Recent content by Samy_A

The following paper from 1951 constructs a continuous function x, for which the convolution x*x is only differentiable in 0. Jarník, V. "Sur le produit de composition de deux fonctions continues." Studia Mathematica 12.1 (1951): 58-64 https://eudml.org/doc/216531

Correct.

You beat me by two months: 3 august will be 36 for me.

What is more expensive: seven discounted memory sticks or five non-discounted-memory sticks? In algebraic terms, the same question would be: which number is larger, 7 * (x-3) or 5x?

Of course. As I said, one of those days ... It could be 2, 3 or 4 people.

So, last last try at 5. For fixed n, let ##P(m)## be the probability that the m'th generation still doesn't include the starter. We have ##P(1)=1, P(2)=1##. For m>2: For starter not to be in generation m+1, he has to satisfy two conditions: not be in generation m (probability P(m)), and not...

Never mind, one of those days. :oldconfused:

Second shot at 5: I'm not sure about terminology. Here I define generation m as the people who are told the rumor during the m'th iteration of the process. EDIT: it wasn't clear in the OP, but from post #87, it appears that generation m consists of all the people who have been told the rumor...

Deleted: Note to myself: read the question before answering.

Is this true forever, or only for the next generation? I mean: say A tells the secret to B. After a few generations, the secret is again told to B. Can he now tell it to A?

That was quick. Anyway, then we are back at post #7. Why did you think that the limit should be 2? Give your detailled reasoning, then someone will hopefully be able to help you find the error(s). Because, as @SammyS wrote, the correct limit is 1.

Let's start with the RHS: ##\displaystyle \ [\lim_{x \rightarrow 1} (2\sin(x-1)/(x-1))]##. As you stated, ##\displaystyle \lim_{x \rightarrow 1} (2\sin(x-1)/(x-1))=2##, so that ##\displaystyle \ [\lim_{x \rightarrow 1} (2\sin(x-1)/(x-1))]=[2]=2##. For the LHS, ##\displaystyle \lim_{x...

Can you explain why you think that ##\displaystyle \lim_{x \rightarrow 1} [2\sin(x-1)/(x-1)] = [\lim_{x \rightarrow 1} (2\sin(x-1)/(x-1))]##?

What you claim is that ##\displaystyle \lim_{x \rightarrow 1} [2\sin(x-1)/(x-1)] = [\lim_{x \rightarrow 1} (2\sin(x-1)/(x-1))]##. But is that claim correct? Is [] a continuous function?

Are there cities in the US with more than one zipcode? If there are, the first dependency is wrong. You would need a table with CityName, CityState, CityZip, CityPK, and then have a foreign key in the Companies table referring to CityPK. (And to fully normalize, you could make two tables, one...