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 Post subject: Calculus
PostPosted: Fri, 19 Nov 2010 15:31:19 UTC 
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I heard this problem from a colleague recently, and I thought her solution was sufficiently clever to merit a calculus-level proposed problem. If you're forcing the answer, or using a calculator/graphing device, you're trying too hard.

The problem:

Find the minimum value of:

|\sin x+\cos x+\tan x+\csc x+\sec x+\cot x|.

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PostPosted: Mon, 22 Nov 2010 11:32:58 UTC 
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 Post subject: Re: Calculus
PostPosted: Tue, 23 Nov 2010 04:17:10 UTC 
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Shadow wrote:
I heard this problem from a colleague recently, and I thought her solution was sufficiently clever to merit a calculus-level proposed problem. If you're forcing the answer, or using a calculator/graphing device, you're trying too hard.

The problem:

Find the minimum value of:

|\sin x+\cos x+\tan x+\csc x+\sec x+\cot x|.


man111,

Please don't post mathematics via images. Edit: Sorry, Shadow said it's OK.

Your final answer looks correct to me.

I like your work down to the line before equation 2.

$E=|t+{2\over t-1}|

I would then find the extrema of $f(t)=t+{2\over t-1},\ \  -\sqrt{2}\le t\le \sqrt{2}, and apply the result to E. Since f(t) has no real zeros, the absolute value will not introduce additional critical points.

$f'(t)=1-{2\over (t-1)^2}

f'(t)=0 \ \implies\  t=1 \pm\sqrt{2}. 1 +\sqrt{2} is not in the domain of f.

f(t) has relative extrema at t=-\sqrt{2},\  1 +\sqrt{2}, \ \sqrt{2}.

f(-\sqrt{2})=2-3\sqrt{2} \implies E=-2+3\sqrt{2}, \mbox{ at } t=-\sqrt{2}.

f(1-\sqrt{2})=1-2\sqrt{2} \implies E=-1+2\sqrt{2}, \mbox{ at } t=1-\sqrt{2}.

f(\sqrt{2})=2+3\sqrt{2} \implies E=2+3\sqrt{2}, \mbox{ at } t=\sqrt{2}.

\min(|\sin x+\cos x+\tan x+\csc x+\sec x+\cot x|)=-1+2\sqrt{2}.


Last edited by Sam Snyder on Tue, 23 Nov 2010 04:43:12 UTC, edited 2 times in total.

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 Post subject: Re: Calculus
PostPosted: Tue, 23 Nov 2010 04:21:32 UTC 
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Sam Snyder wrote:
Shadow wrote:
I heard this problem from a colleague recently, and I thought her solution was sufficiently clever to merit a calculus-level proposed problem. If you're forcing the answer, or using a calculator/graphing device, you're trying too hard.

The problem:

Find the minimum value of:

|\sin x+\cos x+\tan x+\csc x+\sec x+\cot x|.


man111,

Please don't post mathematics via images.

You final answer looks correct to me.

I like your work down to the line before equation 2.

$E=|t+{2\over t-1}|

I would then find the extrema of $f(t)=t+{2\over t-1},\ \  -\sqrt{2}\le t\le \sqrt{2}, and apply the result to E. Since f(t) has no real zeros, the absolute value will not introduce additional critical points

$f'(t)=1-{2\over (t-1)^2}

f'(t)=0 \ \implies\  t=1 \pm\sqrt{2}. 1 +\sqrt{2} is not in the domain of f.

f(t) has relative extrema at t=-\sqrt{2},\  1 +\sqrt{2}, \ \sqrt{2}.

f(-\sqrt{2})=-1-2\sqrt{2} \implies E=1+2\sqrt{2}, \mbox{ at } t=-\sqrt{2}.

f(1-\sqrt{2})=1-2\sqrt{2} \implies E=-1+2\sqrt{2}, \mbox{ at } t=1-\sqrt{2}.

f(\sqrt{2})=2+3\sqrt{2} \implies E=2+3\sqrt{2}, \mbox{ at } t=\sqrt{2}.

\min(|\sin x+\cos x+\tan x+\csc x+\sec x+\cot x|)=-1+2\sqrt{2}.


He can post however he feels like, there's no reason he shouldn't.

Also, the absolute value won't ever introduce new critical points if you're looking at zeros of the derivative, the only complex number of norm 0 is 0.

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 Post subject: Re: Calculus
PostPosted: Tue, 23 Nov 2010 04:37:26 UTC 
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Shadow wrote:

Also, the absolute value won't ever introduce new critical points if you're looking at zeros of the derivative, the only complex number of norm 0 is 0.


But the absolute value can introduce new critical points at zeros of a real function.


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 Post subject: Re: Calculus
PostPosted: Tue, 23 Nov 2010 04:47:43 UTC 
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Sam Snyder wrote:
Shadow wrote:

Also, the absolute value won't ever introduce new critical points if you're looking at zeros of the derivative, the only complex number of norm 0 is 0.


But the absolute value can introduce new critical points at zeros of a real function.


Yes, but if you want to be situation specific, zeros of the function represent the minimum, so "new critical values" would all be moot anyways, just check for zeros of the function. I mean, it's okay to be thorough with what you're doing, but if you're going that far you need to go all the way, otherwise it just sounds like you're looking for a reason to add another post to your total post-count. The most qualitative thing you could have said there was not the lack of zeros making new critical points, but rather if it were zero we would already be done.

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 Post subject: Re: Calculus
PostPosted: Tue, 23 Nov 2010 05:07:33 UTC 
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Shadow wrote:
Sam Snyder wrote:
Shadow wrote:

Also, the absolute value won't ever introduce new critical points if you're looking at zeros of the derivative, the only complex number of norm 0 is 0.


But the absolute value can introduce new critical points at zeros of a real function.


Yes, but if you want to be situation specific, zeros of the function represent the minimum, so "new critical values" would all be moot anyways, just check for zeros of the function.



|(x-1)^2-4| has two critical points in addition to the one (x-1)^2-4 has.


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 Post subject: Re: Calculus
PostPosted: Tue, 23 Nov 2010 05:12:46 UTC 
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Sam Snyder wrote:
Shadow wrote:
Sam Snyder wrote:
Shadow wrote:

Also, the absolute value won't ever introduce new critical points if you're looking at zeros of the derivative, the only complex number of norm 0 is 0.


But the absolute value can introduce new critical points at zeros of a real function.


Yes, but if you want to be situation specific, zeros of the function represent the minimum, so "new critical values" would all be moot anyways, just check for zeros of the function.



|(x-1)^2-4| has two critical points in addition to the one (x-1)^2-4 has.


See the edit to my prior post.

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