University of Florida/Egm6341/s10.team3.aks/HW3

(4)Use Error estimate for Taylor series, composite trapezoidal and composite Simpsons rule

Problem Statement

Use Error estimate for Taylor series, composite trapezoidal and composite Simpson's rule to find n such that

  ${\displaystyle E_{n}=I-I_{n}=O(10^{-6})}$

and compare to numerical results.

Solution

${\displaystyle I=\int _{0}^{1}{\frac {e^{x}-1}{x}}dx}$

Taylor Series

${\displaystyle e^{x}=\sum _{j=0}^{\infty }{\frac {x^{j}}{j!}}\Rightarrow {\frac {e^{x}-1}{x}}=\sum _{j=1}^{\infty }{\frac {x^{j-1}}{j!}}}$

${\displaystyle I_{n}=\int _{0}^{1}f_{n}(x)dx=\int _{0}^{1}\sum _{j=1}^{\infty }{\frac {x^{j-1}}{j!}}dx=\sum _{j=1}^{n}{\frac {x^{j}}{j!j}}|_{0}^{1}=\sum _{j=1}^{n}{\frac {1}{j!j}}}$

${\displaystyle f(x)-f_{n}(x)=R_{n}(x)={\frac {(x-x_{0})^{n}}{(n+1)!}}f^{(n+1)}(\xi )}$ with ${\displaystyle \xi \in [0,x]\,}$ and ${\displaystyle x_{0}=0\,}$

${\displaystyle \Rightarrow f(x)-f_{n}(x)={\frac {x^{n}}{(n+1)!}}f^{(n+1)}(\xi )}$

${\displaystyle E_{n}=I-I_{n}=\int _{0}^{1}\left[f(x)-f_{n}(x)\right]dx=\int _{0}^{1}\underbrace {\frac {x^{n}}{(n+1)!}} _{w(x)}\underbrace {f^{(n+1)}\left(\xi (x)\right)} _{g(x)}dx}$

${\displaystyle \Rightarrow g(\alpha )\int _{0}^{1}w(x)dx}$ for ${\displaystyle \alpha \in [0,1]}$

${\displaystyle E_{n}=\,}$

${\displaystyle \Rightarrow max={\frac {g(\alpha )}{(n+1)!(n+1)}}}$, ${\displaystyle \alpha ={\frac {e}{(n+1)!(n+1)}}\,}$

${\displaystyle \Rightarrow min={\frac {g(\alpha )}{(n+1)!(n+1)}}}$, ${\displaystyle \alpha ={\frac {1}{(n+1)!(n+1)}}\,}$

${\displaystyle n=2,E_{2}\leq .151016\,}$

${\displaystyle n=4,E_{5}\leq .00453\,}$

${\displaystyle n=6,E_{6}\leq 6.2792X10^{-4}\,}$

${\displaystyle n=8,E_{8}\leq 8.42X10^{-6}\,}$

Below are the values from Numerical Analysis of Taylor series from HW_1

${\displaystyle n=2,E_{2}\leq .151016\,}$

${\displaystyle n=4,E_{5}\leq .00453\,}$

${\displaystyle n=6,E_{6}\leq 6.2792X10^{-4}\,}$

${\displaystyle n=8,E_{8}\leq 8.42X10^{-6}\,}$

We are getting same values from both analysis for Taylor series

Trapazoidal Rule

Error for Composite Trapazoidal rule is given by

${\displaystyle \displaystyle \left|E_{n}^{1}\right|\leq {\frac {(b-a)^{3}}{12n^{2}}}M_{2}}$

where ${\displaystyle M_{2}=max\left|f^{(2)}(\zeta )\right|}$ for ${\displaystyle \zeta \varepsilon [a,b]}$

For the given function ${\displaystyle f(x)={\frac {e^{x}-1}{x}}}$, we have

${\displaystyle \displaystyle f^{(2)}(x)={\frac {e^{x}[x^{2}-2x+2]-2}{x^{3}}}}$

For the given interval [0,1] the maximum value of function ${\displaystyle f^{(2)}(x)}$ is achieved at ${\displaystyle x=1}$

${\displaystyle \displaystyle M_{2}=f^{(2)}(x=1)={\frac {e[1+2-2]-2}{1}}=0.71828182}$

${\displaystyle E_{n}^{1}\leq {\frac {(1-0)^{3}}{12n^{2}}}X0.71828182}$

${\displaystyle E_{2}^{1}\leq .014964204}$

${\displaystyle E_{4}^{1}\leq 3.74105X10^{-3}}$

${\displaystyle E_{6}^{1}\leq 1.662689X10^{-3}}$

${\displaystyle E_{8}^{1}\leq 9.3526278X10^{-4}}$

${\displaystyle E_{16}^{1}\leq 2.3381569X10^{-4}}$

${\displaystyle E_{32}^{1}\leq 5.845392X10^{-5}}$

${\displaystyle E_{64}^{1}\leq 1.4613481X10^{-5}}$

${\displaystyle E_{128}^{1}\leq 3.65337026X10^{-6}}$

${\displaystyle E_{256}^{1}\leq 9.1334256X10^{-7}}$

Below are the results from Numerical analysis from HW 1

${\displaystyle E_{2}^{1}\leq 0.010389576}$

${\displaystyle E_{4}^{1}\leq 0.002602468}$

${\displaystyle E_{8}^{1}\leq 0.650935X10^{-3}}$

${\displaystyle E_{16}^{1}\leq 162753X10^{-4}}$

${\displaystyle E_{32}^{1}\leq 4.06892X10^{-5}}$

${\displaystyle E_{64}^{1}\leq 1.0172X10^{-5}}$

${\displaystyle E_{128}^{1}\leq 2.54263X10^{-6}}$

${\displaystyle E_{256}^{1}\leq 6.3528X10^{-7}}$

We can see that we are getting order ${\displaystyle 10^{-6}}$ at n=128 from both the analysis.

Composite Simpsons Rule

The error estimate of the Composite Simpson's rule is given as

${\displaystyle E_{n}^{2}\leq {\frac {(b-a)^{5}}{2880n^{4}}}M_{4}}$

where  ${\displaystyle M_{4}=max\left|f^{(4)}(\zeta )\right|}$ for ${\displaystyle \zeta \varepsilon [a,b]}$

For the given function ${\displaystyle f(x)={\frac {e^{x}-1}{x}}}$, we have

${\displaystyle \displaystyle f^{(4)}(x)={\frac {e^{x}[x^{4}-4x^{3}+12x^{2}-24x]-24}{x^{5}}}}$

The function ${\displaystyle f^{4}(x)}$ has maximum value at ${\displaystyle x=1}$

${\displaystyle \displaystyle M_{4}={\frac {e[1-4+12-24+24]-24}{1}}=0.46453645}$

${\displaystyle E_{n}^{1}\leq {\frac {(1-0)^{5}}{2880n^{4}}}X0.46453645}$

${\displaystyle E_{2}^{2}\leq 1.00810X10^{-5}}$

${\displaystyle E_{4}^{2}\leq 6.300678X10^{-7}}$

Below are the results from Numerical analysis from HW 1

${\displaystyle E_{2}^{2}\leq 1.06514X10^{-4}}$

${\displaystyle E_{4}^{2}\leq 6.76473X10^{-6}}$

We can see that we are getting order ${\displaystyle 10^{-6}}$ at n=4 from both the analysis.