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CHAPTER 4 Taylor Series and Numerical Differentiation
Problem 1 : Consider the function f ( x ) = e sin x. Using the Taylor series expansion about x = 0, evaluate the function to O ( h 2 ) and O ( h 4 ). Plot the TE as a function of h for both the cases. Take h from 0.1 to 0.4. The given function f ( x ) is (1) To plot the TE as a function of h , the Taylor series expansion can be written as (2) The derivatives are evaluated as (3-1) (3-2) (3-3) (3-4) Since โ โ and โ โ , , , , and (^) (4) Therefore, the Taylor series for f ( x ) is
Thus, from Equations (2), (3), (4) and (5), the Taylor series expansion about x = 0 can be written as: i) For O ( h 2 ) ๏ Answer ii) For O ( h 4 ) ๏ Answer And the TE from Equation (2) can be written as i) For O ( h 2 ) ๏ Answer ii) For O ( h 4 ) ๏ Answer With xi = 0 and h = 0.1 ~ 0.4, we can calculate the TE as shown in Table 1 and 2. Table 1 The Taylor series expansion values and (TE)ex as function of h h Exact for O ( h 2 ) (TE)ex for O ( h 4 ) (TE)ex h = 0.1 (^1) 1.1 โ 0.1 1.105 โ 0. h = 0.2 (^1) 1.2 โ 0.2 1.22 โ 0. h = 0.3 (^1) 1.3 โ 0.3 1.345 โ 0.
Problem 2 : 4 .5. The pressure p of a gas is given by the expression log p = 21.6 โ 2420/ T , where T is the temperature in kelvins. Using the exact value of p from this expression, at T = 400 K, and the Taylor series expansion for p ( T ), compute the pressures at 410, 420, and 450 K. Compare these values with the exact ones obtained from the given expression. The given relation between p and T is (1) The Equation (1) can be written as follows: (2) The Equation (2) can be written as (3) where and (^) (4) From the Equation (1), the exact value of p from this expression at T = 400 K is (5) Then, the Taylor series expansion for p ( T ) is given by (6) Now , and so on, can be obtained by as
From the above equations, p can be calculated at 410, 420, and 450 K. (See the Table 1 )
Central differencing formulas O [(ฮ x ) 2 ]: For the first derivative (2-1) For the second derivative (2-2) I. The first and second derivatives of sin x at x = 0 are and (3) Using above equations, the first and second derivatives at x =0 can be calculated (See Table 1 ) Table 1- 1 Forward difference approximations: O [(ฮ x )^2 ] Numerical Exact TE ฮ x (^) (at xf i = 0) fi +1 fi +2 fi +3 f' f" f' f" f' f"
- 2 0 0.1987 0.38942 0.5646 1.0132 โ 0.0079 1 0 โ 0.01315 0. 0 .1 0 0.0998 0.19867 0.2955 1.0033 โ 0.001 1 0 โ 0.00332 0. 0.01 0 0. 010000 0.02 0.02999 1.00003 โ 0.000001 1 0 โ 0.000033 0. Table 1- 2 Central difference approximations O [(ฮ x ) 2 ] Numerical Exact TE ฮ x (^) (at xf i = 0) fi +1 fi โ 1 f' f" f' f" f' f"
- 2 0 0.19867 โ 0.19867 0.9933 5 0 1 0 0.00665 0 0 .1 0 0.09983 โ 0.09983 0.99833 0 1 0 0.00167 0 0.01 0 0.010000 โ 0.010000 0.99998 0 1 0 0.00002 0 II.
The first and second derivatives of sin x at x = 0 can be written as and (^) (4) From Equations (1), (2) and (4), the first and second derivatives at x = 0 can be calculated (See Table 2 ) Table 2- 1 Forward difference approximations: O [(ฮ x )^2 ] Numerical Exact TE ฮ x (^) (at xf i = 0) fi +1 fi +2 fi +3 f' f" f' f" f' f"
- 2 1 1.22140 1.4918 3 1.82213 0.98447 0.9541 6 1 1 0.01553 0.0458 5 0 .1 1 1.10517 1.22140 1.3498 6 0.9964 1 0.98976 1 1 0.0035 59 0.0102 4 0.01 1 1. 01005 1.02020 1.0304 6 0.99997 0.9999 1 1 1 0.000034 0. Table 2- 2 Central difference approximations: O [(ฮ x )^2 ] Numerical Exact TE ฮ x (^) (at xf i = 0) fi +1 fi โ 1 f' f" f' f" f' f"
- 2 1 1.221403 0.818731 1.006680 1.003338 1 1 โ 0.006680 โ 0. 0 .1 1 1.105171 0.904837 1.001668 1.000834 1 1 โ 0.001668 โ 0. 0.01 1 1. 010050 0.990050 1.000017 1.000008 1 1 โ 0.000017 โ 0. As expected, the accuracy improves as the step size (ฮ x ) decreases. Moreover, the central differencing yields more accurate results, as compared to forward differencing. Problem 4 : 4 .12. The measured temperature distribution in a solar energy heating system may be represented by the equation where x is the distance away from the surface being heated and T is the temperature. Compute the temperature gradient dT / dx and the second derivative d^2 T / dx^2 at x = 0. The heat transfer rate is proportional to the temperature gradient at x = 0. The second derivative
Problem 5 : 4 .17. The hot-wire anemometer is an instrument used for measuring velocities or temperatures. If, during its calibration, the output signal E is measured as 0, 1.7, 3.3, and 5.6 volts at velocities V of 0, 1, 1.5, and 2 m/s, obtain the gradient dE / dV at V = 0 m/s, using the polynomial representation of the function E ( V ). Use the polynomial as (1) From the calibration data ( Table 1 ) Table 1: The hot-wire anemometer calibration data E ( V ) [Volts] V [m/s] 0.0 0. 1.7 1. 3.3 1. 5.6 2. Using equation (1) and Table 1, the polynomial value at V=0 is Thus, (2) Using equations (1) and (2), Thus
Solving equations (3-1) to (3-3) (4) With equations (1), (2), and (4), the polynomial representation is (5) From equation (5), the gradient dE / dV is (6) Note that the truncation error for this approximation is of the order of ( dV ) 3
