Computing $int_0^1 prod_x in mathcalX (p_xy + (1-p_x)) dy$
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How would I compute this integral:
$$int_0^1 prod_x in mathcalX (p_xy + (1-p_x)) dy$$
I have tried using the chain rule and not made any headway on that front. Any advice is welcome.
polynomials definite-integrals
edited Aug 14 at 11:13
Blue
43.7k868141
asked Aug 14 at 11:09
Television
13711
add a comment |Â
up vote
0
down vote
favorite
How would I compute this integral:
$$int_0^1 prod_x in mathcalX (p_xy + (1-p_x)) dy$$
I have tried using the chain rule and not made any headway on that front. Any advice is welcome.
polynomials definite-integrals
edited Aug 14 at 11:13
Blue
43.7k868141
asked Aug 14 at 11:09
Television
13711
2
Can you agree that $p_x=partial p/partial x$ and explain what $mathcalX$ here is?
– Fakemistake
Aug 14 at 11:14
@Fakemistake $p_x$ is just a constant. $mathcalX$ is just a set, the cardinality of which governs the degree of the polynomial whose monomials are of the form $(p_xy+(1-p_x))$.
– Television
Aug 14 at 11:16
2
The substitution $u = 1 - y$ simplifies the factors of the integrand to the form $1-p_x u$. The product can then be expanded as a polynomial in $u$, with coefficients expressed as elementary symmetric polynomials of the $p_i$. The integration of the expanded polynomial is straightforward.
– Blue
Aug 14 at 11:55
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How would I compute this integral:
$$int_0^1 prod_x in mathcalX (p_xy + (1-p_x)) dy$$
I have tried using the chain rule and not made any headway on that front. Any advice is welcome.
polynomials definite-integrals
edited Aug 14 at 11:13
Blue
43.7k868141
asked Aug 14 at 11:09
Television
13711
How would I compute this integral:
$$int_0^1 prod_x in mathcalX (p_xy + (1-p_x)) dy$$
I have tried using the chain rule and not made any headway on that front. Any advice is welcome.
polynomials definite-integrals
edited Aug 14 at 11:13
Blue
43.7k868141
asked Aug 14 at 11:09
Television
13711
edited Aug 14 at 11:13
Blue
43.7k868141
edited Aug 14 at 11:13
Blue
43.7k868141
edited Aug 14 at 11:13
Blue
43.7k868141
43.7k868141
asked Aug 14 at 11:09
Television
13711
asked Aug 14 at 11:09
Television
13711
asked Aug 14 at 11:09
Television
13711
13711
2
Can you agree that $p_x=partial p/partial x$ and explain what $mathcalX$ here is?
– Fakemistake
Aug 14 at 11:14
@Fakemistake $p_x$ is just a constant. $mathcalX$ is just a set, the cardinality of which governs the degree of the polynomial whose monomials are of the form $(p_xy+(1-p_x))$.
– Television
Aug 14 at 11:16
2
The substitution $u = 1 - y$ simplifies the factors of the integrand to the form $1-p_x u$. The product can then be expanded as a polynomial in $u$, with coefficients expressed as elementary symmetric polynomials of the $p_i$. The integration of the expanded polynomial is straightforward.
– Blue
Aug 14 at 11:55
add a comment |Â
2
Can you agree that $p_x=partial p/partial x$ and explain what $mathcalX$ here is?
– Fakemistake
Aug 14 at 11:14
@Fakemistake $p_x$ is just a constant. $mathcalX$ is just a set, the cardinality of which governs the degree of the polynomial whose monomials are of the form $(p_xy+(1-p_x))$.
– Television
Aug 14 at 11:16
2
The substitution $u = 1 - y$ simplifies the factors of the integrand to the form $1-p_x u$. The product can then be expanded as a polynomial in $u$, with coefficients expressed as elementary symmetric polynomials of the $p_i$. The integration of the expanded polynomial is straightforward.
– Blue
Aug 14 at 11:55
2
2
Can you agree that $p_x=partial p/partial x$ and explain what $mathcalX$ here is?
– Fakemistake
Aug 14 at 11:14
Can you agree that $p_x=partial p/partial x$ and explain what $mathcalX$ here is?
– Fakemistake
Aug 14 at 11:14
@Fakemistake $p_x$ is just a constant. $mathcalX$ is just a set, the cardinality of which governs the degree of the polynomial whose monomials are of the form $(p_xy+(1-p_x))$.
– Television
Aug 14 at 11:16
@Fakemistake $p_x$ is just a constant. $mathcalX$ is just a set, the cardinality of which governs the degree of the polynomial whose monomials are of the form $(p_xy+(1-p_x))$.
– Television
Aug 14 at 11:16
2
2
The substitution $u = 1 - y$ simplifies the factors of the integrand to the form $1-p_x u$. The product can then be expanded as a polynomial in $u$, with coefficients expressed as elementary symmetric polynomials of the $p_i$. The integration of the expanded polynomial is straightforward.
– Blue
Aug 14 at 11:55
The substitution $u = 1 - y$ simplifies the factors of the integrand to the form $1-p_x u$. The product can then be expanded as a polynomial in $u$, with coefficients expressed as elementary symmetric polynomials of the $p_i$. The integration of the expanded polynomial is straightforward.
– Blue
Aug 14 at 11:55
add a comment |Â
1 Answer
1
active
oldest
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up vote
1
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accepted
I assume $|chi|<infty.$
If we are able to rewrite the product as follows
$$prod_xinchi(p_xy+(1-p_x))=sum_n=0^chia_ny^n$$
then the integral is very easy to compute and it results in
$$int_0^1prod_xinchi(p_xy+(1-p_x))dy=int_0^1sum_n=0^chia_ny^ndy=sum_n=0^chifraca_nn+1.$$
Now the terms $a_n$ are those where you multiply exactly $n$ $p_xy$ together and $|chi|-n$ terms of the form $(1-p_x)$ (look at the degreee). Thus
$$a_n=sum_=nprod_p_xin A p_xcdotprod_p_xnotin A(1-p_x)$$
answered Aug 14 at 11:23
b00n heT
8,08411430
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I assume $|chi|<infty.$
If we are able to rewrite the product as follows
$$prod_xinchi(p_xy+(1-p_x))=sum_n=0^chia_ny^n$$
then the integral is very easy to compute and it results in
$$int_0^1prod_xinchi(p_xy+(1-p_x))dy=int_0^1sum_n=0^chia_ny^ndy=sum_n=0^chifraca_nn+1.$$
Now the terms $a_n$ are those where you multiply exactly $n$ $p_xy$ together and $|chi|-n$ terms of the form $(1-p_x)$ (look at the degreee). Thus
$$a_n=sum_=nprod_p_xin A p_xcdotprod_p_xnotin A(1-p_x)$$
answered Aug 14 at 11:23
b00n heT
8,08411430
add a comment |Â
up vote
1
down vote
accepted
I assume $|chi|<infty.$
If we are able to rewrite the product as follows
$$prod_xinchi(p_xy+(1-p_x))=sum_n=0^chia_ny^n$$
then the integral is very easy to compute and it results in
$$int_0^1prod_xinchi(p_xy+(1-p_x))dy=int_0^1sum_n=0^chia_ny^ndy=sum_n=0^chifraca_nn+1.$$
Now the terms $a_n$ are those where you multiply exactly $n$ $p_xy$ together and $|chi|-n$ terms of the form $(1-p_x)$ (look at the degreee). Thus
$$a_n=sum_=nprod_p_xin A p_xcdotprod_p_xnotin A(1-p_x)$$
answered Aug 14 at 11:23
b00n heT
8,08411430
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I assume $|chi|<infty.$
If we are able to rewrite the product as follows
$$prod_xinchi(p_xy+(1-p_x))=sum_n=0^chia_ny^n$$
then the integral is very easy to compute and it results in
$$int_0^1prod_xinchi(p_xy+(1-p_x))dy=int_0^1sum_n=0^chia_ny^ndy=sum_n=0^chifraca_nn+1.$$
Now the terms $a_n$ are those where you multiply exactly $n$ $p_xy$ together and $|chi|-n$ terms of the form $(1-p_x)$ (look at the degreee). Thus
$$a_n=sum_=nprod_p_xin A p_xcdotprod_p_xnotin A(1-p_x)$$
answered Aug 14 at 11:23
b00n heT
8,08411430
I assume $|chi|<infty.$
If we are able to rewrite the product as follows
$$prod_xinchi(p_xy+(1-p_x))=sum_n=0^chia_ny^n$$
then the integral is very easy to compute and it results in
$$int_0^1prod_xinchi(p_xy+(1-p_x))dy=int_0^1sum_n=0^chia_ny^ndy=sum_n=0^chifraca_nn+1.$$
Now the terms $a_n$ are those where you multiply exactly $n$ $p_xy$ together and $|chi|-n$ terms of the form $(1-p_x)$ (look at the degreee). Thus
$$a_n=sum_=nprod_p_xin A p_xcdotprod_p_xnotin A(1-p_x)$$
answered Aug 14 at 11:23
b00n heT
8,08411430
answered Aug 14 at 11:23
b00n heT
8,08411430
answered Aug 14 at 11:23
b00n heT
8,08411430
answered Aug 14 at 11:23
b00n heT
8,08411430
8,08411430
add a comment |Â
add a comment |Â
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2
Can you agree that $p_x=partial p/partial x$ and explain what $mathcalX$ here is?
– Fakemistake
Aug 14 at 11:14
@Fakemistake $p_x$ is just a constant. $mathcalX$ is just a set, the cardinality of which governs the degree of the polynomial whose monomials are of the form $(p_xy+(1-p_x))$.
– Television
Aug 14 at 11:16
2
The substitution $u = 1 - y$ simplifies the factors of the integrand to the form $1-p_x u$. The product can then be expanded as a polynomial in $u$, with coefficients expressed as elementary symmetric polynomials of the $p_i$. The integration of the expanded polynomial is straightforward.
– Blue
Aug 14 at 11:55