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 1 hour ago
Glorfindel
1811211
asked 16 hours ago
StarBucK
710212
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 1 hour ago
Glorfindel
1811211
asked 16 hours ago
StarBucK
710212
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 1 hour ago
Glorfindel
1811211
asked 16 hours ago
StarBucK
710212
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 1 hour ago
Glorfindel
1811211
asked 16 hours ago
StarBucK
710212
edited 1 hour ago
Glorfindel
1811211
asked 16 hours ago
StarBucK
710212
edited 1 hour ago
Glorfindel
1811211
edited 1 hour ago
Glorfindel
1811211
edited 1 hour ago
Glorfindel
1811211
1811211
asked 16 hours ago
StarBucK
710212
asked 16 hours ago
StarBucK
710212
asked 16 hours ago
StarBucK
710212
710212
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 13 hours 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 $,texttt{!=},$ and $,texttt{=!=},$ operators is very important here. Because the truth value of $,texttt{b != 0},$ could not be determined, therefore the
$,texttt{If},$ expressions were returned unevaluated.
answered 11 hours 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 13 hours 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 13 hours 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 13 hours 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 13 hours ago
Andrew
1,9011115
answered 13 hours ago
Andrew
1,9011115
answered 13 hours ago
Andrew
1,9011115
answered 13 hours 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 $,texttt{!=},$ and $,texttt{=!=},$ operators is very important here. Because the truth value of $,texttt{b != 0},$ could not be determined, therefore the
$,texttt{If},$ expressions were returned unevaluated.
answered 11 hours 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 $,texttt{!=},$ and $,texttt{=!=},$ operators is very important here. Because the truth value of $,texttt{b != 0},$ could not be determined, therefore the
$,texttt{If},$ expressions were returned unevaluated.
answered 11 hours 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 $,texttt{!=},$ and $,texttt{=!=},$ operators is very important here. Because the truth value of $,texttt{b != 0},$ could not be determined, therefore the
$,texttt{If},$ expressions were returned unevaluated.
answered 11 hours 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 $,texttt{!=},$ and $,texttt{=!=},$ operators is very important here. Because the truth value of $,texttt{b != 0},$ could not be determined, therefore the
$,texttt{If},$ expressions were returned unevaluated.
answered 11 hours ago
Somos
2107
edited 1 hour ago
answered 11 hours ago
Somos
2107
answered 11 hours ago
Somos
2107
answered 11 hours ago
Somos
2107
2107
add a comment |
add a comment |
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