Taylor Polynomial

Topics: Calculus - Polynomial - Function

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(definition)

Let $f:\mathbb{R}\to \mathbb{R}$ be a function, with $n$ derivatives.

We define the degree $n$ Taylor polynomial around $a\in \mathbb{R}$ as:

$$P_f(n,a)=\sum _{i=0}^n\frac{f^i(a)}{i!}(x-a{)}^i$$

…where $f^i(a)$ denotes the $i$th derivative of $f$ evaluated at the point $a$. Note that the zeroth derivative $f$ is defined to be $f$ itself, while $0!$ is defined to be $1$.

Example

Let $f(x)=e^x$. We’ll build $P_f(1,0)$ first, then $P_f(4,0)$.

Calculating the derivatives of $f$ is trivial, since the derivative of $e^x$ is always itself:

• $f(x)=e^x$
• $f^{\prime }(x)=e^x$
• $f^{\prime \prime }(x)=e^x$
• $f^{\prime \prime \prime }(x)=e^x$
• etc.

Evaluating the derivatives at $0$ is also trivial:

• $f(0)=1$
• $f^{\prime }(0)=1$
• $f^{\prime \prime }(0)=1$
• $f^{\prime \prime \prime }(0)=1$
• etc.

With all of this, we can now easily build $P_f(1,0)$:

$$P_f(1,0)=1+(x-0)1=1+x$$

…and $P_f(4,0)$:

$$P_f(4,0)=1+1(x-0)+\frac{1}{2!}(x-0{)}^2+\frac{1}{3!}(x-0{)}^3+\frac{1}{4!}(x-0{)}^4$$

$$P_f(4,0)=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\frac{x^4}{4!}$$