Thompson - Computing for Scientists and Engineers (523188), страница 15
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Thus show, by combining terms from the two convergent series, thatthe cosh gathers the even terms of the exponential and the sinh gathers its oddterms, producing for the hyperbolic cosine(3.29)and for the hyperbolic sine(3.30)(b) As an alternative derivation, use relations (2.53) and (2.54) between hyperbolic and circular functions, and assume that the power-series expansions (3.20)and (3.21) can be extended to complex variables, from x to ix. Thus again derive the expansions (3.29) and (3.30). nThe second method shows, in terms of complex variables, where the sign changesbetween the power series for circular and hyperbolic functions originate, since thealternating signs for the former series are canceled by the factors of i 2 = -1 whicharise in forming the latter series.The convergence properties of the cosh and sinh are very much better than thoseof the cos and sin, because the sums in both (3.29) and (3.30) have only positiveterms.
Therefore, given the very rapid convergence of the series for circular functions, the convergence of the series for hyperbolic functions will be even more rapidfor small x. Note, however, as shown in Figure 2.4, that these hyperbolic functions are unbounded rather than periodic, so the restrictions to small arguments thatwe invoked for the circular functions cannot be made so realistically for them.The numerical properties of the Maclaurin expansions for cosh and sinh are explored in Project 3 in Section 3.5. Their coding is essentially that for the cos andsin expansions, apart from the lack of sign reversals between successive terms.72POWER SERIESLogarithms in series expansionsThere are many applications for power series expansions of the natural logarithm,1n. Maclaurin expansions of 1n (x) cannot be made because the logarithm is divergent as x tends to zero, and its first derivative diverges as 1/x .
Instead, one usuallyconsiders the expansion of 1n (1 + x ) about x = 0.Exercise 3.17Show that if one wants an expansion of 1n (a + bx) about x = 0, then an expansion in terms of (1 + u ), where u = bx /a, is sufficient, except that the region of convergence for u differs by the factor b/a relative to that for x, and istherefore smaller if this factor is of magnitude greater than unity.
nA power series expansion of 1n ( 1 + x ) is easiest made directly from Taylor’stheorem, (3.6). The successive derivatives after the first, which is just 1/(1 + x ),are easy to evaluate. They alternate in sign and grow in a factorial manner, but theyare one step behind the Taylor-theorem factorial. Thus you can readily obtain theseries expansion(3.31)This series expansion converges only if |x| < 1, as can be seen by comparison withthe divergent harmonic series.The result (3.31) and its consequences are of considerable interest.Exercise 3.18(a) Show in detail the steps leading to (3.31).(b) Write down the series expansion for 1n (1 - x ) by using (3.31) with thesign of x changed, then subtract the series expansion from the expressions in(3.31). Thereby show that(3.32)(c) Check the parity symmetry of the logarithm on the left side of (3.32) byshowing that it is an odd function of x, and that this result agrees with the symmetry of the right-side expression.
Show also that the simpler expresssion in(3.31) does not have a definite symmetry under sign reversal of x. nAs you can see from the coefficients of the powers of x in (3.31), the logarithmic series has quite poor convergence, mainly relying for its convergence on the decreasing values of the powers of x, rather than on the presence of factorials in thedenominator. We investigate the numerical convergence in more detail as part of thecomputing project in Section 3.5.3.2 TAYLOR EXPANSIONS OF USEFUL FUNCTIONS73The power series expansion of the logarithm may be adequate in limited rangesof the variable, especially if one is willing to interpolate within a table of logarithms.Suppose we have a table in computer memory with entries at t - h, t, t + h, .
. . ,and that we want the interpolated logarithm at some point a distance d from t.Clearly, it is sufficient to consider values of d such that |d/h | < 1/2. How closelyspaced do the table entries have to be in order to obtain a given accuracy, , of theinterpolated values?Exercise 3.19(a) Use the power-series expansion of the logarithm, (3.31), to show that(3.33)in which the error, , can be estimated as the first neglected term, which is(3.34)(b) Show that if the tabulated increments satisfy h/tbe less than 10-7. n0.01, then the error willTable lookup therefore provides an efficient way to obtain values of logarithms overa small range of its argument.
In this example 200 entries would give better thanpart-per-million accuracy. For applications such as Monte Carlo simulations, wherethere is intrinsically much approximating, this accuracy is probably sufficient.Now that we have some experience with handling power series for logarithms, itis interesting to try a more difficult function involving logarithms.Series expansion of x 1n(x)The function x 1n(x) is of interest because, although it has a Maclaurin expansion,this cannot be obtained directly by taking successive derivatives of this function andevaluating them at x = 0, since the derivatives of the logarithm diverge as x 0.You should first be convinced that x 1n(x) is indeed well-behaved near the origin.To do this, try the following exercise.Exercise 3.20-yy(a) Set x = e , with y > 0, so that x 1n( x) = -y/e . Thence argue thatx0 as y, and therefore that x 1n(x)0 from below.(b) Make a graph of x 1n(x) for x in the range 0.05 to 2.00 by steps of 0.05,and check that it agrees with the solid curve in Figure 3.4.
Thus this function iswell-behaved over the given range of X, and is therefore likely to have a seriesexpansion about the origin, at least over a limited range of X. n74POWER SERIESFIGURE 3.4 The function x 1n (x), shown by the solid curve, and its approximations, shownby the dashed curves.The derivation of the series expansion of x 1n(x), examples of which are shownin Figure 3.4, is easiest made by considering the Taylor expansion about x = 1 ofthe logarithm, written in the form(3.35)We also have the geometric series(3.36)If we now multiply the two left-side expressions in (3.35) and (3.36), and their corresponding series on the right sides, then we have a double series in which we cangroup like powers of x as(3.37)This is not the conventional form of a series expansion, which usually has a singlepower, k, of X.
You can easily remedy this by working the following exercise.Exercise 3.21(a) To understand what the series (3.37) looks like, write out the first few powers of x on the right-hand side, in order to see that the coefficients are 1,1 + 1/2, 1 + 1/2 + 1/3, etc.3.2 TAYLOR EXPANSIONS OF USEFUL FUNCTIONS75(b)Set k = m + n in (3.37), then rearrange the summation variables in orderto show that(3.38)where the harmonic coefficients, hk, given by(3.39)are just the sums of the reciprocals of the first k integers. nThus we have an expansion about x = 1 of a series that does not seem to be particularly relevant to our purposes. But we can soon change all that, as follows.If in (3.38) we set y = 1 - x, then write(3.40)By relabeling, yx, we have a power-series expansion.
For any number ofterms in this expansion it coincides with the function at both x = 0 and at x = 1,at both of which points it is zero. The expansion to the first three terms (as the dashed curve) is compared with the function (the solid curve) in Figure 3.4. Note thatthe agreement is fairly poor between the two anchor points, because of the slow convergence of the series. The harmonic series is divergent, so the hk coefficients growsteadily without limit. Only the decreasing powers of x (less than unity) make theseries convergent.To reduce the poor convergence properties, we can play tricks with the logarithmseries. Suppose that we want an expansion that converges rapidly near x = a,where a 1.
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