Thanks for your reply, but I'm sorry I really don't follow.

The only tangent line I can think of is by differentiating the equation to get 2x+1, then plugging in 2 from (2,-3) to get a slope of 5. Then I get a tangent line: y=5x-13. How else can I find a tangent line?

(2,-3) is not on your curve. So you wouldn't plug in x=2. You'd plug in x0, the unknown x value of the point on the curve:


The point of this exercise is that on one hand, you know the slope (in terms of x0) because you differentiated, but you also know the slope just from high school algebra (you have 2 points...). Equate them and solve.

Oh right!!! That makes a lot of sense now. Thanks for your help!!

1. The problem statement, all variables and given/known data

Use a Green's function to solve:

u" + 2u' + u = e^-x

with u(0) = 0 and u(1) = 1 on 0<=x<=1

2. Relevant equations

This from the lecture notes in my course:



3. The attempt at a solution

Solving for the homogeneous equation first:

u" + 2u' + u = 0

From the characteristic equation,

lambda² + 2lambda + 1 = 0

lambda= -1 (repeated root)

Characteristic solution:

u₁(x) = c₁e^-x and u₂ = c₂e^-xx

To satisfy boundary conditions,

u(0) = c₁e⁰ + c₂e⁰(0)
c₁ = 0

Thus take f(x) = e^-xx

and

u(1) = c₁e^-1 + c₂e^-1(1)
get c₁ = e/2 and c₂ =e/2

Thus take g(x) = (1/2)e^-x+1(x+1)

evaluating the wroknskian, W = (1/2)e^-2x+1

I contruct green's function as per the formula provided above and carry out the integral and get final answer of,

u(x) = xe^-x(x² - 3)

which obviously doesn't satisfy the second boundary condition of u(1) = 1.

I found out the solution from Wolfram alpha:

u = (1/2)e^-xx(x+2e-1)

I've also tried many different combinations of f(x) and g(x) but none seen to work

Any help appreciated !

Find the limit as x approaches 0 of sin(x) / x + tan (x)

Attempt:

I've only managed to rearrange to formula to the following

sin (x) / [x + (sin(x) / cos(x))]

How am I supposed to proceed from here? That extra 'x' in the denominator just messes everything up for me.

lim x->0 1/(x/sin(x) + 1/cos(x))=?. I assume you covered the limit as x->0 of sin(x)/x.

So, my quantum teacher doesn't really want to teach us quantum mechanics, he'd rather just give us tough problems that we don't think he knows how to really solve.



So, for part a) I used the substitution



and



to express the final state as,

.

When I overhead him explaining it to some other people, he said that for one photon,



where S is the scattering operator, ks are the polarization vectors and epsilons can be in any direction perpendicular to the polarization vector.

Any help tonight would be great.

Hey thanks for your reply.
I know that limit as x->0 of sin(x) / x = 1.

But how did you manage to rearrange the equation to:
lim x->0 1/(x/sin(x) + 1/cos(x)) ? That's what I'm confused about.

Factor out a sin(x) on bottom to cancel the one on top...

1/(x/sin(x) + 1/cos(x))= 1/(xcosx + sinx/sinxcosx) = sinxcosx/(xcosx + sinx)

= sinx/(x + (sinx/cosx)) = sinx/(x+tanx)

Guys,

I'm new here. And I think this question might be a newb one.

How to use Law of Large Numbers (LLN) to estimate the value of pi=3.1415926

Bit of a (hopefully) quick question - how would one determine the mean curvature of a surface xyz=1?

I have a surface parametrization s = (u,v,(uv)^-1) and the gaussian curvature K = 3(x^-2+y^-2+z^-2)^-2, but I'm drawing a blank on how to calculate the principle curvatures, and through them, the mean curvature.

I've got one definition in my notes, a seperate and different one from wiki, and, well, I'm stumped, I'm just not making any prgoress, as far as I can see the first and second fundamental forms are way to complex (at least for this parametrization), and the question seems to be implying that the mean curvature is an 'easy' step of the process, so I don't know....any help?

Ok i think there is a "conceptual" error here, and some calculation errors.
Lets fix first the former since its more important.

This is how you do an inhomogenous diff equation in general:
Get a solution for the inhomogeneous equation with some method (for example Green's method). Then get a complete system of solutions for the homogeneous equation. Then you add the right combination of the homogenous solutions to the inhom. solution to satisfy conditions.

So for this exercise:
v := (1/2)*x²*e^-x is a solution for the inhom. equation (which you should have gotten using the formula i think).
(What you wrote is not a solution for the inhom. equation).

As you found yourself a complete system of solutions for the hom. equation is:
u₁(x) = e^-x
and u₂ = e^-xx

(compl. system here means thatthese span the vspace of hom solutions)

Now you are looking for the solution in the form of:
v + c₁u₁+c₂u₂

Subsitituting this into the boundary conditions provides you with 2 linear equations for c₁ and c₂, which I am sure you can solve.

Now why did you not get this inhom solution v using Green's method?
Well due to calculations errors i suppose. For example your g does not vanish in 1, like it should.
There might be more errors, you will have to check yourself.

Let me know if something is not clear. Good luck with finishing the problem.

I dont see any tricks here, and calculating this directly from definitions is far from impossible just very ugly.

Step1: you calculate the partial derivatives of s
(so (1,0,-(u^2)v) (0,1,-(v^2)u) )

Step2: calculate the norm vector from that
(that is the vector product of the partial derivatives divided by its norm.
the vector product is something like (u^2v, v^2u, 1))

Step3: calculate 2nd partial derivatives of s, scalar multiply with the norm vector. you get some 2x2 matrix.
Now the eigenvalues should be the principal curvatures.
Mean curvature is their average so that should be the trace of the matrix divided by 2. And gaussian curvature is the product so the determinant of this matrix.

Note: I do not take responsibility for that, I really learnt it a long time ago. Maybe there is some extra constants/signs somewhere. But surely googling around will give you the formulas.

Note2: usually taking a better parametrization makes the calculation cleaner but i can not think of one for this case. (maybe coz i am dumb)

have fun with finishing it

Answer is semi-long and i dont feel like typing it.
Google finds it for you, look for "Buffon's needle problem"
(or simple google "law of large numbers" "estimate pi")

here is one page that explains, though surely there are better sources:
http://www.ds.unifi.it/VL/VL_EN/buffon/buffon2.html

Shoot, sorry, I completely forgot to edit my post and say I'd figured it out...actually, it turns out I figured it out to the minute you posted that :|

But yeah, I ended up just deriving the equations from scratch using the second fundamental forms, turned out the wiki equations were wrong, so that was a fun hour of rather huge equations, which culminated in me finding a nice little formula. And then left me feeling deflated a minute later after I was linked to the formula in my textbook. C'est la vie. Thanks Artagas.

You are welcome but what "wiki equations" did you find to be wrong? If it is some wikipedia thing please link it to me and i will try to verify if it is wrong and fix it if needed.
I mean diff geo is a popular topic these days and half the undergrads rutinely use wikipedia to look up definitions/formulas etc., if there is an error it should be fixed asap.

I sucks at the physics.. so here we go..

Car is traveling 50 kp/h and it stops at distance of 70m. With the same conditions how far is the stopping distance at the speed of 100kp/h.

I have all the formulas but with two unknown variables I'm lost..

I mean do I have to combine two formulas in to one equation?

Like from s=v_0 * t + 1/2at^2 and some other?

Halp!

E:

I don't need the entire solution.. just somebody to nudge me in the right direction thx

okaaay i really step out of my comfort zone trying to answer some classical mechanics question, and might just make myself look ridicolous saying something completely stupid but i will risk it

so as i understand the kinetic energy of the car is proportional to the square of its speed. Now if there is some constant force stopping it (suppose that is hugely unrealistic but i dont know a lot about cars and forces ) then the work done to stop the car is just proportional to the distance traveled.

so if the car travels at double speed then the distance needed to stop is 4x the original distance. So 280m.
Flame away if that is stoopid.

yeah I thought about this but thought it would be too simple.. and we're talking about braking distance here but all the conditions (the acceleration/deceleration) stay the same. But I think you might be right.

Is there anyone that is capable and willing to help me with some basic Statistics problems?

I figured poker players are competent in this subject and I will pay you for your help.

PM if you can help. Thanks!

The problems regard introductory standard deviation concepts, variance, etc.

why dont you just go ahead and post your questions? Then people will be able to decide if they can help.

Because of the equation you posted ill assume the question wants constant acceleration. In which case your second equation needed is v=v_0+at

So I'm given a graph of acceleration (y axis) vs. force (x axis), with three straight lines of varying slopes. What does the slope of each line represent?

So far, all I've been able to come up with is slope = acceleration/force, in other words, slope = 1/mass (but I'm not sure how this is relevant/why the graph wouldn't be force - y-axis, acceleration - x-axis if the slope really represented the mass).

Can anyone point me in the right direction?

Thanks!

From a cursory look it looks like you're right about the slope = 1/mass, but I can provide you an answer to why the y-axis is acceleration and the x-axis is force: any function representing a slop in the xy plane has a dependent and an independent quantity, and by convention the dependent quantity is always expressed on the y-axis, as it's the result, whereas the x-axis is the input.

For example, if you were plotting a displacement/time curve, the time is on the x-axis because that's what you're changing, and the displacement is the result of that change, and a product of the relationship between time and speed.