I remember learning this out of curiosity in the past, but math doesnt stick in my brain unless chemistry is used as a disguise. Where can I read about this method again? Will the wiki page on Gauss get me there efficiently enough?
I found that pretty handy. Came across it looking for a Numberphile video on the subject. I was pretty sure I’d seen one before but didn’t have a lot of luck finding it.
Just wanted to say thanks! Clicking your link made me realize I should try to figure out how to open up YT on PipePipe automatically, so now my phone works better :)
The sum of all natural numbers smaller than and including x is equal to
(x+1)(x/2)
The sum of all even numbers up to and including x is that minus (n/2)² so
[(x+1)(x/2)]-(x/2)²
That would mean the sum of all odd numbers under x is equal to
(x/2)²
or sum of all odd including x (if x is odd of course) is
[(x+1)/2]²
Since the sum of all even numbers up to x is the sum of all numbers minus the sum of all odd numbers.
[-(x/2)² +2x +1]/2 (another way of writing the sum of all evens under and including a number) looks suspiciously polynomial. I want to go further.
Edit:
Interestingly, the sum of all evens under and including x is larger than the sum of all odds under x by half of x. So
[(x+1)(x/2)-(x/2)²]-(x/2) = (x/2)²
So another formula for sum of all evens is
(x/2)²+(x/2)
So,
2[(x/2)²]+(x/2) = (x+1)(x/2)
The left is the sum of all odds plus the sum of all evens under and including x, the right is the original formula we started with, sum of all natural numbers under and including x. Since they both give us the total sum for all natural numbers up to and including x, the left hand side is a different formula giving us the same result!
It is not well defined. Because an order of summation is not given you could just as easily sum pairs of (0,1), (-1,2), (-2,3), (-3,4)... (-x, x+1) and conclude you are constantly adding 1 to your total so it goes to infinity instead
Or do the reverse of (-1,0), (-2,1), ... (-x-1, x) and get that the each pair adds -1 so the sum goes to negative Infinity
Order of the addition sometimes changes infinite sums. Infinitely large things are weird sometimes
It's not split in half, it's reversed, which is why both are equal to S. If you just took half, then 1 + 2 + ... would not equal 100 + 99 + ... and so they both wouldn't be equal to S.
Instead of halving the list, they just reversed it and summed it to 2S later, which they then half. So no typo here.
Oh ok. Yeah fair. I guess I'm used to the simplified version. Where it doesn't use the full list, just splits it in half. This method would also work for odd lists of numbers rather than only even. Makes sense.