Studying Mathematics Thread

[polymath257]

Technically, here, the limit doesn't exist. If the limit *does* exist, the only possibility is what you gave: a removable discontinuity.

And there are many ways a limit can fail to exist:

1. A jump discontinuity (as above)

2. A vertical asymptote (where the limit is infinite). F9x)=1/x does this as x->0.

3. Even the one-sided limits can fail to exist through oscillation. f(x)=sin(1/x) does this as x->0.

Yes, you are correct, the overall limit at X=1 does not exist because the two sides don't match.

Your example 3 is very interesting, I remember it from my lectures a long time ago. They would call it "not well behaved", informally.

That example is used for a large number of other strange examples in upper level courses. So, for example,

f(x)=x*sin(1/x) for x!=0 and f(0)=0 defines a *continuous* function. The limit as x->0 is 0, which is f(0). In essence, the x in front of the sin forces things to go to 0.

Should we start talking about derivatives?

[robvalue]

Ahah, clever!

Sure, I love derivatives! One of my favourite things. It's such an incredibly powerful tool. It was Newton who set them in motion, I believe? (Pun not intended but taken credit for anyway)

[polymath257]

So, first, the basic aspect of derivatives is linear approximation.

Think of it like this. Suppose we have a function f(x) and an x-value x=a. We want to approximate f(x) as well as possible by a linear function y=mx+b. For convenience, we
will rewrite that as y=m(x-a) + b. So, for x values close to x=a, we want f(x) approximately equal to m(x-a) + c. Now, it is easy to plug in and demand that f(a)=m(a-a)+c, so c=f(a).

But what about the slope of that line? Well, the slope of the line through (a, f(a) ) and (x, f(x) ) is {f(x)-f(a)}/{x-a}. Unfortunately, we cannot just plug in x=a for this because we then get 0/0, which is an indefinite limit (such expressions can come out to be anything). So, we take the next best approach and do a limit.

So, if we let m be the *limit* of {f(x)-f(a)}/{x-a} as x->a, then we should get the 'best' straight line approximation to f(x).

if this limit exists (see above discussion), then we say the function f(x) is differentiable at x=a and define f'(a) to be that limit. So, the derivative f'(a) is the slope of the line that best approximates the function f(x) close to x=a.

Now, if this derivative exists, then the function f(x) *has* to be continuous at x=a. Why? Because x-a->0 as x->a, so to get the limit for the derivative, we need f(x)-f(a)->0 also (otherwise the limit would be infinite).

On the other hand, it is *quite* possible for a fun be continuous at a point but not differentiable there. Take the function f(x)=x if x>=0 and f(x)=-x if x<0. This graphs as a 'V' shape with the corner at x=0. And, in fact, the 'derivative from the left' is -1 and the 'derivative from the right' is +1. This means the derivative does not exist (just like jumps for ordinary limits).

Much stranger is the fact that there are continuous functions that do now have derivatives *anywhere*. For the example above, the derivative failed only at x=0. But there are functions (first discovered by Weierstrass) where the derivative exists at no point at all. In some sense, they are continuous by 'all corners'.

Next, we can go to functions of more than one variable, f(x,y) and ask for the 'best' linear approximation in exactly the same way. Only now, the linear approximation is in two variables and it looks like Ax +By +C for constants A,B, and C. if we want the approximation to work at (x,y)=(a,b), it is easier to write the linear function as A(x-a) +B(x-b) +D. Again, just as before, D=f(a,b).

But now, the coefficients A and B turn out to be limits where we only vary x or y (respectively) as we approach (a,b). This leads to the idea of 'partial derivatives'.

Clearly, now, this can be generalized to any number of variables. It turns out that we can even do this for *infinitely* many variables if we are working in a Hilbert or Banach space. In every case, the ultimate idea for derivatives is as linear approximations.

---

Ahah, clever!

Sure, I love derivatives! One of my favourite things. It's such an incredibly powerful tool. It was Newton who set them in motion, I believe? (Pun not intended but taken credit for anyway)

Well, Newton was one of the first. Leibnitz, Fermat, and Bolya contributed to the overall theory.

[Kernel Sohcahtoa]

IMO, when studying math, it is extremely important to not get discouraged when encountering a challenging concept, proof, definition, theorem, etc.  A couple of days ago, I encountered a proof dealing with the power series and the concept of the radius of convergence.  When I got to sentence 3, I was unsure how the author drew a particular conclusion, and as a result, I got stuck and ultimately decided to take a break from it.  However, today I was thinking more clearly and realized that if I took the negation of a particular definition (in this case, the definition of a bounded sequence), then the sentence made much more sense.  Thus, difficult concepts are a part of mathematics: sometimes I'm fortunate enough to be able to wrap my head around them; however, there are times when it is necessary to let some difficult concepts go and make a note of them, so that you don't waste time staring into space when you could be  learning new/important material that comes more easily to you (sometimes covering new material may provide more insight about the challenging concept).  Hence, you can always come back to the difficult stuff later or ask someone who knows more about math for help, especially if the concepts are fundamental to successfully understanding the topic one is studying; there's no shame in asking for help, as it is part of the learning process.

Also, IMO, when studying higher level mathematics, it is crucial that one has a solid understanding of mathematical proofs.  IMO, when taking a course in mathematical proofs for the first time, it is important that students do the following: give themselves ample time to digest the various proof-writing concepts, especially the basic proof-writing techniques; be willing to practice and work various proof exercises, so that they will gain first-hand experience navigating the proof-writing process (this often involves at least a draft of one's thought processes which is eventually harmonized into a coherent final proof); learn how to read condensed proofs via spotting key words that give away basic proof-writing techniques and being able to fill in the missing details via considering how one would get from A to B via theorems, definitions, previous knowledge, etc.; when writing final/condensed proofs, be sure to keep any clarifying notes (especially the analysis of proof, which IMO, is where the proof is actually explained and understood) that explain the fine details of a proof, because these notes will save you the time of having to figure things out again, especially if you haven't reviewed the proof for some time (days, weeks, months) and are somewhat rusty with the concepts, definitions, theorems, etc., that are involved in the proof.  IMO, reading the condensed proof of another author makes one appreciate the value of including an analysis of proof with the final condensed proof.

[robvalue]

I had to learn many proofs by heart for my exams. Something like 40% of the marks were for this. Understanding them makes it a lot easier to remember them of course.

[GrandizerII]

Speaking of proofs, since I am an amateur compared to you guys, I'll do something relatively basic. Let's see how to come up with the quadratic formula (I am doing this now without checking any sources, and it's been more than a year since I last did some serious algebra studying/practices).

ax^2 + bx + c = 0 (starting equation from which to eventually derive the quadratic formula, with coefficient 'a' being non-zero)

x^2 + b/a(x) + c/a = 0 (divide both sides of the equation by 'a')

x^2 + b/a(x) + (b/(2a))^2 - (b/(2a))^2 + c/a = 0 (complete the square and add in the compensatory term to keep the equation true)

x^2 + b/a(x) + (b/(2a))^2 = (b/(2a))^2 - c/a (move all terms not containing 'x' to the right-hand side of the equation)

(x + (b/(2a)))^2 = b^2/(4a^2) - c/a (factor the left-hand side and expand the first term of the right-hand side)

x + b/(2a) = +- sqrt(b^2/(4a^2) - c/a) (reverse the square on the left-hand side in the proper manner)

x + b/(2a) = +- sqrt((b^2-4ac)/(4a^2)) (unify the fractions within the square root)

x + b/(2a) = +- sqrt(b^2-4ac)/(2a) (pull the denominator of the right-hand side fraction out of the square root)

x = -b/(2a) +- sqrt(b^2-4ac)/(2a) (move the second term of left-hand side to the right-hand side)

x = (-b +- sqrt(b^2-4ac))/(2a) (combine the two fractions on the right-hand side, and we finally have the infamous quadratic formula)

[robvalue]

Very nice! Good work, and it shows what I was saying about how understanding helps the memory. It all follows, it's not just meaningless drivel to memorise.

Now do the formula for cubics Just kidding, I didn't see that done until university, and it hurt my poor brain.

Funnily enough, I remember hearing that there is also a formula for power 4 polynomials, but that there cannot be a general formula for power 5 or more. I wonder if my memory is correct there. There's probably some restriction about the kind of formula we're talking about.

Instead, we can use iterative methods and such, for approximate solutions.

[GrandizerII]

Very nice! Good work, and it shows what I was saying about how understanding helps the memory. It all follows, it's not just meaningless drivel to memorise.

Now do the formula for cubics Just kidding, I didn't see that done until university, and it hurt my poor brain.

Funnily enough, I remember hearing that there is also a formula for power 4 polynomials, but that there cannot be a general formula for power 5 or more. I wonder if my memory is correct there. There's probably some restriction about the kind of formula we're talking about.

lol, I don't even know what it's supposed to be like (the cubic formula). But I can imagine it being rather tedious to derive. Maybe a few years from now once I get into studying proofs for real.

[robvalue]

It would appear to the casual onlooker to be the darkest form of magic

Great, I hope you enjoy your further studies!

I was obsessed with maths. At "A level" (age 16-18) I did maths, further maths, physics and chemistry. I chose the last two because they were the closest to further further and further further further maths. I found physics the hardest out of all of them, as I found mechanics to be the hardest maths unit; not because of the maths involved, but the translation of real world problems using physical rules. Then of course, I did maths at university. And then became a maths teacher...

I wonder if I'd be better at it now. Maybe my young brain was too completely abstract. I did spend most of my time in pretend worlds. Then again, I still do. I think I would be better now, I feel I'm more well rounded.

[GrandizerII]

It would appear to the casual onlooker to be the darkest form of magic

Great, I hope you enjoy your further studies!

I was obsessed with maths. At "A level" (age 16-18) I did maths, further maths, physics and chemistry. I chose the last two because they were the closest to further further and further further further maths. I found physics the hardest out of all of them, as I found mechanics to be the hardest maths unit; not because of the maths involved, but the translation of real world problems using physical rules. Then of course, I did maths at university. And then became a maths teacher...

I wonder if I'd be better at it now. Maybe my young brain was too completely abstract. I did spend most of my time in pretend worlds. Then again, I still do. I think I would be better now, I feel I'm more well rounded.

At my age (35), and especially since I didn't have the chance to get into maths seriously when I was younger, I know I would most likely never be able to come up with innovative proofs on my own that others have yet to come up with, as my fluid intelligence should have already started its decline. But it doesn't mean one still can't improve their knowledge on a subject regardless of age, as there's also crystallized intelligence which should continue to increase until perhaps far later on in life.

Yeah, physics does put me off a little bit with the whole translation to the real world thing where things are no longer as precise as they are in the mathematical world, and a lot of assumptions then have to be made (which really bugs me personally). But that's the real world for you.