I never learned calculus in high school and didn't go to college. I'm completely allergic to institutional learning.

Yeah, I know, not having the piece of paper doesn't impress HR departments, but that's not really what I wanted out of life (also, I'm doing pretty well regardless).

I tried to learn calculus from a text book, but they're pretty awful books in general, and particularly bad as far as being tutorials.

I guess I need to encounter a problem that I can't easily solve with conventional math. What are some good, illustrative ones?

Physics. Calculus was invented to solve physics problems. Time, distance, acceleration. Pick some seemingly basic physics problems and try to solve them. If you find them truly interesting, that might motivate you to learn calculus.

Richard Feynman used this book to teach himself calculus:

http://www.amazon.com/exec/obidos/tg/detail/-/0312185480/002-2450025-6129651?v=glance

Completely offtopic, but I caught "Hit Me Baby One More Time" on Much More Music tonight (it's an awesome British show where one-hit retro wonders play their big hit, and also a cover of a new song). Anyways Tommy Two Tone, of 867-5309 fame, was one of the contestants, and they do a little "where are they now". The guy is a flipping GUI designer (or possibly even developer) for the financial industry in Portland.

I don't know why, but I find that very disturbing.

8-6-7-5---3-0-9...eeeee.....ine.....

That's a British show? I know the Emcee was British, but that show aired here months ago. It must have aired there at the same time because there was audience voting.

You're probably right - the voice of it was a Brit so I just presumed, but now that I think about it everything else seemed American.

Out of all of the crap about Richard Feynman I've read, how in the world did I miss the book that was supposed to have been his first calculus book?

Argh. I wish I had this book 10 years ago in high school.

Ordered from half.com for $1.29.

Physics, Aaron? How about to model a savings account?

Lets say I have 1000/year to invest into a savings account that earns 5% return for 10 years. I get as far as:

Let i = interest rate (1.05)
Let x = yearly investment (1000)

p = (((((((((x*i+x)*i+x)*i+x)*i+x)*i+x)*i+x)*i+x)*i+x)*i+x)

There's got to be a way of representing this as a constant time equation. I just have no idea how to do it.

Unless I'm completely off my rocker, I need to figure out the instantaneous rate of change for one year and simply multiply by number of years.

Totally at a loss for how to get there.

If new amounts are invested yearly and interest is compounded yearly, as you have written it, it is an algebraic problem of summing arithmetic and geometric series.

However, if interest were to be compounded continuously (or say, every day), then it would become a problem of calculus.

Calculus happens when the terms to be summed become negligibly small and huge in number compared to the whole.

To expand on the series part of it:

In the first year you invest x, and interest is compounded over the next ten years, so the final growth of that initial x is

x * i^10

where i^10 is 1.05 to the power of 10.

The second year, you invest another x, and this grows to x * i^9.
The third year, x * i^8.
And so on.

Adding them all up and turning it round you have:

Total = x * i + x * i^2 + x * i^3 + ... + x * i^10

This summation is a geometric progression of the form

S = a + ar + ar^2 + ar^3 + ... + ar^(N-1)

where a = x * i, r = i and N = 10.

There is a formula for the sum of a geometric progression that allows S to be computed for any values of a, r and N.

I'll come to that next if you are following so far, and if nobody points out any errors I might have made up to this point.

To find the sum of a geometric progression:

Let S = a + ar + ar^2 + ... + ar^(N-1)

Muliply S by r and write the two series underneath each other

S = a + ar + ar^2 + ... + ar^(N-1)
Sr =    ar + ar^2 + ... + ar^(N-1) + ar^N

Subtract one from the other, and all the middle terms cancel out:

S-Sr = a - ar^N

or

S(1-r) = a(1-r^N)

so

S = a (1 - r^N) / (1 - r)

and that is the formula for the sum of a geometric progression.

One thing about maths that puzzled me is the Fundamental Theorem of 'X'.

What makes a theorem fundamental?

Back on topic, I've not read any of the five books recommended here:
http://books.slashdot.org/books/04/03/04/028253.shtml

Buyer beware :) (Or downloader...)

And from a comment to that /. article http://mathworld.wolfram.com/

"One thing about maths that puzzled me is the Fundamental Theorem of 'X'. What makes a theorem fundamental?"

A "Fundamental Theorem" forms the basis for a whole branch of mathematics.

Mathematics consists of a collection of theorems, many of which are small steps from one to the other. But sometimes there is a big leap to make, where a lot of concepts come together to create something new.

Like the "Fundamental Theorem of Calculus", which states that integration (calculating the area under a function) is the inverse of differentiation (calculating the slope of a function).

This historically connected a new concept, speed, to a classical mathematical concept, area, through an algebraic method of formulamanipulation, and boosted the analysis of many physical and other dynamic problems.

To give a rigid proof of the Fundamental Theorem of Calculus you need to construct difficult notions of collections of division and limits to infinity, but once proven, you can use it to proceed relatively easy into many directions.

Most specialized fields of mathematics have their own Fundamental Theorem, which starts with a definition of a mathematical object and then it proves that the object has a certain basic non trivial property.

Hi Ian,

I expect that one day I might have figured out everything you described, except for this part.


Muliply S by r and write the two series underneath each other

S = a + ar + ar^2 + ... + ar^(N-1)
Sr =  ar + ar^2 + ... + ar^(N-1) + ar^N


What's happening here? Why multiply by r at all and then subtract? Why do you add that final ar^N on the second line?

Huh?

I recommend "Forgotten Calculus" and "Forgotten Algebra". I loved the teaching style of both books.

http://www.amazon.com/exec/obidos/tg/detail/-/0764119982/002-1371354-5321620?v=glance

"What's happening here? Why multiply by r at all and then subtract? Why do you add that final ar^N on the second line?"

The final ar^N comes from multiplying the ar^(N-1) on the line above by r -- note the "missing" a from the bottom sequence, which is now ar.

Muliplying by r and then subtracting gives you:

S - Sr = a + ar - ar + ar^2 - ar^2 + ... + ar^(N-1) - ar^(N-1) + ar^N

So most of the terms cancel out, leaving you with

(S - Sr) = a - ar^N

which is a lot easier to figure out, especially if N is some enormous figure; just pump the numbers in once rather than performing lots of calculations.

Er, that last plus in the expanded add/subtract sequence should obviously be minus. :)

I taught myself calculus with a book entitled, appropriately, Teach Yourself Calculus:
http://www.amazon.com/exec/obidos/tg/detail/-/0071421289/qid=1125156906/sr=8-1/ref=pd_bbs_1/002-6861986-6660856?v=glance&s=books&n=507846

Amazon offers it in a bundle with Teach Yourself Trigonometry, which I suggest getting if your trigonometry is rusty.

Incidentally, I also recommend Teach Yourself Ancient Greek, from the same series, for anyone so inclined.

Hi Michael,

As to why multiply by r at all and then subtract, it is really just a flash of insight, a cunning ploy. If you do it, you get nearly all of the terms to vanish when you do the subtraction. End result, you get a shortcut to the answer, even if there should be a million terms to add up in the original sequence.

In case the algebra is not obvious, consider that x squared times x is x cubed, and that x cubed times x is x to the fourth power. In general, x to the (N-1)th power times x is x to the Nth power. In the case of the sequence:

  a * r = ar
  ar * r = ar^2
ar^2 * r = ar^3
...
ar^(N-1) * r = ar^N

About cunning ploys, consider the old favorite of how do you add up all the numbers from 1 to 1000 without adding them up?

Write the answer S as

S = 1 + 2 + 3 + 4 + 5 + ... + 999 + 1000

Turn the sequence around (cunning ploy) and write it underneath

S =  1 +  2 +  3 +  4 +  5 + ... + 999 + 1000
S = 1000 + 999 + 998 + 997 + 996 + ... +  2 +  1

Adding the two sequences we have

S + S = 2S = 1001 + 1001 + 1001 + 1001 + 1001 + ... + 1001 + 1001

or

2S = 1000 * 1001

or

S = 1000 * 1001 / 2

Expressed in words, this is the old favourite: "Add the first number to the last number, divide by two, and multiply by however many numbers there are."

My columns didn't line up properly in that last post. Here is my experiment at getting it right:

  S =    1 +    2 +    3 +    4 +    5 + ... +  999 + 1000
  S = 1000 +  999 + 998 +  997 +  996 + ... +    2 +    1

2S = 1001 + 1001 + 1001 + 1001 + 1001 + ... + 1001 + 1001

Success! (of a sort)

Wow.

Math is magical.

Some sort of working PRE tag would be great here. These are technical support forums, aren't they?

You might like 'A tour of the Calculus' by David Berlinski. The book is easy to read. Berlinksi is passionate about the subject and it shows in all the pages. He intersperses the appendixes, which has proofs, and writes them too. It's not a dry appendix at the end of the book which we never get to. In one of them he says you might want to skip the proof and go to the fun part, which is the succeeding chapter, but tells us that the proof is the real fun part and goes on explaining it. The language is poetic throughout the book. If you like it you will doubly enjoy the book. If not that part may be a little irking. It won't teach Calculus but will surely help a lot in understanding how Calculus came about and why.