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Verify this statement if f(x) is infinitely differentiable

  1. Sep 17, 2014 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    [itex]f(mx)=f(x) + (m-1)xf'(x)+\dfrac{(m-1)^2}{2!} x^2 f''(x)........[/itex]


    2. Relevant equations
    Taylor's Series


    3. The attempt at a solution
    If I approximate the LHS of the eqn using Taylor's polynomial,
    [itex]f(mx)=f(mx)+mxf'(mx)+\dfrac{(mx)^2f''(mx)}{2!}+........ [/itex]

    But, I'm lost from here. It doesn't resemble RHS at all!
     
  2. jcsd
  3. Sep 17, 2014 #2

    ehild

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    Let be y=m the variable and k=x a constant. Do the Taylor expansion of f(ky) about y=1.

    ehild
     
  4. Sep 17, 2014 #3

    utkarshakash

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    Doing so gives me [itex] f(k)+(ky-k)f'(k)+\dfrac{(ky-k)^2f"(k)}{2!}+......... [/itex]

    which indeed is equal to RHS! Brilliant! But why did you assume x as a constant and m as a variable?
     
    Last edited: Sep 17, 2014
  5. Sep 17, 2014 #4

    haruspex

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    To get f(x) on the right, you need something of the form f(x+t) on the left. What does that make t equal to?
     
  6. Sep 17, 2014 #5

    utkarshakash

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    t=(m-1)x
     
  7. Sep 17, 2014 #6

    ehild

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    because that solves the problem :biggrin: The right side is very much like a Taylor expansion only you have m-1 instead of x-xo and f' is multiplied by x and f" multiplied by x2, as if x was the coefficient of the independent variable, so it must be m and the expansion is about m-1.
     
  8. Sep 17, 2014 #7

    utkarshakash

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    Nice! What about this question
    $$f(\dfrac{x^2}{x+1}) = f(x) - \dfrac{x}{x+1}f'(x)+\left( \dfrac{x}{x+1} \right) ^2 f"(x)+........ $$

    This doesn't look as simple as the previous one. There are alternating + and - signs along with x/x+1 as the coefficient in each term.
     
  9. Sep 17, 2014 #8

    Ray Vickson

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    Write ##x^2/(x+1) = x + t## and figure out the value of ##t##.

    BTW: in both these questions, just having ##f## infinitely differentiable may not be not enough: you might also need ##f## to be analytic. There are infinitely-differentiable functions which do not equal their Maclauren or Taylor series, but that is for functions expanded about a fixed point ##x_0##; this problem is different, so I am not really sure whether or not there are counterexamples in this case. However, if ##f## is known to be analytic you are basically done.
     
    Last edited: Sep 17, 2014
  10. Sep 17, 2014 #9

    utkarshakash

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    But what about the factorials that I get in the denominator in the expansion of f(x+t)?
     
  11. Sep 17, 2014 #10

    Ray Vickson

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    What has that got to do with anything? Of course the Taylor expansion has factorials in the denominators---so what?
     
  12. Sep 17, 2014 #11

    haruspex

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    Looks to me like the question has left out the factorials by mistake.
     
  13. Sep 17, 2014 #12

    utkarshakash

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    Hmm, that might be true. Here's another expression which I'm finding difficult to solve.

    [itex]f(x)=f(0)+xf'(x)-\frac{x^2}{2!}f"(x)+.............. [/itex]

    Here t=0. Now, if I expand it as Taylor's series, I get
    $$f(0)+xf'(0)+x^2/2! f''(0)...........$$
    which does not contain alternating + and - signs as in the original question. Also, the argument of f is 0 for all terms in my case.
     
  14. Sep 18, 2014 #13

    haruspex

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    Try rearranging the given equation
    with f(0) one side and all else the other
     
  15. Sep 18, 2014 #14

    utkarshakash

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    $$f(0) = f(x) -xf'(x)+\frac{x^2}{2!} f''(x) - \frac{x^3}{3!}f'''(x)+........$$

    I still can't figure out what to do next.
     
  16. Sep 18, 2014 #15

    haruspex

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    As before, the standard form is f(x+t) = f(x) + tf'(x) + ... So you need to find substitutions for x and t in there to turn f(x+t) into f(0) etc.
     
  17. Sep 18, 2014 #16

    utkarshakash

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    Thanks!
     
  18. Sep 19, 2014 #17

    HallsofIvy

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    By the way, is the only condition here that f be infinitely differentiable? If that is the case then it does not necessarily follow that the Taylor's series for f itself converges to f. For that you need that f is "analytic".
     
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