Natural Constant e

Bernoulli and the Discovery of Natural Constant

📜 Historical Background

In the late 17th century, Swiss mathematician Jacob Bernoulli (1654-1705) accidentally discovered the natural constant ee while studying compound interest problems in finance. This story begins with basic financial concepts.

💰 Principal, Interest, and Interest Rate

Before discussing compound interest, let’s understand some basic financial concepts:

Basic Financial Concepts
  • Principal: The initial amount of money invested or borrowed, denoted as PP. For example, if you deposit 100 yuan in a bank, that 100 yuan is the principal.
  • Interest: The compensation earned by the principal over a certain period of time when lent or deposited. For example, if the bank gives you an additional 2 yuan after one year, that 2 yuan is the interest.
  • Interest Rate: The ratio of interest to principal, usually expressed as a percentage. For example, an annual interest rate of 5% means that for every 100 yuan of principal, 5 yuan of interest is earned per year.

🔄 The Compound Interest Problem

In the 17th century, bankers and merchants already used compound interest to calculate interest in practice. Compound interest means adding the interest generated in each period to the principal, and then calculating interest together in the next period. This way, “interest begets interest,” and the sum of principal and interest grows faster than simple interest.

Bernoulli posed this classic question:

Problem: If you have 1 unit of principal with an annual interest rate of 100%, how much money will you have after one year?

1️⃣ Compounded Once per Year

If interest is calculated only once a year, the year-end sum is:

A1=1×(1+1)=2A_1 = 1 \times (1 + 1) = 2

2️⃣ Compounded Twice per Year

If interest is calculated twice a year, with each time at 50% rate:

  • Annual rate 100%, compounded twice, so each rate is:

    Each rate=100%2=50%=0.5\text{Each rate} = \frac{100\%}{2} = 50\% = 0.5
  • First calculation (after half year):

    1×(1+0.5)=1.51 \times (1 + 0.5) = 1.5
  • Second calculation (after another half year):

    1.5×(1+0.5)=2.251.5 \times (1 + 0.5) = 2.25
  • In formula form:

    Sum=1×(1+0.5)2=1.52=2.25\text{Sum} = 1 \times (1 + 0.5)^2 = 1.5^2 = 2.25

3️⃣ Compounded n Times per Year

If interest is calculated nn times per year, with each rate at 1/n1/n, the year-end sum is:

An=1×(1+1n)nA_n = 1 \times \left(1 + \frac{1}{n}\right)^n

4️⃣ Compounding Frequency Approaches Infinity: Continuous Compounding

Bernoulli wondered: as the compounding frequency increases (n→∞n \to \infty), what value does the sum approach? He calculated results for different nn values:

Compounding Frequency nSum (1+1/n)n(1+1/n)^n
12.0000
22.2500
102.5937
1002.7048
10002.7169
100002.7181
∞e ≈ 2.71828

As nn approaches infinity, the sum approaches a limit, which is:

e=lim⁡n→∞(1+1n)n≈2.71828e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n \approx 2.71828

5️⃣ Historical Significance

In 1683, Bernoulli first published related calculations. Although the constant was not yet denoted by ee, this was the first appearance of ee.

In the 18th century, the great mathematician Leonhard Euler (1707-1783) systematically studied this constant and first used the letter ee to denote it. Euler proved the irrationality of ee and discovered its profound connections with trigonometric functions and complex numbers (the transcendence of ee was later proved by French mathematician Charles Hermite in 1873). ee gradually became an indispensable important constant in mathematical analysis, probability theory, number theory, and other fields.


Deep Understanding of Natural Constant e

📐 Two Definitions

Definition of Natural Constant e

The natural constant ee can be defined in the following two equivalent ways:

Method 1: Limit Definition (Compound Interest Limit)

e=lim⁡n→∞(1+1n)ne = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n

Method 2: Series Definition (Taylor Expansion)

e=∑n=0∞1n!=1+11+12!+13!+14!+⋯e = \sum_{n=0}^{\infty} \frac{1}{n!} = 1 + \frac{1}{1} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots
SymbolTypePronunciation/NoteMeaning in this article
eeMathematical constantNatural constantEuler’s number, approximately 2.71828
n!n!Mathematical symboln factorialn!=n×(n−1)×⋯×1n! = n \times (n-1) \times \cdots \times 1
∑\sumGreek letterSigmaSummation symbol
lim⁡\limMathematical symbolLimitLimit operation

🔍 Rigorous Proof of Limit Definition

Proof Method 1: Binomial Expansion

Binomial Expansion Proof

We want to prove:

e=lim⁡n→∞(1+1n)ne = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n

Proof Process:

Using the binomial theorem:

(1+1n)n=∑k=0n(nk)(1n)k\left(1 + \frac{1}{n}\right)^n = \sum_{k=0}^{n} \binom{n}{k} \left(\frac{1}{n}\right)^k

Where the binomial coefficient is:

(nk)=n!k!(n−k)!=n(n−1)⋯(n−k+1)k!\binom{n}{k} = \frac{n!}{k!(n-k)!} = \frac{n(n-1)\cdots(n-k+1)}{k!}

Substituting:

(1+1n)n=∑k=0nn(n−1)⋯(n−k+1)k!⋅nk\left(1 + \frac{1}{n}\right)^n = \sum_{k=0}^{n} \frac{n(n-1)\cdots(n-k+1)}{k! \cdot n^k}=∑k=0n1⋅(1−1n)⋯(1−k−1n)k!= \sum_{k=0}^{n} \frac{1 \cdot \left(1 - \frac{1}{n}\right) \cdots \left(1 - \frac{k-1}{n}\right)}{k!}

As n→∞n \to \infty, for fixed kk, each factor (1−jn)→1\left(1 - \frac{j}{n}\right) \to 1, so:

lim⁡n→∞(1+1n)n=∑k=0∞1k!=e\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = \sum_{k=0}^{\infty} \frac{1}{k!} = e

Q.E.D.

Proof

Proof Method 2: Squeeze Theorem

Squeeze Theorem Proof

We can prove the following inequality:

(1+1n)n<e<(1+1n)n+1\left(1 + \frac{1}{n}\right)^n < e < \left(1 + \frac{1}{n}\right)^{n+1}

As nn approaches infinity, both sides approach the same limit ee.

Proof Process:

Let an=(1+1n)na_n = \left(1 + \frac{1}{n}\right)^n. We can prove that the sequence {an}\{a_n\} is monotonically increasing and bounded above.

  • Monotonicity: an+1>ana_{n+1} > a_n (can be proved by AM-GM inequality)
  • Boundedness: an<3a_n < 3 (can be proved by comparison with geometric series)

Therefore {an}\{a_n\} converges, and its limit is ee.

Q.E.D.

Proof

🔢 Derivation of Series Definition

Starting from the limit definition, we can derive the series definition:

e=lim⁡n→∞(1+1n)ne = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n

Using binomial expansion and letting n→∞n \to \infty, we get:

e=1+11!+12!+13!+14!+⋯=∑k=0∞1k!e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots = \sum_{k=0}^{\infty} \frac{1}{k!}

This series converges very quickly! Calculating the first few terms:

  • 1+1=21 + 1 = 2
  • +12=2.5+ \frac{1}{2} = 2.5
  • +16≈2.6667+ \frac{1}{6} \approx 2.6667
  • +124≈2.7083+ \frac{1}{24} \approx 2.7083
  • +1120≈2.7167+ \frac{1}{120} \approx 2.7167
  • +1720≈2.7181+ \frac{1}{720} \approx 2.7181

With only 7 terms, we can achieve accuracy to 3 decimal places!


Amazing Properties of e

⭐ Core Properties

Core Properties of e

1. Base of Natural Logarithm

ln⁡x=log⁡ex\ln x = \log_e x

Natural logarithm has a special position in calculus because its derivative form is the simplest.

2. Derivative Equals Itself

ddxex=ex\frac{d}{dx}e^x = e^x

exe^x is the only real function whose derivative equals itself!

3. Irrational and Transcendental

  • ee is an irrational number (proved by Euler in 1737)
  • ee is a transcendental number (proved by Hermite in 1873)

Transcendental means ee is not a root of any polynomial with rational coefficients, which is fundamentally different from algebraic numbers (like 2\sqrt{2}).

4. Decimal Expansion

e=2.7182818284590452353602874713527…e = 2.7182818284590452353602874713527\ldots

The decimal part is infinite and non-repeating.

🎯 Why Choose e as the Base?

Comparison of Exponential Function Derivatives

For a general exponential function f(x)=axf(x) = a^x, its derivative is:

ddxax=axln⁡a\frac{d}{dx}a^x = a^x \ln a

Only when a=ea = e, ln⁡e=1\ln e = 1, the derivative equals itself:

ddxex=ex⋅ln⁡e=ex⋅1=ex\frac{d}{dx}e^x = e^x \cdot \ln e = e^x \cdot 1 = e^x

This is why in calculus we prefer exponential functions with base ee!

🔗 Continued Fraction Representation

ee also has an elegant continued fraction representation:

e=2+11+12+11+11+14+11+11+⋯e = 2 + \frac{1}{1 + \frac{1}{2 + \frac{1}{1 + \frac{1}{1 + \frac{1}{4 + \frac{1}{1 + \frac{1}{1 + \cdots}}}}}}}

The pattern is: [2;1,2,1,1,4,1,1,6,1,1,8,1,1,10,…][2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, \ldots]

This pattern was discovered by Euler, revealing the internal structural beauty of ee.


Euler’s Formula: The Most Beautiful Formula in Mathematics

🌟 Discovery of Euler’s Formula

Euler's Formula
eix=cos⁡x+isin⁡xe^{ix} = \cos x + i\sin x

Where ii is the imaginary unit, i2=−1i^2 = -1.

Proof

Proof (using Taylor series):

Expand eixe^{ix}, cos⁡x\cos x and sin⁡x\sin x into Taylor series respectively:

eix=1+ix+(ix)22!+(ix)33!+(ix)44!+⋯e^{ix} = 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \cdots=1+ix−x22!−ix33!+x44!+ix55!−⋯= 1 + ix - \frac{x^2}{2!} - i\frac{x^3}{3!} + \frac{x^4}{4!} + i\frac{x^5}{5!} - \cdots=(1−x22!+x44!−⋯ )+i(x−x33!+x55!−⋯ )= \left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots\right) + i\left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\right)=cos⁡x+isin⁡x= \cos x + i\sin x

Q.E.D.

💎 Euler’s Identity

When x=πx = \pi, we get the famous Euler’s Identity:

eiπ+1=0e^{i\pi} + 1 = 0

This formula is known as “the most beautiful formula in mathematics” because it connects the five most important constants in mathematics:

ConstantSymbolMeaning
11Multiplicative identity
00Additive identity
ππ\piPi, geometric constant
eeeNatural constant, analytical constant
iiiImaginary unit, algebraic constant

Physicist Richard Feynman called it “the most remarkable formula in mathematics.”


Practical Applications of e

📈 Continuous Growth and Decay Models

Exponential Growth/Decay Model
P(t)=P0ektP(t) = P_0 e^{kt}
  • P0P_0: Initial value
  • kk: Growth rate (k>0k>0) or decay rate (k<0k<0)
  • tt: Time
Radioactive Decay

Carbon-14 has a half-life of about 5730 years. The decay of radioactive substances follows:

N(t)=N0e−λtN(t) = N_0 e^{-\lambda t}

Where λ=ln⁡2T1/2\lambda = \frac{\ln 2}{T_{1/2}} is the decay constant.

Bacterial Growth

Under ideal conditions, bacterial population grows exponentially:

N(t)=N0ertN(t) = N_0 e^{rt}

Where rr is the growth rate.

🎲 Applications in Probability Theory

  • Poisson Distribution: P(X=k)=λke−λk!P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}
  • Normal Distribution: Probability density function contains e−x2e^{-x^2}
  • Central Limit Theorem: ee appears in limit expressions

🏦 Financial Mathematics

  • Continuous Compounding: A=PertA = Pe^{rt}
  • Option Pricing: Black-Scholes formula uses ee

🔬 Physics and Engineering

  • Newton’s Law of Cooling: T(t)=Tenv+(T0−Tenv)e−ktT(t) = T_{env} + (T_0 - T_{env})e^{-kt}
  • RC Circuit Charging: V(t)=V0(1−e−t/RC)V(t) = V_0(1 - e^{-t/RC})
  • Atmospheric Pressure: P(h)=P0e−h/HP(h) = P_0 e^{-h/H}

Comparison of e and π

Propertyeπ
Definition SourceAnalysis (limits, calculus)Geometry (circle circumference/diameter ratio)
First Discovery17th century (Bernoulli)Ancient (Archimedes, etc.)
Mathematical PropertiesIrrational, transcendentalIrrational, transcendental
Decimal Representation2.71828…3.14159…
Main ApplicationsCalculus, differential equations, probabilityGeometry, trigonometry, physics
Series Expansion∑n=0∞1n!\sum_{n=0}^{\infty} \frac{1}{n!}4(1−13+15−17+⋯ )4\left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots\right)
Euler’s Formulaeiπ+1=0e^{i\pi} + 1 = 0Same, both are unified here

Exercises

Exercise 1: Basic Concepts

Write the limit definition and series definition of ee.

natural constant edefinitionlimitseries

Approach:

The natural constant ee has two equivalent definitions: one as the limit of the compound-interest factor (1+1n)n\left(1 + \frac{1}{n}\right)^n, and one as the sum of the reciprocals of the factorials. Write down both.

Steps:

  1. Limit definition: e=lim⁡n→∞(1+1n)ne = \lim_{n \to \infty} (1 + \frac{1}{n})^n, which is exactly the limit of the compound-interest sum for a principal of 11, nn periods and a rate of 1n\frac{1}{n} per period.
  2. Series definition: e=∑n=0∞1n!=1+11!+12!+13!+⋯e = \sum_{n=0}^{\infty} \frac{1}{n!} = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots.
  3. The series converges, because the ratio of consecutive terms 1n+1→0\frac{1}{n+1} \to 0, so its partial sums increase and stay bounded.
  4. Both definitions describe the same constant e≈2.71828e \approx 2.71828, one as the limit of a growth factor and the other as a term-by-term summable series.

Answer:

Limit Definition: e=lim⁡n→∞(1+1n)ne = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n

Series Definition: e=∑n=0∞1n!=1+11!+12!+13!+⋯e = \sum_{n=0}^{\infty} \frac{1}{n!} = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots

Exercise 2: Property Judgment

Determine whether ee is a rational number and explain why.

natural constant eirrational numbertranscendental number

Approach:

First recall the criterion for a rational number — whether it can be written as a ratio of two integers — and then use the two classical theorems about ee to show that it fails even that criterion.

Steps:

  1. A rational number can be expressed as a ratio of two integers, but ee cannot be expressed this way.
  2. ee is an irrational number (proved by Euler in 1737).
  3. Furthermore, ee is a transcendental number (proved by Hermite in 1873), meaning it is not a root of any polynomial with rational coefficients.
  4. Every transcendental number is irrational, because a rational number is always a root of some polynomial with rational coefficients (for instance px−qpx - q), so ee belongs to a strictly stronger class.

Answer:

ee is not a rational number.

Exercise 3: Basic Calculation

Calculate the values of e0e^0, e1e^1, and ln⁡e\ln e.

natural constant eexponential functionlogarithmic function

Approach:

The first two values follow from the exponent laws; the last one reads ln⁡e\ln e as the logarithm to base ee of the value ee, which the equation e1=ee^1 = e settles at once.

Steps:

  1. By the exponent laws the 00-th power of any non-zero number equals 11, so e0=1e^0 = 1.
  2. The 11-st power of any number equals itself, so e1=ee^1 = e.
  3. By the definition of the logarithm, ln⁡e\ln e is the yy with ey=ee^y = e; since e1=ee^1 = e, we get y=1y = 1, that is ln⁡e=1\ln e = 1.

Answer:

  • e0=1e^0 = 1 (any non-zero number to the power of 0 equals 1)
  • e1=ee^1 = e (any number to the power of 1 equals itself)
  • ln⁡e=1\ln e = 1 (because e1=ee^1 = e)

Exercise 4: Limit Identification (Graduate Exam Type)

Which of the following limits equals ee? (A) lim⁡n→∞(1+1n)n\lim_{n\to\infty} \left(1 + \frac{1}{n}\right)^n (B) lim⁡n→∞(1+1n)2n\lim_{n\to\infty} \left(1 + \frac{1}{n}\right)^{2n} (C) lim⁡n→∞(1+2n)n\lim_{n\to\infty} \left(1 + \frac{2}{n}\right)^n (D) lim⁡n→∞(1+12n)n\lim_{n\to\infty} \left(1 + \frac{1}{2n}\right)^n

natural constant elimitgraduate exam

Approach:

All four options have the shape (1+cn)kn\left(1 + \frac{c}{n}\right)^{kn}; rewriting the exponent as a multiple of nn lets the basic limit lim⁡n→∞(1+1n)n=e\lim_{n \to \infty} (1 + \frac{1}{n})^n = e be applied, after which the four results are compared with ee.

Steps:

  1. (A) lim⁡n→∞(1+1n)n=e\lim_{n\to\infty} \left(1 + \frac{1}{n}\right)^n = e ✓
  2. (B) lim⁡n→∞(1+1n)2n=[lim⁡n→∞(1+1n)n]2=e2\lim_{n\to\infty} \left(1 + \frac{1}{n}\right)^{2n} = [\lim_{n\to\infty} \left(1 + \frac{1}{n}\right)^n]^2 = e^2 ✗
  3. (C) lim⁡n→∞(1+2n)n=lim⁡n→∞[(1+2n)n2]2=e2\lim_{n\to\infty} \left(1 + \frac{2}{n}\right)^n = \lim_{n\to\infty} \left[\left(1 + \frac{2}{n}\right)^{\frac{n}{2}}\right]^2 = e^2 ✗
  4. (D) lim⁡n→∞(1+12n)n=lim⁡n→∞[(1+12n)2n]12=e12=e\lim_{n\to\infty} \left(1 + \frac{1}{2n}\right)^n = \lim_{n\to\infty} \left[\left(1 + \frac{1}{2n}\right)^{2n}\right]^{\frac{1}{2}} = e^{\frac{1}{2}} = \sqrt{e} ✗
  5. Of the four results only (A) equals ee, the others being e2e^2 and e\sqrt{e}, neither of which is ee.

Answer:

(A)

Exercise 5: Transcendental Number Concept

Determine whether ee is a transcendental number and explain why.

natural constant etranscendental numbergraduate exam

Approach:

Recall the definition of a transcendental number and the classical theorem about ee, and then compare the two properties to show that transcendence is the stronger one.

Steps:

  1. A transcendental number is a number that is not a root of any polynomial with rational coefficients.
  2. In 1873, French mathematician Charles Hermite first proved that ee is a transcendental number.
  3. This means there is no polynomial P(x)P(x) with integer coefficients such that P(e)=0P(e) = 0.
  4. This is a stronger property than being irrational: all transcendental numbers are irrational, but not all irrational numbers are transcendental (for example, 2\sqrt{2} is irrational but not transcendental).

Answer:

ee is a transcendental number.

Exercise 6: Series Calculation

Use the series definition to calculate the approximate value of ee, accurate to 3 decimal places.

natural constant eseriesnumerical calculation

Approach:

Add the terms of the series e=∑n=0∞1n!e = \sum_{n=0}^{\infty} \frac{1}{n!} one by one and compare with 2.7182.718 after each addition, until a new term is too small to change the value to three decimal places.

Steps:

  1. Starting from the series e=∑n=0∞1n!e = \sum_{n=0}^{\infty} \frac{1}{n!}, compute and accumulate each term 1n!\frac{1}{n!}; the running totals are listed in the table below.
  2. After accumulating up to n=6n = 6 we already have 2.71812.7181, whose distance from 2.7182.718 is less than 0.00050.0005.
  3. Adding the term 15040≈0.0002\frac{1}{5040} \approx 0.0002 for n=7n = 7 gives 2.71832.7183, and every later term has absolute value below 10−410^{-4}, so the value to three decimal places can no longer change.
  4. Hence, to three decimal places, e≈2.718e \approx 2.718.
n1/n!Cumulative Sum
011
112
21/2 = 0.52.5
31/6 ≈ 0.16672.6667
41/24 ≈ 0.04172.7083
51/120 ≈ 0.00832.7167
61/720 ≈ 0.00142.7181
71/5040 ≈ 0.00022.7183

Answer:

e≈2.718e \approx 2.718 (accurate to 3 decimal places)

Exercise 7: Euler’s Formula

Use Euler’s formula to prove: eiπ+1=0e^{i\pi} + 1 = 0

natural constant eEuler formulacomplex number

Approach:

Substitute x=πx = \pi into Euler’s formula eix=cos⁡x+isin⁡xe^{ix} = \cos x + i\sin x and use the values cos⁡π=−1\cos \pi = -1 and sin⁡π=0\sin \pi = 0.

Steps:

  1. According to Euler’s formula:

    eix=cos⁡x+isin⁡xe^{ix} = \cos x + i\sin x
  2. Let x=πx = \pi:

    eiπ=cos⁡π+isin⁡πe^{i\pi} = \cos \pi + i\sin \pi
  3. Since cos⁡π=−1\cos \pi = -1 and sin⁡π=0\sin \pi = 0:

    eiπ=−1+i⋅0=−1e^{i\pi} = -1 + i \cdot 0 = -1
  4. Rearranging:

    eiπ+1=0e^{i\pi} + 1 = 0

Q.E.D. This is the famous Euler’s identity!

Answer:

eiπ+1=0e^{i\pi} + 1 = 0

Exercise 8: Continuous Compounding

If the principal is 1000 yuan, the annual interest rate is 5%, calculated by continuous compounding, what is the sum after 10 years?

natural constant econtinuous compoundingapplication

Approach:

Continuous compounding is governed by A=PertA = Pe^{rt}; substitute the principal, the rate and the number of years and only the approximation of e0.5e^{0.5} remains to be evaluated.

Steps:

  1. The continuous compounding formula is

    A=PertA = Pe^{rt}
  2. Where:

    • P=1000P = 1000 (principal)
    • r=0.05r = 0.05 (5% annual rate)
    • t=10t = 10 (years)
  3. Calculation:

    A=1000⋅e0.05×10=1000⋅e0.5A = 1000 \cdot e^{0.05 \times 10} = 1000 \cdot e^{0.5}
  4. Since e0.5≈1.6487e^{0.5} \approx 1.6487:

    A≈1000×1.6487=1648.72 yuanA \approx 1000 \times 1.6487 = 1648.72 \text{ yuan}

Answer:

Approximately 1648.72 yuan

Exercise 9: Limit Transformation

Find the limit: lim⁡x→0(1+2x)1x\lim_{x\to 0} (1 + 2x)^{\frac{1}{x}}

natural constant elimitgraduate exam

Approach:

The expression has the indeterminate form 1∞1^{\infty}; substitute t=1xt = \frac{1}{x} so that the exponent becomes tt, and then rewrite the base in the form 1+2t1 + \frac{2}{t} to apply the basic limit defining ee.

Steps:

  1. Let t=1xt = \frac{1}{x}; when x→0x \to 0, t→∞t \to \infty.

    lim⁡x→0(1+2x)1x=lim⁡t→∞(1+2t)t\lim_{x\to 0} (1 + 2x)^{\frac{1}{x}} = \lim_{t\to\infty} \left(1 + \frac{2}{t}\right)^t
  2. Rewrite it as a power of the basic limit:

    =lim⁡t→∞[(1+2t)t2]2= \lim_{t\to\infty} \left[\left(1 + \frac{2}{t}\right)^{\frac{t}{2}}\right]^2
  3. Since lim⁡t→∞(1+2t)t2=e\lim_{t\to\infty}\left(1 + \frac{2}{t}\right)^{\frac{t}{2}} = e:

    =[e]2=e2= [e]^2 = e^2

Answer:

e2e^2

Exercise 10: Differential Equation

Find the particular solution of the differential equation dydx=y\frac{dy}{dx} = y with y(0)=1y(0) = 1.

natural constant edifferential equationderivative

Approach:

The equation is separable: divide by yy, integrate both sides and then use the initial condition to fix the integration constant.

Steps:

  1. Separate variables:

    dyy=dx\frac{dy}{y} = dx
  2. Integrate both sides:

    ln⁡y=x+C\ln y = x + C
  3. From y(0)=1y(0) = 1, we get ln⁡1=0+C\ln 1 = 0 + C, so C=0C = 0.

  4. Therefore:

    ln⁡y=x⇒y=ex\ln y = x \Rightarrow y = e^x

Answer:

y=exy = e^x

This also explains why the derivative of exe^x equals itself!


Summary

📚 Core Points Review

  1. Historical Discovery: Discovered by Bernoulli while studying compound interest (1683), systematically studied and named by Euler (18th century)
  2. Dual Definitions: Limit definition (compound interest) and series definition (Taylor expansion)
  3. Core Properties: Derivative equals itself, base of natural logarithm, irrational number, transcendental number
  4. Euler’s Formula: eix=cos⁡x+isin⁡xe^{ix} = \cos x + i\sin x, connecting different mathematical fields
  5. Wide Applications: Calculus, probability theory, physics, engineering, finance

🔗 Course Connections

This course is closely related to:

  • Exponential Functions (exponential-function.mdx): Foundation of exe^x
  • Logarithmic Functions (logarithmic-function.mdx): Base of ln⁡x\ln x
  • Limits (limits/): Second important limit
  • Series (infinite-series/): Exponential series
  • Derivatives (differential-calculus/): (ex)′=ex(e^x)' = e^x
  • Integrals (integral-calculus/): ∫exdx=ex+C\int e^x dx = e^x + C

🌟 Mathematical Quote

“God created the integers, all else is the work of man.” — Leopold Kronecker

And the discovery of ee tells us: even starting from human-made concepts (like compound interest), we can discover profound and beautiful truths in mathematics.

📖 Further Reading

  • “Euler’s Complete Works”: Understand Euler’s systematic research on ee
  • “Mathematical Constants”: by Steven Finch, deep exploration of various mathematical constants
  • “Mathematical Thought from Ancient to Modern Times”: by Morris Kline, understand the context of mathematical history

Symbols in this Article

SymbolTypePronunciation/NoteMeaning in this article
eeMathematical constantNatural constantEuler’s number, approximately 2.71828
π\piGreek letterPiCircle constant, approximately 3.14159
iiMathematical symbolImaginary uniti2=−1i^2 = -1
ln⁡\lnMathematical symbolNatural logarithmLogarithm with base ee
∑\sumGreek letterSigmaSummation symbol
lim⁡\limMathematical symbolLimitLimit operation
n!n!Mathematical symboln factorialn×(n−1)×⋯×1n \times (n-1) \times \cdots \times 1
∞\inftyMathematical symbolInfinityInfinite quantity
∫\intMathematical symbolIntegralIntegral operation

Greek Letter Quick Reference

UpperLowerEnglish NameMeaning in Text
Ππ\piPiCircle constant (Pi)
Σ∑\sumSigmaSummation symbol
-∞\inftyInfinityInfinity

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