Recent content by kaze

A infinitely long rod, A, with linear mass density, say p, is placed along the z-axis. Another thin rod, B, of length L and mass density, say p_{2}, is placed orthogonally to A on the x-axis from x=a to x=a+L. What is the gravitational force experienced by rod B from rod A? I have solved a...

Both methods show that log(2) equals your series by definition of the Taylor Series, I suppose.

I've just discovered this forum and have almost been addicted to it. I've seen lots of good advice and since I am coming up on some big decisions in the next few months, I thought you guys/gals could give me some of that wonderful advice. I am a starting my senior year this fall at the...

So the general form for a taylor series expanded at a point a is: \sum^{\infty}_{n=0}\frac{f^{(n)}(a)}{n!}(x-a)^n. Since log(0) does not exist, choose a=1 and then expand log to get: \sum^{\infty}_{n=0}\frac{f^{(n)}(a)}{n!}(x-a)^n=log(1)+(x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}-\cdots +\cdots...

in your last post, you are showing algebraically what K needs to be in order to get (3pi)/4. But the K must stay in the equation because sin is a periodic function. Like I said, pi/4, (3pi)/4, (5pi)/4 and (7pi)/4 all work as solutions. But if you add 2pi to any of these they always work. So...

when you solve 6*sin^2(x)-3=0, you get sin(x)=+/- 1/sqrt(2). So, the first answer you gave is x=(pi/4)+2*pi*K. But this isn't all the answers. There are other such x values in [0,2pi) that satisfy sin(x)=+/- 1/sqrt(2) and they are (3pi)/4, (5pi)/4 and (7pi)/4. So you need to modify your...