What is the additive identity property?
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Zero is known as the additive identity because when you add it to anything (say "x"), it doesn't change the value of "x".
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This depends on the specific Lie group. The exponential map for GL^+(n,R) is not surjective (for n=2, [[-1 0][0 -2]] is not an exponential), but is is surjective for GL(n,C). Notice that it cannot be surjective unless the group is connected [GL(n,R) is not]. The map tends not to be injective. For example exp[[0 0][0 0]] and exp[[2pi*i 0][0 0]] are both the identity. Compactness considerations give negative answers for bijectivity for all compact Lie groups.\n\nThe exponential map is bijective on a neighborhood of the identity for every Lie group. The reason is that the differential map is essentially the identity and so the Inverse Function Theorem applies. Cleartly it is both injective and surjective for R^n.
535203
A relatively low cost, easy to operate video signature comparator and a corresponding method by which a known signature is visually compared to a sample signature to determine whether both signatures were made by the same individual. The signator comparator has particular application to forgery detection or to applications for verifying the authenticity of a signature. An electronic circuit is disclosed which displays video signals that are representative of the known and sample signatures. The circuit enhances video image density variations through extreme contrast and brightness adjustment and forces the lighter regions of each signature to white, such that only the dark stroke lines (indicative of those regions having the greatest density where the maker applied the most pressure) are displayed on video monitor. By visually comparing the characteristic density/pressure variations of the known signature to the sample, an observer can more accurately determine whether both signatures originated from the same maker.
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To find the opposite, or the additive inverse, just change the sign to the opposite sign.\nIf it's positive, then make it negative.\nIf it's negative, then make it positive.
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a*(a+b)=a^2+a*b\nis a special case of \ndistributive law of multiplication over addition:\nx*(y+z)=x*y+x*z\n\nReading it from RHS to LHS,\nit is taking the factor a out of the sum.\na^2+ab=a*(a+b)
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Consider the left-hand side(LHS) [cot(x) + 1] / [cot(x) – 1]. We want to prove that this is equal to the right-hand side (RHS), that is, [1 + tan(x)] / [1 – tan(x)]. Thus, we have the following:\n\nLHS = [cot(x) + 1] / [cot(x) + 1]\n        =  { [cot(x) + 1] / [cot(x) – 1] }* { tan(x) / tan(x) }.\n\nNOTE: We multiply tan(x) to cot(x) + 1 and the other tan(x) is multiplied to cot(x) – 1.\n\n        =  { tan(x) [cot(x) + 1] } / { tan(x) [cot(x) – 1] }\n\n        =  { tan(x) cot(x) + tan(x)] } / { tan(x) cot(x) – tan(x)] }, by distributive property of multiplication over addition (subtraction)\n\n        =  { 1 + tan(x)] } / { 1 – tan(x)] }, since tan(x) cot(x) = 1 by reciprocal identity.\n\n        =  RHS.\n\nIndeed, [cot(x) + 1] / [cot(x) – 1] = [1 + tan(x)] / [1 – tan(x)]
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x+0=x is always true. This is the definition of the identity property of addition.
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Its alt+3, and FYI Alt and a number is a way of printing orginal symbols from ASCII text, (first printer data kinda thing)
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sqrt x^2 = x is  not an identity.Its true only 4 positive values of x.\nThe reason is explained as follows:-\n\nConsider x > 0\nthen x^2 > 0 (of course,yes)\nthen sqrt x^2 = ± x.\nsince x > 0(as considered above),\nhence sqrt x^2 = x (for all x > 0).\n\nNow,Consider x < 0\nthen x^2 > 0(of course,yes)\nthen sqrt x^2 = ± x.\nsince x < 0(as considered above),\nhence sqrt x^2 = - x (for all x < 0).\n\nSo in this case,after taking sqrt,we have 2 insert a negative sign 2 balance the equation.\n\nHence,sqrt x^2 = x is not an identity(coz its true only 4 positive values of x).
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an essential and distinguishing attribute of something or someone
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In order to provide you with the theory behind the distributive property, here is a more technical explanation than others are likely to give, but I think you'll be able to follow it:\n\nFirst I need to tell you what an operator is. Suppose we have something like:\n\n5+3=8\n\nThen, "+" would be the operator. Specifically, it's called the addition operator. But if instead we had:\n\n2-1=1\n\nThen "-" would be an operator, and, again, more specifically, it's the subtraction operator.\n\nIn fact, any time we have something which acts between two numbers and produces another number, it's an operator. So, things like multiplication and division are also operators. In fact, we can even make up operators like this:\n\n2o3=10\n\nThere, I made up an "o" operator which happens to produce 10 when you give it 2 and 3. Maybe we would even define the "o" operator like this:\n\naob=2(a+b)\n\nOr, in words, "a operator b equals 2 times the quantity a plus b." So that for 2 and 3 it gives:\n\n2o3=2(2+3)=2x5=10\n\nOk, so now that we know what an operator is, we'll next say the following: operators have properties!\n\nFor instance, the addition operator has the commutative property; i.e. it doesn't matter what order you put the operation in. So, 1+1=1+1. The division operator does not have this property; i.e. 2/1 does not equal 1/2.\n\nThe distributive property is a property that some operators have, while others do not. Actually, the distributive property only comes into play when thinking about combinations of operators. The most common example of two operators that when put together have a distributive property is multiplication and addition. But, in general, an operator is said to have the distributive property if and only if it looks like this:\n\nav(b f c)=avb f avc\n\nwhere a, b, and c are numbers, while v is one operator and f is the other. Here's how it works for multiplication when combined with addition:\n\na(b+c)=ab+ac\n\nFor example,\n\n5(3+1)=5(3) + 5(1)\n\nWhich, of course, is equal to 15+5, or 20.
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simple\n\nintegration = addition \ndifferentiation = division
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A - sign outside (  ) changes all the signs inside the (  ). \nn^2   -3n   +1 \n-4n^2  -n  +5   Stacking the values in columns helps combine the 'like' terms, which have the same letter(s) to the same power(s)\n\n-3n^2 -4n +6
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I can help you with the part of  "Identify the expression that completes the equasion so that it is an identity.\nsin^2x/(1-cosx)"\n\nYou got sin^2x/(1-cosx). Then to complete the equation you multiply[(1+cosx )/(1+cosx)] to the equation.  \n\nThen you get {[(sin^2x)(1+cosx)]/[(1-cosx)(1+cosx)]}...Then by factorization you get that {[(sin^2x)(1+cosx)]/(1-cos^2)]}....The (1-cos^2) is the same as to sin^2x.  \n\nThen you cancel the terms sin^2x/(1-cos^2)...It leaves the equation as following ==> (1+cosx)(1) = (1+cosx)\n\nHope someone helps you with the rest SOrry
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x to the n + y to the n = z to the n where n is an integer? ... damn it's been so long I can't remember for sure. Some Brit solved the proof already.
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I wont give you the entire proof, just give you the key arguments.\n\nknowing the joint distibution of (X, Y) allows to compute the marginal densities of X and Y.\nIndeed, let f_X be the marginal density of X, then:\n\nf_X(x) = &#92;int f(x, y) d_y\n\nf_Y(y) = &#92;int f(x, y) d_x\n\nNow, you might compute E(Y| X = x) = &#92;int y f_{Y|x} (y) d_y\n\nwhere f_{Y| x} is the conditionnal density of Y knowing the event X = x\n\nrecall that f_{Y| x}  = f(x, y) / f_X(x), and you just have to compute the resulting integrals\n\nregards,\n\njfp
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Let s be in S.  Since f is a homomorphism, f(r)=f(1)*f(r), thus \ns*f(r)=s*f(1)*f(r).  We rearrange and factor to get\n[s-s*f(1)]*f(r)=0.  The hypotheses imply that f(r) is not a zero divisor, so we conclude that s=s*f(1).  Similarly using f(r)=f(r)*f(1) we get s=f(1)*s, thus f(1) is an identity element.