Numerical Differentiation Estimates

8. $$\mathrm { f } ( x ) = 2 x ^ { - \frac { 2 } { 3 } } + \frac { 1 } { 2 } x - \frac { 1 } { 3 x - 5 } - \frac { 5 } { 2 } \quad x \neq \frac { 5 } { 3 }$$ The table below shows values of \(\mathrm { f } ( x )\) for some values of \(x\), with values of \(\mathrm { f } ( x )\) given to 4 decimal places where appropriate.

1. Complete the table giving the values to 4 decimal places. The equation \(\mathrm { f } ( x ) = 0\) has exactly one positive root, \(\alpha\). Using the values in the completed table and explaining your reasoning,
2. determine an interval of width one that contains \(\alpha\).
3. Hence use interval bisection twice to obtain an interval of width 0.25 that contains \(\alpha\). Given also that the equation \(\mathrm { f } ( x ) = 0\) has a negative root, \(\beta\), in the interval \([ - 1 , - 0.5 ]\)
4. use linear interpolation once on this interval to find an approximation for \(\beta\). Give your answer to 3 significant figures.

3 The diagram shows the graph of \(y = f ( x )\) for values of \(x\) from 1 to 3.5.
\includegraphics[max width=\textwidth, alt={}, center]{4023e87c-34b1-4abd-9acc-ede5e4d68c7f-03_945_1248_312_244} The table shows some values of \(x\) and the associated values of \(y\).

1. Use the forward difference method to calculate an approximation to \(\frac { \mathrm { dy } } { \mathrm { dx } }\) at \(x = 2\).
2. Use the central difference method to calculate an approximation to \(\frac { \mathrm { dy } } { \mathrm { dx } }\) at \(x = 2\).
3. On the copy of the diagram in the Printed Answer Booklet, show how the central difference method gives the approximation to \(\frac { \mathrm { dy } } { \mathrm { dx } }\) at \(x = 2\) which was found in part (b).
4. Explain whether your answer to part (a) or your answer to part (b) is likely to give a better approximation to \(\frac { \mathrm { dy } } { \mathrm { dx } }\) at \(x = 2\).

7 The value of a function, \(\mathrm { y } = \mathrm { f } ( \mathrm { x } )\), and its gradient function, \(\frac { \mathrm { dy } } { \mathrm { dx } }\), when \(x = 2\), is given in Table 7.1. \begin{table}[h] \end{table}

1. Determine the approximate value of the error when \(f ( 2 )\) is used to estimate \(f ( 2.03 )\). The Newton-Raphson method is used to find a sequence of approximations to a root, \(\alpha\), of the equation \(\mathrm { f } ( x ) = 0\). The spreadsheet output showing the iterates, together with some further analysis, is shown in Table 7.2. \begin{table}[h]
   \captionsetup{labelformat=empty} \caption{Table 7.2}

   	A	B	C	D
   1	r	Xr	difference	ratio
   2	0	12
   3	1	-13.1165572	-25.1165572
   4	2	1.76283279	14.87939004	-0.5924136
   5	3	2.18052157	0.41768878	0.02807163
   6	4	2.182419024	0.001897454	0.00454275
   7	5	2.182419066	\(4.13985 \mathrm { E } - 08\)	\(2.1818 \mathrm { E } - 05\)

   \end{table}
2. 1. Explain what the values in column D tell you about the order of convergence of this sequence of approximations.
   2. Without doing any further calculation, state the value of \(\alpha\) as accurately as you can, justifying the precision quoted.

1 The table shows some values of \(x\), together with the associated values of a function, \(\mathrm { f } ( x )\).

1. Use the information in the table to calculate the most accurate estimate of \(f ^ { \prime } ( 2 )\) possible.
2. Calculate an estimate of the error when \(f ( 2 )\) is used as an estimate of \(f ( 2.05 )\).

6 Table 6.1 shows some values of \(x\) and the associated values of a function, \(y = f ( x )\). \begin{table}[h] \end{table}

1. Explain why it is not possible to use the central difference method to calculate an estimate of \(\frac { \mathrm { dy } } { \mathrm { dx } }\) when \(x = 1\).
2. Use the forward difference method to calculate an estimate of \(\frac { \mathrm { dy } } { \mathrm { dx } }\) when \(x = 1\). A student uses the forward difference method to calculate a series of approximations to \(\frac { \mathrm { dy } } { \mathrm { dx } }\) when \(x = 2\) with different values of the step length, \(h\). These approximations are shown in Table 6.2, together with some further analysis. \begin{table}[h]
   \captionsetup{labelformat=empty} \caption{Table 6.2}

   \(h\)	0.8	0.4	0.2	0.1	0.05	0.025	0.0125	0.00625
   approximation	0.130452	0.138647	0.143381	0.145942	0.147277	0.147959	0.148304	0.148477
   difference		0.008195	0.004734	0.002561	0.001335	0.000682	0.000345	0.000173
   ratio			0.577633	0.541099	0.521186	0.510762	0.505424	0.502723

   \end{table}
3. 1. Explain what the ratios of differences tell you about the order of the method in this case.
   2. Comment on whether this is unusual.
4. Determine the value of \(\frac { \mathrm { dy } } { \mathrm { dx } }\) when \(x = 2\) as accurately as possible. You must justify the precision quoted.

5 You are given that
\(g ( x ) = \frac { \sqrt [ 3 ] { x ^ { x } + 25 } } { 2 }\). Fig. 5.1 shows two values of \(x\) and the associated values of \(\mathrm { g } ( x )\). \begin{table}[h] \end{table}

1. Use the central difference method to calculate an estimate of \(\mathrm { g } ^ { \prime } ( 1.5 )\), giving your answer correct to 3 decimal places. The equation \(x ^ { x } - 8 x ^ { 3 } + 25 = 0\) has two roots, \(\alpha\) and \(\beta\), such that \(\alpha \approx 1.5\) and \(\beta \approx 4.4\).
2. Obtain the iterative formula \(x _ { n + 1 } = g \left( x _ { n } \right) = \frac { \sqrt [ 3 ] { x _ { n } ^ { X _ { n } } + 25 } } { 2 }\).
3. Use your answer to part (a) to explain why it is possible that the iterative formula \(x _ { n + 1 } = g \left( x _ { n } \right) = \frac { \sqrt [ 3 ] { x _ { n } ^ { X _ { n } } + 25 } } { 2 }\) may be used to find \(\alpha\).
4. Starting with \(x _ { 0 } = 1.5\), use the iterative formula to find \(x _ { 1 } , x _ { 2 } , x _ { 3 } , x _ { 4 } , x _ { 5 }\), and \(x _ { 6 }\).
5. Use your answer to part (d) to state the value of \(\alpha\) correct to 8 decimal places. Starting with \(x _ { 0 } = 4.5\) the same iterative formula is used in an attempt to find \(\beta\). The results are shown in Fig. 5.2. \begin{table}[h]

   \(n\)	\(x _ { n }\)
   0	4.5
   1	4.81826433
   2	6.27473453
   3	23.2937196
   4	\(2.0654 \mathrm { E } + 10\)
   5	\#NUM!

   \captionsetup{labelformat=empty} \caption{Fig. 5.2}
   \end{table}
6. Explain why \#NUM! is displayed in the cell for \(x _ { 5 }\).
7. On the diagram in the Printed Answer Booklet, starting with \(x _ { 0 } = 4.5\), illustrate how the iterative formula works to find \(x _ { 1 }\) and \(x _ { 2 }\).
8. Determine what happens when the relaxed iteration \(x _ { n + 1 } = ( 1 - \lambda ) x _ { n } + \lambda g \left( x _ { n } \right)\) is used to try to find \(\beta\) with \(x _ { 0 } = 4.5\), in each of the following cases.
   • \(\lambda = 0.5\)
   • \(\lambda = - 0.4\)

7 Fig. 7.1 shows two values of \(x\) and the associated values of \(\mathrm { f } ( x )\). \begin{table}[h] \end{table}

1. Use the forward difference method to calculate an estimate of the gradient of \(\mathrm { f } ( x )\) at \(x = 3\), giving your answer correct to 4 decimal places. Fig. 7.2 shows some spreadsheet output with additional values of \(x\) and the associated values of \(\mathrm { f } ( x )\). \begin{table}[h]

   \(x\)	3	3.00001	3.0001	3.001	3.01	3.1
   \(\mathrm { f } ( x )\)	6.082763	6.08274	6.082541	6.08054	6.060454	5.848846

   \captionsetup{labelformat=empty} \caption{Fig. 7.2}
   \end{table} These values have been used to produce a sequence of estimates of the gradient of \(\mathrm { f } ( x )\) at \(x = 3\), together with some further analysis. This is shown in the spreadsheet output in Fig. 7.3. \begin{table}[h]

   \(h\)	0.1	0.01	0.001	0.0001	0.00001
   estimate	-2.339165	-2.230883	-2.220532	-2.219501	-2.219398
   difference	0.1082815	0.010352	0.0010307	0.000103
   ratio	0.095602	0.099567	0.0999568

   \captionsetup{labelformat=empty} \caption{Fig. 7.3}
   \end{table} Tommy states that the differences between successive estimates is decreasing so rapidly that the order of convergence of this sequence of estimates is much faster than first order.
2. Explain whether or not Tommy is correct.
3. Use extrapolation to determine the value of the gradient of \(\mathrm { f } ( x )\) at \(x = 3\) as accurately as possible, justifying the precision quoted.
4. Calculate an estimate of the absolute error when \(\mathrm { f } ( 3 )\) is used as an approximation to \(\mathrm { f } ( 3.02 )\).

4 The table shows some values of \(x\) and the associated values of \(\mathrm { f } ( x )\).

1. Calculate four estimates of the derivative of \(\mathrm { f } ( x )\) at \(x = 4\).
2. Without doing any further calculation, state the value of \(f ^ { \prime } ( 4 )\) as accurately as you can, justifying the precision quoted.

4 The table below gives values of a function \(y = \mathrm { f } ( x )\).

1. Calculate three estimates of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) at \(x = 0.4\) using the central difference method.
2. State the value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) at \(x = 0.4\) to an appropriate degree of accuracy. Justify your answer.