S(n+1)=aS(n)+bS(n-1)，答皆兄與１２３問

令a=x+y,b=-xy

We have   S(n+1)=(x+y)S(n)-xyS(n-1)

S(n+1)-xS(n)=y(S(n)-xS(n-1)=...=y^n(S(1)-xS(0))

S(n+1)-yS(n)=...................=x^n(S(1)-yS(0)

(x-y)S(n)=(x^n-y^n)S(1)+(xy^n-yx^n)S(0)

S(n)=(x^n-y^n)/(x-y)S(1) + b(x^(n-1)-y^(n-1))/(x-y)S(0)

These formula are better. Mr.123's skill is good, but ...,this Q should be two parts.

Last week,The founder of Recursive Function in China Prof.Mo passed away at his 94,I miss him.

所有跟帖： 

• Who is Prof.Mo ? -皆兄弟也- ♂ (0 bytes) () 10/18/2011 postreply 22:06:30

• 謝謝！同問PROF MO的中文名? -wxcfan123- ♂ (0 bytes) () 10/18/2011 postreply 22:08:58

• 答皆兄與１２３問:莫紹揆 -jinjing- ♀ (97 bytes) () 10/19/2011 postreply 07:44:42

• 首次聽說莫紹揆。見過王浩，一個有所成就的數理邏輯學家。 -皆兄弟也- ♂ (0 bytes) () 10/19/2011 postreply 10:08:48

• 莫有曆史問題,沒當成院士.數學學科走下坡路,數邏更甚. -jinjing- ♀ (70 bytes) () 10/19/2011 postreply 17:24:15