Mixing
Trying to understand to equal sums.
Why is it that?
$$z + sum_{2 le k le n}^{infty} frac{q^{n-1}}{n-1} k(k-1)p^{n-k} dbinom{2n-k-2}{n-2}z^n = sum_{j, k ge 0}^{infty} frac{q^{j+k-1}}{j+k-1} k(k-1)p^j dbinom{2j+k-2}{j}z^{j+k}. $$
I get that you can just plug $j+k=n$, but where does the $z$ appears in the latter?
summation
asked Nov 20 '18 at 18:50
user528867
133
add a comment |
Why is it that?
$$z + sum_{2 le k le n}^{infty} frac{q^{n-1}}{n-1} k(k-1)p^{n-k} dbinom{2n-k-2}{n-2}z^n = sum_{j, k ge 0}^{infty} frac{q^{j+k-1}}{j+k-1} k(k-1)p^j dbinom{2j+k-2}{j}z^{j+k}. $$
I get that you can just plug $j+k=n$, but where does the $z$ appears in the latter?
summation
asked Nov 20 '18 at 18:50
user528867
133
Note that the sum on the left was over $k$, the one in the right is over $j$ and $k$? Also the binomial coefficient is not obtained just by substitution of$ j+k=n$. It may help if you tell about where does this come from (what context), what does it represent?
– NoChance
Nov 20 '18 at 18:59
The first has index $kge 2$. The second allows $k=0,1$. I'd assume fo k=1 the term is z. At least that what I assume just by skimming.
– fleablood
Nov 20 '18 at 19:03
Page number 3. jstor.org/stable/2322567?seq=3#metadata_info_tab_contents
– user528867
Nov 20 '18 at 19:05
add a comment |
Why is it that?
$$z + sum_{2 le k le n}^{infty} frac{q^{n-1}}{n-1} k(k-1)p^{n-k} dbinom{2n-k-2}{n-2}z^n = sum_{j, k ge 0}^{infty} frac{q^{j+k-1}}{j+k-1} k(k-1)p^j dbinom{2j+k-2}{j}z^{j+k}. $$
I get that you can just plug $j+k=n$, but where does the $z$ appears in the latter?
summation
asked Nov 20 '18 at 18:50
user528867
133
Why is it that?
$$z + sum_{2 le k le n}^{infty} frac{q^{n-1}}{n-1} k(k-1)p^{n-k} dbinom{2n-k-2}{n-2}z^n = sum_{j, k ge 0}^{infty} frac{q^{j+k-1}}{j+k-1} k(k-1)p^j dbinom{2j+k-2}{j}z^{j+k}. $$
I get that you can just plug $j+k=n$, but where does the $z$ appears in the latter?
summation
summation
asked Nov 20 '18 at 18:50
user528867
133
asked Nov 20 '18 at 18:50
user528867
133
asked Nov 20 '18 at 18:50
user528867
133
asked Nov 20 '18 at 18:50
user528867
133
asked Nov 20 '18 at 18:50
user528867
133
133
Note that the sum on the left was over $k$, the one in the right is over $j$ and $k$? Also the binomial coefficient is not obtained just by substitution of$ j+k=n$. It may help if you tell about where does this come from (what context), what does it represent?
– NoChance
Nov 20 '18 at 18:59
The first has index $kge 2$. The second allows $k=0,1$. I'd assume fo k=1 the term is z. At least that what I assume just by skimming.
– fleablood
Nov 20 '18 at 19:03
Page number 3. jstor.org/stable/2322567?seq=3#metadata_info_tab_contents
– user528867
Nov 20 '18 at 19:05
add a comment |
Note that the sum on the left was over $k$, the one in the right is over $j$ and $k$? Also the binomial coefficient is not obtained just by substitution of$ j+k=n$. It may help if you tell about where does this come from (what context), what does it represent?
– NoChance
Nov 20 '18 at 18:59
The first has index $kge 2$. The second allows $k=0,1$. I'd assume fo k=1 the term is z. At least that what I assume just by skimming.
– fleablood
Nov 20 '18 at 19:03
Page number 3. jstor.org/stable/2322567?seq=3#metadata_info_tab_contents
– user528867
Nov 20 '18 at 19:05
Note that the sum on the left was over $k$, the one in the right is over $j$ and $k$? Also the binomial coefficient is not obtained just by substitution of$ j+k=n$. It may help if you tell about where does this come from (what context), what does it represent?
– NoChance
Nov 20 '18 at 18:59
Note that the sum on the left was over $k$, the one in the right is over $j$ and $k$? Also the binomial coefficient is not obtained just by substitution of$ j+k=n$. It may help if you tell about where does this come from (what context), what does it represent?
– NoChance
Nov 20 '18 at 18:59
The first has index $kge 2$. The second allows $k=0,1$. I'd assume fo k=1 the term is z. At least that what I assume just by skimming.
– fleablood
Nov 20 '18 at 19:03
The first has index $kge 2$. The second allows $k=0,1$. I'd assume fo k=1 the term is z. At least that what I assume just by skimming.
– fleablood
Nov 20 '18 at 19:03
Page number 3. jstor.org/stable/2322567?seq=3#metadata_info_tab_contents
– user528867
Nov 20 '18 at 19:05
Page number 3. jstor.org/stable/2322567?seq=3#metadata_info_tab_contents
– user528867
Nov 20 '18 at 19:05
add a comment |
1 Answer
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We start with the right-hand side and obtain
begin{align*}
color{blue}{sum_{j,kgeq 0}}&color{blue}{frac{q^{j+k-1}}{j+k-1}k(k-1)p^jbinom{2j+k-2}{j}z^{j+k}}\
&=sum_{n=0}^inftyleft(sum_{{j+k=n}atop{j,kgeq 0}}frac{q^{j+k-1}}{j+k-1}k(k-1)p^jbinom{2j+k-2}{j}right)z^ntag{1}\
&=sum_{n=0}^inftysum_{k=0}^nfrac{q^{n-1}}{n-1}k(k-1)p^{n-k}binom{2n-k-2}{n-k}z^ntag{2}\
&=sum_{0leq kleq nleq infty}frac{q^{n-1}}{n-1}k(k-1)p^{n-k}binom{2n-k-2}{n-2}z^ntag{3}\
&,,color{blue}{=z+sum_{2leq kleq nleq infty}frac{q^{n-1}}{n-1}k(k-1)p^{n-k}binom{2n-k-2}{n-2}z^n}tag{4}\
end{align*}
and the claim follows.
Comment:
In (1) we reorder the summands by introducing $n=j+k, ngeq 0$.
In (2) we eliminate the index $j$ by substituting $j=n-k$.
In (3) we rewrite the index region and use the binomial identity $binom{p}{q}=binom{p}{p-q}$.
In (4) we observe that the summands with indices $(n,k)in{(0,0),(0,1),(1,0)}$ vanish due to the factor $k(k-1)$, whereas in the case $(n,k)=(1,1)$ the expression $frac{k-1}{n-1}$ cancels, leaving $z$.
answered Nov 21 '18 at 16:48
Markus Scheuer
60.2k455143
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
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We start with the right-hand side and obtain
begin{align*}
color{blue}{sum_{j,kgeq 0}}&color{blue}{frac{q^{j+k-1}}{j+k-1}k(k-1)p^jbinom{2j+k-2}{j}z^{j+k}}\
&=sum_{n=0}^inftyleft(sum_{{j+k=n}atop{j,kgeq 0}}frac{q^{j+k-1}}{j+k-1}k(k-1)p^jbinom{2j+k-2}{j}right)z^ntag{1}\
&=sum_{n=0}^inftysum_{k=0}^nfrac{q^{n-1}}{n-1}k(k-1)p^{n-k}binom{2n-k-2}{n-k}z^ntag{2}\
&=sum_{0leq kleq nleq infty}frac{q^{n-1}}{n-1}k(k-1)p^{n-k}binom{2n-k-2}{n-2}z^ntag{3}\
&,,color{blue}{=z+sum_{2leq kleq nleq infty}frac{q^{n-1}}{n-1}k(k-1)p^{n-k}binom{2n-k-2}{n-2}z^n}tag{4}\
end{align*}
and the claim follows.
Comment:
In (1) we reorder the summands by introducing $n=j+k, ngeq 0$.
In (2) we eliminate the index $j$ by substituting $j=n-k$.
In (3) we rewrite the index region and use the binomial identity $binom{p}{q}=binom{p}{p-q}$.
In (4) we observe that the summands with indices $(n,k)in{(0,0),(0,1),(1,0)}$ vanish due to the factor $k(k-1)$, whereas in the case $(n,k)=(1,1)$ the expression $frac{k-1}{n-1}$ cancels, leaving $z$.
answered Nov 21 '18 at 16:48
Markus Scheuer
60.2k455143
add a comment |
We start with the right-hand side and obtain
begin{align*}
color{blue}{sum_{j,kgeq 0}}&color{blue}{frac{q^{j+k-1}}{j+k-1}k(k-1)p^jbinom{2j+k-2}{j}z^{j+k}}\
&=sum_{n=0}^inftyleft(sum_{{j+k=n}atop{j,kgeq 0}}frac{q^{j+k-1}}{j+k-1}k(k-1)p^jbinom{2j+k-2}{j}right)z^ntag{1}\
&=sum_{n=0}^inftysum_{k=0}^nfrac{q^{n-1}}{n-1}k(k-1)p^{n-k}binom{2n-k-2}{n-k}z^ntag{2}\
&=sum_{0leq kleq nleq infty}frac{q^{n-1}}{n-1}k(k-1)p^{n-k}binom{2n-k-2}{n-2}z^ntag{3}\
&,,color{blue}{=z+sum_{2leq kleq nleq infty}frac{q^{n-1}}{n-1}k(k-1)p^{n-k}binom{2n-k-2}{n-2}z^n}tag{4}\
end{align*}
and the claim follows.
Comment:
In (1) we reorder the summands by introducing $n=j+k, ngeq 0$.
In (2) we eliminate the index $j$ by substituting $j=n-k$.
In (3) we rewrite the index region and use the binomial identity $binom{p}{q}=binom{p}{p-q}$.
In (4) we observe that the summands with indices $(n,k)in{(0,0),(0,1),(1,0)}$ vanish due to the factor $k(k-1)$, whereas in the case $(n,k)=(1,1)$ the expression $frac{k-1}{n-1}$ cancels, leaving $z$.
answered Nov 21 '18 at 16:48
Markus Scheuer
60.2k455143
add a comment |
We start with the right-hand side and obtain
begin{align*}
color{blue}{sum_{j,kgeq 0}}&color{blue}{frac{q^{j+k-1}}{j+k-1}k(k-1)p^jbinom{2j+k-2}{j}z^{j+k}}\
&=sum_{n=0}^inftyleft(sum_{{j+k=n}atop{j,kgeq 0}}frac{q^{j+k-1}}{j+k-1}k(k-1)p^jbinom{2j+k-2}{j}right)z^ntag{1}\
&=sum_{n=0}^inftysum_{k=0}^nfrac{q^{n-1}}{n-1}k(k-1)p^{n-k}binom{2n-k-2}{n-k}z^ntag{2}\
&=sum_{0leq kleq nleq infty}frac{q^{n-1}}{n-1}k(k-1)p^{n-k}binom{2n-k-2}{n-2}z^ntag{3}\
&,,color{blue}{=z+sum_{2leq kleq nleq infty}frac{q^{n-1}}{n-1}k(k-1)p^{n-k}binom{2n-k-2}{n-2}z^n}tag{4}\
end{align*}
and the claim follows.
Comment:
In (1) we reorder the summands by introducing $n=j+k, ngeq 0$.
In (2) we eliminate the index $j$ by substituting $j=n-k$.
In (3) we rewrite the index region and use the binomial identity $binom{p}{q}=binom{p}{p-q}$.
In (4) we observe that the summands with indices $(n,k)in{(0,0),(0,1),(1,0)}$ vanish due to the factor $k(k-1)$, whereas in the case $(n,k)=(1,1)$ the expression $frac{k-1}{n-1}$ cancels, leaving $z$.
answered Nov 21 '18 at 16:48
Markus Scheuer
60.2k455143
We start with the right-hand side and obtain
begin{align*}
color{blue}{sum_{j,kgeq 0}}&color{blue}{frac{q^{j+k-1}}{j+k-1}k(k-1)p^jbinom{2j+k-2}{j}z^{j+k}}\
&=sum_{n=0}^inftyleft(sum_{{j+k=n}atop{j,kgeq 0}}frac{q^{j+k-1}}{j+k-1}k(k-1)p^jbinom{2j+k-2}{j}right)z^ntag{1}\
&=sum_{n=0}^inftysum_{k=0}^nfrac{q^{n-1}}{n-1}k(k-1)p^{n-k}binom{2n-k-2}{n-k}z^ntag{2}\
&=sum_{0leq kleq nleq infty}frac{q^{n-1}}{n-1}k(k-1)p^{n-k}binom{2n-k-2}{n-2}z^ntag{3}\
&,,color{blue}{=z+sum_{2leq kleq nleq infty}frac{q^{n-1}}{n-1}k(k-1)p^{n-k}binom{2n-k-2}{n-2}z^n}tag{4}\
end{align*}
and the claim follows.
Comment:
In (1) we reorder the summands by introducing $n=j+k, ngeq 0$.
In (2) we eliminate the index $j$ by substituting $j=n-k$.
In (3) we rewrite the index region and use the binomial identity $binom{p}{q}=binom{p}{p-q}$.
In (4) we observe that the summands with indices $(n,k)in{(0,0),(0,1),(1,0)}$ vanish due to the factor $k(k-1)$, whereas in the case $(n,k)=(1,1)$ the expression $frac{k-1}{n-1}$ cancels, leaving $z$.
answered Nov 21 '18 at 16:48
Markus Scheuer
60.2k455143
edited Nov 22 '18 at 7:16
answered Nov 21 '18 at 16:48
Markus Scheuer
60.2k455143
answered Nov 21 '18 at 16:48
Markus Scheuer
60.2k455143
answered Nov 21 '18 at 16:48
Markus Scheuer
60.2k455143
60.2k455143
add a comment |
add a comment |
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Note that the sum on the left was over $k$, the one in the right is over $j$ and $k$? Also the binomial coefficient is not obtained just by substitution of$ j+k=n$. It may help if you tell about where does this come from (what context), what does it represent?
– NoChance
Nov 20 '18 at 18:59
The first has index $kge 2$. The second allows $k=0,1$. I'd assume fo k=1 the term is z. At least that what I assume just by skimming.
– fleablood
Nov 20 '18 at 19:03
Page number 3. jstor.org/stable/2322567?seq=3#metadata_info_tab_contents
– user528867
Nov 20 '18 at 19:05