Analysis - Integration Watch

TheEd
Badges: 0
Rep:
?
#1
Report Thread starter 8 years ago
#1

of a continuous function f:[a,b]->R.

How would you do either part here?
0
reply
nuodai
Badges: 14
#2
Report 8 years ago
#2
Let P_n = \{0, \frac{1}{n}, \cdots, \frac{n-1}{n}, 1\} be your partition -- then your upper and lower sums will take a similar form to the RHS of your equation. Then show that the lower and upper sums over this partition tend to the same limit as n \to \infty, and deduce the result.

For the second part, you need to think about what I did in constructing \{0, \frac{1}{n}, \cdots, \frac{n-1}{n}, 1\}, and do a similar thing on the interval [a,b].
reply
TheEd
Badges: 0
Rep:
?
#3
Report Thread starter 8 years ago
#3
(Original post by nuodai)
Let P_n = \{0, \frac{1}{n}, \cdots, \frac{n-1}{n}, 1\} be your partition -- then your upper and lower sums will take a similar form to the RHS of your equation. Then show that the lower and upper sums over this partition tend to the same limit as n \to \infty, and deduce the result.

For the second part, you need to think about what I did in constructing \{0, \frac{1}{n}, \cdots, \frac{n-1}{n}, 1\}, and do a similar thing on the interval [a,b].
I see how you've constructed P_n with a=0 and b=1 but I'm having trouble finding formulae for the upper and lower sums. If the lower sum is defined as follows:

then how would this be calculated for our P_n ?

a=x_0=0<x_1=\frac{1}{n}<\cdots<x  _{n-1}=\frac{n-1}{n}<1=x_n=b
0
reply
DFranklin
Badges: 18
Rep:
?
#4
Report 8 years ago
#4
Surely you have a proof that continuous functions are integrable in your notes?
0
reply
nuodai
Badges: 14
#5
Report 8 years ago
#5
(Original post by TheEd)
I see how you've constructed P_n with a=0 and b=1 but I'm having trouble finding formulae for the upper and lower sums. If the lower sum is defined as follows:

then how would this be calculated for our P_n ?

a=x_0=0<x_1=\frac{1}{n}<\cdots<x  _{n-1}=\frac{n-1}{n}<1=x_n=b
Well you can work out x_{j+1}-x_j fairly easily (the difference between two consecutive terms in this partition is always the same).

You also need to use the continuity of f in here. If x_j \le x \le x_{j+1} and x_{j+1}-x_j = \dfrac{1}{n} then clearly 0 \le x - x_{j} \le x_{j+1}-x_{j} (by subtracting x_{j} in each part of the inequality), and so 0 \le x - x_{j} \le \dfrac{1}{n} and so on... basically the idea is to us limits and the fact that if a_n \to a then f(a_n) \to f(a). Doing this on both the upper and lower sum will give you something identical to what's in the question. Then you can deduce integrability using the result given in the question.
reply
TheEd
Badges: 0
Rep:
?
#6
Report Thread starter 8 years ago
#6
(Original post by DFranklin)
Surely you have a proof that continuous functions are integrable in your notes?
I have this:
0
reply
nuodai
Badges: 14
#7
Report 8 years ago
#7
(Original post by TheEd)
I have this:
That's just the definition of Riemann integrability, not a proof that continuous functions are integrable. In any case, see my above post.
reply
TheEd
Badges: 0
Rep:
?
#8
Report Thread starter 8 years ago
#8
(Original post by nuodai)
That's just the definition of Riemann integrability, not a proof that continuous functions are integrable. In any case, see my above post.
Yeah I just realised and started looking for a proof. I have the theorem that every continuous function is Riemann Integrable but no proof to go with it.
0
reply
DFranklin
Badges: 18
Rep:
?
#9
Report 8 years ago
#9
(Original post by nuodai)
You also need to use the continuity of f in here. If x_j \le x \le x_{j+1} and x_{j+1}-x_j = \dfrac{1}{n} then clearly 0 \le x - x_{j} \le x_{j+1}-x_{j} (by subtracting x_{j} in each part of the inequality), and so 0 \le x - x_{j} \le \dfrac{1}{n} and so on... basically the idea is to us limits and the fact that if a_n \to a then f(a_n) \to f(a). Doing this on both the upper and lower sum will give you something identical to what's in the question. Then you can deduce integrability using the result given in the question.
It's not as simple as that; you also need to use the fact that [0,1] is a closed interval.

For example, 1/x is continuous on (0, 1], but it isn't Riemann integrable.

If you know about uniform continuity (and that continuous functions on closed intervals are uniformly continuous) then it's straightforward to use a proof like yours, otherwise life gets a bit harder.
0
reply
rowzee
Badges: 1
Rep:
?
#10
Report 8 years ago
#10
Integration via Riemann sums looks like a pain in the neck. We did the theory of integration through step functions and regulated functions which seemed altogether easier.
0
reply
DFranklin
Badges: 18
Rep:
?
#11
Report 8 years ago
#11
They're not really that different. It's probably slightly easier to prove something *is* integrable using regulated functions, slightly easier to prove it *isn't* integrable using upper/lower sums.
0
reply
TheEd
Badges: 0
Rep:
?
#12
Report Thread starter 8 years ago
#12
(Original post by nuodai)
Well you can work out x_{j+1}-x_j fairly easily (the difference between two consecutive terms in this partition is always the same).

You also need to use the continuity of f in here. If x_j \le x \le x_{j+1} and x_{j+1}-x_j = \dfrac{1}{n} then clearly 0 \le x - x_{j} \le x_{j+1}-x_{j} (by subtracting x_{j} in each part of the inequality), and so 0 \le x - x_{j} \le \dfrac{1}{n} and so on... basically the idea is to us limits and the fact that if a_n \to a then f(a_n) \to f(a). Doing this on both the upper and lower sum will give you something identical to what's in the question. Then you can deduce integrability using the result given in the question.
Does m_0=\text{inf}\{f(x):0\leq x\leq\frac{1}{n}\}=f(0) and m_1=\text{inf}\{f(x):\frac{1}{n}  \leq x\leq\frac{2}{n}\}=f(\frac{1}{n}  ) and so on... up to m_{n-1}=\text{inf}\{f(x):\frac{n-1}{n}\leq x\leq 1\}=f(\frac{n-1}{n}) ?

If this is right then L(P_n,f)=\frac{1}{n}\displaystyl  e\sum_{j=0}^{n-1}m_j=\frac{1}{n}\left[f(0)+f(\frac{1}{n})+f(\frac{2}{n  })+\cdots +f(\frac{n-1}{n})\right]=\frac{1}{n}\sum_{k=0}^{n-1}f\left(\frac{k}{n}\right)

But I need to have L(P_n,f)=\displaystyle \frac{1}{n} \sum_{k=1}^{n}f\left(\frac{k}{n}  \right) so I need an f(1) instead of an f(0) . What's wrong?
0
reply
DFranklin
Badges: 18
Rep:
?
#13
Report 8 years ago
#13
If you don't know what uniform continuity is and you don't have a proof of this in your notes, there's pretty much no chance you're going to work it out for yourself.

It is not a simple result to prove. Either ask your lecturer, search the internet, or buy a textbook.
0
reply
TheEd
Badges: 0
Rep:
?
#14
Report Thread starter 8 years ago
#14
(Original post by DFranklin)
If you don't know what uniform continuity is and you don't have a proof of this in your notes, there's pretty much no chance you're going to work it out for yourself.

It is not a simple result to prove. Either ask your lecturer, search the internet, or buy a textbook.
Uniform continuity is for any \varepsilon >0 there exists \delta_{\varepsilon} > 0 such that for all x,y \in [0,1]\;\;|x-y|< \delta_{\varepsilon} \Rightarrow |f(x)-f(y)|< \varepsilon
0
reply
DFranklin
Badges: 18
Rep:
?
#15
Report 8 years ago
#15
OK, then use uniform continuity and it comes out in a few lines.

Note that your "proof" above doesn't use continuity *at all*. (You are wrong in thinking that m_0 = f(0). You seem to be confused between \inf \{f(x): 0\leq x\leq\frac{1}{n}\} and f(\inf \{x : 0\leq x\leq \frac{1}{n}\})).
0
reply
TheEd
Badges: 0
Rep:
?
#16
Report Thread starter 8 years ago
#16
(Original post by DFranklin)
OK, then use uniform continuity and it comes out in a few lines.

Note that your "proof" above doesn't use continuity *at all*. (You are wrong in thinking that m_0 = f(0). You seem to be confused between \inf \{f(x): 0\leq x\leq\frac{1}{n}\} and f(\inf \{x : 0\leq x\leq \frac{1}{n}\})).
Ok but what does \inf \{f(x): 0\leq x\leq\frac{1}{n}\} equal? And what does f(\inf \{x: 0\leq x\leq\frac{1}{n}\}) equal?
0
reply
DFranklin
Badges: 18
Rep:
?
#17
Report 8 years ago
#17
The 2nd one equals f(0). The first one equals the minimum value of f(x) for 0 <= x <= 1/n.
0
reply
TheEd
Badges: 0
Rep:
?
#18
Report Thread starter 8 years ago
#18
(Original post by DFranklin)
OK, then use uniform continuity and it comes out in a few lines.
I don't see where to start using uniform continuity. I know


as this was used to deduce the result (*) given in the question. Is this of use?
0
reply
DFranklin
Badges: 18
Rep:
?
#19
Report 8 years ago
#19
How do you know this? (It's not true for general f).

I think you need to post the *whole* question.
0
reply
TheEd
Badges: 0
Rep:
?
#20
Report Thread starter 8 years ago
#20
(Original post by DFranklin)
How do you know this? (It's not true for general f).

I think you need to post the *whole* question.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • University of the Arts London
    MA Performance Design and Practice Open Day Postgraduate
    Thu, 24 Jan '19
  • Brunel University London
    Undergraduate Experience Days Undergraduate
    Sat, 26 Jan '19
  • SOAS University of London
    Postgraduate Open Day Postgraduate
    Sat, 26 Jan '19

Are you chained to your phone?

Yes (58)
19.46%
Yes, but I'm trying to cut back (119)
39.93%
Nope, not that interesting (121)
40.6%

Watched Threads

View All
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise