Question 4

• October 30th 2006, 04:17 AM
ThePerfectHacker
Question 4
This week I will post 2 problems because I am not sure how this is still going to work.

First: Given the diophantine equation,
$x^2+y^3=n$ show that the number of solutions cannot exceed $n$. (Easy)

Second: The Dirchlet function:
Show that the Dirichlet function is not Riemann integratble on $[0,1]$ (Hard)

This Dirchlet function is defined to map rationals into zero and irrationals into one.

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One more thing, can you rate this threads. By selecting the number of stars for it. Whether you liked it or hated it.
• November 6th 2006, 08:32 AM
ThePerfectHacker
I am supprised why nobody responded. Either because the questions are too difficult (which I cannot imagine, the first one was really easy). Or they are just not fun.
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First Problem:
Okay you have,
$x^2+y^3=n$ with $n\geq 1$.
That means,
$1\leq x\leq n^{1/2}$
Because otherwise, since $y\geq 1$ there is no way to make the left hand smaller.
Similarly,
$1\leq y\leq n^{1/3}$
The best case senerio is when all are solutions in that case,
$n^{1/2}\cdot n^{1/3}=n^{5/6}
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Second Problem:

$D(x)=\left\{ \begin{array}{cc}1&\mbox{ if irrational}\\ 0&\mbox{ if rational} \end{array} \right\}$
There are a number of things we need to know.

First, if the function is integratble then the Riemann sum is well-defined as long as the norm sequence limit is zero.

Second, if $\int_a^b f(x)dx$ exists and $c\in (a,b)$ then both the integrals exists and,
$\int_a^b f(x)dx=\int_a^c f(x) dx+\int_c^b f(x) dx$

Third, $1 is irrational therefore, $1/e<1$.

Now we can begin.

The idea is that we are going to assume that the Dirichlet function is Riemann integrable then arrive at a contradiction.

If the Dirichlet function is integrable on $[0,1]$ then we can partition the interval in any way we choose as long as the norm sequence limit is zero. We will use the right end-points. In, that case,
$\int_0^1 D(x)dx=\lim_{n\to \infty}\sum_{k=1}^n D\left( \frac{k}{n} \right)\cdot \frac{1}{n}$
But the trick is to realize that,
$\frac{k}{n}$ is rational!
Thus,
$D\left(\frac{k}{n}\right)=0$
Thus, (limit of zero)
$\int_0^1 D(x) dx=0$.

But, since the Dirchelt integral exists (by assumption) then,
$\int_0^1 D(x)=\int_0^{1/e}D(x) dx+\int_{1/e}^1 D(x)dx$
That means its value is by left-endpoints,
$\lim_{n\to\infty}\sum_{k=1}^n D\left(\frac{k}{n}\right) \cdot \frac{1}{en}$+ $\lim_{n\to\infty}\sum_{k=1}^n D\left(\frac{1}{e}+\frac{k}{n}\right) \cdot \frac{1-1/e}{n}$

But,
$D\left( \frac{k}{n}\right)=0$ for it is rational. Und $D\left(\frac{1}{e}+\frac{k}{n}\right)=1$ for it is irrational.
Thus,
$0+\lim_{n\to\infty}\sum_{k=1}^n 1\cdot \frac{1-1/e}{n}=\lim_{n\to\infty}\frac{n(1-1/e)}{n}=1-1/e\not = 0$
Thus, the limit is not well-defined by partitioning.
Thus, the function is not integrable.

If you are curious how I invented this problem is when I was in Calculus class we were doing integration over surfaces and I asked whether every bounded function (not continous) is integratble. My professor said it has to be discontinous at uncountably many points. So I decided to test whether what he said was true or false by using the Dirichlet function (since it is discontinous at uncountably many points) and he seemed to have been correct. So the credit of this problem should go to my professor.
• November 7th 2006, 03:31 PM
TriKri
For the first problem, how do you know that $y\geq 1$?
• November 7th 2006, 07:00 PM
ThePerfectHacker
Quote:

Originally Posted by TriKri
For the first problem, how do you know that $y\geq 1$?

Because it is a DIOPHANTINE equation. Traditionally they are solved in only the positive integers.
• April 10th 2008, 04:18 PM
arbolis
I didn't try the second exercice, but I think there is an easier way to do it. If you define a partition P in [a,b], with every step equal to $(b-a)/n$. When you calcul the $L(f,P)$, I think its value is 0. Now calcul the $U(f,P)$, it's 1. Thefore the integral is not Riemann integrable. I'm probably wrong, I would apreciate if you tell me so and explaining.
• April 10th 2008, 06:05 PM
ThePerfectHacker
Quote:

Originally Posted by arbolis
I didn't try the second exercice, but I think there is an easier way to do it. If you define a partition P in [a,b], with every step equal to $(b-a)/n$. When you calcul the $L(f,P)$, I think its value is 0. Now calcul the $U(f,P)$, it's 1. Thefore the integral is not Riemann integrable. I'm probably wrong, I would apreciate if you tell me so and explaining.

When I came up with that solution above I did not yet know what you were referring to. (About Supremum and Infimum on intervals). Using that it is much easier.
• October 23rd 2008, 08:37 PM
chiph588@
for the second one, couldn't you just divide $[0,1]$ into a partition $P$ and say that
$\lim_{n \to \infty} \sum_{i=0}^n D(x_{i max})(x_{i+1}-x_{i}) = \lim_{n \to \infty} \sum_{i=0}^n 1(x_{i+1}-x_{i}) = 1$

but $\lim_{n \to \infty} \sum_{i=0}^n D(x_{i min})(x_{i+1}-x_{i}) = \lim_{n \to \infty} \sum_{i=0}^n 0(x_{i+1}-x_{i}) = 0$

Therefore $D(x)$ is not Riemann integrable.
• October 24th 2008, 07:38 AM
arbolis
Quote:

Originally Posted by chiph588@
for the second one, couldn't you just divide $[0,1]$ into a partition $P$ and say that
$\lim_{n \to \infty} \sum_{i=0}^n D(x_{i max})(x_{i+1}-x_{i}) = \lim_{n \to \infty} \sum_{i=0}^n 1(x_{i+1}-x_{i}) = 1$

but $\lim_{n \to \infty} \sum_{i=0}^n D(x_{i min})(x_{i+1}-x_{i}) = \lim_{n \to \infty} \sum_{i=0}^n 0(x_{i+1}-x_{i}) = 0$

Therefore $D(x)$ is not Riemann integrable.

Yes chiph588@, that's exactly what I've done.
• October 24th 2008, 12:19 PM
chiph588@
Whoops! I didn't see you posted that, sorry.
• October 24th 2008, 12:42 PM
arbolis
There is no problem! (Wink)