TVM solve for interest rate, revisited

[Albert Chan]

(06-26-2024 02:31 PM)Albert Chan Wrote:  let Q = 1/q ≈ n*log(2)      → (1+i)^Q ≈ (1+i0)^Q - 1

Previous post showed why ln(2) is there, because RHS is (s-1)
If RHS is (s-2), then ln(3) pops out ... prove is not complete.

Let's try again, assuming n is huge. (i is interest rate, not √(-1))

C = i*n / (1-(1+i)^-n) ≈ i*n / (1-e^(-i*n)) = 1 + (n*i)/2 + (n*i)²/12 - ...

s = (1+i0)^Q = (1+C/n)^Q ≈ (1 + 1/n + i/2 + n/12*i² - ...)^Q

(1+1/n)^Q ≈ ((1+1/n)^n) ^ ln(2) ≈ e ^ ln(2) = 2

We can pull out tiny 1/n term, as correction instead (note: n ≈ Q/ln(2))

RHS = s-1 ≈ 2 * (1 + i/2 + n/12*i² - ...)^Q - 1 ≈ 1 + Q*i + 0.490*(Q*i)² - ...

LHS = (1+i)^Q = 1 + Q*i + Q*(Q-1)/2*i² + ...  ≈ 1 + Q*i + 0.500*(Q*i)² + ...

QED

With Q=n*ln(2), RHS is still smaller.

Assuming positive n, C ≈ i*n / (1-e^(-i*n)) is understimated.
C = i*n / (1-(1+i)^-n) = 1 + (n+1)/2*i + (n²-1)/12*i² - (n²-1)/24*i³ - ...

Assuming positive n, factor 2 also under-estimated.
factor^(1/Q) = (1+C/n) / (1+(C-1)/n) = (1 + 1/(n+C-1)) ≤ (1+1/n)

If the goal is to get factor ≈ 2, take log2 both side:

1/Q = q ≈ log2(1+1/(n+C-1))      → Q ≈ (n+C-.5) * ln2