
{ }
The function z=f(X1,X2)= X1X2 (X1,X2 ≧0） is quasiconcave .
(半凹的嗎?)
We shall now check it. Let u= (u1,u2) and v=(v1,v2)
be any two points in the domain. Then f(u)=u1u2 and f(v)=V1V2
assume that: f(v)≧f(u) or v1v2≧u1u2 (v1 v2 u1 u2≧0) (12.27)
since the partial derivitives of are f1=x2 and f2=x1 12.22式 amounts
to the condition that (請問這邊f的偏導數根下面推出的方程式有何關係?)
f1(u)(v1-u1)+f2(u)(v2-u2) = u2(v1-u1)-u1(v2-u2) ≧0
(為何透過導數可以判定需要都以 u去測度?
以及f1(u)和f2(u)為何等於 u2和u1 ?? 下面12.28的式子怎麼來的)
or upon rearangement ,
u2(v1-u1)≧u1(u2-v2) (12.28)
we need to consider four possiblities regarding the value of u1 and u2
First, if u1=u2=0 then 12.28 is trivially satisfied. Second ,
if u1=0 but u2>0 then 12.28 reduce to the condition u2v1≧0. which is again
(違反的意思嗎?? 那為何可推出使u2 v1不為負成立)
satisfied since u2 and v1 are both nonnegtive.
Third , if u1>0 and u2=0 then (12.28) reduce to the condition 0≧-u1v2,
(也同樣是違反0≧-u1v2嗎 那又為何等是成立?)
which is still satisfied.
Last, suppose that u1 and u2 are both possitive
so that u1 and u2 are both possitive , so that v1 and v2 are also positive.
(這邊比較沒問題)
Subtracting v2u1 from both sides of (12.27), we obtain
v2(v1-u1)≧u1(u2-v2) (12.29)
(這邊還看的懂 只是不知道同時減掉 v2u1的用意??)
Three subpossibilities now present themselves.
(不懂這個單字?? 推測??)
1.If u2=v2 then v1≧u1 .In fact , we should have v1>u1 since (v1,v2) are
distinct points. (不懂?? u2=v2不是會使等號右邊變成0? 那怎麼又推得
v1≧u1 以及distinct point 的 v1>u1? 不是也應該是 v1=u1??
The fact that u2=v2 and v1>u1 implies that condition(12.28 is satisfied)
2.If u2>v2 , then we must also have v1>u1 by(12.29) .Multiplying both sides
of (12.29) by u2/v2 , we get
u2(v1-u1)≧ u2/v2[u1(u2-v2)]>u1(u2-v2) [since u2/v2 >1 ] (12.30)
Thus 12.28 is again satisfied (這邊我看的懂 我比較沒意見)
3. The final subpossiblities is that u2<v2, implying that u2/v2 is a positive
fraction. In this case . the first line of (12.30) still holds.
The second line also holds , but now for a different reason:
(以上我還能夠理解 下面兩句就看不懂 )
a fraction (u2/v2) of a negative number (u2-v2)
is greater than the latter number itself.
請問 整個題目的中心主旨 就是在映證 原本
透過12.27的假設而來的 12.28恆等式嗎 u2(v1-u1)≧u1(u2-v2)
題目太複雜了 又有用到一些特別的英文 煩請高手 感激不盡
--
Anything, without blind insistence, is right.
--
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