The trapezoidal rule applied on ${\int}_{0}^{2}[f(x)]dx$ gives the value 5 and the Midpoint rule gives the value $4$. What value does Simpson's rule give?

letumsnemesislh
2022-07-10
Answered

The trapezoidal rule applied on ${\int}_{0}^{2}[f(x)]dx$ gives the value 5 and the Midpoint rule gives the value $4$. What value does Simpson's rule give?

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Prove: Let$S(n)$ and $T(n)$ be the approximations of a function using n intervals by using Simpson's rule and the Trapezoid rule respectfully.

$S(2n)=\frac{4T(2n)-T(n)}{3}$

$S(2n)=\frac{4T(2n)-T(n)}{3}$

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Show that one extrapolation of the trapezoid rule leads to Simpson's rule.

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How $2n$ numbers come out at the left side while only $n$ numbers at the right side.

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asked 2022-05-10

Let

$F(x)={\int}_{-\mathrm{\infty}}^{x}f(t)dt,$

where $x\in \mathcal{R}$, $f\ge 0$ is complicated (it cannot be integrated analytically).

Can we used the Simpson's rule to approximate this integral, knowing that $f(-\mathrm{\infty})=0$?

$F(x)={\int}_{-\mathrm{\infty}}^{x}f(t)dt,$

where $x\in \mathcal{R}$, $f\ge 0$ is complicated (it cannot be integrated analytically).

Can we used the Simpson's rule to approximate this integral, knowing that $f(-\mathrm{\infty})=0$?

asked 2022-05-13

How to derive the quadrature weights for the trapezoid and the Simpsons rule?

Trapeziod rule:$\underset{a}{\overset{b}{\int}}f(x)$$\approx $$\frac{b-a}{2}$$[f(a)+f(b)]$

Simpson's rule:$\underset{a}{\overset{b}{\int}}f(x)$$\approx $$\frac{b-a}{6}$$[f(a)+4f((a+b)/2)+f(b)]$

Trapeziod rule:$\underset{a}{\overset{b}{\int}}f(x)$$\approx $$\frac{b-a}{2}$$[f(a)+f(b)]$

Simpson's rule:$\underset{a}{\overset{b}{\int}}f(x)$$\approx $$\frac{b-a}{6}$$[f(a)+4f((a+b)/2)+f(b)]$

asked 2022-06-14

The error bound formulas for trapezoidal rule and simpson's rule say that:

Error Bound for the Trapezoid Rule: Suppose that $\begin{array}{l}\text{Error Bound for the Trapezoid Rule: Suppose that}|{f}^{\mathrm{\prime}\mathrm{\prime}}(x)|\le k\text{for some}k\in \mathbb{R}\text{where}\\ a\le x\le b.\text{Then}\\ \phantom{\rule{2em}{0ex}}\left|{E}_{T}\right|\le k\frac{(b-a{)}^{3}}{12{n}^{2}}\\ \text{Error Bound for Simpson's Rule: Suppose that}|{f}^{(4)}(x)|\le k\text{for some}k\in \mathbb{R}\text{where}\\ a\le x\le b.\text{Then}\\ \phantom{\rule{2em}{0ex}}\left|{E}_{S}\right|\le k\frac{(b-a{)}^{5}}{180{n}^{4}}\end{array}$

Using these formulas, is it possible to find functions where Trapezoid Rule is more accurate than Simpson's rule?

Error Bound for the Trapezoid Rule: Suppose that $\begin{array}{l}\text{Error Bound for the Trapezoid Rule: Suppose that}|{f}^{\mathrm{\prime}\mathrm{\prime}}(x)|\le k\text{for some}k\in \mathbb{R}\text{where}\\ a\le x\le b.\text{Then}\\ \phantom{\rule{2em}{0ex}}\left|{E}_{T}\right|\le k\frac{(b-a{)}^{3}}{12{n}^{2}}\\ \text{Error Bound for Simpson's Rule: Suppose that}|{f}^{(4)}(x)|\le k\text{for some}k\in \mathbb{R}\text{where}\\ a\le x\le b.\text{Then}\\ \phantom{\rule{2em}{0ex}}\left|{E}_{S}\right|\le k\frac{(b-a{)}^{5}}{180{n}^{4}}\end{array}$

Using these formulas, is it possible to find functions where Trapezoid Rule is more accurate than Simpson's rule?