FeynmanFtw
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Take the case for the mean:
<br /> \bar{x} = \frac{1}{N} \Big( \sum_{i=1}^Ni \Big)<br />
If we simply use the formula for the sum of a geometric series, we get
\bar{x} = \frac{N}{2} (2a + (N - 1)d)
where a and d both equal 1, so we simply get the result
\bar{x} = \frac{1}{2} (N + 1)
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.
<br /> \bar{x} = \frac{1}{N} (1 + 2 + 3 + ... + N)<br />
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?
<br /> \bar{x} = \frac{1}{N} \Big( \sum_{i=1}^Ni \Big)<br />
If we simply use the formula for the sum of a geometric series, we get
\bar{x} = \frac{N}{2} (2a + (N - 1)d)
where a and d both equal 1, so we simply get the result
\bar{x} = \frac{1}{2} (N + 1)
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.
<br /> \bar{x} = \frac{1}{N} (1 + 2 + 3 + ... + N)<br />
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?