25 KiB
| title | TARGET DECK | FILE TAGS | tags | ||
|---|---|---|---|---|---|
| Sequences | Obsidian::STEM | algebra::sequence |
|
Overview
A sequence is an ordered list of numbers. These are usually represented as a closed formula or a recursive definition.
%%ANKI Basic What is a sequence? Back: An ordered list of values. Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI
Basic
How does sequence 0, 1, 2, \ldots differ from set \mathbb{N}?
Back: Order matters in the former.
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI
Basic
How is sequence a_1, a_2, \ldots more compactly denoted?
Back: (a_n)_{n \geq 1}
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI
Basic
What is the first index of (a_n)_{n \geq 1}?
Back: 1
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI
Basic
What terms make up the expanded form of sequence (a_n)_{n \geq 0}?
Back: a_0, a_1, a_2, \ldots
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI
Basic
What term refers to the subscripts in sequence a_0, a_1, a_2, \ldots?
Back: Indices.
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI Basic How do functions and sequences relate to one another? Back: Every sequence is equivalent to a function mapping index to value. Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI
Basic
How is sequence a_0, a_1, a_2, \ldots interpreted as a function?
Back: As a function that assigns n \mapsto a_n.
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI Basic What two methods allow exact specification of a sequence? Back: Closed formulas and recursive definitions. Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI
Basic
What is a closed formula for sequence e.g. (a_n)_{n \geq 0}?
Back: A formula that can be computed using a fixed number of "well-known" standard operations, independent of n.
Reference: Ronald L. Graham, Donald Ervin Knuth, and Oren Patashnik, Concrete Mathematics: A Foundation for Computer Science, 2nd ed (Reading, Mass: Addison-Wesley, 1994).
END%%
%%ANKI
Cloze
{n!} is a closed formula even though equivalent expression {\Pi_{k=1}^n k} is not.
Reference: Ronald L. Graham, Donald Ervin Knuth, and Oren Patashnik, Concrete Mathematics: A Foundation for Computer Science, 2nd ed (Reading, Mass: Addison-Wesley, 1994).
END%%
%%ANKI
Basic
Why is \Pi_{k=1}^n k not considered a closed formula?
Back: The number of "well-known" standard operations (i.e. multiplications) scale with n.
Reference: Ronald L. Graham, Donald Ervin Knuth, and Oren Patashnik, Concrete Mathematics: A Foundation for Computer Science, 2nd ed (Reading, Mass: Addison-Wesley, 1994).
END%%
%%ANKI
Basic
Why is n! considered a closed formula?
Back: Because the factorial operation is considered a "well-known" standard operation.
Reference: Ronald L. Graham, Donald Ervin Knuth, and Oren Patashnik, Concrete Mathematics: A Foundation for Computer Science, 2nd ed (Reading, Mass: Addison-Wesley, 1994).
END%%
%%ANKI Basic What two parts make up a recursive definition of a sequence? Back: The recurrence relation and an initial condition. Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI Basic What distinguishes a recurrence relation and a recursive definition? Back: A recursive definition includes a recurrence relation along with an initial condition. Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI
Basic
How is sequence (a_n)_{n \geq 0} = 1, 2, 4, 8, \ldots represented as a closed formula?
Back: a_n = 2^n
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI
Basic
How is sequence (a_n)_{n \geq 0} = 1, 2, 4, 8, \ldots represented as a recursive definition?
Back: a_n = 2 \cdot a_{n-1} with a_0 = 1.
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI Cloze It is harder to find {terms of a sequence} with a recursive definition than closed formula. Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI Basic Which is considered easier to find: closed formulas or recursive definitions? Back: Recursive definitions. Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI Basic What general strategy is typically employed when finding a closed formula for a sequence? Back: First find a recursive definition. Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
Recurrence Relations
To solve a recurrence relation means to find a closed form for the relation (with respect to initial conditions).
%%ANKI
Basic
What is the recurrence relation for the Fibonacci sequence?
Back: F_n = F_{n-1} + F_{n-2}
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI Basic What does it mean to solve a recurrence relation? Back: To find a closed formula satisfying the relation and initial conditions. Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
Telescoping
Telescoping refers to the property of summations in which consecutive terms cancel out. We can use telescoping to solve recurrences of form a_n = a_{n-1} + f(n) by noticing that: \begin{align*} a_1 - a_0 & = f(1) \ a_2 - a_1 & = f(2) \ \vdots \ a_n - a_{n-1} & = f(n) \ \hline a_n - a_0 & = \sum_{k=1}^n f(n) \end{align*}
%%ANKI Basic What does it mean for a sum to be telescoping? Back: Pairs of consecutive terms in the summation cancel each other out. Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI Basic What imagery is invoked by the term "telescoping" with respect to a sum? Back: A collapsing telescope. Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI Basic What summands typically remain after evaluating a telescoping sum? Back: The first and the last. Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI
Basic
What is the result of the following after observing telescoping? (2 - 1) + (3 - 2) + \cdots + (100 - 99) + (101 - 100)
Back: -1 + 101
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI
Basic
How is the following rewritten to highlight telescoping? (2 - 1) + (3 - 2) + \cdots + (100 - 99) + (101 - 100)
Back: (-1 + 2) + (-2 + 3) + \cdots + (-99 + 100) + (-100 + 101)
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI
Basic
What is the result of the following? \sum_{n=1}^N (a_n - a_{n-1})
Back: a_N - a_0
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI
Basic
What property is used to quickly verify the following identity? \sum_{n=1}^N (a_n - a_{n-1}) = a_N - a_0$$
Back: This is a telescoping sum.
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI
Basic
Schematically show how telescoping can be used to solve a_n = a_{n-1} + f(n).
Back: \begin{align*}
a_1 - a_0 & = f(1) \
& \vdots \
a_n - a_{n-1} & = f(n) \
\hline
a_n - a_0 & = \sum_{k=1}^n f(k)
\end{align*}$$Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI
Basic
What is the closed formula of recurrence a_n = a_{n-1} + f(n)?
Back: a_n = a_0 + \sum_{k=1}^n f(k)
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI
Basic
What summation property can be used to derive the closed formula of a_n = a_{n-1} + f(n)?
Back: Telescoping.
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
Iteration
Iteration refers to the expansion of terms, starting at the initial conditions, in the hope of discovering a pattern. It is more general than #Telescoping is. Consider a_n = a_{n-1} + f(n) again. We solve with iteration like so: \begin{align*} a_1 & = a_0 + f(1) \ a_2 & = (a_0 + f(1)) + f(2) \ \vdots \ a_n & = (\cdots(a_0 + f(1)) + f(2)) + \cdots) + f(n) \ \hline a_n & = a_0 + \sum_{k=1}^n f(n) \end{align*}
%%ANKI Basic What does it mean to solve a recurrence relation using iteration? Back: Repeatedly expand terms, starting at the initial conditions, to discover a pattern. Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI
Basic
What is the result of "iterating" the following recursive definition twice? $a_n = 3a_{n-1} + 2$
Back: \begin{align*}
a_1 & = 3(a_0) + 2 \
a_2 & = 3(3(a_0) + 2) + 2
\end{align*}
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI
Basic
Schematically show how iteration can be used to solve a_n = a_{n-1} + f(n).
Back: \begin{align*}
a_1 & = a_0 + f(1) \
& \vdots \
a_n & = (\cdots((a_0 + f(1)) + f(2)) + \cdots) + f(n) \
\hline
a_n & = a_0 + \sum_{k=1}^n f(k)
\end{align*}
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI
Basic
How fast does the number of terms grow when iterating a_n = 3a_{n-1} + 2?
Back: Linearly.
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI
Basic
How fast does the number of terms grow when iterating a_n = 2a_{n-1} + 3a_{n-2}?
Back: Exponentially (the number of terms double each iteration).
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI
Basic
Why should you avoid using iteration to solve a_n = 2a_{n-1} + 3a_{n-2}?
Back: The number of terms grows unwieldy very quickly.
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
%%ANKI Basic When solving recurrences, is telescoping or iteration a more general technique? Back: Iteration. Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
END%%
Characteristic Roots
When encountering linear homogeneous recurrence relations with constant coefficients, we can use the characteristic root technique to solve. We demonstrate with a quadratic characteristic polynomial, though this technique generalizes to higher-order polynomials as well.
Given recurrence relation a_n + \alpha a_{n-1} + \beta a_{n-2} = 0, the characteristic polynomial is r^2 + \alpha r + \beta. If r_1 and r_2 are distinct roots of the characteristic polynomial, then the solution to the recurrence relation is $a_n = ar_1^n + br_2^n$
where a and b are determined by the initial conditions. If the characteristic polynomial only has one root r, the solution is instead $a_n = ar^n + bnr^n$
%%ANKI Basic The characteristic root technique only works when solving what kind of recurrence relation? Back: Linear homogeneous recurrence relations with constant coefficients. Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf. Tags: algebra::polynomial
END%%
%%ANKI Basic What does "linear" refer to in "linear homogeneous recurrence relations with constant coefficients"? Back: The recurrence relation is a linear combination of previous terms. Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf. Tags: algebra::polynomial
END%%
%%ANKI Basic What does "homogeneous" refer to in "linear homogeneous recurrence relations with constant coefficients"? Back: Every term in the relation is a multiple of previous terms. Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf. Tags: algebra::polynomial
END%%
%%ANKI
Basic
Why isn't a_n = 2a_{n-1} + 3a_{n-2} a linear homogeneous recurrence relation with constant coefficients?
Back: It is.
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
Tags: algebra::polynomial
END%%
%%ANKI
Basic
Why isn't a_n = 2a_{n-1} \cdot 3a_{n-2} a linear homogeneous recurrence relation with constant coefficients?
Back: It is non-linear.
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
Tags: algebra::polynomial
END%%
%%ANKI
Basic
Why isn't a_n = 2a_{n-1} + 3 a linear homogeneous recurrence relation with constant coefficients?
Back: It is non-homogeneous.
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
Tags: algebra::polynomial
END%%
%%ANKI
Basic
Why isn't a_n = 2a_{n-1} + na_{n-2} a linear homogeneous recurrence relation with constant coefficients?
Back: It has a nonconstant coefficient.
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
Tags: algebra::polynomial
END%%
%%ANKI Basic What is the most famous linear homoegeneous recurrence relation with constant coefficients? Back: The Fibonacci sequence's recurrence relation. Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf. Tags: algebra::polynomial
END%%
%%ANKI
Basic
Why might we guess the solution to e.g. a_n = a_{n-1} + 6a_{n-2} is geometric?
Back: Every step of iteration multiplies a previous iteration by 6.
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
Tags: algebra::polynomial
END%%
%%ANKI
Basic
How is a_n = a_{n-1} + 6a_{n-2} factored to yield its characteristic polynomial?
Back: r^{n-2}(r^2 - r - 6) = 0
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
Tags: algebra::polynomial
END%%
%%ANKI
Basic
What is the characteristic equation of a_n = a_{n-1} + 6a_{n-2}?
Back: r^2 - r - 6 = 0
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
Tags: algebra::polynomial
END%%
%%ANKI
Basic
What is the characteristic polynomial of a_n = a_{n-1} + 6a_{n-2}?
Back: r^2 - r - 6
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
Tags: algebra::polynomial
END%%
%%ANKI
Basic
What is the characteristic polynomial of a_n + \alpha a_{n-1} + \beta a_{n-2} = 0?
Back: r^2 + \alpha r + \beta
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
Tags: algebra::polynomial
END%%
%%ANKI
Basic
What is the characteristic equation of a_n + \alpha a_{n-1} + \beta a_{n-2} = 0?
Back: r^2 + \alpha r + \beta = 0
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
Tags: algebra::polynomial
END%%
%%ANKI
Basic
What is the characteristic equation of a_n = \alpha a_{n-1} + \beta a_{n-2}?
Back: r^2 - \alpha r - \beta = 0
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
Tags: algebra::polynomial
END%%
%%ANKI
Basic
Given recurrence a_n = \alpha a_{n-1} + \beta a_{n-2}, what guess is used to derive the concept of a characteristic polynomial?
Back: The guessing of a geometric solution, e.g. r^n.
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
Tags: algebra::polynomial
END%%
%%ANKI
Basic
What distinguishes the characteristic polynomial from the characteristic equation of a recurrence relation?
Back: The latter sets the characteristic polynomial equal to 0.
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
Tags: algebra::polynomial
END%%
%%ANKI Basic Using the quadratic characteristic root technique, what distinguishes the form of the closed solution? Back: The number of distinct roots of the characteristic polynomial. Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf. Tags: algebra::polynomial
END%%
%%ANKI
Basic
Suppose the characteristic polynomial of a_n = \alpha a_{n-1} + \beta a_{n-2} has distinct roots r_1 and r_2. What is its solution?
Back: a_n = ar_1^n + br_2^n where a and b are determined by initial conditions.
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
Tags: algebra::polynomial
END%%
%%ANKI
Basic
Suppose the characteristic polynomial of a_n = \alpha a_{n-1} + \beta a_{n-2} has single root r. What is its solution?
Back: a_n = ar^n + bnr^n where a and b are determined by initial conditions.
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
Tags: algebra::polynomial
END%%
%%ANKI
Basic
How many distinct roots does the characteristic polynomial of the following have? $a_n = \alpha a_{n-1} + \beta a_{n-2} = ar^n + bnr^n$
Back: One.
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
Tags: algebra::polynomial
END%%
%%ANKI
Basic
How many distinct roots does the characteristic polynomial of the following have? $a_n = \alpha a_{n-1} + \beta a_{n-2} = ar_1^n + br_2^n$
Back: Two.
Reference: Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
Tags: algebra::polynomial
END%%
Bibliography
- Oscar Levin, Discrete Mathematics: An Open Introduction, 3rd ed., n.d., https://discrete.openmathbooks.org/pdfs/dmoi3-tablet.pdf.
- Ronald L. Graham, Donald Ervin Knuth, and Oren Patashnik, Concrete Mathematics: A Foundation for Computer Science, 2nd ed (Reading, Mass: Addison-Wesley, 1994).