Questions and Answers #47 Fractions

Questions and Answers Sheet 47 Fractions Question #1 2 = 2 / (2 / (2 / ())) ? Can we say that We have  = 1+ (with  1  1 = 1+ 1+ 1 1 = 1+  = 1 1+ 1+ 1  being the Golden Ratio) Which gives us the confirmed infinite fraction 1  = 1+ 1 1+ 1+ 1 We also have 2= 2 2 2 = = = 2 2 2 2 2 2 So by analogy we can deduce that 2= 2 2 2 2 The sequence (an )n + such that a1 = 2, an+1 = 2 gives lim an = 2 , so indeed the representation n→+ an should be correct. 2 , it is 1 1 1 2 = 1 + ( 2 − 1) = 1 + = 1+ = 1+ = 1 1 1+ 2 2+ 2+ 1 1+ 2 2+ 1+ 2 Everywhere on the Internet that I see a continued fraction of Why haven't I seen the representation 2 = 2 / (2 / (2 / (2 / ))) mentioned, and see the above used instead? Is something wrong about my representation? I would say it isn't mentioned because it is not useful: you cannot approximate 2 using the representation, unlike you can using the above one (which you can do by removing n any member). Answer: You have: 2 := 2 2 2 2 = 2 2  2 2 2 2 Which is not a very helpful definition. Question #2 Fraction Sum Series This question was asked in (selection) IMO for 8th graders. 1/ 2 +1/ 6 +1/12 +1/ 20 +1/ 30 +1/ 42 +1/ 56 +1/ 72 +1/ 90 +1/110 +1/132 I have noticed that it can be written as 1/ (1  2) + 1/ (2  3) + 1/ (3  4) + 1/ (4  5).... + 1/ (11 12) However I don't know how to continue.. Answer: Hint: Your sum can be written in the following form: 11 1  n(n + 1) n =1 Apply partial fraction decomposition to A B 1 + to write it in the form and see what happens n n +1 n(n + 1) as you simplify. The series should telescope. I.e. adjacent terms will cancel with one another, leaving you with only the 1 1 1 = − . The series then looks like n(n + 1) n n + 1 1 1 1 − + − 11 11 12 first and last terms not cancelled. You have 1 1 1 1 1 1 1 − + − + − + − 1 2 2 3 3 4 4 Question #3 Express w1/ w for w = 2+ 3 in the simplest form with a rational denominator 5− 3 Express the following in the simplest form (with a rational denominator) i) w ii) 1 w I'm confused about (ii) question pls help me. Answer: If w = 1 5− 3 2+ 3 = , then If you can do i, you should be able to do ii. The process is the w 2+ 3 5− 3 same. Question #4 Calculate X of a math problem I am trying to learn some more math and I got stuck on this: 0.2 = 140 X How do I calculate X? EDIT Sorry I meant to calculate 28 = 140 X So that X = 0.2 but how do I calculate it? Answer: 0.2 = 140 X 0.2  X = 140  X X  0.2 = 140 X 1 1  0.2  = 140 X  140 140 0.2  =X 140 0.2 10 2 1 X= = = . 140 10 1400 700  P.S. 1 : Multiplying the both sides of 0.2  X = 140  X X i.e. 0.2 = 140 X . P.S. 2 : Multiplying the both sides of 0.2 = 140 by X gives us X 28 = 140 X by X gives us 28  X = 140  X X i.e. 28 = 140 X . Then, multiplying the both sides by 28  1 gives us 140 1 1 = 140 X  140 140 i.e. 28 = X. 140 Then, we have X= 28 28 1 = = = 0.2. 140 28  5 5 Question #5 Prove that a b c d 2 + + +  . b + 2c + 3d c + 2d + 3a d + 2a + 3b a + 2b + 3c 3 I was thinking of trying a  b  c since the inequality is cyclic. I am not sure how to use this, though, or if this would help simplify the inequality. It also doesn't look like I can really use AM-GM or CauchySchwarz. Answer: We need only note that 𝑎 𝑏 𝑐 𝑑 + + + 𝑏 + 2𝑐 + 3𝑑 𝑐 + 2𝑑 + 3𝑎 𝑑 + 2𝑎 + 3𝑏 𝑎 + 2𝑏 + 3𝑐 = 𝑎2 𝑏2 𝑐2 𝑑2 + + + 𝑎𝑏 + 2𝑎𝑐 + 3𝑎𝑑 𝑏𝑐 + 2𝑑𝑏 + 3𝑎𝑏 𝑐𝑑 + 2𝑎𝑐 + 3𝑏𝑐 𝑎𝑑 + 2𝑏𝑑 + 3𝑐𝑑 ≥ (𝑎 + 𝑏 + 𝑐 + 𝑑)2 4(𝑎𝑏 + 𝑎𝑐 + 𝑎𝑑 + 𝑏𝑐 + 𝑏𝑑 + 𝑐𝑑) = 𝑎2 + 𝑏 2 + 𝑐 2 + 𝑑 2 + 2(𝑎𝑏 + 𝑎𝑐 + 𝑎𝑑 + 𝑏𝑐 + 𝑏𝑑 + 𝑐𝑑) 4(𝑎𝑏 + 𝑎𝑐 + 𝑎𝑑 + 𝑏𝑐 + 𝑏𝑑 + 𝑐𝑑) 2 (𝑎𝑏 + 𝑎𝑐 + 𝑎𝑑 + 𝑏𝑐 + 𝑏𝑑 + 𝑐𝑑) + 2(𝑎𝑏 + 𝑎𝑐 + 𝑎𝑑 + 𝑏𝑐 + 𝑏𝑑 + 𝑐𝑑) ≥3 4(𝑎𝑏 + 𝑎𝑐 + 𝑎𝑑 + 𝑏𝑐 + 𝑏𝑑 + 𝑐𝑑) = 2 3 Question #6 Absolute values in logarithms in a solution of differential equation How have the moduli signs disappeared in the following step: 1 ( ln | g + kv | − ln | g + ku |) = −t k Therefore  g + kv  ln   = −kt g + ku   gk and u are positive constants. t is time, v is velocity. Context: the above calculations are from solving the equation dv / dt = − g − kv given that v = u when t = 0 , and that ug and k are positive constants. Answer: g + kv need not be positive, but it will have the same sign as g + ku , because the solutions cannot cross the equilibrium at v = − g / k . Hence, the quotient inside the logarithm is positive. 1 ( ln | g + kv | − ln | g + ku |) = −t at all; there is a cleaner way of solving k − kt this ODE. Namely, one goes from ln | g + kv |= − kt + C (with indefinite constant C ) to g + kv = Ae , Ideally, you would not arrive at where A takes the place of  e C Question #7 If the chance of an event was 1/128 and increased by 20%, what is the new chance? So I have something that has a 1/128 chance of occurring, let's say. Suddenly, the chances of that thing happening are increased by 20%. How is that fraction written? Would you multiply 1/128 by 6/5 (yielding 3/320), or would you take 128, multiply it by 4/5, and then invert it (1/102)? This isn't for homework, I'm merely curious. Answer: 6 / 5 = 1.2 is the correct approach. This increases a value by 20% . Multiplying by 4 / 5 decreases a value by 20% , but inverting that is not the same as increasing by 20% . It results in 4 multiplying by 5 / 4 , which is an increase of 25% . On top of that, 128  = 102.4, not 102 5 Multiplying by Question #8 How do you solve B and C for 𝐴= s −1 s A Bs + C = + 2 ? 2 s +1 s +1 s +1 s +1 𝑠2 − 𝑠 1 − (−1) |𝑠=−1 = =1 2 𝑠 +1 1+1 Answer: We have s −1 s A Bs + C = + 2 2 s +1 s +1 s +1 s +1 (1) Multiplying (1) by s and making s →  gives 1= A+ B from which Making B=0 s = 0 in (1) gives 0 = A+C from which C = −1 Question #9 How does this seemingly-trivial simplification work? In a section on inductive proofs in the book Modelling Computing Systems: Mathematics for Computer Science (Muller, Struth) there is a simplification that is assumed to be trivial, but that I can't figure out. It occurs in this step: ? ( k + 1)( k + 2)( k + 3) k (k + 1)(k + 2) + (k + 1)(k + 2) = 3 3 How does one get from the first expression to the second? Answer: k (k + 1)(k + 2) k (k + 1)(k + 2) + 3(k + 1)( k + 2) + (k + 1)(k + 2) = 3 3 (k + 1)(k + 2)(k + 3) = 3 Question #10 How do I solve this fraction addition problem? 2 13 5 1 4 + −9 yeilds result of −5 but WolframAlpha says the answer is −5 9 18 18 2 Answer: In −9 1 the negative sign applies to the whole quantity. Thus the calculation should be 2 2 1 38 19 76 171 95 5 4 + (−9 ) = + (− ) = − = − = −5 9 2 9 2 18 18 18 18 Question #11 Adding fractions (to find an equal) without known denominator The question is "Find v ": 1 1 1 = + 20 30 v I have no idea what I'm really doing so if someone could explain in a somewhat easy-to-understand way I'd really appreciate it! Answer: Hint: 1 1 1 = − v 20 30 then find the inverse of 1 . v Question #12 How do you add two fractions? I have a fraction I am trying to solve. I know the answer already, as Wolfram says it is 143 . 300 The fraction is: 5 3 + = ? 12 50 Please explain why and how your method works. Answer: Null has given you a good way. Here's a way without worrying about the LCM: a c ad + bc + = b d bd In the example, 5 3 (5)(50) + (12)(3) 286 143 + = = = 12 50 (12)(50) 600 300 The price of not worrying about the LCM is that you get an answer, that isn't in lowest terms, so you have the extra step at the end of reducing the fraction. Question #13 How to prove that ln12 is irrational witout using the change of base rule? ln18 I have to show that ln12 is irrational by using change of base rule. ln18 At the beginning I have proved that ln r is irrational for any rational rr  1 . Then using this we can say that ln12 and ln18 are irrational. But from here it is difficult for me to show that the fraction is irrational knowing that both the numerator and the denominator are irrational. Answer: You can't prove it using that the quotient of irrationals is irrational for the simple reason that the statement is false. You may instead use a different strategy: suppose ln12 2 ln 2 + ln 3 a = = ln18 ln 2 + 2 ln 3 b for positive and coprime integers a and b. Then ln12 2 ln 2 + ln 3 a = = ln18 ln 2 + 2 ln 3 b that becomes (2b − a ) ln 2 = (2a − b) ln 3 which tells you that ln 3 / ln 2 is rational as well. By the change of base rule, this is the same as saying that log 2 3 is rational, so log 2 3 = h k for positive integers h and k. Therefore log 2 3 = h k Or 3 = 2 that's impossible because of unique factorization of integers. k h Question #14 Find x in log x = (log x) 2 2 I couldn't find x. Answer: log x 2 = (log x)2 2log x = (log x)2 log x(log x − 2) = 0log x = 0 → x = 1log x = 2 → x = 102 Question #15 How to calculate value of expressions when a = 22 Round the answer to three significant figures: 77 3a for this one I am not sure if I do 77 77 = 1.17 or = 56 3(22) 3(22) Sorry if this is written in a confusing way. Answer: 77 means you will just replace a in the formula with c 22 . 3a Therefore, 77 or, when rounded to 3 significant figures, 1.17 3  22 Question #16 If x + y + z = 0 , prove that x2 y2 z2 + + =1 2 x 2 + yz 2 y 2 + zx 2 z 2 + xy A problem in my homework had asked me: When x + y + z = 0 , evaluate x2 y2 z2 + + 2 x 2 + yz 2 y 2 + zx 2 z 2 + xy Without too much difficulty, one can see that the value should be 1 using ( x, y, z ) = (1, 0, −1) I decided to use x = − y − z , which turned out not to be as difficult as initially thought. However, would someone care to enlighten me to some other methods of doing this? Answer: HINT: 2 x 2 + yz = 2( y + z ) 2 + yz = (2 y + z )( y + 2 z ) Now 2 y + z = x + y + z + y − x = y − x Question #17 How to calculate the Integer portion of a fraction using only +, -,  and *? I made something in excel that calculates the days left until a given date, and from that how many weeks were left. I had it so that 9 days displayed as 1.2 using this formula:  A  A   B − B B  A    +  10 B Where B is the "base" you are counting in and A is the number you are trying to count. This all works and the question is purely for personal interest only. So, Is there a way to calculate the Integer portion of a fraction only using the operators +, -,  and *? Answer: You can take advantage of the fact that modulo is the same as the quotient operator for positive values and then sign correct the result. Not the most elegant solution but something like this should work.  | A| −| A|%| B |  , for sgn(A) = sgn(B)   |B|   A   where sgn is the signum function.  B  =  | A | − | A | % | B | − − 1, for sgn(A)  sgn(B)    |B| Question #18 Partial Fraction Decomposition with complex poles I have a function which I'd like to perform partial fraction decomposition on, to allow easier inverse laplace transform. F ( s) = 1 s ( s + 140s + 104 ) 2 I begin with finding the poles s = 0, s = −70  j 10 51 To which I then try putting F ( s ) in this form: F ( s) = A B B + + s s − 70 + j 10 51 s − 70 − j 10 51 A B B Because one unknown ( F ( s) = + ) is just the complex + s s − 70 + j 10 51 s − 70 − j 10 51 conjugate of B, I only need to find out what A, and B is. A = F (0) = 10−4 B = F (−70 + j 10 51)  6.31514 109 + j  6.44274 109 Where the last step was done in Mathematica. Answer, according to several other people, is supposed to be B = −70 + j  71.41 , which looks a lot nicer, but I'm not sure HOW they got to that answer. Answer: Note that the partial fraction expansion of F ( s ) is: F ( s) = where A B B + + s s − s0 s − s0 s0 = −70 + i10 51 and s0 = s0 = −70 − i10 51 are the complex-conjugate roots of s 2 + 140 s + 10 4 Note that we have A = lim sF ( s ) s →0 = 10−4 and B = lim( s − s0 ) F ( s ) s → s0 = 1 s0 ( s0 − s0 ) = −70 − i10 51 i 20 51104 = −51 + i 7 51 1020000 Question #19 What does Pi equal to What is the approximation of pi in a fraction form. I am very curious to know what it is. I have been seeing pi almost everywhere. Answer: The idea here is that you can compute the continued fraction of achieve an approximation of  = 3.141592653589  and then truncate it somewhere to 1  = 3+ 1 7+ 1 15 + 1 1+ 1 292 + 1 1+ 1 1+ 1+ 1 2+ The further along you truncate it, the more accurate your approximation. If we cut it off pretty early (at 7), we get the classic   3+ 22 approximation: 7 1 22 = = 3.1428571428571 7 7 If we cut it off later (say at 292), we get a better approximation:  1  3+ 1 7+ 1 15 + 1+ 1 292 1  3+ 1 292 15 + 293 1  3+ 293 7+ 4687 4687  3+ 33102 103993  = 3.1415926530119 33102 7+  You can use this method to get a rational number that is as close to π as you need. Question #20 Is there an example to demonstrate why 1 equals 2? (1 / 2) Answer: I know you are looking for something about pizzas but I think fractions in general work better if you work with a definition more akin to the axiomatic definition. I mean what is 1  in terms of pizzas? 1 as the number, which when multiplied by x, gives me 1. This works fine for pizzas x 1 because, e.g. 2  = 1 two half slices is the whole. 2 I would define Now whatever 1 1 is, if I multiply it by I get one so that 1 2 2 1 1  = 1... 1 2 2 but this is nothing other than two... (or multiply both sides of the equation on the left by two).