# Proofs

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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)?
Last edited by Student 999; 1 month ago
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1 month ago
#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)?
flaurie_gcx might be able to help?
thanks 0
1 month ago
#3
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 for which , this should be easy.

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

First hint

But

Then

To finish

It is true if you restrict to a closed interval*, eg. if you ask only for 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 is arbitrary and cosine can be swapped for any other continuous function. This is the Weierstrass Approximation Theorem.

the * makes it work essentially because
Last edited by _gcx; 1 month ago
1
#4
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 for which , this should be easy.

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

First hint

But

Then

To finish

It is true if you restrict to a closed interval*, eg. if you ask only for 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 is arbitrary and cosine can be swapped for any other continuous function. This is the Weierstrass Approximation Theorem.

the * makes it work essentially because
0
1 month ago
#5
(Original post by Student 999)
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 , (for "asymptotic equivalence", basically they're close with very large, but this is formulated precisely - means as ) ie. as . " as " 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 )

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 changing. Basically if , you're looking at . But it's a bit clearer to consider the graph of , since is equivalent to . Your second paragraph is correct, my point is that you can't compress the graph by adding a constant, so you can't pick so that falls between and , (the system of equations , has no solutions) because the coefficient of cosine is 1 (or -1, doesn't really matter)
Last edited by _gcx; 1 month ago
0
#6
(Original post by _gcx)
This wouldn't be acceptable at a university level, the symbol you'd really want to use is , (for "asymptotic equivalence", basically they're close with very large, but this is formulated precisely - means as ) ie. as . " as " 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 )

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 changing. Basically if , you're looking at . But it's a bit clearer to consider the graph of , since is equivalent to . Your second paragraph is correct, my point is that you can't compress the graph by adding a constant, so you can't pick so that falls between and , (the system of equations , 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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