Recent content by Another1

Question \[ \int dx_1dx_2...dx_d e^{(x^2_1+x^2_2+...+x^2_d)^{r/2}} = \frac{\pi ^{d/2}(d/r)!}{(d/2)!} \] How can I derive this answer?

How can I find Cauchy principal value. of this integral \[ n(x) = \int_{a}^{b} \frac{d \omega}{\omega ' ^2 - x^2} \] Where $ a<x<b $ I case $a = 0, b = 3, x = 1$ We get \[ n(1) = \int_{0}^{3} \frac{d \omega}{\omega ' ^2 - 1^2} = −0.3465735902799727 \] The result shown is the Cauchy...

\[ \int_{0}^{\inf} \frac{e^{-\frac{(x-a)^2}{b}}}{x^2-c^2} dx\] or \[ \int_{0}^{constant} \frac{e^{-\frac{(x-a)^2}{b}}}{x^2-c^2} dx\] maybe application Residue theorem integral ? because this problem same the kramers kronig relation?

$\dot{r} $ mean full derivative of r by dt because \[ r=r(q_1,...,q_n,t) \] and \[ q_n = q_n(t) \] any $q_n$ as function of time so $\dot{r}$is formed by taking the derivative with respect to dt for $( q_1,...,q_n,t )$

\[ \frac{\partial \dot{r}}{\partial \dot{q_k}} = \frac{\partial r}{\partial q_k} \] where \[ r = r(q_1,...,q_n,t \] solution \[ \frac{dr }{dt } = \frac{\partial r}{\partial t} + \sum_{i} \frac{\partial r}{\partial q_i}\frac{\partial q_i}{\partial t}\] \[ \dot{r} = \frac{\partial r}{\partial t}...

from problem I find \[ r = r_0 + At \] \[ x_0 = 3 + 2t\] \[ y_0 = -1 - 2t\] \[ z_0 = 1 + t\] and \[ A = (2,-2,1)\] but i don't understand What is the distance of closest approach? someone tell me to a formula please.

I select \[ y_1= c_1x^r\] and \[ y_2= c_2x^s\] so \[ (x+1)x^2y_1'' + xy_1' +(x+1)^3y_1 = 0\] and \[ (x+1)x^2y_2'' + xy_2' +(x+1)^3y_2 = 0\] i find first and second deriative of y1 and y2 I get two equations \[r(r-1)(x+1) +rx - (x+1)^3 = 0 \] \[s(s-1)(x+1) +sx - (x+1)^3 = 0 \]

in problem b from \[ y_1y_2 = c \] so I was able to specify that \[ y_1 = c_1x^2 \] abd \[ y_2 = c_2x^{-2} \] Correspond to \[ y_1y_2 = c_1c_2 = c = constant \] then I can find \[ y_1', y_1'', y_2',y_2'' \] So. I can solve \[2p_1p_2 +p_2' = 0\] But in problem C, I have no idea, so I assign \[...

y'' = uv'' +2u'v'+ u''v so y''+ Vy = uv'' +2u'v'+ u''v + Vuv = 0

y''(x)+y'(x)+F(x)=0 Pleas me a idea

Why first terms equal to zero ? ? ?

(a+b)^{a+b}=a^a+y ; sorry i am edited a^b to a^a Suppose we know a and b. y in the term of a, b?

x_0\cos(\phi) = 2.78 x_0\sin(\phi)=2.78 \left( \frac{\gamma^2/2}{ \sqrt{10-\frac{\gamma^2}{4}}} \right) x_0e^{-15\gamma} \cos\left(30\sqrt{10-\frac{\gamma^2}{4}}-\phi\right)=1 I don't know awsner of \phi , x_,\gamma

I don't understand, please ckeck Let V=\Bbb{R}^2 and {u=(u_1,u_2), v=(v_1.v_2)}\in\Bbb{R}^2 , {k}\in \Bbb{R} define of operation u\oplus v = (u_1+v_1,u_2+v_2) and k \odot u =(2ku_1,2ku_2) check V is vector over field \Bbb{R} ? ________________________________________________________________ I...