EDIT n millioner:
Beviset er i noen innlegg under. Uff, dette ble en rotete tråd fra min side
Bevis for alternativ definisjon av Zeta-funksjonen.
Moderatorer: Vektormannen, espen180, Aleks855, Solar Plexsus, Gustav, Nebuchadnezzar, Janhaa
Pokker. Tenkte meg noe slikt. Vil det si at du ikke kan ta grensen på naturlige tall at all? Hvis en kan jeg si atCharlatan skrev:Du kan nok ikke induktere mhp n her. n går mot uendelig når man trekker grensen så det gir ingen mening å anta at grenseverdien stemmer for en gitt n. Induksjonsvariabelen du eventuelt må bruke er s.
[tex]\zeta(s)=\lim_{n\to\infty}\sum_{t=1}^n t^s,[/tex]
og så vise at
[tex]\sum_{t=1}^n k^{-s}=\frac{a_n}{n!^s},[/tex]
så må vel også grenseverdien være den samme?
EDIT:
Det jeg har kommet frem til er forøvrig at
[tex]\zeta(s)=\lim_{n\to\infty} n!^{-s}\sum_{k=0}^{n-1}\left(k!n^{\underline{n-k-1}}\right)^s[/tex]
Ja, det stemmer.Pokker. Tenkte meg noe slikt. Vil det si at du ikke kan ta grensen på naturlige tall at all? Hvis en kan jeg si at
[tex]\zeta(s)=\lim_{n\to\infty}\sum_{t=1}^n t^s,[/tex]
og så vise at
[tex]\sum_{t=1}^n k^{-s}=\frac{a_n}{n!^s},[/tex]
så må vel også grenseverdien være den samme?
EDIT:
Her er hele skiten. Beklager at det er på engelsk, men jeg skrev det på engelsk og da ble det slik. Ja, også var det slik at jeg setter pris på både stor og liten pirking og omvelting
Let's first look at the recurrence
[tex]\{a_n\};& \\a_1=1;& \\a_n=n^{s}a_{n-1}+(n-1)!^s&.[/tex]
Then
[tex]\zeta(s)=\lim_{n\rightarrow\infty}\frac{a_n}{n!^s}.[/tex]
\begin{proof}To show that this is correct, we first prove that [tex]\sum_{k=1}^nk^{-s}=a_n.[/tex]
We do this by induction: Assume this proposition is true for all [tex]1,\dots,n.[/tex] We try for [tex]n \rightarrow n+1[/tex] and find
[tex]\sum_{1 \leq k \leq n}k^{-s}+\frac{1}{(n+1)^{s}}=\frac{a_n}{n!^s}+\frac{1}{(n+1)^{s}}\\=\frac{(n+1)!^sa_n+n!^s}{(n+1)!^s}\\=\frac{a_{n+1}}{(n+1)!^s}[/tex]
Now, since the two expressions are equivalent, and [tex]\zeta(s)=\lim_{n\to\infty}\sum_{k=1}^nk^{-s},[/tex] we can conclude that [tex]\zeta(s)=\lim_{n\rightarrow\infty}\frac{a_n}{n!^s}.[/tex]
\end{proof}
Now, we will want to find a solution to [tex]\{a_n\}[/tex], and we shall show that
[tex]\sum_{0\leq k < n} \left(k!n^{\underline{n-k-1}}\right)^s,[/tex]
is a solution to the recurrence where [tex]n^{\underline m}\equiv n(n-1)\dots(n-m+1), n^{\underline 0}=1.[/tex]
By subsitution the identity then becomes
[tex]\zeta(s)=\lim_{n\to\infty} n!^{-s}\sum_{0\leq k < n}\left(k!n^{\underline{n-k-1}}\right)^s.[/tex]
\begin{proof}To show that this is indeed the Riemann-[tex]\zeta[/tex] function, let us calculate its difference with respect to [tex]n.[/tex] In other words, we let
[tex]f(n)=\sum_{0\leq k < n} \left(\frac{k!n^{\underline{n-k-1}}}{n!}\right)^s.[/tex]
Here the constant term (with regards to the sum) is absorved for clarity. The difference of [tex]f,[/tex] written [tex]\Delta f,[/tex] is defined to be [tex]f(n+1)-f(n).[/tex] We have
[tex]\Delta f = \sum_{0\leq k < n+1} \left(\frac{k!(n+1)^{\underline{n-k}}}{(n+1)!}\right)^s-\sum_{0\leq k < n} \left(\frac{k!n^{\underline{n-k-1}}}{n!}\right)^s\\=\left(\frac{n!(n+1)^{\underline{\cancel{n-n}}}}{(n+1)!}\right)^s+\sum_{0\leq k < n} \left(\frac{\cancel{(n+1)}k!n^{\underline{n-k-1}}}{\cancel{(n+1)}n!}\right)^s -\sum_{0\leq k < n} \left(\frac{k!n^{\underline{n-k-1}}}{n!}\right)^s\\=(n+1)^{-s}+\cancel{\sum_{0\leq k < n} \left(\frac{k!n^{\underline{n-k-1}}}{n!}\right)^s -\sum_{0\leq k < n}\left(\frac{k!n^{\underline{n-k-1}}}{n!}\right)^s},[/tex]
which is conclusive because there is no independent constant term in either representation of [tex]\zeta.[/tex] That is, there is no term [tex]C \neq 0[/tex] so that
[tex]\lim_{h\to0} f(x+h)-f(x)=C. [/tex] Wait a minute, I don't think this last part makes much sense. Then I'm not sure how to filter out any constants.
\end{proof}
Her er hele skiten. Beklager at det er på engelsk, men jeg skrev det på engelsk og da ble det slik. Ja, også var det slik at jeg setter pris på både stor og liten pirking og omvelting
Let's first look at the recurrence
[tex]\{a_n\};& \\a_1=1;& \\a_n=n^{s}a_{n-1}+(n-1)!^s&.[/tex]
Then
[tex]\zeta(s)=\lim_{n\rightarrow\infty}\frac{a_n}{n!^s}.[/tex]
\begin{proof}To show that this is correct, we first prove that [tex]\sum_{k=1}^nk^{-s}=a_n.[/tex]
We do this by induction: Assume this proposition is true for all [tex]1,\dots,n.[/tex] We try for [tex]n \rightarrow n+1[/tex] and find
[tex]\sum_{1 \leq k \leq n}k^{-s}+\frac{1}{(n+1)^{s}}=\frac{a_n}{n!^s}+\frac{1}{(n+1)^{s}}\\=\frac{(n+1)!^sa_n+n!^s}{(n+1)!^s}\\=\frac{a_{n+1}}{(n+1)!^s}[/tex]
Now, since the two expressions are equivalent, and [tex]\zeta(s)=\lim_{n\to\infty}\sum_{k=1}^nk^{-s},[/tex] we can conclude that [tex]\zeta(s)=\lim_{n\rightarrow\infty}\frac{a_n}{n!^s}.[/tex]
\end{proof}
Now, we will want to find a solution to [tex]\{a_n\}[/tex], and we shall show that
[tex]\sum_{0\leq k < n} \left(k!n^{\underline{n-k-1}}\right)^s,[/tex]
is a solution to the recurrence where [tex]n^{\underline m}\equiv n(n-1)\dots(n-m+1), n^{\underline 0}=1.[/tex]
By subsitution the identity then becomes
[tex]\zeta(s)=\lim_{n\to\infty} n!^{-s}\sum_{0\leq k < n}\left(k!n^{\underline{n-k-1}}\right)^s.[/tex]
\begin{proof}To show that this is indeed the Riemann-[tex]\zeta[/tex] function, let us calculate its difference with respect to [tex]n.[/tex] In other words, we let
[tex]f(n)=\sum_{0\leq k < n} \left(\frac{k!n^{\underline{n-k-1}}}{n!}\right)^s.[/tex]
Here the constant term (with regards to the sum) is absorved for clarity. The difference of [tex]f,[/tex] written [tex]\Delta f,[/tex] is defined to be [tex]f(n+1)-f(n).[/tex] We have
[tex]\Delta f = \sum_{0\leq k < n+1} \left(\frac{k!(n+1)^{\underline{n-k}}}{(n+1)!}\right)^s-\sum_{0\leq k < n} \left(\frac{k!n^{\underline{n-k-1}}}{n!}\right)^s\\=\left(\frac{n!(n+1)^{\underline{\cancel{n-n}}}}{(n+1)!}\right)^s+\sum_{0\leq k < n} \left(\frac{\cancel{(n+1)}k!n^{\underline{n-k-1}}}{\cancel{(n+1)}n!}\right)^s -\sum_{0\leq k < n} \left(\frac{k!n^{\underline{n-k-1}}}{n!}\right)^s\\=(n+1)^{-s}+\cancel{\sum_{0\leq k < n} \left(\frac{k!n^{\underline{n-k-1}}}{n!}\right)^s -\sum_{0\leq k < n}\left(\frac{k!n^{\underline{n-k-1}}}{n!}\right)^s},[/tex]
which is conclusive because there is no independent constant term in either representation of [tex]\zeta.[/tex] That is, there is no term [tex]C \neq 0[/tex] so that
[tex]\lim_{h\to0} f(x+h)-f(x)=C. [/tex] Wait a minute, I don't think this last part makes much sense. Then I'm not sure how to filter out any constants.
\end{proof}
Sist redigert av edahl den 24/06-2009 11:34, redigert 2 ganger totalt.
Vent litt! Jeg har et skikkelig og algebraisk bevis for identiteten nå:edahl skrev:Heh, det har jeg ikke tenkt på før så jeg har faktisk aldri brukt det for 'induksjon'Markonan skrev:Verbformen til induksjon blir forresten å indusere.
Men holder beviset da?
Observer at [tex]\frac{m!}{n!}=n!^{\underline{n-m}},\text{for n < m}[/tex] og at [tex]n^{\underline{n-k-1}} = \frac{(k+1)!}{n!}[/tex]
Dermed er
[tex]\lim_{n\to\infty}\sum_{0\leq k < n}\left(\frac{k!n^{\underline{n-k-1}}}{n!}\right)^s = \lim_{n\to\infty} \sum_{0 \leq k < n}\left(\frac{ \frac{\cancel{k!}(k+1)} {\cancel{k!}\cancel{n!}} }{\cancel{n!}}\right)^s = \lim_{n\to\infty}\sum_{0 \leq k < n}\left(\frac{1}{k+1}\right)^s=\lim_{n\to\infty}\sum_{1 \leq k \leq n} \left( \frac{1}{k} \right)^s = \zeta(s).[/tex]
Hvis DEN ikke holder vet jeg ikke hva gjør
EDIT: Tråd nr. 100 ..!