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Courtney's question at Yahoo! Answers regarding Error Bound for approximation methods

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MarkFL

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Feb 24, 2012
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Here is the question:

Consider the definite integral ∫_0^1〖sin⁡(x^2 )dx〗. How large must n be to guarantee that:?

1. |∫_0^1〖sin⁡(x^2 )dx〗- M_n |< .00001
2. |∫_0^1〖sin⁡(x^2 )dx〗- T_n |< .00001
3. |∫_0^1〖sin⁡(x^2 )dx〗- S_n |< .00001
I guess I'm mostly confused on how to find n. Thanks for all your help.
I have posted a link there to this topic so the OP can see my work.
 
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MarkFL

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Feb 24, 2012
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Hello Courtney,

We are given:

\(\displaystyle I=\int_0^1\sin\left(x^2 \right)\,dx\)

For the definitions of the Error Bound for the 3 methods, we identify:

\(\displaystyle a=0,\,b=1,\,f(x)=\sin\left(x^2 \right)\)

1.) The Error Bound $E_n$ for the Midpoint Rule is:

If there exists a number $M>0$ such that $\left|f''(x) \right|\le M$ for all $x$ in $[a,b]$, then:

\(\displaystyle E_n\le\frac{M(b-a)^3}{24n^2}\)

Using the function given, we find:

\(\displaystyle f''(x)=2\cos\left(x^2 \right)-4x^2\sin\left(x^2 \right)\)

Here is a plot of $y=\left|f''(x) \right|$ on the given interval:

courtney1.jpg

We see that:

\(\displaystyle f''(1)=4\sin(1)-2\cos(1)\ge f''(x)\) for all $x$ in the interval. Thus, we want:

\(\displaystyle \frac{\left(4\sin(1)-2\cos(1) \right)(1-0)^3}{24n^2}<0.00001\)

\(\displaystyle \frac{\left(2\sin(1)-\cos(1) \right)}{12n^2}<\frac{1}{100000}\)

or

\(\displaystyle n^2>\frac{25000}{3}\left(2\sin(1)-\cos(1) \right)\approx9521.99719789711\)

\(\displaystyle 97^2<9521.99719789711<98^2\)

Hence, by taking $n\ge98$ we obtain the desired accuracy.

2.) The Error Bound $E_n$ for the Trapezoidal Rule is:

If there exists a number $M>0$ such that $\left|f''(x) \right|\le M$ for all $x$ in $[a,b]$, then:

\(\displaystyle E_n\le\frac{M(b-a)^3}{12n^2}\)

Using the results of 1.) we see that we want:

\(\displaystyle \frac{\left(4\sin(1)-2\cos(1) \right)(1-0)^3}{12n^2}<0.00001\)

\(\displaystyle \frac{\left(2\sin(1)-\cos(1) \right)}{6n^2}<\frac{1}{100000}\)

or

\(\displaystyle n^2>\frac{50000}{3}\left(2\sin(1)-\cos(1) \right)\approx19043.99439579422\)

\(\displaystyle 137^2<19043.99439579422<138^2\)

Hence, by taking $n\ge138$ we obtain the desired accuracy.

3.) The Error Bound $E_n$ for Simpson's Rule is:

If there exists a number $M>0$ such that $\left|f^{(4)}(x) \right|\le M$ for all $x$ in $[a,b]$, then:

\(\displaystyle E_n\le\frac{M(b-a)^5}{180n^4}\)

Using the function given, we find:

\(\displaystyle f^{(4)}(x)=4\left(\left(4x^3-3 \right)\sin\left(x^2 \right)-12x^2\cos\left(x^2 \right) \right)\)

Here is a plot of $y=\left|f^{(4)}(x) \right|$ on the given interval along with the absolute maximum:

courtney2.jpg

Thus, we want:

\(\displaystyle \frac{28.42851540309637345267676583(1-0)^5}{180n^4}<0.00001\)

\(\displaystyle \frac{28.42851540309637345267676583}{180n^4}<\frac{1}{100000}\)

or

\(\displaystyle n^4>15793.61966838687414037598101\bar{6}\)

\(\displaystyle 11^4<15793.61966838687414037598101\bar{6}<12^4\)

Hence, by taking $n\ge12$ we obtain the desired accuracy.