Analysis - Integration Watch

DFranklin
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#21
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#21
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...
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TheEd
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#22
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#22
(Original post by 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})
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DFranklin
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#23
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#23
By the definitions of sup and inf, basically.
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TheEd
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#24
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(Original post by 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?
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DFranklin
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#25
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#25
I would define g:[0,1]->R by g(x) = f(a+x(b-a)).
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TheEd
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#26
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(Original post by 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?
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DFranklin
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#27
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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.
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TheEd
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#28
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(Original post by 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?
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