Investigation 2 - Continuous Compound Interest | 探究二：连续复利

📚 Investigation 2 - Continuous Compound Interest | 探究二：连续复利

In many financial scenarios, interest is not applied just once a year but rather multiple times within a year. As the compounding frequency increases, the final amount grows slightly due to the effect of earning interest on interest more often. This investigation explores what happens when compounding becomes so frequent that we can think of it as happening continuously, every single instant. Underpinning this idea is one of the most important constants in mathematics: e. You will investigate the behaviour of the compound interest formula as the number of compounding periods tends to infinity and discover how continuous compound interest provides a natural model for growth processes.

在许多金融场景中，利息并非每年只计一次，而是一年中多次计息。随着复利频率的增加，由于更频繁地赚取利滚利，最终金额会略微增长。本探究将探索当复利变得如此频繁，以至于可以认为它在每一个瞬间连续发生时的情况。支撑这一思想的是数学中最重要的常数之一：e。你将研究复利公式在计息次数趋向无穷大时的行为，并发掘连续复利如何为增长过程提供一个自然的模型。

1. Understanding Compound Interest | 理解复利

rest from previous periods. The basic discrete compound interest formula is A = P(1 + r/n)^nt, where P is the principal, r is the annual nominal interest rate (as a decimal), n is the number of compounding periods per year, and t is the time in years.

复利意味着利息不仅根据初始本金计算，还根据之前期间累积的利息进行计算。基本的离散复利公式为 A = P(1 + r/n)^nt，其中 P 为本金，r 为年名义利率（以小数表示），n 为每年计息次数，t 为时间（以年为单位）。

For example, if you invest 1000 dollars at an annual rate of 5 % compounded quarterly, after 3 years the amount would be A = 1000(1 + 0.05/4)^{4 × 3} = 1000(1.0125)¹² r than with annual compounding because interest is credited four times a year, and each time it begins earning additional interest sooner.

例如，如果你以 5% 的年利率按季复利投资 1000 美元，3 年后的金额为 A = 1000(1 + 0.05/4)^{4 × 3} = 1000(1.0125)¹² ≈ 1160.75 美元。这个数值比按年复利更高，因为利息每年计入四次，每次都能更早地开始产生额外利息。

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2. Discrete Compounding Examples | 离散复利示例

Let us fix a principal of 1 dollar, an annual rate of 100 % (r = 1), and a time period of 1 year. The formula simplifies to A = (1 + 1/n)ⁿ. As n changes, the final amount tells us how much 1 dollar grows under different compounding frequencies. This setup is ideal for investigating the behaviour of the expression as n varies.

我们固定本金为 1 美元，年利率为 100%（r = 1），时间为 1 年。公式简化为 A = (1 + 1/n)ⁿ。随着 n 变化，最终金额告诉我们 1 美元在不同复利频率下的增长情况。这种设定非常适合研究该表达式随 n 变化的行为。

The table below shows the values of (1 + 1/n)ⁿ for selected values of n. Observe how the amount increases slowly and appears to approach a certain limit.

下表列出了选定 n 值时 (1 + 1/n)ⁿ 的值。观察金额如何缓慢增加，并似乎趋近于某个极限。

n	(1 + 1/n)n
1	2
2	2.25
4	2.44140625
12	2.61303529
365	2.71456748
1000	2.71692393
10000	2.71814593
100000	2.71826824

nise as the famous mathematical constant e. This experiment suggests that limₙ→∞ (1 + 1/n)ⁿ = e.

数值似乎逼近 2.71828...，你可能认出这就是著名的数学常数 e。这个实验表明 limₙ→∞ (1 + 1/n)ⁿ = e。

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3. Increasing the Compounding Frequency | 增加复利频率

Now consider a more realistic rate, say r = 0.05 (5 % semi-annually (n=2), quarterly (n=4), monthly (n=12), daily (n=365), and hourly (n=8760).

现在考虑一个更实际的利率，比如 r = 0.05（5%），本金 P = 1000，时间 t = 10 年。我们可以计算各种复利频率下的金额：每年（n=1）、每半年（n=2）、每季度（n=4）、每月（n=12）、每日（n=365）和每小时（n=8760）。

Using the formula A = 1000(1 + 0.05/n)¹⁰ⁿ amount increases with n, the increments become smaller and the sequence seems to converge.

使用公式 A = 1000(1 + 0.05/n)¹⁰ⁿ，我们得到如下近似金额：n=1 时，A=1628.89；n=2 时，A=1638.62；n=4 时，A=1643.62；n=12 时，A=1647.01；n=365 时，A=1648.66；n=8760 时，A=1648.72。注意，虽然金额随着 n 增加而增加，但增量越来越小，且数列似乎收敛。

This convergence hints that there is a limiting value representing what would happen if interest were compounded infinitely many times per year. Such a scenario is called continuous compounding.

这种收敛性暗示存在一个极限值，表示如果利息每年无限次复利将会发生的情况。这种情景被称为连续复利。

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4. Investigating the Limit | 探究极限

riting the expression. For a fixed principal P, rate r, and time t, we have A = P(1 + r/n)^nt. Let us set m = n/r, so that n = m r. Then (1 + r/n)^nt = [(1 + 1/m)^m]^rt. As n → ∞, m → ∞, and (1 + 1/m)^m → e.

我们可以通过改写表达式来分析极限。对于固定的本金 P、利率 r 和时间 t，有 A = P(1 + r/n)^nt。令 m = n/r，则 n = m r。于是 (1 + r/n)^nt = [(1 + 1/m)^m]^rt。当 n → ∞ 时，m → ∞，并且 (1 + 1/m)^m → e。

Therefore, the limiting amount is A = Pe^rt. This is the continuous compound interest formula. It shows that under continuous compounding, the growth is exponential with constant e, but the exponent depends on both the rate and time.

因此，极限金额为 A = Pe^rt。这就是连续复利公式。它表明，在连续复利下，增长是指数型的，以 e 为底常数，但指数取决于利率和时间。

l limit that defines e. To be convinced, you can test numerically: for r=0.05 and t=10, Pe^rt = 1000 e^0.05×10 = 1000 e⁰.5 ≈ 1648.72, which matches the trend of increasing n perfectly.

该推导运用了定义 e 的基本极限。为确信起见，你可以进行数值检验：对于 r=0.05，t=10，Pe^rt = 1000 e^0.05×10 = 1000 e⁰.5 ≈ 1648.72，这与增加 n 的趋势完全吻合。

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5. The Number e Emerges | 数字 e 的出现

ional and arises naturally in the study of growth and decay. In the context of compound interest, e appears as the factor by which a unit principal grows under 100 % continuous interest for one unit of time. It is the limit of (1 + 1/n)ⁿ as n becomes infinitely large.

常数 e ≈ 2.718281828 是一个无理数，自然地出现在增长与衰减的研究中。在复利背景下，e 代表单位本金在 100% 连续利率下每单位时间的增长因子。它是当 n 无限增大时 (1 + 1/n)ⁿ 的极限。

nential functions, logarithms, and differential equations. In fact, the function A(t) = Pe^rt dA/dt = rA, meaning the rate of change of the amount is proportional to the amount itself.

认识 e 有助于将复利与更广泛的数学概念联系起来，如指数函数、对数和微分方程。事实上，函数 A(t) = Pe^rt 满足微分方程 dA/dt = rA，这意味着数量的变化率与数量本身成正比。

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6. Deriving the Continuous Compound Interest Formula | 推导连续复利公式

Starting from A = P(1 + r/n)^nt r)t = m(rt). The expression inside the bracket is 1 + r/(mr) = 1 + 1/m. So A = P[(1 + 1/m)^m]^rt.

从 A = P(1 + r/n)^nt 出发，我们使用代换 m = n/r。那么 n = mr，指数变为 (mr)t = m(rt)。括号内的表达式为 1 + r/(mr) = 1 + 1/m。因此 A = P[(1 + 1/m)^m]^rt。

Now take the limit as n → ∞ (hence m → ∞). The quantity (1 + 1/m)^m tends to e. By properties of limits, the entire expression tends to P e^rt. This is why continuous compounding yields A = Pe^rt.

现在取 n → ∞（因此 m → ∞）时的极限。量 (1 + 1/m)^m 趋近于 e。根据极限的性质，整个表达式趋近于 P e^rt。这就是为什么连续复利得到 A = Pe^rt。

The formula is elegant and eliminates the discrete compounding variable n, allowing us to model growth as a continuous process. It is widely used in finance, biology, and physics.

该公式简洁雅致，消去了离散复利变量 n，使我们能够将增长模拟为连续过程。它被广泛应用于金融、生物学和物理学。

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7. Continuous Compounding Formula A = Pe^rt | 连续复利公式 A = Pe^rt

When interest is compounded continuously, the future value A after t years is given by:

A = Pe^rt

当利息连续复利时，t 年后的未来值 A 由下式给出：

A = Pe^rt

Here r is the annual nominal interest rate expressed as a decimal, and e is the base of the natural logarithm. The exponent rt represents the product of rate and time, which can be any real number. The principal P grows by the factor e^rt.

这里 r 是以小数表示的年名义利率，e 是自然对数的底数。指数 rt 表示利率与时间的乘积，可以是任意实数。本金 P 以因子 e^rt 增长。

mple, with P = 2500, r = 0.03, and t = 4, the continuously compounded amount is A = 2500 e^0.12 htly larger than the corresponding daily compounding amount, demonstrating the maximum possible accumulation under the given nominal rate.

例如，当 P = 2500，r = 0.03，t = 4 时，连续复利金额为 A = 2500 e^0.12 ≈ 2500 × 1.1275 = 2818.75。这略高于相应的每日复利金额，展示了在给定名义利率下可能的最大积累值。

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8. Effective Annual Rate for Continuous Compounding | 连续复利的实际年利率

ding, the EAR can be found by setting t=1 and considering the growth factor. Since A = P e^r after one year, the effective annual rate is EAR = e^r - 1.

实际年利率（EAR）可用于比较不同的复利方法。对于连续复利，可通过设 t=1 并考虑增长因子求得实际年利率。一年后 A = P e^r，因此实际年利率为 EAR = e^r - 1。

If the nominal rate is 6 % compounded continuously, then EAR = e⁰.06 - 1 ≈ 1.0618365 - 1 = 0.0618365, or about 6.18 %. In contrast, daily compounding at the same nominal rate yields EAR = (1 + 0.06/365)³⁶⁵ - 1 ≈ 0.0618313. The difference is tiny but mathematically significant.

若名义利率为 6% 连续复利，则 EAR = e⁰.06 - 1 ≈ 1.0618365 - 1 = 0.0618365，约 6.18%。相比之下，相同名义利率下每日复利的 EAR = (1 + 0.06/365)³⁶⁵ - 1 ≈ 0.0618313。差异虽微小，但在数学上却是重要的。

This shows that continuous compounding gives the highest possible effective rate for a given nominal rate, making it a useful upper bound in financial mathematics.

这表明，对于给定的名义利率，连续复利给出了最高的可能实际利率，使其成为金融数学中有用的上界。

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9. Graphical Illustration | 图形说明

Plotting the function A(t) = Pe^rt reveals smooth exponential growth. On the same axes, we could plot discrete compounding values for different n at selected time points. The discrete points lie slightly below the continuous curve, and the gap narrows as n increases. This visualisation reinforces the idea that continuous compounding is the limiting case.

绘制函数 A(t) = Pe^rt 的图像可展示平滑的指数增长。在同一坐标轴上，我们可以标出不同 n 在选定时间点的离散复利值。离散点位于连续曲线的稍下方，且随着 n 增加间距缩小。这一可视化强化了连续复利是极限情形这一概念。

The continuous curve has a constant proportional growth rate: its derivative is A' , growth occurs in jumps at the end of each period, but as periods become infinitesimally small, the function approaches the differentiable exponential curve.

连续曲线具有恒定的比例增长率：其导数为 A'(t) = r A(t)。在离散复利中，增长在每个周期末跳跃发生，但当周期无限变小，该函数逼近可微的指数曲线。

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10. Real-World Applications and Limitations | 实际应用与局限

Continuous compounding is used in theoretical finance to price derivatives, in the Black-Scholes model for option pricing, and in calculating the present value of perpetuities. It also models natural processes such as population growth, radioactive decay, and continuously reinvested dividends.

连续复利被用于理论金融中为衍生品定价、期权定价的 Black-Scholes 模型，以及计算永续年金的现值。它还模拟自然过程，如人口增长、放射性衰变以及连续再投资的红利。

However, in everyday banking, truly continuous compounding is not offered because interest is credited at discrete intervals. The daily, monthly, or quarterly compounding is the practical norm. Even so, continuous compounding provides a simple and elegant approximation that is often used in economic models.

然而，在日常银行业务中，并不提供真正的连续复利，因为利息是在离散的时间间隔内计入的。每日、每月或每季复利是实际的标准做法。即便如此，连续复利提供了一个简单而优雅的近似，常用于经济模型。

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11. Solving Problems with Continuous Compounding | 使用连续复利解决问题

ed for an investment to double under continuous compounding, set A = 2P and solve: 2P = Pe^rt → 2 = e^rt → ln 2 = rt → t = (ln 2)/r. This is a form of the 'Rule of 72' but more precise.

要求连续复利下投资翻倍所需的时间，可设 A = 2P 求解：2P = Pe^rt → 2 = e^rt → ln 2 = rt → t = (ln 2)/r。这是“72法则”的一种更精确形式。

For example, at r = 4 % ves 72/4 = 18 years, which is close but slightly overestimates for continuous compounding.

例如，当 r = 4% 时，t = ln(2)/0.04 ≈ 0.6931/0.04 = 17.33 年。近似的 72 法则给出 72/4 = 18 年，很接近，但对连续复利而言略微高估。

Similarly, we can solve for the principal or the rate if other variables are known, using natural logarithms and exponentiation. These techniques are standard in IB mathematics and build familiarity with e and ln.

类似地，如果已知其他变量，可以使用自然对数和指数运算求解本金或利率。这些技巧在 IB 数学中是标准的，并能帮助你熟悉 e 和 ln。

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12. Summary and Reflection | 总结与反思

investigation, we have seen how the formula for discrete compound interest A = P(1 + r/n)^nt transitions to the continuous compounding formula A = Pe^rt nstant e emerges as the fundamental growth factor, and the effective annual rate becomes e^r - 1. Continuous compounding represents the mathematical idealisation of constantly reinvesting interest.

通过此探究，我们看到了离散复利公式 A = P(1 + r/n)^nt 如何通过取 n → ∞ 的极限转变为连续复利公式 A = Pe^rt。常数 e 作为基本增长因子出现，实际年利率变为 e^r - 1。连续复利代表着不间断再投资利息的数学理想化。

Exploring this topic deepens understanding of limits, exponential functions, and the natural logarithm. It also demonstrates how a simple practical question, 'What happens if we compound more often?', leads to one of the most important discoveries in mathematics. The logical progression from arithmetic to calculus is mirrored in the journey from discrete to continuous interest.

探索这一主题加深了对极限、指数函数和自然对数的理解。它还展示了如何从一个简单的实际问题——“如果更频繁地复利会怎样？”——引向数学中最重要的发现之一。从离散到连续利息的历程，反映了从算术到微积分的逻辑演进。

Published by TutorHao | Mathematics Revision Series | aleveler.com

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