We want to multiply series S1=(A+B+C+D+E+&c) by series S2=(P+Q+R+S+T+&c). The result is another series S2=(α+β+γ+δ+ε+&c).
Here, the order in which we keep the multiplied terms is essential. Not only helps to avoid forgetting a term, but also provides the key to what follows:
(A+B+C+D+E&c)·(P+Q+R+S+T&c) =
= AP + BP + CP + DP + EP + &c +
+ AQ + BQ + CQ + DQ + &c +
+ AR + BR + CR + &c +
+ AS + BS + &c +
+ AT + &c
+________________________________
α + β + γ + δ + ε + &c
If we keep this order, multiplying series is really easy. Example of multiplication:
S1 = a + x/2 + x²/(3a) + x³/(4a²) + x⁴/(5a³) + &c S2 = a - x/3 + x²/(5a) - x³/(7a²) + x⁴/(9a³) + &c.
Then we build the ordered form:
a² + ax/2 + x²/3 + x³/(4a) + x⁴/(5a²) + &c
- ax/3 - x²/6 - x³/(9a) - x⁴/(12a²) + &c
+ x²/5 + x³/(10a)+ x⁴/(15a²) + &c
- x³/(7a) - x⁴/(14a²) + &c
+ x⁴/(9a²) + &c
+ -----------------------------------------
a² + ax/6 + 11x²/30+121/(1260a)+281x⁴/(1260a²) + &c
The previous ordered form is very useful again for division of series. Before, we had S1·S2=S3 and now we are given S3 and S2 and we want S1 = S3/S2. Again we write the same order:
(A+B+C+D+E&c)·(P+Q+R+S+T&c) =
= AP + BP + CP + DP + EP + &c +
+ AQ + BQ + CQ + DQ + &c +
+ AR + BR + CR + &c +
+ AS + BS + &c +
+ AT + &c
+________________________________
α + β + γ + δ + ε + &c
And now we see how the first term is just A = α/P. Now, this value can be plugged in the next column, so that we can isolate B = (β-AQ)/P. Then, the last two terms can be used in the third column to get C=(γ-AR-BQ)/P, and so on. A summary of division:
A = α/P B = (β - AQ)/P C = (γ - AR - BQ)/P D = (δ - CQ - BR - AS)/P .
But there is no need to memorise this. Only the order of terms is enough to easily consider column by column.
We can do the previous example but now in reverse.
S3 = α + β + γ + δ + ε + &c S3 = a² + ax/6 + 11x²/30+121/(1260a)+281x⁴/(1260a²) + &c S2 = P + Q + R + S + T + &c S2 = a + x/2 + x²/(3a) + x³/(4a²) + x⁴/(5a³) + &c
Then, we want S1 = S3 / S2:
A = a B = x/6 - x/2 = -x/3 C = 11x²/(30a) - x²/(3a) + x²/(6a) = x²/(5a) D = 121/1260 - x³/(10a²) + x³/(9a²) - x³/(4a²) = -x³/(7a²)
and so on.
To obtain the reciprocal of a series, just apply division in which the series at the numerator is just 1.
It is just the division with α=1 and β=γ=δ=ε=...=0. Our formulae become:
A = 1 / P B = (- AQ) / P C = (- AR - BQ) / P D = (- CQ - BR - AS) / P E = (- DQ - CR - BS - AT) / P &c
Example: find the reciprocal of
S1 = P + Q + R + S + T + &c S1 = a + x/2 + x²/(3a) + x³/(4a²) + x⁴/(5a³) + &c
Then,
A = 1/a B = -x/(2a²) C = -x²(3a³) + x²/(4a³) = -x²/(12a³) D = x³/(24a⁴) + x³/(6a⁴) - x³/(4a⁴) = -x³/(24a⁴) E = etc = -19x⁴/(720a⁵) &c
This case is also trivial, since it is the multiplication of a series by itself. We simply do
P=A, Q=B, R=C, S=D, T=E, &c
The order is
S1² = A² + AB + AC + AD + AE + &c
+ AB + B² + BC + BD + &c
+ AC + BC + C² + &c
+ AD + BD + &c
+ + AE + &c
------------------------------------
S1² = α + β + γ + δ + ε + &c
Notice how this can be rewritten as
S1² = A² + 2AB + 2AC + 2AD + 2AE + 2AF + 2AG + &c
+ B² + 2BC + 2BD + 2BE + 2BF + &c
+ C² + 2CD + 2CE + &c
+ D² + &c
This case is the inverse of the previous one. We recall the previous result as
S1² = A² + 2AB + 2AC + 2AD + 2AE + 2AF + 2AG + &c
+ B² + 2BC + 2BD + 2BE + 2BF + &c
+ C² + 2CD + 2CE + &c
+ + D² + &c
--------------------------------------------------
S2 = P + Q + R + S + T + U + V + &c
So we are given S2 and we want S1, and then
A = P^(1/2) B = Q / (2A) C = (R - B²) / (2A) D = (S - 2BC) / (2A) E = (T - 2BD -C²) / (2A) &c
Example:
sqrt(1 - x/2 + x²/8 - x³/16 + &c)
P Q R S
where we get
A = 1 B = -x/4 C = x²/16 - x²/32 = x²/32 D = -x³/32 + x³/128 = -3x³/128 &c
The cube of a series consists in multiplying a series by itself two times. Let's do the cube of the series S1 = A + B + C + D + &c. But we know the square of it is S2² = A² + (2AB) + (2AC + B²) + (2AD + 2BC) + (2AE + 2BD + C²) + &c. Then, by multiplying this by S1 again, we get
S1³ = A³ + 3A²B + 3AB² + 3A²D + 3AC² + 3BC² + &c
+ 3A²C + 6ABC + 3B²C + 3B²D + &c
+ B³ + 6ABD + 6ACD + &c
+ 3A²E + 6ABE + &c
+ 3AF² + &c
Here, just take the inverse of the cube operation.
Once we have seen some particular cases, we may want to deal with the generic multinomial case. It is attributed to Leibniz in the 1670's, although he did not publish it. It was published by De Moivre (Philosophical Transactions, nº 230).
The multinomial theorem is not an easy thing to apply.
Let's consider
(az + bz² + cz³ + dz⁴ + &c)ᵏ
where k can be an integer number or not. De Moivre provides some very complicated formulas. There are three things that can happen here:
Let's develop the first and the third points.
Let's have
(a + b + c)³
In this case we have the multinomial coefficient
3! / (j1! j2! j3!) defined as B[3,{j1,j2,j3}]
where j's are the exponents in each term.
Example: abc term has j1=j2=j3=1. Then we get the following expansion:
B[3,{3,0,0}]·a³ + B[3,{2,1,0}]·a²·b + B[3,{2,0,1}]·a²·c +
B[3,{0,3,0}]·b³ + B[3,{1,2,0}]·a·b² + B[3,{0,2,1}]·b²·c +
B[3,{0,0,3}]·c³ + B[3,{1,0,2}]·a·c² + B[3,{0,1,2}]·b·c² +
+B[3,{1,1,1}]·a·b·c
This is quite simple, but it only works for a positive
integer exponent.
Let's have the following terms to be raised to exponent m:
([1] + [2] + [3] + [4] + [5] + ...)^m
where [n] is the nth term of the series. The result will be written as R1 + R2 + R3 + R4 + ...
Begin by preparing two papers, indicated here:
movable paper ............. | etc | etc | | | | | [6] |(5m- | | | | | [5] |(4m- | | | | | [4] |(3m- | | | | | [3] |(2m- | | | | | [2] |( m- | +-----+-----+ | [1] | +-----+-----+ fixed paper +-----+-----------------------------+ | 0 )| [1]^m | | | | | 1 )| | | | | | 2 )| | | | | | 3 )| | | | | | 4 )| | | | | | 5 )| | | | | | etc | | .....................................
Now set the initial position of the papers:
Bring [2] beside 0:
.............
| etc | etc |
| | |
| [6] |(5m- |
| | |
| [5] |(4m- |
| | |
| [4] |(3m- | Here we read
| | | [2]·(m-0)·R1 and we / [1] = R2
| [3] |(2m- | so we write it on the next line
| | |+-----+-----------------------------+
| [2] |( m- || 0 )| [1]^m = R1 |
+-----+-----+| | |
| [1] || 1 )| [2]·(m-0)·R1 / [1] = R2 |
+-----+-----+| | |
| 2 )| |
| | |
| 3 )| |
| | |
| 4 )| |
| | |
| 5 )| |
| | |
| etc | |
.....................................
Now move to the next position of the papers: bring the left down.
.............
| etc | etc |
| | |
| [6] |(5m- |
| | | Now we read two matching lines:
| [5] |(4m- | [3]·(2m-0)·R1 |
| | | [2]·( m-1)·R2 | we add both and divide
| [4] |(3m- | by /2[1]
| | |+-----+-----------------------------+
| [3] |(2m- || 0 )| [1]^m = R1 |
| | || | |
| [2] |( m- || 1 )| [2]·m·R1/[1] = R2 |
+-----+-----+| | |
| [1] || 2 )|([3]2mR1 + [2](m-1)R2)/(2[1])|
+-----+-----+| | |
| 3 )| |
| | |
| 4 )| |
| | |
| 5 )| |
| | |
| etc | |
.....................................
Next position of the papers: repeat process.
.............
| etc | etc | Next term has three parts and /(3[1])
| | |
| [6] |(5m- | ([4]3mR1 + [3](2m-1)R2 + [2](m-2)R3)/(3[1])=
| | |
| [5] |(4m- | = R4
| | |+-----+-----------------------------+
| [4] |(3m- || 0 )| [1]^m = R1 |
| | || | |
| [3] |(2m- || 1 )| [2]mR1/[1] = R2 |
| | || | |
| [2] |( m- || 2 )|([3]2mR1 + [2](m-1)R2)/(2[1])=R3 |
+-----+-----+| | |
| [1] || 3 )| R4 |
+-----+-----+| | |
| 4 )| |
| | |
| 5 )| |
| | |
| etc | |
.....................................
and so on. Notice the terms in parentheses can be negative!
We must iterate until we get Rj = 0. Then we stop.
These exercises can be performed now by using different techniques learned here. The idea is to reduce everything to a series consisting of simple terms.