Limit Sum Riemann: Solve Homework Problem

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Homework Help Overview

The discussion revolves around calculating a limit using Riemann sums, specifically focusing on the expression involving the tangent function. The subject area is calculus, particularly the concept of definite integrals and their relation to Riemann sums.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster expresses uncertainty about how to start the problem and seeks guidance. Some participants explain the concept of Riemann sums and how they relate to definite integrals, suggesting the identification of the function and limits of integration. Others question the values of 'a' and 'b' and the form of the function involved.

Discussion Status

Participants are actively engaging in clarifying the concepts of Riemann sums and definite integrals. Some guidance has been provided regarding the identification of the function and the limits, but there remains a lack of consensus on the specific values for 'a' and 'b', as well as the form of the integrand.

Contextual Notes

There is some confusion regarding the identification of the limits of integration and the function to be integrated, with participants expressing differing interpretations of the problem setup.

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Homework Statement



Calculate the limit with Riemann.

Homework Equations



[tex]\displaystyle\lim_{n \to{+}\infty}{\displaystyle\frac{pi}{4}\cdot{} \displaystyle\sum_{k=0}^n{tan^2(\displaystyle\frac{k\cdot{} pi}{4n})\cdot{}\displaystyle\frac{1}{n}}}[/tex]

The Attempt at a Solution



I don't know how to start this problem...

Help me please, thank you.
 
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"With Riemann" means that you treat this as a Riemann sum- which is used to define the definite integral.

If we divide the x-axis, from x= a to x= b, into n intervals then each interval has length [itex]\Delta x= (b- a)/n[itex]. And, if we take the value of the integrand, f, in each interval at the left end of that interval, we have f((b-a)k/n) as the height of the rectangle we are forming on that interval so its area is f((b-a)k/n)(b-a)/n and the whole arera is <br /> [tex]\displaystyle (b- a)\sum_{k=0}^n f((b-a)k/n)\cdot\frac{1}{n}[tex] The limit turns that into an integral.<br /> <br /> So you need to indentify f(x), a, and b in this particular sum, and integrate.[/tex][/tex][/itex][/itex]
 
Explain me in this case please, is the first time I have to do one problem like this. Thank you
 
Halls' already explained it, you just have to put in a bit of effort into trying to understand what he said. Some of his code didn't appear properly so I'll put it up again for you.

HallsofIvy said:
"With Riemann" means that you treat this as a Riemann sum- which is used to define the definite integral.

If we divide the x-axis, from [itex]x= a[/itex] to [itex]x= b[/itex], into [itex]n[/itex] intervals then each interval has length [itex]\Delta x= (b-a)/n[/itex]. And, if we take the value of the integrand, [tex]f(x)[/tex], in each interval at the left end of that interval, we have [tex]f\left(\frac{(b-a)k}{n}\right)[/tex] as the height of the rectangle we are forming on that interval so its area is [tex]f(\frac{(b-a)k}{n})\left(\frac{b-a}{n}\right)[/tex] and the whole area is
[tex]\displaystyle (b- a)\sum_{k=0}^n f\left( \frac{(b-a)k}{n}\right)\cdot\frac{1}{n}[/tex]
If we take the limit as the number of rectangles [itex]n \to \infty[/itex] this becomes [tex]\int^b_a f(x) dx[/tex].
So you need to indentify f(x), a, and b in this particular sum, and integrate.

So carry you this idea for an easy (and common) case, [tex]\int^1_0 f(x) dx[/tex]. You want to find an expression for Riemann sums for that integral. If you do the same idea as what Halls described already, you should get [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].
 
I'm sorry but I don't understand, how i know a and b? I'm sorry I'm lost.

I have the idea.. b is [tex]pi / 4[/tex] and [tex]a = 0[/tex] ?

And f(x) = [tex]tan^2(k*pi/4n)[/tex] ? ?
 

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