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Compound interest
Why 30 years at 5% earn you not €1,500 but €3,322 — and where the difference is made.
C_final = C × (1 + t)^nProblem / motivation
You invest €1,000 at 5% a year and do not touch it for 30 years. How much do you think you get back? Careful, there is a trap: the “intuitive” answer is wrong, but most people are also wrong about the size of their error.
The spontaneous reasoning is that of SIMPLE interest: 5% of €1,000, i.e. €50 a year, hence €1,500 of interest over 30 years — and a final capital of €2,500, starting capital included. That is already a first trap: €1,500 is the interest, not the sum you get back.
But that is not how an investment works: each year the interest joins the capital and earns interest in its turn. This is COMPOUNDING. This course will not merely announce the spectacular result: it will build the formula year by year, put an exact figure on what compounding adds compared with simple interest, show WHERE that extra comes from, and end with the three questions every saver asks — how long to double, what waiting costs, and what all this is worth once inflation is taken out.
Assumptions
In your view, which assumptions are needed for this model to hold? Jot down your ideas — no lead is wrong, this is your worksheet.
Your worksheet is still empty. Go for it: propose at least one idea.
0 idea(s) proposedFormalization
Start with €1,000 at 5%. After a year, you have the capital PLUS the interest: 1,000 + 1,000 × 0.05. Taking 1,000 out as a factor, that is 1,000 × (1 + 0.05) = 1,000 × 1.05 = €1,050. There is where the (1 + t) comes from: it is not a convention, it is “the capital plus its interest”, factorised. The second year the same operation starts again — but on €1,050 this time: 1,050 × 1.05 = €1,102.50. Then 1,102.50 × 1.05 = €1,157.63. Note the difference from simple interest: the first year earns €50, the second €52.50, the third €55.12.
Let us write the third year without going through the stages: 1,000 × 1.05 × 1.05 × 1.05. The factor 1.05 comes back as many times as there are years — which is exactly what an EXPONENT records: 1,000 × 1.05³. After n years, C × (1 + t)ⁿ. Careful not to confuse this with a multiplication: (1 + t) raised to the power 30 is 4.32, whereas (1 + t) multiplied by 30 would be 31.5 — the first gives €4,322, the second €31,500, which makes no sense at all. A power is a repetition of multiplications, never a multiplication.
The formula also answers the reverse question. Doubling means reaching a factor of 2: we look for the n such that (1 + t)ⁿ = 2. The exact answer comes from a logarithm — n = ln 2 / ln(1 + t) — but there is a famous mental shortcut, the RULE OF 72: divide 72 by the rate in per cent. At 7%, 72 ÷ 7 ≈ 10.3 years, where the exact computation gives 10.24. Where does that 72 come from? From ln 2 ≈ 0.693, i.e. 69.3 in per cent: it is rounded up to 72 because it divides cleanly by 2, 3, 4, 6, 8, 9 and 12, and because that rounding offsets the approximation for common rates. The shortcut is excellent between 4% and 10%, and degrades at the extremes: at 15% it announces 4.8 years against 4.96 in reality.
One last trap, the one that makes people misapply the formula to real products. If an account advertises 5% a year but compounds every MONTH, you do not compute with 5% and 1 period: you compute with 5/12% and 12 periods. Result: (1 + 0.05/12)¹² = 1.05116, i.e. 5.116% actually earned over the year instead of 5%. The difference looks tiny, but it is the same mechanism as the whole course — repeated for 30 years, it adds about €146 (€4,467.74 instead of €4,321.94). Keep the rule: the rate and the exponent must always speak of the SAME period. Click each term:
Solving / calculation
Let us run the €1,000 at 5% over 30 years all the way: the computation, the honest comparison with simple interest, the place where the extra is produced, then the three questions a saver really asks.
- The first three years, by hand (the narrative, step 1)
1,000 × 1.05 = 1,050; 1,050 × 1.05 = 1,102.50; 1,102.50 × 1.05 = 1,157.63the same factor, repeated — hence a power - The 30 years in one go
1,000 × 1.05^30, with 1.05^30 = 4.3219C_final = €4,321.94, of which €3,321.94 is interest - The honest comparison: the same 30 years in SIMPLE interest
€50 a year × 30 years = €1,500 of interest, plus the €1,000 at the start€2,500 — that really is a final capital, not €1,500 - What compounding really adds
4,322 − 2,500 (or, on the interest alone: 3,322 − 1,500)€1,822 — there is the “interest on interest” - Where that extra is produced
interest in year 1: 1,000 × 5%; in year 15: 1,000 × 1.05^14 × 5%; in year 30: 1,000 × 1.05^29 × 5%€50, then €99.00 in year 15, then €205.81 in year 30 — more than four times the first - The price of waiting: starting 10 years later, at 40 instead of 30
1,000 × 1.05^20€2,653 instead of €4,322 — 39% less for ten years of delay - How long to double, at 5%?
rule of 72: 72 ÷ 5 = 14.4 years (exact: ln 2 / ln 1.05 = 14.21 years)≈ 14 years — so the capital doubles twice in 30 years, and a little more - And in purchasing power? We deflate by inflation (2% a year)
4,322 ÷ 1.02^30€2,386 of today's money — the real gain is twice as modest as the headline gain
Keep four things, in this order. The formula is not a convention: (1 + t) is “the capital plus its interest” factorised, and the exponent is merely the repetition of that factor. Compounding adds €1,822 here compared with simple interest — not €2,800, a figure you get by wrongly subtracting an amount of interest from a capital. That extra is produced LATE: the thirtieth year earns €206 where the first earned €50 and the fifteenth €99, which explains why ten years of delay cost 39% of the result. And above all, those €4,322 are a nominal figure: once inflation is taken out, only €2,386 of today's purchasing power is left. Compounding works both ways — for your savings, and for prices.
compound: C × (1 + t)^n · simple: C × (1 + t × n) · real: compound ÷ (1 + inflation)^nThree things to try. One: keep comparing the two columns, compound and simple — the gap is invisible in the early years and becomes enormous after twenty. Two: set the duration to 10 years then to 20, and see that the result does far more than double. Three: raise inflation and watch the real gain melt while the headline gain does not budge.
Economic interpretation
Three readings: why time counts more than the rate, what compounding does when it works against you, and the rounding error that costs dear over thirty years.
Compare two savers. The first invests €1,000 at 5% for 30 years and gets €4,322. The second, cleverer, secures 7% — but starts ten years later: 1,000 × 1.07^20 = €3,870. Two extra points of return do not make up for ten years of delay. The reason lies in where the gain is produced: in our example the last year earns €206 and the first €50. It is the final years that carry the result, and those are precisely the ones you lose by starting late. That is also why a small sum invested early often beats a large sum invested late.
Beware of a widespread shortcut: compounding does NOT erode a debt, it inflates it — unpaid interest joins the principal owed and earns interest in its turn. That is exactly what the course “The dynamics of public debt” describes with its snowball effect, and it is the mechanism of an overdraft or a revolving credit. What erodes a debt is something else: INFLATION, which lightens the real value of a sum fixed in advance. Two opposite mechanisms, often confused because both involve debt and time.
To go from an advertised return to a real return, people commonly subtract inflation: 5% minus 3% is “about 2%”. Over one year the approximation is excellent. Over thirty compounded years it is not: the exact computation deflates by dividing, (1.05/1.03)^30, and gives €1,781 of purchasing power, against €1,811 with the shortcut — a real gain overstated from 781 to 811 €, i.e. nearly 4%. The lesson is consistent with the rest of the course: what is negligible over one period stops being so when repeated. The course “Do your savings beat inflation?” sets out that exact division.
Limits / critiques
Exercises
€1,000 is invested at 10% a year for 2 years, interest reinvested. What is the final capital (in €)?
At 6% a year, in how many years does a capital double, according to the rule of 72?
An account advertises 6% a year but compounds every HALF-YEAR (3% every six months). What is the final capital for €1,000 after one year (in €)?
Two savers invest €1,000. One gets 5% for 30 years, the other 7% but starts only 10 years later (so 20 years of investment). Which ends up with more?
The same €1,000 at 10% for 2 years, but in SIMPLE interest: what is the final capital (in €)?
True or false: €1,000 invested at 5% earns MORE THAN TWICE as much interest over 10 years as over 5 years.
Another investment reaches €10,000 after 20 years, inflation having run at 3% a year over the period. What are those €10,000 worth in today's purchasing power (in €)?
True or false: compounding interest gradually lightens the weight of a debt.