The ancients mastered synthetic (axiomatic, pure) geometry, which is performed with a compass and a straight-edge. A straight-edge with regularly spaced markings is called a ruler, but synthetic geometry does not need any marking, just the ability to draw straight lines as long as you wish.
From Descartes, geometry is also approached from an algebraic point of view, by using formulae and (usually Cartesian) coordinates. This is called analytic geometry.
Both are very powerful and should be learned by any mathematician and physicist. Newton seems to have no favourite choice between them, since he masters and greatly develops both of them.
Newton call species to what we call polynomials.
A species and a series is the same thing. An infinite series is the same as an infinite polynomial.
A decimal expression of a number can also be thought as a species.
For example, 1.2345 can be expressed as:
1.2345 = 1·X⁰ + 2·X⁻¹ + 3·X⁻² + 4·X⁻³ + 5·X⁻⁴ ,
where X = 10.
So the same approach you have with decimal numbers to add, subtract, multiply, divide and extract roots you can apply it to species and infinite series.
Before entering into the contents of Newton's book, we should review some important concepts.
Let's have the number x=0.13131313...
So now we multiply by 100 to get
100·x = 13.131313... = 13 + x .
This is a great application of the infinity idea. Even if we multiply by 100, the infinitely repeating decimals go on forever, so we obtain x again added to 13. Then,
99·x = 13 => x = 13/99 .
A little more difficult example, with repeating decimals appearing after some non-repeating decimals have appeared:
x = 21.25613131313 = 21.256 + 0.000131313...
Multiply by 1000 so that we get 0.131313 alone. We always want the repeating decimals to begin at the first decimal position.
1000·x = 21256 + 0.131313...
Then solve the 0.131313 as before ---> 13/99.
And then
1000·x = 21256 + 13/99 .
x = 21256/1000 + 13/99000 .
x = (21256·99 + 13) / 99000 .
x = 2104357 / 99000 .
Instead of the word base, in this book you may find the word "scale".
When we say we work in base 10 or in scale 10, 10 is the *root* of the scale or base.
We are TOO used to base 10. We use it because (most of us) have 10 fingers in our hands. But this is by no means the only possible base, neither the most clever. We are also used to see chunks of binary data, with 0's and 1's. See how for decimal base we need ten symbols: 0,1,2,3,4,5,6,7,8,9 and for binary (base-2) we need just two symbols: 0,1.
For base-7 we need 0,1,2,3,4,5,6 and for base 16 we need 0,1,2,3,4,5,6,7,8,9... but wait, we need 6 more symbols. Do we invent new number symbols? No, we just use letters, so we continue with A,B,C,D,E,F. In base 50 you may run out of usual letter symbols as well.
When we multiply two numbers, like 89·174=15486, it is assumed everything is in base 10. Not a difficult operation, but requires memorising (in elementary school) many rules, the multiplication tables. Memory is a great thing to exercise, but there is a much clever way to deal with numbers than the decimal system.
We could also multiply 59·AE=3C7E, which is the same, in base 16 (hexadecimal), as before. But you could say: not happy with the multiplication tables for base 10, should we learn them also with 16 symbols? Not at all, because 16 is a power of 2, and all bases which are powers of 2 are way more clever than bases that are not powers of 2, like base 10.
Let's show how we could learn in school in an ideal world.
First, we need to expand each hex *digit* into a binary number, by having the easy translation:
+--------+-------------+---------------------------+ | binary | hexadecimal | how to know it: | +--------+-------------+---------------------------+ | 0000 | 0 | just 0 | +--------+-------------+---------------------------+ | 0001 | 1 | 2⁰ | +--------+-------------+---------------------------+ | 0010 | 2 | power of 2: 2¹ | +--------+-------------+---------------------------+ | 0011 | 3 | 1 + 2 | +--------+-------------+---------------------------+ | 0100 | 4 | power of 2: 2² | +--------+-------------+---------------------------+ | 0101 | 5 | 4 + 1 | +--------+-------------+---------------------------+ | 0110 | 6 | 4 + 2 | +--------+-------------+---------------------------+ | 0111 | 7 | 4 + 2 + 1 | +--------+-------------+---------------------------+ | 1000 | 8 | power of 2: 2³ | +--------+-------------+---------------------------+ | 1001 | 9 | 8 + 1 | +--------+-------------+---------------------------+ | 1010 | A | 8 + 2 | +--------+-------------+---------------------------+ | 1011 | B | 8 + 2 + 1 | +--------+-------------+---------------------------+ | 1100 | C | 8 + 4 | +--------+-------------+---------------------------+ | 1101 | D | 8 + 4 + 1 | +--------+-------------+---------------------------+ | 1110 | E | 8 + 4 + 2 | +--------+-------------+---------------------------+ | 1111 | F | trivial (last one) | +--------+-------------+---------------------------+
But ideally, when you are used to work with them, you see the binary chunk and already identify it with its hex digit. For example, you see 1100 and already say C. The third column, as you practise more and more, becomes less and less relevant.
We want to multiply 59·AE, so we immediately we translate each hex digit into a 4 digit binary chunk. Instead of 5 we must see 0101. Instead of 9 we see 1001. Instead of A we see 1010. Instead of E we see 1110. Writing one on top on the other we get:
01011001 10101110
Now we apply the multiplication rules. Look how simple they become!
1·0 = 0·1 = 0·0 = 0 1·1 = 1 .
Also the addition rules:
0 + 0 = 0 1 + 0 = 0 + 1 = 1 1 + 1 = 0 (carry 1) .
Armed with such simple rules, let's multiply these two numbers:
01011001
x 10101110
_________________
00000000
01011001
01011001
01011001 notice how it is either a chain of 0
00000000 or a copy of the top number
01011001 1+1 = 0, carry 1 on next
00000000 1+1+1 = 1 and carry 1 on next
01011001 1+1+1+1 = 0 and carry 1 two places away
|| | || 1+1+1+1+1 = 1 --> 2 places away
+_____|__________ 1+1+1+1+1+1 = 0 --> 3 places away
useful to place sticks for carried 1's
11110001111110
The final number is 0011 1100 0111 1110, which translated back to hex digits is = 3C7E.
Of course, you may be tempted to translate 3C7E to base 10. How do we do this? We know the result must be 15486.
It is useful to write 15486 in explicit decimal powers, because a number with many digits is in fact the abbreviation of a species!
15486 = 1·X⁴ + 5·X³ + 4·X² + 8·X¹ + 6·X⁰ ,
where X is the (Roman) 10. On the other hand, 3C7E (in base 16) can be expanded as a species as well:
3C7E = 3·H³ + C·H² + 7·H¹ + E·H⁰ ,
where, as X was 10 in base 10, H is ... 16? Be careful, since 16 is a number written in base 10. In base 10, growing numbers from 0 are 1,2,3,4,5,6,7,8,9 and once we run out of symbols, we start with two digit numbers: 10,11,12,13,14, etc. See how the last digit runs the same cycle again, only to be accompanied by a first digit that is now 1.
In other bases we do the same. In base 3, we count 0,1,2 and then we run out of symbols, so we continue with 10,11,12 and then 20,21,22, etc. In base 16, 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F and ... 10. The first two digit symbol is always 10.
So H means 10 in base 16.
But we want to be able to convert from base 16 to base 10. So a direct way would be to substitute H by 16, C by 12 and E by 14:
3·16³ + 12·16² + 7·16¹ + 14·16⁰ = 15486.
And thus we recover the base 10 expression. What if we know want to go back to base 16 (on another base, in fact)? We write again the species for the number in base 10:
15486 = 1·X⁴ + 4·X³ + 4·X² + 8·X¹ + 6·X⁰ ,
We could again use a brute force to express X in hex terms, where X = A, and do
1·A⁴ + 4·A³ + 4·A² + 8·A¹ + 6·A⁰ ,
which should be enough. But then you would need to perform all these powers, multiplications and additions. There must be a better way.
Begin by computing the relevant powers: 16⁰ = 1, 16¹ = 16, 16² = 256, 16³ = 4096, 16⁴ = no need, since it is bigger than our number. Then, our highest power is 16³ = 4096. We divide 15486 / 4096 but only to extract the integer part, which is much easier. You can rapidly see how 4 is too much, and that 3 is our number. This means that 3 is our first digit, so we can start building
15486 = 3·16³ + ...
Now we subtract 15486 - 3·16³ = 3198. We divide this by the next power, 16² = 256, seeking only the integer. We get 12, so we update our result to
15486 = 3·16³ + 12·16² ...
We subtract 12·16² and get 126. Time to divide this by 16 and get the integer part, which is 7. We update to
15486 = 3·16³ + 12·16² + 7·16¹ + ...
We subtract 7·16 and get 14. And clearly, 14 divided by 16⁰ is just 14, so our final update is
15486 = 3·16³ + 12·16² + 7·16¹ + 14·16⁰ .
Now it is only a matter to rewrite the powers of 16 as powers of H and the coefficients in hex notation: 3 is still 3, and 7 is still 7, but 12 is C and 14 is E, so
3·H³ + C·H² + 7·H¹ + E·16⁰ = 3C7E.
Bases are fascinating. For example, we could write a number in base π if we wished. What would 2 be like in base π? We would explore powers of π first. We see that π¹ is already too big for 2, so we explore lower powers, like π⁰ =1, or π⁻¹ ≈ 0.31831, and so on. So our first division is 2 divided by π⁰, which gives 2. This means that our first digit is 2. And subtracting 2 from 2 gives 0, so we are done. Conclusion, 2 in base π is written as 2. And π in base π would be.. you already know this: just 10.
Another fascinating aspect is how the base is related to the periodicity of decimals. We know that in base 10, 1/3 can be written as 0.3333... But what if we rewrite 1/3 to another base? Will we get periodic decimals in the new base? Let's try base 9.
First, we see how our highest power of 9 must be 9⁻¹, so we must divide 1/3 by 1/9, which gives us 3. We begin building our number as
3·9⁻¹ + ...
Then we must subtract this from our number: 0.3333... - 3·9⁻¹ = 0. This means we are finished, so 1/3 in base 9 is written just as 0.3. No infinite decimals, just a single one, and not repeating itself!
What would happen in base 8? The first division gives an integer part of 2, which is our first digit.
Then, 1/3 - 2·8⁻¹ = 0.08333... so we need to divide this number by 8⁻², which has 5 as integer part. For now, our number is 0.25, but we must continue. Subtracting we get 0.005208333..., and this divided by 8⁻³ gives an integer part of 2, which is our next digit, so our number is for now 0.252. Next step gives again 5, so that we learn that 1/3 in base 8 is 0.252525...
In conclusion, a (rational) number showing periodic decimals in one base can exhibit a finite number of decimals in another base, and vice versa.
There is a lot more to be discovered about bases. For example:
When expressing a number as a series, there are two problems we can have.
First, we can know the coefficients and the root of the series but ignoring the *aggregate*, i.e. the result of the sum of all terms.
1/1² + 1/2² + 1/3² + 1/4² + ... = ?
Second, we can know the result of the sum (the aggregate) and the coefficients but ignore the root:
5x⁴ + 3x³ + 7x = 53070 .
Here 53070 is the aggregate, the coefficients are 5, 3, 7 and the root is x.
So, solving a species (a polynomial) is just a problem of finding the base in which a number is written!
Imagine we have the equation
4x⁴ + 2x³ + 3x² + 0x¹ + 2x⁰ + 4x⁻¹ + 3x⁻² = 2827.92 .
So looking for the result of the equation, x, is looking in which base the number 2827.92 (base 10) appears as 42302.43. Solving the equation we find x=5.
So far we have only used integer powers of the root. For example 23 = 2·X¹ + 3·X⁰ where X = 10.
But couldn't we directly express whatever number (like 23) not as a sum of integer powers of X but as a single term?
This would require 23 = Xⁿ. If we can solve for n, we would be able to represent 23 in base X (10) with a single term.
In this example, n = logarithm in base 10 of 23 = log₁₀(23) ≈ 1.36173... (How to calculate this will be seen much later.)
Isn't this fantastic? Logarithms are justified as single term expansions of a number given a root.
See how digits of numbers is a series representation. See how base/scale is the root of a series. See how logarithms are the exponents of a single term series. Suddenly, everything seems to be a series!
Let a + bc + cx² + dx³ + ex⁴ + &c be a series, finite or infinite. Let (p/q + x) be a factor to which multiply the previous series. We multiply and set (p/q + x) =0, so
ap/q + (bp+aq)·x/q + (cp+bq)·x²/q + (dp+cq)·x³/q + &c = 0 .
As we know that x = -p/q makes the factor (p/q + x) = 0, we can substitute to get
ap/q - (bp+aq)·p/q² + (cp+bq)·p²/q³ - (dp+cq)·p³/q⁴ + ... = 0 .
We can multiply all by q/p and rewrite as
a = (bp+aq)/q - (cp+bq)·p/q² + (dp+cq)·p²/q³ - ...
You can think "what a weird way to build a series!". And it is true. Why choosing this method instead of another? The answer is that there are many methods of building a series, but not for all of them we are able to calculate the supplement. Continue reading to see why this form is special.
Now, it is important to see that this series is always equal to a. Imagine we take two terms to the series: then bp/q + a - cp²/q² - bp/q = a - cp²/q². If we add the third term, we get a - cp²/q² + dp³/q³ + cp²/q² = a + dp³/q³. And so on. Each new term will cancel part of the previous, in such a way that we always get a plus/minus something, where this something becomes smaller and smaller (if p/q < 1). So it is clear that the aggregate of the series is a.
If we stop at whatever number of steps, as we have said, we will get a ± something, or just a + something if we allow the something to be whatever sign.
The negative version of this 'something' is called the *supplement* of the series. So the form of this series is designed as to be able to calculate the supplement. In other words, the supplement is what you need to add to get the exact aggregate of the series.
If the series is diverging, the supplement will grow. If the series is converging, the supplement will decrease.
So, a partial sum plus the supplement always give the exact aggregate. Also, if the series is convergent, we may neglect the supplement at some point and assume the partial sum as a good approximation of the aggregate.
We need now to keep our promise and calculate the supplement. If we take 1 term, the partial sum is a+bp/q, so the supplement is -bp/q. If we take 2 terms, the partial sum is a - cp²/q², so the supplement is +cp²/q². If we take three terms, the supplement is -dp³/q³. It is easy to see that for n terms, the supplement is (-1)ⁿ·l·(p/q)ⁿ, where l is the (n+1)th coefficient in the (a+bx+cx²+...) series.
Although the method seems quite abstract, consider for example a=1, b=2, c=3 and so on. And p=q=1. Then
a = (bp+aq)/q - (cp+bq)·p/q² + (dp+cq)·p²/q³ - ...
becomes
1 = (2+1) - (3+2) + (4+3) - ... = 3 - 5 + 7 - 9 + 11 - ...
See how this series gives (if grouping every pair of terms), -2 -2 -2 ... which seems to diverge!
Let's choose now a=1, b=1/2, c=1/3, d=1/4, &c, with p=q=1. Then we get
1 = (1+1/2) - (1/2+1/3) + (2/3+1/4) - ... = = 3/2 + 5/6 + 7/12 + 9/20 + ...
Notice how the numerators grow like 3, 5, 7, 9, 11, &c while denominators grow by adding 4, 6, 8, 10, &c. How non-trivial is that the result of such series gives 1!
Another method to build a series in which is the supplement is known is to consider two progressions, A,B,C,D, &c and a,b,c,d, &c. The idea is to express A as the result of a series.
We begin by assuming that A is equal to a plus a supplement, so we are building the supplement ad hoc! A = a + p, where p is the supplement. Then, it is clear that p = (A-a).
In order to build a series from it, instead of adding the whole supplement p (a thing that would stop the series), we add only a part of it. The fraction that we will take of p is given by b/B. So we call add as second term in the series just (b/B)·p = (b/B)·(1-a/A)·A, with the supplement being now q = (1-b/B)p.
The process continues for as many steps as we wish.
Another way to see this is by beginning with a tautology, and then start inflating it. Notice how each time we complicate the expression instead of simplifying it:
A = A
A = a + A - a = a + A(1-a/A) = a + p, with supplement p = A(1-a/A).
A = a + p·(1 + b/B - b/B) = a + A(1-a/A)b/B + A(1-a/A)(1-b/B)
|___p___| |_______q______|
A = a + A(1-a/A)b/B + A(1-a/A)(1-b/B)c/C + A(1-a/A)(1-b/B)(1-c/C).
|___p__| |_____q_______| |________r___________|
The supplements are p, q , r, &c.
Let's see the power of such method with some examples. Take A=B=C=...=2 and a=b=c=...=1. Then we get
2 = 1 + 1/2 + 1/2² + 1/2³ + &c .
To me, it's riveting how by such simplistic construction we end up obtaining such a wonderful result. And this is just one out of infinite possibilities!
We will present another method to build a series that will offer almost miraculous results.
We propose two series that are going to be added to give 1:
1 - a/A - b/B - c/C - d/D - e/E - &c
+ + a/A + b/B + c/C + d/D + e/E + &c
----------------------------------------------------
1 = (A+a)/A + (Ab-Ba)/(AB) + (Bc-Cb)/(BC) + (Cd-Dc)/(CD) + &c.
Instead of adding vertically, which trivially would cancel everything except 1, just add *obliquely*. For example, 1+a/A, then -a/A + b/B, and so on.
The method is as silly as it gets, but the results are as powerful as it can become.
Make A=2, B=3, C=4, D=5, &c and a=b=c=---=1 and see how we get
1 = 3/2 - 1/(2·3) - 1/(3·4) - 1/(4·5) - 1/(5·6) ,
which can be rewritten as
1/2 = 1/(2·3) + 1/(3·4) + 1/(4·5) + 1/(5·6) + &c.
Imagine you were presented with such a sum. We would never guess that the aggregate could be such an elegant 1/2.
Let's bring yet another "silly" method of building a series with 1 as a result. Quite similar to the previous one.
1 + a/A + b/B + c/C + d/D + e/E + &c
+ - a/A - b/B - c/C + d/D - e/E - &c
----------------------------------------------------
1 = (A-a)/A + a(B-b)/(AB) + ab(C-c)/(ABC) + abc(D-d)/(ABCD) + &c.
And now let's explore one out of its infinite possibilities: A=B=C=...=3 and a=b=c=...=1. We get
1/2 = 1/3 + 1/3² + 1/3³ + 1/3⁴ + &c.
Or even better, do A=2, B=3, C=4, &c and a=b=c=...=1. Then we obtain
1 = 1/2 + 2/(2·3) + 3/(2·3·4) + 4/(2·3·4·5) + &c .
Do you recognize these denominators? They are factorials! We can write
1 = 1/2! + 2/3! + 3/4! + 4/5! + &c.
Isn't this amazing? Welcome to the world of series, probably the most powerful tools of all mathematics.