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#?

2 Answers
Mar 5, 2017

Answer:

#{ (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.

Mar 5, 2017

Answer:

#{ (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) :}#