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Splitting of a sum involving a ceiling function
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I want to calculate a sum involving a ceiling function in its argument, by splitting the sum into several sums, and replacing the ceiling function by its evaluated value within the intervals of these separate sums.
Example with two sums for ceiling values of $2n+1$ and $2n+2$:
$$sum_m=1^2n+1 b^-biglceil2sqrtn^2+mbigrceil = sum_m=1^n b^-2n-1 +sum_m=n+1^2n+1 b^-2n-2$$
A more general case would be (for ceiling values of $an+1$ until $an+a$):
$$sum_m=1^2n+1 b^-biglceil asqrtn^2+mbigrceil = sum_m=1^? b^-an-1 + sum_m=?^? b^-an-2 + cdots + sum_m=?^2n+1 b^-an-a$$
The problem in this case is, that I don't know what would be the integral summation limits? I could state them by some abstract floor and ceiling functions, but this would introduce floor and ceiling expressions again into the calculation, that is exactly the thing that I would like to circumvent. I thought about the possibility of 'extending' the whole summation interval in some way, according to the choice of $a$, but the upper limit $2n+1$ is connected to the square root part of the expression, since ultimately the expression should be summed again by an outer sum of $sum_n = 0^infty$.
Thanks for any ideas.
summation ceiling-function
edited Jul 25 at 16:20
Xander Henderson
13k83150
asked Jul 25 at 16:15
Iridium
113
add a comment |Â
up vote
0
down vote
favorite
I want to calculate a sum involving a ceiling function in its argument, by splitting the sum into several sums, and replacing the ceiling function by its evaluated value within the intervals of these separate sums.
Example with two sums for ceiling values of $2n+1$ and $2n+2$:
$$sum_m=1^2n+1 b^-biglceil2sqrtn^2+mbigrceil = sum_m=1^n b^-2n-1 +sum_m=n+1^2n+1 b^-2n-2$$
A more general case would be (for ceiling values of $an+1$ until $an+a$):
$$sum_m=1^2n+1 b^-biglceil asqrtn^2+mbigrceil = sum_m=1^? b^-an-1 + sum_m=?^? b^-an-2 + cdots + sum_m=?^2n+1 b^-an-a$$
The problem in this case is, that I don't know what would be the integral summation limits? I could state them by some abstract floor and ceiling functions, but this would introduce floor and ceiling expressions again into the calculation, that is exactly the thing that I would like to circumvent. I thought about the possibility of 'extending' the whole summation interval in some way, according to the choice of $a$, but the upper limit $2n+1$ is connected to the square root part of the expression, since ultimately the expression should be summed again by an outer sum of $sum_n = 0^infty$.
Thanks for any ideas.
summation ceiling-function
edited Jul 25 at 16:20
Xander Henderson
13k83150
asked Jul 25 at 16:15
Iridium
113
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to calculate a sum involving a ceiling function in its argument, by splitting the sum into several sums, and replacing the ceiling function by its evaluated value within the intervals of these separate sums.
Example with two sums for ceiling values of $2n+1$ and $2n+2$:
$$sum_m=1^2n+1 b^-biglceil2sqrtn^2+mbigrceil = sum_m=1^n b^-2n-1 +sum_m=n+1^2n+1 b^-2n-2$$
A more general case would be (for ceiling values of $an+1$ until $an+a$):
$$sum_m=1^2n+1 b^-biglceil asqrtn^2+mbigrceil = sum_m=1^? b^-an-1 + sum_m=?^? b^-an-2 + cdots + sum_m=?^2n+1 b^-an-a$$
The problem in this case is, that I don't know what would be the integral summation limits? I could state them by some abstract floor and ceiling functions, but this would introduce floor and ceiling expressions again into the calculation, that is exactly the thing that I would like to circumvent. I thought about the possibility of 'extending' the whole summation interval in some way, according to the choice of $a$, but the upper limit $2n+1$ is connected to the square root part of the expression, since ultimately the expression should be summed again by an outer sum of $sum_n = 0^infty$.
Thanks for any ideas.
summation ceiling-function
edited Jul 25 at 16:20
Xander Henderson
13k83150
asked Jul 25 at 16:15
Iridium
113
I want to calculate a sum involving a ceiling function in its argument, by splitting the sum into several sums, and replacing the ceiling function by its evaluated value within the intervals of these separate sums.
Example with two sums for ceiling values of $2n+1$ and $2n+2$:
$$sum_m=1^2n+1 b^-biglceil2sqrtn^2+mbigrceil = sum_m=1^n b^-2n-1 +sum_m=n+1^2n+1 b^-2n-2$$
A more general case would be (for ceiling values of $an+1$ until $an+a$):
$$sum_m=1^2n+1 b^-biglceil asqrtn^2+mbigrceil = sum_m=1^? b^-an-1 + sum_m=?^? b^-an-2 + cdots + sum_m=?^2n+1 b^-an-a$$
The problem in this case is, that I don't know what would be the integral summation limits? I could state them by some abstract floor and ceiling functions, but this would introduce floor and ceiling expressions again into the calculation, that is exactly the thing that I would like to circumvent. I thought about the possibility of 'extending' the whole summation interval in some way, according to the choice of $a$, but the upper limit $2n+1$ is connected to the square root part of the expression, since ultimately the expression should be summed again by an outer sum of $sum_n = 0^infty$.
Thanks for any ideas.
summation ceiling-function
edited Jul 25 at 16:20
Xander Henderson
13k83150
asked Jul 25 at 16:15
Iridium
113
edited Jul 25 at 16:20
Xander Henderson
13k83150
edited Jul 25 at 16:20
Xander Henderson
13k83150
edited Jul 25 at 16:20
Xander Henderson
13k83150
13k83150
asked Jul 25 at 16:15
Iridium
113
asked Jul 25 at 16:15
Iridium
113
asked Jul 25 at 16:15
Iridium
113
113
add a comment |Â
add a comment |Â
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