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Thread: Extracurricular Elementary Mathematics Marathon

  1. #51
    -insert title here- Paradoxica's Avatar
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    Re: Extracurricular Elementary Mathematics Marathon

    Quote Originally Posted by seanieg89 View Post




    This establishes the claim for all n larger than 6.

    Verifying the inequality for the five values of n between 2 and 6 with a calculator or by hand completes the proof.
    Wow... you're so bothered to type up all that ... My solution consisted of creating an intermediate inequality and inducting on that.
    If I am a conic section, then my e = ∞

    Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.

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    Re: Extracurricular Elementary Mathematics Marathon

    Quote Originally Posted by Paradoxica View Post
    Wow... you're so bothered to type up all that ... My solution consisted of creating an intermediate inequality and inducting on that.
    Just went with the immediate idea that popped into my head, didn't take long to tex/compute.

    If you want a weaker constant than 1/sqrt(3) (like 1/sqrt(2) for instance) then the same method as above gets you immediate results without needing to go several terms into the product before doing the telescoping.

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    Re: Extracurricular Elementary Mathematics Marathon

    Quote Originally Posted by seanieg89 View Post
    Just went with the immediate idea that popped into my head, didn't take long to tex/compute.

    If you want a weaker constant than 1/sqrt(3) (like 1/sqrt(2) for instance) then the same method as above gets you immediate results without needing to go several terms into the product before doing the telescoping.
    The intermediate inequality you induct on is placing 1/sqrt(3n+1) between the two. It may seem arbitrary but the problem is that the right hand side doesn't shrink fast enough for an inductive proof to be able to compare the two sides.
    If I am a conic section, then my e = ∞

    Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.

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    Supreme Member seanieg89's Avatar
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    Re: Extracurricular Elementary Mathematics Marathon




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    Re: Extracurricular Elementary Mathematics Marathon


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    Re: Extracurricular Elementary Mathematics Marathon

    You should probably specify the domain of the functional equation you posted Paradoxica. It it the reals?


    Assuming this is the case (and in fact this working is also valid if the domain is the integers, rationals or complex numbers), then:

    6/

    Let x=0 and let y=-f(0)+z, so

    f(z)=f(f(0)+y)=y=-f(0)+z.

    This implies that all solutions are of the form f(z)=z+c for some constant c.

    However, if we apply this formula to our original functional equation, we get:

    x+y+2c=x+y.

    Hence c must be zero, and the only solution to the functional equation is the identity function f(x)=x.
    Last edited by seanieg89; 7 Feb 2016 at 6:31 PM.

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    Re: Extracurricular Elementary Mathematics Marathon

    Quote Originally Posted by Paradoxica View Post




    Edit: done above

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    Supreme Member seanieg89's Avatar
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    Re: Extracurricular Elementary Mathematics Marathon

    Lols, mega procrastination on my behalf. Am supposed to be brushing up on something for a supervisor meeting. These questions are fun though.

  9. #59
    -insert title here- Paradoxica's Avatar
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    Re: Extracurricular Elementary Mathematics Marathon

    Quote Originally Posted by Paradoxica View Post
    And we are done!

    someone else post next, I've had my due run for the time being.
    If I am a conic section, then my e = ∞

    Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.

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    Supreme Member seanieg89's Avatar
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    Re: Extracurricular Elementary Mathematics Marathon

    1. Does there exist a non-constant polynomial P(x) with integer coefficients such that P(k) is prime for every positive integer k?

    Justify your response.
    Last edited by seanieg89; 7 Feb 2016 at 9:12 PM.

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    Senior Member KingOfActing's Avatar
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    Re: Extracurricular Elementary Mathematics Marathon

    Quote Originally Posted by seanieg89 View Post
    1. Does there exist a non-constant polynomial P(x) with integer coefficients such that P(k) is prime for every positive integer k?

    Justify your response.
    I'm not sure if this works but:

    Suppose there exists such a polynomial. P(0) =/= 0, as otherwise the x | P(x). Suppose P(0) = p for some prime number p. Then P(p) = ap^n + bp^n-1 + ... + p which is divisible by p, and is such a composite number. (1 is not a prime number, right?)
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    Re: Extracurricular Elementary Mathematics Marathon

    Quote Originally Posted by KingOfActing View Post
    (1 is not a prime number, right?)
    Correct, 1 is not prime.

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    Re: Extracurricular Elementary Mathematics Marathon

    Quote Originally Posted by KingOfActing View Post
    I'm not sure if this works but:

    Suppose there exists such a polynomial. P(0) =/= 0, as otherwise the x | P(x). Suppose P(0) = p for some prime number p. Then P(p) = ap^n + bp^n-1 + ... + p which is divisible by p, and is such a composite number. (1 is not a prime number, right?)
    Being divisible by p does not imply that you are a composite number. Why can't P(p)=p itself?

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    Senior Member KingOfActing's Avatar
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    Re: Extracurricular Elementary Mathematics Marathon

    Quote Originally Posted by seanieg89 View Post
    Being divisible by p does not imply that you are a composite number. Why can't P(p)=p itself?
    I think I'm grasping at straws here, but:

    P(kp) must be divisble by p, so that it either equals p or is composite. Since we claimed P(x) is never composite, P(kp) = p for all k. That is, P(kp) - p is the zero polynomial, and hence P(x) must be constant.
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    Re: Extracurricular Elementary Mathematics Marathon

    Quote Originally Posted by KingOfActing View Post
    I think I'm grasping at straws here, but:

    P(kp) must be divisble by p, so that it either equals p or is composite. Since we claimed P(x) is never composite, P(kp) = p for all k. That is, P(kp) - p is the zero polynomial, and hence P(x) must be constant.
    Bingo , back yourself more!

    Will let someone else post one now.

  16. #66
    -insert title here- Paradoxica's Avatar
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    Re: Extracurricular Elementary Mathematics Marathon

    Quote Originally Posted by seanieg89 View Post
    Justify your response.
    This isn't English.

    Here, we say "Prove"

    Anyway, I digress.

    If I am a conic section, then my e = ∞

    Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.

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    Re: Extracurricular Elementary Mathematics Marathon

    Quote Originally Posted by Paradoxica View Post
    This isn't English.

    Here, we say "Prove"
    Condescending much? Let people choose their own words. Clarity and precision of semantic content is far more important in mathematical writing than such choices in wording. Especially in a forum post of a single question lol.

    I did not write this question as "Prove X" simply because I wanted students to approach the question without knowing for sure whether X was true or false.

  18. #68
    -insert title here- Paradoxica's Avatar
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    Re: Extracurricular Elementary Mathematics Marathon

    Quote Originally Posted by seanieg89 View Post
    Condescending much? Let people choose their own words. Clarity and precision of semantic content is far more important in mathematical writing than such choices in wording. Especially in a forum post of a single question lol.

    I did not write this question as "Prove X" simply because I wanted students to approach the question without knowing for sure whether X was true or false.
    You could also say "prove your result" or "prove or disprove"

    I see the second one a lot in Putnam questions.
    If I am a conic section, then my e = ∞

    Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.

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    Supreme Member seanieg89's Avatar
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    Re: Extracurricular Elementary Mathematics Marathon

    Quote Originally Posted by Paradoxica View Post
    You could also say "prove your result" or "prove or disprove"
    I could. I could also say: "Justify your answer by way of proof", or countless variations of this.

    I made a choice, and the meaning was unambiguous. You will be frequently frustrated in future studies if such variations in wording bother you.

  20. #70
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    Re: Extracurricular Elementary Mathematics Marathon

    Quote Originally Posted by seanieg89 View Post
    Condescending much? Let people choose their own words. Clarity and precision of semantic content is far more important in mathematical writing than such choices in wording. Especially in a forum post of a single question lol.

    I did not write this question as "Prove X" simply because I wanted students to approach the question without knowing for sure whether X was true or false.
    I think Paradoxica might have been being tongue-in-cheek when he said that comment. Or at least, I thought he was being tongue-in-cheek when I first saw the comment (thought he was just jokingly taking a jab at English maybe), not sure now upon seeing his reply to your post I quoted.
    Last edited by InteGrand; 8 Feb 2016 at 1:39 AM.

  21. #71
    what is that?It is Cowpea RealiseNothing's Avatar
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    Re: Extracurricular Elementary Mathematics Marathon

    Quote Originally Posted by InteGrand View Post
    I think Paradoxica might have been being tongue-in-cheek when he said that comment. Or at least, I thought he was being tongue-in-cheek when I first saw the comment (thought he was just jokingly taking a jab at English maybe), not sure now upon seeing his reply to your post I quoted.
    Nah he was defs srs lol.
    Carrotsticks likes this.

  22. #72
    Supreme Member seanieg89's Avatar
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    Re: Extracurricular Elementary Mathematics Marathon

    Also, there are only finitely many solutions to that functional equation I believe.

    It is clear that if a solution vanishes somewhere it vanishes everywhere (let y be a root to make the RHS vanish and then vary x so the LHS vanishing tells you that f is identically zero).

    Let us assume now that f is a nonvanishing solution.

    Applying the functional equation tells us:

    f(x+y+z)=f(x+(y+z))=f(x)f(y+z)f(xy+xz)=f(x)f(y)f(z )f(yz)f(xy)f(xz)f(x^2yz).

    On the other hand, if we group the terms differently, we obtain f(x+y+z)=f(y+(x+z))=f(x)f(y)f(z)f(yz)f(xy)f(xz)f(x y^2z).

    Setting z=1 and dividing out the factors (which are nonzero!) tells us f(x^2y)=f(xy^2) for all real x and y.

    But for nonzero a,b we can always solve x^2y=a, xy^2=b. (x^3=a^2/b, y^3=b^2/a).

    This implies that f must be constant on R \ {0}. Moreover, by setting x = y =/= 0 in the functional equation, we obtain that this constant must satisfy x^3=x, which has only three solutions.

    Similarly, setting x=y=0 tells us that the value at the origin must also solve this same cubic.

    So we have narrowed down possible solutions to the functional equation to a finite set of functions and we are done.

    (It would not take much additional effort to find ALL solutions to the functional equation if one so desired.)

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    Supreme Member seanieg89's Avatar
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    Re: Extracurricular Elementary Mathematics Marathon

    Here's a cute one:

    Find all solutions f: R\{2/3}->R to:

    504x-f(x)=f(2x/(3x-2))/2.
    Last edited by seanieg89; 11 Feb 2016 at 2:18 PM.

  24. #74
    -insert title here- Paradoxica's Avatar
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    Re: Extracurricular Elementary Mathematics Marathon

    Quote Originally Posted by seanieg89 View Post
    Here's a cute one:

    Find all solutions f: R->R to:

    If I am a conic section, then my e = ∞

    Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.

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    Supreme Member seanieg89's Avatar
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    Re: Extracurricular Elementary Mathematics Marathon

    I do not believe your first solution is correct, neither is the statement that if f(x) is a solution then f(2x/(3x-2)) is one too.

    This would be the case if the functional equation was symmetric in these two unknowns, but the factor of 1/2 stops that from happening.

    A quick way to verify that this solution doesn't work without plugging into the functional equation directly is by considering the limit L of f(x)/x as x->0. The functional equation implies that L must be 1008 if it exists, whereas the limit in your first proposed solution is -1008.

    The second solution is the unique solution and you are done after solving the simultaneous equations. (Because all of the previous lines are quantified over all x, you can actually be sure that this is the unique solution after reaching the line f(x)=blah).


    (I obviously cooked up the numbers so that 2016 featured in the unique solution.)

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