# determining if it is convergent or divergent + taylor series

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Hi, a good question.

You would use the $1/(x^{p})$ integrals (integral p-test) to show that functions of the form $P(x)/Q(x)$ converge or diverge, for two polynomials P(x) and Q(x). What you do is look at the highest exponent in P(x) and Q(x) and the difference of the two. If the difference is more than one, then typically $\int _{a}^{\infty }{\frac {P(x)}{Q(x)}}\,dx$ converges. If the difference is 1 or less you typically face a divergent integral. To finish your argument you want to show that $P(x)/Q(x) , for $p>1$ in the convergent case, and $P(x)/Q(x)>C/(x^{p})$ , for $p\leq 1$ in the divergent case. Last step is then the integral comparison test.

As for Taylor series, if you are asked for the full series you need to write out every term, hence use Sigma notation. Sometimes you are just asked for the first few terms, then you can avoid the Sigma notation, or course. Spotting the pattern to be able to write the series in Sigma notation takes some practice. But there are a few building blocks that are useful: If the sign changes for every summand, use $(-1)^{n}$ . If only even numbers show up you use $2n$ , for odd numbers use either $2n+1$ or $2n-1$ . You should always convince yourself that your try at the Sigma notation is correct by plugging in n=0, 1, 2. It is a little trial-and-error after all.

Hope this helps. Good luck!

Bernhard Konrad03:57, 7 April 2012

Thanks so much Bernhard! Your tips on the taylor series is definitely helpful!

Can you also check my post for question for the math 103 2011 exam questions 4.a please :)

Jt05:02, 7 April 2012

Did just that. My pleasure, Jt :)

Bernhard Konrad05:07, 7 April 2012

Also, I found out that you're an instructor for the Math 103 course this semester and was wondering if you're going to have any review session before the final exam. thanks!

Jt05:06, 7 April 2012

Yes, I am the instructor for Section 206 this winter. I have a review session on Tuesday, April 10th at 10am in GEOG 200. You are welcome to join! Not sure how many seats this room offers though. In doubt I'd have to give priority to students from my section, of course.

Bernhard Konrad05:42, 7 April 2012

Bernhard, I have a few questions that i hope you can help me clarify...

1. for finding the normalizing constant, is it integral from a to b of the function = A or integral from a to b of the function TIMES A = 1 and then solve for A?

2. In the problem sets, 7.11 i was wondering why it shouldn't be a permutation since you technically can't choose if you want a head or tail. Q7.11: how many ways are there to get 3 times H and 2 times T by tossing a fair coin 5 times? What is the probability of getting 3 heads in 5 fair coin tosses?

3. for an unfair object, like in APRIL 2011 FINAL EXAM number 2, if it's twice as likely and there's in total 6 possibilities, does that mean you add an extra probability....like hence it results 7 and not 6? because it says the probability for a 6 is 2/7 and for other numbers it's 5/7.

4. What is the use of the normalized binomial in CH7? z= (x-np)/standard deviation

5. is random walker or hardy weinberg genetics going to be on the exam? if so is random walker just the same concept of tossing a coin, just that now it's a person walking? and is there shortcuts to drawing the genotype table so you can solve for the probabilities quicker?

Sorry for all these questions!

Jt21:29, 7 April 2012

Hello Jt, since this is a wiki you don't need to address your questions to me. Anyone is free to answer them.

Let's see:

1. The integral from a to b over the function needs to equal 1. Hence A = 1/integral. See, e.g. here.

2. I'm not sure if I get where your confusion comes from. But a coin toss is the classical example of a Bernoulli distribution. Here p = 0.5, hence q = 1-p = 0.5. If you let Head be a success, then the probability of 3 successes is $C(5,3)p^{3}q^{2}=C(5,3)(1/2)^{3}(1/2)^{2}=C(5,3)(1/2)^{5}=10/32,$ as asserted.

3. I'm not sure which question you are referring to. But if the question is something like it is twice as likely that roll a 6 than any other number you do the following: Let a be the probability of any other number. Then 2a is the probability of a 6. The total probability needs to be 1. With this information you can find a: a+a+a+a+a+2a = 1, i.e. a = 1/7 and thus prob(6) = 2/7.

4. The normalized binomial is needed to justify the use of a continuous random variable (with Gaussian as pdf) as an approximation to a discrete random variable (with binomial pdf). So in this sense it motivates chapter 8. In real life this is extremely useful in [Statistical hypothesis testing, but we didn't go into details here so I don't think you should worry about it.

5. Of course I can't comment on your first question. But yes, this is an example that, mathematically, it is the same as tossing a coin. But the application can we very different and easily applicable to a real world problem. Same with Hardy Weinberg genetics, since this relies on the binomial distribution (and is hence mathematically the same as a coin toss) you can apply the theoretical results on the binomial to genetics.

Hope this helps. Your questions are very welcome :)

Bernhard Konrad22:02, 7 April 2012

thanks so much for the clarification! :D

Jt22:17, 7 April 2012