Question

Consider the tables created using an initial investment of $1,000 and quarterly compounding of interest.

Table A represents the function that models the total amount of one investment, a(x), based on the annual interest rate, x, as a percent.

Table B represents the function that models the interest rate, r(d), as a percent, based on the total amount at the end of the investment, d.

Answer 1

Assume the investment runs for one year with quarterly compounding and initial principal P = $1,000.

Table A (amount as a function of annual rate x, given as a percent)
- Quarterly rate = (x/100)/4 = x/400.
- Amount after one year: a(x) = 1000*(1 + x/400)^4, where x is in percent.

Table B (annual rate as a function of final amount d, returned as a percent)
- Solve d = 1000*(1 + x/400)^4 for x:
(d/1000)^{1/4} = 1 + x/400
x = 400*((d/1000)^{1/4} - 1).
- So r(d) = 400*((d/1000)^{1/4} - 1) (percent).

Domains/notes
- Need 1 + x/400 > 0, so x > −400 (practically x will be a nonnegative interest rate).
- d must be > 0; for positive principal, d > 0 and the formulas above give the inverse.

Example rows
- x = 2% -> a(2) = 1000*(1.005)^4 ≈ $1,020.15
- x = 5% -> a(5) = 1000*(1.0125)^4 ≈ $1,050.95
- x = 8% -> a(8) = 1000*(1.02)^4 ≈ $1,082.43

Inverse checks
- r(1082.43) = 400*((1082.43/1000)^{1/4} − 1) ≈ 8%.