TVM solve for interest rate, revisited

[Albert Chan]

(06-10-2022 08:25 PM)Albert Chan Wrote:  g = n*x / ((1+x)^n-1) = 1 - (n-1)/2*x + (n²-1)/12*x² - (n²-1)/24*x³ + ...

For integer n, g is decaying if n>0, growing if n<0 (= compounding factor C)
Note: if n=1, g=1; if n=-1, g=1+x ⇒ For n=±1, f is a straight line.

g ≥ 1 - (n-1)/2*x            // x=0 tangent line, but inequality holds for other x.

We relied on above identity to prove g'' > 0 ⇒ f'' has same sign throughout.

It is easy to show with plots (pick any n>1), but harder to proof.
(Perhaps I missed something simple ...)

I had placed this question of proving it here (response welcome)

I managed to proof it, using XCAS.

Let R = 1+x > 0, to prove:

g = n*(R-1)/(R^n-1) ≥ 1 - (n-1)/2*(R-1)

→ Or, n ≥ (1 - (n-1)/2*(R-1)) * (R^n-1)/(R-1)      // note: multiplied factor > 0

(RHS, R)' = (1 - n*(n+1)/2*R^(n-1) + (n²-1)*R^n - n*(n-1)/2*R^(n+1)) / (R-1)^2

If n > 1, numerator have 3 sign changes, 1 or 3 positive roots (Descartes rules of signs)
Taking taylor series, we showed that it had triple equal roots, at x=0

XCAS> M := (1 - n*(n+1)/2*R^(n-1) + (n²-1)*R^n - n*(n-1)/2*R^(n+1))
XCAS> series(M(R=1+x), x, 0, 4, polynom)

(n/6-n^3/6)*x^3 + (-n/4+n^2/8+n^3/4-n^4/8)*x^4

2 roots get cancelled by denominator, x^2, we have only 1 extremum, at x=0
Taking limit at x=0, we have

LHS = g(0) = 1
RHS = 1 - (n-1)/2*0 = 1         → LHS = RHS

For x≠0, two sides don't touch. Example, pick x=∞

LHS = g(∞) = 0 (decay to nothing)
RHS = 1 - (n-1)/2*∞ = -∞      → LHS > RHS

For x > -1, LHS ≥ RHS      QED