# LessonAmortization Formula

#### MarkFL

##### La Villa Strangiato
Staff member
Moderator
Math Helper
Let $$P$$ = monthly payment, $$A$$ = amount borrowed, $$i$$ = monthly interest rate, and $$n$$ = the number of payments.

Also, let $$\displaystyle D_n$$ be the debt amount after payment $$n$$.

Consider the recursion:

(1) $$\displaystyle D_{n}=(1+i)D_{n-1}-P$$

(2) $$\displaystyle D_{n+1}=(1+i)D_{n}-P$$

Subtracting (1) from (2) yields the homogeneous recursion:

$$\displaystyle D_{n+1}=(2+i)D_{n}-(1+i)D_{n-1}$$

whose associated auxiliary equation is:

$$\displaystyle r^2-(2+i)r+(1+i)=0$$

$$\displaystyle \left(r-(1+i)\right)\left(r-1\right)=0$$

Thus, the closed-form for our recursion is:

$$\displaystyle D_n=k_1(1+i)^n+k_2$$

Using initial values, we may determine the coefficients $$\displaystyle k_i$$:

$$\displaystyle D_0=k_1+k_2=A$$

$$\displaystyle D_1=k_1(1+i)+k_2=(1+i)A-P$$

Solving this system, we find:

$$\displaystyle k_1=\frac{Ai-P}{i},\,k_2=\frac{P}{i}$$ and so we have:

$$\displaystyle D_n=\left(\frac{Ai-P}{i}\right)(1+i)^n+\left(\frac{P}{i}\right)=\frac{(Ai-P)(1+i)^n+P}{i}$$

Now, equating this to zero, we can solve for P:

$$\displaystyle \frac{(Ai-P)(1+i)^n+P}{i}=0$$

$$\displaystyle (Ai-P)(1+i)^n+P=0$$

$$\displaystyle (P-Ai)(1+i)^n=P$$

$$\displaystyle P\left((1+i)^n-1\right)=Ai(1+i)^n$$

$$\displaystyle P=\frac{Ai(1+i)^n}{(1+i)^n-1}$$

$$\displaystyle P=\frac{Ai}{1-(1+i)^{-n}}$$

harpazo

#### harpazo

##### Pure Mathematics
Let $$P$$ = monthly payment, $$A$$ = amount borrowed, $$i$$ = monthly interest rate, and $$n$$ = the number of payments.

Also, let $$\displaystyle D_n$$ be the debt amount after payment $$n$$.

Consider the recursion:

(1) $$\displaystyle D_{n}=(1+i)D_{n-1}-P$$

(2) $$\displaystyle D_{n+1}=(1+i)D_{n}-P$$

Subtracting (1) from (2) yields the homogeneous recursion:

$$\displaystyle D_{n+1}=(2+i)D_{n}-(1+i)D_{n-1}$$

whose associated auxiliary equation is:

$$\displaystyle r^2-(2+i)r+(1+i)=0$$

$$\displaystyle \left(r-(1+i)\right)\left(r-1\right)=0$$

Thus, the closed-form for our recursion is:

$$\displaystyle D_n=k_1(1+i)^n+k_2$$

Using initial values, we may determine the coefficients $$\displaystyle k_i$$:

$$\displaystyle D_0=k_1+k_2=A$$

$$\displaystyle D_1=k_1(1+i)+k_2=(1+i)A-P$$

Solving this system, we find:

$$\displaystyle k_1=\frac{Ai-P}{i},\,k_2=\frac{P}{i}$$ and so we have:

$$\displaystyle D_n=\left(\frac{Ai-P}{i}\right)(1+i)^n+\left(\frac{P}{i}\right)=\frac{(Ai-P)(1+i)^n+P}{i}$$

Now, equating this to zero, we can solve for P:

$$\displaystyle \frac{(Ai-P)(1+i)^n+P}{i}=0$$

$$\displaystyle (Ai-P)(1+i)^n+P=0$$

$$\displaystyle (P-Ai)(1+i)^n=P$$

$$\displaystyle P\left((1+i)^n-1\right)=Ai(1+i)^n$$

$$\displaystyle P=\frac{Ai(1+i)^n}{(1+i)^n-1}$$

$$\displaystyle P=\frac{Ai}{1-(1+i)^{-n}}$$
Very impressive work.

MarkFL