# What's the shape of this “addsTo” function …?

Note that in this combinatronics question,

How many lists of 100 numbers (1 to 10 only) add to 700?

For an array of 100 numbers, each 1 to 10 inclusive, and the total is T - how many such arrays exist?

In fact due to the ASTOUNDING answers there, we now know

for T=700 the answer is 1.2 e92

Being curious I now wonder about the shape of this result for various T.

Sadly, all I know are the points T<100, T=100, T=700 (thanks, MSE!), and T=1000.

Many questions arise, is there a T that makes a maximum, is it the same all over except the ends, is it bumpy or erratic ... does it make any difference if T is prime, even, etc ... and, since 19^92 is pretty small and there's only 600 Ts, in fact, is T=700 just freakishly small for some reason (what reason?) ...... or?

• The rough shape is a bell curve with center at $550$ and symmetric around that point. – Thomas Andrews Sep 11 '14 at 14:42
• Yeah, I didn't read the linked-to question, so I didn't realize I duplicated a lot of work, so just made the comment. Basically, if you pick a random integer from $1$ to $10$ 100 times, the the probability of getting $T$ is $\frac{1}{10^{100}}$ times your count. And the general tendency of repeating processes like this is to get a bell curve. – Thomas Andrews Sep 11 '14 at 14:45
• It's $$\sum_{j=0}^{45} (-1)^j \binom{100}{j}\binom{549-10j}{99}$$ Need some time to find a program to calculate it. – Thomas Andrews Sep 11 '14 at 14:51
• The approximation of the distribution as a bell curve appeared as @mjqxxxx's answer to your other problem. One can perhaps obtain asymptotic corrections to this with more sophisticated analysis. – Semiclassical Sep 12 '14 at 16:32
• ah, i didn't notice that, thanks semi. still it would be interesting to know the max etc - still thanks – Fattie Sep 12 '14 at 16:35

According to the Central Limit Theorem, when you add a large number of independent, identically distributed random variables (each variable can be a sum of other random variables), the distribution of the sum tends to a normal distribution whose mean is the sum of the means of the individual distributions, and whose variance is the sum of the variances of the individual distributions.

For example, if we sum $1$ number ($1$-$10$), the distribution looks like

$\hspace{3cm}$

If we sum $2$ numbers ($1$-$10$), the distribution looks like

$\hspace{3cm}$

If we sum $3$ numbers ($1$-$10$), the distribution starts to looks like a normal distribution

$\hspace{3cm}$

If we sum $10$ numbers ($1$-$10$), the distribution looks even more like a normal distribution

$\hspace{3cm}$

Now lets look at the sum of $100$ numbers ($1$-$10$)

$\hspace{3cm}$

The number of ways for $100$ numbers ($1$-$10$) to sum to $700$, which was computed in this answer, is pointed to by the arrow. Note that the range of the sum is $9n$ where $n$ is how many numbers are added; however, the standard deviation is $\sqrt{8.25n}$. Thus, the relative spread gets smaller; that is, the distribution becomes narrower about the mean as $n$ get bigger.

The maximum number of ways to sum to $T$ is at the mean of the distribution; that is, at $T=550$, where the number of ways is $$13868117806391314648666325510838589167047653141664\\4888545033078503482282975641730091720919340564340$$ which is approximately $1.3868117806391314649\times10^{98}$.

• Wow - you guys are amazing. I'm so dense I didn't even realise, of course the question is just equivalent to looking at the random distribution. one question - you see the final graph, in fact the answer to the question ........ – Fattie Sep 13 '14 at 9:30
• ...... would we still call that "a bell curve," a normal distribution, since it's so incredibly pointy? As the answer explains, for 1 or 2 numbers, I guess you'd say, "it is not" a Normal distribution. Well now, when you get to very high numbers (300, 500, 10,000) again does it become "not" a Normal distribution? Or, is it still perfectly a "bell curve" but just very pointy? Is bell curvey -ness subjective? Or conversely, just by definition, is every result here officially a normal distribution (even the 1 or 2 cases)? Cheers!!!!! – Fattie Sep 13 '14 at 9:31
• @JoeBlow: A scaled bell curve looks pointy, and that is what this is: a scaled bell curve. If scaled so that the standard deviation is $1$, its total sum is $1$, and translated so that the mean is $0$, it would look very close to a normal distribution bell curve. Scaled this way would map the total range of possibilities to the whole real line. – robjohn Sep 13 '14 at 9:39
• Gotchya, I'll have to think about that. If I'm not mistaken, in your graphs, indeed they all start and end at the zero points (so, for "10" it's 10-->100, and for "100" it's 100-->1000 etc). And the scale heights are the same so there's no "trick". But it seems to me they are radically different: the shapes of "100" and "10million" would be terribly different right? Recall I was asking "is 700 just freakishly small for some reason?" Indeed, that is the reason: there is no such "freakish smallness" in the (say) "10" or "15" case when you are still equally near the middle .... but ... – Fattie Sep 13 '14 at 9:53
• ... indeed this function becomes "amazingly pointy" as the count gets higher. I'm thinking like a video game programmer, if something depended on that shape, play would be spectacularly different with the "10" shape versus the "500" shape. (You'd almost always, versus never, run in to the goblin or whatever.) Anyways - I'm wasting your time - I'll go take a refresher course in normal curves THANK YOU SO MUCH. Do you often provide such good answers that nobody else even has to bother? ;-) – Fattie Sep 13 '14 at 9:55


Khan Academy is a … Next lesson. The powers don’t need to be “2” all the time. Played 0 times. Thew following steps will be useful to simplify any radical expressions. Test - I. Note that every positive number has two square roots, a positive and a negative root. The standard way of writing the final answer is to place all the terms (both numbers and variables) that are outside the radical symbol in front of the terms that remain inside. Test - II . When the radical is a cube root, you should try to have terms raised to a power of three (3, 6, 9, 12, etc.). Square root, cube root, forth root are all radicals. Show all your work to explain how each expression can be simplified to get the simplified form you get. 72 36 2 36 2 6 2 16 3 16 3 48 4 3 A. APTITUDE TESTS ONLINE. Simplifying Radicals Worksheet … Multiply all numbers and variables outside the radical together. Exponents and power. To play this quiz, please finish editing it. For this problem, we are going to solve it in two ways. It must be 4 since (4)(4) =  42 = 16. Step 1 : If you have radical sign for the entire fraction, you have to take radical sign separately for numerator and denominator. Section 6.3: Simplifying Radical Expressions, and . 2) Product (Multiplication) formula of radicals with equal indices is given by To read our review of the Math Way -- which is what fuels this page's calculator, please go here . An algebraic expression that contains radicals is called a radical expression An algebraic expression that contains radicals.. We use the product and quotient rules to simplify them. by lsorci. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Simplifying Radical Expressions DRAFT. However, I hope you can see that by doing some rearrangement to the terms that it matches with our final answer. For the number in the radicand, I see that 400 = 202. For example, These types of simplifications with variables will be helpful when doing operations with radical expressions. So, , and so on. Our mission is to provide a free, world-class education to anyone, anywhere. The number 16 is obviously a perfect square because I can find a whole number that when multiplied by itself gives the target number. Well, what if you are dealing with a quotient instead of a product? applying all the rules - explanation of terms and step by step guide showing how to simplify radical expressions containing: square roots, cube roots, . In this tutorial, the primary focus is on simplifying radical expressions with an index of 2. Equivalent forms of exponential expressions. Next, express the radicand as products of square roots, and simplify. 5 minutes ago. #1. . Improve your math knowledge with free questions in "Simplify radical expressions with variables I" and thousands of other math skills. Then, it's just a matter of simplifying! A perfect square number has integers as its square roots. Section 6.4: Addition and Subtraction of Radicals. Notice that each group of numbers or variables gets written once when they move outside the radical because they are now one group. Radical expressions can often be simplified by moving factors which are perfect roots out from under the radical sign. Quotient Property of Radicals. But there is another way to represent the taking of a root. . Additional simplification facilities for expressions containing radicals include the radnormal, rationalize, and combine commands. More so, the variable expressions above are also perfect squares because all variables have even exponents or powers. Please click OK or SCROLL DOWN to use this site with cookies. In this tutorial, the primary focus is on simplifying radical expressions with an index of 2. Example 11: Simplify the radical expression \sqrt {32} . \left(\square\right)^{'} \frac{d}{dx} \frac{\partial}{\partial x} \int. Homework. Therefore, we have √1 = 1, √4 = 2, √9= 3, etc. Another way to solve this is to perform prime factorization on the radicand. TRANSFORMATIONS OF FUNCTIONS. Variables with exponents also count as perfect powers if the exponent is a multiple of the index. The answer must be some number n found between 7 and 8. We hope that some of those pieces can be further simplified because the radicands (stuff inside the symbol) are perfect squares. You can use the same ideas to help you figure out how to simplify and divide radical expressions. Solving Radical Equations Method 1: Perfect Square Method -Break the radicand into perfect square(s) and simplify. The goal is to show that there is an easier way to approach it especially when the exponents of the variables are getting larger. are called conjugates to each other. If the denominator is not a perfect square you can rationalize the denominator by multiplying the expression by an appropriate form of 1 e.g. Example 1. The first law of exponents is x a x b = x a+b. However, it is often possible to simplify radical expressions, and that may change the radicand. First we will distribute and then simplify the radicals when possible. Simplifying Radical Expressions Date_____ Period____ Simplify. Simplifying Radical Expressions. However, the best option is the largest possible one because this greatly reduces the number of steps in the solution. 16 x = 16 ⋅ x = 4 2 ⋅ x = 4 x. no fractions in the radicand and. The properties we will use to simplify radical expressions are similar to the properties of exponents. A radical expression is composed of three parts: a radical symbol, a radicand, and an index. So which one should I pick? The key to simplify this is to realize if I have the principal root of x over the principal root of y, this is the same thing as the principal root of x over y. All that you have to do is simplify the radical like normal and, at the end, multiply the coefficient by any numbers that 'got out' of the square root. To simplify a radical, factor the radicand (under the radical) into factors whose exponents are multiples of the index. Scientific notations. Multiply all numbers and variables inside the radical together. Adding and Subtracting Radical Expressions, That’s the reason why we want to express them with even powers since. Test to see if it can be divided by 4, then 9, then 25, then 49, etc. Simplify radical expressions using the product and quotient rule for radicals. Here it is! To help me keep track that the first term means "one copy of the square root of three", I'll insert the "understood" "1": Don't assume that expressions with unlike radicals cannot be simplified. Example 2: to simplify ( 3. . The symbol is called a radical sign and indicates the principal square root of a number. Product Property of n th Roots. How to Simplify Radicals with Coefficients. For the numerical term 12, its largest perfect square factor is 4. These two properties tell us that the square root of a product equals the product of the square roots of the factors. To simplify this sort of radical, we need to factor the argument (that is, factor whatever is inside the radical symbol) and "take out" one copy of anything that is a square. Simplifying logarithmic expressions. Then express the prime numbers in pairs as much as possible. If found, they can be simplified by applying the product and quotient rules for radicals, as well as the property a n n = a, where a is positive. Let's look at to help us understand the steps involving in simplifying radicals that have coefficients. To simplify complicated radical expressions, we can use some definitions and rules from simplifying exponents. By multiplying the variable parts of the two radicals together, I'll get x 4 , which is the square of x 2 , so I'll be able to take x 2 out front, too. So we expect that the square root of 60 must contain decimal values. What rule did I use to break them as a product of square roots? Always look for a perfect square factor of the radicand. Part A: Simplifying Radical Expressions. This calculator simplifies ANY radical expressions. You may use your scientific calculator. Improve your math knowledge with free questions in "Simplify radical expressions" and thousands of other math skills. Picking the largest one makes the solution very short and to the point. Recall that the Product Raised to a Power Rule states that $\sqrt[n]{ab}=\sqrt[n]{a}\cdot \sqrt[n]{b}$. These properties can be used to simplify radical expressions. Use rational exponents to simplify radical expressions. Finish Editing. Pairing Method: This is the usual way where we group the variables into two and then apply the square root operation to take the variable outside the radical symbol. Step 4: Simplify the expressions both inside and outside the radical by multiplying. This type of radical is commonly known as the square root. Type your expression into the box under the radical sign, then click "Simplify." Notice that the square root of each number above yields a whole number answer. Here are the search phrases that today's searchers used to find our site. Perfect Squares 1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 324 400 625 289 = 2 = 4 = 5 = 10 = 12 Simplifying Radicals Simplifying Radical Expressions Simplifying Radical Expressions A radical has been simplified when its radicand contains no perfect square factors. Vertical translation. Remember, the square root of perfect squares comes out very nicely! Let’s deal with them separately. The paired prime numbers will get out of the square root symbol, while the single prime will stay inside. nth roots . Code to add this calci to your website Just copy and paste the below code to your webpage where you want to display this calculator. And it checks when solved in the calculator. We need to recognize how a perfect square number or expression may look like. Play. Radical Expressions are fully simplified when: –There are no prime factors with an exponent greater than one under any radicals –There are no fractions under any radicals –There are no radicals in the denominator Rationalizing the Denominator is a way to get rid of any radicals in the denominator Mathematics. If found, they can be simplified by applying the product and quotient rules for radicals, as well as the property $$\sqrt[n]{a^{n}}=a$$, where $$a$$ is positive. The radicand contains no fractions. Let’s do that by going over concrete examples. Type any radical equation into calculator , and the Math Way app will solve it form there. Simplification of expressions is a very useful mathematics skill because, it allows us to change complex or awkward expression into more simple and compact form. If and are real numbers, and is an integer, then. Simplify a Term Under a Radical Sign. And we have one radical expression over another radical expression. Example 3: Simplify the radical expression \sqrt {72} . Simplify the expression: It is possible that, after simplifying the radicals, the expression can indeed be simplified. You can do some trial and error to find a number when squared gives 60. The solution to this problem should look something like this…. Simplify expressions with addition and subtraction of radicals. For example, the sum of $$\sqrt{2}$$ and $$3\sqrt{2}$$ is $$4\sqrt{2}$$. Live Game Live. After doing some trial and error, I found out that any of the perfect squares 4, 9 and 36 can divide 72. x^{\circ} \pi. In order to simplify radical expressions, you need to be aware of the following rules and properties of radicals 1) From definition of n th root(s) and principal root Examples More examples on Roots of Real Numbers and Radicals. Mathplanet is licensed by Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 Internationell-licens. The roots of these factors are written outside the radical, with the leftover factors making up the new radicand. +1) type (r2 - 1) (r2 + 1). Procedures. As long as the powers are even numbers such 2, 4, 6, 8, etc, they are considered to be perfect squares. Share practice link. Example 10: Simplify the radical expression \sqrt {147{w^6}{q^7}{r^{27}}}. Example 1: Simplify the radical expression \sqrt {16} . Simplify each of the following. Algebraic expressions containing radicals are very common, and it is important to know how to correctly handle them. Radical Expressions and Equations Notes 15.1 Introduction to Radical Expressions Sample Problem: Simplify 16 Solution: 16 =4 since 42 =16. You will see that for bigger powers, this method can be tedious and time-consuming. We have to consider certain rules when we operate with exponents. Variables in a radical's argument are simplified in the same way as regular numbers. This quiz is incomplete! Practice: Evaluate radical expressions challenge. Think of them as perfectly well-behaved numbers. Play this game to review Algebra I. Simplify. The product of two conjugates is always a rational number which means that you can use conjugates to rationalize the denominator e.g. Adding and Subtracting Radical Expressions The calculator presents the answer a little bit different. To find the product of two monomials multiply the numerical coefficients and apply the first law of exponents to the literal factors. Remember the rule below as you will use this over and over again. Simplifying Radical Expressions with Variables When radicals (square roots) include variables, they are still simplified the same way. . The following are the steps required for simplifying radicals: Start by finding the prime factors of the number under the radical. Step 1 : Decompose the number inside the radical into prime factors. This is easy to do by just multiplying numbers by themselves as shown in the table below. That is, we find anything of which we've got a pair inside the radical, and we move one copy of it out front. Step 3 : A radical expression is said to be in its simplest form if there are, no perfect square factors other than 1 in the radicand, $$\sqrt{16x}=\sqrt{16}\cdot \sqrt{x}=\sqrt{4^{2}}\cdot \sqrt{x}=4\sqrt{x}$$, $$\sqrt{\frac{25}{16}x^{2}}=\frac{\sqrt{25}}{\sqrt{16}}\cdot \sqrt{x^{2}}=\frac{5}{4}x$$. Looks like the calculator agrees with our answer. Perfect cubes include: 1, 8, 27, 64, etc. $$\frac{x}{4+\sqrt{x}}=\frac{x\left ( {\color{green} {4-\sqrt{x}}} \right )}{\left ( 4+\sqrt{x} \right )\left ( {\color{green}{ 4-\sqrt{x}}} \right )}=$$, $$=\frac{x\left ( 4-\sqrt{x} \right )}{16-\left ( \sqrt{x} \right )^{2}}=\frac{4x-x\sqrt{x}}{16-x}$$. Recall that the Product Raised to a Power Rule states that $\sqrt[x]{ab}=\sqrt[x]{a}\cdot \sqrt[x]{b}$. Discovering expressions, equations and functions, Systems of linear equations and inequalities, Representing functions as rules and graphs, Fundamentals in solving equations in one or more steps, Ratios and proportions and how to solve them, The slope-intercept form of a linear equation, Writing linear equations using the slope-intercept form, Writing linear equations using the point-slope form and the standard form, Solving absolute value equations and inequalities, The substitution method for solving linear systems, The elimination method for solving linear systems, Factor polynomials on the form of x^2 + bx + c, Factor polynomials on the form of ax^2 + bx +c, Use graphing to solve quadratic equations, Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 Internationell-licens. There is a rule for that, too. Introduction. ACT MATH ONLINE TEST. no perfect square factors other than 1 in the radicand. Example 14: Simplify the radical expression \sqrt {18m{}^{11}{n^{12}}{k^{13}}}. Radical expressions are expressions that contain radicals. Evaluating mixed radicals and exponents. Actually, any of the three perfect square factors should work. Start studying Algebra 5.03: Simplify Radical Expressions. In the same way we know that, $$\sqrt{x^{2}}=x\: \: where\: \: x\geq 0$$, These properties can be used to simplify radical expressions. You factor things, and whatever you've got a pair of can be taken "out front". The first rule we need to learn is that radicals can ALWAYS be converted into powers, and that is what this tutorial is about. Simplifying radical expression. Save. Comparing surds. Remember that getting the square root of “something” is equivalent to raising that “something” to a fractional exponent of {1 \over 2}. Quantitative aptitude. Simplifying Radical Expressions. Search phrases used on 2008-09-02: Students struggling with all kinds of algebra problems find out that our software is a life-saver. Factoring to Solve Quadratic Equations - Know Your Roots; Pre-Requisite 4th, 5th, & 6th Grade Math Lessons: MathTeacherCoach.com . We can add or subtract radical expressions only when they have the same radicand and when they have the same radical type such as square roots. no perfect square factors other than 1 in the radicand $$\sqrt{16x}=\sqrt{16}\cdot \sqrt{x}=\sqrt{4^{2}}\cdot \sqrt{x}=4\sqrt{x}$$ no … If we combine these two things then we get the product property of radicals and the quotient property of radicals. In the next a few examples, we will use the Distributive Property to multiply expressions with radicals. You can use the same ideas to help you figure out how to simplify and divide radical expressions. 0. Extended Keyboard; Upload; Examples; Random; Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. We know that The corresponding of Product Property of Roots says that . −1)( 2. . By quick inspection, the number 4 is a perfect square that can divide 60. If you would like a lesson on solving radical equations, then please visit our lesson page . Rationalize Radical Denominator - online calculator Simplifying Radical Expressions - online calculator SIMPLIFYING RADICAL EXPRESSIONS INVOLVING FRACTIONS. Determine the index of the radical. Example 12: Simplify the radical expression \sqrt {125} . To find the product of two monomials multiply the numerical coefficients and apply the first law of exponents to the literal factors. Now for the variables, I need to break them up into pairs since the square root of any paired variable is just the variable itself. Sometimes radical expressions can be simplified. This lesson covers . But before that we must know what an algebraic expression is. Simplify … The radicand contains both numbers and variables. Simplifying Expressions – Explanation & Examples. Simplifying simple radical expressions Simplifying Radicals Practice Worksheet Awesome Maths Worksheets For High School On Expo In 2020 Simplifying Radicals Practices Worksheets Types Of Sentences Worksheet . Enter the expression here Quick! Some of the worksheets below are Simplifying Radical Expressions Worksheet, Steps to Simplify Radical, Combining Radicals, Simplify radical algebraic expressions, multiply radical expressions, divide radical expressions, Solving Radical Equations, Graphing Radicals, … Once you find your worksheet(s), you can either click on the pop-out icon or download button to print or download … $$\sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}\cdot {\color{green} {\frac{\sqrt{y}}{\sqrt{y}}}}=\frac{\sqrt{xy}}{\sqrt{y^{2}}}=\frac{\sqrt{xy}}{y}$$, $$x\sqrt{y}+z\sqrt{w}\: \: and\: \: x\sqrt{y}-z\sqrt{w}$$. Dividing Radical Expressions 7:07 Simplify Square Roots of Quotients 4:49 Rationalizing Denominators in Radical Expressions 7:01 To simplify this radical number, try factoring it out such that one of the factors is a perfect square. 1) 125 n 5 5n 2) 216 v 6 6v 3) 512 k2 16 k 2 4) 512 m3 16 m 2m 5) 216 k4 6k2 6 6) 100 v3 10 v v 7) 80 p3 4p 5p 8) 45 p2 3p 5 9) 147 m3n3 7m ⋅ n 3mn 10) 200 m4n 10 m2 2n 11) 75 x2y 5x 3y 12) 64 m3n3 Aptitude test online. Print; Share; Edit; Delete; Report an issue; Host a game. Dividing Radical Expressions. We just have to work with variables as well as numbers Example 13: Simplify the radical expression \sqrt {80{x^3}y\,{z^5}}. Let’s simplify this expression by first rewriting the odd exponents as powers of an even number plus 1. \int_{\msquare}^{\msquare} \lim. Example 8: Simplify the radical expression \sqrt {54{a^{10}}{b^{16}}{c^7}}. Play this game to review Algebra II. The simplify/radical command is used to simplify expressions which contain radicals. Simplifying radical expressions calculator. 0. Compare what happens if I simplify the radical expression using each of the three possible perfect square factors. To simplify radicals, rather than looking for perfect squares or perfect cubes within a number or a variable the way it is shown in most books, I choose to do the problems a different way, and here is how. Multiplication tricks. Below is a screenshot of the answer from the calculator which verifies our answer. If you have square root (√), you have to take one term out of the square root for every two same terms multiplied inside the radical. Thus, the answer is. “Division of Even Powers” Method: You can’t find this name in any algebra textbook because I made it up. A radical expression is composed of three parts: a radical symbol, a radicand, and an index. Starting with a single radical expression, we want to break it down into pieces of “smaller” radical expressions. Separate and find the perfect cube factors. Here are the steps required for Simplifying Radicals: Step 1: Find the prime factorization of the number inside the radical. Recognize a radical expression in simplified form. 10-2 Lesson Plan - Simplifying Radicals (Members Only) 10-2 Online Activities - Simplifying Radicals (Members Only) ... 1-1 Variables and Expressions; Solving Systems by Graphing - X Marks the Spot! The symbol is called a radical sign and indicates the principal square root of a number. To simplify radical expressions, look for factors of the radicand with powers that match the index. Example 1: to simplify ( 2. . Improve your math knowledge with free questions in "Simplify radical expressions" and thousands of other math skills. When simplifying, you won't always have only numbers inside the radical; you'll also have to work with variables. Example 5: Simplify the radical expression \sqrt {200} . Multiplying Radical Expressions. If the term has an even power already, then you have nothing to do. Example 2: Simplify the radical expression \sqrt {60}. Horizontal translation. For example the perfect squares are: 1, 4, 9, 16, 25, 36, etc., because 1 = 12, 4 = 22, 9 = 32, 16 = 42, 25 = 52, 36 = 62, and so on. Step 2 : We have to simplify the radical term according to its power. This type of radical is commonly known as the square root. One way to think about it, a pair of any number is a perfect square! Edit. Start studying Algebra 5.03: Simplify Radical Expressions. A perfect square is the product of any number that is multiplied by itself, such as 81, which is the product of 9 x 9. no radicals appear in the denominator of a fraction. Be sure to write the number and problem you are solving. WeBWorK. Simplifying Radical Expressions . For instance, x2 is a p… Keep in mind that you are dealing with perfect cubes (not perfect squares). For the case of square roots only, see simplify/sqrt. Algebra 2A | 5.3 Simplifying Radical Expressions Assignment For problems 1-6, pick three expressions to simplify. These properties can be used to simplify radical expressions. Example 6: Simplify the radical expression \sqrt {180} . Level 1 $$\color{blue}{\sqrt5 \cdot \sqrt{15} \cdot{\sqrt{27}}}$$ $5\sqrt{27}$ $30$ $45$ $30\sqrt2$ ... More help with radical expressions at mathportal.org. 25 16 x 2 = 25 16 ⋅ x 2 = 5 4 x. 1) Simplify. Practice. Radical expressions come in many forms, from simple and familiar, such as$\sqrt{16}$, to quite complicated, as in $\sqrt[3]{250{{x}^{4}}y}$. There is a rule for that, too. COMPETITIVE EXAMS. To multiply radicals, you can use the product property of square roots to multiply the contents of each radical together. Procedures. For example, the square roots of 16 are 4 and … Otherwise, check your browser settings to turn cookies off or discontinue using the site. The radicand contains no factor (other than 1) which is the nth or greater power of an integer or polynomial. Simplify #2. simplifying radical expressions. Step 2 : If you have square root (√), you have to take one term out of the square root for every two same terms multiplied inside the radical. If you have radical sign for the entire fraction, you have to take radical sign separately for numerator and denominator. The main approach is to express each variable as a product of terms with even and odd exponents. Fantastic! Simplify any radical expressions that are perfect squares. 9th - University grade . Algebra 2A | 5.3 Simplifying Radical Expressions Assignment For problems 1-6, pick three expressions to simplify. Radical expressions are written in simplest terms when. Step 2 : We have to simplify the radical term according to its power. +1 Solving-Math-Problems Page Site. Topic Exercises. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Example 7: Simplify the radical expression \sqrt {12{x^2}{y^4}} . Rationalizing the Denominator. Going through some of the squares of the natural numbers…. Square roots are most often written using a radical sign, like this, . We use cookies to give you the best experience on our website. Take a look at our interactive learning Quiz about Simplifying Radical Expressions , or create your own Quiz using our free cloud based Quiz maker. In addition, those numbers are perfect squares because they all can be expressed as exponential numbers with even powers. Multiplying Radical Expressions However, the key concept is there. \sum. The simplest case is when the radicand is a perfect power, meaning that it’s equal to the nth power of a whole number. Symbol is called a radical sign, then click  simplify simplify radical expressions and are real numbers, and other tools! Polynomials and radical expressions, look for factors of the factors is a screenshot of three! Factor of the first law of exponents is x a x b = x a+b solution! For this problem should look something like this… use Polynomial Multiplication to multiply expressions with radicals such one. 200, the primary focus is on simplifying radical expressions based on the radicand no longer has perfect! Also perfect squares comes out of the square root to simplify the radical ) into factors whose exponents are of. That the square root, cube root, forth root are all radicals then 9, then you have sign... At to help you figure out how to multiply radicals, you wo n't always have numbers! Because I can find a perfect square factor we expect that the square of. Next, express the prime numbers in pairs as much as possible powers that the., games, and an index of 2 power of an integer, then click  simplify expressions. Quotient rule for radicals want to express it as some even power plus 1 apply! Expressions, we want to express it as some even power plus 1 further because! Radical into prime factors the target number only left numbers are prime be taken  front! Mathplanet is licensed by Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 Internationell-licens of a product equals the product quotient... Factorization of the radicand, and is an easier way to solve this is easy to do when radicals square. Squares comes out of the exponent properties radicand contains no factor ( other than 1 in same. Operations with radical expressions Online calculator to simplify this expression by first rewriting the odd powers as even plus. Find out that any of the exponent properties squares of the three perfect square factors to! The most important step in understanding and mastering algebra sign ( square root of radical... Radicand contains no factor ( other than 1 ) ( r2 + 1 ) with even powers picking largest. No perfect square is Polynomial Multiplication to multiply expressions with an index simplified. 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When a “normal” fraction contains fractions in either the numerator or denominator or both, then we consider it to be a complex fraction. This algebra video tutorial explains the process of simplifying complex numbers or imaginary numbers. Simplifying Expressions Containing Complex Numbers Because i really is a radical, and because we do not want to leave radicals in the denominators of fractions, we do not want to leave any complex numbers in the denominators of fractions. When you want … Example 2: Simplify the complex fraction below. Please click OK or SCROLL DOWN to use this site with cookies. The overall LCD of the denominators is \color{red}6x. Dividing Complex Numbers Write the division of two complex numbers as a fraction. In this case, the above expression is a complex fraction with 1/2 as the numerator and 1/6 as the denominator. Simplifying Complex Expressions Calculator. Solutions Graphing Practice ; Geometry beta; Notebook Groups Cheat Sheets; Sign In; Join; Upgrade; Account Details Login Options Account Management Settings … If either numerator or denominator consists of several numbers, these numbers must be combined into one number. Example 3: Simplify the complex fraction below. The line or slash in that separates the numerator and the denominator in a fraction represents division. Addition and Subtraction . We have got a great deal of high-quality reference information on subjects varying from geometry to math . If you observe, the complex denominator is already in the form that we want – having one fractional symbol. The goal of this lesson is to simplify complex fractions. Ti 89 titanium Laplace download, ti-84 instructions log, log function ti-89, graph inequality calculator, second order nonhomogeneous differential equations x+2, free automatic algebra answers, calculator for multipling whole numbers with fractions. Thus, by finding the square root of both sides, you get: Therefore – 8 is the only possible value of the complex fraction. o Recognize and simplify complex fractions . Similarly do this in … In order to simplifying complex numbers that are ratios (fractions), we will rationalize the denominator by multiplying the top and bottom of the fraction by i/i. Preview: Input Expression: Simplify expression. -+ * / ^. Method II consists in multiplying the numerator and denominator by the LCD of all fractional expressions first. If the bakery utilized 1/2 of a bag of baking flour on that day. Our next step would be to transform the complex numerator into a “simple” or single fraction. Free fraction worksheets 2 simplifying fractions equivalent fractions fractions mixed numbers. In this tutorial the instructor shows how to simplify complex fractions. Example 1: to simplify $(1+i)^8$ type (1+i)^8 . This is great! For this problem, we are going to use Method 2 only. Then, the number of batches of cakes produced by the bakery on the day. Calculate the batches of cakes manufactured by the bakery on that day. The proper fraction is the one where numerator is greater than the denominator, while the improper fraction is the one where denominator is greater than the numerator. Employ the division rule by multiplying the top of the fraction by the reciprocal of the bottom. After going over a few examples, you should realize that Method 2 is much better than Method 1 because almost always it takes fewer steps to get to the final answer. Many times the mini-lesson will not be enough for you to start working on the problems. The line or slash in that separates the numerator and the denominator in a fraction represents division. In the case you have service with algebra and in particular with Simplifying Complex Fractions Calculator or elementary algebra come pay a visit to us at Algebra-help.org. Dividing positive and negative numbers. In fact, complex fractions in which the numerator and denominator both contain a single fraction are usually fairly easy to solve. Method I consists in simplifying the numerator and denominator first. Fraction answers are provided in reduced form (lowest terms). The bakery used 1/2 of a bag of baking flour on a certain day. Then reduce if possible. Basic instructions for the worksheets What we have in mind is to show how to take a complex number and simplify it. Each bag will hold 1/12 pound of trail mix. A complex fraction containing a variable is known as a complex rational expression. Example 4: Simplify the complex fraction below. A complex fraction is a fraction that contains another fraction. If either numerator or denominator consists of several numbers, these numbers must be combined into one number. Practice: Signs of expressions. From simplifying complex numbers with exponents to numerical, we have got every part discussed. Both the numerator and denominator of the complex fraction are already expressed as single fractions. Learn more Accept. Fraction answers are provided in reduced form (lowest terms). Then simplify if possible. Home Simplifying Complex Fractions Fractions Complex Fractions Fractions, Ratios, Money, Decimals and Percent Fraction Arithmetic Fractions Worksheet Teaching Outline for Fractions … Observe that the LCD of all the denominators is just \color{red}12x. Multiply both the numerator and the denominator of the complex fraction by the LCD. It may look a bit intimidating at first; however, if you pay attention to details, I guarantee you that it is not that bad. Simplify the fraction its lowest terms possible. 2 4 4 5 There are two methods used to simplify such kind of fraction. In this tutorial the instructor shows how to simplify complex fractions. Practice: Multiplying negative numbers . If you would like to see the other way of simplifying complex fractions you can refer to the textbook. Simplify the result to the lowest terms possible. 3/(1/2) is a complex fraction whereby, 3 is the numerator and 1/2 is the denominator. Example 5: Simplify the complex fraction below. If either numerator or denominator consists of several numbers, these numbers must be combined into one number. Use this to multiply through the top and bottom expressions. Simple, yet not quite what we had in mind. Complex Fractions – Explanation & Examples A fraction is made up of two parts: a numerator and a denominator; the number above the line is the numerator and the number below the line is the denominator. Create here an unlimited supply of worksheets for simplifying complex fractions — fractions where the numerator, the denominator, or both are fractions/mixed numbers. Simplify complex fractions that contain several different mathematical operations; In Multiply and Divide Mixed Numbers and Complex Fractions, we saw that a complex fraction is a fraction in which the numerator or denominator contains a fraction. Amount of trail mix each bag holds = 1/12 pound. Create single fractions in both the numerator and denominator, then follow by dividing the fractions. This type of fraction is also known as a compound fraction. The complex number calculator is also called an imaginary number calculator. $$i \text { is defined to be } \sqrt{-1}$$ From this 1 fact, we can derive a general formula for powers of $$i$$ by looking at some examples. Simplifying Fractions and Complex Fractions Therefore, take the reciprocal of the denominator, Thus, the number of batches of cakes manufactured by the bakery = 3, Simplify the complex fraction: (2 1/4)/(3 3/5). It is used to represent how many parts we have out of the total number of parts. Why a negative times a negative makes sense. How many cups scoops can fill the chicken feeder? Use this as the common multiplier for both top and bottom expressions. A complex fraction is the same as a normal fraction, but has a fraction problem in the numerator and a fraction problem in the denominator. We simplified complex fractions by rewriting them as division problems. Rationalizing Complex Numbers In this unit we will cover how to simplify rational expressions that contain the imaginary number, "i". Use this Complex Fractions Calculator to do math and add, subtract, multiply and divide complex fractions. Equivalent fractions worksheets fraction equivalent fractions equivalent fractions are equal to each other two fractions are equal if they represent the also complex fractions worksheet Simplify Complex Fractions 1 . In this method, we want to create a single fraction both in the numerator and denominator. The worksheets are meant for the study of rational numbers, typically in 7th or 8th grade math (pre-algebra and algebra 1). Simplifying complex expressions The following calculator can be used to simplify ANY expression with complex numbers. Quiz: Simplifying Fractions and Complex Fractions Decimals Signed Numbers (Positive Numbers and Negative Numbers) A complex fraction is a fraction that contains another fraction. Like last week at the Java Hut when a customer asked the manager, Jobius, for a 'simple cup of coffee' and was given a cup filled with coffee beans. Find the LCD of the entire problem, that is, the LCD of the top and bottom denominators. The worksheets are … Simplifying Expressions Containing Complex Numbers Because i really is a radical, and because we do not want to leave radicals in the denominators of fractions, we do not want to leave any complex numbers in the denominators of fractions. Find the least common denominator (LCD) of all fractions appearing within the complex fraction. 4. Multiplying and dividing negative numbers. Table 1 $\text{ Table 1} \\ \begin{array}{ccc|c} \hline Expression & & Work & Result \\\hline \red{i^ \textbf{2}} & … Then simplify if possible. Example 1: to simplify$(1+i)^8$type (1+i)^8. Example 1: Simplify the complex fraction below. For this problem, we are going to use Method 1 only. The complex number calculator is able to calculate complex numbers when they are in their algebraic form. The first thing to do is arrive at a simple fraction both in the numerator and the denominator. Imaginary numbers are based on the mathematical number $$i$$. This website uses cookies to ensure you get the best experience. Amount of baking floor used to make a batch of cakes = 1/6 of a bag. Simplifying Fractions and Complex Fractions In the denominator, we will … Now that we have developed a solid foundation regarding what fractions are as well as some different types of fractions, we can now turn to application of the basic arithmetic operations (addition, subtraction, multiplication, and division) to fractions. Complex Fractions Worksheet & fractions good to start with perhaps then move on to more. . Before I can cancel anything off, I need to simplify that top parentheses, because it has a negative exponent on it. Finish off by canceling out common factors to get the final answer. Simplify complex fractions that contain several different mathematical operations In Multiply and Divide Mixed Numbers and Complex Fractions, we saw that a complex fraction is a fraction in which the numerator or denominator contains a fraction. I can't cancel off, say, the a 's, because that a 4 isn't really on top. A complex fraction can be defined as a fraction in which the denominator and numerator or both contain fractions. In cases that involve simple numbers, addition and subtraction of fractions is easy … In complex fractions either or both the numerator and the denominator contain fractions or mixed numbers. After doing so, we can expect the problem to be reduced to a single fraction which can be simplified as usual. The following calculator can be used to simplify ANY expression with complex numbers. Apply the division rule of fractions by multiplying the numerator by the reciprocal or inverse of the denominator. Complex fractions aren't necessarily difficult to solve. When possible reduce, simplify and convert to mixed numbers, any final fraction results. This lesson is also about simplifying. There are two methods used to simplify such kind of fraction. This means we have to work a bit on the complex numerator. In this method of simplifying complex fractions, the following are the procedures: Generate a single fraction both in the denominator and the numerator. Multiplying numbers with different signs. Simplify complex fractions worksheets grade 7 305805 worksheet. Capacity of the chicken feeder = 9/10 of a cup of grains. Reduce all fractions when possible. If necessary, simplify the numerator and denominator into single fractions. The complex number online calculator, allows to perform many operations on complex numbers. There is another type of fraction called Complex Fraction, which we will see below. Why a negative times a negative is a positive. Another kind of fraction is called complex fraction, which is a fraction in which the numerator or the denominator contains a fraction. In the case you have service with algebra and in particular with Simplifying Complex Fractions Calculator or elementary algebra come pay a visit to us at Algebra-help.org. Obviously, this problem would require us to do that first before we perform division. (3/7)/9 is also a complex fraction with 3/7 and 9 as the numerator and denominator respectively. Generate a single fraction both in the denominator and the numerator. In complex fractions either or both the numerator and the denominator contain fractions or mixed numbers. Convert mixed numbers to improper fractions. Let’s look at some of the key steps for each simplification method: In this method of simplifying complex fractions, the following are the procedures: This is the easiest method of simplifying complex fractions. The above expression is a complex fraction, therefore, change the division as multiplication and take reciprocal of the fraction in denominator. Then simplify if possible. Simplifying Complex Fractions When a “normal” fraction contains fractions in either the numerator or denominator or both, then we consider it to be a complex fraction. 1. A chicken feeder can hold 9/10 of a cup of grains. If each piece the wire is 1/12 of the wire, how many pieces of the wire can Kelvin cut? For example. Here are the steps for this method: Kelvin cuts 3/4 meters of a wire into smaller pieces. The next step to do is to apply division rule by multiplying the numerator by the reciprocal of the denominator. We use cookies to give you the best experience on our website. Practice: Simplify complex fractions. Start by finding the Least Common Multiple of al the denominator in the complex fractions. A complex fraction is the same as a normal fraction, but has a fraction problem in the numerator and a fraction problem in the denominator. Simplify the fraction its lowest terms possible. Sometimes, we can take things too literally. $$i \text { is defined to be } \sqrt{-1}$$ From this 1 fact, we can derive a general formula for powers of $$i$$ by looking at some examples. Simplifying complex fractions. I can either move the whole parentheses down, square, and then simplify; or else I can take the negative-square through first, and then move things up or down. Add the fractions in the numerator, and subtract the ones in the denominator. Some examples of complex fractions are: 6 7 3 3 4 5 8 x 2 5 6 To simplify a complex fraction, remember that the fraction bar means division. Given that 3/10 of a cup grains fills the feeder, therefore the number of scoops can be found by dividing 9/10 by 3/10. If the feeder is being filled by scoop that only holds 3/10 of a cup of grains. Looking at the denominators \large{x} and \large{x^2}, its LCD must be \large{x^2} Multiply the top and bottom by this LCD. Since the LCD of 3y and 6y is just \textbf{6y}, we will now multiply the complex numerator and denominator by this LCD. This type of fraction is also known as a compound fraction. We have got a great deal of high-quality reference information on subjects varying from geometry to math By using this website, you agree to our Cookie Policy. It is […] To do this take the lcm of the denominators of the fractions in the numerator and simplify it. Use this Complex Fractions Calculator to do math and add, subtract, multiply and divide complex fractions. The types of numerator and denominator determines the type of fraction. It helps kids to work better in operating fractions comparing fractions creating equivalent fractions and more. Free Algebra Tutorials! Algebra often involves simplifying expressions, but some expressions are more confusing to deal with than others. The worksheets are meant for the study of rational numbers, typically in 7th or 8th grade math (pre-algebra and algebra 1). Simplify complex fractions by multiplying each term by the least common denominator. All expression will be simplified as much as possible show help ↓↓ examples ↓↓ ^. Then, the total length of a wire is 3/4 meters. Here are the steps required for Adding and Subtracting Rational Expressions: Step 1: Simplify the rational expression in the numerator of the original problem by adding or subtracting the fractions as necessary. Simplify any Complex Fraction Enter any fraction into the two text boxes and we will show you all work using the LCM method and the "flip" method. simplify x2 + 4x − 45 x2 + x − 30 simplify x2 + 14x + 49 49 − x2 simplify 6 x − 1 − 3 x + 1 simplify 5x 6 + 3x 2 Home: Miscellaneous Equations: Operations with Fractions: Undefined Rational Expressions: Inequalities: … Start by multiplying the numerator of the complex fraction by the reciprocal of its denominator. Employ the division rule by multiplying the top of the fraction by the reciprocal of the bottom. We can multiply by i/i because it is equal to one and won't change the value of the fraction. The problem requires you to apply the FOIL method (multiplication of two binomials) and a simple factorization of trinomial. The LCD in the numerator is (x + 3)(x – 1). There are two methods used to simplify complex fractions. 3. Calculate the possible value of x in the following complex fraction. This is the currently selected item. What is an imaginary number anyway? … Start by converting the top and bottom into improper fractions: Find the reciprocal of the denominator and change the operator: Multiply the numerators and denominators separately: The numerator and denominator of the fraction have a common factor number 9, simplify the fraction to the lowest terms possible. Analysis of this question results in complex fractions: The problem is solved by finding the reciprocal of the denominator, and in this case, it is 3/10. Remember to distribute the negative (or subtraction) sign between the two fractions. Create a single fraction in the numerator and denominator. Imaginary numbers are based on the mathematical number $$i$$. Complex Fractions – Explanation & Examples. To see extra written explanation next to the work, click on the "verbose mode" here. Otherwise, check your browser settings to turn cookies off or discontinue using the site. This algebra video tutorial explains how to simplify complex fractions especially those with variables and exponents - positive and negative exponents. 5. We simplified complex fractions by rewriting them as division problems. A bakery uses 1/6 of a bag of baking flour in a batch of cakes. You need to see someone explaining the … A fraction is made up of two parts: a numerator and a denominator; the number above the line is the numerator and the number below the line is the denominator. Step 1: Simplify the rational expression in the numerator of the original problem by subtracting the fractions. Ti 89 titanium Laplace download, ti-84 instructions log, log function ti-89, graph inequality calculator, second order nonhomogeneous differential equations x+2, free automatic algebra answers, calculator for multipling whole numbers with fractions. Complex Numbers''Simplifying roots of negative numbers video Khan Academy May 13th, 2018 - What are the complex numbers We re asked to simplify the principal square root of negative because we have a negative 52 here inside of the radical' 'miL2872X Ch07 469 550 9 29 06 18 13pm Page 469 IA May 12th, 2018 - Radicals And Complex Numbers 469 In This Chapterwe Study Radical Expressions This … There are two methods. Multiply the both the numerator and denominator of the complex fraction by this L.C.M. Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step. Are coffee beans even chewable? Simplifying Fractions and Complex Fractions. (3/4)/(9/10) is another complex fraction with 3/4 as the numerator and 9/10 as the denominator. Come to Mathfraction.com and read and learn about matrices, long division and many other algebra subject areas . 2. The complex symbol notes i. Worksheets for complex fractions Create here an unlimited supply of worksheets for simplifying complex fractions — fractions where the numerator, the denominator, or both are fractions/mixed numbers. working... Polynomial … Video Lesson . Example 2: to simplify$\dfrac{2+3i}{2-3i}\$ type (2+3i)/(2-3i). The first thing to do is arrive at a simple fraction both in the numerator and the denominator. Complex numbers involve the quantity known as i , an “imaginary” number with the property i = √−1.If you have to simply an expression involving a complex number, it might seem daunting, but it’s quite a simple process once you learn the basic rules. Video Tutorial on Simplifying Imaginary Numbers. Multiply this LCD to the numerator and denominator of the complex fraction.

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# Number of points of intersections, no of parts of chords inside circle

$n$ points ($n>1$) are taken on the circumference of a circle. Through any two of them a chord is drawn. No three chords intersect at one point inside the circle. i) Find how many points of intersections of these chords are inside the circle? ii) Find how many parts do these chords divide the circle?

I know one solution is to make a graph and use Euler's formula $v-e+f=2$. But that idea I would have never come up with. Is there any other way to approach this?

• The number of separate parts inside the circle is a sequence that starts $1,2,4,8,16,\ldots$. Can you guess the next term? – Arthur Sep 2 '16 at 20:14
• 32. How do you prove formally that the sequence gives the parts inside the circle? – Amrita Sep 2 '16 at 20:52
• It's actually 31. I was deliberately trying to throw you off. I don't know how to prove it, but Euler characteristic seems like a not-too-bad idea. – Arthur Sep 2 '16 at 20:54
• The number of points of intersection inside the circle is $\binom n4$, since each set of four points on the circle determines a pair of intersecting chords. – bof Aug 3 '19 at 12:08

I do not think there is a better approach to find the number of parts (part ii) comparing to the Euler's formula. But to use it one should first find the number of vertices and edges, and for this task I do not see how the Euler's formula can help.

For simplicity we disregard the circle arcs and will consider only the graph consisting of line segments connecting the points.

Let $$s$$ be a segment connecting two points of the circle. The points split the circle in two parts. Let the number of points in one of the parts be $$k$$. Then the number of points in the other part is $$(n-k-2)$$. Every segment which connects a point lying in one part with a point lying in the other part will intersect the segment $$s$$, whereas any segment connecting two points lying in the same part will not intersect $$s$$. Therefore the overall number of intersection points of the segment $$s$$ with other segments is $$k(n-2-k)$$ and the overall number of intersection points on all segments drawn from each of the considered circle points is: $$v''_n=\sum_{k=0}^{n-2}k(n-2-k)=\frac{(n-1)(n-2)(n-3)}6=\binom{n-1}3\tag1$$ and the overall number of the intersection points inside the circle is $$v'_n=\frac n4 v''_n=\binom n4,\tag2$$ where factor 4 in the denominator accounts for the fact that while summing over all $$n$$ circle points every internal intersection point is counted 4 times.

To compute the overall number of vertices we shall add the number of the circle points to the obtained result: $$v_n=v'_n+n=\frac{n(n+1)(n^2-7n+18)}{24}.\tag3$$

The number of edges can be computed similarly. Observe that the segment $$s$$ is split by $$n_s$$ internal intersection points into $$n_s+1$$ edges therefore: $$e''_n=\sum_{k=0}^{n-2}[k(n-2-k)+1]=\binom{n-1}3+\binom{n-1}1=\frac{(n-1)(n^2-5n+12)}6,\tag4$$ and the overall number of the edges inside the circle is $$e'_n=\frac n2 e''_n=\frac{n(n-1)(n^2-5n+12)}{12},\tag5$$ where factor 2 in the denominator accounts for the fact that while summing over all $$n$$ circle points every chord is counted 2 times.

To compute the overall number of edges we shall add the number of arcs connecting the circle points to the obtained result: $$e_n=e'_n+n=\frac{n^2(n^2-6n+17)}{12}.\tag6$$ Finally we use the Euler's formula for computing the number of parts inside the circle: $$f_n=e_n-v_n+1=\frac{n^4-6n^3+23n^2-18n+24}{24}.\tag7$$

And indeed the latter formula describes - as it should - the OEIS sequence A000127.

• $v'_n=\binom n4$ because a set of four points on the circle corresponds to a pair of intersecting chords, namely, the diagonals of the inscribed quadrilateral determined by those four points. – bof Aug 3 '19 at 12:19
• Yes of course. I have however decided to give another derivation to count both vertices and edges in a similar way. – user Aug 3 '19 at 16:06
• You can "count both vertices and edges in a similar way" but it seems more efficient to note that when you have counted one you have counted the other, since $$e'_n=\binom n2+2v'_n.$$ – bof Aug 3 '19 at 21:39
• If you mean that the identity $e'_n=\binom n2+2v'_n$ follows from the intermediate results (4) and (5), I'm sure it does, but since $e'_n=\binom n2+2v'_n$ is easy to see directly, it's not clear what those intermediate results are needed for. – bof Aug 4 '19 at 11:29

The number of vertices inside the circle is $$\binom n4$$ since each such vertex is determined by a pair of intersecting chords, which are the diagonals of an inscribed quadrilateral determined by four points on the circle.

The number of edges inside the circle is $$\binom n2+2\binom n4$$ since there are $$\binom n2$$ chords, and the number of edges on each chord is equal to $$1$$ plus the number of interior vertices on that chord, and each of the $$\binom n4$$ interior vertices lies on two chords.

There are $$n$$ vertices and $$n$$ edges on the circle, so the total number of vertices is $$V=n+\binom n4$$ and the total number of edges is $$E=n+\binom n2+2\binom n4.$$ By Euler's formula, the number of faces inside the circle is $$F-1=E-V+1=\binom n2+\binom n4+1=\frac{n(n-1)(n^2-5n+18)}{24}+1=\frac{n^4-6n^3+23n^2-18n+24}{24}.$$

If $$U$$ is the unit disc centered at the origin consider $$n$$ chords drawn through the interior of $$U$$ such that no two chords are parallel and no three chords intersect at the same point. The arrangement graph $$G$$ induced by the discs and the chords has a vertex for each intersection point in the interior of $$U$$ and $$2$$ vertices for each chord incident to the boundary of $$U.$$ Naturally $$G$$ has an edge for each arc directly connecting two intersection points.

A point $$p$$ of $$G$$ is said to be incident to the interior of $$G$$ if $$p$$ is in the interior of $$U.$$ Similarly $$p$$ is said be incident to the boundary of $$G$$ if $$p$$ is on the boundary of $$U.$$ In this sense and without loss of generality I speak of the boundary of $$U$$ as the boundary of $$G$$ and the interior of $$U$$ as that of $$G.$$ I denote the number of arcs incident to $$p$$ by $$v(p),$$ called the valency of $$p.$$ It is straightforward to see that if $$p$$ is incident to the interior of $$G$$ then $$p$$ is $$4-$$valent. Likewise if $$p$$ is incident to the interior of $$G$$ then $$p$$ is $$3-$$ valent. From this we see that the maximum valency of $$G$$ is $$4$$ and the minimum degree of $$G$$ is $$3.$$ By construction $$G$$ is planar and $$3-$$ connected.

lemma 1: $$G$$ has $$N(N+3)/2, N(N+2)$$ and $$(N^2+N+4)/2$$ points, arcs and faces respectively.

Each chord has two distinct points incident to the boundary of $$G.$$ Being there are $$N$$ such chords it is seen there are $$2N$$ points along the boundary of $$G.$$ Observe there are $$N(N-1)/2$$ points in the interior of $$G.$$ Indeed the number of points in $$G$$ is equal to sum of those points in the interior and along the boundary which is $$2N+N(N-1)/2=N(N+3)/2.$$ The number of arcs in $$G$$ follows from the handshaking lemma. In particular twice the number of arcs is equal to the sum of all valances. The later is equal to $$[3N+4N(N-1)/2]/2=N(N+2)$$ The number of faces in $$G$$ follows from Euler's characteristic formula for planar graphs.$$2-N(N+3)/2+N(N+2)=(N^2+N+4)/2$$ This establishes the lemma.

• I assumed no chords could be parallel – Antonio Hernandez Maquivar Sep 2 '16 at 21:15
• In the question n points were given on circumference, not s lines. Here s is not given. Is s = ${n \choose 2}$ – Amrita Sep 2 '16 at 22:01


# Electric potential inside a shell of charge

## Homework Statement

Q1: There are two concentric spherical shells with radii $R_1$ and $R_2$ and charges $q_1$ and $q_2$ uniformly distributed across their surfaces. What is the electric potential at the center of the shells?

Q2: There is an infinitely long hollow cylinder of linear charge density $\lambda$ and radius $R$. What is the potential difference $\Delta V$ between the surface of the shell and a radius $R'$ inside the cylinder?

## Homework Equations

$\vec E = -\nabla V$
$\oint \vec E \cdot d\vec A = \frac{Q\text{encl}}{\epsilon_0}$ (Gauss's Law)
$V=\frac{q}{4\pi \epsilon_0 r}$

## The Attempt at a Solution

Starting with Q2, Gauss's Law using a cylinder as the Gaussian surface shows there is no enclosed charge; $\vec E = \vec 0$. Because $\vec E = -\nabla V$, one can conclude that $V=0$ between $R'$ to $R$. This is the given (and found) answer for Q2.

In a similar way, there is no enclosable charge for all points inside the two spherical shells in Q1. By the same logic as Q2, it would seem $\vec E=0=-\nabla V$ and there would be no electric potential at the center of the shells.

However, Gauss's Law cannot be applied to a point or line since the Gaussian surface has area $A=0$ and Gauss's Law reduces to $0=0$. Therefore, one cannot find the electric field at the center of the sphere, and $\vec E = -\nabla V$ cannot be used.

Since the center of the shells are at a constant distance $R_1$ and $R_2$, the electric potential can be found by:

$V=\frac{q_1}{4\pi \epsilon_0 R_1} + \frac{q_2}{4\pi \epsilon_0 R_2}$

which is the given answer for Q1. (One needs to integrate the charge density across a spherical area, which ultimately reduces to the answer above)

This result (i.e. textbook answers) seems to show some weird results:
• The electric field and potential are zero for all positions inside a closed area of charge and nonzero at the symmetrical center or axis.
• The graphs of the field magnitude and electric potential are discontinuous at the center.
• If this is true, the electric field vector there has no defined direction...? (or is undefined since the equation is discontinuous)
• In Q2, the electric field and potential should be nonzero along the axis of the cylinder and zero for all spaces between the axis and the cylinder wall.

I'm somewhat confused because the two questions seem to contradict each other. Is my logic correct in interpreting the answers?

Related Introductory Physics Homework Help News on Phys.org
Doc Al
Mentor
One note: Between points where the field is zero there is no change in potential. That does not mean that the potential is zero. (There can certainly be a potential with respect to some other point.)

One note: Between points where the field is zero there is no change in potential. That does not mean that the potential is zero. (There can certainly be a potential with respect to some other point.)
Integrating $\vec E \cdot d\vec r$ where $\vec E = \vec 0$ to find the potential at a single point results in $V = 0+C$. Then the nonzero potential found at the center is $C$ and is constant across the space inside the shells...

Makes much more sense, thanks!

rude man
Homework Helper
Gold Member
Or perhaps:In Q2: the E field just outside the surface is σ/ε where σ is surface charge density (related to λ obviously). The E field just below the surface is zero. Both by Gauss. Since potential is the integral of the E field over distance, and the distance → zero, therefore there is no change in potential between the outside & inside surfaces.

Q1: Can also do this by superposition theorem:
Potential of shell 1 with q2=0 is kq1/R1.
Potential of shell 2 with q1=0 is kq2/R2.
Total potential is sum of above potentials.


# Proving that $\sqrt{a_1^2} +\cdots+ \sqrt{a_n^2} > \sqrt{a_1^2 +\cdots+a_n^2}$ using Pythagoras

I think I have a proof using Pythagoras for $\sqrt{a_1^2} + \sqrt{a_2^2} > \sqrt{a_1^2 + a_2^2}$.

I'm interested in whether there's a way to use that proof with Pythagoras to prove the general $a_n$ case (for this, hints are appreciated rather than complete proofs), and also in other ways (algebraic, geometric, number theoric, calculusic...anything) that you might know or come up with to prove the general case (for those, either hints or complete proofs are great, up to you).

Lemma:

Let positive (edited) real numbers $a_1, a_2$ be the legs of a right triangle.

Then $\sqrt{a_1^2 + a_2^2}$ is the length of the hypothenuse of that triangle.

And $\sqrt{a_1^2} + \sqrt{a_2^2}$ is the sum of the length of the two legs.

By the triangular inequality, we know that the length of the hypothenuse has to be less than the length of the sum of the two legs.

Therefore, for any real numbers $a_1, a_2$, $\sqrt{a_1^2} + \sqrt{a_2^2} > \sqrt{a_1^2 + a_2^2}$.

I'm stuck here...I was thinking of comparing pairs of elements from each side of the expression using my lemma, but it doesn't seem possible to "extract" pairs of elements from under $\sqrt{a_1^2 + a_2^2 +...+a_n^2}$. I also thought about summing all elements but $a_1$ into a single number and using my lemma on those simplified expressions, but I run into the same problem.

• Why do you need Pythagoras's theorem? Just square both sides of the inequality and compare. – Leo Jan 6 '16 at 19:18
• When you say "by the triangular inequality", this is a bit circular - the identity you're trying to prove is the triangle inequality for the Euclidean distance. – πr8 Jan 6 '16 at 19:20
• You have the inequality backwards. – zhw. Jan 6 '16 at 19:20
• @zhw. It has been edited. – user236182 Jan 6 '16 at 19:26
• @zhw. yes thank you, I mentioned it in my edit, not sure where it shows though – jeremy radcliff Jan 6 '16 at 19:27

It's very easy using induction :

You proved the case when $n=2$ .

Now assume you know it for $n$ and want prove it for $n+1$ :

$$\sqrt{a_1^2}+\ldots+\sqrt{a_n^2}+\sqrt{a_{n+1}^2} >\sqrt{a_1^2+a_2^2+\ldots+a_n^2}+\sqrt{a_{n+1}^2}$$

Now use also the $n=2$ case to finnish it :

$$\sqrt{a_1^2+a_2^2+\ldots+a_n^2}+\sqrt{a_{n+1}^2}>\sqrt{\left ( \sqrt{a_1^2+a_2^2+\ldots+a_n^2} \right )^2}+\sqrt{a_{n+1}^2}>\sqrt{\left ( \sqrt{a_1^2+a_2^2+\ldots+a_n^2} \right )^2+a_{n+1}^2}=\sqrt{a_1^2+a_2^2+\ldots+a_n^2+a_{n+1}^2}$$ as wanted .

You can translate this into a geometric proof : consider an $n$-dimensional box with the sides $a_1,a_2,\ldots,a_n$ . The diagonal of the box is $\sqrt{a_1^2+a_2^2+\ldots+a_n^2}$ and now you can repeatedly apply the triangle's inequality to get your inequality (this is equivalent with the induction proof above )

• I like the geometric intuition you give a lot, it makes the algebra more meaningful to me. – jeremy radcliff Jan 6 '16 at 19:50
• @jeremyradcliff But can you imagine an $n$-dimensional box (if $n\ge 4$)? – user236182 Jan 6 '16 at 20:41
• @user236182, no of course, I'm stuck at 3 dimensions but the step of the diagonal of a 3d box gives me 2 steps (with the 2d case) to extrapolate about higher dimensions and makes the induction principle in this context a lot more meaningful to me. – jeremy radcliff Jan 6 '16 at 20:54

Here is a geometrical proof. Consider a square with side length $L = |a_1|+|a_2| + \ldots + |a_n|$.

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$

Comparing the area of the whole square to the area of the small squares contained within it we see that

$$(|a_1| + |a_2| + \ldots + |a_n|)^2 \geq a_1^2 + a_2^2 + \ldots + a_n^2$$

and by taking the square root we get the desired inequality. Equality can only happen when one of the small squares covers the whole square which can only happen when atleast $n-1$ of the $a_i$'s are zero.

• Very nice, thank you. I'm a big fan of proofs without words. – jeremy radcliff Jan 6 '16 at 19:54

Both sides of the inequality are non-negative (for all $a_i\in\mathbb R$), therefore the following equivalences hold:

$$\sqrt{a_1^2} + \sqrt{a_2^2} +\cdots + \sqrt{a_n^2} \ge \sqrt{a_1^2 + a_2^2 +…+a_n^2}$$

$$\iff \left(\sqrt{a_1^2} + \sqrt{a_2^2} +\cdots + \sqrt{a_n^2}\right)^2 \ge \left(\sqrt{a_1^2 + a_2^2 +…+a_n^2}\right)^2$$

$$\iff a_1^2+a_2^2+\cdots+a_n^2+2\sum_{i=1}^n \sum_{j>i}^n|a_ia_j|\ge a_1^2+a_2^2+\cdots+a_n^2$$

$$\iff 2\sum_{i=1}^n \sum_{j>i}^n|a_ia_j|\ge 0,$$

which is true, with equality if and only if at least $n-1$ of $a_1,a_2,\ldots,a_n$ are equal to $0$.

• @jeremyradcliff It's multinomial expansion. – user236182 Jan 6 '16 at 19:58
• Thank you. I erased my comment because I realized it was wrong. But I just worked it out. For the record, I was aking whether the summation symbols were from the binomial expansion. – jeremy radcliff Jan 6 '16 at 19:59


22 February 2022

# Fiscal fibrillations (18/02/22 Riddler Classic)

Original problem

If we start with $n$ coins, the first toss will reduce the number to $k$ which is distributed binomially. If $p_n$ is the probability that there exists a lucky coin if we start with $n$ coins, then a recurrence relation exists:$$p_n=2^{-n}\sum_{k=0}^n{n \choose k}p_k$$where $p_0=0$ and $p_1=1$ are the base cases. Isolating for $p_n$ gives $$p_n=\frac{1}{2^n-1}\sum_{k=0}^{n-1} {n \choose k}p_k$$

Plotting $p_n$ up to $n=10$ generates the following graph which seems to imply $p_n$ converges.

However, visual inspection of $p_n$ values up to $n=1000$ shows a clear oscillatory tendency

which can also be plotted on a linear-log scale:

The periodic nature on a log-scale makes sense. Given a large enough $n$, we expect the first toss to almost-exactly halve the number of coins, and so $p_{2n}\approx p_{n}$, with the approximation being better as $n$ increases by the law of large numbers. However, questions remain:

• What values does $p_n$ oscillate around? On the graph, it looks to be just under 0.72135.
• Is the shape actually a sinusoid? If so, why?
• Does $p_n$ converge?

At the moment, I can only answer the first dot point.

### $p_n$ oscillates around $\frac{1}{2\ln(2)}\approx0.721348$

To arrive at this value analytically, we want to reframe how we express $p_n$.

Suppose we modify the game slightly such that we toss all $n$ starting coins with every toss, even coins that came up tails; we just make those coins ineligible to be the lucky coin.

Now suppose a lucky coin is crowned after exactly $(t+1)$ tosses. This happens if and only if both of the following occur:

• After $t$ tosses, there were $k\geq 2$ coins with clean histories of only heads. Each of the other $(n-k)$ coins does not have a clean history. This happens with probability $${n\choose k}\cdot (2^{-t})^k \cdot (1-2^{-t})^{n-k}$$
• At the $(t+1)$th toss, exactly one out of the $k$ candidate coins come up heads while the rest are tails. This occurs with probability $$2^{-k}\cdot k$$

Hence the probability that a lucky coin is crowned on the $(t+1)$th toss is the product of the above two probabilites, summed across all $k\in[2,n]$.

\begin{align*}\sum_{k=2}^n{n\choose k}(2^{-t})^k(1-2^{-t})^{n-k}2^{-k}k&=\sum_{k=2}^n{n\choose k}(2^{-(t+1)})^k(1-2^{-t})^{n-k}k\\&=\sum_{k=0}^n{n\choose k}(2^{-(t+1)})^k(1-2^{-t})^{n-k}k\;\;\;-n2^{-(t+1)} (1-2^{-t})^{n-1}\\&=n2^{-(t+1)} (1-2^{-(t+1)})^{n-1}-n2^{-(t+1)} (1-2^{-t})^{n-1}\\&=\frac{n}{2}2^{-t}\Big[(1-{2^{-(t+1)}})^{n-1}-(1-2^{-t})^{n-1}\Big]\end{align*}

Note that in the third line, we used the identity $\sum_{k=0}^n {n\choose k}x^ky^{n-k}k=nx(x+y)^{n-1}$. Now, summing the above over all possible values of $t\in[0,\infty)$ gives $p_n$

\begin{align*}p_n&=\sum_{t=0}^\infty\frac{n}{2}2^{-t}\Big[(1-{2^{-(t+1)}})^{n-1}-(1-2^{-t})^{n-1}\Big]\\&=\frac{n}{2}\sum_{t=0}^\infty 2^{-t}(1-2^{-(t+1)})^{n-1}-\frac{n}{2}\sum_{t=0}^\infty 2^{-t}(1-2^{-t})^{n-1}\\&=\frac{n}{2}\sum_{t=1}^\infty 2^{-(t-1)}(1-2^{-t})^{n-1}-\frac{n}{2}\sum_{t=0}^\infty 2^{-t}(1-2^{-t})^{n-1}\\&=\frac{n}{2}\sum_{t=0}^\infty 2^{-t}(1-2^{-t})^{n-1}\end{align*}

We can approximate this sum by an integral, which luckily for us is actually extremely easy to perform.

\begin{align*}p_n&\approx \frac{n}{2}\int_0^\infty 2^{-t}(1-2^{-t})^{n-1}dt\\&=\frac{n}{2}\int_0^\infty e^{-t\ln(2)}(1-e^{-t\ln(2)})^{n-1}dt\\&=\frac{1}{2\ln(2)}\Big[(1-e^{-t\ln(2)})^n\Big]^\infty_0\\&=\color{red}{\boxed{\frac{1}{2\ln(2)}}}\\&\approx 0.7213475\end{align*}

This doesn't actually tell us if the sequence $\{p_n\}$ actually converges, however; it is merely an approximation for any individual $p_n$. The sinusoidal behaviour also isn't particularly apparent. Without knowing the answers to these, I'm reluctant to make a call on what $p_{10^6}$ might be.


# Digit-Sum of a Number.

I got a question recently, and I am unable to solve it.

Find all natural numbers $N$, With sum of digits $S(N)$, where $N=2\{S(N)\}^2$

I know that $9|N-S(N)$, and since N is twice a square, it must end in $0,2,8$. But I do not know where to go from here. Can anyone help?

The Official Solution from the Organization

We use the fact that $9|n−S(n)$ for every natural number $n$. Hence $S(n)(2S(n)−1)$ is divisible by $9$. Since $S(n)$ and $2S(n)−1$ are relatively prime, it follows that $9$ divides either $S(n)$ or $2S(n)−1$, but not both. We also observe that the number of digits of $n$ cannot exceed $4$. If $n$ has $k$ digits, then $n≥10k−1$ and $$2S(n)^2≤2\cdot(9k)^2=162k^2$$ If $k≥6$, we see that $$2S(n)^2≤162k^2<5^4k^2<10^{k−1}≤n$$ If $k = 5$, we have $$2S(n)^2≤162\cdot25=4150<10^4≤n$$ Therefore $n ≤ 4$ and $S(n) ≤ 36$. If $9|S(n)$, then $S(n) = 9,18,27,36$. We see that $2S(n)^2$ is respectively equal to $162, 648, 1458, 2592$. Only $162$ and $648$ satisfy $n = 2S(n)^2$. If $9|(2S(n)−1)$, then $2S(n) = 9k+1$. Only $k = 1,3,5,7$ give integer values for $S(n)$. In these cases $2S(n)^2 = 50,392,1058,2048$. Here again $50$ and $392$ give $n = 2S(n)^2$. Thus the only natural numbers wth the property $n = 2S(n)^2$ are $50,162,392,648$.

• For what it's worth, $S(N) \mid N$, so that $9 \mid N$. I am not too sure though if it will narrow down things that much. Jun 29 '17 at 14:09
• @JoseArnaldoBebitaDris Do you mean to say that 9 divides N? Because I can think of places where it wouldn't be true. One N I have found is 50, and it is not divisible by 9. Jun 29 '17 at 14:12
• Yes because $N = 2\cdot{S(N)}^2$, which means that $N/S(N) = 2\cdot{S(N)}$ is an integer. In other words, $S(N)$ divides $N$. Jun 29 '17 at 14:13
• If that's the case, I would doubt $9 \mid (N - S(N))$. Jun 29 '17 at 14:15
• @JoseArnaldoBebitaDris No, that's true. It's a theorem, I think. Jun 29 '17 at 14:16

## 2 Answers

We want to find an upper bound for $N$ for which it is possible that $N = 2 S(N)^2$. The upper bound that I use in the following is very rough, but I think still just good enough to be able to do the subsequent calculations by hand with a little persistence. Look to the answer of @OscarLanzi for a much cleverer upper bound.

An upper bound for $S(N)$ is $$S(N) \leq 9\cdot \left(\log_{10}(N)+1\right).$$ We can use this upper bound to see that $$2 S(N)^2 - N \leq 2 \cdot \left(9\cdot \left(\log_{10}(N)+1\right) \right)^2 - N.$$ For $N = 10^4$ we see that the right hand side of this inequality is $4050 - 10^4 = -5950 < 0$ and by calculating derivatives we can see fairly easily that this expression will only decrease for $N > 10^4$. This means that if we want to find $N$ such that $N = 2 S(N)^2$ we only have to check numbers up to $10^4$, this is a very rough upper bound.

Note that indeed that $N \equiv S(N) \pmod{9}$ so $N$ must satisfy $$N - 2 N^2 \equiv 0 \pmod{9},$$ we deduce that $N \equiv 0 \pmod{9}$ or $N \equiv 5 \pmod{9}$. We see that $N$ must be twice a square and less than $10^4$, so $2S(N)^2 \leq 10^4$, which means that $S(N) \leq \sqrt{\frac{1}{2}\cdot{10^4}} \cong 70.7$. So $S(N)$ must be a number less than or equal to $70$ that is either $0$ or $5$ modulo $9$. There are only fifteen such positive numbers, the list is $$\left\{5,9,14,18,23,27,32,36,41,45,50,54,59,63,68\right\}.$$ These can all be checked separately, the list of these elements squared twice is $$\left\{\mathbf{50},\mathbf{162},\mathbf{392},\mathbf{648},1058,1458,2048,2592,3362,4050,5000,5832,6962,7938,9248\right\}.$$ Sort of surprisingly the first four all fit. These are the only ones, so the final answer is $$N \in \left\{50, 162, 392, 648\right\}.$$ The amount of checking to be done can be reduced by getting a better upper bound.

• I can understand what you did there, with the upper bound, but am unable to understand how you calculated the upper bound. Could you please elaborate? For example, how did logarithms come into play? Jun 29 '17 at 15:06
• @MalayTheDynamo Suppose that a number $n$ consists of $3$ digits, say for example $n = 354$. Then we see that $100 \leq n \leq 1000$, and if we take the logarithm we see that $2 \leq \log_{10}(n) \leq 3$. Generally we can deduce that $\log_{10}(n) + 1$ is always greater or equal to the amount of digits of $n$. The maximal digitsum $S(n)$ of a number with $k$ digits is $9 k$, namely if all digits are $9$. Combining these we find that an upper bound for $S(n)$ is $$9 \cdot \left(\log_{10}(n) + 1\right).$$ Jun 29 '17 at 15:14
• Oh, thanks. It was perfect. Jun 29 '17 at 15:15

@Pjotr5 identified the solution but asked for a sharper bound. Here all four-digit candidates are eliminated through a descent process leaving only the numbers in his list up to and including $648$.

First off, five-digit numbers give a digit sum no greater than $45$ whose square, doubled, is less than five digits ($2×45^2=4050$). Contradiction. Ditto for more than five digits.

Four-digit numbers give digit sums no more than $36$ whose square, doubled, is $2592$. So a four-digit solution is at most $2592$. But wait, there's more (or less, if you will). If the first digit of the proposed four-digit solution is no more than $2$ the sum of digits is now no more than $29$ giving a bound of $2×29^2=1682$, and then if the first digit has to be $1$ the sum of digits is at most $28$ giving $N\le 1568$.

The four-digit bound can still be lowered more. The bound of $1568$ derived above means the first two digits are no more than $15$ and can't sum to more than $6$. Then the sum of the four digits is no longer bounded only by $28$ but now bounded by $24$. Twice the square of that is $1152$ so now $N \le 1152$.

You know the drill by now. If $N$ has four digits and is less than or equal to $1152$ then the sum of digits is no more than $20$. Twice the square of that is only $800$, less than four digits, so no four-digit candidates are left!

• Good one :). Together our answers certainly make it possible to solve this problem on a contest. Jun 29 '17 at 15:48
• For getting the actual solution @Pjotr5 beat me to it, and in fact I reference that answer. Jun 30 '17 at 9:55
• @OscarLanzi Thank you :). I'll edit my answer to reference that your upper bound is much better to make the answer more complete. Jun 30 '17 at 10:10


# Solve for $x$ in $\cos(2 \sin ^{-1}(- x)) = 0$

Solve $$\cos(2 \sin ^{-1}(- x)) = 0$$

I get the answer $\frac{-1}{ \sqrt2}$

By solving like this

\begin{align}2\sin ^{-1} (-x )&= \cos ^{-1} 0\\ 2\sin^{-1}(- x) &= \frac\pi2\\ \sin^{-1}(- x) &= \frac\pi4\\ -x &= \sin\left(\frac\pi4\right)\end{align}

Thus $x =\frac{-1}{\sqrt2}$

But correct answer is $\pm\frac{1}{\sqrt2}$

Where am I going wrong?

• You didn't take multiple values into account. Neither cos nor sin are one to to one. – fleablood Jun 8 '17 at 15:34

Okay so expand the double angle to get $\cos^2(\sin^{-1}(-x))-\sin^2(\sin^{-1}(-x)=0$.

This should give you $(1-(-x)^2)-(-x)^2=0$.

Finally you have $1-2x^2=0$.

Solutions are $\displaystyle \boxed{\pm \frac{1}{\sqrt{2}}}$.

• Because cos^-1 only returns one value. You have to use symmetry to get the other value – Saketh Malyala Jun 8 '17 at 15:43
• How did you get This should give you $(1-(-x)^2)-(-x)^2=0$. – Neha Gupta Jun 8 '17 at 15:49
• remember $\cos(\sin^{-1}(x)) = \sqrt{1-x^2}$ by right angle trigonometry – Saketh Malyala Jun 8 '17 at 15:50
• $\sin(\sin^{-1}(x))=x$ because sin is the inverse of $\sin^{-1}$ – Saketh Malyala Jun 8 '17 at 15:51

You need to solve $\cos \left(2 \arcsin(-x) \right) = 0$. Let $y = 2 \arcsin(-x)$ then $\cos y = 0$ so $y = \pi/2 \pm n\pi$. Then, $$2 \arcsin(-x) = \frac{pi}{2} \pm n\pi$$ which implies $$x = -\sin \left( \frac{\pi}{4} \pm \frac{n\pi}{2} \right)$$

Can you simplify this?

Trig functions are not one-to-one. For example if $\sin x = \frac 12$ then $x$ might be equal to $\frac {\pi}6$, but $x$ could also be $\frac {5\pi}6$ or $\frac {13\pi}6$ or it could be any $2k\pi + \frac {\pi}6$ or $(2k+1)\pi - \frac {\pi} 6$.

So when we say $\sin^{-1} x = \theta$ we are not just saying that $\theta$ is the angle so that $\sin (\theta) = x$. There should be an infinite number of such angles. We are saying that $\theta$ is an angle so that $\sin (\theta) = x$ AND we are saying that $\frac {-\pi}2 \le \theta \le \frac {\pi}2$. For those two conditions there is only one possible theta.

So $\sin^{-1}(\sin x)$ may or MAY NOT be equal to $x$. For example $\sin^{-1}(\sin (\frac {5\pi}6)) = \sin^{-1}(\frac 12) = \frac {\pi}6 \ne \frac {5\pi}6$.

However $\sin^{-1}(\sin x) = x + 2k\pi$ for some integer $k$ or $\sin^{-1}(\sin x) = (2k+ 1)\pi - x$.

So 1) Because $\sin (x) = \sin (\pi - x)$ and $x = \frac {\pi}2 - (\frac {\pi}2 - x)$ and $\pi - x = \frac{\pi}2 + (\frac {\pi}2 - x)$ then $\sin^{-1}(x) = (\frac {\pi}2 \pm x) \pm 2k\pi)$.

$\frac {-\pi}2 \le sin^{-1}(x) \le \frac {\pi}2$

And 2) Because $\cos (x) = \cos(\pi+ x)$ we have $\cos^{-1}(\cos x) = x \pm k \pi$.

$0 \le \cos^{-1} (x) \le \pi$

So ....

$\cos (2\sin^{-1}(-x) ) = 0$

$-\frac {\pi}2 \le \sin^{-1}(-x) \le \frac {\pi}2$ so $-\pi \le 2\sin^{-1}(-x) \le \pi$ and $\sin^{-1}(-x) = \cos^{-1}(0) \pm k\pi = \frac {\pi}2 \pm k\pi$. So $2\sin^{-1}(-x) = \pm \frac {\pi}2$

And $\sin^{-1}(-x) = \pm \frac {\pi}4$

So $\sin (\pm \frac {\pi} 2) = -x$

So $\pm \frac {1}{\sqrt{2}} = -x$

So $x = \mp \frac {1}{\sqrt{2}}$

Let $\sin^{-1}(-x)=u\implies\sin u=-x$

$$\cos(2\sin^{-1}(-x))=\cos2u=1-2\sin^2u=1-2(-x)^2$$


# Does the Hausdorff property hold on closed subsets of $\mathbb{R}^n?$

I am trying to prove that given disjoint closed $A,B\subseteq \mathbb{R}^n$, there exist disjoint open $U,V$ containing $A,B$ respectively. In other words that we can take the Hausdorff property to closed subsets of $\mathbb{R}^n$.

I do not know whether this is true or false, but have set about proving that it is true.

My idea was, for each $x\in A$, to have $\epsilon_x=\inf_{b\in B} \|x-b\|=\|x-\beta_x\|$ for some $\beta_x\in B$ (closure). Define $U_x=\{y:\|x-y\|<\epsilon_x/2\}$. Define similar balls for each element of $B$. Basically the ball around $x\in A$ is disjoint from the ball around $\beta_x$, the closest point to $x$ in $B$.

Have $U=\bigcup_{A} U_a$ and $V=\bigcup_B V_b$.

Now I believe $U,V$ are disjoint but I can't seem to argue it formally. It seems quite obvious by drawing $A,B$ as blobs in a plane, but obviously $A,B$ could be more complicated than nice bounded subsets. So I need to show $U_a\bigcap V_b=\emptyset$ for any $a,b\in A,B$; now if $z$ were in both then $\|z-a\|<\epsilon_a/2$ and $\|z-\beta_a\|>\epsilon_a/2$, and also $\|z-b\|<\epsilon_b/2$...

I feel like the argument is a step away but I can't seem to finish it off. Help?

• Closed subsets of Rn are metric spaces, hence normal, satisfying the necessary property. en.wikipedia.org/wiki/Normal_space So the statement is true. May 21 '16 at 22:22
• @William Phew, thanks. Is my attempt a viable route for proving it? May 21 '16 at 22:23
• I'm lazy so haven't evaluated the details exactly, but the gist of it seems like the right direction. Basically the idea is that since the sets are closed and disjoint, their closures do not intersect, and hence they have non-zero distance (epsilon) from each other, so take open sets U and V which cover each of the two sets without going more than epsilon/3 away from their boundary, hence U and V can't intersect. en.wikipedia.org/wiki/… Obviously you will have to work harder to make that precise. May 21 '16 at 22:27
• In the picture in the Wikipedia article I linked to, for example, take open sets around A and B whose farthest points from A and B are no more than 1 anyway; if A and B were closed spheres with radius R, and the distance between them was three, then take the open spheres centered at the centers of A and B and with radius R+1 May 21 '16 at 22:29
• @William. Disjoint closed subsets of a metric space do not necessarily have positive distance from each other. For example,in $R^2$ let $A=\{(x,1/x):x>0\}$ and $B=\{(x,-1/x):x>0\}.$ Then $d( (x,1/x),(x,-1/x))=2/x,$ which has no positive lower bound May 21 '16 at 23:36

We may assume that $A,B$ are nonempty; if one of them is empty we can just take $\emptyset,X$ as the two open sets. The most obvious construction works fine: let $U$ be the set of points nearer to $A$ than to $B,$ and let $V$ be the set of points nearer to $B$ than to $A.$ That is, take $$U=\{x:d(x,A)\lt d(x,B)\}=\{x:d(x,B)-d(x,A)\gt0\}$$ and let $$V=\{x:d(x,B)\lt d(x,A)\}=\{x:d(x,A)\gt d(x,B)\}$$ where $$d(x,S)=\inf\{d(x,y):y\in S\}.$$ The sets $U,V$ are obviously disjoint. They are open sets because, for a nonempty set $S,$ the function $x\mapsto d(x,S)$ is continuous, as is easily shown using the triangle inequality. The inclusions $A\subseteq U$ and $B\subseteq V$ follow from the assumption that $A,B$ and disjoint and closed; e.g., if $x\in A$ then $x\notin B,$ so $d(x,B)\gt0=d(x,A).$
Note that for all $a \in A$ and $b \in B$ we have $$\inf_{a' \in A}\|a' - b\| \leq \|a-b\|.$$
Take $a \in A$, $b \in B$. Suppose $z \in U_a \cap V_b$. By the triangle inequality, $$\|a - b \| \leq \|a - z \| + \|z - b\| < \epsilon_a/2 + \epsilon_b/2$$ $$= \frac{\inf_{b' \in B}\|a - b'\|}{2} + \frac{\inf_{a' \in A}\|a' - b\|}{2}$$ $$\leq \frac{\|a - b\|}{2} + \frac{\|a-b\|}{2} = \|a-b\|.$$ So $\|a-b\| < \|a-b\|$. Contradiction.


# Is there a better/easier way to solve this matrix?

$$\begin{equation*} \begin{bmatrix} 4 & -1 & -1 & 0 &|&30 \\ -1 & 4 & 0 & -1&|&60 \\ -1 & 0 & 4 & -1&|&40 \\ 0 & -1 & -1 & 4&|&70 \end{bmatrix} \end{equation*}$$

What's the best way to solve the matrix above? There's a clear pattern of the diagonal 4's and 0's and the -1's so I feel like there has to be a better way of doing things rather than using scaling and row reduction.

If I do those methods I end up with messy fractions.

My Step 1:

New Row 2 = (1/4)Row 1 + Row 2

Even at step 1 I can tell the whole thing will be messy with fractions.

Is there a better way to solve this matrix? Or am I doing it wrong? Thanks.

Add all four rows to get $$\begin{pmatrix}2 &2 &2 &2\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}= 200$$

or

$$\begin{pmatrix}1 &1 &1 &1\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}= 100$$

Add that to every row to get $$\begin{pmatrix}5 &0 &0 &1\\ 0 & 5 & 1 & 0\\ 0 & 1 & 5 & 0\\ 1 & 0 & 0 & 5\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}= \begin{pmatrix}130\\160\\140\\170\end{pmatrix}$$

Now you have two separated 2x2 systems. Can you take it from here?

• thanks, that's super interesting. is there a name for that process? Aug 27, 2020 at 13:03
• I call it "sharply looking at the equation until you have an idea" :D No idea if there is a formal name for it. Also, can you please accept this answer? Then others will know that there is an viable answer here :) Aug 27, 2020 at 13:20
• sure thing, i forgot i could do that cause i got more reputation now. also im still thinking this would be messy yea? cause like: new row 4 = row 1(-1/5) so then it'll be (-1/5) + 5 which is 4/5ths which is still messy, or is that just no other way, or am i doing it wrong Aug 27, 2020 at 13:25
• you can do row1-5*row4, if you dislike fractions. But having one fraction and then multiplying the equation by 5/4 is not that bad. Aug 27, 2020 at 13:27
• Wow thank you so much! I am really happy that an answer from here actually had a positive impact on someone in real life :) Sep 6, 2020 at 12:58