Chapter 3.6, Problem 64E

Solution

To justify: The total interest paid on a $15-\text{year}$ mortgage is less than half of the total interest paid on a $30-\text{year}$ mortgage with the same loan amount and $\text{APR}$ .

True.

Given information: $15-\text{year}$ mortgage is less than half of the total interest paid on a $30-\text{year}$ mortgage.

Calculation:

To calculate the periodical payment $R$ ,

  $\begin{array}{c}pv=a\\ R\frac{1-{\left(1+i\right)}^{-n}}{i}=A\\ R=\frac{Ai}{1-{(1+i)}^{-n}}\end{array}$

The total amount paid in interest is,

  $\begin{array}{c}nR-A=n\frac{Ai}{1-{(1+i)}^{-n}}-A\\ =A\left(\frac{ni}{1-{(1+i)}^{-n}}-1\right)\end{array}$

Let’s define,

  $I(n)=A\left(\frac{ni}{1-{(1+i)}^{-n}}-1\right)$

From the question it is known that,

  $I(180)<\frac{1}{2}I(360)$

Substituting the formula in above equation

  $A\left(\frac{180i}{1-{(1+i)}^{-180}}-1\right)<\frac{1}{2}A\left(\frac{360i}{1-{(1+i)}^{-360}}-1\right)$

Divide by $A$ on both sides to get,

  $\left(\frac{180i}{1-{(1+i)}^{-180}}-1\right)<\frac{1}{2}\left(\frac{360i}{1-{(1+i)}^{-360}}-1\right)$

Multiply by $\text{2}$ on both sides to get,

  $\left(\frac{360i}{1-{(1+i)}^{-180}}-2\right)<\left(\frac{360i}{1-{(1+i)}^{-360}}-1\right)$

Rearrange,

  $\left(\frac{360i}{1-{(1+i)}^{-180}}-2\right)-\left(\frac{360i}{1-{(1+i)}^{-360}}-1\right)<1$

Graph:

To find the graph for the following function,

  $f(x)=\frac{360x}{1-{(1+x)}^{-180}}-\frac{360x}{1-{(1+x)}^{-360}}$

  

Interpretation:

It is observed that the above graph the inequality is true for all values of $i\in (0,1)$ .