How do you solve the system $- 5 = - 64 a + 16 b - 4 c + d$ $-5 = -64a + 16b - 4c + d$ , $- 4 = - 27 a + 9 b - 3 c + d$ $-4 = -27a + 9b - 3c + d$ , $- 3 = - 8 a + 4 b - 2 c + d$ $-3 = -8a + 4b - 2c + d$ , $4 = - a + b - c + d$ $4 = -a + b - c + d$ ?

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How do you solve the system $- 5 = - 64 a + 16 b - 4 c + d$ $-5 = -64a + 16b - 4c + d$ , $- 4 = - 27 a + 9 b - 3 c + d$ $-4 = -27a + 9b - 3c + d$ , $- 3 = - 8 a + 4 b - 2 c + d$ $-3 = -8a + 4b - 2c + d$ , $4 = - a + b - c + d$ $4 = -a + b - c + d$ ?

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Answer 1 · 2017-03-05T10:23:44

${ (a=1), (b=9), (c=27), (d=23) :}$

Explanation:

Given:

${ (-5 = -64a+16b-4c+d), (-4 = -27a+9b-3c+d), (-3 = -8a+4b-2c+d), (4 = -a+b-c+d) :}$

Consider the function:

$f(x) = ax^3+bx^2+cx+d$

Note that the given system of equations is equivalent to:

${ (f(-4) = -5), (f(-3) = -4), (f(-2) = -3), (f(-1) = 4) :}$

Since the sampling points are at equal intervals of size $1$ , we can conveniently find a formula for a cubic function satisfying this system by examining differences...

Start with the original values:

$color(blue)(-5), -4, -3, 4$

Write down the sequence of differences between successive terms:

$color(blue)(1), 1, 7$

Write down the sequence of differences between successive terms:

$color(blue)(0), 6$

Write down the sequence of differences between successive terms:

$color(blue)(6)$

Then we can write down a formula for $f(x)$ using the initial term of each of these sequences (see footnote):

$f(x) = color(blue)(-5)/(0!)+color(blue)(1)/(1!)(x+4)+color(blue)(0)/(2!)(x+4)(x+3)+color(blue)(6)/(3!)(x+4)(x+3)(x+2)$

$color(white)(f(x)) = -5+x+4+x^3+9x^2+26x+24$

$color(white)(f(x)) = x^3+9x^2+27x+23$

So:

${ (a=1), (b=9), (c=27), (d=23) :}$

$color(white)()$
Footnote

Note that:

$color(purple)(1/(0!))_(color(white)(1/1))$ is a constant function taking the value $1$ when $x = -4$

$color(purple)(1/(1!)(x+4))_(color(white)(1/1))$ is a linear function taking the value $0$ when $x = -4$ and $1$ when $x = -3$

$color(purple)(1/(2!)(x+4)(x+3))_(color(white)(1/1))$ is a quadratic function taking the value $0$ when $x = -4$ or $x = -3$ and the value $1$ when $x = -2$

$color(purple)(1/(3!)(x+4)(x+3)(x+2))_(color(white)(1/1))$ is a cubic function taking the value $0$ when $x = -4$ , $x = -3$ or $x = -2$ and the value $1$ when $x = -1$

So as we add suitable multiples of these functions in turn, we get a sequence of polynomials of increasing degree that match each of the sample points in turn. The suitable multiples are the differences we found.

Answer 2 · 2017-03-05T12:08:17

${ (a=1), (b=9), (c=27), (d=23) :}$

Explanation:

Given:

${ (-5 = -64a+16b-4c+d), (-4 = -27a+9b-3c+d), (-3 = -8a+4b-2c+d), (4 = -a+b-c+d) :}$

Write in matrix form:

$((-64, 16, -4, 1, -5),(-27, 9, -3, 1, -4),(-8, 4, -2, 1, -3),(-1, 1, -1, 1, 4))$

Perform some row operations to make the left hand side into a $4xx4$ identity matrix. These may not be optimal, but this is a sequence I found...

Subtract $3xx"row"2$ from $"row"1$ to get:

$((17, -11, 5, -2, 7),(-27, 9, -3, 1, -4),(-8, 4, -2, 1, -3),(-1, 1, -1, 1, 4))$

Add $2xx"row"1$ to $"row"2$ to get:

$((17, -11, 5, -2, 7),(7, -13, 7, -3, 10),(-8, 4, -2, 1, -3),(-1, 1, -1, 1, 4))$

Add $2xx"row"3$ to $"row"1$ to get:

$((1, -3, 1, 0, 1),(7, -13, 7, -3, 10),(-8, 4, -2, 1, -3),(-1, 1, -1, 1, 4))$

Add $"row"1 + "row"3$ to $"row"2$ to get:

$((1, -3, 1, 0, 1),(0, -12, 6, -2, 8),(-8, 4, -2, 1, -3),(-1, 1, -1, 1, 4))$

Add $8xx"row"1$ to $"row"3$ to get:

$((1, -3, 1, 0, 1),(0, -12, 6, -2, 8),(0, -20, 6, 1, 5),(-1, 1, -1, 1, 4))$

Add $"row"1$ to $"row"4$ to get:

$((1, -3, 1, 0, 1),(0, -12, 6, -2, 8),(0, -20, 6, 1, 5),(0, -2, 0, 1, 5))$

Subtract $"row"2$ from $"row"3$ to get:

$((1, -3, 1, 0, 1),(0, -12, 6, -2, 8),(0, -8, 0, 3, -3),(0, -2, 0, 1, 5))$

Subtract $"row"3+"row"4$ from $"row"2$ to get:

$((1, -3, 1, 0, 1),(0, -2, 6, -6, 6),(0, -8, 0, 3, -3),(0, -2, 0, 1, 5))$

Divide $"row"2$ by $-2$ to get:

$((1, -3, 1, 0, 1),(0, 1, -3, 3, -3),(0, -8, 0, 3, -3),(0, -2, 0, 1, 5))$

Subtract $4xx"row"4$ from $"row"3$ to get:

$((1, -3, 1, 0, 1),(0, 1, -3, 3, -3),(0, 0, 0, -1, -23),(0, -2, 0, 1, 5))$

Add $"row"3$ to $"row"4$ to get:

$((1, -3, 1, 0, 1),(0, 1, -3, 3, -3),(0, 0, 0, -1, -23),(0, -2, 0, 0, -18))$

Multiply $"row"3$ by $-1$ and divide $"row"4$ by $-2$ to get:

$((1, -3, 1, 0, 1),(0, 1, -3, 3, -3),(0, 0, 0, 1, 23),(0, 1, 0, 0, 9))$

Permute rows $2$ , $3$ and $4$ to get:

$((1, -3, 1, 0, 1),(0, 1, 0, 0, 9),(0, 1, -3, 3, -3),(0, 0, 0, 1, 23))$

Subtract $"row"2$ from $"row"3$ to get:

$((1, -3, 1, 0, 1),(0, 1, 0, 0, 9),(0, 0, -3, 3, -12),(0, 0, 0, 1, 23))$

Divide $"row"3$ by $-3$ to get:

$((1, -3, 1, 0, 1),(0, 1, 0, 0, 9),(0, 0, 1, -1, 4),(0, 0, 0, 1, 23))$

Add $"row"4$ to $"row"3$ to get:

$((1, -3, 1, 0, 1),(0, 1, 0, 0, 9),(0, 0, 1, 0, 27),(0, 0, 0, 1, 23))$

Add $3xx"row"2 - "row"3$ to $"row"1$ to get:

$((1, 0, 0, 0, 1),(0, 1, 0, 0, 9),(0, 0, 1, 0, 27),(0, 0, 0, 1, 23))$

We can now read off $a, b, c, d$ from the last column as:

${ (a=1), (b=9), (c=27), (d=23) :}$

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How do you solve the system $- 5 = - 64 a + 16 b - 4 c + d$ $-5 = -64a + 16b - 4c + d$ , $- 4 = - 27 a + 9 b - 3 c + d$ $-4 = -27a + 9b - 3c + d$ , $- 3 = - 8 a + 4 b - 2 c + d$ $-3 = -8a + 4b - 2c + d$ , $4 = - a + b - c + d$ $4 = -a + b - c + d$ ?

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Explanation:

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How do you solve the system -5 = -64a + 16b - 4c + d−5=−64a+16b−4c+d-5 = -64a + 16b - 4c + d, -4 = -27a + 9b - 3c + d−4=−27a+9b−3c+d-4 = -27a + 9b - 3c + d, -3 = -8a + 4b - 2c + d−3=−8a+4b−2c+d-3 = -8a + 4b - 2c + d, 4 = -a + b - c + d4=−a+b−c+d4 = -a + b - c + d?

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Explanation:

Explanation:

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How do you solve the system $- 5 = - 64 a + 16 b - 4 c + d$ $-5 = -64a + 16b - 4c + d$ , $- 4 = - 27 a + 9 b - 3 c + d$ $-4 = -27a + 9b - 3c + d$ , $- 3 = - 8 a + 4 b - 2 c + d$ $-3 = -8a + 4b - 2c + d$ , $4 = - a + b - c + d$ $4 = -a + b - c + d$ ?