= Quadratic Taylor majorant of a reciprocal quadratic
{title2=$\frac{x^2-4x+5}{4}-\frac1{1+x^2}=\frac{(x-1)^4}{4(1+x^2)}$}
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 $p\geq f$ 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 $f'''(x)=24x(1-x^2)/(1+x^2)^4<0$ for $x>1$. The exact identity is stronger: it controls the sign globally and shows the fourth-order contact at the expansion point.
\Image[/past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2018/continuous-mathematics-taylor-majorant.png]
{title=The quadratic Taylor polynomial globally majorizes the reciprocal quadratic}
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