bloodhound
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Posts posted by 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?
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pretty sweet
[ce]H2SO4 + Al[/ce]
maybe you should add some stuff for the commonly used arrows as well.
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Just nitpicking: [math]\sin(x) \leq 1[/math].
Rest of it is fine though
just nitpicking.. only when the sin is constrained to R
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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.
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yeah sorry, i should have made it more clear. Its quite easy to visualise partial derivatives when you have a function of two variables.
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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.
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well, that how the partial derivative defined. It's just an ordinary derivate where one of the variables is taken to be a constant.
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I was just saying if some vectors span R^3 they dont have to be necessarily linearly independent.
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The columns of [math]A[/math] span [math]\mathbb{R} ^3[/math'] if and only if they are 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.
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well. just take another example then... [math](1-\sqrt{2})+\sqrt{2}=1[/math]
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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.
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First of all I hate having multiple servers... I am mostly on Rizon with nick bhound.
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and you can share files as well!
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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.
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What exactly is this washer method?
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what level does this have to be.
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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.
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yeah. this been answered before.
http://www.scienceforums.net/forums/showthread.php?t=3903
heres my solution.
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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.
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i thought the orbits were just described by probabilitiy densities. [but i might be wrong as i am not a physicist.]
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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]
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then i guess adobe premier is the way to go.
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Bush's response was equally clear: If you do' date=' you'll be provoking me and the Congress at your peril."
[/quote']
what exactly is this threat.
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why do mpegs when u can do avi.
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