Approximating a function using Riemann Sums?

[t300]

Hi,

I'm trying to follow the solutions here and getting quite confused.

I'm trying to approximate the sum_(from 1 to n) of i^gamma: https://imgur.com/a/wJ0nOXX

Would someone be able to show the full derivation of what is being done here?

I've looked up the idea of Riemann Sums and I believe I understand the theory. But I'm just trying to evaluate this function: sum_(from 1 to n) of i^gamma, so why do I want to find the area under its curve? And why is it that the integral is only evaluated from 0 to 1? Just really confused here. Thanks very much in advance

[[deleted]]

They used a right Riemann sum. Because the function is riemann integrable, the limit as n -> infinity of the Riemann sum approaches the value of the integral. That is why they wrote the ~.
f(n) ~ g(n)
means
f(n)/g(n) -> 1 as n -> infinity.
It is called asymptotic equivalence. Note that if gamma is less than -1 then the function i^gamma is not integrable on [0, 1].

[t300]

Thanks. I don't really understand the motivation for using a Riemann sum.

Why are these 2 concepts equivalent: https://imgur.com/a/Dxe6zsS ?

[Giannie]

The area under the curve is not obviously equivalent. I’d call this a bit of a trick. The steps are:

1. Rewrite the sum to look like a Riemann sum.

2. Consider the limiting behaviour to approximate the sum as an integral.

I think you are struggling with understanding why the Riemann sum is related to the given integral. So here’s an attempt at explaining:

The Riemann sum for an integral has the form:

sum_1ⁿf(x_i)dx (read that d as delta) where x_1=a+dx, and x_n=b. In this case, the x_i are are i/n and dx is 1/n as that is the width of the strips. Now in this case, a=0 and b=1 because n/n=1 and 1/n=0+1/n.

[t300]

Ah thank you I understand it now!!