Student 999
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#1
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#1
Is my explanation for part ii lacking, could a simple proof by counterexample be enough?

For part iii, currently the only idea in my mind is considering the Taylor expansion of cos(x) to an infinite degree as polynomial P(x)?Name:  Screenshot 2022-06-25 at 19.42.48.png
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Size:  87.6 KBName:  IMG_213F668ED53C-1.jpeg
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Last edited by Student 999; 1 month ago
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_Rusty_
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#2
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#2
(Original post by Student 999)
Is my explanation for part ii lacking, could a simple proof by counterexample be enough?

For part iii, currently the only idea in my mind is considering the Taylor expansion of cos(x) to an infinite degree as polynomial P(x)?Name:  Screenshot 2022-06-25 at 19.42.48.png
Views: 14
Size:  87.6 KBName:  IMG_213F668ED53C-1.jpeg
Views: 14
Size:  105.9 KB
flaurie_gcx might be able to help?
thanks
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_gcx
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For ii your reasoning is not correct since cosine is not injective: if cos(x) = cos(y) you could have x = -y. To establish it's not equal for all theta you just need to find a \theta for which \cos(\sin \theta) \ne \sin(\cos \theta), this should be easy.

iii is false. I'm not sure what level you're supposed to answer it at.

First hint
Spoiler:
Show


Note that any non-constant polynomial P goes to \infty or -\infty as x \to \infty.

But
Spoiler:
Show


cosine is bounded between -1 and 1 so the inequality will be broken for sufficiently large x for any non-constant polynomial P

Then
Spoiler:
Show


This leaves only constant polynomials P.

To finish

Spoiler:
Show


Note that \cos x - c would have to fall between -10^{-6} and 10^{-6}. But subtracting a constant does not compress the cosine wave, it only translates it up or down. More precisely, \cos x - c will always oscillate between -1 - c and 1 - c, and this will always have width 2 and not 2 \times 10^{-6}.

It is true if you restrict x to a closed interval*, eg. if you ask only for 0 \le x \le 1 for example and taking the first something terms of the Taylor expansion works for it. You would have to look up what the remainder of a taylor series is for this. In this case the number 10^{-6} is arbitrary and cosine can be swapped for any other continuous function. This is the Weierstrass Approximation Theorem.

the * makes it work essentially because
Spoiler:
Show
you can no longer say "for sufficiently large x", since there is an upper limit to what x can be.
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Student 999
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#4
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#4
Name:  IMG_831538CF5082-1.jpeg
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Size:  55.7 KB Would this be correct, also for 'To finish' part is that sort of an explanation by contradiction as in cos(x) - C shifts the graph/y values up or down hence the inequality won't be true for values of big values of C just how the polynomial P acts for different values of x?

Essentially hinting that it's effect does not change the difference between the maximum and minimum values so -1-c<= cos(x)-c<=1-c unlike something like kcos(x) where you can compress the graph to match a certain range of values.
(Original post by _gcx)
For ii your reasoning is not correct since cosine is not injective: if cos(x) = cos(y) you could have x = -y. To establish it's not equal for all theta you just need to find a \theta for which \cos(\sin \theta) \ne \sin(\cos \theta), this should be easy.

iii is false. I'm not sure what level you're supposed to answer it at.

First hint
Spoiler:
Show


Note that any non-constant polynomial P goes to \infty or -\infty as x \to \infty.

But
Spoiler:
Show


cosine is bounded between -1 and 1 so the inequality will be broken for sufficiently large x for any non-constant polynomial P

Then
Spoiler:
Show


This leaves only constant polynomials P.

To finish

Spoiler:
Show


Note that \cos x - c would have to fall between -10^{-6} and 10^{-6}. But subtracting a constant does not compress the cosine wave, it only translates it up or down. More precisely, \cos x - c will always oscillate between -1 - c and 1 - c, and this will always have width 2 and not 2 \times 10^{-6}.

It is true if you restrict x to a closed interval*, eg. if you ask only for 0 \le x \le 1 for example and taking the first something terms of the Taylor expansion works for it. You would have to look up what the remainder of a taylor series is for this. In this case the number 10^{-6} is arbitrary and cosine can be swapped for any other continuous function. This is the Weierstrass Approximation Theorem.

the * makes it work essentially because
Spoiler:
Show
you can no longer say "for sufficiently large x", since there is an upper limit to what x can be.
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_gcx
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(Original post by Student 999)
Name:  IMG_831538CF5082-1.jpeg
Views: 11
Size:  55.7 KB Would this be correct, also for 'To finish' part is that sort of an explanation by contradiction as in cos(x) - C shifts the graph/y values up or down hence the inequality won't be true for values of big values of C just how the polynomial P acts for different values of x?

Essentially hinting that it's effect does not change the difference between the maximum and minimum values so -1-c<= cos(x)-c<=1-c unlike something like kcos(x) where you can compress the graph to match a certain range of values.
This wouldn't be acceptable at a university level, the symbol you'd really want to use is \sim, (for "asymptotic equivalence", basically they're close with x very large, but this is formulated precisely - f \sim g means f(x)/g(x) \to 1 as x \to \infty) ie. P(x) - \cos(x) \sim P(x) as x \to \infty. "P(x) - \cos(x) \to P(x) as x \to \infty" is a bit sloppy beyond A-level. (I'm not particularly happy with it being used at A-level even, I'd prefer sticking to \approx)

Your conclusion is unclear, you say "hence diverges" but you haven't said anything about the inequality. Also you should make clear that P is not constant.

I'm not sure why you're talking about c changing. Basically if P(x) = c, you're looking at |P(x) - \cos x| = |c - \cos(x)|. But it's a bit clearer to consider the graph of \cos x - c, since |x| &lt; c is equivalent to -c &lt; x &lt; c. Your second paragraph is correct, my point is that you can't compress the graph by adding a constant, so you can't pick c so that \cos x - c falls between -10^{-6} and 10^{-6}, (the system of equations -1 - c = -10^{-6}, 1 - c = 10^{-6} has no solutions) because the coefficient of cosine is 1 (or -1, doesn't really matter)
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Student 999
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#6
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#6
(Original post by _gcx)
This wouldn't be acceptable at a university level, the symbol you'd really want to use is \sim, (for "asymptotic equivalence", basically they're close with x very large, but this is formulated precisely - f \sim g means f(x)/g(x) \to 1 as x \to \infty) ie. P(x) - \cos(x) \sim P(x) as x \to \infty. "P(x) - \cos(x) \to P(x) as x \to \infty" is a bit sloppy beyond A-level. (I'm not particularly happy with it being used at A-level even, I'd prefer sticking to \approx)

Your conclusion is unclear, you say "hence diverges" but you haven't said anything about the inequality. Also you should make clear that P is not constant.

I'm not sure why you're talking about c changing. Basically if P(x) = c, you're looking at |P(x) - \cos x| = |c - \cos(x)|. But it's a bit clearer to consider the graph of \cos x - c, since |x| &lt; c is equivalent to -c &lt; x &lt; c. Your second paragraph is correct, my point is that you can't compress the graph by adding a constant, so you can't pick c so that \cos x - c falls between -10^{-6} and 10^{-6}, (the system of equations -1 - c = -10^{-6}, 1 - c = 10^{-6} has no solutions) because the coefficient of cosine is 1 (or -1, doesn't really matter)
Yeah your third paragraph is what I was trying to get at minus my own confusion haha, in terms of the notation that you talked about should I focus on learning the basics of university 'standard notation' to help me get a better understanding of maths or continue trying to solve questions and wait to learn it when I'm actually at uni? If the former is there a website/book that you would specifically recommend? Thanks for your time
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