• Support PF! Buy your school textbooks, materials and every day products Here!

Any good calculus places to start?

  • #1

Main Question or Discussion Point

Hello, I'm new here! Before I ask, I shall introduce myself. I am Lazernugget! I am probably younger than most people here, yet I am dedicated to sciences and mathematics, so I joined. I am young, as I said, But I am reading a thick trigonometry book, understand very advanced math, (But not all the symbols like: [tex]\partial[/tex] or [tex]\int[/tex] So) I may not get all the formulas you may show me. I know a BIT of calculus, but that is where my question come in:

Would you mind showing some basic calculus functions, and then explain it for me? I may ask for more advanced ones afterwords. Then I am curious to know what these are for:

[tex]\int[/tex] [tex]\oint[/tex] [tex]\sum[/tex] [tex]\nabla[/tex] [tex]\partial[/tex] and [tex]\otimes[/tex] are? Then use them in a simple mathematical equation.

Thanks! Bye!
 

Answers and Replies

  • #2
berkeman
Mentor
57,296
7,276
Hello, I'm new here! Before I ask, I shall introduce myself. I am Lazernugget! I am probably younger than most people here, yet I am dedicated to sciences and mathematics, so I joined. I am young, as I said, But I am reading a thick trigonometry book, understand very advanced math, (But not all the symbols like: [tex]\partial[/tex] or [tex]\int[/tex] So) I may not get all the formulas you may show me. I know a BIT of calculus, but that is where my question come in:

Would you mind showing some basic calculus functions, and then explain it for me? I may ask for more advanced ones afterwords. Then I am curious to know what these are for:

[tex]\int[/tex] [tex]\oint[/tex] [tex]\sum[/tex] [tex]\nabla[/tex] [tex]\partial[/tex] and [tex]\otimes[/tex] are? Then use them in a simple mathematical equation.

Thanks! Bye!
Welcome to the PF. This is a reasonable place to start:

http://en.wikipedia.org/wiki/Calculus

And there are links at the end out to other sources of information.

I'm going to move this thread to Academic Guidance for now.
 
  • #3
Ah yes, thanks, I'll try, but Wikipedia can be...long drawn...

While I try, could someone explain the symbols I typed above?
 
  • #4
348
0
It's awesome to see someone so motivated at such a young age. I hope that enthusiasm stays with you. Regarding your question, there are a plethora of resources out there for learning basic calculus. If you prefer learning from a textbook, I like M. Kline's "Calculus." Of course, there's also the gold standard: Stewart's "Calculus." There are also youtube videos and lectures posted on itunes you might want to check out.

Regarding your next question on the specific symbols: the first and the third are something you'll see in your study of basic calculus. The next second, fourth and fifth you'll see in more advanced calculus and the last is the tensor product operator and you probably won't see it for quite a while. Unfortunately, without knowing calculus, it would be impossible to really describe those symbols.
 
Last edited:
  • #5
Nabeshin
Science Advisor
2,205
16
[tex]\int[/tex] Integral. An infinite sum.

[tex]\oint[/tex] Integration along a closed surface.

[tex]\sum[/tex] Summation.

[tex]\nabla[/tex] Del, a derivative operator.

[tex]\partial[/tex] Partial derivative. Distinguished from a total derivative in that you only differentiate with respect to one out of many variables.

[tex]\otimes[/tex] Tensor product.

If none of that makes sense, then good it shouldn't. Really, you shouldn't focus on the symbols though. What's more important is that you understand what's behind them, so just be patient and you'll get there eventually.
 
  • #6
Sooo...um, Wouldn't an infinite sum just be infinity? (Integral)

How do you use those symbols? Thanks...
 
  • #7
445
0
Sooo...um, Wouldn't an infinite sum just be infinity? (Integral)
http://www.msstate.edu/dept/abelc/math/integrals.html [Broken]
 
Last edited by a moderator:
  • #8
18
0
Sooo...um, Wouldn't an infinite sum just be infinity? (Integral)

How do you use those symbols? Thanks...
An integral is just another way of finding area. i.e. the space underneath a curve/line/whatever. it's useful for more advanced calculations like volume of a solid with a weird shape and many, many more applications.


I wouldn't worry too much about these things. Math builds up to that stuff. You'll learn it eventually through practice and application problems.
 
  • #9
901
2
Sooo...um, Wouldn't an infinite sum just be infinity? (Integral)

How do you use those symbols? Thanks...
It's good to know the difference between and integral and an anti-derivative. An anti-derivative is an indefinite integral thus you aren't finding the area, definite integral has a boundary and therefore you are finding the area.
Anti-derivative [tex]\int f(x)dx[/tex]
Integral: [tex]\int_a^b \ f(x)dx[/tex] bounded by [tex]a,b[/tex]. Finding the area is described by the fundamental theorem of calculus:
[tex]F(b)-F(a)[/tex] where [tex]F(x)[/tex] is the integral evaluated at the point [tex]x[/tex]
Del Operator ([tex]\nabla[/tex]): [tex]\nabla = \frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z} +...[/tex]
Partial derivative ([tex]\partial_{x...n}[/tex]): Is an operator in multi-variable calculus in which you differentiate with respect to one variable and treat the others as constants.
Tensor Product ([tex]\otimes[/tex]): Don't worry about this right now as it shouldn't be learned until you have a grasp of the other branches. A tensor would be identified by [tex]T_{ij}[/tex].
Summation : [tex]\sum[/tex] is essentially a series, what I mean by that is if you have [tex]\sum_{n=1}^{3} 2^n = 2^1 + 2^2 + 2^3 = 14[/tex]

I almost forgot! This is a great place to start learning maths: www.khanacademy.org
 
Last edited:
  • #10
724
0
Sooo...um, Wouldn't an infinite sum just be infinity? (Integral)
You have to study the basic stuff before this can make sense. In short, an integral is an infinite sum of infinitely small numbers. Look at the Riemman sum method for approximating the area under a curve. You can think of integration as a Riemman sum with infinitely thin rectangles.
 
  • #11
201
0
To understand calculus you need to understand the problem faced without it. Let's say you have a graph of a straight horizontal line between two points on the x-axis. You know how to calculate the area under it because it's just a square. You can do this as well for a slanted line, but you don't know how to do this for curves. It's very difficult to do this geometry without calculus to find the area. So instead you put a bunch of little squares under the curve to try to approximate or roughly guess how much there is under the curve by adding up the area of each square. The squares don't fit perfectly under the curves because they have jagged edges, so you put slimmer squares under it and try to fit it until you do it so much that the slim squares become so thin that it just looks like the squares filled the curve perfectly. These very-very slim squares are infinitesimally thin so you need an infinite summation of all of these squares and this is called the integral.

The other fundamental problem is the slope of a curve at a spot at only one point. You know how to calculate the slope between two points on a curve, but when you put these two points closer and closer you can't calculate them because eventually the difference between them is zero and the formula for slope no longer works. So instead you keep calculating what the slope is when it is really really close and you find the a general trend to do this and you call it the derivative. These derivatives can be proven with limits but I don't want to get into that.

The signs for these the derivate is the infinitesimal change in x over y (which is the slope of the regular line if it wasnt infinitesimal) known as [tex]\frac{dy}{dx}[/tex]. The integral is the slim width of the square which is the [tex]dx[/tex] times the height of it which is the function evaluated at that point in x and from the start to the end of the sum from a to b (on the x axis) which is [tex]\int_{a}^{b} f(x) dx[/tex]. Derivatives give the slope of the tangent line and integrals give areas of curves.

If you have the infinite sums of the infinitely small what do you get? The integral. It turns out that these two concepts, the derivative and the integral, are related by the fundamental theorems of calculus.

Now that you understand the problems I hope you can start reading a calculus book. Wikipedia is not a good place to start. Seem interesting?
 
Last edited:
  • #12
47
0
Pauls Online Math Notes has been very helpful in my quest to master calculus. His Calculus I notes are available http://tutorial.math.lamar.edu/Classes/CalcI/CalcI.aspx" [Broken]. I believe somebody else on this thread mentioned khanacadamy.org, but I would like to verify that person's claim that it is a valuable learning tool.

I understand where your curiosity comes from with all these new symbols. I was thinking the same way just a few months ago. I wasted probably a month just trying to figure out what all the symbols meant rather than learning what they mean and how to use them one at a time, which is what I have been doing these past few months. Comparing the two study strategies I've used, the latter has been much, MUCH more effective for me; and I assume it would be more effective for most individuals also.
 
Last edited by a moderator:
  • #13
mathwonk
Science Advisor
Homework Helper
10,901
1,064
it takes a long time just to understand any one of the symbols you wrote down. You could spend almost a lifetime just on the integral. I am 68 years old and still studying its basic properties, (and it is not my specialty). Just find a book you can understand and start in.

Trigonometry is one of the hardest subjects to do well in the beginning since even defining the basic functions cosine and sine requires a function of arc length in terms of x and y coordinates, and then inverting it. This process cannot be done carefully until one has integral calculus in hand, so in a sense trigonometry is more sophisticated than calculus, and comes later. Elementary presentations of trigonometry necessarily do things in an imprecise way.

after you have had calculus and learned to make sense of "infinite sums" you can define

cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! ±....... which also shows why the function is "even".
 
  • #14
mathwonk
Science Advisor
Homework Helper
10,901
1,064
here are the notes from the first day of one of my calculus classes, explaining what "real" numbers are, the numbers that are used in beginning calculus.


Day One, 2200 sp 2000 Real Numbers
Real numbers are in one one correspondence with points on the line, and are used especially to measure lengths. A real number is represented symbolically as a finite or infinite decimal, with the finite decimals being the ones ending in all zeroes. The finite decimals have another expansion ending in all 9’s, (except for the number zero = 0.000000..... which has only that one expansion).
Whole numbers or integers, are those real numbers corresponding to decimals with all zeroes after the decimal point. Rational numbers, or quotients of whole numbers, are a larger subclass of real numbers, corresponding to decimals whose expansions eventually repeat after some point (e.g. 3.4762989898......., where the repeating part is the ......98.....). A real number whose decimal expansion never repeats nor terminates in all zeroes (a form of repeating) is an “irrational” real number, since it cannot be written as a quotient of two whole numbers. (One can see by looking carefully at the result of the division process that dividing two whole numbers always results in an eventually repeating decimal. As a student I had trouble proving this, but try a few examples.)
Since calculators contain only a finite number of decimal places, they cannot represent most real numbers, and thus the answers they give to most questions involving real numbers are almost always (slightly) wrong. Since real numbers are so complicated to describe explicitly, it is not easy even to perform familiar operations on them such as adding them and multiplying them. Some form of limit process is necessary even for these simple operations to be done correctly.
But it is easy to add and multiply finite decimals, and every real number can be represented as closely as desired by a sequence of approximations by finite decimals. Thus one way to define sums of two positive real numbers is to approximate the summands by finite decimals, add the approximations, and then define the real sum as the unique number which is approximated by the sequence of approximate sums.
I.e. add the integer parts of the two summands and take this as the first approximation to the sum, then add the approximations which also include the “tenths” parts and take this as the second approximation. Then add the two approximations which include the hundredths parts and call this the third approximation. This process never stops but it gives an infinite sequence of approximations to the answer, and these approximations determine the answer.
I.e. the answer can be approximated in this way to any degree of accuracy (which is better than a calculator can do) and we can thus describe the actual sum theoretically as “the smallest real number which is not smaller than any of the approximating finite sums. Every problem in this course is solved by some such “limit”, i.e. infinite approximation, process.
 
  • #15
mathwonk
Science Advisor
Homework Helper
10,901
1,064
here is day 2:


Day 2, 2200, jan.13, 2000. Least upper bounds
Recall how to describe the real number defined by an infinite decimal in terms of the sequence of finite decimals in its expansion. We said that the real number represented by an infinite decimal, is the smallest real number which is not smaller than any of its finite decimal approximations. This is the first example of the important concept of “least upper bound”. We define an “upper bound” of a set S of real numbers as a number L which is not smaller than any number in S, and a "least upper bound" as an upper bound such that no smaller number is an upper bound.
I.e. L is a least upper bound (lub) for the set of numbers in S if and only if:
1) L is an upper bound for numbers in S, i.e. for every number x in S, x ≤ L, and
2) No number smaller than L is an upper bound for all numbers in S,
i.e. for any number M < L, there is some number x in S with x > M.
Ex: find the l.u.b. of the slopes of all lines joining points of form (1,1) and (x,x^2) where x < 1.
Solution: By the two point formula, the slope of the line joining (1,1) and (x,x^2), is (x^2-1)/(x-1) = x+1, if x < 1. Thus the least number larger than all numbers of form 1+x with x < 1, is 2. I.e. property 1) holds since if x < 1, then 1+x < 2. And property 2) holds since x < 1 implies 1+x < 2. Then if M is any number less than 2, there is an x with x < 1, and 1+x > M. I.e. M < 2 so M-1 < 1, so just take x greater than M-1, and less than 1, i.e. M-1 < x < 1. Then M < 1+x < 2, as desired.

Thus the slopes of the secant lines to the curve y = x^2, joining points (1,1) and (x,x^2) for x<1, have form 1+x, and the smallest number larger than all these slopes shopuld be the slope of the tangent line to the curve y = x^2 at the point (1,1). Namely this slope should be 2. I.e. from looking at the graph of the curve y = x^2, near the point (1,1), we see that the tangent line at (1,1) should have slope larger than the slopes of any of the secant lines joining points (x,x^2), and (1,1), but the slope of these secants becomes arbitrarily near the slopes of the tangent line. Thus the tangent line's slope should be the smallest number larger than the slope of every such secant line, namely 2. Similarly, if we take the point (a,a^2) on the curve y = x^2, the secant lines joining points (a,a^2) and (x,x^2) for x < 2, have slope a+x where x < a. Thus the lub of these slopes is 2a. Hence the slope of the tangent line to y = x^2 at (a,a^2) should be 2a.

Exercise: Show the slope of the tangent line to th curve y = x^3 at (1,1) is 3, and at (a,a^3) is 3a^2.
 
  • #16
mathwonk
Science Advisor
Homework Helper
10,901
1,064
now you know how to use differential calculus and limits to define precisely the slope of the tangent line to a parabola. Thus you have already started to learn calculus.
 
  • #17
201
0
Note that mathwonk's notes are compact and summarize whole chapters of calculus books. Therefore, it is harder to digest without visuals, examples, etc. Get your parents to buy you an introductory calculus book, or find someone in your family with a higher education and ask them if they can teach you.
 
  • #18
mathwonk
Science Advisor
Homework Helper
10,901
1,064
i tried to make my notes at most one page per lecture, so they comprise a less than 45 page differential calculus book. They are quite hard and slow to read, but i hope that someone with patience can actually learn more from those 2 pages I posted than from whole chapters of some books. But that person must accept that they should be read very slowly. Also, i am here to answer questions on them. Probably they should be taken as thought provoking starts for discussion with friends.
 
  • #19
724
0
You did supremum on your second day? That's insane.
 
  • #20
mathwonk
Science Advisor
Homework Helper
10,901
1,064
well i may be insane, but my position is that we are brainwashed into thinking certain simple ideas are hard by the way they are presented. ask a class why .9999.... is supposed to be equal to 1. the simplest way to say this i know of is to say that 1 is the smallest number larger than all the finite decimals of form .999......9. calling it the "sup" just makes it sound hard.

if we expect people to use words like "limit" we need them to begin with easier ideas. what you call "sup" is just the limit of an increasing sequence. i.e. general limits are much harder than "sup" 's. I claim that you can teach anything you actually understand, but maybe not so well if you just parrot formal definitions. of course this one page of notes is a very short version of a one hour lecture full of examples.

I have also taught euler characteristics to 2nd graders, quite successfully, not of course by defining them in terms of homology groups, but by handing out colored cardboard polyhedrons.

I also suffer from the belief that some of my students are very bright and will benefit from being told what is actually going on behind the jargon. Usually some of them prove me right, and they deserve all i can give them. i do not penalize the others for not getting it. I try to teach ambitiously and grade generously.
 
Last edited:
  • #21
724
0
I never said it was hard. I just thought it may be a little dense for the second day of calculus. :)

I like teachers who reveal the inner workings to those ready to learn them. Not many do though, for fear of confusing the crap out of those not ready

While self-studying it, I've noticed a lot of calculus stuff that I learned in Alg II. I'm taking it now, *so far*, the only difference between Alg II(HS algebra) and Calculus is that leaving off the remainder of long division of polynomials has a new name, "removing discontinuity," and it all has to have "lim" in from of it. From my self-studying, I know there's plenty to be covered that wasn't in Alg II, but I'm speaking in terms of so far.
 
Last edited:
  • #22
mathwonk
Science Advisor
Homework Helper
10,901
1,064
you seem like a sharp guy tyler, and i apologize for throwing out these 2 minute synopses of a one hour lecture, but here is day three, using least upper bounds again to prove the area formula for a circle. of course this is just reinforcement of an hour lecture with lots of pictures and entertaining singing and dancing (just kidding). sometimes kids (i.e. it happened at least once) come up after this lecture and say this is the first time they have ever understood why this formula is true. are you a high school student? you seem very advanced for that. I think I'll try tossing you some of my more creative stuff on limits, if it will copy here, but i doubt it will.


Day 3, 2200, jan.18, 2000. Area of circle as a least upper bound
Recall the definition of the “least upper bound” of a set S of real numbers, is “the smallest real number not smaller than any number in the set S”. Then 1 is the least upper bound of the infinite set of real numbers of form {.9, .99, .999, .9999, ...........}, since 1 is larger than all these numnbers, and yet these numbers eventually become larger than any number less than 1.
Indeed this is what it means to say the infinite decimal .9999999..... represents (i.e. equals) the number 1, since the real number represented by an infinite decimal is the least upper bound of the set of all finite decimals obtained by truncating the infinite decimal at all finite stages.
E.g. the real number represented by the infinite decimal .333333....., is by definition the least upper bound of the sequence of finite decimals {.3, .33, .333, .3333, .......}, and equals 1/3.
Recall that area is measured in “square units” and that the area of a plane region is intuitively the number of unit squares that will fit inside the region. We make sense of the area of regions such as rectangles, parallelograms, triangles, and polygons, by cutting up a certain number of unit squares and reassembling them to fit inside the region.
Then consider a region such as a circle, where no amount of cutting and reassembling can ever make a finite number of unit squares fit perfectly inside the region. One way to make sense of the concept of area here is to approximate the area of the circle by the areas of an infinite number of simpler regions for which we understand area better, such as inscribed polygons. We define the area of a circle as the least upper bound of the areas of all (convex) polygons which can be inscribed in the circle.

We will use the Pythagorean “distance formula” to measure the length of a line segment and we define the length of a circle to be the least upper bound of the lengths (i.e. perimeters) of all polygons which can be inscribed in the circle. We define the number π to be the ratio π = C/2r of the circumference to twice the radius of any circle, (accepting that this ratio is the same for all circles). We argue that the area of a circle of radius r is πr^2 as follows: the area of the circle is the lub of the areas of all inscribed polygons. Looking only at regular polygons, i.e. ones with all sides the same length, seems acceptable. Then we see by dividing the polygon up into triangles that the area of the polygon is half the product of the perimeter of the polygon times the common height of each triangle, (where that common height is the distance from the center of the polygon to the center of any one edge of the polygon).

We agree the lub of this product of increasing factors, is the product of the lub’s of the factors. Thus the area of the circle is 1/2 the product of the lub of the heights of the triangles times the lub of the perimeters of the polygons. But the lub of the heights of the triangles is the radius of the circle, and the lub of the perimeters of the polygons is the circumference of the circle. Thus the lub of the areas of the inscribed polygons is 1/2 the product of the radius of the circle times its circumference, i.e. A = (1/2)(2πr)(r) = πr^2.
 
  • #23
mathwonk
Science Advisor
Homework Helper
10,901
1,064
here is a detailed actual lecture from the beginning of the theory of area.
 

Attachments

  • #24
724
0
Thanks. I'm going to keep a local copy. :)

Yeah, I'm a junior in high school. I took Pre-Calulus last semester(with a 100, I might add :P), and am taking Calculus at the local community college this semester.

Recall the definition of the “least upper bound” of a set S of real numbers, is “the smallest real number not smaller than any number in the set S”. Then 1 is the least upper bound of the infinite set of real numbers of form {.9, .99, .999, .9999, ...........}, since 1 is larger than all these numnbers, and yet these numbers eventually become larger than any number less than 1.
I'm going to have to call circular argument here. You are assuming 1 <= .999... to prove 1 = .999... :)

Instead, take a simple property of the reals: There are infinitely many members between any two members. What's a number between .999... and 1? There is none. Therefore, by the contrapositive of the aforementioned property, .999... = 1.

EDIT: I just noticed you never explicitly said that it was a proof .999... = 1. But, I get the feeling it was implied, no?
 
  • #25
201
0
What happened to the simple old 1=(1/3)*3=(.333...)*3=.999...?
 

Related Threads on Any good calculus places to start?

  • Last Post
Replies
10
Views
3K
Replies
5
Views
2K
  • Last Post
Replies
5
Views
2K
Replies
5
Views
276
Replies
4
Views
2K
Replies
6
Views
1K
Replies
3
Views
7K
  • Last Post
Replies
20
Views
5K
  • Last Post
Replies
9
Views
3K
Top