Proof of the Linear Independence of {1, log(a/(a+1)) and log((a+1)/(a+2))} for Every Positive Integer a and Corresponding Explicit Linear Independence Measures By Shalosh B. Ekhad Theorem : For any positive integer a, the three numbers {1, log(a/(a+1)), log((a+1)/(a+2))} are linearly independent. In other words there exists a positive number, nu, such that if q,p1,p2 are integers and Q=max(|q|,|p1|,|p2|) Q>=Q0, where Q0 is a sufficiently large number, then |q+p1*log(a/(a+1))+p2*log((a+1)/(a+2))|>1/Q^nu . Furthermore, an upper bound for nu is as follows if a is even then nu can be taken as / / 6 5 4 3 2 |2 ln(2) + ln(3) - ln(|108 a + 648 a + 1440 a + 1440 a + 614 a + 76 a - 1 \ \ 12 11 10 9 8 + (11664 a + 139968 a + 730944 a + 2177280 a + 4072464 a 7 6 5 4 3 2 + 4959360 a + 3940188 a + 1985832 a + 592933 a + 90676 a + 4974 a / (2/3) / 1/3\ \ / | + 52 a + %1 + 1) / %1 |/(a (a + 2))) - ln(a (a + 2)) - 2| / |ln( / / / / \ / |/ 1944 I || \\ 6 5 20 4 20 3 307 2 19 -1/54 I a - 1/9 I a - -- I a - -- I a - ---- I a - ---- I a + 1/5832 I 81 81 2916 1458 / 1/2 \ \ (1/3) | 3 | (2/3) 1/2 2 | %1 + |1/11664 I - -----| %1 + (I + 3 ) (a + 2 a + 1/6) / \ 11664/ 2 (a + 1) / 8 7 6 5 3733 4 385 3 4411 2 19 \ |a + 8 a + 51/2 a + 41 a + ---- a + --- a + ---- a + --- a + 1/1944| \ 108 27 1944 972 / \ \ | / 1/3 | | / (%1 a (a + 2))) - 2 ln(2) + ln(a (a + 2)) + 2| / / / 1/2 %1 := 6 3 a (a + 2) 2 2 4 3 2 3 4 1/2 (-(1 + 2 a) (2 a + 3) (9 a + 36 a + 74 a + 76 a + 2) (a + 1) ) + 2 / 16 15 14 13 24757 12 11 1259712 (a + 1) |a + 16 a + 115 a + 490 a + ----- a + 8018/3 a \ 18 2382761 10 1166365 9 19318909 8 1146325 7 25771639 6 + ------- a + ------- a + -------- a + ------- a + -------- a 648 324 7776 972 69984 817135 5 8937953 4 96245 3 1205 2 19 \ + ------ a + ------- a + ------ a + ------ a - ------ a - 1/1259712| 11664 1259712 314928 209952 314928 / and in Maple format (2*ln(2)+ln(3)-ln((108*a^6+648*a^5+1440*a^4+1440*a^3+614*a^2+76*a-1+(11664*a^12 +139968*a^11+730944*a^10+2177280*a^9+4072464*a^8+4959360*a^7+3940188*a^6+ 1985832*a^5+592933*a^4+90676*a^3+4974*a^2+52*a+(6*3^(1/2)*a*(a+2)*(-(1+2*a)^2*( 2*a+3)^2*(9*a^4+36*a^3+74*a^2+76*a+2)^3*(a+1)^4)^(1/2)+1259712*(a+1)^2*(a^16+16 *a^15+115*a^14+490*a^13+24757/18*a^12+8018/3*a^11+2382761/648*a^10+1166365/324* a^9+19318909/7776*a^8+1146325/972*a^7+25771639/69984*a^6+817135/11664*a^5+ 8937953/1259712*a^4+96245/314928*a^3+1205/209952*a^2-19/314928*a-1/1259712))^(2 /3)+1)/(6*3^(1/2)*a*(a+2)*(-(1+2*a)^2*(2*a+3)^2*(9*a^4+36*a^3+74*a^2+76*a+2)^3* (a+1)^4)^(1/2)+1259712*(a+1)^2*(a^16+16*a^15+115*a^14+490*a^13+24757/18*a^12+ 8018/3*a^11+2382761/648*a^10+1166365/324*a^9+19318909/7776*a^8+1146325/972*a^7+ 25771639/69984*a^6+817135/11664*a^5+8937953/1259712*a^4+96245/314928*a^3+1205/ 209952*a^2-19/314928*a-1/1259712))^(1/3))/a/(a+2))-ln(a*(a+2))-2)/(ln(1944*I/(6 *3^(1/2)*a*(a+2)*(-(1+2*a)^2*(2*a+3)^2*(9*a^4+36*a^3+74*a^2+76*a+2)^3*(a+1)^4)^ (1/2)+1259712*(a+1)^2*(a^16+16*a^15+115*a^14+490*a^13+24757/18*a^12+8018/3*a^11 +2382761/648*a^10+1166365/324*a^9+19318909/7776*a^8+1146325/972*a^7+25771639/ 69984*a^6+817135/11664*a^5+8937953/1259712*a^4+96245/314928*a^3+1205/209952*a^2 -19/314928*a-1/1259712))^(1/3)*((-1/54*I*a^6-1/9*I*a^5-20/81*I*a^4-20/81*I*a^3-\ 307/2916*I*a^2-19/1458*I*a+1/5832*I)*(6*3^(1/2)*a*(a+2)*(-(1+2*a)^2*(2*a+3)^2*( 9*a^4+36*a^3+74*a^2+76*a+2)^3*(a+1)^4)^(1/2)+1259712*(a+1)^2*(a^16+16*a^15+115* a^14+490*a^13+24757/18*a^12+8018/3*a^11+2382761/648*a^10+1166365/324*a^9+ 19318909/7776*a^8+1146325/972*a^7+25771639/69984*a^6+817135/11664*a^5+8937953/ 1259712*a^4+96245/314928*a^3+1205/209952*a^2-19/314928*a-1/1259712))^(1/3)+(1/ 11664*I-1/11664*3^(1/2))*(6*3^(1/2)*a*(a+2)*(-(1+2*a)^2*(2*a+3)^2*(9*a^4+36*a^3 +74*a^2+76*a+2)^3*(a+1)^4)^(1/2)+1259712*(a+1)^2*(a^16+16*a^15+115*a^14+490*a^ 13+24757/18*a^12+8018/3*a^11+2382761/648*a^10+1166365/324*a^9+19318909/7776*a^8 +1146325/972*a^7+25771639/69984*a^6+817135/11664*a^5+8937953/1259712*a^4+96245/ 314928*a^3+1205/209952*a^2-19/314928*a-1/1259712))^(2/3)+(I+3^(1/2))*(a^2+2*a+1 /6)*(a+1)^2*(a^8+8*a^7+51/2*a^6+41*a^5+3733/108*a^4+385/27*a^3+4411/1944*a^2+19 /972*a+1/1944))/a/(a+2))-2*ln(2)+ln(a*(a+2))+2) On the other hand, if a is odd and >=3 then nu can be taken as / / 6 5 4 3 2 |ln(3) - ln(|108 a + 648 a + 1440 a + 1440 a + 614 a + 76 a - 1 + ( \ \ 12 11 10 9 8 7 11664 a + 139968 a + 730944 a + 2177280 a + 4072464 a + 4959360 a 6 5 4 3 2 + 3940188 a + 1985832 a + 592933 a + 90676 a + 4974 a + 52 a / (2/3) / 1/3\ \ / | + %1 + 1) / %1 |/(a (a + 2))) - ln(a (a + 2)) - 2| / |ln(1944 I / / / / \ / |/ || \\ 6 5 20 4 20 3 307 2 19 -1/54 I a - 1/9 I a - -- I a - -- I a - ---- I a - ---- I a + 1/5832 I 81 81 2916 1458 / 1/2 \ \ (1/3) | 3 | (2/3) 1/2 2 | %1 + |1/11664 I - -----| %1 + (I + 3 ) (a + 2 a + 1/6) / \ 11664/ 2 (a + 1) / 8 7 6 5 3733 4 385 3 4411 2 19 \ |a + 8 a + 51/2 a + 41 a + ---- a + --- a + ---- a + --- a + 1/1944| \ 108 27 1944 972 / \ \ | / 1/3 | | / (%1 a (a + 2))) + ln(a (a + 2)) + 2| / / / 1/2 %1 := 6 3 a (a + 2) 2 2 4 3 2 3 4 1/2 (-(1 + 2 a) (2 a + 3) (9 a + 36 a + 74 a + 76 a + 2) (a + 1) ) + 2 / 16 15 14 13 24757 12 11 1259712 (a + 1) |a + 16 a + 115 a + 490 a + ----- a + 8018/3 a \ 18 2382761 10 1166365 9 19318909 8 1146325 7 25771639 6 + ------- a + ------- a + -------- a + ------- a + -------- a 648 324 7776 972 69984 817135 5 8937953 4 96245 3 1205 2 19 \ + ------ a + ------- a + ------ a + ------ a - ------ a - 1/1259712| 11664 1259712 314928 209952 314928 / and in Maple format (ln(3)-ln((108*a^6+648*a^5+1440*a^4+1440*a^3+614*a^2+76*a-1+(11664*a^12+139968* a^11+730944*a^10+2177280*a^9+4072464*a^8+4959360*a^7+3940188*a^6+1985832*a^5+ 592933*a^4+90676*a^3+4974*a^2+52*a+(6*3^(1/2)*a*(a+2)*(-(1+2*a)^2*(2*a+3)^2*(9* a^4+36*a^3+74*a^2+76*a+2)^3*(a+1)^4)^(1/2)+1259712*(a+1)^2*(a^16+16*a^15+115*a^ 14+490*a^13+24757/18*a^12+8018/3*a^11+2382761/648*a^10+1166365/324*a^9+19318909 /7776*a^8+1146325/972*a^7+25771639/69984*a^6+817135/11664*a^5+8937953/1259712*a ^4+96245/314928*a^3+1205/209952*a^2-19/314928*a-1/1259712))^(2/3)+1)/(6*3^(1/2) *a*(a+2)*(-(1+2*a)^2*(2*a+3)^2*(9*a^4+36*a^3+74*a^2+76*a+2)^3*(a+1)^4)^(1/2)+ 1259712*(a+1)^2*(a^16+16*a^15+115*a^14+490*a^13+24757/18*a^12+8018/3*a^11+ 2382761/648*a^10+1166365/324*a^9+19318909/7776*a^8+1146325/972*a^7+25771639/ 69984*a^6+817135/11664*a^5+8937953/1259712*a^4+96245/314928*a^3+1205/209952*a^2 -19/314928*a-1/1259712))^(1/3))/a/(a+2))-ln(a*(a+2))-2)/(ln(1944*I/(6*3^(1/2)*a *(a+2)*(-(1+2*a)^2*(2*a+3)^2*(9*a^4+36*a^3+74*a^2+76*a+2)^3*(a+1)^4)^(1/2)+ 1259712*(a+1)^2*(a^16+16*a^15+115*a^14+490*a^13+24757/18*a^12+8018/3*a^11+ 2382761/648*a^10+1166365/324*a^9+19318909/7776*a^8+1146325/972*a^7+25771639/ 69984*a^6+817135/11664*a^5+8937953/1259712*a^4+96245/314928*a^3+1205/209952*a^2 -19/314928*a-1/1259712))^(1/3)*((-1/54*I*a^6-1/9*I*a^5-20/81*I*a^4-20/81*I*a^3-\ 307/2916*I*a^2-19/1458*I*a+1/5832*I)*(6*3^(1/2)*a*(a+2)*(-(1+2*a)^2*(2*a+3)^2*( 9*a^4+36*a^3+74*a^2+76*a+2)^3*(a+1)^4)^(1/2)+1259712*(a+1)^2*(a^16+16*a^15+115* a^14+490*a^13+24757/18*a^12+8018/3*a^11+2382761/648*a^10+1166365/324*a^9+ 19318909/7776*a^8+1146325/972*a^7+25771639/69984*a^6+817135/11664*a^5+8937953/ 1259712*a^4+96245/314928*a^3+1205/209952*a^2-19/314928*a-1/1259712))^(1/3)+(1/ 11664*I-1/11664*3^(1/2))*(6*3^(1/2)*a*(a+2)*(-(1+2*a)^2*(2*a+3)^2*(9*a^4+36*a^3 +74*a^2+76*a+2)^3*(a+1)^4)^(1/2)+1259712*(a+1)^2*(a^16+16*a^15+115*a^14+490*a^ 13+24757/18*a^12+8018/3*a^11+2382761/648*a^10+1166365/324*a^9+19318909/7776*a^8 +1146325/972*a^7+25771639/69984*a^6+817135/11664*a^5+8937953/1259712*a^4+96245/ 314928*a^3+1205/209952*a^2-19/314928*a-1/1259712))^(2/3)+(I+3^(1/2))*(a^2+2*a+1 /6)*(a+1)^2*(a^8+8*a^7+51/2*a^6+41*a^5+3733/108*a^4+385/27*a^3+4411/1944*a^2+19 /972*a+1/1944))/a/(a+2))+ln(a*(a+2))+2) When a=1, then nu can be taken as 20.018720479288865664 -------------------------------------- Proof: Consider the two integrals 1 + 2 a / (2 n) 2 2 n 2 2 n | x (x - (1 + 2 a) ) (x - (2 a + 3) ) E1(n) = | -------------------------------------------- dx | 2 2 2 (2 n + 1) / (x - (1 + 2 a) (2 a + 3) ) 0 2 a + 3 / (2 n) 2 2 n 2 2 n | x (x - (1 + 2 a) ) (x - (2 a + 3) ) E2(n) = | -------------------------------------------- dx | 2 2 2 (2 n + 1) / (x - (1 + 2 a) (2 a + 3) ) 0 It is readily seen that we can write a + 1 E1(n) = A1(n) + B(n) ln(-----) a + 2 a E2(n) = A2(n) + B(n) ln(-----) a + 1 For THREE sequences of RATIONAL numbers A1(n), A2(n), B(n) . Note the crucial fact that the coefficient of log((a+1)/(a+2)) in E1(n) and the coefficient of log(a/(a+1)) in E2(n) are identical, but of course, A1(n), A2(n) are different. Using the amazing Almkvist-Zeilberger algorithm (using a different Maple package called EKHAD.txt), it emerges as that both E1(n) and E2(n), and hence also the sequences of rational numbers A1(n), A2(n), B(n) satisfy the third order linear recurrence equation 10 10 6 4 2 -65536 (1 + 2 a) (2 a + 3) (2 n + 1) (n + 2) (n + 1) (a + 1) (72 a n 4 3 2 4 3 2 2 3 + 312 a n + 288 a n + 306 a + 1248 a n + 592 a n + 1224 a 2 2 2 2 + 2638 a n + 608 a n + 2721 a + 2780 a n + 16 n + 2994 a + 76 n + 84) 10 10 6 8 4 F(n) - 32768 (n + 2) (1 + 2 a) (2 a + 3) (a + 1) (1152 a n 8 3 7 4 8 2 7 3 6 4 + 7296 a n + 9216 a n + 15744 a n + 58368 a n + 24736 a n 8 7 2 6 3 5 4 8 + 13536 a n + 125952 a n + 157824 a n + 19392 a n + 3672 a 7 6 2 5 3 4 4 7 + 108288 a n + 343664 a n + 129792 a n - 58880 a n + 29376 a 6 5 2 4 3 3 4 6 + 298080 a n + 298656 a n - 366696 a n - 192768 a n + 81306 a 5 4 2 3 3 2 4 5 + 272448 a n - 781708 a n - 1242528 a n - 212544 a n + 76572 a 4 3 2 2 3 4 4 - 683910 a n - 2766768 a n - 1387680 a n - 86144 a n - 208845 a 3 2 2 3 4 3 - 2530584 a n - 3140560 a n - 566784 a n - 2176 n - 797508 a 2 2 3 2 - 2925704 a n - 1294656 a n - 14688 n - 941568 a - 1218784 a n 2 - 34128 n - 397272 a - 32504 n - 10680) F(n + 1) + 32768 (2 n + 3) 10 10 6 10 4 10 3 (1 + 2 a) (2 a + 3) (a + 1) (15552 a n + 114048 a n 9 4 10 2 9 3 8 4 + 155520 a n + 291168 a n + 1140480 a n + 708480 a n 10 9 2 8 3 7 4 10 + 294048 a n + 2911680 a n + 5211216 a n + 1935360 a n + 87516 a 9 8 2 7 3 6 4 + 2940480 a n + 13371240 a n + 14318208 a n + 3414240 a n 9 8 7 2 6 3 + 875160 a + 13634412 a n + 37089600 a n + 25436832 a n 5 4 8 7 6 2 + 3841344 a n + 4162158 a + 38503776 a n + 66639432 a n 5 3 4 4 7 6 + 28806336 a n + 2567728 a n + 12293424 a + 70623576 a n 5 2 4 3 3 4 6 + 76247088 a n + 19357000 a n + 882112 a n + 23665824 a 5 4 2 3 3 2 4 + 82288848 a n + 51647804 a n + 6677536 a n + 110880 a n 5 4 3 2 2 3 4 + 28697760 a + 56507480 a n + 17921456 a n + 843252 a n + 1216 a n 4 3 2 2 3 4 + 20273811 a + 19793312 a n + 2274902 a n + 9640 a n - 32 n 3 2 2 3 2 + 7233612 a + 2529538 a n + 27084 a n - 248 n + 934680 a + 31300 a n 2 10 10 - 680 n + 11928 a - 768 n - 288) F(n + 2) - 32768 (1 + 2 a) (2 a + 3) 6 4 2 4 (2 n + 5) (2 n + 3) (n + 3) a (a + 1) (a + 2) (72 a n + 168 a n 3 2 4 3 2 2 3 2 2 + 288 a n + 66 a + 672 a n + 592 a n + 264 a + 1454 a n + 608 a n 2 2 + 675 a + 1564 a n + 16 n + 822 a + 44 n + 24) F(n + 3) = 0 but of course with different initial conditions. Using the Poincare lemma, we can use it to deduce the growth of E1(n), E2(n), and the common growth of A1(n),A2(n), B(n). We approximate the above linear recurrence equation by a CONSTANT coefficients linear recurrence obtained by taking the leading term in n. The indicial polynomial, in N, (after normalizing the constant term to be 1) 4 3 2 1 + (4 a + 16 a - 11 a - 54 a - 34) N 6 5 4 3 2 2 + (-108 a - 648 a - 1440 a - 1440 a - 614 a - 76 a + 1) N 3 + a (a + 2) N and in Maple format 1+(4*a^4+16*a^3-11*a^2-54*a-34)*N+(-108*a^6-648*a^5-1440*a^4-1440*a^3-614*a^2-76*a+1)*N^2+a*(a+2)*N^3 Let C1(a) be the largest root, C2(a), the second largest root, and C3(a) the smallest rule of the above equation that can be expressed explicitly thanks to Cardano. It follows from the Poincare lemma that, ignoring polynomial corrections that disappear at the end A1(n),A2(n),B(n)=OMEGA(C1(a)^n) E1(n)=OMEGA(C2(a)^n), E2(n)=OMEGA(C3(a)^n) or the other way E2(n)=OMEGA(C3(a)^n), E1(n)=OMEGA(C2(a)^n) Note that C1(a) is big and C2(a), C3(a) are small. We now need three divisibility lemmas that we leave to the reader Lemma 1: B(n)*lcm(1..2*n) is always an integer Lemma 2: A1(n)*K1(a)^n*lcm(1..2*n) is always an integer (or at worst has bounded denominator) where K1(a)= a+2 if a is odd, and K1(a)=a/2+1 if a is even Lemma 3: A2(n)*K2(a)^n*lcm(1..2*n) is always an integer (or at worst has bounded denominator) where K2(a)= a if a is odd, and K2(a)=a/2 if a is even Let K(a)=K1(a)*K2(a) In other words, K(a)=a(a+2) if a is odd, and K(a)=(a/2)(a/2+1) if a is even Multiplying the equations a + 1 E1(n) = A1(n) + B(n) ln(-----) a + 2 a E2(n) = A2(n) + B(n) ln(-----) a + 1 by K(a)^n*lcm(1..2*n), and letting E1new(n)=K(a)^n*lcm(1..2*n)*E1(n), E2ne\ w(n)=K(a)^n*lcm(1..2*n)*E2(n) We have E1new(n)=A1new(n)+Bnew(n)*log((a+1)/(a+2)) E2new(n)=A2new(n)+Bnew(n)*log((a/(a+1))) where A1new(n), A2new(n), Bnew(n) are INTEGER sequences Recall the fact that lcm(1..n) is asymptotic to e^n and hence lcm(1..2*n) is asymptotic to e^(2n) . We now that A1new(n), A2new(n), Bnew(n) are, up to polynomial factors that disappear at the wash A1new(n),A2new(n),Bnew(n)=OMEGA((C1(a)*e^2*K(a))^n ) E1new(n), E2new(n) =OMEGA((C3(a)*e^2*K(a))^n Finally, using Lemma 3 in Salikhov's paper, that is due to M. Hata (Acta Arith. 63(4), 335-349 (1993)), We get that one can take nu to be ln(C1(a)) + ln(K(a)) + 2 - ------------------------ ln(C2(a)) + ln(K(a)) + 2 plugging for C1(a), C2(a) their values given by the Cardano formulas and K(a)=a(a+2) if a is odd, and K(a)=(a/2)(a/2+1) if a is even we get the expressions in the statement of the theorem. QED. Finally let's plot the independce measure in terms of a, from a=1 to a=1000. The higher curve is for the odd case + HHHHHHHHHHHHHH 35+ HHHHHHHHHHHHHHHHHHH + HHHHHHHHHHHHHH + HHHHHHHHHH 30+ HHHHHHH + HHHHH + HHHH 25+ HHH + HHH + HH 20+HH +H +H HHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH 15*H HHHHHHHHHHHHHHHHHHHH * HHHHHHHHHH * HHHHH 10* HH *H *H 5 * *--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+ 200 400 600 800 1000