Recent content by ianb

18 replies? I'm really grateful to you guys. I realize I need to take a chill pill here and work things out right away. There's a toastmasters group around my place as well. Thanks folks, I'm going to read through everything posted again.

Dunno, I get stuck and fail to articulate what I try to say. People tell me how well I write and all that (mind you I'm no good at all, that's just what they say), but words just don't roll out of my tongue in the worst of moments (while participating in class, socialising, etc.). I speak...

I'm not following the format because this isn't really a homework question -- I'm just studying for my finals and encountered a few graphs in the electricity unit, and want a few things cleared up. So in a current-voltage graph, the slope is 1/R, and power (according to my teacher) is the...

Haha, wow, I could actually make sense of all these theorems - awesome. I have once attempted an analysis text and was overwhelmed with the terminology that was apparently essential to calculus - point-set topology and metric spaces? Etc.

Thanks, mindscrape. I suppose the book could have misprinted the answer. One more thing, though, if you don't mind: When the switch is closed, I calculated two parallel circuits. Are these two circuits, in turn, parallel to each other, or in series to each other (like I did above)?

Homework Statement http://img514.imageshack.us/img514/1223/questionju1.jpg Homework Equations Parallell: \frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} Series: R_{eq} = R_{1} + R_{2} The Attempt at a Solution Part 1 - When the switch is not closed: R_{eq_2} = R +...

Wow. I'm embarrassed to say the least. You, sir, rock my world.

I'm sorry, I should have been clearer. Message edited accordingly.

Hm. What if the lower bound is zero, I don't think it does. Look at the substitution method, and evaluate it from 0 to 1.7. You will still get a difference of 0.25 than if you compare it with the other method. Of course, you might say that the constants do indeed cancel even with 0 set as the...

Ah, my apologies. I was rushing through latex'ing my equations (still a newbie)--should have double checked it.

Well I can certainly show that, just expand it. But the issue arises when you evaluate the area, say from 0 to 1.7, and using either ways give different answers. In fact, the answers differ by 0.25. Thoughts?

simple integral -- but a question of methodology \int \frac{(x-1)}{2} dx First, try substitution. Let u = x - 1, and du = dx\int \frac{u}{2} du \frac{1}{2} \int u \ du \frac{1}{2} \times \frac{u^{2}}{2} \frac{1}{2} \times \frac{(x-1)^2}{2} \frac{(x-1)^{2}}{4} + C Or we could do this...

Thanks for commenting. I have room for another book (Amazon gift certificate), and I'm thinking of "https://www.amazon.com/dp/038790459X/?tag=pfamazon01-20". I also have Rudin's book, which turned out to be an overly ambitious choice. As you can see, I'm a beginner, and I'm approaching abstract...

Is Polya's How To Solve It book a suitable introduction to proof writing? I sort of just ordered it and I'm not sure I've made the right choice here.

Wow, okay, then I guess we could say F_(T,2)(0) + F_(T,1) * sin50(.30) + F_(T,1) * cos50(.30) = 10(.15) but that will give F_(T,1) = 3.55, which is incorrect. Of course, I could have made something wrong there but there is a catch somewhere that I probably wasn't taught before.