I'm not quite sure I understand your explanation. Also, if I integrate using those values, won't I be left with x's in my final answer? It looks as if I'll end up with complex numbers as well.
Homework Statement
A cylinder has a diameter of 2 inches. One end is cut perpendicular to the side of the cylinder and the other side is cut at an angle of 40 degrees to the side. The length at the longest point is 10 inches. Find the volume of the sample.
I believe this is what it would...
Homework Statement
I have an algorithm that implements Gaussian elimination. According to the text, with some modification of the indices and their in the loops, I should be able to have this algorithm perform Gauss-Jordan elimination. I also have to reduce the matrix to reduced row-echelon...
Ok, so I've found v3 to be \frac{\sqrt{6}}{6}.
I plugged this back into v_{1} + v_{2} -2v_{3} = 0 to get v_{2} = \frac{\sqrt{6}}{3} - v_{1}.
I plugged this into v_{1}^{2} + v_{2}^{2} + v_{3}^2 = 1 and ended up with 2v_{1}^{2} - \frac{2\sqrt{6}}{3}v_{1} -\frac{1}{6} = 0.
From this, I found...
Yes, you're right. I had forgotten about that. Even using that, though, I am still unsure of the next steps I should take. Is the previous work I showed applicable? Was I going in the wrong direction?
Homework Statement
The problem states to find a unit vector that is orthogonal to \left\langle1, 1, -2\right\rangle, forms an angle of \frac{\pi}{4} with \left\langle1, 1, 1\right\rangle and has v1 > 0.
Homework Equations
cos\theta = \frac{\vec{u}\bullet\vec{v}}{|\vec{u}||\vec{v}|}...
Apologies everyone, it is actually \mu \approx m.
Also, I do see how I could turn LCKurtz suggestion into a geometric series: m * \frac{1}{1-(- \frac{m}{M})} is the sum of the geometric series: \Sigma^{\infty}_{0} (-1)^nm(\frac{m}{M})^n
Homework Statement
\mu = \frac{mM}{m+M}
a. Show that \mu = m
b. Express \mu as m times a series in \frac{m}{M}
Homework Equations
\mu = \frac{mM}{m+M}
The Attempt at a Solution
I am having trouble seeing how to turn this into a series. How can I look at the given function...
Ok, ok, one more question!
After using the ratio test to find the radius of convergence, I say that |x| < 1, so -1 < |x| < 1. For the interval of convergence, I found that the series diverges for -1 and converges for 1. For k = -100, I found that x = \frac{1}{e^{100}} - 1 = -1. Now, -1 is...
Great! Thanks so much for the help. It's really important to me that I understand exactly why something is being done and you've been a huge help with that. Now I just have to see if I can find these values of x :tongue:.
I'm a bit rusty on my rules for ln's and e's, but I'll give it a shot. I think I understand the general premise now. I take the power series, differentiate it and then find a formula for it. If I integrate that formula, that will go back and give me the formula for the original series, right...