Taylor polynomial ($T_{n,a}(x)={\sum }_{k=0}^n{f}^{(k)}(a)(x-a{)}^k/k!$)

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.

The quadratic Taylor polynomial of $f(x)=1/(1+x^2)$ at $x=1$ is $p(x)=(x^2-4x+5)/4$. Multiplying $p-f$ by $4(1+x^2)$ gives $(x-1{)}^4$, proving the exact displayed identity. Thus $p\ge f$ for every real $x$, with equality only at one. In particular, a claim that $f>p$ to the right of one has the wrong sign. The Lagrange remainder also proves this direction there, since $f^{\prime \prime \prime }(x)=24x(1-x^2)/(1+x^2{)}^4<0$ for $x>1$. The exact identity is stronger: it controls the sign globally and shows the fourth-order contact at the expansion point.

The multivariate Taylor polynomial of total degree $p$ at $x_0$ is

For positive $a,b$, fixed real $p,q$, and $f(a,b)=a^p{b}^q$, change the inputs to $a(1+h),b(1+j)$ with $h,j$ small. The ratio of the new value to the old one is exactly $(1+h{)}^p(1+j{)}^q$. Applying the binomial series to each factor and multiplying proves the displayed Taylor expansion. Positive inputs and $1+h,1+j>0$ make arbitrary real exponents well defined. A percentage change of $k$ in $a$ means $h=k/100$, not $h=k$. The mixed term $pqhj$ is required even though there is no such term in either one-variable expansion separately.