Recent content by jellicorse

OK, thanks for the tip, Adithyan...

Hi, I was wondering whether anyone could tell me how to deal with this manipulation, which I am unable to see. a_1=\frac{F}{(-m_1\omega^2+K_1+K_3)-K_3a_2} a_2= \frac{K_3a_1}{(-m_2\omega^2 +K_2+K_3)} Starting with a_1...

OK, thanks Chogg...

Hi, I am trying to follow an introductory problem in my book for which no solutions are provided and have got stuck. I was wondering whether anyone could tell me how to go about this problem and where I am going wrong. The problem starts: Consider the eqquations: y_1= x_1+2x_2...

Not sure why it wasn't working; I've just typed the URL in again and I think it's OK now.

Thanks for the hints Silversonic. I will go back through it again but I think I can see how this works out now! (I just changed the limits because I though it wasn't necessary to integrate beyond \frac{T}{2} because the function there is zero.)

I have been trying to follow how the complex Fourier coefficients are obtained; the reference I am using is at www.thefouriertransform.com. However I am unable to follow the author's working exactly and wondered if anyone could help me see where I am going wrong. First, I understand that the...

Oh, I see now. Thanks a lot, it all makes perfect sense now!

Thanks a lot, Mark... I am getting closer but have one problem. Having done it several times over, the same thing keeps happening: Using the product rule (uv)' = u'v + uv' I get one extra term. I'll type it piece by piece for clarity: \frac{\partial}{\partial u}( u \frac{\partial...

This is the part that I don't understand. The first part is OK...

Sorry. z =f (x,y), with x=\frac{1}{2}(u^2-v^2) and y =uv

I am not quite sure how \frac{\partial}{\partial u}\left(\frac{\partial z}{\partial u}\right) =\frac{\partial}{\partial u}\left( u \frac{\partial z}{\partial x}+v\frac{\partial z}{\partial y} \right) comes to \frac{\partial z}{\partial x} + u\frac{\partial}{\partial u}\left(\frac{\partial...

Thanks a lot Astronuc. I'll bear that in mind for future posts.

Oh, cheers for letting me know! I'll see if I can contact a moderator...

I think you have to select "show image in new tab" and it should work...