Questions and Answers #2 Rational Functions

Questions and Answers Sheet 2 Rational Functions Question #1 Does it follow that r(x)+p(x) is a rational function (and not a polynomial)? Suppose 𝑟(𝑥) ≠ 0 is a rational function (and not a polynomial) p(x) is a polynomial function of 𝑥 ∈ 𝑅𝑅 . Answer: Yes. Suppose that r(x)+p(x)=s(x) were a polynomial. Then also r(x)=s(x)=p(x) would be a polynomial, hence r(x) is a polynomial, contradiction. Question #2 Does there exists some simple criteria to know when the primitive of a rational function of CC[z] is still a rational function? In fact my question is more about the stability of this property. Let P and Q two co' polynomials and let A and B two co' polynomials such that 𝐴 𝑃 ′ 𝑃′ 𝑄 − 𝑃𝑄′ =( ) = 𝐵 𝑄 𝑄2 Then considering a pertubation 𝐴𝜉 of A (in the sense the roots of 𝐴𝜉 converge to the one of A with the same multiplicity as 𝜉 goes to zero.): does 𝐴𝜉 𝐵 admit a primitive which is rational function? Answer: One simple criterion is the following. If the denominator of f has squarefree factorization 𝑄1 𝑄22 … 𝑄𝑛𝑛 then f has a partial fraction expansion 𝑃1 𝑄1 + ⋯+ 𝑃𝑛 𝑛 𝑄𝑛 , which is the derivative of a rational function ⇔ each ⇔ 𝑄|𝑊(𝑃, 𝑄′, (𝑄^2)′, … , (𝑄^{𝑖 − 1})′) , where W denotes the Wroian. Recall that the squarefree factorization of a polynomial over a field of characteristic 0 may be quickly computed by gcds. Question #3 Find a rational function 𝑓: 𝑅𝑅 → 𝑅𝑅 with range f(RR)=[0,1] (Thus 𝑓(𝑥) = 𝑃(𝑥) 𝑄(𝑥) for all 𝑥 ∈ 𝑅𝑅 for suitable polynomials P and Q, where Q has no real root). I have no clue where to begin. Answer: The image of the rational function 𝑥 → 𝑥 1+𝑥 2 is a compact interval. Compose it with a suitable affine function to obtain a rational function with image [0,1]. Question #4 Find divisor of rational function 𝑓 = (𝑤0 , 𝑤1 ) on a surface 𝑋 = {𝑤0 𝑤1 − 𝑤2 𝑤3 } ⊂ 𝑃𝑃3 How should I to take into account this surface? 𝑃 𝑄𝑖 is Answer: I interpret your question as asking about the rational function 𝑓 = 𝑤1 𝑤0 on the surface X. The divisor of zeros is 𝑑𝑖𝑣(𝑓)0 = 𝐷2 + 𝐷3 where 𝐷2 is the line 𝑤1 = 𝑤2 = 0 and 𝐷3 is the line 𝑤1 = 𝑤3 = 0. This can be checked on the affine chart 𝑤0 = 1. The divisor of poles is ÷ (𝑓)∞ = 𝐸2 + 𝐸3 where 𝐸2 is the line 𝑤0 = 𝑤2 = 0 and 𝐸3 is the line 𝑤0 = 𝑤3 = 0. This can be checked on the affine chart 𝑤1 = 1. The locus of indeterminacy of f consists of the two points 𝐷3 ∩ 𝐸3 = [0: 0: 1: 0] and 𝐷2 ∩ 𝐸2 = [0: 0: 0: 1] And finally the answer to your question is 𝑑𝑖𝑣(𝑓) = 𝑑𝑖𝑣(𝑓)0 − 𝑑𝑖𝑣(𝑓)∞ = 𝐷2 + 𝐷3 − 𝐸2 − 𝐸3 A picture The lines 𝐷2 , 𝐸3 are in one of the two rulings of the quadric X and the lines 𝐷3 , 𝐸2 on the other one. You might try to draw these four lines as the sides of a square and see where the indeterminacy points lie. Question #5 For a course on Galois theory, we proved the fundamental theorem of symmetric polynomials, which states that every symmetric polynomial can be uniquely written as a polynomial in the elementary symmetric polynomials. As an exercise we were asked to prove that every symmetric rational function also is a rational function in the elementary symmetric polynomials. For this I have the following (partial) proof. Let K be a field, let 𝑓𝜖𝐾(𝑇1 , … , 𝑇𝑛 ) be a symmetric function. We can write 𝑓 = 𝑔 ℎ with g,h co'. Let 𝜎𝜖𝑆𝑛 , then 𝜎(𝑔) 𝜎(ℎ) 𝑔 𝑔 ℎ ℎ = 𝜎 ( ) = , so 𝜎(𝑔)ℎ = 𝜎(ℎ)𝑔. Since g and h are co', it follows that 𝑔|𝜎(𝑔) , and since the total degrees of the polynomials are the same it follows there exists some 𝑘𝜖𝑒𝐾 × such that 𝜎(𝑔) = 𝑘𝑔. Similarly for h. Of course I would like to conclude from this that we must have 𝜎(𝑔) = 𝑔 since 𝜎 doesn't change the coefficients, but I realised the argument is more subtle than that, since this is not necessarily true. For example, with 𝐾 = 𝐹𝐹7 and n=3 we have (321) ⋅ (𝑋1 𝑋2 + 2𝑋2 𝑋3 + 4𝑋3 𝑋1 ) = 2 ⋅ (𝑋1 𝑋2 + 2𝑋2 𝑋3 + 4𝑋3 𝑋1 ). So, does anyone know how to proceed? Ofcourse, once you have 𝜎(𝑔) = 𝑔 and 𝜎(ℎ) = ℎ for all 𝜎𝜖𝑆𝑛 the result follows from the fundamental theorem of symmetric polynomials. As a sidenote: I know there's a proof using the fundamental theorem of Galois theory, but I would like to finish a proof using this approach. Answer: Let 𝑓𝜖𝑄(𝑇1 , … , 𝑇𝑛 ) be a symmetric rational function, then we can write 𝑓 = 𝑔 ℎ with 𝑔, ℎ𝜖𝑍[𝑇1 , … , 𝑇𝑛 ]. If h is a symmetric polynomial, then g=hf is symmetric too. By the fundamental theorem of symmetric polynomials, f is a rational function in the elementary symmetric polynomials. If h is not a symmetric polynomial, consider ℎ′ = ∏𝜎𝜖 𝑆𝑛 𝜎ℎ. Then hh' is a symmetric polynomial by {𝑒} construction. By writing 𝑓 = 𝑔 ℎ = 𝑔ℎ′ ℎℎ′ , we reduce ourselves to the previous case. Question #6 For each of the following functions, determine if it is a polynomial, rational function, both, or neither ℎ(𝑥) = 3𝑥 −9 + 6𝑥 −4 − 3 a. Polynomial function only b. Rational function only c. Both polynomial and rational function 𝑗(𝑥) = (8𝑥 2 − 5√𝑥 + 6) 2 a. Polynomial function only b. Rational function only c. Both polynomial and rational function Answer: ℎ(𝑥) = 3𝑥 −9 + 6𝑥 −4 − 3 = 3 6 3 + 6𝑥 9 − 3𝑥 9 + 4−3= 9 𝑥 𝑥 𝑥9 Rational function only Question #7 For the function, determine if it is a polynomial, rational function, both, or neither 𝑔(𝑥) = 4𝑥 5 +𝑥 2 −8𝑥 9 +2√𝑥+9 a. Polynomial function only b. Rational function only c. Both polynomial and rational function Answer: d. Neither polynomial nor rational function Question #8 From the definition of rational function 𝑓(𝑥) = so for the function 𝑓(𝑥) = 𝑥 −2 +3 𝑥−5 𝑝(𝑥) , where P(x) and q(x) are polynomials and 𝑞(𝑥) ≠ 0 𝑞(𝑥) by the definition f(x) isnt rational since the numerator is not polynomial but by multiplying both numerator and denominator by 𝑥 2 we get 𝑓(𝑥) = 3𝑥 2 +1 𝑥 3 −5𝑥 2 which is rational and can we say that both functions are equal at every point? Answer: They are not equal at every point. They are equal where they are both defined, on the intersection of their domains, which is all real numbers except 0 and 5. (So they are not equal at 0 or at 5 since they aren't even defined there.) Question #9 Given a rational function 𝑝(𝑥) 𝑞(𝑥) , what would the 𝑥 𝑘 coefficient of this rational function mean (in particular for the negative k's). Is there some sort of expansion 𝑝(𝑥) 𝑞(𝑥) 𝑘 = ∑∞ −∞ 𝑎𝑘 𝑥 where one would say 𝑎𝑘 is the coefficient of 𝑥 𝑘 ? How would you find the coefficient? A very simple rational function would be find the 𝑥 −1 (𝑥−1) . How would you (𝑥−2)(𝑥−3) coefficient of this? Answer: There is no such expansion that converges everywhere. To see this, consider what happens to both sides of your equation as 𝑥 → 0 and 𝑥 → ∞. You can, however, expand 𝑝(𝑥) 𝑞(𝑥) as different series that converge around different values of x. For instance, near 0 you have (𝑥−1) 1 1 𝑘 2 1 𝑘 2 2 3 3 𝑘 = ∑∞ 𝑘=0 [ ( ) − ( ) ] 𝑥 , (𝑥−2)(𝑥−3) which you can derive by using partial fractions and the formula for geometric series. Similarly expanding 𝑝(𝑦 −1 ) 𝑞(𝑦 −1 ) 𝑝(𝑦 −1 ) 𝑞(𝑦 −1 ) with respect to y will give you a power series in 𝑥 −1 valid as 𝑥 → ∞: 1 2 6 3 1−3𝑦 =− + 1 − 1 1 2 1−2𝑦 1 2 1 1 2 1 6 3 2 6 3 2 ∞ 𝑘 𝑘 𝑘 𝑘 𝑘 −𝑘 = − + ∑∞ 𝑘=0 [ 3 − 2 ] 𝑦 = − + ∑𝑘=0 [ 3 − 2 ] 𝑥 So in some sense you could say that the coefficient of 𝑥 −1 is 1, near infinity. Question #10 For the function, determine if it is a polynomial, rational function, both, or neither 𝑓(𝑥) = (6𝑥 5 − 4)2 a. Polynomial function only b. Rational function only c. Both polynomial and rational function Answer: c. Both polynomial and rational function