Solve a double summation in R
2
Is there any way of solving the following sum in R:
r
edited Nov 20 '18 at 18:18
m0nhawk
15.6k83262
asked Nov 20 '18 at 18:18
Nasir AbbasNasir Abbas
186
add a comment |
2
Is there any way of solving the following sum in R:
r
edited Nov 20 '18 at 18:18
m0nhawk
15.6k83262
asked Nov 20 '18 at 18:18
Nasir AbbasNasir Abbas
186
add a comment |
2
2
2
Is there any way of solving the following sum in R:
r
edited Nov 20 '18 at 18:18
m0nhawk
15.6k83262
asked Nov 20 '18 at 18:18
Nasir AbbasNasir Abbas
186
Is there any way of solving the following sum in R:
r
r
edited Nov 20 '18 at 18:18
m0nhawk
15.6k83262
asked Nov 20 '18 at 18:18
Nasir AbbasNasir Abbas
186
edited Nov 20 '18 at 18:18
m0nhawk
15.6k83262
asked Nov 20 '18 at 18:18
Nasir AbbasNasir Abbas
186
edited Nov 20 '18 at 18:18
m0nhawk
15.6k83262
edited Nov 20 '18 at 18:18
m0nhawk
15.6k83262
edited Nov 20 '18 at 18:18
m0nhawk
15.6k83262
15.6k83262
asked Nov 20 '18 at 18:18
Nasir AbbasNasir Abbas
186
asked Nov 20 '18 at 18:18
Nasir AbbasNasir Abbas
186
asked Nov 20 '18 at 18:18
Nasir AbbasNasir Abbas
186
186
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
4
You can calculate this without any for loops:
double_sum <- function(j) {
sum(sapply(1:j, function(i) sum(1/i:j))^2) / j^2
}
And then calculate for each j:
> sapply(1:50, outer_sum)
[1] 1.00000000 0.62500000 0.46296296 0.36979167 0.30866667 0.26527778 0.23279883 0.20753348 0.18729669 0.17071032
[11] 0.15686052 0.14511659 0.13502879 0.12626754 0.11858565 0.11179403 0.10574549 0.10032374 0.09543562 0.09100565
[21] 0.08697198 0.08328344 0.07989737 0.07677785 0.07389447 0.07122127 0.06873600 0.06641942 0.06425487 0.06222779
[31] 0.06032545 0.05853663 0.05685142 0.05526106 0.05375773 0.05233445 0.05098496 0.04970367 0.04848551 0.04732591
[41] 0.04622074 0.04516625 0.04415901 0.04319591 0.04227410 0.04139098 0.04054415 0.03973142 0.03895077 0.03820032
Or something strange (built upper triangle matrix for coefficient and then sum rows and results):
mat_sum <- function(j) {
d <- outer(rep(1, j), 1:j, FUN="/")
d[lower.tri(d)] <- 0
sum(rowSums(d)^2) / j^2
}
And benchmarks:
> s <- 1:100
> microbenchmark::microbenchmark(for_sum=sapply(s, sumfun), double_sum=sapply(s, double_sum), mat_sum=sapply(s, mat_sum))
Unit: milliseconds
expr min lq mean median uq max neval
for_sum 9.601222 10.261159 11.996525 10.774037 11.894962 30.56077 100
double_sum 6.075801 6.678923 8.787946 7.373223 8.697266 21.37783 100
mat_sum 7.809572 8.770058 13.766358 10.190758 18.500802 46.18336 100
answered Nov 20 '18 at 18:40
m0nhawkm0nhawk
15.6k83262
Manually I checked, the second value of sum (for j=2) is 0.375. There is something wrong with the code.
– Nasir Abbas
Nov 21 '18 at 5:31
In Wolfram Mathematica I'm getting the same result, as in R. The exact value forj=2is5/8. Check the Wolfram|Alpha.
– m0nhawk
Nov 21 '18 at 16:10
Please, recheck your calculations, you took the square of each term, but you need to take the square of the whole inner sum. Here is the result forj=2:(1/2 + 1/4)^2 + (1/4)^2 = 5/8, and you calculated:(1/2)^2 + (1/4)^2 + (1/4)^2 = 3/8.
– m0nhawk
Nov 21 '18 at 16:41
add a comment |
2
sumfun <- function(j) {
res <- 0
for(i in 1:j) {
temp <- 0
for(k in i:j) {
temp <- temp + 1/(k*j)
}
res <- res + temp^2
}
return(res)
}
sapply(1:50, sumfun)
[1] 1.00000000 0.62500000 0.46296296 0.36979167 0.30866667 0.26527778 0.23279883 0.20753348 0.18729669 0.17071032 0.15686052
[12] 0.14511659 0.13502879 0.12626754 0.11858565 0.11179403 0.10574549 0.10032374 0.09543562 0.09100565 0.08697198 0.08328344
[23] 0.07989737 0.07677785 0.07389447 0.07122127 0.06873600 0.06641942 0.06425487 0.06222779 0.06032545 0.05853663 0.05685142
[34] 0.05526106 0.05375773 0.05233445 0.05098496 0.04970367 0.04848551 0.04732591 0.04622074 0.04516625 0.04415901 0.04319591
[45] 0.04227410 0.04139098 0.04054415 0.03973142 0.03895077 0.03820032
answered Nov 20 '18 at 18:32
Dan YDan Y
3,7211627
Manually I checked, the second value of sum (for j=2) is 0.375. There is something wrong with the code.
– Nasir Abbas
Nov 21 '18 at 5:33
Check your work. The double sum for J=2 is (1/2 + 1/4)^2 + 1/4^2 = 10/16 = 0.625.
– Dan Y
Nov 21 '18 at 5:45
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
4
You can calculate this without any for loops:
double_sum <- function(j) {
sum(sapply(1:j, function(i) sum(1/i:j))^2) / j^2
}
And then calculate for each j:
> sapply(1:50, outer_sum)
[1] 1.00000000 0.62500000 0.46296296 0.36979167 0.30866667 0.26527778 0.23279883 0.20753348 0.18729669 0.17071032
[11] 0.15686052 0.14511659 0.13502879 0.12626754 0.11858565 0.11179403 0.10574549 0.10032374 0.09543562 0.09100565
[21] 0.08697198 0.08328344 0.07989737 0.07677785 0.07389447 0.07122127 0.06873600 0.06641942 0.06425487 0.06222779
[31] 0.06032545 0.05853663 0.05685142 0.05526106 0.05375773 0.05233445 0.05098496 0.04970367 0.04848551 0.04732591
[41] 0.04622074 0.04516625 0.04415901 0.04319591 0.04227410 0.04139098 0.04054415 0.03973142 0.03895077 0.03820032
Or something strange (built upper triangle matrix for coefficient and then sum rows and results):
mat_sum <- function(j) {
d <- outer(rep(1, j), 1:j, FUN="/")
d[lower.tri(d)] <- 0
sum(rowSums(d)^2) / j^2
}
And benchmarks:
> s <- 1:100
> microbenchmark::microbenchmark(for_sum=sapply(s, sumfun), double_sum=sapply(s, double_sum), mat_sum=sapply(s, mat_sum))
Unit: milliseconds
expr min lq mean median uq max neval
for_sum 9.601222 10.261159 11.996525 10.774037 11.894962 30.56077 100
double_sum 6.075801 6.678923 8.787946 7.373223 8.697266 21.37783 100
mat_sum 7.809572 8.770058 13.766358 10.190758 18.500802 46.18336 100
answered Nov 20 '18 at 18:40
m0nhawkm0nhawk
15.6k83262
Manually I checked, the second value of sum (for j=2) is 0.375. There is something wrong with the code.
– Nasir Abbas
Nov 21 '18 at 5:31
In Wolfram Mathematica I'm getting the same result, as in R. The exact value forj=2is5/8. Check the Wolfram|Alpha.
– m0nhawk
Nov 21 '18 at 16:10
Please, recheck your calculations, you took the square of each term, but you need to take the square of the whole inner sum. Here is the result forj=2:(1/2 + 1/4)^2 + (1/4)^2 = 5/8, and you calculated:(1/2)^2 + (1/4)^2 + (1/4)^2 = 3/8.
– m0nhawk
Nov 21 '18 at 16:41
add a comment |
4
You can calculate this without any for loops:
double_sum <- function(j) {
sum(sapply(1:j, function(i) sum(1/i:j))^2) / j^2
}
And then calculate for each j:
> sapply(1:50, outer_sum)
[1] 1.00000000 0.62500000 0.46296296 0.36979167 0.30866667 0.26527778 0.23279883 0.20753348 0.18729669 0.17071032
[11] 0.15686052 0.14511659 0.13502879 0.12626754 0.11858565 0.11179403 0.10574549 0.10032374 0.09543562 0.09100565
[21] 0.08697198 0.08328344 0.07989737 0.07677785 0.07389447 0.07122127 0.06873600 0.06641942 0.06425487 0.06222779
[31] 0.06032545 0.05853663 0.05685142 0.05526106 0.05375773 0.05233445 0.05098496 0.04970367 0.04848551 0.04732591
[41] 0.04622074 0.04516625 0.04415901 0.04319591 0.04227410 0.04139098 0.04054415 0.03973142 0.03895077 0.03820032
Or something strange (built upper triangle matrix for coefficient and then sum rows and results):
mat_sum <- function(j) {
d <- outer(rep(1, j), 1:j, FUN="/")
d[lower.tri(d)] <- 0
sum(rowSums(d)^2) / j^2
}
And benchmarks:
> s <- 1:100
> microbenchmark::microbenchmark(for_sum=sapply(s, sumfun), double_sum=sapply(s, double_sum), mat_sum=sapply(s, mat_sum))
Unit: milliseconds
expr min lq mean median uq max neval
for_sum 9.601222 10.261159 11.996525 10.774037 11.894962 30.56077 100
double_sum 6.075801 6.678923 8.787946 7.373223 8.697266 21.37783 100
mat_sum 7.809572 8.770058 13.766358 10.190758 18.500802 46.18336 100
answered Nov 20 '18 at 18:40
m0nhawkm0nhawk
15.6k83262
Manually I checked, the second value of sum (for j=2) is 0.375. There is something wrong with the code.
– Nasir Abbas
Nov 21 '18 at 5:31
In Wolfram Mathematica I'm getting the same result, as in R. The exact value forj=2is5/8. Check the Wolfram|Alpha.
– m0nhawk
Nov 21 '18 at 16:10
Please, recheck your calculations, you took the square of each term, but you need to take the square of the whole inner sum. Here is the result forj=2:(1/2 + 1/4)^2 + (1/4)^2 = 5/8, and you calculated:(1/2)^2 + (1/4)^2 + (1/4)^2 = 3/8.
– m0nhawk
Nov 21 '18 at 16:41
add a comment |
4
4
4
You can calculate this without any for loops:
double_sum <- function(j) {
sum(sapply(1:j, function(i) sum(1/i:j))^2) / j^2
}
And then calculate for each j:
> sapply(1:50, outer_sum)
[1] 1.00000000 0.62500000 0.46296296 0.36979167 0.30866667 0.26527778 0.23279883 0.20753348 0.18729669 0.17071032
[11] 0.15686052 0.14511659 0.13502879 0.12626754 0.11858565 0.11179403 0.10574549 0.10032374 0.09543562 0.09100565
[21] 0.08697198 0.08328344 0.07989737 0.07677785 0.07389447 0.07122127 0.06873600 0.06641942 0.06425487 0.06222779
[31] 0.06032545 0.05853663 0.05685142 0.05526106 0.05375773 0.05233445 0.05098496 0.04970367 0.04848551 0.04732591
[41] 0.04622074 0.04516625 0.04415901 0.04319591 0.04227410 0.04139098 0.04054415 0.03973142 0.03895077 0.03820032
Or something strange (built upper triangle matrix for coefficient and then sum rows and results):
mat_sum <- function(j) {
d <- outer(rep(1, j), 1:j, FUN="/")
d[lower.tri(d)] <- 0
sum(rowSums(d)^2) / j^2
}
And benchmarks:
> s <- 1:100
> microbenchmark::microbenchmark(for_sum=sapply(s, sumfun), double_sum=sapply(s, double_sum), mat_sum=sapply(s, mat_sum))
Unit: milliseconds
expr min lq mean median uq max neval
for_sum 9.601222 10.261159 11.996525 10.774037 11.894962 30.56077 100
double_sum 6.075801 6.678923 8.787946 7.373223 8.697266 21.37783 100
mat_sum 7.809572 8.770058 13.766358 10.190758 18.500802 46.18336 100
answered Nov 20 '18 at 18:40
m0nhawkm0nhawk
15.6k83262
You can calculate this without any for loops:
double_sum <- function(j) {
sum(sapply(1:j, function(i) sum(1/i:j))^2) / j^2
}
And then calculate for each j:
> sapply(1:50, outer_sum)
[1] 1.00000000 0.62500000 0.46296296 0.36979167 0.30866667 0.26527778 0.23279883 0.20753348 0.18729669 0.17071032
[11] 0.15686052 0.14511659 0.13502879 0.12626754 0.11858565 0.11179403 0.10574549 0.10032374 0.09543562 0.09100565
[21] 0.08697198 0.08328344 0.07989737 0.07677785 0.07389447 0.07122127 0.06873600 0.06641942 0.06425487 0.06222779
[31] 0.06032545 0.05853663 0.05685142 0.05526106 0.05375773 0.05233445 0.05098496 0.04970367 0.04848551 0.04732591
[41] 0.04622074 0.04516625 0.04415901 0.04319591 0.04227410 0.04139098 0.04054415 0.03973142 0.03895077 0.03820032
Or something strange (built upper triangle matrix for coefficient and then sum rows and results):
mat_sum <- function(j) {
d <- outer(rep(1, j), 1:j, FUN="/")
d[lower.tri(d)] <- 0
sum(rowSums(d)^2) / j^2
}
And benchmarks:
> s <- 1:100
> microbenchmark::microbenchmark(for_sum=sapply(s, sumfun), double_sum=sapply(s, double_sum), mat_sum=sapply(s, mat_sum))
Unit: milliseconds
expr min lq mean median uq max neval
for_sum 9.601222 10.261159 11.996525 10.774037 11.894962 30.56077 100
double_sum 6.075801 6.678923 8.787946 7.373223 8.697266 21.37783 100
mat_sum 7.809572 8.770058 13.766358 10.190758 18.500802 46.18336 100
answered Nov 20 '18 at 18:40
m0nhawkm0nhawk
15.6k83262
edited Nov 20 '18 at 19:19
answered Nov 20 '18 at 18:40
m0nhawkm0nhawk
15.6k83262
answered Nov 20 '18 at 18:40
m0nhawkm0nhawk
15.6k83262
answered Nov 20 '18 at 18:40
m0nhawkm0nhawk
15.6k83262
15.6k83262
Manually I checked, the second value of sum (for j=2) is 0.375. There is something wrong with the code.
– Nasir Abbas
Nov 21 '18 at 5:31
In Wolfram Mathematica I'm getting the same result, as in R. The exact value forj=2is5/8. Check the Wolfram|Alpha.
– m0nhawk
Nov 21 '18 at 16:10
Please, recheck your calculations, you took the square of each term, but you need to take the square of the whole inner sum. Here is the result forj=2:(1/2 + 1/4)^2 + (1/4)^2 = 5/8, and you calculated:(1/2)^2 + (1/4)^2 + (1/4)^2 = 3/8.
– m0nhawk
Nov 21 '18 at 16:41
add a comment |
Manually I checked, the second value of sum (for j=2) is 0.375. There is something wrong with the code.
– Nasir Abbas
Nov 21 '18 at 5:31
In Wolfram Mathematica I'm getting the same result, as in R. The exact value forj=2is5/8. Check the Wolfram|Alpha.
– m0nhawk
Nov 21 '18 at 16:10
Please, recheck your calculations, you took the square of each term, but you need to take the square of the whole inner sum. Here is the result forj=2:(1/2 + 1/4)^2 + (1/4)^2 = 5/8, and you calculated:(1/2)^2 + (1/4)^2 + (1/4)^2 = 3/8.
– m0nhawk
Nov 21 '18 at 16:41
Manually I checked, the second value of sum (for j=2) is 0.375. There is something wrong with the code.
– Nasir Abbas
Nov 21 '18 at 5:31
Manually I checked, the second value of sum (for j=2) is 0.375. There is something wrong with the code.
– Nasir Abbas
Nov 21 '18 at 5:31
In Wolfram Mathematica I'm getting the same result, as in R. The exact value for
j=2 is 5/8. Check the Wolfram|Alpha.– m0nhawk
Nov 21 '18 at 16:10
In Wolfram Mathematica I'm getting the same result, as in R. The exact value for
j=2 is 5/8. Check the Wolfram|Alpha.– m0nhawk
Nov 21 '18 at 16:10
Please, recheck your calculations, you took the square of each term, but you need to take the square of the whole inner sum. Here is the result for
j=2: (1/2 + 1/4)^2 + (1/4)^2 = 5/8, and you calculated: (1/2)^2 + (1/4)^2 + (1/4)^2 = 3/8.– m0nhawk
Nov 21 '18 at 16:41
Please, recheck your calculations, you took the square of each term, but you need to take the square of the whole inner sum. Here is the result for
j=2: (1/2 + 1/4)^2 + (1/4)^2 = 5/8, and you calculated: (1/2)^2 + (1/4)^2 + (1/4)^2 = 3/8.– m0nhawk
Nov 21 '18 at 16:41
add a comment |
2
sumfun <- function(j) {
res <- 0
for(i in 1:j) {
temp <- 0
for(k in i:j) {
temp <- temp + 1/(k*j)
}
res <- res + temp^2
}
return(res)
}
sapply(1:50, sumfun)
[1] 1.00000000 0.62500000 0.46296296 0.36979167 0.30866667 0.26527778 0.23279883 0.20753348 0.18729669 0.17071032 0.15686052
[12] 0.14511659 0.13502879 0.12626754 0.11858565 0.11179403 0.10574549 0.10032374 0.09543562 0.09100565 0.08697198 0.08328344
[23] 0.07989737 0.07677785 0.07389447 0.07122127 0.06873600 0.06641942 0.06425487 0.06222779 0.06032545 0.05853663 0.05685142
[34] 0.05526106 0.05375773 0.05233445 0.05098496 0.04970367 0.04848551 0.04732591 0.04622074 0.04516625 0.04415901 0.04319591
[45] 0.04227410 0.04139098 0.04054415 0.03973142 0.03895077 0.03820032
answered Nov 20 '18 at 18:32
Dan YDan Y
3,7211627
Manually I checked, the second value of sum (for j=2) is 0.375. There is something wrong with the code.
– Nasir Abbas
Nov 21 '18 at 5:33
Check your work. The double sum for J=2 is (1/2 + 1/4)^2 + 1/4^2 = 10/16 = 0.625.
– Dan Y
Nov 21 '18 at 5:45
add a comment |
2
sumfun <- function(j) {
res <- 0
for(i in 1:j) {
temp <- 0
for(k in i:j) {
temp <- temp + 1/(k*j)
}
res <- res + temp^2
}
return(res)
}
sapply(1:50, sumfun)
[1] 1.00000000 0.62500000 0.46296296 0.36979167 0.30866667 0.26527778 0.23279883 0.20753348 0.18729669 0.17071032 0.15686052
[12] 0.14511659 0.13502879 0.12626754 0.11858565 0.11179403 0.10574549 0.10032374 0.09543562 0.09100565 0.08697198 0.08328344
[23] 0.07989737 0.07677785 0.07389447 0.07122127 0.06873600 0.06641942 0.06425487 0.06222779 0.06032545 0.05853663 0.05685142
[34] 0.05526106 0.05375773 0.05233445 0.05098496 0.04970367 0.04848551 0.04732591 0.04622074 0.04516625 0.04415901 0.04319591
[45] 0.04227410 0.04139098 0.04054415 0.03973142 0.03895077 0.03820032
answered Nov 20 '18 at 18:32
Dan YDan Y
3,7211627
Manually I checked, the second value of sum (for j=2) is 0.375. There is something wrong with the code.
– Nasir Abbas
Nov 21 '18 at 5:33
Check your work. The double sum for J=2 is (1/2 + 1/4)^2 + 1/4^2 = 10/16 = 0.625.
– Dan Y
Nov 21 '18 at 5:45
add a comment |
2
2
2
sumfun <- function(j) {
res <- 0
for(i in 1:j) {
temp <- 0
for(k in i:j) {
temp <- temp + 1/(k*j)
}
res <- res + temp^2
}
return(res)
}
sapply(1:50, sumfun)
[1] 1.00000000 0.62500000 0.46296296 0.36979167 0.30866667 0.26527778 0.23279883 0.20753348 0.18729669 0.17071032 0.15686052
[12] 0.14511659 0.13502879 0.12626754 0.11858565 0.11179403 0.10574549 0.10032374 0.09543562 0.09100565 0.08697198 0.08328344
[23] 0.07989737 0.07677785 0.07389447 0.07122127 0.06873600 0.06641942 0.06425487 0.06222779 0.06032545 0.05853663 0.05685142
[34] 0.05526106 0.05375773 0.05233445 0.05098496 0.04970367 0.04848551 0.04732591 0.04622074 0.04516625 0.04415901 0.04319591
[45] 0.04227410 0.04139098 0.04054415 0.03973142 0.03895077 0.03820032
answered Nov 20 '18 at 18:32
Dan YDan Y
3,7211627
sumfun <- function(j) {
res <- 0
for(i in 1:j) {
temp <- 0
for(k in i:j) {
temp <- temp + 1/(k*j)
}
res <- res + temp^2
}
return(res)
}
sapply(1:50, sumfun)
[1] 1.00000000 0.62500000 0.46296296 0.36979167 0.30866667 0.26527778 0.23279883 0.20753348 0.18729669 0.17071032 0.15686052
[12] 0.14511659 0.13502879 0.12626754 0.11858565 0.11179403 0.10574549 0.10032374 0.09543562 0.09100565 0.08697198 0.08328344
[23] 0.07989737 0.07677785 0.07389447 0.07122127 0.06873600 0.06641942 0.06425487 0.06222779 0.06032545 0.05853663 0.05685142
[34] 0.05526106 0.05375773 0.05233445 0.05098496 0.04970367 0.04848551 0.04732591 0.04622074 0.04516625 0.04415901 0.04319591
[45] 0.04227410 0.04139098 0.04054415 0.03973142 0.03895077 0.03820032
answered Nov 20 '18 at 18:32
Dan YDan Y
3,7211627
edited Nov 20 '18 at 18:44
answered Nov 20 '18 at 18:32
Dan YDan Y
3,7211627
answered Nov 20 '18 at 18:32
Dan YDan Y
3,7211627
answered Nov 20 '18 at 18:32
Dan YDan Y
3,7211627
3,7211627
Manually I checked, the second value of sum (for j=2) is 0.375. There is something wrong with the code.
– Nasir Abbas
Nov 21 '18 at 5:33
Check your work. The double sum for J=2 is (1/2 + 1/4)^2 + 1/4^2 = 10/16 = 0.625.
– Dan Y
Nov 21 '18 at 5:45
add a comment |
Manually I checked, the second value of sum (for j=2) is 0.375. There is something wrong with the code.
– Nasir Abbas
Nov 21 '18 at 5:33
Check your work. The double sum for J=2 is (1/2 + 1/4)^2 + 1/4^2 = 10/16 = 0.625.
– Dan Y
Nov 21 '18 at 5:45
Manually I checked, the second value of sum (for j=2) is 0.375. There is something wrong with the code.
– Nasir Abbas
Nov 21 '18 at 5:33
Manually I checked, the second value of sum (for j=2) is 0.375. There is something wrong with the code.
– Nasir Abbas
Nov 21 '18 at 5:33
Check your work. The double sum for J=2 is (1/2 + 1/4)^2 + 1/4^2 = 10/16 = 0.625.
– Dan Y
Nov 21 '18 at 5:45
Check your work. The double sum for J=2 is (1/2 + 1/4)^2 + 1/4^2 = 10/16 = 0.625.
– Dan Y
Nov 21 '18 at 5:45
add a comment |
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