Kombinatorik - Beweis zu Partitionen?

13/02/2011 - 17:10 von JW | Report spam
Hallo,

es sei Pi[n,k](k[1],k[2],...,k[n]) die Menge der Zahl-Partitionen der Ganzen
Zahl n mit k=k[1]+k[2]+...+k[n] Teilen.

Hat jemand eine Idee, wie ich folgende Aussage beweisen kann?

Sum{i=0..n}Sum{j=0..k}(Sum{Pi[i,j](r[1],r[2],...,r[i])}a[1]^r[1]*a[2]^r[2]*...*a[i]^r[i]/(1!^r[1]r[1]!*2!^r[2]r[2]!*...*i!^r[i]r[i]!))*(Sum{Pi[n-i,k-j](s[1],s[2],...,s[n-i])}b[1]^s[1]*b[2]^s[2]*...*b[n-i]^s[n-i]/(1!^s[1]s[1]!*2!^s[2]s[2]!*...*(n-i)!^s[n-i]s[n-i]!))
=Sum{Pi[n,k](k[1],k[2],...,k[n])}Sum{t[1]=0..k[1]}Sum.{t[2]=0..k[2]}...Sum{t[n]=0..k[n]}binomial(k[1],t[1])f[1]^t[1]g[1]^(k[1]-t[1])*binomial(k[2],t[2])f[2]^t[2]g[2]^(k[2]-t[2])*...*binomial(k[n],t[n])f[n]^t[n]g[n]^(k[n]-t[n]))/(1!^k[1]k[1]!*2!^k[2]k[2]!*...*n!^k[n]k[n]!)

bzw. dort wo es geht ohne eckige Klammern:

Sum{i=0..n}Sum{j=0..k}(Sum{Pi[i,j](r1,r2,...,r[i])}a1^r1*a2^r2*...*a[i]^r[i]/(1!^r1r1!*2!^r2r2!*...*i!^r[i]r[i]!))*(Sum{Pi[n-i,k-j](s1,s2,...,s[n-i])}b1^s1*b2^s2*...*b[n-i]^s[n-i]/(1!^s1s1!*2!^s2s2!*...*(n-i)!^s[n-i]s[n-i]!))
=Sum{Pi[n,k](k1,k2,...,kn)}Sum{t1=0..k1}Sum.{t2=0..k2}...Sum{t[n]=0..k[n]}(binomial(k1,t1)f1^t1g1^(k1-t1)*binomial(k2,t2)f2^t2g2^(k2-t2)*...*binomial(k[n],t[n])f[n]^t[n]g[n]^(k[n]-t[n]))/(1!^k1k1!*2!^k2k2!*...*n!^k[n]k[n]!)


in LATEX:
\[\sum_{i=0}^{n}\sum_{j=0}^{k}\left(\sum_{\pi_{i,j}(r_{1},r_{2},...,r_{i})}\frac{a_{1}^{r_{1}}\cdot
a_{2}^{r_{2}}\cdot ...\cdot a_{i}^{r_{i}}}{1!^{r_{1}}r_{1}!\cdot
2!^{r_{2}}r_{2}!\cdot ...\cdot
i!^{r_{i}}r_{i}!}ight)\cdot\left(\sum_{\pi_{n-i,k-j}(s_{1},s_{2},...,s_{n-i})}\frac{b_{1}^{s_{1}}\cdot
b_{2}^{s_{2}}\cdot ...\cdot b_{n-i}^{s_{n-i}}}{1!^{s_{1}}s_{1}!\cdot
2!^{s_{2}}s_{2}!\cdot ...\cdot (n-i)!^{s_{n-i}}s_{n-i}!)}ight)\]
\\\[=\sum_{\pi_{n,k}(k_{1},k_{2},...,k_{n})}\sum_{t_{1}=0}^{k_{1}}\sum_{t_{2}=0}^{k_{2}}...\sum_{t_{n}=0}^{k_{n}}\frac{\left(^{k_{1}}_{t_{1}}ight)a_{1}^{t_{1}}b_{1}^{(k_{1}-t_{1})}\cdot\left(^{k_{2}}_{t_{2}}ight)a_{2}^{t_{2}}b_{2}^{(k_{2}-t_{2})}\cdot
...\cdot\left(^{k_{n}}_{t_{n}}ight)a_{n}^{t_{n}}b_{n}^{(k_{n}-t_{n})}}{1!^{k_{1}}k_{1}!\cdot
2!^{k_{2}}k_{2}!\cdot ...\cdot n!^{k_{n}}k_{n}!}\]

in MAPLE:
Sum(Sum(Sum(a[1]^r[1]*a[2]^r[2].`...`.a[i]^r[i]/(`1`!^r[1]*r[1]!*2!^r[2]*r[2]!.`...`.i!^r[i]*r[i]!),[r[1],r[2],`...`,r[i]]=Pi[i,j](r[1],r[2],`...`,r[i]))*Sum(b[1]^s[1]*b[2]^s[2].`...`.b[n-i]^s[n-i]/(`1`!^s[1]*s[1]!*2!^s[2]*s[2]!.`...`.(n-i)!^s[n-i]*s[n-i]!),[s[1],s[2],`...`,s[n-i]]=Pi[n-i,k-j](s[1],s[2],`...`,s[n-i])),j=0..k),i=0..n)=Sum(Sum(Sum(`...`.Sum((binomial(k[1],t[1])*a[1]^t[1]*b[1]^(k[1]-t[1])*binomial(k[2],t[2])*a[2]^t[2]*b[2]^(k[2]-t[2]).`...`.(binomial(k[n],t[n])*a[n]^t[n]*b[n]^(k[n]-t[n])))/(`1`!^k[1]*k[1]!*2!^k[2]*k[2]!.`...`.(n!^k[n]*k[n]!)),t[n]=0..k[n]),t[2]=0..k[2]),t[1]=0..k[1]),[k[1],k[2],`...`,k[n]]=Pi[n,k](k[1],k[2],`...`,k[n]));

Bin kein Mathematiker.

Danke!
 

Lesen sie die antworten

#1 wernertrp
13/02/2011 - 19:07 | Warnen spam
On 13 Feb., 17:10, "JW" wrote:
Hallo,

es sei Pi[n,k](k[1],k[2],...,k[n]) die Menge der Zahl-Partitionen der Ganzen
Zahl n mit k=k[1]+k[2]+...+k[n] Teilen.

Hat jemand eine Idee, wie ich folgende Aussage beweisen kann?

Sum{i=0..n}Sum{j=0..k}(Sum{Pi[i,j](r[1],r[2],...,r[i])}a[1]^r[1]*a[2]^r[2]*­...*a[i]^r[i]/(1!^r[1]r[1]!*2!^r[2]r[2]!*...*i!^r[i]r[i]!))*(Sum{Pi[n-i,k-j­](s[1],s[2],...,s[n-i])}b[1]^s[1]*b[2]^s[2]*...*b[n-i]^s[n-i]/(1!^s[1]s[1]!­*2!^s[2]s[2]!*...*(n-i)!^s[n-i]s[n-i]!))
=Sum{Pi[n,k](k[1],k[2],...,k[n])}Sum{t[1]=0..k[1]}Sum.{t[2]=0..k[2]}...Sum{­t[n]=0..k[n]}binomial(k[1],t[1])f[1]^t[1]g[1]^(k[1]-t[1])*binomial(k[2],t[2­])f[2]^t[2]g[2]^(k[2]-t[2])*...*binomial(k[n],t[n])f[n]^t[n]g[n]^(k[n]-t[n]­))/(1!^k[1]k[1]!*2!^k[2]k[2]!*...*n!^k[n]k[n]!)

bzw. dort wo es geht ohne eckige Klammern:

Sum{i=0..n}Sum{j=0..k}(Sum{Pi[i,j](r1,r2,...,r[i])}a1^r1*a2^r2*...*a[i]^r[i­]/(1!^r1r1!*2!^r2r2!*...*i!^r[i]r[i]!))*(Sum{Pi[n-i,k-j](s1,s2,...,s[n-i])}­b1^s1*b2^s2*...*b[n-i]^s[n-i]/(1!^s1s1!*2!^s2s2!*...*(n-i)!^s[n-i]s[n-i]!))
=Sum{Pi[n,k](k1,k2,...,kn)}Sum{t1=0..k1}Sum.{t2=0..k2}...Sum{t[n]=0..k[n]}(­binomial(k1,t1)f1^t1g1^(k1-t1)*binomial(k2,t2)f2^t2g2^(k2-t2)*...*binomial(­k[n],t[n])f[n]^t[n]g[n]^(k[n]-t[n]))/(1!^k1k1!*2!^k2k2!*...*n!^k[n]k[n]!)

in LATEX:
\[\sum_{i=0}^{n}\sum_{j=0}^{k}\left(\sum_{\pi_{i,j}(r_{1},r_{2},...,r_{i})}­\frac{a_{1}^{r_{1}}\cdot
a_{2}^{r_{2}}\cdot ...\cdot a_{i}^{r_{i}}}{1!^{r_{1}}r_{1}!\cdot
2!^{r_{2}}r_{2}!\cdot ...\cdot
i!^{r_{i}}r_{i}!}ight)\cdot\left(\sum_{\pi_{n-i,k-j}(s_{1},s_{2},...,s_{n­-i})}\frac{b_{1}^{s_{1}}\cdot
b_{2}^{s_{2}}\cdot ...\cdot b_{n-i}^{s_{n-i}}}{1!^{s_{1}}s_{1}!\cdot
2!^{s_{2}}s_{2}!\cdot ...\cdot (n-i)!^{s_{n-i}}s_{n-i}!)}ight)\]
\\\[=\sum_{\pi_{n,k}(k_{1},k_{2},...,k_{n})}\sum_{t_{1}=0}^{k_{1}}\sum_{t_{­2}=0}^{k_{2}}...\sum_{t_{n}=0}^{k_{n}}\frac{\left(^{k_{1}}_{t_{1}}ight)a_­{1}^{t_{1}}b_{1}^{(k_{1}-t_{1})}\cdot\left(^{k_{2}}_{t_{2}}ight)a_{2}^{t_­{2}}b_{2}^{(k_{2}-t_{2})}\cdot
...\cdot\left(^{k_{n}}_{t_{n}}ight)a_{n}^{t_{n}}b_{n}^{(k_{n}-t_{n})}}{1!­^{k_{1}}k_{1}!\cdot
2!^{k_{2}}k_{2}!\cdot ...\cdot n!^{k_{n}}k_{n}!}\]

in MAPLE:
Sum(Sum(Sum(a[1]^r[1]*a[2]^r[2].`...`.a[i]^r[i]/(`1`!^r[1]*r[1]!*2!^r[2]*r[­2]!.`...`.i!^r[i]*r[i]!),[r[1],r[2],`...`,r[i]]=Pi[i,j](r[1],r[2],`...`,r[i­]))*Sum(b[1]^s[1]*b[2]^s[2].`...`.b[n-i]^s[n-i]/(`1`!^s[1]*s[1]!*2!^s[2]*s[­2]!.`...`.(n-i)!^s[n-i]*s[n-i]!),[s[1],s[2],`...`,s[n-i]]=Pi[n-i,k-j](s[1],­s[2],`...`,s[n-i])),j=0..k),i=0..n)=Sum(Sum(Sum(`...`.Sum((binomial(k[1],t[­1])*a[1]^t[1]*b[1]^(k[1]-t[1])*binomial(k[2],t[2])*a[2]^t[2]*b[2]^(k[2]-t[2­]).`...`.(binomial(k[n],t[n])*a[n]^t[n]*b[n]^(k[n]-t[n])))/(`1`!^k[1]*k[1]!­*2!^k[2]*k[2]!.`...`.(n!^k[n]*k[n]!)),t[n]=0..k[n]),t[2]=0..k[2]),t[1]=0..k­[1]),[k[1],k[2],`...`,k[n]]=Pi[n,k](k[1],k[2],`...`,k[n]));

Bin kein Mathematiker.

Danke!



Diese Mathematik führt zu Augenschàden.

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