The Official Math/Physics/Whatever Homework questions thread - Page 25

It's not very difficult to actually compute what a_n is, and from there it is trivial to compute the limit of a_{n+1}/a_n. This way you don't have to worry about proving that the limit actually exists.

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03-01-2010 , 01:31 PM

#603

furyshade

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Quote:

Originally Posted by blah_blah

maybe i'm not understanding, do you mean just calculate the ratio for higher order terms?

perhaps i'm missing something to do with how to use generating functions. we didn't really go over it in class, he just gave us a handout on it and put it on the homework so i could be missing something really easy.

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03-01-2010 , 02:38 PM

#604

Myrmidon7328

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Join Date: Dec 2007 Posts: 2,133

Sorry for the poor notation, but say I have a discrete random process X, with n outcomes, X_1,...X_n, each with probability P(X_1),...,P(X_n), respectively.

E[X] = sum[P(X_i)X_i]

How do I calculate the variance? Is it just

Var[X] = sum[P(X_i)(X_i - E[X])^2]
?

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03-02-2010 , 02:40 AM

#605

blah_blah

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Join Date: Feb 2007 Posts: 1,659

Or \sum_i P(X_i)X_i^2-(\sum_i P(X_i)X_i)^2.

In general, E[f(X)] = \sum_i P(X_i)f(X_i) -- in this case, f(x) = (x-E(X))^2.

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03-02-2010 , 02:43 AM

#606

blah_blah

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Join Date: Feb 2007 Posts: 1,659

Quote:

Originally Posted by furyshade

You have the recursion a_n-a_{n-1}-6a_{n-2} = 0. The characteristic polynomial of this recursion is r^2-r-6r=0. This has roots r_1,r_2. The solution to the recursion is a_n = c(r_1)^n+d(r_2)^n (where c,d can be determined by the boundary conditions a_0=0,a_1=1).

I forget how to do this by generating functions though.

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03-04-2010 , 05:10 PM

#607

ColeW123

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Probability Question!

If five numbers are selected at random from the set {1,2,3,...,20}, hwat is the probability that their minimum is larger than 5?

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03-04-2010 , 05:25 PM

#608

Wyman

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Well, this should be
(the probability that the first number you pick is larger than 5) *
(the probability that the second number you pick is larger than 5) *
(the probability that the third number you pick is larger than 5) *
(the probability that the fourth number you pick is larger than 5) *
(the probability that the fifth number you pick is larger than 5)

What do you get for those?

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03-04-2010 , 05:38 PM

#609

ColeW123

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15^5/20^5 ? seems too easy...

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03-04-2010 , 05:41 PM

#610

Wyman

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That would be correct if your five numbers could be (6,6,6,6,6), for instance.

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03-04-2010 , 05:43 PM

#611

ColeW123

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I guess I think it's just strange because the section of the book is Combinations, but I'm not using them for the problem.

Last edited by ColeW123; 03-04-2010 at 05:52 PM.

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03-04-2010 , 05:46 PM

#612

Wyman

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Quote:

Originally Posted by ColeW123

I guess I think it's just strange because the section of the book is Combinatorics, but I"m not using them for the problem.

Yes. What I'm saying is that you shouldn't get (15/20)^5 unless you can pick the number 6 five times. You can't, right? You have to pick 5 random numbers, and you don't put the number you picked back each time.

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03-04-2010 , 05:52 PM

#613

ColeW123

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Would it be (15 choose 5) / (20 choose 5) if I were to use combinations?

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03-04-2010 , 06:03 PM

#614

Wyman

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Yes. This is the number of 5-element subsets of {6,...,20} divided by the number of 5-element subsets of {1,...,20}.

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03-04-2010 , 06:16 PM

#615

ColeW123

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Quote:

Originally Posted by Wyman

Yes. This is the number of 5-element subsets of {6,...,20} divided by the number of 5-element subsets of {1,...,20}.

Thank you very much sir. Seems too easy but I guess it makes sense.

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03-04-2010 , 06:18 PM

#616

Wyman

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Don't worry. It gets harder.

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03-04-2010 , 06:34 PM

#617

ColeW123

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Okay, I'm officially stuck on this Conditional Probability problem...

Suppose that 28 crayons, of which four are red, are divided random among Jack, Marty, Sharon, and Martha (seven each). If Sharon has exactly one red crayon, what is the probability that Marty has the remaining 3?

P(Martha has 1 red) = P(B) = (4 choose 1)*(24 choose 6) / (28 choose 7)

I know the formula says P(A|B) = P(AB)/P(B)

How do I find P(AB)?

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03-04-2010 , 11:38 PM

#618

Banzai-

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You don't imo. That formula would be used to find P(AB). P(A|B) is much easier, and does not require knowledge of P(B). Just think of it as a new problem with 3 people and 21 crayons, of which 3 are red. Then find probability Marty gets all 3.

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03-05-2010 , 12:32 AM

#619

vetiver

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Logic class. Supposed to "simplify" sentence 5 into "negation normal form", meaning no superfluous use of negatives. My work is shown in line 6, with my final answer being the bottom line:

¬ = "it is false that"
∨ = "or" (meaning 1 or more is true)
∧ = "and"

I've run through this many times and can't find where I went wrong - do you see a flaw in my reductions?

EDIT: missing parentheses at the end, but it's still wrong so just assume that it's there.

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03-05-2010 , 01:34 AM

#620

vetiver

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^ nevermind, figured it out, misplaced sign.

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03-05-2010 , 05:45 AM

#621

ANONN123

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reviewing some fluid dynamics from last year and i'm drawing a complete blank (and i got an A in this class, lol)

just a point in the right direction would be cool

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03-06-2010 , 05:19 AM

#622

tarheeljks

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any recommendations for a complex analysis book?

Quote:

Originally Posted by Myrmidon7328

late to the party here but

Var[X]= E[(X-µ)²]= E[X²] - (E[X])²

in your notation it's

Var[X] = sum[P(X_i)(X_i)²] - (E[X])^2

= sum[P(X_i)(X_i)²] - (sum[P(X_i)(X_i])^2

what you have originally sort of looks like sample variance which is

(1/N)*sum(1,N)[(x_i-µ)²]

Last edited by tarheeljks; 03-06-2010 at 05:49 AM.

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03-06-2010 , 05:45 AM

#623

slipstream

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Quote:

Originally Posted by tarheeljks

any recommendations for a complex analysis book?

I loved "Complex Analysis" by Stein & Shakarchi. It's the third in a four-book series which seeks to highlight the interconnectedness between different parts of analysis. It's the only book I've ever used for the subject, so I can't comment on how it differs from other books, but I found the material very clearly presente and really elegant. A more canonical choice is Ahlfors.

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03-06-2010 , 06:33 AM

#624

blah_blah

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Join Date: Feb 2007 Posts: 1,659

What he wrote is fine, as

\sum P(x_i)(x_i-E[X])^2 = \sum P(x_i)x_i^2-2E[X]\sum P(x_i)x_i+\sum P(x_i)E[X]^2 = E[X^2]-2E[X]E[X]+E[X]^2 = E[X^2]-(E[X])^2

re: a complex analysis text, I second Stein and Shakarchi as the best option, and Ahlfors as a good secondary reference. I would put them in this order regardless of price, but the fact that Stein and Shakarchi is a third of the price, more modern, and covers more material makes the choice trivial.