The Official Math/Physics/Whatever Homework questions thread - Page 214

LondonBroil · 08-01-2012, 03:55 PM

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?

Wyman · 08-01-2012, 04:13 PM

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).

LondonBroil · 08-01-2012, 05:22 PM

Very simple when you explain it like that!

la6ki · 08-03-2012, 02:09 AM

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?

roundr2000blufedme · 08-05-2012, 11:36 AM

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.

non-self-weighter · 08-05-2012, 10:22 PM

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.

non-self-weighter · 08-05-2012, 10:39 PM

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.

Wyman · 08-05-2012, 10:41 PM

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)

non-self-weighter · 08-05-2012, 11:00 PM

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.

Chasqui · 08-05-2012, 11:19 PM

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?

Wyman · 08-06-2012, 12:04 AM

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.

Alex Wice · 08-06-2012, 05:42 AM

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.

Wyman · 08-06-2012, 06:49 AM

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.

non-self-weighter · 08-06-2012, 10:32 AM

Proving the series is convergent is easy. That's not what I wanted help with.

Wyman · 08-06-2012, 11:20 AM

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].