Ms17 - Modelling growth, decay and finance with sequences

Ms17 - Modelling growth, decay and finance with sequences

Turn repeated changes into models with a clear time index. You will compare simple and compound interest, find integer-time thresholds, and build savings and loan models whose predictions respect the timing of transactions.

Model equal changes or equal multipliers

Three deposits of 8080 pounds add up to 240240 pounds, but can moving their dates change the interest even when the rate stays fixed? We will compare deposits before and after monthly interest. First, decide what kind of change each interval represents.

You already know

An arithmetic sequence adds a constant difference; a geometric sequence multiplies by a constant ratio. A p%p\% decrease retains the fraction 1−p/1001-p/100. A recurrence gives the next value from the current one.

A 20%20\% fall needs a 25%25\% rise to recover: 0.8(1.25)=10.8(1.25)=1. Equal percentage rises and falls do not undo each other, because they use different starting amounts.

Models indexed from time zero

For initial value X0X_0 and non-negative integer nn,

Xn=X0+ndorXn=X0qnX_n=X_0+nd\qquad\text{or}\qquad X_n=X_0q^n

These represent constant addition dd or constant multiplication qq per interval.

Not in the formula sheet: learn it.

Index zero is the value before any interval has passed. A fitted model is a prediction under assumptions, not a promise that the same change will continue indefinitely.

Worked example 1: a fixed loss

Question. A tank contains 500500 litres initially and loses 5050 litres per day. Model its contents after two days and state a limit of the model.

Plan. Subtract a fixed amount for each completed day.

Xn=500−50nfixed daily lossX2=500−100=400two days\begin{aligned} X_n&=500-50n && \text{fixed daily loss} \\ X_2&=500-100=400 && \text{two days} \end{aligned}

Why restrict the time range? Negative contents after day ten would have no physical meaning.

Answer. 400400 litres; the model can apply only while water remains, with no refilling and the assumed constant loss.

Check. Its first values are 500,450,400500,450,400 litres.

Change only the loss rule to 10%10\% of the current contents each day. The first day's loss is still 5050 litres.

Worked example 2: a fixed percentage loss

Question. With 500500 litres initially, model a 10%10\% daily loss and find the contents after two days.

Plan. Retain 90%90\% each day, so multiply twice by 0.90.9.

Xn=500(0.9)nretention factorX2=500(0.9)2=405two intervals\begin{aligned} X_n&=500(0.9)^n && \text{retention factor} \\ X_2&=500(0.9)^2=405 && \text{two intervals} \end{aligned}

Why is more water left? The second loss is only 10%10\% of 450450, rather than another 5050 litres.

Answer. 405405 litres, assuming a fixed daily loss fraction with no additions.

Check. 500500 becomes 450450, then 405405.

Values 80,100,12080,100,120 at equal intervals have constant differences; 80,100,12580,100,125 have constant ratios. Check several changes, compare the prediction with later observations, and state when a rule stops being plausible.

Watch out

Wrong: a 10%10\% loss uses multiplier 0.10.1. Right: 0.10.1 is the fraction lost; 0.90.9 remains. Do not confuse the change with the new value.

Change the daily fraction for a fresh model.

Interest changes the principal or the whole balance

The word “interest” is not enough to choose a model: you need to know which balance is used in the calculation.

Simple and compound interest

Simple interest uses the initial principal for every interval. Compound interest uses the balance at the start of the current interval, including interest already credited. For example, at 3%3\%, interest on 100100 pounds is initially 33 pounds under either rule, but the next compound calculation uses 103103 pounds.

Let ii be the decimal interest rate per interval and PP the initial principal. Assume no deposits, withdrawals, fees or tax, and a fixed interest rule.

Interest balances

After nn completed intervals,

Bn=P(1+ni)for simple interest,Bn=P(1+i)nfor compound interestB_n=P(1+ni)\quad\text{for simple interest},\qquad B_n=P(1+i)^n\quad\text{for compound interest}

The interval of ii must match the count nn.

Not in the formula sheet: learn it.

The distinction is addition of equal interest amounts versus multiplication by an equal growth factor.

Worked example 3: the arithmetic balance

Question. Find the balance of 30003000 pounds after four years at 3%3\% annual simple interest.

Plan. Calculate one year's interest on the principal, then add four equal amounts.

annual interest=3000(0.03)=90initial principalB4=3000+4(90)=3360four additions\begin{aligned} \text{annual interest}&=3000(0.03)=90 && \text{initial principal} \\ B_4&=3000+4(90)=3360 && \text{four additions} \end{aligned}

Why is 9090 used each time? The principal stays the reference amount under simple interest.

Answer. 33603360 pounds.

Check. The total interest is 12%12\% of 30003000, or 360360 pounds.

Change only the rule to annual compounding, keeping the principal, rate and term.

Worked example 4: the geometric balance

Question. Find the compound balance and its excess over the simple balance after four years.

Plan. Multiply four times by 1.031.03, then compare the unrounded balances.

B4=3000(1.03)4=3376.52643compound modelB4−3360=16.52643extra interest\begin{aligned} B_4&=3000(1.03)^4=3376.52643 && \text{compound model} \\ B_4-3360&=16.52643 && \text{extra interest} \end{aligned}

Why does compounding gain more? Earlier interest earns further interest in subsequent years.

Answer. Balance 3376.533376.53 pounds; an extra 16.5316.53 pounds, to pennies.

Check. The first two compound balances are 30903090 and 3182.703182.70, already slightly above the corresponding simple balances.

A stated monthly rate of 1%1\% gives effective annual growth 1.0112−1≈0.1268251.01^{12}-1\approx0.126825, or about 12.68%12.68\%. It is not exactly 12%12\%. An effective annual rate cannot be divided by twelve without checking the rate convention. Retain full precision, then round a requested final money amount to pennies.

Watch out

Wrong: pair a monthly rate with a yearly exponent. Right: two years contains 2424 monthly intervals. Check whether the question states a monthly rate or an annual effective rate.

Change the principal and compare both interest rules at the same rate and term.

Logarithms locate an integer-time threshold

A continuous exponent calculation locates a boundary; the discrete time index decides which interval is allowed. A short logarithm rule makes this manageable before the later logarithms lessons.

For positive zz, log⁡z\log z is the exponent satisfying 10log⁡z=z10^{\log z}=z. Since this function increases, taking logarithms preserves an inequality between positive arguments.

Power rule for logarithms

For q>0q>0,

log⁡(qn)=nlog⁡q\log(q^n)=n\log q

Indeed, q=10log⁡qq=10^{\log q} gives qn=10nlog⁡qq^n=10^{n\log q}. Natural logarithms may be used consistently instead.

Not in the formula sheet: learn it.

For growth, 600(1.08)n>900600(1.08)^n>900 gives n>log⁡1.5/log⁡1.08≈5.26845n>\log1.5/\log1.08\approx5.26845. The first whole interval is 66: the values at 55 and 66 are about 881.597881.597 and 952.125952.125. Do not round the boundary to the nearest integer.

Worked example 5: a decay threshold

Question. Determine the first non-negative integer nn for which 1200(0.85)n<6001200(0.85)^n<600.

Plan. Take logarithms after dividing by 12001200, and account for a negative divisor.

0.85n<0.5normalisenlog⁡0.85<log⁡0.5positive argumentsn>log⁡0.5log⁡0.85≈4.26502negative divisor reverses sign\begin{aligned} 0.85^n&<0.5 && \text{normalise} \\ n\log0.85&<\log0.5 && \text{positive arguments} \\ n&>\frac{\log0.5}{\log0.85}\approx4.26502 && \text{negative divisor reverses sign} \end{aligned}

Why reverse the inequality? log⁡0.85<0\log0.85<0.

Answer. The first permitted interval is n=5n=5.

Check. At 44, the value is 626.4075626.4075; at 55, it is 532.446375532.446375, so the first crossing is between them.

Change only the direction of the requested inequality: now find the last interval above the same boundary.

Worked example 6: the last interval before decay crosses

Question. Determine the last non-negative integer nn for which 1200(0.85)n>6001200(0.85)^n>600.

Plan. Solve the reversed comparison and select the greatest integer satisfying it.

nlog⁡0.85>log⁡0.5divide then take logsn<4.26502…divide by a negative number\begin{aligned} n\log0.85&>\log0.5 && \text{divide then take logs} \\ n&<4.26502\ldots && \text{divide by a negative number} \end{aligned}

Why choose 44? The model decreases, and the question asks for the last interval before dropping below 600600.

Answer. n=4n=4.

Check. The same adjacent values are above and below 600600, respectively.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account