3 ms·
I'm going to try to get through all of them That's inspiring. I've been working through them all so far, albeit slowly. It's worth noting that later chapters h
by gruseom 15y ago
I'm going to try to get through all of them
That's inspiring. I've been working through them all so far, albeit slowly. It's worth noting that later chapters have fewer exercises. The one I'm looking forward to most, Chapter 8, has less than a third as many as Chapter 1 does. Presumably they get less mechanical as well. I'd be content if I find enough time to "pay my dues" the slow way through the earlier chapters. When I was younger I used to scorn the mechanical stuff. Now I'm taking pleasure in it. We'll see if that lasts.
For those who don't intend to do all of them, here are some exercises I found interesting:
1.2: 3, 10, 11, 14, 15, 23
1.3: 4, 14, 17, 18 (that's about as far as I've got so far)
(Also, see my comments above if you don't have a way to look up those exercises by number - and especially if you have the Indian edition.)
- lepton 15y agoOK, these were interesting. Are we ready to share thoughts on them? For 1.2.10, I got a single equation involving the three unknowns--does that sound right?
- gruseom 15y agoLet's talk in detail about 1.2 and hold off on 1.3 until tptacek confirms. Re 1.2.10, I think that's right. Because the x-coords grow by the same amount as you go from one point to the next, the y-coords must as well, giving y3-y2 = y2-y1.
- mechanical_fish 15y agoYes, this is correct. (Though I haven't looked to see which problem this is, I can pretty much tell it's that one.) If we're not through 1.3 yet I'm not far behind! Will try to plug away on my 1.2 problems in between bugfixes.
- gruseom 15y agoI'm glad to hear that. I've been hoping that a group pace would emerge. If it's slower than we planned, that's ok as long as it still overcomes friction.
- lepton 15y agoGood. That's a really geometric understanding, which is helpful. I saw it as somewhat analogous to the back-substitution after lower-triangulation; using y=mx + b and starting with y_0 = b, then y_1 = m + y_0.