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### Messages - Andrew Hardy

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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
$$f' = \overset{\circ}f ' + ((f(a-0) -(f(a+0))\delta(x-a)$$

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##### 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,

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

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

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##### 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

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##### 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 »

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

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