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Interest.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=0.5cm]{geometry} \begin{document} \text{\large Interest} \begin{align*} \text{Sequences and Series:}\\ \text{AP:}\quad&T_n=a+(n-1)\:d\:,\quad S_n=\tfrac{\:1\:}{2}n\big[2a+(n-1)d\big]=\tfrac{\:1\:}{2}\:n\:(a+l)\\ \text{GP:}\quad&T_n=ar^{n-1}\:,\quad S_n=\frac{a(1-r^n)}{1-r}=\frac{a-lr}{1-r}\:,\quad S_\infty=\frac{a}{1-r}\:\:\text{where }|r|<1\\ \\ \text{Common Notations:}\quad&\text{(All are positive.)}\\ &\text{$P$: Principle}\\ &\text{$R$: Interest Rate}\\ &\text{$I$: Interest}\\ &\text{$N$ or $n$: Number of Periods}\\ &\text{$M$: Payment per Period}\\ &\text{$A_n$: Balance at the \it end \rm of the Period $n$}\\ \\ \text{Simple Interest:}\quad&\text{APs with}\quad a=P+PR\:,\quad d=PR\\ &I=PR\:,\quad A_{k+1}=A_k+I=A_k+PR\\ &A_0=P\quad\text{($A_0$ is not in the sequence, which starts from $A_1$.)}\\ &A_1=A_0+PR=P+PR=P(1+R)\\ &A_2=A_1+PR=(P+PR)+PR=P+PR\cdot 2=P(1+R\cdot 2)\\ &A_3=A_2+PR=(P+PR\cdot 2)+PR=P+PR\cdot 3=P(1+R\cdot 3)\\ &\ldots\\ &\boxed{A_n=P(1+Rn)}\\ \\ \text{Compound Interest:}\quad&\text{GPs with}\quad a=P(1+R)\:,\quad r=1+R\\ &I=A_k R\:,\quad A_{k+1}=A_k+I=A_k+A_k R=A_k(1+R)\\ &A_0=P\quad\text{($A_0$ is not in the sequence, which starts from $A_1$.)}\\ &A_1=A_0(1+R)=P(1+R)\\ &A_2=A_1(1+R)=\left[P(1+R)\right](1+R)=P(1+R)^2\\ &A_3=A_2(1+R)=\left[P(1+R)^2\right](1+R)=P(1+R)^3\\ &\ldots\\ &\boxed{A_n=P(1+R)^n}\qquad\text{For depreciation: }\boxed{A_n=P(1-R)^n}\\ \\ &\text{If each period is divided into $h$ parts, over which the interest is compounded, then}\\ &A_n=P\left(1-\tfrac{R}{h}\right)^{nh}=P\left(1-\tfrac{Rn}{nh}\right)^{nh}=P\left(1-\tfrac{Rn}{m}\right)^m\:,\quad\text{where $m=nh$ .}\\ &\lim_{h\to\infty}A_n=\lim_{m\to\infty}P(1+\tfrac{Rn}{m})^m=P\:e^{Rn}\:.\quad\text{In this case, $n$ becomes a continuous measurement of time $T$.}\\ &A_T=P\:e^RT\:,\quad A_0=P\:,\quad\text{The Interest}\quad I=A_T-P=A_T-A_0=\left[\frac{dA}{dt}\right]_0^T=\int_0^T PR\:e^Rt\:dt\:.\\ \therefore\quad&\boxed{I=\int_0^T PR\:e^Rt\:dt}\:,\quad\text{where $I$ is the interest compounded instantaneously;}\\ &\quad\text{$P$ is the principle; $R$ is interest rate per period; and $T$ is the number of periods.}\\ \end{align*} \begin{align*} \text{Regular Payments}\quad&\text{(with compound interest)}\\ &\text{Starting with a zero balance, if a payment of $M$ is made at the \it beginning \rm of each period,}\\ &\text{what is the balance at the \it end \rm of the Nth period?}\\ \text{Method I:}\quad&\text{Let $T_k$ be what the payment made at the \it beginning \rm of the ``k''th period is worth}\\ &\text{at the \it end \rm of the Nth period (when the investment matures).}\\ T_k=&M(1+R)^{n-k+1}\\ A_n=&\sum_{k=1}^n T_k=T_1+T_2+\ldots+T_{n-1}+T_n\\ =&M(1+R)^n+M(1+R)^{n-1}+\ldots+M(1+R)^2+M(1+R)\\ =&M(1+R)\cdot\left[(1+R)^{n-1}+(1+R)^{n-2}+\ldots+(1+R)+1\right]\\ =&M(1+R)\cdot\frac{(1+R)^n-1}{(1+R)-1}\\ \therefore\quad&\boxed{A_n=M\cdot\frac{(1+R)\times\left[(1+R)^n-1\right]}{R}}\ldots\text{\small how much your investment is worth if you invest $M$ per period for $n$ periods.}\\ &\boxed{M=A_n\cdot\frac{R}{(1+R)\times\left[(1+R)^n-1\right]}}\ldots\text{\small how much to invest per period if you want to receive $A_n$ after $n$ periods}\\ \\ \text{Method II:}\quad&\text{Let $A_k$ be the balance at the \it end \rm of the ``k''th period (which is what the bank statement would show).}\\ A_1=&M(1+R)\\ A_2=&(A_1+M)(1+R)=\left[M(1+R)+M\right]\cdot(1+R)=M\left[(1+R)^2+(1+R)\right]\\ A_3=&(A_2+M)(1+R)=\left(M\left[(1+R)^2+(1+R)\right]+M\right)\cdot(1+R)=M\left[(1+R)^3+(1+R)^2+(1+R)\right]\\ &\ldots\\ A_n=&M\cdot\left[(1+R)^n+(1+R)^{n-1}+\ldots+(1+R)\right]\\ =&M(1+R)\cdot\left[(1+R)^{n-1}+(1+R)^{n-2}+\ldots+(1+R)+1\right]\\ &\quad\text{which will give the same result as in Method I.}\\ \\ \text{Paying Off a Loan}\quad&\text{(with compound interest)}\\ \text{Method I:}\quad&\text{Let's consider investing by regular payments aiming at receiving an amount at the end of $n$ periods to pay}\\ &\text{off a loan of $P$. The investment calculation differs from the ``Regular Payments'' formula as the payment is}\\ &\text{made at the \it end \rm of the period, i.e. no payment at the beginning of the first period, but an extra payment}\\ &\text{is made at the end of the last. Still $n$ payments of $M$ have been made but the amount is $[M(1+R)^n-M]$}\\ &\text{less to compensate the interest for that ``late'' payment. The $Loan=Investment$ formula becomes:}\\ &P(1+R)^n=M\cdot\frac{(1+R)\times\left[(1+R)^n-1\right]}{R}-[M(1+R)^n-M]=M\cdot\frac{(1+R)^n-1}{R}\\ \therefore\quad&\boxed{M=\frac{PR(1+R)^n}{(1+R)^n-1}=\frac{PR}{1-\frac{1}{(1+R)^n}}}\ldots\text{\small how much to pay per period to pay off a loan of $P$ in $n$ periods}\\ \\ \text{Method II:}\quad&\text{Let $B_k$ be the loan balance at the \it end \rm of the ``k''th period (which is what the bank statement would show).}\\ B_1=&P(1+R)-M\\ B_2=&B_1(1+R)-M=\left[P(1+R)-M\right]\cdot(1+R)-M=P(1+R)^2-M(1+R)-M\\ B_3=&B_2(1+R)-M=\left[P(1+R)^2-M(1+R)-M\right]\cdot(1+R)-M=P(1+R)^3-M(1+R)^2-M(1+R)-M\\ &\ldots\\ B_n=&P(1+R)^n-M(1+R)^{n-1}-M(1+R)^{n-2}-\ldots-M(1+R)-M=0\quad\text{(Last period balance must be zero.)}\\ &P(1+R)^n=M\cdot\frac{(1+R)^n-1}{(1+R)-1}=M\cdot\frac{(1+R)^n-1}{R}\\ &\quad\text{which will give the same result as in Method I.}\\ \end{align*} \end{document}