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1 数学分析wsj习题5.41

$$$ p_n(x)=\frac{1}{2^nn!}\frac{d^n}{dx^n}(x^2-1)^n\\ prove:\ p_n(1)=1 \\ proof\ 1:\\ \frac{d^n}{dx^n}(x^2-1)^n=n!\sum_{k=1}^n\binom{n}{k}\binom{2k}{n}x^{2k-n}(-1)^{n-k}\\ x=1\\ \sum_{k}\binom{n}{k}\binom{2k}{n}(-1)^{n-k}\\\ =\sum_{k}\binom{n}{k}\binom{2k}{n}(-1)^{n-k}\\ =\sum_{k,j}\binom{n}{k}\binom{k}{j}\binom{k}{n-j}(-1)^{n-k}\\ =\sum_{k,j}\binom{n}{j}\binom{n-j}{k-j}\binom{k}{n-j}(-1)^{n-k}\\ =\sum_{k,j,l}\binom{n}{j}\binom{n-j}{k-j}\binom{k-j}{l}\binom{j}{n-j-l}(-1)^{n-k}\\ =\sum_{k,j,l}\binom{n}{j}\binom{n-j}{l}\binom{n-j-l}{k-j-l}\binom{j}{n-j-l}(-1)^{n-k}\\ =\sum_{k,j,l}\binom{n}{j}\binom{n-j}{l}\binom{j}{n-j-l}(-1)^{n+j+l}(-1)^{k-j-l}\binom{n-j-l}{k-j-l}\\ =\sum_{k,j,l}\binom{n}{j}\binom{n-j}{l}\binom{j}{n-j-l}(-1)^{n+j+l}[j+l==n]\\ =\sum_{j}\binom{n}{j}\\ =2^n $$$ $$$ prove \ 2: \sum_{k}\binom{n}{k}\binom{2k}{n}(-1)^{n-k}\\ =\sum_{k}\binom{k}{n-k}\binom{2k}{k}(-1)^{n-k}\\ =[x^n]\sum_{k}\binom{2k}{k}(1+x)^kx^k(-1)^{n-k} \ \ (ps:\sum_k\binom{2k}{k}x^k=\sqrt{\frac{1}{1-4x}})\\ =[x^n]\sqrt{\frac{1}{1+4x(1+x)}}(-1)^n\\ =[x^n]\frac{1}{1+2x}(-1)^n\\ =2^n $$$

but finally it occurred to me that ......

$$$ \frac{1}{2^nn!}\frac{d^n}{dx^n}(x^2-1)^n\\ =\frac{1}{2^nn!}\frac{d^n}{dx^n}((x+1)^n(x-1)^n)\\ apply \ leibniz \ formula $$$

2

$$$ prove \\\sum_k{\binom{n}{k}\frac{(-1)^k}{m+k+1}}=\frac{m!n!}{(m+n+1)!} $$$

I found there was no suitable tool for it

so I had to look into Concrete Mathematics -- Gosper!

现学

proof 1:

$$$ S(n)=t(n,k)=\binom{n}{k}\frac{(-1)^k}{m+k+1}\\ \hat{t}(n,k)=\beta_0(n)t(n,k)+\beta_1(n)t(n+1,k)=T(k+1)-T(k)\\ T(k) \ has \ the \ form \ T(k)=\frac{r(k)s(k)t(k)}{p(k)} \\ \frac{t(n+1,k)}{t(n,k)}=\frac{n+1}{n+1-k}\\ so \ let \ p(n,k)=\beta_0(n)(n+1-k)+(n+1)\beta_1(n)\\ now\ \hat{t}(n,k)=p(n,k)\frac{t(n,k)}{n+1-k}\\ \bar{t}(n,k)=\frac{t(n,k)}{p(n,k)}\\ \bar{p}(n,k)=\frac{\hat{p}(n,k)}{p(n,k)}\\ \frac{\bar{t}(n,k+1)}{\bar{t}(n,k)}=\frac{\bar{p}(n,k+1)q(n,k)}{\bar{p}(n,k)r(n,k+1)} =\frac{-(n+1-k)(m+k+1)}{(k+1)(m+k+2)}\\ \bar{p}(n,k)=1\\ q(n,k)=(n+1-k)(m+k+1)\\

r(n,k)=-k(m+k+1)\\ (n+1-k)\beta_0(n)+(n+1)\beta_1(n)=(m+k+1)((n+1-k)s(n,k+1)+ks(n,k))\\ consider \ \ the \ \ coefficient \ \ of \ \ k\\ the \ \ degree \ \ of \ \ s \ \ must \ \ be \ \ 0\\ hence \ s(n,k)=1 \ \ \beta_0(n)=-n-1 \ \ \beta_1(n)=m+n+2\\ S(n+1)/S(n)=(n+1)/(m+n+2) $$$

proof 2

$$$ \Delta^n(\frac{1}{x})\\ =\Delta^n((1-x)^{\underline{-1}})=(-1)^n(x-1)^{\underline{-n-1}}=(-1)^n\frac{n!}{x(x+1)...(x+n)} $$$

on the other hand we can expand

$$$ \Delta^n=(E-1)^n=..... $$$

proof 3

then I observe the form

it is similar to a identity that haunted me for more than a year

it is about the connection between the recurrence and the inclusive-exclusive expression of stirling number of second kind

$$$ \sum_{i=1}^m \frac{\binom{m}{i}(-1)^{m-i}}{1-ix}=\frac{x^m \ m!}{\prod_{i=1}^m(1-ix)} $$$

the left hand side is inclusive-exclusive

the right of obtained by the recurrence

it is easy(?) to see this is equivalent to the probem

thus we get a combinatorial demonstration

ps : whether x is an integer is not important since 考虑多项式在无限个点值为零则为零

proof 4

but in fact if we consider starting from the right hand side

Partial Fraction Expansion

$$$ F(x)=\frac{x!}{(x+n+1)!}=\frac{1}{(x+n+1)(x+n)...(x+1)} $$$

convert it into partial fraction

$$$ F(x)=\sum_{k=1}^{n+1}\frac{b_k}{x+k}\\ b_k=\lim_{x\to-k}(x+k)F(x)=\frac{1}{(n+1-k)!(k-1)!(-1)^k}\\ F(x)=\frac{1}{n!}\sum_{k=1}^{n+1}\binom{n}{k-1}\frac{(-1)^{k-1}}{x+k}\\ $$$

then set x=m.