# Taylor's Theorem/One Variable

## Theorem

Let $f$ be a real function which is:

- of differentiability class $C^n$ on the closed interval $\closedint a x$

and:

- at least $n + 1$ times differentiable on the open interval $\openint a x$.

Then:

\(\ds \map f x\) | \(=\) | \(\ds \frac 1 {0!} \map f a\) | ||||||||||||

\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds \frac 1 {1!} \paren {x - a} \map {f'} a\) | |||||||||||

\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds \frac 1 {2!} \paren {x - a}^2 \map {f''} a\) | |||||||||||

\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds \cdots\) | |||||||||||

\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds \frac 1 {n!} \paren {x - a}^n \map {f^{\paren n} } a\) | |||||||||||

\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds R_n\) |

where $R_n$ (sometimes denoted $E_n$) is known as the **error term** or **remainder**, and can be presented in one of $2$ forms:

- Lagrange Form

- $R_n = \dfrac {\map {f^{\paren {n + 1} } } \xi} {\paren {n + 1}!} \paren {x - a}^{n + 1}$

for some $\xi \in \openint a x$.

- Cauchy Form

- $R_n = \dfrac {\map {f^{\paren {n + 1} } } \xi \paren {x - \xi}^n} {n!} \paren {x - a}$

for some $\xi \in \openint a x$.

## Proof

### Integral Version

This proof requires $f^{\paren n}$ to be absolutely continuous on $\closedint a x$, so that the Fundamental Theorem of Calculus holds.

Except at the end when the Mean Value Theorem is invoked, differentiability of $f^{\paren n}$ need not be assumed, since absolute continuity implies:

- differentiability almost everywhere
- the validity of the Fundamental Theorem of Calculus

provided the integrals involved are understood as Lebesgue integrals.

Consequently, the integral form of the remainder holds with this particular weakening of the assumptions on $f$.

We first prove Taylor's Theorem with the integral remainder term.

The Fundamental Theorem of Calculus states that:

- $\ds \int_a^x \map {f'} t \rd t = \map f x - \map f a$

which can be rearranged to:

- $\ds \map f x = \map f a + \int_a^x \map {f'} t \rd t$

Now we can see that an application of Integration by Parts yields:

\(\ds \map f x\) | \(=\) | \(\ds \map f a + x \map {f'} x - a \map {f'} a - \int_a^x t \map {f''} t \rd t\) | $u = \map {f'} t$ and $\d v = \d t$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \map f a + \int_a^x x \map {f''} t \rd t + x \map {f'} a - a \map {f'} a - \int_a^x t \map {f''} t \rd t\) | $\ds \int_a^x x \map {f''} t \rd t = x \map {f'} x - x \map {f'} a$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \map f a + \paren {x - a} \map {f'} a + \int_a^x \paren {x - t} \map {f''} t \rd t\) | factoring out some common terms |

Another application yields:

- $\ds \map f x = \map f a + \paren {x - a} \map {f'} a + \frac 1 2 \paren {x - a}^2 \map {f''} a + \frac 1 2 \int_a^x \paren {x - t}^2 \map {f'''} t \rd t$

By repeating this process, we may derive Taylor's theorem for higher values of $n$.

This can be formalized by applying the technique of Principle of Mathematical Induction.

So, suppose that Taylor's theorem holds for a $n$, that is, suppose that:

\(\ds \map f x\) | \(=\) | \(\ds \map f a\) | ||||||||||||

\(\ds \) | \(+\) | \(\ds \frac {\map {f'} a} {1!} \paren {x - a}\) | ||||||||||||

\(\ds \) | \(+\) | \(\ds \cdots\) | ||||||||||||

\(\ds \) | \(+\) | \(\ds \frac {\map {f^{\paren n} } a} {n!} \paren {x - a}^n\) | ||||||||||||

\(\ds \) | \(+\) | \(\ds \int_a^x \frac {\map {f^{\paren {n + 1} } } t} {n!} \paren {x - t}^n \rd t\) | $*$ |

We can rewrite the integral using Integration by Parts.

A primitive of $\paren {x - t}^n$ as a function of $t$ is given by $\dfrac {-\paren {x - t}^{n + 1} } {n + 1}$.

So:

\(\ds \) | \(\) | \(\ds \int_a^x \map {\frac {f^{\paren {n + 1} } } t} {n!} \paren {x - t}^n \rd t\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds -\intlimits {\map {\frac {f^{\paren {n + 1} } } t} {\paren {n + 1} n!} \paren {x - t}^{n + 1} } a x + \int_a^x \frac {\map {f^{\paren {n + 2} } } t} {\paren {n + 1} n!} \paren {x - t}^{n + 1} \rd t\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \frac {\map {f^{\paren {n + 1} } } a} {\paren {n + 1}!} \paren {x - a}^{n + 1} + \int_a^x \frac {\map {f^{\paren {n + 2} } } t} {\paren {n + 1}!} \paren {x - t}^{n + 1} \rd t\) |

The last integral can be solved immediately, which leads to

- $R_n = \dfrac {\map {f^{\paren {n + 1} } } \xi} {\paren {n + 1}!} \paren {x - a}^{n + 1}$

$\blacksquare$

### Proof using Cauchy Mean Value Theorem

An alternative proof, which holds under milder technical assumptions on the function $f$, can be supplied using the Cauchy Mean Value Theorem.

Let $G$ be a real-valued function continuous on $\closedint a x$ and differentiable with non-vanishing derivative on $\openint a x$.

Let:

- $\map F t = \map f t + \dfrac {\map {f'} t} {1!} \paren {x - t} + \dotsb + \dfrac {\map {f^{\paren n} } t} {n!} \paren {x - t}^n$

By the Cauchy Mean Value Theorem:

- $(1): \quad \dfrac {\map {F'} x} {\map {G'} \xi} = \dfrac {\map F x - \map F a} {\map G x - \map G a}$

for some $\xi \in \openint a x$.

Note that the numerator:

- $\map F x - \map F a = R_n$

is the remainder of the Taylor polynomial for $\map f x$.

On the other hand, computing $\map {F'} \xi$:

- $\map {F'} \xi = \map {f'} \xi - \map {f'} \xi + \dfrac {\map {f''} \xi} {1!} \paren {x - \xi} - \dfrac {\map {f''} \xi} {1!} \paren {x - \xi} + \dotsb + \dfrac {\map {f^{\paren {n + 1} } } t} {n!} \paren {x - \xi}^n = \dfrac {\map {f^{\paren {n + 1} } } \xi} {n!} \paren {x - \xi}^n$

Putting these two facts together and rearranging the terms of $(1)$ yields:

- $R_n = \dfrac {\map {f^{\paren {n + 1} } } \xi} {n!} \paren {x - \xi}^n \dfrac {\map G x - \map G a} {\map {G'} \xi}$

which was to be shown.

This page has been identified as a candidate for refactoring.Find a way of neatly handling the two different remainder formsUntil this has been finished, please leave
`{{Refactor}}` in the code.
Because of the underlying complexity of the work needed, it is recommended that you do not embark on a refactoring task until you have become familiar with the structural nature of pages of $\mathsf{Pr} \infty \mathsf{fWiki}$.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Refactor}}` from the code. |

Note that the Lagrange Form of the Remainder comes from taking $\map G t = \paren {x - t}^{n + 1}$ and the given Cauchy Form of the Remainder comes from taking $\map G t = t - a$.

$\blacksquare$

### Proof using Rolle's Theorem directly

Yet another proof for Lagrange Form of the Remainder can be constructed applying Rolle's theorem directly $n$ times; this proof might be easier to visualize geometrically.

Let the function $g$ be defined as:

- $\map g t = \map {R_n} t - \dfrac {\paren {t - a}^{n + 1} } {\paren {x - a}^{n + 1} } \map {R_n} x$

Then:

- $\map {g^{\paren k} } a = 0$

for $k = 0, \dotsc, n$, and $\map g x = 0$.

Apply Rolle's Theorem successively to $g, g', \dotsc, g^{\paren n}$.

Then there exist:

- $\xi_1, \ldots, \xi_{n + 1}$

between $a$ and $x$ such that:

- $\map {g'} {\xi_1} = 0, \map {g''} {\xi_2} = 0, \ldots, \map {g^{\paren {n + 1} } } {\xi_{n + 1} } = 0$

Let $\xi = \xi_{n + 1}$.

Then:

- $0 = \map {g^{\paren {n + 1} } } \xi = \map {f^{\paren {n + 1} } } \xi - \dfrac {\paren {n + 1}!} {\paren {x - a}^{n + 1} } \map {R_n} x$

and the formula for $\map {R_n} x$ follows.

$\blacksquare$

## Also see

- Mean Value Theorem: Taylor's Theorem when $n = 0$

## Source of Name

This entry was named for Brook Taylor.

## Sources

- 1998: David Nelson:
*The Penguin Dictionary of Mathematics*(2nd ed.) ... (previous) ... (next): Entry:**Taylor's theorem** - 2008: David Nelson:
*The Penguin Dictionary of Mathematics*(4th ed.) ... (previous) ... (next): Entry:**Taylor's theorem** - 2014: Christopher Clapham and James Nicholson:
*The Concise Oxford Dictionary of Mathematics*(5th ed.) ... (previous) ... (next): Entry:**Taylor's Theorem**