Questions and Answers #4 Rational Functions

Questions and Answers Sheet 4 Rational Functions Question #1 I need to write the power series: ∑∞ 𝑛=1 1 (𝑥−3)2𝑛−1 1 − (𝑥−2)2𝑛−1 I need to write it as a rational function. Answer: In general ∑∞ 𝑛=1 1 𝑢2𝑛−1 = 𝑢 𝑢2 −1 . Your expression becomes 𝑥−3 (𝑥−3)2 −1 𝑥−2 − (𝑥−2)2 −1 = (𝑥−3)((𝑥−2)2 −1)−(𝑥−2)((𝑥−3)2 −1) ((𝑥−3)2 −1)((𝑥−2)2 −1) You could simplify it. Note that |x-3|>1 and |x-2|>1 required. Question #2 I thought it was interesting that and 1 (𝑢2 −2𝑢−1)2 𝑢2 +1 (𝑢2 −2𝑢−1)2 has the very simple integral − 𝑢 𝑢2 −2𝑢−1 but both of 𝑢2 2 (𝑢2 −𝑤𝑢−1) are very complicated (the transcendental parts cancel each other though). So my question is how do I check by looking at a rational function whether or not it's a derivative of a rational function? For example 1 (𝑥 2 +1)2 isn't but 𝑥 (𝑥 2 +1)2 is. What should I do? Answer: Examine the poles of your function (in the complex plane). If all residues are zero, you are in good shape. Question #3 I want a function that passes through the (x, y) points (3,0),(4,-1),(6,3) and (7,2). Using Lagrange interpolation I get − 𝑥 3 −15𝑥 2 +70𝑥−102 2 but using this interpolation method for rational functions I get the much simpler (𝑥−3) . This is a case where (𝑥−5) the rational function interpolation is a lot simpler than polynomial interpolation. Is there a method to get (𝑥−3) (𝑥−5) given the points above, a sort of interpolation method that uses rational functions? Answer: 𝑓(𝑥) = − 𝑔(𝑥) = 𝑥 3 −15𝑥 2 +70𝑥−102 2 𝑥−3 𝑥−5 pass through points (3,0), (4, −1), (6,3), (7,2). But what does happen if you want to interpolate close to 𝑥 = 5? 𝑓(5) = 1 but 𝑔(5± ) = ±∞. This makes a small problem, isn't ? Question #4 I wanted to find a function f(x) that has the following properties: 1.extreme values at x=-1 and x=0 and x=1 2.x -axis is asymptote No other asymptotes or extrema. I found the following two by integrating and trial and error: 𝑓1 (𝑥) = 𝑥 2 𝑒 −𝑥 2 𝑓2 (𝑥) = 2 ln(𝑥 2 + 1) + ln(𝑥 4 − 𝑥 2 + 1) How can I find one that is actually a rational function? Answer: If the x-axis has to be an asymptote, you need a denominator of higher degree than the numerator. In order to get the extreme values where you want them, suggest a rational function with sufficient parameters, e.g. suggest a function of the form: 𝑓(𝑥) = 𝑎𝑥 2 +𝑏𝑥+𝑐 𝑥 4 +1 Note that taking 𝑥 4 + 1 instead of 𝑥 3 + 1 (which is of sufficiently high degree as well) ensures no vertical asymptote is added. Finding f'(x) and requiring 𝑓 ′ (−1) = 𝑓 ′ (0) = 𝑓 ′ (1) = 0 yields b=c=0 suggesting that any function with nonzero parameter a of the following form will do: 𝑑(𝑥) = 𝑎𝑥 2 𝑥4 + 1 You can verify that indeed, taking e.g. a=1, this function satisfies your requirements. Question #5 I was trying to convert a rational function R(X) to a polynomial of type 𝑅|𝑋, 𝑋 −1 | but I failed. I searched internet and 𝑅|𝑋, 𝑋 −1 | has its own name!: "Laurent polynomials". a. Every Laurent polynomial is a combination of rational functions. Is every Laurent polynomial possible to be equal to a single rational function? b. Are there methods to convert rational functions to Laurent polynomials? How are they? Answer: a. Yes. A Laurent polynomial is an expression of the type 𝑎𝑛 𝑋 𝑛 + 𝑎𝑛+1 𝑋 𝑛+1 + ⋯ + 𝑎𝑛+𝑘 𝑋 𝑛+𝑘 , (1) with 𝑛 ∈ 𝑍 and 𝑘 ∈ 𝑁 . This is a polynomial (and therefore a rational function) is 𝑛 ≥ 0 . Otherwise (1) = 𝑎𝑛 + 𝑎𝑛+1 𝑋 𝑛 + ⋯ + 𝑎𝑛+𝑘 𝑋 𝑘 𝑋 −𝑛 which is a rational function. 2. No. For instance, 1 1+𝑋 is a rational function, but you can't express it as an element of 𝑅[𝑋, 𝑋 −1 ] . Question #6 I'm searching the antiderivative of rational functions: 1) ∫ 1+𝑥 √2𝑥+1 𝑑𝑥 For this one we have 𝑡 = √2𝑥 + 1 then 𝑠𝑡 = 1 √2𝑥+2 𝑑𝑥 but then I do not see the way to compute the antiderivative. Same thing for 2) ∫ 1 𝑥+√𝑥+1 What are the methods to find antiderivative of rational functions like these ones? 1 1 1 2 2 2 For (1): Try the substitution u=2x+1. Then 𝑑𝑢 = 2𝑑𝑥 ⇒ 𝑑𝑢 = 𝑑𝑥, and 𝑥 = (𝑢 − 1) ⇒ 𝑥 + 1 = (𝑢 − 1) + 1 1 = (𝑢 + 1) 2 Then your integral becomes 1+𝑥 1 ∫ √2𝑥+1 𝑑𝑥 = 4 ∫ 1 𝑢+1 1 𝑢2 𝑑𝑢 = 1 4 1 − ∫ (𝑢2 + 𝑢 2 ) 𝑑𝑢 Answer: 1 1 2 2 For (1): Try the substitution u=2x+1 . Then 𝑑𝑢 = 2𝑑𝑥 ⇒ 𝑑𝑢 = 𝑑𝑥, and 𝑥 = (𝑢 − 1) ⇒ 𝑥 + 1 = 1 2 1 (𝑢 − 1) + 1 = (𝑢 + 1) 2 Then your integral becomes ∫ 1+𝑥 1 𝑢+1 1 𝑑𝑥 = ∫ 1 𝑑𝑢 = 4 4 √2𝑥 + 1 𝑢2 1 1 − ∫ (𝑢2 + 𝑢 2 ) 𝑑𝑢 Question #7 I'm trying to find out under which conditions does a sum of a rational function ∑ 𝑓(𝑛) converges, with 𝑓(𝑥) = 𝐴(𝑥) 𝐵(𝑥) , with deg 𝐵 > deg 𝐴 and ∀𝑋, 𝐵(𝑋) ≠ 0. Answer: As long as convergence is concerned, assume that we're summing ∑∞ 𝑛=𝑁 𝑓(𝑛) where N is large enough so that 𝐵(𝑛) ≠ 0 for 𝑛 ≥ 𝑁 and f(n) is always well-defined. If 𝑓(𝑛) ≤ 0 for large n then consider -f(n) instead. We have 𝑓(𝑛) ∼ 𝐶 𝑛deg 𝐴 𝑛deg 𝐵 = 𝐶. 1 𝑛deg 𝐵 −𝑛deg 𝐴 so that by comparison this series is convergent precisely when deg 𝐵 = deg 𝐴 > 1, that is deg 𝐵 ≥ deg 𝐴 + 2. Question #8 𝑓(𝑥) If f is quadratic function such that f(0)=1 and ∫ 2(𝑥+1)3 𝑑𝑥 is a rational function, find the value of f'(0). 𝑥 I already tried solving this question by using general quadratic equation 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 and then using partial fraction method but it became very complicated. Answer: We have that b=f'(0) and c=f(0)=1. Moreover, by using partial fraction method we get 𝑎𝑥 2 +𝑏𝑥+1 𝑥 2 (𝑥+1)3 𝐴 𝐵 𝑥 𝑥2 = + + 𝐶 𝑥+1 𝐷 𝐸 + (𝑥+1)2 + (𝑥+1)3 . If the integral is a rational function, i.e. a ratio of polynomials, then A=0 and C=0 because the integrals of those terms yield logarithms. Hence 𝑎𝑥 2 + 𝑏𝑥 + 1 = 𝐵(𝑥 + 1)3 + 𝐷𝑥 2 (𝑥 + 1) + 𝐸𝑥 2 , that is (𝐵 + 𝐷)𝑥 3 + (3𝐵 + 𝐷 + 𝐸 − 𝑎)𝑥 2 + (3𝐵 − 𝑏)𝑥 + (𝐵 − 1) = 0. Are you able to find b? Question #9 If I have an analytic function of a complex variable, I can write a Taylor series and in some cases can truncate the high powers to obtain a good approximation over some part of the function's domain. I would like to be able to generate rational functions (quotients of polynomials, as in 𝑛(𝑥) 𝑑(𝑥) where n and d are polynomial functions) that play a similar role. Is there a general way to do this? Answer: To see how it arises, first note that in a rational function of degree N in the numerator and degree M in the denominator, 𝑅(𝑥) = 𝑛 ∑𝑛=0 𝑁 𝑎𝑛 𝑥 1+∑𝑚=1 𝑏𝑚𝑥 𝑚 𝑀 , where a 1 has been taken out of the sum in the denominator to eliminate a redundancy in the coefficients, there are M+N+1 free parameters: 𝑎0 , 𝑎1 , … 𝑎𝑁 , along with 𝑏1 , … , 𝑏𝑀 . So, we assert that the rational function should match the function we are approximating all the way up to M+Nth order around some point 𝑥0 : 𝑓(𝑥0 ) = 𝑅(𝑥0 ) 𝑑𝑓 𝑑𝑥 (𝑥0 ) = 𝑑2 𝑓 𝑑𝑥 2 𝑑𝑅 𝑑𝑥 (𝑥0 ) = 𝑑 𝑀+𝑁 𝑑𝑥 𝑀+𝑁 (𝑥0 ) 𝑑2𝑅 𝑑𝑥 2 (𝑥0 ) = (𝑥0 ) 𝑑 𝑀+𝑁 𝑅 𝑑𝑥 𝑀+𝑁 (𝑥0 ) It seems natural that this would uniquely specify what R must be, and indeed it does. Question #10 In general, how does one determine if a rational function is regular? I have the particular problem of determining in which points of the circle 𝑉(𝑥 2 + 𝑦 2 − 1) ⊆ 𝐴2 is the rational function 𝛼 = 𝑦−1 𝑥 regular? Answer: To say that 𝜙 = 𝑦−1 𝑥 is regular on the circle means that there exist polynomials 𝑝(𝑋, 𝑌), 𝑞(𝑋, 𝑌)𝜖𝑘[𝑋, 𝑌] such that: 𝑌 − 1 = 𝑋𝑝(𝑋, 𝑌) + 𝑞(𝑋, 𝑌)(𝑥 2 + 𝑌 2 − 1)𝜖𝑘[𝑋, 𝑌] But substituting X=0 in that equality yields 𝑌 − 1 = 𝑞(0, 𝑌)(𝑌 2 − 1)𝜖𝑘[𝑌] which is impossible since the left hand side has degree 1 whereas the right hand side is zero or has degree ≥ 2 . Hence the rational function 𝜙 is not regular. Since 𝜙 is not regular but is clearly regular at all points of the circle different from P=(0,-1), it follows that P is the only point where 𝜙 is not regular, i.e. the only pole of 𝜙.