Recent content by Petek

Your argument looks good* (we assume that c > 0). You've shown that if |x| < c, then -c < x < c. To show the statements are equivalent, use a similar argument to show that -c < x < c implies that |x| < c. *Except you didn't cover the case x = 0, which is very simple. If you're writing this...

Instead of saying that the absolute value of a number is its distance from zero, I suggest that you use the following definition: If ##x \geq 0## then ##|x| = x##. If ##x < 0## then ##|x| = -x##. Then consider the two cases in which either ##x \geq 0## or ##x < 0##.

I got the same result as you in Excel (most current version). I also plugged the equation into WolframAlpha and got the same result (both results for Korifice).

Thanks. Did not know that.

Both the link following the Members item and SEARCH now work as expected. My post count wasn't updated, but I'll assume that stat isn't updated in real time. Thanks Greg!

If I select USER from the main forum page, the link after Messages returns only two of my posts. This action previously returned all of my posts. Using the forum's SEARCH function to search for my posts gives the same result. Has there been a recent change to Search?

The expression ##f(x) \ll g(x)## uses what is called Vinogradov notation. It means that there exists a constant ##C > 0## such that ##|f(x)| \leq Cg(x)## where it's usually understood that ##x \rightarrow \infty##. The Vinogradov notation ##f(x) \ll g(x)## is the same as the so-called Big O...

Have you considered posting your question on MathOverflow? That site caters to research-level problems. It has a tag for questions about Bessel functions, so you might find some advice there.

What if ##x = 0.101001000100001000001 \cdots##? (with the number of zeros between ones increasing by one each time). That number is clearly irrational, but the set S contains numbers of the form ##0.10 \cdots, 0.0010 \cdots, 0.00010 \cdots## (letting k equal a suitable power of 10) and so on...

Perhaps I should flesh out my argument, since I wasn't assuming that k and n were fixed. Your lemma states, in effect, that given any ##\epsilon > 0##, there exist integers k and n (depending on ##\epsilon##) such that ##|k\pi - n| < \epsilon##. I argue that, if this is true, then there exist...

Is this statement true? Here's a heuristic argument casting doubt: Let ##x = \pi##, for simplicity. We can assume, wlog, that ##\epsilon < 1##, Then the expression becomes ##|k\pi - n| < \epsilon < 1##. That implies that ##n = [k\pi]##, the greatest integer less than ##k\pi##. Thus, ##|k\pi -...

What about using the partial convergents of the continued fraction for ##\pi##?

@anemone Welcome to PF!. I think I solved some of your challenge problems on MHB.

The proposition states that a certain property holds for any five consecutive integers. You have only shown that it holds for five specific integers. You need to generalize your argument.

The course description appears to cover topics from functional analysis. However, doesn't the instructor have a public email address? Even if on sabbatical, he might recommend a suitable book.