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Name of the following summation: $sum_a=b^inftybinoma-1b-1x^a-b=(1-x)^-b$
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I was proofing a formula when I meet a summation that I culdn't solve.
After some efforts and investigations I've successfully recognized it in its generalized formula:
$$sum_a=b^inftybinoma-1b-1x^a-b=(1-x)^-b$$
that I saw online in a list of knowed series.
I've searched for a long time now, but I can't find information about it, even it's name, can you help me please?
I would like to prove it.
sequences-and-series
asked Jul 14 at 21:48
Fabio
258
add a comment |Â
up vote
0
down vote
favorite
I was proofing a formula when I meet a summation that I culdn't solve.
After some efforts and investigations I've successfully recognized it in its generalized formula:
$$sum_a=b^inftybinoma-1b-1x^a-b=(1-x)^-b$$
that I saw online in a list of knowed series.
I've searched for a long time now, but I can't find information about it, even it's name, can you help me please?
I would like to prove it.
sequences-and-series
asked Jul 14 at 21:48
Fabio
258
1
You can find it as a special case of Binomial series here. Also, on a side note, changing index variable so that it starts from $0$ instead of $b$ is helpful to see it more clearly
– Rumpelstiltskin
Jul 14 at 21:54
@Adam In the special case paragraph of the link it says $$sum_k=0^inftybinomk+betakz^k=frac1(1-z)^beta+1$$ but In my case I have to proof $$sum_k=0^inftybinomk+betabetaz^k=frac1(1-z)^beta+1$$ Do you think that is still a Binomial series?
– Fabio
Jul 14 at 22:14
1
Surely, if $beta$ is a non-negative integer (recall that then $binomk+betabeta=binomk+betak+beta-beta = binomk+betak $)
– Rumpelstiltskin
Jul 14 at 22:38
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I was proofing a formula when I meet a summation that I culdn't solve.
After some efforts and investigations I've successfully recognized it in its generalized formula:
$$sum_a=b^inftybinoma-1b-1x^a-b=(1-x)^-b$$
that I saw online in a list of knowed series.
I've searched for a long time now, but I can't find information about it, even it's name, can you help me please?
I would like to prove it.
sequences-and-series
asked Jul 14 at 21:48
Fabio
258
I was proofing a formula when I meet a summation that I culdn't solve.
After some efforts and investigations I've successfully recognized it in its generalized formula:
$$sum_a=b^inftybinoma-1b-1x^a-b=(1-x)^-b$$
that I saw online in a list of knowed series.
I've searched for a long time now, but I can't find information about it, even it's name, can you help me please?
I would like to prove it.
sequences-and-series
asked Jul 14 at 21:48
Fabio
258
asked Jul 14 at 21:48
Fabio
258
asked Jul 14 at 21:48
Fabio
258
asked Jul 14 at 21:48
Fabio
258
258
1
You can find it as a special case of Binomial series here. Also, on a side note, changing index variable so that it starts from $0$ instead of $b$ is helpful to see it more clearly
– Rumpelstiltskin
Jul 14 at 21:54
@Adam In the special case paragraph of the link it says $$sum_k=0^inftybinomk+betakz^k=frac1(1-z)^beta+1$$ but In my case I have to proof $$sum_k=0^inftybinomk+betabetaz^k=frac1(1-z)^beta+1$$ Do you think that is still a Binomial series?
– Fabio
Jul 14 at 22:14
1
Surely, if $beta$ is a non-negative integer (recall that then $binomk+betabeta=binomk+betak+beta-beta = binomk+betak $)
– Rumpelstiltskin
Jul 14 at 22:38
add a comment |Â
1
You can find it as a special case of Binomial series here. Also, on a side note, changing index variable so that it starts from $0$ instead of $b$ is helpful to see it more clearly
– Rumpelstiltskin
Jul 14 at 21:54
@Adam In the special case paragraph of the link it says $$sum_k=0^inftybinomk+betakz^k=frac1(1-z)^beta+1$$ but In my case I have to proof $$sum_k=0^inftybinomk+betabetaz^k=frac1(1-z)^beta+1$$ Do you think that is still a Binomial series?
– Fabio
Jul 14 at 22:14
1
Surely, if $beta$ is a non-negative integer (recall that then $binomk+betabeta=binomk+betak+beta-beta = binomk+betak $)
– Rumpelstiltskin
Jul 14 at 22:38
1
1
You can find it as a special case of Binomial series here. Also, on a side note, changing index variable so that it starts from $0$ instead of $b$ is helpful to see it more clearly
– Rumpelstiltskin
Jul 14 at 21:54
You can find it as a special case of Binomial series here. Also, on a side note, changing index variable so that it starts from $0$ instead of $b$ is helpful to see it more clearly
– Rumpelstiltskin
Jul 14 at 21:54
@Adam In the special case paragraph of the link it says $$sum_k=0^inftybinomk+betakz^k=frac1(1-z)^beta+1$$ but In my case I have to proof $$sum_k=0^inftybinomk+betabetaz^k=frac1(1-z)^beta+1$$ Do you think that is still a Binomial series?
– Fabio
Jul 14 at 22:14
@Adam In the special case paragraph of the link it says $$sum_k=0^inftybinomk+betakz^k=frac1(1-z)^beta+1$$ but In my case I have to proof $$sum_k=0^inftybinomk+betabetaz^k=frac1(1-z)^beta+1$$ Do you think that is still a Binomial series?
– Fabio
Jul 14 at 22:14
1
1
Surely, if $beta$ is a non-negative integer (recall that then $binomk+betabeta=binomk+betak+beta-beta = binomk+betak $)
– Rumpelstiltskin
Jul 14 at 22:38
Surely, if $beta$ is a non-negative integer (recall that then $binomk+betabeta=binomk+betak+beta-beta = binomk+betak $)
– Rumpelstiltskin
Jul 14 at 22:38
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
It's a form of
the generalized binomial theorem.
It can be rewritten like this:
$beginarray\
sum_a=b^inftybinoma-1b-1x^a-b
&=sum_a=0^inftybinoma+b-1b-1x^a\
endarray
$
From the generalized binomial theorem:
$beginarray\
(1-x)^-b
&=sum_n=0^infty binom-bn (-x)^n\
&=sum_n=0^infty dfracprod_k=0^n-1(-b-k)n! (-1)^nx^n\
&=sum_n=0^infty dfrac(-1)^nprod_k=0^n-1(b+k)n! (-1)^nx^n\
&=sum_n=0^infty dfracprod_k=b^n+b-1(k)n! x^n\
&=sum_n=0^infty dfrac(n+b-1)!(b-1)!n! x^n\
&=sum_n=0^infty binomn+b-1n x^n\
endarray
$
and this is the same
with $n$ instead of $a$.
answered Jul 14 at 22:36
marty cohen
69.3k446122
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
It's a form of
the generalized binomial theorem.
It can be rewritten like this:
$beginarray\
sum_a=b^inftybinoma-1b-1x^a-b
&=sum_a=0^inftybinoma+b-1b-1x^a\
endarray
$
From the generalized binomial theorem:
$beginarray\
(1-x)^-b
&=sum_n=0^infty binom-bn (-x)^n\
&=sum_n=0^infty dfracprod_k=0^n-1(-b-k)n! (-1)^nx^n\
&=sum_n=0^infty dfrac(-1)^nprod_k=0^n-1(b+k)n! (-1)^nx^n\
&=sum_n=0^infty dfracprod_k=b^n+b-1(k)n! x^n\
&=sum_n=0^infty dfrac(n+b-1)!(b-1)!n! x^n\
&=sum_n=0^infty binomn+b-1n x^n\
endarray
$
and this is the same
with $n$ instead of $a$.
answered Jul 14 at 22:36
marty cohen
69.3k446122
add a comment |Â
up vote
2
down vote
accepted
It's a form of
the generalized binomial theorem.
It can be rewritten like this:
$beginarray\
sum_a=b^inftybinoma-1b-1x^a-b
&=sum_a=0^inftybinoma+b-1b-1x^a\
endarray
$
From the generalized binomial theorem:
$beginarray\
(1-x)^-b
&=sum_n=0^infty binom-bn (-x)^n\
&=sum_n=0^infty dfracprod_k=0^n-1(-b-k)n! (-1)^nx^n\
&=sum_n=0^infty dfrac(-1)^nprod_k=0^n-1(b+k)n! (-1)^nx^n\
&=sum_n=0^infty dfracprod_k=b^n+b-1(k)n! x^n\
&=sum_n=0^infty dfrac(n+b-1)!(b-1)!n! x^n\
&=sum_n=0^infty binomn+b-1n x^n\
endarray
$
and this is the same
with $n$ instead of $a$.
answered Jul 14 at 22:36
marty cohen
69.3k446122
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
It's a form of
the generalized binomial theorem.
It can be rewritten like this:
$beginarray\
sum_a=b^inftybinoma-1b-1x^a-b
&=sum_a=0^inftybinoma+b-1b-1x^a\
endarray
$
From the generalized binomial theorem:
$beginarray\
(1-x)^-b
&=sum_n=0^infty binom-bn (-x)^n\
&=sum_n=0^infty dfracprod_k=0^n-1(-b-k)n! (-1)^nx^n\
&=sum_n=0^infty dfrac(-1)^nprod_k=0^n-1(b+k)n! (-1)^nx^n\
&=sum_n=0^infty dfracprod_k=b^n+b-1(k)n! x^n\
&=sum_n=0^infty dfrac(n+b-1)!(b-1)!n! x^n\
&=sum_n=0^infty binomn+b-1n x^n\
endarray
$
and this is the same
with $n$ instead of $a$.
answered Jul 14 at 22:36
marty cohen
69.3k446122
It's a form of
the generalized binomial theorem.
It can be rewritten like this:
$beginarray\
sum_a=b^inftybinoma-1b-1x^a-b
&=sum_a=0^inftybinoma+b-1b-1x^a\
endarray
$
From the generalized binomial theorem:
$beginarray\
(1-x)^-b
&=sum_n=0^infty binom-bn (-x)^n\
&=sum_n=0^infty dfracprod_k=0^n-1(-b-k)n! (-1)^nx^n\
&=sum_n=0^infty dfrac(-1)^nprod_k=0^n-1(b+k)n! (-1)^nx^n\
&=sum_n=0^infty dfracprod_k=b^n+b-1(k)n! x^n\
&=sum_n=0^infty dfrac(n+b-1)!(b-1)!n! x^n\
&=sum_n=0^infty binomn+b-1n x^n\
endarray
$
and this is the same
with $n$ instead of $a$.
answered Jul 14 at 22:36
marty cohen
69.3k446122
answered Jul 14 at 22:36
marty cohen
69.3k446122
answered Jul 14 at 22:36
marty cohen
69.3k446122
answered Jul 14 at 22:36
marty cohen
69.3k446122
69.3k446122
add a comment |Â
add a comment |Â
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1
You can find it as a special case of Binomial series here. Also, on a side note, changing index variable so that it starts from $0$ instead of $b$ is helpful to see it more clearly
– Rumpelstiltskin
Jul 14 at 21:54
@Adam In the special case paragraph of the link it says $$sum_k=0^inftybinomk+betakz^k=frac1(1-z)^beta+1$$ but In my case I have to proof $$sum_k=0^inftybinomk+betabetaz^k=frac1(1-z)^beta+1$$ Do you think that is still a Binomial series?
– Fabio
Jul 14 at 22:14
1
Surely, if $beta$ is a non-negative integer (recall that then $binomk+betabeta=binomk+betak+beta-beta = binomk+betak $)
– Rumpelstiltskin
Jul 14 at 22:38