math.tex

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\title{Math Notes for Compound Interest}
\author{}
\date{}

\begin{document}
\maketitle

\section{Basic Formula for Debt Owed}

First let's calculate the debt after a time period $t$ given a principal $P$, an
interest rate $r$, a compound frequency $n$, a cyclic payment $p$, and a payment
frequency $m$.
It is assumed that $n \geq m$.

\begin{align*}
    a &= 1 + \frac{r}{n}\\
    D_0 &= P\\
    D_1 &= aP\\
    D_2 &= a^2 P\\
        &\vdots\\
    D_{n/m} &= a^{n/m} P - p\\
    D_{n/m+1} &= a (a^{n/m} P - p)\\
    D_{n/m+2} &= a^2 (a^{n/m} P - p)\\
            &\vdots\\
    D_{2n/m} &= a^{n/m} (a^{n/m} P - p) - p\\
             &= a^{2n/m} P - a^{n/m} p - p\\
             &= a^{2n/m} P - (a^{n/m} + 1) p\\
           &\vdots\\
    D_{nt} &= a^{nt} P - p \sum_{i = 0}^{mt - 1} a^{in/m}\\
\end{align*}

Remember that we can find the solution to the finite exponential series as so
\begin{align*}
    x &= \sum_{i = 0}^{n - 1} a^i\\
    a x &= \sum_{i = 1}^{n} a^i\\
    a x - a^n + 1 &= \sum_{i = 0}^{n - 1} a^i\\
    a x - a^n + 1 &= x\\
    (1 - a) x &= 1 - a^n\\
    x &= \frac{1 - a^n}{1 - a}
\end{align*}

Plugging this in gives us our debt calculation.
\begin{align*}
    D_{nt} &= a^{nt} P - p \frac{1 - a^{mt (n/m)}}{1 - a^{n/m}}\\
           &= a^{nt} P - p \frac{1 - a^{nt}}{1 - a^{n/m}}
\end{align*}


\section{Calculating Parameters from Known Values}

For the following we will always assume $D_{nt} = 0$.

\subsection{Principal}

\begin{align*}
    0 &= D_{nt}\\
    0 &= a^{nt} P - p \frac{1 - a^{nt}}{1 - a^{n/m}}\\
    P &= \frac{a^{-nt} - 1}{1 - a^{n/m}} p
\end{align*}


\subsection{Payments}

From here we can calculate the minimum payment needed to pay off all debts in
a certain time frame.
\begin{align*}
    P &= \frac{a^{-nt} - 1}{1 - a^{n/m}} p\\
    p &= \frac{1 - a^{n/m}}{a^{-nt} - 1} P
\end{align*}

Taking the limit of $t$ we can also find the minimum payment needed to maintain
debt.
\begin{align*}
    p_\text{min} &= \lim_{t \to \infty} \frac{1 - a^{n/m}}{a^{-nt} - 1} P\\
                 &= \frac{1 - a^{n/m}}{-1} P\\
                 &= (a^{n/m} - 1) P
\end{align*}


\subsection{Time}

We can also calculate how long it will take to pay off all debts.
\begin{align*}
    p &= \frac{1 - a^{n/m}}{a^{-nt} - 1} P\\
    (a^{-nt} - 1) p &= (1 - a^{n/m}) P\\
    a^{-nt} - 1 &= (1 - a^{n/m}) \frac{P}{p}\\
    a^{-nt} &= (1 - a^{n/m}) \frac{P}{p} + 1\\
    -nt &= \log_a\del{(1 - a^{n/m}) \frac{P}{p} + 1}\\
    t &= -\frac{\log_a\del{(1 - a^{n/m}) \frac{P}{p} + 1}}{n}\\
\end{align*}


\section{Recommended Payment}

\begin{align*}
    T &= mtp\\
    \dod{T}{p} &= -\delta\\
    mt &= -\delta\\
    t &= -\frac{\delta}{m}\\
    p &= \frac{1 - a^{n/m}}{a^{n\delta/m} - 1} P
\end{align*}


\section{Meeting a Specific Return on Investment}

Say that we want to only pay a specific amount of return on a debt.
For instance we only want to pay \$1200 back on a \$1000 dollar loan.
We can say $G$ is what we're willing to pay back, where $G > P$.
We then have

\begin{align*}
    mtp &= G\\
    t &= \frac{G}{mp}\\
    p &= \frac{1 - a^{n/m}}{a^{-nt} - 1} P\\
      &= \frac{1 - a^{n/m}}{a^{-nG/mp} - 1} P
\end{align*}

This is easily solvable, but we can use Newton's method to find an answer using
a calculator.
\begin{align*}
    f(p) &= \frac{1 - a^{n/m}}{a^{-nG/mp} - 1} P\\
    f'(p) &= -\frac{1 - a^{n/m}}{(a^{-nG/mp} - 1)^2} P \del{-\frac{nG \ln(a)}{mp} a^{-nG/mp}} \del{-\frac{1}{p^2}}\\
          &= -\frac{1 - a^{n/m}}{p^2 (a^{-nG/mp} - 1)^2} \del{\frac{nG \ln(a)}{mp} a^{-nG/mp}} P\\
          &= \frac{a^{n/m} - 1}{p^2 (a^{-nG/mp} - 1)^2} \del{\frac{nG \ln(a)}{mp} a^{-nG/mp}} P\\
    p_{n + 1} &= p_n - \frac{f(p_n)}{f'(p_n)}
\end{align*}

\end{document}