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Polynomial & trig question

  1. Mar 12, 2013 #1
    Could anyone help me out with this:

    Which of the following statements are true and which are false? Justify your answers.

    iii) There exists a polynomial P such that [itex] |P(x) - \cos(x)| \leq 10^{-6} [/itex]

    I've tried to thinking about it, and it seems as though it is false, because [itex] |cos(x)| \leq 1 [/itex] and the polynomial can go up to an extremely high power so instinctively it can't be true. However, I have no idea on how to prove this (or if it's correct).

    I've also tried to look at the expansion of cos(x), leading me to *again* think it's false, but again, I'm not sure how to go about proving it.
     
  2. jcsd
  3. Mar 12, 2013 #2

    Dick

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    Try to show any nonconstant polynomial becomes unbounded as x->infinity. The largest term in the polynomial will be the one with the highest power. Show that dominates the other terms as x->infinity.
     
  4. Mar 12, 2013 #3

    Simon Bridge

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    Is there a difference between "justify" and "prove"?

    Note: the polynomial expansion would be exact for an infinite number of terms. What happens for finite terms?
     
  5. Mar 12, 2013 #4
    I'm not really sure how to show it, I mean it makes sense just thinking about it. As x -> infinity then surely any non constant polynomial will also tend to infinity, I have no idea how to go about proving this unfortunately. Any chance you could start me off?

    Also - The definition of a polynomial: is x^0 a polynomial?
    I'm not sure if there is a difference, but in the footnotes of the question the author states that we should, if true, show a proof that it's true, if false, then show by counter-example or proof that it is false.

    I'm not quite sure what you mean that the polynomial expansion would be exact for an infinite number of terms, and I don't see how this relates to the question.

    apologies for my lack of knowledge,
     
  6. Mar 12, 2013 #5

    Dick

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    Yes, constants are polynomials. But they don't approximate cos(x) very well. To show other polynomials suppose your polynomial is ##p(x)=a_n x^n + a_{n-1} x^{n-1} + ... +a_0##. Define ##f(x)=\frac{p(x)}{a_n x^n}##. What's the limit of f(x) as x->infinity?
     
  7. Mar 12, 2013 #6

    Simon Bridge

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    If P(x)=cos(x) then P(x) would be the "exact polynomial expansion" of cos(x) [also called a power series expansion] ... it would have form: ##P(x)=\sum_{i=0}^n a_i x^i## ... as Dick explains. How big does n have to be for P(x)=cos(x)? Infinite right?
    At finite values of n it won't be "exact" but it is still an approximate expansion. (It is possible to have an exact expansion with finite n - just not for cos(x): do you see why?)

    But, in your case, you just have to get P(x) very close to cos(x) ... within 10-6 in fact... (i.e. it needn't be exact) and your polynomial needs to have a finite n.

    If you just needed "justification" in the weak sense, then you just needed to point out that any finite order polynomial will become arbitrarily large or small for |x| very big while cos(x) is always in [-1,1]. This is why I asked that question.

    Instead - you are faced with proving that ... which Dick is helping you with.
    Enjoy :)
     
  8. Mar 12, 2013 #7

    Dick

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    I agree with Simon. "any fule kno" that nonconstant polynomials aren't bounded. Just pointing that out might be proof enough. Not that the proof is hard, though.
     
  9. Mar 13, 2013 #8
    Would f(x) also tend to infinity?
    because cos(x) is an infinite expansion?

    Yes I understand this. If x is small then any finite order polynomial will also be small, however , for small values of x, cos(x) will be closer to 1, so wont satisfy the inequality in the range, so the statement is false.
     
  10. Mar 13, 2013 #9

    Simon Bridge

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    Hmmm... you should try it then. I'm a big fan of hard experience.
    Consider the 2nd order polynomial ##P(x)=1-x^2## : for small x, P(x) is about 1... same as cos(x).

    Whatever x is - what is the largest value possible for cos(x)? What is the smallest value?
    How about a polynomial - as x gets very big or very negative - what happens to a polynomial?
     
  11. Mar 13, 2013 #10

    Dick

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    Noo. Take a simple example. p(x)=2x^2-x+3. f(x)=(2x^2-x+3)/(2x^2)=1-1/(2x)+3/(2x^2). What's the limit of that as x->infinity?
     
  12. Mar 20, 2013 #11
    what is the largest value possible for cos(x)? 1
    How about a polynomial - as x gets very big or very negative - what happens to a polynomial?: It gets very big or very small

    1

    I don't see how this relates to the question though.

    Overall, I understand why it's false, any polynomial will not hold for all values of x, as they will rather get very big, or very small, hence there is not a polynomial which satisfies the inequality. I don't think I need to prove it for this question, but out of curiosity, how would I go about do so?

    Thank you for your patience.
     
  13. Mar 20, 2013 #12

    Dick

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    It's related to the proof I was suggesting in post 5. If you put ##f(x)=\frac{p(x)}{a_n x^n}##, I would say that goes to 1 as x->infinity for any polynomial. That was just an example.
     
  14. Mar 21, 2013 #13

    Simon Bridge

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    ... what you are saying is that no matter what the value of x, -1 ≤ cos(x) ≤ 1, but for any finite order polynomial, P(x) must be very much bigger than 1 or very much smaller than -1 for some values of x.

    Can you not see how this relates to the original wording of the problem back in post #1?
    Oh well... over to you Dick.
     
  15. Mar 27, 2013 #14
    I thought this was very rude... I'm just trying to get some help.

    I've finished the problem and managed to prove it, sorry if I did not do it as fast as you Simon Bridge, but I am just learning.

    thank you for all your help, both of you.
     
  16. Mar 27, 2013 #15

    Simon Bridge

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    That was not supposed to be rude so my apologies for seeming so.

    There is a limit to what sort of help you can get here - if I had told you, in so many words, how what I was saying related to the original question, like you asked, then that would have amounted to doing your homework for you. If you could not get that connection, then I would be unable to help you.
    Logically I should step aside in favor of someone better qualified - that was Dick, who has been doing this much longer than me and may know a better approach. I'd watch what he did and learn.

    What would be helpful now would be if you showed us the solution/method you eventually came up with.
    That way others would be able to benefit from your experience.
     
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