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 6 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 6 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 6 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 6 hours ago
StarBucK
688211
edited 2 hours ago
bbgodfrey
44.2k858109
asked 6 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 6 hours ago
StarBucK
688211
asked 6 hours ago
StarBucK
688211
asked 6 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 3 hours ago
Andrew
1,8561115
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 $,texttt{!=},$ and $,texttt{=!=},$ operators is very important here. Because the truth value of $,texttt{b != 0},$ could not be determined, therefore the
$,texttt{If},$ statements were returned unevaluated.
answered 1 hour ago
Somos
1607
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 3 hours ago
Andrew
1,8561115
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 3 hours ago
Andrew
1,8561115
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 3 hours ago
Andrew
1,8561115
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 3 hours ago
Andrew
1,8561115
answered 3 hours ago
Andrew
1,8561115
answered 3 hours ago
Andrew
1,8561115
answered 3 hours ago
Andrew
1,8561115
1,8561115
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 $,texttt{!=},$ and $,texttt{=!=},$ operators is very important here. Because the truth value of $,texttt{b != 0},$ could not be determined, therefore the
$,texttt{If},$ statements were returned unevaluated.
answered 1 hour ago
Somos
1607
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 $,texttt{!=},$ and $,texttt{=!=},$ operators is very important here. Because the truth value of $,texttt{b != 0},$ could not be determined, therefore the
$,texttt{If},$ statements were returned unevaluated.
answered 1 hour ago
Somos
1607
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 $,texttt{!=},$ and $,texttt{=!=},$ operators is very important here. Because the truth value of $,texttt{b != 0},$ could not be determined, therefore the
$,texttt{If},$ statements were returned unevaluated.
answered 1 hour ago
Somos
1607
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 $,texttt{!=},$ and $,texttt{=!=},$ operators is very important here. Because the truth value of $,texttt{b != 0},$ could not be determined, therefore the
$,texttt{If},$ statements were returned unevaluated.
answered 1 hour ago
Somos
1607
edited 25 mins ago
answered 1 hour ago
Somos
1607
answered 1 hour ago
Somos
1607
answered 1 hour ago
Somos
1607
1607
add a comment |
add a comment |
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