Taylor's theorem

Taylor's theorem expresses a sufficiently differentiable function as a finite Taylor polynomial plus a Taylor remainder. One convenient version assumes $f$ is $n+1$ times continuously differentiable on the segment between $a$ and $x$. Thenfor some intermediate point $\xi$, when $x\ne a$. Thus a bound $|f^{(n+1)}|\le M$ on that segment gives $|R_{n,a}(x)|\le M|x-a{|}^{n+1}/(n+1)!$. The intermediate point usually need not be computed: bounding the derivative throughout the segment is enough. This is why local approximation can yield reliable numerical bounds. A finite Taylor polynomial does not require an infinite Taylor series to represent the function.

Historically, the Royal Society holds Brook Taylor's letter of 26 July 1712 to John Machin. Its catalogue connects his work on Kepler's second law with the expansion later published in the 1715 Methodus incrementorum directa et inversa. This is a documented route from an orbital calculation to a general approximation method, rather than a claim that every earlier use of such expansions began with Taylor. See Royal Society archive, letter MS/82/2.

For a smooth real function satisfying the displayed equation with positive integer $r$, differentiating makes $f^{(j+r)}=f^{(j)}$. On any compact interval, all derivatives are therefore bounded by the maximum of the bounds of $f,f^{\prime },\dots ,f^{(r-1)}$. The Taylor formula with integral remainder consequently has error at most $M{R}^{n+1}/(n+1)!$, which tends to zero uniformly for $|x|\le R$. Hence the function equals its globally convergent Taylor series. For $r=3$ and initial derivatives $(1,0,0)$, it is ${\sum }_{k\ge 0}{x}^{3k}/(3k)!$.

For a continuously differentiable vector-valued function whose domain contains the segment from $x$ to $x+h$, apply the fundamental theorem of calculus componentwise to $t\mapsto F(x+th)$. The resulting integral matrix is valid even when there is no common intermediate point giving a vector mean-value formula. For a score function, it integrates the Hessian matrix along the segment to a candidate estimate.

If $f$ has $k+1$ continuous derivatives, its Taylor remainder about $a$ is

The integral remainder in Taylor theorem for $\log (1-t)$ gives the exact finite identityFor $0\le x\le t<1$, $(t-x)/(1-x)\le t$, so the nonnegative remainder is at most $t^{N+1}/(1-t)$ and tends to zero. At $t=1/2$ it is at most $2^{-N}$, proving both convergence to $\log 2$ and an explicit error bound without assuming the logarithmic power series beforehand.

A Taylor expansion about a point consists of successive derivative terms there, together with either a finite remainder or an infinite Taylor series when that series represents the function.

For a scalar function with continuous third derivatives near $a$, apply the one-dimensional Taylor theorem to $g(t)=f(a+th)$. The chain rule gives $g^{\prime }(0)=Df(a)h$ and $g^{\prime \prime }(0)=h^T{H}_f(a)h$, proving the displayed formula. Here $H_f$ is the Hessian matrix. The finite polynomial part is the Multivariate Taylor polynomial; the remainder controls its accuracy.

For a fixed unit direction $v$, the multivariable Taylor expansion gives the displayed leading error. Equal directional derivatives in two directions do not imply equal approximation errors: the linear changes depend on the gradient, while the quadratic errors depend on the Hessian matrix. For example, $f(x,y)=x+y+2x^2+y^2$ at the origin has equal derivatives in the two coordinate directions but errors $2s^2$ and $s^2$.

A Peano zero through order $r$ means that the function admits the zero Taylor polynomial with remainder $o(|t{|}^r)$ at $x_0$. This does not assert the existence of ordinary iterated derivatives on a neighborhood. For example, multiplying $t^{r+1}$ by a bounded continuous function with dense points of nondifferentiability preserves this estimate while destroying neighborhood differentiability.

A parabolic approximation retains terms through second order in a local Taylor expansion. Near the closest point of a circle of radius $a$ and its tangent line, the separation is $x^2/(2a)+O(x^4/a^3)$.

For a polynomial $p$ of degree at most $m$, the Taylor formula is exact:

If $f$ is $n$ times differentiable between $a$ and $x$, then for some $\xi$ strictly between them,

The Taylor remainder in this form evaluates one higher derivative at an intermediate point $\xi$ between the expansion point $a$ and $x$. If $|f^{(n+1)}|\le M$ on that interval, then $|R_n(x)|\le M|x-a{|}^{n+1}/(n+1)!$. This bounds truncation error without knowing the particular intermediate point. The differentiability hypotheses are those of the Taylor theorem with Lagrange remainder.

The Taylor remainder is the difference between a function and a finite Taylor polynomial. Bounds on higher derivatives or their Hölder continuity control this error.

If $f^{(m-1)}$ is locally absolutely continuous, then its Taylor remainder can be written as an integral involving the weak derivative $f^{(m)}$. This form permits norm bounds when the highest derivative is only integrable.

For a function with derivatives of every order at $a$, its Taylor series is the power seriesIts partial sums are Taylor polynomials. At $a=0$ it is called a Maclaurin series. Forming the coefficients, proving that its partial sums form a convergent sequence, and proving that their limit equals $f(x)$ are separate tasks. A smooth function provides all the coefficients, but a real analytic function is additionally represented locally by the resulting power series.

For the exponential function, every derivative of $e^x$ at zero equals one, givingHere equality holds for every real $x$: the Taylor remainder is bounded by $e^{|x|}|x{|}^{N+1}/(N+1)!$, which tends to zero for each fixed $x$. At $x=1$, the degree-five Taylor polynomial is $163/60\approx 2.71666667$, and the absolute difference from $e$ is less than $e/720$. The factorial denominator eventually overwhelms every fixed power, allowing local derivative information to reconstruct this function throughout the real line.

The geometric series shows why the region of validity matters. For $f(x)=1/(1-x)$, the Taylor series at zero is ${\sum }_{k\ge 0}{x}^k$, with radius of convergence one. For $x\ne 1$, its finite Taylor polynomial satisfies the exact identityAt $x=1/2$ and $N=5$, the remainder is $1/32$. At $x=0.9$, the same degree-five polynomial has value $4.68559$ and remainder $5.31441$, although the Taylor series still converges to $10$ there. Convergence of the infinite expression does not guarantee accuracy from a particular small number of terms. For $|x|>1$, even the terms fail to tend to zero, so this power series cannot represent the function. The standard expansions and the conditions for convergent series are discussed in OpenStax, working with Taylor series.

A stronger caution comes from the flat functionIt is a smooth function with $h^{(n)}(0)=0$ for every $n$, so every Taylor polynomial at zero is zero and its Taylor series converges everywhere to zero. Nevertheless $h(1/2)=e^{-4}>0$. Off zero, repeated differentiation produces a polynomial in $1/x$ times $e^{-1/x^2}$; the exponential decay makes each required limit at zero vanish. This explains how all the local coefficients can miss a positive function arbitrarily close to the centre. Real analyticity supplies the additional condition that rules out this behavior.

The coefficient of $(z-z_0{)}^n$ in the Taylor series of a function at $z_0$. The displayed formula defines it whenever the indicated derivative exists. For a holomorphic function on a disk, the Cauchy integral formula also gives $a_n=(2\pi i{)}^{-1}\oint f(z)(z-z_0{)}^{-n-1}dz$ along a circle enclosing the center within that disk.

A Taylor series centred at zero. Its coefficients are the derivatives there divided by the corresponding factorials.