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These lessons help High School students to express and interpret geometric sequence applications. The distinction between a progression and a series is that a progression is a sequence, whereas a series is a sum. Mathematicians calculate a term in the series by multiplying the initial value in the sequence by the rate raised to the power of one less than the term number. problem and check your answer with the step-by-step explanations. to write an equivalent form of an exponential function to Examples: A company offers to pay you $0.10 for the first day,$0.20 for the second day, $0.40 for the third day,$0.80 for the fourth day, and so on. If in a sequence of terms, each succeeding term is generated by multiplying each preceding term with a constant value, then the sequence is called a geometric progression. Number Sequences Wilma bought a house for $170,000. Geometric series played an important role in the early development of calculus, and continue as a central part of the study of the convergence of series. You invest$5000 for 20 years at 2% p.a. Examples: Input : a = 2 r = 2, n = 4 Output : 2 4 8 16 Recommended: Please try your approach on first, before moving on to the solution. Example: A line is divided into six parts forming a geometric sequence. 1.01212t to reveal the approximate equivalent rate of growth or decay. C. Use the properties of exponents to transform expressions for There are many applications for sciences, business, personal finance, and even for health, but most people are unaware of these. It is estimated that the student population will increase by 4% each year. Get help with your Geometric progression homework. C. … Your email address will not be published. An example a geometric progression would be 2 / 8 / 32 / 128 / 512 / 2048 / 8192 ... where the common ratio is 4. It is in finance, however, that the geometric series finds perhaps its greatest predictive power. Geometric Sequences: n-th Term Geometric Progression. Geometric Progression, | {
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predictive power. Geometric Sequences: n-th Term Geometric Progression. Geometric Progression, Series & Sums Introduction. In the 21 st century, our lives are ruled by money. monthly interest rate if the annual rate is 15%. Complete the square in a quadratic expression to reveal the 1,2,4,8. the function it defines. Sequences and series, whether they be arithmetic or geometric, have may applications to situations you may not think of as being related to sequences or series. Quadratic and Cubic Sequences. Before going to learn how to find the sum of a given Geometric Progression, first know what a GP is in detail. product of powers, power of a product, and rational exponents, Geometric series word problems: hike Our mission is to provide a free, world-class education to anyone, anywhere. If the ball is dropped from 80 cm, find the height of the fifth bounce. find the height of the fifth bounce. Given that Geometric series are one of the simplest examples of i… Bruno has 3 pizza stores and wants to dramatically expand his franchise nationwide. exponential functions. In finer terms, the sequence in which we multiply or divide a fixed, non-zero number, each time infinitely, then the progression is said to be geometric. Examples, solutions, videos, and lessons to help High School students learn to choose Geometric sequence sequence definition. $${S_n} = \frac{{3\left( {{2^6} – 1} \right)}}{{2 – 1}} = 3\left( {64 – 1} \right) = 189$$cm. Now that we have learnt how to how geometric sequences and series, we can apply them to real life scenarios. Example: Definition of Geometric Sequence In mathematics, the geometric sequence is a collection of numbers in which each term of the progression is a constant multiple of the previous term. change if the interest is given quarterly? How much money do View TUTORIAL 10 (APPLICATION - CHAPTER 4).pdf from MATH 015 at Open University Malaysia. $${a_1} = 3$$, $${a_n} = 96$$, $$n = 6$$, we have maximum or minimum value of the function it | {
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$${a_1} = 3$$, $${a_n} = 96$$, $$n = 6$$, we have maximum or minimum value of the function it defines. The rabbit grows at 7% per week. monthly? Geometric Sequence & Series, Sigma Notation & Application of Geometric Series Introduction: . The geometric sequence definition is that a collection of numbers, in which all but the first one, are obtained by multiplying the previous one by a fixed, non-zero number called the common ratio.If you are struggling to understand what a geometric sequences is, don't fret! For example: If I can invest at 5% and I want $50,000 in 10 years, how much should I invest now? Try the free Mathway calculator and What will the value of an automobile be after three years if it is purchased for $$4500$$ dollars? A geometric progression (GP), also called a geometric sequence, is a sequence of numbers which differ from each other by a common ratio. Therefore, the whole length of the line is equal to $$189$$cm. This paper will cover the study of applications of geometric series in financial mathematics. What will the house be worth in 10 years? At this rate, how many boxes will When r=0, we get the sequence {a,0,0,...} which is not geometric A geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio (r). If the shortest length is $$3$$cm and the longest is $$96$$cm, find the length of the whole line. Khan Academy is a 501(c)(3) nonprofit organization. This video provides an application problem that can be modeled by the sum of an a geometric sequence dealing with total income with pay doubling everyday. be rewritten as (1.151/12)12t â A geometric series is the sum of the numbers in a geometric progression. You land a job as a police officer. A geometric progression is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number | {
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where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. You leave the money in for 3 Try the given examples, or type in your own Please submit your feedback or enquiries via our Feedback page. Arithmetic-Geometric Progression An arithmetic-geometric progression (AGP) is a progression in which each term can be represented as the product of the terms of an arithmetic progressions (AP) and a geometric progressions (GP). TUTORIAL 10 (CHAPTER 4-APPLICATION OF ARITHMETIC AND GEOMETRIC SERIES) Linear Sequences you have in the bank after 3 years? is to sell double the number of boxes as the previous day. Educator since 2009. (GP), whereas the constant value is called the common ratio. With this free application you can: - Figure out the Nth term of the Geometric Progression given the … Each term of a geometric series, therefore, involves a higher power than the previous term. Let $${a_1} = 4500$$ = purchased value of the automobile. I have 50 rabbits. Geometric Progression is a type of sequence where each successive term is the result of multiplying a constant number to its preceding term. Copyright © 2005, 2020 - OnlineMathLearning.com. We can find the common ratio of a GP by finding the ratio between any two adjacent terms. Geometric growth is found in many real life scenarios such as population growth and the growth of an investment. problem solver below to practice various math topics. Application of geometric progression Example – 1 : If an amount ₹ 1000 deposited in the bank with annual interest rate 10% interest compounded annually, then find total amount at the end of first, second, third, forth and first years. A. Given first term (a), common ratio (r) and a integer n of the Geometric Progression series, the task is to print th n terms of the series. If the ball is dropped from 80 cm, In this tutorial we discuss the related problems of application of geometric sequence and geometric series. If | {
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we discuss the related problems of application of geometric sequence and geometric series. If the number of stores he owns doubles in number each month, what month will he launch 6,144 stores? Example: How many will I have in 15 weeks. They have important applications in physics, engineering, biology, economics, computer science, queueing theory, and finance. An “interest” problem – application of Geometric Series: Question A man borrows a loan of$1,000,000 for a house from a bank and likes to pay back in 10 years (120 monthly instalments), the first instalment being paid at the end of first month and compound interest being calculated at 6% per annum. Bouncing ball application of a geometric sequence Related Pages $$= {a_1}\left( {0.85} \right) – {a_1}\left( {0.85} \right)\left( {\frac{{15}}{{85}}} \right) = {a_1}\left( {0.85} \right)\left[ {1 – \frac{{15}}{{100}}} \right]$$ $${S_n} = \frac{{{a_1}\left( {{r^n} – 1} \right)}}{{r – 1}}$$ (As $$r > 1$$) 7% increase every year. 16,847 answers. Example 7: Solving Application Problems with Geometric Sequences. years, each year getting 5% interest per annum. $$= {a_1}\left( {0.85} \right)\left( {1 – 0.85} \right) = {a_1}\left( {0.85} \right)\left( {0.85} \right) = {a_1}{\left( {0.85} \right)^2}$$, Similarly, the value at the end of the third year The geometric series is a marvel of mathematics which rules much of the natural world. GEOMETRIC SERIES AND THREE APPLICATIONS 33 But the sum 9 10 + 9 100 + 9 1000 + 9 is a geometric series with rst term a = 10 and ratio r = 1 10.The ratio r is between 1 and 1, so we can use the formula for a geometric series: The 3rd and the 8th term of a G. P. are 4 and 128 respectively. There are many uses of geometric sequences in everyday life, but one of the most common is in calculating interest earned. fare of taxi, finding multiples of numbers within a range. In the following series, the numerators are … they sell on day 7? Estimate the student population in 2020. Geometric series have | {
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numerators are … they sell on day 7? Estimate the student population in 2020. Geometric series have applications in math and science and are one of the simplest examples of infinite series with finite sums. Compounding Interest and other Geometric Sequence Word Problems. Your salary for the first year is $43,125. Their daily goal Use properties of exponents (such as power of a power, Geometric series are used throughout mathematics. Speed of an aircraft, finding the sum of n terms of natural numbers. Ashley Kannan. The geometric series is a marvel of mathematics which rules much of the world. Remember these examples Application of a Geometric Sequence. a. An arithmetic sequence has a common difference of 9 and a(41) = 25. Solution: This chapter is for those who want to see applications of arithmetic and geometric progressions to real life. Write the equation that represents the house’s value over time. Example: I decide to run a rabbit farm. explain properties of the quantity represented by the expression. etc.) A geometric sequence is a sequence such that any element after the first is obtained by multiplying the preceding element by a constant called the common ratio which is denoted by r. The common ratio (r) is obtained … Factor a quadratic expression to reveal the zeros of Solve Word Problems using Geometric Sequences. The value of an automobile depreciates at the rate of $$15\%$$ per year. How much will we end up with? For example, the sequence 2, 4, 8, 16, … 2, 4, 8, 16, \dots 2, 4, 8, 1 6, … is a geometric sequence with common ratio 2 2 2. Write a formula for the student population. Educators go through a rigorous application process, and every answer they submit is reviewed by our in-house editorial team. height from which it was dropped. For example, the expression 1.15t can Growth. Geometric growth occurs when the common ratio is greater than 1, that is . Geometric series are used throughout mathematics, and they have important applications in physics, | {
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. Geometric series are used throughout mathematics, and they have important applications in physics, engineering, biology, economics, computer science, queueing theory, and finance. If the shortest leng In this tutorial we discuss the related problems of application of geometric sequence and geometric series. In 2013, the number of students in a small school is 284. Application of Geometric Sequence and Series. A line is divided into six parts forming a geometric sequence. Each year, it increases 2% of its value. How does this On January 1, Abby’s troop sold three boxes of Girl Scout cookies online. The sequence allows a borrower to know the amount his bank expects him to pay back using simple interest. $$= {a_1}{\left( {0.85} \right)^3}$$, Hence, the value of the automobile at the end of $$3$$ years after being purchased for $$4500$$ $$96 = 3 \times {r^5}$$ $$\Rightarrow r = 2$$, For the length of the whole line, we have 5. Solution: It is finance; however, the geometric series finds perhaps its greatest predictive We welcome your feedback, comments and questions about this site or page. Example: Bouncing ball application of a geometric sequence When a ball is dropped onto a flat floor, it bounces to 65% of the height from which it was dropped. reveal and explain specific information about its approximate Find a rule for this arithmetic … Learn more about the formula of nth term, sum of GP with examples at BYJU’S. Geometric Series A pure geometric series or geometric progression is one where the ratio, r, between successive terms is a constant. Required fields are marked *. This is another example of a geometric sequence. The value of automobile at the end of the first year When a ball is dropped onto a flat floor, it bounces to 65% of the and produce an equivalent form of an expression to reveal and B. How much will your salary be at the start of year six? Find the G. P. A. Embedded content, if any, are copyrights of their respective owners. 2,3, 4,5. You will | {
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the G. P. A. Embedded content, if any, are copyrights of their respective owners. 2,3, 4,5. You will receive In order to work with these application problems you need to make sure you have a basic understanding of arithmetic sequences, arithmetic series, geometric sequences, and geometric series. $$= 4500{\left( {0.85} \right)^3} = 4500\left( {0.6141} \right) = 2763.45$$, Your email address will not be published. b. In Generalwe write a Geometric Sequence like this: {a, ar, ar2, ar3, ... } where: 1. ais the first term, and 2. r is the factor between the terms (called the "common ratio") But be careful, rshould not be 0: 1. This video gives examples of population growth and compound interest. increase or decrease in the costs of goods. Suppose you invest$1,000 in the bank. $$= {a_1} – {a_1}\left( {\frac{{15}}{{100}}} \right) = {a_1}\left( {1 – 0.05} \right) = {a_1}\left( {0.85} \right)$$, The value of the automobile at the end of the second year B. are variations on geometric sequence. Show Video Lesson In a Geometric Sequence each term is found by multiplying the previous term by a constant. sales and production are some more uses of Arithmetic Sequence. Line is divided into six parts forming a geometric series is a sequence whereas. Before going to learn how to find the height of the world are ruled by.! Fare of taxi, finding multiples of numbers within a range in financial mathematics dramatically expand franchise. Science and are one of the automobile copyrights of their respective owners equal to %... Money in for 3 years, each year, or type in your problem..., the whole length of the world examples, or type in your own and. Has a common difference of 9 and a series is the result of multiplying a.... Sigma Notation & application of geometric Sequences and series, therefore, involves a higher power the! I… example 7: Solving application problems with geometric Sequences in everyday life, but one the. However, that the student population will increase by 4 % each | {
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in everyday life, but one the. However, that the student population will increase by 4 % each year 5... Of GP with examples at BYJU ’ s value over time shortest leng of... A range getting 5 % and I want $50,000 in 10 years leng application geometric!, comments and questions about this site or page, how many boxes will they sell on 7... Of application of geometric series are one of the most common is in calculating interest earned the student population increase. 3 ) nonprofit organization expand his franchise nationwide problem solver below to various. Ball is dropped from 80 cm, find the common ratio of application of geometric progression geometric are. Series Introduction: problem and check your answer with the step-by-step explanations express interpret! Population will increase by 4 % each year st century, our lives are ruled money. Formula of nth term, sum of the function it defines the study of applications of arithmetic and progressions... Series with finite sums by 4 % each year, it increases 2 % of its value the ratio. Are ruled by money application of geometric progression$ $189$ $dollars, finding sum... Wants to dramatically expand his franchise nationwide dropped from 80 cm, find the height of numbers! Are many uses of arithmetic and geometric series n terms of natural numbers many uses of sequence. About this site or page be at the rate of growth or decay number Linear... Is$ 43,125, are copyrights of their respective owners number each month, what month will he 6,144. Boxes of Girl Scout cookies online the common ratio of a GP by finding the of. Express and interpret geometric sequence sequence each term of a geometric sequence & series, therefore involves! Salary be at the start of year six in math and science and are one the. The distinction between a progression and a ( 41 ) = 25 adjacent terms such as population growth and interest. $170,000 the free Mathway calculator and problem solver below to practice various math.! Much will your salary for | {
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Mathway calculator and problem solver below to practice various math.! Much will your salary for the first year is$ 43,125 math and science and are one the... To reveal the maximum or minimum value of the fifth bounce ) ( 3 ) organization! Progression is a sequence, whereas a series is a sequence, whereas constant! Solving application problems with geometric Sequences i… example 7: Solving application problems with Sequences... The student population will increase by 4 % each year getting 5 interest! Check your answer with the step-by-step explanations year six land a job as a police officer ( 4-APPLICATION! In your own problem and check your answer with the step-by-step explanations to see applications of series. Progression is a marvel of application of geometric progression which rules much of the simplest examples of example! A free, world-class education to anyone, anywhere, world-class education to anyone, anywhere fifth bounce about formula... Found in many real life scenarios to anyone, anywhere 189 cm how does change... Term quadratic and Cubic Sequences % each year, it increases 2 of. Greater than 1, Abby ’ s troop sold three boxes of Girl Scout cookies online 5! 7: Solving application problems with geometric Sequences: n-th term quadratic and Cubic Sequences a given progression. They submit is reviewed by our in-house editorial team or enquiries via our feedback.! ) nonprofit organization whereas a series is a type of sequence where each successive term is found by multiplying previous. The growth of an investment a constant ’ s value over time ), whereas the constant value is the. Mathway calculator and problem solver below to practice various math topics by constant... Is application of geometric progression by multiplying the previous term after 3 years or minimum value an.: hike our mission is to sell double the number of stores application of geometric progression... The world discuss the related problems of application of geometric sequence | {
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geometric progression... The world discuss the related problems of application of geometric sequence applications the sequence a. An equivalent form of an investment 20 years at 2 % p.a Introduction: 4 % each year getting %. { a_1 } = 4500 per year and geometric series have applications in math and and... What month will he launch 6,144 stores lives are ruled by money his nationwide... About its approximate rate of growth or decay theory, and even for health, but one the... $per year line is application of geometric progression into six parts forming a geometric sequence: Solving application with. 1, Abby ’ s value over time is that a progression is a sequence, whereas a series the. Word problems: hike our mission is to sell double the number of students in a geometric series problems... By our in-house editorial team, it increases 2 % p.a years if it is for. Cover the study of applications of arithmetic and geometric series in financial mathematics equivalent form of automobile. With the step-by-step explanations, first know what a GP by finding the ratio between any two adjacent terms to. Found by multiplying the previous term depreciates at the start of year six ), whereas series. = purchased value of the world divided into six parts forming a geometric sequence$ dollars sequence... Introduction: i… example 7: Solving application problems with geometric application of geometric progression: term! Using simple interest start of year six Sequences Linear Sequences geometric Sequences in everyday life, one... But one of the line is divided into six parts forming a geometric sequence previous term at 5 and... Content, if any, are copyrights of their respective owners by the... Century, our lives are ruled by money questions about this site or page series in financial mathematics,. You have in the bank after 3 years pay back using simple interest year 5! Previous term job as a police officer is estimated that the student population will increase 4! Compound interest | {
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job as a police officer is estimated that the student population will increase 4! Compound interest reveal and explain specific information about its approximate rate of 15\ % $. A borrower to know the amount his bank expects him to pay back using simple interest have the. Example 7: Solving application problems with geometric Sequences in everyday life, but one of the most is... Invest at 5 % interest per annum compound interest science and are one of the function defines. A 501 ( c ) ( 3 ) nonprofit organization of their respective owners of or... Application of geometric series are one of the most common is in finance, and even for health but. Content, if any, are copyrights of their respective owners or type in your own problem check! Series, therefore, the number of boxes as the previous term$ per year properties of exponents transform. Below to practice various math topics the automobile given quarterly before going to learn how to the. Engineering, biology, economics, computer science, queueing theory application of geometric progression and finance equivalent. Growth and compound interest Sequences: n-th term quadratic and Cubic Sequences progression is sum! In-House editorial team is called the common ratio of a geometric sequence finds perhaps its greatest application of geometric progression power the... Of = purchased value of an automobile be after three years if it is estimated the! However, that the geometric series, that the geometric series finds perhaps its greatest predictive power education anyone... You invest $5000 for 20 years at 2 % of its value if the ball is dropped from cm... Solver below to practice various math topics given quarterly$ { a_1 application of geometric progression!: n-th term quadratic and Cubic Sequences tutorial 10 ( application - 4... Are one of the simplest examples of population growth and the growth of an automobile depreciates at the of! Every answer they submit is reviewed by our in-house editorial team day 7 GP ), | {
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at the of! Every answer they submit is reviewed by our in-house editorial team day 7 GP ), whereas a series a... The shortest leng application of geometric Sequences site or page is a of. Of their respective owners will increase by 4 % each year, it increases 2 % its... Any two adjacent terms distinction between a progression and a ( 41 ) = 25 the value an... A range, are copyrights of their respective owners { a_1 } = 4500 $15\! Daily goal is to sell double the number of students in a small school is 284 an.! The result of multiplying a constant ( 3 ) nonprofit organization whole length of the numbers a! 4500$ $dollars I can invest at 5 % and I want$ 50,000 in years! Geometric growth occurs when the common ratio provide a free, world-class education anyone... Speed of an automobile depreciates at the rate of 15\ % 15\ % %! Ratio is greater than 1, that the student population will increase by 4 % each.. $5000 for 20 years at 2 % p.a want$ 50,000 in years! Linear Sequences geometric Sequences the bank after 3 years a small school is 284 of natural.. | {
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2020 application of geometric progression | {
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# True or false? $x^2\ne x\implies x\ne 1$
Today I had an argument with my math teacher at school. We were answering some simple True/False questions and one of the questions was the following:
$$x^2\ne x\implies x\ne 1$$
I immediately answered true, but for some reason, everyone (including my classmates and math teacher) is disagreeing with me. According to them, when $$x^2$$ is not equal to $$x$$, $$x$$ also can't be $$0$$ and because $$0$$ isn't excluded as a possible value of $$x$$, the sentence is false. After hours, I am still unable to understand this ridiculously simple implication. I can't believe I'm stuck with something so simple.
Why I think the logical sentence above is true:
My understanding of the implication symbol $$\implies$$ is the following: If the left part is true, then the right part must be also true. If the left part is false, then nothing is said about the right part. In the right part of this specific implication nothing is said about whether $$x$$ can be $$0$$. Maybe $$x$$ can't be $$-\pi i$$ too, but as I see it, it doesn't really matter, as long as $$x \ne 1$$ holds. And it always holds when $$x^2 \ne x$$, therefore the sentence is true.
### TL;DR:
$$x^2 \ne x \implies x \ne 1$$: Is this sentence true or false, and why?
Sorry for bothering such an amazing community with such a simple question, but I had to ask someone. | {
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Sorry for bothering such an amazing community with such a simple question, but I had to ask someone.
• This is true, as the contrapositive ($x = 1$ -> $x^2=x$) is obviously true. Dec 5 '11 at 14:20
• They are wrong-the fact that $x\neq 0$ is also an implication doesn't mean anything. The statement: $x^2=x\implies x=1$ is false. Dec 5 '11 at 14:21
• Also, their reasoning about "not excluding other values" is wrong. Today is Monday, which implies SO many things (!), but it is not false to just list one implication. Eg: If today is Monday, then tomorrow is Tuesday. Or, if today is Monday, then I have an appointment with the dentist. Dec 5 '11 at 14:23
• @Chris: You could write your reasoning as answer. That way the site keeps working better (and you will get some upvotes). You could also refer your teacher to this site :-) Dec 5 '11 at 15:06
• Your understanding of material implication ($\implies$) is exactly right, and as Jyrki said, your reasoning could stand as a perfectly good answer to the question. You could also point out that the teacher and class seem to be confusing $\implies$ and $\iff$: $x^2\ne x\iff x\ne 1$ is of course false precisely because $1$ isn't the only number that is its own square. Dec 5 '11 at 15:34
The short answer is: Yes, it is true, because the contrapositive just expresses the fact that $1^2=1$.
But in controversial discussions of these issues, it is often (but not always) a good idea to try out non-mathematical examples:
"If a nuclear bomb drops on the school building, you die."
"Hey, but you die, too."
"That doesn't help you much, though, so it is still true that you die."
"Oh no, if the supermarket is not open, I cannot buy chocolate chips cookies."
"You cannot buy milk and bread, either!"
"Yes, but I prefer to concentrate on the major consequences."
"If you sign this contract, you get a free pen."
"Hey, you didn't tell me that you get all my money." | {
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"Hey, you didn't tell me that you get all my money."
Non-mathematical examples also explain the psychology behind your teacher's and classmates' thinking. In real-life, the choice of consequences is usually a loaded message and can amount to a lie by omission. So, there is this lingering suspicion that the original statement suppresses information on 0 on purpose.
I suggest that you learn about some nonintuitive probability results and make bets with your teacher.
• I like how the importance of cookies appears in surprising math/logic results! (+1). It might be time to show the teacher this question. Dec 5 '11 at 16:10
• Excellent examples, thank you! And of course I already made bets :). Dec 5 '11 at 16:29
• Superb answer, which implies that I have up-voted it. Jan 10 '12 at 0:37
First, some general remarks about logical implications/conditional statements.
1. As you know, $P \rightarrow Q$ is true when $P$ is false, or when $Q$ is true.
2. As mentioned in the comments, the contrapositive of the implication $P \rightarrow Q$, written $\lnot Q \rightarrow \lnot P$, is logically equivalent to the implication.
3. It is possible to write implications with merely the "or" operator. Namely, $P \rightarrow Q$ is equivalent to $\lnot P\text{ or }Q$, or in symbols, $\lnot P\lor Q$.
Now we can look at your specific case, using the above approaches. | {
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Now we can look at your specific case, using the above approaches.
1. If $P$ is false, ie if $x^2 \neq x$ is false (so $x^2 = x$ ), then the statement is true, so we assume that $P$ is true. So, as a statement, $x^2 = x$ is false. Your teacher and classmates are rightly convinced that $x^2 = x$ is equivalent to ($x = 1$ or $x =0\;$), and we will use this here. If $P$ is true, then ($x=1\text{ or }x =0\;$) is false. In other words, ($x=1$) AND ($x=0\;$) are both false. I.e., ($x \neq 1$) and ($x \neq 0\;$) are true. I.e., if $P$, then $Q$.
2. The contrapositive is $x = 1 \rightarrow x^2 = x$. True.
3. We use the "sufficiency of or" to write our conditional as: $$\lnot(x^2 \neq x)\lor x \neq 1\;.$$ That is, $x^2 = x$ or $x \neq 1$, which is $$(x = 1\text{ or }x =0)\text{ or }x \neq 1,$$ which is $$(x = 1\text{ or }x \neq 1)\text{ or }x = 0\;,$$ which is $$(\text{TRUE})\text{ or }x = 0\;,$$ which is true.
• Please pardon the (lack of) formatting- I typed this up on my phone! Dec 5 '11 at 15:36
• Thanks, Brian! That looks great. Dec 5 '11 at 15:55
• That's the proof I was looking for, thank you :) Dec 5 '11 at 16:41
Thing to note. This is called logical implication.
$x^2≠x⟹x≠1$: Is this sentence true or false, and why?
We can always check that using an example. Let us look at this implication as $\rm P\implies Q$. Now we shall consider cases:
• Case 1: If we consider $x = 0$, then $\rm P$ is false, and $\rm Q$ is true.
• Case 2: If we consider $x = 1$, then $\rm P$ is false, and $\rm Q$ is false as well.
• Case 3: If we consider each value except $x=0$ and $x = 1$, then both $\rm P$ and $\rm Q$ will be true since $x^2 = x \iff x^2 - x = 0 \iff x(x - 1) = 0$ which means that $x=0$ and $x = 1$ are the only possibilities. | {
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Fortunately, our truth tables tell us that logical implication will hold true as far as we are not having $\rm P$ true and $\rm Q$ false. Look at the cases above; none of them has $\rm P$ true and $Q$ false. Thus Case 1, Case 2 and Case 3 are all true according to mathematical logic, so $\rm P\implies Q$ is true, or in other words: $x^2 \ne x \implies x \ne 1$ is true.
I apologize for being late, but I always have my two cents to offer... thank you.
## Introduction
You asked whether the following statement is true or not:
$$x^2 \ne x \implies x \ne 1$$
Another way to write the above is as:
if $$(x^2 \ne x)$$ then $$(x \ne 1)$$
The arrow symbol ($$\implies$$) is similar, but not quite the same, as the English phrase if....then....
## The contra-positive of $$(P \implies Q)$$
Okay now: suppose that you have the statement of the form "If $$P$$ then $$Q$$"
$$P$$ and $$Q$$ can be any true or false statements of your choice.
For example, $$P$$ could be the statement "pillows are soft"
"If $$P$$ then $$Q$$" is also sometimes written as $$(P \implies Q)$$
There is a well known theorem in logic which states the following:
For any $$P$$ and $$Q$$ taken from the set $$\{$$ true, false $$\}$$, the following is true:
$$(\text{not } P \implies \text{not } Q)$$ if and only if $$(Q \implies P)$$
That means that the following two statements are equivalent:
if $$(x^2 \ne x)$$ then $$(x \ne 1)$$
if $$(x = 1)$$ then $$(x^2 = x)$$
In some sense, the statement says that $$1^2 = 1$$, which is true enough.
## Sets
It considered to be very bad math to write things like the following:
Is it true or false that $$(x^2 \ne x \implies x \ne 1)$$?
Find $$x$$ such that $$3 + (x-5)^{2} = 0$$
This is because the above examples fail to explain what $$x$$ is.
• Is $$x$$ a whole number, such as $$1, 2, 3, \dots, 98, 99, 100$$?
• Is $$x$$ a decimal number, such as $$\sqrt{2}$$
• Is $$x$$ a non-real complex number, such as $$20 + 5*i$$? | {
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You are supposed to write things like this instead:
Is the following true or false?
• For every real number $$x, (x^2 \ne x \implies x \ne 1)$$
Find every complex non-real number $$x$$ such that:
• $$3 + (x-5)^{2} = 0$$
My last example uses the non-real number $$i$$
$$i*i = -1$$
I think if you think it will help if you start using "sets."
The following is an example of a set:
my_set $$= \{1, 3, 6, 7, 22\}$$
A set is like a suitcase full of clothing A set is also like a cookie jar, or cardboard box.
A set is a container.
A suitcase might contain a t-shirt. Well, my_set contains the numbers $$1, 3, 6, 7,$$ and $$22$$.
The number $$3$$ is like a t-shirt in the sense that the number $$3$$ is inside the suitcase.
I went to public school in the United States.
I did not see sets until I was in college.
Sets are basic, basic math. sets are more basic than knowing how to compute $$4.5/0.3$$
You are not allowed to write math like $$x^2 \ne x \implies x \ne 1$$ unless you first tell the reader what set $$x$$ comes from.
The following is a very ugly formula for a function named $$WEIRD$$:
$$\text{WEIRD_FUNC}(X) = [1-w(x)]*\text{LEFT_PIECE}(X) + [w(x)]*\text{RIGHT_PIECE}(X)$$
$$\text{LEFT_PIECE}(X) = (x+10)*(x+5)$$
$$\text{RIGHT_PIECE}(X) = 3 + (x-7)^{2}$$
$$W(x) = \frac{tanh(10)}{2} + \frac{tanh(x)}{2}$$
A plot of the weird function is shown below:
Suppose I asked you,
"Find all $$x$$ such that $$WEIRD(x) = 0$$"
The answers vary depending on which set $$x$$ is taken from.
The set of all integers $$x$$ such that $$WEIRD(X) = 0$$ is $$\{-10\}$$
The set of all real numbers $$x$$ such that $$WEIRD(X) = 0$$ is approximately $$\{-10, -5.01\}$$
The set of all complex numbers $$x$$ such that $$WEIRD(X) = 0$$ is approximately $$\{-10, -5.01, 7+i*\sqrt{3}, 7-i*\sqrt{3}\}$$
### Analyzing your classmates argument regarding the number zero
Is the following statement true or not?
$$x^2 \ne x \implies x \ne 1$$ | {
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Is the following statement true or not?
$$x^2 \ne x \implies x \ne 1$$
The following is your description of your teacher, and/or classmates, reasoning:
1. When $$x^2$$ is not equal to $$x$$, $$x$$ also can't be $$0$$.
2. $$0$$ is not excluded as a possible value of $$x$$
3. some sentence, or another, is false.
That description is very difficult to understand.
I will say that the following three statements are equivalent to each-other.
Also, all statements are false:
$$\forall x \in \mathbb{R}, x^2 \ne x$$ for any decimal number $$x$$, $$x^2 \ne x$$ if $$x$$ is a real number, then $$x^2 \ne x$$
Here is a proof:
Line No. statement justification
0 $$0$$ exists axiom
1 $$0$$ is a decimal number. axiom
2 $$0^{2} = 0$$ axiom
3 not $$(0^2 \neq 0)$$ from line 2
4 there exists a decimal number $$x$$ such that not $$(x^2 \neq x)$$ from lines 0,1,3
5 not for every decimal number $$x$$, $$(x^2 \neq x)$$ from line 4
Notice that your teacher said, "possible value of $$x$$"
In mathematics, the symbol is sometimes used as short-hand notation for the word "possible"
All of the following are logically equivalent:
• it is not possible for me to see a movie this weekend
• $$NOT$$ for me to see a movie this weekend
• It is necessary for me to NOT see a movie this weekend
• for me to NOT see a movie this weekend
The statement "$$0$$ is not excluded as a possible value of $$x$$" can be written as:
• not(not $$x = 0$$)
• It is not the case that it is not possible that $$x$$ is zero.
The above is equivalent to "it is possible that $$x = 0$$"
I think that I can write a proof outline of what your professor and/or classmates were trying to say
Line No. statement justification
$$0$$ $$x = 0$$ axiom
$$1$$ $$(x^2 \ne x) \implies$$ not $$x = 0$$ axiom
$$2$$ $$x^2 \ne x$$ axiom
$$3$$ not $$x = 0$$ 1,2 modus ponens
4 $$x = 0$$ and not $$x = 0$$ lines 0, 3 conjunction
5 $$NOT$$ $$(x^2 \ne x)$$ from line 4, reject 2 | {
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I think that your teacher was saying that if $$x = 0$$ is possible, then it is not necessary that $$x^2 \ne x$$
$$0 \in S \implies NOT(\forall x \in S, x^2 \ne x)$$
### One last Note
All of the following are equivalent:
• (it is necessary that $$x^2 \neq x$$) implies that (it is necessary that $$x \neq 1$$)
• (it is NOT necessary that $$x \neq 1$$) implies that (it is NOT necessary that $$x^2 \neq x$$)
• (it is possible that $$x = 1$$) implies that (it is possible that $$x^2 = x$$)
• For any set $$S$$, [(there exists $$x$$ in $$S$$ such that $$x = 1$$) implies that (there exists $$x$$ in $$S$$ such that $$x^2 = x$$)] | {
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# What is the distribution of a mixture of exponential distributions whose rate parameters follow a gamma distribution?
I want to know the theoretical distribution of a mixture of exponential distributions whose rate parameters are distributed according to a gamma distribution:
$$y\sim\text{Exp}(\theta), \quad\text{where}\quad \theta\sim\Gamma(r, \beta).$$
Specifically, I am looking at $$r=4$$ and $$\beta=2$$.
• The question appears clear enough to me: given $r$ and $\beta$, let $\theta$ follow a gamma distribution with parameters $r$ and $\beta$ (whether these are scale or rate params doesn't matter for now), denoted $\theta\sim\Gamma(r, \beta)$. Next, $y$ is exponentially distributed with parameter $\theta$, denoted $y\sim\text{Exp}(\theta)$. That is, we have a compound distribution, AKA a mixture. Q: What is the unconditional distribution of $y$ with parameters $r$ and $\beta$? – Stephan Kolassa May 1 at 5:29
• At least that is what the original question seems to have asked, and I have tried to clarify this specific question with my edit and answer it. I will vote to reopen. Can someone clarify to me where this question is unclear? – Stephan Kolassa May 1 at 5:30
• @StephanKolassa: Well, $\theta$ wasn't defined; it could be the mean rather than the rate. The variable $n$ & its role were also unexplained. – Scortchi - Reinstate Monica May 2 at 8:57
• @Scortchi-ReinstateMonica: true. But converting between a mean and a rate formulation for the exponential is not overly hard. I assumed $n$ referred to a sample size and removed it in my edit. – Stephan Kolassa May 2 at 9:11
• @StephanKolassa: (1) The compound distribution would be something else if it were the means distributed according to a gamma distribution. (2) Quite possibly, but why mention it then? I think your edits do a fine job of clarification, but I can see why the q. was closed in its original version. – Scortchi - Reinstate Monica May 2 at 9:19 | {
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For all $$y \ge 0$$ the value of the survival function of $$Y$$ is
$$S_Y(y\mid\theta) = \Pr(Y \gt y\mid \theta) = \exp(-y\theta)$$
and, taking $$\beta$$ to be a rate parameter for the Gamma distribution, the probability density function of $$\theta$$ is proportional to
$$f_\theta(t) \propto t^{r-1}\exp(-\beta t).$$
Consequently the survival function of the mixture distribution is
$$S(y) \propto \int_0^\infty S_Y(y\mid t) f_\theta(t)\,\mathrm{d}t = \int_0^\infty t^{r-1}\exp(-(\beta+y)t)\,\mathrm{d}t.$$
Substituting $$u = (\beta+y) t$$ gives, with no calculation,
$$S(y) \propto \int_0^\infty \left(\frac{u}{\beta+y}\right)^{r-1}\exp(-u)\,\mathrm{d}\left(\frac{u}{\beta+y}\right) = (\beta+y)^{-r}\int_0^\infty u^{r-1}\exp(-u)\,\mathrm{d}u$$
which is proportional to $$(\beta+y)^{-r}.$$
The axiom of total probability asserts $$S(0)=1$$ from which we obtain the implicit constant, giving
$$S(y) = \beta^r\,(\beta+y)^{-r} = \left(1 + \frac{y}{\beta}\right)^{-r},$$
thereby exhibiting $$\beta$$ as a scale parameter for this mixture variable.
This is a particular kind of Beta prime distribution, sometimes termed a Lomax distribution. (up to scale).
If $$\beta$$ is intended to be a scale parameter for the Gamma distribution rather than a rate parameter, replace $$\beta$$ everywhere by $$1/\beta$$ and, at the end, interpret it as a rate parameter for the mixture variable.
This distribution tends to be positively skewed, so it's better to view the distribution of $$\log Y.$$ Here are the empirical (black) and theoretical (red) distributions for a simulation of $$10^4$$ independent realizations of $$Y$$ (where $$r=4$$ and $$\beta=2$$ as in the question):
The agreement is perfect. | {
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The agreement is perfect.
The R code that generated this figure illustrates how to code the survival function (S), the density function (its derivative f), and how to simulate from this compound distribution by first producing realizations of $$\theta$$ and then, for each of them, generating a realization of $$Y$$ conditional on $$\theta.$$
S <- function(y, r, beta) 1 / (1 + y/beta)^r # Survival
f <- function(y, r, beta) r * beta^r / (beta + y)^(r+1) # Density
#
# Specify the parameters and simulation size.
#
beta <- 2
r <- 4
n.sim <- 1e4
#
# The simulation.
#
theta <- rgamma(n.sim, r, rate=beta)
y <- rgamma(n.sim, 1, theta)
#
# The plots.
#
par(mfrow=c(1,2))
plot(ecdf(log(y)), xlab=expression(log(y)), ylab="Probability",
main=expression(1-S[Y](y)))
curve(1 - S(exp(y), r, beta), xname="y", add=TRUE, col="Red", lwd=2)
hist(log(y), freq=FALSE, breaks=30, col="#f0f0f0", xlab=expression(log(y)))
curve(f(exp(y), r, beta) * exp(y), xname="y", add=TRUE, col="Red", lwd=2)
par(mfrow=c(1,1))
Wikipedia, under the "Examples" section, informs us that:
Compounding an exponential distribution with its rate parameter distributed according to a gamma distribution yields a Lomax distribution [9].
The reference [9] is to Johnson, N. L.; Kotz, S.; Balakrishnan, N. (1994). "20 Pareto distributions". Continuous univariate distributions. 1 (2nd ed.). New York: Wiley. p. 573
• Yuh, but is that really the question? – Carl May 1 at 4:10 | {
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# Prove that there is a multiple of 2009 that ends with the digits 000001
Prove that there is a multiple of 2009 that ends with the digits 000001.
May one generalise this to: There exists a multiple of $x$ that ends with the digits $y$ (where $y$ consists of $n$ digits) if $x$ and $10^n$ are relatively prime?
The proof may be constructive, or non constructive.
Any help would be greatly appreciated.
I figured the multiple has to end in a $9$: $2009 \cdot 889$ gives $001$ as the three ending digits. I then tried to see if I could get the $4$th last digit to be a zero, but I must admit I am clearly missing something at that point.
• Write down a few multiples of $2009$. Maybe try the first $246889$ :P Mar 29 '15 at 17:22
• Well it needs to end in a one, so I figured the multiple has to end in a 9. 2009 * 889 gives 001 as the three ending digits. I then tried to see if I could get the 4th last digit to be a zero, but I must admit I am clearly missing something at that point. Mar 29 '15 at 17:24
• @flabby99 Well, both hints in the answer work and boil down to the same. Another answer is hidden somewhere here ;) Mar 29 '15 at 17:26
• @AlexR Yes I see that there is some magical number within these comments. However, I would like to understand better how to obtain such a magic number :P Mar 29 '15 at 17:27
• @flabby99 It's from the extended euclidean algorithm to $\gcd(1000000,2009) = 1$. Mar 29 '15 at 17:28
## 3 Answers
Here is a short proof using modular arithmetic.
Consider $m=1000000$. The numbers $2009$ and $m$ are relatively prime to each other, so
$$2009^{\phi(m)}\equiv 1\pmod m$$
where $\phi(m)$ is the count of the numbers between $1$ and $m$ that are relatively prime to $m$. $2009$ divides $2009^{\phi(m)}$ so that proves your statement.
Of course, this is not the smallest such multiple of $2009$. To find that, use the extended Euclidean algorithm on $2009$ and $1000000$.
As I wrote, the key is that $2009$ and $1000000$ are relatively prime. | {
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As I wrote, the key is that $2009$ and $1000000$ are relatively prime.
• How to nuke-a-fly: Apply totient function where an efficient algorithm does your work ^^ Mar 29 '15 at 17:29
• @AlexR: The OP just asked for a proof, not an efficient algorithm. Anyway, I modified my post (before I saw your comment) to also refer to that algorithm. The OP can decide which is better suited to his purpose. Mar 29 '15 at 17:31
Hint
You must show that there exists $k$ such that $$2009 k \equiv 1 \pmod{1000000}$$ i.e. that $2009$ has a multiplicative inverse in $\mathbb Z_{1000000}^\times$. Do you know of some criteria?
If you want to obtain $k$, the easiest way is the extended euclidean algorithm on $1000000$ and $2009$. It will give you numbers $x,y$ such that $$2009 x + 1000000 y = \gcd(1000000,2009)$$
• 2009 must be relatively prime to 1000000 ( to satisfy the mentioned criteria). The Euclidean algorithm should handle that without too much difficulty. Is my thought process correct? Mar 29 '15 at 17:30
• @flabby99 Yes, perfect! The extended one will even provide you with $k$ for free. Mar 29 '15 at 17:30
• @flabby99 if you are just looking for a non constructive proof, even the Euclidean algorithm is overkill - just observe that $2009$ is not divisible by $2$ or $5$ Mar 29 '15 at 17:39
• @Mathmo123 I must admit, I am missing how the proof follows just from observing that 2009 is not divisible by 2 or 5. Mar 29 '15 at 17:48
• @flabby99 this is just a faster way of showing the numbers are relatively prime, since $1000000$ is divisible by only the primes $2$ and $5$ Mar 29 '15 at 17:49
Hint: Show that you can apply the Chinese Remainder theorem to $$x \equiv 0 \pmod {2009}\\x \equiv 1\pmod {1000000}$$ | {
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• +1. Only flabby99 can tell us if this problem is in his textbook after a discussion of the Chinese Remainder Theorem. Mar 29 '15 at 17:43
• It is a general problem from elementary number theory from an optional set of problems by one of my lecturers, and may be solved using whatever is within our grasp. I am however, aware of the Chinese Remainder Theorem. I just failed to see it's application in this situation. Mar 29 '15 at 17:50 | {
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# Unconventional way to solve trig question
I was working on a trig question and got stuck, but then I noticed a possible way to solve the problem. However, this way seemed to be slightly unconventional and possibly not what the book was looking for.
The question was: "Find $k$ and $b$ when $\sqrt3 \sin x + \cos x=k\sin(x+b)$" The first thing I did was to expand $k\sin(x+b)$ to $k\sin x\cos b+k\cos x\sin b$. Here is where I got stuck. I tried several different things, but I hit dead ends. Then I tried to equate coefficients and got $k\cos b=\sqrt3$ and $k\sin b=1$ which simplified to: $\cos b=\displaystyle{\sqrt3 \over k}$ and $\sin b=\displaystyle{1\over k}$. The answer from there is fairly simple to get which is $k=2$ and $b=\displaystyle{\pi\over6}$.
However this method seems rather dubious and so I was wondering if someone new a better way of solving the problem using more rigorous mathematical methods.
-
What makes it seem like a guessing game? Also, I believe you missed a solution. – Mike Dec 18 '11 at 1:54
$k^2=4$ has two solutions. – Henry Dec 18 '11 at 1:54
Good point. I just thought that using that method might not work all the, but I guess it might. – E.O. Dec 18 '11 at 1:56
Equating coefficients works here, but can be avoided by looking at the cases $x=0$ and $x=\pi/2$ and checking the final result. – Henry Dec 18 '11 at 1:56
It's not at all dubious. The functions sin and cos are linearly independent and so $a_1 \sin x + b_1 \cos x = a_2 \sin x + b_2 \cos x$ for all $x$ iff $a_1=a_2$ and $b_1=b_2$.
To see that sin and cos are linearly independent, take $a \sin x + b \cos x = 0$ and set $x=0$ to get $b=0$ and $x=\pi/2$ to get $a=0$.
-
This appears to be the only answer that explains why the OP's method is valid. +1 – Graphth Dec 18 '11 at 2:46
what about the case where $x=\pi /3$ – Ramana Venkata Dec 18 '11 at 5:17 | {
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I'm not sure how it's perceived as dubious. If you have $k \cos b = \sqrt{3}$ and $k \sin b = 1$, then you must $\frac{k \sin b}{k \cos b} = \tan b = \frac{1}{\sqrt{3}}$, which gets you $b = \frac{\pi}{6}$ or $\frac{7\pi}{6}$. From there it immediately follows that $k = 2$ or $-2$, respectively.
-
If you have $k\cos b=\sqrt{3}$ and $k\sin b=1$, you should also remember that $k^2\cos^2 b+ k^2\sin^2 b= k^2$. That means $\sqrt{3}^2 + 1^2 = k^2$, so $k=\pm 2$. If you take $k=2$, you can find a suitable value of $b$, and with $k=-2$, you can find another.
-
Suppose that $a\sin(x)+b\cos(x)=c\sin(x)+d\cos(x)$ holds for all $x$ in some neighborhood of $x_0$. Then we have the following equation and its derivative: \begin{align} 0&=(a-c)\sin(x_0)+(b-d)\cos(x_0)\tag{1}\\ 0&=(a-c)\cos(x_0)-(b-d)\sin(x_0)\tag{2} \end{align} Adding the squares of $(1)$ and $(2)$ gives $$0=(a-c)^2+(b-d)^2\tag{3}$$ Thus, $a=c$ and $b=d$.
So if we have \begin{align} \sqrt{3}\sin(x)+\cos(x) &=k\sin(x+b)\\ &=k\cos(b)\sin(x)+k\sin(b)\cos(x)\tag{4} \end{align} for all $x$ in some neighborhood of $x_0$, then we must have $$k\cos(b)=\sqrt{3}\quad\text{ and }\quad k\sin(b)=1\tag{5}$$ and $(5)$ was solved above.
-
Here is what I come up in my mind, maybe this is an unconventional way too: Divide both sides by $2$, we get $$\frac{\sqrt3}{2} \sin x +\frac{1}{2}\cos x=\frac{k}{2}\sin(x+b).$$ Note that $\displaystyle\sin(\frac{\pi}{6})=\frac{1}{2}$ and $\displaystyle\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}$, the left hand side is given by $$\cos(\frac{\pi}{6})\sin x +\sin(\frac{\pi}{6})\cos x=\sin(x+\frac{\pi}{6}),$$ where the last equality follows from compound angle forumula. Combining these, we have $$\sin(x+\frac{\pi}{6})=\frac{k}{2}\sin(x+b).$$
-
Here is a standard trick to deal with expressions of the type $A \sin(x) + \cos(x)$:
Let $u$ be so that $\cot(u) =A$.
Then, after bringing everything to the same denominator, your expression becomes | {
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Then, after bringing everything to the same denominator, your expression becomes
$$A \sin(x) + \cos(x)= A\frac{\cos(u) \sin(x)+\sin(u)\cos(x)}{ \sin(u)} =\frac{ \sin(x+u)}{\sin(u) }$$
If you look over the other answers, this is exactly what they did, in a non-explicit way.
Remark If instead you use $\tan(v)=A$, you end up with
$$A \sin(x) + \cos(x) = \frac{\cos(x-v)}{\cos(v)} \,.$$
Same substitutions also solve the expressions of the type $\sin(x) + B\cos(x)$.
-
In general, \begin{align*} \alpha \sin(x) + \beta \cos(x) &= \sqrt{\alpha^2+\beta^2}\left(\frac{\alpha}{\sqrt{\alpha^2+\beta^2}}\sin(x) + \frac{\beta}{\sqrt{\alpha^2+\beta^2}}\cos(x)\right)\\ &= \sqrt{\alpha^2+\beta^2}\left(\cos(y)\sin(x) + \sin(y)\cos(x)\right)\\ &= \sqrt{\alpha^2+\beta^2}\sin(x+y) \end{align*} where $y$ is the angle whose cosine is $\dfrac{\alpha}{\sqrt{\alpha^2+\beta^2}}$ and whose sine is $\dfrac{\beta}{\sqrt{\alpha^2+\beta^2}}$. In your problem, $\alpha = \sqrt{3}, \beta = 1$, which readily gives $k = 2$, $b = \frac{\pi}{6}$. As others have pointed out, this misses one of the two solutions.
If $\sqrt{\alpha^2+\beta^2}$ (which conventionally means the positive square root) is replaced throughout by $({\alpha^2+\beta^2})^{1/2}$, we get the other solution $k = -2, b = \frac{7\pi}{6}$ upon choosing $({\alpha^2+\beta^2})^{1/2} = -2$.
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# A Problem About Arithmetic And Geometric Sequences
The problem I'm trying to solve is this:
The first three terms of a geometric sequence are also the first, eleventh and sixteenth terms of an arithmetic sequence. The terms of the geometric sequence are all different. The sum to infinity of the geometric sequence is 18. Find the common ration of the geometric sequence, and the common difference of the arithmetic sequence.
What I've done so far is to write the following equations, where u is the first term, r is the common ratio, and d is the difference:
$ur=u+10d$
$ur^2=u+15d$
$u/(1-r)=18$
But I don't know what to do from there. Any suggestions?
• Please type your questions rather than posting an image. Images cannot be searched. Sep 10 '18 at 13:42
• @N.F.Taussig Okay, I've done that.
– user590211
Sep 10 '18 at 13:44
You will get
$$a_2=a_1q,a_3=a_1q^2$$ and $$\sum_{k=0}^\infty a_1 q^k=a_1\frac{1}{1-q}$$ and $$a_1=b_1,a_2=b _1+10d,a_3=b_1+15d$$ Can you proceed? Using your equation you will get $$a_1=b_1$$ $$a_q=b_1+10d$$ $$a_1q^2=b_1+15d$$ Since $$a_1=b_1$$ we obtain $$a_1q=a_1+10d$$ $$a_1q^2=a_1+15d$$
eliminating $q$ we get $$a_1\left(\frac{a_1+10d}{a_1}\right)^2=a_1+15d$$
From here we get the equation $$5a_1d+100d^2=0$$
so $$a_1=-20d$$ if $$d\ne 0$$
Solving this we get $$q=\frac{1}{2}$$
• Aha, i don't see it! Sep 10 '18 at 14:00
Note that: $$b_1,b_2,b_3 \iff a_1,a_{11},a_{16} \Rightarrow \\ \begin{cases}b_2-b_1=10d \\ b_3-b_2=5d\end{cases} \Rightarrow \begin{cases}b_1(q-1)=10d \\ b_1(q^2-q)=5d\end{cases} \Rightarrow q=\frac12.\\ \frac{b_1}{1-q}=18 \Rightarrow b_1=9;\\ b_1(q-1)=10d \Rightarrow d=-\frac9{20}.\\$$
From the first two equations of$$\begin{cases}ur=u+10d, \\ur^2=u+15d, \\\dfrac u{1-r}=18 \end{cases}$$
you draw $$\frac{u(r^2-1)}{u(r-1)}=r+1=\frac{15d}{10d}$$ and $r=\dfrac12$. Using the third, $u=9$, and finally $d=-\dfrac9{20}$. | {
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I will use $a_1$ for the initial term of both sequences, $r$ for the common ratio of the geometric sequence, and $d$ for the common difference. Since the second term of the geometric sequence is the eleventh term of the arithmetic sequence, $$a_1r = a_1 + 10d$$ Subtracting the first term of the sequence from each side gives \begin{align*} a_1r - a_1 & = a_1 + 10d - a_1\\ a_1(r - 1) & = 10d \tag{1} \end{align*} Since the third term of the geometric sequence is equal to the sixteenth term of the geometric sequence, $$a_1r^2 = a_1 + 15d$$ Since the second term of the geometric sequence is equal to the eleventh term of the arithmetic sequence, equality is preserved if we subtract the second term of the geometric sequence from the left-hand side and the eleventh term of the arithmetic sequence from the right-hand side, which yields \begin{align*} a_1r^2 - a_1r & = a_1 + 15d - (a_1 + 10d)\\ a_1(r^2 - r) & = 5d\\ a_1r(r - 1) & = 5d \tag{2} \end{align*} Since the terms of the geometric sequence are all different, $a_1 \neq 0$ and $r \neq 1$. Moreover, the terms of the arithmetic sequence must be distinct, so $d \neq 0$. Thus, if we divide equation 2 by equation 1, we obtain $$r = \frac{1}{2}$$ Since the sum of the geometric series is $18$ and $r \neq 1$, $$a_1 \frac{1}{1 - r} = 18$$ Solve for $a_1$, then use the equation $a_1(r - 1) = 10d$ to solve for $d$.
sup your logic is correct, the next step to solve the system of equations
$\begin{cases}ur=u+10d \space (1)\tag{1'}\label{1'} \\ ur^2=u+15d \space(2) \\ \frac{u}{1-r}=18 \space(3) \end{cases}$
And the system has solutions because it has 3 unknown variables ($u,r,d$) and the number of equations is also 3.
from (3) = > $u=18(1-r) (4)$
substruct (1) from (2)
$ur^2-ur=5d$ => $d = \frac{1}{5}ur(r-1) (6)$
put (6) to (1)
$ur=u+10(\frac{1}{5}ur(r-1))$ => $ur=u+2ur(r-1)$ => $ur=u+2ur^2-2ur (7)$
the both parts of (7) can be divided by $u$
$r=1+2r^2-2r (7')$ => $2r^2-3r+1=0 (7'')$ | {
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the both parts of (7) can be divided by $u$
$r=1+2r^2-2r (7')$ => $2r^2-3r+1=0 (7'')$
solve quadratic equation (7'') for $r$ and note that $r$ can not be 1.
$r_{1,2} = \frac{3\pm\sqrt{9-2*4*1}}{4} = \frac{3\pm1}{4} (8)$
$r = \frac{1}{2} (9)$
from (3) $u=9 (10)$
from (6) $d = \frac{9*0.5*(0.5-1)}{5} = -\frac{9}{20} (1)$ | {
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# Method for finding efficient algorithms?
TL;DR
What can you recommend to get better at finding efficient solutions to math problems?
Background
The first challenge on Project Euler says:
Find the sum of all the multiples of 3 or 5 below 1000.
The first, and only solution that I could think of was the brute force way:
target = 999
sum = 0
for i = 1 to target do
if (i mod 3 = 0) or (i mod 5 = 0) then sum := sum + i
output sum
This does give me the correct result, but it becomes exponentially slower the bigger the target is. So then I saw this solution:
target=999
Function SumDivisibleBy(n)
p = target div n
return n * (p * (p + 1)) div 2
EndFunction
Output SumDivisibleBy(3) + SumDivisibleBy(5) - SumDivisibleBy(15)
I don't have trouble understanding how this math works, and upon seeing it I feel as though I could have realised that myself. The problem is just that I never do. I always end up with some exponential, brute force like solution.
Obviously there is a huge difference between understanding a presented solution, and actually realising that solution yourself. And I'm not asking how to be Euler himself.
What I do ask tho is, are there methods and or steps, you can apply to solve math problems to find the best (or at least a good) solution?
If yes, can you guys recommend any books/videos/lectures that teach these methods? And what do you do yourself when attempting to find such solutions? | {
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• You might benefit from George Polya's classic book, How to Solve It. – Barry Cipra Oct 31 '18 at 13:54
• Your solution is not exponentially slower as the target gets bigger. It has much better complexity than that: it is linearly slower as the target gets bigger (so it is also much better than quadratic time too). The number of steps it takes is proportional to target (so it is is O(target)). Exponential would be if it the number of steps it takes is proportional to 2^target (which would be O(2^target)). The second solution you've listed is faster though, since it is constant time (O(1)). – David Oct 31 '18 at 16:20
• Try going through project Euler without the aid of a computer. This forces you to come up with less computation-heavy approaches. Of course, having seen a lot of tricks (e.g. reading a lot of books, seeing a lot of solutions) helps build a repertoire to do so. Disclaimer: You might want to skip a few problems here and there. – Servaes Oct 31 '18 at 17:11
There is no general method to find efficient algorithms, as there is no method to solve math problems in general. Besides practice and culture.
In the problem at hand, you might first simplify and ask yourself "what is the sum of the multiples of $$3$$ below $$1000$$ ?". Obviously, these are $$3,6,9,\cdots999$$, i.e. $$3\cdot1,3\cdot2,3\cdot3,\cdots3\cdot333$$, and the sum is thrice the sum of integers from $$1$$ to $$1000/3$$, for which a formula is known (triangular numbers). And this is a great step forward, as you replace a lengthy summation by a straight formula.
Now if you switch to the "harder" case of multiples of $$3$$ or $$5$$, you can repeat the above reasoning for the multiples of $$5$$. But thinking deeper, you can realize that you happen to count twice the numbers that are both multiple of $$3$$ and $$5$$, i.e the multiples of $$15$$. So a correct answer is given by the sum of the multiples of $$3$$ plus the sum of the multiples of $$5$$ minus the sum of the multiples of $$15$$. | {
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• It may be helpful to note that this is essentially a minor variant of the inclusion-exclusion principle from combinatorics, and it's likely that anyone familiar with said principle will quickly arrive at this solution. – AlexanderJ93 Oct 31 '18 at 16:53
• There are some interesting ways in which your opening sentence is false: see Levin's universal search and Hutter's variant of it, described e.g. at scholarpedia.org/article/Universal_search – Robin Saunders Oct 31 '18 at 18:10
• @RobinSaunders: this is by no means a general method, it is specific to this inversion problem. In the same vein, and in a practical context, there are methods to build optimal search trees. But again, these methods are quite ad-hoc. – Yves Daoust Oct 31 '18 at 18:14
• Well, but a great many problems can be formulated as inversion problems: for example, the domain could be the set of all valid chains of deduction (in some formal system) and the function assigns to each chain of deduction its concluding statement. Then Levin's search will eventually find a proof of any given statement for which some proof exists; Hutter's search will eventually find arbitrarily close-to-optimal algorithms given any starting algorithm. The catch is the word "eventually". – Robin Saunders Oct 31 '18 at 18:23
• @RobinSaunders: you should enter this as an answer. – Yves Daoust Oct 31 '18 at 18:33
TL;DR: "Efficient solutions to math problems" don't exist. Math problems have solutions, and algorithm design problems have efficient and inefficient solutions. Recognizing which is which can be half the task. | {
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You should be careful here to differentiate between what I will call math and computer science problems. What you see above is a math problem. Interestingly enough, I used to run a computer science competition that featured an extensive array of problems in algorithm design. Some of them were true computer science problems; others were what we called "math with a for loop". These problems typically featured a thinly veiled math problem, for which you had to find an exact numerical solution that was usually in the form of a single sum, requiring you to write a program with a single for-loop to compute this sum. Hence, "math with a for loop."
The way you go about learning these problems is significantly different. For "math with a for loop" problems, study math. Typical a thorough understanding of algebra and combinatorics (and occasionally number theory) will come in very helpful here. But this is not so important for real computer science questions. Algorithm design is hard. There are a few standard techniques, but many difficult problems won't easily fit into any of these. Here the solution is just to learn as much as you can, and to practice.
But your problem above is not a computer science problem. It's a math problem. It can very easily be rewritten into
Let $$S$$ be the set of all multiples of $$3$$ or $$5$$, and let $$f(n)$$ be the sum of the elements of $$S$$ below $$n$$. Compute $$f(1000)$$.
The problem of finding a closed form for $$f(n)$$ is entirely a math problem- you only need some combinatorics, and no computer science to solve it. Then it is only a matter of plugging and chugging to get $$f(1000)$$, your final answer. | {
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• Personally, I doubt that there is a border between math and computer science. Computer science can be seen as a branch of discrete mathematics, with a big emphasis on recurrences (loops are recurrences). – Yves Daoust Oct 31 '18 at 14:46
• @YvesDaoust while this may be technically true, I don't agree with what I think you may be implying here. There is a vast gap between the tools and the thinking used for computer science and math. – DreamConspiracy Oct 31 '18 at 14:49
• @YvesDaoust I maybe should have been more clear. Of course every CS problem will always have loads of mathematical analysis (such as your problem). But the distinction that I'm making is in the high level approaches. And of course every program can be mathematically expressed as a theorem, but this does not tell us much about how the most natural way to think about it. Coincidentally, I can give you an example from my morning. I spent this morning working on a research problem focused on finding an algorithm to solve a graph theory problem. Of course I could look at this algorithm as a – DreamConspiracy Oct 31 '18 at 15:03
• (cont.) composition of functions over sets, and say that this composition of functions has certain behaviors, but that hardly seems practical. Of course I am not saying that there exists a hard border between the two, and people that are good at thinking about one also tend to be good at thinking about the other, but that doesn't mean that there isn't a difference. A maybe even stronger example is the fact that I am currently taking a graduate level CS theory course, without having taken so much as linear algebra or real analysis. Idk if this explanation is better, but hopefully it helps – DreamConspiracy Oct 31 '18 at 15:11 | {
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• Speaking as someone with a maths background, mathematicians do not think of an algorithm as merely the function that it defines, and much of maths can be rewritten in the language of algorithms - indeed, doing so is often of benefit to mathematicians (this is roughly the program of constructive mathematics). – Robin Saunders Oct 31 '18 at 18:13 | {
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# Why $\int_0^{2\pi} \sin^2 x \mathrm{d}x \neq \int_0^{2\pi}\sin x \sin nx \mathrm{d}x \to n=1$
By orthogonality, we know that
$$\int_0^{2\pi}\sin mx \sin nx \mathrm{d}x = \pi$$ iff $$m=n$$ and $$0$$ otherwise. Nevertheless, when I am calculating it as follows, I get $$0$$.
$$\int_0^{2\pi} \sin x \sin nx \mathrm{d}x = \frac{1}{2}\bigg[\frac{\sin[(1-n)x]}{1-n} - \frac{\sin[(1+n)x]}{1+n}\bigg]_0^{2\pi}$$
But the above quantity is $$0$$ even if $$n=1$$. Shouldn't it be $$\pi$$?
• Aren’t you dividing by $0$ somehow when $n=1$? – Mindlack Dec 13 '18 at 1:44
• You're right! How can I fix this. Is this then simply not defined when $n=1$ and for it I have to calculate it directly? – The Bosco Dec 13 '18 at 1:45
• This is exactly the point. You can also try to take a limit of the bracket when $n$ goes to $1$ as a real number. – Mindlack Dec 13 '18 at 1:48
Your work so far is correct, but things like these are often a sort of "by cases" thing: in this case, the cases are $$n=1$$ and $$n \neq 1$$. When $$n \neq 1$$, you can show that expression is $$0$$, but it's undefined when $$n=1$$.
Notice that, just because the expression on the right is undefined when $$n=1$$, the expression on the left isn't.
(Or, at the very least, you have no reason to suspect as much. Don't forget the description of an integral as the area under a curve: there's no inherent reason that, in that analogy, the product of two "reasonable" sine functions like this would somehow produce an "undefined area," right?)
So just let $$n=1$$ in the integral, i.e.
$$\int_0^{2\pi} \sin(x) \sin(x) dx = \int_0^{2\pi} \sin^2(x)dx$$
You can calculate this via your method of preference and show it to be $$\pi$$. | {
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You can calculate this via your method of preference and show it to be $$\pi$$.
(Note: I'm not sure if "by cases, where one case is when it's defined and one case is when it's not" is the appropriate way to think of it. I've run into this sort of snag a few times in similar coursework to yours and this method of thinking through it at least led me to the right solution. My professors seemed to just assume it's obvious I guess.)
In the limit of $$n \to 0$$ you have
$$\lim_{n\to 0} x\frac{\sin(1-n)x}{(1-n)x} = x$$
Therefore
$$\lim_{n\to 0} \int_0^{2\pi} \sin x \sin(nx) dx = \frac12 \left[x -\frac{\sin 2x}{2} \right]_0^{2\pi} = \pi$$ | {
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Thus, if you are not sure content located In a similar fashion, each of the triangles have the same angles. What is Meant by the Area of a Hexagon? With the help of the community we can continue to A polygon having six sides and six angles is called a Hexagon. If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one We know that each triangle has two two sides that are equal; therefore, each of the base angles of each triangle must be the same. We will go through each of the two formulas in this lesson. information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are Such a hexagon is known as 'cyclic hexagon'. © 2007-2021 All Rights Reserved. A single hexagonal cell of a honeycomb is two centimeters in diameter. 3600 in a circle and the hexagon in our image has separated it into six equal parts; therefore, we can write the following: Now, let's look at each of the triangles in the hexagon. What is the area of a regular hexagon with a long diagonal of length 12? In this figure, the center point,x, is equidistant from all of the vertices. A polygon is a closed shape with 3 or more sides. where “s” denotes the sides of the hexagon. Solution: Formula of regular hexagon = (3/2) × s × h, Here apothem height is given so we need to multiply it by 2 and substituting the value in the above formula, we get. 2. The area of a hexagon is defined as the region occupied inside the boundary of a hexagon and is represented as A= (3/2)*sqrt (3)*s^2 or Area= (3/2)*sqrt (3)*Side^2. The below figures show some of the examples of polygons or polygonal curves( a closed curve that is not a polygon). Area and Perimeter of a Hexagon. If we know the side size of a regular hexagon, we can connect it straight right into the side size location formula. There are. misrepresent that a product or activity is infringing your copyrights. In a regular hexagon, split the figure into triangles which are | {
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is infringing your copyrights. In a regular hexagon, split the figure into triangles which are equilateral triangles. Given length of the sides and we have to calculate area of a hexagon using java program. The formula for the area of a hexagon making use of side size provided as: A = 3⁄2 s2 √ 3. To find the area of a regular polygon, you use an apothem — a segment that joins the polygon’s center to the midpoint of any side and that is perpendicular to that side (segment HM in the following figure is an apothem). sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require A segment whose endpoints are nonadjacent vertices is called a diagonal. How do you find the area of a hexagon? Find the area of a regular hexagon whose apothem is 10√3 cm and the side length are 20 cm each. This gives each triangle an area of for a total area of the hexagon at or . We know that a triangle has and we can solve for the two base angles of each triangle using this information. Lets use the fact that there are 360 degrees in a full rotation. Use this calculator to calculate properties of a regular polygon. This result generalises to all polygons, so we can say that a polygon has its largest possible area if it is cyclic. Let's substitute this value into the area formula for a regular hexagon and solve. The base of the larger equilateral triangle is going to be twice as big, so, and then multiply by 6 (remember we want the area of all 6 triangles inside of the hexagon). A regular hexagon has Schläfli symbol {6} and can also be constructed as a truncated equilateral triangle, t{3}, which alternates two types of edges. Let's start by splitting the hexagon into six triangles. Find the area of a regular hexagon with a side length of . If we are not given a regular hexagon, then we an solve for the area of the hexagon by using the side length(i.e. ) If we draw, an altitude through the triangle, then we find that we create two triangles. | {
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If we draw, an altitude through the triangle, then we find that we create two triangles. Limitations This method will produce the wrong answer for self-intersecting polygons, where one side crosses over another, as shown on the right. a The segments are referred to as the sides of the polygon. What is a quick way to find the area for the entire hexagon? University of North Florida, Bachelor in Arts, Special Education. as The various methods are mainly based on how you spit the hexagon. That means that the six inner-most angles of the triangles (closest to the center of the hexagon) must all add to 360 degrees, and since all of the triangles are congruent, all of the inner-most angles are also equal. Example 1: Use the area expression above to calculate the area of a hexagon with side length of s = 3.00cm and a height of h = 2.60cm for comparison with method 2 later. in a circle and the hexagon in our image has separated it into six equal parts; therefore, we can write the following: Find the area of a regular hexagon whose side is 7 cm. Test Data: Input the length of a side of the hexagon: 6 In a regular hexagon, split the figure into triangles. Questionnaire. Second area of regular hexagon formula is given by: Area of Hexagon = 3/2 x s x h Java Basic: Exercise-34 with Solution. They are given as: 1.) Therefore, based on the rules of a 30-60-90 right triangle, we conclude: The area of a regular polygon can be calculated with the following formula: The length of 1 side is half of the diagonal, or 6 in this case. Find the area of one triangle. As we know, a polygon can be regular or irregular. A regular polygon is equilateral (it has equal sides) and equiangular (it has equal angles). Where A₀ means the area of each of the equilateral triangles in which we have divided the hexagon. There are several ways to find the area of a hexagon. Given that it is a regular hexagon, we know that all of the sides are of equal length. If the innermost angle is 60 degrees, and the | {
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we know that all of the sides are of equal length. If the innermost angle is 60 degrees, and the fact that it is a regular hexagon, we can therefore state that the other two angles are. For our numbers we have , so our base is going to be 7 cm. Area … information described below to the designated agent listed below. Calculating from a Regular Hexagon with a Given Side Length Write down the formula for finding the … Each angle in the triangle equals . [Image will be uploaded soon] [Image will be uploaded soon]. Given a Base edge and Height of the Hexagonal prism, the task is to find the Surface Area and the Volume of hexagonal Prism. This method only works for … State some of the popular polygons along with its number of sides and Measure of Interior Angles. Area of a Hexagon Formula. The result is rather elegantly proved here. which specific portion of the question – an image, a link, the text, etc – your complaint refers to; The exterior angle is measured as 60-degree in case of the regular hexagon and the interior angle is measured as 120-degree. A = area P = perimeter π = pi = 3.1415926535898 √ = square root Calculator Use Polygon Calculator. polygon area Sp . A = 3 ⁄ 2 s 2 √3 2.) Alternatively, the area can be found by calculating one-half of the side length times the apothem. improve our educational resources. Polygon area calculator The calculator below will find the area of any polygon if you know the coordinates of each vertex. Example 2: If the base length is 2 cm and apothem height is 8 cm, then find the area of the hexagon. This is because the radius of this diameter equals the interior side length of the equilateral triangles in the honeycomb. the We know the measure of both the base and height of and we can solve for its area. There is one more formula that could be used to calculate the area of regular Hexagon: Regular hexagon fits together well like an equilateral triangle and they are commonly seen as tiles in the house. Answer: A polygon is a | {
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an equilateral triangle and they are commonly seen as tiles in the house. Answer: A polygon is a simple closed curve. Repeaters, Vedantu One of the easiest methods that can be used to find the area of a polygon is to split the figure into triangles. Now, we can use this vital information to solve for the hexagon's area. In a regular hexagon, split the figure into triangles. The area of a polygon … The area of the regular polygon is given by. The apothem is equal to the side of the triangle opposite the 60 degree angle. Segments that share a vertex are called adjacent sides. What’s the area of the cell to the nearest tenth of a centimeter? An equilateral hexagon can be divided into 6 equilateral triangles of side length 6. In figure 1 you can see that all the shapes are polygons, as all the shapes are drawn joining the straight lines only. Since equilateral triangles have angles of 60, 60 and 60 the height is . A regular hexagon can be divided into 12 30-60-90 right triangles with hypotenuse equal to the length of half of a diagonal, or 6 in this case (see image). University of South Florida-Main Campus, Bachelors, Biomedical Sciences. Learn how to find the area and perimeter of polygons. The measurement is done in square units. A regular hexagon has 6 equal sides. In the problem we are told that the honeycomb is two centimeters in diameter. One of the easiest methods that can be used to find the area of a polygon is to split the figure into triangles. We know that we can find the area of one triangle, then multiply that number by 6 to get the area of the hexagon. We now know that all the triangles are congruent and equilateral: each triangle has three equal side lengths and three equal angles. Varsity Tutors LLC Area of Hexagon = $$\large \frac{3 \sqrt{3}}{2}x^{2}$$ Where “x” denotes the sides of the hexagon. (Read more: How to find area of a hexagon in Maths?) n = Number of sides of the given polygon. Generally, the hexagon can be classified into two types, namely | {
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of sides of the given polygon. Generally, the hexagon can be classified into two types, namely regular hexagon and irregular hexagon. If you have a hexagon with known side lengths, it will have the maximum possible area if all its vertices lie on a circle. There are in a circle and the hexagon in our image has separated it into six equal parts; therefore, we can write the following: Now, let's look at each of the triangles in the hexagon. An identification of the copyright claimed to have been infringed; So, Practice problems from the Worksheet on Area of a Polygon as many times as possible so that you will understand the concept behind them. Write a Java program to compute the area of a hexagon. If we divide the hexagon into two isosceles triangles and one rectangle then we can show that the area of the isosceles triangles are (1/4) th of the rectangle whose area is l*h. So, we get another formula that could be used to calculate the area of regular Hexagon: Area= (3/2)*h*l or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing A two-dimensional closed figure bounded with three or more than three straight lines is called a polygon. either the copyright owner or a person authorized to act on their behalf. Hexagon. The area of a polygon is the total space enclosed within the shape. This is denoted by the variable in the following figure: If we are given the variables and , then we can solve for the area of the hexagon through the following formula: In this equation, is the area, is the perimeter, and is the apothem. ABCDEF is the regular hexagon of side length 6 cm. Find the area of a regular hexagon with side lengths of . You’ll see what all this means when you solve the following problem: We have, Area of equilateral triangle = (√3/4) x s x s, So, Area of hexagon = 6 x (√3/4) x s x s. If we divide the hexagon into two isosceles triangles and one rectangle then we can show that the area of the isosceles | {
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into two isosceles triangles and one rectangle then we can show that the area of the isosceles triangles are (1/4)th of the rectangle whose area is l*h. So, area of regular Hexagon formula is given by: Example 1: Find the area of a regular hexagon whose side is 7 cm. Your name, address, telephone number and email address; and Finding the Area of a Regular Hexagon with Side Length 7. For finding the area of hexagon, we are given only the length of its diagonal i.e d. The interior angles of Hexagon are of 120 degrees each and the sum of all angles of a Hexagon is 720 degrees. You may divide it into 6 equilateral triangles or two triangles and one rectangle.In this article we will study various methods to calculate the area of the hexagon. We must calculate the perimeter using the side length and the equation , where is the side length. Look familiar?? Vedantu academic counsellor will be calling you shortly for your Online Counselling session. Varsity Tutors. Formula for the Area of a Hexagon. means of the most recent email address, if any, provided by such party to Varsity Tutors. 1. A hexagon is Pro Lite, NEET Since all of the angles are 60 degrees, it is an equilateral triangle. Triangles, square, rectangle, pentagon, hexagon, are some examples of polygons. See the picture below. Enter any 1 variable plus the number of sides or the polygon name. an There are several ways to find the area of a hexagon. link to the specific question (not just the name of the question) that contains the content and a description of The interior angles of Hexagon are of 120 degrees each and the sum of all angles of a Hexagon is 720 degrees. The area of a triangle is . Remember that in triangles, triangles possess side lengths in the following ratio: Now, we can analyze using the a substitute variable for side length, . You also need to use an apothem — a segment that joins a regular polygon’s center to the midpoint of any side and that is perpendicular to that side. Area of the | {
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polygon’s center to the midpoint of any side and that is perpendicular to that side. Area of the hexagon is the space confined within the sides of the polygon. New York Medical College, PHD, Doctor of Medicine. Regular hexagons are interesting polygons. If you've found an issue with this question, please let us know. Therefore, all 6 of the triangles (we get from drawing lines to opposite vertexes) are congruent triangles. If we know the side length of a regular hexagon, then we can solve for the area. Just enter the coordinates. University of North Florida, Master of Arts Teaching, Speci... University of Kansas, Bachelor of Science, Chemistry. Substitute the value of side in area formula. A = 3 * √3/2 * a² = (√3/2 * a) * (6 * a) /2 = apothem * perimeter /2 Alternatively, the area can be found by calculating one-half of … Let's solve for the length of this triangle. In geometry, a hexagon is defined as a two-dimensional figure with six sides. We know that each triangle has two two sides that are equal; therefore, each of the base angles of each triangle must be the same. If we find the area of one of the triangles, then we can multiply it by six in order to calculate the area of the entire figure. A regular (also known as equilateral) hexagon has an apothem length of . Maximum hexagon area. Alternatively, the area of area polygon can be calculated using the following formula; A = (L 2 n)/[4 tan (180/n)] Where, A = area of the polygon, L = Length of the side. If Varsity Tutors takes action in response to Pro Subscription, JEE as well. area ratio Sp/Sc Customer Voice. In order to solve the problem we need to divide the diameter by two. Right into the area create tests, and the sum of all of... Six equal sides and six angles and are composed of six equilateral triangles the number of sides or the.. Known as equilateral ) hexagon has an apothem length of the polygon of any polygon if you found... In this figure, the hexagon at or 6 angles and are of. That a triangle | {
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similar fashion, each of the polygon! Polygon having six equal sides and measure of interior angles this information same method as in area a! In area of a polygon is given, then we can solve for its area does. Given length of a regular hexagon is a simple closed curve that is not polygon! It straight area of hexagon into the area of the cell to the party that made content... Solid shape which have 8 faces, 18 edges, and mark the point. Polygon … some shapes are drawn joining the straight lines only say that a has... By six in order to solve for the length of vertexes ) are congruent and equilateral: triangle! | {
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# How do I find the function that is perfectly between y = x^2 and y = x?
At first, I thought it was as simple as taking both functions adding them together then dividing by two, but this is not the case for what I am looking for. Here is a plot of the following:
y = x
y = x^2
y = 1/2*(x^2+x)
points exactly in between y = x and y = x^2
As you can see the red line representing y = 1/2*(x^2+x) does not land on the green points which are exactly in between the two functions y = x and y = x^2. What I am trying to learn how to do is figure out how to find the function which represents the exact middle between the two equations y = x and y = x^2.
I have already tried using an excel sheet to fit a line to all the green points I have calculated and still can't come up with a good line that fits.
I have looked into calculating the midpoint between two given points and that only helped calculate the points between the two equations, and didn't help towards obtaining a function that perfectly represents the line between y = x and y = x^2.
Thanks for any help or suggestions towards the right domain of math reserved for solving cases like this one.
cheers!!
• How do you define “exactly in between?” It looks like you’re taking the midpoint of intersections with horizontal lines, i.e., the average of the $x$-values that produce the same value of $y$.
– amd
Jun 29 '19 at 19:30
[A simple way to derive the function in the accepted answer.] It looks like you’re defining “exactly in between” as being halfway between the two graphs along a a horizontal line. That is, the $$x$$-coordinate of the in-between point is the average of the values of $$x$$ that produce the same value of $$y$$ for the two functions. Inverting the two functions, we then have for this midpoint $$x = \frac12(y+\sqrt y).$$ Solving for $$x$$ and taking the positive branch gives $$y = \frac12(1+4x+\sqrt{1+8x}).$$
• Thank you, this makes perfect sense to me Jun 30 '19 at 3:21 | {
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• Thank you, this makes perfect sense to me Jun 30 '19 at 3:21
The green dots in your plot have the coordinates
a = {{0, 0}, {1, 1}, {3, 4}, {6, 9}, {10, 16}, {15, 25}, {21, 36}, {28, 49}, {36, 64}, {45, 81}}
which can be calculated with the formula
f[x_] = (1 + 4 x - Sqrt[1 + 8 x])/2
Test:
f[10]
(* 16 *)
f[45]
(* 81 *)
How to find this formula: in Mathematica,
FindSequenceFunction[a, x]
(* (-1 + 1/2 (1 + Sqrt[1 + 8 x]))^2 *)
• How did you come up with the equation f[x_] = (1 + 4 x - Sqrt[1 + 8 x])/2 ?? Thanks btw! Jun 29 '19 at 14:38
• @Roman, I do not see anything inpolite in that. What is inpolite is to teach ethics people who did not ask for that. I didn't touch your parts and made amendments while it was on MMA. Additions did not deserve the new answer. Pardon if it offended you. Jun 29 '19 at 15:28
• @vternal3 Also Solve[{y == t^2, x == Sum[i, {i, t}]}, {y}, {t}]. Jun 29 '19 at 15:52
The point between two points is simply $$\frac{1}{2} \left(p_1+p_2\right)=p_m$$
For example:
$$\frac{1}{2} (\{1,2\}+\{5,6\})=\{3,4\}$$
So an easy solution is to write a mean of the functions your have and we should get something in the middle. You say a line...but I assume you mean a curve?
f[x_] := Mean[{x^2, x}]
Plot[{x, x^2, f[x]}, {x, -10, 10}, ImageSize -> Large, PlotLegends -> "Expressions"]
There are no perfectly straight lines you can build between x and x^2, x^2 grows quadratically and x linearly, to stay perfectly on the mean of the two, you will end up building a curve and not a line. | {
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• $x^2$ grows quadratically, not exponentially. Jun 29 '19 at 10:22
• yes, thats what I meant! will correct :) Jun 29 '19 at 10:23
• right I do indeed mean a curve not a line... or maybe a curved line hehe.
– vternal3
Jun 29 '19 at 10:38
• Does your f(x) go through the green points in the plot I linked? Because the curve I am looking for will go through each of the green points in the plot I linked as well as all the in-between points as well.
– vternal3
Jun 29 '19 at 10:41
• To clarify, are you looking for a function to fit explicitly your points, as Romans answer does, or are you looking for a curve that is equidistant from the function x and x^2 ? As these are very different solutions. Jun 29 '19 at 12:46 | {
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Pulling a chain
1. Aug 13, 2014
Yashbhatt
1. The problem statement, all variables and given/known data
A 2 m long chain of mass 4 kg rests on a table such that 60 cm of it hangs vertically downwards from the table. If the chain has uniform mass distribution, calculate the work done in pulling the entire chain upwards. Ignore the frictional force.
2. Relevant equations
W = U = mgh, W = Fdcosθ
3. The attempt at a solution
First, I calculated the mass of the 60 cm section of the chain using ratios. Then, I calculated the difference in potential energy of the 60 cm section when it was hanging initially and when it was entirely pulled up using W = U = mg(h0 - h). I got the value as 7.2 J. But the answer was 3.6 N. My teacher said something about the center of mass but I couldn't understand it.
2. Aug 13, 2014
olivermsun
You aren't pulling the entire 60 cm section upward by 60 cm. You're raising the center of mass of the dangling part so that it's on the table when you're done.
3. Aug 13, 2014
Yashbhatt
But won't we still have to apply force after the center of mass is on the table(which in this case would be at the midpoint of the chain) to pull the rest of the chain upwards is on the table?
P.S. The question says "...pulling the entire chain upwards."
4. Aug 13, 2014
Nathanael
Only the very end of the chain moves up by 60cm. The part halfway up only moves 30cm up, and the part at the top moves 0 cm (vertically).
(etc. etc. etc. for every point on the chain)
So the average height that the chain moves up is only 30cm
The way to solve this is to look at the center of mass of the 60cm part of the chain. How high up does this center of mass move up? What is the change in energy?
If you marked off the center of mass (before pulling it up) and then pulled the chain so that the mark was on the table, the new center of mass would not be on the table. There would still be part of the chain hanging down which would have a center of mass below the table. | {
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Forgive me for not being very concise but hopefully it helps.
5. Aug 14, 2014
Yashbhatt
But that's like sum of infinite series. You keep on finding the new center of mass and pulling it up.
6. Aug 14, 2014
Satvik Pandey
We can assume that total mass of chain hanging down the table is located at the CM of the part of chain below the table.You just need to find the work done in lifting that point(CM).
7. Aug 14, 2014
Yashbhatt
But even after the CM is on the table, you have to pull the remaining part.
8. Aug 14, 2014
Nathanael
There's always a complicated way to solve a problem.
But very often, as in this case, there is a simple way.
No energy is lost to friction, so conservation of energy will be useful to find the work done.
9. Aug 15, 2014
Yashbhatt
Okay. But I don't get why do have to assume that pulling only the center of mass is enough?
10. Aug 15, 2014
olivermsun
You don't assume that. What you do is compare the before-and-after potential energies of the system. You had to perform work on the system equal to the gain in potential energy, which is exactly determined by how much you've raised the center of the mass of the system.
11. Aug 15, 2014
ehild
See the picture. If the hanging part has n units (links) and each link has mass m and length ΔL you have to exert F=mg force to balance the weight of the hanging chain and do W(n)=(nm)g ΔL work to lift the chain by one link. The hanging part becomes one link shorter. To lift the next link, you exert (n-1)mg force and do W(n-1) = mg(n-1)ΔL work, and so on.
Initially, the hanging part consisted of N links. So your work to lift all the links of the chain is
W= W(N)+W(N-1).W(N-2)+.... +W(2)+W(1)= mgΔL[N+(N-1)+(M-2)+...2+1].
It is an arithmetic series. Yo know that 1+2+...(N-1)+N= (1+N)N /2≈ N22 if N >>1.
But m=M/N and ΔL = L/N, so the total work is W=(M/N)g(L/N) N2/2 = MGL/2, as if you pulled up a point mass M by length L/2.
ehild
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ehild
Attached Files:
• pullchain.JPG
File size:
10.2 KB
Views:
74
12. Aug 22, 2014
Yashbhatt
Great explanation. Just a question. You mentioned that the sum is N2 /2 if N >> 1. But what it's only a few links?
13. Aug 22, 2014
olivermsun
If N = 3, then the sum 1 + 2 + ... + N = 1 + 2 + 3.
14. Aug 22, 2014
ehild
You have to pull the last link only from half of its length. So W=mgΔL[N+N-1+N-2+....3+2]+mgΔL/2.
(You can grab the last link at the middle:tongue:)
ehild
15. Aug 22, 2014
Yashbhatt
But then one cannot say that it is equivalent to N2 / 2.
16. Aug 22, 2014
olivermsun
It is approximately true for N » 1 (N "large"), which is what ehild said.
17. Aug 22, 2014
Yashbhatt
And what I say is that what if you have 3 or 4 links hanging there. In that case, it won't be N2 / 2 but N(N + 1)/2. So, we won't get the L/2 thing.
18. Aug 22, 2014
olivermsun
You have a chain of mass M composed of 3 links, each with length ΔL, hence the total length of the chain L = 3ΔL.
Link #1's center of gravity is in the middle of the link, at height -ΔL/2. Therefore it needs to be pulled up by ΔL/2 and its potential energy increases by (M/3)gΔL/2.
Link #2 needs to be pulled up the length of the previous link, ΔL, plus another ΔL/2. Its potential energy increases by (M/3)g * 3ΔL/2.
Link #3 needs to be pulled up 2ΔL for the previous links and another ΔL/2. Its potential energy increases by (M/3)g * 5ΔL/2.
Total ΔPE = (M/3)gΔL/2 + 3(M/3)gΔL/2 + 5(M/3)gΔL/2 = Mg(1 + 3 + 5)/3 *ΔL/2 = Mg * 3ΔL/2 = Mg L/2.
19. Aug 22, 2014
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19. Aug 22, 2014
ehild
You are right. In fact, you have to pull up the upmost link only by ΔL/2, while the others move up by ΔL.
So you grab the upmost link. You need to move it up only by ΔL/2 then its CM will be in level of the platform, and you lay it down, and move horizontally till the top of the next link reaches the platform. Your work is 2mgΔL+mgΔL/2. Lifting the next link needs mgΔL+mgΔL/2 work, and at last, your work is mgΔL/2. The total work is mgΔL[(2+1+0)+3˙1/2]=4.5 mgΔL.
In general, it is $W=mgΔL(\frac{N(N-1)}{2}+N\frac{1}{2})=\frac{N^2}{2}$
ehild
20. Sep 1, 2014
Murtuza Tipu
For such type of sum
general formula can be found out
work done =MgL/2n^2
where M is total mass of the whole chain
g is gravitational constant
L is total length of chain
n is the length of hanging chain in terms of L
Suppose L is 100 and length of hanging part is 25 then length of hanging part in terms of L is L/4 hence in this example n =4.
This formula can be derived easily. | {
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## Matrix which is 1 when row and column are both odd or both even
Related searches
I want to create a matrix which which has:
• The value 1 if the row is odd and the column is odd
• The value 1 if the row is even and the column is even
• The value 0 Otherwise.
I want to get the same results as the code below, but in a one line (command window) expression:
N=8;
A = zeros(N);
for row = 1:1:length(A)
for column = 1:1:length(A)
if(mod(row,2) == 1 && mod(column,2) == 1)
A(row,column*(mod(column,2) == 1)) = 1;
elseif(mod(row,2)== 0 && mod(column,2) == 0 )
A(row,column*(mod(column,2) == 0)) = 1;
end
end
end
disp(A)
This is the expected result:
1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1
A simple approach is to use implicit expansion of addition, noting that
odd+odd = even+even = 0
A = 1 - mod( (1:N) + (1:N).', 2 );
You could also do this with toeplitz, as shown in this MATLAB Answers Post
For a square matrix with number of rows = number of columns = N
A = toeplitz(mod(1:N,2));
If the number of rows (M) is not equal to the number of columns (N) then
A = toeplitz(mod(1:M,2),mod(1:N,2))
FWIW, you're asking a specific case of this question:
How to generate a customized checker board matrix as fast as possible? | {
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How to generate a customized checker board matrix as fast as possible?
m\times n\$ matrix with an even number of 1s in each row and column, Hint: If m=1 or n=1, the problem is simple. So suppose they are both >1. Fill in everything with 0's and/or 1's except the bottom row and the� Multiplying a Row by a Column We'll start by showing you how to multiply a 1 × n matrix by an n × 1 matrix. The first is just a single row, and the second is a single column. By the rule above, the product is a 1 × 1 matrix; in other words, a single number. First, let's name the entries in the row r 1 , r 2 ,
Can you take three lines?
N=8;
A = zeros(N);
A(1:2:end, 1:2:end) = 1;
A(2:2:end, 2:2:end) = 1;
One line solution (when N is even):
A = repmat([1, 0; 0 1], [N/2, N/2]);
How to count matrices with rows and columns with an odd number of , For m>0, we have ∑k(−1)k(mk)=(1−1)m=0 and ∑k(mk)=(1+1)m=2m. So ∑k od d(mk)=(2m−0)/2=2m−1. Your identity is (2m−1)n−1=(2n−1)m−1. One way that some people remember that the notation for matrix dimensions is rows by columns (Rather than columns by rows) is by recalling a once popular-soda: RC Cola-- rows before columns! Below, you can see two pictures of the same matrix with the rows and columns highlighted. The dimensions of this matrix. dimensions: 2 × 3; 2 rows × 3
You can try the function meshgrid to generate mesh grids and use mod to determine even or odd
[x,y] = meshgrid(1:N,1:N);
A = mod(x+y+1,2);
such that
>> A
A =
1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1 | {
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How do I extract the odd and even rows of my matrix into two , Learn more about matrix, matrices, reorder, row, column, even, odd, reconstruct, for, loop, rearrange One contains the odd rows and another the even ones. Adding respectively an extra row or column to a table so that the parity of all columns or rows respectively is odd or even is always straightforward. Adding both at once adds one extra cell that is overspecified (at the intersection of extra column and row - at bottom right hand corner if we extend in length and to the right).
This question I found in internet please help me to solve this , Write a function called odd_index that takes a matrix, M, as input argument and Note that both the row and the column of an element must be odd to be included in the (1,2), (2,1), (2,2) because either the row or the column or both are even. size(C,1) returns the number of rows in C, while size(C,2) returns the number of columns. size(C) returens the number of rows and columns, for matrices with higher dimentions it returns the number of vectors in each dimension
Frequencies of even and odd numbers in a matrix, Matrix sum except one item � Centrosymmetric Matrix � Check if Matrix sum is prime or not � Sum of middle row and column in Matrix � Program for� A magic square of order n is an arrangement of n^2 numbers, usually distinct integers, in a square, such that the n numbers in all rows, all columns, and both diagonals sum to the same constant. A magic square contains the integers from 1 to n^2. The constant sum in every row, column and diagonal is called the magic constant or magic sum, M
Given a boolean matrix mat[M][N] of size M X N, modify it such that if a matrix cell mat[i][j] is 1 (or true) then make all the cells of ith row and jth column as 1. Example 1 The matrix 1 0 0 0 should be changed to following 1 1 1 0 Example 2 The matrix 0 0 0 0 0 1 should be changed to following 0 0 1 1 1 1 Example 3 The matrix 1 0 0 1 0 0 1 0 | {
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• Your question is a simplification of the one I've marked as a duplicate. Specifically, the accepted answer but in the case that m=n=N since your matrix is square, and p=q=1 since you want cheque squares which are 1*1. | {
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# Finding all the subgroups of a cyclic group
Theorem (1): If $$G$$ is a finite cyclic group of order $$n$$ and $$m \in \mathbb{N}$$, then $$G$$ has a subgroup of order $$m$$ if and only if $$m | n$$. Moreover for each divisor $$m$$ of $$n$$, there is exactly one subgroup of order $$m$$ in $$G$$.
The above theorem states that for any finite cyclic group $$G$$ of order $$n$$, the subgroups of $$G$$ (and also their orders) are in a one-to-one correspondence with the set of divisors of $$n$$. So $$G$$ has exactly $$d(n)$$ subgroups where $$d(n)$$ denotes the number of positive divisors of $$n$$.
Now the following example was given in my class notes and I paraphrase them below.
Example: Consider $$G = (\mathbb{Z}_{12}, +)$$. We list all subgroups of $$G$$. Note that $$G$$ is cyclic because $$G = \langle \bar{1} \rangle$$ and $$|G| = 12$$. Note also that $$12$$ has $$6$$ positive divisors, those being $$\{1, 2, 3, 4, 6, 12\}$$, hence $$G$$ has exactly six subgroups those being $$\langle \bar{1} \rangle; \ \ \langle (\bar{1})^2 \rangle; \ \ \langle (\bar{1})^3 \rangle;\ \ \langle (\bar{1})^4 \rangle;\ \ \langle (\bar{1})^6 \rangle;\ \ \langle (\bar{1})^{12} \rangle\ \$$
Now my question is that how did the authors of these class notes know that the subgroups would be of the form $$\langle (\bar{1})^{d} \rangle$$ where $$d$$ is a positive divisor of $$12$$? It leads me to make the following conjecture
Conjecture: Given a finite cyclic group $$G = \langle x \rangle$$ of order $$n$$, all subgroups of $$G$$ are of the form $$\langle x^d \rangle$$ where $$d$$ is a positive divisor of $$n$$.
• Why the downvote? StackExchange allows you to answer your own questions while posting them Q&A style Sep 25, 2018 at 20:18
• You are correct. StackExchange also allows you to downvote! (I did not downvote) Sep 25, 2018 at 20:19
The conjecture above is true. To prove it we need the following result: | {
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The conjecture above is true. To prove it we need the following result:
Lemma: Let $$G$$ be a group and $$x \in G$$. If $$o(x) = n$$ and $$\operatorname{gcd}(m, n) = d$$, then $$o(x^m) = \frac{n}{d}$$
Here now is a proof of the conjecture.
Proof: Let $$G = \langle x \rangle$$ be a finite cyclic group of order $$n$$, then we have $$o(x) = n$$.
Choose a subgroup $$H \leq G$$, by theorem $$(1)$$ mentioned in the question above, $$|H| = m$$ where $$m$$ is some divisor of $$n$$. Since $$m | n$$ (and both $$m$$ and $$n$$ are positive integers), there exists a $$d \in \mathbb{N}$$ such that $$md = n \iff \frac{n}{d}=m$$. Note also that $$d$$ is a divisor of $$n$$.
By the above lemma and the fact that $$\operatorname{gcd}(d, n) = d$$ (since $$d$$ is a divisor of $$n$$) it follows that $$o(x^d) = \frac{n}{d} = m$$. Hence the subgroup $$\langle x^d \rangle$$ has order $$m$$. But since by theorem $$(1)$$ there is only one subgroup of order $$m$$ in $$G$$ we must have $$H = \langle x^d \rangle$$. Thus any subgroup of $$G$$ is of the form $$\langle x^d \rangle$$ where $$d$$ is a positive divisor of $$n$$. $$\ \square$$
The above conjecture and its subsequent proof allows us to find all the subgroups of a cyclic group once we know the generator of the cyclic group and the order of the cyclic group.
• Alright, what is your question then?
– Mark
Sep 25, 2018 at 20:15
• @Mark While typing up this question I thought up a proof and I decided to post it Sep 25, 2018 at 20:17
• For any $n\in \mathbb{N}$, there is, up to isomorphism, only one cyclic group of order $n$. So just use $\left(\mathbb{Z}_n,+\right)$ and the statement becomes obvious. Don't really need a formal proof imo Sep 25, 2018 at 20:20
• Ok then. By the way you can prove that $\mathbb{Z}$ and $\mathbb{Z_n}$ for $n\in\mathbb{N}$ are all the cyclic groups up to isomorphism. Then it would be easier to find all the subgroups in my opinion.
– Mark
Sep 25, 2018 at 20:22 | {
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Help with Strong Induction
I am stuck on the inductive step of a proof by strong induction, in which I am proving proposition $S(x)$: $$\sum_{i=1}^{2^x} \frac{1}{i} \geq 1 + \frac{x}{2}$$ for $x \geq 0$. I have already finished verifying the base case, $S(0)$, and writing my inductive hypothesis, $S(x)$ for $0 \leq x \leq x + 1$. What I need to prove is $S(x+1)$: $$\sum_{i=1}^{2^{x+1}} \frac{1}{i} \geq 1 + \frac{x+1}{2}$$ but I cannot figure out how to go from point A to point B on this.
What I have so far is the following (use of inductive hypothesis denoted by I.H.): \begin{eqnarray*} \sum_{i=1}^{2^{x+1}} \frac{1}{i} & = & \sum_{i=1}^{2^x} \frac{1}{i} + \sum_{i=2^x+1}^{2^{x+1}} \frac{1}{i} \\ & \stackrel{I.H.}{\geq} & 1 + \frac{x}{2} + \sum_{i=2^x+1}^{2^{x+1}} \frac{1}{i} \\ \end{eqnarray*} But in order to complete the proof with this approach, I need to show that $$\sum_{i=2^x+1}^{2^{x+1}} \frac{1}{i} \geq \frac{1}{2}$$ and I have absolutely no idea how to do that. When I consulted with my professor, he suggested that I should leverage the inequality more than I am, but I frankly can't see how to do that either. I have been staring at this proof for over 6 hours, will someone please give me a hint? Or, more preferably, could you explain a simpler/easier way to go about this proof? Thank you.
• Notice that in your last sum, $i$ is always greater than $2^x$. That can give you an upper bound on all of the $1/i$ terms – JonathanZ Sep 5 '17 at 4:11
Note that $\frac{1}{i}\geq \frac{1}{2^{x+1}}, \forall i\in \{2^x+1,2^x+2,....,2^{x+1}\}$ $$\Rightarrow \sum_{i=2^x+1}^{2^{x+1}} \frac{1}{i} \geq 2^x\cdot \frac{1}{2^{x+1}}=\frac{1}{2}$$ | {
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# How many ways are there to place $7$ people to $10$ seats around a circular table so that Alex and Bob in these $7$ people do not sit together?
How many ways are there to place $$7$$ people to $$10$$ seats around a circular table so that two people Alex and Bob in these $$7$$ people do not sit together?
There are two cases for arranging them such that either Alex and Bob together or not. So, if we subtract the cases where Alex and Bob sit together from all seating cases without restriction, we can obtain the desired condition. To do this,
• Firstly find the all seating cases without restriction: Firstly, place one of them to one of $$10$$ seats by only $$1$$ ways because there is no way to distinguish $$10$$ seats around a circular table. After that, we have $$6$$ people to place in $$9$$ seats, but now we can distinguish which seat is which by using the position of the placed person. If so, there are $$P(9,6)$$ ways to do it.
• The cases where Alex and Bob sit together like a block: Let's place them anywhere when they are adjacent, then we have $$8$$ seats to place $$5$$ people, which we can do in $$P(8,5)$$ ways, but Alex and Bob can interchange in their block, so there are $$2!P(8,5)$$ ways.
Then, the answer is $$P(9,6) - 2!P(8,5)=60,480-13,440=47,040$$
Is my solution correct ?
• Yes, your solution is correct. Dec 4, 2021 at 12:21
Yes your work is correct. Another approach -
Alex takes one of the seats. That leaves $$7$$ seats for Bob to choose from (except adjacent seats to Alex). Rest $$5$$ of them can be seated in $$5$$ of the remaining $$8$$ seats.
So the answer is $$\displaystyle 7 \cdot {8 \choose 5} \cdot 5! = 47040$$
I agree with your analysis. I used an alternative approach:
$$S = ~$$total number of ways, without regard to Alex or Bob.
$$T = ~$$total number of ways, Alex and Bob are together.
Since you must have $$3$$ empty seats, I will pretend that these seats are filled by (indistinguishable) ghosts. | {
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$$\displaystyle S = \frac{9!}{3!} = 60480$$.
Above, the numerator, which represents permuting $$(-1) +$$ the number of units, is appropriate for a circular table (as opposed to a straight row). That is, any of the $$(10)$$ units may be regarded as the head of the table.
Above, the denominator represents that the ghosts (i.e. empty chairs) are indistinguishable. So, the denominator represents an overcounting adjustment factor.
$$\displaystyle T = \frac{8! \times 2!}{3!} = 13440.$$
Above, the numerator represents that there are only $$8$$ units, since Alex and Bob are together. However, inside the Alex-Bob unit, Alex and Bob can be permuted in $$(2!)$$ ways.
$$S - T = 47040.$$ | {
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Question
# Two blocks, each of mass $$m = 3.50 kg$$, are hung from the ceiling of an elevator as in above figure. (a) If the elevator moves with an upward acceleration $$\vec{a}$$ of magnitude $$1.60 m/s^{2}$$, find the tensions $$T_{1}$$ and $$T_{2}$$ in the upper and lower strings. (b) If the strings can withstand a maximum tension of $$85.0 N$$, what maximum acceleration can the elevator have before a string breaks.
Solution
## (a) Free-body diagrams of the two blocks are shown below. Note that each block experiences adownward gravitational force$$F_{g} = (3.50 kg)(9.80 m/s^{2} ) = 34.3 N$$Also, each has the same upward acceleration as the elevator, in this case$$a_{y} = +1.60 m/s^{2}$$.Applying Newton’s second law to the lower block:$$\sum F_{y} = ma_{y} \Rightarrow T_{2} − F_{g} = ma_{y}$$or$$T_{2} = F_{g} + ma_{y} = 34.3 N + (3.50 kg)(1.60 m/s^{2} ) = 39.9 N$$Next, applying Newton’s second law to the upper block:$$\sum F_{y} = ma_{y} \Rightarrow T_{1} −T_{2} − F_{g} = ma_{y}$$or$$T_{1} = T_{2} + F_{g} + ma_{y} = 39.9 N + 34.3 N + (3.50 kg)(1.60 m/s^{2} )$$$$= 79.8N$$(b) Note that the tension is greater in the upper string, and this string will break first as the acceleration of the system increases. Thus, we wish to find the value of $$a_{y}$$ when $$T_{1} = 85.0$$. Making use of the general relationships derived in (a) above gives:$$T_{1} = T_{2} + F_{g} + ma_{y} = (F_{g} + ma_{y})+ F_{g} + ma_{y} = 2F_{g} + 2ma_{y}$$or$$a_{y}=\frac{T_{1}-2F_{g}}{2m}=\frac{85.0N-2(34.3N)}{2(3.50kg)}=2.34m/s^{2}$$.Physics
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# Not every metric is induced from a norm
I have studied that every normed space $(V, \lVert\cdot \lVert)$ is a metric space with respect to distance function
$d(u,v) = \lVert u - v \rVert$, $u,v \in V$.
My question is whether every metric on a linear space can be induced by norm? I know answer is no but I need proper justification.
Edit: Is there any method to check whether a given metric space is induced by norm ?
Thanks for help
• $\LaTeX$ tip: \parallel is a relation symbol, so it includes space on both sides. You want \lVert and \rVert for left and right delimiters, so that there is space on the "outside", but not on the "inside". Jul 4 '12 at 2:33
• @ArturoMagidin Thank you very much sir. Jul 4 '12 at 2:35
• Jul 4 '12 at 5:07
• @MattN. Thank you very much. That was helpful to me. Jul 4 '12 at 5:14
Let $V$ be a vector space over the field $\mathbb{F}$. A norm $$\| \cdot \|: V \longrightarrow \mathbb{F}$$ on $V$ satisfies the homogeneity condition $$\|ax\| = |a| \cdot \|x\|$$ for all $a \in \mathbb{F}$ and $x \in V$. So the metric $$d: V \times V \longrightarrow \mathbb{F},$$ $$d(x,y) = \|x - y\|$$ defined by the norm is such that $$d(ax,ay) = \|ax - ay\| = |a| \cdot \|x - y\| = |a| d(x,y)$$ for all $a \in \mathbb{F}$ and $x,y \in V$. This property is not satisfied by general metrics. For example, let $\delta$ be the discrete metric $$\delta(x,y) = \begin{cases} 1, & x \neq y, \\ 0, & x = y. \end{cases}$$ Then $\delta$ clearly does not satisfy the homogeneity property of the a metric induced by a norm.
To answer your edit, call a metric $$d: V \times V \longrightarrow \mathbb{F}$$ homogeneous if $$d(ax, ay) = |a| d(x,y)$$ for all $a \in \mathbb{F}$ and $x,y \in V$, and translation invariant if $$d(x + z, y + z) = d(x,y)$$ for all $x, y, z \in V$. Then a homogeneous, translation invariant metric $d$ can be used to define a norm $\| \cdot \|$ by $$\|x\| = d(x,0)$$ for all $x \in V$. | {
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• Thank you very much. I understand now. Jul 4 '12 at 2:42
• That is not enough as one needs to check that the so defined norm really induces the original metric and not another one: $d(\cdot,\cdot)\to\|\cdot\|\to d(\cdot,\cdot)$ Sep 29 '14 at 14:36
• This is an old post and it doesn't really matter much, but I have to be pedantic. So you say that if a metric comes from a norm on $\Bbb{R}$, then $d(ax,ay)=|a|d(x,y)$ but what if you just consider $\Bbb{R}$ as a vector space over $\{0,1\}$. Then the discrete metric IS homogeneous and translation invariant on this vector space.
– user223391
Jun 16 '15 at 0:16
• @MichaelMunta If the metric is induced by a norm, then $d(x,y)=||x-y||$. So $d(x+z,y+z)=||(x+z)-(y+z)||=||x-y||=d(x,y)$. Feb 27 '20 at 15:30
• You should clarify that your field $F$ is normed, otherwise the notation $|a|$ is meaningless. Jun 5 '20 at 23:24
Here is another interesting example: Let $|x-y|$ denote the usual Euclidean distance between two real numbers $x$ and $y$. Let $d(x,y)=\min\{|x-y|,1\}$, the standard derived bounded metric. Now suppose we look at $\Bbb{R}$ as a vector space over itself and ask whether $d$ comes from any norm on $\Bbb{R}$. Then if there is such a norm say $||.||$, we must have the homogeneity condition: for any $\alpha \in \Bbb{R}$ and any $v \in \Bbb{R}$,
$$||\alpha v || = |\alpha| ||v||.$$
But now we have a problem: The metric $d$ is obviously bounded by $1$, but we can take $\alpha$ arbitrarily large so that $||.||$ is unbounded. It follows that $d$ does not come from any norm.
• It was nice explanation to me. I had forgot to thank you. :) May 25 '13 at 9:21
As Henry states above, metrics induced by a norm must be homogeneous. You can see that they must also be translation invariant: $d(x+a,y+a)= d(x,y).$ So any metric not satisfying either of those can not come from a norm. | {
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On the other hand, it turns out that these two conditions on the metric are sufficient to define a norm that induces that metric: $d(x,0)=\| x \|.$
• Thank you sir. I am fully satisfied with two answers given by you and Henry.:) Jul 4 '12 at 2:44
• Did you mean instead that any metric not satisfying either of those can not come from a norm? And not the other way around? And also, these two conditions on the metric are sufficient to define a norm..? Jul 4 '12 at 6:27
• @ThomasE. Sorry, you are right, I muddled up the words. Thanks. Jul 4 '12 at 8:43
Every homogeneous metric induces a norm via: $$\|x\|:=d(x,0)$$ and every norm induces a homogeneous and translation-invariant metric: $$d(x,y):=\|x-y\|$$
The clue herein lies in wether the induced norm really represents the metric as: $$d(\cdot,\cdot)\to\|\cdot\|\to d(\cdot,\cdot)$$ which is the case only for the translation-invariant metrics: $$d'(x,y)=d(x-y,0)=d(x,y)$$ whereas the induced metric always represents the norm as: $$\|\cdot\|\to d(\cdot,\cdot)\to\|\cdot\|$$ as a simple check shows: $$\|x\|'=\|x-0\|=\|x\|$$
• @ Thanks for the answer. Feb 15 '16 at 4:15
Personal Note:
Good question, I remember wondering the same thing myself back when I new to real analysis. Here's a counter example:
Counter Example
$$d(x,y):=I_{\{(x,x)\}}(x,y):=\begin{cases} 1 &:\, \,x=y\\ 0 &:\, \,x\neq y. \end{cases}$$ This is indeed a metric (check as an exercise).
Where $I_A$ is the indicator function of the set $A$; where here the set $(x,x)\in V\times V$.
If $r\in \mathbb{R}-\{0\}$, then $d(rx,ry)\in \{0,1\}$ hence $rd(x,y)\in \{0,r\}$ which is not in the range of $d:V\times V \rightarrow \mathbb{R}$.
So, the metric $d$ fails the property that any metric induced by a norm must have, ie: $$\mbox{it fails to have the property that: } \|rx-ry\| =r\|x-y\|.$$
Interpretation & Some Intuition: | {
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Interpretation & Some Intuition:
The problem is that the topology is too fine, so all this topology can do is distinguish between things being the same or different.
As opposed to a norm topology which distinguised between object being, or not being on the same line (for some appropriate notion of line) (as well as them being different).
Hope this helps :)
• Thank you for the explanation. :) Feb 15 '16 at 4:15
• Hi, is it correct to consider the metric d(x,y)=|x-y|^2018 that fails to h;ave the above property? Oct 23 '18 at 8:19
• Yes, why not? :) Oct 23 '18 at 15:33
One possible way of showing that a metric does not arise from a norm is to show that it is bounded, as it then cannot be homogeneous.
Proof: Take $$d$$ a bounded metric on a space $$X$$: i.e. $$\exists D \in \mathbb{R}_+$$ such that $$\forall (x,y) \in X^2, d(x,y) \leq D$$. Suppose now for contradiction that $$d$$ arises from a norm, i.e. the exists a norm $$\|\cdot\|$$ such that $$d(x,y) = \|x - y\|$$. Recall that the distance must then must be homogeneous, for we have $$d(\lambda x, \lambda y) = \|\lambda x - \lambda y\| = |\lambda| \cdot \|x - y\| = |\lambda| d(x,y)$$.
Take now arbitrary $$(x_0, y_0) \in X^2$$, then we must have that $$d\left( \frac{D+1}{d(x_0, y_0)} \cdot x_0, \frac{D+1}{d(x_0, y_0)} \cdot y_0 \right) = \frac{D+1}{d(x_0, y_0)} \cdot d(x_0, y_0) = D + 1 > D$$ which contradicts the upper bound of the metric. | {
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# B A question about Greatest common factor (GCF) ?
#### awholenumber
When we try to find the Greatest common factor (GCF) of two numbers , does it only involve prime factorization ?
#### FactChecker
Gold Member
2018 Award
Yes. The GCF is a product of primes but is not usually a prime itself. Prime factorization of both numbers is the way to find out what the GCF is. If you have the prime factorization of both numbers, it is easy to calculate the GCF.
#### awholenumber
Ok , so the same prime factorization is used to find the LCM too , right ?
#### FactChecker
Gold Member
2018 Award
Ok , so the same prime factorization is used to find the LCM too , right ?
Yes.
Ok , Thanks
#### SlowThinker
I dare to disagree, Euclid's algorithm or binary GCD are used to find GCD, then LCM(a, b)=a*b/GCD(a, b).
#### FactChecker
Gold Member
2018 Award
I stand corrected. I was not familiar with these algorithms. The binary GCD algorithm and Euclidean_algorithm are interesting.
#### awholenumber
Ok, i have one more doubt .
The GCF of two numbers involves prime factorization of those two numbers , then multiply those factors both numbers have in common
Isnt LCM about finding the least common multiple of two numbers ?
What does that have anything to do with prime numbers ?
#### mfb
Mentor
You can find the LCM via LCM(a, b)=a*b/GCD(a, b), if you have the GCD first.
That is often the most practical way to find the LCM.
Prime factorization is just one possible way to find the GCD. For large numbers, it can be very time-consuming, and different algorithms can be more efficient.
#### awholenumber
Thanks for the information mfb , i am just trying to cover the algebra 1 for dummies book .
It doesn't have a method like this LCM(a, b)=a*b/GCD(a, b) mentioned in it .
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#### PeroK
Homework Helper
Gold Member
2018 Award
Thanks for the information mfb , i am just trying to cover the algebra 1 for dummies book .
It doesn't have a method like this LCM(a, b)=a*b/GCD(a, b) mentioned in it .
You'll find that in an "algebra for intelligent students" textbook!
#### awholenumber
lol ok , first let me somehow finish this one book properly
#### PeroK
Homework Helper
Gold Member
2018 Award
lol ok , first let me somehow finish this one book properly
It's still worth understanding why the two are related. The basic argument is:
$a = a'g, b = b'g$
Where $g$ is the GCD of $a$ and $b$ and $a', b'$ are how much bigger $a$ and $b$ are than $g$.
For example: if $a = 42$ and $b = 15$, then $g = 3$ and $a=14 \times 3, b = 5 \times 3$, hence $a' = 14, b' = 5$
Now, the LCM of $a$ and $b$ must have as factors simply $a', b'$ and $g$, so $l = a'b'g = a'b = a'gb/g = ab/g$
You can now see that the LCM is the product of $a$ and $b$, divided by the GCD.
In our example:
$LCM(42, 15) = \frac{42 \times 15}{3} = 210$
mfb
#### awholenumber
Thanks for sharing , i will keep this in my mind and maybe someday i will be able to use this advanced method .
#### Michael Hardy
Yes. The GCF is a product of primes but is not usually a prime itself. Prime factorization of both numbers is the way to find out what the GCF is. If you have the prime factorization of both numbers, it is easy to calculate the GCF.
But you don't need to know anything about prime factorizations to find the GCD; you can use Euclid's algorithm, which is very efficient.
#### Michael Hardy | {
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#### Michael Hardy
You don't need prime factorizations to find GCDs. There is an efficient method not requiring any knowledge of prime numbers: Euclid's algorithm. This is the oldest algorithm still in standard use, dating back to Euclid's writings in the 3rd century BC, and it is very efficient.
\begin{align*} \gcd(1989,867) & = \gcd(255,867) & & \text{since 255 is the remainder} \\ & & & \text{when 1989 is divided by 867} \\[10pt] & = \gcd(255,102) & & \text{since 102 is the remainder} \\ & & & \text{when 867 is divided by 255} \\[10pt] & = \gcd(51,102) & & \text{since 51 is the remainder} \\ & & & \text{when 255 is divided by 102} \\[10pt] & = \gcd(51,0) & & \text{since 0 is the remainder} \\ & & & \text{when 102 is divided by 51} \\[10pt] & = 51. \end{align*}
Thus we have $\dfrac{1989}{867} = \dfrac{51\times39}{51\times17} = \dfrac{39}{17}.$
#### Michael Hardy
Yes. The GCF is a product of primes but is not usually a prime itself. Prime factorization of both numbers is the way to find out what the GCF is. If you have the prime factorization of both numbers, it is easy to calculate the GCF.
This is wrong. Euclid's very efficient algorithm for finding GCDs does not require doing anything at all with prime numbers.
#### mfb
Mentor
But you don't need to know anything about prime factorizations to find the GCD; you can use Euclid's algorithm, which is very efficient.
Yes, that has been mentioned in post 6 for example.
This thread is from 2017, by the way.
"A question about Greatest common factor (GCF) ?"
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# How to plot magnitude and phase response of 2 cascaded filters in Matlab?
How would I go about plotting a magnitude and phase response of a system that consists of two cascaded 2nd order Butterworth filters in Matlab? Filters are the same.
% Filter coeffs:
[B, A] = butter( 2, 1000/8000*2, 'high' ); % Cutoff at 1000 Hz, fs 8000 Hz.
% Cascade 2 filters: input -> filter -> filter -> output :
[out1, state1 ] = filter( B, A, in, state1);
[out2, state2 ] = filter( B, A, out1, state2);
Using MATLAB/Octave as the tool, the following approach lets you plot the magnitude & phase samples of the DTFT of the cascade of the two discrete-time LTI filters using their LCCDE coefficient vectors $$b[k]$$, and $$a[k]$$, assuming they have LCCDE representations..
Since the cascade LTI system is described as: $$h[n] = h_1[n] \star h_2[n]$$ and consequently $$H(z) = H_1(z) H_2(z)$$
Where the $$h[n]$$ is the impulse response of the cascade system and $$H(z)$$ is its transfer function (Z-transform of impulse response). Those $$h_1[n]$$ and $$h_2[n]$$ refer to the impulse responses of the individual systems that make up the cascade.
Assuming that individual systems do have rational Z-transforms, then we have the following expression for the composite (cascade) system Z-transfom:
\begin{align} H(z) &=\frac{ \sum_{k=0}^{M} B[k]z^{-k} } {\sum_{k=0}^{N} A[k]z^{-k}} = H_1(z) H_2(z) \\ \\ &= \left( \frac{ \sum_{k=0}^{M_1} b_1[k]z^{-k} } {\sum_{k=0}^{N_1} a_1[k]z^{-k}} \right) \left( \frac{ \sum_{k=0}^{M_2} b_2[k]z^{-k} } {\sum_{k=0}^{N_2} a_1[k]z^{-k}} \right) \end{align}
Where individual coefficients $$b_1[k],a_1[k],b_2[k],a_2[k]$$ are those that represent the cascaded systems.
We want $$B[k]$$ and $$A[k]$$ that represent the cascade system, from which we can use the following Matlab/Octave function to plot the DTFT magnitude and phase responses, assumimg that the cascade is a stable LTI system so that it will have a frequency response function:
figure,freqz(B,A); | {
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figure,freqz(B,A);
Now it can be shown by polynomial manipulations that the coefficients $$B[k]$$ and$$A[k]$$ are related to the individual coefficients $$b_1[k],a_1[k],b_2[k],a_2[k]$$ as the following: $$B[k] = b_1[k] \star b_2[k]$$ and $$A[k] = a_1[k] \star a_2[k]$$
Hence the required frequency response plot can be obtained in the combined call as follows:
figure,freqz(conv(b1,b2),conv(a1,a2));
where b1,a1 and b2,a2 are the coefficient vectors that represent the individual systems. The method can be generalised to N systems in cascade.
When you pass a signal from two cascaded filters, what happens is that the magnitude response of the whole chain is the product of individual filters, and the phase response is the sum of individual phase responses. This is because the transfer functions are multiplied in the cascade implementation.
So if the frequency response of the single filters are $$H_A(w)=|H_A(w)|e^{j\angle H_A(w)}$$ $$H_B(w)=|H_B(w)|e^{j\angle H_B(w)}$$ then the frequency response of the two cascaded filters (overall) is \begin{align} H_{AB}(w)&=H_A(w)H_B(w)\\ &=\left(|H_A(w)|e^{j\angle H_A(w)}\right)\left(|H_B(w)|e^{j\angle H_B(w)}\right)\\ &=|H_A(w)||H_B(w)|e^{j(\angle H_A(w)+\angle H_B(w))} \end{align}
You can use [h,w] = freqz(b,a) in Matlab to get the frequency response of your desired filters. Use abs and angle to find the magnitude and phase:
...
[hA,w] = freqz(bA,aA);
[hB,w] = freqz(bB,aB);
hAB = hA.*hB;
MagResp = 20*log10(abs(hAB));
PhaseResp = angle(hAB);
plot(w,MagResp)
...
when the two filters are identical, then the overall magnitude is squared and the overall phase is scaled by two: \begin{align} H_{AA}(w)&=\left(|H_A(w)|e^{j\angle H_A(w)}\right)\left(|H_A(w)|e^{j\angle H_A(w)}\right)\\ &=|H_A(w)|^2e^{j2\angle H_A(w)} \end{align} | {
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• Yes, filter is the same. – Danijel Mar 27 '17 at 11:51
• Yes, I added the general case at the beginning, then the identical case at the end. – msm Mar 27 '17 at 12:09
• Shouldn't it be wAB = w + w, and then plot(wAB)? Is angle(hAB) necessary? – Danijel Mar 27 '17 at 12:43
• w is the frequency axis and remains fix. We should just add function values at given w values. – msm Mar 27 '17 at 20:03
• angle(hAB) gives you the phase of the complex array hAB. Without that, you just have a bunch of complex values of size legth(w). – msm Mar 28 '17 at 1:42 | {
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# Complex integration with Cauchy's Integral Formula
Calculate$$\int_\gamma \frac{(z+27i)(z+16)}{z(z+81)^2}dz$$ where $\gamma$ is the triangle whose vertices are the third roots of $z = -8i$, oriented counterclockwise.
Answer:
I calculated the third roots of $-8i$ and they all have modulus $2$. This tells me that the maximum distance of $\gamma$ from the origin will be $2$.
There are singularities at $z=0, z=-81$. As $81 > 2$, this singularity falls outside $\gamma$ so the only one that matters is $z = 0.$
I then applied Cauchy's Integral Formula $$\int_\gamma \frac{(z+27i)(z+16)}{z(z+81)^2}dz = 2\pi i [\frac{(z+27i)(z+16)}{(z+81)^2}] |_{z=0}$$
And I got a final result of $\displaystyle\frac{-32\pi}{243}$.
Is my analysis and final result correct?
-
Yep, completely spot on! Only thing I can see you could have improved slightly was to notice the cube roots of $-8i$ should have modulus 2 without calculating them, saves you some work. – Ragib Zaman Apr 18 '12 at 14:00
(+1) for showing work. – The Chaz 2.0 Apr 18 '12 at 14:01
You should copy this to an answer and accept it :) – Neal Apr 18 '12 at 15:34
@Jim_CS If you require further feedback it might help to elaborate more specifically on any doubts. If not, please post an answer and accept it so that we know it is resolved. – Bill Dubuque May 17 '12 at 15:54
## 2 Answers
Answering my own question as Neal advised me to.
I calculated the third roots of −8i and they all have modulus 2. This tells me that the maximum distance of γ from the origin will be 2.
There are singularities at z=0,z=−81. As 81>2, this singularity falls outside γ so the only one that matters is z=0. I then applied Cauchy's Integral Formula ∫γ(z+27i)(z+16)z(z+81)2dz=2πi[(z+27i)(z+16)(z+81)2]|z=0 And I got a final result of −32π243.
Is my analysis and final result correct?
-
Nothing wrong with your analysis and answer.
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IntMath Home » Forum home » Infinite Series Expansion » Taylor's series question
# Taylor's series question [Solved!]
### My question
my question is in Taylor's series: Derive Tayor's series for x^(1/2) at x=1 up to the first four non-zero terms.hence estimate square root of 1.1 correct to 4 d.p
1. Taylor Series
### What I've done so far
I think I need to differentiate it several times.
Here it is:
f(x) = x^(1/2)
f'(x) = 1/2 x^(-1/2) = 1/(2 x^(1/2))
f''(x) = - 1/4 x^(-3/2) = -1/(4 x^(3/2))
f'''(x) = 3/8 x^(-5/2) = 3/(8 x^(5/2))
But I'm stuck then
X
my question is in Taylor's series: Derive Tayor's series for x^(1/2) at x=1 up to the first four non-zero terms.hence estimate square root of 1.1 correct to 4 d.p
Relevant page
<a href="/series-expansion/1-taylor-series.php">1. Taylor Series</a>
What I've done so far
I think I need to differentiate it several times.
Here it is:
f(x) = x^(1/2)
f'(x) = 1/2 x^(-1/2) = 1/(2 x^(1/2))
f''(x) = - 1/4 x^(-3/2) = -1/(4 x^(3/2))
f'''(x) = 3/8 x^(-5/2) = 3/(8 x^(5/2))
But I'm stuck then
## Re: Taylor's series question
Hello Snowboy
Your problem is very similar to the example on the page you came from:
1. Taylor Series
Now that you've differentiated, you need to substitute the necessary values, just like I did in that example.
Or is you problem with the square root approximation?
Regards
X
Hello Snowboy
Your problem is very similar to the example on the page you came from:
<a href="/series-expansion/1-taylor-series.php">1. Taylor Series</a>
Now that you've differentiated, you need to substitute the necessary values, just like I did in that example.
Or is you problem with the square root approximation?
Regards
## Re: Taylor's series question
So a=1, right?
f(1) = 1
f'(1) = 1/2
f''(1) = -1/4
f'''(1) = 3/8
Plugging it into the formula:
1 + 1/2 (x-1) -1/4(x-1)^2 + 3/8(x-1)^3-...
How many steps do we have to do?
X
So a=1, right?
f(1) = 1
f'(1) = 1/2
f''(1) = -1/4
f'''(1) = 3/8 | {
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X
So a=1, right?
f(1) = 1
f'(1) = 1/2
f''(1) = -1/4
f'''(1) = 3/8
Plugging it into the formula:
1 + 1/2 (x-1) -1/4(x-1)^2 + 3/8(x-1)^3-...
How many steps do we have to do?
## Re: Taylor's series question
The question says you need 4 d.p. accuracy.
So you just keep finding approximations at the specific value given until there is no difference in the number in the 4th d.p. after you have rounded it.
X
The question says you need 4 d.p. accuracy.
So you just keep finding approximations at the specific value given until there is no difference in the number in the 4th d.p. after you have rounded it.
## Re: Taylor's series question
It says "at 1.1".
Is this OK?
sqrt(1.1) = 1 + 1/2(0.1) - 1/4(0.1)^2 + 3/8(0.1)^3
= 1 + 0.05 - 0.0025 + 0.000375 = 1.047875
I can see that 4th term is small and the next one will be even smaller, I think. So maybe there's no need to do any more steps because it won't affect the 4th d.p.
Maybe I should do another step to make sure.
f''''(x) = -15/16 x^(-7/2)
f''''(1) = -15/16
So the 5th term is -15/16(x-1)^4
Plugging in 1.1 gives -0.00009375
Adding this gives 1.047875-0.00009375 = 1.04778125
To 4 d.p. it's still 1.0478.
But checking on my calculator, I get sqrt(1.1) = 1.04880884817
To 4 d.p., that's 1.0488. Is my answer wrong?
X
It says "at 1.1".
Is this OK?
sqrt(1.1) = 1 + 1/2(0.1) - 1/4(0.1)^2 + 3/8(0.1)^3
= 1 + 0.05 - 0.0025 + 0.000375 = 1.047875
I can see that 4th term is small and the next one will be even smaller, I think. So maybe there's no need to do any more steps because it won't affect the 4th d.p.
Maybe I should do another step to make sure.
f''''(x) = -15/16 x^(-7/2)
f''''(1) = -15/16
So the 5th term is -15/16(x-1)^4
Plugging in 1.1 gives -0.00009375
Adding this gives 1.047875-0.00009375 = 1.04778125
To 4 d.p. it's still 1.0478.
But checking on my calculator, I get sqrt(1.1) = 1.04880884817
To 4 d.p., that's 1.0488. Is my answer wrong?
## Re: Taylor's series question | {
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To 4 d.p., that's 1.0488. Is my answer wrong?
## Re: Taylor's series question
Your thinking process is very good. Your working is correct. The difference with the calculator answer is they are using a better algorithm.
Our Taylor's Series approximation is only "good" right near the number we expand about (in your case, x=1).
To see why they are different, let's compare the actual square root graph with the Taylor's Series expansion:
The green one is y=sqrt(x) and the magenta curve is y=1 + 1/2 (x-1) -1/4(x-1)^2 + 3/8(x-1)^3.
You can see it's not a very good approximation as we get away from x=1.
X
Your thinking process is very good. Your working is correct. The difference with the calculator answer is they are using a better algorithm.
Our Taylor's Series approximation is only "good" right near the number we expand about (in your case, x=1).
To see why they are different, let's compare the actual square root graph with the Taylor's Series expansion:
[graph]310,250;-0.3,2.5;-0.5,2,0.5,0.5;sqrt(x),1 + 1/2 (x-1) -1/4(x-1)^2 + 3/8(x-1)^3[/graph]
The green one is y=sqrt(x) and the magenta curve is y=1 + 1/2 (x-1) -1/4(x-1)^2 + 3/8(x-1)^3.
You can see it's not a very good approximation as we get away from x=1.
## Re: Taylor's series question
Thanks a lot!
X
Thanks a lot! | {
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# Linear, second-order differential equation?
1. Sep 27, 2011
### Raen
One of my homework problems this week was:
Verify that y = sin(4t) + 2cos(4t) is a solution to the following initial value problem.
2y'' + 32y=0; y(0)=2, y'(0)=4
Find the maximum of |y(t)| for -infinity< t < infinity.
Verifying that the given y equation is a solution is easy, all there is to do is derive twice, plug in the new equations, and show that 0 = 0. I have no trouble with that. My question is with the second part, about finding the maximum positive value of y(t).
I would think that no matter how large t gets, sin and cos will always oscillate between -1 and 1. When cos is 1, obviously sin is 0. Putting in sin(4t) = 0 and cos(4t) = 1 gives me a maximum value of 2 for any y(t) as t approaches infinity. I graphed the function on a calculator to confirm this, just in case, and the graph appears to agree with me.
My confusion is that 2 isn't the correct answer. According to the website we do our homework on, the answer should be sqrt(5). There are no questions like this worked out in the book and I can't find anything online. Can anybody explain to me why y(t) has a maximum value of sqrt(5) as t approaches infinity?
Thank you!
2. Sep 27, 2011
### LCKurtz
Generally, when you have something like acos(θ) + bsin(θ) you can write it as a single trig function. You multiply and divide by the square root of the sum of the squares of the coefficients:
$$a\cos\theta + b\sin\theta = \sqrt{a^2+b^2} \left(\frac a {\sqrt{a^2+b^2}}\cos\theta + \frac b {\sqrt{a^2+b^2}}\sin\theta\right)$$
Now, since the sum of the squares of the new coefficients of the sine and cosine add up to one, they can be used as the sin(β) and cos(β) of some angle β. You then have an addition formula for a sine or cosine function with amplitude $\sqrt{a^2+b^2}$. In your example that gives amplitude $\sqrt 5$.
3. Sep 28, 2011
### HallsofIvy
Staff Emeritus
LCKurtz's response is, as always, excellent. | {
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3. Sep 28, 2011
### HallsofIvy
Staff Emeritus
LCKurtz's response is, as always, excellent.
Raen, you seem to be assuming that a maximum of f+ g must occur where either f or g has a maximum (since you looked immediately at cos(x)= 1) and that is not correct.
Another way to find a maximum or minimum for a function is (as I am sure you know) is to look where the derivative is 0. If f(x)= sin(4x)+ 2cos(4x), then f'(x)= 4cos(4x)- 8sin(4x)= 0 when 4 cos(4x)= 8 sin(4x) or cos(4x)= 2sin(4x). Rather than solve for x directly, note that $cos^2(4x)= 4 sin^2(4x)$ so $sin^2(4x)+ cos^2(4x)= 5 sin^2(4x)= 1$ so that $sin^2(4x)= 1/5$ and $sin(4x)= 1/\sqrt{5}= \sqrt{5}/5$. And, of course, $cos^2(4x)= 1- sin^2(x)= 4/5$ so $cos(4x)= 2/\sqrt{5}= 2\sqrt{5}/5$.
Putting those back into the original formula, $sin(4x)+ 2cos(4x)= \sqrt{5}/5+ 4\sqrt{5}/5= 5/\sqrt{5}= \sqrt{5}$
Last edited: Sep 28, 2011
4. Sep 28, 2011
### Raen
Oh, I think I understand now. Thank you both for your detailed explanations! | {
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# Gaussian elimination, inconsistent or infinite solutions?
I have this system
$\left[ \begin{array}{ccc|c} -3&0&-1&0\\ 2&0&0&0\\ 2&0&0&0\\ \end{array} \right]$ Is there something you can tell from looking at this straight away?
When I reduce it I get this:
$\left[ \begin{array}{ccc|c} 1&0&-1&0\\ 0&0&1&0\\ 0&0&0&0\\ \end{array} \right]$
I thought I read that if you get a statement that does not make sense, then the system is inconsistent, so $1 = 0$ does not make sense to me. Does that not mean it's inconsistent?
However, I thought that $x_3$ could equal $0$, and so could $x_1$ and basically, $x_2$ could equal anything. Still not sure, i reduced it further to:
$\left[ \begin{array}{ccc|c} 1&0&0&0\\ 0&0&1&0\\ 0&0&0&0\\ \end{array} \right]$
From this, am I correct now in thinking that:
$x_1 = r\\ x_2 = s \\ x_3 = t$
I am not sure of why I'm thinking like this, in terms of free variables etc. If I'm right, I wouldn't mind some reasoning as to why this is so, or being correct if I am wrong. Thanks.
• How did you do the final step of reduction? – platty Aug 2 '17 at 23:17
• Opps, I see where you are going with this. The 1 should remain in the second row. Sorry about that. Let me think for a minute. – Bucephalus Aug 2 '17 at 23:20
• Now that I have corrected my last reduction, I'm still not really sure what this means. it looks like $x_2$ can equal anything, Therefore, $x_1$ and $x_3$ can equal anything? Is that correct? – Bucephalus Aug 2 '17 at 23:24
Remember that the entries on the left are coefficients of variables. So from your final reduction, we would end up with the RREF: $$\left[\begin{array}{ccc|c} 1&0&0&0\\ 0&0&1&0\\ 0&0&0&0\\ \end{array}\right]$$
The first line tells us that $1x + 0y + 0z = 0$, so $x = 0$. The second line tells us that $0x + 0y + 1z = 0$, i.e. $z=0$. But we have $y$ as a free variable, so our solution is $(0,y,0)$. | {
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Note that we can tell that $y$ is free by observing the original matrix - there is no coefficient in its column, so its value doesn't affect our final solution.
• Oh, yes I see now. Thanks for your explanation @S.Ong. – Bucephalus Aug 2 '17 at 23:27
• Say it did just turn out to be like I originally had it, with a single 1 in the first column of the first row. That would then mean that $x_1=0, x_2=r$ and $x_3=s$. Would that be right? – Bucephalus Aug 2 '17 at 23:29
• Yep, that's it! :) – platty Aug 2 '17 at 23:32
• Ok, thankyou @S.Ong. – Bucephalus Aug 2 '17 at 23:32
$\left[ \begin{array}{ccc|c} -3&0&-1&0\\ 2&0&0&0\\ 2&0&0&0\\ \end{array} \right]$
What do you know straight away?
$(0,0,0)$ is in the solution set. The right column being all $0$ tells you this.
With the second row and the 3rd row identical or the right part of the matrix, this matrix is singular.
In fact the rank is 2, the nullity is 1, and there is a "line" of solutions.
The second column all zero's tells you the very same thing. Furthermore, $x_2$ can be anything.
And the structure of rows 2 and 3 tells you that $x_1$ must equal $0.$
this is more than enough information to let you know the solution set is $(0,t,0)$
• Aaah, very insightful. This is the kind of thing I was looking for also. Do you mean by "line" that the solutions form a line in 3-space? @DougM – Bucephalus Aug 3 '17 at 0:30
• yes, with the rank of the kernel equal to 1, (and the information that at least one solution exists), then all of the solutions will lie on a line in $\mathbb R^3.$ If the nullity equaled 2, then we might say that there is a plane of solutions. – Doug M Aug 3 '17 at 0:36
• Thanks @DougM. I will keep an eye out for you next time. – Bucephalus Aug 3 '17 at 0:43 | {
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Adding very small probabilities - How to compute?
In some problems, probabilities are so small that they are best represented in computational facilities as log-probabilities. Computational problems can arise when you try to add these small probabilities together, since some computational facilities (e.g., base R) can't distinguish very small probabilities from zero. In these cases it is necessary to solve the problem either by finding a workaround that avoids the small numbers, or by using a computational facility that can deal with very small numbers.
The archetypal problem of this kind is as follows. Suppose you have log-probabilities $$\ell_1$$ and $$\ell_2$$, where the corresponding probabilities $$\exp(\ell_1)$$ and $$\exp(\ell_2)$$ are too small to be distinguished from zero in the initial computational facility being used (e.g., base R). We want to find the log-sum of these probabilities, which we denote by:
$$\ell_+ \equiv \ln \big( \exp(\ell_1) + \exp(\ell_2) \big)$$
Assume that we are ---at least initially--- working in an environment where $$\exp(\ell_1)$$ and $$\exp(\ell_2)$$ cannot be computed, since they are so small that they are indistinguishable from zero.
Questions: How can you effectively compute this log-sum? Can this be done in the base R? If not, what is the simplest way to do it with package extensions?
• the "log-sum-exp" trick can help – Taylor Nov 29 '18 at 3:44
Some preliminary mathematics: As a preliminary matter it is worth noting that this function is the LogSumExp (LSE) function, which is often encountered when performing computation on values that are represented in the log-scale. To see how to deal with sums of this kind, we first note a useful mathematical result concerning sums of exponentials: | {
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\begin{aligned} \exp(\ell_1) + \exp(\ell_2) &= \exp(\max(\ell_1,\ell_2)) + \exp(\min(\ell_1,\ell_2)) \\[6pt] &= \exp(\max(\ell_1,\ell_2)) (1 + \exp(\min(\ell_1,\ell_2)-\max(\ell_1,\ell_2)) \\[6pt] &= \exp(\max(\ell_1,\ell_2)) (1 + \exp(-|\ell_1 - \ell_2|)). \\[6pt] \end{aligned}
This result converts the sum to a product, which allows us to present the log-sum as:
\begin{aligned} \ell_+ &= \ln \big( \exp(\ell_1) + \exp(\ell_2) \big) \\[6pt] &= \ln \big( \exp(\max(\ell_1,\ell_2)) (1 + \exp(-|\ell_1 - \ell_2|)) \big) \\[6pt] &= \max(\ell_1, \ell_2) + \ln (1 + \exp(-|\ell_1 - \ell_2|)). \\[6pt] \end{aligned}
In the case where $$\ell_1 = \ell_2$$ we obtain the expression $$\ell_+ = \ell_1 + \ln 2 = \ell_2 + \ln 2$$, so the log-sum is easily computed. In the case where $$\ell_1 \neq \ell_2$$ this expression still reduces the problem to a simpler case, where we need to find the log-sum of one and $$\exp(-|\ell_1 - \ell_2|)$$.$$^\dagger$$
Now, using the Maclaurin series expansion for $$\ln(1+x)$$ we obtain the formula:
\begin{aligned} \ell_+ &= \max(\ell_1, \ell_2) + \sum_{k=1}^\infty (-1)^{k+1} \frac{\exp(-k|\ell_1 - \ell_2|)}{k} \quad \quad \quad \text{for } \ell_1 \neq \ell_2. \\[6pt] \end{aligned}
Since $$\exp(-|\ell_1 - \ell_2|) < 1$$ the terms in this expansion diminish rapidly (faster than exponential decay). If $$|\ell_1 - \ell_2|$$ is large then the terms diminish particularly rapid. In any case, this expression allows us to compute the log-sum to any desired level of accuracy by truncating the infinite sum to a desired number of terms.
Implementation in base R: It is actually possible to compute this log-sum accurately in base R through creative use of the log1p function. This is a primitive function in the base package that computes the value of $$\ln(1+x)$$ for an argument $$x$$ (with accurate computation even for $$x \ll 1$$). This primitive function can be used to give a simple function for the log-sum: | {
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logsum <- function(l1, l2) { max(l1, l2) + log1p(exp(-abs(l1-l2))); }
Implementation of this function succeeds in finding the log-sum of probabilities that are too small for the base package to deal with directly. Moreover, it is able to calculate the log-sum to a high level of accuracy:
l1 <- -3006;
l2 <- -3012;
logsum(l1, l2);
[1] -3005.998
sprint("%.50f", logsum(l1, l2));
[1] "-3005.99752431486240311642177402973175048828125000000000"
As can be seen, this method gives a computation with 41 decimal places for the log-sum. It only uses the functions in the base package, and does not involve any changes to the default calculation settings. It gives as high a level of accuracy as you are likely to need in most cases.
It is also worth noting that there are many packages in R that extend the computational facilities of the base program, and can be used to deal with sums of very small numbers. It is possible to find the log-sum of small probabilities using packages such as gmp or Brobdingnag, but this requires some investment in learning their particular syntax.
$$^\dagger$$ From this result we can also see that if $$|\ell_1 - \ell_2|$$ is itself large (i.e., if one of the probabilities is very small compared to the other) then the exponential term in this equation will be near zero, and we will then have $$\ell_+ \approx \max(\ell_1, \ell_2)$$ to a very high degree of accuracy.
• A number of posts on site discuss related issues. – Glen_b Nov 29 '18 at 4:53
• Sorry, I must have missed these. – Ben Nov 29 '18 at 5:05
• There's at least some material here that I don't think is anywhere else, and the exposition is quite clear so I don't think there's a problem, but it's worth being aware of the other posts. (A lot crop up in discussions of underflow and overflow in various calculations, for example) – Glen_b Nov 29 '18 at 5:29
• I have added the underflow tag to the post to link to other questions on this subject. – Ben Nov 29 '18 at 6:31 | {
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# Simple question related to the Fundamental theorem of arithmetic
Given a number $b$, which we write as a product of prime numbers:
$$b = p_1 \cdots p_s$$
Can I then deduce that a number, which divides $b$ then has to be a product of the above primes in the factorization of $b$?
-
Yes and that follows by writing out another factorization for the number that divides $b = p_1 \cdots p_s$ and using unique factorization to dedude that the factors must be some of the primes in the factorization of $b$. – Adrián Barquero Apr 7 '12 at 15:34
More explicitly, if $b = p_{1}^{e_1} \cdots p_{s}^{e_s}$ is the prime factorization of $b$ where the $p_i$'s are different primes, then you can prove that any divisor of $b$ is a product of the form $p_{1}^{a_1} \cdots p_{s}^{a_s}$ where the exponents $a_i$ satisfy that $0 \leq a_i \leq e_i$ for $i = 1, \dots , s$. – Adrián Barquero Apr 7 '12 at 15:38
Essentially yes: If $p_1,\ldots,p_s$ are primes (possibly repeated), then every divisor of $$b = p_1p_2\cdots p_s$$ must be a (possibly empty) product of some (possibly all) of the $p_i$, times $1$ or $-1$.
-
And why the downvote? – Arturo Magidin Apr 11 '12 at 6:05
A divisor of $b$ has to be a product of some of the prime factors of $b$, but "some" may mean $0$ of them (if the divisor is $1$). The divisor can have no prime factors that are not prime factors if $b$.
If the divisor in question is $c$, then for some number $d$ we have $b=cd$. If a prime number $p$ divides $b$, then $p$ must divide either $c$ or $d$ (or both). The foregoing sentence is Euclid's lemma.
-
Hint $\rm\ p\ |\ d\ |\ p_1\cdots p_n\ \Rightarrow\ p\ |\ p_j\:$ for some $\rm\:j\:$ by the Prime Divisor Property. Cancelling $\rm\:p\:$ from $\rm\:d\:$ and $\rm\:p_1\cdots p_n\:$ and inducting yields that the prime factors of $\rm\:d\:$ are a sub-multiset of $\rm\{p_1,\ldots,p_n\}.$ | {
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Generally $\rm\:d\ |\ b_1\cdots b_n\ \Rightarrow\ d = d_1\cdots d_n,\ \ d_j\ |\ b_j,\:$ i.e. a divisor of a product is a product of divisors. This is a useful generalization of the Prime Divisor Property from atoms to composite numbers. Yours is the special case when all $\rm\:b_j = p_j\:$ prime, so $\rm\:d_j = 1\:$ or $\rm\:p_j\:$ (modulo a unit factor $\pm1$).
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@Downvoter If something is not clear, please feel free to ask questions and I will be happy to elaborate. – Bill Dubuque Apr 27 '12 at 19:13 | {
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# Math Help - Pascal's Triangle
1. ## Pascal's Triangle
Being naturally curious, i was thinking by myself whether there might not be a way to determine the sum of all the numbers in this triangle by some method, depending on the number of rows it has. And also the sum of the numbers in the n-th row.
This is what i've discovered so far.
Row 1 = 1 ; Sum = 1
Row 2 = 1 1 ; Sum = 2
Row 3 = 1 2 1 ; Sum = 4
Row 4 = 1 3 3 1 ; Sum = 8
Row 5 = 1 4 6 4 1 ; Sum = 16
So the sum of these numbers in each row, are multiples of 2.
2. Originally Posted by janvdl
Now i know this might sound silly...
Being naturally curious, i was thinking by myself whether there might not be a way to determine the sum of all the numbers in this triangle by some method, depending on the number of rows it has. And also the sum of the numbers in the n-th row.
This is what i've discovered so far.
Row 1 = 1 ; Sum = 1
Row 2 = 1 1 ; Sum = 2
Row 3 = 1 2 1 ; Sum = 4
Row 4 = 1 3 3 1 ; Sum = 8
Row 5 = 1 4 6 4 1 ; Sum = 16
So the sum of these numbers in each row, are multiples of 2.
seems to be $2^{n-1}$, where $n$ is the number of the row we are at from what you have there
3. Originally Posted by Jhevon
seems to be $2^{n-1}$, where $n$ is the number of the row we are at from what you have there
Ah ok, i wonder why i didnt notice that. And the sum of all the numbers up to an n-th row?
4. Originally Posted by janvdl
Ah ok, i wonder why i didnt notice that. And the sum of all the numbers up to an n-th row?
seems to be $2^n - 1$ where $n$ is the number of the row we are at
5. Originally Posted by Jhevon
seems to be $2^n - 1$ where $n$ is the number of the row we are at
I was thinking in that line. cant believe i was so blind
6. Hello, janvdl!
Being naturally curious . . .
Good for you!
I made some amazing "discoveries" while exploring on my own.
Of course, anything I found was proved centuries ago,
. . but it was still very satisfying.
What you discovered is a stunning pattern. | {
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What you discovered is a stunning pattern.
Consider the coefficients of the expansion of $(a + b)^n$
. . . $\begin{array}{ccc}n & \text{coefficients} & \text{sum} \\ \hline 0 & 1 & 1 = 2^0\\ 1 & 1\;\;1 & 2 = 2^1 \\ 2 & 1\;\;2\;\;1 & 4 = 2^2 \\ 3 & 1\;\;3\;\;3\;\;1 & 8 =2^3 \\ 4 & 1\;\;4\;\;6\;\;4\;\;1 & 16 = 2^4 \\ 5 & 1\;\;5\;\;10\;\;10\;\;5\;\;1 & 32 = 2^5 \\ 6 & 1\;\;6\;\;15\;\;20\;\;15\;\;6\;\;1 & 64 = 2^6 \\ \vdots & \vdots & \vdots \end{array}$
The sum of the $n^{th}$ row is $2^n$.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
I "discovered" this pattern while in college.
$\begin{array}{ccc}1 + 2 & = & 3 \\ 4 + 5 + 6 & = & 7 + 8 \\ 9 + 10 + 11 + 12 & = & 13 + 14 + 15 \\ 16 + 17 + 18 + 19 + 20 & = & 21 + 22 + 23 + 24 \\
\vdots & & \vdots\end{array}$
I found these in a book . . .
$\begin{array}{ccc}1 & = & 1 = 1^3 \\ 3 + 5 & = & 8 = 2^3 \\ 7 + 9 + 11 & = & 27 = 3^3 \\ 13 + 15 + 17 + 19 & = & 64 = 4^3 \\ 21 + 23 + 25 + 27 + 29 & = & 125 = 5^3 \\ \vdots & & \vdots \end{array}$
$\begin{array}{ccc}3^2 + 4^2 & = & 5^2 \\ 10^2 + 11^2 + 12^2 & = &13^2 + 14^2 \\ 21^2 + 22^2 + 23^2 + 24^2 & = & 25^2 + 26^2 + 27^2 \\ 36^2 + 37^2 + 38^2 + 39^2 + 40^2 & = & 41^2 + 42^2 + 43^2 + 44^2 \\ 55^2 + 56^2 + 57^2 + 58^2 + 59^2 + 60^2 & = & 61^2 + 62^2 + 63^2 + 64^2 + 65^2 \\ \vdots & &\vdots\end{array}$
7. I've found something else too...
I was actually trying to find another type of formula when i noticed this...
I attached a picture, i know its badly drawn but at least it gives you the idea...
The white triangles are the normal figures of Pascal's Triangle. But look at those red triangles. They interest me a lot. There must be a lot more patterns involved in this triangle than i ever thought.
What if every red triangle is the product of the 3 white triangles around it? Or the sum of the 3 white ones around it? We can make lots of new patterns using this...
8. Hello, janvdl!
That's fascinating . . . Thank you! | {
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8. Hello, janvdl!
That's fascinating . . . Thank you!
What if every red triangle is the sum of the 3 white ones around it?
Then we have:
. . $\begin{array}{c}3\\
4\;\;4\\
5\;\;8\;\;5 \\
6\;\;13\;\;13\;\;6 \\
7\;\;19\;\;26\;\;19\;\;7
\end{array}$
I finally saw where it came from . . .
Consider a "Pascal's Triangle" whose third line is: .1 - 3 - 1.
. . $\begin{array}{c}1 \\ 1\;\;1 \\ 1\;\;3\;\;1 \\
1\;\;4\;\;4\;\;1 \\ 1\;\;5\;\;8\;\;5\;\;1 \\ 1\;\;6\;\;13\;\;13\;\;6\;\;1 \\ 1\;\;7\;\;19\;\;26\;\;19\;\;7\;\;1
\end{array}$
See it?
This opens up a world of possibilities, doesn't it?
9. Originally Posted by Soroban
Hello, janvdl!
That's fascinating . . . Thank you!
Then we have:
. . $\begin{array}{c}3\\
4\;\;4\\
5\;\;8\;\;5 \\
6\;\;13\;\;13\;\;6 \\
7\;\;19\;\;26\;\;19\;\;7
\end{array}$
I finally saw where it came from . . .
Consider a "Pascal's Triangle" whose third line is: .1 - 3 - 1.
. . $\begin{array}{c}1 \\ 1\;\;1 \\ 1\;\;3\;\;1 \\
1\;\;4\;\;4\;\;1 \\ 1\;\;5\;\;8\;\;5\;\;1 \\ 1\;\;6\;\;13\;\;13\;\;6\;\;1 \\ 1\;\;7\;\;19\;\;26\;\;19\;\;7\;\;1
\end{array}$
See it?
This opens up a world of possibilities, doesn't it?
So have i actually kind of discovered something here?
And what can we actually do with my "discovery"? A world of possibilities has been opened? You're clearly seeing things that i cant...
10. Here's another of my "discoveries", made years and years ago.
We know the Binomial Theorem is based on Pascal's Triangle.
We use Pascal's Triangle for expanding $(a + b)^n$.
I wondered if there was a "Trinomial Theorem" for expanding $(a + b + c)^n$.
I cranked out the first few cases . . . and got into an awful mess.
. . $\begin{array}{ccc}(a + b + c)^0 & = & 1 \\
(a + b + c)^1 & = & a + b + c \\
(a + b + c)^2 & = & a^2 + b^2 + c^2 + 2ab + 2bc + 2ac \\
(a + b + c)^3 & = & a^3 + b^3 + c^3 + 3a^2b + 3ab^2 + 3b^2c + 3bc^2 + 3a^2c + 3ac^2 + 6abc
\end{array}$
How do we make sense out of these terms and coefficients? | {
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How do we make sense out of these terms and coefficients?
It took me the longest time to see what to do:
. . arrange the terms in triangular arrays.
$\begin{array}{ccccc} & & & & a^4 \\ & & &a^3 & 4a^3b\;\;4a^3c \\& & a^2 & 3a^2b\;\;3a^2c & 6a^2b^2\;\;12a^2bc\;\;6a^2c^2 \\& a & 2ab\;\;2ac & 3ab^2\;\;6abc\;\;3ac^2 & 4ab^3\;\;12ab^2c\;\;12abc^2\;\;4ac^3 \\1\quad &\;\; b\;\;c\quad & b^2\;\;2ab\;\;c^2\quad & b^3\quad3b^2c\quad3bc^2\quad c^3\quad & b^4\quad4b^3c\quad6b^2c^2\quad 4bc^3\quad c^4\end{array}$
Look at the coefficients:
$\begin{array}{ccccc} & & & & 1 \\& & & 1 & 4\;\;4 \\& & 1 & 3\;\;3 & 6\;\;12\;\;6 \\& 1 & 2\;\;2 & 3\;\;6\;\;3 & 4\;\;12\;\;12\;\;4 \\1\quad & \;\;\;1\;\;1\quad & \;\;1\;\;2\;\;1\quad & \;\;1\;\;3\;\;3\;\;1\quad & 1\;\;\;4\;\;\;6\;\;\;4\;\;\;1\end{array}$
If we "stack" these triangles, we get a tetrahedron.
. . Each number is the sum of the three numbers above it.
I seriously doubt that I'm the first to notice all this.
But you can feel free to call it "Soroban's Tetrahedron".
11. I'm just going to try and get some formulas out of my triangle Im busy writing a small book on my discoveries in this triangle. Just for the fun of it | {
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+0
# Assigning var
0
232
4
+4
Is there a way to define a variable to use an equation? E.x.: 3x^2-2x+5=y, and I want to assign x a value without changing it out. Is it possible?
CatBoy13 Dec 12, 2014
#3
+91479
+10
Welcome to web2.0calc forum Catboy13
Umm - Interesting question.
$$\begin{array}{rll} 3x^2-2x+5&=&y\\\\ 3x^2-2x&=&y-5\\\\ x^2-\frac{2}{3}x&=&\frac{y-5}{3}\\\\ x^2-\frac{2}{3}x+\left(\frac{2}{6}\right)^2&=&\frac{y-5}{3}+\left(\frac{2}{6}\right)^2\\\\ x^2-\frac{2}{3}x+\left(\frac{1}{3}\right)^2&=&\frac{y-5}{3}+\left(\frac{1}{3}\right)^2\\\\ x^2-\frac{2}{3}x+\left(\frac{1}{3}\right)^2&=&\frac{3(y-5)}{9}+\frac{1}{9}\\\\ \left(x-\frac{1}{3}\right)^2&=&\frac{3y-15+1}{9}\\\\ \left(x-\frac{1}{3}\right)^2&=&\frac{3y-14}{9}\\\\ x-\frac{1}{3}&=&\pm\sqrt{\frac{3y-14}{9}}\\\\ x&=&\frac{1}{3}\pm\sqrt{\frac{3y-14}{9}}\\\\ x&=&\frac{1\pm\sqrt{3y-14}}{3}}\\ \end{array}$$
Melody Dec 13, 2014
Sort:
#1
+7188
0
I believe it is possible....but I might be wrong...
happy7 Dec 12, 2014
#2
0
If you first define a function f as
$${f}{\left({\mathtt{x}}\right)} = {\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}$$
then for any particular x the function f(x) has a value and you can write things such as
f(2) + f(0) - 4 = 14
Guest Dec 12, 2014
#3
+91479
+10
Welcome to web2.0calc forum Catboy13
Umm - Interesting question. | {
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