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Function equal up to nth order, diffbility

  1. Jul 4, 2007 #1
    1. The problem statement, all variables and given/known data
    Two functions f,g:R->R are equal up to nth order at if

    lim h-->0 [f(a+h) - g(a+h) / hn ]= 0

    Show that f is differentiable at a iff there is a function g(x)=b+m(x-a) such that f and g are equal up to first order at a.

    2. Relevant equations

    f is differentiable at a if the following limit exists:

    lim h-->0 [f(a+h) - f(a) / h]


    3. The attempt at a solution

    Direction 1:
    =>

    Assume f is differentiable. Show there exists a function g(x)=b+m(x-a) such that f and g are equal up to first order at a.

    So, lim h-->0 [f(a+h) - f(a) / h] exists and equals some L.

    =>lim h-->0 [f(a+h) - f(a) - g(a+h) + g(a+h)/ h]
    =>lim h-->0 [(f(a+h)- g(a+h)) - f(a) + g(a+h)/ h]

    this implies that we want to show

    lim h-->0 [g(a+h) - f(a)/h = L]

    I'm not sure where to go from here.


    Direction 2:
    <=

    Assume that lim h-->0 [f(a+h) -g(a+h) / h = 0]
    show lim h-->0 [f(a+h) - f(a) / h] is equal to some constant.


    Take lim h-->0 [f(a+h) -g(a+h) / h = 0]:

    => lim h-->0 [f(a+h) -g(a+h) -f(a) + f(a) - g(a) + g(a) / h = 0]

    => lim h-->0 [(f(a+h)-f(a)) -(g(a+h) - g(a)) + (f(a) - g(a))/ h = 0]

    but lim h-->0 [f(a+h)-f(a)/h] is just f'
    and lim h-->0 [g(a+h) - g(a)/h] is -m
    so, we just need to show that
    f(a) = g(a) and then lim h-->0 [f(a) - g(a)/h] = 0 and f' = m
    Not sure how to do that.

    That's all.

    Thanks for your help.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 4, 2007 #2
    There exists g(x)=m+b(x-a) such that f and g are equal to the first order, so you need to find such g (i.e., the value of m and b) according to the diff'bility of f.
     
  4. Jul 5, 2007 #3
    for both parts?
     
  5. Jul 5, 2007 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No, kekido said "such that f and g are equal to the first order" so he is talking about "if there exist g= m(x-a)+b such that f and g are equal to first order, then f is differentiable at x=a."

    However, the theorem as stated is not true. For example, if f(x)= 2x if x is not 0, 1 if x= 0, and g(x)= 2x, we have f(0+h)= 2h for all non-zero h, g(0+h)= 2h for all h so
    [tex]lim_{x\rightarrow a}\frac{f(0+h)-g(0+h)}{h}= 0[/itex]
    but f is not differentiable at n.

    You have to have some statement about what happens at x= a.
     
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