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Finite Differences By Daniela G. Ambrosano Finite Differences Finite differences is a method to find the polynomial that could generate a sequence, assuming the sequence can be generated by a polynomial. Finding Successive Differences 15 Find successive difference between terms of sequence; the difference sequence. Term #n f(n) 1 5 2 15 15 - 5 = 10 3 33 33 - 15 = 18 4 59 59 – 33 =26 5 93 93 – 59 = 34 Finding differences, continued… Continue finding differences between the terms of difference sequences until the difference sequence is a constant. Term#n f(n) 1 5 2 15 15 - 5 = 10 3 33 33 - 15 = 18 18 - 10 = 8 4 59 59 - 33 = 26 26 - 18 = 8 5 93 93 - 59 = 34 34 - 26 = 8 Continuing on… The second difference in the sequence is constant so we need a 2nd degree polynomial. General Form: an bn c 2 Why? The number of difference sequences is the power of the polynomial that generated the original sequence. This is because it takes that many derivatives to get to a constant. Generating a system (n+1) equations in (n+1) variables The variables in (n+1) are the coefficients of the general polynomial. To generate a system (n+1) equations in (n+1) variables, we subtract the various term numbers to equal the corresponding term values. For n = 1 an bn c a(1 ) b(1) c 5 2 2 Polynomials 2 For n = 1 a(1 ) b(1) c 5 For n=2 For n = 3 a(2 ) b(2) c 15 2 a(3 ) b(3) c 33 2 Solve the System of Equations Solve the system of equations to find the coefficients/variables in two ways: *substitution *elimination Eliminating C the 2nd equation for n=2 and subtract from it the 1st equation for n=1 Take 4a 2b c 15 (a b c 5) 3a b 10 take the 3rd equation for n=3 and subtract from it the 2nd equation for n=2 9a 3b c 33 Now (4a 2b c 15) 5a b 18 Eliminate b and Solve for a Now we use the new equations 3a + b = 10 and 5a + b = 18, subtract them and then solve for a 5a b 18 (3a b 10) 2a 8 a4 Substituting Now we can substitute a = 4 in 3a + b = 10 to find the value of b 3(4) b 10 b 2 Then substitute the values of a = 4 and b = -2 into the 1st equation a + b + c = 5 to find the value of c 4 (2) c 5 c3 What is f(n)? Now that we know the values of a, b, and c we can substitute them into the original 2nd degree polynomial that will generate the difference sequences. f (n) 4n (2)n 3 2 Check Polynomial Plug in a value for n to check if the polynomial generated is correct Ex: n = 2 f (2) 4(2 ) (2) 3 2 16 (4) 3 15 Original Sequence Table Term # n 1 2 3 4 5 f(n) 5 15 33 59 93 The End