1. Complements are used in digital computers to simplify subtraction and logical manipulations. 1's complement inverts all bits of a binary number. 2's complement inverts all bits and adds 1 to the least significant bit.
2. Subtraction using 1's complement involves taking the 1's complement of the subtrahend, adding it to the minuend, and handling any carry. Addition using 1's complements involves taking the complement of negative numbers before adding.
3. Subtraction using 2's complement involves taking the 2's complement of the subtrahend and adding it to the minuend. The result is negative if there is no carry out, and positive if there is a carry out.
This document discusses the z-transform, which is a tool for analyzing discrete-time signals and systems analogous to the Laplace transform for continuous-time signals. It defines the z-transform, describes its region of convergence and pole-zero plot, lists some common z-transform pairs, and discusses how the location of poles and zeros in the z-plane relates to the stability of a system. Examples are provided to illustrate key concepts like finding the z-transform of sequences and determining the region of convergence.
Basic theorems and properties of boolean algebraHanu Kavi
The document discusses Boolean algebra laws and theorems that are used to simplify Boolean expressions and logic. Some of the key laws and theorems covered include the associative, distributive, commutative, absorption, and duality laws. De Morgan's theorems are also explained, which relate to taking the complement of a sum or product of variables. Truth tables are used to demonstrate De Morgan's theorems. The overall purpose is to provide some of the fundamental laws and theorems of Boolean algebra that can be applied to simplify logical expressions.
The document discusses various properties of signals including:
- Analog signals can have an infinite number of values while digital signals are limited to a set of values.
- Phase describes the position of a waveform relative to a reference point in time.
- Total energy and average power of continuous and discrete signals can be calculated through integrals and sums.
- Periodic, even, odd, exponential, and sinusoidal signals are described.
- Unit impulse and step signals are defined for both discrete and continuous time domains.
- A signal's frequency spectrum shows the collection of component frequencies and bandwidth is the range of these frequencies.
Digital Data, Digital Signal | Scrambling TechniquesBiplap Bhattarai
Digital signal is a sequence of discrete, discontinuous voltage pulses.
Each pulse is a signal element.
Binary data are transmitted by encoding the bit stream into signal elements.
In the simplest case, one bit is represented by one signal element.
- E.g., 1 is represented by a lower voltage level, and 0 is represented by a higher voltage level
This document discusses the design of finite impulse response (FIR) filters. It begins by describing the basic FIR filter model and properties such as filter order and length. It then covers topics such as linear phase response, different filter types (low-pass, high-pass, etc.), deriving the ideal impulse response, and filter specification in terms of passband/stopband edges and ripple levels. The document concludes by outlining the common FIR design method of windowing the ideal impulse response, describing popular window functions, and providing a step-by-step example of designing a low-pass FIR filter using the Hamming window.
The document discusses minimizing Boolean expressions using Karnaugh maps. It explains that Karnaugh maps provide a graphical way to simplify logic circuits by grouping adjacent 1s in the map. The steps for minimization using Karnaugh maps are outlined, including drawing the map, entering values, forming the largest possible groups of 1s, and selecting the fewest groups needed to cover all 1s. Rules for grouping such as group size and overlap are also covered.
1. Complements are used in digital computers to simplify subtraction and logical manipulations. 1's complement inverts all bits of a binary number. 2's complement inverts all bits and adds 1 to the least significant bit.
2. Subtraction using 1's complement involves taking the 1's complement of the subtrahend, adding it to the minuend, and handling any carry. Addition using 1's complements involves taking the complement of negative numbers before adding.
3. Subtraction using 2's complement involves taking the 2's complement of the subtrahend and adding it to the minuend. The result is negative if there is no carry out, and positive if there is a carry out.
This document discusses the z-transform, which is a tool for analyzing discrete-time signals and systems analogous to the Laplace transform for continuous-time signals. It defines the z-transform, describes its region of convergence and pole-zero plot, lists some common z-transform pairs, and discusses how the location of poles and zeros in the z-plane relates to the stability of a system. Examples are provided to illustrate key concepts like finding the z-transform of sequences and determining the region of convergence.
Basic theorems and properties of boolean algebraHanu Kavi
The document discusses Boolean algebra laws and theorems that are used to simplify Boolean expressions and logic. Some of the key laws and theorems covered include the associative, distributive, commutative, absorption, and duality laws. De Morgan's theorems are also explained, which relate to taking the complement of a sum or product of variables. Truth tables are used to demonstrate De Morgan's theorems. The overall purpose is to provide some of the fundamental laws and theorems of Boolean algebra that can be applied to simplify logical expressions.
The document discusses various properties of signals including:
- Analog signals can have an infinite number of values while digital signals are limited to a set of values.
- Phase describes the position of a waveform relative to a reference point in time.
- Total energy and average power of continuous and discrete signals can be calculated through integrals and sums.
- Periodic, even, odd, exponential, and sinusoidal signals are described.
- Unit impulse and step signals are defined for both discrete and continuous time domains.
- A signal's frequency spectrum shows the collection of component frequencies and bandwidth is the range of these frequencies.
Digital Data, Digital Signal | Scrambling TechniquesBiplap Bhattarai
Digital signal is a sequence of discrete, discontinuous voltage pulses.
Each pulse is a signal element.
Binary data are transmitted by encoding the bit stream into signal elements.
In the simplest case, one bit is represented by one signal element.
- E.g., 1 is represented by a lower voltage level, and 0 is represented by a higher voltage level
This document discusses the design of finite impulse response (FIR) filters. It begins by describing the basic FIR filter model and properties such as filter order and length. It then covers topics such as linear phase response, different filter types (low-pass, high-pass, etc.), deriving the ideal impulse response, and filter specification in terms of passband/stopband edges and ripple levels. The document concludes by outlining the common FIR design method of windowing the ideal impulse response, describing popular window functions, and providing a step-by-step example of designing a low-pass FIR filter using the Hamming window.
The document discusses minimizing Boolean expressions using Karnaugh maps. It explains that Karnaugh maps provide a graphical way to simplify logic circuits by grouping adjacent 1s in the map. The steps for minimization using Karnaugh maps are outlined, including drawing the map, entering values, forming the largest possible groups of 1s, and selecting the fewest groups needed to cover all 1s. Rules for grouping such as group size and overlap are also covered.
EC Binary Substraction using 1's Complement,2's ComplementAmberSinghal1
The binary number system represents all data as combinations of 0s and 1s. It is used in computer systems, where digits are combined to form binary numbers like 1001 or 11000110. A digit 0 or 1 in a binary number is called a bit. For example, 1001 is a 4-bit binary number and 11000110 is an 8-bit binary number. There are different methods for performing binary subtraction, including using 1's complement or 2's complement operations. With 1's complement subtraction, the bits of the number being subtracted are flipped before adding. With 2's complement subtraction, 1 is added to the 1's complement before adding.
The z-transform provides a method to analyze discrete-time signals and systems using complex variable theory. It is defined as the summation of a sequence multiplied by z to the power of the time index from negative infinity to positive infinity. The region of convergence consists of values of z where this summation converges. It is determined by the locations of the zeros and poles of the z-transform function. Examples show how different sequences lead to different regions of convergence bounded by these zeros and poles.
Gauss Forward And Backward Central Difference Interpolation Formula Deep Dalsania
This PPT contains the topic called Gauss Forward And Backward Central Difference Interpolation Formula of subject called Numerical and Statistical Methods for Computer Engineering.
The document describes the Modified Booth's Algorithm for binary multiplication of negative numbers. It uses bit pair recoding of the multiplier and defines a recoding table. As an example, it shows the step-by-step binary multiplication of -13 x -7 using bit pair recoding of the multiplier, multiplication according to the recoding table, and addition of partial products to get the final result of 91.
Introducing Data Types & Operators
Skills gained:
1- Familiarity with data types
2- Modeling Memories
3- More on Expressions & Operators
This is part of VHDL 360 course
This document provides information about sequences and series in mathematics. It defines sequences, limits of sequences, convergence and divergence of sequences, infinite series, tests to determine convergence of series like the divergence test, limit comparison test, ratio test, root test, and power series. Examples of applying these concepts to specific series are also included.
A demultiplexer is a device that takes a signal containing multiple data streams and reconstructs the original separate streams. It works in reverse of a multiplexer, which combines multiple streams into one signal. Demultiplexers come in various types depending on the number of select lines, which determine how many output streams they can separate the combined signal into, from 1-to-2 up to 1-to-16. Common applications of demultiplexers include communication systems and converting serial to parallel signals.
The document discusses proofs by contraposition. It explains that a statement of the form "if p then q" can be proven by showing its contrapositive "if not q then not p" is true. It provides examples of proofs using this method, including proving if n^2 is even then n is even, if m + n is even then m and n have the same parity, and if 3n + 2 is odd then n is odd. Homework exercises are provided applying this proof technique.
The document discusses different number systems:
- Decimal uses base 10 with digits 0-9
- Binary uses base 2 with digits 0-1
- Octal uses base 8 with digits 0-7
- Hexadecimal uses base 16 with digits 0-9 and A-F
It provides methods to convert between decimal, binary, octal, and hexadecimal numbers.
DIGITAL COMMUNICATION: ENCODING AND DECODING OF CYCLIC CODE ShivangiSingh241
Cyclic codes are a type of linear code where any cyclic shift of a codeword is also a codeword. This allows for efficient encoding and decoding using shift registers.
Encoding of cyclic codes can be done by dividing the message polynomial by the generator polynomial, with the remainder becoming the parity bits. Encoding circuits use shift registers with feedback to efficiently perform this division. Decoding uses the syndrome, which is computed by shifting the received word into a syndrome register. A decoder then attempts to match the syndrome to an error pattern, correcting errors one symbol at a time by shifting the syndrome and received word simultaneously.
1. Generating functions can be used to represent sequences as power series and solve recurrence relations.
2. Common examples of generating functions are presented for various sequences like the constant sequence {1,1,...}, the sequence of powers of 2, and binomial coefficients.
3. The process of using generating functions to solve recurrence relations involves rewriting the relation, multiplying by x^n and summing, identifying the generating function, and extracting the nth term.
Sequential circuits are circuits whose outputs depend not only on present inputs but also on past inputs or states. There are two types: synchronous use a clock signal to synchronize state changes, asynchronous can change state at any time. Common memory elements are flip-flops including RS, D, JK, and T flip-flops. The master-slave flip-flop construction using two flip-flops avoids unpredictable states by separating the sampling and output functions.
This document discusses predicates and quantifiers in predicate logic. It begins by explaining the limitations of propositional logic in expressing statements involving variables and relationships between objects. It then introduces predicates as statements involving variables, and quantifiers like universal ("for all") and existential ("there exists") to express the extent to which a predicate is true. Examples are provided to demonstrate how predicates and quantifiers can be used to represent statements and enable logical reasoning. The document also covers translating statements between natural language and predicate logic, and negating quantified statements.
Convolution discrete and continuous time-difference equaion and system proper...Vinod Sharma
This document discusses discrete time signals and discrete time convolution. Discrete time convolution describes the output of a linear, time-invariant system when its input is one discrete signal and its impulse response is another. The output is the sum of each input sample multiplied by the corresponding flipped and shifted impulse response. This allows characterization of a system by its impulse response. The document provides examples of discrete time convolution and discusses properties such as the number of samples in the convolved output depending on the lengths of the input and impulse response signals.
What are Flip Flops and Its types. What are Flip Flops and Its types. What are Flip Flops and Its types. What are Flip Flops and Its types. What are Flip Flops and Its types. What are Flip Flops and Its types. What are Flip Flops and Its types.
This document discusses four basic concepts: sets, functions, relations, and binary operations. It provides definitions and examples of each concept. Sets are collections of distinct objects. A function relates each element in its domain to a unique element in its codomain. Relations define connections between elements in two sets. Binary operations are mathematical operations that combine two operands, like addition and multiplication, and have properties such as commutativity and distributivity.
The Laplace transform allows solving differential equations using algebra by transforming differential operators into algebraic operations. It transforms a function of time (f(t)) into a function of a complex variable (F(s)), allowing differential equations describing systems to be solved for variables of interest. Key properties include linearity, time and frequency shifting, and relationships between derivatives, integrals, and the Laplace transform that enable solving differential equations by taking the transform, performing algebra, and applying the inverse transform.
Mathematical Induction
CMSC 56 | Discrete Mathematical Structure for Computer Science
October 18, 2018
Instructor: Allyn Joy D. Calcaben
College of Arts & Sciences
University of the Philippines Visayas
This document summarizes key aspects of the discrete Fourier transform (DFT). It defines the DFT, provides the formula for calculating it, and explains that the DFT transforms a discrete-time signal from the time domain to the frequency domain. It also outlines several important properties of the DFT, including linearity, shift property, duality, symmetry, and circular convolution. Examples are provided to illustrate duality and symmetry. References for further information on the discrete Fourier transform are also included.
1) The document discusses solving linear recurrence relations, which are relations where each term is a linear combination of previous terms.
2) It provides methods for solving both homogeneous relations, where the right side is only previous terms, and non-homogeneous relations, where the right side also includes an additional term.
3) The key steps are finding the characteristic equation, which determines the possible solutions, then using the initial conditions to find the specific solution. Examples are worked through to demonstrate the process.
The document discusses recurrence relations and their applications. It begins by defining a recurrence relation as an equation that expresses the terms of a sequence in terms of previous terms. It provides examples of recurrence relations and their solutions. It then discusses solving linear homogeneous recurrence relations with constant coefficients by finding the characteristic roots and obtaining an explicit formula. Applications discussed include financial recurrence relations, the partition function, binary search, and the Fibonacci numbers. It concludes by discussing the case when the characteristic equation has a single root.
EC Binary Substraction using 1's Complement,2's ComplementAmberSinghal1
The binary number system represents all data as combinations of 0s and 1s. It is used in computer systems, where digits are combined to form binary numbers like 1001 or 11000110. A digit 0 or 1 in a binary number is called a bit. For example, 1001 is a 4-bit binary number and 11000110 is an 8-bit binary number. There are different methods for performing binary subtraction, including using 1's complement or 2's complement operations. With 1's complement subtraction, the bits of the number being subtracted are flipped before adding. With 2's complement subtraction, 1 is added to the 1's complement before adding.
The z-transform provides a method to analyze discrete-time signals and systems using complex variable theory. It is defined as the summation of a sequence multiplied by z to the power of the time index from negative infinity to positive infinity. The region of convergence consists of values of z where this summation converges. It is determined by the locations of the zeros and poles of the z-transform function. Examples show how different sequences lead to different regions of convergence bounded by these zeros and poles.
Gauss Forward And Backward Central Difference Interpolation Formula Deep Dalsania
This PPT contains the topic called Gauss Forward And Backward Central Difference Interpolation Formula of subject called Numerical and Statistical Methods for Computer Engineering.
The document describes the Modified Booth's Algorithm for binary multiplication of negative numbers. It uses bit pair recoding of the multiplier and defines a recoding table. As an example, it shows the step-by-step binary multiplication of -13 x -7 using bit pair recoding of the multiplier, multiplication according to the recoding table, and addition of partial products to get the final result of 91.
Introducing Data Types & Operators
Skills gained:
1- Familiarity with data types
2- Modeling Memories
3- More on Expressions & Operators
This is part of VHDL 360 course
This document provides information about sequences and series in mathematics. It defines sequences, limits of sequences, convergence and divergence of sequences, infinite series, tests to determine convergence of series like the divergence test, limit comparison test, ratio test, root test, and power series. Examples of applying these concepts to specific series are also included.
A demultiplexer is a device that takes a signal containing multiple data streams and reconstructs the original separate streams. It works in reverse of a multiplexer, which combines multiple streams into one signal. Demultiplexers come in various types depending on the number of select lines, which determine how many output streams they can separate the combined signal into, from 1-to-2 up to 1-to-16. Common applications of demultiplexers include communication systems and converting serial to parallel signals.
The document discusses proofs by contraposition. It explains that a statement of the form "if p then q" can be proven by showing its contrapositive "if not q then not p" is true. It provides examples of proofs using this method, including proving if n^2 is even then n is even, if m + n is even then m and n have the same parity, and if 3n + 2 is odd then n is odd. Homework exercises are provided applying this proof technique.
The document discusses different number systems:
- Decimal uses base 10 with digits 0-9
- Binary uses base 2 with digits 0-1
- Octal uses base 8 with digits 0-7
- Hexadecimal uses base 16 with digits 0-9 and A-F
It provides methods to convert between decimal, binary, octal, and hexadecimal numbers.
DIGITAL COMMUNICATION: ENCODING AND DECODING OF CYCLIC CODE ShivangiSingh241
Cyclic codes are a type of linear code where any cyclic shift of a codeword is also a codeword. This allows for efficient encoding and decoding using shift registers.
Encoding of cyclic codes can be done by dividing the message polynomial by the generator polynomial, with the remainder becoming the parity bits. Encoding circuits use shift registers with feedback to efficiently perform this division. Decoding uses the syndrome, which is computed by shifting the received word into a syndrome register. A decoder then attempts to match the syndrome to an error pattern, correcting errors one symbol at a time by shifting the syndrome and received word simultaneously.
1. Generating functions can be used to represent sequences as power series and solve recurrence relations.
2. Common examples of generating functions are presented for various sequences like the constant sequence {1,1,...}, the sequence of powers of 2, and binomial coefficients.
3. The process of using generating functions to solve recurrence relations involves rewriting the relation, multiplying by x^n and summing, identifying the generating function, and extracting the nth term.
Sequential circuits are circuits whose outputs depend not only on present inputs but also on past inputs or states. There are two types: synchronous use a clock signal to synchronize state changes, asynchronous can change state at any time. Common memory elements are flip-flops including RS, D, JK, and T flip-flops. The master-slave flip-flop construction using two flip-flops avoids unpredictable states by separating the sampling and output functions.
This document discusses predicates and quantifiers in predicate logic. It begins by explaining the limitations of propositional logic in expressing statements involving variables and relationships between objects. It then introduces predicates as statements involving variables, and quantifiers like universal ("for all") and existential ("there exists") to express the extent to which a predicate is true. Examples are provided to demonstrate how predicates and quantifiers can be used to represent statements and enable logical reasoning. The document also covers translating statements between natural language and predicate logic, and negating quantified statements.
Convolution discrete and continuous time-difference equaion and system proper...Vinod Sharma
This document discusses discrete time signals and discrete time convolution. Discrete time convolution describes the output of a linear, time-invariant system when its input is one discrete signal and its impulse response is another. The output is the sum of each input sample multiplied by the corresponding flipped and shifted impulse response. This allows characterization of a system by its impulse response. The document provides examples of discrete time convolution and discusses properties such as the number of samples in the convolved output depending on the lengths of the input and impulse response signals.
What are Flip Flops and Its types. What are Flip Flops and Its types. What are Flip Flops and Its types. What are Flip Flops and Its types. What are Flip Flops and Its types. What are Flip Flops and Its types. What are Flip Flops and Its types.
This document discusses four basic concepts: sets, functions, relations, and binary operations. It provides definitions and examples of each concept. Sets are collections of distinct objects. A function relates each element in its domain to a unique element in its codomain. Relations define connections between elements in two sets. Binary operations are mathematical operations that combine two operands, like addition and multiplication, and have properties such as commutativity and distributivity.
The Laplace transform allows solving differential equations using algebra by transforming differential operators into algebraic operations. It transforms a function of time (f(t)) into a function of a complex variable (F(s)), allowing differential equations describing systems to be solved for variables of interest. Key properties include linearity, time and frequency shifting, and relationships between derivatives, integrals, and the Laplace transform that enable solving differential equations by taking the transform, performing algebra, and applying the inverse transform.
Mathematical Induction
CMSC 56 | Discrete Mathematical Structure for Computer Science
October 18, 2018
Instructor: Allyn Joy D. Calcaben
College of Arts & Sciences
University of the Philippines Visayas
This document summarizes key aspects of the discrete Fourier transform (DFT). It defines the DFT, provides the formula for calculating it, and explains that the DFT transforms a discrete-time signal from the time domain to the frequency domain. It also outlines several important properties of the DFT, including linearity, shift property, duality, symmetry, and circular convolution. Examples are provided to illustrate duality and symmetry. References for further information on the discrete Fourier transform are also included.
1) The document discusses solving linear recurrence relations, which are relations where each term is a linear combination of previous terms.
2) It provides methods for solving both homogeneous relations, where the right side is only previous terms, and non-homogeneous relations, where the right side also includes an additional term.
3) The key steps are finding the characteristic equation, which determines the possible solutions, then using the initial conditions to find the specific solution. Examples are worked through to demonstrate the process.
The document discusses recurrence relations and their applications. It begins by defining a recurrence relation as an equation that expresses the terms of a sequence in terms of previous terms. It provides examples of recurrence relations and their solutions. It then discusses solving linear homogeneous recurrence relations with constant coefficients by finding the characteristic roots and obtaining an explicit formula. Applications discussed include financial recurrence relations, the partition function, binary search, and the Fibonacci numbers. It concludes by discussing the case when the characteristic equation has a single root.
This document discusses recurrence relations and their solutions. It begins by defining recurrence relations and providing examples of recurrence relations modeling real-world problems. It then explains that linear homogeneous recurrence relations of degree k can be solved by finding the characteristic roots of the characteristic equation. For relations of degree two, the solution is a linear combination of the characteristic roots raised to successive powers. The document provides examples of solving specific recurrence relations, including deriving an explicit formula for the Fibonacci numbers.
pdf of recurrence relation which is used in statisticspranjaltarte
This document discusses recurrence relations and their solutions. It begins by defining recurrence relations and providing examples of recurrence relations modeling real-world problems. It then explains that linear homogeneous recurrence relations of degree k can be solved by finding the characteristic roots of the characteristic equation. For relations of degree two, the solution is a linear combination of the characteristic roots raised to successive powers. The document provides examples of solving specific recurrence relations, including deriving an explicit formula for the Fibonacci numbers.
The document describes recursive sequences defined by initial conditions and a recurrence relation. It provides examples of linear homogeneous recurrence relations with constant coefficients, including their characteristic equations and solutions. It notes that any linear combination of solutions to a linear homogeneous recurrence is also a solution, and discusses finding general solutions when the characteristic equation has repeated roots.
The document discusses recurrence relations and methods for solving them. It defines a recurrence relation as an equation that expresses the terms of a sequence in terms of previous terms. It provides examples of homogeneous and non-homogeneous recurrence relations. For homogeneous relations, it describes guessing a solution of the form T(n)=x^n and finding the characteristic equation. For non-homogeneous relations, it explains adding the non-recursive term to the characteristic equation. It then works through examples of solving both homogeneous and non-homogeneous recurrence relations.
Impact of Linear Homogeneous Recurrent Relation Analysisijtsrd
A linear homogeneous recurrence relation of degree k with constant coefficients is a recurrence. A recurrence relation is an equation that recursively defines a sequence or multidimensional array of values, once one or more initial terms are given each further term of the sequence or array is defined as a function of the preceding terms. Thidar Hlaing "Impact of Linear Homogeneous Recurrent Relation Analysis" Published in International Journal of Trend in Scientific Research and Development (ijtsrd), ISSN: 2456-6470, Volume-3 | Issue-5 , August 2019, URL: https://www.ijtsrd.com/papers/ijtsrd26662.pdfPaper URL: https://www.ijtsrd.com/computer-science/other/26662/impact-of-linear-homogeneous-recurrent-relation-analysis/thidar-hlaing
The document discusses recurrence relations and methods for solving them. It covers:
- Recurrence relations define problems where the solution is defined in terms of smaller instances of the same problem.
- Methods for solving recurrence relations include the iterative method, substitution method, recursion tree method, and Master's method.
- Examples are provided to demonstrate applying these methods, such as expanding the recurrence iteratively until a pattern emerges or applying the Master's method.
The document discusses recurrence relations and their applications. It can be summarized as follows:
1. A recurrence relation expresses the terms of a sequence in terms of previous terms, providing a rule to generate all terms. It does not specify initial values.
2. Recurrence relations are useful for modeling many real-world phenomena like financial growth, partitions of sets, and algorithm analysis.
3. Linear homogeneous recurrence relations of constant coefficients can be solved by finding the characteristic roots from the characteristic equation. The general solution is a linear combination of terms with the roots.
The document discusses recursive algorithms and recurrence relations. It provides examples of solving recurrence relations for different algorithms like Towers of Hanoi, selection sort, and merge sort. Recurrence relations define algorithms recursively in terms of smaller inputs. They are solved to find closed-form formulas for the running time of algorithms.
The document discusses solving recurrence relations through the following steps:
1) Write the recurrence relation and initial conditions as a characteristic equation.
2) Find the solutions to the characteristic equation.
3) Use the solutions to write the general solution as a linear combination with constants.
4) Determine the constants using the initial conditions.
Two examples are provided to demonstrate solving both regular and similar solutions to linear homogeneous recurrence relations.
The document discusses techniques for solving homogeneous and inhomogeneous recurrence equations. It provides examples of solving recurrence equations by finding the characteristic equation, determining the roots, and using the general solution formula. It also discusses special cases like having repeated roots in the characteristic equation and divide-and-conquer recurrence relations.
LINEAR RECURRENCE RELATIONS WITH CONSTANT COEFFICIENTSAartiMajumdar1
This document discusses linear recurrence relations with constant coefficients. It covers homogeneous solutions, particular solutions, and the total solution. It also discusses solving recurrence relations using generating functions. Key points:
- Homogeneous solutions are found by solving the characteristic equation.
- Particular solutions are found for homogeneous and non-homogeneous equations.
- The total solution is the sum of the homogeneous and particular solutions.
- Generating functions can also be used to solve recurrence relations.
The document provides an introduction to the binomial theorem. It begins by discussing binomial coefficients through the Pascal's triangle. It then derives an explicit formula for binomial coefficients using factorials. Finally, it states the binomial theorem and provides examples of using it to expand algebraic expressions and estimate numerical values.
This document provides an introduction to z-transforms, which are a major mathematical tool for analyzing discrete systems including digital control systems. It discusses sequences and difference equations, which are foundational topics for understanding z-transforms. Specifically, it defines what a sequence is, distinguishes between first and second order difference equations, and explains how sequences can be shifted to the left or right. It also introduces key concepts like causal sequences and the unit step sequence that are important for z-transform analysis.
The document discusses different methods for solving recurrence equations:
1. Substitution method involves guessing a solution and using induction to prove it is correct.
2. Iterative method expands the recurrence into a summation and evaluates it.
3. Master method handles recurrences of the form T(n)=aT(n/b)+f(n). It provides 3 cases to bound the solution.
4. Recursion tree method models the recurrence as a tree to track problem sizes and costs at each level. Summing costs across levels provides the total solution.
This document discusses limiting ratios of generalized recurrence relations. It begins by introducing recurrence relations and defining a generalized recurrence relation with a parameter m. It then analyzes several special cases for values of m=2,3,4,5,10. For each case it determines the characteristic equation and uses numerical methods to find the positive real root, which represents the limiting ratio. It is shown that as m increases, the limiting ratios approach 2. The document formally proves this result with Theorem 1, showing that the limiting ratio λ of the generalized recurrence relation converges to 2 as m approaches infinity. In conclusion, the paper has generalized the Fibonacci recurrence relation and analyzed the behavior of limiting ratios for this more generalized case.
Building API data products on top of your real-time data infrastructureconfluent
This talk and live demonstration will examine how Confluent and Gravitee.io integrate to unlock value from streaming data through API products.
You will learn how data owners and API providers can document, secure data products on top of Confluent brokers, including schema validation, topic routing and message filtering.
You will also see how data and API consumers can discover and subscribe to products in a developer portal, as well as how they can integrate with Confluent topics through protocols like REST, Websockets, Server-sent Events and Webhooks.
Whether you want to monetize your real-time data, enable new integrations with partners, or provide self-service access to topics through various protocols, this webinar is for you!
The Rising Future of CPaaS in the Middle East 2024Yara Milbes
Explore "The Rising Future of CPaaS in the Middle East in 2024" with this comprehensive PPT presentation. Discover how Communication Platforms as a Service (CPaaS) is transforming communication across various sectors in the Middle East.
What to do when you have a perfect model for your software but you are constrained by an imperfect business model?
This talk explores the challenges of bringing modelling rigour to the business and strategy levels, and talking to your non-technical counterparts in the process.
Nashik's top web development company, Upturn India Technologies, crafts innovative digital solutions for your success. Partner with us and achieve your goals
Everything You Need to Know About X-Sign: The eSign Functionality of XfilesPr...XfilesPro
Wondering how X-Sign gained popularity in a quick time span? This eSign functionality of XfilesPro DocuPrime has many advancements to offer for Salesforce users. Explore them now!
WMF 2024 - Unlocking the Future of Data Powering Next-Gen AI with Vector Data...Luigi Fugaro
Vector databases are transforming how we handle data, allowing us to search through text, images, and audio by converting them into vectors. Today, we'll dive into the basics of this exciting technology and discuss its potential to revolutionize our next-generation AI applications. We'll examine typical uses for these databases and the essential tools
developers need. Plus, we'll zoom in on the advanced capabilities of vector search and semantic caching in Java, showcasing these through a live demo with Redis libraries. Get ready to see how these powerful tools can change the game!
How GenAI Can Improve Supplier Performance Management.pdfZycus
Data Collection and Analysis with GenAI enables organizations to gather, analyze, and visualize vast amounts of supplier data, identifying key performance indicators and trends. Predictive analytics forecast future supplier performance, mitigating risks and seizing opportunities. Supplier segmentation allows for tailored management strategies, optimizing resource allocation. Automated scorecards and reporting provide real-time insights, enhancing transparency and tracking progress. Collaboration is fostered through GenAI-powered platforms, driving continuous improvement. NLP analyzes unstructured feedback, uncovering deeper insights into supplier relationships. Simulation and scenario planning tools anticipate supply chain disruptions, supporting informed decision-making. Integration with existing systems enhances data accuracy and consistency. McKinsey estimates GenAI could deliver $2.6 trillion to $4.4 trillion in economic benefits annually across industries, revolutionizing procurement processes and delivering significant ROI.
Stork Product Overview: An AI-Powered Autonomous Delivery FleetVince Scalabrino
Imagine a world where instead of blue and brown trucks dropping parcels on our porches, a buzzing drove of drones delivered our goods. Now imagine those drones are controlled by 3 purpose-built AI designed to ensure all packages were delivered as quickly and as economically as possible That's what Stork is all about.
Boost Your Savings with These Money Management AppsJhone kinadey
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1. CS103A Handout 23
Winter 2002 February 22, 2002
Solving Recurrence Relations
Introduction
A wide variety of recurrence problems occur in models. Some of these recurrence relations can be solved
using iteration or some other ad hoc technique. However, one very important class of recurrence relations
can be explicitly solved in a systematic way. These are the recurrence relations that express the terms of a
sequence as a linear combination of previous terms.
Definition: A linear homogeneous recurrence relation of degree k with constant coefficients
is a recurrence relation of the form:
an = c1an− 1 + c2an−2 + L+ ckan−k
where c1,c 2 ,c 3 ,K,c k are real numbers, and ck ≠ 0 .
The recurrence relation in the definition is linear since the right-hand side is a sum of multiple of the
previous terms of the sequence. The recurrence relation is homogeneous since no terms of the recurrence
relation fail to involve a previous term of the sequence in some way. The coefficients of the terms of the
sequence are all constants, rather than functions that depend on n. The degree is k because an is expressed
in terms of the previous k terms of the sequence. A consequence of strong induction is that a sequence
satisfying the recurrence relation in the definition is uniquely determined by this recurrence relation and
the k initial condition:
an = c1an− 1 + c2an−2 + L+ ckan−k
a0 = C0
a1 = C1
KK
ak −1 = Ck −1
Solving the Little Monsters
The basic approach for solving linear homogeneous recurrence relations is to look for solutions of the form
an = r
n
, wherer is a constant. Note that an = r
n
is a solution to the recurrence if and only if:
r
n
= c1r
n−1
+ c2r
n− 2
+ c3r
n−3
+ L+ ckr
n−k
When both sides of the equation are divided by r
n−k
and the right-hand side is subtracted from the left,
we obtain the equivalent equation:
r
k
− c1r
k− 1
− c 2r
k − 2
− c3r
k − 3
−L − ck −1r − ck = 0
2. Consequently, the sequence with an = r
n
is a solution if and only if r is a solution to this last equation,
which is called the characteristic equation of the recurrence relation. The solutions of this equation are
called characteristic roots of the recurrence relation. As we’ll see, these characteristic roots can be used to
give an explicit formula for all the solutions of the recurrence relation. First, we should develop results
that deal with linear homogeneous recurrence relations with constant coefficients of degree two. The
results for second order relations can be extended to solve higher-order equations. The mathematics
involved in proving everything is really messy, so even though I’ll refer to it in lecture, I don’t want to
place it in here, because you might incorrectly think you’re responsible for knowing the proof, and you’re
not.
Let’s turn our undivided attention to linear homogeneous recurrence relations of degree two. First, consider
the case where there are two distinct characteristic roots.
A Fact Without Proof: Let c1 and c2 be real numbers. Suppose that r
2
− rc1 − c2 = 0 has two
distinct roots r1 and r2 . Then the sequence an{ } is a solution of the
recurrence relation an = c1an− 1 + c2an−2 if and only if an = b1r1
2
+ b2r2
2
for n = 0,1,2,3,4,5,K where b1 and b2 are constants determined by
the initial conditions of the recurrence relation.
Problem: Find an explicit solution to the tiling recurrence relation we developed
last time. You may recall that it went a-something like this:
Tn = Tn− 1 + Tn− 2
T0 = 1
T1 = 1
Well, we surmise that the solution will be a linear combination or terms having the form Tn = r
n
.
Following the rule from above, we just plug in Tn = r
n
into out recurrence relation and see what constraints
are placed on r . Let’s listen in:
Tn = Tn− 1 + Tn− 2
guess that Tn = r
n ⇒
r
n
= r
n− 1
+ r
n− 2
r
2
= r + 1
r2
− r − 1= 0
r1 =
1+ 5
2
;r2 =
1 − 5
2
Well, so the guess that Tn = r
n
is a good one provided that r be equal to one of the two roots above. In
fact, any linear combination of
1+ 5
2
n
and
1− 5
2
n
will also satisfy the recurrence if you ignore the
initial conditions for a moment—that is, Tn = b1
1+ 5
2
n
+ b2
1− 5
2
n
for any constant coefficients as
factors. It’s only when you supply the initial conditions that b1 and b2 are required to adopt a specific
value. In fact, the initial conditions here dictate that:
2
3. b1 + b2 = 1
b1
1 + 5
2
+ b2
1− 5
2
= 1
And after solving for b1 and b2 do we finally arrive at the unique solution to our recurrence problem:
Tn =
5 + 5
10
1 + 5
2
n
+
5 − 5
10
1 − 5
2
n
How very satisfying.
Problem: Find a closed form solution to the bit string problem modeled in our last
handout.
Welllllll, the recurrence relation, save the initial conditions, is exactly the same as it that for the tiling
problem; that translates to a solution of the same basic form. But the fact that the initial conditions are
different hints that the constants multiplying the individual terms:
note same form, but possibly
different constants
The different initial conditions place different constraint on the values of b1 and b2 do we. Now the
following system of equations must be solved:
b1 + b2 = 1
b1
1 + 5
2
+ b2
1− 5
2
= 2
We end up with something like this as our solution in this case:
Bn =
5 + 3 5
10
1 + 5
2
n
+
5− 3 5
10
1− 5
2
n
Problem: Solve the recurrence relation given below
Jn = Jn−2
J0 = 3J1 = 5
You may be asking, “Jerry, where in the world is the Jn− 1 term?” Relax—it’s in there—it’s just that its
multiplying coefficient is zero, so I didn’t bother writing it in. It’s still a homogeneous equation of degree
two, so I solve it like any other recurrence of this type.
Bn = b1
1+ 5
2
n
+ b2
1− 5
2
n
3
4. Jn = Jn−2
J0 = 3J1 = 5
⇒
Jn = Jn−2
r2
= 1
r
2
− 1 = 0
r1 = 1;r2 = − 1
Jn = b1 1( )n
+ b2 −1( )n
= b1 + b2 −1( )n
Funny little twist—dropping the 1( )n
, since it’s transparent to us anyway.
b1 + b2 = 5
b1 − b2 =
5
3
b1 =
10
3
,b 2 =
5
3
⇒ Jn =
10
3
+
5
3
−1( )n
Solving Coupled Recurrence Relations
The first recurrences handout included examples where solutions involved coupled recurrence relations.
The circular Tower of Hanoi Problem, the n × 3 tiling problem, the 2 × 2 × npillar problem, and the
Martian DNA problem were all complex enough to require (or at least benefit from) the invention of a
second counting problem and a second recurrence variable. Remember the 2 × 2 × n pillar recurrence? If not,
here it is once more:
Sn =
1
2
2Sn−1 + Sn− 2 + 4Tn−1
n = 0
n = 1
n ≥ 2
Tn =
0
1
Sn−1 + Tn−1
n = 0
n = 1
n ≥ 2
Because S and T are defined in terms of one another, it’s possible to use the second recurrence of the two to
eliminate all occurrences or T from the first. It takes a algebra and a little ingenuity, but when we rid of
the second variable, we are often left with a single recurrence relation which can be solved like other
recurrences we’ve seen before. The above system of recurrences reduces to a single linear, homogeneous,
constant-coefficient equation once we get rid of T. Don’t believe me? Read on, Thomas.
Problem: Solve the above system of recurrences for Sn by eliminating Tn and
solving the linear equation that results.
I want to completely eliminate all traces ofT and arrive at (what will turn out to
be) a (cubic) recurrence relation for just Sn , and then solve it. First things first: We
want to get rid of theTn−1 term from the recurrence relation for Sn , and to do that, I make the neat little
observation that the second equality below follows from the first by replacing n by n-1:
Sn = 2Sn−1 + Sn− 2 + 4Tn−1
Sn−1 = 2Sn− 2 + Sn−3 + 4Tn−2
Subtracting the second equation from the first, we get:
4
5. Sn − Sn−1 = 2Sn− 1 + Sn− 2 + 4Tn−1( ) − 2Sn−2 + Sn− 3 + 4Tn− 2( )
= 2Sn−1 − Sn−2 − Sn− 3 + 4Tn− 1 − 4Tn− 2
Sn = 3Sn−1 − Sn−2 − Sn− 3 + 4Tn−1 − 4Tn− 2
= 3Sn−1 − Sn−2 − Sn− 3 + 4 Tn− 1 − Tn− 2( )
Believe it or not, this is progress, because I have something very interesting to say about Tn−1 − Tn− 2 —it’s
always equal to Sn− 2 . Just look at the crafty manipulations I work up from theTn recurrence relation:
Tn = Sn−1 + Tn−1
Tn−1 = Sn− 2 + Tn− 2
Tn−1 − Tn− 2 = Sn− 2 + Tn− 2 − Tn−2
= Sn− 2
How crafty! Now I can eradicate any mention ofT from the Sn recurrence relation, and I do so like-a this:
Sn = 3Sn−1 − Sn− 2 − Sn−3 + 4 Tn−1 − Tn− 2( )
= 3Sn−1 − Sn− 2 − Sn−3 + 4Sn−2
= 3Sn−1 + 3Sn− 2 − Sn−3
Therefore, an uncoupled recurrence relation for Sn can be expressed as follows:
Sn =
1
2
2S1 + S0 + 4T1 = 9
3Sn−1 + 3Sn− 2 − Sn−3
n = 0
n = 1
n = 2
n ≥ 3
Admittedly, that was a lot of work, but this is pretty exciting, because we’re on the verge of taking what
is clearly a homogeneous, constant-coefficient equation and coming up with a closed form solution. As
always, guess a solution of the form Sn = a
n
and substitute to see what values of a make everything work
out regardless of the initial conditions. More algebra:
Sn S n = a
n = 3Sn−1 + 3Sn− 2 − Sn−3 Sn =a n
a
n
= 3a
n−1
+ 3a
n− 2
− a
n−3
a3
= 3a2
+ 3a − 1
0 = a
3
− 3a
2
− 3a + 1
0 = a + 1( ) a
2
− 4a + 1( )
0 = a + 1( ) a − 2 + 3( ) a − 2 − 3( )
a = − 1, 2 ± 3
That means that the general solution to our recurrence (ignoring the initial conditions) is
Sn = x 2 + 3( )
n
+ y 2 − 3( )
n
+ z −1( )n
, where x, y, and z can be any real numbers. Boundary conditions
require that:
5
6. S0 = x + y+ z = 1
S1 = 2 + 3( )x + 2 − 3( )y − z = 2
S2 = 7 + 4 3( )x + 1 − 4 3( )y + z = 9
This is a system of three linear equations for three unknowns, and it admits exactly one solution. Solving
for x, y, and z is a matter of algebra, and after all that algebra is over, we converge on a closed formula of
Sn =
1
6
2 + 3( )
n+1
+
1
6
2 − 3( )
n+1
+
1
3
−1( )n
, which is
1
6
2 + 3( )
n+1
rounded to the nearest integer. Neat!
Problem: Revisit the Martian DNA problem, and show that the number of valid
Martian DNA strands of length n is given as an + bn = F3n + 2 (yes, the Fibonacci
number!)
Let’s bring back the recurrence that defined a and b.
an =
0 n = 0
2 n = 1
an−1 + 2bn−1 n ≥ 2
bn =
1 n = 0
3 n = 1
2an− 1 + 3bn−1 n ≥ 2
Eliminate one of the recurrence terms in order to get a closed solution (though this time we’ll ultimately
need to solve for both variables, because you’re interested in their sum.)
an = an−1 + 2bn− 1
bn = 2an−1 + 3bn−1
Notice that the first one tells us something about an − an−1 , so let’s subtract a shifted version of the second
equation from the original to get at something that’s all b and no a.
bn = 2an− 1 + 3bn−1
bn−1 = 2an− 2 + 3bn−2
bn − bn−1 = 2 an−1 − an− 2( ) + 3bn−1 − 3bn− 2
bn = bn−1 + 2 an−1 − an− 2( )+ 3bn−1 − 3bn− 2
The first equation tells us that an−1 − an− 2 = 2bn− 2 . Substitution yields
bn = bn−1 + 4bn− 2 + 3bn−1 − 3bn− 2
= 4bn−1 + bn− 2
The characteristic equation here is r
2
− 4r − 1 = 0 ; bn = c1 2 + 5( )
n
+ c2 2 − 5( )
n
. Because b0 = 1 and
b1 = 3 , we determine c1 , c2 by solving the following two equations:
6
7. c1 + c2 = 1
2 + 5( )c1 + 2 − 5( )c2 = 3
⇒ c1 =
5 + 5
10
, c2 =
5 − 5
10
.
bn =
5 + 5
10
2 + 5( )
n
+
5 − 5
10
2 − 5( )
n
.
You might think you have to do the same thing for a, and in theory you’re right in that you need to solve
it, but we more or less have. Because
bn = 2an−1 + 3bn−1
we actually know that
bn = 2an−1 + 3bn−1
2an−1 = bn − 3bn−1
an−1 =
1
2
bn −
3
2
bn− 1
an =
1
2
bn+1 −
3
2
bn
an + bn =
1
2
bn+1 −
3
2
bn + bn
=
1
2
bn+1 −
1
2
bn
Recall that we’re really just interested in an + bn , and since it can be defined just in terms of the bn, we get:
an + bn =
1
2
bn+1 −
3
2
bn + bn
=
1
2
bn+1 −
1
2
bn
=
1
2
5 + 5
10
2 + 5( )
n+1
+
5 − 5
10
2 − 5( )
n+1
−
1
2
5 + 5
10
2 + 5( )
n
+
5 − 5
10
2 − 5( )
n
=
1
2
5 + 5
10
2 + 5( )− 1( )2 + 5( )
n
+
1
2
5 − 5
10
2 − 5( )− 1( ) 2 − 5( )
n
=
1
2
5 + 5
10
1 + 5( ) 2 + 5( )
n
+
1
2
5 − 5
10
1 − 5( ) 2 − 5( )
n
=
1
2
10 + 6 5
10
2 + 5( )
n
+
1
2
10 − 6 5
10
2 − 5( )
n
=
5 + 3 5
10
2 + 5( )
n
+
5 − 3 5
10
2 − 5( )
n
That’s typically the closed-form you’d leave it in, but we also want to show that an + bn = F3n + 2 . Here
goes:
7
8. F3n + 2 =
1
5
1+ 5
2
3n +2
−
1
5
1− 5
2
3n + 2
=
1
5
1+ 5
2
2
1+ 5
2
3n
−
1
5
1− 5
2
2
1− 5
2
3n
=
1
5
6 + 2 5
4
1+ 5
2
3
n
−
1
5
6 − 2 5
4
1− 5
2
3
n
=
6 + 2 5
4 5
2 + 5( )
n
−
6 − 2 5
4 5
2 + 5( )
n
=
5
5
6 + 2 5
4 5
2 + 5( )
n
−
5
5
6 − 2 5
4 5
2 + 5( )
n
=
10 + 6 5
20
2 + 5( )
n
+
10 − 6 5
20
2 + 5( )
n
=
5 + 3 5
10
2 + 5( )
n
+
5 − 3 5
20
2 + 5( )
n
Therefore, an + bn = F3n + 2 , and we rejoice.
Problem: What about the Circular Tower of Hanoi recurrence? Why can’t we solve
that one as easily?
Well, you probably already noticed that both the pillar and the DNA recurrences
didn’t have any inhomogeneous terms anywhere. That’s not the case with the
Circular Tower of Hanoi recurrence, so while we are certainly invited to eliminate one of the variables
from the set of recurrences, there’s not much hope for solving it like we did for Martian DNA.
Qn =
0;
2Rn− 1 + 1
if n = 0
if n > 0
Rn =
0;
Qn + Qn−1 + 1
if n = 0
if n > 0
Clearly, it’s a piece of cake to define Qn in terms of previous Qs, but these 1s just aren’t going to go away.
Qn = 2Rn− 1 + 1
= 2 Qn−1 + Qn− 2 + 1( )+ 1
= 2Qn− 1 + 2Qn− 2 + 3
so that
Qn =
0
1
2Qn + 2Qn−1 + 3
if n = 0
if n = 1
if n > 1
We’re sorta bumming because of that inhomogeneous 3.
There is a trick to solving this one, but it’s not something I’m formally going to require you to know, since
there’s little pedagogical value in memorizing tricks. For those of you with nothing to do on a Friday
8
9. night, try substituting Qn = An−1 − 1 into the above system (making sure to adjust those base cases as
well.) Solve for An , then take whatever you got there and subtract 1 from it to get Qn .
General Approach to Solving all Linear First-Order Recurrences
There is a general technique that can reduce virtually any recurrence of the form
anTn = bnTn− 1 + cn
to a sum. The idea is to multiply both sides by a summation factor, sn , to arrive at
snanTn = snbnTn− 1 + sncn . This factor sn is chosen to make snbn = sn− 1an−1 . Then if we write Sn = snanTn
we have a sum-recurrence for Sn as:
Sn = Sn−1 + sncn
Hence Sn = s0a0T0 + sk ck
1≤ k ≤n
∑ = s1b1T0 + sk ck
1≤ k ≤n
∑ and the solution to the original recurrence becomes
Tn =
1
snan
s1b1T0 + skck
1≤k ≤ n
∑
.
So, you ask: How can we be clever enough to find the perfect sn ? Well, the relation sn = sn− 1an−1 bn can
be unfolded by repeated substitution for the si to tell us that the fraction:
sn =
an− 1an− 2an− 3 Ka1
bnbn− 1bn− 2 Kb2
(or any convenient constant multiple of this value) will be a suitable summation factor.
Problem: Solve Vn =
5 n =0
nVn− 1 + 3⋅n! n ≥ 1
by finding the appropriate summation
factor.
Derivation of the solution just follows protocol. You're specifically told to use
summation factors here, so we should do that. Notice here that an = 1 and bn = n, so that the summation
factor should be chosen as:
sn =
1n− 1
n!
=
1
n!
V0 = 5 for sure, but the recurrence formula Vn = nVn− 1 + 3⋅n! becomes
1
n!
Vn =
1
n − 1( )!
Vn−1 + 3. Let
Tn =
1
n!
Vn for the time being, just so we can solve an easier recurrence relation. We tackle Tn = Tn− 1 + 3,
and wouldn't you know it: Tn = 3n + T0 = 3n +
1
0!
V0 = 3n + 5 . Tn =
1
n!
Vn . We need Vn = n!Tn , so therefore
we have that Vn = n! 3n + 5( ). Woo!
9
10. Problem: Solve Cn =
0 n =0
n + 1 +
2
n
Ck
0≤k ≤n− 1
∑ n ≥ 1
by finding the appropriate
summation factor.
Let’s write out the recurrences for Cn and Cn−1 a little more explicitly.
Cn = n + 1 +
2
n
Ck
0≤k ≤n− 1
∑
Cn−1 = n +
2
n − 1
Ck
0≤k ≤n− 2
∑
Subtracting the second one from the first one is a good idea, but only after the multiply through by a factor
that’ll make each of the summations equal to each other:
Cn = n + 1 +
2
n
Ck
0≤k ≤n− 1
∑
n − 1
n
Cn−1 =
n − 1
n
n +
n − 1
n
2
n − 1
Ck
0≤k ≤n− 2
∑
= n − 1 +
2
n
Ck
0≤k ≤n− 2
∑
Subtracting the second one from the first, we arrive at:
Cn −
n − 1
n
Cn− 1 = n + 1 +
2
n
Ck
0≤ k≤ n−1
∑
− n − 1+
2
n
Ck
0 ≤k ≤n− 2
∑
= 2 +
2
n
Cn−1
Cn =
n + 1
n
Cn−1 + 2
nCn = n + 1( )Cn− 1 + 2n
If we choose a summation factor of sn =
an− 1an− 2an− 3 Ka1
bnbn− 1bn− 2 Kb2
=
n − 1( ) n − 2( ) n − 3( )K1
n + 1( )n n − 1( )K3
=
2
n n + 1( )
and multiply
through, we arrive at:
2
n n + 1( )
nCn =
2
n n + 1( )
n + 1( )Cn−1 +
2
n n + 1( )
2n
2
n + 1( )
Cn =
2
n
Cn− 1 +
4
n + 1( )
Cn
n + 1
=
Cn−1
n
+
2
n + 1
10
11. If we let n + 1( )Dn = Cn, then we finally arrive at an equation that looks reasonable: Dn = Dn−1 +
2
n + 1
.
Repeated substitution yields the following:
Dn = Dn−1 +
2
n + 1
= Dn−2 +
2
n
+
2
n + 1
= Dn−3 +
2
n − 1
+
2
n
+
2
n + 1
= Dn−4 +
2
n − 2
+
2
n − 1
+
2
n
+
2
n + 1
M
= D0 + 2
1
k + 1
1≤k ≤n
∑
= 2 Hn+1 − 1( )
= 2 Hn +
1
n + 1
− 1
= 2 Hn −
n
n + 1
Recall that Cn = n + 1( )Dn, so that Cn = 2 n + 1( ) Hn −
n
n + 1
= 2 n + 1( )Hn − 2n.
11