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Chapter 7: Problem 4

Find the general solution of \(x_{n+2}-2 x_{n+1}+2 x_{n}=0\).

Short Answer

Expert verified
The general solution is \(x_n = C_1 (1+i)^n + C_2 (1-i)^n\).

Step by step solution

01

Find the Characteristic Equation

We start by assuming a solution of the form \(x_n = r^n\). Substituting this into the recurrence relation \(x_{n+2} - 2x_{n+1} + 2x_n = 0\), we get:\(r^{n+2} - 2r^{n+1} + 2r^n = 0\).We can divide through by \(r^n\) (assuming \(r eq 0\)) to obtain the characteristic equation:\[r^2 - 2r + 2 = 0.\]
02

Solve the Characteristic Equation

To solve the quadratic \(r^2 - 2r + 2 = 0\), we use the quadratic formula:\[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},\]where \(a = 1\), \(b = -2\), and \(c = 2\).Substituting these values, we get:\[r = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 8}}{2} = \frac{2 \pm \sqrt{-4}}{2}.\]Since the discriminant \(b^2 - 4ac = -4\) is negative, we have complex roots:\[r = \frac{2 \pm 2i}{2} = 1 \pm i.\]
03

Write the General Solution

The roots \(r_1 = 1+i\) and \(r_2 = 1-i\) are complex conjugates, which allows us to express the general solution using these roots. The general solution to the recurrence relation, when complex roots are present, is given by:\[x_n = C_1 (1+i)^n + C_2 (1-i)^n.\]This solution represents the linear combination of the terms with the coefficients \(C_1\) and \(C_2\), which are determined by any initial conditions if provided.

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Key Concepts

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

Characteristic Equation
When dealing with recurrence relations, a powerful approach to finding solutions is to determine the characteristic equation.
This is a polynomial equation that arises when assuming a solution of the form \(x_n = r^n\). Substituting this into the recurrence relation simplifies it significantly. In our exercise, the recurrence relation is \(x_{n+2} - 2x_{n+1} + 2x_n = 0\).
  • Set \(x_n = r^n\)
  • The recurrence relation transforms to \(r^{n+2} - 2r^{n+1} + 2r^n = 0\).
  • After dividing by \(r^n\), assuming \(r eq 0\), we get the characteristic equation \(r^2 - 2r + 2 = 0\).

The characteristic equation is a quadratic equation, and solving it will reveal the roots, which are key to finding the general solution.
Quadratic Formula
To solve the characteristic equation, you often need the quadratic formula, a fundamental tool for finding the roots of quadratic equations. A quadratic equation has the form \(ax^2 + bx + c = 0\). In our case, the characteristic equation is \(r^2 - 2r + 2 = 0\).
The quadratic formula is given by:
  • \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Identifying the coefficients from our equation:
  • \(a = 1\)
  • \(b = -2\)
  • \(c = 2\)
Substitute these into the formula:
  • \(r = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 2}}{2}\)
This simplifies to:
  • \(r = \frac{2 \pm \sqrt{4 - 8}}{2} = \frac{2 \pm \sqrt{-4}}{2}\)
Since the discriminant \(b^2 - 4ac\) is negative, the quadratic formula reveals that the equation has complex roots.
Complex Roots
Complex roots arise when the discriminant (\(b^2 - 4ac\)) in the quadratic formula is negative. This means there is no real square root, and the roots are complex numbers. In our case, the discriminant is \(-4\).
The roots of our characteristic equation are computed as:
  • \(r = \frac{2 \pm \sqrt{-4}}{2} = \frac{2 \pm 2i}{2}\)
This simplifies to:
  • \(r_1 = 1 + i\)
  • \(r_2 = 1 - i\)
These are complex conjugates. When recurrence relations have complex roots, the general solution is a linear combination of terms involving these roots:
  • \(x_n = C_1 (1+i)^n + C_2 (1-i)^n\)
Here, \(C_1\) and \(C_2\) are constants determined by initial conditions (if provided). Complex roots contribute oscillatory behaviors, which often appear in the solutions to recurrence relations with complex roots.

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Most popular questions from this chapter

Let \(v_{1}, \ldots, v_{n}\) be a basis of a vector space \(V\), and let \(\rho\) be a permutation of \(\\{1, \ldots, n\\}\). Let \(\alpha: V \rightarrow V\) be the unique linear map that satisfies \(\alpha\left(\sum_{j} \lambda_{j} v_{j}\right)=\sum_{j} \lambda_{j} v_{\rho(j)}\). Show that \(\alpha\) is an isomorphism of \(V\) onto itself.

By assuming that \(x_{n}=(-1)^{n}(\lambda n+\mu)\) find a solution of the difference equation \(x_{n+2}+(n+2) x_{n+1}+(n+2) x_{n}=0\).

Let \(V=\mathbb{R}\), and consider \(V\) as a vector space over \(\mathbb{R}\). Show that if \(U\) is a subspace of \(V\) then either \(U=\\{0\\}\) or \(U=V\).

Let \(V\) be a vector space over \(\mathbb{F}\) and let \(v_{1}, \ldots, v_{m}\) be any vectors in \(V\). Show that the set \(V_{0}\) of all linear combinations of \(v_{1}, \ldots, v_{m}\) is a vector space over \(\mathbb{F}\); it is the vector space generated by \(v_{1}, \ldots, v_{m}\).

Let \(S\) be the vector space of solutions of the differential equation \(\ddot{x}+4 x=0 .\) Show that \(\operatorname{dim}(S)=2\), and that \(\alpha: S \rightarrow S\) defined by \(\alpha(x)=\dot{x}\) is an isomorphism of \(S\) onto itself.
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