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Suppose \(a_{0}=0, a_{1}=1\), and \(a_{k+1}=3 a_{k}+4 a_{k-1}\) for all \(k \geq 1\). Use methods of linear algebra to find the formula for \(a_{k}\).

Short Answer

Expert verified
The formula for \(a_k\) given the linear recurrence relation \(a_{k+1} = 3a_k + 4a_{k-1}\) and initial conditions \(a_0 = 0\) and \(a_1 = 1\) is: \[a_k = \frac{1}{5} \cdot 4^k - \frac{1}{5} \cdot (-1)^k\]

Step by step solution

01

Given the recurrence relation \(a_{k+1} = 3a_k + 4a_{k-1}\), we can identify that this is a linear recurrence relation with constant coefficients due to the linear relationship between the terms. #Step 2: Find the characteristic polynomial #

We need to find the characteristic polynomial associated with the recurrence relation. In the case of the recurrence relation \(a_{k+1} = 3a_k + 4a_{k-1}\), the polynomial is: \[r^2 - 3r - 4\] #Step 3: Find the roots of the polynomial #
02

Now, we will find the roots of the polynomial found in step 2. To do this, we will factor the polynomial: \[(r - 4)(r + 1)\] Thus, the roots of the polynomial will be \(r = 4\) and \(r = -1\). #Step 4: Write down the general solution for the linear recurrence relation #

The general solution for a linear homogenous recurrence relation is \[a_k = c_1 \cdot 4^k + c_2 \cdot (-1)^k\] Where \(c_1\) and \(c_2\) are constants that we need to find using the given initial conditions. #Step 5: Solve for constants \(c_1\) and \(c_2\) using the initial conditions #
03

With the initial conditions \(a_0 = 0\) and \(a_1 = 1\), we can find the values of \(c_1\) and \(c_2\) by plugging them into the general solution: For \(k = 0\) (initial condition): \(a_0 = c_1 \cdot 4^0 + c_2 \cdot (-1)^0 = 0\), meaning: \(c_1 + c_2 = 0\) For \(k = 1\) (initial condition): \(a_1 = c_1 \cdot 4^1 + c_2 \cdot (-1)^1 = 1\), meaning: \(4c_1 - c_2 = 1\) Solving this system of equations, we find that \(c_1 = \frac{1}{5}\) and \(c_2 = -\frac{1}{5}\). #Step 6: Write down the final formula for \(a_k\) #

Now that we have found the values for \(c_1\) and \(c_2\), we can write the final formula for \(a_k\): \[a_k = \frac{1}{5} \cdot 4^k - \frac{1}{5} \cdot (-1)^k\] This is the formula for \(a_k\) given the linear recurrence relation and initial conditions.

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Key Concepts

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

Characteristic Polynomial
In the study of linear recurrence relations, the characteristic polynomial plays a significant role. It acts like a bridge between the recurrence form and its solution. A characteristic polynomial is derived from a linear recurrence relation by replacing each term with powers of a common variable, typically denoted by \(r\). For example, if you have a recurrence relation of the form \(a_{k+1} = 3a_k + 4a_{k-1}\), the corresponding characteristic polynomial is obtained as \(r^2 - 3r - 4\).

This polynomial allows us to determine the behavior of the sequence defined by the recurrence relation. It is crucial to formulating the general solution of the sequence, as it provides the roots we need to construct the solution. In summary, the characteristic polynomial is the cornerstone for solving linear recurrence relations.
Roots of Polynomial
Once we have the characteristic polynomial, finding its roots is the next step. The roots give us essential "building blocks" for constructing the general solution of the recurrence relation.

To find these roots, we need to factor the polynomial. For instance, factoring the polynomial \(r^2 - 3r - 4\) results in \((r-4)(r+1)\). From this factorization, it's clear that the roots are \(r = 4\) and \(r = -1\).

These roots are key because each root corresponds to a component in the general solution formula. The nature of these roots affects how the solution behaves, especially when there are repeated roots or complex roots. In our case, both roots are real and distinct, which simplifies the solution.
General Solution for Recurrence Relations
With the roots of the characteristic polynomial in hand, we can write the general solution for the recurrence relation. The general solution for a linear homogeneous recurrence relation looks like a sum of terms involving powers of the roots.

If the roots are \(r_1\) and \(r_2\), the solution generally takes the form \(a_k = c_1 r_1^k + c_2 r_2^k\), where \(c_1\) and \(c_2\) are constants to be determined using initial conditions.

For example, given the roots \(4\) and \(-1\), the general solution is \(a_k = c_1 \, 4^k + c_2 \, (-1)^k\). These expressions tell us how every term in the sequence is constructed. These formulas are versatile, working for any initial conditions that provide the specific values we need for \(c_1\) and \(c_2\).
Initial Conditions in Recurrence Relations
Initial conditions are the specific values of the first few terms in a sequence, crucial for determining exact constants for the general solution. Given initial conditions allow us to solve for the unknown constants \(c_1\) and \(c_2\) in the general solution.

In our example, we were given \(a_0 = 0\) and \(a_1 = 1\). Using these, we can set up equations: for \(k = 0\), the equation becomes \(c_1 + c_2 = 0\), and for \(k = 1\), it becomes \(4c_1 - c_2 = 1\).

Solving these equations yields \(c_1 = \frac{1}{5}\) and \(c_2 = -\frac{1}{5}\). With these constants, the exact formula for \(a_k\) is \(a_k = \frac{1}{5} \, 4^k - \frac{1}{5} \, (-1)^k\). Initial conditions thus anchor the abstract solution to specific numerical values, defining the sequence uniquely.

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Most popular questions from this chapter

Consider the linear transformation \(T: \mathcal{M}_{n \times n} \rightarrow \mathcal{M}_{n \times n}\) defined by \(T(X)=X^{\top}\). Find its eigenvalues and the corresponding eigenspaces. (Hint: Consider the equation \(X^{\top}=\) \(\lambda X\).)

We say an \(n \times n\) matrix \(N\) is nilpotent if \(N^{r}=\mathrm{O}\) for some positive integer \(r\). a. Show that 0 is the only eigenvalue of \(N\). b. Suppose \(N^{n}=\mathrm{O}\) and \(N^{n-1} \neq \mathrm{O}\). Prove that there is a basis \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{n}\right\\}\) for \(\mathbb{R}^{n}\) with respect to which the matrix for \(N\) becomes (Hint: Choose \(\mathbf{v}_{1} \neq \mathbf{0}\) in \(\mathbf{C}\left(N^{n-1}\right)\), and then define \(\mathbf{v}_{2}, \ldots, \mathbf{v}_{n}\) appropriately to end up with this matrix representation. To argue that \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{n}\right\\}\) is linearly independent, you might want to mimic the proof of Theorem 2.1.)

Let \(a, b, c \in \mathbb{R}\), and let \(f\left(x_{1}, x_{2}\right)=a x_{1}^{2}+2 b x_{1} x_{2}+c x_{2}^{2}\). a. The Spectral Theorem tells us that there exists an orthonormal basis for \(\mathbb{R}^{2}\) with respect to whose coordinates \(\left(y_{1}, y_{2}\right)\) we have $$ f\left(x_{1}, x_{2}\right)=\bar{f}\left(y_{1}, y_{2}\right)=\lambda y_{1}^{2}+\mu y_{2}^{2} . $$ Show that the \(y_{1} y_{2}\)-axes are obtained by rotating the \(x_{1} x_{2}\)-axes through an angle \(\alpha\), where $$ \cot 2 \alpha=\frac{a-c}{2 b} . $$ Determine the type (ellipse, hyperbola, etc.) of the conic section \(f\left(x_{1}, x_{2}\right)=1\) from \(a, b\), and \(c\). (Hint: Use the characteristic polynomial to eliminate \(\lambda^{2}\) in your computation of \(\tan 2 \alpha\).) b. Use the formula for \(\bar{f}\) above to find the maximum and minimum of \(f\left(x_{1}, x_{2}\right)\) on the unit circle \(x_{1}^{2}+x_{2}^{2}=1\).

Show that the eigenvalues of an upper (or lower) triangular matrix are its diagonal entries.

Show that if \(\lambda\) is the only eigenvalue of a symmetric matrix \(A\), then \(A=\lambda I\).

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