47202 利用微積分得到$1^k+2^k+\cdots+n^k$的求和公式

一、求和函數的定義

\begin{align} 1+2+\cdots+n=\,&\frac{n^2+n}2, \label{1}\\ 1^2+2^2+\cdots+n^2=\,&\frac{2n^3+3n^2+n}6, \label{2} \end{align}

\begin{align} 1^3+2^3+\cdots+n^3=\frac{n^4+2n^3+n^2}4. \label{3} \end{align}

$$1^k+2^k+\cdots+n^k=(n\ \hbox{的}\ k+1\ \hbox{次多項式}). \qquad\hbox{(註一)}$$

$$F_k (0)=0^k=0,\quad F_k (1)=1^k=1,\quad F_k (2)=1^k+2^k,\ldots,F_k (n)=1^k+2^k+\cdots+n^k .$$

$$F_k (0)=0,\ \hbox{並且}\ F_k (n)-F_k (n-1)=n^k,\quad n\ge 1.$$

$$F_k (0)=0,\qquad F_k (x)-F_k (x-1)=x^k.$$

\begin{align} F_1 (x)=\,&\frac{x^2+x}2, \label{4}\\ F_2 (x)=\,&\frac{2x^3+3x^2+x}6, \label{5}\\ F_3 (x)=\,&\frac{x^4+2x^3+x^2}4. \label{6} \end{align}

二、求和函數的存在性和唯一性

\begin{align*} &F_k'(x)-F_k'(x-1)=\,kx^{k-1},\\ {\hbox{或}} &\frac 1k F_k'(x)-\frac 1k F_k'(x-1)=\,x^{k-1}. \end{align*}

$$\frac 1k F_k' (x)=F_{k-1} (x)+c_k.$$

$$\int_0^x F_{k-1} (t)dt+c_k x=G(x).$$

$$G' (x)=F_{k-1} (x)+c_k=\frac 1k F_k' (x),$$

$$F_k (x)=k G(x)=k\Big(\int_0^x F_{k-1} (t)dt+c_k x\Big).$$

$$1=F_k (1)=k\Big(\int_0^1 F_{k-1} (t)dt+c_k \Big),$$

$$c_k=\frac 1k -\int_0^1 F_{k-1} (t)dt.$$

$$F_k (x)=k\Big(\int_0^x F_{k-1} (t)dt+c_k x\Big),$$

$$c_k=\frac 1k -\int_0^1 F_{k-1} (t)dt.$$

\begin{align*} \frac 1k [F_k (x)-F_k (x-1)] =&\int_0^x F_{k-1} (t)dt+c_k x-\int_0^{x-1} F_{k-1} (t)dt-c_k (x-1)\\ =&\int_0^x F_{k-1} (t)dt -\int_0^{x-1} F_{k-1} (t)dt+c_k\\ =&\int_0^x F_{k-1} (t)dt -\int_0^{x-1} F_{k-1} (t)dt+\frac 1k -\int_0^1 F_{k-1} (t)dt\\ =&\int_1^x F_{k-1} (t)dt -\int_0^{x-1} F_{k-1} (t)dt+\frac 1k.\\ {\hbox{但 }} &\hskip 10pt\int_0^{x-1} F_{k-1} (t)dt =\int_1^x F_{k-1} (t-1)dt,\quad\hbox{(註三)}\\ \hbox{原式}=&\int_1^x F_{k-1} (t)dt -\int_1^x F_{k-1} (t-1)dt+\frac 1k\\ =&\int_1^x t^{k-1} dt+\frac 1k=\dfrac{x^k}k-\frac 1k +\frac 1k =\dfrac{x^k}k, \end{align*}

\begin{align*} F_1 (x)=\,&\frac 12 (x^2+x),\quad c_2=\frac 12-\int_0^1 F_1 (t)dt =\frac 1{12},\hskip 5cm~\\ \frac 12 F_2 (x)=&\int_0^x F_1 (t)dt+c_2 x=\frac 16 x^3+\frac 14 x^2+\frac 1{12} x,\\ \hbox{得}\quad F_2 (x)=\,&\frac 13 x^3+\frac 12 x^2+\frac 16 x \qquad \hbox{(見式 \eqref{5})}. \end{align*}

\begin{align*} F_2 (x)=\,&\frac 13 x^3+\frac 12 x^2+\frac 16 x,\qquad c_3=\frac 13-\int_0^1 F_2 (t)dt =0,\hskip 3.5cm~\\ \frac 13 F_3 (x)=&\int_0^x F_2 (t)dt+0 \cdot x=\frac 1{12} x^4+\frac 16 x^3+\frac 1{12} x^2,\\ \hbox{得}\quad F_3 (x)=\,&\frac 14 x^4+\frac 12 x^3+\frac 14 x^2\qquad \hbox{(見式 \eqref{6})}. \end{align*}

\begin{align*} \hbox{假設}\quad F_{k-1} (x)=\,&\dfrac 1k x^k+(\hbox{低次項}),\\ \hbox{則}\qquad\ \frac 1k F_k (x)=&\int_0^x F_{k-1} (t)dt+c_k x=\int_0^x(\frac 1k t^k+\cdots)dt+c_k x\hskip 3cm~\\ =\,&\frac 1{k (k+1)} x^{k+1}+(\hbox{低次項}), \end{align*}

$$\dfrac{dh(x-1)}{dx}=h'(x-1).$$

$$\int_a^b h(x)dx=\int_{a+1}^{b+1}h(x-1)dx;$$