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For $f$ sufficiently differentiable on the interval between $a$ and $x$, Taylor's theorem with its first $n$ terms readswhere $\xi$ lies between $a$ and $x$. This is the Taylor theorem with Lagrange remainder. For example, $|f^{(n)}|\le M$ on that interval gives $|R_n(x)|\le M|x-a{|}^n/n!$. The finite formula does not assume that an infinite Taylor series converges to the function; that further conclusion requires the remainders to tend to zero.

Taylor's theorem gives a finite Taylor polynomial of degree $n$, consisting of the first $n+1$ terms, and a remainder:In the Taylor theorem with Lagrange remainder form, assuming the required derivatives exist on an interval containing $a$ and $x$,for some $\xi$ strictly between $a$ and $x$ when $x\ne a$; at $x=a$ the remainder is zero. In particular, if $|f^{(n+1)}|\le M$ on that segment, then $|R_{n+1}(x)|\le M|x-a{|}^{n+1}/(n+1)!$. This is a finite expansion; it does not assume equality with an infinite Taylor series.

For $f(x,y)=y$ and $\beta =1/2$, the explicit midpoint method multiplies the current value by $R=1+h+h^2/2$. Starting from one, induction givesLet $A=e^h$ and $B=R$. The Taylor theorem with Lagrange remainder for the exponential function givesThe difference of powers identity gives $A^n-B^n=(A-B){\sum }_{k=0}^{n-1}{A}^{n-1-k}{B}^k$. Every summand lies between one and $A^{n-1}$. Since $nh=x_n$, the global error therefore satisfiesThe same inequalities hold at $n=0$, where the error vanishes. At the fixed endpoint $x_N=1$, the lower bound $h^2/6$ prevents any higher order of a numerical method, while the upper bound is $e{h}^2/6$. Thus the convergence order is exactly two.

Let a nonnegative kernel for density estimation $K$ have integral one, zero first moment and finite second moment ${\mu }_2(K)$. If a probability density function $f$ has bounded second derivative, its kernel density estimator has pointwise bias at most $h^2{\mu }_2(K)\Vert f^{\prime \prime }{\Vert }_{\infty }/2$. Apply the Taylor theorem with Lagrange remainder to $f(x-hu)$ and integrate: the first-order term vanishes and the remainder is bounded by $h^2{u}^2\Vert f^{\prime \prime }{\Vert }_{\infty }/2$. For $K=1_{[-1/2,1/2]}$, ${\mu }_2(K)=1/12$.

A Taylor polynomial replaces a function near a chosen point by a polynomial with matching derivatives. If $f$ has derivatives through order $n$ at $a$, defineThe factorial denominator makes $T_{n,a}^{(j)}(a)=f^{(j)}(a)$ for $0\le j\le n$. This determines a unique polynomial of degree of a polynomial at most $n$. Order zero matches the value, order one also matches the tangent line, and order two also incorporates the second derivative. Matching more local information is useful because polynomials are easy to evaluate, differentiate, and integrate. The distinction between the finite Taylor polynomial and the infinite Taylor series is developed in OpenStax, Taylor and Maclaurin series.

For a computed example, let $f(x)=\sqrt{1+x}$ and expand at zero. Its first three derivatives giveAt $x=0.1$, the linear approximation is $1.05$ and the quadratic approximation is $1.04875$, compared with $\sqrt{1.1}\approx 1.048808848$. The quadratic correction captures the downward bending of the square root and reduces the absolute difference from about $0.00119115$ to $0.0000588482$.

The Taylor remainder turns this convenient approximation into a controlled calculation. On $[0,0.1]$, $f^{\prime \prime \prime }(x)=3/[8(1+x{)}^{5/2}]$ is positive and at most $3/8$. The Taylor theorem with Lagrange remainder therefore givesConsequently $1.04875<\sqrt{1.1}\le 1.0488125$ follows without using the displayed decimal evaluation as evidence. The Taylor polynomial and the Taylor remainder play different roles: the former supplies an answer that is easy to compute, while the latter specifies what accuracy is justified. Farther from the expansion point, the same low-degree polynomial need not remain accurate.

Suppose a probability density function and its third derivative are bounded. A square-integrable unit-integral kernel for density estimation with first two moments zero and ${\mu }_3=\int |u{|}^3|K(u)|du<\infty$ has bias of a kernel density estimator at most $\Vert f^{(3)}{\Vert }_{\infty }{\mu }_3{h}^3/6$, by Taylor theorem with Lagrange remainder and moment cancellation. Independence gives variance at most $\Vert f{\Vert }_{\infty }\Vert K{\Vert }_2^2/(nh)$. The Cauchy-Schwarz inequality bounds the expected centred absolute deviation by the square root of this variance. Adding the bias proves the inequality uniformly in $x$. The choice $h=n^{-1/7}$ gives expected absolute error $O(n^{-3/7})$.