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-#LyX 2.4 created this file. For more info see https://www.lyx.org/
-\lyxformat 620
-\begin_document
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-\index Index
-\shortcut idx
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-\end_header
-
-\begin_body
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc4[HM20]
-\end_layout
-
-\end_inset
-
-(
-\emph on
-Sums of powers.
-\emph default
-) When 
-\begin_inset Formula $f(x)=x^{m}$
-\end_inset
-
-,
- the high-order derivatives of 
-\begin_inset Formula $f$
-\end_inset
-
- are all zero,
- so Euler's summation formula gives an 
-\emph on
-exact
-\emph default
- value for the sum
-\begin_inset Formula 
-\[
-S_{m}(n)=\sum_{0\leq k<n}k^{m}
-\]
-
-\end_inset
-
-in terms of Bernoulli numbers.
- (It was the study of 
-\begin_inset Formula $S_{m}(n)$
-\end_inset
-
- for 
-\begin_inset Formula $m=1,2,3,\dots$
-\end_inset
-
- that led Bernoulli and Seki to discover those numbers in the first place.) Express 
-\begin_inset Formula $S_{m}(n)$
-\end_inset
-
- in terms of Bernoulli 
-\emph on
-polynomials
-\emph default
-.
- Check your answer for 
-\begin_inset Formula $m=0$
-\end_inset
-
-,
- 1,
- and 2.
- (Note that the desired sum is performed for 
-\begin_inset Formula $0\leq k<n$
-\end_inset
-
- instead of 
-\begin_inset Formula $1\leq k<n$
-\end_inset
-
-;
- Euler's summation formula may be applied with 0 replacing 1 throughout.)
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-For 
-\begin_inset Formula $k\geq0$
-\end_inset
-
-,
- we have 
-\begin_inset Formula $f^{(k)}(x)=m^{\underline{k}}x^{m-k}$
-\end_inset
-
-.
- If 
-\begin_inset Formula $k<m$
-\end_inset
-
-,
- this is nonzero except that 
-\begin_inset Formula $f^{(k)}(0)=0$
-\end_inset
-
-;
- if 
-\begin_inset Formula $k=m$
-\end_inset
-
-,
- 
-\begin_inset Formula $f^{(m)}(x)\equiv1$
-\end_inset
-
-,
- and if 
-\begin_inset Formula $k>m$
-\end_inset
-
-,
- 
-\begin_inset Formula $f^{(k)}(x)\equiv0$
-\end_inset
-
-.
- Using Euler's summation formula,
- when 
-\begin_inset Formula $m\geq1$
-\end_inset
-
-,
-\begin_inset Formula 
-\[
-S_{m}(n)=\sum_{0\leq k<n}k^{m}=\int_{0}^{n}x^{m}\text{d}x+\sum_{k=1}^{m}\frac{B_{k}}{k!}m^{\underline{k-1}}n^{m-k+1}=\frac{n^{m+1}}{m+1}+\sum_{k=1}^{m}\frac{B_{k}}{m-k+1}\binom{m}{k}n^{m-k+1}.
-\]
-
-\end_inset
-
-Then,
-\begin_inset Formula 
-\[
-S'_{m}(n)=n^{m}+\sum_{k=1}^{m}B_{k}\binom{m}{k}n^{m-k}=\sum_{k}B_{k}\binom{m}{k}n^{m-k}=B_{m}(x),
-\]
-
-\end_inset
-
-which means that
-\begin_inset Formula 
-\[
-\int_{0}^{n}B_{m}=(S_{m}(n)-S_{m}(0))=S_{m}(n)
-\]
-
-\end_inset
-
- and therefore,
-\begin_inset Formula 
-\[
-S_{m}(n)=\int_{0}^{n}B_{m}.
-\]
-
-\end_inset
-
-Now we check the solution,
- and in particular we check that this works for 
-\begin_inset Formula $m=0$
-\end_inset
-
- as well:
-\begin_inset Formula 
-\begin{align*}
-\int_{0}^{n}B_{0} & =\int_{0}^{n}\text{d}x=n-0=n, & S_{0}(n) & =\sum_{0\leq k<n}1=n;\\
-\int_{0}^{n}B_{1} & =\int_{0}^{n}(x-\tfrac{1}{2})\text{d}x=\frac{n^{2}-n}{2}, & S_{1}(n) & =\sum_{0\leq k<n}k=\frac{n(n-1)}{2};\\
-\int_{0}^{n}B_{2} & =\int_{0}^{n}(x^{2}-x+\tfrac{1}{6})\text{d}x=\frac{2n^{3}-3n^{2}+n}{6},
-\end{align*}
-
-\end_inset
-
-and we check the last result we use induction.
- For 
-\begin_inset Formula $n=0$
-\end_inset
-
-,
- 
-\begin_inset Formula $S_{2}(0)=0$
-\end_inset
-
-.
- For 
-\begin_inset Formula $n\geq0$
-\end_inset
-
-,
-\begin_inset Formula 
-\begin{align*}
-S_{2}(n) & =\frac{1}{6}\left(2(n-1)^{3}-3(n-1)^{2}+(n-1)\right)+(n-1)^{2}\\
- & =\frac{1}{6}\left(2n^{3}\cancel{-6n^{2}}\cancel{+6n}\cancel{-2}-3n^{2}\cancel{+6n}\cancel{-3}+n\cancel{-1}\cancel{+6n^{2}}\cancel{-12n}\cancel{+6}\right)=\int_{0}^{n}B_{2}.
-\end{align*}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc9[M25]
-\end_layout
-
-\end_inset
-
-Find the asymptotic value of 
-\begin_inset Formula $\binom{2n}{n}$
-\end_inset
-
- with a relative error of 
-\begin_inset Formula $O(n^{-3})$
-\end_inset
-
-,
- in two ways:
-\end_layout
-
-\begin_layout Enumerate
-via Stirling's approximation;
-\end_layout
-
-\begin_layout Enumerate
-via exercise 1.2.6–47 and Eq.
- 1.2.11.1–(16).
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer
-\end_layout
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Enumerate
-We have,
-\begin_inset Formula 
-\[
-\binom{2n}{n}=\frac{(2n)!}{n!^{2}},
-\]
-
-\end_inset
-
-so 
-\begin_inset Formula 
-\begin{multline*}
-\ln\binom{2n}{n}=\ln(2n)!-2\ln(n!)=\\
-=(2n+\tfrac{1}{2})\ln(2n)-2n+\sigma+\frac{1}{24n}-(2n+1)\ln n+2n-2\sigma-\frac{1}{6n}+O(n^{-3})=\\
-=(2n+\tfrac{1}{2})(\ln n+\ln2)-(2n+1)\ln n-\sigma-\frac{1}{8n}+O(n^{-3})=\\
-=(2n+\tfrac{1}{2})\ln2-\frac{\ln n}{2}-\sigma-\frac{1}{8n}+O(n^{-3}),
-\end{multline*}
-
-\end_inset
-
-and therefore
-\begin_inset Formula 
-\[
-\binom{2n}{n}=\exp(\cdots)=\frac{2^{2n}}{\sqrt{\pi n}}\left(1-\frac{1}{8n}+\frac{1}{128n^{2}}+O(n^{3})\right).
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Enumerate
-\begin_inset Note Greyedout
-status open
-
-\begin_layout Plain Layout
-(I had to look up the solution.)
-\end_layout
-
-\end_inset
-
-Exercise 1.2.6–47,
- when 
-\begin_inset Formula $r=-1/2$
-\end_inset
-
-,
- the first identity simplifies to
-\begin_inset Formula 
-\[
-(-1)^{k}\binom{-\frac{1}{2}}{k}=(-1)^{k}\binom{-k-1}{k}\Bigg/4^{k}=\binom{2k}{k}\Bigg/4^{k}.
-\]
-
-\end_inset
-
-Thus,
- using Eq.
- 1.2.6–(21),
-\begin_inset Formula 
-\begin{multline*}
-\binom{2n}{n}=4^{n}(-1)^{n}\binom{-\frac{1}{2}}{n}=4^{n}\binom{n-\frac{1}{2}}{n}=4^{n}\frac{\Gamma(n+\frac{1}{2})}{\Gamma(n+1)\Gamma(\frac{1}{2})}=4^{k}\frac{\Gamma(n+\frac{1}{2})}{n\Gamma(n)\sqrt{\pi}}=\\
-=\frac{4^{n}n^{\overline{1/2}}}{n\sqrt{\pi}}=\frac{2^{2n}}{n\sqrt{\pi}}\left(\stirla{1/2}{1/2}n^{1/2}+\stirla{1/2}{-1/2}n^{-1/2}+\stirla{1/2}{-3/2}n^{-3/2}+O(n^{-5/2})\right).
-\end{multline*}
-
-\end_inset
-
-Using Eqs.
- 1.2.6–(48),
- (49),
- and (57),
-\begin_inset Formula 
-\begin{align*}
-\stirla{1/2}{1/2} & =1,\\
-\stirla{1/2}{-1/2} & =\binom{1/2}{2}=\frac{\frac{1}{2}(-\frac{1}{2})}{2}=-\frac{1}{8},\\
-\stirla{1/2}{-3/2} & =\binom{1/2}{4}+2\binom{3/2}{4}=\frac{-\frac{15}{16}}{24}+2\frac{\frac{9}{16}}{24}=\frac{3}{16\cdot24}=\frac{1}{128},
-\end{align*}
-
-\end_inset
-
-so in the end,
-\begin_inset Formula 
-\[
-\binom{2n}{n}=\frac{2^{2n}}{\sqrt{\pi n}}\left(1-\frac{1}{8n}+\frac{1}{128n^{2}}+O(n^{-3})\right).
-\]
-
-\end_inset
-
-
-\end_layout
-
-\end_body
-\end_document