Mortgage Rates: truth about The Real Rate vs. The Advertised Rate

Real Rate vs. Advertised Rate

The mortgage industry cheats consumers by advertising a lower interest rate, than what is actually charged. If $x$ is the advertised rate, then the real rate you end up paying is
$y = (1+\frac{x}{12})^{12} - 1$
e.g. taking $x = 0.03375$ for a 3.375% interest rate, the real rate is
$y = (1+\frac{0.03375}{12})^{12} - 1 = 0.03428$ or 3.428%

How?

If $y$ = the real rate of interest per year, then to calculate the monthly rate of interest $r$ , we need to solve $(1+y)^t = (1+r)^{12t}$ . This gives
$r = (1+y)^{\frac{1}{12}} - 1$
Whereas if $x$ is the advertised rate, then the monthly rate of interest is calculated as $\frac{x}{12}$ (refer to the wikipedia article where it says: Since the quoted yearly percentage rate is not a compounded rate, the monthly percentage rate is simply the yearly percentage rate divided by 12). So
$\frac{x}{12} = (1+y)^{\frac{1}{12}} - 1$
or,
$y = (1+\frac{x}{12})^{12} - 1$

Monthly payment Calculation:
Let $p$ = amount borrowed from the bank a.k.a. loan amount
$r$ = monthly rate of interest
$z$ = monthly payment
At end of Month 1:
Amount Due = $p(1+r)$
You pay $z$
Balance $p_1 = p(1+r) - z$

At end of Month 2:
Amount Due = $p_1(1+r)$
You pay $z$
Balance $p_2 = p_1(1+r) - z$

At end of $n$ -th month (when loan is paid in full):
Amount Due = $p_{n-1}(1+r)$
You pay $z$
Balance $p_n = p_{n-1}(1+r) - z = 0$
This can be written as
$p(1+r)^n - \sum_{i=0}^{i=n-1}(1+r)^{i}z = 0$
which can be solved to give
$z = pr\frac{(1+r)^n}{(1+r)^n-1}$
Total payments over $n$ months, $T = nz$
Total Interest paid to the bank $I = T - p = p(nr\frac{(1+r)^n}{(1+r)^n-1} - 1)$
is directly proportional to $p$ => the more you borrow, the more interest you will end up paying.
First Order Approximation of $z$ : Assume $r \ll 1$ , then $z = p(r+\frac{1}{n})$