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Prove f(x) is zero in range [-1,1]

  1. Nov 28, 2017 #1
    1. The problem statement, all variables and given/known data
    A function f(x) is continuous in the range [-1,1]
    f(x) = 2xf(x2 - 1)

    Prove that f(x) = 0 everywhere in range [-1,1]

    2. Relevant equations
    I don't know how to proceed.
    By putting values as x = 0, I got f(0) = 2*0 = 0.
    f(1) = 2*1*f(0) = 0.

    And I also get f(negative values) = f(positive values). So it's like a cosine wave.

    But how to prove that the values are equal to zero.

    3. The attempt at a solution
    How to proceed? Continuous means value of f(x) when I approach from left side = value of f(x) when I approach from right side. Not sure how to imply it here.
     
  2. jcsd
  3. Nov 28, 2017 #2

    BvU

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    How ? And: if you are right, and if you can also get the opposite, you have it !
     
  4. Nov 28, 2017 #3
    No, I think i wrote wrong.
    Putting x = -1,
    I get f(-1) = 2(-1)*f((-1)2 -1 )
    which is f(-1) = -2f(0)
    and f(1) = 2f(0)

    f(0) = 0*f(-1) = 0.
    So f(0) = 0
    So f(-1) = 0 and so is f(1) from underlined parts.

    So f(x) = 0 for x = -1, 0, 1. But how to prove for other values of x?

    when x = 0.5. I get f(0.5) = 2*0.5 f(0.25 - 1) = f(0.75)
    and when I put x = 0.75 I get
    f(0.75) = 2*0.75 f(0.5625-1) = 1.5f(-0.4375)

    I don't know how to prove that all these f(x) are equal to zero.
     
  5. Nov 28, 2017 #4
    Although I goofed up there, how would I have it if I proved this. why would f(x) be zero if f(-x) = f(x).
    A cosine wave is continuous in range -90 to + 90 but it's not always zero, and it satisfies condition f(-x) = f(x)
     
  6. Nov 28, 2017 #5

    PeroK

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    This looks tricky. What happens if you start from any point in ##x_1 \in [-1, 1]## and generate a recursive sequence by ##x_{n+1} = x_n^2 - 1##?
     
  7. Nov 28, 2017 #6
    What's a recursive sequence?

    I tried with mean value theorem.
    a = -1, b = 1
    there is a point c between a and b such that f'(c) = [ f(b) - f(a) ] / [b - a]
    this is zero since f(-1) = f(1) = 0
    So since f'(c) is zero, we get a constant value and since one constant value is zero we can say that f(x) = constant = 0.
     
  8. Nov 28, 2017 #7

    PeroK

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    It doesn't say that ##f## is differentiable. But, another idea is to use the fact that a continuous function attains its max and min on a closed interval.

    Regarding my previous idea. If you start with a number. E.g. ##x_1 = 1/2##, then ##x_2 = x_1^2 - 1 = -3/4## and ##x_3 = x_2^2 - 1 = -7/16## and so on. This sequence may converge to a point ##x## where ##x^2 - 1 = x## (if there is such a point).

    These are just ideas. I haven't spotted the solution yet, so they may not help!
     
    Last edited: Nov 28, 2017
  9. Nov 28, 2017 #8
    Simple proof:

    Go to the point where x²-1=x and call it P12. (Should be -0.6180339887498949 and 1.6180339887498949)
    f(P12)=2P12 f(P12)
    1/2=P12 except if f is zero. And because P12 is clearly not 1/2, f must be zero.

    Edit: Wait im stupid. That says only that f(P12) must be zero...
     
  10. Nov 28, 2017 #9

    fresh_42

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    So maybe integration is an idea. What about ##F(x):= \int_{-1}^x f(y)dy\,##? With ##g(x)=x^2-1## we get easier conditions plus differentiability. However, I haven't found the trick, yet.
     
  11. Nov 28, 2017 #10

    PeroK

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    Do you think it might be false?

    The condition on the interval ##[0, 1]## is irrelevant. If you can satisfy the condition on ##[-1, 0]## that would do it.
     
  12. Nov 28, 2017 #11

    fresh_42

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    What could be false? I only thought continuity invites to be integrated to ##F'(x)=f(x)## with ##F(0)=F(-1)=F(1)=0## and ##F=F\circ g## if I didn't make a mistake. These are easier to handle.
    Sure, but do we have it on one of them? I'm looking for an argument without iterations.
     
  13. Nov 28, 2017 #12

    PeroK

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    That there might be a continuous function that meets the criteria?
     
  14. Nov 28, 2017 #13

    fresh_42

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    I don't get it, sorry. Since it is supposed to result in ##f(x)=0## it'll be both, continuous and differentiable. I think that this is the only solution, I just don't see the killer argument. I simply thought, that we can get differentiability for free by an integration.
     
  15. Nov 28, 2017 #14

    Ray Vickson

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    Iteration plus integration will do the job. Let
    $$ F(x) = \int_0^x f(t) \, dt,$$
    so that we get
    $$F(x) = F(x^2-1) - F(-1).$$
    Since ##f## is an odd function, ##F## is even; thus ##F(-1) = F(1)##, so we have ##F(x) = F(x^2-1) - F(0)##. Using ##F(0)=0## we have ##F(1) = -F(1)##, so ##F(1) =0.## Therefore, we have
    $$F(x) = F(x^2 - 1).$$

    Now, for any ##x_0 \in (-1,1)## the iterative scheme ##x_1 = x_0^2-1, x_2 = x_1^2 - 1, \ldots## converges to the negative solution of ##x = x^2-1##, which is ##\tau \equiv (1-\sqrt{r})/2 \doteq -0.61803##. (We can see this by looking at the "cobweb diagram" of the iteration scheme ##x_{n+1} = x_n^2 - 1.##). Since ##F(x_{n+1}) = F(x_n)## for all ##n \geq 0## we have ##F(x_0) = \lim_{n \to \infty} F(x_n ) = F(\tau)## for any ##x_0##. In other words, ##F(x) = \text{constant}.##
     
  16. Nov 28, 2017 #15

    Delta²

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    My idea is that ##f(x)=2xf(x^2-1)=2x2(x^2-1)f((x^2-1)^2-1)=...=2^np_n(x)f(q_n(x))## where ##p_n## and ##q_n## polynomials of degree ##2^n-1## and ##2^n##. If we can prove that ##\lim_{n->\infty}q_n(x)=0## then I think we got it.
     
    Last edited: Nov 28, 2017
  17. Nov 28, 2017 #16

    PeroK

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    I found that this sequence converges alternatingly to ##0## and ##-1##.
     
  18. Nov 28, 2017 #17

    Delta²

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    That's only if you start with 0, 1 or -1. for any other x in (-1,1) it converges to -0.618...=##\frac{1-\sqrt{5}}{2}##.

    Now I see clearly that the polynomial ##q_n(x)## in post #15 converges to -0.618.. for any x in (-1,0) or (0,1) and f(-0.618...)=0.
     
  19. Nov 28, 2017 #18

    PeroK

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    Can you prove that? Try starting at 0.5, for example, and see what happens.
     
  20. Nov 28, 2017 #19

    Ray Vickson

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    Yes, you are right. However, the basic limiting argument still holds: ##F(x_0) = F(x_n) ## for all ##n##, and for very large ##n## we have that ##x_n## is either ##0## or ##-1## (alternating). But since ##F(0) = F(-1) = 0## we have ##F(x_0) = 0##. That is for any ##x_0 \in (-1,1)## different from ##(1-\sqrt{5})/2.##
     
    Last edited: Nov 28, 2017
  21. Nov 28, 2017 #20

    Delta²

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    Well I ve got to eat my hat, seems you are right. But still the sequence ##f(q_n(x))## converges alternatingly to f(0) or f(-1) and since they are equal to 0, converges to zero.
     
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