Homework 1 collaboration area
Feel free to toss around ideas here. Feel free to form teams to toss around ideas. Feel free to create your own workspace for your own team. --Steve Bell
By the way, I sign my name at the end of a post by typing three tildes in a row, and the Rhea puts my name in for me. - Steve Bell
Here is my favorite formula:
$ f(a)=\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z-a}\ dz. $
This is a test formula:
$ A \vec x= \vec b $ - Eun Young
People have noted an error in the solutions in the back of the book. This from Jeff R.:
For Section 7.2 problem 29, they have the profit vector as
[85, 62, 30]
but the problem defines it as
[35, 62, 30]
which obviously gives a different answer. Just wanted to make you aware of the mistake. (Thanks, Jeff. Steve Bell )
Question from a student :
I have a question on the Kirchoff law problem, section 7.3 problem 18. On the loop portion of the defining equations, relative to a clockwise direction, I find the equation of the right loop to be -12 I2 + 8 I3 = -24. Is this correct to assume the I2 term is negative due to the counterclockwise flow of I2, as with the voltage term?
Answer from Eun Young :
Yes, it's correct and it's same as $ 12 I_2 -8 I_3 = 24. $
Remark from Steve Bell :
I remember from my engineering days that engineers make one convention about the sign of the increase in voltage around a loop in Kirchoff's Laws and physicists make the opposite convention, so this might be a cultural thing. The important thing to understand here is the math.
Question from a student :
That equation is the same thing I got, I just took a different direction for my KVL around the loop (e.g. $ 12 I_2 - 8 I_3 = 24 $). However, I don't fully understand what your question is. I do however have a question on p. 287 #12, 14, and 15. Is the book looking for a rigorous proof or just an example of this property? --Ryan Russon 18:48, 25 August 2013 (UTC)
Answer from Eun Young :
When a problem asks you to show or prove something, you need to provide proof. When a problem asks you to disprove something, you need to give an example. Hence, you need to prove #12, 14, and 15. There is a theorem about rank in sec 7.4. With this theorem, you can show the properties in #12, 14, and 15 easily.
Remark from Steve Bell :
Perhaps the word "proof" is overkill here. I would like you to be able to explain in words why something is a general fact, much like I do in class. It doesn't need to feel like a mathematical proof, but it does need to be convincing.
From --Ryan Russon
Thanks for the clarification. I was on my way to working on a very rigorous proof and then wondered if all the work was necessary.
Question from a student :
I have a question on the last 2 homework problems in section 7.4 #32, 34. How exactly do we determine if something is a vector space? Do all vectors have to be linearly independent? I know in class we went over a couple of test to determine if vectors are a subspace. Does that come into play? Thank you. From --Katie Mathews
Answer from Eun Young :
In #32, we basically want to see if $ \{ (v_1, v_2, v_3 ) \in R^3 | 3v_1 - 2 v_2 + v_3=0 \text{ and } 4 v_1 + 5v_2=0 \} $ is a vector space. Note that this set is a subset of a vector space $ R^3 $. So, you just need to show whether it is a subspace or not. #34 is similar. In general, to determine if a set with operations is a vector space, you need to check 8 axioms. You can find those on [1].
Remark from Steve Bell:
Yes, you'll find those vector space axioms in section 7.9. The good news for us now is that we don't need to check all those axioms because our sets are subsets of $ {\mathbb R}^n $. We only need to check two things.
1. Is the set closed under addition?
2. Is the set closed under multiplication?
That's the Subspace Test.
In #34, if all the components of the vectors in the space have to have absolute value one, then multiplying by certain constants would take you out of the space. For example, if you multiplied by 2, you'd get a vector that has absolute value 2 in each component. So the space is not closed under scalar multiplication, and is therefore NOT a vector space.
Perhaps an easier way to see that it is not a vector space is to note that the zero vector is not in the space. (If you multiply a vector by c=0, it would need to be in the space. That's why the zero vector has to be there.)
