Hi! I'm looking for a good intro to calculus book, and i was wondering if you guys had any suggestions. Thanks, piirx
There is a book called "A Tour of the Calculus" - it is a blue colored book.. I do not know the name of the author. I read this while in AP Calc. just for fun... I enjoyed it. Also, try going into Barnes&Noble and browsing the math section... there should be some good books there.
Hah, oh Calculus, how I miss thee. Really, you are in for some ridiculous stuff. Not that I think it is unnecessary, but just wait until you start doing the first derivatives (you guys know, the real long way, I don't remember the formula). Anyways, it takes around 6 minutes per derivative at first, mostly because you have to write A TON. About 2 days later they will start teaching you how to do the same thing in 2 seconds.... jerks.
Jerks?!?!? How so? You seem to me to be the type of person that does not want to understand math... you just want to know a bunch of rules. That is why people like you hate math... you do not understand it... so you get frustrated. The rules are derived from the slow process you do in the first few days... you jerk. By the way... the the derivative is: lim(delta_x -> 0) [f(x + delta_x) - f(x)]/delta_x Try deriving the power rule from it... or the derivative of sin(x) from it. Just curious if you can do it without reference to other material. Please Register or Log in to view the hidden image! I appoligize if I am taking your message the wrong way........
Calculus Book That Tour of Calculus book is an entertaining book and I actually enjoyed reading it. Although, as an intro to Calculus, I think the author is not very articulate and sometimes you loose track with what his point is. There is another book I would recommend, "Calculus Made Easy" by Silvanus Thompson and Martin Gardner. Although, one nice thing with the Tour of Calculus book, he gives you an insight to calculus without overloading you with math. Check them both out at your local bookstore. CJ
Just for fun, the derivative of sin(x) from first principles: f'(x) = lim(delta_x -> 0) [f(x + delta_x) - f(x)]/delta_x = lim(dx -> 0) (sin(x+dx) - sin(x))/dx = lim(dx -> 0) (sin x cos dx + sin dx cos x - sin x)/dx = lim (dx -> 0) [sin x (cos dx - 1)/dx + cos x (sin dx)/dx] = cos x The last step follows from the fact that cos x = 1 + x<sup>2</sup>/2 + ... so cos dx - 1 = (dx)<sup>2</sup> + ... Therefore sin x (cos dx - 1)/dx ~= sin x dx -> 0 as dx -> 0 Also sin x = x - ... So sin(dx)/dx -> 1 as dx -> 0 Ain't calculus fun?
Fun Fun Haha, man, havent done a derivative like that in years. Reminds me when we had to add up all the stupid rectangles at different heights along some exponential curve. Remember the good old Reinmanns Sums? It all teaches you the fundamentals though, and its important. Math can be fun if you understand it.
Hi James, I am not sure if mathematicians would like your notation (something like cos(dx) is lickely to give them a heart attack), but I enjoyed reading it Please Register or Log in to view the hidden image! Bye! Crisp
i object to the fact that you use a taylor series for cos x, which depends on knowing the derivative of cos x, to calculate the derivative of cos x. that is circular. it s a nice way to guess the derivatives, but it ain t first principles.
Well here is how I derived it at first: dsin(x)/dx = lim(h->0) [sin(x - h) - sin(x)]/h = lim(h->0) [sin(x)cos(h) - sin(h)cos(x) - sin(x)]/h = lim(h->0) [-sin(h)cos(x)]/h = -cos(x) Now, the - comes from assumptions in the derivation, which can be easily fixed. sin(pi/2) = 1. sin(pi/2 - |some small number|) gives a value less than 1. Therefore to the left, dy/dx is negative. cos(pi/2) = 0 cos(pi/2 - |the same small number|) = some negative. Because cos(x) is negative in (pi/2, 3pi/2), the sign is positive. I am sure a mathematician would have a heart attack at this one, too. lol. I know another way to derive it similar to James R's way, but I used properties of limits for trig functions.
lethe: Fair enough. No Taylor series allowed. Then let's do it another way. I got down to here: f'(x) = lim (dx -> 0) [sin x (cos dx - 1)/dx + cos x (sin dx)/dx] Now, I need to establish two things: (1) lim(x -> 0) (sin x)/x = 1 and (2) lim(x -> 0) (cos x - 1)/x = 0 ----------- Proof of (1): sin x < x < tan x Dividing by sin x: 1 < x/sin x < 1/cos x Now, lim(x->0) 1 = 1 and lim(x->0) 1/cos x = 1. So, using the sandwich rule: lim(x->0) x/sin x = 1 Therefore lim(x->0) (sin x)/x = 1 ....... QED. ------------ Proof of (2): (2) lim(x -> 0) (cos x - 1)/x = 0 cos x = 1 - 2sin<sup>2</sup>x so (cos x - 1)/x = -2(sin<sup>2</sup>x)/x = -2 sin x (sin x)/x Therefore: lim(x->0) -2 sin x ((sin x)/x) = (lim(x->0) -2 sin x)(lim(x->0) (sin x)/x) = 0 × 1 = 0....... QED. Happy now?
I'd like to point out that both identities are pretty easy to derive intuitively.. the first by examining the relationship between x and sin x on a unit circle as x -> 0, and the second by examining a graph of (cos x) - 1 as x -> 0. I like the neat formal proofs too, of course.
Ummmmmmmmmmm... Sorry about that.... Stupid algebra... that negative will disappear. Here is the correction: dsin(x)/dx = lim(h->0) [sin(x - h) - sin(x)]/(-h) = lim(h->0) [sin(x)cos(h) - sin(h)cos(x) - sin(x)]/(-h) = lim(h->0) [-sin(h)cos(x)]/(-h) = cos(x) :bugeye:
yes, good show. i guess i will add that in my experience, in a rigorous analysis setting, the easiest way to define sine, cosine, exp, and logarithm is in terms of their differential equations, rather than in terms of some picture you draw on a circle, which makes analytic proofs more difficult. in that setting, it is trivial to calculate their derivatives from first principles. the sine function has cosine as its derivative <i>by definition</i>. however, this is not a thread on the foundations of analysis, this is a thread for calc I, and that derivation you gave is exactly what i was looking for.
Personally, and I have seen many Calculus books in my day, but "The Joy of Calculus" is one of the most well written math books that I have ever read. Not sure who its by though.
I'm starting a business related calc class two weeks from today and reading this thread scared me to death. Please Register or Log in to view the hidden image!
