2. PHANTOM GRAPHS PART 2 Philip Lloyd Epsom Girls Grammar School Web site: www.phantomgraphs.weebly.comwww.phantomgraphs.weebly.com

3. Now we will consider: y = x (x – 3) 2 As before, any horizontal line (or plane) should cross this graph at 3 places because any equation of the form : x 3 – 6x 2 + 9x = “a constant ”, has 3 solutions. (1, 4 ) 3

4. The maximum point is at (1, 4) and the minimum point is at (3, 0). We now realise that the phantom curves will start to appear at the turning points (1, 4) and (3, 0) If x 3 – 6x 2 + 9x = 5 then x = 4.1 and 0.95 ± 0.6 i If x 3 – 6x 2 + 9x = 6 then x = 4.2 and 0.90 ± 0.8 i If x 3 – 6x 2 + 9x = 7 then x = 4.3 and 0.86 ± 0.9 i So the left hand phantom is leaning to the left. 3 (1,4) 1

5. The right hand phantom will be appearing at the minimum point when x = 3, y = 0. If x 3 – 6x 2 + 9x = – 1 then x = –0.1 and 3.05 ± 0.6 i If x 3 – 6x 2 + 9x = – 2 then x = –0.2 and 3.1 ± 0.8 i If x 3 – 6x 2 + 9x = – 3 then x = –0.3 and 3.14 ± 0.9 i So the right hand phantom is leaning to the right. 3

8. AUTOGRAPH VERSION. y = x(x - 3)² y = x(x - 3)²

9. Solutions of cubic crossing 3 planes Solutions of cubic crossing 3 planes AUTOGRAPH VERSION. Solutions of cubic crossing 3 planes Solutions of cubic crossing 3 planes Notice how this typical cubic will cross any horizontal plane exactly 3 times. Notice how this typical cubic will cross any horizontal plane exactly 3 times. This shows how any equation of the form y= x 3 + ax 2 + bx + c always has 3 solutions. This shows how any equation of the form y= x 3 + ax 2 + bx + c always has 3 solutions.

10. A very interesting case is the hyperbola : y 2 – x 2 = 1 or y 2 = x 2 + 25 25 25 If x = 0 then y = ± 5 And, if x is large then y 2 ≈ x 2 producing asymptotes y = ± x

11. The graph of y 2 = x 2 + 25 for real x values is : 5 - 5 What x values produce y values between -5 and +5 ?

12. y 2 = x 2 + 25 If y = 4 then 16 = x 2 + 25 and – 9 = x 2 so x = ± 3i Similarly if y = 3 then 9 = x 2 + 25 so x = ± 4i And if y = 0 then 0 = x 2 + 25 so x = ± 5i

13. These points should seem very familiar as points on a CIRCLE of radius 5 units. (0, 5) ( ± 3 i, 4) ( ± 4 i, 3) ( ± 5 i, 0) The circle has imaginary x values but REAL y values. This phantom circle is in the plane at right angles to the hyperbola and joining its two halves.

14. The graph of y 2 = x 2 + 25 for real y values is : 5 - 5

17. y² = x² + 25 y² = x² + 25 AUTOGRAPH VERSION. y² = x² + 25 y² = x² + 25

18. SOME AFTERTHOUGHTS………………(not for Year 13 students) Consider the graph y = 2x 2 = 2 + 2 x 2 – 1 x 2 – 1 This has a horizontal asymptote y = 2 and two vertical asymptotes x = ± 1 -1 1 2 y x What x values can produce y values from y = 0 to y = 2 ?

19. If y = 1 then 2x 2 = 1 x 2 – 1 So 2x 2 = x 2 – 1 and x 2 = – 1 Producing x = ± i If y = 1.999 then 2x 2 = 1.999 x 2 – 1 So 2x 2 = 1.999x 2 – 1.999 and 0.001x 2 = – 1.999 Producing x 2 = – 1999 x  ± 45 i This is a “phantom graph” which approaches the horizontal asymptotic plane y = 2 and is at right angles to the x, y plane, resembling an upside down normal distribution curve… Side view at right angles 2 to the x, y plane. 0 2

21. AUTOGRAPH VERSION. Asymptotic plane Asymptotic plane

22. What about TRIGONOMETRICAL GRAPHS? If y = cos(x) what about y values > 1 and < - 1 ? - π/2 π/2 π 3π/2 1 Solving cos( x ) = 2 seems impossible ! but with some intuition, we could try to find out what cos( i ) is equal to.

23. Using cos(x) = 1 – x 2 + x 4 – x 6 + x 8 -.. …. 2! 4! 6! 8! Let’s find cos(± i ) = 1 + 1 + 1 + 1 + 1..  1.54 (ie > 1 ) 2! 4! 6! 8! Similarly cos(± 2i) = 1 + 4 + 16 + 64 + …  3.8 2! 4! 6! Also find cos(π + i) = cos(π) cos(i) – sin(π) sin(i) = – 1 × cos(i) – 0  – 1.54 These results imply that the cosine graph also has its own “phantoms” in vertical planes at right angles to the usual x, y graph, emanating from each max/min point. NB cos(i) = cosh(1)

24. -π/2 π/2 π 3π/2 2π

25. AUTOGRAPH VERSION. y = cos(x) y = cos(x)

26. Consider the exponential function y = e x 1 x (real) y (real) We need to be able to find x values which produce negative y values. How can we find x if e x = – 1 ? Using the expansion for e x = 1 + x + x 2 + x 3 + x 4 + x 5 +.. …. 2! 3! 4! 5! We can find e x i = 1 + xi + (xi) 2 + (xi) 3 + (xi) 4 + (xi) 5 +.. …. 2! 3! 4! 5! = (1 – x 2 + x 4 – x 6 + x 8 +… ) + i ( x – x 3 + x 5 – x 7 +...) 2! 4! 6! 8! 3! 5! 7! = cos x + i sin x

27. If we are to get REAL y values then using e xi = cos x + i sin x, we see that the imaginary part, sin x, has to be zero. This only occurs when x = 0, π, 2π, 3π,… (or generally nπ) e πi = cosπ + isinπ = – 1 + 0i, e 2πi = cos2π + isin2π = + 1 + 0i e 3πi = cos3π + isin3π = – 1 + 0i, e 4πi = cos4π + isin4π = + 1 + 0i Now consider y = e X where X = x + 2nπ i (ie even numbers of π) ie y = e x + 2nπ i = e x × e 2nπ i = e x × 1 = e x Also consider y = e X where X = x +(2n+1)π i ( ie odd numbers of π) ie y = e x + (2n+1)π i = e x × e (2n+1)π i = e x × – 1 = – e x

28. This means that the graph of y = e X consists of parallel identical curves occurring at X = x + 2nπi = x + even N os of πi and, upside down parallel identical curves occurring at X = x + (2n + 1)πi = x + odd N os of πi

29. y (real) x (real) x (unreal) Graph of y = e X where X = x + nπi πiπi 2πi2πi -πi-πi -2πi 1 1 1

30. x (real) y (real) x (unreal) Graph of y = e X where X = x + nπi π 2 π2 π - π - 2 π

31. AUTOGRAPH VERSION. y = exp(x) y = exp(x)

32. I call this THE ALPHA GRAPH y 2 = x(x – 3) 2 1 2 3 4 x y321y321

33. y 2 = x(x – 3) 2 Using a technique from previous graphs: 4 1 4 I choose an x value such as x = 4, calculate the y value, ie y = 4 then solve the equation x(x – 3) 2 = 4 already knowing one factor is (x – 4) ie If x(x – 3) 2 = 4 then x 3 – 6x 2 + 9x – 4 = 0 so (x – 4)(x 2 – 2x + 1) = 0 factorising (x – 4)(x – 1) 2 = 0 x = 4 or 1

34. SIMILARLY : If x = 5, y 2 = 20 and y = ±4.5 or ±2√5 So x(x – 3) 2 = 20 x 3 – 6x 2 + 9x – 20 = 0 (x – 5)(x 2 – x + 4) = 0 x = 5 or ½ ± 1.9i If x = 6, y 2 = 54 and y = ±7.3 So x(x – 3) 2 = 54 x 3 – 6x 2 + 9x – 54 = 0 (x – 6)(x 2 + 9) = 0 x = 6 or ± 3i If x = 7, y 2 = 112 and y = ±10.58 So x(x – 3) 2 = 112 x 3 – 6x 2 + 9x – 112 = 0 (x – 7)(x 2 + x + 16) = 0 x = 7 or – ½ ± 4i

35. 642642 1 2 3 4

36. y² = x(x - 3)² AUTOGRAPH VERSION.

37. This is a more familiar graph: y = x 2 x – 1 - 1 1 2 x y4321y4321

38. y = x 2 x – 1 If y = any real number c then x 2 = c x – 1 which produces a quadratic equation: x 2 = cx – c So for values of c from 0 to 4 we will get complex conjugate solutions. y4321y4321 -1 1 2 x

39. y = x 2 x – 1 If y = 0 x = 0 If y = 1 x = 1 ± √3i 2 2 If y = 2 x = 1 ± i If y = 3 x = 3 ± √3i 2 2 If y = 4 x = 2 These points produce the phantom “oval” shape as shown in the next picture. y4321y4321 -1 1 2 x

40. 1 2 3 5432154321

41. y = x²/(x - 1) AUTOGRAPH VERSION.

42. Consider an apparently “similar” equation but with a completely different “Phantom”. y = x 2 = x 2 (x – 1)(x – 4) x 2 – 5x + 4

43. AUTOGRAPH VERSION. Curve with 2 vertical asymptotes

44. y = x 3 /(x 2 – 1)

45. AUTOGRAPH VERSION. y = x³/(x² - 1)

46. y = x 4 /(x 2 – 1)

47. y = x^4/(x² - 1) AUTOGRAPH VERSION.

48. THE END Please check out the Web site: www.phantomgraphs.weebly.com

49. Not all graphs have phantoms! Consider the equation : y = x x 2 – 1

50. Briefly, if we consider any real y value, for example y = 1 then if we solve : x = 1 we just get a simple quadratic x 2 – 1 equation to solve which has 2 solutions. Notice that ANY horizontal line will always cross this curve twice already. There are no “missing” values which is the usual sign that there are “phantoms”.

51. The algebra for this case is as follows… Let z = x + iy = x + iy (x + iy) 2 – 1 (x 2 – y 2 – 1) + 2xyi = x(x 2 – y 2 – 1) + 2xy 2 + i ( y(x 2 – y 2 – 1 ) – 2x 2 y (x 2 – y 2 – 1) 2 + 4x 2 y 2 = x 3 + xy 2 – x + i( -y(x 2 + y 2 + 1) ) (x 2 – y 2 – 1) 2 + 4x 2 y 2 So if Im(z) = 0 then either y = 0 producing the original curve OR x 2 + y 2 + 1 = 0 and since this has no real solutions there are no phantoms.