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Error Analysis in Taylor Polynomials

We can find the Taylor series of any differentiable function $f\left(x\right)=\sum _{n=0}^{\infty }{c_{n}\left(x-a\right)^{n}}$ , where $c_{n}={\tfrac {f^{\left(n\right)}\left(a\right)}{n!}}$ .

We can approximate the function with a finite polynomial by taking the Nth degree Taylor Polynomial of $f$ at $x=a$ :

T_{N}\left(x\right)=\sum _{n=0}^{N}{{\frac {f^{\left(n\right)}\left(a\right)}{n!}}\,\left(x-a\right)^{n}}

Remainder

Everything else past the finite polynomial that we chopped off:

{\begin{aligned}R_{N}\left(x\right)&=\sum _{n=N+1}^{\infty }{{\frac {f^{\left(n\right)}\left(a\right)}{n!}}\,\left(x-a\right)^{n}}\\&=f\left(x\right)-T_{N}\left(x\right)\end{aligned}}

We can estimate $R_{N}$ ; this shows how far off we are at any point on the given interval (of $x$ ) when we stop at the Nth degree Taylor Polynomial:

Alternating Series

{\begin{aligned}\left|R_{N}\left(x\right)\right|&\leq a_{N+1}\\&\leq {\frac {f^{\left(N+1\right)}\left(a\right)}{\left(N+1\right)!}}\,\left|x-a\right|^{N+1}\end{aligned}}

Taylor's Inequality

If $\left|f^{\left(N+1\right)}\left(x\right)\right|\leq M$ on a given interval (in other words, if we can get an upper bound on the derivative), then

\left|R_{N}\left(x\right)\right|\leq {\frac {M}{\left(N+1\right)!}}\,\left|x-a\right|^{N+1}

Note: given; do not memorize

Example

Use a 3rd degree Taylor polynomial at $a=1$ to approximate $f\left(x\right)={\sqrt {x}}$ on the interval [0.9, 1.1] and determine the accuracy of your results using the remainder theorem

$n$	$f^{\left(n\right)}\left(x\right)$	$f^{\left(n\right)}\left(1\right)$	${\frac {f^{\left(n\right)}\left(1\right)}{n!}}\,\left(x-1\right)^{n}$
0	${\sqrt {x}}$	1	1
1	${\frac {1}{2}}\,x^{-1/2}$	${\frac {1}{2}}$	${\frac {1/2}{1!}}\,\left(x-1\right)^{1}$
2	$-{\frac {1}{4}}\,x^{-3/2}$	$-{\frac {1}{4}}$	$-{\frac {1/4}{2!}}\,\left(x-1\right)^{2}$
3	${\frac {3}{8}}\,x^{-5/2}$	${\frac {3}{8}}$	${\frac {3/8}{3!}}\,\left(x-1\right)^{3}$

T_{3}\left(x\right)=1+{\frac {1}{2}}\,\left(x-1\right)-{\frac {1}{8}}\,\left(x-1\right)^{2}+{\frac {1}{16}}\,\left(x-1\right)^{3}

$N=3$
$a=1$
$x\in \left[0.9,\,1.1\right]$ $x\in \left[0.9,\,1.1\right]$ . . . Which value of $x$ $x$ maximizes the error bound?
- $M=\left|f^{\left(4\right)}\left(x\right)\right|$ (max. of next derivative on the interval of $x$ )
- M_max occurs at $x=0.9$ : ${\frac {15/16\,\left(0.9\right)^{7/2}}{4!}}\,\left|0.9-1\right|^{4}$