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])
$\[<br /> f(x) = \left\{ \begin{array}{ccc} 0 & 0 \le x < 1/2 \\ 1 & 1/2 \le x \le 1 \end{array} \right. \]$

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 $x_j \ge 1/2, x_{j-1} < 1/2$
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.

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

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

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:
$U(f,P)=\left\{\begin{array}{ccc}<br /> \sum\limits_{j=1}^{n} (0)(1/n)=0 & \text{if} & j< n/2\\<br /> \sum\limits_{j=1}^{n} (1)(1/2n)= & \text{if} & j\geq n/2<br /> \end{array}\right.$
and
$L(f,P)=\left\{\begin{array}{ccc}<br /> \sum\limits_{j=1}^{n} (0)(1/n)=0 & \text{if} & j \le n/2\\<br /> \sum\limits_{j=1}^{n} (1)(1/2n+1)= & \text{if} & j > n/2<br /> \end{array}\right.$
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.