f:[0,1]-R is a contious function s.t intregation from 0 to 1 fg=0 for any step function g.Show f=0 in [0,1]?
475920
Homework?\n\nSuppose f is not identically 0. Then there is an interval where f stays away from 0. The g be the step function of this interval to get a contradiction.
395570
You can't!  \n\nDefine a recursive sequence as follows:\n\nx_{n+1} = 1/2(x_n - 1)\n\nThen for any x_0 you plug in for x, you can determine the values of all x_i (i nonnegative integer).  The general formula for this sequence is as follows:\n\nx_n = (1/2)^n * (x_0 - n)\n\nNote then that if you plug in a rational number, all numbers in this sequence will be rational while if you plug in an irrational number, all numbers in this sequence will be irrational.\n\nTherefore, all such sequences that share a common element MUST both have a rational seed, x_0 so that they only contain rational numbers, or both have an irrational seed x_0, so that they only contain irrational numbers.\n\nWithout the condition of continuity then, the function g defined by g(x) = 1 if x is rational and g(x) = 0 if x is irrational is a nonconstant function that will satisfy the given condition.
49739
i'm not sure what you mean by the 2nd method.  can you elaborate?  the first way sounds correct.\n\nEDIT:  okay, i think i see the problem the way you've phrased it in terms of the unit step functions.  they don't describe the same function as the original piecewise f(t).\n\nconsider your equation in terms of the step functions:\nthe first term is equal to t everywhere.  the second term t*1(t-T) is equal to zero when t<T, and is equal to t when t>T.  the third term T*1(t-T) is equal to zero when t<T, and is equal to T when t>T.\n\naltogether the way you have it written, \nfor t<0, f(t) = t-0+0 = t,\nfor 0<t<T, f(t) = t-0+0 = t,\nfor t>T, f(t) = t-t+T = T.\n\nyou've already said this stuff so far, but i'd like to point out that this is not the original function you specify, which is equal to 0 when t<0, not t.  what you should have is a thing that is zero until t=0, then ramps up at a slope of 1 until t=T, then is equal to T thereafter.  what you have now in terms of the step functions is just a constant ramp up to T, all the way from t = negative infinity.\n\np.s. the - sign inside the unit step function *is* a subtraction.  1(t) is defined as 0 when t<0, and 1 when t>0.  subtracting T shifts the step over to t=T.  it's just like f(x) = x^2 is a parabola with a minimum at x=0, but f(x-2) is the same parabola shifted to the right by 2 units.\n\np.p.s this means, i think that the 2nd term should be t*(1(0)-1), instead of -t*1(t-T).  then this describes the original piecewise function.
1099079
Yes. Even if you define the sequence with F(1)=F(2)=1, you can still work backward to get\nF(0)=0\nF(-1)=1\nF(-2)=-1\nF(-3)=2\nF(-4)=-3\nThe negative fibonacci sequence is identical to the positive one except that it alternates between positive and negative numbers. Also, every other number is not even, but every third number is even. If you define F(0)=0, F(1)=1, then look at every F(n) where n is a multiple of 3.\nThere's also an interesting equation that gives all values in the Fibonacci sequence, even the ones that aren't integers. I don't feel like typing it here (it wouldn't be readable anyway), but it's at http://en.wikipedia.org/wiki/Fibonacci_number#Closed_form_expression
126194
d=(-infinity, +infinity)\nr= [-1,1]
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Let f(z) = x+iy\n\nNow, When |z| = 1, f(z) is in R, this means y = 0\n\nBut z = x1+iy1\n\nFor f(x1+iy1) = x, the function must map every point in the complex plane to the real axis.\n\nNow on diffentiating f(z) w.r.t. z, we have z'f'(z).\n\nThis quantity is 0. Thus f(z) is a constant.
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No\n\n\n\nf(3+1)=f(4)=2(4)+1=9\n\nbut\n\nf(3) + f(1)= [2(3)+1] + [2(1)+1] = 7 + 3 = 10
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1) [1/x] <= 1/x, for all x>0\n2) [1/x] = 1/x, iff 1/x is an integer, so x=1/n for some integer n.\n3) [1/x] < 1/x, for all x != 1/n for any n (x>0).\n4) So, f(x) = [x*[1/x]] = [x*1/x] = [1] = 1 for all x=1/n for some integer n.\n5) Also, x*[1/x] < x*1/x = 1 for all x != 1/n for any n (x>0),\nthen f(x) = [x*[1/x]] = [<1] = 0 for all x != 1/n for any n (x>0).\n\nlim f(x)(x->a) = b, iff for ALL sequences {a_n->a} f(a_n)->b.\n\nLet's take 2 sequences:\n{a_n = 1/n} f(a_n)=1 for all n (see 4), so f(a_n)->1.\n{b_n = 2/(2*n+1)} f(b_n)=0 for all n (see 5), so f(b_n)->0.\nEasy to see that {a_n}->0, {b_n}->0.\n\nThus, there is NO limit of f(x) when x->0.\n\nPS: "iff" means "if and only if"
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f(-2) = 5. What's so difficult about this question?
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Break this down into three functions depending on the value of x:\n\nWhen x<-1 you have f(x) = (-x-1)(2-x) because (x+1) and (x-2) are negative.  This is the same as f(x) = (x+1)(x-2)\n\nWhen -1<=x<2 you have f(x) = (x+1)(2-x).\nWhen x>=2 then f(x) = (x+1)(x-2).\n\nSo f(x) = (x+1)(x-2) = x^2-x-2 outside -1<=x<2,\nand f(x) = -x^2+x+2 in -1<=x<2.\n\nf'(x) = 2x-1 outside -1<=x<2, and f'(x) = 1-2x in -1<=x<2.
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First subsitute -1 wherever you see x\n\nf(-1) = 4(-1) - 3\n      = -4 - 3\n      = -7
918808
wouldnt it be like 3<x<-3 f(x)\nso like x would be 1 so it would be just f if you really want to simply that far done.
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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.
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Consider the function G(x) := 1 + integral (from 0 to x) of f(t) dt.  It is easy to see that G(0) = 1.  It is an easy exercise to show that it is well defined.  And by the fundamental theorem of calculus, it satisfies the last desired condition.
221721
Product rule:\n\n(fg)' = (f ')g + f(g ')\n\nSo,\n\n(-2x^2 * ln(x))' = (-4x * ln(x)) + (-2x^2 * 1/x)\n\nMultiply and clean up:\n\n-4xln(x) - 2x
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this means\n\nx^3 = -1\nor x^3+1 = 0\n\nf(x) = x^3+1\n\nf(-1) = 0\nso one root is -1\n\nfor the other root by division\n(x+1)(x^2-x+1)\nx^2-x+1 = 0 => x= (1+/-sqrt(3)i)/2
1082751
1)   f(x) = x+1/x =(x^2+1)/x  \n      g(x) = x-2/x =(x^2-2)/x\n(f+g)(x) = (x^2+x^2+1-2)/x = (2*x^2-1)/x = 2*x -1/x\n\n2)  f(x) = -2/x\n     g(x) = (x+1) / (x-3)\n(f+g)(x) = -2/x + (x+1)/(x-3) = {-2*(x-3)+x*(x+1)}/x*(x-3)\n              = (-2x+6+x^2-x)/(x^2-3x) =(x^2-x+6)/x^2-3x