Recent content by certainly

It burns through just about everything, except teflon, right? Can you describe the extra laboratory precautions that you had to take when handling that stuff ? And (if you are at liberty to) what you used it for ?

Yeah, your right. I didn't actually do the integral, I only saw the answer and thought, this might take some time.

In regards to post #9:- I'm not entirely sure what you're trying to do here. Did you have a look at the links I provided in post #8. I think your trying to come up with a value for ##\int_0^{\pi} \frac{1}{1+\alpha cos(x)} dx## for ##|\alpha|<1##using ##\int_0^{\pi} \frac{1}{\alpha-cos(x)}...

See this, example 3. Now let's do that daunting looking integral. Like I said before, this is a good problem. First write ##\int_0^{\pi} \frac{1}{\alpha}-\frac{1}{(\alpha)(1+\alpha cos(x))} dx## for the original integral. However ##\int \frac{1}{1+\alpha cos(x)} dx## itself is not nice, in-fact...

Hmmmm...I see your point of view. [Edit:- I am, more or less self taught. So sometimes there exist these small gaps in my knowledge, that make me look like a complete beginner :-)]

But there are also the floor and ceiling functions which have constant 2nd derivatives, where they are continuous and also the sawtooth function (the fractional part of ##x##) which will have a constant 2nd derivative when ##x## is not an integer.

Oh! I see, thanks!

But analytically speaking aren't we first taking the absolute value and then squaring it, so it shouldn't be a polynomial right ? (I wasn't sure of this, that is why I said "I think")

##|x|^2## is not a polynomial (I think), and it has a constant second derivative.

There is a good deal of material. For instance the corollary to a theorem of Fermat :- "Any prime ##p## of the form ##4k+1## can be represented uniquely as the sum of 2 squares. Or (related) Lagrange's theorem:- "Any positive integer ##n## can be written as the sum of 4 squares, some of which...

Ah, yes! silly error on my part. Edited post accordingly.

Yes, this is correct. But ##a## and ##b## don't need to be prime. As long as, ##gcd(a,b)=1## and ##a|n## and ##b|n## then, ##ab|n##.

Actually, if 2 and 3 divide a number ##n##, then since ##gcd(2,3)=1##, ##n## also divides ##2\times 3=6##.

Did I scare you off ? I assumed you were familiar with some number theory terminology. If you have any questions, please ask.

Please...spend...a...little...more...time...writing...your...posts...a...lot...more...clearly.