(11C) TVM for HP-11C

[Albert Chan]

(12-03-2020 08:53 PM)Albert Chan Wrote:  Iterations were setup with Newton's method.
But, denominator is not quite f'(i), resulting in better correction.

More important reason for modified denominator is safety.
Note: below proofs required Bernoulli inequality, i>-1, n ≤ 0, or n ≥ 1: \(\quad(1+i)^n ≥ 1 + n\,i\)

Quote:Case FV=0: a*i/(1-k) = 1       → f(i) = a*i - (1-k) = 0       → f'(i) = a - k*n/(1+i)

If we use f'(i) for Newton's denominator, and f'(i) ≤ 0, it is likely it would converge to trivial solution, i=0

Replacing with modified slope, (1-k)/i - k*n/(1+i), we remove this problem.

Proof:
Let K = 1/k = (1+i)^n ≥ 1 + n*i       → (K-1)/i ≥ n

Modified slope = \(\large {1-k \over i} - {k\,n \over 1+i}
= k \left({K-1 \over i} - {n \over 1+i}\right)
≥ k \left(n - {n \over 1+i}\right) = {k\,n\,i \over i+1} \)

If i > 0, n ≥ 1, then modified slope will be positive.

Quote:Case PV=0: a*i/(K-1) = 1       → f(i) = a*i - (K-1) = 0       → f'(i) = a - K*n/(1+i)

Same issue as above. But, this time we wanted negative "slope".
Replacing with modified slope, (K-1)/i - K*n/(1+i), we remove this problem.

Proof:

Modified slope = \(\large{(K-1)(1+i)\;-\;K\,n\,i \over i\,(1+i)}
= {K\over i\,(1+i)} \normalsize (1+(1-n)\,i - (1+i)^{1-n} )\)

Applying Bernoulli inequality, last term is non-positive.

If i > 0, n ≥ 2, then modified slope will be negative.