Wikipedia:Reference desk/Archives/Mathematics/2010 April 11

= April 11 =

Expressing an integer as a sum of powers of a prime
Hello all. This question is part of a proof that I am reading from a book. Essentially we have an integer i which has a prime p in its prime factorization. Now the book says let $$\mu(i)$$ be the largest integer such that $$\frac{i}{p^{\mu(i)}}$$ is an integer. I take it to mean that that $$\mu(i)$$ is the highest power of p in the prime factorization of i. Now the book states that $$i=\sum_{j\ge \mu(i)} a_jp^j$$ where $$a_j$$ 's are positive integers. I take it that it is implicitly assumed that this sum is finite, but I don't understand how to prove this statement formally. The next statement is that $$1\le a_{\mu(i)}<p$$. I don't quite follow this one either. Can anyone please help me out? Thanks.-Shahab (talk) 06:34, 11 April 2010 (UTC)


 * The infinite sum might make sense if the book is describing p-adic numbers - the emphasis on a prime base suggests this is plausible. Alternatively, if $$j\ge \mu(i)$$ is a misprint for $$j\le \mu(i)$$ then the book is just saying that i has an expansion in base p with $$\mu(i)$$ digits $$a_j$$, in which the most significant digit $$a_{\mu(i)}$$ is non-zero. Gandalf61 (talk) 08:31, 11 April 2010 (UTC)
 * The $$a_j$$'s are positive or nonnegative? If the latter then it makes perfect sense. All the large powers will have zero coefficient, and this is just the expansion in base p. The $$j\ge\mu(i)$$ just means that all the small powers are unnecessary because i is divisible by $$p^{\mu(i)}$$. All of them are $$0\le a_j<p$$ and also, $$a_{\mu(i)}$$ can't be zero because then i would be divisible by $$p^{\mu(i)+1}$$. -- Meni Rosenfeld (talk) 10:45, 11 April 2010 (UTC)


 * Meni, what do you mean by saying that "The $$j\ge\mu(i)$$ just means that all the small powers are unnecessary because i is divisible by $$p^{\mu(i)}$$". Don't the small powers contribute to the sum? Isn't it better to say $$j\le\mu(i)$$. Thanks-Shahab (talk) 11:04, 11 April 2010 (UTC)
 * There aren't any small powers. If a number is divisible by $$10^3$$ then its last 3 digits in base 10 are 0. For example, $$98000=9\cdot10^4+8\cdot10^3$$, no powers of 10 smaller than $$10^3$$ involved. Exactly this also happens for expansion to base p. Saying $$j\le\mu(i)$$ would be incorrect. -- Meni Rosenfeld (talk) 11:15, 11 April 2010 (UTC)
 * Ok, thanks. But shouldn't the base be $$p^{\mu(i)}$$? And what does Gandalf61 mean when he says that it might be a misprint? Because the book does have a lot of those!-Shahab (talk) 11:20, 11 April 2010 (UTC)
 * Why should the base be $$p^{\mu(i)}$$? Of course we can expand the number in any base we want. For whatever reason, the book needs to use the base-p expansion. One of the things it uses to learn about the properties of this expansion (in particular, the power in which it starts) is our knowledge of $$\mu(i)$$.
 * I think Gandalf61 hasn't thought this through before he said it might be a misprint. -- Meni Rosenfeld (talk) 11:46, 11 April 2010 (UTC)
 * Indeed. Meni is correct - if all $$a_j$$ beyond some point are 0 then the sum is just describing the base p expansion of i in which least significant $$\mu(i)$$ digits are zero, and first non-zero digit is $$a_{\mu(i)}$$. Gandalf61 (talk) 10:45, 12 April 2010 (UTC)

how do you distribute poison?
Question:

A garage has 2 vans and 3 cars which can be hired out for day at a time. Requests for the hire of a van follow a Poison distribution with a mean of 1.5 requests per day and requests for the hire of a car fiollow an independent Poison distribution with a mean of 4 requests per day.

Find the probability that on a particular day there is at least one request for a van and at least two requests for a car, given that there are a total of four requests on that day.

I tried:

There are 2 possibilities: 2 vans and 2 cars or 1 van and 3 cars. So I found poisonpdf(1.5,2), poisonpdf(4,2), poisonpdf(1.5,1) and poisonpdf(4,3) but how do I combine them?

Similar question: In the morning, the number of people entering a bank per minute has a Poison distribution with mean 3, and independently, the number of people leaving the bank per minute has a Poison distribution with mean 2.

Find the probability that during a particular minute in the morning, exactly one person will enter the bank given that the total number of people entering and leaving the bank is exactly 4.

Does that mean 3 people left the bank? poisonpdf(3,1) and poisonpdf(2,3)? —Preceding unsigned comment added by 59.189.218.247 (talk) 10:38, 11 April 2010 (UTC)
 * Distributing poison is illegal. I can help you distribute Poisson though :)
 * Let's start with the first question. Are there any additional sections to this question? Because the number of vehicles in the garage is irrelevant to the question asked - perhaps it needs to be interpreted in some special way.
 * I'll answer the question as written. This is a conditional probability problem - you need to divide the probability that $$v\ge1,\ c\ge2,\ v+c=4$$ by the probability that $$v+c=4$$.
 * To calculate the first probability, note that the probability of the union of two mutually exclusive events is the sum of probabilities, and the probability of the intersection of two independent events is the product of their probabilities. -- Meni Rosenfeld (talk) 11:04, 11 April 2010 (UTC)

LOL OMG MY SPELLING FAIL! For the first question, that is the third part. The first two parts are:

i) Find the probability that not all requests for the hire of a van can be met on any particular day. [SOLVED: 1 - poissoncdf(1.5,2)=0.191]

ii) Find the least number of vans that the garage should have so that, on any particular day, the probability that a request for the hire of a van for that day has to be refused is less than 0.1. [SOLVED using graphic calculator, table of poissoncdf(1.5,X)>0.9, answer is 3.]

I think that the probability that v + c = 4 would be poissonpdf(5,4), correct? Which events are mutually exclusive? I know the requests for vans and requests for cars are independent. THANKS! —Preceding unsigned comment added by 59.189.218.247 (talk) 11:32, 11 April 2010 (UTC)
 * The events that are mutually exclusive (and together exhaust the requirement $$v\ge1,\ c\ge2,\ v+c=4$$) are: The first is v=2, c=2. The second is v=1, c=3. So you need to sum the probabilities of the first and the second. To find the first, for example, you need to multiply $$P(v=2)P(c=2)$$. -- Meni Rosenfeld (talk) 11:51, 11 April 2010 (UTC)

Solved! THANKS! May come back with more questions soon, hehe.