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For Loop and Sum
Why should I avoid the For loop in Mathematica?How to solve a problem in relative motion?FindFit giving weird values on a logistic equationReplace notebooks and pens with Mathematica?Module that prints first m positive odd integers and returns the square root of the sum of themTrouble defining function for homeworkVariable Increments in a TableWhat is p[a] forInfinite While loopAn efficient way to translate a triple for-loop from Matlab to MathematicaPlotting gradient and newton directions for parametric system of nonlinear ODEs
$begingroup$
I am requested to do a sum of the first fifty positive even numbers by using For loop.
The For loop that I created is
For[natNum = 2, natNum <= 50, natNum = natNum + 2, Print[natNum]]
The output of this is a list of numbers from 2 to 50 and they are increasing by 2. Now I need these numbers to be sum so I created
Sum[For[natNum = 2, natNum <= 50, natNum = natNum + 2, Print[natNum]]]
However, it is not a correct input. I am not sure if I should write the Sum inside the For loop.
homework
edited 23 mins ago
MarcoB
36.6k556112
asked 3 hours ago
Laura FigueroaLaura Figueroa
61
$endgroup$
add a comment |
$begingroup$
I am requested to do a sum of the first fifty positive even numbers by using For loop.
The For loop that I created is
For[natNum = 2, natNum <= 50, natNum = natNum + 2, Print[natNum]]
The output of this is a list of numbers from 2 to 50 and they are increasing by 2. Now I need these numbers to be sum so I created
Sum[For[natNum = 2, natNum <= 50, natNum = natNum + 2, Print[natNum]]]
However, it is not a correct input. I am not sure if I should write the Sum inside the For loop.
homework
edited 23 mins ago
MarcoB
36.6k556112
asked 3 hours ago
Laura FigueroaLaura Figueroa
61
$endgroup$
2
$begingroup$
If this is a class assignment, I'd suggest withdrawing from the class.Forloops are almost never a good way to do anything in Mathematica.
$endgroup$
– John Doty
2 hours ago
$begingroup$
If you want the first 50 positive even numbers, I believe that is {2, 4, ..., 100}. The sum is (2+100) + (4+98) + ... + (50 + 52) = 102 * 25 = 2550.
$endgroup$
– mjw
1 hour ago
$begingroup$
Total@Range[2, 2*50, 2]would be pretty direct. Note also thatPrintprints its arguments, but does not return them (in fact, it returnsNull).
$endgroup$
– MarcoB
17 mins ago
add a comment |
$begingroup$
I am requested to do a sum of the first fifty positive even numbers by using For loop.
The For loop that I created is
For[natNum = 2, natNum <= 50, natNum = natNum + 2, Print[natNum]]
The output of this is a list of numbers from 2 to 50 and they are increasing by 2. Now I need these numbers to be sum so I created
Sum[For[natNum = 2, natNum <= 50, natNum = natNum + 2, Print[natNum]]]
However, it is not a correct input. I am not sure if I should write the Sum inside the For loop.
homework
edited 23 mins ago
MarcoB
36.6k556112
asked 3 hours ago
Laura FigueroaLaura Figueroa
61
$endgroup$
I am requested to do a sum of the first fifty positive even numbers by using For loop.
The For loop that I created is
For[natNum = 2, natNum <= 50, natNum = natNum + 2, Print[natNum]]
The output of this is a list of numbers from 2 to 50 and they are increasing by 2. Now I need these numbers to be sum so I created
Sum[For[natNum = 2, natNum <= 50, natNum = natNum + 2, Print[natNum]]]
However, it is not a correct input. I am not sure if I should write the Sum inside the For loop.
homework
homework
edited 23 mins ago
MarcoB
36.6k556112
asked 3 hours ago
Laura FigueroaLaura Figueroa
61
edited 23 mins ago
MarcoB
36.6k556112
asked 3 hours ago
Laura FigueroaLaura Figueroa
61
edited 23 mins ago
MarcoB
36.6k556112
edited 23 mins ago
MarcoB
36.6k556112
edited 23 mins ago
MarcoB
36.6k556112
36.6k556112
asked 3 hours ago
Laura FigueroaLaura Figueroa
61
asked 3 hours ago
Laura FigueroaLaura Figueroa
61
asked 3 hours ago
Laura FigueroaLaura Figueroa
61
61
2
$begingroup$
If this is a class assignment, I'd suggest withdrawing from the class.Forloops are almost never a good way to do anything in Mathematica.
$endgroup$
– John Doty
2 hours ago
$begingroup$
If you want the first 50 positive even numbers, I believe that is {2, 4, ..., 100}. The sum is (2+100) + (4+98) + ... + (50 + 52) = 102 * 25 = 2550.
$endgroup$
– mjw
1 hour ago
$begingroup$
Total@Range[2, 2*50, 2]would be pretty direct. Note also thatPrintprints its arguments, but does not return them (in fact, it returnsNull).
$endgroup$
– MarcoB
17 mins ago
add a comment |
2
$begingroup$
If this is a class assignment, I'd suggest withdrawing from the class.Forloops are almost never a good way to do anything in Mathematica.
$endgroup$
– John Doty
2 hours ago
$begingroup$
If you want the first 50 positive even numbers, I believe that is {2, 4, ..., 100}. The sum is (2+100) + (4+98) + ... + (50 + 52) = 102 * 25 = 2550.
$endgroup$
– mjw
1 hour ago
$begingroup$
Total@Range[2, 2*50, 2]would be pretty direct. Note also thatPrintprints its arguments, but does not return them (in fact, it returnsNull).
$endgroup$
– MarcoB
17 mins ago
2
2
$begingroup$
If this is a class assignment, I'd suggest withdrawing from the class.
For loops are almost never a good way to do anything in Mathematica.$endgroup$
– John Doty
2 hours ago
$begingroup$
If this is a class assignment, I'd suggest withdrawing from the class.
For loops are almost never a good way to do anything in Mathematica.$endgroup$
– John Doty
2 hours ago
$begingroup$
If you want the first 50 positive even numbers, I believe that is {2, 4, ..., 100}. The sum is (2+100) + (4+98) + ... + (50 + 52) = 102 * 25 = 2550.
$endgroup$
– mjw
1 hour ago
$begingroup$
If you want the first 50 positive even numbers, I believe that is {2, 4, ..., 100}. The sum is (2+100) + (4+98) + ... + (50 + 52) = 102 * 25 = 2550.
$endgroup$
– mjw
1 hour ago
$begingroup$
Total@Range[2, 2*50, 2] would be pretty direct. Note also that Print prints its arguments, but does not return them (in fact, it returns Null).$endgroup$
– MarcoB
17 mins ago
$begingroup$
Total@Range[2, 2*50, 2] would be pretty direct. Note also that Print prints its arguments, but does not return them (in fact, it returns Null).$endgroup$
– MarcoB
17 mins ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Anyone asking you to write For loops in Mathematica for such a problem is a dolt. Nevertheless, here's how you might do that:
rslt = 0;
For[i = 0, i <= 50, i += 2, rslt += i];
rslt
And here's how someone with some familiarity with Mathematica might write it
Plus @@ Range[25]*2
Now, spend the time you were going to waste writing a For loop by reading the answers to this question Why should I avoid the For loop in Mathematica?
answered 2 hours ago
High Performance MarkHigh Performance Mark
528411
$endgroup$
add a comment |
$begingroup$
The first fifty positive integers can be computed with Table:
Q = 0; (* Initialization)
T = Table[Q + k, {k, 2 Range[50]}]
We can sum up the values of T with Sum:
Sum[T[[k]], {k, Range[50]}]
2550
answered 1 hour ago
mjwmjw
3566
$endgroup$
$begingroup$
Perhaps better / more idiomatic than yourSumapproach would be usingTotal[T]orPlus@@T.
$endgroup$
– MarcoB
18 mins ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Anyone asking you to write For loops in Mathematica for such a problem is a dolt. Nevertheless, here's how you might do that:
rslt = 0;
For[i = 0, i <= 50, i += 2, rslt += i];
rslt
And here's how someone with some familiarity with Mathematica might write it
Plus @@ Range[25]*2
Now, spend the time you were going to waste writing a For loop by reading the answers to this question Why should I avoid the For loop in Mathematica?
answered 2 hours ago
High Performance MarkHigh Performance Mark
528411
$endgroup$
add a comment |
$begingroup$
Anyone asking you to write For loops in Mathematica for such a problem is a dolt. Nevertheless, here's how you might do that:
rslt = 0;
For[i = 0, i <= 50, i += 2, rslt += i];
rslt
And here's how someone with some familiarity with Mathematica might write it
Plus @@ Range[25]*2
Now, spend the time you were going to waste writing a For loop by reading the answers to this question Why should I avoid the For loop in Mathematica?
answered 2 hours ago
High Performance MarkHigh Performance Mark
528411
$endgroup$
add a comment |
$begingroup$
Anyone asking you to write For loops in Mathematica for such a problem is a dolt. Nevertheless, here's how you might do that:
rslt = 0;
For[i = 0, i <= 50, i += 2, rslt += i];
rslt
And here's how someone with some familiarity with Mathematica might write it
Plus @@ Range[25]*2
Now, spend the time you were going to waste writing a For loop by reading the answers to this question Why should I avoid the For loop in Mathematica?
answered 2 hours ago
High Performance MarkHigh Performance Mark
528411
$endgroup$
Anyone asking you to write For loops in Mathematica for such a problem is a dolt. Nevertheless, here's how you might do that:
rslt = 0;
For[i = 0, i <= 50, i += 2, rslt += i];
rslt
And here's how someone with some familiarity with Mathematica might write it
Plus @@ Range[25]*2
Now, spend the time you were going to waste writing a For loop by reading the answers to this question Why should I avoid the For loop in Mathematica?
answered 2 hours ago
High Performance MarkHigh Performance Mark
528411
answered 2 hours ago
High Performance MarkHigh Performance Mark
528411
answered 2 hours ago
High Performance MarkHigh Performance Mark
528411
answered 2 hours ago
High Performance MarkHigh Performance Mark
528411
528411
add a comment |
add a comment |
$begingroup$
The first fifty positive integers can be computed with Table:
Q = 0; (* Initialization)
T = Table[Q + k, {k, 2 Range[50]}]
We can sum up the values of T with Sum:
Sum[T[[k]], {k, Range[50]}]
2550
answered 1 hour ago
mjwmjw
3566
$endgroup$
$begingroup$
Perhaps better / more idiomatic than yourSumapproach would be usingTotal[T]orPlus@@T.
$endgroup$
– MarcoB
18 mins ago
add a comment |
$begingroup$
The first fifty positive integers can be computed with Table:
Q = 0; (* Initialization)
T = Table[Q + k, {k, 2 Range[50]}]
We can sum up the values of T with Sum:
Sum[T[[k]], {k, Range[50]}]
2550
answered 1 hour ago
mjwmjw
3566
$endgroup$
$begingroup$
Perhaps better / more idiomatic than yourSumapproach would be usingTotal[T]orPlus@@T.
$endgroup$
– MarcoB
18 mins ago
add a comment |
$begingroup$
The first fifty positive integers can be computed with Table:
Q = 0; (* Initialization)
T = Table[Q + k, {k, 2 Range[50]}]
We can sum up the values of T with Sum:
Sum[T[[k]], {k, Range[50]}]
2550
answered 1 hour ago
mjwmjw
3566
$endgroup$
The first fifty positive integers can be computed with Table:
Q = 0; (* Initialization)
T = Table[Q + k, {k, 2 Range[50]}]
We can sum up the values of T with Sum:
Sum[T[[k]], {k, Range[50]}]
2550
answered 1 hour ago
mjwmjw
3566
answered 1 hour ago
mjwmjw
3566
answered 1 hour ago
mjwmjw
3566
answered 1 hour ago
mjwmjw
3566
3566
$begingroup$
Perhaps better / more idiomatic than yourSumapproach would be usingTotal[T]orPlus@@T.
$endgroup$
– MarcoB
18 mins ago
add a comment |
$begingroup$
Perhaps better / more idiomatic than yourSumapproach would be usingTotal[T]orPlus@@T.
$endgroup$
– MarcoB
18 mins ago
$begingroup$
Perhaps better / more idiomatic than your
Sum approach would be using Total[T] or Plus@@T.$endgroup$
– MarcoB
18 mins ago
$begingroup$
Perhaps better / more idiomatic than your
Sum approach would be using Total[T] or Plus@@T.$endgroup$
– MarcoB
18 mins ago
add a comment |
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2
$begingroup$
If this is a class assignment, I'd suggest withdrawing from the class.
Forloops are almost never a good way to do anything in Mathematica.$endgroup$
– John Doty
2 hours ago
$begingroup$
If you want the first 50 positive even numbers, I believe that is {2, 4, ..., 100}. The sum is (2+100) + (4+98) + ... + (50 + 52) = 102 * 25 = 2550.
$endgroup$
– mjw
1 hour ago
$begingroup$
Total@Range[2, 2*50, 2]would be pretty direct. Note also thatPrintprints its arguments, but does not return them (in fact, it returnsNull).$endgroup$
– MarcoB
17 mins ago