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 08-01-2012, 03:55 PM #3196 Pooh-Bah     Join Date: Jan 2003 Location: The one who is knocking Posts: 4,766 Re: The Official Math/Physics/Whatever Homework questions thread Friend sent this email out this morning: I have 6 acronyms. There are a total of 63 (?) combinations of these 6. What are they? RA AS PsA PS CD UC For example: RA AS PsA PS CD UC RA, AS, PsA, PS, CD, UC RA, AS, PsA, PS, CD RA, AS RA, PsA RA, PS RA, CD RA, UC RA, AS, PS RA, AS, CD RA, AS, UC RA, AS, PsA RA, PS, CD RA, PS, PsA RA, PS, UC RA, CD, UC RA, CD, PsA RA, UC, PsA RA, AS, PsA, CD RA, AS, PsA, UC RA, AS, PsA, PS RA, AS, PS, CD RA, AS, PS, UC Are there actually 63 or is she mistaken? Isn't this a permutation problem?
 08-01-2012, 04:13 PM #3197 Carpal \'Tunnel     Join Date: Mar 2007 Location: Redoubling with gusto Posts: 10,693 Re: The Official Math/Physics/Whatever Homework questions thread There are 64 if you count the empty set. In each set, each of the 6 elements is either in or out. So we have 2^6 = 64 ways to do that. To make it more concrete, call the 6 guys A, B, C, D, E, and F. And let each set be represented by a string of 0s and 1s telling whether each element is in the set. i.e. 000000 = none of them are in the set 111111 = all of them are in the set 100000 = just A 010000 = just B 010001 = B & F etc There are 64 such strings, including 000000 (since these strings are binary representations of the numbers 0 thru 63).
 08-01-2012, 05:22 PM #3198 Pooh-Bah     Join Date: Jan 2003 Location: The one who is knocking Posts: 4,766 Re: The Official Math/Physics/Whatever Homework questions thread Very simple when you explain it like that!
 08-03-2012, 02:09 AM #3199 veteran     Join Date: Oct 2010 Posts: 2,617 Re: The Official Math/Physics/Whatever Homework questions thread I'm mostly stuck on this: z = x^2 + 2*y^2 x=rcosθ y=rsinθ (∂z/∂x)θ=? In case the characters don't appear properly, what's the partial derivative of z with respect to x with theta held constant?
 08-05-2012, 11:36 AM #3200 enthusiast     Join Date: Jul 2009 Posts: 65 Re: The Official Math/Physics/Whatever Homework questions thread These questions are regarding trigonometric identities. First prove 1+2sin2x+cos2x/2+sin2x-2cos2x=cotx I have played around with using different double angles for the cos2x, but I seem to cancel everything. Second Prove tan(x-y) +tany= sec(x-y)sinxsecy On the LHS I have used the sum and difference angle for tanx-tany/1+tanxtany +tany. I have attempted squaring the tanx and tany, and using another identity , but I am note sure if it is legal. Any help is greatly appreciated.
08-05-2012, 10:22 PM   #3201

Join Date: May 2008
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by roundr2000blufedme These questions are regarding trigonometric identities. First prove 1+2sin2x+cos2x/2+sin2x-2cos2x=cotx I have played around with using different double angles for the cos2x, but I seem to cancel everything. Second Prove tan(x-y) +tany= sec(x-y)sinxsecy On the LHS I have used the sum and difference angle for tanx-tany/1+tanxtany +tany. I have attempted squaring the tanx and tany, and using another identity , but I am note sure if it is legal. Any help is greatly appreciated.
Could you show your work? People here could probably quickly and easily identify any mistakes.

 08-05-2012, 10:39 PM #3202 adept     Join Date: May 2008 Location: increasing potential GDP Posts: 1,170 Re: The Official Math/Physics/Whatever Homework questions thread I have this Series question that I'm sure I could answer easily with another test (rather than the divergence test, because I can see it will fail), and I'm sure the limit of this term approaches 0, but I can't seem to calculate it. Determine whether the series is convergent or divergent: $\sum\limits_{n = 1}^\infty {\frac{1 + 2^n}{3^n}$Suppose first I wanted to try the divergence test (again, I know it will fail, but this just made me think about how to take the limit of this term). $\lim_{n \to \infty}{\frac{1 + 2^n}{3^n}= \frac{\infty}{\infty}$ So, I use l'Hopital's rule: $\lim_{n \to \infty}{\frac{1 + 2^n}{3^n}= \lim_{n \to \infty}\frac{ln(2)2^n}{ln(3)3^n} = \frac{ln(2)}{ln(3)}\lim_{n \to \infty}{\frac{2^n}{3^n}$ Stuck! Must I set this term equal to y, and take the log of both sides, and attack the problem that way? Or is there some simple rule or simple algebra I'm forgetting here? Thank you.
 08-05-2012, 10:41 PM #3203 Carpal \'Tunnel     Join Date: Mar 2007 Location: Redoubling with gusto Posts: 10,693 Re: The Official Math/Physics/Whatever Homework questions thread Ratio test Edit: also you did the divergence test wrong. Edit2: pull a 3^n out of all terms (or a 2^n, same difference)
 08-05-2012, 11:00 PM #3204 adept     Join Date: May 2008 Location: increasing potential GDP Posts: 1,170 Re: The Official Math/Physics/Whatever Homework questions thread Thank you, Wyman. How did I do the divergence test wrong? If a = (1 + 2^n)/(3^n), it was my understanding that the divergence test states that if lim of a as n approaches infinity != 0, then the sum diverges. Is it my understanding or execution that is incorrect? Thank you.
08-05-2012, 11:19 PM   #3205

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Quote:
 Originally Posted by roundr2000blufedme These questions are regarding trigonometric identities. First prove 1+2sin2x+cos2x/2+sin2x-2cos2x=cotx I have played around with using different double angles for the cos2x, but I seem to cancel everything. Second Prove tan(x-y) +tany= sec(x-y)sinxsecy On the LHS I have used the sum and difference angle for tanx-tany/1+tanxtany +tany. I have attempted squaring the tanx and tany, and using another identity , but I am note sure if it is legal. Any help is greatly appreciated.
Are there any parentheses in the first problem that you forgot to include?

What do you mean by cancel everything?

Regarding the idea of squaring both sides: if the square of two reals m,n are equal does that guarantee that m,n are equal?

08-06-2012, 12:04 AM   #3206
Carpal \'Tunnel

Join Date: Mar 2007
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by non-self-weighter Thank you, Wyman. How did I do the divergence test wrong? If a = (1 + 2^n)/(3^n), it was my understanding that the divergence test states that if lim of a as n approaches infinity != 0, then the sum diverges. Is it my understanding or execution that is incorrect? Thank you.
Your understanding is fine. Execution (getting infinity/infinity) leaves something to be desired. See my edits for a hint.

08-06-2012, 05:42 AM   #3207
aka Double Ice

Join Date: Jun 2007
Posts: 4,563
Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by non-self-weighter Determine whether the series is convergent or divergent: $\sum\limits_{n = 1}^\infty {\frac{1 + 2^n}{3^n}$
I think Wyman suggested the ratio test, but I would try to show that the partial sums are bounded and increasing -- should be a one liner.

 08-06-2012, 06:49 AM #3208 Carpal \'Tunnel     Join Date: Mar 2007 Location: Redoubling with gusto Posts: 10,693 Re: The Official Math/Physics/Whatever Homework questions thread Yes, or one could claim that this sum is less than the sum of something like ( 2^n +2^n)/3^n, and use what we know about geometric series. Many ways to get at this -- he may not know that monotone bounded sequences are convergent.
 08-06-2012, 10:32 AM #3209 adept     Join Date: May 2008 Location: increasing potential GDP Posts: 1,170 Proving the series is convergent is easy. That's not what I wanted help with.
 08-06-2012, 11:20 AM #3210 Carpal \'Tunnel     Join Date: Mar 2007 Location: Redoubling with gusto Posts: 10,693 Re: The Official Math/Physics/Whatever Homework questions thread Oh. NSW, sorry I admit I only read half your post. At the point you got stuck, you just observe that 2^n/3^n = (2/3)^n --> 0 So the divergence test fails [but that doesn't mean that the series converges! You should use an alternate test (comparison or ratio come to mind) to prove that].

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