Piecewise Function Integration

[wany]

Homework Statement

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

Homework Equations

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

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 [itex]x_j, x_{j-1}[/itex] are < 1/2 then both inf and sup are 0;
if both [itex]x_j, x_{j-1}[/itex] are >= 1/2 then both inf and sup are 1;
so now it is possible for one case to be [itex]x_j \ge 1/2, x_{j-1} < 1/2[/itex]
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 [tex]M_j(f)[/tex], i.e.

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

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

 

[wany]

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

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:
[itex]
U(f,P)=\left\{\begin{array}{ccc}
\sum\limits_{j=1}^{n} (0)(1/n)=0 & \text{if} & j< n/2\\
\sum\limits_{j=1}^{n} (1)(1/2n)= & \text{if} & j\geq n/2
\end{array}\right.
[/itex]
and
[itex]
L(f,P)=\left\{\begin{array}{ccc}
\sum\limits_{j=1}^{n} (0)(1/n)=0 & \text{if} & j \le n/2\\
\sum\limits_{j=1}^{n} (1)(1/2n+1)= & \text{if} & j > n/2
\end{array}\right.
[/itex]
So now viewing as the limit goes to infinity, [itex]\lim_{n \rightarrow \infty}\sum\limits_{j=1}^{n} (1)(1/2n+1)=0[/itex]
and [itex]\lim_{n \rightarrow \infty}\sum\limits_{j=1}^{n} (1)(1/2n)=0[/itex]

 

[micromass]

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

[tex]\sum_{j=1}^n{M_j(f)\Delta x_j[/tex]

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

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

[tex] M_1(f)\Delta x_1+M_2(f)\Delta x_2+M_3(f)\Delta x_3[/tex]

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

[tex]\Delta x_2+\Delta x_3=2/3[/tex]

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
[itex]\lceil \frac{n+1}{2} \rceil[/itex] values that have the sup of 1.

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

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

So case 1: n is odd: so we have [itex]\frac{n-1}{2}*\frac{1}{n}=\frac{n-1}{2n}=\frac{1}{2}-\frac{1}{2n}[/itex]
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 [itex]\frac{n-1-1}{2}*\frac{1}{n}=\frac{n-2}{2n}=\frac{1}{2}-\frac{2}{2n}[/itex]
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.