0:00 / 0:00

Taylor Polynomial Remainder

Since Taylor Polynomials are only evaluated to a finite nn value, they are approximations. Thus, there is some error when using a Taylor Polynomial. The error is the difference in the exact answer and the Taylor polynomial:
Rn(x)=f(x)Tn(x)\boxed{R_n(x)=f(x)-T_n(x)}
where nn is "evaluated at infinity”. Clearly, we cannot evaluate n at \infty. However, we have the following result:

Taylor’s Remainder Theorem

The error of the nth degree Taylor Polynomial approximation, Rn(x)R_n(x) ,is given by
Rn(x)=f(n+1)(c)(n+1)!(xa)n+1\boxed{R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}}

where cc is an unknown number between xx and aa.

Bounding Error

To find a bound on the error RnR_n, we can maximize f(n+1)(c)f^{(n+1)}(c) If we let
M=max{f(n+1)(c)} \boxed{M=|\text{max}\{f^{(n+1)}(c)\}|}
then
Rn(x)M(n+1)!xan+1\boxed{\lvert R_n(x)\rvert\leq \frac{M}{(n+1)!}\lvert x-a\rvert^{n+1}}


PAGE BREAK

Procedure for Taylor Remainder

  1. Determine the function f(x)f(x), the degree of the approximation nn, and the point at which it is centered about aa.
  2. Find a point cc, between aa and the xx in question, that maximizes f(n+1)(c)f^{(n+1)}(c) .
  3. Use the equation above to compute a bound on the error.
0:00 / 0:00

Example: Taylor Polynomial Remainder

A function f(x)f(x) has a 3rd derivative equal to 9(1x) \dfrac{9}{(1-x)}. The second degree Taylor Polynomial T2(x)centered at a=0T_2(x)\, \, \text{centered at }\,a=0 is used to approximate f(0.1)f(0.1). Find the upper bound for the error given by the Taylor Polynomial.


Rn(x)M(n+1)!xan+1  ,M=|max{f(n+1)(c)}\displaystyle \lvert R_n(x)\rvert\leq \frac{M}{(n+1)!}\lvert x-a\rvert^{n+1} \ \ , M=\text{|max}\{f^{(n+1)}(c)\}|

R2(x)M3!x3 R_2(x) \leq \left|\dfrac{M}{3!}\right||x|^3


f(x)=91xf'''(x) = \dfrac{9}{1-x}

a=0 ,x=0.1    0c0.1a=0 \ , x=0.1 \implies 0\le c \le 0.1

f(c)=91cas c  increases ,1c  decreases     f(c) increases    choose c=0.1 to maximize f(x) on [0,0.1]M=f(0.1)=910.1=10R2(x)103!x3=106x3=53x3R2(x)53x3R2(0.1)53(0.1)3R2(0.1)1600\begin{array}{l} f'''(c) = \dfrac{9}{1-c} \\ \\ \text{as } c \ \text{ increases }, 1-c \ \text{ decreases }\implies f'''(c) \text{ increases} \\ \\ \implies \text{choose } c=0.1 \ \text{to maximize } f'''(x) \text{ on } [0,0.1] \\ \\ M=f'''(0.1) = \dfrac{9}{1-0.1}=10 \\ \\ \\ R_2(x) \leq \left|\dfrac{10}{3!}\right||x|^3 =\left|\dfrac{10}{6}\right||x|^3=\dfrac{5}{3}|x|^3 \\ \\ R_2(x) \leq \dfrac{5}{3}|x|^3 \\ \\ R_2(0.1) \leq \dfrac{5}{3}(0.1)^3 \\ \\ \boxed{R_2(0.1) \leq \dfrac{1}{600}} \end{array}

Thus the upper bound for the error is 1600\dfrac{1}{600}.

Practice: Taylor Polynomial Remainder

Use the remainder theorem check if the following inequality is true or false:

ln(89)19\displaystyle\ln\left(\frac{8}{9}\right)\le-\frac{1}{9}

Let f(x)=ln(1x2)f(x)=\ln(1-x^2).
Extra Practice