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

Problem Statement

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

  $E_n=I-I_n=O(10^{-6})$

and compare to numerical results.

Solution

$I=\underset{0}{\overset{1}{\int }}\frac{e^x-1}{x}dx$

Taylor Series

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

$I_n=\underset{0}{\overset{1}{\int }}{f}_n(x)dx=\underset{0}{\overset{1}{\int }}\sum _{j=1}^{\mathrm{\infty }}\frac{x^{j-1}}{j!}dx=\sum _{j=1}^n\frac{x^j}{j!j}{|}_0^1=\sum _{j=1}^n\frac{1}{j!j}$

$f(x)-f_n(x)=R_n(x)=\frac{(x-x_0{)}^n}{(n+1)!}{f}^{(n+1)}(\xi )$ with $\xi \in [0,x]$ and $x_0=0$

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

$E_n=I-I_n=\underset{0}{\overset{1}{\int }}[f(x)-f_n(x)]dx=\underset{0}{\overset{1}{\int }}\underset{w(x)}{\underbrace{\frac{x^n}{(n+1)!}}}\underset{g(x)}{\underbrace{f^{(n+1)}(\xi (x))}}dx$

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

$E_n=$

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

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

$n=2,E_2\le .151016$

$n=4,E_5\le .00453$

$n=6,E_6\le 6.2792X10^{-4}$

$n=8,E_8\le 8.42X10^{-6}$

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

$n=2,E_2\le .151016$

$n=4,E_5\le .00453$

$n=6,E_6\le 6.2792X10^{-4}$

$n=8,E_8\le 8.42X10^{-6}$

We are getting same values from both analysis for Taylor series

Trapazoidal Rule

Error for Composite Trapazoidal rule is given by

$\left|E_n^1\right|\le \frac{(b-a{)}^3}{12n^2}{M}_2$

where $M_2=max|f^{(2)}(\zeta )|$ for $\zeta \epsilon [a,b]$

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

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

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

$M_2=f^{(2)}(x=1)=\frac{e[1+2-2]-2}{1}=0.71828182$

$E_n^1\le \frac{(1-0{)}^3}{12n^2}X0.71828182$

$E_2^1\le .014964204$

$E_4^1\le 3.74105X10^{-3}$

$E_6^1\le 1.662689X10^{-3}$

$E_8^1\le 9.3526278X10^{-4}$

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

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

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

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

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

Below are the results from Numerical analysis from HW 1

$E_2^1\le 0.010389576$

$E_4^1\le 0.002602468$

$E_8^1\le 0.650935X10^{-3}$

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

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

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

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

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

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

Composite Simpsons Rule

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

$E_n^2\le \frac{(b-a{)}^5}{2880n^4}{M}_4$

where  $M_4=max|f^{(4)}(\zeta )|$ for $\zeta \epsilon [a,b]$

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

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

The function $f^4(x)$ has maximum value at $x=1$

$M_4=\frac{e[1-4+12-24+24]-24}{1}=0.46453645$

$E_n^1\le \frac{(1-0{)}^5}{2880n^4}X0.46453645$

$E_2^2\le 1.00810X10^{-5}$

$E_4^2\le 6.300678X10^{-7}$

Below are the results from Numerical analysis from HW 1

$E_2^2\le 1.06514X10^{-4}$

$E_4^2\le 6.76473X10^{-6}$

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