Why are the Maclaurin coefficients of $e^x$ equal to $1/n$! for all n?

Conditions

• The function is $f(x) = e^x$.
• The expansion is centered at $x=0$ (Maclaurin series).
• The general Taylor coefficient formula $c_n = f^{(n)}(a) / n!$ is applied.

Reasoning, step by step

1. Recall the general formula for the $n$-th coefficient of a Maclaurin series: $c_n = f^{(n)}(0) / n!$.
2. Identify the property of the exponential function: its $n$-th derivative is $f^{(n)}(x) = e^x$ for all $n$.
3. Evaluate the $n$-th derivative at the center $x=0$: $f^{(n)}(0) = e^0 = 1$.
4. Substitute this value into the coefficient formula: $c_n = 1 / n!$.
5. Conclude that the Maclaurin series is $e^x = \sum_{n=0}^{\infty} x^n / n!$.

Example

The script states: 'The exponential has all derivatives equal to itself. At zero each derivative equals one, so its Maclaurin coefficients are $1/n$!.'

Common misconceptions

• Assuming the derivatives of $e^x$ cycle or change form, which would result in alternating signs or different coefficients.
• Forgetting to evaluate the derivative at the center point $x=0$ before dividing by $n!$.

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Connected concepts

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