Let's follow Newton dividing a²/(b+x) in the same manner we have proceeded before. We consider that in (b+x) it is b what dominates, being x comparatively small.
a²/b - a²x/b² + a²x²/b³ - a²x³/b⁴ + &c
_______________________________________
b+x ) a²
a² + a²x/b
-____________
0 - a²x/b
- a²x/b -a²x²/b²
-__________________
0 + a²x²/b²
a²x²/b² + a²x³/b³
-__________________
0 - a²x³/b³ --> remainder after 3 terms.
Conclusion: for b > x we have a²/(b+x) = a²/b - a²x/b² + a²x²/b³ - a²x³/b⁴ + &c. I don't think a² in the numerator was necessary at all. We could get rid of it, and also multiply all by b so that we get x/b < 1 as a single variable:
1/(1+x/b) = 1 - x/b + (x/b)² - (x/b)³ + (x/b)⁴ - &c.
What if is x who dominates over b? Then we take the division 1/(1+b/x) and we get
1/(1+b/x) = 1 - b/x + (b/x)² - (b/x)³ + (b/x)⁴ - &c.
Newton proposes to divide 1/(1+x²) now. We consider x² < 1.
1 - x² + x⁴ - x⁶ + &c
________________________
1+x² ) 1
1 + x²
-______
0 - x²
- x² - x⁴
-___________
0 + x⁴
x⁴ + x⁶
-___________
0 - x⁶ --> remainder after 3 terms.
If x² dominates over 1, then
1/x² - 1/x⁴ + 1/x⁶ - &c
__________________________
x²+1 ) 1
1 + 1/x²
-______
0 - 1/x²
- 1/x² - 1/x⁴
-_____________
0 + 1/x⁶ ---> remainder after 2 terms.
We are proposed (2x¹⸍² - x³⸍²) / (1 + x¹⸍² - 3x). We suppose that x < 1 so 1 dominates.
2x¹⸍² - 2x
__________________________
(1 + x¹⸍² - 3x) ) 2x¹⸍² - x³⸍²
2x¹⸍² + 2x - 6x³⸍²
-____________________
0 - 2x + 5x³⸍²
We stop the process here to recall that the interim dividend now is -2x + 5x³⸍² and that we need to divide -2x by 1. We continue:
2x¹⸍² - 2x + 7x³⸍² - &c
__________________________
(1 + x¹⸍² - 3x) ) 2x¹⸍² - x³⸍²
2x¹⸍² + 2x - 6x³⸍²
-____________________
0 - 2x + 5x³⸍²
- 2x - 2x³⸍² + 6x²
-______________________-
0 + 7x³⸍² - 6x² ---> remainder after 2 terms.
We can apply this knowledge to particular numbers. For example, in a²/(b+x) we can set a=1, b=6 and x=1, so that
1/(6+1) = 1/6 - 1/6² + 1/6³ - 1/6⁴ + &c = 1/7 .
If b=8, a=1 and x=-1, we arrive to
1/7 = 1/8 + 1/8² + 1/8³ + 1/8⁴ + &c.
I find this fascinating.
To force powers of 10 we set a=1, b=10 and x=-3. Then
1/7 = 1/10 + 3/100 + 9/10³ + 27/10⁴ + 31/10⁵ + &c.
What a way to get the decimal expansion of 1/7 ! This does not directly give the digits, but with simple addition we can find 0.1 + 0.03 + 0.009 + 0.0027 + 0.00031 = 0.14201, which already approximates quite well the result (~0.142857).
Ready for a riveting application of division as series? One can change division into multiplication, which is a lot faster!
Consider the division 1/(b+x) and if you later want another numerator, like a², just multiply by it. So let's be concerned by the reciprocal of a number like (b+x).
After the 1st term of the series, which is 1/b, the remainder is -x/b, so that the supplement is (remainder/divisor) equal to -x/(b²+xb). This means that 1/(b+x) is exactly 1/b - x/(b²+xb).
Now, the trick: multiply all by by x/b: we get x/(b²+xb) = x/b² - x²/(b³+xb²). As you can see, by this simple multiplication we obtain an expression for the supplement as the sum of the 2nd term of the series plus the new supplement.
If we multiply all by x²/b² now, we will get an expression for this new supplement, expressed as the sum of the 3rd AND 4th terms of the series plus another supplement after it.
Notice that we had 1/b + s1 (supplement 1). Then, after multiplying by x/b, we have 1/b - x/b² + s2. But upon multiplying by (x²/b²) we have now two new terms, so 1/b - x/b² + x²/b³ - x³/b⁴.
Another step, multiplying by (x/b)⁴ and we will get FOUR new terms. Notice how each step doubles the number of terms! We just multiply and get an accelerated output for division.
This process is even more amazing when we apply to numbers. Say we want to calculate 1/29.
We first need some decimal figures with regular division. In this case, it is optimal if we get a remainder made of a single digit.
0.03448
________
29 ) 100000
87|||
-___v||_
130||
116||
-___v|_
140|
116|
-___v_
240
232
-____
8
We get 0.03448 as quotient and 8 as remainder. And of course, 8/29 as supplement. This means we can write as a full result 1/29 = 0.03448[8/29].
What I put into brackets is the supplement referred to the last digit we write. Usually we don't write the brackets but just the fraction in a smaller font size. However, I prefer to write it between brackets to avoid confusion, and I think it is better for plain text.
This notation is very convenient, because compare the two fractions that appear in 1/29 = 0.03448[8/29]. If we multiply the 1/29 by 8, the result could be plugged into the brackets! Let's do it.
8·(1/29) is a process that must multiply both the 0.3448 and the supplement (recall that D/d = q + s). Then, we do by hand 0.03448·8 and get 0.27584. Also, 8·8/29 = 64/29.
Then, 8/29 = 0.27584[64/29] but the supplement should be expressed as a proper fraction, so notice how [64/29] = 0[64/29] = 2[6/29]. After this, 8/29 = 0.27586[6/29].
And now the magic of this notation allows to just *concatenate* this 8/29 to our previous 1/29. We can write 1/29 = 0.0344827586[6/29]. With a single multiplication we have doubled the number of decimals!.
Now we have 1/29 against 6/29 in the same expression. Why not multiplying by 6? 0.0344827586[6/29] = 0.2068965517[7/29], so we concatenate 1/29 = 0.03448275862068965517[7/29].
Of course, this multiplication is tedious, but imagine to get all these decimals by division!
The number 1/29 is rational and sooner or later we should find some periodicity. But it has not appeared yet, so why not multiplying by 7? We get 7/29 = 0.24137931034482758620[20/29]. And we concatenate it to obtain 1/29 = 0.0344827586206896551724137931034482758620[20/29].
Try that with a calculator, or even with a computer and see whether you get the same result! Maybe you get a surprise.
Of course, now there is no point in further multiplying, because the periodicity has already appeared.
This notation may be familiar to you, but it was invented (I think) by Newton himself.
Instead of 1/x², we can write x⁻², or instead of 1/x³ we can write x⁻³.
Instead of √x we may write x¹⸍², or instead of √x³ we may write x³⸍². This is visually convenient, but not semantically, since the Unicode symbol ⸍ does not mean division but "Right Raised Omission Bracket". We could also write x^(3/2) or x**(3/2) and that's all, but sometimes it is more convenient to use superscripts. For people using screen readers, please forgive me for using ⸍ with the meaning of superscript division!
Instead of a²/x - a²b/x² + a²b²/x³ + &c we can write a²x⁻¹ - a²b²x⁻² + a²b²x⁻³ + &c.
Square roots in plain text (with a single line) are painful, since the symbol √ does not extend its arm beyond a single character. Perhaps √(a² - x²) is clear enough. Although it is enough as sqrt(a²-x²).
However, we can now write (a²-x²)¹⸍² as well.
For a cubic root like ∛(a²-x²) we can write (a²-x²)¹⸍³.
"We may not improperly distinguish powers into affirmative or negative, integral and fractional."
This is such a unification! It is so absurd to have different symbols for what is essentially the same.