Comprehensive Guide to Simple Interest, Compound Interest, and Annuities

Simple and Compound Interest Foundations

  • Interest calculations are divided into two main categories: simple interest and compound interest.

  • Simple Interest Formula:      A=P×r×tA = P \times r \times t

    • AA = Total amount / balance

    • PP = Principal (initial deposit or loan amount)

    • rr = Annual interest rate

    • tt = Time in years

  • Solving for the rate rr algebraically in simple interest is straightforward:      r=AP×tr = \frac{A}{P \times t}

  • Compound Interest Formula:      A = P\times\begin{pmatrix}1 + \frac{r}{m}\right)^{m t}

    • AA = Account balance after tt years

    • PP = Principal / initial deposit

    • rr = Annual interest rate

    • mm = Number of compounding periods per year

    • tt = Time in years

  • Alternative textbook notation:      A=P(1+I)nA = P(1 + I)^n

    • I=rmI = \frac{r}{m} (interest rate per compounding period)

    • n=mtn = m t (total number of compounding periods)

  • While textbook notation simplifies fractions into variables II and nn, using A = P\times\begin{pmatrix}1 + \frac{r}{m}\right)^{m t} directly is recommended to avoid incorrectly plugging in the unadjusted annual rate rr.

  • Solving for variables like rr or tt algebraically in compound interest is significantly more difficult than in simple interest because it requires undoing powers and fractions. Graphing tools and calculators are essential for solving these equations.

Graphing and Functional Notation for Compound Interest

  • Compound interest functions produce curved, exponential graphs, unlike the straight lines produced by simple interest models.

  • Functional Notation:

    • Account balance as a function of time is written as A(t)A(t).

    • Example: A(3)A(3) represents the balance at 3 years, A(7)A(7) at 7 years, and A(9)A(9) at 9 years along a financial timeline.

  • Example: Calculating Monthly Compounded Interest

    • Deposit: P=7500P = 7500

    • Annual Interest Rate: r=5.2%=0.052r = 5.2\% = 0.052

    • Compounding Frequency: Monthly (m=12m = 12

    • Model:          A(t) = 7500\times\begin{pmatrix}1 + \frac{0.052}{12}\right)^{12 t}

    • Graphing Setup on TI Calculators:

    • Enter formula into Y-editor: Y_1 = 7500 \times \begin{pmatrix}1 + \frac{0.052}{12}\right)^{(12 x)}

    • Enclose the power expression inside parentheses: (12 x).

    • Window settings: Set X-range from 00 to 1515. Set Y-minimum to −10-10 to make the x-axis visible under the curve.

Comparing Investment Accounts and Graph Intersections

  • Investment Comparison Example:

    • Account A: A(t)=10000e0.047tA(t) = 10000 e^{0.047 t}

    • Account B: B(t) = 10000\times\begin{pmatrix}1 + \frac{0.047}{365}\right)^{365 t}

  • Balance after t=20t = 20 years (365×20=7300365 \times 20 = 7300 compounding periods):

    • Both accounts yield approximately \text{\\$25,590.27}, with Account B returning roughly 12c12\text{c} more than Account A due to discrete daily compounding frequency.

  • Graphing Behavior:

    • Curves for similar continuous and daily compounding accounts appear virtually identical on standard graphing windows.

    • To observe separation between two closely aligned interest curves, extreme zoom parameters are required.

  • Finding Differences Between Accounts:

    • To find when the difference between two account balances reaches a specific threshold (e.g., \text{\\$500}):          A(t)−B(t)=500A(t) - B(t) = 500

    • Solution via Graphing:

    • Set Y1=A(x)−B(x)Y_1 = A(x) - B(x)

    • Set Y2=500Y_2 = 500

    • Adjust window parameters: Set xmax⁡=100x_{\max} = 100.

    • Execute calculator intersection function (2nd TRACE -> 5: intersect).

    • Result: The account balances differ by \text{\\$500} at t=23.21t = 23.21 years.

Continuous Compound Interest and Transcendental Numbers

  • Continuous Compounding Formula:      A=PertA = P e^{r t}

    • PP = Initial principal

    • rr = Annual interest rate

    • tt = Time in years

    • ee = Euler's number (≈2.71828…\approx 2.71828\dots)

  • Transcendental Numbers:

    • π≈3.14159…π \approx 3.14159\dots

    • Official world record for memorizing digits of ππ: 70,00070,000 decimal places (verified by official checkers working in shifts).

    • Unofficial claimed record: 111,700111,700 decimal places by Akira from Japan (unverified due to absence of official checkers).

    • e≈2.71828…e \approx 2.71828\dots

    • Official world record for memorizing digits of ee: 14,00014,000 decimal places.

  • Example: Continuous Compounding Calculation

    • Deposit: P=3000P = 3000

    • Rate: r=6.7%=0.067r = 6.7\% = 0.067

    • Time: t=12t = 12 years

    • Calculation:          A = 3000 e^{0.067 \times 12} = 3000 e^{0.72} \approx \text{\\$6,163.30}

  • Estimating Growth Thresholds (Doubling Investment to \text{\\$6,000}):

    • Equation:          6000=3000e0.067t  ⟹  2=e0.067t6000 = 3000 e^{0.067 t} \implies 2 = e^{0.067 t}

    • Graphing Method: Graph Y1=3000e0.067xY_1 = 3000 e^{0.067 x} and Y2=6000Y_2 = 6000; intersection occurs at t=11.55t = 11.55 years.

    • Algebraic Method (using logarithms):          ln⁡(2)=0.067t\ln(2) = 0.067 t

         t=ln⁡(2)0.067≈11.55 yearst = \frac{\ln(2)}{0.067} \approx 11.55\,\text{years}

Mathematical Derivation of ee and Indeterminate Forms

  • Derivation from Compounding Periods:

    • As the number of compounding periods mm approaches infinity (m→∞m \rightarrow \infty):          \lim_{m \rightarrow \infty} P\times\begin{pmatrix}1 + \frac{r}{m}\right)^{m t} = P e^{r t}

    • Evaluating \begin{pmatrix}1 + \frac{r}{\infty}\right)^{\infty} yields \begin{pmatrix}1 + 0\right)^{\infty} = 1^{\infty}.

  • Classification of Mathematical Expressions:

    • Undefined Forms: Division by zero (x0\frac{x}{0}) cannot be evaluated.

    • Indeterminate Forms: Expressions in calculus that can evaluate to various values depending on context:

    • 00\frac{0}{0}

    • 1∞1^{\infty}

    • ∞∞\frac{\infty}{\infty}

Future Value of an Annuity

  • Annuities are structured financial streams categorized into two types: saving money over time or paying back a loan.

  • Future Value Formula (Savings / Investments):      FV = PMT \times \left[ \frac{\begin{pmatrix}1 + \frac{r}{m}\right)^{m t} - 1}{\frac{r}{m}} \right]

    • FVFV = Future value of all payments combined with accumulated interest

    • PMTPMT = Regular payment / deposit amount made each period

    • rr = Annual interest rate

    • mm = Number of payments/compounding periods per year

    • tt = Total time in years

  • Example 1: Regular Monthly Savings

    • Monthly Deposit (PMTPMT): \text{\\$250}

    • Annual Interest Rate (rr): 4.5%=0.0454.5\% = 0.045

    • Frequency (mm): Monthly (1212

    • Duration (tt): 1515 years (mt=180m t = 180 total payments)

    • Calculation:          FV = 250 \times \left[ \frac{\begin{pmatrix}1 + \frac{0.045}{12}\right)^{180} - 1}{\frac{0.045}{12}} \right] = \text{\\$64,103.67}

    • Total Amount Deposited (Bookkeeping Calculation):          \text{Total Principal} = 180\,\text{payments} \times \text{\\250} = \text{\\45,000.00}

    • Total Interest Earned:          Interest=Final Balance−Total Deposited\text{Interest} = \text{Final Balance} - \text{Total Deposited}

         \text{Interest} = \text{\\64,103.67} - \text{\\45,000.00} = \text{\\$19,103.67}

  • Example 2: Retirement Savings Calculation

    • Monthly Deposit (PMTPMT): \text{\\$400}

    • Annual Interest Rate (rr): 7%=0.077\% = 0.07

    • Compounding Frequency (mm): Monthly (1212

    • 10-Year Accumulation (mt=120m t = 120 payments):          FV = 400 \times \left[ \frac{\begin{pmatrix}1 + \frac{0.07}{12}\right)^{120} - 1}{\frac{0.07}{12}} \right] = \text{\\$69,233.92}

    • 30-Year Accumulation (mt=360m t = 360 payments):          FV = 400 \times \left[ \frac{\begin{pmatrix}1 + \frac{0.07}{12}\right)^{360} - 1}{\frac{0.07}{12}} \right] = \text{\\$487,988.40}

    • Graphing Annuity Growth:

    • Window parameters: x∈[0,35]x \in [0, 35].

    • The shape is non-linear and exponential due to continuous compounding of accumulated interest on recurring deposits.

Comparing Investment Options

  • Example Problem (\text{\\$8,000} principal invested over 2525 years):

    • Option A: 4.9%4.9\% compounded monthly (m=12m = 12          A_1(25) = 8000\times\begin{pmatrix}1 + \frac{0.049}{12}\right)^{300} = \text{\\$27,165.49}

    • Option B: 4.7%4.7\% compounded continuously          A_2(25) = 8000 e^{0.047 \times 25} = 8000 e^{1.175} = \text{\\$25,951.14}

    • Conclusion: Option A (4.9%4.9\% monthly) is superior to Option B (4.7%4.7\% continuous) despite continuous compounding, because the higher nominal interest rate outweighs the continuous compounding frequency.

Present Value of an Annuity and Mortgages

  • Present Value Formula (Loans / Mortgages Payback):      PV = PMT \times \left[ \frac{1 - \begin{pmatrix}1 + \frac{r}{m}\right)^{-m t}}{\frac{r}{m}} \right]

    • PVPV = Present value / loan amount borrowed

    • PMTPMT = Regular periodic payment

    • rr = Annual interest rate

    • mm = Number of payments per year

    • tt = Loan duration in years

  • Impact of Federal Reserve Rate Adjustments:

    • Rate adjustments by the Federal Reserve (e.g., a quarter-point or 0.25%0.25\% change) significantly impact overall repayment totals on long-term loans like 30-year mortgages.

  • Solving for Payment Amount (PMTPMT) using Calculator Shortcuts:

    • Re-arranging algebraically requires dividing PVPV by a complex fraction expression.

    • Calculator Technique: Multiply the present value PVPV by the bracketed formula raised to the negative one power (−1-1):          PMT = PV \times \left[ \frac{1 - \begin{pmatrix}1 + \frac{r}{m}\right)^{-m t}}{\frac{r}{m}} \right]^{-1}

  • Example: 30-Year Mortgage Calculation

    • Loan Amount (PVPV): \text{\\$5,000,000}

    • Rate (rr): 6.5%=0.0656.5\% = 0.065

    • Term (tt): 3030 years (mt=360m t = 360 payments)

    • Payment Calculation:          PMT = 5000000 \times \left[ \frac{1 - \begin{pmatrix}1 + \frac{0.065}{12}\right)^{-360}}{\frac{0.065}{12}} \right]^{-1} \approx \text{\\$31,600.00\,per month}