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The first few terms of a series \(u_{n}\), starting with \(u_{0}\), are \(1,2,2,1,6,-3\). The series is generated by a recurrence relation of the form $$ u_{n}=P u_{n-2}+Q u_{n-4} $$ where \(P\) and \(Q\) are constants. Find an expression for the general term of the series and show that the series in fact consists of two other interleaved series given by $$ \begin{array}{r} u_{2 m}=\frac{2}{3}+\frac{1}{3} 4^{m} \\ u_{2 m+1}=\frac{7}{3}-\frac{1}{3} 4^{m} \end{array} $$ for \(m=0,1,2, \ldots\)

Short Answer

Expert verified
For the series \(u_{n}\): \(P = 5\) and \(Q = -4\). Interleaved sequences are: \(u_{2 m} = \frac{2}{3} + \frac{1}{3} 4^{m}\) and \(u_{2 m+1} = \frac{7}{3} - \frac{1}{3} 4^{m}\).

Step by step solution

01

- Identify the recurrence formula and initial variables

Given the series \(u_{0} = 1, u_{1} = 2, u_{2} = 2, u_{3} = 1, u_{4} = 6, u_{5} = -3\) and the relation \(u_{n} = P u_{n-2} + Q u_{n-4}\), set up equations using the known terms.
02

- Set up the equations

Substitute the known series terms into the recurrence relation to form equations: \(u_{4} = P u_{2} + Q u_{0}\). This gives \(6 = P \cdot 2 + Q \cdot 1\) leading to \(6 = 2P + Q\). Similarly for \(u_{5} = P u_{3} + Q u_{1}\), this gives \(-3 = P \cdot 1 + Q \cdot 2\), leading to \(-3 = P + 2Q\).
03

- Solve for P and Q

Solve the system of linear equations: \6 = 2P + Q\ \-3 = P + 2Q\. Multiplying the second equation by 2 gives \(-6 = 2P + 4Q\). Subtracting from the first equation \6 - (-6) = (2P + Q) - (2P + 4Q)\ yields \12 = -3Q\, giving \(Q = -4\). Substituting back to find \(P\), \6 = 2P - 4\ which simplifies to \10 = 2P\, giving \(P = 5\).
04

- Verify the recurrence relation

Use \(P = 5\) and \(Q = -4\) to verify the recurrence relation with the given terms. This relation is valid based on the provided terms.
05

- Identify interleaved series

Notice that the series has two interleaved subsequences: \(u_{2m}\) and \(u_{2m+1}\). Recognize that \(u_{2m}\) occurs at even indices (0,2,4...) and \(u_{2m+1}\) at odd indices (1,3,5...).
06

- Find the expressions for interleaved series

Form the equations based on known values and pattern recognition for these series. Thus: for \(u_{2m} = \frac{2}{3} + \frac{1}{3}4^m\), verify: \(u_{0} = 1, u_{2} = 2, u_{4} = 6\). For the other \(u_{2m+1} = \frac{7}{3} - \frac{1}{3}4^m\), verify: \(u_{1} = 2, u_{3} = 1, u_{5} = -3\).

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Key Concepts

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

linear algebra
Linear algebra is a branch of mathematics concerning vector spaces and linear mappings between such spaces. It is fundamental for handling systems of linear equations. In this exercise, you saw how we used a recurrence relation which is essentially a linear equation, to relate terms of the series. By setting up equations using the given terms of the series, we were able to form a system of linear equations:
\[6 = 2P + Q\]
\[-3 = P + 2Q\]
We solved these equations using basic algebraic techniques, highlighting the core of linear algebra. This step is crucial because it allows us to find the constants that define the relationship between the terms in our series.
series
A series is a sequence of numbers in which the terms are related to each other by a specific formula or rule. In our problem, the series is defined by the recurrence relation \(u_{n} = P u_{n-2} + Q u_{n-4}\) and initial terms. By inspecting the provided terms \(u_{0} = 1, u_{1} = 2, u_{2} = 2, u_{3} = 1, u_{4} = 6, u_{5} = -3\), we can determine the form of the series:
\[u_{2m} = \frac{2}{3} + \frac{1}{3}4^m\]
\[u_{2m+1} = \frac{7}{3} - \frac{1}{3}4^m\]
These formulas describe two interleaved series, one for even indices and one for odd indices. Recognizing this helps us understand the pattern and progression of the series terms more clearly.
pattern recognition
Pattern recognition is the ability to observe patterns, quickly identify, and often predict future occurrences based on these observations. In this exercise, recognizing that the series terms could be split into two interleaved sub-series was crucial. This was done by looking at the terms and noting:
  • Even indexed terms \(u_{2m}\): \(u_{0} = 1, u_{2} = 2, u_{4} = 6\)
  • Odd indexed terms \(u_{2m+1}\): \(u_{1} = 2, u_{3} = 1, u_{5} = -3\)
By analyzing these patterns, we were able to write the general expressions for both subsequences, accurately describing the whole series.
system of equations
A system of equations is a set of multiple equations used to solve for multiple variables. In our task, we had to find the constants \(P\) and \(Q\) in the recurrence relation \(u_{n} = P u_{n-2} + Q u_{n-4}\). The equations formed were:
\[6 = 2P + Q\]
\[-3 = P + 2Q\]
By solving this system, we found that \(P = 5\) and \(Q = -4\). This step involved standard techniques from linear algebra and algebra, emphasizing how systems of equations are solved to find unknown values in a mathematical context. This understanding is applicable not only in series analysis but in various mathematical and real-world problems.

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Most popular questions from this chapter

The theory of bent beams shows that at any point in the beam the 'bending moment' is given by \(K / \rho\), where \(K\) is a constant (that depends upon the beam material and cross-sectional shape) and \(\rho\) is the radius of curvature at that point. Consider a light beam of length \(L\) whose ends, \(x=0\) and \(x=L\), are supported at the same vertical height and which has a weight \(W\) suspended from its centre. Verify that at any point \(x(0 \leq x \leq L / 2\) for definiteness) the net magnitude of the bending moments, (bending moment \(=\) force \(\times\) perpendicular distance) due to the weight and support reactions, evaluated on either side of \(x\), is \(W x / 2\). If the beam is only slightly bent, so that \((d y / d x)^{2} \ll 1\), where \(y=y(x)\) is the downward displacement of the beam at \(x\), show that the beam profile satisfies the approximate equation $$ \frac{d^{2} y}{d x^{2}}=-\frac{W x}{2 K} $$ By integrating this equation twice and using physically imposed conditions on your solution at \(x=0\) and \(x=L / 2\), show that the downward displacement at the centre of the beam is \(W L^{3} /(48 K)\).

The function \(f(t)\) satisfies the differential equation $$ \frac{d^{2} f}{d t^{2}}+8 \frac{d f}{d t}+12 f=12 e^{-4 t} $$ For the following sets of boundary conditions determine whether it has solutions, and, if so, find them: (a) \(f(0)=0, \quad f^{\prime}(0)=0, \quad f(\ln \sqrt{2})=0\) (b) \(f(0)=0, \quad f^{\prime}(0)=-2, \quad f(\ln \sqrt{2})=0 .\)

The two functions \(x(t)\) and \(y(t)\) satisfy the simultaneous equations $$ \begin{aligned} &\frac{d x}{d t}-2 y=-\sin t \\ &\frac{d y}{d t}+2 x=5 \cos t \end{aligned} $$ Find explicit expressions for \(x(t)\) and \(y(t)\), given that \(x(0)=3\) and \(y(0)=2\). Sketch the solution trajectory in the \(x y\)-plane for \(0 \leq t<2 \pi\), showing that the trajectory crosses itself at \((0,1 / 2)\) and passes through the points \((0,-3)\) and \((0,-1)\) in the negative \(x\)-direction.

Two unstable isotopes \(A\) and \(B\) and a stable isotope \(C\) have the following decay rates per atom present: \(A \rightarrow B, 3 \mathrm{~s}^{-1} ; A \rightarrow C, 1 \mathrm{~s}^{-1} ; B \rightarrow C, 2 \mathrm{~s}^{-1}\). Initially a quantity \(x_{0}\) of \(A\) is present and none of the other two types. Using Laplace transforms, find the amount of \(C\) present at a later time \(t\).

Find the general expression for the \(u_{n}\) satisfying $$ u_{n+1}=2 u_{n-2}-u_{n-1} $$ with \(u_{0}=u_{1}=0\) and \(u_{2}=1\), and show that they can be written in the form $$ u_{n}=\frac{1}{5}-\frac{2^{n / 2}}{\sqrt{5}} \cos \left(\frac{3 \pi n}{4}-\phi\right) $$ where \(\tan \phi=2\).

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