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MATHEMATICAL NOTES. 301

We shall denote Napier's logarithms by Log, and logarithms to base eby log.

Now,even

supposingthat we know

nothing aboutthe

mode of constructionof Napier's system, but have only his tables, we can determine the constantsa and b if we know only the Logs of two distinct numbers.

We find from the tables

Log 107=0 and Log 9995000=5001-25.

Hence we have 7a log 10+ b=0,and a log 9995000+ b= 5001-25.

Therefore a(7 log 10- log 9995000)= -5001-25.

The coefficient of a= - log '9995=-000500125, as is easily calculated fromthe logarithmic series.

Hence a= - 107, exactly, so far as the figures go.Then b= - 7a log 10= 7 x 107x 2-302585093= 161180956-5.The relationship between Napier's logarithms and Hyperbolic logarithms

is thereforeLog x= 10 (log 107- log x)

=161 180957- 107 ogx.Then Logl= 107 og 107= 161 80 956'5.This may be verified by taking examples from the tables.*

Thus Log 2909=81 425 681 and Log 29092=Log8462281= 1 669 663;

therefore Log 1 =2 Log 2909- Log 29092= 161 181699.

Again, by interpolating, Log 11636 = 67 562 746,

Log 100 =115 129 512,

Log 1163600= 21 510 657;

therefore Log 1= Log 11636+ Log 100- Log 1163600

=161 181601. D. M. Y. SOMMERVILLE.

484. [E.5; V.a.A.] Further remarks on the integralsin

dx.

I have been asked for my opinion of the difficulty of Prof. Dixon's proof,

as compared with those of the proofs I discussed in my previous note on thissubject (Gazette,v. p. 98). The conciseness and ingenuity of the proof noone could question. Dr. Bromwichand Mr.Whipplehave also sent me proofsthat I had not seen or forgot to mention, and it seems appropriatethat Ishould attempt to assign marks, according to the principles I then stated,to all three. Before I do so, however, I should like to mention two generalprinciplessuggested by Dr. Bromwichas comments on my system of marking,as I am myself of opinion that their acceptancemakes my method fairerandeasier to apply.

*I have no access to a copy of Napier's tables, but on calculating the valuesof Napier's logarithms from the

correspondinghyperboliclogarithms,it

appearsthat

these numbersshould be corrected as follows:Log2909 =81 425 310 Log11636 = 67 562366Log29092= 1 669663 Log100 = 115 129255

Log1163600= 21 510664The hyperboliclogarithmsof 2, 3 and 10 which are used here I have obtainedfrom

Adams'paper in Proc. R. Soc. (1878)27, p. 92. The value of log2909I have calculatedfromthe formulalog2909=log 3000+ log (1- in). The valuesare

log2= '6931471806 log10 =2'30258 50930

log3=1 09861 22887 log2909= 7'9755646585

MATHEMATICAL NOTES. 301

We shall denote Napier's logarithms by Log, and logarithms to base eby log.

Now,even

supposingthat we know

nothing aboutthe

mode of constructionof Napier's system, but have only his tables, we can determine the constantsa and b if we know only the Logs of two distinct numbers.

We find from the tables

Log 107=0 and Log 9995000=5001-25.

Hence we have 7a log 10+ b=0,and a log 9995000+ b= 5001-25.

Therefore a(7 log 10- log 9995000)= -5001-25.

The coefficient of a= - log '9995=-000500125, as is easily calculated fromthe logarithmic series.

Hence a= - 107, exactly, so far as the figures go.Then b= - 7a log 10= 7 x 107x 2-302585093= 161180956-5.The relationship between Napier's logarithms and Hyperbolic logarithms

is thereforeLog x= 10 (log 107- log x)

=161 180957- 107 ogx.Then Logl= 107 og 107= 161 80 956'5.This may be verified by taking examples from the tables.*

Thus Log 2909=81 425 681 and Log 29092=Log8462281= 1 669 663;

therefore Log 1 =2 Log 2909- Log 29092= 161 181699.

Again, by interpolating, Log 11636 = 67 562 746,

Log 100 =115 129 512,

Log 1163600= 21 510 657;

therefore Log 1= Log 11636+ Log 100- Log 1163600

=161 181601. D. M. Y. SOMMERVILLE.

484. [E.5; V.a.A.] Further remarks on the integralsin

dx.

I have been asked for my opinion of the difficulty of Prof. Dixon's proof,

as compared with those of the proofs I discussed in my previous note on thissubject (Gazette,v. p. 98). The conciseness and ingenuity of the proof noone could question. Dr. Bromwichand Mr.Whipplehave also sent me proofsthat I had not seen or forgot to mention, and it seems appropriatethat Ishould attempt to assign marks, according to the principles I then stated,to all three. Before I do so, however, I should like to mention two generalprinciplessuggested by Dr. Bromwichas comments on my system of marking,as I am myself of opinion that their acceptancemakes my method fairerandeasier to apply.

*I have no access to a copy of Napier's tables, but on calculating the valuesof Napier's logarithms from the

correspondinghyperboliclogarithms,it

appearsthat

these numbersshould be corrected as follows:Log2909 =81 425 310 Log11636 = 67 562366Log29092= 1 669663 Log100 = 115 129255

Log1163600= 21 510664The hyperboliclogarithmsof 2, 3 and 10 which are used here I have obtainedfrom

Adams'paper in Proc. R. Soc. (1878)27, p. 92. The value of log2909I have calculatedfromthe formulalog2909=log 3000+ log (1- in). The valuesare

log2= '6931471806 log10 =2'30258 50930

log3=1 09861 22887 log2909= 7'9755646585

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7/29/2019 Hardy 1916


THE MATHEMATICAL GAZETTE.

Thefirstis, that whena proofinvolves two inversionsresemblingone anotherboth in principleand detail, then the mark for the second should be reduced

by one-half, e.g.from 10 to 5.

The second rests on the following considerations.Let us consider, for clearness,the case of an integration of a series over a

finite range. Then certain integrationsstand out by themselvesas peculiarlyeasy to justify, viz. those in which the series is uniformly and absolutelyconvergent in virtue of " Weierstrass'sM-Test " (Bromwich, Infinite Series,p. 113 and also p. 207). Such a case is that involved in

7rcosnOde 1-f,

n7"

o 1 - 2pos 0 +p 1 1{+ 2 pkcosklO cosnOdO=

-21o-2pcos 0 +jp2 l-p-^o 1-p.0

(where, of course, Ip I< 1). Cases in which the series is uniformlybut not

absolutely convergent are intrinsically more difficult, depending as they doon oneorother of the tests calledby Dr. Bromwich" Abel's"and " Dirichlet's"

(I.c.pp. 113-4),and, in the long run,on " Abel's Lemma" (p. 54). An exampleof this kind is the integrationin what I called in my formernote, "Mr. Berry'ssecond proof."

A similar distinction is to be made in regardto proofs of the continuity ofan infinite series, and, mutatis mutandis, proofs of the legitimacy of other

types of inversions, although it cannot, of course, always be presented in soclear-cut a form.

Dr. Bromwichsuggests that 15 marks at least should be assigned for anyinversion of the more difficulttype, and I am disposedto adopt this suggestion.The 10 or 15 marksare, of course,

onlyintended as a

roughstandardby which

to make a beginning,and would generally be modifiedin any particularcase.On looking through my former note, in the light of these remarks, I am

disposed to make the following alterations in my original marks.

Proof 2.* Assign 25 instead of 20 for the inversion of the integrations,makingthe total 37 instead of 32. Also add 5 to the total for Proof5, makingit 45.

Proof 4. Add 5 marksin respectof the first inversion,makingthe total 45.

Proof 7. Deduct say 28, leaving 80.Dr. Bromwich'sproof, mentioned in the last footnote of my former note

(which we may call Proof 8), I should now mark at about 50 instead of 45.In so far as it is concernedwith this particular integral, it differslittle from

Proof 4. I proceed to consider the three new proofs.Proof 9 (Dr. Bromwich'ssecondproof). We have

714f f7r r iQ~d

e-xsinOcos(xos )dO= RLxe ]d0

-=2

(1 -xsin 0-2 cos20+ 3 sin30+... dO

1 x3 x5

2 -+3.3! 5.5!'

1 rx sin tdt.=2 Jo t

But the original integral is numerically less than

enuerdep(t ,(. ex) < 2x

* The numbers are those assignedto the proofs in my final list (Gazette,v. p. 98).

302





7/29/2019 Hardy 1916


MATHEMATICAL NOTES.

and the result followson makingn tend to oo. This proofis due to Schl6milch,

Utbungsbucher h6herenAnalysis, p. 173. It is to be observed that it and

Proofs 2 and 5 are really variations of the same theme, and closely connectedwith the proof by contour integration. This being so, it is clear that the

straightforward proof by contour integration should have a better markthan either; and I deduct 2 from the marks for Proofs 3 and 6, which I had

perhaps marked a little severely.As both integrations of a series in Proof 9 are of the simplest type, I give

10 marks for the first integration, 5 for the second, 5 for the summation ofthe trigonometricalseries and 10 for the final step in the argument, making30 in all. I add 5 more because the proof has a certain artificiality, thoughless than Mr.Michell's,as contrasted with such proofs as 1 and 4; the totalis therefore 35.

Proof 10 (Mr. Whipple's proof). This depends on the transformations

' sin x d .lim sinnh_ lim ( 1I dx = limAh -=iEm =

(-l)=h-- v0 nh h- A)-

This proof has the great merit of being absolutely natural and straight-forward n idea; unfortunately,however,the inversion s one of a very difficultcharacter.* Also the assumption of a knowledge of the sum of the trigo-nometrical series is a rather serious matter. I cannot give less than30+15=45 marks. At the same time, the proof seems to me a most

interesting and suggestive one.Proof 11 (Prof. Dixon's proof).t This proof I findharderto mark than any.

It is exceedingly concise, and it avoids all inversions of whatever kind in the

most ingeniousway; its difficultiesare, indeed, of a kind which rather bafflemy rules. Rules or no rules, I cannot regard it as on a par with Proof 1,nor, I think, as being really quite as simple as 3 or 9. At a rough guess, letus say 36. As it is absurd to try to make very fine distinctions, I alter themarks of 9 and 2 from 35 and 37 to 36 each also. My final list, then, worksout as follows:

1. Proof 1 (Mr. Berry's second) - - - 28.2. Proof 3 (Mr. Berry's orm of his third) - 32.

[Proof 9 (Dr. Bromwich's second) - - - 36.3. Proof 11 (Prof. Dixon's) - - - - 36.

Proof 2 (My form of Mr. Michell's) - - 36.

6. Proof 6 (Ordinary orm of Mr. Berry'sthird) 40.(Proof 4 (Mr. Berry'sfirst) - - - 45.

7. Proof 5 (Mr. Michell's original) - - - 45.

tProof 10 (Mr. Whipple's) - - - - 45.10. Proof 8 (Dr. Bromwich'sfirst) - - - 50.11. Proof 7 (Prof. Nansen's) - - - - 80.

As I remarked before, Proofs 7 and 8, especially the first, are heavilypenalised because they include proofs of a great deal of extraneous matter.As regards absence of serious theoretical difficulty, I think 9 is possibly the

simplest of all. I regret the low position of 4 and 10; each is a good proof,proceeding to its goal by a straight and natural route, but unfortunately

meeting with difficult theoretical obstacles. Finally, I may point out thatProof 1 is sometimes presentedin a somewhat differentform (see Bromwich,Infinite Series, p. 468), in which the integration of a series may be made to

depend upon an "M-test," and which may possibly deserve a slightly lowermark. G. H. HARDY.

*See a paper by Dr. Bromwich and myself in the Quarterly Journal, vol. xxxix.p. 222; also Cesaro,AlgebraischeAnalysis, p. 699.

t Gazette,vi. p. 225.

303



