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Great I'll take a look at those! Thanks!

I've got a tricky computational statistics problem and I was wondering if anyone could help me solve it. Okay, so in your left pocket is a penny and in your right pocket is a dime. On a fair toss, the probability of showing a head is p for the penny and d for the dime. You randomly chooses a...

Hey guys, I'm trying to find a conditional distribution based on the following information: ##Y|u Poisson(u \lambda)##, where ##u~Gamma( \phi)## and ##Y~NegBinomial(\frac{\lambda \phi}{1+ \lambda \phi}, \phi^{-1})## I want to find the conditional distribution ##u|Y## Here's what I've got so...

Hey guys, I was wondering if you could help me out with a question I've got, I really don't know where to go or really where to start! Here's the question: Let S be a subspace of a finite dimensional vector space V. Show that there exists a Linear Mapping L: V → V such that the kernel of L is...

Oh I think I just got it... \sum^{n=1}_{infinity}(1/2)(1/2)^{n-1}=\frac{1/2}{1-(1/2)}=1 durr

Ok, but there's this example in my textbook: \sum^{n=1}_{infinity}(1/2)^{n}=\frac{1/2}{1-(1/2)}=1 "The series is a geometric series with a=1/2 and r=1/2" I'm confused as to how a=1/2

Oh! Everything would cancel out except a - ar^n S - r*S = a - ar^n ... S = a(1 - r^n)/(1-r) which explains why -1<r<1 since the limit n->inf r^n would diverge above 1

So far in class to find the sum we've just been comparing everything to the geometric series or using a bound on the error, so no

But judging by what everyones been saying, I'm assuming if n=1, then the sum is a/(1-r) and if it starts at 0 you have to multiply everything by r, making the actual sum ar/(1-r) EDIT: n=1 to infinity of course

I was hoping someone could help me understand how to find the actual sum of the geometric series, whether it starts at n=0, n=1 or n=N or whatever!

Keep your panties on ladies, I was making a pun

No idea how to figure it out, want to help me through it instead of being a dick, Dick?

Ok I just looked at another question and it does have n=1 to infinity, but it still defines the sum like: (1/4)/(1-(1/4))

how about n=1->infinity? The textbook doesn't say, it just says that's the sum =\ I'm assuming its n=1->inf though

So the textbook is wrong? What is the actual sum of the geometric series then?

Ok.. so why does my textbook say the sum is a/(1-r) when it's actually ar/(1-r) ?

Not following :S

Well, for the question's sake let's say we start at n=1, why is there a difference if we start at n=2?

I'm confused about the sum of the geometric series: \sum ar^{n-1} = \frac{a}{1-r} when |r|<1 but if you have a series like: \sum (1/4)^{n-1} the sum is: \frac{1/4}{1-(1/4)} should't it be \frac{1}{1-(1/4)} because there is no a value?

I ended up using trial and error and I ended up with N=4. I'm not sure if there's a better way to find M, but I made a list of derivatives and found it that way and that worked! Thanks!

I must have copied down the formula wrong, woops! Im still not understanding.. I can find the error when finding an approximation for the series, but I don't understand how to find an approximation of the series when I have an error...

[PLAIN]http://img600.imageshack.us/img600/1210/11096142.png [Broken] Hey I was wondering if you guys could help me out with this question... I think I have the right power series: = \frac{1}{1-x} + \frac{x}{1-x} = (1+x+x^{2}+x^{3}+...)(x+x^{2}+x^{3}+x^{4}+...) =...

Ohh that's interesting! I guess I'll learn all that soon enough, thanks :)

Ok, I understand that, but what's the difference between conditional convergence and absolute convergence? I always thought it was just 1s and 0s, it either converges or not, but conditional convergence is in between? Does is converge slower or something?

So I have this series: \sum^{infinity}_{n=3}(-1)^{n-1}\frac{ln(n)}{n} And I'm trying to use the AST to find out if it converges or not. First of all, I'm stuck trying to show that ln(n)/n is decreasing... But then after that. I'm assuming I can compare it with 1/n to show that it...

Thanks :)

pi/2, so by the test for divergence it must diverge! Ohh

I'm sure it doesn't, but how do I find out? \sum^{infinity}_{n=1} arctan (n) I thought about using the integral test, but it's not decreasing. Any hints? Could I somehow use proof by induction to show that its an increasing function?

Ohhh! I get it, thanks! About part c)... distance = velocity x time, so would I just multiply my equation by t? I can't seem to wrap my head around this stuff

Hold up, you still get 0... e^0 = 1, 1 - 1 = 0, mg0 = 0, 0/k = 0

Ha, see? stupid mistake :P

[PLAIN]http://img204.imageshack.us/img204/11/60272414.png [Broken] Hey guys, I'm stuck on this question, part b) I figured out a) and v(t) for V(0) = 0 ends up being mg - mge-kt/m k (Sorry, latex was being difficult) But then when I try to figure out the limit...

Thanks a lot guys! So what about b)? will V0 approach infinity because the top approaches infinity faster than the bottom?

So C = (-mgR^2) / (h+R) ?

I'm having trouble with part a) of this question... [PLAIN]http://img69.imageshack.us/img69/5815/98157006.png [Broken] So I started off by solving the DE above a), and I've gotten it down to: \frac{1}{2} m v^{2} = \frac{mgR^{2}}{(x + R)} + C I can tell I'm getting close, but I'm a...

So I've gotten the integral that I'm doing now down to: int(cos(x)/sin^2(x) dx) I looked it up on one of those online integral calculators to get me on the right track, and the answer is: -1/sin(x) It seems so simple, what am I missing?

Oh sorry, the term is x^{3} not x^{2}

\int x^{2} \sqrt{4+x^{2}} dx I've already subbed in: x = 2tan\theta dx = 2sec^{2}\theta d\theta and I've gotten down to: 16 \int tan^{3}\theta sec^{3}\theta d\theta But now I have noo idea what to do! Can someone give me a hint?

Awesome! Thanks a lot for your help, you really helped me understand it instead of just throwing an answer at me

Ohh, I get it, so my equation will be F'(2x)*'(2x), meaning the final answer would be -4x*cos(2x) !

I think I get it! F'(4) is 0, so the answer is -F'(x) = -2x*cos(2x) ? Where am I doing something with the chain rule though?

I'm still not seeing what you're saying... You can differentiate F(4)-F(2x), it would be F'(4)-F'(2x), meaning the answer is what I said first? 4*cos(4) - 2x*cos(2x) ? Ahh all this theory stuff makes my head spin :cry:

Oh, I guess I was reading FTC part II wrong =P so I would have to integrate t*cos(t), making the answer... dy/dx[4*sin(4) + cos(4)] - [2x*sin(2x) + cos(2x)] ?

(Not quite sure how to use Latex so I print screened it :tongue:) [PLAIN]http://img87.imageshack.us/img87/5991/calc1.png [Broken] So I've been staring at this question, and I think I might have it but I'm not 100% sure, is the answer just t*cos t |42x = (4)*cos(4) - (2x)*cos(2x) ? Or...