Science:Math Exam Resources/Courses/MATH105/April 2013/Question 01 (j)

MATH105 April 2013

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Question 01 (j)
Short-Answer Questions. Put your answer in the box provided but show your work also. Each question is worth 3 marks, but not all questions are of equal difficulty. Find a bound for the error in approximating $\displaystyle \int _{1}^{5}{\frac {1}{x}}\,dx$ using Simpson’s rule with $n=4$. Do not write down Simpson’s rule approximation $\displaystyle S_{4}$.

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Hint
Given $n$, an even integer, and an upper bound $K\geq |f^{(4)}(x)|$ for all $x$ in the interval $[a,b]$, the error that comes from using Simpson's Rule is as follows: $|I-S_{n}|\leq {\frac {K(b-a)^{5}}{180\cdot n^{4}}}$ where I is the integral in question. How would you find K?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Given $n$, an even integer, and an upper bound $K\geq |f^{(4)}(x)|$ for all $x$ in the interval $[a,b]$, the error that comes from using Simpson's Rule is as follows: $|I-S_{n}|\leq {\frac {K(b-a)^{5}}{180\cdot n^{4}}}$ where $I=\int _{1}^{5}{\frac {1}{x}}\,dx$. We know that $\displaystyle a=1$, $\displaystyle b=5$ and $\displaystyle n=4$. To find an upper bound on the fourth derivative, we first compute that ${\begin{aligned}f(x)&={\frac {1}{x}}\\f'(x)&=-{\frac {1}{x^{2}}}\\f''(x)&=2{\frac {1}{x^{3}}}\\f'''(x)&=-6{\frac {1}{x^{4}}}\\f''''(x)&=24{\frac {1}{x^{5}}}\end{aligned}}$ As the fourth derivative $\displaystyle f''''(x)={\frac {24}{x^{5}}}$ is a decreasing function in x, its maximum occurs at the smallest possible value, when $\displaystyle x=1$. Hence a value of K is given by $\displaystyle K=24$ Hence, we have a bound on the error is given by $\left|\int _{1}^{5}{\frac {1}{x}}\,dx-S_{n}\right|\leq {\frac {24(5-1)^{5}}{180\cdot 4^{4}}}={\frac {24\cdot 4^{5}}{180\cdot 4^{4}}}={\frac {8}{15}}$

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