Paradox: 1 – 1/2 and 1/3 – 1/4 and . . . = 0(simonsingh.net)
simonsingh.net
Paradox: 1 – 1/2 and 1/3 – 1/4 and . . . = 0
http://www.simonsingh.net/Simpsons_Mathematics/1-12-13-14-0/
5 comments
Yeah that's what I thought the article was going to be about. By re-arranging terms you can make it convert to any real number.
And 1 + 2 + 3 + 4 + ⋯ = ...
where is the easiest place on the web to graph this?
What do you want to graph?
The first expression, A1, is:
You can graph these sorts of things at Wolfram Alpha, or even just type the expression into Google and often you'll get something useful. Here's an example:
https://news.ycombinator.com/item?id=9393392
The first expression, A1, is:
sum_{x=1}^{oo} 1/x
S(n) = sum_{x=1}^n 1/x
You can graph these sorts of things at Wolfram Alpha, or even just type the expression into Google and often you'll get something useful. Here's an example:
https://news.ycombinator.com/item?id=9393392
Thanks, in my messing around with WA, I wasnt getting a chart using the limit function, but now i realized I wanted the summation. I havent done much graphing online.
TBH I was just curious how the graphs looked taking the positives and negatives independently.
sum_{x=1}^{oo} 1/x
vs
sum_{x=1}^{oo} (-1)1/x
vs
sum_{x=1}^{oo} (-1)1/(x+1)
TBH I was just curious how the graphs looked taking the positives and negatives independently.
sum_{x=1}^{oo} 1/x
vs
sum_{x=1}^{oo} (-1)1/x
vs
sum_{x=1}^{oo} (-1)1/(x+1)
I suspect you have problems there with asterisks for multiplication. Edit it to use dot instead of star, then I might be able to reply sensibly.
8pts, 6 comments and already the site's buckled.
Sometimes this place makes me think that not putting a caching proxy in front of your blog should be a crime.
Sometimes this place makes me think that not putting a caching proxy in front of your blog should be a crime.
A1 and A2 have different limits.
It will be clear if the upper limit is changed to something other than infinity.
Lets say A1 is sum of 1/x, x=1 to 10.
For A1 and A2 to be equal, A2 will have limits x=1 to 5.
Thus, you cannot strike of sum of (1/x) on both sides in the equation, A1=D.
> A1 and A2 have different limits. It will be clear
> if the upper limit is changed to something other
> than infinity.
1 + 1/2 + 1/4 + 1/8 + 1/16 + ...
1/2 + 1/8 + 1 + 1/32 + 1/128 + 1/4 + ...
If your argument were valid then these would have different limits, because if you cut them both off after some number of terms the totals will always be different. However, these two series in fact have the same limit.
And in fact A1 and A2 have the same limit, so that's not where the problem lies.
>For A1 and A2 to be equal, A2 will have limits x=1 to 5.
I should have added 'For A1 and A2 to be equal' at the start. If you want them to be equal, they need to have different limits and if they have same limits, they are not equal.
I should have added 'For A1 and A2 to be equal' at the start. If you want them to be equal, they need to have different limits and if they have same limits, they are not equal.
I'm finding that completely impossible to parse.
In particular:
In particular:
If you want them to be equal, they need
to have different limits and if they have
same limits, they are not equal.
sum_{x=0}^oo (-1)^x (1/x)
sum_{y=0}^oo [ 1/(2y) - 1/(2y+1) ]
[deleted]
Exactly. If you use different variable at two sides..say 'x' and 'n' , it will be more clear.
http://en.wikipedia.org/wiki/Riemann_series_theorem