That makes things awkward.

n!/3^n = 1/3 * 2/3 * 3/3 * 4/3 * 5/3 * ... * n/3 -> infinity so 3^n/n! goes to 0.

The ratio test just tells you it converges since a_n/a_(n-1) = -3/n -> 0, but since it alternates sign, it can only converge to 0.

How about: for n > 4, |a_n| < 3/4 * |a_{n-1}|, so |a_n| < (3/4)^(n-4) * |a_4|, which clearly goes to 0.

Question on my midterm today:

You know that {a_n} is increasing and bounded on top (though you don't have to prove either of those), with a_1=2 and a_(n+1)=(2+a_n)^(1/2). Find lim a_n as n -> infinity.

Doesn't the way that a_1 and a_(n+1) are defined imply the following:
a_1=2
a_2=(2+2)^(1/2) = 2
a_3=(2+2)^(1/2) = 2
a_4=(2+2)^(1/2) = 2
etc
etc

So that the limit is 2? The part that confuses me, and makes me think that I must be thinking of this wrong, is that we were to assume that the sequence was increasing. It seems to me that if a_1=2, and if the above computations are correct, then its always the case that 2=<a_n=<2. But if that is the case how can it be said that the sequence is increasing?

The limit is 2, but a sequence can be said to be increasing or monotonically increasing even if all the values are the same. It was news to me too. So increasing means the same as non-decreasing, which is odd since decreasing can also mean that all the values are the same, and that would mean that a non-decreasing sequence can be decreasing. Or maybe wiki is wrong. Anyway, the terms strictly or weakly can be used to clarify this.

Ha, mathworld says monotone increasing is the same as strictly increasing and means the values cannot be the same. That disagrees with wiki.

Interesting. Here is the definition from my textbook:

My calculus book disagrees with both wiki and mathworld. It says that increasing and decreasing cannot have the values the same, but monotone can have them the same. That's how I've always understood it.

So your monotonic agrees with mathworld. My book agrees with your definition of increasing and decreasing, but it says it can be monotonic if it is increasing (>), decreasing (<), non-increasing (<=), or non-decreasing (>=). That doesn't disagree with yours, it just adds 2 more conditions.

Monotone "should" include non-increasing and non-decreasing to get the broadest possible application of the monotone convergence theorem.

Increasing/decreasing "should" be strict by default because of the existence of non-increasing and non-decreasing as perfectly logical descriptors.

At least, that's how I would do it if I ran the world of nomenclature.

If you go by wiki's definition, a sequence could be both increasing and decreasing. Their default is weakly rather than strictly.

Also, it explicitly states that non-decreasing doesn't mean not decreasing, and non-increasing doesn't mean not increasing.

This is funny from Mathworld:


MONOTONE

Another word for monotonic.


MONOTONE INCREASING

Always increasing; never remaining constant or decreasing. Also called strictly increasing.


MONOTONE DECREASING

Always decreasing; never remaining constant or increasing. Also called strictly decreasing.


MONOTONIC SEQUENCE

A sequence {a_n} such that either (1) a_(i+1)>=a_i for every i>=1, or (2) a_(i+1)<=a_i for every i>=1.

does anyone have wicked mad skills with R?

Or are you just kinda good with it in general?

Please send pm. Help greatly appreciated.

(excel might work too, but I doubt it)

Need help sort of automating a sequence and sorting/organizing data that's output.

Bruce, your question about the mcnuggets from the other thread is the kind of question I find myself upset that I don't intuitively know how to do. If I could, I'd like to have you critique my thought process on the problem.

You can't order exactly 1 mcnugget or 2 or 3 mcnuggets given the quantities you've said they come in. Then it makes sense that there is a set of numbers beyond those that I've mentioned which you can not exactly purchase given the quantities you've mentioned. The goal is to find the largest number in that set.

My first intuition was to just extend the sequence above:
1,2,3,4,5,7,8,10,11,13,14,16,17,19...

But as I do so I think two things (1) for all I know this list might go on forever, (2) it gets harder to ensure that I have not made a mistake the larger the numbers get. Next I wonder if these numbers have a property that can be described in a general way, IE is there a function which models numbers in our set, and can we find the max value of that function using the methods I've learned in calc.

Then I decide maybe I'll get a feel for what must be true of the numbers in that set: The numbers can not be divisible by 6, 9, or 20. But because I know that you can order 15 nuggets, and 15 isn't divisible by those numbers, there must be other limiters as well. The obvious ones are that numbers in the set cant be equal to divisible by (6+9), (6+20), or (20+9). But then I find myself thinking that I don't know enough about the different ways to combine numbers to make sure I rule out all the cases I need to.

Then I find myself thinking again, I don't even know if there is a finite number that can't be ordered, so maybe I'm wasting my time. Then I think, I wonder what Bruce would do.

Then I think that it might be that we are looking for the largest number in the set of numbers that isn't divisible by 3, 4, or 5, but then realize that 29 isn't divisible by 3,4,5 but that it couldn't be a part of our set because we can order that many nuggets.

Hint: Every number is either a multiple of 3, 1 more than a multiple of 3, or 2 more than a multiple of 3.

I have Z = (X1* ----* Xn)^n, where X1 thru Xn all have mean u and sd o, all are SIID.

I need to find P(Z > a) for a given a, n, u, and o.

What should I do? I had to get the distribution of Z^n earlier, which I got was normal with mean u^n and variance (o^2 + u^2)^n - u^(2*n).

Im thinking distribution of Z is lognormal, and i think i may need to use a t-distribution similar to normal - thats where im stuck

I haven't been able to make the connection yet.

If you only use 6 and 9, you only cover multiples of 3, but you cover all multiples of 3 except for 3 itself. If you only use 6's, you cover every even multiple of 3: 6,12,18,.... If you use one 9 and the rest 6's, you cover every odd multiple of 3: 9, 15, 21... So to cover the other numbers, consider what happens when you use one or more 20s.

Switching gears to physics homework:

Looking through my solutions manual I have a couple of questions which seem to result from poor trig recollection.

s(x,t)=6nm*cos(kx+(3000rads/sec)t+ϕ) We want to know how long it takes a molecule related to the soundwave expressed by the above function to travel from 2nm to -2nm.

They make the following generalization t=0, x=0, s=0 and then conclude that the phase is -pi/2. They then say that this makes the function 6nm*sin(ωt). I see where pi/2 comes from, but what I don't understand is why it's negative pi/2 and why the function switches to sin.

This is messed up. The mean and variance you got u^n and (o^2 + u^2)^n - u^(2*n) are the mean and variance of X1*---*Xn, not Z^n. That's because for independent r.v.s, the expected value of a product is a product of expected values since the covariance is 0. So the mean is u^n. Then for the variance, we know that

E[(X1*---*Xn)^2] = E(X1^2) *---*E(Xn^2) = (o^2+u^2)^n

so the variance is

(o^2+u^2)^n - u^(2*n).

So I think you mean those are the mean and variance of Z, and Z = (X1*---*Xn). But this distribution is not normal. In the limit for large n, it will become lognormal. For small values of n, it is complicated and depends on n and the distribution of X. If X is normal, for n=2, the distribution is a modified Bessel function of the 2nd kind. For n=3 it is a Meijer G-function. Now assuming you have large enough n that this is lognormal, raising a lognormal variable to the nth power will produce another lognormal random variable, and this will multiply the mean by n and the variance by n^2. So whether you want P(Z<a) or P(Z^n) < a, you would use the cumulative lognormal distribution with the appropriate mean and variance. Clearly if you know the distribution for Z^n, then P(Z<a) = P(Z^n < a^n). The cummulative lognormal distribution can be expressed in terms of the error function as

1/2 + 1/2 * erf[(ln(x) - u) / sqrt(2*sigma^2)]

where u and sigma^2 are the mean and variance of ln(x) which you computed, not the original u and o^2. This is like using the cumulative normal distribution when the r.v.s are normal which is equivalent to the t-distribution when sigma is known. You only need the t-distribution when sigma is not known and you are using an estimate of sigma from your data.

Oops, typo. It was

Z = (X1* ----* Xn)^(1/n).

Thanks.

Try plotting sin(x) and cos(x) on the same axes.

edit: and cos is an even function, so it's both pi/2 and -pi/2. Plotting both functions should make this clearer

OK, so you got the mean and variance of Z^n which is just the product, and if we're assuming the product is lognormal, then Z will be lognormal with a mean of 1/n * u^n and a variance of 1/n^2 * [(o^2 + u^2)^n - u^(2*n)]. Those other distributions I mentioned are only when X is normal. The lognormal will hold as long as n is large enough that the sum of the logs is normal by the central limit theorem, so we would have to assume that. Of course you can use the cumulative standard normal distribution and lookup ln(a) after you subtract the mean and divide by the standard deviation.

From memory I know that they are the same wave except that the origins are different. I am assuming that a shift of -pi/2 makes the cos function a sin function. This would explain why there is some connection, but my real confusion stems more from why we move from a perfectly good function that gives the displacement of a particle to a dfferent function. What necessitated that?

Why is it helpful to consider the function at t=0 x=0 s=0 and then make the switch to sin?