# Tag Info

0

It seems that I've found the solution: Let's express the "line of sources" as superposition of $\delta$-functions. To obtain it we use parametrization of the line: $x = x(p), y = y(p)$, where $p \in [0; L]$ (L -- is length of the line). So we can write down the expression: $$\int_0^L \delta(x - x(p))\delta(y-y(p))q(x(p), y(p), t)\,dp$$ Also it is clear ...

0

You should carry on. You are on the right track. Substitute $L(y)=Y(s)$, use the initial conditions, solve for $Y(s)$ and then find the inverse Laplace transform.

4

Find max: $x^2 + y^2 = 1 \implies x, y\in[-1,1] \implies x^2 \ge x^3, y^2\ge y^3 \implies 1 = x^2+y^2\ge x^3 + y^3$ Follow that, find min: It is easy to see that $x^3 + y^3$ min $\iff$ $x, y < 0$ and $|x^3 + y^3|$ max Cleaner: $$x^2 \ge |x|^3, y^2\ge |y|^3 \implies 1= x^2 + y^2 \ge|x^3|+|y^3| \ge|x^3 +y^3|$$ $$\iff x^3+y^3 \in[-1,1]$$

0


3

Hint Prove that $$\frac{d^k}{dx^k}\sin x=\sin\left(x+k\frac{\pi}{2}\right)$$

2

Hint For the second series discuss the different cases: $x,y>1$ and $x=1,y>1$ and $x<1,y>1$ and $x>1, y=1$ etc For the first series you can use the ratio test which fits better with the factorial. Added: For the second series let's show the case $x>1,y<1$: we have $$\left(\frac{1+x^k}{1+y^k}\right)^{1/k}\sim_\infty ... 1 The notation \frac {d^{950}}{dx^{950}} \sin x usually means , take the 950th derivative of \sin x , i.e., differentiate the function \sin x \;950 times. But notice that the derivatives are periodic, i.e., they repeat after a certain number of times. d/dx(\sin x)=\cos x; d/dx(\cos x)=-\sin x \dots Then 950 goes on cycles of 4 a certain number ... 0 I do not like the proof via squeeze theorem that \sin x < x < \tan x because how do you prove that x<\tan x? One can prove that \sin x < x as the line is the shortest distance between two points. But what kind of geometric argument do you use to justify that x<\tan x? I do not see of one. So I believe it is not possible to prove this ... 0 The basic concept of inverse trigonometry is as follows - Arctan is same as tan^-1 with in the specified values. tan^-1 x + tan ^-1 y = tan ^ -1 (x+ y/1−xy) if xy<1 π+ tan ^ -1 (x+y/1−xy) if xy>1 and x>0 and y>0 -π + tan ^ -1 (x + y/1-xy) if xy>1 and x<0 and y This concept will ... 0 Hint: Try s = e^\frac{-u^2}{2}, and dv = \frac{1}{u^2} 1 Another reason is that the proof of the Mean Value Theorem uses Rolle's Theorem, which does not require differentiability on a closed interval [a,b] but an open interval (a,b). 2 The only difference is that the usual way applies to more functions. There are functions which are continuous on a closed interval and differentiable on the open interval, but not differentiable on the closed interval, and we'd like to apply the mean value theorem to them as well. Since the theorem is true when we only assume differentiability on the open ... 3 Hint. Take the logarithm of your term and since n is large, use a first order Taylor expansion. Multiplied by the exponent, you will find 0 as the limit of the logarithm, then 1 for the limit. 4 Hint: rewrite your limit as$$\lim_{n\to\infty}\left[\left(1-\frac{\sqrt {2t}}{n}\right)^{-n/2}\left(1+\frac{\sqrt {2t}}{n}\right)^{-n/2}\right]$$and use the product rule. 4 Hints:$$(1)\; \left(1-\frac{2t}{n^2}\right)^{-n/2}=\left(\left[\left(1-\frac{2t}{n^2}\right)^{n^2}\right]^{-1/2}\right)^{1/n}(2)\;\text{ For any function}\;\;f(n)\;\;s.t.\;\;\lim_{n\to\infty}f(n)=\infty\;,\;\;\lim_{n\to\infty}\left(1+\frac x{f(n)}\right)^{f(n)}=e^x(3)\;\lim_{n\to\infty}a_n=a>0\implies\lim_{n\to\infty}\sqrt[n]{a_n}=1

Top 50 recent answers are included