Using Riemann Upper sums to solve limits

In summary, the conversation discusses the challenges of using the Riemann definition of the integral to solve limit-series computations. The speaker is able to see the limit-series from the given integral, but struggles to find the integral from the limit-series. They ask if there is a general method to find the function from k/n, but it seems that the most common method is through inspection. The conversation ends with the speaker expressing disappointment that this is the case.
  • #1
Gib Z
Homework Helper
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I often see people use the Riemann definition of the integral to solve a certain limit-series computation, but they usually just skip a step that I can follow one way but not the other. Given the integral, I can see the limit-series that comes from it, but when trying to find the integral from the limit-series I have a problem.

For example, over the interval a=0 and b=1, [tex]\int^1_0 f(x) dx = \lim_{n\to \infty} \frac{1}{n} \sum_{k=1}^{n} f\left(\frac{k}{n}\right)[/tex] using Riemann Upper sums.

So basically If required to somehow work out [tex]f(k/n) = something[/tex], so that I can re-express it in terms of f(x). So i guess the question is, does anyone know a general method to spot the function of k/n, as to find f(x) ?
 
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  • #2
I don't believe is any, rather than pure inspection.

In most examples I've seen, the sum can be worked to be on the most common form

[tex]\lim_{\Delta P\rightarrow 0}\sum_{n=0}^\infty f(x_i^*)(x_{i+1}-x_i),[/tex]

where [itex]P=\{x_0,...,x_n,...\}[/itex] and [itex]x_i^* \in (x_i,x_{i+1})[/itex], so its easier to spot the [itex]f[/itex].
 
  • #3
Damn I was hoping that wouldn't be the answer :( Thanks
 

FAQ: Using Riemann Upper sums to solve limits

1. What are Riemann Upper sums?

Riemann Upper sums are a method of approximating the value of a limit by dividing the interval [a, b] into smaller subintervals and taking the maximum value of the function within each subinterval. This approach is based on the Riemann sum, which is a numerical method for estimating the area under a curve.

2. How do you use Riemann Upper sums to solve limits?

To use Riemann Upper sums to solve a limit, you first divide the interval [a, b] into smaller subintervals of equal width. Then, you take the maximum value of the function within each subinterval and multiply it by the width of the subinterval. Finally, you add all of these values together to get an approximation of the limit.

3. What is the benefit of using Riemann Upper sums to solve limits?

The benefit of using Riemann Upper sums is that it provides a more accurate approximation of the limit compared to other methods, such as the midpoint rule or the trapezoidal rule. This is because it takes into account the maximum values of the function within each subinterval, rather than just the average value.

4. Are there any limitations to using Riemann Upper sums to solve limits?

One limitation of using Riemann Upper sums is that it can be time-consuming and labor-intensive, especially when calculating the maximum value of the function within each subinterval. Additionally, it may not always provide an exact solution to the limit, but rather an approximation.

5. In what situations is it most useful to use Riemann Upper sums to solve limits?

Riemann Upper sums are most useful when dealing with continuous functions over a closed interval. They are also helpful in cases where the function is not easily integrable, as it provides a way to approximate the area under the curve. Additionally, Riemann Upper sums can be used to solve limits that involve infinite series or improper integrals.

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