/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q4E In Example 1, page 126, the impr... [FREE SOLUTION] | 91影视

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Chapter 3: Q4E (page 130)

In Example 1, page 126, the improved Euler鈥檚 method approximation to \({\bf{e}}\) with step size \({\bf{h}}\)was shown to be \({\bf{\;}}{\left( {{\bf{1 + h + }}\frac{{{{\bf{h}}^{\bf{2}}}}}{{\bf{2}}}} \right)^{\frac{{\bf{1}}}{{\bf{h}}}}}\).First prove that the error \({\bf{e - \;}}{\left( {{\bf{1 + h + }}\frac{{{{\bf{h}}^{\bf{2}}}}}{{\bf{2}}}} \right)^{\frac{{\bf{1}}}{{\bf{h}}}}}\)approaches zero as \({\bf{h}} \to 0\). Then use L鈥橦opital鈥檚 rule to show the \(\mathop {{\bf{lim}}}\limits_{{\bf{h}} \to 0} \frac{{{\bf{error}}}}{{{{\bf{h}}^{\bf{2}}}}}{\bf{ = }}\frac{{\bf{e}}}{6} \approx 0.45305\).Compare this constant with the entries in the last column of Table 3.5.

Short Answer

Expert verified

By the use L鈥橦opital鈥檚 rule to show the \(\mathop {\lim }\limits_{h \to 0} \frac{{error}}{{{h^2}}} = \frac{e}{6} \approx 0.45305\). The closest value with the last value from table 3.5 for \(h = {10^{ - 3}}\,\,is\,\,0.45271\).

Step by step solution

01

Important hint.

For getting the solution apply L鈥檋opital rule.

02

Prove that the error \({\bf{e - \;}}{\left( {{\bf{1 + h + }}\frac{{{{\bf{h}}^{\bf{2}}}}}{{\bf{2}}}} \right)^{\frac{{\bf{1}}}{{\bf{h}}}}}\) approaches zero as \({\bf{h}} \to 0\).As  \(\mathop {\lim }\limits_{h \to 0} {\left( {1 + h} \right)^{\frac{1}{h}}} = e\)By using this formula

\(\begin{array}{c}\mathop {\lim }\limits_{h \to 0} \left( {e - {{\left( {1 + h + \frac{{{h^2}}}{2}} \right)}^{\frac{1}{h}}}} \right)\\ = \mathop {\lim }\limits_{h \to 0} e - \mathop {\lim }\limits_{h \to 0} {\left( {1 + h + \frac{{{h^2}}}{2}} \right)^{\frac{1}{h}}}\\ = e - \mathop {\lim }\limits_{h \to 0} {\left( {{{\left( {1 + h + \frac{{{h^2}}}{2}} \right)}^{\frac{1}{{h + \frac{{{h^2}}}{2}}}}}} \right)^{\frac{{h + \frac{{{h^2}}}{2}}}{h}}}\\ = e - \mathop {\lim }\limits_{h \to 0} {e^{\frac{{h + \frac{{{h^2}}}{2}}}{h}}}\end{array}\)

\(\begin{array}{c} = e - {e^1}\\ = 0\end{array}\)

03

Find the result by applying L’Hopital’s rule.

\(\begin{array}{c} = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {e - \;{{\left( {1 + h + \frac{{{h^2}}}{2}} \right)}^{\frac{1}{h}}}} \right)}^'}}}{{{{({h^2})}^'}}}\\ = \mathop {\lim }\limits_{h \to 0} {\frac{{\left( { - \;{{\left( {1 + h + \frac{{{h^2}}}{2}} \right)}^{\frac{1}{h}}}} \right)}}{{2h}}^'}\end{array}\)

04

Evaluate the result by the derivative of part in step 2.

\(\begin{array}{c}{\left( { - \;{{\left( {1 + h + \frac{{{h^2}}}{2}} \right)}^{\frac{1}{h}}}} \right)^'} = - {\left( {{e^{\left( {\frac{1}{h}in{{\left( {1 + h + \frac{{{h^2}}}{2}} \right)}^{\frac{1}{h}}}} \right)}}} \right)^'}\\ = - {e^{\left( {\frac{1}{h}in{{\left( {1 + h + \frac{{{h^2}}}{2}} \right)}^{\frac{1}{h}}}} \right)}}\left( {\frac{{ - \ln \left( {1 + h + \frac{{{h^2}}}{2}} \right)}}{{{h^2}}} + \frac{{2(h + 1)}}{{h(2 + 2h + {h^2}}}} \right)\\ = \left( {\frac{{ - \ln \left( {1 + h + \frac{{{h^2}}}{2}} \right)}}{{{h^2}}} + \frac{{2(h + 1)}}{{h(2 + 2h + {h^2}}}} \right){\left( {1 + h + \frac{{{h^2}}}{2}} \right)^{\frac{1}{h}}}\end{array}\)

05

Now Apply derivative part in step 2.

\( = \mathop {\lim }\limits_{h \to 0} {\frac{{\left( { - \;{{\left( {1 + h + \frac{{{h^2}}}{2}} \right)}^{\frac{1}{h}}}} \right)}}{{2h}}^'}\)

\(\begin{array}{c} = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {\frac{{ - \ln \left( {1 + h + \frac{{{h^2}}}{2}} \right)}}{{{h^2}}} + \frac{{2(h + 1)}}{{h(2 + 2h + {h^2})}}} \right){{\left( {1 + h + \frac{{{h^2}}}{2}} \right)}^{\frac{1}{h}}}}}{{2h}}\\\frac{e}{2}.\mathop {\lim }\limits_{h \to 0} - \frac{{\left( {\frac{{ - \ln \left( {1 + h + \frac{{{h^2}}}{2}} \right)}}{{{h^2}}} + \frac{{2(h + 1)}}{{h(2 + 2h + {h^2})}}} \right)}}{h}\\\frac{e}{2}\mathop {\lim }\limits_{h \to 0} \left( { - \frac{{2h(h + 1) - \ln (\frac{{{h^2}}}{2} + h + 1)({h^2} + 2h + 2)}}{{{h^3}({h^2} + 2h + 2)}}} \right)\\ = \frac{e}{2}\mathop {\lim }\limits_{h \to 0} \frac{{ - h - {h^2} + \left( {1 + h + \frac{{{h^2}}}{2}} \right).\ln \left( {1 + h + \frac{{{h^2}}}{2}} \right)}}{{{h^3} + {h^4} + \frac{{{h^5}}}{2}}}\end{array}\)

By solving this and apply as \(h \to 0\).

\(\begin{array}{l} = \frac{e}{6}\\ \approx 0.45305\end{array}\)

06

Find the values of different values of \({\bf{h}}\).

For \(h = 1\) then \(\frac{{{\rm{error}}}}{{{h^2}}} = 0.21828\)

For \(h = {10^{ - 1}}\,\,{\rm{then}},\,\,\frac{{{\rm{error}}}}{{{h^2}}} = 0.42010\)

For \(h = {10^{ - 2}}\,{\rm{then}}\,\,\frac{{{\rm{error}}}}{{{h^2}}} = 0.44966\)

For \(h = {10^{ - 3}}\,{\rm{then}}\,\,\frac{{{\rm{error}}}}{{{h^2}}} = 0.45271\)

Therefore, the closest value with the last value from table 3.5 for \(h = {10^{ - 3}}\,{\rm{is}}\,\,0.45271\).

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