Symmetric Functions

Excerpts

Symmetric Functions
A rational function of the independent variables $x_1, x_2, \dotsc, x_n$ is said to be *symmetric* in them if it is unaltered by the interchange of any two of the variables.
Symmetric Functions
In general, if $t$ is a rational function of $x_1, \dotsc, x_n, \Sigma t$ denotes the sum of $t$ and all of the distinct functions obtained from $t$ by permutations of the variables; such a $\Sigma$-function (read *sigma function*) is symmetric in $x_1, \dotsc, x_n$.
Symmetric Functions
In particular, $\Sigma \alpha = \alpha + \beta + \gamma$, $\Sigma \alpha\beta$, and $\alpha\beta\gamma$ are called the three *elementary symmetric functions* of $\alpha$, $\beta$, $\gamma$.
Symmetric Functions
$(\Sigma \alpha)^2 = \Sigma \alpha^2 + 2\Sigma \alpha\beta$, whence $\Sigma \alpha^2 = p^2 - 2q$.

Equations

Symmetric Functions
\Sigma \alpha = \alpha + \beta + \gamma
The sigma sum of the three variables alpha, beta, gamma is their plain sum.
Symmetric Functions
E_1 = \Sigma x_1
The first elementary symmetric function E_1 is the sum of the variables.
Symmetric Functions
E_2 = \Sigma x_1x_2
The second elementary symmetric function E_2 is the sum of all products of the variables taken two at a time.
Symmetric Functions
E_n = x_1x_2 \dotsm x_n
The n-th elementary symmetric function E_n is the product of all n variables.
Symmetric Functions
x^n + c_1x^{n-1} + c_2x^{n-2} + \dotsb + c_n = 0
The general polynomial equation of degree n with leading coefficient unity, whose roots are alpha_1, ..., alpha_n.
Symmetric Functions
x^n - E_1 x^{n-1} + E_2 x^{n-2} - \dotsb + (-1)^n E_n = 0
The equation whose roots are x_1, ..., x_n has coefficients given by the signed elementary symmetric functions of those roots.
Symmetric Functions
rx_1^2 + rx_2^2 + sx_1 + sx_2 \equiv r(E_1^2 - 2E_2) + sE_1
For two variables the given symmetric polynomial equals the same expression written in E_1 and E_2 with the coefficients r and s.
Symmetric Functions
f(x) \equiv (x - \alpha_1)(x - \alpha_2) \dotsm (x - \alpha_n)
A polynomial f(x) of degree n equals the product of linear factors x minus each of its roots.
Symmetric Functions
f'(x) \equiv \frac{f(x)}{x - \alpha_1} + \frac{f(x)}{x - \alpha_2} + \dotsb + \frac{f(x)}{x - \alpha_n}
The derivative of a polynomial in factored form is the sum of f(x) divided by each linear factor.
Symmetric Functions
s_k = \Sigma \alpha_1^k
The sum of the k-th powers of the roots is denoted s_k.
Symmetric Functions
s_1 = -c_1
The sum of the roots equals minus the coefficient c_1 of x^{n-1}.
Symmetric Functions
s_2 = c_1^2 - 2c_2
The sum of the squares of the roots equals c_1 squared minus twice c_2.
Symmetric Functions
s_k + c_1 s_{k-1} + c_2 s_{k-2} + \dotsb + c_{k-1} s_1 + kc_k = 0
For k up to n, the power sums satisfy a recursion in the coefficients of the equation, with the k-th term weighted by k.
Symmetric Functions
s_n + c_1 s_{n-1} + c_2 s_{n-2} + \dotsb + c_{n-1} s_1 + nc_n = 0
The last of Newton's identities, obtained by substituting each root into the equation and adding.
Symmetric Functions
s_k + c_1 s_{k-1} + c_2 s_{k-2} + \dotsb + c_n s_{k-n} = 0
For k greater than n the power sums satisfy the recursion with the n coefficients of the equation.
Symmetric Functions
x^2 + px + q \equiv (x - \alpha)(x - \beta)
A monic quadratic equals the product of its linear factors with roots alpha and beta.
Symmetric Functions
1 + py + qy^2 \equiv (1 - \alpha y)(1 - \beta y)
The quadratic after the substitution x = 1/y, multiplied by y squared, factors into the two linear factors in y.
Symmetric Functions
\frac{-p - 2qy}{1 + py + qy^2} \equiv \frac{\alpha}{1 - \alpha y} + \frac{\beta}{1 - \beta y}
The logarithmic-type derivative of the quadratic in y equals the sum of the two geometric-type fractions for its roots.
Symmetric Functions
\frac{1}{1 - r} \equiv 1 + r + r^2 + \dotsb + r^{k-1} + \frac{r^k}{1-r}
The finite geometric expansion of 1/(1 - r), with its remainder term r^k/(1 - r).
Symmetric Functions
\frac{\alpha}{1 - \alpha y} + \frac{\beta}{1 - \beta y} = s_1 + s_2 y + \dotsb + s_k y^{k-1} + \frac{\phi y^k}{1 + py + qy^2}
The sum of the two root fractions, expanded to k terms, has coefficients equal to the power sums s_1, ..., s_k.
Symmetric Functions
s_k = k\sum (-1)^{i+j} \frac{(i+j-1)!}{i!j!} p^iq^j
For the quadratic x^2 + px + q, the k-th power sum is given by a finite sum over i and j with i + 2j = k.
Symmetric Functions
s_k = k\sum_{j=0}^K (-1)^j \frac{(k-j-1)!}{(k-2j)!j!} p^{k-2j} q^j
The k-th power sum of the roots of x^2 - px + q is an explicit finite sum in p and q, with K the largest integer not exceeding k/2.
Symmetric Functions
s_k = k\sum (-1)^{r_1 + \dotsb + r_n} \frac{(r_1 + \dotsb + r_n-1)!}{r_1! \dotsm r_n!} c_1^{r_1} \dotsm c_n^{r_n}
The k-th power sum of the roots of a general equation is a sum over non-negative integer sets r_1, ..., r_n with r_1 + 2r_2 + ... + nr_n = k.
Symmetric Functions
x^k + \left(\frac{q}{x}\right)^k = c
With the roots x and q/x of x^2 - px + q, the equation in x states that the k-th powers sum to the constant c.
Symmetric Functions
\Sigma \alpha_1^a \alpha_2^b = \frac{1}{m} (s_a s_b - s_{a+b})
The sigma sum of products of two different powers of the roots equals half or the whole of s_a s_b minus s_{a+b}, according to whether a equals b.
Symmetric Functions
\Sigma \alpha_1^4 \alpha_2^3 \alpha_3^2 = s_2s_3s_4 - s_2s_7 - s_3s_6 - s_4s_5 + 2s_9
The sigma sum of a term in three roots with exponents 4, 3, 2 is expressed in the power sums s_k.
Symmetric Functions
\Sigma x_1^2x_2x_3x_4 = E_1E_4 - 5E_5
The sigma sum of x_1^2 x_2 x_3 x_4 is expressed through the elementary symmetric functions E_1, E_4 and E_5.
Symmetric Functions
\Sigma \frac{\alpha^2 + \beta^2}{\alpha + \beta} = \frac{2q^2 - 2p^2q + 4pr}{pq - r}
For the cubic x^3 + px^2 + qx + r with roots alpha, beta, gamma, the sum of the three fractions equals (2q^2 - 2p^2 q + 4pr) divided by (pq - r).

Problems

Exercise Page129

The data holds no problems for this exercise yet.

Exercise Page133

Exercise Page133, problem 1, p. 133
$\Sigma \dfrac{\beta\gamma + \alpha^2}{\beta + \gamma}$, % [** PP: Added ,]
Printed answer:
- $\dfrac{p^4 - 3p^2q + 5pr + q^2}{r - pq}$.
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- other: not a kind the checker handles (p**4 - 3*p**2*q + 5*p*r + q**2)/(r - p*q)
Exercise Page133, problem 10, p. 133
$\alpha\beta + \alpha\gamma$, $\alpha\beta + \beta\gamma$, $\alpha\gamma + \beta\gamma$.
Printed answer:
- $y = q+r/x$.
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- other: not a kind the checker handles Eq(y, q + r/x)
Exercise Page133, problem 11, p. 133
$\dfrac{2\alpha - 1}{\beta + \gamma - \alpha}$, etc.
Printed answer:
- $x = \dfrac{1-py}{2+2y}$.
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- other: not a kind the checker handles Eq(x, (1 - p*y)/(2 + 2*y))
Exercise Page133, problem 12, p. 133
$\dfrac{\beta\gamma + 3\alpha^2}{\beta + \gamma - 2\alpha}$, etc.
Printed answer:
- $y = \dfrac{4x^2 + px + q}{-3x-p}$, see §112.
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- other: not a kind the checker handles Eq(y, (4*x**2 + p*x + q)/(-3*x - p))
Exercise Page133, problem 13, p. 133
$\Sigma\dfrac{\beta^2 + \gamma^2 + \delta^2}{\beta + \gamma + \delta}$.
Printed answer:
- $\dfrac{2q(p^3 + 2pq - r)}{p^2q - pr + s} - 5p$, see Ex. 17.
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- other: not a kind the checker handles 2*q*(p**3 + 2*p*q - r)/(p**2*q - p*r + s) - 5*p
Exercise Page133, problem 14, p. 133
$\Sigma\dfrac{\beta\gamma + \beta\delta + \gamma\delta}{\beta + \gamma + \delta - 3}$.
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- (none printed)
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Exercise Page133, problem 15, p. 133
Prove that if $y_1$, $y_2$, $y_3$ are the roots of $y^3 + py + q = 0$, the equation with the roots $z_1 = (y_2 - y_3)^2$, $z_2 = (y_1 - y_3)^2$, $z_3 = (y_1 - y_2)^2$ is % z^3 + 6pz^2 + 9p^2 z + 4p^3 + 27q^2 = 0. Hints: since $z_1 = \Sigma y_1^2 - 2y_2y_3 - y_1^2 = -2p + 2q/y_1 - y_1^2$, etc., we set $z = -2p + 2q/y - y^2$. By the given equation, $y^2 + p + q/y = 0$. Thus the desired substitution is $z = -p + 3q/y$, $y = 3q/(z + p)$.
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Exercise Page133, problem 16, p. 133
Hence find the discriminant of the reduced cubic equation. % %
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Exercise Page133, problem 17, p. 133
If $x_1, \dotsc, x_n$ are the roots of $f(x)=0$, show that 1x_1 - c = -f’(c)f(c). Hint: $x_1 - c = y_1, \dotsc, x_n - c = y_n$ are the roots of f(c+y) = f(c) + yf’(c) + y^2( )+ = 0, as shown by Taylor’s theorem. Or we may employ $(5)$ below % [** PP: Added ‘below’] for $x = c$.
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Exercise Page133, problem 2, p. 133
$\Sigma \dfrac{3\beta\gamma - 2\alpha^2}{\beta + \gamma - \alpha}$.
Printed answer:
- $\dfrac{(5p^2-12q)(p^2-4q)}{4(p^3 - 4pq + 8r)} - \dfrac{13}{4}p$.
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- other: not a kind the checker handles (5*p**2 - 12*q)*(p**2 - 4*q)/(4*(p**3 - 4*p*q + 8*r)) - 13*p/4
Exercise Page133, problem 3, p. 133
Why would the use of $\beta\gamma = -r/\alpha$ complicate Exs. 1, 2? Verify that = -r = f() - r = ^2 + p + q.
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Exercise Page133, problem 4, p. 133
Why would you use $\beta\gamma = -r/\alpha$ in finding $\Sigma \dfrac{\beta^2 + \gamma^2}{\beta\gamma + c}$?
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Exercise Page133, problem 5, p. 133
Find $\Sigma (\beta + \gamma)^2$.
Printed answer:
- $2p^2-2q$.
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- other: not a kind the checker handles 2*p**2 - 2*q
Exercise Page133, problem 6, p. 133
Find $\Sigma (\alpha + \beta - \gamma)^3$.
Printed answer:
- $24r-p^3$.
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- other: not a kind the checker handles 24*r - p**3
Exercise Page133, problem 7, p. 133
Find $\smash{\Sigma \left(\dfrac{\beta - \gamma}{\beta + \gamma}\right)^2}$.
Printed answer:
- $\dfrac{3p^2q^2 - 4p^3r - 4q^3 - 2pqr - 9r^2}{(r - pq)^2}$.
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- other: not a kind the checker handles (3*p**2*q**2 - 4*p**3*r - 4*q**3 - 2*p*q*r - 9*r**2)/(r - p*q)**2
Exercise Page133, problem 8, p. 133
Find a necessary and sufficient condition on the coefficients that the roots, in some order, shall be in harmonic progression. Hint: If $\dfrac{1}{\alpha} + \dfrac{1}{\gamma} = \dfrac{2}{\beta}$, then $\dfrac{-3r}{q} - \beta = 0$, and conversely. Hence the condition is (-3rq - ) (-3rq - ) (-3rq - ) = f(-3rq) = 0.
Printed answer:
- $27r^2 - 9pqr + 2q^3 = 0$.
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- other: not a kind the checker handles Eq(27*r**2 - 9*p*q*r + 2*q**3, 0)
Exercise Page133, problem 9, p. 133
Find the cubic equation with the roots $\beta\gamma - \dfrac{1}{\alpha}$, $\alpha\gamma - \dfrac{1}{\beta}$, $\alpha\beta - \dfrac{1}{\gamma}$. Hint: since these are $(-r - 1)/\alpha$, etc., make the substitution $(-r - 1)/x = y$.
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- (none printed)
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Exercise Page136

Exercise Page136, problem 1, p. 136
For a cubic equation, $s_4 = c_1^4 - 4c_1^2 c_2 + 4c_1 c_3 + 2c_2^2$.
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- (none printed)
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Exercise Page136, problem 2, p. 136
For an equation of degree $\geqq 4$, $s_4 = c_1^4 - 4c_1^2 c_2 + 4c_1 c_3 + 2c_2^2- 4c_4$.
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- (none printed)
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Exercise Page136, problem 3a, p. 136
Find $s_2$, $s_3$, $s_4$, $s_5$ for $x^2 - px + q = 0$.
Printed answer:
- $s_2 = p^2 - 2q$,
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- other: not a kind the checker handles p**2 - 2*q
Exercise Page136, problem 3b, p. 136
Find $s_2$, $s_3$, $s_4$, $s_5$ for $x^2 - px + q = 0$.
Printed answer:
- $s_3 = p^3 - 3pq$,
unverified: no computed check settled this one (yet)
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- other: not a kind the checker handles p**3 - 3*p*q
Exercise Page136, problem 3c, p. 136
Find $s_2$, $s_3$, $s_4$, $s_5$ for $x^2 - px + q = 0$.
Printed answer:
- $s_4 = p^4 - 4p^2q + 2q^2$,
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles p**4 - 4*p**2*q + 2*q**2
Exercise Page136, problem 3d, p. 136
Find $s_2$, $s_3$, $s_4$, $s_5$ for $x^2 - px + q = 0$.
Printed answer:
- $s_5 = p^5 - 5p^3q + 5pq^2$.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles p**5 - 5*p**3*q + 5*p*q**2
Exercise Page136, problem 4, p. 136
Find $s_k$ for $x^5 - 3 = 0$.
Printed answer:
- $s_{5n} = 5·3^n$, $s_k = 0$ if $k$ is not divisible by $5$.
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- other: not a kind the checker handles Piecewise((5*3**(k/5), Mod(k, 5) == 0), (0, True))
Exercise Page136, problem 5a, p. 136
Find $s_2$, $s_3$, $s_6$, $s_7$ for $x^5 - px + q = 0$.
Printed answer:
- All zero.
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How it was checked
- other: not a kind the checker handles 0
Exercise Page136, problem 5b, p. 136
Find $s_2$, $s_3$, $s_6$, $s_7$ for $x^5 - px + q = 0$.
Printed answer:
- All zero.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles 0
Exercise Page136, problem 5c, p. 136
Find $s_2$, $s_3$, $s_6$, $s_7$ for $x^5 - px + q = 0$.
Printed answer:
- All zero.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles 0
Exercise Page136, problem 5d, p. 136
Find $s_2$, $s_3$, $s_6$, $s_7$ for $x^5 - px + q = 0$.
Printed answer:
- All zero.
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How it was checked
- other: not a kind the checker handles 0

Exercise Page140

Exercise Page140, problem 1, p. 140
For the quadratic $x^2 - px + q = 0$ write out the expressions for $s_2$, $s_3$, $s_4$, $s_5$ given by $(19)$, and compare with those obtained from Newton’s identities (Ex. 3, §106).
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- (none printed)
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Exercise Page140, problem 2, p. 140
Find $s_4$ for a quartic equation by Waring’s formula.
Printed answer:
- See Ex. 2, p. 136.
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Exercise Page140, problem 3, p. 140
For $k=5$, $(20)$ becomes De Moivre’s quintic $p^5 - 5qp^3 + 5q^2p = c$. Solve it by radicals for $p$.
Printed answer:
- $\epsilon^j \sqrt[5]{\frac{1}{2}c + \sqrt{Q}} + \epsilon^{5-j} \sqrt[5]{\frac{1}{2}c - \sqrt{Q}}$, $Q = \frac{1}{4}c^2 - q^5$($j=0$, $1$, $2$, $3$, $4$).
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- solve: the printed answer does not match the problem epsilon**j*(c/2 + sqrt(c**2/4 - q**5))**Rational(1,5) + epsilon**(5-j)*(c/2 - sqrt(c**2/4 - q**5))**Rational(1,5)
Exercise Page140, problem 4, p. 140
Solve $(20)$ by radicals when $k=7$.
Printed answer:
- $\epsilon^j \sqrt[7]{\frac{1}{2}c + \sqrt{Q}} + \epsilon^{7-j} \sqrt[7]{\frac{1}{2}c - \sqrt{Q}}$, $Q = \frac{1}{4}c^2 - q^7$($j=0$, $1,\dotsc, 6$).
unverified: no computed check settled this one (yet)
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- solve: no printed answer to check epsilon**j*(c/2 + sqrt(c**2/4 - q**7))**Rational(1,7) + epsilon**(7-j)*(c/2 - sqrt(c**2/4 - q**7))**Rational(1,7)

Exercise Page141

Exercise Page141, problem 1, p. 141
$\Sigma \alpha_1^2 \alpha_2^2$.
Printed answer:
- $c_2^2 - 2c_1c_3 + 2c_4$.
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- other: not a kind the checker handles c_2**2 - 2*c_1*c_3 + 2*c_4
Exercise Page141, problem 2, p. 141
$\Sigma \alpha_1^3 \alpha_2$.
Printed answer:
- $c_1^2c_2 - 2c_2^2 - c_1c_3 + 4c_4$.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles c_1**2*c_2 - 2*c_2**2 - c_1*c_3 + 4*c_4
Exercise Page141, problem 3, p. 141
$\Sigma \alpha_1^2 \alpha_2 \alpha_3$.
Printed answer:
- $c_1c_3 - 4c_4$.
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How it was checked
- other: not a kind the checker handles c_1*c_3 - 4*c_4
Exercise Page141, problem 4, p. 141
$\Sigma \alpha_1^2 \alpha_2^2 \alpha_3^2$.
Printed answer:
- $c_3^2 - 2c_2c_4$.
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How it was checked
- other: not a kind the checker handles c_3**2 - 2*c_2*c_4
Exercise Page141, problem 5, p. 141
If $a\geqq b > c > 0$, prove that _1^a _2^b _3^c = 1m (s_a s_b s_c - s_a s_b+c - s_b s_a+c - s_c s_a+b + 2s_a+b+c), where $m = 1$ if $a > b$, $m = 2$ if $a = b$.
Printed answer:
- (none printed)
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- other: not a kind the checker handles
Exercise Page141, problem 6, p. 141
$\Sigma \alpha_1^a \alpha_2^b \alpha_3^b = \frac{1}{2}(s_a s_b^2 - s_as_{2b} - 2s_b s_{a+b} + 2s_{a+2b})$, $a > b > 0$.
Printed answer:
- (none printed)
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- other: not a kind the checker handles
Exercise Page141, problem 7, p. 141
$\Sigma \alpha_1^a \alpha_2^a \alpha_3^a = \frac{1}{6}(s_a^3 - 3s_a s_{2a} + 2s_{3a})$, $a > 0$.
Printed answer:
- (none printed)
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- other: not a kind the checker handles

Exercise Page142

Exercise Page142, problem 1, p. 142
$\Sigma \alpha_1^2 \alpha_2 \alpha_3$.
Printed answer:
- $c_1c_3 - 4c_4$ if $n>3$, $c_1c_3$ if $n=3$.
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- other: not a kind the checker handles c1*c3 - 4*c4
Exercise Page142, problem 10, p. 142
$\Sigma \dfrac{\beta}{\alpha} = \Sigma \dfrac{\beta + \gamma + \delta}{\alpha} = \Sigma \dfrac{-p - \alpha}{\alpha} = -4 - p \Sigma \frac{1}{\alpha}$.
Printed answer:
- $-4 + pr/s$.
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How it was checked
- other: not a kind the checker handles -4 + p*r/s
Exercise Page142, problem 11, p. 142
$\Sigma \dfrac{\beta}{\alpha^2}$. Use $\Sigma \dfrac{1}{\alpha}·\Sigma \dfrac{\beta}{\alpha} = \Sigma \dfrac{\beta}{\alpha^2} + 3\Sigma \dfrac{1}{\alpha} + 2\Sigma \dfrac{\gamma}{\alpha\beta}$.
Printed answer:
- $(rs - pr^2 + 2pqs)/s^2$.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles (r*s - p*r**2 + 2*p*q*s)/s**2
Exercise Page142, problem 12i, p. 142
Express $\Sigma \alpha_1^a \alpha_2^b \alpha_3^c \alpha_4^d$ in terms of the $s_k$ when (*i*) $a>b>c>d>0$, and (*ii*) when $a=b=c=d$.
Printed answer:
- $s_a s_b s_c s_d - \Sigma s_a s_b s_{c+d} + 2\Sigma s_a s_{b+c+d} + \Sigma s_{a+b} s_{c+d} - 6s_{a+b+c+d}$.
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Exercise Page142, problem 12ii, p. 142
Express $\Sigma \alpha_1^a \alpha_2^b \alpha_3^c \alpha_4^d$ in terms of the $s_k$ when (*i*) $a>b>c>d>0$, and (*ii*) when $a=b=c=d$.
Printed answer:
- $\tfrac{1}{24}(s_a^4 - 6s_a^2s_{2a} + 8s_as_{3a} + 3s_{2a}^2 - 6s_{4a})$.
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Exercise Page142, problem 13i, p. 142
By solving the first $k$ of Newton’s identities $(10)$ as a system of linear equations, find an expression in the form of a determinant (*i*) for $s_k$ in terms of $c_1, \dotsc, c_k$, and (*ii*) for $c_k$ in terms of $s_1, \dotsc, s_k$.
Printed answer:
- s_k = - | arraycccccc 1 & 0 & 0 & …& 0 & c_1 c_1 & 1 & 0 & …& 0 & 2c_2 c_2 & c_1 & 1 & …& 0 & 3c_3 c_3 & c_2 & c_1 & …& 0 & 4c_4 [2]6 c_k-1 &c_k-2 &c_k-3 & …& c_1 & kc_k array|, s_3 = - vmatrix 1 & 0 & c_1 c_1 & 1 & 2c_2 c_2 & c_1 & 3c_3 vmatrix, where all but the last term in the main diagonal is $1$, and all terms above the diagonal are zero except those in the last column. If $k>n$, we must take $c_j =0 \quad (j>n)$.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page142, problem 13ii, p. 142
By solving the first $k$ of Newton’s identities $(10)$ as a system of linear equations, find an expression in the form of a determinant (*i*) for $s_k$ in terms of $c_1, \dotsc, c_k$, and (*ii*) for $c_k$ in terms of $s_1, \dotsc, s_k$.
Printed answer:
- k! c_k = - | arraycccccc 1 & 0 & 0 & …& 0 & s_1 s_1 & 2 & 0 & …& 0 & s_2 s_2 & s_1 & 3 & …& 0 & s_3 [2]6 s_k-1 & s_k-2 & s_k-3 & …& s_1 & s_k array|, 3! c_3 = - vmatrix 1 & 0 & s_1 s_1 & 2 & s_2 s_2 & s_1 & s_3 vmatrix.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page142, problem 14, p. 142
One set of $n$ numbers is a mere rearrangement of another set if $s_1, \dotsc, s_n$ have the same values for each set.
Printed answer:
- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page142, problem 2, p. 142
$\Sigma \alpha_1^2 \alpha_2^2 \alpha_3$.
Printed answer:
- $3c_1c_4 - c_2c_3 - 5c_5$.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles 3*c1*c4 - c2*c3 - 5*c5
Exercise Page142, problem 3, p. 142
[0pt][l]$\Sigma \alpha_1^2 \alpha_2^2 \alpha_3 \alpha_4$.
Printed answer:
- $c_2c_4 - 4c_1c_5 + 9c_6$.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles c2*c4 - 4*c1*c5 + 9*c6
Exercise Page142, problem 4, p. 142
$\Sigma \alpha_1^2 \alpha_2^2 \alpha_3^2$.
Printed answer:
- $c_3^2 - 2c_2c_4 + 2c_1c_5 - 2c_6$.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles c3**2 - 2*c2*c4 + 2*c1*c5 - 2*c6
Exercise Page142, problem 5, p. 142
$\alpha^2$, $\beta^2$, $\gamma^2$.
Printed answer:
- $y^3 - (p^2 - 2q)y^2 + (q^2 - 2pr)y - r^2 = 0$.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles y**3 - (p**2 - 2*q)*y**2 + (q**2 - 2*p*r)*y - r**2
Exercise Page142, problem 6, p. 142
$\alpha\beta$, $\alpha\gamma$, $\beta\gamma$.
Printed answer:
- $y^3 - qy^2 + pry - r^2 = 0$.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles y**3 - q*y**2 + p*r*y - r**2
Exercise Page142, problem 7, p. 142
$\dfrac{2}{\alpha}$, $\dfrac{2}{\beta}$, $\dfrac{2}{\gamma}$.
Printed answer:
- $ry^3 + 2qy^2 + 4py + 8 = 0$.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles r*y**3 + 2*q*y**2 + 4*p*y + 8
Exercise Page142, problem 8, p. 142
$\alpha^2 + \beta^2$, $\alpha^2 + \gamma^2$, $\beta^2 + \gamma^2$.
Printed answer:
- Eliminate $x$ by $y = s_2 - x^2$.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page142, problem 9, p. 142
$\alpha^2 + \alpha\beta + \beta^2$, etc.
Printed answer:
- Use $p^2 - q + px = y$.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles