Why does the Taylor series for $e^x$ centered at 0 become sum $x^n/n$!?

Conditions

• $f(x)=e^x$
• center $a=0$

Reasoning, step by step

1. Start with the Taylor coefficient formula $c_n = f^{(n)}(a) / n$!.
2. Specialize to the example function and center: $f(x) = e^x$ and $a = 0$.
3. Evaluate the nth derivative of $e^x$ at $x=0$. Since $d^n/dx^n(e^x) = e^x$, the value is $e^0 = 1$.
4. Substitute the evaluated derivative into the coefficient formula to get $c_n = 1 / n$!.
5. Write the final power series by substituting $c_n$ into the sum: $e^x$ = sum ($1/n$!) $x^n$ = sum $x^n/n$!.

Example

The instructor explains the key simplification for the exponential function: differentiating $e^x$ does not change it. Thus the nth derivative is still $e^x$, and evaluating at the center 0 gives $e^0=1$. The board simplifies to $e^x$=sum_{$n=0$}^infty $x^n/n$!.

Common misconceptions

• Assuming that 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!.

Watch the explanation

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