A Course of Modern Analysis by E. T. Whittaker, G. N. Watson

By E. T. Whittaker, G. N. Watson

A process sleek research - moment variation - An Unabridged, thoroughly Revised Printing Of the 1st variation, With Additions (Riemann Integration, crucial Equations, Riemann-Zeta functionality) And Corrections, And in line with All next versions (at publication), retailer For The twenty third bankruptcy facing Ellipsoidal Harmonics And Lame's Equation - This version Has Been Digitally Enlarged, to incorporate The Decimal approach Of Paragraphing, With Appendix, record Of Quoted Authors, And accomplished normal Index.

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Can we say f is Riemann integrable on [a, b] and lim Ina,b fn(x)dx n-'O° 6 Ja f (x)dx? The following exercises show that the answer in general is in the negative. 3. Show that each fn is (i) Let fn(x) = Riemann integrable and lim fn(x) = f (x) = 0 V x E (0, 1], but n-+oo 1 1 fdx does not converge to Ifol (ii) Let {rl, r2, f(x)dx. JO ... } be an enumeration of the rationals in [0, 1]. , 1 0 if x E {rl,r2,... , rn}, if x E [0,1] \ {r1, r2, ... , rn}. 1. Riemann integration 40 Show that {f}>i is a convergent sequence of Riemann integrable funclim fn(x) is a bounded function which is not Riemann tions and f (x) integrable.

At k(V + 1)x, = 1, 2, ... , thfunction f(mx)/m2 has a jump of magnitude 1/2n2 from the right and 1/2n2 from the left. ) Hence R is discontinuous at x if x = m/2k, m and 2k being relatively prime. We note that such points are dense in R. Finally, we show that R(x) is integrable. Let e > 0 be any real number. Then D, x E [a, b] w(R, x) > EJ _ {x E [a, b] x = m/2k, m and 2k relatively prime and 7r2/8k2 > e}. 1. Riemann integration 34 Thus DE is at most a finite set, and hence is a null set. 16, R is integrable.

Riemann integration 12 Thus a2 = sup{L(Pn,f)} n>1 2 ia sup{L(P,f)} _ f(x)dx P < infjU(Pj)j < infjU(Pnjf)j a2 Hence f is Riemann integrable and a fn (iii) Let f : [a, b] f (x)dx = 2 R be defined by f(x):=< 0 1 if x is a rational, a < x < b, if x is an irrational, a < x < b. It is easy to see that for any partition P of [a, b], U(P, f) = (b - a) and L(P, f) = 0. Hence 6 = 0< (b - a) = ff(x)dx 6 ff(x)dx. Thus f is not Riemann integrable. In general, it is not easy to compute the upper and the lower integrals of a function and verify its integrability.

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