Concepts in Calculus


by mbrmbrg
Tags: calculus, concepts
mbrmbrg
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#1
Aug16-06, 03:23 PM
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Anyone know a good text that covers the conceptual aspects of calculus? I can do the math, but I like to understand conceptually what's going on.

The current bug in my ear is indeterminate forms (limits of 0xinfty, 0/0, 1^infty, etc.). I see that they are indeterminate (heck, I do the problems!) but I want to know why.
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Werg22
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Aug16-06, 05:06 PM
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It depends on what you are looking for. If you want a rigorous classic text, or just a generally proof free run of the mill text. If you want the latter, there are plenty of books. If you want the former, I suggest

Introduction to Calculus and Analysis, by Richard Courant and Fritz John

It's the kind of book you always come back to when you need explanation or proof. It takes time to read, but you will certainly have a very solid base afterwise.
neurocomp2003
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Aug16-06, 06:26 PM
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Intro to Calc, Stewart.
Else you might want to pick up a computability book. I think its much clearer to learn infinities in Computability.

StatusX
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Aug16-06, 06:58 PM
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Concepts in Calculus


Quote Quote by mbrmbrg
The current bug in my ear is indeterminate forms (limits of 0xinfty, 0/0, 1^infty, etc.). I see that they are indeterminate (heck, I do the problems!) but I want to know why.
Do you want to know why we can't assign definite values to expressions like 0/0 and 1^infinity? That's just because these expressions are shorthand for limit expressions that don't all have the same limit. For example, say we have:

[tex]\lim_{x \rightarrow a} f(x) = L [/tex]

[tex]\lim_{x \rightarrow a} g(x) = K [/tex]

where a might be infinity. Now if L and K are finite and non-zero, then we can safely say things like:

[tex]\lim_{x \rightarrow a} f(x)/g(x) = L/K [/tex]

[tex]\lim_{x \rightarrow a} f(x)^{g(x)} = L^K [/tex]

However, if we try to extend these expressions to cases where one of L or K is zero or infinite, we run into some problems. For example, if f(x)=x, g(x)=x2, and a=0, then L=K=0, but how are we to interpret 'L/K' or 'K/L'. If you evaluate the corresponding limits, you find that the one corresponding to L/K is 0 while the one corresponding to K/L is infinity. But they both have the form '0/0'. It turns out that we can't extend the above expressions to cover all the cases where one or both of L or K is zero or infinity in a way that is correct for all possible limit expressions of each form.

If you want to go a little deeper and find the reason for this, you just have to understand that what we are essentially doing is looking at:

[tex] \lim_{x \rightarrow a, y \rightarrow b} h(x,y) [/tex]

for different functions h(x,y), such as x/y and xy. These functions are not defined at certain points (eg, x/y is not defined along y=0), including at infinite values of x and y. In some of these cases (like x/y at x=0 and y=infinity), you can assign these functions values in such a way that the function becomes continuous at this point (ie, the limit of the function as you approach one of these points is just the value you've assigned to the function at that point). But at others (like x/y at x=y=0), you cannot do this, for the limits depend on precisely how you approach the points. 'How you approach the points' is determined by f(x) and g(x) above, and for these indefinite types, you have to work out the limit in each individual case.
VietDao29
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Aug17-06, 01:54 AM
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Or you can look at it in an easy way.
0 * everything = 0 (assuming that the everything is not infinity), right? e.g: 0 * 5 = 0, 0 * 10100 = 0
[tex]\infty * \mbox{everything} = \infty , \ \mbox{everything} \neq 0[/tex]. If you multiply a number (not 0) by a very very great number, what do you get?
So, if you multiply one number that tends to 0, and another that tends to infinity. What do you get? 0, or infinity, or something in between?
---------
* 0 / b = 0, for all b not 0, and b is a constant.
* If b tends to 0, a is a constant, then a / b tends to infinity.
So what's a / b if a, and b both tends to 0?
---------
* If a is a constant and a > 1, then if b tends to infinity, then ab tends to infinity, right?
* If a is a constant and a < 1, then if b tends to infinity, then ab tends to 0.
So what if, a tends to 1, and b tends to infinity? Is it 0, or infinity, or something in between?
Werg22
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Aug17-06, 04:13 PM
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It's very important to see limits from an analytical point of view, rather than "philosophical". The concept of a limit is purely definitional.

Say you have function f(x). You want to know the limit as x tends to y. If you can write f(x) in the form of

f(x) = C + e

Where C is a constant, and e is a value that can be made as small as possible, as long as x gets closer to y.

By definition,

lim f(x) = C
x ->y
loom91
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Aug20-06, 12:58 PM
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Quote Quote by StatusX
Do you want to know why we can't assign definite values to expressions like 0/0 and 1^infinity? That's just because these expressions are shorthand for limit expressions that don't all have the same limit.
I can't agree with that. 1/0 is undefined not because the llimit doesn't exist. 1/0 is undefined because we did not define it when we defined the operation of division and it turns out there is no consistent way of defining it without incurring serious costs. By 1/0 one means 1 divided by 0, not a shorthand for the limit of 1/x as x tends to 0.

Molu
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Aug21-06, 03:00 PM
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Quote Quote by loom91
I can't agree with that. 1/0 is undefined not because the llimit doesn't exist. 1/0 is undefined because we did not define it when we defined the operation of division and it turns out there is no consistent way of defining it without incurring serious costs. By 1/0 one means 1 divided by 0, not a shorthand for the limit of 1/x as x tends to 0.
You could be stubborn and say that 1/0 must mean 1 divided by 0, and since such an operation is undefined, 1/0 is a meaningless symbol. Or you could think of it as standing for a certain class of limit expressions, as I described above, and the symbol becomes useful again. The same goes for other expressions like 0/0, 0^0, etc. The symbols just sit there gathering dust otherwise.

But more importantly, rember that limits are already used to extend these operations to all real numbers. How would you interpret [itex]\pi^e[/itex]? Well, we originally defined [itex]x^y[/itex] only when y is a positive integer. We extend its domain so as to preserve the rules for multiplying exponents, first to the negative integers by defining [itex]x^{-y}=1/x^y[/itex], and then to rational numbers by defining root extractions as the inverse of the corresponding power, ie, [itex]{(x^y)}^{1/y}=x[/itex], and then defining [itex]x^{y/z}={(x^y)}^{1/z}[/itex]. But what about aribtrary real numbers y? We can only interpret this by using successively better rational approximations of y, evaluation the exponent using our above defintion at these approximations, and then defining [itex]x^y[/itex] as the limit as the rational approximations approach y.

So, in fact, [itex]x^y[/itex] really is only defined on the real numbers in terms of limits. [itex]x/y[/itex] is a little more subtle, because the real numbers can be defined algebraically as a field without referring to their analytic properties, and if this is your approach, it is true that 1/0 is just meaningless. But it is perfectly valid, and in some cases necessary, to look at these operations from an analytic point of view, and as defined in terms of classes of limit expressions.

But that's just my understanding. Maybe an expert can do a better job answering this question.


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