高専の数学解説

第9章 9.12

(1)

Im,n=nm+n+1Im,n−1I_{m,n} = \frac{n}{m+n+1} I_{m,n-1}

Im,n=∫01xm(1−x)ndx=[1m+1xm+1(1−x)n]01−∫011m+1xm+1{−n(1−x)n−1}dxI_{m,n} = \int_0^1 x^m(1-x)^n dx = \left[\frac{1}{m+1}x^{m+1}(1-x)^n\right]_0^1 - \int_0^1 \frac{1}{m+1}x^{m+1}\{-n(1-x)^{n-1}\}dx

=nm+1∫01xm(x−1+1)(1−x)n−1dx=nm+1∫01{xm(1−x)n−1−xm(1−x)n}dx= \frac{n}{m+1}\int_0^1 x^m(x-1+1)(1-x)^{n-1}dx = \frac{n}{m+1}\int_0^1 \{x^m(1-x)^{n-1} - x^m(1-x)^n\}dx

=nm+1{∫01xm(1−x)n−1dx−∫01xm(1−x)ndx}=nm+1(Im,n−1−Im,n)= \frac{n}{m+1}\left\{\int_0^1 x^m(1-x)^{n-1}dx - \int_0^1 x^m(1-x)^n dx\right\} = \frac{n}{m+1}\left(I_{m,n-1} - I_{m,n}\right)

(1+nm+1)Im,n=nm+1Im,n−1\left(1 + \frac{n}{m+1}\right)I_{m,n} = \frac{n}{m+1}I_{m,n-1}

m+1+nm+1Im,n=nm+1Im,n−1Im,n=nm+n+1Im,n−1\frac{m+1+n}{m+1}I_{m,n} = \frac{n}{m+1}I_{m,n-1} \qquad I_{m,n} = \frac{n}{m+n+1}I_{m,n-1}

(2)

Im,n=∫01xm(1−x)ndx=[−1n+1(1−x)n+1xm]01+∫011n+1(1−x)n+1mxm−1dxI_{m,n} = \int_0^1 x^m(1-x)^n dx = \left[-\frac{1}{n+1}(1-x)^{n+1}x^m\right]_0^1 + \int_0^1 \frac{1}{n+1}(1-x)^{n+1} m x^{m-1} dx

=mn+1∫01(1−x)n(1−x)xm−1dx=mn+1∫01{xm−1(1−x)n−xm(1−x)n}dx=mn+1(Im−1,n−Im,n)= \frac{m}{n+1}\int_0^1 (1-x)^n(1-x)x^{m-1}dx = \frac{m}{n+1}\int_0^1 \{x^{m-1}(1-x)^n - x^m(1-x)^n\}dx = \frac{m}{n+1}\left(I_{m-1,n} - I_{m,n}\right)

(1+mn+1)Im,n=mn+1Im−1,nIm,n=mm+n+1Im−1,n\left(1 + \frac{m}{n+1}\right)I_{m,n} = \frac{m}{n+1}I_{m-1,n} \qquad I_{m,n} = \frac{m}{m+n+1}I_{m-1,n}

m=0m=0 n=0n=0 のとき

I0,0=∫01x0(1−x)0dx=∫01dx=1I_{0,0} = \int_0^1 x^0(1-x)^0 dx = \int_0^1 dx = 1

このとき Im,n=m! n!(m+n+1)!I_{m,n} = \dfrac{m!\,n!}{(m+n+1)!} が成り立っている

よって(1)、(2)で求めた式から

Im,n=1×⋯×mnm+n+1=m! n!(m+n+1)!I_{m,n} = 1 \times \cdots \times \frac{mn}{m+n+1} = \frac{m!\,n!}{(m+n+1)!} が全ての自然数 m,nm,n で成り立つことが分かる

解説: r31bn1z