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PHANTOM GRAPHS PART 1.

PHANTOM GRAPHS PART 1. Philip Lloyd Epsom Girls Grammar School Web site: www.phantomgraphs.weebly.com. We often say that “Solutions of a quadratic are where the graph crosses the x axis ”. y = x 2 – 4x + 3 y = x 2 – 2x + 1 y = x 2 – 2x + 2

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PHANTOM GRAPHS PART 1.

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  1. PHANTOM GRAPHSPART 1. Philip Lloyd Epsom Girls Grammar School Web site: www.phantomgraphs.weebly.com

  2. We often say that “Solutions of a quadratic are where the graph crosses the x axis”. y = x2 – 4x + 3 y = x2 – 2x + 1 y = x2 – 2x + 2 1 3 1 1 Crosses twice Crosses once Does not cross at all so x2 – 4x + 3 = 0 so x2 – 2x + 1 = 0 so x2 – 2x + 2 = 0 has 2 (real) solutions has 1 (real)solution has no (real) solutions (or 2 equal solutions)

  3. But then we say that this last equation has “complex” or “imaginary” solutions. We can find where y actually CAN equal zero by “completing the square” : x2 – 2x + 2 = 0 x2 – 2x = – 2 x2 – 2x + 1 = – 2 + 1 (x – 1)2 = – 1 x – 1 = ± i x = 1 + i and x = 1 – i

  4. Clearly, the graph does not actually cross the x axis at these points. It does not cross the x axis at all! So, physically, wherearethese “imaginary” solutions?

  5. It is probably better to consider a simpler case using just y = x2 For positivey values : we get the usual points… (±1, 1) (±2, 4) (±3, 9) etc But we can also find negative, real y values even though the graph does not seem to exist under the x axis: If y = – 1 then x2 = – 1 and x = ±i If y = – 4 then x2 = – 4 and x = ±2i If y = – 9 then x2 = – 9 and x = ±3i 9 4 1 -3 -2 -1 1 2 3

  6. The big breakthrough is to change from an x AXIS….… Real y axis Real x axis

  7. The big breakthrough is to change from an x AXIS….… to an x PLANE ! Real y axis Unreal x axis Real x axis Complex xplane or “Argand plane” (Instead of just an x axis)

  8. This produces another parabola underneath the usual y = x2 but at right angles to it! (A sort of phantom parabola “hanging” from the usual y = x2 graph)

  9. THE GRAPH OF y = x2 with REAL y VALUES. Real y Normal parabola x (real) x (imaginary) Phantom parabola at RIGHT ANGLES to the normal one.

  10. AUTOGRAPH VERSION.

  11. Going back to y = x2 – 2x + 2 ……. This can be written as: y = (x – 1)2 + 1 Normally we say the MINIMUM VALUE of y is 1 But the REAL Minimum value of y is not 1! We just showed y can be 0 ! (ie when x = 1 + i and x = 1 – i ) (1,1)

  12. In fact y can equal any real value! Suppose y = – 3 So (x – 1)2 + 1 = – 3 (x – 1)2 = – 4 x = 1 + 2i and x = 1 – 2i Similarly if y = – 8 (x – 1)2 + 1 = – 8 (x – 1)2 = –9 x = 1 + 3i and x = 1 – 3i

  13. In fact there is NO MINIMUM REAL y VALUE because all complex x values of the form x = 1 ± K iwill actually produce more REAL VALUES of y to – ∞. These values are all in the same PLANE at right angles to the basic graph. No other complex x values will produce real y values. The result is another parabola “hanging” from the vertex of the normal graph.

  14. So, instead of saying … “Solutions of quadratics are where the graph crosses the x AXIS” we should now say … “Solutions of quadratics are where the graph crosses the x PLANE”.

  15. AUTOGRAPH VERSION.y = (x - 1)² + 1

  16. y = (x – 6)2 + 1 y = (x – 2)2 y = (x + 4)(x + 2

  17. AUTOGRAPH VERSION.3 parabolas http://autograph-maths.com/activities/philiplloyd/phantom.html

  18. Now consider y = x4 For positive y values we get the usual points (±1, 1), (±2, 16), (±3, 81) but equations involving x4 such as : x4 = 1 or x4 = 16 have 4 solutions not just 2. (This is called the Fundamental Theorem of Algebra.)

  19. If y = 1, x4 = 1 so using De Moivre’s Theorem: r4cis 4θ = 1cis (360n) r = 1 and 4θ = 360n θ = 0, 90, 180, 270 x1 = 1 cis 0 = 1 x2 = 1 cis 90 = i x3 = 1 cis 180 = – 1 x4 = 1 cis 270 = – i

  20. If y = 16, x4 = 16 so using De Moivre’s Theorem: r4cis 4θ = 16cis (360n) r = 2 and 4θ = 360n θ = 0, 90, 180, 270 x1 = 2 cis 0 = 2 x2 = 2 cis 90 = 2i x3 = 2 cis 180 = – 2 x4 = 2 cis 270 = – 2i

  21. This means y = x4 has another phantom part at right angles to the usual graph. y Phantomgraph Usual graph Unreal x Real x

  22. BUT WAIT, THERE’S MORE !!! The y values can also be negative. If y = –1, x4 = –1 Using De Moivre’s Theorem: r4cis 4θ = 1cis (180 + 360n) r = 1 and 4θ = 180 + 360n θ = 45 + 90n x1 = 1 cis 45 x2 = 1 cis 135 x3 = 1 cis 225 x4 = 1 cis 315

  23. Similarly, if y = –16, x4 = –16 Using De Moivre’s Theorem: r4cis 4θ = 16cis (180 +360n) r = 2 and 4θ = 180 + 360n θ = 45 + 90n x1 = 2 cis 45 x2 = 2 cis 135 x3 = 2 cis 225 x4 = 2 cis 315

  24. This means that the graph of y = x4 has TWO MORE PHANTOMSsimilar to the top two curves but rotated through 45 degrees.

  25. AUTOGRAPH VERSION.y = x^4

  26. Consider a series of horizontal planes cutting this graph at places such as : y = 81 So we are solving the equation: x4 = 81 The result is a series of very familiar Argand Diagrams which we have never before associated with cross sections of a graph.

  27. Unreal x Solutions of x4 = 81 Real x -3 -2 -1 1 2 3

  28. Unreal x Solutions of x4 = 16 Real x -3 -2 -1 1 2 3

  29. Unreal x Solutions of x4 = 1 Real x -3 -2 -1 1 2 3

  30. Unreal x Solutions of x4 = 0.0001 Real x -3 -2 -1 1 2 3

  31. Unreal x Solutions of x4 = 0 Real x -3 -2 -1 1 2 3

  32. Unreal x Solutions of x4 = –0.0001 Real x -3 -2 -1 1 2 3

  33. Unreal x Solutions of x4 = –1 Real x -3 -2 -1 1 2 3

  34. Unreal x Solutions of x4 = –16 Real x -3 -2 -1 1 2 3

  35. Unreal x Solutions of x4 = –81 Real x -3 -2 -1 1 2 3 ,

  36. y = x4 ordinary form of the equation z = (x + iy)4 form of the equation for “Autograph” z = x4 + 4x3yi + 6x2y2i2 + 4xy3i3 + y4i4 z = x4+ 4x3yi – 6x2y2– 4xy3i+ y4 Re(z) = x4 – 6x2y2 + y4 Im(z)= 4yx(x2 – y2) If Im(z) = 0 then y = 0 or x = 0 or y = ±x Subs y = 0 and Re(z) = x4 ( basic curve ) Subs x = 0 and Re(z) = y4( top phantom ) Subs y = ±x and Re(z) = – 4 x4 ( bottom 2 phantoms) z x y

  37. Next we have y = x3 Equations with x3 have 3 solutions. If y = 1 then x3 = 1 so r3cis 3θ = 1cis (360n) r = 1 θ = 120n = 0, 120, 240 x1 = 1 cis 0 x2 = 1 cis 120 x3 = 1 cis 240

  38. Similarly, if y = 8 then x3 = 8 so r3cis 3θ = 8cis (360n) r = 2 θ = 120n = 0, 120, 240 x1 = 2 cis 0 x2 = 2 cis 120 x3 = 2 cis 240

  39. Also y can be negative. If y = –1, x3 = –1 r3cis 3θ = 1cis (180 +360n) r = 1 and 3θ = 180 + 360n θ = 60 + 120n x1 = 1 cis 60 x2 = 1 cis 180 x3 = 1 cis 300

  40. The result is THREE identical curves situated at 120 degrees to each other!

  41. AUTOGRAPH VERSION.y = x³

  42. Again consider a series of horizontal Argand planes cutting this graph at places such as : y = 27 So we are solving the equation: x3 = 27 The result is a series of very familiar Argand Diagrams which we have never before associated with cross sections of a graph.

  43. Unreal x Solutions of x3 = 27 Real x -3 -2 -1 1 2 3

  44. Unreal x Solutions of x3 = 8 Real x -3 -2 -1 1 2 3

  45. Unreal x Solutions of x3 = 1 Real x -3 -2 -1 1 2 3

  46. Unreal x Solutions of x3 = 0.001 Real x -3 -2 -1 1 2 3

  47. Unreal x Solutions of x3 = 0 Real x -3 -2 -1 1 2 3

  48. Unreal x Solutions of x3 = – 0.001 Real x -3 -2 -1 1 2 3

  49. Unreal x Solutions of x3 = –1 Real x -3 -2 -1 1 2 3

  50. Unreal x Solutions of x3 = –8 Real x -3 -2 -1 1 2 3

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