Question

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

Answer 1

`{ (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

`{ (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) :}`