Formulæ in Algebra

Involution.
${(a + b)}^{2}$	$= a^{2} + 2 a b + b^{2}$ .
${(a - b)}^{2}$	$= a^{2} - 2 a b + b^{2}$ .
${(a + b)}^{3}$	$= a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$ .
${(a - b)}^{3}$	$= a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}$ .
${(a + b)}^{n}$	$= a^{n} + n . a^{n - 1} b + &c.$ , where the index of `a` continually decreases by unity, and that of `b` increases, and where the coefficient of each term is formed from the preceding by the rule “Multiply together the coefficient and the index of `a` and divide by the place of the term.”
${(a - b)}^{n}$	$= a^{n} - n . a^{n - 1} b + &c.$ , where each term is formed as in the former case, and the signs are alternately + and −.
Resolution into rational factors.
(1) $a^{2} + b^{2}$	cannot be resolved.
(2) $a^{2} - b^{2}$	$= (a + b) . (a - b)$ .
If `n` be an odd prime,
(3) $a^{n} + b^{n}$	$= (a + b) . (a^{n - 1} - a^{n - 2} b + \dots)$ .
(4) $a^{n} - b^{n}$	$= (a - b) . (a^{n - 1} + a^{n - 2} b + \dots)$ .
If `T` be a power of 2,
(5) $a^{T} + b^{T}$	cannot be resolved.
(6) $a^{T} - b^{T}$	can be resolved by form (2).
If `N` contain odd prime factors only,
(7) $a^{N} + b^{N}$	””(3).
(8) $a^{N} - b^{N}$	””(4).
If `M` be even and contain an odd prime factor,
(9) $a^{M} + b^{M}$	””(3).
(10) $a^{M} - b^{M}$	””(2) or (4).
If `x`, `y` be commensurable,
(11) $a^{x} + b^{y}$ }	can be put into one of the first four forms,
(12) $a^{x} - b^{y}$	according to the nature of the G. C. M.
Rationalising binomial surds.
If `r` be odd,
(1) $\sqrt[r]{a} + \sqrt[r]{b}$ is submultiple of	$a + b$
(2) $\sqrt[r]{a} - \sqrt[r]{b}$ is ”	$a - b$
If `s` be even,
(3) $\sqrt[s]{a} + \sqrt[s]{b}$ is submultiple of	$a - b$
(4) $\sqrt[s]{a} - \sqrt[s]{b}$ is ”	”
Quadratic equation (one Variable).
If $x^{2} - p x + q = 0$ ,
the values of the coefficients, in terms of the roots, are	$p =$ sum of roots
	$q =$ product of roots.
If $A x^{2} + B x + C = 0$ ,
the roots are	$x = \frac{- B \pm \sqrt{B^{2} - 4 A C}}{2 A}$ .
and the test for
(1) roots equal with opposite signs	$B = 0$ .
(2) roots real	$B^{2} - 4 A C ≮ 0$ .
(3) ” imaginary	” $< 0$ .
(4) ” identical	” $= 0$ .
(5) ” rational	” $≮ 0$ , and is a square.
Simultaneous Equations (two Variables).
If $A_{1} x + B_{1} y + C_{1} = 0$ ,
If $A_{2} x + B_{2} y + C_{2} = 0$ ,
the roots are	$\frac{x}{\| \begin{matrix} B_{1}, & C_{1} \\ B_{2}, & C_{2} \end{matrix} \|} = \frac{- y}{\| \begin{matrix} A_{1}, & C_{1} \\ A_{2}, & C_{2} \end{matrix} \|}$
	$= \frac{1}{\| \begin{matrix} A_{1}, & B_{1} \\ A_{2}, & B_{2} \end{matrix} \|}$ .
and test for Equations being
(1) consistent	$\| \begin{matrix} A_{1} & B_{1} \\ A_{2} & B_{2} \end{matrix} \| \neq 0$ .
(2) inconsistent	$\| \begin{matrix} A_{1} & B_{1} \\ A_{2} & B_{2} \end{matrix} \| = 0$ , and either
	$\| \begin{matrix} A_{1} & C_{1} \\ A_{2} & C_{2} \end{matrix} \|$ or $\| \begin{matrix} B_{1} & C_{1} \\ B_{2} & C_{2} \end{matrix} \| \neq 0$ .
(3) identical	$‖ \begin{matrix} A_{1} & B_{1} & C_{1} \\ A_{2} & B_{2} & C_{2} \end{matrix} ‖ = 0$ .
Summation of Series.
Arithmetical.
If $a =$ first term,
If $b =$ common difference,
If $n =$ number of terms,
If $l =$ last term,
If $S =$ sum;
the formula connecting
(1) `a`, `b`, `n`, `S`	$S = (2 a + \bar{n - 1} . b) . \frac{n}{2}$ .
(2) `a`, `l`, `S`	$S = (a + l) . \frac{n}{2}$ .
Geometrical.
If $a =$ first term,
If $r =$ common ratio,
If $n =$ number of terms,
If $S =$ sum;
the formula connecting
(1) `a`, `r`, `n`, `S`	$S = a . \frac{r^{n} - 1}{r - 1}$ .
(2) `a`, `r`, `S` (where `n` is infinite	$S = \frac{a}{1 - r}$ .
Arithmetical Mean, &c.
If `α`, `β`, be two terms having one between them,
(1) the Arithmetical Mean	$\frac{α + β}{2}$ .
(2) the Geometrical Mean	$\sqrt{α β}$ .
(3) the Harmonical Mean	$\frac{2 α β}{α + β}$ .
Binomial Theorem.
$(a \pm b)^{n}$	$= a^{n} \pm n a^{n - 1} b + \frac{n . (n - 1)}{2} . a^{n - 2} b^{2} \pm \frac{n . (n - 1) . (n - 2)}{3} . a^{n - 3} b^{3} + \dots$
the sum of the coefficients	$= 2^{n}$ .
Permutations and Combinations.
Permutation of `n` things
(1) taken `r` together	$n . (n - 1) \dots (n - r + 1)$ .
(2) taken all together	$n$ .
Combinations of `n` things taken `r` together	$\frac{n . (n - 1) \dots (n - r + 1)}{r}$ .
Permutations of `n` things taken all together, when there are `p` of one kind, `q` of another kind, &c.	$\frac{n}{p . q \dots}$ .
Total number of combinations of `n` things	$= 2^{n} - 1$ .
Interest, Discount, &c.
If $P =$ Principal,
If $r =$ interest of £1 for 1 year,
If $n =$ number of years,
If $I =$ total interest,
If $M =$ amount,
If $D =$ discount,
If $A =$ annuity,
Simple interest:
formula connecting `P`, `r`, `n`, `I`	$I = P n r$ .
”” `P`, `r`, `n`, `M`	$M = P + P n r$ .
”” `M`, `r`, `n`, `D`	$D = \frac{M n r}{1 + n r}$ .
Compound interest:
formula connecting `P`, `r`, `n`, `M`	$M = P (1 + r)^{n}$ .
Terminable annuities:
formula connecting `A`, `r`, `n`, `M`	$M = A . \frac{(1 + r)^{n} - 1}{r}$ .
”” `A`, `r`, `n`, `V`	$V = A . \frac{(1 + r)^{n} - 1}{r . (1 + r)^{n}}$ .
Perpetual annuity:
formula connecting `A`, `r`, `V`	$V = \frac{A}{r}$ .
Deferred terminable annuity:
If $d =$ number of years for which it is deferred,
If $n =$ number of years for which it is to continue,
formula connecting `A`, `r`, `d`, `n`, `V`	$V = A . \frac{(1 + r)^{n} - 1}{r . (1 + r)^{d + n}}$ .
Deferred perpetual annuity:
formula connecting `A`, `r`, `d`, `V`	$V = A . \frac{1}{r . (1 + r)^{d}}$ .
Logarithms.
If `a`, `b`, &c., = the bases used
`M`, `N`, &c., = any numbers,
(1) $log 1$	$= 0$ .
(2) $log a$	$= 1$ .
(3) ${log}_{a} M \times N \times &c.$	${log}_{a} M + {log}_{a} N + &c.$
(4) ${log}_{a} \frac{M}{N}$	$= {log}_{a} M - {log}_{a} N$ .
(5) ${log}_{a} M^{x}$	$x . {log}_{a} M$ .
(6) ${log}_{b} M$ (in terms of logs to base `a`)	$\frac{{log}_{a} M}{{log}_{a} b}$ .
Exponential and Logarithmic Series.
`e` (the Napierian base)	$= 1 + 1 + \frac{1}{2} + \frac{1}{3} + \dots ad infin.$
	= 2.7182, &c.
$a^{x}$ (in a series of ascending powers of `x`)	$= 1 + ({log}_{e} a) . x + \frac{({log}_{e} a)^{2}}{2} . x^{2} + \dots$
hence $e^{x}$	$= 1 + x + \frac{x^{2}}{2} + \dots$
${log}_{e} (1 + x)$	$= x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + \dots$
$\frac{1}{{log}_{e} 10}$	$= . 43429, &c.$
${log}_{10} (n + 1)$	$= {log}_{10} n + \frac{2}{{log}_{e} 10} . (\frac{1}{2 n + 1} + \frac{1}{3 (2 n + 1)^{3}} + \frac{1}{5 (2 n + 1)^{5}} + \dots)$ .