# Upper and lower sum

1. Apr 26, 2012

### Uniquebum

Let $f:[0,1]->ℝ$ and $f(x) = \frac{1}{n}$ when $x=\frac{1}{n²}, n=1,2,...$ and 0 in other case.

Define such spacing/interval $D$ that $S_D-s_D < \frac{1}{100}$. Here $S_D$ refers to the upper sum and $s_D$ the the lower sum.

Now, i'm not sure how to approach this really since the sum is defined so that the values of x jump a different interval every time. Any help would be appreciated. Hopefully i managed to explain the problem clearly enough.

Last edited: Apr 26, 2012
2. Apr 26, 2012

### SammyS

Staff Emeritus
That looks like a typo.

Is $\displaystyle f(x) = \frac{1}{x}$, when $\displaystyle x=\frac{1}{n^2}, n=1,2,...$ and 0 in otherwise.

... or perhaps ... $\displaystyle f(x) = x$, when $\displaystyle x=\frac{1}{n^2}, n=1,2,...$ and 0 in otherwise.

Are you doing this with Riemann sums ? ... and ... Do the intervals need to be equal in size?

3. Apr 26, 2012

### Uniquebum

It actually is $f(x)=\frac{1}{n}$. Yea, i'm trying to approach this with Riemann sums and the intervals don't need to be equal in size as far as i understand.

4. Apr 26, 2012

### Uniquebum

Right, I gave it a wild shot and this is what i came up with.

The upper sum is defined as
$\sum_{k=1}^{n}(x_{k+1}-x_k)*sup(f(x))$

Now let's choose such width for the Riemann sum quadrilateral that
$0.5*(\frac{1}{n} - \frac{1}{n+1})= \frac{1}{2n(n+1)} = r_n$.

Lets place each quadrilateral so that the supremus (=the point 1/n) is in the middle. This way the infimum will always be 0 and the lower sum $s_D=0$.

Calculating the upper sum gives
$\sum_{k=1}^{n}(x_{k+1}-x_k)*sup(f(x)) = \sum_{k=1}^{n}(2*r_n*\frac{k}{n})$
because $\frac{k}{n}, k=1,...,n$ goes through all the supremums.
So
$S_D = \sum_{k=1}^{n}(2*r_n*\frac{k}{n}) = 2*r_n\sum_{k=1}^{n}k$
which with a little work leads into
$S_D = \frac{1}{2n}$ since $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$

Now $S_D-s_d=\frac{1}{2n}-0=\frac{1}{2n}<\frac{1}{100}$ when $n\geq 50$.

Whether that's right or wrong is beyond me...

5. Apr 26, 2012

### Uniquebum

I'm actually starting to wonder whether it might be a typo by my professor.

Since it seems the function would only get one value if defined the way i stated in the original post. $\frac{1}{n}$ right?... or?