4 ms·

This is one of those domain-specific areas in which Wolfram is awesome: http://www.wolframalpha.com/input/?i=sum+of+n+from+1+to+365 http://www.wolframalpha.com/

by m0th87 17y ago

This is one of those domain-specific areas in which Wolfram is awesome: http://www.wolframalpha.com/input/?i=sum+of+n+from+1+to+365 http://www.wolframalpha.com/input/?i=sum+of+n+from+1+to+365

patio11 17y ago
We can't all be Gauss but, then again, we're not in fourth grade anymore, either. You can sum over any range trivially: average of the endpoints times length of the range. The average of 1 and 365 is 183. There are 365 elements between 1 and 365 (both inclusive). 183 * 365 = 66795. Yay, right answer. (There is a formula. I have a poor memory for formulas but am good at transforming English into them, so I just remember the English gloss.) I'm decently impressed that Wolfram was able to spit out something equivalent to the right formula with only a little massaging of the inputs from me: http://www.wolframalpha.com/input/?i=sum+of+n+from+n%3Da+to+n%3Db http://www.wolframalpha.com/input/?i=sum+of+n+from+n%3Da+to+...
- deleted 17y ago
  [deleted]
- m0th87 17y ago
  I was referring to the fact that Wolfram actually understood the input, not that it answered a trivial math question :)
  - mahmud 17y ago
    (loop for x from 1 to 365 summing x) ==> 66795
three14 17y ago
There's a famous possibly true story about Gauss having to sum the numbers 1 to 100 as a child. Most versions tell it that he noticed that 1 + 100 = 2 + 99 = 3 + 98 and so on, yielding a general formula of n * (n + 1) / 2, so you could probably skip Wolfram for this one. (Explanation: you have n numbers, on average each is equal to the first + the last divided by 2) Unless you want it for this: http://www.wolframalpha.com/input/?i=sum+of+n+from+1+to+n http://www.wolframalpha.com/input/?i=sum+of+n+from+1+to+n
Raphael 17y ago
Preaching to the choir. I used it all day for physics homework.
seldo 17y ago
I knew there must be a formula but I totally just broke out a spreadsheet and auto-generated the sequence and then auto-summed it. Math/brain fail.
- RyanMcGreal 17y ago
  Alternately: fast repetition of trivial operations win.
icey 17y ago
A REPL is nice too: (reduce + (range 366))
- deleted 17y ago
  [deleted]
- Torn 17y ago
  Python: reduce (operator.add, range(366))
  - tome 17y ago
    or sum(range(366))
    - deleted 17y ago
      [deleted]
    - Torn 17y ago
      didn't even think of that one! :)
      - karanbhangui 17y ago
        or plain math: n(n+1)/2
        mahmud 17y ago
        How quickly can you generalize the plain math formula to these sequences: http://www.research.att.com/njas/sequences/index.html http://www.research.att.com/njas/sequences/index.html
    - deleted 17y ago
      [deleted]
    - mahmud 17y ago
      (sum (range 1 365)) assuming you defined these already; I have them in my utils library ;-) (defun range (start end) (loop for x from start to end collecting x)) (defun sum (sequence) (reduce #'+ sequence))

by m0th87 17y ago

This is one of those domain-specific areas in which Wolfram is awesome: http://www.wolframalpha.com/input/?i=sum+of+n+from+1+to+365 http://www.wolframalpha.com/input/?i=sum+of+n+from+1+to+365

patio11 17y ago

We can't all be Gauss but, then again, we're not in fourth grade anymore, either. You can sum over any range trivially: average of the endpoints times length of the range. The average of 1 and 365 is 183. There are 365 elements between 1 and 365 (both inclusive). 183 * 365 = 66795. Yay, right answer. (There is a formula. I have a poor memory for formulas but am good at transforming English into them, so I just remember the English gloss.) I'm decently impressed that Wolfram was able to spit out something equivalent to the right formula with only a little massaging of the inputs from me: http://www.wolframalpha.com/input/?i=sum+of+n+from+n%3Da+to+n%3Db http://www.wolframalpha.com/input/?i=sum+of+n+from+n%3Da+to+...

deleted 17y ago
[deleted]
m0th87 17y ago
I was referring to the fact that Wolfram actually understood the input, not that it answered a trivial math question :)
- mahmud 17y ago
  (loop for x from 1 to 365 summing x) ==> 66795

three14 17y ago

There's a famous possibly true story about Gauss having to sum the numbers 1 to 100 as a child. Most versions tell it that he noticed that 1 + 100 = 2 + 99 = 3 + 98 and so on, yielding a general formula of n * (n + 1) / 2, so you could probably skip Wolfram for this one. (Explanation: you have n numbers, on average each is equal to the first + the last divided by 2) Unless you want it for this: http://www.wolframalpha.com/input/?i=sum+of+n+from+1+to+n http://www.wolframalpha.com/input/?i=sum+of+n+from+1+to+n

Raphael 17y ago

Preaching to the choir. I used it all day for physics homework.

seldo 17y ago

I knew there must be a formula but I totally just broke out a spreadsheet and auto-generated the sequence and then auto-summed it. Math/brain fail.

RyanMcGreal 17y ago
Alternately: fast repetition of trivial operations win.

icey 17y ago

A REPL is nice too: (reduce + (range 366))

deleted 17y ago
[deleted]
Torn 17y ago
Python: reduce (operator.add, range(366))
- tome 17y ago
  or sum(range(366))
  - deleted 17y ago
    [deleted]
  - Torn 17y ago
    didn't even think of that one! :)
    - karanbhangui 17y ago
      or plain math: n(n+1)/2
      - mahmud 17y ago
        How quickly can you generalize the plain math formula to these sequences: http://www.research.att.com/njas/sequences/index.html http://www.research.att.com/njas/sequences/index.html
  - deleted 17y ago
    [deleted]
  - mahmud 17y ago
    (sum (range 1 365)) assuming you defined these already; I have them in my utils library ;-) (defun range (start end) (loop for x from start to end collecting x)) (defun sum (sequence) (reduce #'+ sequence))