Jeff Bivin -- LZHS

Finite Differences

By: Jeffrey Bivin

Lake Zurich High School

jeff.bivin@lz95.org

Updated 12-5-08

Jeff Bivin -- LZHS

Determine the degree of the function given the set of finite differences.

x 1 2 3 4 5 6 7 8 9

f(x) 0 13 32 57 88 125 168 217 272

13 19 25 31 37 43 49 55

6 6 6 6 6 6 6

1st

2nd

Jeff Bivin -- LZHS

Now, identify the function of degree 2.

f(x) = ax2 + bx + c

f(2) = a(2)2 + b(2) + c = 4a + 2b + c = 13

f(1) = a(1)2 + b(1) + c = 1a + 1b + c = 0

f(3) = a(3)2 + b(3) + c = 9a + 3b + c = 6

For a 2nd degreefunction, how manypoints do we need to write the equation? 3

We will use this system to solve for

a, b, & c.

Jeff Bivin -- LZHS

Now, identify the function of degree 2.

f(x) = ax2 + bx + c

f(2) = a(2)2 + b(2) + c = 4a + 2b + c = 13

f(1) = a(1)2 + b(1) + c = 1a + 1b + c = 0

f(3) = a(3)2 + b(3) + c = 9a + 3b + c = 6

6139

13124

0111

RREF

a = 3b = 4c = -7

f(x) = ax2 + bx + c

f(x) = 3x2 + 4x - 7

Jeff Bivin -- LZHS

Determine the degree of the function given the set of finite differences.

x 1 2 3 4 5 6 7 8 9

f(x) -4 -7 38 227 704 1661 3338 6023 10052

-3 45 189 477 957 1677 2685 4029

48 144 288 480 720 1008 1344

1st

2nd

96 144 192 240 288 3363rd

48 48 48 48 484th

Jeff Bivin -- LZHS

Now, identify the function of degree 4.

f(x) = ax4 + bx3 + cx2 + dx + e

f(2) = 16a + 8b + 4c + 2d + e = -7

f(1) = 1a + 1b + 1c + 1d + e = -4

f(3) = 81a + 27b + 9c + 3d + e = 38

For a 4th degreefunction, how manypoints do we need towrite the equation? 5

We will use this system to solve for

a, b, & c.

f(4) = 256a + 64b + 16c + 4d + e = 227

f(5) = 625a + 125b + 25c + 5d + e = 704

Jeff Bivin -- LZHS

Solve the system:

7041525125625

227141664256

381392781

7124816

411111

a = 2b = -4c = -2

16a + 8b + 4c + 2d + e = -7

1a + 1b + 1c + 1d + e = -4

81a + 27b + 9c + 3d + e = 38

256a + 64b + 16c + 4d + e = 227

625a + 125b + 25c + 5d + e = 704

RREF

d = 1e = -1

f(x) = ax4 + bx3 + cx2 + dx + e

f(x) = 2x4 – 4x3 – 2x2 + x – 1

Jeff Bivin -- LZHS

Determine the degree of the function

4, 7, 10, 13, 16, 19, 22, 25, 28

3, 3, 3, 3, 3, 3, 3, 31st difference

Jeff Bivin -- LZHS

Jeff Bivin -- LZHS

Now, write the linear model

4, 7, 10, 13, 16, 19, 22, 25, 28

f(1) f(2)

(1, 4)

(2, 7)

313

1247

m

)1(34 xy334 xy13 xy

Jeff Bivin -- LZHS

Jeff Bivin -- LZHS

Determine the degreeof the function

-1, 0, 5, 14, 27, 44, 65, 90, 119

1, 5, 9, 13, 17, 21, 25, 291st difference

4, 4, 4, 4, 4, 4, 4 2nd difference

Jeff Bivin -- LZHS

Jeff Bivin -- LZHS

Now write the quadratic model

-1, 0, 5, 14, 27, 44, 65, 90, 119

f(1) f(2) f(3)

cbnannf 2)(

1)1()1()1( 2 cbacbaf

024)2()2()2( 2 cbacbaf

539)3()3()3( 2 cbacbaf

Jeff Bivin -- LZHS

Jeff Bivin -- LZHS

Now write the quadratic model

-1, 0, 5, 14, 27, 44, 65, 90, 119

f(1) f(2) f(3)

cbnannf 2)(

1)1()1()1( 2 cbacbaf

024)2()2()2( 2 cbacbaf

539)3()3()3( 2 cbacbaf

Jeff Bivin -- LZHS

5139

0124

1111

RREF

a = 2

b = -5

c = 2

252)( 2 nnnf

Jeff Bivin -- LZHS

Determine the degreeof the function

1, 10, 47, 130, 277, 506, 835, 1282, 1865

9, 37, 83, 147, 229, 329, 447, 583

28, 46, 64, 82, 100, 118, 136

18, 18, 18, 18, 18, 183rd difference

2nd difference

1st difference

Jeff Bivin -- LZHS

Jeff Bivin -- LZHS

Now write the quadratic model

f(1) f(2) f(3)

dcnbnannf 23)(

1)1()1()1()1( 23 dcbadcbaf

1, 10, 47, 130, 277, 506, 835, 1282, 1865f(4)

10248)2()2()2()2( 23 dcbadcbaf

473927)3()3()3()3( 23 dcbadcbaf

13041664)4()4()4()4( 23 dcbadcbaf

Jeff Bivin -- LZHS

Jeff Bivin -- LZHS

Now write the quadratic model

f(1) f(2) f(3)

dcnbnannf 23)(

1)1()1()1()1( 23 dcbadcbaf

1, 10, 47, 130, 277, 506, 835, 1282, 1865f(4)

10248)2()2()2()2( 23 dcbadcbaf

473927)3()3()3()3( 23 dcbadcbaf

13041664)4()4()4()4( 23 dcbadcbaf

130141664

4713927

101248

11111

RREF

a = 3 b = -4 c = 0 d = 2 243)( 23 nnnf