Recent content by seaglespn

Thanks for your help @nazzard... :smile: Cheers!

\[ \mathop {\lim }\limits_{n \to \infty } \frac{{1 - q^{n + 1} }}{{1 - q}} = \frac{1}{{1 - q}} \] ? Sorry about double post... my refresh is kinda slow :smile:

\[ \mathop {\lim }\limits_{n \to \infty } \frac{{1 - q^{n + 1} }}{{1 - q}} = \frac{1}{{1 - q}} = ? \]

Ok, I have done the math, and I end up with the correct answer, after I wasn't so sure about the : \[ b_n = b_1 \frac{{q^n - 1}}{{q - 1}} \] Where the power of q must be the TOTAL number of elements... Sorry, my mistake... :smile: The sum thends to a constant... but that might be a...

Some sums, don't sum up :) I have a problem that require some math tricks, and after I tried to solve it myself I looked at the answer and I don't understand how this is done : \[ \sum\limits_{k = 0}^n {\left( {\frac{2}{5}} \right)^k } + \sum\limits_{k = 0}^n {\left( {\frac{3}{5}}...

This is what I've told earlier... I just forgot to mention that there must EXIST the limit in f(x), where x is the point we're checking out... So please don't get me wrong... but there was a single point to correct me, you wasn't supose to totaly disagree with me, I wasn't after that sure...

If I was knowing the answer I wasn't asking for help. :smile: And I know what means "differentiable", I said that I don't know what else I've got to prove other than that that function in continuous in order to demonstrate it's differentiability. So can anyone show me the correct solution to...

In 0 f=0 f'=1; so the angle made by the tangent is 45. As far as I can tell, and see, the function is continuous, I can't find or remember a larger definition on differentiability... I don't know what else has a function to be in order to be differentiable... help ?! :smile:

Well, I don't know how to derivate |x| , but I thought that first I need to discuss |x| , and thus to create two functions, and after that I need to see if first limit from left (x>0) equals limit in 0 and also equals limit from rigth( in case x<0 ), if this is correct, then I do the derivate...

prefix mega means 10^6 over the basic unit. prefix giga means 10^9 over the basic unit. prefix terra means 10^12 ... so there are 1.000.000.000.000 flops in a terra flop 1.000.000.000 flops in a giga flop and so on... These prefixes can be applied to any unit, not just to flops(i.e. MegaByte...

Thank you guys for another promt ,very needed , help. :smile:

well, if x<0 then |x|=-x, else if x>=0 |x|=x... So is never negative, always possitive, a function witch draws a line, twice :), so this was the big deal? :D Now the question is seams easy to me... the denominator is never 0, because |x| cannot be -1, so that function is always differentiable...

I have f(x) = x / (1 + |x|) The task is to study f's derivability and calculate f'. I don't have any experience in this type of derivates, pls help :smile: .

Thanks again to all of you, I just finished the problem, and another one (same type), and I totally understood the tehnique, thx!

@HallsofIvy : Thanks a lot! Now I can get the things started :).