If and summation : why I have the indice 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 it returns me an a[m] ? m is the indice 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 hours ago
bbgodfrey
44.2k858109
asked 5 hours ago
StarBucK
688211
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 it returns me an a[m] ? m is the indice 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 hours ago
bbgodfrey
44.2k858109
asked 5 hours ago
StarBucK
688211
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 it returns me an a[m] ? m is the indice 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 hours ago
bbgodfrey
44.2k858109
asked 5 hours ago
StarBucK
688211
Consider the following code :
Sum[If[b != 0, a[m], 0], {m, 1, 4}]
(* 4 If[b != 0, a[m], 0] *)
Why it returns me an a[m] ? m is the indice 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 hours ago
bbgodfrey
44.2k858109
asked 5 hours ago
StarBucK
688211
edited 2 hours ago
bbgodfrey
44.2k858109
asked 5 hours ago
StarBucK
688211
edited 2 hours ago
bbgodfrey
44.2k858109
edited 2 hours ago
bbgodfrey
44.2k858109
edited 2 hours ago
bbgodfrey
44.2k858109
44.2k858109
asked 5 hours ago
StarBucK
688211
asked 5 hours ago
StarBucK
688211
asked 5 hours ago
StarBucK
688211
688211
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 hours ago
Andrew
1,8461115
add a comment |
The other answer by Andrew technically answers your question, but,
perhaps what you 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 $texttt{!=}$ and $texttt{=!=}$ is very important here. Because the truth value of $,texttt{b != 0},$ could not be determined, the
$,texttt{If},$ statement was returned unevaluated.
answered 24 mins ago
Somos
1507
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 hours ago
Andrew
1,8461115
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 hours ago
Andrew
1,8461115
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 hours ago
Andrew
1,8461115
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 hours ago
Andrew
1,8461115
answered 2 hours ago
Andrew
1,8461115
answered 2 hours ago
Andrew
1,8461115
answered 2 hours ago
Andrew
1,8461115
1,8461115
add a comment |
add a comment |
The other answer by Andrew technically answers your question, but,
perhaps what you 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 $texttt{!=}$ and $texttt{=!=}$ is very important here. Because the truth value of $,texttt{b != 0},$ could not be determined, the
$,texttt{If},$ statement was returned unevaluated.
answered 24 mins ago
Somos
1507
add a comment |
The other answer by Andrew technically answers your question, but,
perhaps what you 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 $texttt{!=}$ and $texttt{=!=}$ is very important here. Because the truth value of $,texttt{b != 0},$ could not be determined, the
$,texttt{If},$ statement was returned unevaluated.
answered 24 mins ago
Somos
1507
add a comment |
The other answer by Andrew technically answers your question, but,
perhaps what you 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 $texttt{!=}$ and $texttt{=!=}$ is very important here. Because the truth value of $,texttt{b != 0},$ could not be determined, the
$,texttt{If},$ statement was returned unevaluated.
answered 24 mins ago
Somos
1507
The other answer by Andrew technically answers your question, but,
perhaps what you 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 $texttt{!=}$ and $texttt{=!=}$ is very important here. Because the truth value of $,texttt{b != 0},$ could not be determined, the
$,texttt{If},$ statement was returned unevaluated.
answered 24 mins ago
Somos
1507
edited 1 min ago
answered 24 mins ago
Somos
1507
answered 24 mins ago
Somos
1507
answered 24 mins ago
Somos
1507
1507
add a comment |
add a comment |
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