Piecewise Function Integration

[wany]

Homework Statement

Find formulas for the upper and lower sums of f on P_n, and use them to compute the value of \int_0^1f(x)dx.
P_n:=\{\frac{j}{n}:j=0,1,...,n\} (a partition of [0,1])
\[<br /> f(x) = \left\{ \begin{array}{ccc} 0 &amp; 0 \le x &lt; 1/2 \\ 1 &amp; 1/2 \le x \le 1 \end{array} \right. \]

Homework Equations

U(f,P)=\sum\limits_{j=1}^{n} M_j(f)\Delta x_j and
L(f,P)=\sum\limits_{j=1}^{n} M_j(f)\Delta x_j
where M_j=sup f([x_{j-1},x_j]) and m_j=inf f([x_{j-1},x_j])
if \lim_{n \rightarrow \infty} L(f,P_n)=\lim_{n \rightarrow \infty} U(f,P_n) then this equals \int_0^1f(x)dx

The Attempt at a Solution

So it is easy to see that this function is bounded on [0,1]. So now we can break this up into the different partitions, but now is where I run into a problem. It is finding the inf and the sup of each interval:
so obviously if both x_j, x_{j-1} are < 1/2 then both inf and sup are 0;
if both x_j, x_{j-1} are >= 1/2 then both inf and sup are 1;
so now it is possible for one case to be x_j \ge 1/2, x_{j-1} &lt; 1/2
in which case sup =1/2 and inf =0.

I am stuck from this point. Any help would be appreciated.

 

[micromass]

Can you tell us how exactly you are stuck?
It seems that you have found the good expression for the M_j(f), i.e.

M_j(f)=\left\{\begin{array}{ccc}<br /> 0 &amp; \text{if} &amp; j&lt; n/2\\<br /> 1 &amp; \text{if} &amp; j\geq n/2<br /> \end{array}\right.

and something analogous for m_j(f). Now you just need to plug those things in in U(f,P) and L(f,P)...

 

[wany]

so <br /> m_j(f)=\left\{\begin{array}{ccc}<br /> 0 &amp; \text{if} &amp; j \leq n/2\\<br /> 1 &amp; \text{if} &amp; j &gt; n/2<br /> \end{array}\right.<br />
So now
U(f,P)=\left\{\begin{array}{ccc}<br /> \sum\limits_{j=1}^{n} (0)(j/n)=0 &amp; \text{if} &amp; j&lt; n/2\\<br /> \sum\limits_{j=1}^{n} (1)(j/n)= &amp; \text{if} &amp; j\geq n/2<br /> \end{array}\right.
and
L(f,P)=\left\{\begin{array}{ccc}<br /> \sum\limits_{j=1}^{n} (0)(j/n)=0 &amp; \text{if} &amp; j \le n/2\\<br /> \sum\limits_{j=1}^{n} (1)(j/n)= &amp; \text{if} &amp; j &gt; n/2<br /> \end{array}\right.

So now solving I am not sure if I am correct. Are the sums not supposed to be taken from j=1 to n in such cases?

 

[wany]

wait so would it actually be:
<br /> U(f,P)=\left\{\begin{array}{ccc}<br /> \sum\limits_{j=1}^{n} (0)(1/n)=0 &amp; \text{if} &amp; j&lt; n/2\\<br /> \sum\limits_{j=1}^{n} (1)(1/2n)= &amp; \text{if} &amp; j\geq n/2<br /> \end{array}\right.<br />
and
<br /> L(f,P)=\left\{\begin{array}{ccc}<br /> \sum\limits_{j=1}^{n} (0)(1/n)=0 &amp; \text{if} &amp; j \le n/2\\<br /> \sum\limits_{j=1}^{n} (1)(1/2n+1)= &amp; \text{if} &amp; j &gt; n/2<br /> \end{array}\right.<br />
So now viewing as the limit goes to infinity, \lim_{n \rightarrow \infty}\sum\limits_{j=1}^{n} (1)(1/2n+1)=0
and \lim_{n \rightarrow \infty}\sum\limits_{j=1}^{n} (1)(1/2n)=0

 

[micromass]

No, what you wrote is incorrect.
You have the sum

\sum_{j=1}^n{M_j(f)\Delta x_j

and you are supposed to plug in the values for M_j(f). But for every M_j(f) you have a different value.

Lets take the example with n=3, then we got

M_1(f)\Delta x_1+M_2(f)\Delta x_2+M_3(f)\Delta x_3

We got that M_1(f)=0 and M_2(f)=M_3(f)=1, thus the above expression yields:

\Delta x_2+\Delta x_3=2/3

Now you just need to do the same thing with n arbitrary (I strongly suggest you do the cases n=4 and n=5 first, then you will see the general case).

 

[wany]

Ok so I think I got the general case, but I am not certain. So I did it for n=4,n=5,n=6, and got 3/4,3/5, and 2/3 respectively. So basically I think that depending on n, there will be
\lceil \frac{n+1}{2} \rceil values that have the sup of 1.

So case 1: n is odd: so we have \frac{n+1}{2}*\frac{1}{n}=\frac{n+1}{2n}=\frac{1}{2}+\frac{1}{2n}
so taking this limit as n goes to infinity we get that it is 1/2.
So case 2: n is even: so we have that \frac{n+1+1}{2}*\frac{1}{n}=\frac{n+2}{2n}=\frac{1}{2}+\frac{2}{2n}
so taking this limit as n goes to infinity we get that it is 1/2.

Now for the lower bound, it is \lfloor \frac{n-1}{2} \rfloor

So case 1: n is odd: so we have \frac{n-1}{2}*\frac{1}{n}=\frac{n-1}{2n}=\frac{1}{2}-\frac{1}{2n}
so taking this limit as n goes to infinity we get that it is 1/2.
So case 2: n is even: so we have that \frac{n-1-1}{2}*\frac{1}{n}=\frac{n-2}{2n}=\frac{1}{2}-\frac{2}{2n}
so taking this limit as n goes to infinity we get that it is 1/2.

Is this correct?

 

[micromass]

That seems to be good!

 

[wany]

Thanks a lot, this was very helpful.