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Expressing Sequences Explicitly

Expressing Sequences Explicitly. By: Matt Connor Fall 2013. Pure Math Analysis Calculus and Real Analysis Sequences. Sequence- A list of numbers or objects in a specific order 1,3,5,7,9,..... Finite Sequence- contains a finite number of terms 2,4,6,8

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Expressing Sequences Explicitly

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  1. Expressing Sequences Explicitly • By: Matt Connor • Fall 2013

  2. Pure Math • Analysis • Calculus and Real Analysis • Sequences

  3. Sequence- A list of numbers or objects in a specific order • 1,3,5,7,9,..... • Finite Sequence- contains a finite number of terms • 2,4,6,8 • Infinite Sequence- contains an infinite number of terms • 2,4,8,16, ........

  4. Arithmetic Sequence- add or subtract a constant to get from one term to the next • 88, 77, 66, 55,....... • Geometric Sequence- multiply or divide by a common ratio to get from one term to the next • 6, 12, 24, 48,........

  5. Recursive Formula- formula for a sequence that relates the previous term(s) to find the new one. • ex: An = A(n-1)+ 4 • Explicit Formula- formula that finds any term in the sequence without knowing any other terms. • ex: An = 1+ 2(n-1) • all you need to know is n

  6. Arithmetic Sequences • General Forms • Recursive formula • An = A(n-1) + d • Explicit formula • An = A1 + d(n-1)

  7. Geometric Sequences • General Forms • Recursive formula • An = r(An-1) • Explicit formula • An = A1 (rn-1)

  8. What about sequences that are not arithmetic or geometric? • This means they do not have a common constant or ratio • These are commonly called Fibonacci-type • The difficult thing about these is finding an explicit formula

  9. Fibonacci Sequence Explicit Formula • Now we will go through deriving an explicit formula for the Fibonacci Sequence • We know the relational formula is • An = An−1 + An−2 • We guess an explicit formula of the form An =Cxn and plug it in to the relational equation and get • Cxn= Cxn−1 + Cxn−2

  10. Cxn = Cxn−1 + Cxn−2 this will always simplify to an equation with the same coefficients as the relational equation, • x2 = x + 1 • Then we collect the terms on one side to use the quadratic formula. • x2 −x−1=0

  11. The quadratic formula gives us x=(1/2)(1±√5) • Therefore: An= B((1/2)(1+√5))n + C((1/2)(1-√5))n • Next we use the first two Fibonacci numbers to find two equations representing B and C • A0=1 and A1=1

  12. This gives us two equations for B and C • B+C=1 and • B(1/2)(1+√5) + C(1/2)(1-√5)=1 • Then we simplify the second equation we have • (B + C) + (B - C)√5 = 2 and since our first equation tells us that B+C=1 we can replace that.

  13. 1 + (B-C)√5 = 2 • We then further simplify this to get the second of our two equations • B+C=1 and B-C=1/√5 • If we add these two equations and simplify we can then solve for B • B= (√5+1)/(2√5)

  14. And then insert the value of B to find the value of C • C=(√5-1)/(2√5) • One More Step!!

  15. If we replace the B and C in our equation for An • This is Binet’s formula, an explicit formula for finding the nth Fibonacci number. An=

  16. As you have seen finding an explicit formula for the nth term in a Fibonacci-type sequence is much more difficult. • . . . . . but they are possible to find!

  17. Resources • http://www.kenston.k12.oh.us/khs/academics/math/AA_11-3A_geometric_sequences_explicit.pdf • http://www.kenston.k12.oh.us/khs/academics/math/AA_11-2A_arithmetic_sequences.pdf • http://www.geom.uiuc.edu/~demo5337/s97b/fibonacci.html • http://faculty.mansfield.edu/hiseri/MA1115/1115L30.pdf

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