The constant known as e
Table of Contents Calculating e Approximating e^x Understanding Euler's Identity What is e e e
To grasp the concept of e e e 100 100 100 100 100% 100
A year later, your investment doubles to 200 200 200 2 2 2 50 50% 50 150 150 150
( 1 , 5 ) 2 = 2 , 25 (1,5)^2 = 2,25 ( 1 , 5 ) 2 = 2 , 25 Notice how it increases. Let's break down the investment further into quarters, each with a 25% return. The growth now becomes to:
( 1 , 25 ) 4 = 2 , 441 (1,25)^4 = 2,441 ( 1 , 25 ) 4 = 2 , 441 What if the investment were even more frequent, say monthly? This is the total growth factor at the end of the year:
( 1 + 1 12 ) 12 = 2 , 613 (1 + \frac{1}{12})^{12} = 2,613 ( 1 + 12 1 ) 12 = 2 , 613 The formula for calculating the growth factor is thus:
( 1 + 1 n ) n (1 + \frac{1}{n})^n ( 1 + n 1 ) n Does this mean the growth factor increases infinitely? Not quite. Eventually, it approaches a constant value, known as e e e e e e
lim n → ∞ ( 1 + 1 n ) n \lim\limits_{n \to \infty} (1 + \frac{1}{n})^n n → ∞ lim ( 1 + n 1 ) n But, how do we calculate this? Well, this is where binomial expansion comes to place. Remember the formula for binomial expansion:
( a + b ) n = ∑ k = 0 n ( n k ) a n − k b k (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k ( a + b ) n = k = 0 ∑ n ( k n ) a n − k b k Where ( n k ) \binom{n}{k} ( k n )
n ! ( k ! ) ( n − k ) ! \frac{n!}{(k!)(n-k)!} ( k !) ( n − k )! n ! Now let's plug our definition for e e e
e = lim n → ∞ ∑ k = 0 n ( n ! ( k ! ) ( n − k ) ! ) 1 n − k ( 1 n ) k = ∑ k = 0 n lim n → ∞ n ! ( k ! ) ( n − k ) ! ⋅ 1 n k = ∑ k = 0 n lim n → ∞ 1 k ! ⋅ n ! ( n k ) ( n − k ) ! \begin{split}
e &= \lim\limits_{n \to \infty} \sum_{k=0}^{n} (\frac{n!}{(k!)(n-k)!}) 1^{n-k} (\frac{1}{n})^k \\
&= \sum_{k=0}^{n} \lim\limits_{n \to \infty} \frac{n!}{(k!)(n-k)!} \cdot \frac{1}{n^k}\\
&= \sum_{k=0}^{n} \lim\limits_{n \to \infty} \frac{1}{k!} \cdot \frac{n!}{(n^k)(n-k)!}
\end{split} e = n → ∞ lim k = 0 ∑ n ( ( k !) ( n − k )! n ! ) 1 n − k ( n 1 ) k = k = 0 ∑ n n → ∞ lim ( k !) ( n − k )! n ! ⋅ n k 1 = k = 0 ∑ n n → ∞ lim k ! 1 ⋅ ( n k ) ( n − k )! n ! Now, we can focus on this part of the equation:
n ! ( n k ) ( n − k ) ! = n ! ( n − k ) ! ⋅ 1 n k \frac{n!}{(n^k)(n-k)!} = \frac{n!}{(n-k)!} \cdot \frac{1}{n^k} ( n k ) ( n − k )! n ! = ( n − k )! n ! ⋅ n k 1 The fraction n ! ( n − k ) ! \frac{n!}{(n-k)!} ( n − k )! n ! ( n ) ( n − 1 ) ( n − 2 ) ⋯ ( n − k + 1 ) (n)(n-1)(n-2)\cdots(n-k+1) ( n ) ( n − 1 ) ( n − 2 ) ⋯ ( n − k + 1 ) n − k n-k n − k n n n n n n
lim n → ∞ n ! ( n − k ) ! ⋅ 1 n k = lim n → ∞ n n ⋅ n − 1 n ⋯ n − k + 1 n = 1 ⋅ 1 ⋯ 1 = 1 \begin{split}
\lim\limits_{n \to \infty} \frac{n!}{(n-k)!} \cdot \frac{1}{n^k} &= \lim\limits_{n \to \infty} \frac{n}{n} \cdot \frac{n-1}{n} \cdots \frac{n-k+1}{n} \\
&= 1 \cdot 1 \cdots 1 \\
&= 1
\end{split} n → ∞ lim ( n − k )! n ! ⋅ n k 1 = n → ∞ lim n n ⋅ n n − 1 ⋯ n n − k + 1 = 1 ⋅ 1 ⋯ 1 = 1 From here we can incorporate it to our binomial expansion:
e = ∑ k = 0 n lim n → ∞ n ! ( n k ) ( n − k ) ! ⋅ 1 k ! e = ∑ k = 0 ∞ 1 k ! \begin{split}
e &= \sum_{k=0}^{n} \lim\limits_{n \to \infty} \frac{n!}{(n^k)(n-k)!} \cdot \frac{1}{k!} \\
e &= \sum_{k=0}^{\infty} \frac{1}{k!}
\end{split} e e = k = 0 ∑ n n → ∞ lim ( n k ) ( n − k )! n ! ⋅ k ! 1 = k = 0 ∑ ∞ k ! 1 Getting the value of e e e e x e^x e x x = 1 x=1 x = 1
Let us try to get the approximation of e x e^x e x lim n → ∞ ( 1 + 1 n ) n x \lim\limits_{n \to \infty} (1 + \frac{1}{n})^{nx} n → ∞ lim ( 1 + n 1 ) n x
e x = lim n → ∞ ∑ k = 0 n ( n x k ) 1 n x − k ( 1 n ) k = lim n → ∞ ∑ k = 0 n ( n x ! ( k ! ) ( n x − k ) ! ) ( 1 n ) k = ∑ k = 0 n lim n → ∞ n x ! ( k ! ) ( n x − k ) ! ⋅ 1 n k = ∑ k = 0 n lim n → ∞ 1 k ! ⋅ n x ! ( n k ) ( n x − k ) ! \begin{split}
e^x &= \lim\limits_{n \to \infty} \sum_{k=0}^{n} \binom{nx}{k} 1^{nx-k}(\frac{1}{n})^k \\
&= \lim\limits_{n \to \infty} \sum_{k=0}^{n} (\frac{nx!}{(k!)(nx-k)!})(\frac{1}{n})^k \\
&= \sum_{k=0}^{n} \lim\limits_{n \to \infty} \frac{nx!}{(k!)(nx-k)!} \cdot \frac{1}{n^k}\\
&= \sum_{k=0}^{n} \lim\limits_{n \to \infty} \frac{1}{k!} \cdot \frac{nx!}{(n^k)(nx-k)!}
\end{split} e x = n → ∞ lim k = 0 ∑ n ( k n x ) 1 n x − k ( n 1 ) k = n → ∞ lim k = 0 ∑ n ( ( k !) ( n x − k )! n x ! ) ( n 1 ) k = k = 0 ∑ n n → ∞ lim ( k !) ( n x − k )! n x ! ⋅ n k 1 = k = 0 ∑ n n → ∞ lim k ! 1 ⋅ ( n k ) ( n x − k )! n x ! Just like with n ! ( n − k ) ! \frac{n!}{(n-k)!} ( n − k )! n ! n x ! ( n x − k ) ! \frac{nx!}{(nx-k)!} ( n x − k )! n x ! n n n n x − k nx-k n x − k n x nx n x
lim n → ∞ n x ! ( n x − k ) ! ⋅ 1 n k = lim n → ∞ n x n ⋅ n x − 1 n ⋯ n x − k + 1 n = x ⋅ x ⋯ x = x k \begin{split}
\lim\limits_{n \to \infty} \frac{nx!}{(nx-k)!} \cdot \frac{1}{n^k} &= \lim\limits_{n \to \infty} \frac{nx}{n} \cdot \frac{nx-1}{n} \cdots \frac{nx-k+1}{n} \\
&= x \cdot x \cdots x \\
&= x^k
\end{split} n → ∞ lim ( n x − k )! n x ! ⋅ n k 1 = n → ∞ lim n n x ⋅ n n x − 1 ⋯ n n x − k + 1 = x ⋅ x ⋯ x = x k Plugging it back to the sum, we get:
e x = ∑ k = 0 n lim n → ∞ n x ! ( n k ) ( n x − k ) ! ⋅ 1 k ! = ∑ k = 0 ∞ x k k ! \begin{split}
e^x &= \sum_{k=0}^{n} \lim\limits_{n \to \infty} \frac{nx!}{(n^k)(nx-k)!} \cdot \frac{1}{k!} \\
&= \sum_{k=0}^{\infty} \frac{x^k}{k!}
\end{split} e x = k = 0 ∑ n n → ∞ lim ( n k ) ( n x − k )! n x ! ⋅ k ! 1 = k = 0 ∑ ∞ k ! x k To derive e x e^x e x
e x = ∑ k = 0 ∞ x k k ! = x 0 0 ! + ∑ k = 1 ∞ x k k ! = 1 + ∑ k = 1 ∞ x k k ! d e x d x = 0 + ∑ k = 1 ∞ k ⋅ x k − 1 k ! = ∑ k = 1 ∞ x k − 1 ( k − 1 ) ! = ∑ k = 0 ∞ x k k ! = e x \begin{split}
e^x &= \sum_{k=0}^{\infty} \frac{x^k}{k!}\\
&= \frac{x^0}{0!} + \sum_{k=1}^{\infty} \frac{x^k}{k!}\\
&= 1 + \sum_{k=1}^{\infty} \frac{x^k}{k!}\\
\frac{de^x}{dx} &= 0 + \sum_{k=1}^{\infty} \frac{k \cdot x^{k-1}}{k!} \\
&= \sum_{k=1}^{\infty} \frac{x^{k-1}}{(k-1)!}\\
&= \sum_{k=0}^{\infty} \frac{x^{k}}{k!}\\
&= e^x
\end{split} e x d x d e x = k = 0 ∑ ∞ k ! x k = 0 ! x 0 + k = 1 ∑ ∞ k ! x k = 1 + k = 1 ∑ ∞ k ! x k = 0 + k = 1 ∑ ∞ k ! k ⋅ x k − 1 = k = 1 ∑ ∞ ( k − 1 )! x k − 1 = k = 0 ∑ ∞ k ! x k = e x Understanding Euler's Identity
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