Prove the following trigonometric identities, showing all work:

1+tan2AtanA=sec2A
1+cot2AcotA=csc2AcotA

sin5A+sin2A-sinA=sin2A(2cos3A+1)
cos3A+sin2A-sin4A=cos3A(1-2sinA)
sinA-sin2A+sin3A=sin2A(2cosA-1)
sin3A-sinA-sin5A=sin3A(1-2cos2A)
(sin2A)/(1+cos2A)=tanA
tanA+cotA=2csc2A

I was misrembering series as sequence, and trying to say "sum of sequence" (when the series is already the sum of the sequence). Sorry. The second one, if it were a sequence, converges to 0, which isn't any more difficult/interesting than seeing that the first one goes to infinity. The second one, as a series, is more interesting than the first because the terms of the sequence actually do go to 0, so there's a chance that it could converge.

It might be-the sum of bn is not bounded for |z|>1/e either. I don't know if you consider it implicit that z is so constrained.

Lol, np. Regarding the summation-by-parts, I meant that for z ∈ℂ with |z| = 1/e, then we have for each partial sum B_K of ∑b_n:

|B_K| = |∑(ez)ⁿ|
.......= |(1 - (ez)^K+1)/(1-ez)| if z != 1/e
.......≤ (1 + |ez|^K+1)/|1-ez|
.......= 2/|1-ez] since |z|=1/e

And since c_n = e^-nnⁿ/n! looks to be a monotonically decreasing sequence which goes to 0 by Stirling's formula, then the series ∑b_nc_n converges by an application of summation by parts.

Does that seem right?

i think i know a) but not sure about b)

15 students are to be sorted into 3 groups of 5, how many ways to choose if a) it doesn't matter which group is which and b) it does matter

for a) i think it is (15nCr3)^3, for b) i'm not sure, i feel it is (15nCr3)^3*(3nCr1)^3 but not confident

Yes I was just confirming what you said.

Also I see there are two power series you are considering, so this works at the radius of convergence for one of them (even if z takes complex values, not just real).

For the other power series, where at the radius of convergence, the terms goes to zero, (with magnitude like n^{-1/2}), you could use Limit comparison test for one of IOC, Alternating series test for the other and Dirichlet's test if z takes complex values.

http://en.wikipedia.org/wiki/Convergence_tests
http://en.wikipedia.org/wiki/Alternating_series_test
http://en.wikipedia.org/wiki/Dirichlet%27s_test

PS I see above you linked to

http://en.wikipedia.org/wiki/Summation_by_parts

which I guess would lead to the same place as Dirichlet's test.

For a philosophy(Metaphysics) paper: does anyone have an opintion on linguistically motivated platonic realism vs. Wilfrid Sellar's metalinguistic nominalism?

just realized i wrote part a) totally wrong, now my theory is

(15nCr5)*(10nCr5)*(5nCr5)

and part b) would be

(15nCr5)*(10nCr5)*(5nCr5)*(3nPr3)

One way to look at this is that we have a 15 letter word with 5 a's, 5 b's, and 5 c's, and you want to know how many ways you can rearrange the letters (i.e. group matters) Then divide by 3! (number of ways to relabel the groups) to get the #of ways when group doesn't matter.

So it should be 15!/(5!)^3 and 15!/[(5!)^3 3!].

The first is (15c5) * (10c5), and the second is 15C5 * 10C5 / 3!

i don't think i understand how using combinations groups matter. the way i thought of it was a 15 card deck, the first hand is 15c5, the second is 10c5, the third is 5c5 then each of the resulting combinations could be permuted 3p3 ways to find then number of cases where groups matter. i figure this logic is wrong i would just like to know where i am screwing up

You're multiplying by 3! when you've already chosen the order of the groups implicitly.

ah, so you are saying that in getting rid of the problem of group by giving an arbitrary order i group them, i then need to divide out 3! to remove my implied grouping. i think i get it, might need to think about it a bit more.

15C5 is the number of ways to choose 5 people to be in GROUP 1.

What's the quadratic mean of a Poisson Process?

I went to wikipedia and checked out a few library books about probability theory, but I'm really confused about how to calculate it and get a non-zero answer. I'm not asking for a solution, but I'm totally stumped right at the beginning.

Physics midterm take home test... Forgot a lot of stuff...More to come most likely

1. A stone is thrown upward with an initial velocity of 15 m/s
a) What will its maximum height be?
b) What is its velocity at 1.8 seconds?

2. A 125 Kg crate slides down a ramp at a constant 3m/s. The ramp is inlined at 16.7 degrees.
a) What is the normal force?
b) What is the coefficient of of kinetic friction?

3. Three forces act on a box. The first pulls with a force of 10N east, the second at an angle of 35 degrees South of West, the third 18 N north. What is the magnitude and angle of the resultant force?

It's been six years since I took physics, but I believe the first question can be solved with the equation velocity = initial velocity + acceleration*time

The height will be a maximum when velocity reaches zero.
So: 0=15m/s + -9.8m/s[superscript]2[/superscript]*time
Then, -15m/s= -9.8m/s[superscript]2[/superscript]*time
Then, (-15m/s)/(-9.8m/s[superscript]2[/superscript])=time
So the time that the whatever hits maximum height is at -15/-9.8= ~1.5s

Then you can plug that into the equation height=initial height + initial velocity*time + 1/2(acceleration*time)

So: Height = 0 + 15 m/s*1.5s + 1/2(-9.8m/s[superscript]2[/superscript])*1.5s[superscript]2[/superscript]
Height = 22.5 + 11.025
= 33.525m

Not sure if you are supposed to use -9.8m/s[superscript]2[/superscript] or -10 as gravity. I used -9.8 for the first calculation, which gave me a lot of decimal places for time that it takes to reach maximum height. I simply plugged in 1.5s to get the final answer, so you should just work out the problem yourself given what I typed to get an exact answer.

to figure out the velocity of the ball is at 1.8s you can just plug 1.8s, gravity, and initial velocity into the first equation I gave.

velocity = 15m/s + -9.8m/s[superscript]2[/superscript]*1.8s
= -2.64 m/s

I know this is a convoluted response. I forgot how to solve the second problem.

The third problem is just adding vectors... which I really don't want to do through typing.

Someone lemme know if I'm ******ed.

Sorry that superscripts don't work... should still make sense though, I hope.

So here is what I have so far:

The quadratic variation is the sum of all [N_t_(i+1)-N_t_i]^2, with the max |N_t_(i+1)-N_t_i| --> 0.

Since poisson distributions are independent, and we want to find N(t), partition the interval [0,t] to n subintervals. Let h = max |N_t_(i+1)-N_t_i|. So, since I'm taking the limit as h becomes arbitarily tiny, I can rewrite this sum as n*(E(N_h))^2 right? But then, I get n*(lam*(1\n))^2, which goes to zero. This can't be right, so I must have made at least one error here right?

EDIT: In my OP, I should have said quadratic variation obv.

An airplane takes off at 10:00 AM and reaches a cruising speed of 225 kph at 10:45 AM. It cruises at this speed until noon, then decelerates for 35 min. to 100 kph. It circles for 30 min., then decelerates until it lands 15 min. later. What is its average speed?

An arrow is shot horizontally 1.5m above the ground . It lands 5.55m away. With what total velocity does t hit the ground.

Easy proof I can't solve, because I haven't taken proofs lol.

n(n-1)/2 My teacher called it the handshake problem where n is the number of people.


I get how to explain it but how do I solve it?
I.E. the n(n-1) is showing how many handshakes happen, but it's doubling the number because you don't count Bob shaking Dan's hand and Dan shaking Bob's hand twice so you have to divide by 2.

This is the same as nC2 (n choose 2). It says how many ways can you take n people, and pick two of them, so that order doesn't matter.

Anyways, the intuition is that there are n people in the beginning. Then, there are n-1 people that he can shake hands with. But, say A and B are two people. You're counting A shakes B, and B shakes A as well. So you divide by 2, because order doesn't matter.

I'm new to proofs and always thought it had to involve numbers, but I can just explain in words?

You can explain it in words, but its probably stronger to say that its equivalent to nC2, and then show that the nC2 formula yields what you want.

Regarding the vertical component of the arrow's motion, u = 0, s = 1.5 and a = 9.8. From s = ut + (1/2)at^2 we get that t = sqrt(1.5/4.9). From v = u + at we get that v = 9.8t.

Then regarding horizontal motion, a = 0 and t = sqrt(1.5/4.9), so 5.55 = ut. Since v = u you now know the final horizontal velocity.

The final speed is the square root of the squares of the final vertical and horizontal velocities.

Hey everyone

I'm 99% certain that I'm doing this correctly, but my friend had this example in his lecture notes and it had a different answer than me.

Determine the area of the region enclosed by 4x + y^2 = 12 and x=y.

My answer is in the spoiler, I would appreciate some help with this.



Thanks.