Questions and Answers #11 Logarithms

Questions and Answers Sheet 11 Logarithms Question #1 ln(1+3𝑥 ) Find lim 𝑥→−∞ ln(1+2𝑥 ) Answer: HINT: 𝑧 ≤ log(1 + 𝑧) ≤ 𝑧(1) 𝑧+1 for 𝑧 ≻ 1. Then, use the squeeze theorem. NOTE 1: We can prove the inequalities in (1) using standard tools, which don't require the use of derivatives. We need only to note that 𝑒 𝑥 ≥ 1 + 𝑥(2) For x>-1 , we take the log of both sides to reveal log(1 + 𝑥) ≤ 𝑥 𝑧 which is the right-hand side inequality of (1). Likewise, if we make the substitution 𝑥 = − (𝑧+1) in (2), we obtain 𝑧 𝑒 −(1+𝑧) ≥1− 𝑧 𝑧+1 = 1 𝑧+1 whereupon rearranging yields for 𝑧 ≻ 1 𝑧 log(1 + 𝑧) ≥ 𝑧+1 as was to be shown. NOTE 2: We can prove the inequality given in (2) using the limit definition of 𝑒 𝑥 𝑥 𝑛 𝑒 𝑥 = lim (1 + ) 𝑛→∞ 𝑛 𝑥 𝑛 I showed in THIS ANSWER, without the use of derivatives, that (1 + ) is a monotonically increasing 𝑛 function of n (If x<0 , this is true for n>|x| ). 𝑥 𝑛 Then, we have 𝑒 𝑥 ≥ (1 + ) ≥ 1 + 𝑥 𝑛 where this last inequality follows from Bernoulli's inequality. Question #2 2 Proof that 3 < log(2) < Positive integrals 1 2𝑥(1−𝑥)2 ∫0 1+𝑥 2 𝑑𝑥 = 𝜋 − 3 and 1 𝑥 4 (1−𝑥)4 ∫0 1+𝑥 2 𝑑𝑥 = prove that 3<𝜋< 22 7 22 7 − 𝜋 7 10 Is there a similar argument for the following log(2) inequality? 2 7 < log(2) < 3 10 Answer: 2 7 There are positive integrals that relate log(2) to its first four convergents: 0,1, , . 3 10 1 2𝑥 = log(2) ∫0 1+𝑥 2 1 (1−𝑥)2 ∫0 1+𝑥 2 𝑑𝑥 = 1 − log(2) 1 𝑥 2 (1−𝑥)2 ∫0 1+𝑥 2 𝑑𝑥 1 𝑥 4 (1−𝑥)2 7 ∫0 1+𝑥 2 𝑑𝑥 = 10 − log(2) Therefore, 1 𝑥 2 (1−𝑥)2 − ∫0 2 1+𝑥 2 1 𝑥 4 (1−𝑥)2 < 0 < ∫0 − log(2) < 0 < 3 2 < log(2) < 3 7 7 10 1+𝑥 2 𝑑𝑥 − log(2) 10 A similar set is available with denominators (1+x) : 1 1 ∫0 𝑑𝑥 = log(2) 1+𝑥 1 𝑥 ∫0 1+𝑥 𝑑𝑥 = 1 − log(2) 1 1 𝑥 2 (1−𝑥) 2 ∫ 1+𝑥 𝑑𝑥 = log(2) − 3 2 0 1 1 𝑥 5 (1−𝑥) 7 ∫ 1+𝑥 𝑑𝑥 = 10 − log(2) 2 0 and series versions are given by 2 1 log(2) − = ∑∞ 𝑘=1 (2𝑘+1)(2𝑘+2)(2𝑘+3) 3 7 10 1 − log(2) = ∑∞ 𝑘=2 (2𝑘+2)(2𝑘+3)(2𝑘+4) Question #3 Information on the sum ∑∞ 𝑛=1 = log 𝑛 𝑛! Answer: Using Dobii's formula for Bell numbers, we have 1 𝐵(𝑛) = ∑∞ 𝑘=0 Hence, ∑∞ 𝑘=1 𝑒 𝑑 𝑑𝑛 log 𝑘 𝑘! 𝑘𝑛 𝑘! 1 𝑘 𝑛 log 𝑘 𝑒 𝑘! 𝐵(𝑛) = ∑∞ 𝑘=2 hence, = 𝐵0′ 𝑒 Note that the first term (k=1) is 0 . Question #4 How can I prove that 𝑥 − 𝑥2 2 < ln(1 + 𝑥) Answer: Hint: Prove that ln(1 + 𝑥) − 𝑥 + 𝑥2 2 is strictly increasing for x>0 edit: to see why this isn't a complete proof, consider 𝑥 2 − 1 for x>0. It's strictly increasing; does that show that 𝑥 2 > 1 ? I hope not, because it's not true! Question #5 Prove an algorithm for logarithmic mean 𝑎0 −𝑏0 lim 𝑎𝑛 = lim 𝑏𝑛 = 𝑛→∞ ln 𝑎0 −ln 𝑏0 𝑛→∞ Answer: You have several exact identities, such as 𝑎𝑛 + 𝑏𝑛 𝑎𝑛+1 − 𝑏𝑛+1 = ⇒ 𝑎𝑛 − 𝑏𝑛 = 2−𝑛 (𝑎0 − 𝑏0 ), 2 𝑎𝑛+1 𝑏𝑛+1 𝑎𝑛 =√ 𝑏𝑛 ⟹ 𝑎𝑛 𝑎 = ( 0) 𝑏𝑛 2−𝑛 𝑏0 The first tells us that if one of the sequences has a limit the other has the some limit. In combination bn can be eliminated to get, via mean value theorem, 𝑎𝑛 = 2−𝑛 (𝑎0 −𝑏0 ) 2−𝑛 𝑏 1−( 0 ) = 𝑎0 𝑎0 −𝑏0 ln 𝑎0 −ln 𝑏0 +𝑂(2−𝑛 ) Question #6 Solve −𝐵 ln 𝑦 − 𝐴𝑦 ln 𝑦 + 𝐴𝑦 − 𝐴 = 0 for y Answer: Hint: 0 = −𝐵 log 𝑦 − 𝐴𝑦 log 𝑦 + 𝐴𝑦 − 𝐴 = log 1 𝑦𝐵 + log 1 𝑦 𝐴𝑦 + 𝐴𝑦 + 𝐴 = log 1 𝑦 𝐴𝑦+𝐵 + 𝐴𝑦 + 𝐴 This doesn't look, as you say, like having a closed form but perhaps the above, together or not with xavier's comment (Lambert function), can give you some good approximation. Question #7 1 ln(− ln 𝑥)𝑙𝑖 2 𝑥 A closed form for ∫0 𝑥 𝑑𝑥 Answer: I will start from Vladimir's result and my calculation. Let ∞ 2𝛤(𝑠) 𝐹(𝑠) = ∫0 𝑧 𝑠−1 𝐸𝑖(−𝑧)2 𝑑𝑧 = 𝑠 1 𝑢𝑠−1 ∫02 1−𝑢 𝑑𝑢. Then we have ∞ ∫0 𝑧 𝑠−1 𝐸𝑖(−𝑧)2 log 𝑧 𝑑𝑧 = 𝐹 ′ (𝑠) 1 2𝛤(𝑠) 𝑠 𝑠 = 𝐹(𝑠) (𝜓0 (𝑠) − ) + 1 ∫02 𝑢𝑠−1 log 𝑢 1−𝑢 𝑑𝑢 . Plugging s=1 to both sides, 𝐼 = 𝐹(1)(𝜓0 (1) − 1) + 2 ∫ 1 log 𝑢 20 1−𝑢 𝑑𝑢 1 = −2(𝛾 + 1) log 2 + 2[𝐿𝑖2 (1 − 𝑢)]20 = −𝜁(2) − log 2 2 − 2(𝛾 + 1) log 2 as desired. Here, 𝑧 log(1−𝑡) 𝐿𝑖2 (𝑧) = − ∫0 𝑡 𝑑𝑡 is the dilogarithm and we exploited the special values 𝐿𝑖2(1) = 𝜁(2), 1 1 1 Li2 ( ) =  (2) − log 2 2 2 2 2 Question #8 Prove 1 𝑥+1 < ln ( 𝑥+1 𝑥 1 1 𝑥 𝑥+1 ) < 0.5 ( + ) Answer: Let us recall a property of integrals, so simple one could think nothing good could come out of it, and yet: In the present case, one may start from the identity log ( (𝑥+1) 𝑥 𝑥+1 ) = ∫𝑥 𝑢(𝑡)𝑑𝑡 with 𝑢(𝑡) = 1 𝑡 and note that 1 𝑢(𝑡) ≥ (𝑥+1) for every t in the interval (x, x+1). This should give one inequality. For the other inequality, one may prove that 𝑣(𝑠) = 𝑢(𝑥 + 1/2 + 𝑠) + 𝑢(𝑥 + 1/2 − 𝑠) is an increasing function of s for s 1 in (0,1/2) and note that log((𝑥 + 1)/𝑥) = ∫ 𝑣(𝑠)𝑑𝑠. 20 Can you reach that point and finish the proof from there? Question #9 Prove by Mean Value Theorem 𝑥 1+𝑥 < ln(1 + 𝑥) < 𝑥 for 𝑥 > 0 Answer: By the mean value theorem, given x>0 , there exists 𝑐 ∈ (0, 𝑥) such that 𝑓(𝑥) − 𝑓(0) = 𝑓 ′ (𝑐)𝑥, i.e., 𝑥 ln(1 + 𝑥) = . Since 0 8 = 23 , 2 log3 5  3  2 log 2 3 hence log3 5  log 2 3. (ii) Since 3 < 4 = 22 and 11 > 9 = 32 , log 2 3 < 2 < log 3 11. Question #15 0 ln(1+𝑥) Show that ∫1 1 ln 𝑥 1 𝑑𝑥 = − ∫0 2 𝑥 1−𝑥 𝑑𝑥 without actually evaluating both integrals Answer: HINT: Note that we have 1 1 log(𝑥) ∫ 𝑑𝑥 2 0 1−𝑥 2 1𝑥 log(𝑥 ) 𝑑𝑥 1−𝑥 2 𝑥→𝑥 2 ∫0 = ⏞ 1 1 1 = ∫ log(𝑥) ( − ) 𝑑𝑥 1 − 𝑥 1 + 𝑥 0 Can you finish now? Question #16 Logarithmic Series Evaluation I was trying to generate a direct formula for this series but I am not sure whether it is possible to do so. 1 ln(1) + 2 ln(2) + 3 ln(3) + 4 ln(4) + ⋯ + (𝑛 − 1) ln(𝑛 − 1) + 𝑛 ln(𝑛) Answer: By partial summation: 1 ∑𝑛𝑘=1 𝑘 log 𝑘 = ((𝑛 + 1), (2)) log 𝑛 − ∑𝑛−1 𝑘=1 ((𝑘 + 1), (2)) log (1 + ) 𝑘 and using the approximation 1 1 1 𝑘 𝑘 2𝑘(𝑘+1) log (1 + ) = − − 1 6𝑘(𝑘+1)2 1 + 𝑂 ( 4) we get: 𝑘 6𝑛2 +6𝑛+1 ∑𝑛𝑘=1 𝑘 log 𝑘 = 12 log 𝑛 − 𝑛2 4 + 𝑂(1). In the O(1)-term hides the logarithm of the Glaisher-Kinkelin constant. Question #17 1 Integral of ∫ 𝑑𝑥 log 𝑥 Answer: If you don't like special functions symbols: 1 ∫ log 𝑥 𝑑𝑥 = ∫ ln 10 ln 𝑥 𝑑𝑥 Let 𝑢 = ln 𝑥, Then 𝑥 = 𝑒 𝑢 𝑑𝑥 = 𝑒 𝑢 𝑑𝑢 ∴∫ ln 10 =∫ = 𝑑𝑥 ln 𝑥 𝑒 𝑢 ln 10 𝑑𝑢 𝑢 𝑢𝑛 ln 10 ∫ ∑∞ 𝑛=0 𝑛!𝑢 𝑑𝑢 = ∫ ∑∞ 𝑛=0 = ∫( ln 10 𝑢 𝑢𝑛−1 ln 10 𝑛! + ∑∞ 𝑛=1 𝑑𝑢 𝑢𝑛−1 ln 10 = ln 1 0 ln 𝑢 + ∑∞ 𝑛=1 ) 𝑑𝑢 𝑛! 𝑢𝑛 ln 10 𝑛!𝑛 +𝐶 = ln 1 0 ln ln 𝑥 + ∑ 𝑛 = 1∞ (ln 𝑥)𝑛 ln 10 𝑛!𝑛 +𝐶 Question #18 Evaluate lim log(1+𝑦) 𝑦 𝑦→0 without LHR or Taylor series Answer: lim log(1+𝑦) 𝑦 𝑦→0 1 1 = lim log(1 + 𝑦)𝑦 = log lim (1 + 𝑦)𝑦 = log 𝑒 = 1 𝑦→0 𝑦→0 Question #19 How to prove monotonicity of this function? Let a>0. How to prove that the function: 𝑓(𝑥) = 𝑎𝑥−1 log(𝑎𝑥) ⋅ log 𝑥 𝑥−1 , is monotonic (depending on a<1 or a>1 ). I know that we can calculate the derivative and determine its sign, but this needs much of calculation. I'm wondering if we can decide the monotonicity using a simple trick. Answer: Consider g: → g (t ) = ln et − 1 t (where the value at 0 is defined by continuity). Then from ln 𝑓 (𝑥) = 𝑔(ln 𝑥 + ln 𝑎) − 𝑔(ln 𝑥) we know that 𝑔′′ (𝑡) = 1 ln 𝑎 is monotone for all 𝑎 ≠ 1 if g is convex. But g 's second derivative 𝑡2 cosh 𝑡−1− 2 𝑡 2 (cosh 𝑡−1) is nonnegative. Admittedly, since ∞ 𝑔(𝑡) = 𝑡 + ∑ 𝑛=1 𝐵𝑛 𝑡 𝑛 𝑛 𝑛! is itself an important elementary function, using the second derivative to reduce the result to inequalities of other elementary functions seems unsatisfactory. Maybe there's a more intrinsic route? Question #20 Express ln 𝑥 𝑦 2 as a sum, difference or multiple of logarithms. Answer: Given, ln 𝑥 𝑦 2 On simplification, we get ⇒ ln 𝑥 𝑦 2 = ln 𝑥 + ln 𝑦 2 [∵ ln(𝑥𝑦) = ln 𝑥 + ln 𝑦] = ln 𝑥 + 2 ln 𝑦 [∵ ln(𝑥 𝑦 ) = 𝑦 ln 𝑥]