Supremum, infimum, maximum and minimum

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gn17
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Hi tsr,

A=(2^n)/((2^n)+1)cos((npi)/2) such that n is an element of natural numbers. Which of the following statements is true? a)modulus of minA=modulus of maxA b)supA=infA c)modulus of supA=modulus of infA d)modulus of minA=modulus of supA

I got the max to be 1, minA to be -1, supA to be 1 and infA to be -1. So according to my answers modulus of all the above answers except b) make sense but there is only one correct answer so which one is right and how do i get to it?

Also i worked out this answer from trial and error, is there a quicker way of doing this since this was a non-calculator question.
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Smaug123
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(Original post by gn17)
Hi tsr,

A=(2^n)/((2^n)+1)cos((npi)/2) such that n is an element of natural numbers. Which of the following statements is true? a)modulus of minA=modulus of maxA b)supA=infA c)modulus of supA=modulus of infA d)modulus of minA=modulus of supA

I got the max to be 1, minA to be -1, supA to be 1 and infA to be -1. So according to my answers modulus of all the above answers except b) make sense but there is only one correct answer so which one is right and how do i get to it?

Also i worked out this answer from trial and error, is there a quicker way of doing this since this was a non-calculator question.
Could you clarify - do you mean A = \dfrac{2^n}{2^n+1} \cos(\dfrac{n \pi}{2})?
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gn17
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(Original post by Smaug123)
Could you clarify - do you mean A = \dfrac{2^n}{2^n+1} \cos(\dfrac{n \pi}{2})?
Yes
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Smaug123
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(Original post by gn17)
Yes
So for n odd, \cos(\frac{n \pi}{2}) = 0; for n even, \cos(\frac{n \pi}{2}) = (-1)^{n/2}. Clearly the sequence increases in modulus with n, because the fraction term tends to 1, and 1 is an upper bound on the modulus of A. Moreover, it is the supremum of the modulus. However, it is not attained for any n, so the maximum does not exist.

The inf is -1, similarly, and the minimum doesn't exist.
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gn17
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(Original post by Smaug123)
So for n odd, \cos(\frac{n \pi}{2}) = 0; for n even, \cos(\frac{n \pi}{2}) = (-1)^{n/2}. Clearly the sequence increases in modulus with n, because the fraction term tends to 1, and 1 is an upper bound on the modulus of A. Moreover, it is the supremum of the modulus. However, it is not attained for any n, so the maximum does not exist.

The inf is -1, similarly, and the minimum doesn't exist.
But when e.g n=80, I do get A to be 1 and when n is 82 I get A to be -1.
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davros
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(Original post by gn17)
But when e.g n=80, I do get A to be 1 and when n is 82 I get A to be -1.
How can you possibly get A to be 1? Are you doing some sort of estimation with a calculator or something?

You'll never get \dfrac{2^n}{2^n + 1} = 1 for any finite n.
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gn17
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(Original post by davros)
How can you possibly get A to be 1? Are you doing some sort of estimation with a calculator or something?

You'll never get \dfrac{2^n}{2^n + 1} = 1 for any finite n.
I just realised my mistake, yes I did use a calculator and I didn't know that a calculator would round off the answer, I thought even the calculator would not give us 1 or -1 for any value of n, thanks for pointing that out. I know it was a non-calc question but I wasn't able to work it out without the calc, turns I out I couldn't do it with the calc as well!
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gn17
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(Original post by davros)
How can you possibly get A to be 1? Are you doing some sort of estimation with a calculator or something?

You'll never get \dfrac{2^n}{2^n + 1} = 1 for any finite n.
Can I also ask how you know that it will never give us 1? Is it because if we take the limit as n tends to infinity we get 1?
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Smaug123
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(Original post by gn17)
I just realised my mistake, yes I did use a calculator and I didn't know that a calculator would round off the answer, I thought even the calculator would not give us 1 or -1 for any value of n, thanks for pointing that out. I know it was a non-calc question but I wasn't able to work it out without the calc, turns I out I couldn't do it with the calc as well!
It returns an answer correct to 16dp or whatever. Once the value of the expression is greater than 0.9999999999999999, it will return 1.000000000000000.
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gn17
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(Original post by Smaug123)
It returns an answer correct to 16dp or whatever. Once the value of the expression is greater than 0.9999999999999999, it will return 1.000000000000000.
Yh, just found out, thanks for your help
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james22
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(Original post by gn17)
Can I also ask how you know that it will never give us 1? Is it because if we take the limit as n tends to infinity we get 1?
The only way it can be 1 is if 2^n/(2^n+1)=1 so 2^n=1+2^n so 1=0, which is a contradiction so it will never equal 1.
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astroturf
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(Original post by gn17)
I just realised my mistake, yes I did use a calculator and I didn't know that a calculator would round off the answer, I thought even the calculator would not give us 1 or -1 for any value of n, thanks for pointing that out. I know it was a non-calc question but I wasn't able to work it out without the calc, turns I out I couldn't do it with the calc as well!
let me give one simple advice: take your calculator home and forget about it (assuming you are studying maths at any serious uni)
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Smaug123
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(Original post by astroturf)
let me give one simple advice: take your calculator home and forget about it (assuming you are studying maths at any serious uni)
Well said - I haven't used a calculator in all my one-and-three-quarter years.
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Noble.
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(Original post by Smaug123)
Well said - I haven't used a calculator in all my one-and-three-quarter years.
I can't even think of an undergrad topic where a calculator would be any use...
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Smaug123
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(Original post by Noble.)
I can't even think of an undergrad topic where a calculator would be any use...
There was one question in a numerical analysis example sheet which asked you to simulate a five-decimal-place computer. By hand. Performing approximately twenty additions, ten multiplications and five divisions.
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davros
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(Original post by Smaug123)
There was one question in a numerical analysis example sheet which asked you to simulate a five-decimal-place computer. By hand. Performing approximately twenty additions, ten multiplications and five divisions.
And your problem with that is?
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Smaug123
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(Original post by davros)
And your problem with that is?
You sadist!
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james22
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(Original post by Noble.)
I can't even think of an undergrad topic where a calculator would be any use...
Some more advanced stats modules use one I think.
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davros
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(Original post by Smaug123)
You sadist!
No calculators in the Tripos!
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BlueSam3
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(Original post by gn17)
Can I also ask how you know that it will never give us 1? Is it because if we take the limit as n tends to infinity we get 1?
You're overthinking it a lot. It's entirely possible to have a sequence that tends to one and that has values including one (the sequence (1,1,1,...) being an obvious example, and alternating terms between this and any other sequence tending to one will give you some non-trivial examples). It isn't, however, possible to have a strictly increasing sequence tending to one that takes the value 1 for some n.

Whilst that is a reason why \frac{2^n}{2^n+1}\neq 1 for all n, it's far from the most obvious reason. The obvious reason is that a fraction is equal to one if and only if the numerator and denominator are equal, and it should be obvious that 2^n \neq 2^n+1.
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