- #1
FeynmanFtw
- 46
- 4
Take the case for the mean:
[tex]
\bar{x} = \frac{1}{N} \Big( \sum_{i=1}^Ni \Big)
[/tex]
If we simply use the formula for the sum of a geometric series, we get
[itex]\bar{x} = \frac{N}{2} (2a + (N - 1)d)[/itex]
where a and d both equal 1, so we simply get the result
[itex]\bar{x} = \frac{1}{2} (N + 1)[/itex]
What I've been trying to do is to get the same result by being more rigorous, in that I've attempted to expand the series, i.e.
[tex]
\bar{x} = \frac{1}{N} (1 + 2 + 3 + ... + N)
[/tex]
and replace the terms with the appropriate expressions using N, for example 1 and 2 would be N-(N-1) and N-(N-2) respectively, and so forth. Unfortunately I just keep going round in circles and never achieve the correct result.
Am I wasting my time? Or have I simply not seen the next step?
[tex]
\bar{x} = \frac{1}{N} \Big( \sum_{i=1}^Ni \Big)
[/tex]
If we simply use the formula for the sum of a geometric series, we get
[itex]\bar{x} = \frac{N}{2} (2a + (N - 1)d)[/itex]
where a and d both equal 1, so we simply get the result
[itex]\bar{x} = \frac{1}{2} (N + 1)[/itex]
What I've been trying to do is to get the same result by being more rigorous, in that I've attempted to expand the series, i.e.
[tex]
\bar{x} = \frac{1}{N} (1 + 2 + 3 + ... + N)
[/tex]
and replace the terms with the appropriate expressions using N, for example 1 and 2 would be N-(N-1) and N-(N-2) respectively, and so forth. Unfortunately I just keep going round in circles and never achieve the correct result.
Am I wasting my time? Or have I simply not seen the next step?