#C01 · Calculus & Analysis · Lv.2View in 3D Mind-Map

Limits, Continuity & Euler's Number e

Jacob Bernoulli & Augustin-Louis Cauchy · 1683 CE

e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n \approx 2.718281828

\lim_{x \to 0} \frac{\sin x}{x} = 1,\quad \frac{d}{dx}e^x = e^x

Formalizes infinite approach via limits and defines Euler's number e ≈ 2.71828 from continuous compounding—the unique base whose exponential function is its own derivative.

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Compounding n

\left(1 + \frac{1}{100}\right)^{100} = 2.704814 \approx e = 2.718282

(1 + 1/n)^n 근삿값

2.704814

자연상수 극한 e

2.718282

수렴 오차 |e - 근사|

1.35e-2

Prerequisite Path (6 STEP)

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Click steps in order to follow the 3D neon path from basics to target.

STEP 1#A01Linear Equation & Balance Axiom

ax + b = 0 \implies x = -\frac{b}{a}

STEP 2#A02Polynomial Expansion & Factoring

(a+b)^2 = a^2 + 2ab + b^2

STEP 3#A03Laws of Exponents & Radicals

a^m \cdot a^n = a^{m+n}

STEP 4#A06Logarithm Laws & Change of Base

\log_a(xy) = \log_a x + \log_a y

STEP 5#A08Binomial Theorem & Pascal's Triangle

(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k

STEP 6#C01Limits, Continuity & Euler's Number e

e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n \approx 2.718281828

🚀 Next Unlocks (Derived Formulas) (4)

#C02Derivative Definition & Instantaneous Rate #C07Euler's Formula & Euler's Identity #P06Normal (Gaussian) Distribution & Z-Score #P10Softmax Function & Stochastic Gradient Descent

Geometric & Intuitive Insight

연이율 100%를 1년에 한 번 주면 2배가 되지만, 매달·매일·매초로 쪼개어 이자에 이자를 연속으로 붙이면 무한히 커지지 않고 정확히 2.7182818…(자연상수 e)에 수렴합니다.

Step-by-Step Derivation

Step 1.1단계: 이항정리로 (1 + 1/n)^n = ∑ (nCk / n^k) = 1 + 1 + (1 - 1/n)/2! + (1 - 1/n)(1 - 2/n)/3! + … 전개
Step 2.2단계: n → ∞ 극한을 취하면 각 괄호가 1로 수렴하여 e = ∑ (1 / k!) = 1 + 1 + 1/2! + 1/3! + 1/4! + …
Step 3.3단계: 이를 이용해 (e^x)' = lim ((e^(x+h) - e^x)/h) = e^x · lim ((e^h - 1)/h) = e^x 가 증명됩니다.

AI & Coding Real-World Use

Deep learning activations (Sigmoid, Softmax, GELU) and continuous compounding e^(rt) in quant finance

#Sigmoid & Softmax#Continuous Compounding#Calculus Foundation

const approxE = (n: number) => Math.pow(1 + 1 / n, n);

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