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(a) By solving Eq. (9) numerically show that the real part of the complex roots changes sign when \(r \cong 24.737\). (b) Show that a cubic polynomial \(x^{3}+A x^{2}+B x+C\) has one real zero and two pure imaginary zeros only if \(A B=C\). (c) By applying the result of part (b) to Eq. (9) show that the real part of the complex roots changes sign when \(r=470 / 19\).

Short Answer

Expert verified
#Answer# (a) By solving Eq. (9) numerically, we can find out the value of r when the real part of the complex roots changes sign is close to 24.737. (b) A cubic polynomial x³ + Ax² + Bx + C has one real zero and two pure imaginary zeros if A*B = C. (c) Applying the result from part (b) to Eq. (9) and using numerical methods to find the value of r, we can verify that the real part of the complex roots changes sign when r = 470/19.

Step by step solution

01

(a) Numerical solution of Eq. (9)

To solve Eq. (9) numerically and find out the value of \(r\) when the real part of the complex roots changes sign, we need to use a numerical method such as the bisection method, Newton-Raphson method, or any other suitable method to find the root of the equation. Once the value of r is found, we can verify if it is close to 24.737.
02

(b) Condition for pure imaginary zeros of a cubic polynomial

To find the condition for a cubic polynomial to have one real zero and two pure imaginary ('9) numerically satisfy given float roots(condition \(r=470/19\)), compute the real and imaginary parts accordingly, and check if real parts of conditions change signs considering the numerical error tolerance. When the real part of complex roots changes sign and the conditions above are met, it can be concluded that \(r=470/19\), proving the result.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complex Roots
Complex roots are solutions to polynomial equations that involve the square root of negative numbers. These roots have a real part and an imaginary part. We deal with complex numbers using the format: \(a + bi\), where \(a\) is the real part and \(bi\) is the imaginary part.
Often in mathematics, we encounter equations that have these types of solutions, especially when dealing with polynomials of degree greater than or equal to two.
  • The concept of complex roots is essential when discussing polynomial roots, where not all solutions are neatly plotted on a real number line.
  • Understanding how these roots behave is vital for numerical methods, which can predict or analyze the behavior of complex systems.
Complex roots often occur in conjugate pairs, meaning if \(a + bi\) is a root, then \(a - bi\) is also a root, especially in real-coefficient polynomials. Being able to determine when and how these roots change can be crucial for specific mathematical applications, like in the exercise we discuss here.
Cubic Polynomials
Cubic polynomials are equations of the form \(x^3 + Ax^2 + Bx + C\). These are important because they represent the simplest form of polynomial equations that can have complex, real, and/or mixed types of roots.
  • A cubic equation will always have three roots, as given by the Fundamental Theorem of Algebra, considering multiplicity and complex numbers.
  • The nature of these roots can be all real, two complex and one real, or all three complex.
In part (b) of our original exercise, it is shown that for a cubic polynomial to have one real and two pure imaginary (complex) roots, the condition \(AB=C\) must be satisfied. This condition indicates a specific relationship among the coefficients that dictates the root types. Verifying this may require computation of discriminants or using other criteria that indicate the change from real to complex roots or vice versa.
Newton-Raphson Method
The Newton-Raphson method is a powerful technique for finding successively better approximations to the roots (or zeros) of a real-valued function. This method is particularly advantageous for its speed and efficiency.
  • The method uses an iterative process to approach a single root. Starting with an initial guess, it refines this guess using the formula: \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\).
  • This technique is especially useful for handling more complex roots when compared to simpler methods like the Bisection Method.
The key to success with the Newton-Raphson method is choosing an initial guess close to the actual root, ensuring convergence to the correct answer. Additionally, there is a need to compute both the function and its derivative, making it more suited to problems where these are easily accessible or computable. In the context of solving the cubic polynomial from the exercise, this method could help locate precise values where root conditions change.
Bisection Method
Unlike Newton-Raphson, the Bisection Method is a more basic numerical method for finding roots of a continuous function. This method works by repeatedly dividing an interval in half and then selecting the subinterval in which the root must lie. This is effective because it relies on the Intermediate Value Theorem.
  • The process is straightforward: if a function \(f\) changes signs over an interval \([a, b]\), then a root exists between \(a\) and \(b\).
  • By repeatedly halving the interval and choosing the subinterval in which the sign change occurs, one can home in on a root to within a desired precision.
This method is simple but guarantees success as long as the function meets the continuity criteria and initially has different signs at the ends of the interval. This conscientious approach is especially useful in contexts where a robust, though slower, method is preferred to more complex iterations like Newton-Raphson. It's commonly used for locating sign changes, important in assessing the values of \(r\) where complex root conditions alter.

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Most popular questions from this chapter

an autonomous system is expressed in polar coordinates. Determine all periodic solutions, all limit cycles, and determine their stability characteristics. $$ d r / d t=r(1-r)(r-2), \quad d \theta / d t=-1 $$

Assuming that the trajectory corresponding to a solution \(x=\phi(t), y=\psi(t),-\infty0\) such that \(\phi\left(t_{0}+T\right)=x_{0}, \psi\left(t_{0}+T\right)=y_{0} .\) Show that \(x=\Phi(t)=\phi(t+T)\) and \(y=\Psi(t)=\psi(t+T)\) is a solution and then use the existence and uniqueness theorem to show that \(\Phi(t)=\phi(t)\) and \(\Psi(t)=\psi(t)\) for all \(t .\)

(a) Find all the critical points (equilibrium solutions). (b) Use a computer to draw a direction field and portrait for the system. (c) From the plot(s) in part (b) determine whether each critical point is asymptotically stable, stable, or unstable, and classify it as to type. $$ d x / d t=(2+x)(y-x), \quad d y / d t=(4-x)(y+x) $$

In this problem we show how small changes in the coefficients of a system of linear equations can affect a critical point that is a center. Consider the system $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{0} & {1} \\ {-1} & {0}\end{array}\right) \mathbf{x} $$ Show that the eigenvalues are \pmi so that \((0,0)\) is a center. Now consider the system $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{\epsilon} & {1} \\ {-1} & {\epsilon}\end{array}\right) \mathbf{x} $$ where \(|\epsilon|\) is arbitrarily small. Show that the eigenvalues are \(\epsilon \pm i .\) Thus no matter how small \(|\epsilon| \neq 0\) is, the center becomes a spiral point. If \(\epsilon<0,\) the spiral point is asymptotically stable; if \(\epsilon>0,\) the spiral point is unstable.

Consider the eigenvalues given by equation ( 39 ). Show that $$\left(\sigma_{1} X+\sigma_{2} Y\right)^{2}-4\left(\sigma_{1} \sigma_{2}-\alpha_{1} \alpha_{2}\right) X Y=\left(\sigma_{1} X-\sigma_{2} Y\right)^{2}+4 \alpha_{1} \alpha_{2} X Y$$ Hence conclude that the eigenvalues can never be complex-valued.

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