Recent content by incognitO

First write the area of the anular region from the disk with radius 1 centered in x=6 to x=y^2+1, then integrate over the whole height of the figure (a drawing will make it clearer). EDIT: Hah, that's exactly what you did... assuming your calculation is good, then ur all set!

its because \log e=1... look at it this way... \begin{array}{c}a^x=y(x) \\ \\x \log a=\log y(x) \\ \\ \dfrac{d}{dx}(x \log a)=\dfrac{d}{dx}\log y(x) \\ \\ \log a =\dfrac{y'(x)}{y(x)} \\ \\ y(x) \log a =y'(x)\end{array} wich means \frac{d}{dx}a^x=a^x \log a e is not different from...

(\nabla S)^2=|\nabla S|^2 \qquad \hbox{the square gradient} you can establish an equivalent system of ode's for your nonlinear problem and then aswer the questions for existence, unicity and solvability... for more details check first chapter of Fritz John book.

Also, basic courses are pretty much the same in every university in the world, and i bet you every good mathematitian, no matter where in the world was born or formed ,has read the classical texts.

in series solutions with recurrence relations as the type mentioned by Dr Transport, is costumary to choose a_0=1, a_1=0 to find the first independent solution and then a_0=0 and a_1=1 to find the second independent solution. In the case of your Bessel function, this is not necesary science...

The above are functions present in Hamiltonian systems, which are a big subject of study for mathematitians too... Specially in P.D.E. EDIT: Not to mention Calculus of Variations.

think of it... in classical mechanics, you can have functions like the action L which is defined in the form L=L(x,y,z,t), but x=x(t), y=y(t), z=z(t), so \frac{d L}{d t}=\frac {\partial L}{\partial x} \dot{x}(t)+\frac {\partial L}{\partial y} \dot{y}(t)+\frac {\partial L}{\partial z}...

How about this... |f(u_1)-f(u_0)| \le H(u_0,u_1) |u_1-u_0| where H(u_0,u_1)=\left\{\begin{array}{cc} \dfrac{2}{(1+u_0)^3} & \hbox{if} \quad u_0<u_1 \\ & \\ \dfrac{2}{(1+u_1)^3} & \hbox{if} \quad u_0>u_1 \end{array}\right. So f(u) \in C^{0,1} in -1<u_0,u_1 \le 0, just not uniformly...

I don't understand what you mean. The worst behavior of the bound is when u_1 and u_0 are close to -1, but that should reflect in the bounding constant H(u_0) right? Let me put it this way (maybe I am saying stupid things but, that woulndt be new :P ) If -a <u \le 0 and 0<a<1 , then...

just insert the series J_{\frac{3}{2}} on your O.D.E. The only way the equality will hold is if all the therms on the O.D.E. are zero. That will give you a condition over the a_n's.

Hello guys, I am trying to prove that the function f(u)=-\frac{1}{(1+u)^2} is Hölder continuous for -1<u \le 0 but I am stuck. Here is what I have done: If |u_1-u_0| \le \delta then \left|-\frac{1}{(1+u_1)^2}+\frac{1}{(1+u_0)^2}\right| \le...