Mixing
$sum_N=6^inftyfrac6NbinomN-15p^6(1-p)^N-6$
Clash Royale CLAN TAG#URR8PPP
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Find the sum: $sumlimits_N=6^inftyfrac6NbinomN-15p^6(1-p)^N-6$
I could not find a way to manipulate $binomN-15$ to get any suitable form here. Note that $binomN-15p^6(1-p)^N-6$ are probability masses of negative binomial distribution.
probability probability-distributions summation negative-binomial
edited Jul 16 at 18:21
David G. Stork
7,6912929
asked Jul 16 at 18:03
Stat_prob_001
18710
add a comment |Â
up vote
1
down vote
favorite
Find the sum: $sumlimits_N=6^inftyfrac6NbinomN-15p^6(1-p)^N-6$
I could not find a way to manipulate $binomN-15$ to get any suitable form here. Note that $binomN-15p^6(1-p)^N-6$ are probability masses of negative binomial distribution.
probability probability-distributions summation negative-binomial
edited Jul 16 at 18:21
David G. Stork
7,6912929
asked Jul 16 at 18:03
Stat_prob_001
18710
1
Look here at the special cases of Binomial series. Then integrate once
– Rumpelstiltskin
Jul 16 at 18:12
It seems that it doesn´t exist a closed form. See here what wolfram alpha figured out.
– callculus
Jul 16 at 18:33
1
@callculus It looks like the first right parenthesis belongs at the end of the whole expression. See here
– saulspatz
Jul 16 at 18:38
@saulspatz Thanks for the correction. I haven´t noticed the misinterpretation of the parenthesis by w.a .
– callculus
Jul 16 at 18:41
Mathematica gives: $$frac60 p^6 log (p)-(p-1) p (p (p (p (137 p-163)+137)-63)+12)10 (p-1)^6$$ which suggests that manipulation will be rather difficult.
– David G. Stork
Jul 16 at 18:49
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Find the sum: $sumlimits_N=6^inftyfrac6NbinomN-15p^6(1-p)^N-6$
I could not find a way to manipulate $binomN-15$ to get any suitable form here. Note that $binomN-15p^6(1-p)^N-6$ are probability masses of negative binomial distribution.
probability probability-distributions summation negative-binomial
edited Jul 16 at 18:21
David G. Stork
7,6912929
asked Jul 16 at 18:03
Stat_prob_001
18710
Find the sum: $sumlimits_N=6^inftyfrac6NbinomN-15p^6(1-p)^N-6$
I could not find a way to manipulate $binomN-15$ to get any suitable form here. Note that $binomN-15p^6(1-p)^N-6$ are probability masses of negative binomial distribution.
probability probability-distributions summation negative-binomial
edited Jul 16 at 18:21
David G. Stork
7,6912929
asked Jul 16 at 18:03
Stat_prob_001
18710
edited Jul 16 at 18:21
David G. Stork
7,6912929
edited Jul 16 at 18:21
David G. Stork
7,6912929
edited Jul 16 at 18:21
David G. Stork
7,6912929
7,6912929
asked Jul 16 at 18:03
Stat_prob_001
18710
asked Jul 16 at 18:03
Stat_prob_001
18710
asked Jul 16 at 18:03
Stat_prob_001
18710
18710
1
Look here at the special cases of Binomial series. Then integrate once
– Rumpelstiltskin
Jul 16 at 18:12
It seems that it doesn´t exist a closed form. See here what wolfram alpha figured out.
– callculus
Jul 16 at 18:33
1
@callculus It looks like the first right parenthesis belongs at the end of the whole expression. See here
– saulspatz
Jul 16 at 18:38
@saulspatz Thanks for the correction. I haven´t noticed the misinterpretation of the parenthesis by w.a .
– callculus
Jul 16 at 18:41
Mathematica gives: $$frac60 p^6 log (p)-(p-1) p (p (p (p (137 p-163)+137)-63)+12)10 (p-1)^6$$ which suggests that manipulation will be rather difficult.
– David G. Stork
Jul 16 at 18:49
add a comment |Â
1
Look here at the special cases of Binomial series. Then integrate once
– Rumpelstiltskin
Jul 16 at 18:12
It seems that it doesn´t exist a closed form. See here what wolfram alpha figured out.
– callculus
Jul 16 at 18:33
1
@callculus It looks like the first right parenthesis belongs at the end of the whole expression. See here
– saulspatz
Jul 16 at 18:38
@saulspatz Thanks for the correction. I haven´t noticed the misinterpretation of the parenthesis by w.a .
– callculus
Jul 16 at 18:41
Mathematica gives: $$frac60 p^6 log (p)-(p-1) p (p (p (p (137 p-163)+137)-63)+12)10 (p-1)^6$$ which suggests that manipulation will be rather difficult.
– David G. Stork
Jul 16 at 18:49
1
1
Look here at the special cases of Binomial series. Then integrate once
– Rumpelstiltskin
Jul 16 at 18:12
Look here at the special cases of Binomial series. Then integrate once
– Rumpelstiltskin
Jul 16 at 18:12
It seems that it doesn´t exist a closed form. See here what wolfram alpha figured out.
– callculus
Jul 16 at 18:33
It seems that it doesn´t exist a closed form. See here what wolfram alpha figured out.
– callculus
Jul 16 at 18:33
1
1
@callculus It looks like the first right parenthesis belongs at the end of the whole expression. See here
– saulspatz
Jul 16 at 18:38
@callculus It looks like the first right parenthesis belongs at the end of the whole expression. See here
– saulspatz
Jul 16 at 18:38
@saulspatz Thanks for the correction. I haven´t noticed the misinterpretation of the parenthesis by w.a .
– callculus
Jul 16 at 18:41
@saulspatz Thanks for the correction. I haven´t noticed the misinterpretation of the parenthesis by w.a .
– callculus
Jul 16 at 18:41
Mathematica gives: $$frac60 p^6 log (p)-(p-1) p (p (p (p (137 p-163)+137)-63)+12)10 (p-1)^6$$ which suggests that manipulation will be rather difficult.
– David G. Stork
Jul 16 at 18:49
Mathematica gives: $$frac60 p^6 log (p)-(p-1) p (p (p (p (137 p-163)+137)-63)+12)10 (p-1)^6$$ which suggests that manipulation will be rather difficult.
– David G. Stork
Jul 16 at 18:49
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
What we want to find:
$$sum_k=0^infty frac6k+6binomk+5kp^6(1-p)^k =S(p) $$
What we know from Binomial series:
$$sum_k=0^infty binomk+5kz^k =frac1(1-z)^6 $$
Some manipulation, and integration:
$$int_0^1-xsum_k=0^infty binomk+5kz^k+5 dz =sum_k=0^infty frac1k+6binomk+5k(1-x)^k+6 = int_0^1-x fracz^5(1-z)^6dz = frac300x^4-300x^3+200x^2-75x+1260x^5+ln(x)-frac13760 $$
After substitution, and some multiplying by constants:
$$S(p) =frac300p^5-300p^4+200p^3-75p^2+12p10(1-p)^6+frac6p^6ln(p)(1-p)^6-frac137p^610(1-p)^6 $$
answered Jul 16 at 18:50
Rumpelstiltskin
1,424315
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
What we want to find:
$$sum_k=0^infty frac6k+6binomk+5kp^6(1-p)^k =S(p) $$
What we know from Binomial series:
$$sum_k=0^infty binomk+5kz^k =frac1(1-z)^6 $$
Some manipulation, and integration:
$$int_0^1-xsum_k=0^infty binomk+5kz^k+5 dz =sum_k=0^infty frac1k+6binomk+5k(1-x)^k+6 = int_0^1-x fracz^5(1-z)^6dz = frac300x^4-300x^3+200x^2-75x+1260x^5+ln(x)-frac13760 $$
After substitution, and some multiplying by constants:
$$S(p) =frac300p^5-300p^4+200p^3-75p^2+12p10(1-p)^6+frac6p^6ln(p)(1-p)^6-frac137p^610(1-p)^6 $$
answered Jul 16 at 18:50
Rumpelstiltskin
1,424315
add a comment |Â
up vote
4
down vote
accepted
What we want to find:
$$sum_k=0^infty frac6k+6binomk+5kp^6(1-p)^k =S(p) $$
What we know from Binomial series:
$$sum_k=0^infty binomk+5kz^k =frac1(1-z)^6 $$
Some manipulation, and integration:
$$int_0^1-xsum_k=0^infty binomk+5kz^k+5 dz =sum_k=0^infty frac1k+6binomk+5k(1-x)^k+6 = int_0^1-x fracz^5(1-z)^6dz = frac300x^4-300x^3+200x^2-75x+1260x^5+ln(x)-frac13760 $$
After substitution, and some multiplying by constants:
$$S(p) =frac300p^5-300p^4+200p^3-75p^2+12p10(1-p)^6+frac6p^6ln(p)(1-p)^6-frac137p^610(1-p)^6 $$
answered Jul 16 at 18:50
Rumpelstiltskin
1,424315
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
What we want to find:
$$sum_k=0^infty frac6k+6binomk+5kp^6(1-p)^k =S(p) $$
What we know from Binomial series:
$$sum_k=0^infty binomk+5kz^k =frac1(1-z)^6 $$
Some manipulation, and integration:
$$int_0^1-xsum_k=0^infty binomk+5kz^k+5 dz =sum_k=0^infty frac1k+6binomk+5k(1-x)^k+6 = int_0^1-x fracz^5(1-z)^6dz = frac300x^4-300x^3+200x^2-75x+1260x^5+ln(x)-frac13760 $$
After substitution, and some multiplying by constants:
$$S(p) =frac300p^5-300p^4+200p^3-75p^2+12p10(1-p)^6+frac6p^6ln(p)(1-p)^6-frac137p^610(1-p)^6 $$
answered Jul 16 at 18:50
Rumpelstiltskin
1,424315
What we want to find:
$$sum_k=0^infty frac6k+6binomk+5kp^6(1-p)^k =S(p) $$
What we know from Binomial series:
$$sum_k=0^infty binomk+5kz^k =frac1(1-z)^6 $$
Some manipulation, and integration:
$$int_0^1-xsum_k=0^infty binomk+5kz^k+5 dz =sum_k=0^infty frac1k+6binomk+5k(1-x)^k+6 = int_0^1-x fracz^5(1-z)^6dz = frac300x^4-300x^3+200x^2-75x+1260x^5+ln(x)-frac13760 $$
After substitution, and some multiplying by constants:
$$S(p) =frac300p^5-300p^4+200p^3-75p^2+12p10(1-p)^6+frac6p^6ln(p)(1-p)^6-frac137p^610(1-p)^6 $$
answered Jul 16 at 18:50
Rumpelstiltskin
1,424315
answered Jul 16 at 18:50
Rumpelstiltskin
1,424315
answered Jul 16 at 18:50
Rumpelstiltskin
1,424315
answered Jul 16 at 18:50
Rumpelstiltskin
1,424315
1,424315
add a comment |Â
add a comment |Â
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1
Look here at the special cases of Binomial series. Then integrate once
– Rumpelstiltskin
Jul 16 at 18:12
It seems that it doesn´t exist a closed form. See here what wolfram alpha figured out.
– callculus
Jul 16 at 18:33
1
@callculus It looks like the first right parenthesis belongs at the end of the whole expression. See here
– saulspatz
Jul 16 at 18:38
@saulspatz Thanks for the correction. I haven´t noticed the misinterpretation of the parenthesis by w.a .
– callculus
Jul 16 at 18:41
Mathematica gives: $$frac60 p^6 log (p)-(p-1) p (p (p (p (137 p-163)+137)-63)+12)10 (p-1)^6$$ which suggests that manipulation will be rather difficult.
– David G. Stork
Jul 16 at 18:49