Mixed Recursion, Part 2 Molly, at age 25, has decided it’s not too early to plan for retirement already. She wants to consider how much money she could have at age 60 if she begins investing now. She plans to deposit $5,000 into an account that pays 4.8% interest, compounded monthly, and she will add another $200 at the end of each month.

Mixed Recursion, Part 2 We could use a recurrence relation for this problem, and then use a spreadsheet to help us calculate how much money will be in Molly’s account by age 60. Can you see any disadvantages of a spreadsheet in this case?

Mixed Recursion, Part 2 Think about how big the spreadsheet would become. We’re going to be looking at a period of 35 years, and our recurrence relation applies to each month. Our spreadsheet would have 35(12), or 420 rows. Do you want to scroll through 420 rows to get to the answer?

Mixed Recursion, Part 2 Think about how big the spreadsheet would become. We’re going to be looking at a period of 35 years, and our recurrence relation applies to each month. Our spreadsheet would have 35(12), or 420 rows. Do you want to scroll through 420 rows to get to the answer? What if there were a quicker way?

Mixed Recursion, Part 2 You lucky dogs, you! There is indeed a quicker way to get to the punch line of this investment problem. Instead of using a recurrence relation, we’ll look for a closed-form solution. Hey, it worked for arithmetic and geometric recursions, so why not for a mixed recursion?

Mixed Recursion, Part 2 Closed-form Solution For a Mixed Recursion For a recurrence relation t n = at n-1 + b The closed-form solution is t n = (t 1 – p)(a n-1 ) + p where p is the fixed point.

Mixed Recursion, Part 2 For our example with Molly, the recurrence relation is B n = 1.004B n so a = 1.004, and we need to find p. p = 1.004p =.004p p = -50,000

Mixed Recursion, Part 2 Now we have the terms we need to make our closed-form solution: B n = (5,220 – (-50,000))(1.004 n-1 ) – 50,000 B n = (55,220)(1.004 n-1 ) – 50,000 (Note that B 1 is the value at the end of the first month: 5,000(1.004) = 5,220.)

Mixed Recursion, Part 2 Now we can determine how much money Molly will have after 35 years, or 420 months. B n = (55,220)(1.004 n-1 ) – 50,000 B 420 = (55,220)( ) – 50,000 Any estimates for how much it will be?

Mixed Recursion, Part 2 Now we can determine how much money Molly will have after 35 years, or 420 months. B n = (55,220)(1.004 n-1 ) – 50,000 B 420 = (55,220)( ) – 50,000 = $244,118.32

Mixed Recursion, Part 2 Molly might also be interested in knowing when her balance will reach a given amount, say $100,000. We can do that. B n = (55,220)(1.004 n-1 ) – 50, ,000 = (55,220)(1.004 n-1 ) – 50, ,000 = (55,220)(1.004 n-1 ) = n-1

Mixed Recursion, Part n-1 = We need to get to the exponent, which means we need to use logarithms. (n – 1)log(1.004) = log(2.7164) (n – 1) = [log(2.7164)]/[log(1.004)] n – 1 = n = So it will take about 252 months, or 21 years.