Finding the Best Textbook for Introductory Calculus Refresher

[denverdoc]

Homework Statement

just looking for opinions regarding the best text for introductory calculus as in refresher--I used Taylor eons ago, which i recall as being pretty descent.

Homework Equations

The Attempt at a Solution

 

[neutrino]

Seeing that you have nearly 600 posts, I'm sure that you know that this is not the place to post this question. You can search the Calc and Beyond and the Science books reviews forums for many threads on this topic.

 

[denverdoc]

thanks, Neutrino; frankly I wasn't sure where to go; PF is a very big place and I have only begun to scratch the surface.

 

[nicktacik]

I like Stewart...

 

[ZioX]

egad! Stay away from Stewarts. His definition of integration is not well defined. And I do mean in the mathematical sense.

I guess it's not bad for an intro text.

 

[denverdoc]

stewart seems to be the most controversial--on amazon it seems to get 5 starts or 1-2.

 

[Jumpy Lee]

stewart is good

your just looking for a refesher right?

get a 2nd ed stewart book

it is going to be pretty cheap.

 

[SeReNiTy]

There is no better introduction calculus text then calculus by spivak.

 

[Werg22]

Courant's text is one of the deepest out there, but it is time consuming.

 

[ZioX]

For those interested, Stewarts' defines the integral as riemann sums as the the mesh of the partitions (largest interval of the partition) goes to zero. The problem with this is if you look at the identity function on [0,1] of the rationals. You can always have the end points of your partitions be rationals, and then letting the mesh go to zero gives you 1. However, you can also always let the end points of your partitions be irrational, letting the mesh go to zero will yield 0.

 

[Werg22]

For those interested, Stewarts' defines the integral as riemann sums as the the mesh of the partitions (largest interval of the partition) goes to zero. The problem with this is if you look at the identity function on [0,1] of the rationals. You can always have the end points of your partitions be rationals, and then letting the mesh go to zero gives you 1. However, you can also always let the end points of your partitions be irrational, letting the mesh go to zero will yield 0.

I'm afraid I do not see what you mean. A very important result in calculus is that the Riemann sum of any repartition of an interval [a, b] as the largest interval goes to zero will always yield the same value. This is due to the fact that any function defined as integrable is also uniformly continuous (in a closed interval, if we chose an appropriate h, then f(x + h) - f(x) < k, where k is any value we wish and x is any value in the in the interval). Your argument with irrational numbers does not hold. Irrational numbers are given all the arithmetic properties of rationals, hence making Riemann summation consistent. Also, the integral of the identity function over [0, 1] is 1/2, not 1. This said, Stewart is certainly a masterful mathematician, and I deem his calculus textbook beyond reproach.

 

[morphism]

I think ZioX meant to say the indicator -not identity- function (of the rationals), that is the function which is 1 on the rationals and 0 otherwise. Although I don't really follow his argument. It seems to me that in both cases you'd get 0.

 

[Data]

For those interested, Stewarts' defines the integral as riemann sums as the the mesh of the partitions (largest interval of the partition) goes to zero. The problem with this is if you look at the identity function on [0,1] of the rationals. You can always have the end points of your partitions be rationals, and then letting the mesh go to zero gives you 1. However, you can also always let the end points of your partitions be irrational, letting the mesh go to zero will yield 0.

The indicator function of the rationals is not Riemann integrable on any interval of positive length, precisely because the limit as the mesh fineness goes to zero of the Riemann sums doesn't exist. I don't actually have a copy of Stewart's book (it was the assigned text for my first year calculus course, but I never bothered to open it and I can't find it now!), but my second-year vector calculus text (and every other text I've seen) defines the Riemann integral in a way that is certainly equivalent (unless he does something very silly!).

 

[Werg22]

I think ZioX meant to say the indicator -not identity- function (of the rationals), that is the function which is 1 on the rationals and 0 otherwise. Although I don't really follow his argument. It seems to me that in both cases you'd get 0.

But in this case the integral cannot be defined since the function g(x, h) = f(x+h) - f(x) would have an infinite number of jump discontinuity on any closed interval - the concept of a limit does not apply to the Riemann sum of this function because the Riemann sum has, how would you say, a "bad behavior".

Edit: Oops, just saw Data's post. That makes mine just a repetition of what has already been said.

 

[denverdoc]

Guys, thanks,

I just want to reacquaint myself with the tools of the trade, I have a good diffy text (kreyzig) but some of the simple stuff i haven't used for 20+ years. Just looking for a good cookbook with enuf theory to understand what's being assumed. I know the forum has to be carful re endorsement, else there should be some stickies. Weakest area on PF was book reviews, which is where Neutrino suggested I look.So I'm back til we get moved.

 

[mathwonk]

whatever you can learn from. i suggest beginning with taylor, the one you are used to, and then branching out to spivak, or something else yoiu find challenging, and which adds something to what you know.

 

[ZioX]

My point was that this is how stewarts defines his integrals, as far as I remember. As the mesh goes to zero and it comes out to some number, then that is the definite integral. But that is not a well-defined definition, as my indicator function shows. If, however, you're integrating a continuous function and you get some number, then it is Riemann integrable with that limit.

 

[Data]

It's not defined for every function, but then again neither is the Lebesgue integral, which is essentially the best that we can do on the reals. It is a correct definition for the Riemann integral (a function is Riemann integrable iff that limit exists!). I don't know of any introductory calculus book that defines the Lebesgue integral before the Riemann integral.

 

[ZioX]

You know, I just realized that I was correcting Stewarts definition. Stewarts defines integration as the limit of the riemann sums as the number of partitions goes to infinity.