Eg:At which values of z is modulus of z continuous\nAnswer: modulus of z is continuous at all z not equal to 0\n\nBUT I DONT UNDERSTAND WHY
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Firstly, modulus of z is continuous at ALL z, also for z=0. Cite from attached article 1: "The real absolute value function is continuous everywhere. It is differentiable everywhere except for x = 0."\n\nMay be you mean differentiable, not continuous?\nOr you are talking about continuity, but mean "sign" instead of "modulus" ? Sign(x), which is 1 for all x>0, -1 for all x<0 and 0 for x=0 is continuous at all x except 0.\n\nIn both cases, look at the definition (see attached article 2 for continuity, 3 for differentiability.\n\nBriefly, countinuity of function f(x) in point x_0 mean than any sequence, which limit is x_0, have another limit f(x_n) equal to f(x_0).\nNow you can easily check continuity of modulus and sign.\n\nDifferentiability is a bit more complex, but attached articles have enoygh information.\n\nHope it helps.
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A function relates each of its inputs to exactly one output. That is for each value of x in the domain, there must be one and only one y.\n1) 2) are OK for each x in R there is exactly one y (and vice versa)\n3) Not OK.  Take y=-2 and y=2. Both have x=4\n4) OK. There are x-values that lead to the same y, but that is allowed.\n6) Not OK. Take x=2, y>4. There are an infinite number of y-values.
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The background is , as you rightly say , very technical in the mathematical sense. It starts with developments in Differential geometry ( Marsden Morse) , leading to the catastrophy theory of Renee Thom. Essentially if you have a 'smooth' function ( i.e. differentiable several times ) such as ax^i+by^j+cz^k=0 , then this is actually a prototype for a whole range of functions if you vary a,b and c.if you , perversly , analyse the functions with respect to the a,b and c , there are often areas where a sudden discontinuity occurs.so , for instance your equation may have a solution  for x where a=5 say. if you change a to 5.000001 , then the solution for x is suddenly very different. A way to picture this is to look at the situation where the equation is that of a Corkscrew. for a given x and z , there are many values of y which fit the equation(one for each turn of the screw). I could go on here but hopefully this is enough to get you started.Thanks for the question.
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I'm guessing you looked on wikipedia cause all those confusing terms were there.  Don't worry, I didn't understand it either :-)\n\nThe idea of the Intermediate Value Theorem is that for any interval [a,b] on any continous function f(x), there is a point somewhere in that interval where f(x) = c for any value of c such that c is between f(a) and f(b).  If you think about what a continuous function looks like and try to come up with an exception, you'll probably be able to understand intuitively why it's true.  I don't know that you can formally prove it without all the technical terms, but you should be able to prove in informally.  Basically think about what a continuous function is.  For a function to be continuous on the interval [a,b], you have to be able to trace the graph from a to b without taking your pencil off the paper.  So if there is a point c between f(a) and f(b) for which f(x) does not equal c for at least one value of x between a and b it means there's a hole somewhere.  If there's a hole, the function is not continuous.
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Use the maximum modulus principle. The maximum modulus of f(z) on the ball {z:|z|<=r} has to occur on the boundary and only on the boundary if the function is non-constant.
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I know what you are talking about.\n\nIts in Linear Algebra.\n\nThis system can be expressed quite compactly by writing Ax=b, where A is an m x n  matrix whose elements are the coefficients, aij, in the linear system. We use A to denote both the linear map and the matrix; note that A is linear map from R^n to R^m . The column vector of unknowns is x which belongs to R^n and the right-side column vector is b which belongs to R^n . When b=0, the system is said to be homogeneous.\n\nFirst consider the homogeneous system Ax=0. The solution set of this system is the null space of A, denoted null A. The question is whether or not the solution set includes more than the zero vector. We have seen that if A is not injective (one-to-one), then the null space contains more than {0} . Furthermore, we have seen that if m<n, then A cannot be injective. Putting these observations together, we see that a homogeneous system with more unknowns than equations (n>m) must have nonzero solutions. This makes sense: there are not enough equations (constraints) to determine a solution uniquely.\n\nOops you asked about "nonhomogenus solutions". We will get there now. :)\n\nNow consider the nonhomogeneous system Ax=b. In order for a solution to exist, b must be a linear combination of the columns of A (with weights given by the components of x). Using familiar terminology, this means that b must lie in the range of A = R^m. Introducing new terminology that is used with matrices, this means that b must lie in the column space of A, which is the span of the columns of A. If , then there is at least one solution for all choices of b belong to R^m. This means that if A is surjective (onto), then there are solutions for all . Conversely, if n<m, then A is not surjective and no solution exists for some choices of b. We conclude that for a nonhomogeneous system with more equations than unknowns (m>n), no solution exists for some choices of b. This also makes sense: it is possible that there are too many equations (constraints) to determine a solution. \n\nWhat about m=n?\nA special case arises when m=n. In this case, if the columns of A are linearly independent, they span R^m and all b belong to R^m  lie in the column space of A (including b=0). Thus, all vectors b admit a solution, x, whose components give the weights in expressing b as a linear combination of the columns of A. Because this representation is unique, the system Ax=b has a unique solution for all b belong to R^m , in the case that m=n and the columns of A are linearly independent. \n\nThus, \n\nAx=b has solutions for ALL b. It does NOT depend on the subspace of R^n.\n\nRegards
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You ask a number of questions here.  I will first say that the Mandelbrot set does not really have any direct practical value.\n\nThe mandelbrot set like other fractals is interesting, in that they are calculated out in iterations, and the number of iterations used determines whether or not a point is in the set.  To give an example, you know that if you define a function 2=x+y, you take any two values for x and y and see if they fit that function.  x=2, y=3 is not a point that fits this function, whicle x=0.5, y=1.5 does fit this function.\n\nWith the Mandelbrot set, you cannot look at a point and see if it fits the function.  The way the madelbrot set is formed is to place imaginary numbers on one axis, real numbers on the other, and define the points as follows:\n\nfor any complex number z, such that z=z^2 + 1, iterated over and over does not ever exceed 2, that number z is part of the set.\n\nThere is only one way to determine if a number z is part of the set and that is to say, we are going to do a certain number of iterations, and if z<2 at the end of our calculations, that z is part of the set, otherwise it is not.  This means that whether or not a complex number is part of the madelbrot set depends on the number of iterations you use in your inclusion criteria for z.  There is no absolute answer to the question "Is z a part of the madelbrot set?"\n\nSO yes the Mandelbrot set does go on forever, and the more iterations used in calculating it, the more detailed the fractal gets.
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f(x) = sinx on [0,pi/2] is 1-1 on this interval, therefore there can be no mode, since all values occur only once.\n\nThe mean is equal to the area under f(x) = sinx on this inteval divided by the length of the interval.  In this case, it is 1/[pi/2-0] = 1/(pi/2) = 2/pi.\n\nThe mode is the value where half the area is below this value and half is above it.  Therefore, we want to know for what value of a is Integral from 0 to a of sinx = 1/2.  When you integrate and evaluate, you get cosa = 1/2, which means a = pi/3, so pi/3 is the median.
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(I am assuming you mean that neither a nor b are 0).  Since p is prime, you are really working in the multiplicative group G of nonzero elements of the field Zp, the integers modulo p.  Thus, for b^x = a (mod p) to have a solution,  the integer a must be in the subgroup generated by b.\n\nIt is well known that G is cyclic.  Thus, for b^x = a (mod p) to have a solution for any integer a<p, then b must be a generator of G.\n\nIf there is a solution, there will be infinitely many solutions.  This is because b^(p-1) = 1 (mod p) for any nonzero integer b<p (the order of G is p-1).  Hence, if b^x = a (mod p) and n is any multiple of p-1, then b^(nx) = a (mod p).  However, the solution is unique modulo the order of the element b.
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thanks 4 voting my answer as best but i didnt understand what u said\no divided by any no is always zero and with exception of zero where i guess it's not defined(aint that also zero iam not sure)
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uniform continuity implies continuity, but not conversely.  The first link seems to help quit a bit, the wikipedia not so much so, though it does have a good example with the details missing.\n\n\nhttp://www.maths.nott.ac.uk/personal/jff/G12RAN/pdf/Uniform.pdf#search='uniform%20continuity'\n\nhttp://en.wikipedia.org/wiki/Uniform_continuity
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This is an excellent question.\n\nWhen dealing with "non-routine" functions, you must go back to the definition.  In the case of a piecewise function like this, you need to check that \n\nf'(x)=&#92;lim_{h &#92;to 0} (f(x+h)-f(x))/h \n\nexists from both the left and right directions.\n\nFor your problem, from the left &#92;lim_{h &#92;to 0-} (f(2+h)-f(x))/h=2 and from the right\n&#92;lim_{h &#92;to 0+} (f(2+h)-f(x))/h=5.\n\nSince the left and right limits are not equal, the function is not differentiable at x=2.  (Note that if the "+3" we omitted, it would be differentiable and some would say that we should have verified continuity first...)\n\nChallenge:\n1.  f(x)=0, x rational\n x, x irrational\n\nis continuous only at x=0.\n\n2.  f(x)=0, x rational\n x^2, x irrational\n\nis differentiable (and continuous) only at x=0.\n\nGood luck! :)
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The problem is to show that f is identically 0 using the density of the rational numbers. I would strongly suggest to the asker that they do their own homework at this level. If you don't understand the questions or techniques, you will have a *lot* of trouble later on.
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The definition of continuity that the preimage of any open set is open, is elegant but not directly amenable to intuition in my opinion. Maybe an easier way of putting it (logically equivalent) is the following. It's much closer to the the usually concept of continuity but applies to any topological spaces:\n\nA function f:X-->Y is continuous if and only if it preserves limit point relationships, that is: If A is any subset of X and p a limit point of A, then f(p) is a limit point, or point, of f(A).\n\nIf every open set S containing f(p) has an open preimage f^(-1)(S), then p being the limit point of some set A means that points of A are contained in f^(-1)(S) --> points of f(A) are contained in S. The converse is also easily proved so this is an equivalent definition of continuity.\n\nThe concept of a limit point p of a set A is relatively easy to grasp. It just means there's no open neighborhood of p which doesn't contain points of A - there's no finite margin separating p from points of A.\n\nIf f is continuous, for any point p we can typically construct a sequence A having p as its unique limit point, then the definition of continuity tells us that the sequence of image points under f is f(p).
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Hint:\nPick y and z different. Let \nf(x)=d(x,y)-d(x,z).\nThen f(y)<0 and f(z)>0. Show that the range of f is uncountable.
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Rational:  it makes sense to divide 0 by another number and get 0\nReal:  0 is a real number\nIrrational:  No, that would be the case if it were 5/0\nInteger:  Any whole positive or negative number, and/or 0 \nWhole:  0 is not fractional\n\nTherefore:  0 is a real, whole, rational integer
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To extend the above two discussions and bring the matter to a conclusion, more or less.\n\nLet us define a sequence by T0 = 1, Tn = x^Tn-1.  Call the limit of this sequence as n -> infinity  as f(x).\n\nFor any given y = f(x), we see that:\n\n     lim Tn = lim Tn-1 = y\n>>\n    lim Tn = lim(x^Tn-1)=x^limTn-1\n>>\n     y = x^y\n>> \n    x = y ^ (1/y)\n\nFor y >= 1:\nThis can easily be seen to satisfy the  special cases of y = 1, 2 and 4.\n\nx, as a function of y has a maximum at y = e.~ 2.718, which leads to x~1.447.\n\nFor values of  x > e^(1/e),  f(x) does not exist; and the sequence T cannot have a limit > e.\n\nFor y < 1, we can similarly see that limits < 1/e are not possible.  The smallest x for which the limit exists is (1/e)^e ~ 0.066, giving a limit 1/e ~ 0.368