• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13
  14. 14
    14
  15. 15
    15
  16. 16
    16
  17. 17
    17

Shadow Functions Maths IB HL Portfolio

Extracts from this document...

Introduction

June 25

2012


Shadow Functions

Polynomials represent a large area of mathematics. The word itself is a combination of the Greek poly and Latin binomium, meaning "many" and "binomial".

A polynomial function is in the form:

image00.png

where

 is a non-negative integer and image01.png

are constants.

We define roots or zeros of a polynomial the x-values for which these equal to zero;

Real roots of a polynomial are graphically represented by their y-intercepts, but how can complex roots be identified?

In this task, I will investigate the method of shadow functions and their generators, which help identify the real and imaginary components of complex zeros from key points along the x-axis.

To this effect, I will make use of the software Geogebra, a graphing program.

Part A: Quadratic Polynomials

Quadratic polynomials are in the form image02.png

 where

 and

.

Their roots can be calculated by using the so-called discriminant image02.png

.

And are calculated to be

image03.png

Let us consider the quadratic function:

image04.png

, where

Where the vertex has coordinates image02.png

We can find the zeros of this quadratic by expanding the function:

image05.png

Where the discriminant  image01.png

,

thus image02.png

has two complex roots:  

image06.png

The "shadow function" to image02.png

is another quadratic image02.png

 which shares the same vertex but has opposite concavity and two real roots.

image07.png

image01.png

image08.png

Let us proceed and use various values for

 and

...read more.

Middle

Giving

image02.png

Allowing us to express image02.png

in terms of  image02.png

 and image02.png

.

image04.png

image01.png

image10.png

image11.png

image11.png

I will now investigate similar cubic functions to see if the relationship between image11.png

 and image02.png

 remains the same.

Let us consider

image09.png

 and

image09.png

Graphed:

image21.png

The shadow generating function is given by the equation

,

image02.png

Expressing image02.png

in terms of  image02.png

 and image02.png

:

image09.png

image02.png

image02.png

image11.png

Which is the same relationship as with our previous cubic. Let us see if this pattern holds true with another similar cubic.

image09.png

with shadow function

image09.png

Graphed:

image22.png

The shadow generating function can be read to be

,

image02.png

Again, expressing image02.png

in terms of  image02.png

 and image02.png

:

image09.png

image02.png

image02.png

image11.png

It seems that, just like in the case of the quadratics, the relationship between image02.png

, image02.png

and image02.png

 is given by the equation:

image10.png

Let us try to prove this statement, by generalising the functions image02.png

, image02.png

and image02.png

.

We have:

image04.png

image04.png

.

These two cubics intersect where image11.png

,

image01.png

image01.png

image02.png

The points of intersection are thus

 and image02.png

.

And the equation of the line passing through both points image02.png

.

Where image13.png

image02.png

Giving image05.png

The relationship between image02.png

, image02.png

and image02.png

:

image04.png

image10.png

image10.png

image10.png

image11.png

My assumption has been proven correct.

The relationship between image02.png

, image02.png

and image02.png

 in the case of cubic polynomials is exactly the same as for quadratics. Perhaps, the zeros of image02.png

 can also be used in the same way to graphically determine the real and imaginary components of the complex zeros of image02.png

.

Taking into consideration the function image09.png

The roots are

 and

, as proven in part A.

...read more.

Conclusion

image02.png

 will still work.

To show this I will use the function

image04.png

 with roots

 and

.

The shadow function will have roots

 and

 with equation:

image04.png

Graphed:

image26.png

As we have two pairs of conjugate roots in this case, we will also have to draw two circles, with centre A and C (representing the

 and

 parts of the real roots under the form of

 and

) and radius

 and

 respectively. The lines passing through A and C and parallel to the y-axis intercept the circles at E, F and G, H. The new points represent, as expected, the complex roots

 and

 when interpreted on the Argand plane.

Conclusion:

In this assessment, I have investigated and proven the relationship between quadratics and cubics with complex roots and their shadow-functions: image10.png

.  I also showed that, when knowing the equation of the shadow function, you can easily, graphically identify the complex roots of the original function, with my "circle"-method.

However, the case of quartics is more complicated. First of all, there are two different cases to consider:

The first case with two complex roots  allowed me to define the shadow-generating function as image15.png

, maintaining the above stated relationship.

But for the second case with four complex roots I failed to identify the shadow-generating function.

Yet, both cases still permitted me to graphically determine the complex roots of the original function from key-points along the x-axis of the shadow-function.

...read more.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Maths essays

  1. Investigating Quadratic functions

    This is because y=4x� is narrower than y=2x�. Now, I would like to graph the family y=ax� with a coefficient between zero and one. Again, the position of the two functions and its vertex are the same. Their openings are also facing up which is probably because the coefficient is positive, however I can't fully conclude yet.

  2. Math Portfolio: trigonometry investigation (circle trig)

    in trial to verify the conjecture, the value of sin turn out to be positive while the values of cos and tan turn out to be negative. When we put a random angle from quadrant 3, the range of -180<?<-90, -164� in trial to verify the conjecture, the value of

  1. Math SL Circle Portfolio. The aim of this task is to investigate positions ...

    8 This pattern between the length of and (when is held constant) develops an exponential relationship (' ? ). Therefore, the general statement to represent this would be: ' = (n= r) To check the validity of the general statement, other values for r were also calculated.

  2. Ib math HL portfolio parabola investigation

    * This can also be expressed as (observed from the cubic equation on page 18): f(x) = axn - a (r1+ r2+r3...rn) xn-1 +.... + (-1)n a (r1r2r3...rn) I am now going to use mathematical induction and try and prove that the sum of the roots of a polynomial of degree n is Let us assume that p (k)

  1. Population trends. The aim of this investigation is to find out more about different ...

    My model isn't similar to what most of the data is like, the initial there points which are close to a date of the previous set of data are in a good range from my curve, however the points from the year 2000 onwards are very distant and aren't a prediction of what could happen according to my model.

  2. MAths HL portfolio Type 1 - Koch snowflake

    I have used another free online software in order to simulate this. Verifying the values using the diagrams: Number of sides: From the diagrams above we can see that each successive iteration has transformed each side into four different sides and of a smaller length.

  1. Moss's Egg. Task -1- Find the area of the shaded region inside the two ...

    Thus, AC2 = 33 + 32 = 18, BC2 = 32 + 32 = 18 AC = = , BC = = The perimeter of triangle ABC is the sum of AB + AC + BC P = + + 6 = + 6 14.5 cm * Note: We reject

  2. Gold Medal heights IB IA- score 15

    The c value affects the horizontal shift which is calculated after knowing all over variables then solving for c. Lastly, the d value represents the vertical shift, which can be calculated by finding the average point of heights. Algebraically approaching the function Amplitude (a)

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work