bloodhound

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Well... yea... i am back. Done me exams. One of the modules was a bitch, and it dragged my average down. hovering around 70% at the moment. Didn't really put much effort into it. I normally start revising the night before the exam which has generaly been my style and has got me through so far. don't know if I should change it. But with 6 exams with 4 in 4days maybe i took a toll. Well anyway I am starving as usual. Hopefully I will start posting again. So... yea... how's everyone been doing?

pretty sweet [ce]H2SO4 + Al[/ce] maybe you should add some stuff for the commonly used arrows as well.

just nitpicking.. only when the sin is constrained to R

My domestic news comes from BBC. I catch up international news on Yahoo. And for analysis I have subscription to The Economist. For tech stuff and reviews I got to tomshardware.

yeah sorry, i should have made it more clear. Its quite easy to visualise partial derivatives when you have a function of two variables.

Just use the definition... you just have to show that for every positive epsilon, you can find a delta st. if the distance between x and 1 is greater than 0 and less than delta, then the distance between sqrt(x) and 1 is less than epsilon.

well, that how the partial derivative defined. It's just an ordinary derivate where one of the variables is taken to be a constant.

I was just saying if some vectors span R^3 they dont have to be necessarily linearly independent.

Hmm... I am not sure about that... I think it spans iff every element of R^3 can be written as a linear combination of the column vectors. A set of vectors which span a space and on top of that are linearly independent is the basis set.

well. just take another example then... [math](1-\sqrt{2})+\sqrt{2}=1[/math]

I usually idle on irc 24/7 its just that mirc becomes so cluttered with all the channels and server open , i can't be boetthered to connect to yet another server.

First of all I hate having multiple servers... I am mostly on Rizon with nick bhound.

and you can share files as well!

algebracus, you did the same thing i did. but i gave up after the 4th wave of elimination as i couldn't be bothered anymore.

What exactly is this washer method?

what level does this have to be.

Just general info: A point of inflection is a point where a curve crosses its tangent. So a point of inflection doesnt have to be a stationary point. But a stationary point which is neither a maximum or a minimum has to be by default a point of inflection. For the case where the second derivative is equal to 0 at the stationary point, one of the ways forward is to look at the power series (taylor series) of the function and how it behaves close to the stationary point.

yeah. this been answered before. http://www.scienceforums.net/forums/showthread.php?t=3903 heres my solution.

i assume you meant [sinx][exp(-sinx)]. just do the most obvious thing which is to find its derivative. which comes out to be [cosx]exp[-sinx] - [sinx][cosx]exp[-sinx]. to find the stationary points , find points such that the derivative at the point is zero. so setting that to 0 we get [cosx]exp[-sinx] - [sinx][cosx]exp[-sinx] = 0 iff cosx - sinx[cosx] = 0 as the exponential is never 0. so we have either cosx = 0 or sinx = 1 which gives just the same arithmetic sequence of solutions namely (pi/2 + n*pi), where n ranges over the integers.

i thought the orbits were just described by probabilitiy densities. [but i might be wrong as i am not a physicist.]

well, the first flaw lies in you taking squares. you have written[math]e^{(i\pi)^{2}}[/math] , when it should be [math](e^{i\pi})^2[/math] but then, you might say [math](e^{i\pi})^{2}=1[/math] [math](e^{2i\pi})=1[/math] [math]2i\pi = \ln(1) = 0[/math] and how does that work? its just like dave said, taking logarithms of complex numbers is a dodgy proposition. Lets just say it is not always true that [math]\ln{e^{z}}=z[/math] if you are working with complex variables. [edit] why isn't the latex showing up?[/edit]

then i guess adobe premier is the way to go.

why do mpegs when u can do avi.