What happens to the other terms in the Taylor series when taking the nth derivative at $x=a$?

Conditions

• The function is infinitely differentiable at a.
• n is a non-negative integer.
• The derivative is evaluated exactly at the center point $x=a$.

Reasoning, step by step

1. Identify the terms to the left of the nth term, which have degrees strictly less than n.
2. Apply the nth derivative to these lower-order terms; they become exactly zero.
3. Identify the terms to the right of the nth term, which have degrees greater than n.
4. Apply the nth derivative to these higher-order terms; they retain at least one factor of (x-a).
5. Evaluate the resulting expressions at $x=a$; the factor (a-a) becomes zero, causing the entire term to vanish.
6. Conclude that only the nth term survives the nth differentiation and evaluation at $x=a$.

Example

The video explains that terms to the left vanish because differentiating a polynomial of degree m, m times, where $m < n$, results in 0. Terms to the right vanish at $x=a$ because they contain at least one (x-a) term, so at $x=a$, the nth derivative vanishes.

Common misconceptions

• Assuming that higher-order terms become zero immediately upon differentiation, rather than only when evaluated at $x=a$.
• Thinking that lower-order terms leave a constant remainder after n differentiations.

Watch the explanation

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