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Calculus Made Easy (1910)

calculusmadeeasy.org

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Re: Calculus Made Easy (1910)

#71

Post by zonethundery »

dopkew wrote: post #69
dopkew wrote: post #68

In chapter 2: > Let us think of x as a quantity that can grow by a small amount so as to become x+dx, where dx is the small increment added by growth. The square of this is x2+2x⋅dx+(dx)^2. The second term is not negligible because it is a first-order quantity; while the third term is of the second order of smallness, being a bit of, a bit of x^2. It seems to me that the third term is actually a bit of a bit of x, ra…

And in chapter 2: >Now if, for such a purpose, we regard 1/1,000,000 (or one millionth) as a small quantity, then 1/1,000,000 of 1/1,000,000, that is 1/1,000,000,000,000 (or one billionth) .. 1/1,000,000,000,000 is actually one trillionth

From wikipedia[0]: A billion is a number with two distinct definitions:

1,000,000,000, i.e. one thousand million, or 109 (ten to the ninth power), as defined on the short scale. This is now the meaning in both British and American English.

Historically, in British English, 1,000,000,000,000, i.e. one million million, or 1012 (ten to the twelfth power), as defined on the long scale. This is one thousand times larger than the short scale billion, and equivalent to the short scale trillion.

[0] https://en.wikipedia.org/wiki/Billion

Re: Calculus Made Easy (1910)

#72

Post by fesoliveira »

mettamage wrote: post #62

I notice that I stumble over math over small but important details. I understand the big ideas, but then at chapter 4 in the book it says: y+dy = (x+dx)^-2 is equal to x^−2 * (1 + dx/x)^−2 [1] To me (not that strong at math) this isn't apparent at all. I have a couple of options here: 1. Spend a couple of hours fiddling around and trying to figure out the answer. 2. Hopefully find some app. 3. Ask a friend. Regarding…

From your other post, you might have been a bit tired due to sleep deprivation, so don't be too hard on yourself for not getting it ;)

As for the expression, you just have to expand it:

  y+dy = (x+dx)^-2 
       = 1 / (x+dx)^2 
       = 1 / (x^2 + 2xdx + d^2x) 
       = 1 / [x^2 * (1 + 2dx/x + d^2x / x^2)] 
       = 1 / [x^2 * (1 + dx/x)^2]
       = [x^2 * (1 + dx/x)^2]^-1
       = x^-2 * (1 + dx/x)^-2
It looks like a bit of a jump at first, but he just skipped the expansion of the expression. When I see this kind of thing, it helps me to just mess around with both start and end to see if I can find a way to get from one to the other.

Re: Calculus Made Easy (1910)

#73

Post by [deleted] »

userbinator wrote: post #53
qwerty456127 wrote: post #30

Earlier quoted context omitted.

Indeed. They never explained this at school and we were just memorizing and applying formulae having no idea of the meaning. So many years have passed an I've only realized what does the d actually mean some weeks ago and now I see this book explaining it this easy at the very beginning written in 1910!

I've seen a similar effect in various other topics too, and propose the saying "the closer a subject is to its infancy, the clearer it will be taught". Two examples of this come to mind. The first is in the automotive industry, with early videos such as https://news.ycombinator.com/item?id=15122031 (and older discussion at https://news.ycombinator.com/item?id=8513209 ) as well as the detailed yet straightforward expl…

[deleted]

Re: Calculus Made Easy (1910)

#74

Post by seanjamz »

xelxebar wrote: post #61
aeneasmackenzie wrote: post #23

I like Calculus Made Easy because it uses informal infinitesimals. You can make these fully rigorous if you want and they're a much more intuitive technique than epsilon-delta.

Is it really all that intuitive, though? I mean, where does (dx)^2 = 0 come from?? Usually people say that, well, since dx is already small, then (dx)^2 is really really small, so for magical reasons it's okay to pretend that it's zero. I mean, if we're willy-nilly ignoring small things, why can't we ignore the already "infinitely small" dx? Personally, I always found hand-waving such an infinitesimal explanation to…

One thing that's problematic with this approach is the assumption that dx is a small constant. Its not, it represents a limit, specifically a value approaching 0. Look at the quantity (x + dx)^2. By expanding the terms you get x^2 + 2xdx + dx^2. Look at the last two terms, which both involve a dx. Lets look at how these compare to each other by putting the last term over the middle term so we get (dx^2 / 2xdx). Since we are in a limit, consider the value of this as dx approaches 0. You can cancel one of the dx's so you have (dx / 2x), and now you can clearly see that this limit will be 0. What this last limit shows, is that the last term is infinitely smaller than the middle term as dx shrinks. Which is why it gets "pretended" to be 0 in some math and physics classes. This is no approximation though, and can be carried along in your calculations if you choose to keep it.

Re: Calculus Made Easy (1910)

#76

Post by perl4ever »

qwerty456127 wrote: post #44
romwell wrote: post #35

Earlier quoted context omitted.

Hey, there's more! First, why "d"? Well, "d" is for "difference". As in: as x changes from x_1 to x_2, the difference (x_2 - x_1) -- when it's very small. But wait, there's more! The commonly used symbols for finite difference like that is the Greek letter Delta: Δ For a list of values x_1, x_2, x_3, x_4,.. we write Δx_i = (x_i - x_{i-1}). That is, Δx_i is the i'th change . (Side note: an airline had a marketing slog…

Do you realize this comment of yours is worth more than years of studying (high school + Bc.)? Perhaps schools are better at your location but the ones I've attended never explained this, just write that down, this equals that, memorize and f-ck you. Thanks G-d we have the Internet and people like you nowadays...

I remember when I had calculus in school and the textbook was a huge book weighing like five pounds...and my mom's calculus textbook from the 1950s was this tiny little thin book. But they hadn't added anything to the subject of calculus since then!
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