University of Florida/Egm6341/s10.team2.niki/HW3

Error is defined as

${\displaystyle \displaystyle E_{n}=I-I_{n}=\int _{a}^{b}[f(x)-f_{n}(x)]dx}$

For the Taylor series, from the discussion on p6-4, we know that the error is nothing but the remainder of the Taylor series integrated over the given interval i.e

${\displaystyle \displaystyle \int _{a}^{b}R_{n+1}(x)=\int _{a}^{b}[{\frac {(x-x_{0})^{n+1}}{(n+1)!}}f^{(n+1)}(\xi )]}$ where ${\displaystyle \xi \varepsilon [a,b]}$

(1 p2-3)

For the given function the error is

${\displaystyle \displaystyle E_{n}=\int _{0}^{1}\underbrace {\frac {x^{n}}{(n+1)!}} _{w(x)}\underbrace {e^{\xi (x)}} _{g(x)}dx}$

(2)

Using the Integral Mean Value Theorem, we have

${\displaystyle \displaystyle E_{n}=e^{\xi (x=\alpha )}\int _{0}^{1}{\frac {x^{n}}{(n+1)!}}dx}$

(3)

integrating we get,

${\displaystyle \displaystyle E_{n}=e^{\xi (x=\alpha )}{\frac {1}{(n+1)!(n+1)}}}$

(4)

This function has a minimum when ${\displaystyle \alpha =0}$ and maximum when ${\displaystyle \alpha =1}$ thus we have

${\displaystyle \displaystyle {\frac {1}{(n+1)!(n+1)}}\leq E_{n}\leq {\frac {e}{(n+1)!(n+1)}}}$

(5)

Setting the upper bound of the error to ${\displaystyle 10^{-6}}$ we have

${\displaystyle \displaystyle \ E_{n}\leq {\frac {e}{(n+1)!(n+1)}}=10^{-6}}$

(6)

Solving this equation for n we get

${\displaystyle \displaystyle (n+1)!(n+1)=e*10^{6}\Rightarrow n\approx 7.92=7}$

(A)

From the discussion on page 16-3 we have

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

(1)

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

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

(2)

Evaluating over the given interval it is seen that the function ${\displaystyle f^{(2)}(x)}$ has maximum value at ${\displaystyle x=1}$

${\displaystyle \displaystyle M_{2}=f^{(2)}(x=1)={\frac {e[1+2-2]-2}{1}}=0.718282}$

(3)

Setting the error to the ${\displaystyle 10^{-6}}$ and solving for ${\displaystyle n}$ we get

${\displaystyle \displaystyle {\frac {(1-0)^{3}}{12n^{2}}}(0.718282)=10^{-6}\Rightarrow n\approx 244.657=245}$

(B)

We have from p 17-2, the error estimate of the Composite Simpson's rule as

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

(1)

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

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

(2)

Evaluating over the given interval it is seen that the function ${\displaystyle f^{(2)}(x)}$ has maximum value at ${\displaystyle x=1}$

${\displaystyle \displaystyle M_{4}=f^{(4)}(x=1)={\frac {e[1-4+12-24+24]-24}{1}}=9(e)-24=0.464536}$

(3)

Setting the error to the ${\displaystyle 10^{-6}}$ and solving for ${\displaystyle n}$ we get

${\displaystyle \displaystyle {\frac {(1-0)^{5}}{2880n^{4}}}(0.464536)=10^{-6}\Rightarrow n\approx 3.35735=3}$

(c)

Given below is the data from the numerical evaluation of the given function using Composite Trapezoidal Rule.

First we plot a semilog graph for the data. The graph is shown below:

The graph is not a straight line which implies that the relationship between ${\displaystyle {\boldsymbol {h}}}$ and ${\displaystyle {\boldsymbol {log(error)}}}$ is not exponential. Hence we plot a log-log graph as below:

This is seen to be linear. A straight line if fitted to the data the equation of which is given above. From the discussion above, we see that the slope of line is 2.1447 which is very close to the analytical value of 2.

Given below is the data obtained from HW1 for the COmposite Simpsons Rule.

Plotting a semilog graph of ${\displaystyle {\boldsymbol {log(error)}}}$ against ${\displaystyle {\boldsymbol {h}}}$ we see that it is non-linear as in the case of the Composite Trapezoidal Rule.

Thus, plotting the log-log graph as below,

we see that the slope of the line is 4.0881 which is very close to the analytically determined value of 4.

Given below is the data for the Taylor series method. Eventhough ${\displaystyle {\boldsymbol {h}}}$ is not used in the Taylor series method,defining ${\displaystyle {\boldsymbol {h}}}$ based on the number of terms of the series, it is seen that the error and h are not related exponentially or by a power relation i.e. both semilog and log log plots are not linear.

--Egm6341.s10.team2.niki 08:10, 17 February 2010 (UTC)