查看“︁代數/本書課文/求和/差分變換”︁的源代码

[[代數/本書課文/求和/組合數求和|組合數求和]]中由<math>p(k)=\sum_{j=0}^{m} C^{k-a}_{j}\Delta^j p(a)</math>得到多項式的求和結果。<ref>{{cite book|author=Károly Jordán|year=1950|title=Calculus of Finite Differences|url=http://crypto.cs.mcgill.ca/~simonpie/webdav/ipad/EBook/Math%C3%A9matique/Calculus/Math%20-%20Calculus%20of%20Finite%20Differences.pdf}}</ref>

{{Robox|theme=3|title=证明：<math>p(k)=\sum_{j=0}^{m} C^{k-a}_{j}\Delta^j p(a)</math>}}
設<math>p(k)=\sum_{j=0}^{m} a_j C^{k-a}_{j}=a_0+a_1 C^{k-a}_{1}+a_2 C^{k-a}_{2}+\dots+a_m C^{k-a}_{m}</math>

<math>p(a)=a_0</math>

<math>\Delta C^k_l=C^{k+1}_l-C^k_l=C^k_{l-1}</math>

<math>\Delta^j p(k)=a_j+a_{j+1} C^{k-a}_{1}+a_{j+2} C^{k-a}_{2}+\dots+a_m C^{k-a}_{m-j}</math>

<math>\Delta^j p(a)=a_j</math>

<math>p(k)=\sum_{j=0}^{m} C^{k-a}_{j}\Delta^j p(a)</math>

{{Robox/Close}}

除此之外，還可以利用變換<math>\sum_{k=0}^\infty u_k v_k x^k=\sum_{k=0}^\infty \frac{\Delta^k u_0 x^k}{k!}\frac{d^k}{dx^k}(\sum_{l=0}^\infty v_l x^l)</math>求和，可設<math>u_k=p(k)</math>為多項式使和式的項數有限。（<math>\Delta^{m+1} u_k=0</math>）<ref>{{cite book|author=Murray Spiegel|title=Schaum's Outline of Calculus of Finite Differences and Difference Equations|year=1971|url=https://lpuguidecom.files.wordpress.com/2016/10/difference-equation-e-book.pdf}}</ref>

{{Robox|theme=3|title=证明：<math>\sum_{k=0}^\infty u_k v_k x^k=\sum_{k=0}^\infty \frac{\Delta^k u_0 x^k}{k!}\frac{d^k}{dx^k}(\sum_{l=0}^\infty v_l x^l)</math>}}

<math>\frac{d^k}{dx^k}(\sum_{l=0}^\infty v_l x^l)=\sum_{l=k}^\infty l(l-1)\dots (l-k+1)v_l x^{l-k}=\sum_{l=k}^\infty \frac{l!}{(l-k)!}v_l x^{l-k}</math>

<math>\sum_{k=0}^\infty \frac{\Delta^k u_0 x^k}{k!}\frac{d^k}{dx^k}(\sum_{l=0}^\infty v_l x^l)=\sum_{k=0}^\infty \sum_{l=k}^\infty C^l_k \Delta^k u_0 v_l x^l=\sum_{l=0}^\infty \sum_{k=0}^l C^l_k \Delta^k u_0 v_l x^l=\sum_{l=0}^\infty v_l x^l (\sum_{k=0}^l C^l_k \Delta^k u_0) =\sum_{l=0}^\infty u_l v_l x^l=\sum_{k=0}^\infty u_k v_k x^k</math>

{{Robox/Close}}

{{ExampleRobox|title=例子：<math>\sum_{k=0}^\infty u_k x^k=\sum_{k=0}^\infty \frac{\Delta^k u_0 x^k}{(1-x)^{k+1}}</math>}}

<math>v_k=1,\frac{d^k}{dx^k}(\sum_{l=0}^\infty v_l x^l)=\frac{d^k}{dx^k}(\sum_{l=0}^\infty x^l)=\frac{d^k}{dx^k}(\frac{1}{1-x})=\frac{k!}{(1-x)^{k+1}}</math>

<math>\sum_{k=0}^\infty u_k x^k=\sum_{k=0}^\infty \frac{\Delta^k u_0 x^k}{k!}\frac{d^k}{dx^k}(\sum_{l=0}^\infty v_l x^l)=\sum_{k=0}^\infty \frac{\Delta^k u_0 x^k}{(1-x)^{k+1}}</math>

{{Robox/Close}}

{{ExampleRobox|title=例子：<math>\sum_{k=0}^\infty \frac{u_k x^k}{k!}=e^x(\sum_{k=0}^\infty \frac{\Delta^k u_0 x^k}{k!})</math>}}

<math>v_k=\frac{1}{k!},\frac{d^k}{dx^k}(\sum_{l=0}^\infty v_l x^l)=\frac{d^k}{dx^k}(\sum_{l=0}^\infty \frac{x^l}{l!})=\frac{d^k}{dx^k}e^x=e^x</math>

<math>\sum_{k=0}^\infty \frac{u_k x^k}{k!}=\sum_{k=0}^\infty \frac{\Delta^k u_0 x^k}{k!}e^x=e^x(\sum_{k=0}^\infty \frac{\Delta^k u_0 x^k}{k!})</math>
{{Robox/Close}}


==參考資料==
{{reflist}}