(低羅沒路嘟 Cá tháng Tư)

命嗔得界紹典各伴沒排曰衛總分，差分。帝享應役使用"chữ tượng hình" 帝發展"tư duy" 立程，命嗔法得曰排呢朋"chữ Nôm" 台位字國語。

於立11，12 各伴諸得學衛各概念道咸吧辟分。各概念呢固忍性質窒葉，執眾些丹澗化役性算假治懶一，𡮈一，面辟，世辟，。。。。

相自，台概念差分吧總分共固忍性質葉相自道咸，辟分吧侞世執些性總沒隔丹澗欣。

1。差分

些諸別：

~f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}~

由欺性總，~h~ 羅數元年欺 ~h \rightarrow 0~ 時 ~h~ 止固世羅 ~1~ 或 ~-1~. 帝丹澗化，些𩲡 ~h=1~ 帝固定義差

~\Delta(f(x)) = f(x+1) - f(x)~

~\Delta(f(x))~ 讀羅差分𧵑 ~f~ 在點 ~x~，或論欣羅差分𧵑 ~f(x)~

些固性質差𧵑道咸: ~(x^m)' = mx^{m-1}~

些仕此性差分𧵑 ~x^3~ 覘差分料固世固性質相自能空。

~\Delta(x^3) = (x+1)^3 - x^3 = 3x^2 + 3x + 1~

結果呢空葉吧易汝如 ~(x^3)' = 3x^2~, 位丕欺爫役某差分，些勤類累承特別噲羅累承減﹝能群噲羅皆承減﹞

~x^{\underline{m}} = x(x-1)(x-2)...(x-m+1)~

些𧡊低羅辟𧵑 ~m~ 承數減寅 ~x~, ~x-1~, ..., ~x-m+1~. 既號 ~x^{\underline{m}}~ 讀羅 ~x~ 帽 ~m~ 減

些此性差分𧵑 ~x^{\underline{m}}~

~\Delta(x^{\underline{m}}) = (x+1)^{\underline{m}} - x^{\underline{m}} = (x+1)x(x-1)(x-2)...(x-m+2) - x(x-1)(x-2)...(x-m+1) = x(x-1)(x-2)...(x-m+2)[(x+1) - (x-m+1)]~

~\Delta(x^{\underline{m}}) = mx(x-1)(x-2)...(x-m+2) = mx^{\underline{m-1}}~

丕些固公識葉吧易汝 ~\boxed{\Delta(x^{\underline{m}}) = mx^{\underline{m-1}}}~

於𨕭眾些某定義累承減某數帽揚。些仕定義累承減朱數帽空揚如差帝性質棟哈於𨕭中。

~x^{\underline{0}} = 1~

~x^{\underline{-m}} = \frac{1}{(x+1)(x+2)(x+3)...(x+m)}~ (~m > 0~)

差欺定義得累承減朱每數元，些徐固得公識人台累承減

~\boxed{x^{\underline{n+m}} = x^{\underline{n}}(x-n)^{\underline{m}}}~

同時，些共固公識差﹝種公識開展二識Newton﹞

~(x+y)^{\underline{n}} = \sum_{k=0}^n C_n^k x^{\underline{k}}y^{\underline{n-k}}~

沒性質葉女𧵑道咸羅 ~(e^x)' = e^x~. 料固沒咸 ~f(x)~ 鬧課滿 ~\Delta f(x) = f(x)~ 能空？

些固： ~\Delta f(x) = f(x) \Rightarrow f(x+1) - f(x) = f(x) \Rightarrow f(x+1) = 2f(x)~

易𧡊 ~f(x) = 2^x~ 羅沒咸課滿 ~f(x+1) = 2f(x)~, 吧位世 ~\boxed{\Delta(2^x) = 2^x}~

役性差分咸帽共可丹澗

~\Delta(c^x) = c^{x+1} - c^x \Rightarrow \boxed{\Delta(c^x) = (c-1)c^x}~

2。總分

些諸別：

~\int_a^b f(x) dx = \lim_{n \rightarrow \infty} \sum_{i=1}^{n-1} (x_{i+1} - x_i)f(x_i)~

某 ~x_i = a + \frac{i}{n+1}(b-a)~ (tức ~x_0, x_1, ..., x_n, x_{n+1}~ 羅各點中斷 ~[a,b]~ 隔調饒)

﹝帝易想象吧汝公識呢，流意 ~\sum_{i=1}^{n-1} (x_{i+1} - x_i)f(x_i)~ 羅總面辟各刑字日用帝約量面辟噽𠁑途是咸數 ~f(x)~﹞

些禮敢興慈概念辟分帝定義總分如差：

~\sum_a^b f(x) \delta x = f(a) + f(a+1) + ... f(b-1) = \sum_{k=a}^{b-1} f(k)~

流意總止豸慈 ~a~ 典 ~b-1~ 帝總分固各性質種辟分差：

• ~\sum_a^b f(x) \delta x + \sum_b^c f(x) \delta x = \sum_a^c f(x) \delta x~ (種 ~\int_a^b f(x)dx + \int_b^c f(x)dx = \int_a^c f(x)dx~)
• 裊 ~g(x) = \Delta(f(x))~ 時
  • ~\sum_a^b g(x) \delta x = f(x)]_a^b = f(b) - f(a)~ (種裊 ~g(x) = f'(x)~ 時 ~\int_a^b g(x) dx = f(x)]_a^b~)
  • 些既號 ~\sum g(x) \delta x = f(x) + C~ (種元咸)

瘴明性質裊 ~g(x) = \Delta(f(x))~ 時 ~\sum_a^b g(x) \delta x = f(x)]_a^b = f(b) - f(a)~

些固：

~\sum_a^b g(x) \delta x = \sum_{k=a}^{b-1} g(x) = \Delta(f(a)) + \Delta(f(a+1)) + \Delta(f(a+2)) + ... + \Delta(f(b-1))~

~\sum_a^b g(x) \delta x = (f(a+1) - f(a)) + (f(a+2) - f(a+1)) + (f(a+3) - f(a+2)) + ... + (f(b) - f(b-1))~

~\sum_a^b g(x) \delta x = f(b) - f(a)~

慈各公識性差分於𨕭，些易陽固得各公識性總分差：

• ~\sum x^{\underline{m}} \delta x = \frac{x^{\underline{m+1}}}{m+1} + C~ (中欺 ~m \ne -1~)
• ~\sum c^x \delta x = \frac{c^x}{c-1} + C~ (中欺 ~c \ne 1~). 欺 ~c = 1~ 時 ~\sum c^x \delta x = \sum 1 \delta x = x + C~

悲徐些勤所公識朱 ~\sum x^{\underline{m}} \delta x~ 欺 ~m = -1~. 於門解辟，些別

~\int x^{-1} dx = \ln |x| + C~

相自，些共固

~\sum x^{\underline{-1}} = H_x + C~ 某 ~H_x = \sum_{1 \leq k \leq x} \frac{1}{k}~

調呢朱𧡊 ~H_x~ 固堯性質種某咸 ~\ln~. 倘些共諸瘴明得 ~|H_n - \ln n| < 1~ 欺 ~n~ 都懶.

3. 應用丹澗𧵑總分

察排算性 ~\square_n = \sum_{0 \leq k \leq n} k^2~

話𥆾時固𡲈如排算𨕭空連觀之典差分，總分位各性質於𨕭止停朱累承減𠹲空拜累承平常。雖然，些固世轉 ~k^2~ 朋總各累承減朋公識 ~k^2 = k^{\underline{2}} + k^{\underline{1}}~

些固：

~\sum_{0 \leq k \leq n} k^2 = \sum_0^{n+1} k^2 \delta k = \sum_0^{n-1} (k^{\underline{2}} + k^{\underline{1}}) \delta k~

~\boxed{\sum_{0 \leq k \leq n} k^2 = [\frac{k^{\underline{3}}}{3} + \frac{k^{\underline{2}}}{2}]_0^{n+1}}~

慈低些止役開展羅囉沒公識朱 ~\square_n~

相自，些固世用公識 ~k^3 = k^{\underline{3}} + 3k^{\underline{2}} + k^{\underline{1}}~ 帝性 ~C_n = \sum_{0 \leq k \leq n} k^3~ 如差：

~C_n = \sum_0^{n+1} (k^{\underline{3}} + 3k^{\underline{2}} + k^{\underline{1}}) \delta k~

~C_n = [\frac{k^{\underline{4}}}{4} + k^{\underline{3}} + \frac{k^{\underline{2}}}{2}]_0^{n+1}~

爫世鬧帝些尋得各係數中各公識 ~k^2 = k^{\underline{2}} + k^{\underline{1}}~ 吧 ~k^3 = k^{\underline{3}} + 3k^{\underline{3}} + k^{\underline{1}}~. 俱者利羅些使用數Sterling 類I ﹝蓋呢共種役些使用數祖合能三角Pascal 帝尋係數𧵑各數巷中開展 ~(a+b)^n~ 丕﹞。

4. 差分𧵑沒辟，總分曾焚

察差分𧵑沒辟台咸 ~u~ 吧 ~v~. 些仕尋隔性 ~\Delta(uv)~ 慈 ~u, v, \Delta(u), \Delta(v)~

~\Delta(uv) = u(x+1)v(x+1) - u(x)v(x) = u(x+1)v(x+1) + u(x+1)v(x) - u(x+1)v(x) - u(x)v(x)~

﹝役共除 ~u(x+1)v(x)~ 棟𦠘路爫求綏𡧲台數巷﹞

~\Delta(uv) = u(x+1)[v(x+1) - v(x)] + v(x)[u(x+1) - u(x)]~

~\Delta(uv) = u(x+1)\Delta(v) + v(x)\Delta(u)~

帝公識呢葉欣，些定義 ~Eu(x) = u(x+1)~ (~E~ 羅曰悉𧵑慈"Extra" ﹝添﹞中㗂英﹞，些固世曰論公識呢又如差：

~\boxed{\Delta(uv) = Eu\Delta(v) + v\Delta(u)}~

中索，公識呢羅 ~\boxed{\Delta(uv) = u\Delta(v) + Ev\Delta(u)}~. 墨油台公識呢空對稱仍些固世瘴明眾朋饒。

於門解辟，公識性道咸𧵑沒辟執些固得公識性辟分曾焚。相自，些共固公識性總分曾焚如差：

~\sum u\Delta(v) = uv - \sum Ev\Delta(u)~

種如公識性辟分曾焚，些常使用得公識性總分曾焚裊中表識勤性，固沒焚性差分易吧沒焚性總分易。

圍裕 1: 性 ~\sum_{0 \leq k \leq n} k2^k~

頭先，些性 ~\sum x2^x \delta x~

達 ~u(x) = x \Rightarrow \Delta(u) = 1~, ~\Delta(v) = 2^x \Rightarrow v = 2^x \Rightarrow Ev = 2^{x+1}~

欺帝： ~\sum x2^x \delta x = x2^x - \sum 2^{x+1} \delta x = x2^x + 2^{x+1} + C~

位世麻 ~\sum_{0 \leq k \leq n} k2^k = \sum_0^{n+1} x2^x \delta x = [x2^x + 2^{x+1}]_0^{n+1}~

圍裕 2: 性 ~\sum_0^n H_x \delta x~

些𧡊 ~H_x~ 羅沒焚易性差分，位世年些達

~u(x) = H_x \Rightarrow \Delta(u) = x^{\underline{-1}}~

~\Delta(v) = 1 \Rightarrow v = x^{\underline{1}} = x\Rightarrow Ev = (x+1)^{\underline{1}}~

些固： ~\sum H_x \delta x = xH_x - \sum (x+1)^{\underline{1}}x^{\underline{-1}} \delta x~

~\sum H_x \delta x = xH_x - \sum 1 \delta x~

~\sum H_x \delta x = xH_x + x + C~

慈帝易陽性得 ~\boxed{\sum_0^n H_x \delta x = nH_n + n}~

圍裕 3: 性 ~\sum_0^n xH_x \delta x~

達：

~u(x) = H_x \Rightarrow \Delta(u) = x^{\underline{-1}}~

~\Delta(v) = x \Rightarrow v = \frac{x^{\underline{2}}}{2} \Rightarrow Ev = \frac{(x+1)^{\underline{2}}}{2}~

些固：

~\sum xH_x \delta x = \frac{x^{\underline{2}}}{2}H_x - \sum \frac{(x+1)^{\underline{2}}x^{\underline{-1}}}{2} \delta x~

~\sum xH_x \delta x = \frac{x^{\underline{2}}}{2}H_x - \frac{1}{2} \sum x \delta x~

~\sum xH_x \delta x = \frac{x^{\underline{2}}}{2}H_x - \frac{x^{\underline{2}}}{4} + C~

慈帝易陽性得 ~\boxed{\sum_0^n xH_x \delta x = \frac{n^{\underline{2}}}{2}H_n - \frac{n^{\underline{2}}}{4}}~

5. 總合見識

~\Delta(x^{\underline{m}}) = mx^{\underline{m-1}} \Rightarrow \sum x^{\underline{m}} \delta x = \frac{x^{\underline{m+1}}}{m+1} + C~

(特別： ~\Delta(H_x) = x^{\underline{-1}} \Rightarrow \sum x^{\underline{-1}} \delta x = H_x + C~)

~\Delta(c^x) = (c-1)c^x \Rightarrow \sum c^x \delta x = \frac{c^x}{c-1} + C~

(特別：~\Delta(2^x) = 2^x \Rightarrow \sum 2^x \delta x = 2^x + C~)

~\Delta(cf) = c\Delta(f) \Rightarrow \sum cf = c \sum f~ (~c~ 羅恆數)

~\Delta(f + g) = \Delta(f) + \Delta(g) \Rightarrow \sum (f + g) = \sum f + \sum g~

~\Delta(fg) = f\Delta(g) + Eg\Delta(f) \Rightarrow \sum f\Delta(g) = fg - \sum Eg\Delta(f)~