Is There a Polynomial That Meets the Given Criteria?

[phospho]

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 |P(x) - \cos(x)| \leq 10^{-6}

I've tried to thinking about it, and it seems as though it is false, because |cos(x)| \leq 1 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.

 

[Dick]

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 |P(x) - \cos(x)| \leq 10^{-6}

I've tried to thinking about it, and it seems as though it is false, because |cos(x)| \leq 1 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.

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.

 

[Simon Bridge]

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?

 

[phospho]

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.

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?

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?

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,

 

[Dick]

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?

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?

 

[Simon Bridge]

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 :)

 

[Dick]

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.

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.

 

[phospho]

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?

Would f(x) also tend to infinity?

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?)

because cos(x) is an infinite expansion?

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 :)

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 won't satisfy the inequality in the range, so the statement is false.

 

[Simon Bridge]

because cos(x) is an infinite expansion?

Hmmm... you should try it then. I'm a big fan of hard experience.

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 won't satisfy the inequality in the range, so the statement is false.

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?

 

[Dick]

Would f(x) also tend to infinity?

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?

 

[phospho]

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?

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

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?

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.

 

[Dick]

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.

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.

 

[Simon Bridge]

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

... 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.

 

[phospho]

... 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.

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.

 

[Simon Bridge]

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.