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Math Help - Problem 33

  1. #1
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    Problem 33

    Let f(x) be a function on the number line.
    It is given that f(2+x)=f(2-x) \mbox{ and }f(7+x)=f(7-x) and 0 is a zero of f(x). What is the least number of zeros must f(x) have on the interval -2007\leq x\leq 2007.
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    The function is symmetric about x=2 and x=7, so all necessary zero are found by reflecting x=0 repeatedly about x=2 and x=7; if R_2 is a reflection about x=2 and R_7 about x=7, then we obtain a sequence of zeros R_2, \ R_7R_2, \ R_2R_7R_2, ... or R_7, \ R_2R_7, \ R_7R_2R_7, .... If R_2 is our first reflection, then we obtain the following sequence of zeros a_0=0, \ a_1=4, \ a_2 = 10, \ a_3 = -6, \ a_4 = 20 and in general a_{2m}=10m, \ a_{2m-1} = 14-10m, giving us 200+202 = 402 zeros in the interval [-2007, 2007]. If R_7 is our first reflection, then we get the sequence a_{2k}=-10k, \ a_{2k-1} = 10k+4, which yields another 402 zeros. A quick check shows that the two sequences have no terms in common. Hence at least 804 zeros total in [-2007, 2007].
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    Your solution seems correct.
    ----

    This was a slighy modified problem from an AIME question.

    We know f(0) is a zero.
    Thus,
    f(4) is a zero.
    Thus,
    f(10) is a zero.
    Thus,
    f(-6) is a zero.
    Thus,
    f(20) is a zero.
    Thus,
    f(-16) is a zero.
    Thus,
    f(30) is a zero.
    ....
    The pattern is clear.

    Similarly we get that,
    f(14) is a zero.
    Thus,
    f(-10) is a zero.
    Thus,
    f(24) is a zero.
    Thus,
    f(-20) is a zero.
    ...
    The pattern is clear.

    Look at chart below.

    RED: 4,-6,-16,...
    BLUE: 14,24,34,...
    GREEN: 10,20,30,...
    GREY: -10,-20,-30,...

    These are all arithmetic sequences.
    Now we count how many are in this inteveral + 1 (because we never counted zero yet).

    And also, these sequences are disjoint so we do not count the same thing twice.
    Attached Thumbnails Attached Thumbnails Problem 33-picture30.gif  
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    Oops I forgot to count zero
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  5. #5
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    The operation "reflection about 2" is the map s: x \mapsto 4-x

    The operation "reflection about 7" is the map t: x \mapsto 14-x

    Starting with x=0 we can obtain the zeroes of f by repeatedly applying s,t in any combination. We have


    t \circ s (x)= 14-(4-x)=10+x

    Moreover s^2=t^2=1, the identity map, so we can consider only alternating sequences of s,t. If we compose (t \circ s) with itself n times then (t \circ s)^n (x) = 10+(10+(10+...+(10+x)...)= 10n+x (where the exponent means composition - careful!)

    So now we have three possible types of different sequences of composition:
    (t \circ s)^n\:  (x) = 10n+x
    s\circ (t \circ s)^n\:  (x) = 4-(10n+x)
    (t \circ s)^n \circ s\:  (x) = 10n+(4-x)

    Since x=0 is our starting point, the zeroes of f are of the form 10n, 10n+4, -10n+4 for all n \in \mathbb N. It is easy to see that they are almost in bijection with the multiples of 5 on the interval [-2007,2007]. Counting carefully we get 2\Big\lfloor\frac{2007}{5}\Big\rfloor+2=804 zeroes.
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