Mathematics

Associated with Math.SE; for both general discussion & math qu...
Jul 20, 2015 00:19
Can anyone help me with this? I got the answer of 3 but it was incorrect. math.stackexchange.com/questions/1367029/…
Jul 10, 2015 01:08
Great, thanks! @JMoravitz
Jul 10, 2015 01:05
Yep! :) @Fargle, and thanks for the tip @JMoravitz!
Jul 10, 2015 01:04
Art of Problem Solving! @Fargle
Jul 10, 2015 01:02
I got the answer (2). Thank you! :)
Jul 10, 2015 00:58
What I have so far is $\frac{2}{\omega^4+\omega} + \frac{2}{\omega^3+\omega^2}$.
Jul 10, 2015 00:55
Thanks, I'll try it! @SamuelYusim
Jul 10, 2015 00:41
I am having some trouble with this problem: Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega \neq 1$. Find
$\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3}$, (precalculus, roots of unity). Does anyone have any hints? I tried rearranging it to cancel some terms out in the fractions but it ended up becoming a large, messy fraction.
Jul 9, 2015 03:59
Maybe edit it and state "I attempted _________________ and now I am stuck. Any hints?" @abandon
Jul 9, 2015 03:55
@abandon: They want you to add some more details, like what you have solved so far on the problem or what part you are stuck at, etc.,
Jul 9, 2015 03:46
@anon: no, just precalculus
Jul 9, 2015 02:10
Can anyone help me with this problem? Roots of unity. "Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega \neq 1$. Find
\[\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3}"
Apr 7, 2015 22:10
I got the answers $k = 4, 9$ for this problem: Find all real numbers $k$ for which there exists a nonzero, 2-dimensional vector ${v}$ such that
$\begin{pmatrix} 2 & 12 \\ 2 & -3 \end{pmatrix} {v} = k {v}$, but the way I obtained these answers seem a bit too easy. Is $k=4, 9$ correct?
Apr 6, 2015 23:03
@TedShifrin: Okay, that's what I thought. Thanks!
Apr 6, 2015 23:03
For a parallelogram where the vertices are in vectors, the area is $|v_1*w_2 − v_2*w_1|$. math.arizona.edu/~calc/Text/Section13.4.pdf. (page 1). I'm a bit confused on what the bars denote. Is it the norm? Absolute value?
Feb 5, 2015 16:57
@DanielFischer: Cool! Thanks for all of the help!
Feb 5, 2015 16:56
@DanielFischer Therefore...it is zero?
Feb 5, 2015 16:55
@DanielFischer: We are left with $-(e^{\frac{\pi}{10}i} + e^{-\frac{\pi}{10}i})^6$. In the parentheses, this simplifies down to $\cos (\frac{\pi}{10}) + i \sin (\frac{\pi}{10}) + \cos (\frac{\pi}{10}) - i \sin (\frac{\pi}{10})$. This is where the imaginary parts cancel out.
Feb 5, 2015 16:52
@Ramanewbie: Did you try squaring the entire equation?
Feb 5, 2015 16:50
@DanielFischer I'm assuming it is 0 because the imaginary parts cancelled each other out.
Feb 5, 2015 16:47
@DanielFischer: Oh! Is it 0?
Feb 5, 2015 16:46
@DanielFischer: It is to find the imaginary part.
Feb 5, 2015 16:45
@DanielFischer: Are you referring to -(2)^{64}?
Feb 5, 2015 16:44
@DanielFischer: $\frac{\pi}{10}$ is equivalent to $18$ degrees. How would it factor out into a nice, whole integer?
Feb 5, 2015 16:43
@DanielFischer: Does the answer factor out into a nice, whole integer?
Feb 5, 2015 16:41
@DanielFischer: Does that mean, when simplified, it is: $-(2 \cos (\frac{\pi}{10}))^{6}$?
Feb 5, 2015 16:38
@DanielFischer: Would that be $e^{\pi*i}$?
Feb 5, 2015 16:37
@DanielFischer: So something like $(\cos^{6}(\frac{\pi}{6}) + i \sin^{6} (\frac{\pi}{6}))(64 \cos^{6} (\frac{\pi}{10})$??
Feb 5, 2015 16:34
@DanielFischer I reduced that down to $$(\cos (\frac{\pi}{6}) + i \sin (\frac{\pi}{6}))(2 \cos (\frac{\pi}{10})$$. What do I do next?
Feb 5, 2015 16:25
@bolbteppa: What is '?
Feb 5, 2015 16:24
....everyone mysteriously goes quiet O_O
Feb 5, 2015 16:23
I've noticed that whenever I enter this room......
Feb 5, 2015 16:21
Can anyone help me with this? "Find the imaginary part of $(\cos12^\circ+i\sin12^\circ+\cos48^\circ+i\sin48^\circ)^6$."
Feb 5, 2015 16:21
Oh wow ^
Feb 5, 2015 15:13
@Studentmath: Who are you asking?
Feb 5, 2015 15:11
Wouldn't the answer to this: math.stackexchange.com/questions/1134379/… be 70? But I tried 70 and it was incorrect.
Feb 4, 2015 04:43
How would I Find the real part of $(\sqrt{3} - i)^{2011}$?
Feb 4, 2015 03:42
I don't understand what the answer is hinting at. Can anyone help? math.stackexchange.com/questions/1132786/…
Feb 4, 2015 02:09
Feb 3, 2015 20:46
@Axoren: Would "k" be $-3+2\sqrt{6}, -3-2\sqrt{6}$, then?
Feb 3, 2015 20:44
@Axoren: Would it be: $||[1, k+3]|| = \sqrt{1^2+(k+3)^2}$?
Feb 3, 2015 20:43
@Axoren Alright, then
Feb 3, 2015 20:42
@Axoren: What if there are two vectors, then? ||[2, k] - [1, -3]|| ? Would that be simplified to: ||[2,k]|| - ||[1, -3]||?
Feb 3, 2015 20:40
@Axoren: That means if there is $(x,y)^T = \sqrt{x^2+y^2}$?
Feb 3, 2015 20:38
@Axoren: What are the || || on each side of the matrices? I haven't used matrices in a very long time.
Feb 3, 2015 20:34
@Axoren: What is the transpose, then?
Feb 3, 2015 20:33
[text=link]math.stackexchange.com/questions/1132334/… . Can anyone clarify what the answer meant?
Jan 28, 2015 00:47
@Jmoravitz: Got it now. Thanks.
Jan 28, 2015 00:42
I used z = a+bi , w = c+di but I got the messiest expression
Jan 28, 2015 00:41
How did you get that? @JMoravitz?