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Questions on functions and supremums and infimums.

  1. Nov 1, 2006 #1
    1)let g:B->A f:A->B be functions, and gof is the identity function, prove/disprove the following:
    g=f^-1;f is onto;f is 1-1;g is onto;g is 1-1.
    iv'e showed that the first the third and the last are correct, and the others are wrong, am i right?

    2)prove that f(x)=cos(x)+sqrt(3)sin(x) is a bounded function and find its supremum and infimum.

    well to prove that it's a bounded function was easy, it's bounded by:
    -1-sqrt(3)<=f(x)<=1+sqrt(3)
    now to find the supremum and infimum, the supremum is 2, and i have proven it this way:
    first we need to show that it's an upper bound, i.e cos(x)+sqrt(3)sinx<=2
    we have 2-sqrt(3)sin(x)>=1 and we can suppose that cos(x) is positive, bacause the upper bound should be greater also from the least value of f(x) which can be obtained when cos(x)<0, so we can suppose that cos(x)>0 (is this assumption valid?) then we have sqrt(1-sin^2x)<=2-sqrt(3)sin(x) 1-sin^2x<=4+3sin^2x-4sqrt(3)sin(x)
    4sin^2x-4sqrt(3)sin(x)+3>=0
    2sin^2x-sqrt3)^2>=0 which is ofcourse correct.

    now we need to show that for every c>=f(x) 2<c, (in order to show that's least upper bound), so we suppose that c<2, then c>=cos(x)+sqrt(3)sin(x)
    again we assume cos(x)>0 so we have c^2-2csqrt(3)sin(x)+3sin^2(x)>=1-sin^2x 4sin^2x-2sqrt3sin(x)+c^2-1>=0 [/tex]\delta=4*3c-16c^2+16=-4c^2+16<=0[/tex] so we have (2-c)(2+c)<=0 c>=2 or c<=-2
    but this clearly contradicts our assumptions, is my appraoch correct?


    p.s
    just something im interested about, how do you prove that 0.123456789101112.... (all the natural numbers) is irrational?
     
  2. jcsd
  3. Nov 1, 2006 #2

    Hurkyl

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    It is known that the decimal expansion of any rational number is eventually repeating.
     
  4. Nov 1, 2006 #3
    what about the other two questions?
     
  5. Nov 1, 2006 #4

    HallsofIvy

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    In order that a function HAVE an inverse, it must be both 1-1 and onto, and its inverse must be both 1-1 and onto. However, consider this example: A= {a, b}, B= {x,y,z} . f(a)= x, f(b)= y. g(x)= a, g(y)= b, g(z)= b. then gof(a)= g(x)= a, gof(b)= g(y)= b so gof is the identity function. Is g= f-1? What about the other questions?

    Actually, in this case, the sup and inf are the maximum and minimum values so taking the derivative and setting it equal to 0 is a good start.
    You can also note that the "root mean square" of the coefficients, 1 and [itex]\sqrt{3}[/itex] is [itex]\sqrt{1+ 3}= 2[/itex]. Factoring out a 2, you have [itex]f(x)= 2(\frac{1}{2}cos(x)+ \frac{\sqrt{3}}{2}sin(x))[/itex]. Now use the fact that sin(x+ a)= sin(a)cos(x)+ cos(a)sin(x). Here, we have [itex]sin(a)= \frac{1}{2}[/itex] and [itex]cos(a)= \frac{\sqrt{3}}{2}[/itex] so [itex]a= \frac{\pi}{6}[/itex].
    [tex]f(x)= 2 sin(x+ \frac{\pi}{6})[/tex]
    Aren't the max and min (sup and inf) of that obvious?
     
  6. Nov 1, 2006 #5
    In order that a function HAVE an inverse, it must be both 1-1 and onto, and its inverse must be both 1-1 and onto. However, consider this example: A= {a, b}, B= {x,y,z} . f(a)= x, f(b)= y. g(x)= a, g(y)= b, g(z)= b. then gof(a)= g(x)= a, gof(b)= g(y)= b so gof is the identity function. Is g= f-1? What about the other questions?
    no bacuse we dont have a reverse image for f(g(z)) doesnt equal z.
    so it means we dont have it so still i think that f isnt onto, but it's 1-1, cause:
    if f(x1)=y1=y2=f(x2) in B, then x1=g(f(x1))=g(f(x2))=x2.
    from the example you showed clearly g isnt one to one, but it's onto, so i guess i need to prove that g is onto.
    then for every x in A there exists y in B such that g(y)=x when y=f(x).
    thanks, halls.
     
  7. Nov 1, 2006 #6
    halls, am i right in what i wrote?
    im having second thoughts about it. (-:
     
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