Muzzafarath

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About Muzzafarath

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  1. What's the worse thing a gf has done to you?

    Quote:Original post by Promit Quote:Original post by ukdeveloper Because people these days are twisted and morally corrupt. Any person in their 70s or 80s will tell you this, and how things have changed.Exactly. Society nowadays is fucked up. First of all they let those damn blacks use all the same facilities as white people, and they even go on golf courses. Wouldn't have seen that back in the old days. And women are just so damn uppity, voting and wearing pants and generally abusing all the privileges they've been given in this world. They just don't respect their male betters anymore. Then you've got these poor blue collar workers unionizing and making demands of the rich folk who are way too good for that kind of treatment. Even the music nowdays, first they had this rock n roll and now the hip hop and rap, it's corrupting our youth. We're not even trying to send those goddamn commies to jail anymore, and the Japanese are allowed to roam the streets freely. Between that and letting colored kids into our schools, how is there any hope at all? Morals just aren't what they used to be in the good ol' 30s. Moron. There are even women vocalists these days. Good heavens.
  2. Any knowledgeable complex math people?

    I'm afraid you'll need to throw away everything you've done. The only way forward for you would be to prove that lim(z->z0) Imz = Imz0, which is essentially the original problem. The Pythagorean theorem says that |z - z_0|^2 = |Re(z) - Re(z_0)|^2 + |Im(z) - Im(z_0)|^2. Use this to find an inequality which relates |Re(z) - Re(z_0)| and |z - z_0|.
  3. Linear Algebra Question

    You could prove that zero is not an eigenvalue of BA - I. If (BA - I)x = 0, then you'd find that AB has a certain eigenvalue, which can't be...
  4. Plane representation

    Seeing as all those forms could be negative, they're all wrong. The first one comes closest to being correct, though. Consider a plane ax + by + cz + d = 0, with N = (a, b, c) normalized. Let P = (p, q, r) be any point. Let Q be the point on the plane which is closest to P. It should be geometrically clear that Q - P = kN for some real k (that is, Q - P must be perpendicular to the plane). Then Q = kN + P = (ka + p, kb + q, kc + r) but Q also lies on the plane, so we must have a(ka + p) + b(kb + q) + c(kc + r) + d = 0 <=> ka^2 + ap + kb^2 + bq + kc^2 + cr + d = 0 <=> k(a^2 + b^2 + c^2) + ap + bq + cr + d = 0 <=> k = -(ap + bq + cr + d). (Remember that a^2 + b^2 + c^2 = 1 since (a, b, c) was assumed to be normalized). Hence the distance between P and the plane is |Q - P| = |kN| = |k| * |N| = |k| * 1 = |ap + bq + cr + d|. Quote: Shouldn't the distance between the two be 2.0 (unsigned that is)? No. The point (-1, 0, 0) lies on the plane x + 1 = 0. The distance must therefore be 0.
  5. Plane Equation

    That's odd - I've never seen anyone make that distinction (and it seems totally unnecessary). See for example Mathworld or PlanetMath. Quote: Orthogonality means an angle of 90° between the vectors. I could not see how an angle is defined anyway if one or both of the vectors is of length 0. It is very easy to get around this problem - stop thinking about things so concretely.
  6. Plane Equation

    Quote: This is true in many cases but not in general. To be true both original vectors (a) must not be of zero length, and (b) must not be co-linear. You have isolated the sentence from its original context, and perhaps the context have decribed a situation where that conditions were fulfilled. It's true in those cases as well (if one of the above conditions are met, the cross product in question will be the zero vector, which is orthogonal to ALL vectors).
  7. working across the pond

    I would assume that checking with the foreign government would work a lot better.
  8. math functions..

    But then again, self-sufficiency (in the sense that you can find out things by looking them up) is a trait you'd want to develop.
  9. 2=1

    Quote:Original post by Trap how do you get a/a=1 for a complex a? Surely that's not the problem - every non-zero complex number has a multiplicative inverse (if z = a + bi, then z^(-1) = a/(a^2 + b^2) - bi/(a^2 + b^2) (well-defined unless a = b = 0)). It follows immediately that x/x is well-defined for all non-zero complex x, and is equal to 1.
  10. What exactly is the Dot Product?

    Quote: RPTD's formal definition. The specific definition you refer to is just one consequence of that lemma. The OP's questiion was pretty ambigious, though RPTD didn't give the full definition anyway, any function that takes two vectors and spits out a real number is not an inner product. There's more to it than that. And I assure you that the definition of the "usual" dot product in R^3 is not the consequence of any lemma (then it wouldn't be a definition). Quote: No, it's the only way of multiplying a vector with another vector. Multiplying a vector by a scalar is different, and I'm pretty sure that's not called a dotproduct. I see. So I guess "weighted dot products" (that's probably not what they're called in English) don't exist then... *edit* Quote: So, umm, doesn't this only back up what I was saying? No. It says that the "usual" dot product is /A/ way to multiply two vectors. Not /the only/ way.
  11. What exactly is the Dot Product?

    Quote: a dot product is an operation between two elements of a vector space Rn (vector space of dimension n). Nitpick: you can define inner products on other vector spaces than R^n, and there are other n-dimensional spaces than R^n. Quote: at the most fundamental level, that's what it "really is." What on earth could be more fundamental than the actual definition (which was given in the first reply)?
  12. Geometry question

    You could use arctan, but you could also use arcsin or arccos (as you know all the lengths involved).
  13. trig symbols

    Tangent: I read in some book by Richard Feynman (the physicist), that when he "discovered" trigonometry as a child, he made up his own symbols for the trigonometric functions, but he abandoned them since no one else used them.
  14. were does order of operations come from?

    Quote: Which brings up the question, could you derive equations for things like velocity, acceleration, etc, where it would always give the correct outcome if you did addition before multiplication? You'd just have to place the parantheses at different places.
  15. Quote:Original post by Vasant56 finally.. simple calculus: A sum of two numbers is 4. Their product is 1, what's the sum of their squares (should be 14). I did this: x + y = 4 (y = 4-x) xy = 1 x(4-x) = 1 differentiate: 4 - 2x = 0 x = 2 (wrong) thanks for the help.. Think about it for a moment. You want to find a number x such that x(4-x) = 1, you're not claiming that x(4-x) = 1 for all x (within an interval). Thus, in this case, it makes absolutely no sense to differentiate both sides and expect a sensible answer. This method "kind of" works for certain polynomial equations though, if a polynomial p has a root q of multiplicity at least 2, then q will also be a root of p' (should be easy to prove with the fundamental theorem of algebra and the product rule).