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Adjusting from cookbook math

CGiuliano

Lowly Undergrad
Joined
4/19/09
Messages
234
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28
So, I just started analysis and.... yeahhh. I can already tell this is nothing like the "cookbook" calculus that I'm used to. If anyone can shed some insight on this question, please do so. Thanks in advance

Let An (A sub n) denote the area of a regular 2^n-sided polygon inscribed in a unit circle. (Assume n>=2.) Use your knowledge from trigonometry and calculus, prove that the sequence is monotone and bounded above, and give its limit.

Here is where I am.

Area of unit circle = pi(r)^2 = pi(1)^2; A = pi.

Now I know that An would be bounded by the area of the circle i.e. pi. So it is bounded above. Also, I know that by definition of transcribed, the radius would be fixed. It follows that as n increases, the area of An increases. This means the An is monotone (increasing). Since it is bounded above and monotone increasing the limit must be pi (Theorem in my book).


Basically, I solved the problem with logic, but I don't know how to prove this with calculus. If anyone can see a way, please share. Oh yeah, I also wrote up a formula for the area of An: An = (2^n/2) sin(2pi/2).

Thanks again
 
Now I know that An would be bounded by the area of the circle i.e. pi. So it is bounded above. Also, I know that by definition of transcribed, the radius would be fixed. It follows that as n increases, the area of An increases. This means the An is monotone (increasing). Since it is bounded above and monotone increasing the limit must be pi (Theorem in my book).


Basically, I solved the problem with logic, but I don't know how to prove this with calculus. If anyone can see a way, please share. Oh yeah, I also wrote up a formula for the area of An: An = (2^n/2) sin(2pi/2).
Click to expand...

You have to prove the intuitively plausible statement that as n increases, the area increases. Then you have shown that the sequence is both bounded and monotone. If your formula for the area of An is correct, you can show that it increases as n increases (but show how you derived the formula, and use maybe an induction argument to prove that it increases as n increases). Once you have done this, you can assert that being a bounded and montone sequence, it must have a least upper bound -- which will be the limit of the sequence. But you have to demonstrate that pi -- while an upper bound, as you have argued -- is a least upper bound. To do this you have to show that the shaded area that is everything in the circle not contained in An converges to 0 as n increases (using terminology, it's a "null sequence"). Then you're done.
 
I'll help you get started on this one. First of all, your formula is wrong (probably a typo).

An = 2^n*1/2*sin(2pi/2^n) ... the important thing here is the 2^n in the denominator in the sine function

So I'm going to show that this sequence converges to pi using the squeeze theorem, and hopefully give you some motivation for the rest of the problem in the process.

Now we know that the Taylor series of sin(x), around x=0, is sin(x) = x - x^3/3! + x^5/5! +...

Therefore, we know the following inequality holds for x >= 0 : x - x^3/3! <= sin(x) <= x

Therefore, since 2pi/2^n>=0 for all n, we have:
(2pi/2^n) - (2pi/2^n)^3/3! <= sin(2pi/2^n) <= 2pi/2^n,
which then we can turn into:
2^n*1/2*[(2pi/2^n) - (2pi/2^n)^3/3!] <= 2^n*1/2*sin(2pi/2^n) <= 2^n*1/2*(2pi/2^n)

Now here's the fun part. Let's do the RHS first:
2^n*1/2*(2pi/2^n) = pi, which obviously goes to infinity as n->infinity as it is constant (this also shows that the sequence is bounded above by pi)
Now for the LHS:
2^n*1/2*[(2pi/2^n) - (2pi/2^n)^3/3!] = pi - 2pi/(3*4^n) -> pi as n->infinity

Which means that we've proven that the limit of our sequence is indeed pi by the squeeze theorem.

Now you just have to show it's monotone!
 
Very helpful, thanks so much. :tiphat::smt023:)

bbw- I think the book hints at proof by induction.

financeguy- I just worked it out on paper and that is some pretty creative work.
 
why thank you.. its been a while..

just so you know, proving monotonicity will generally involve induction.. its just the nature of the beast

in this case, i would think about the property sin(2x)=2sin(x)cos(x) :)
 
I'm sorry can you explain this:

"Therefore, we know the following inequality holds for x >= 0 : x - x^3/3! <= sin(x) <= x"

Thanks.
 
CGiuliano said:
"Therefore, we know the following inequality holds for x >= 0 : x - x^3/3! <= sin(x) <= x"
Click to expand...

Both inequalities follow from the power series expansion of sin x (= x - x^3/3! + x^5/5! - x^7/7! + ...). To see the first inequality, observe that x^5/5! - ... is a positive number; to see the second, observe that -x^3/3! + ... is a negative number (try proving these by yourself).
 
financeguy said:
why thank you.. its been a while..

just so you know, proving monotonicity will generally involve induction.. its just the nature of the beast

in this case, i would think about the property sin(2x)=2sin(x)cos(x) :)
Click to expand...
If a further hint is required, then you might try devising a recursion formula for (A_{n+1}) in terms of (A_n).

Although it may not be in the spirit of the question, it looks to me like there is a way that doesn't require a Taylor series argument. Consider:
(\lim_{x \rightarrow \infty}\frac{1}{2}x sin(\frac{2 \pi}{x}))
If this limit exists, then the limit
(\lim_{n \rightarrow \infty}\frac{1}{2}2^n sin(\frac{2 \pi}{2^n}))
also exists, and they are equal.

The first expression can be worked on using the things you learn in a first calculus class--L'Hospital's Rule being the nicest of these tools. It's easiest to transform the expression a bit:
(\frac{1}{2} \lim_{y \rightarrow 0}\frac{sin(2 \pi y)}{y})
Again, if this limit exists, then the one above expressed in terms of (x) also does, and they are equal. The expression in (y) is a (\frac{0}{0}) limit, so we can use L'Hospital:
(\frac{1}{2} \lim_{y \rightarrow 0}\frac{sin(2 \pi y)}{y} =)
(\frac{1}{2} \lim_{y \rightarrow 0}\frac{2 \pi cos(2 \pi y)}{1} =)
(\frac{1}{2} 2 \pi = \pi)

This fact, combined with the facts that (1) the sequence is monotonic, and (2) the value (A_2), for example, is less than (\pi), allow you to back into the fact that this limit is also a LUB.
 
Just came back to it now and I think I'm close-- don't give me the answer!-- I'll get back to you in a minute.

---------- Post added at 02:54 PM ---------- Previous post was at 02:26 PM ----------

I'll add to this as it comes:

An+1 = An/cos(pi/2^n)

cos(pi/2^n) <= 1

A2*cos(pi/2^n) <= A2

A2 <= A2/cos(pi/2^n) = A3

A2 <= A3

An <= An+1

!!!!!!!!!!!!!!!
 
if that recursion formula is correct, then all you need is An+1 = An/cos(pi/2^n), cos(pi/2^n) <= 1, and a base (i.e. just show A3>A2) and you're done.

that second part doesnt look right.. looks like a circular argument.. you have to assume that first statement A0 <= pi is true for all n (i.e. that 0 was arbitrary) in order to prove that its true for all n.. in reality, A0 is not arbitrary right? to show boundedness, you really just use inequalities in the general An case.. just say An<=something by changing a term of An. that's really how you do it
 
I actually solved it backwards from the way it is displayed.. And wait, didn't I already show A1>A0? I need to do it again for A3 and A2?

Thanks for all your help.
 
Here's boundess;

(\log(A_{n+1})=\sum_{k=2}^n-\log(\cos(\frac{\pi}{2^k}))),
we have (\log(\cos(\frac{\pi}{2^k}))=\log(1-(1-\cos(\frac{\pi}{2^k})))=\log(1-(1-1+\mathcal{O}_0(2^{-2k})))=\log(1-\mathcal{O}_0(2^{-2k}))=-\mathcal{O}_0(2^{-2k})),
so we have (\log(A_{n+1})=\sum_{k=2}^n\mathcal{O}_0(2^{-2k})\leq \sum_{k=2}^{\infty}\mathcal{O}_0(2^{-2k})<\infty)
 
this is his first analysis class.. jeez i dont even know what Mr Doe is talking about

that may be right, but i wouldnt get confused looking at that solution.. PM me CGiuliano, we'll talk about this problem
 
Sn = Sum from k=1 to n of Ck*cos (kt), where Ck = 1/2^k.

Prove Sn >= .7 for 0 <= t <= .1.

This question is killing me :wall.

Here is what I've got so far:

I estimated cos kt > .95 for 0 <= kt <= .3 which hits the first three terms... now I need to find a lower bound for all the terms, and I think (from previous examples) I'm going to have to use the difference form of the triangle inequality...? Any guidance would be greatly appreciated. Thanks.
 
Just an idea...

(S_n(t)=\sum_{k=1}^n\frac{1}{2^k}\cos(kt)=Re(\sum_{k=1}^n\frac{1}{2^k}e^{itk})=Re(\sum_{i=1}^k(\frac{e^{it}}{2})^k)),

now you know that;

(\frac{1-z^{n+1}}{1-z}-1=\sum_{k=1}^nz^k), so we have

(\sum_{k=1}^n(\frac{e^{it}}{2})^k=\frac{1-\frac{e^{(n+1)it}}{2^{n+1}}}{1-\frac{e^{it}}{2}}-1), now just multiply and divide the first expression on the right side by (((1-\frac{\cos(t)}{2})+i\frac{\sin(t)}{2})) (you are doing this to get rid of the complex terms in the denominator). Now, forget about the complex terms, you should get a clear analytic expression of your sum and then try to do some simple estimate (I'm not going to do that ;)).
 
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