※ 引述《a606155123 (冷氣團團長)》之銘言:
: 2. 條件:△^1f(x)=f(x+1)-f(x) and △^mf(x)=△(△^m-1f(x)) m>2
: Show that△^n X^n=N!
暴力吧...
△ x^n = nx^(n-1) + 比 n-1 次小的項 = nx^(n-1) + O(x^(n-2))
觀察做了 △ 運算之後, 多項式次數會嚴格變少,
然後到 △ x = (x+1)-x = 1, 常數項 △ 1 = 0
所以用歸納法吧: △ x = 1,
假設 △^k x^k = k! 成立, 且 △^k x^m = 0 for m<k
則 △^(k+1) x^(k+1) = △^k ( △ x^(k+1) )
= △^k [ (k+1)x^k + O(x^(k-1)) ]
(k-1<k, by 遞迴假設) = (k+1) △^k x^k = (k+1)!
同時驗證當 m<k+1
△^(k+1) x^m = △^k (△x^m) = △^k (mx^(m-1) + O(x^(m-1)))
(同樣由遞迴假設 m<k+1 => m-1<k) = 0
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