How can I truly learn Calculus without going through the basic steps?

[ahsanxr]

Alright so I should start off by saying that I could be defined as the conventional "math whiz". That is, I have no problems at all doing complex algebra problems or complex and tedious integrals and derivatives where there is little concept involved. Let's just say that I've done really well on any Math class I've ever taken and I've taken some rather advanced stuff for a first year college student. But the thing is, I don't really feel that I've actually truly "learned" a lot of stuff I have studied, which is something that has been discomforting me lately, especially since I'm shooting for a math major. It would be really sad if at the end of my major all I know is how to reduce something complex including all kinds of notations, variables and constants into something simple without really knowing what I have done.

Furthermore, I've heard how hard a math major becomes for students who are good at solving for a number but when things like proof-writing start because they are no longer doing the kind of math they thought they were so good at.

So I would like to start "truly" learning what is all that stuff that I've been studying for some time now but I would like to do that without going over the baby steps students have to go through when they are learning these kinds of things for the first time (so I wouldn't want to read a textbook). Any recommendations about what I should be doing to, what books to read or what sites etc I should be using would be extremely helpful. I am particularly interested in Calculus because I'm pretty confident about my conceptual understanding of the material learned before that.

So basically I need your help to guide me towards mathematical enlightenment.

 

[scorpion990]

Sorry to tell you, but:
"In mathematics you don't understand things. You just get used to them." - Neumann

You are suffering from something that should really be more common in our universities... Half of the people out there follow equations blindly, while the other half are afraid to admit that they don't truly understand what is going on.

To answer your question... You are not going to truly "understand calculus" by reading calculus I/II textbooks. You're going to need to study set theory, number theory, and most importantly, real analysis. Many of the things that are bothering you will be answered if you carefully study these three subjects. All of them are axiomatic in nature and are fairly theoretical. As a math major, you're not going to be able to escape these. In time, you will learn to enjoy them!

For now, just enjoy your education. If you are in a decent program, they will "ease" you into formal mathematics. You'll probably take a linear algebra course (which includes the formal axioms of vector spaces) and a discrete mathematics course (which will ease you into number theoretic and set theory stuff). In time, you will be able to tackle more advanced topics like real analysis.

Good luck!

 

[ahsanxr]

Well I might have not been very clear in my OP but what I mean to say is that I don't really want an understanding touching on the deep, theoretical and axiomatic parts of Calculus. Instead what I'm looking for is an explanation of why we use what we use. For example I know why the derivative represents the slope (which in turn is the rate of change of one variable with respect to another) because I read up on the definition of the derivative and saw some examples. But I still don't clearly know why integration over a certain region represents the area under the curve (for 2 dimensional x and y functions). Also things like why the product rule works and why we can find integration for volumes etc.

 

[TylerH]

Think of differentiation as an infinitely small difference(like, the difference between the value of the function as the variable increases by an infinitesimal) and integration as an infinitely large sum. The area is just the addition of all the values of f(x), from a to b, summed. Integration for volumes is like finding the area for each infinitely small slice of a 3d function, and then infinitely summing the areas of all the slices to find the volume.

The easiest way to learn to understand any of the rules, like product rule, is to derive them. https://secure.wikimedia.org/wikipedia/en/wiki/Product_rule#Proof_of_the_product_rule

 

[chiro]

I echo the above poster recommending getting into an analysis course.

One view of what calculus does is that it allows us to calculate any kind of measure in any dimension where there is non-linearity.

It might seem trivial but its very powerful. Before calculus whenever we wanted to calculate things like length of a curve, area or volume of some arbitrary shape or even do some analysis on non-euclidean geometry, without calculus, we wouldn't have the machinery to analyze these problems.

In high school we are taught to calculate things that have linear components like for example a trapezium or triangle or cube of some prism, but if you have some side that was non-linear (ie some curve), then you would be stuck, and with calculus you could in fact find the appropriate measure.

 

[scorpion990]

Well I might have not been very clear in my OP but what I mean to say is that I don't really want an understanding touching on the deep, theoretical and axiomatic parts of Calculus. Instead what I'm looking for is an explanation of why we use what we use. For example I know why the derivative represents the slope (which in turn is the rate of change of one variable with respect to another) because I read up on the definition of the derivative and saw some examples. But I still don't clearly know why integration over a certain region represents the area under the curve (for 2 dimensional x and y functions). Also things like why the product rule works and why we can find integration for volumes etc.

Understanding why an integral represents the area under a curve is actually fairly straightforward. Convince yourself that you can estimate the area under a curve as a Riemann Sum:
http://en.wikipedia.org/wiki/Riemann_sum

A 1-dimensional integral is just a limit of these sums. The reason why a double integral represents a volume is really the same. Convince yourself that you can write the volume under a surface as a double sum. A double integral is just the limit of this expression.

However, the more important (and interesting) question is why you can compute integrals in terms of antiderivatives. That is, why is the fundamental theorem of calculus true? For the double integral problem, the corresponding question is, why can you use Fubini's theorem to compute the double integral as two single integrals? The answers to these questions, in addition to why the product rule works, lies in the study of real analysis.

 

[Fredrik]

Instead what I'm looking for is an explanation of why we use what we use.

Such things will gradually become clearer as you keep studying. I don't think there's a book you can read that just explains those things to you.

But I still don't clearly know why integration over a certain region represents the area under the curve (for 2 dimensional x and y functions).

This isn't the best way to think about it. I suggest this instead: The integral defines what we mean by the "area" of a non-rectangular region. If you study Riemann's or Darboux's definitions of the integral, you should see why this is a natural generalization of the definition of the area of a rectangle. The theorems derived with this definition as the starting point shouldn't be thought of as revealing "facts" about the god-given concept of area, but as verification that we have successfully defined the "area" of a non-rectangular region in a way that gives it the properties we wanted that concept to have before we chose a definition.

Also things like why the product rule works...

Just do the proof over and over, until you don't need to look at the book anymore, until you're able to do the whole thing from start to finish in your head with your eyes closed. Will you still feel that you don't know "why" it holds when you can do that?

 

[AlephZero]

You are already well on the way to mathematical enlightenment. Why? Because you have figured out that you "don't understand everything". Many people with "good educations" never progressed that far.

I would suggest if you really want to "understand" mathematics, read books by great mathematicians or philosophers of mathematics. Hilbert, Bertrand Russell, and G H Hardy were all excellent writers as well as great mathematicians. But you need to make your own journey of discovery, not follow a plan that somebody else gives you...

Of course you need to read your course textbooks as well, to spoon-feed you the stuff you need to pass your exams. But reading them is just a quick short-cut to getting adegree certificate, not the way to get deep understanding of anything.

 

[paulfr]

Integration
I teach Calculus
Mathematicians will frown on what I am about to say, but ...

Integration is just [ or can be thought of as ] MULTIPLICATION when one of the factors is changing. Recall that Calculus is the Mathematics of Change.

Think of first integrating y = 3 from 0 to 4
The area is that of a single rectangle, 3 units high and 4 units long.
It is just a simple multiplication

Now make y = f(x) more general.
You are still multiplying, it is just a bit more complicated and involves approximations and limits to approximations.

Hope that helps

 

[scorcher863]

it's funny I'm taking calc2 and i was just wondering this same question on my own. i searched for a concpetual explanation for integration on google, and was led to this page lol.

there has to be some layman's explanation for why if you evaluate at some x the function that results from anti-deriving some f(x), you get the area under the curve of f(x) up to the value of the evaluated x.

i honestly see no reason why this must be logically true.

 

[scorcher863]

i think i sort of get it. get two graphs -- one of any graph, and one of it's derivative.

take the derivative graph and while pointing at some point along the function, ask yourself, "if i were to deviate from this graph from this point on (ie. move your finger somewhere higher or lower than the graph), what would that change correspond to in the original graph?"

the answer is that, if i deviate lower than the graph at a given point moving forward*** then the slope of the original function will be less steep, and thus from the given point on the original function will climb less higher. how high the original function climbs represents the area of under the derivative.

***moving lower will decrease the area under the curve, and moving higher will increase it.

i recommend using the graph of y=2x and y'=2 to start (it's not a curve, but if you draw a curve with your imagiation by "deviating" you can see the corresponding change in the y value of the y=2x function in your imagination too.

 

[Fredrik]

there has to be some layman's explanation for why if you evaluate at some x the function that results from anti-deriving some f(x), you get the area under the curve of f(x) up to the value of the evaluated x.

i honestly see no reason why this must be logically true.

I don't know about a layman's explanation, but we can do a sloppy version of the rigorous proof, which ignores some of the details. (Those details are of course crucial when we want the proof to be rigorous). The definition of the derivative says that

[tex]F'(x)=\lim_{h\rightarrow 0}\frac{F(x+h)-F(h)}{h}[/tex]

This means that when h is small,

[tex]F'(x)\approx\frac{F(x+h)-F(h)}{h}[/tex]

which implies

[tex]F'(x)h\approx F(x+h)-F(h)[/tex]

Suppose that f=F'. You should know that the integral is approximately equal to a Riemannn sum:

[tex]\int_a^b f(x)dx=\sum_{i=0}^{n-1} f(x_i) \Delta x_i=\sum_{i=0}^{n-1} F'(x_i) \Delta x_i=\sum_{i=0}^{n-1} (F(x_i+\Delta x_i)-F(x_i))[/tex]

Now recall that [itex]\Delta x_i=x_{i+1}-x_i[/itex], which implies that [itex]x_i+\Delta x_i=x_{i+1}[/itex]. This means that the above is

[tex]=\sum_{i=0}^{n-1} (F(x_{i+1})-F(x_i))=F(x_1})-F(x_0)+F(x_2)-F(x_1)+\cdots+F(x_n)-F(x_{n-1})[/tex]

[tex]=F(x_n)-F(x_0)=F(b)-F(a)[/tex]

 

[l'Hôpital]

http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Geometric_intuition

That should provide a little intuition on the matter. Not a rigorous proof, but it gives the idea.

 

[scorcher863]

http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Geometric_intuition

That should provide a little intuition on the matter. Not a rigorous proof, but it gives the idea.

this is a perfect explanation for me, thanks!

 

[mathwonk]

I don't get it scorcher. That wiki page is exactly what appears in virtually every calculus book in the world. do todays students only read online works? do their brains go blank when they lift a big book with print in it? I admit it is nice to have a short version with big type, but otherwise it is exactly the same explanation found everywhere.

 

[mathwonk]

the whole point is to understand the fundamental theorem of calculus. i.e. the definir=tion of a riemann integral by riemann sums makes it pretty obvious that the limit is the area. then the whole point is why that area function can be computed as an antiderivative. there are two steps: 1) to see geometrically why "the derivative of the area function is the height function (if the height function is continuous)". (I wear a T-shirt with this sentence on it to my class on test day).

2) the fact that two functions with the same derivative differ by a constant, so ANY function whose derivative is the height function must be a constant plus the area function. This is the deep part of the theorem although it sounds easier.

But it is frustrating that the OP said he did not want any deep theoretical explanation, since this understanding can ONLY be gained from a PROOF (i.e. from the theory) of these statements. When someone says "I want to understand but I don't want to see any theory, nor anything deep", I don't know what to say to them, except maybe "trust me, that's the way to go, but think about what the theory means, and ask for motivation."

I.e. it is possible, and preferable, to present theory with understanding.

 

[dBrandon/dC]

Personally, what has helped me to understand calculus is an understanding of physics. You get real-world application of calculus in physics, and it can help you actually see what's going on with your formulas. Take the simple example of velocity and distance. When we differentiate distance with respect to time, we get velocity - the instantaneous rate of change of distance. And differentiating velocity gives us acceleration - the rate of change of velocity. Make sense? So let's say you have a curve - a plot with millage on the y-axis and time on the x-axis. If this plot was linear (like y = x), you could see that as your time progresses, your millage accumulates. If we differentiate that, we would end up with a straight line, since the rate of change is constant. But that line also shows us how much millage we accumulate per unit time. The higher the line (in the y direction), the more millage accumulated per unit of time. Of course, the height of the line represents the slope of the first chart, which is, again, the millage accumulated over time. So what relationship does the second chart (of velocity) have with the first chart (of distance)? The steeper the slope of the first, the higher the line of the second, which also means the greater the area gain under the line in the second as time progresses. That's why integration works. Mathematically, it gives us the area under the curve. But physically, it gives us an increasing area as time progresses, and that area is directly proportional to the millage accumulation from the first chart. We're using the example of velocity here, but the principle holds true for every other case, as well.

 

[Entropee]

I was too lazy to read all the previous comments so forgive me if someone has already posted this. There is a book called "calculus made easy" and from what I have heard it is very good at getting you to actually understand why things are they way they are and so on and so fourth. It however does not go super in depth into problems, which is good since you already know how to do all of them. Its a smallish book so it shouldn't be too much of a problem to read. Get back to me if you end up reading it, Good luck man! P.S. your not alone with your little "problem".

 

[ahsanxr]

I was too lazy to read all the previous comments so forgive me if someone has already posted this. There is a book called "calculus made easy" and from what I have heard it is very good at getting you to actually understand why things are they way they are and so on and so fourth. It however does not go super in depth into problems, which is good since you already know how to do all of them. Its a smallish book so it shouldn't be too much of a problem to read. Get back to me if you end up reading it, Good luck man! P.S. your not alone with your little "problem".

Thanks a lot man. I recently got this book called "Basic Training in Mathematics: A Fitness Program for Science Students" by Yale Physics professor R. Shankar. The reason why I got it was that I've listened to quite a few of his lectures on youtube and found his way of explanation really intuitive and conceptual, so I figured his book would be the same too. Secondly, the book attempts to cover a lot including Calc I-III, some Complex numbers, some Linear Algebra and some Diff Eqs. The reason why I like this is that like you said, I don't need to actually learn how to solve the problems since I'm pretty good at that, I just want a brief conceptual understanding of every topic which hopefully that book will give me. If it doesn't end up being what I'm expecting it to, I'll look into your book too.

 

[Entropee]

Yeah for sure man, if you want you can just search it on amazon.com and read the comments. I'm really excited to start calculus this summer quarter. Only problem is summer quarter is 6 weeks long as opposed to a normal 11 week quarter, so they rush through everything. So I'm trying to teach myself calculus before summer so I can destroy the class without too much trouble. If you have any ideas I'd also love to hear them haha

 

[ahsanxr]

Yeah for sure man, if you want you can just search it on amazon.com and read the comments. I'm really excited to start calculus this summer quarter. Only problem is summer quarter is 6 weeks long as opposed to a normal 11 week quarter, so they rush through everything. So I'm trying to teach myself calculus before summer so I can destroy the class without too much trouble. If you have any ideas I'd also love to hear them haha

Dude, thing is IMO the Calc I-III sequence is insanely easy. If you're looking to do well in those classes (disregarding TRULY learning it, which would take some time I think) then just look at the examples in the book and practice problems. Eventually like all computation based math (i.e non-proof based math) you will catch on to a pattern and will have no problems in doing most of the problems. As for understanding and getting and intuition behind what you're doing, that can't really be done in a rushed semester I think and you would have to learn it by yourself. Eventually if you're a math major then you'll take Basic Real Analysis which is all about (from what I heard) where calculus comes from and doing calculus proofs.

 

[Entropee]

I think you're probably right. I was just worried calc 1 in particular would be too rushed for me.