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Student 999

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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)?

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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_Rusty_

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#2

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#2

(Original post by

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)?

**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)?

thanks

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_gcx

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#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

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

First hint

But

Spoiler:

Show

Then

To finish

Spoiler:

Show

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

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Student 999

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#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.

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

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

**_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

Spoiler:

Show

Then

To finish

Spoiler:

Show

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

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_gcx

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#5

(Original post by

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.

**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.

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

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Student 999

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#6

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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)

**_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)

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