what is tellegen's theorem? state?
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For any linear network which may be linear or non linear, active or passive, time varying or time invarient, the algebraic sum of powers delievered by all branches at any instant is zero. All currents and voltages in that nework must satisfy Kirchoff's laws
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Im not entirely sure if this is correct but, I know that a relation is a partial ordering if it is reflexive, transitive and anti-symmetric.  \n\nSo to satisfy the reflexive part we need that x>=x, which is trivially satisfied by f(x)>=f(x).  For the transitive part we need that if x>=y and y>=z then x>=z, which will be true if for f(x)>=f(y) and f(y)>=f(z) then f(x)>=f(z).\nFor the anti-symmetric part we need that if x>=y and y>=x then x=y, so when f(x)>=f(y) and f(y)>=f(x) then x=y\n\nSo this would suggest that the necessary and sufficient conditions on f are that: (1) when f(x)>=f(y) and f(y)>=f(z) then f(x)>=f(z).  (2) if f(x)>=f(y) and f(y)>=f(x) then x=y.  So f would appear to be one to one.\n\nI hope this helps
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Hi!\nI'm suprised nobody used the easiest way that i will demonstrate now.\n\nLets assume further that x is positive real.\n\nWe start with a simple fact that ANY squares are nonnegative. The idea is we "choose" a square intelligently to suit our purpose. Also not that x does NOT equal zero.\n\nBy experience, \nI choose the square ( x^(1/2) - 1 / [ x^(1/2)] )^2.\nIt must be nonnegative for whatever positive real values of x.\nThus,\n\n    ( x^(1/2) - 1 / [ x^(1/2)] )^2 >= 0\n\nExpanding and regrouping, we get,\n\n    x - 2 + 1/x >= 0 \n      \nor equivalently,\n\n      x + 1/x >= 2\n\nThus, minimum is 2. \nWith equality when x^(1/2) = 1 / [ x^(1/2)], or\n x = 1.\n\nDone.\n\nAlternative Method.\n\nOne can use two variable AM - GM,\nFor any two positive reals m and n, we have\n\n ( m + n )/ 2 >= (mn)^(1/2)\n\nWith equality if and only if m = n.\n\nJust set m = x and n = 1/x, to get,\n\n( x + 1/x )/ 2 >= 1\n\nor  x + 1/x >= 2\n\nAgain, the minimum values is 2 with equality if and only if x = 1 .\n\nRegards
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Hint: If an analytic function is 0 on a set with its accumulation point, then the function is identically 0.\n\nNow, either f or g has to have this property since at each point one or the other is 0.
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http://en.wikipedia.org/wiki/Liskov_substitution_principle click the link to know in detail.
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do you *really* think you're going to get help in set theory from Yahoo Answers?\n\ntry the Wikipedia Reference Desk (http://en.wikipedia.org/wiki/Wikipedia:Reference_desk ).
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Pythagoras Theorem asserts that for a right triangle with short sides of length a and b and long side of length c \na2 + b2 = c2 \n\nor c2-a2 = b2 \n\nhere is a link with all the necessary thing you have to know about this theorem . I hope it helps , just read it carefully and everything is going to be fine :)
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Godel's theorem is a rather technical result about axiom systems. It is widely misunderstood, so be careful of popular explanations of it.\n\nThe set-up is as follows:\n\nAn axiom system consists of a collection of symbols and a collection of axioms (which are strings of symbols) and rules of deduction (which are ways of taking strings of symbols and obtaining another string of symbols). For Godel's result, there should only be countably many symbols (usually there are only finitely many in practice) and it should be possible to decide whether a string is an axiom via a computer program. There are also rule for deciding which strings are 'well formed'.\n\nMost of mathematics can be done in the context of axiom systems of this sort. For example, Zormelo-Fraenkl set theory is of this type.\n\nGodel's result says that for any axiom system which is strong enough to model the natural numbers there is a well formed formula which can neither be proved nor disproved from the axiom system.\n\nEssentially, Godel's result shows that there is no reasonable axiom system for the natural numbers that can answer every question that arises in that system. For example, with set theory, it is known that the axiom of choice can neither be proved nor disproved from just the ZF axioms. If the axiom of choice is added, it is known that the continuum hypothesis can neither be proved nor disproved from the new axiom system.\n\nProbably, the best way to interpret Godel's result is to say that axiom systems are inherently limited in their ability to answer questions that arise in those systems. If they are strong enough, there will be questions that can't be answered in the system. That doesn't mean the questions are unanswerable because considerations outside the system (aesthetics, for example) may dictate an answer.
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There are many proofs but with fallacies. Please visit:\nhttp://www.math.toronto.edu/mathnet/falseProofs/first1eq2.html\n\nhttp://www.math.toronto.edu/mathnet/falseProofs/second1eq2.html
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let |f|=k (some constant).let u,v be the real & img parts of 'f'.\n            the we have |f|=u^2+v^2, u^2+v^2=k^2\n      differentiating partially,wrt to x&y ,and \n\n    USING CAUCHY-REI EQ'S FROM TWO RESULTING EQ'S \n\nu finally get     (u^2+v^2)(du/dx)=0.\n                 &  (u^2+v^2)(du/dy)=0. \nfrom above two eq's   we have \n                     k^2=u^2+v^2=0  implies u=0&v=0\n\n                  so 'f' is constant.
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Hint: Let C=X&#92;O and let f(x)=dist(x,C). The other way is trivial.
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That's kind of a tall order, mentioning every single definition and theorem related to vector spaces and subspaces.  I suggest you ask your professor for help on this one and read the textbook if you need a thorough explanation.\n\nBut anyway, the main idea is: suppose you have a subset W of a vector space V.  Then W is a subspace of V if W itself is a vector space when using the operations inherited from V.\n\nThen there's this theorem which basically says the same thing:\n\nLet V be a vector space and W a subset of V.  Then W is a vector subspace if and only if\n\n1) W is nonempty,\n2) for every x in W and scalar c, cw is in W, and\n3) for every x, y in W, x + y is also in W.\n\nExamples:\n\nA) Let P3 be the polynomials of degree at most 3.  Then P2 (which is the set of polynomials of degree at most 2) is a vector subspace of P3.\n\nB) Let P3 be as above, and let W = { f in P3 : f(7)=0 }.  Then W is a subspace of P3.\n\nC) Consider R^4 = { (x,y,z,w) : x,y,z,w in R }.  Let U = { (x,y,z,w) in R^4 : x + y = 0, z = w }.  Then U is a subspace of R^4.\n\nNonexamples:\n\nA) Consider R^2 = { (x,y) : x,y in R } and D = { (x,y) : x^2 + y^2 <= 1 }.  So R^2 is the real plane and D is the disk of radius 1 centered at the origin.  D is not a subspace of R^2 since it is not closed under addition or scalar multiplication.  For instance (0,1) is in D, but (0,1) + (0,1) = (0,2) is not.  Also, if |c| > 1, then c (0,1) is not in D as well.\n\nB) Consider P3 as above.  Then the subset S = { f in P3 : f(7) = 1 } is not a subspace.  I'll let you figure out why.\n\n\nYou can verify that all these are subspaces or not using the theorem above.  I hope this helps you out.  Cheers!
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Do you mean Chebychev's theorem of proof of Bertrand's postulate on prime number distributions, or Chebychev's theorem concerning standard deviations? \n\nEdit your question to clarify and I'll answer more...\n\nOk, Chebychev's theorem states that at least 1-(1/k^2) of a distribution lies within k standard deviations of the mean.  The reason k must be more than 1 is that at that low of a standard deviation exactly what type of distribution you have becomes important.  Chebychev's theorem works for any distribution, so it can be used to assess general population size, at least within a fairly wide range, with less analysis using just the sample standard deviation.  So, it works for sparse data, non-normal distributions, and so on, within its limits of only holding for calculations of what lies within a range larger than one standard deviation.  \n\nSo, for example, you can take a small subset from a set of data with a mean of 50, and a *sample* standard deviation  of 8, and from Chebychev's theorem know that if you were to take a larger sample, at least 75% (1-1/2^2) would lie within the range of 44 to 66, that is, 50+/-16.
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You might try reading Whitehead and Russell's Principia Mathematica.  I can't promise it will help, as I haven't read it, but I've been told that this issue is addressed.
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Rolle's theorem states that if a function f is continuous on a closed interval [a,b] and differentiable on the open interval (a,b), and f(a) = f(b) then there is some number c in the open interval (a,b) such that\n\nf '(c) = 0. \nIntuitively, this means that if a smooth curve is equal at two points then there must be a stationary point somewhere between them. All the assumptions are necessary. For example, if f(x) = |x|, the absolute value of x, then we have that f(-1) = f(1), but there is no x between -1 and 1 for which f '(x) = 0. This is because that function, although continuous, is not differentiable.\n\nThe theorem was first stated in India by Bhaskara in 1150, and later by Michel Rolle in 1691.\n\nRolle's Theorem is used in proving the mean value theorem, which eliminates the requirement that f(a) = f(b).\n\nI just covered that in my calc class!!\n\nTry this for more:\nmathworld.wolfram.com/RollesTheorem.html
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The Golbach Conjecture is kind of special.It cannot have proof....but in an estrange way.  It has been proved that it's a semidecidable proposition.\n\nA decidable proposition is a proposition that cannot  be proved as wrong or right. We are thinking here, in propositions of arithmetic, and Kurt Gödel proved that those propositions (the decidable) exists in arithmetic so arithmetic is an incomplete theory.  (That's incredible man!, it changed the way we see mathematics).\n\nA semidecidable proposition is a proposition that cannot be proved to be decidable or not ... complex,  but interesting. If you could prove that  Goldbach conjecture is true or false you were proving that it's decidable. So, it doesn't have known proof (even if it is impossible to prove that it cannot be proved). \n\nYou can look in:\n\n\nhttp://en.wikipedia.org/wiki/Decidability\n\nI suggest to see Gödel proof.  It´s simple but impressive!
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Richard Bellman made this principle in regards to a two-stage optimization to get V(t+s, x, φ)\n\n   Using any U up to a time t, with a controller U changes to an optimal one.   At time t+s we get a performance J(t,x,V(s,∙,φ),U).  If you take the supremum with respect to U you get V(t+s, x, φ)