4 ms·

The question is not well defined. If m is the degree of the polynomial, you want ∀ i ∈ {0, ..., m}, c_i ≥ 0 and ∃ i ∈ {1, ..., m}, c_i > 0. Otherwise the solut

by nanis 8y ago

The question is not well defined. If m is the degree of the polynomial, you want ∀ i ∈ {0, ..., m}, c_i ≥ 0 and ∃ i ∈ {1, ..., m}, c_i > 0. Otherwise the solution is either trivial or not unique. > Do you find this interesting? Depends. For a pedagogical tool in 5th to 8th grades, maybe. There may be some reason it is a compelling exercise which I am not seeing though. So, you might want to explain why you find this interesting enough to post it twice.

ColinWright 8y ago
I find it odd that you think the question is not well defined. Choose a polynomial with non-negative integer coefficients. Note that under the usual conventions, the polynomials x^2+1 and x^2+0x+1 are the same. Does that cover the situation you're worried about? Seriously, I don't know anyone who would say that x^2+1 and x^2+0x+1 are different, so maybe I'm not seeing what you're worried about. However, given that, does it answer your objection? As to why I think it's interesting, I think it's slightly surprising that given k=6 and p(k)=665 then the polynomial is uniquely identified as 3x^3+2x+5. Do you not find that surprising? If you don't find it surprising, how would you explain it?
- nanis 8y ago
  > I find it odd that you think the question is not well defined. You must assume that at least some of the coefficients of the x terms are non-zero. > Do you not find that surprising? No. > how would you explain it? I think it is up to you to explain why this is interesting. Without the condition that k is greater than all the coefficients, the solution is clearly non-unique. Given that, the assumption of the coefficients are integers, and that the solution is not a constant polynomial, the statement becomes equivalent to "division is a function".
  - ColinWright 8y ago
    >> I find it odd that you think the question is not well defined. > You must assume that at least some of the coefficients of the x terms are non-zero. That turns out not to be the case. >> Do you not find that surprising? > No. Interesting. >> how would you explain it? > I think it is up to you to explain why this is interesting. So far, many people who are clearly intelligent seem to miss that it works at all, let alone how it works. You appear so far to be among them. As I've said elsewhere, I appreciate that you are trying to poke holes in it, but it really is complete and correct as stated. > Without the condition that k is greater than all the coefficients, the solution is clearly non-unique. Well, yes. That's why having k greater than all the coefficients is a requirement. That's why your example in another comment elsewhere is not an example of the puzzle failing - it does not satisfy the condition that k must exceed all the coefficients. > Given that, the assumption of the coefficients are integers, and that the solution is not a constant polynomial, the statement becomes equivalent to "division is a function". I don't see why this is true - perhaps you could expand on it? Why is the problem statement equivalent to saying "Division is a function."? So now I don't understand what you're trying to say or do. The question as stated is fine - an example that doesn't satisfy the conditions isn't a counter-example. So at least one of us is confused. So let me state clearly, rather than having a bitty back'n'forth. You pick a polynomial p(x) with non-negative integer coefficients. You choose an integer k that's larger than all the coefficients of p(x). You tell me k and p(k). In return, I tell you p(x). So: Are you claiming this doesn't work? Are you claiming it's trivial to accomplish? Are you claiming it's obvious why it works? I'm no longer sure what you're claiming, so I thought I'd ask clearly and plainly.
    - nanis 8y ago
      > Are you claiming this doesn't work? No, I am not. I misinterpreted the "larger than all coefficients" because I was sure that the condition that at least one of the coefficients other than the constant term had to be positive. You pointed out my mistake, you are correct, thank you. > Are you claiming it's trivial to accomplish? It is easy to state what you need to do. Find largest a such that k^a ≤ p(k). Find p(k) / k^a using integer division. Because division is a function, there is unique quotient and unique remainder. The quotient is your first coefficient. Take the remainder, repeat.
      - ColinWright 8y ago
        Right - that's one way of doing it. It's not the way I first thought of, and the other way I have has a connection to something else that I think makes it all kinda nice. That solution is elsewhere in the thread, but here it is in rot13: jevgrcxvaonfrxnaqgurqvtvgfnergurpbrssvpvragf I'm pleased we've got it sorted, though. Thanks for your responses, and I'd be happy to hear more thoughts.

by nanis 8y ago

ColinWright 8y ago

I find it odd that you think the question is not well defined. Choose a polynomial with non-negative integer coefficients. Note that under the usual conventions, the polynomials x^2+1 and x^2+0x+1 are the same. Does that cover the situation you're worried about? Seriously, I don't know anyone who would say that x^2+1 and x^2+0x+1 are different, so maybe I'm not seeing what you're worried about. However, given that, does it answer your objection? As to why I think it's interesting, I think it's slightly surprising that given k=6 and p(k)=665 then the polynomial is uniquely identified as 3x^3+2x+5. Do you not find that surprising? If you don't find it surprising, how would you explain it?

nanis 8y ago
> I find it odd that you think the question is not well defined. You must assume that at least some of the coefficients of the x terms are non-zero. > Do you not find that surprising? No. > how would you explain it? I think it is up to you to explain why this is interesting. Without the condition that k is greater than all the coefficients, the solution is clearly non-unique. Given that, the assumption of the coefficients are integers, and that the solution is not a constant polynomial, the statement becomes equivalent to "division is a function".
- ColinWright 8y ago
  >> I find it odd that you think the question is not well defined. > You must assume that at least some of the coefficients of the x terms are non-zero. That turns out not to be the case. >> Do you not find that surprising? > No. Interesting. >> how would you explain it? > I think it is up to you to explain why this is interesting. So far, many people who are clearly intelligent seem to miss that it works at all, let alone how it works. You appear so far to be among them. As I've said elsewhere, I appreciate that you are trying to poke holes in it, but it really is complete and correct as stated. > Without the condition that k is greater than all the coefficients, the solution is clearly non-unique. Well, yes. That's why having k greater than all the coefficients is a requirement. That's why your example in another comment elsewhere is not an example of the puzzle failing - it does not satisfy the condition that k must exceed all the coefficients. > Given that, the assumption of the coefficients are integers, and that the solution is not a constant polynomial, the statement becomes equivalent to "division is a function". I don't see why this is true - perhaps you could expand on it? Why is the problem statement equivalent to saying "Division is a function."? So now I don't understand what you're trying to say or do. The question as stated is fine - an example that doesn't satisfy the conditions isn't a counter-example. So at least one of us is confused. So let me state clearly, rather than having a bitty back'n'forth. You pick a polynomial p(x) with non-negative integer coefficients. You choose an integer k that's larger than all the coefficients of p(x). You tell me k and p(k). In return, I tell you p(x). So: Are you claiming this doesn't work? Are you claiming it's trivial to accomplish? Are you claiming it's obvious why it works? I'm no longer sure what you're claiming, so I thought I'd ask clearly and plainly.
  - nanis 8y ago
    > Are you claiming this doesn't work? No, I am not. I misinterpreted the "larger than all coefficients" because I was sure that the condition that at least one of the coefficients other than the constant term had to be positive. You pointed out my mistake, you are correct, thank you. > Are you claiming it's trivial to accomplish? It is easy to state what you need to do. Find largest a such that k^a ≤ p(k). Find p(k) / k^a using integer division. Because division is a function, there is unique quotient and unique remainder. The quotient is your first coefficient. Take the remainder, repeat.
    - ColinWright 8y ago
      Right - that's one way of doing it. It's not the way I first thought of, and the other way I have has a connection to something else that I think makes it all kinda nice. That solution is elsewhere in the thread, but here it is in rot13: jevgrcxvaonfrxnaqgurqvtvgfnergurpbrssvpvragf I'm pleased we've got it sorted, though. Thanks for your responses, and I'd be happy to hear more thoughts.