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

[sponsoredwalk]

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)?

 

[mrbohn1]

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.

 

[g_edgar]

Thomas is in BIG TROUBLE if he actually says $x \to \infty$ ... but I think it is just sponsoredwalk who is mistaken ...

 

[sponsoredwalk]

That answered my question perfectly thanks, I was a bit confused because the i was in the subscript.

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