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Taylor Theorem

Level 1 - Math I (Physics) topic page in Power Series.

Principle

Taylor's theorem explains why Taylor polynomials approximate a function. It says that a sufficiently differentiable function equals its Taylor polynomial plus a remainder term.

Taylor theorem form
\[f(x)=P_N(x)+R_N(x)\]

The theorem is the bridge between formal derivative matching and controlled approximation. In physics, this is what justifies keeping only the leading terms in a small-parameter model.

Notation

\(P_N(x)\)
Taylor polynomial of degree N about c
\(R_N(x)\)
remainder after degree N
\(f^{(N+1)}\)
derivative one order higher than the polynomial
\(c\)
centre of the Taylor approximation
\(\xi\)
some point between c and x in the Lagrange remainder form

The exact value of \(\xi\) is usually not known. The useful part is that it lies between the centre and the input, which lets us bound the derivative there.

Method

Step 1: Build the Taylor polynomial

Compute derivatives at \(c\) up to order \(N\), then form \(P_N(x)\).

Step 2: Write the remainder form

One common Level 1 form is the Lagrange remainder:

Lagrange remainder
\[R_N(x)=\frac{f^{(N+1)}(\xi)}{(N+1)!}(x-c)^{N+1}\]

where \(\xi\) lies between \(c\) and \(x\).

Step 3: Use a bound if accuracy is needed

If \(|f^{(N+1)}(t)|\le M\) between \(c\) and \(x\), then

Start from Lagrange remainder
\[R_N(x)=\frac{f^{(N+1)}(\xi)}{(N+1)!}(x-c)^{N+1}\]
Take absolute values
\[|R_N(x)|=\frac{|f^{(N+1)}(\xi)|}{(N+1)!}|x-c|^{N+1}\]
Use derivative bound
\[|f^{(N+1)}(\xi)|\le M\]
Error bound
\[|R_N(x)|\le\frac{M}{(N+1)!}|x-c|^{N+1}\]

Rules

Taylor polynomial
\[P_N(x)=\sum_{n=0}^{N}\frac{f^{(n)}(c)}{n!}(x-c)^n\]
Function plus error
\[f(x)=P_N(x)+R_N(x)\]
Remainder bound
\[|R_N(x)|\le\frac{M}{(N+1)!}|x-c|^{N+1}\]

The bound gets smaller when \(|x-c|\) is small or when the factorial in the denominator dominates the derivative growth.

Examples

Question
Use Taylor's theorem to describe \(e^x\) after the linear term about
\[0\]
Answer
The linear Taylor polynomial is
\[P_1(x)=1+x\]
Taylor's theorem gives
\[e^x=1+x+R_1(x)\]
where
\[R_1(x)=e^{\xi}x^2/2!\]
for some \(\xi\) between
\[0\]
and \(x\).

Checks

  • Taylor's theorem includes an error term; it is not just a polynomial formula.
  • The Lagrange remainder uses a derivative one order higher than the polynomial.
  • Error bounds depend on an interval between \(c\) and \(x\).
  • A small input displacement is what makes local approximation powerful.

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Taylor Series

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Remainders