Questions and Answers #65 Quadratic Function and Equation

Questions and Answers Sheet 65 Quadratic Function and Equation Question #1 Region of the coefficients of a quadratic equation that cause the roots of it to be in the unit disk From Simon Haykin's Adaptive Filter Theory: consider the characteristic equation is 1 + 𝑎1 𝑧 −1 + 𝑎2 𝑧 −2 = 0, then for the roots to be inside the unit circle (or in the unit disk), the coefficients of the quadratic must satisfy the following conditions: −1 ≤ 𝑎1 + 𝑎2 , −1 ≤ 𝑎2 − 𝑎1 , and −1 ≤ 𝑎2 ≤ 1. I could figure this out some of the way. For complex roots it follows that the coefficients lie in the region defined by the intersection of 4𝑎2 > 𝑎12 and 𝑎2 < 1. For real roots, I could prove that |𝑎2 | < 1 , and obviously 4𝑎2 ≤ 𝑎12 . But I cannot figure out how the line boundaries of the region can be derived. Quadratic Function and Equation Answer: From the quadratic equation 𝑧= −𝑎1 2 𝑎 2 ± √( 1 ) − 𝑎2 2 For the greater real root and the upper boundary on the real line  1 1+ a1 2 −a1 a + ( 1 ) 2 − a2 2 2 a  + ( 1 ) 2 − a2 2 a 1 + a1 + ( 1 ) 2 2 a  ( 1 ) 2 − a2 2 a1 + a2  −1 For the lesser real root and the lower boundary on the real line −1 −1 + a1 2  − a1 a − ( 1 ) 2 − a2 2 2 a  − ( 1 ) 2 − a2 2 a −1 + a1 + ( 1 ) 2 2 a  ( 1 ) 2 − a2 2 −1  a2 − a1 For the complex roots being inside the unit circle − a1 a  i a2 − ( 1 ) 2 | 2 2 1 | 1 a a  ( 1 ) 2 + a2 − ( 1 ) 2 2 2 1  a2 12  ( a2 ) 2 1 | a2 | −1  a2  1 Question #2 Root of an equation Given: (5 − 𝑦)2 = (1 − 𝑦)2 Answer: Solve: 𝐴2 − 𝐵2 = (𝐴 + 𝐵)(𝐴 − 𝐵) = 0 ⟺ 𝐴 + 𝐵 = 0 ∨ 𝐴 − 𝐵 = 0 and in your equation the case A-B leds to no solution and we only keep the solution for the case A+B=0 . Question #3 Roots of a quadratic equation. 2 Given: 𝑎(𝑓(𝑥)) + 𝑏𝑓(𝑥) + 𝑐 = 0. Answer: 2 The equation 𝑎(𝑓(𝑥)) + 𝑏(𝑓(𝑥)) + 𝑐 = 0 is "quadratic" only if f(x) is itself a linear function. If u is a solution to 𝑎𝑢2 + 𝑏𝑢 + 𝑐 = 0 (there might be two values of u) then any solution to f(x)=u is solution to 2 𝑎(𝑓(𝑥)) + 𝑏(𝑓(𝑥)) + 𝑐 = 0. How many that is depends on the function, f. Question #4 Roots of hyperbolic functions I am having difficulty with understanding one part of the answer to the following: Find conditions on a, b, and c which are necessary and sufficient to ensure that the equations 𝑎 cos ℎ 𝑥 + 𝑏 sin ℎ 𝑥 = 𝑐 has exactly one root. Answer: If you let 𝑒 𝑥 = 𝑦, you end with the equation (𝑎 + 𝑏)𝑦 2 − 2𝑐𝑦 + (𝑎 − 𝑏) = 0 So, first we need two roots Δ = 𝑏 2 + 𝑐 2 − 𝑎2 ≥ 0 and one of them must be negative since 𝑦 > 0 so 𝑎−𝑏 𝑎+𝑏 <0 that is to say 𝑎2 − 𝑏 2 < 0. Question #5 S and G are roots of 𝑥 2 − 3𝑥 + 1 = 0. Find equation whose roots are 1 𝑆−2 and 1 𝐺−2 . Answer: Note 𝑆𝐺 = 1, 𝑆 + 𝐺 = 3 . Then 1 1 𝑆−2 𝐺−2 1 𝑆−2 + = 1 𝐺−2 1 𝑆𝐺−2(𝑆+𝐺)+4 = = −1 𝑆+𝐺−4 𝑆𝐺−2(𝑆+𝐺)+4 =1 Thus, the equation with roots 1 𝑆−2 and 1 𝐺−2 is 𝑥 2 − 𝑥 − 1 = 0 Question #6 Set 𝑓(𝑥) = 𝐴𝑥 2 + 𝐵 𝑥 + 𝐶 and 𝑔(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 with 𝐴 × 𝑎 ≠ 0, 𝐴, 𝑎𝐵, 𝑏, 𝐶, 𝑐 ∫ 𝑅 satisfies: |𝑓(𝑥)| ≥ |𝑔(𝑥)|∀𝑥 ∈ 𝑅 Prove that |𝐵2 − 4𝐴𝐶| ≥ |𝑏 2 − 4𝑎𝑐| Answer: Step 1 The extrema value of g(x) is 𝑏 2 −4𝑎𝑐 4𝑎 Wrong claim that you made: The smallest value of |g(x)| need not be | 𝑏 2 −4𝑎𝑐 4𝑎 | E.g. Consider 𝑔(𝑋) = (𝑥 − 1)(𝑥 + 1). Clearly the smallest value of |g(x)| is o. Whereas | 𝑏 2 −4𝑎𝑐 4𝑎 | is the absolute value of the extrema of g(x), so this is equal to |-1|=1 Step 2 You need 𝛿𝑔 ≥ 0 in order to conclude that "The smallest value of |g(x)| is | 𝑏 2 −4𝑎𝑐 4𝑎 | '' for your proof to work. Note that the case of 𝛿𝑔 ≥ 0 is pretty simple to deal with. Question #7 Set 𝑓(𝑥) = 𝑥 2 + 𝑎𝑥 + 𝑏. Prove max{ |𝑓(−1)|, |𝑓(0)|, |𝑓(1)|} ≥ 1 2 Answer: Let max{ |𝑓(−1)|, |𝑓(0)|, |𝑓(1)|} = 𝑘. Thus, by your work 2 = (−2𝑏) + (1 − 𝑎 + 𝑏) + (1 + 𝑎 + 𝑏) ≤ ≤ 2|𝑏| + |1 − 𝑎 + 𝑏| + |1 − 𝑎 + 𝑏| ≤ 2𝑘 + 𝑘 + 𝑘 = 4𝑘, which gives 𝑘≥ 1 2 Question #8 Show that if a quadratic has rational coefficients and 𝑥 + 𝑦√𝑧 is a root, then 𝑥 − 𝑦√𝑧 must also be a root. Answer: Maybe try using the following relationships of the roots and coefficients of a quadratic polynomial: Let a, b, 𝑐 ∈ 𝑄. 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 𝑎(𝑥 − 𝑥1 )(𝑥 − 𝑥2 ) = 𝑎(𝑥 2 − (𝑥1 + 𝑥2 ) + 𝑥1 ⋅ 𝑥2 ) = 𝑎𝑥 2 − 𝑎(𝑥1 + 𝑥2 )𝑥 + 𝑎 ⋅ 𝑥1 ⋅ 𝑥2 . From here we get the following system of equalities: b  x + x = − 1 2  a   x x = c  1 2 a 𝑏 Notice that if 𝑥1 = 𝑥 + √𝑧, then 𝑥1 + 𝑥2 = 𝑥 + √𝑧 + 𝑥2 = − ∈ 𝑄, then we can solve for 𝑥2 : 𝑎 𝑏 𝑥2 = (− − 𝑥) − √𝑧 = 𝑚 − √𝑧, where 𝑚 ∈ 𝑄 𝑎 Now let's use the second relationship we proved: 𝑥1 ⋅ 𝑥2 = (𝑥 + √𝑧)(𝑚 − √𝑧) = (𝑥𝑚 − 𝑧) + (𝑚 − 𝑥)√𝑧 = 𝑐 𝑎 ∈𝑄 If 𝑚 − 𝑥 ≠ 0 we could solve for √𝑧 to get that it must be rational, but this contradicts our assumption that it is irrational, hence m=x , which implies 𝑥2 = 𝑚 − √𝑧 = 𝑥 − √𝑧 Question #9 Show that √−1 − √−1 − √−1 … can be expressed in the form 𝛼 or 𝛼 2 , Hence prove 1 (𝑎 + 𝑏𝛼 + 𝑐𝛼 2 )(𝑎 + 𝑏𝛼 2 + 𝑐𝛼) = [(𝑎 − 𝑏)2 + (𝑏 − 𝑐)2 + (𝑐 − 𝑎)2 ] 2 Answer: The LHS is 𝑎2 + 𝑏 2 𝛼 3 + 𝑐 2 𝛼 3 + 𝑎𝑏(𝛼 2 + 𝛼) + 𝑎𝑐(𝛼 2 + 𝛼) + 𝑏𝑐(𝛼 4 + 𝛼) Use the fact that 𝛼 4 = 𝛼. So The LHS becomes 1 1 2 2 𝑎2 + 𝑏 2 + 𝑐 2 − 𝑎𝑏 − 𝑎𝑐 − 𝑏𝑐 = [𝑎2 + 𝑏 2 − 2𝑎𝑏 + 𝑏 2 + 𝑐 2 − 2𝑏𝑐 + 𝑎2 + 𝑐 2 − 2𝑎𝑐] = [(𝑎 − 𝑏)2 + (𝑏 − 𝑐)2 + (𝑎 − 𝑐)2 ] Question #10 Show that 𝑥 − 𝑥 2 = 1 3 has no real solutions. Answer: I mean the easiest way is to use the discriminant (as mentioned in the first comment). But hey, why use the quadratic equation when you can just complete the square (i.e. derive the quadratic equation): 1 1 1 x 2 − x + = 0( x − ) 2 + + ? 3 2 3 1 1 = 0x2 − x + + + ? = 0 → ? 4 3 1 1 1 1 = − ( x − )2 + − 4 2 3 4 1 1 = 0( x − ) 2 + 2 12 1 1 = 0( x − ) 2 = − 2 12 A real value cannot be squared (positive or negative) to give a negative result, so this has no real solutions. If we inspect 𝑏 2 − 4𝑎𝑐, we find: 1 − 1 2 4 1 = −  0 → no real solutions 3 3 1 2 (2 (𝑥 − )) = (2𝑥 − 1)2 = 4 (𝑥 − ) 2 2 So we can use that and just multiply each side by 4 (2𝑥 − 1)2 = − 1 3