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Logarithmic series

  1. Jun 30, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]x=\frac{1+\frac{f(x+1)}{1!}+\frac{f^{2}(x+1)}{2!}+\frac{f^{3}(x+1)}{3!}+........}{1+\frac{f(x)}{1!}+\frac{f^{2}(x)}{2!}+\frac{f^{3}(x)}{3!}+...}[/tex]
    f(x) is a twice differentaible equation.
    1. Find the possible values of x when;
    f(x+0.5)<f(x-0.5)
    2. Find the possible values of x when;
    f(|x|+e-1)<f(|x|+e-2)+1

    3. The attempt at a solution
    On simplifying i get:
    ln x= f(x+1)- f(x)
    The only other data given is that f(x) is twice differentiable, which just means that when the function is expressed it will have two distinct constants, right? How shall I proceed?
     
  2. jcsd
  3. Jun 30, 2008 #2

    HallsofIvy

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    Why would "twice differentiable" say anything about constants?
     
  4. Jun 30, 2008 #3
    The order of differential equation is equal to number of distinct arbitrary constants. Right?
    If not, tell me how can I do these?

    I have shown everything I knew! Please help!!
     
  5. Jul 1, 2008 #4

    HallsofIvy

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    There is no differential equation here!

    x+ 1/2= x- 1/2+ 1 so f(x+ 1/2)= f((x-1/2)+1). Saying that f(x+1/2)< f(x- 1/2) means that f((x-1/2)+ 1)< f(x- 1/2) or f(y+ 1)< f(y) with y= x- 1/2. So what is ln(y) and what does that tell you about x?
     
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