Mixing
Can someone clearly explain the algebra involved in this sum and integral?
$begingroup$
We have :
$$int frac{1}{x} sum_{m=1}^infty frac{(x y)^m}{1-y^m}
dx = sum_{m=1}^infty frac{y^m}{1-y^m} int x^{m-1}dx= sum_{m=1}^infty frac{y^m}{1-y^m} frac{x^m}{m} = sum_{m=1}^infty frac{(x y)^m}{(1-y^m)m}$$
From Sotiris' answer to a previous question here, based off: Jjacqueline's answer to a similar question here.
I think I could follow and maybe improve this if I understood a bit more about how we swap out the integral and the summation signs in the line I mentioned above. I get up to: $$sum_{m=1}^infty int{frac{y^m}{1-y^m} x^{m-1}}dx$$ How does $$int frac{1}{x} sum_{m=1}^infty frac{(x y)^m}{1-y^m} dx$$ come from that? I ask because I don't understand why multiple integrations would be needed in the case of the $frac{1}{m!}$ hypothetical I asked Sotiris about in the first link of this question, knowing the partial derivative of $frac{x^m}{m!}$ with respect to $m$ is this.
integration sequences-and-series gamma-function
asked Jan 27 at 2:13
user3108815user3108815
569
$endgroup$
add a comment |
$begingroup$
We have :
$$int frac{1}{x} sum_{m=1}^infty frac{(x y)^m}{1-y^m}
dx = sum_{m=1}^infty frac{y^m}{1-y^m} int x^{m-1}dx= sum_{m=1}^infty frac{y^m}{1-y^m} frac{x^m}{m} = sum_{m=1}^infty frac{(x y)^m}{(1-y^m)m}$$
From Sotiris' answer to a previous question here, based off: Jjacqueline's answer to a similar question here.
I think I could follow and maybe improve this if I understood a bit more about how we swap out the integral and the summation signs in the line I mentioned above. I get up to: $$sum_{m=1}^infty int{frac{y^m}{1-y^m} x^{m-1}}dx$$ How does $$int frac{1}{x} sum_{m=1}^infty frac{(x y)^m}{1-y^m} dx$$ come from that? I ask because I don't understand why multiple integrations would be needed in the case of the $frac{1}{m!}$ hypothetical I asked Sotiris about in the first link of this question, knowing the partial derivative of $frac{x^m}{m!}$ with respect to $m$ is this.
integration sequences-and-series gamma-function
asked Jan 27 at 2:13
user3108815user3108815
569
$endgroup$
1
$begingroup$
Hint: Treat the variable $y$ as a constant with respect to the variable $x$.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 27 at 3:15
add a comment |
$begingroup$
We have :
$$int frac{1}{x} sum_{m=1}^infty frac{(x y)^m}{1-y^m}
dx = sum_{m=1}^infty frac{y^m}{1-y^m} int x^{m-1}dx= sum_{m=1}^infty frac{y^m}{1-y^m} frac{x^m}{m} = sum_{m=1}^infty frac{(x y)^m}{(1-y^m)m}$$
From Sotiris' answer to a previous question here, based off: Jjacqueline's answer to a similar question here.
I think I could follow and maybe improve this if I understood a bit more about how we swap out the integral and the summation signs in the line I mentioned above. I get up to: $$sum_{m=1}^infty int{frac{y^m}{1-y^m} x^{m-1}}dx$$ How does $$int frac{1}{x} sum_{m=1}^infty frac{(x y)^m}{1-y^m} dx$$ come from that? I ask because I don't understand why multiple integrations would be needed in the case of the $frac{1}{m!}$ hypothetical I asked Sotiris about in the first link of this question, knowing the partial derivative of $frac{x^m}{m!}$ with respect to $m$ is this.
integration sequences-and-series gamma-function
asked Jan 27 at 2:13
user3108815user3108815
569
$endgroup$
We have :
$$int frac{1}{x} sum_{m=1}^infty frac{(x y)^m}{1-y^m}
dx = sum_{m=1}^infty frac{y^m}{1-y^m} int x^{m-1}dx= sum_{m=1}^infty frac{y^m}{1-y^m} frac{x^m}{m} = sum_{m=1}^infty frac{(x y)^m}{(1-y^m)m}$$
From Sotiris' answer to a previous question here, based off: Jjacqueline's answer to a similar question here.
I think I could follow and maybe improve this if I understood a bit more about how we swap out the integral and the summation signs in the line I mentioned above. I get up to: $$sum_{m=1}^infty int{frac{y^m}{1-y^m} x^{m-1}}dx$$ How does $$int frac{1}{x} sum_{m=1}^infty frac{(x y)^m}{1-y^m} dx$$ come from that? I ask because I don't understand why multiple integrations would be needed in the case of the $frac{1}{m!}$ hypothetical I asked Sotiris about in the first link of this question, knowing the partial derivative of $frac{x^m}{m!}$ with respect to $m$ is this.
integration sequences-and-series gamma-function
integration sequences-and-series gamma-function
asked Jan 27 at 2:13
user3108815user3108815
569
asked Jan 27 at 2:13
user3108815user3108815
569
edited Jan 27 at 3:07
user3108815
asked Jan 27 at 2:13
user3108815user3108815
569
asked Jan 27 at 2:13
user3108815user3108815
569
asked Jan 27 at 2:13
user3108815user3108815
569
569
1
$begingroup$
Hint: Treat the variable $y$ as a constant with respect to the variable $x$.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 27 at 3:15
add a comment |
1
$begingroup$
Hint: Treat the variable $y$ as a constant with respect to the variable $x$.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 27 at 3:15
1
1
$begingroup$
Hint: Treat the variable $y$ as a constant with respect to the variable $x$.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 27 at 3:15
$begingroup$
Hint: Treat the variable $y$ as a constant with respect to the variable $x$.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 27 at 3:15
add a comment |
1 Answer
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$begingroup$
I somehow forgot that the integral of polynomial summands is just the integral of the sum of the terms.
i.e:
$$if f(i) = a_i,$$ and $$g(i) = a_ix^i= f(i)x^i,$$
then $$sum_{i=1}^nbigr(int a_ix^i dxbigl) =int a_1x dx+int a_2x^2 dx + int a_3x^3 dx+...+int a_nx^n dx$$ $$ =int a_1x + a_2x^2 + a_3x^3 +...+a_nx^n dx$$
$$= int (sum_{i=1}^{n}f(i)x^i) dx$$
$$= int bigl(sum_{i=1}^{n}g(i)bigr) dx$$
answered Jan 27 at 9:25
user3108815user3108815
569
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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$begingroup$
I somehow forgot that the integral of polynomial summands is just the integral of the sum of the terms.
i.e:
$$if f(i) = a_i,$$ and $$g(i) = a_ix^i= f(i)x^i,$$
then $$sum_{i=1}^nbigr(int a_ix^i dxbigl) =int a_1x dx+int a_2x^2 dx + int a_3x^3 dx+...+int a_nx^n dx$$ $$ =int a_1x + a_2x^2 + a_3x^3 +...+a_nx^n dx$$
$$= int (sum_{i=1}^{n}f(i)x^i) dx$$
$$= int bigl(sum_{i=1}^{n}g(i)bigr) dx$$
answered Jan 27 at 9:25
user3108815user3108815
569
$endgroup$
add a comment |
$begingroup$
I somehow forgot that the integral of polynomial summands is just the integral of the sum of the terms.
i.e:
$$if f(i) = a_i,$$ and $$g(i) = a_ix^i= f(i)x^i,$$
then $$sum_{i=1}^nbigr(int a_ix^i dxbigl) =int a_1x dx+int a_2x^2 dx + int a_3x^3 dx+...+int a_nx^n dx$$ $$ =int a_1x + a_2x^2 + a_3x^3 +...+a_nx^n dx$$
$$= int (sum_{i=1}^{n}f(i)x^i) dx$$
$$= int bigl(sum_{i=1}^{n}g(i)bigr) dx$$
answered Jan 27 at 9:25
user3108815user3108815
569
$endgroup$
add a comment |
$begingroup$
I somehow forgot that the integral of polynomial summands is just the integral of the sum of the terms.
i.e:
$$if f(i) = a_i,$$ and $$g(i) = a_ix^i= f(i)x^i,$$
then $$sum_{i=1}^nbigr(int a_ix^i dxbigl) =int a_1x dx+int a_2x^2 dx + int a_3x^3 dx+...+int a_nx^n dx$$ $$ =int a_1x + a_2x^2 + a_3x^3 +...+a_nx^n dx$$
$$= int (sum_{i=1}^{n}f(i)x^i) dx$$
$$= int bigl(sum_{i=1}^{n}g(i)bigr) dx$$
answered Jan 27 at 9:25
user3108815user3108815
569
$endgroup$
I somehow forgot that the integral of polynomial summands is just the integral of the sum of the terms.
i.e:
$$if f(i) = a_i,$$ and $$g(i) = a_ix^i= f(i)x^i,$$
then $$sum_{i=1}^nbigr(int a_ix^i dxbigl) =int a_1x dx+int a_2x^2 dx + int a_3x^3 dx+...+int a_nx^n dx$$ $$ =int a_1x + a_2x^2 + a_3x^3 +...+a_nx^n dx$$
$$= int (sum_{i=1}^{n}f(i)x^i) dx$$
$$= int bigl(sum_{i=1}^{n}g(i)bigr) dx$$
answered Jan 27 at 9:25
user3108815user3108815
569
edited Jan 30 at 16:16
answered Jan 27 at 9:25
user3108815user3108815
569
answered Jan 27 at 9:25
user3108815user3108815
569
answered Jan 27 at 9:25
user3108815user3108815
569
569
add a comment |
add a comment |
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$begingroup$
Hint: Treat the variable $y$ as a constant with respect to the variable $x$.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 27 at 3:15