If and summation: why do I have the index of the summation in the final result?
Consider the following code:
Sum[If[b != 0, a[m], 0], {m, 1, 4}]
(* 4 If[b != 0, a[m], 0] *)
Why does it return me a[m]? m is the index of the summation, it shouldn't appear in the result.
Here I expect to have:
If[b != 0, a[1], 0]+If[b != 0, a[2], 0]+If[b != 0, a[3], 0]+If[b != 0, a[4], 0]
I don't understand this behavior
(my example here is to understand this behavior, my specific problem is more complicated but the problem is the same as this one showed here).
summation
edited 2 days ago
Glorfindel
1851211
asked Dec 31 '18 at 1:44
StarBucK
740213
add a comment |
Consider the following code:
Sum[If[b != 0, a[m], 0], {m, 1, 4}]
(* 4 If[b != 0, a[m], 0] *)
Why does it return me a[m]? m is the index of the summation, it shouldn't appear in the result.
Here I expect to have:
If[b != 0, a[1], 0]+If[b != 0, a[2], 0]+If[b != 0, a[3], 0]+If[b != 0, a[4], 0]
I don't understand this behavior
(my example here is to understand this behavior, my specific problem is more complicated but the problem is the same as this one showed here).
summation
edited 2 days ago
Glorfindel
1851211
asked Dec 31 '18 at 1:44
StarBucK
740213
add a comment |
Consider the following code:
Sum[If[b != 0, a[m], 0], {m, 1, 4}]
(* 4 If[b != 0, a[m], 0] *)
Why does it return me a[m]? m is the index of the summation, it shouldn't appear in the result.
Here I expect to have:
If[b != 0, a[1], 0]+If[b != 0, a[2], 0]+If[b != 0, a[3], 0]+If[b != 0, a[4], 0]
I don't understand this behavior
(my example here is to understand this behavior, my specific problem is more complicated but the problem is the same as this one showed here).
summation
edited 2 days ago
Glorfindel
1851211
asked Dec 31 '18 at 1:44
StarBucK
740213
Consider the following code:
Sum[If[b != 0, a[m], 0], {m, 1, 4}]
(* 4 If[b != 0, a[m], 0] *)
Why does it return me a[m]? m is the index of the summation, it shouldn't appear in the result.
Here I expect to have:
If[b != 0, a[1], 0]+If[b != 0, a[2], 0]+If[b != 0, a[3], 0]+If[b != 0, a[4], 0]
I don't understand this behavior
(my example here is to understand this behavior, my specific problem is more complicated but the problem is the same as this one showed here).
summation
summation
edited 2 days ago
Glorfindel
1851211
asked Dec 31 '18 at 1:44
StarBucK
740213
edited 2 days ago
Glorfindel
1851211
asked Dec 31 '18 at 1:44
StarBucK
740213
edited 2 days ago
Glorfindel
1851211
edited 2 days ago
Glorfindel
1851211
edited 2 days ago
Glorfindel
1851211
1851211
asked Dec 31 '18 at 1:44
StarBucK
740213
asked Dec 31 '18 at 1:44
StarBucK
740213
asked Dec 31 '18 at 1:44
StarBucK
740213
740213
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
This works:
Sum[If[b != 0, a[m] // Evaluate, 0], {m, 1, 4}]
From the documentation:
If evaluates only the argument determined by the value of the
condition.
$ $
You can use Evaluate to override HoldFirst etc. attributes of built-in
functions.
answered 2 days ago
Andrew
1,9011115
add a comment |
The other answer by Andrew technically answers your question, but,
maybe what you really wanted was
Sum[If[b =!= 0, a[m], 0], {m, 1, 4}]
which returns
a[1] + a[2] + a[3] + a[4]
The difference between the != and =!= operators is very important here. Because the truth value of b != 0 could not be determined, therefore the
If expressions were returned unevaluated.
I suggest moving the If expression out of the summation. That is use
If[b =!= 0, Sum[a[m], {m, 1, 4}], 0]
instead. This simplifies the code and makes it easier to understand.
answered 2 days ago
Somos
2107
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
This works:
Sum[If[b != 0, a[m] // Evaluate, 0], {m, 1, 4}]
From the documentation:
If evaluates only the argument determined by the value of the
condition.
$ $
You can use Evaluate to override HoldFirst etc. attributes of built-in
functions.
answered 2 days ago
Andrew
1,9011115
add a comment |
This works:
Sum[If[b != 0, a[m] // Evaluate, 0], {m, 1, 4}]
From the documentation:
If evaluates only the argument determined by the value of the
condition.
$ $
You can use Evaluate to override HoldFirst etc. attributes of built-in
functions.
answered 2 days ago
Andrew
1,9011115
add a comment |
This works:
Sum[If[b != 0, a[m] // Evaluate, 0], {m, 1, 4}]
From the documentation:
If evaluates only the argument determined by the value of the
condition.
$ $
You can use Evaluate to override HoldFirst etc. attributes of built-in
functions.
answered 2 days ago
Andrew
1,9011115
This works:
Sum[If[b != 0, a[m] // Evaluate, 0], {m, 1, 4}]
From the documentation:
If evaluates only the argument determined by the value of the
condition.
$ $
You can use Evaluate to override HoldFirst etc. attributes of built-in
functions.
answered 2 days ago
Andrew
1,9011115
answered 2 days ago
Andrew
1,9011115
answered 2 days ago
Andrew
1,9011115
answered 2 days ago
Andrew
1,9011115
1,9011115
add a comment |
add a comment |
The other answer by Andrew technically answers your question, but,
maybe what you really wanted was
Sum[If[b =!= 0, a[m], 0], {m, 1, 4}]
which returns
a[1] + a[2] + a[3] + a[4]
The difference between the != and =!= operators is very important here. Because the truth value of b != 0 could not be determined, therefore the
If expressions were returned unevaluated.
I suggest moving the If expression out of the summation. That is use
If[b =!= 0, Sum[a[m], {m, 1, 4}], 0]
instead. This simplifies the code and makes it easier to understand.
answered 2 days ago
Somos
2107
add a comment |
The other answer by Andrew technically answers your question, but,
maybe what you really wanted was
Sum[If[b =!= 0, a[m], 0], {m, 1, 4}]
which returns
a[1] + a[2] + a[3] + a[4]
The difference between the != and =!= operators is very important here. Because the truth value of b != 0 could not be determined, therefore the
If expressions were returned unevaluated.
I suggest moving the If expression out of the summation. That is use
If[b =!= 0, Sum[a[m], {m, 1, 4}], 0]
instead. This simplifies the code and makes it easier to understand.
answered 2 days ago
Somos
2107
add a comment |
The other answer by Andrew technically answers your question, but,
maybe what you really wanted was
Sum[If[b =!= 0, a[m], 0], {m, 1, 4}]
which returns
a[1] + a[2] + a[3] + a[4]
The difference between the != and =!= operators is very important here. Because the truth value of b != 0 could not be determined, therefore the
If expressions were returned unevaluated.
I suggest moving the If expression out of the summation. That is use
If[b =!= 0, Sum[a[m], {m, 1, 4}], 0]
instead. This simplifies the code and makes it easier to understand.
answered 2 days ago
Somos
2107
The other answer by Andrew technically answers your question, but,
maybe what you really wanted was
Sum[If[b =!= 0, a[m], 0], {m, 1, 4}]
which returns
a[1] + a[2] + a[3] + a[4]
The difference between the != and =!= operators is very important here. Because the truth value of b != 0 could not be determined, therefore the
If expressions were returned unevaluated.
I suggest moving the If expression out of the summation. That is use
If[b =!= 0, Sum[a[m], {m, 1, 4}], 0]
instead. This simplifies the code and makes it easier to understand.
answered 2 days ago
Somos
2107
edited 13 hours ago
answered 2 days ago
Somos
2107
answered 2 days ago
Somos
2107
answered 2 days ago
Somos
2107
2107
add a comment |
add a comment |
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