This discussion is now closed.

Check out other Related discussions

Analysis - Integration

Well, that completely changes everything (sigh). Can you please make sure you post sufficient context in future? It was impossible to answer your question properly given what you originally posted.

For Q2(a), note that

\displaystyle L(P_n, f) \leq \frac{1}{n}\sum_1^n f(\frac{k}{n}) \leq R(P_n, f)

, and the result follows.

I have to say, I don't understand how you can have answered Q1 correctly and not been able to do Q2, but still...

Reply 21

TheEd

OP

DFranklin

Well, that completely changes everything (sigh). Can you please make sure you post sufficient context in future? It was impossible to answer your question properly given what you originally posted.

For Q2(a), note that

\displaystyle L(P_n, f) \leq \frac{1}{n}\sum_1^n f(\frac{k}{n}) \leq R(P_n, f)

, and the result follows.

I have to say, I don't understand how you can have answered Q1 correctly and not been able to do Q2, but still...

How do you know that

\displaystyle L(P_n, f) \leq \frac{1}{n}\sum_{k=1}^{n} f(\frac{k}{n}) \leq U(P_n, f)

?

By this inequality:

\displaystyle\displaystyle\lim_{n \rightarrow \infty}L(P_n, f) \leq \displaystyle\lim_{n \rightarrow \infty}\frac{1}{n}\sum_{k=1}^{n} f(\frac{k}{n}) \leq \displaystyle\lim_{n \rightarrow \infty}U(P_n, f)

If

\displaystyle\lim_{n \rightarrow \infty}L(P_n,f)=\lim_{n \rightarrow \infty}U(P_n,f)

then

\displaystyle\lim_{n \rightarrow \infty}L(P_n,f)=\lim_{n \rightarrow \infty}\frac{1}{n}\sum_{k=1}^{n} f(\frac{k}{n})=\lim_{n \rightarrow \infty}U(P_n,f)

. So by the result from Q1,

\displaystyle\int_0^1 f=\lim_{n \rightarrow \infty}\frac{1}{n}\sum_{k=1}^{n} f(\frac{k}{n})

Reply 22

DFranklin

18

By the definitions of sup and inf, basically.

Reply 23

TheEd

OP

DFranklin

By the definitions of sup and inf, basically.

For 2(b) I don't have such an inequality to go by as there is no answer printed in the question so how do I find a formulae for [a,b]->R instead of [0,1]->R?

Reply 24

DFranklin

18

I would define g:[0,1]->R by g(x) = f(a+x(b-a)).

Reply 25

TheEd

OP

DFranklin

I would define g:[0,1]->R by g(x) = f(a+x(b-a)).

If you let

P_n=\{a,b,2b-a,3b-2a,\cdots,nb+a(1-n)\}

then

m_{j+1}-m_j=b-a

If we define

g:[0,1]\rightarrow\mathbb{R}

by

g(x)=f(a+x(b-a))

Working out

L(P_n,g)

m_0=\text{inf}\{f(a+x(b-a)):a\leq x \leq b\}

m_1=\text{inf}\{f(a+x(b-a)):b\leq x \leq 2b-a\}

...

Surely I have to calculate these this time to find a formulae? But what are these equal to?

Reply 26

DFranklin

18

It's no different from what you did before.

I give up. This is basically me doing the question for you, without even having the full question in front of me, or knowing what your course has told you can be assumed.

Reply 27

TheEd

OP

DFranklin

It's no different from what you did before.

I give up. This is basically me doing the question for you, without even having the full question in front of me, or knowing what your course has told you can be assumed.

I should have been dividing by n in my last post.

Anyway, would

\displaystyle\lim_{n \rightarrow \infty}\frac{b-a}{n}\sum_{k=1}^{n} f\left(a+\frac{k(b-a)}{n}\right)

be the formula?

Analysis - Integration

Related discussions

Latest

Trending

Trending

Articles for you