[數學]求值?

[數學]求值?

Searchtruth 於 星期三 八月 06, 2003 7:20 pm


設a,b,x,y屬於R  ,  ax+by=3  ,  ax2+by2=7
, ax3+by3=16 , ax4+by4=42
則ax5+by5之值為何?

Searchtruth
訪客
 

scsnake 於 星期三 八月 06, 2003 10:56 pm


用Mathematica硬算...............20

scsnake
訪客
 

Searchtruth 於 星期三 八月 06, 2003 11:07 pm


... ...無言...
我今天放上來ㄉ題目就是今年台中一中校內數學競賽有獎徵答...此題是B卷~共有3卷~

Searchtruth
訪客
 

bugzpodder 於 星期四 八月 07, 2003 10:32 pm


this is recursive relation.
x,y are the roots of:
u^2-(x+y)u+xy=0

therefore
a1=3
a2=7
a3=16
a4=42
and a_n=(x+y)a_(n-1)-xya_(n-2)  
16=7(x+y)-3xy
42=16(x+y)-7xy

solve for x+y and xy, then caculate a5 using formula

bugzpodder

 
文章: 87
註冊時間: 2003-07-30

Searchtruth 於 星期四 八月 07, 2003 10:36 pm


完ㄌ..看不懂...>"<

Searchtruth
訪客
 

bugzpodder 於 星期四 八月 07, 2003 10:44 pm


homogeneous recurrsive relations.  i dont know it in chinese.  convince yourself of identity:

ax^(n+2)+by^(n+2)=(x+y)(ax^(n+1)+by^(n+1))-xy(ax^n+by^n)

bugzpodder

 
文章: 87
註冊時間: 2003-07-30

bugzpodder 於 星期四 八月 07, 2003 11:00 pm


bugzpodder 寫到:this is recursive relation.
x,y are the roots of:
u^2-(x+y)u+xy=0

therefore
a1=3
a2=7
a3=16
a4=42
and a_n=(x+y)a_(n-1)-xya_(n-2)  
16=7(x+y)-3xy
42=16(x+y)-7xy

solve for x+y and xy, then caculate a5 using formula

i made a sign error, i've fixed it
anyways, to solve it:
16*16=16*7(x+y)-48xy
42*7=16*7(x+y)-49xy
42*7-16*16=xy
xy=-38
x+y=-14
a5=ax^5+by^5=(x+y)a4-xya3=(-14)*42-(-38)*16=20

bugzpodder

 
文章: 87
註冊時間: 2003-07-30




代數學