How Do You Calculate (x_i - x_i_-_1) in Summation Notation?

sponsoredwalk
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Okay I've seen how crazy Riemann sums can get in real analysis and I've noticed a heirarchy of notation.

The Stewart/Thomas etc... kinds of books use;

\lim_{x \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x

Where;

\Delta x = \frac{b - a}{n} and x_i = a + i\Delta x

Then the books like Apostol and Bartle's real analysis use;

\lim_{x \to \infty} \sum_{i=1}^{n} f(x_i) (x_i - x_i_-_1)and what I'd like to know is how to calculate the (x_i - x_i_-_1) for some equation like;

f(x) = x² integrated from 2 to 8. I can do the Δx = (b - a)/n version fine but how do you work the newer notation?

in f(x_i) (x_i - x_i_-_1) I would assume f(x_i) would use any endpoint, i.e. the right endpoint being a + iΔx but how do you make sense of the (x_i - x_i_-_1)?
 
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As you say, they both essentially mean the same thing. To see this, just substitute (i.\frac{b - a}{n} for x_i and (i-1).\frac{b - a}{n} for x_{i-1} in the latter notation, and simplify.

x_i just means the "value you get after adding i\frac{b-a}{n} to a ".

I'm not sure if this answers your question.
 
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Thomas is in BIG TROUBLE if he actually says x \to \infty ... but I think it is just sponsoredwalk who is mistaken ...
 
That answered my question perfectly thanks, I was a bit confused because the i was in the subscript.

Yeah that's actually \lim_{n \to \infty} :redface:
 
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