Trying to understand to equal sums.
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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 11 hours ago
user528867
83
add a comment |
up vote
0
down vote
favorite
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 11 hours ago
user528867
83
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
11 hours ago
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
11 hours ago
Page number 3. jstor.org/stable/2322567?seq=3#metadata_info_tab_contents
– user528867
11 hours ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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 11 hours ago
user528867
83
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 11 hours ago
user528867
83
asked 11 hours ago
user528867
83
asked 11 hours ago
user528867
83
asked 11 hours ago
user528867
83
asked 11 hours ago
user528867
83
83
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
11 hours ago
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
11 hours ago
Page number 3. jstor.org/stable/2322567?seq=3#metadata_info_tab_contents
– user528867
11 hours ago
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
11 hours ago
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
11 hours ago
Page number 3. jstor.org/stable/2322567?seq=3#metadata_info_tab_contents
– user528867
11 hours ago
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
11 hours ago
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
11 hours ago
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
11 hours ago
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
11 hours ago
Page number 3. jstor.org/stable/2322567?seq=3#metadata_info_tab_contents
– user528867
11 hours ago
Page number 3. jstor.org/stable/2322567?seq=3#metadata_info_tab_contents
– user528867
11 hours ago
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
11 hours ago
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
11 hours ago
Page number 3. jstor.org/stable/2322567?seq=3#metadata_info_tab_contents
– user528867
11 hours ago