Appendix lecture 9: Extra terms.

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Terence Simmons
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1 Appendi lecture 9: Etra terms The Hyperolic method has een used to prove that d n = log + 2γ + O /2 n This can e used within the Convolution Method to prove Theorem There eists a constant C such that n X 2 ωn = ζ 2 X log X + C X + O X /2 log X Solution Recall that 2 ω = d µ 2 so n X 2 ωn = a X µ 2 a = a X µ 2 a X/a d X a log X a + 2γ X a + O Recall that µ 2 a is non-zero only if a = m 2, say Thus X /2 a 2 ωn = X µ m m log X 2 m + 2γ X /2 X 2 m + O 2 m 2 n X m 2 X 2 The error here is O X /2 = O X /2 log X m m X /2 For the first term we use a trick seen earlier for dealing with the logarithm: m 2 X µ m X m 2 log X m 2 = X m 2 X µ m m 2 X m 2 dt X t = X m 2 t µ m dt m 2 t 3 In this integrand the sum converges asolutely so we complete it to infinity,

2 m 2 t µ m m 2 = = m= µ m m 2 µ m = m 2 ζ 2 + O m 2 m>t /2 m>t /2 + ε t, 4 ζ 2 where ε t = O /t /2 Inserted into 3 this gives m 2 X µ m X m 2 log X m 2 = X X Since ε t = O /t /2 the integral dt ζ 2 + ε t t = X log X +X ζ 2 ε t dt t converges and so is a constant we will denote y C 0 Then X Hence ε t dt t = C 0 m 2 X X X ε t dt t ε t dt t = C dt 0 + O = C X t 3/2 0 + O X /2 µ m X m 2 log X m 2 = X log X ζ 2 + C 0 X + O X /2 The result 4 with t = X /2 will deal with the remaining term in 2 to give 2 ωn = n X ζ 2 X log X + C 0X + O X / γ X ζ 2 + O X /2 +O X /2 log X 2

3 This gives the stated result with C = C 0 + 2γ ζ 2 Question Generalise the previous result and prove that there eists a constant D k such that n d µ k n = ζ k log + D k + O /2 for k 3 So there is no log term in the error when k 3, and in fact the log in the error of Theorem can e removed y the use of the Hyperolic method Solution Start as aove µ k a d n d µ k n = a = a µ k a /a a log a + 2γ a + O /2 = µ m m log k m + 2γ /2 k m + O k m k m k For the first term we proceed as a 5 µ m m log = k m k m k = dt ζ k + ε k t t ζ k log + ε k t dt t, where ε k t /t /k Because the integral converges it can e completed to C k = ε k t dt t, 3

4 say, with an error Hence m k ε k t dt t dt t 2 /k /k µ m m log k m = k ζ k log + C k + O, /k The second term in 5 is 2γ µ m m k m /k = 2γ ζ k + O /k k While the error in 5 is O /2 m k/2 m /k The sum here converges since k 3, and so is ounded y a constant Comining these three results we get the stated result with D k = C k + 2γ ζ k 4

5 Appendi lecture 9: Additional terms for n d n 2 Let g n = d n 2 where d is the divisor function In Prolem Sheet 4 we have g = µ 2 = d Q 2 Can the Hyperolic Method e used to improve previous results on n d n2? To do so need to first prove two lemmas Lemma 2 There eists a constant C d say for which d a a = 2 log2 U + 2γ log U + C d + O Proof By Partial Summation and, U /2 d a a = U = U d a + U a t d a dt t 2 U log U + 2γ U + O U /2 U + t log t + 2γ t + ε t dt t 2 where ε t t /2 This gives the stated result with C d = 2γ + ε t dt t 2 Lemma 3 There eists a constant C Q say, for which Q 2 = ζ 2 log V + C Q + O V /2 Proof By Partial Summation and the sums of square-free numers 5

6 Q 2 = V = V Q 2 + V t ζ 2 V + O V /2 V + Q 2 dt t 2 dt ζ 2 t + ε t t 2 where ε t t /2 This gives the stated result with C d = ζ 2 + ε t dt t 2 Theorem 4 There eist constants c and c 2 such that n d n 2 = 2ζ 2 log2 + c log + c 2 + O 3/4 log Proof With U and V to e chosen, the Hyperolic method gives n X d n 2 = n X d Q 2 n = and By the Lemmas aove, d a Q 2 X/a Q 2 = a X/ d a = d a Q 2 Q 2 + X/a d a Q 2 d a ζ 2 a + O Q 2 log + 2γ + O a a X/ d a /2, 6 /2 7 6

7 Errors The first error, in 6 is, y partial summation /2 d a = /2 d a + a /2 U /2 2 U a t d a dt t 3/2 /2 U U log U + U t log t dt /2 2 t 3/2 /2 U /2 log U The second error, in 7 is, again y partial summation, /2 Q 2 = /2 Q /2 V /2 2 + V Q 2 dt 2 t 3/2 t /2 V V + V t dt /2 2 t 3/2 /2 V /2 The idea would e to equate these errors, which means demanding U = V Since UV = this means that U = V = /2, when the two errors aove are 3/4 log An error will also arise from the first term in 6, d a = ζ 2 a ζ 2 2 log2 U + 2γ log U + C d ut this is only O 3/4 + O, U /2 7

8 The first term in 7 is Q 2 log = Q 2 log V + log V = log V = log V Q 2 V + t ζ 2 log V + C Q + O Q 2 dt t V /2 8 where η t /t /2 The integral here equals where 2ζ 2 log2 V + C Q log V + D + O D = V dt + ζ 2 log t + C Q + η t t η t dt t, 9 V /2 The errors in 8 and 9 are O log /V /V /2 and O /V /2 which are 3/4 log The second term in 7 is 2γ Q 2 and the error is again 3/4 = 2γ Finally, the last term in the Hyperolic Method is ζ 2 log V + C Q + O, V /2 d a Q 2 = U log U + 2γ U + O U /2 ζ 2 V + O V /2 The error from this is O U /2 V + UV /2 log U which is O 3/4 log y the choice of U and V 8

9 Main Terms The main terms are scattered throughout the aove epressions The main terms will not depend on the choice of U and V, the reason for introducing U and V are that good choices for them will give a good ound on the error term Terms with two logarithms: ζ 2 2 log2 U + log V = 2ζ 2 ζ 2 log V + 2ζ 2 log2 V log 2 U + 2 log V log V + log2 V = log 2 U + 2 log U log V + log 2 V since UV = 2ζ 2 = 2ζ 2 log U + log V 2 = = 2ζ 2 log2 Terms with one logarithm: log UV 2 2ζ 2 ζ 2 2γ log U + C Q log V + C Q log V + 2γ log V U log U ζ 2 ζ 2 V = ζ 2 2γ log U + log V + C Q = ζ 2 2γ log UV + C Q log V V 2γ = + C Q log ζ 2 log V + log V log V U ζ 2 log V + log U ζ 2 9

10 Terms with no logarithm: = C d ζ 2 + D + 2γ C 2γ Q + UV ζ 2 Cd ζ 2 + D + 2γ C 2γ Q + ζ 2 0

11 Appendi Level 9: The Improvement of n d 3 n If we attempted to improve d 3 n = 2 log2 + O log, n y using the improved estimate for n d n from in the Convolution Method, we would get d 3 n = m log m + 2γ /2 m + O m n m The error term here is /2 m m/2 We can get a far smaller error y using the Hyperolic Method Lemma 5 There eists a constant C 2 such that log n n = log 2 log2 + C 2 + O n Proof Writing the logarithm as an integral an interchanging the summation and integration gives log n n = n dt n n t = dt t n Split this sum as n t<n n = n n n t n t<n

12 Then dt t Since t<n n = = n dt n t log + γ + O n t dt n t dt t dt log t + γ + ε t, t = log 2 log2 + O ε t dt t + where ε t /t, ε t dt t the result follows with ε t dt t dt t 2, C 2 = ε t dt t Lemma 6 There eists a constant C d such that n X d n n Solution Left to student Theorem 7 = 2 log2 X + 2γ log X + C d + O d 3 n = 2 X log2 X + 3γ X log X n X X /2 + C d + C 2 + 2γ 2 3γ + X + O X 2/3 log X 2

13 Proof With U and V to e chosen, the Hyperolic method gives d 3 n = d n = d a + d a 0 n X n X X/a a X/ d a The first term on the right hand side equals [ ] X d a a = X d a a + O d a a + O d a = X = X 2 log2 U + 2γ log U + C d + O U /2 +O U log U, 2 having used Lemma 6 The second term on the right hand side of 0 equals d a = X X log X + 2γ X /2 + O a X/ = X log X/ y + 2γ X + O X /2 /2 3

14 So a X/ d a = X log X log + 2γ X + O X /2 V /2 = X log X log V + γ + O V log V X 2 log2 V + C 2 + O V + 2γ X The errors from these two terms in 0 are log V + γ + O +O X /2 V /2 XU /2, U log U, XV and X /2 V / V We attempt to minimise these errors y choosing U and V to equalise them So try XV = X /2 V /2, ie V = X /3 Yet UV = X so this means U = X 2/3 It can e checked that all errors are then X 2/3 log X With this choice of U and V the term in is d a = U log U + 2γ U + O U /2 V + O 6 The error here is O U log U + U /2 V which is again X 2/3 log X given the choice of U and V 4

15 The main terms from 2, 3 and 5 are X 2 log2 U + 2γ log U + C d + X log X log V + γ 7 X 2 log2 V + C 2 + 2γ X log V + γ V U log U + 2γ U = X 2 log2 U + log X log V 2 log2 V +X 2γ log U + γ log X + 2γ log V log U +X C d + C 2 + 2γ γ + 2γ, having used UV = X Note that these main terms should not depend on the choice of U and V, the mean value n X d 3 n will have the same main terms however they are calculated What we are doing in this method is not to calculate eactly the error for our result on this mean value ut to ound this error So a clever choice of U and V will give a etter ound on the error To check that the main terms do not depend on the choice of U and V consider first terms with two logarithms, dropping the X factor, 2 log2 U + log X log V 2 log2 V = 2 log2 U + log UV log V 2 log2 V = 2 log2 U + log U + log V log V 2 log2 V = 2 log2 U + log U log V + 2 log2 V = log U + log V 2 2 = 2 log2 UV = 2 log2 X 5

16 Also, for term with one logarithm again dropping the X factor 2γ log U + γ log X + 2γ log V log U = 2γ log U + log V + γ log X = 3γ log X All this comines to give the result quoted Note though that now we have justified the claim that it does matter what the choice is for U and V you will always get the same main terms, you can go ack to 7 and choose, say, U = and V = X when we get = X 2 log2 U + log X log V 2 log2 V +X 2γ log U + γ log X + 2γ log V log U +X C d + C 2 + 2γ γ + 2γ = X log X log X 2 log2 X + X γ log X + 2γ log X +X C d + C 2 + 2γ γ + 2γ = 2 X log2 X + 3γ X log X + X C d + C 2 + 2γ γ + 2γ 6

= 1 2 x (x 1) + 1 {x} (1 {x}). [t] dt = 1 x (x 1) + O (1), [t] dt = 1 2 x2 + O (x), (where the error is not now zero when x is an integer.

= 1 2 x (x 1) + 1 {x} (1 {x}). [t] dt = 1 x (x 1) + O (1), [t] dt = 1 2 x2 + O (x), (where the error is not now zero when x is an integer. Problem Sheet,. i) Draw the graphs for [] and {}. ii) Show that for α R, α+ α [t] dt = α and α+ α {t} dt =. Hint Split these integrals at the integer which must lie in any interval of length, such as [α,