What is Rolle's theorem?
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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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Here's a copy of the actual paper. Good luck understanding it:\n\nhttp://math.stanford.edu/~lekheng/flt/wiles.pdf
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It is Parseval's theorem you seek. "The total energy contained in a waveform x(t) summed across all of time t is equal to the total energy of the waveform's Fourier Transform X(f) summed across all of its frequency components f."\n\nFor a periodic signal, take the energy in each frequency for one period.  Sum those and you have the total energy.  The maximum amplitude during the period would be if all of that energy occured at once (i.e. an  impulse).  Convert that to dBm or DBW and you have your answer in dB.\n\nLook at the last equation in the link in "sources".
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this is the mean value theorem
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The DeMoivre Laplace theorem is a particular case of the Central Limit Theorem.... it easier to prove the CLT with advanced probability than to prove DMLT. \n\nBoth of the proofs and in Durret:   "Probability Theory and its applications", and also in Protter and Jacod: "Probability Essentials"
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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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First, define a new function g(x)=f(x)-x. f(x) has a fixed point iff ∃x: g(x)=0. By the fundamental theorem of calculus:\ng(x) = g(0) + [0, x]∫g'(a) da\n\nLet c be sup(|f'(x)|). Then f'(x)≤c<1, and so g'(a)≤c-1<0. So for x>0:\ng(x) < g(0) + [0, x]∫c-1 da = g(0) + x(c-1).\nAnd for x<0:\ng(x) > g(0) + [0, x]∫c-1 da = g(0) + x(c-1).\n\nHowever, [x→∞]lim g(0) + x(c-1) = -∞ (since c-1 is negative), and since g(x) is bounded above by this function for x>0, [x→∞]lim g(x) must also be -∞. Conversely, [x→-∞]lim g(0) + x(c-1) = ∞, and since g(x) is bounded below by this functon for x<0, [x→-∞]lim g(x) is ∞. But this means that g(x) is positive for some negative value of x and negative for some positive value of x, thus by the intermediate value theorem, ∃x: g(x)=0 and therefore f(x) has a fixed point. To see that it's the only one, consider that if g(x) had two zeros then by Rolle's theorem there would exist a point where g'(x) is zero, which contradicts the fact that g'(x) is strictly negative everywhere.\n\nNote that this proof remains valid if c is merely the supremum of f'(x), instead of |f'(x)|.\n\nFor the counterexample with |f'(x)|<1 (but not sup(|f'(x)|)<1), consider f(x)={1/x+x if x≥1, 2 if x<1}. This function is everywhere differentiable ([x→1-]lim f(x) = [x→1+]lim f(x) = 2, and [x→1-]lim f'(x) = [x→1+]lim f'(x) = 0), and ∀x, |f'(x)|<1, but f(x) has no fixed points.
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Its the Pythagorean Theorem. It applies to right angle triangles. \nIt states that a^2 + b^2 = c^2\nWhere a and b represent the lengths of legs of the triangle and c is the Hypotenuse.\nThe Hypotenuse is the side that is opposite the right angle (also the longest side). Whereas the legs are the two sides that meet at the right angle.
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Suppose f : D → R has a limit at x0. Prove that | f | : D → R has a limit at x0 and that limx→x0 | f(x) | = | limx→x0 f(x) |.\n\n\nLet limx→x0 f(x) = c\n\nNow | f(x) | = f(x) for f(x) ≥ 0 = c\n= -f(x) for f(x) < 0 = -c\n\nSo limx→x0 | f(x) |  \n\n= limx→x0 f(x) for f(x) ≥ 0 = c\nlimx→x0 -f(x) for f(x) <0 = - limx→x0 f(x) x < 0 = -c\n\n= |c|\n\n= | limx→x0 f(x) | .................... QED
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it is RAYLEIGHS THEORY
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because of the converse of the intercept theorem.the intercepts theorem is as follows\nif a set of parallel lines make equal intercept on any transversal they will make equal intercepts on any other transversal\nlet ABCD be a trpezium with AB paralle DC\nlet E and F be the mid points of AD and BC\nAE/ED=BF/FC =1/2\nnow by the converse of the intercept theorem\nEF must be paralle to AB
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Clairaut's theorem states that if the second partial derivatives of a real function exist and are continuous at a given point, then the order of differentiation in a second order mixed partial derivative does not matter at that point (i.e., d^2(z)/dxdy = d^2(z)/dydx at that point.)\n\nOf course, if the second derivatives of the function exist and are continuous over the entire domain of the function, then the order of differentiation does not matter at any point.\n\nSee source.
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Bayes's Theorem is a simple mathematical formula used for calculating conditional probabilities. It figures prominently in subjectivist or Bayesian approaches to epistemology, statistics, and inductive logic. Subjectivists, who maintain that rational belief is governed by the laws of probability, lean heavily on conditional probabilities in their theories of evidence and their models of empirical learning. Bayes's Theorem is central to these enterprises both because it simplifies the calculation of conditional probabilities and because it clarifies significant features of subjectivist position. Indeed, the Theorem's central insight — that a hypothesis is confirmed by any body data that its truth renders probable — is the cornerstone of all subjectivist methodology.\n\n\n Bayes's Theorem.\n \n PE(H) = [P(H)/P(E)] PH(E) \n\nIn an unfortunate, but now unavoidable, choice of terminology, statisticians refer to the inverse probability PH(E) as the "likelihood" of H on E. It expresses the degree to which the hypothesis predicts the data given the background information codified in the probability P.\n\n                                                               Bayes's Theorem (2nd form)\n \n PE(H) = P(H)PH(E) / [P(H)PH(E) + P(~H)P~H(E)] \n\nIn this guise Bayes's theorem is particularly useful for inferring causes from their effects since it is often fairly easy to discern the probability of an effect given the presence or absence of a putative cause. For instance, physicians often screen for diseases of known prevalence using diagnostic tests of recognized sensitivity and specificity. The sensitivity of a test, its "true positive" rate, is the fraction of times that patients with the disease test positive for it. The test's specificity, its "true negative" rate, is the proportion of healthy patients who test negative. If we let H be the event of a given patient having the disease, and E be the event of her testing positive for it, then the test's specificity and sensitivity are given by the likelihoods PH(E) and P~H(~E), respectively, and the "baseline" prevalence of the disease in the population is P(H). Given these inputs about the effects of the disease on the outcome of the test, one can use it to determine the probability of disease given a positive test.\n\nSpecial Forms of Bayes's Theorem\n\n\nBayes's Theorem can be expressed in a variety of forms that are useful for different purposes. One version employs what Rudolf Carnap called the relevance quotient or probability ratio (Carnap 1962, 466). This is the factor PR(H, E) = PE(H)/P(H) by which H's unconditional probability must be multiplied to get its probability conditional on E. Bayes's Theorem is equivalent to a simple symmetry principle for probability ratios.\n\nProbability Ratio Rule. \n PR(H, E) = PR(E, H) \n\nThe term on the right provides one measure of the degree to which H predicts E. If we think of P(E) as expressing the "baseline" predictability of E given the background information codified in P, and of PH(E) as E's predictability when H is added to this background, then PR(E, H) captures the degree to which knowing H makes E more or less predictable relative to the baseline: PR(E, H) = 0 means that H categorically predicts ~E; PR(E, H) = 1 means that adding H does not alter the baseline prediction at all; PR(E, H) = 1/P(E) means that H categorically predicts E. Since P(E)) = PT(E)) where T is any truth of logic, we can think of as telling us that.\n The probability of a hypothesis conditional on a body of data is equal to the unconditional probability of the hypothesis multiplied by the degree to which the hypothesis surpasses a tautology as a predictor of the data.\n\nAnother useful form of Bayes's Theorem is the Odds Rule. In the jargon of bookies, the "odds" of a hypothesis is its probability divided by the probability of its negation:
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Rolling friction : http://webphysics.davidson.edu/faculty/dmb/PY430/Friction/rolling.html\n\nSliding friction: http://www.mtsu.edu/~phys2010/Lectures/Part_2__L6_-_L11/Lecture_10/Friction/friction.html
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I had to refresh myself on the method of Lagrange multipliers. The following theorem is paraphrased from a problem in Spivak's "Calculus on Manifolds", an excellent calculus text for mathematicians.\nLet A be a subset of R^n. If g:A -> R^m is differentiable and has rank m on the kernel of g, and if f:R^n -> R is also differentiable and the extremum of f on the kernel of g occurs at a, then there exist m real multipliers L1, ..., Lm such that df(a) = (L1)d(g1)(a) + ... + (Lm)d(gm)(a). (d is the differential operator)\nThe proof is not that difficult, but a concise one would require the introduction of too much machinery, so we'll leave it as something to look up.\nIf g has only one component function (if g:A -> R), then the above is very simple with only one multiplier. If n=2, the kernel of g is just a curve in R^2 and the graph of f is a surface in R^3. In this case, you can visualize the constraint on the surface as a curve in the surface of f whose projection on R^2 considered as a subset of R^3 is the kernel of g.\nIn your case, we want to minimize the distance function f(x, y, z) = Sqrt[x^2 + y^2 + z^2] over the two surfaces above, or equivalently the kernel of the differentiable function g(x, y, z) = (2y + 4z - 5, 4(x^2 + y^2) - z^2) (g is thus a map from R^3 -> R^2, so m = 2. The kernel of g is trivially the intersection of the two surfaces. You should show that the rank of g is 2 whenever g(x, y, z) = 0 to validate the use of Lagrange multipliers). \nWe simplify this by noting that the points corresponding to the minimum of f over the kernel of g coincides with the points corresponding to the minimum of [f(x, y, z)]^2 = F(x, y, z) over the kernel of g which is easier to deal with. The theorem tells us we will have to deal with two Lagrange multipliers, one for each component of g. We thus have the system of 5 equations:\n2x = (L1)*0 + (L2)*8x\n2y = (L1)*2 + (L2)*8y\n2z = (L1)*4 + (L2)*(-2z)\ng(x, y, z) = (2y + 4z - 5, 4(x^2 + y^2) - z^2) = (0, 0)\nThese are 5 equations in 5 unknowns, so it is easily solved. Note that the actual values of the multipliers are seldom important.\n\nIn case you are not familiar with the differential operator, if f(x, y, z) = x^2 + y^2 + z^2 + xy, then df(x, y, z) is a covector (actually a cotangent bundle in more formal terms) such that df(x, y, z) = 2x dx + 2y dy + 2z dz. It is similar to notation for implicit differentiation but has a much more rigorous foundation and is used in differential geometry. The dx, dy and dz terms are treated as basis vectors, thus the above is a 3-dimensional vector, normally called the gradient of f in multivariable calculus, where it is treated as if it were a vector in R^3: (2x, 2y, 2z). You should now see where the first 3 equations come from.
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Please use the search function. This question was asked 5 days ago:\n\nA proper polar coordinate system (one that assigns each point in the plane a unique coordinate) will always leave out a ray of points. For example, the standard polar system that identifies the open set (0, oo) X (0, 2Pi) with R^2 - {0} by the function f(r, t) = (rcos(t), rsin(t)) ( = (x, y) ) does not correspond to any points on the positive x-axis and necessarily leaves out the origin (there is no way to include the origin uniquely). This is usually glossed over in a basic calculus or physics course as one actually uses an atlas of polar coordinate charts (this atlas never includes the origin, but at least covers the plane minus the origin); you use whichever chart is convenient for the problem at hand. Some definitions will disregard the need for a homeomorphism and include the positive x-axis by considering a domain that is not open, (0, oo)X[0, 2Pi)
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That is the central limit theorem.  Each distribution can be uniform, but the distributions themselves are normally distributed in their range, their value, their frequency.\nRead the attached link for more details.