Recent content by mihajovics

Thx, I'll try. For example: Since: $F_{n-1} = F_{n} + F_{n-2}$ Then: $F_{2i} \cdot F_{2i-1} = F_{2i}^2 - F_{2i} \cdot F_{2i-2}$ Now use: $F_{n+1} \cdot F_{n-1} = F_{n}^2 + (-1)^n$ ... We'll see :) EDIT: why isn't the latex code showing properly? :(

Homework Statement Write the following expression in a simpler form: $\sum_{1}^{n} F_{2i} \cdot F_{2i-1}$ It doesn't have to be closed-form, probably something on the line of: $\sum_{0}^{n} F_{i}^{2} = F_{n} \cdot F_{n+1}$ (We define the sequence the ususal way, starting the indexing from 0...

Try this book: https://www.amazon.com/dp/0521636361/?tag=pfamazon01-20 Good luck!

If you don't know what potential energy or kinetic energy is, just solve it with your original idea. Also read http://en.wikipedia.org/wiki/Kinetic_energy and http://en.wikipedia.org/wiki/Potential_energy :smile: Otherwise... kinetic energy + potential energy = constant because energy is...

I think you don't have enough information. The crucial piece of the puzzle that's missing is how long the 50N force was applied? 0.01 seconds? 0.1 seconds? It makes a difference. Without that, you can't determine the change in momentum, from which you may deduce the starting velocity v0...

Homework Statement You have a thin lens with a concave and a convex side (let's say with radii r and R). You also have 2 mirrors, a convex one, that fits perfectly into the lens' concave side, and a concave one, that fits perfectly onto the lens' convex side (so the corresponding radii of...

Hi! If I understand correctly, then sin\left(\sqrt{n}\right) is what you are looking for. The proof is based on the idea sin(a)-sin(b) < a-b and sqrt(n+1)-sqrt(n) converges to 0. I originally bumped this post, because I need help with a similar problem, where not an+1-an converges to 0...