If you think about the relative deviation from the original amortisation process, you simply want to know the value $\frac{|R-R'|}{P}$, i.e.:
$$
\sum_{j=0}^M(1+i)^{t_j}\cdot (1+i)^{-\tau_j}-(1+i)^{t_j},
$$
Thus:
$$
\sum_{j=0}^M(1+i)^{t_j}((1+i)^{-\tau_j}-1).
$$
Since you say that the maximum deferral date is one month, i.e. $0\leq\tau_j\leq1, \forall j>0$.
Hence:
$$
\sum_{j=0}^M(1+i)^{t_j}((1+i)^{-\tau_j}-1)\geq0,
$$
and
$$
\sum_{j=0}^M(1+i)^{t_j}((1+i)^{-\tau_j}-1)\leq\sum_{j=0}^M(1+i)^{t_j}((1+i)^1-1)=
$$
$$
=i\sum_{j=0}^M(1+i)^{t_j}=i\cdot\frac{R}{P}.
$$
Thus:
$$
\frac{R'-R}{P}\leq i\frac{R}{P},
$$
Therefore:
$$
\frac{R'-R}{R}\leq i.
$$
I guess this is what you can prove under the hypothesis you stated above.
Okay:
Formula: $n = -(1/30) \cdot \ln(1+b/p(1-(1+i)^{30})) / \ln(1+i)$
First problem
n = 40
b = 7500
p = 250
i = APR/365; we will solve for APR
plug those in
$ 40 = -(1/30) \cdot \ln(1+7500/250(1-(1+APR/365)^{30})) / \ln(1+APR/365) =-(1/30) \cdot \ln(1+30(1-(1+APR/365)^{30}))/ \ln(1+APR/365)$
multiply both side by 30 and then by $ \ln(1+APR/365)$
$1200 = -\ln(1+30(1-(1+APR/365)^{30}))/ \ln(1+APR/365)$
$1200\ln(1+APR/365) = \ln(1+30(1-(1+APR/365)^{30}))$
We get rid of the $ln$s by rising e to both powers. (Trust me, the e's will vanish.)
$e^{1200\ln(1+APR/365)} = e^{-\ln(1+30(1-(1+APR/365)^{30}))} $
$(e^{\ln(1+APR/365)})^{1200} = (e^{\ln(1+30(1-(1+APR/365)^{30}))})^{-1} $
$(1+APR/365)^{1200} = (1+30(1-(1+APR/365)^{30}))^{-1}= \frac{1}{(1+30(1-(1+APR/365)^{30}))} $
Okay, there's no way in heck I'm going to deal with 1200 roots. So I'm going to cheat.
By the binomial theorem we know $(1 + x)^n = 1 + nx + n(n-1)/2*x^2 + ....$ and if x is very small all the x^i terms will become so small as to be insignificant. So I can estimate $$(1 + x)^n = 1 + nx$ (but only for very small values of x.) Now APR/365 will be a very small number. So I will approximate $(1 + APR/365)^{1200} = 1 + 1200APR365$ and $(1 + APR/365)^{30} = 1 + 30APR365$. So:
$1+1200APR/365 = \frac{1}{(1+30(1-(1+30APR/365)))} = \frac{1}{1+30(-30APR/365)}= \frac{1}{1-900APR/365}$
Multiply both sides by $(1-900APR/365)$
$(1+1200APR/365)(1-900APR/365) = 1$
Pour yourself a stiff drink and expand:
$1 + 1200APR/365 - 900APR/365 - 1200*900(APR/365)^2 = 1$
$1200*900(APR/265)^2 = 300APR/365$
We will assume APR isn't 0 so we divide both sides by $300APR/365$ to get
$1200*900(APR/365)^2/(300APR/365) = 1$
$360000APR/365 = 1$
so
APR = 365/360000 = 0.00101388888888888888888888888889 = 1.01%
Unless I made an error which I almost certainly did. SHEESH
Best Answer
Another way to look at the problem that might make the following answer make a little more sense is to instead try to find roots to the polynomial
$$(x+1)^{12}-10x(x+1)^{12}-1.$$
We will need to be careful to omit the value $-1$ from consideration as the function you have given is not defined at $x=-1$.
In general, polynomials of degree $d\ge 5$ are not solvable, i.e. there is no clean way to write out all their roots in terms of addition, subtraction, multiplication, or taking $n^{th}-$roots of rational numbers. While there are specific cases where an algorithm exists, since your given polynomial has degree $13$, I'd be very surprised to see an algorithm to explicitly determine all the roots.
HOWEVER, if you are willing to put in a little bit of elbow grease, Newton's Method for finding roots of polynomials often offers a relatively quick way of finding values very close to being a root. The algorithm is as follows:
Let $x_0$ be any value, preferably one that you suspect is close to being a root.
Let $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$, and repeat until satisfied.
The general idea of Newton's Method is that, looking at a very small range of inputs, the tangent line of a function at a point is a pretty good estimate of the function. Thus, for a very small range the line $$ y-f(x_0) = f'(x_0)(x-x_0) $$ is a good estimate for our polynomial $f$. A root will be where $y=0$, so we find the root of our tangent line is (after moving terms around) $$ x =x_0 -\frac{f(x_0)}{f'(x_0)} = x_1. $$ As we iterate further through the algorithm, our estimates get more and more refined.
In your particular example, using $$ f=(x+1)^{12}-10x(x+1)^{12}-1 \qquad x_0 = \frac{1}{10} $$ the first five estimates for Newtons Method are $$ x_1 = 0.0681369 \qquad x_2 = 0.0471467 \qquad x_3 = 0.0352813 \qquad x_4 = 0.0303015 \qquad x_5=0.0292729 $$
Hopefully, this helps! Thanks for the awesome question :D !!