what are absolute value functions??
553751
Definition of Absolute Value of x, |x|\nIf x >=0, then |x| = x\nIf x < 0, then |x| = -x
1183259
the if function uses 3 parameters:\n\n1) boolean (i.e. either true or false) variable or expression\n\n2) variable to use if the boolean value is true\n\n3) variable to use if the boolean value is false\n\nfor example in the following expression\n\n   if(x>3,x+3,x-3)\n\nif x > 3, then the function returns x+3, otherwise it returns x-3\n\n    if x = 0, since 0>3 is false, then the function returns x-3 which would be -3\n    if x = 4, since 4>3 is true, then the function returns x+3 which would be 7
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x(abs[x- (x^2)] (-1=2)\n\n:> peace\n.
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0, pi/2, pi, 3pi/2, 2pi\n\nall values where 2x = a muliple of pi
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Generally the idea is:\n\nDo Until Decimal = 0\n    If (Decimal Mod 2) Then\n        Let Binary = "1" & Binary\n    Else\n        Let Binary = "0" & Binary \n    End If\n    Let Decimal = Decimal &#92; 2\nLoop\n\n(Decimal is a number, Binary is a string variable)
764395
Hey, firstly I need this corrected: f(x)= Ix^2=5x+6I, because there should not be an equal sign after the power of 2.... I can't do this w/o knowing... if it is subtraction than I will tell you that from what I rembember about this stuff is that absolute value functions are continuous but not differentiable I think.... Because at the point, the f(c) equals the f(x) which means it is cont. but the derivative of f(x) approaching the vertex from the left side does not equal the derivative of the f(x) from the right side...whcih means it is NOT differentiable.... sorry I can't remember more buddy
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Domain ~ x values\nRange ~ y values\n\nYou have to work the function and figure them out.
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I think you mean that: Given a figure (function in the form of a curve), and the value of ordinate (y), does a value of abscissa (x) exist corresponding to that y?  \n\nIf the figure is an open curve, like y=x or y=ax^3, the answer is yes.\n\nBut there may be many other cases. For the curve y=x^2, which is a parabola, x may assume any real value; but y is always non-negative.\n\nAgain, x^2+y^2=a^2, is a circle and a closed figure; x can take up values between -a and +a. y also takes values in the same range.\n\nFor this curve, given a value of y, within the admissible range of -a to +a, there are two values of x. \nFor y=+a or y=-a, there will be only one value of x; and for y>a or y<-a, there will be no value of x.\n\nFinally, for periodic curves like y=sin x, for admissible values of y (between -1 and +1) there exist infinitely many values of x.\n\nSo, you see there is not one answer but many depending on the nature of the figure!
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Well there are the obvious values that are known of arctan. But if your constant K doesn't fall into these categories I'm not sure how you would do this without taylor expansion.\n\nTry this link, it gives the well known values on tan(x) so reading it backwards gives the arctan values
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http://www.math.niu.edu/~beachy/aaol/frames_index.html
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If the value in an absolute value is positive, it remains the same. If it is negative, it is multiplied by -1.\n\nEither a<x, a = x, or a > x. (I think this is called trichotomy?)\n\nIf a = x, then both sides equal 0.\n\nIf a > x, then a-x is positive, and x - a is negative, so...\na - x =  -1(x - a)\n\na - x If a > x, then x-a is positive, and a - x is negative, so...\n-1(a - x) =  x - a\nx - a = x - a\n\na - x = a - x\n\nQED
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Also, remember that if the value inside the absolute value bars is negative, it converts to positive.  So to solve an equation like |x+3| = 5, you really have to solve TWO equations:  x+3=5 or x+3= -5.\n\nIf the absolute value portion of the equation is not by itself, you must make it so first.  For example:\n\n|2x + 1| + 3x = 5       First subtract 3x from both sides:\n\n|2x + 1| = 5 - 3x      Then split into 2 equations:\n\n2x+1=5-3x  or 2x+1 = - (5-3x) [which would be -5 + 3x]\n\nsolve each equation to get the 2 values of x
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#define TRUE  1\n#define FALSE 0\n\n#define ESC      27\n#define F2      188\n#define UP      200\n#define FWD     205\n#define DN      208\n#define BS      203\n\nint getkey(void);\nint keyhit(void);\nvoid curr_cursor(int *, int *);\nvoid cursor(int, int);\nvoid hidecursor(void);\nvoid unhidecursor(void);\nvoid savecursor(void);\nvoid restorecursor(void);\nvoid set_cursor_type(unsigned);\n#define normalcursor() set_cursor_type(0x0607)\n\n\nlook at this code to understand how each function works.\n\nhttp://202.179.135.4/data/DDJ/articles/1991/9103/9103i/9103i.htm#00c1_0013
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You might find this paper useful\nhttp://arxiv.org/abs/math-ph/0106027
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Consider if a= 1, b =-1, then abs(a/b) = 1 and a/b = -1\n\n-1 != 1\nabs(a/b) != a/b\n\n(!= means NOT EQUAL TO)
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A pretty trivial one would be f(x) = absolute value of x and g(x) = -x\n\nf(g(x)) = abs (-x) = x\n\nbut g(f(x)) = - (abs x) = -x
2024
Case 1: f(x) := x-5*y; (y is parameter)\n\nL{f(x)} = int(exp(-s*x)*f(x),x=0..infinity);\nL{f(x)} = int(exp(-s*x)*(x-5*y),x=0..infinity);\nL{f(x)} = 1/s^2-5*y/s;\n\nCase 2: f(y) := x-5*y; (x is parameter)\n\nL{f(y)} = int(exp(-s*y)*f(y),y=0..infinity);\nL{f(y)} = int(exp(-s*y)*(x-5*y),y=0..infinity);\nL{f(y)} = x/s-5/s^2;