Messages

16

Web Bonus Problems / Re: Week 13 -- BP1

« on: April 04, 2018, 06:33:35 PM »

b)
$$ f'(\theta) = -f(\theta') = - \int_{-\infty}^{\infty} f(x)\theta'(x)$$
$$ -\int_{-\infty}^a f\theta'(x) + \int_a^{\infty}  f\theta'(x) $$
if I understand your notation correctly
$$=  \overset{\circ}f '(\theta) + f (\theta) \Big|_{-\infty}^a - f (\theta) \Big|_a^\infty $$
$$ = \overset{\circ}f ' -  \lim{x\to a-}f(x)\theta(a) + \lim{x\to a+}f(x)\theta(a) $$
$$ = \overset{\circ}f ' + ((f(a-0) -(f(a+0))(\theta(a)) $$
where this is defined in the sense of distributions
$$ f' =  \overset{\circ}f '  + ((f(a-0) -(f(a+0))\delta(x-a) $$

17

Web Bonus Problems / Re: Week 13 -- BP1

« on: April 04, 2018, 01:35:49 PM »

a) we know $$ \int_{-\infty}^{\infty} f(x)\delta(x) = f(0) $$
then derivatives of distributions $$ \int_{-\infty}^{\infty} f(x)\theta'(x) = - \int_{-\infty}^{\infty} f(x)'\theta(x)  $$ 

Now $$ - \int_{-\infty}^{\infty} f'(x)\theta(x) = - \int_{0}^{\infty} f'(x)\theta(x) = -f(\infty)+ f(0) = f(0)=  \int_{-\infty}^{\infty} f(x)\delta(x)  $$
therefore $$ \theta'(x) = \delta(x) $$

if I figure out B I will update, otherwise best of luck to someone faster than me

18

Quiz-B / Re: Quiz-B P1

« on: April 04, 2018, 12:58:00 PM »

Via the Energy Equation,

$$ \frac{ R R'^2}{\sqrt{1+R'^2}}  -  \sqrt{(1+R'^2)}R = c $$
$$ R(R'^2-R'^2-1) = c \sqrt{(1+R'^2)} $$

$$ R = -c  \sqrt{(1+R'^2)}  $$
$$ R^2 = c^2(1+(\frac{dR}{dz})^2) $$
$$R^2 - c^2 = c^2(\frac{dR}{dz})^2 $$
$$ c\frac{dR}{dz} = \sqrt{R^2 - c^2} $$
$$ dz = \frac{cdR}{\sqrt{R^2 - c^2}}  $$
$$ z = c  \text{ arcosh}(R/c) + b $$

19

Quiz-B / Re: Quiz-B P1

« on: April 04, 2018, 11:56:26 AM »

I think I fixed my EL equations, but I still see no explicit dependence on z

20

Quiz-B / Re: Quiz-B P2

« on: April 04, 2018, 11:41:15 AM »

Dr. Ivrii,

I believe I updated the induction to be more explicit

21

Quiz-B / Re: Quiz-B P2

« on: April 04, 2018, 11:25:54 AM »

We have the definition that $$  (\partial f)(\phi) = -(f)(\partial\phi) $$

so $$ ( x^k \delta^{(1)}(x) ) = -k  x^{(k-1) }\delta^{(1-1)} (x) $$
assume this holds true for  n-1
 $$ x^{(k+1)} δ^{(n-1)}(x)=(k-1)! (−1)^{(k-1)}δ^{(n-1−k)}(x) $$
via induction  and by the definition $ x^{k+1} δ^{(n-1)}(x) = - x^k δ^{(n)}(x) $exactly, how? and what you make induction with respect to? V.I.
 $$ x^k δ^{(n)}(x)=  k!(−1)^kδ^{(n−k)}(x) $$
 I have to add the condition $$ n\geq k $$ because if n was less than, I'd have a negative derivative which is undefined.

22

Quiz-B / Re: Quiz-B P1

« on: April 04, 2018, 11:12:31 AM »

answer to bonus question

My Lagrangian is $$ \sqrt{(1+R'^2)}R $$
so my Euler Lagrangian equations are
 
  $$ \frac{\partial L}{\partial R'} = \frac{ R R'}{\sqrt{1+R'^2}} $$
$$ \frac{\partial L}{\partial R} = \sqrt{1+R'^2} $$
$$  \frac{\partial L}{\partial z}  = \frac{R R''} {\sqrt{1+R'^2}}  + \frac{R'^2}{\sqrt{1+R'^2}} - \frac{R'^2R''R}{(1+R'^2)^{3/2}}$$



I'm not sure how to get to the solution you're looking for though. The argument isn't explicitly in the Lagrangian so I don't know how I'm going to integrate.

23

Quiz-5 / Re: Quiz5 T0101

« on: March 11, 2018, 04:53:24 PM »

Considering the property that if $ h(x) $ is of the form $ f(x) \cdot g(x) $ that the Fourier transformation is the convolution, $  \hat h(x) = \hat f(x) \ast \hat g(x) $ that gives us the same answer? So these properties of paired functions that we're memorizing are all special cases of this general fact about convolutions? 

24

Quiz-1 / Re: Q1-T5102-P2

« on: January 25, 2018, 09:04:06 AM »

$$ dt/1 = dx/(t^2 +1) $$
$$ (t^2+1)dy = dx $$
$$ t^3 + t = x + C $$
$$ C = t^3 +t - x  $$
$$ U = \phi ( t^3 +t - x) $$

I can't sketch here, but they should roughly look like cubics. They'll be steeper near the origin because of +t