bloodhound

I think that with enough practice, most of the standard differentials and integrals will be stored into ur brain. just like writing, or multiplying simple number, or walking, its gonna be natural. Loads of Practice is what i would recommend.

i think getting another book just for linear algebra, instead of relying on one chapter in the quantum mech book would be a good idea.

oh yeah, forgot about that. i have been used to using algebra of limits in addition and subtraction. dont think i have ever used ALG in limit of products

for the first two. just use L'Hopital's Rule twice so for a) lim = 10/3 b)lim = 1/2 still doing c) i was lucky these didnt come up in my analysis exam

Ah, i see J'Dona, u must be a physicist. as u used j instead of i. oh yeah, dave, can u corrent my script. i just wanna see what it looks like

yes but it seems that if u put $stuff$ the output will be text instead of maths for example [math]I am Bloodhound[/math] compared to [math]$I am Bloodhound[/math]

I don't understand what you mean by the proof for the discriminant. I assume u want to show that discriminant is [math]b^2-4ac[/math] Take a general quadratic [math]ax^2+bx+c[/math] and equate it to [math]0[/math] [math]ax^2+bx+c=0[/math] [math]x^2+\tfrac{b}{a}x+\tfrac{c}{a}=0[/math] Completeting the square, we get, [math](x+\frac{b}{2a})^2-\frac{b^2}{4a^2}+\frac{c}{a}=0[/math] [math](x+\frac{b}{2a})^2}=\frac{b^2}{4a^2}-\frac{c}{a}[/math] [math]x+\frac{b}{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}[/math] [math]x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}[/math] [math]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/math] the [math]b^2-4ac[/math] is defined to be the discriminant. the second part of the questions is obvious from here [math]$If $b^2>4ac$ then$ \sqrt{b^2-4ac}>0[/math] [math]$Making the roots of the quadratic $x_1,x_2$ real[/math] The distinct bit also can be seen clearly from here. I will leave it to the author of the question to write up the proper solution.

why isn't this script coming out properly? <math>$Let$ {v_1,v_2,...,v_k}$be the basis set for the vector space$V$which are linearly independent, then it is possible to construct another basis set${e_1,e_2,...,e_k}$from the original basis set, such that the basis vectors in the new basis set are all orthogonal to each other, i.e.$<e_i,e_j>=0$for$ i≠j$ where$<,>$ is the inner product defined on the Vector Space$V$ </math>

just testing [math]ds^2=dt^2-(dx^2+dy^2+dz^2)[/math] [math]\int_a^{\infty} f(x)dx=\lim_{L\to\infty}\int_a^{L}f(x)dx[/math] [math]$Let$ {v_1,v_2,...,v_k}$be the basis set for the vector space$V$which are linearly independent, then it is possible to construct another basis set${e_1,e_2,...,e_k}$from the original basis set, such that the basis vectors in the new basis set are all orthogonal to each other, i.e.{:content:}lt;e_i,e_j>=0$for$ i≠j$ where{:content:}lt;,>$ is the inner product defined on the Vector Space$V$ [/math]

dont know if this constitutes a rigorous proof, but here we go anyway Let us consider the function [math]y=\cos(x)+i\sin(x)[/math] Continuing to treat [math]i[/math] like any other number, we have, by differentiation, [math]\frac{dy}{dx}=-\sin(x)+i\cos(x)=i(\cos(x)+i\sin(x))[/math] Hence, [math]\frac{dy}{dx}=iy[/math] [math]i\frac{dx}{dy}=\frac{1}{y}[/math] [math]ix=\log(y)+c[/math] But when x= 0 , y = 1 , So c = 0 [math]ix=\log(y)[/math] [math]y=e^{ix}[/math] Therefore [math]y=\cos(x)+i\sin(x)=e^{ix}[/math] Another way to "show" this, is to use taylor expansions of /cosx and /sinx, form the linear combination [math]\cos(x)+i\sin(x)[/math] and show that its equivalent to the taylor expansion of [math]e^{ix}[/math], where [math]i[/math] is taken to be just another number. For the second part of the question, you just plug in x=pi in the identity.

no, thats wrong. the value of n in [math]s_{xy}[/math] and [math]s_{xx}[/math] is the total number of pairs of observations. ie.[math](x_i,y_i)[/math]

wooooooooooooo. free till the end of september. out of 6 exams, i have had only 1 utterly bad one. rest of them were good - excellent. have to clean up my room of all the scrap paper i used to practise. and finally eat something

there is such thing as chaos in certain iterations. I am not sure if that qualifies as random. maybe someone else here can elaborate

I dont know how mine came about. it just sounds cool and i have been using that username for anything i register for. I loved Hellsing and Cowboy Bebop

It depends what means u trisect an angle by. If ur using a compass and a straight edge, then its impossible to trisect a general angle, but certain angles like pi/2 and pi are. In addition, trisection of an arbitrary angle can be accomplished using a marked ruler check http://mathworld.wolfram.com/AngleTrisection.html

yeah , and that as well

I am Bloodhound. I am a MAths Student at university in UK. Got exams tmr, Stats and MAthematical structures day after. so much to learn. I loved Hellsing and Cowboy Bebop by the way

there are actually two dents. the bigger one made by my Middle knuckle and the smaller one by the one on the right of the middle. The bigger one has approx dimensions. of 2 cm in diameter and about 4-5 mm in deep at its deepest the smaller has diameter of about 1.5cm and about 2-3 mm deep. But u have to see it to really appreciate how big it is.

No, an IQ test measures the ability of a person to do IQ tests. There maybe some correlation between actual "intelligence" and the IQ score, but IQ test should not be use alone to judge how clever a person is. By the way I loved Hellsing and Cowboy Bebop

is that ur pic on the avatar?? if it is then u are proper fit!!! *whistles"

i have a pretty strong fist. i usually go round punching walls and stuff. I made a massive dent on my flats plaster. wish i had a camera. my wrist was out for a week. simple things like turning the door handle or turning the tap on was virtually impossible

i thought relativity was about inertial frames. i dont see anything in motion, or different frames of reference in that flash

dave, thats the first thing that came into my mind as well, but it takes hell of a job to expand A2-B2 the way i did it was to show that A-B is divisible by 9 and A+B is divisible by 11 i wonder if there are any better ways to work round it. i was about to use modular arithmetic but i am rubbish at that.