text stringlengths 1 2.12k | source dict |
|---|---|
we can use this method
$$\sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b}\tag{1}$$
where $\psi$ is digamma function.
now we can write
$\sum_{n=1}^\infty\dfrac{1}{(2n-1)(2n+1)}= \frac{1}{4}\sum_{n=1}^\infty\dfrac{1}{(n+\frac{1}{2})(n+\frac{3}{4})}=\frac{1}{2}$
since $\psi(\frac{1}{2})-\psi(\frac{3}{2})=-2$ | {
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The region $R$ is a rectangle having the vertices: (0,1),\,(1,1),\,(1,-1),\,(0,-1) One way we can compute the volume is: ...
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MarkFL replied to a thread logs in Pre-Algebra and Algebra
Okay, we are given: \ln\left(x-\frac{1}{x}-3\right)=2 Converting from logarithmic to exponential form, we have: x-\frac{1}{x}-3=e^2 ...
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As for the $\LaTeX$, look under the "Calculus/Analysis" section of our Quick $\LaTeX$ tool. You will find it, which gives the code: ...
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MarkFL replied to a thread The Fonz and Geometry in Chat Room
Yes, and I was just pointing out that there can be other reasons a student fails besides lack of innate ability to understand the material, or an...
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MarkFL replied to a thread The Fonz and Geometry in Chat Room
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• August 21st, 2017, 12:26 | {
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• August 21st, 2017, 12:26
Here is this week's POTW: ----- If a quadrilateral is circumscribed about a circle, prove that its diagonals and the two chords joining the...
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• August 21st, 2017, 12:22
Suppose \frac{3}{2}\le x \le 5. Prove that 2\sqrt{x+1}+\sqrt{2x-3}+\sqrt{15-3x}<2\sqrt{19}. Congratulations to MarkFL for his correct...
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• August 20th, 2017, 21:11
my solution: with transformation $y=tan(x),dy=sec^2(x)dx$ I = \int_{0}^{\dfrac{\pi}{4}}\left(\tan x + \cot x \right)\left ( \dfrac{\tan x}{1 +...
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• August 20th, 2017, 10:02
$-2+3 ln2$
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• August 20th, 2017, 01:29
MarkFL posted a visitor message on Peter's profile
Hey Peter! (Wave) I edited your post to remove the duplicate content. Sorry for the late reply, I was busy "powering through" a tedious 3 hour...
• August 19th, 2017, 22:48
to find the value of $f(\alpha)=-17+21\sqrt 3 i$ and $f(\beta)=-17-21\sqrt 3i$ seemed time-consuming do you have a better way to get them ?
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• August 18th, 2017, 20:20
yes a miscalculation found the answer is 43524 the solution has been edited what a shame ! my poor calculation
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• August 18th, 2017, 11:01
$f(x)=x^3+20x-17$ $f(r)=r^3+20r-17=0---(1)$ ($r$ is a root of $f$) $f(r+1)=(r+1)^3+20(r+1)-17=3r^2+3r+21$ from $(1):$...
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• August 18th, 2017, 09:04
my solution: for $r_1^3+20r_1-17=r_2^3+20r_2-17=r_3^3+20r_3-17=0$ Using Vieta's formulas $r_1+r_2+r_3=0---(1)$ $r_1r_2+r_2r_3+r_3r_1=20---(2)$...
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• August 17th, 2017, 05:03
my solution: let $A=8x^2-2xy^2$ $B= 6y$ $C= 3x^2+3x^3y^2$ from $A,B$ we have $xy^2+3y-4x^2=0-----(1)$ from $B,C$ we have...
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• August 17th, 2017, 02:10
Small quibble...axis of symmetry is: t=-\frac{1}{105} So, since the parabola opens up, the vertex is a minimum, so the minimum distance will be...
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• August 17th, 2017, 00:59
For a parabola of the form: f(x)=ax^2+bx+c We know the axis of symmetry is at: x=-\frac{b}{2a} So, for:
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### 16 Visitor Messages
1. Hi Albert,
I was meant to reply to the PM but it seems to me your account here has exceeded your stored private messages quota and hence you cannot receive new message(s).
Anyway, I just wanted to thank you for reply and guess what, I just solved that challenge! Hurray! Haha...
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2. Hi Albert,
I am aware that to editing our own post of less than 24 hours old (which I think is a good thing rather than keep adding responses by posting new follow-up posts to make mess of the look of one particular thread) is encouraged and not frowned upon, but I wanted to also let you know I have come to realize of your latest edit/update to my latest challenge only when I was about to reply to it. If by any chance I missed reading it, that would be unfair to you because your newest edited post deserves a credit, imho.
Thus in the future if you want to do a major edit to your previous post, I would recommend you to instead add another reply beneath your previous post, does that sound good to you, Albert?
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3. Hi Albert, congratulations on winning the MHB Challenges Award!
4. Hi Albert,
I wanted to say that I like your profile .sig, Little Prince was a story I read almost 2 years ago but it still touches my heart. It was the best story I ever have read in my life.
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5. Hello, me again Albert!
Here is what our administrator Jameson has said regarding the issue:
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So, when it happens again, try reloading the page, and if after several times the error still occurs, capture a screen shot and email the image to me.
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Mark.
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6. Hello again Albert,
The staff here would really like to help, and it would be informative for use if you could get a screen shot of the error and email it to me, and I will forward it to the staff so we can see exactly what the error looks like at your end.
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7. Hello Albert,
I have reported your problem to the staff here, and I can assure you they are a very knowledgeable and involved team of administrators whose primary goal is that our member's problems get addressed promptly.
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8. Hello Albert,
First I wanted to ask you if the issue you had yesterday was resolved...I waited for your email of the problem description but never got it. I want to help you get to the bottom of the problem if I can.
Second, I edited your latest post of the solution to the minimization problem. We try not to link to other math forums unless we are showing where a problem comes from if posted by someone else, to give credit to the OP. Since this problem is yours, you should post the solution at both places rather than at one, and then linking to that at the other.
Your contributions here are appreciated!
Best Regards,
Mark.
9. Hi Albert! Please use titles which represent the nature of your question. I have renamed your following threads:
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2) http://www.mathhelpboards.com/f28/bi...-squares-3198/ (Original title: mission impossible)
10. Hi again! You have asked in your thread >>here<< how to move a thread if you accidentally posted in a wrong sub-forum. The answer is you can't do that. However you can report the post by clicking on the little triangle (shown below) and writing down your request. I have moved your thread to the Challenge Questions and Puzzles sub-forum.
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# Alternating sum of squares of binomial coefficients
I know that the sum of squares of binomial coefficients is just $${2n}\choose{n}$$ but what is the closed expression for the sum $${n\choose 0}^2 - {n\choose 1}^2 + {n\choose 2}^2 + \cdots + (-1)^n {n\choose n}^2$$?
• Can you do it with generating functions? – Nikhil Ghosh Aug 8 '12 at 2:33
• I thought I had something clever. I have deleted my post until I have a chance to think on the case when $n$ is even. $n$ being odd still yields 0, unless I am totally mistaken. – Emily Aug 8 '12 at 2:41
• Wolfram|Alpha gives this closed form. – joriki Aug 8 '12 at 2:45
• I don't really understand why combinatorial proof went more or less unnoticed (while standard application of generating functions is heavily upvoted). – Grigory M Nov 30 '13 at 13:28
$$(1+x)^n(1-x)^n=\left( \sum_{i=0}^n {n \choose i}x^i \right)\left( \sum_{i=0}^n {n \choose i}(-x)^i \right)$$
The coefficient of $$x^n$$ is $$\sum_{k=0}^n {n \choose n-k}(-1)^k {n \choose k}$$ which is exactly your sum.
On another hand:
$$(1+x)^n(1-x)^n=(1-x^2)^n=\left( \sum_{i=0}^n {n \choose i}(-1)^ix^{2i} \right)$$
Thus, the coefficient of $$x^n$$ is $$0$$ if $$n$$ is odd or $$(-1)^{\frac{n}2}{n \choose n/2}$$ if $$n$$ is even.
• ty fixed it. I hope that was the only one :) – N. S. Aug 8 '12 at 2:51
Here's a combinatorial proof.
Since $\binom{n}{k} = \binom{n}{n-k}$, we can rewrite the sum as $\sum_{k=0}^n \binom{n}{k} \binom{n}{n-k} (-1)^k$. Then $\binom{n}{k} \binom{n}{n-k}$ can be thought of as counting ordered pairs $(A,B)$, each of which is a subset of $\{1, 2, \ldots, n\}$, such that $|A| = k$ and $|B| = n-k$. The sum, then, is taken over all such pairs such that $|A| + |B| = n$. | {
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Given $(A,B)$, let $x$ denote the largest element in the symmetric difference $A \oplus B = (A - B) \cup (B - A)$ (assuming that such an element exists). In other words, $x$ is the largest element that is in exactly one of the two sets. Then define $\phi$ to be the mapping that moves $x$ to the other set. The pairs $(A,B)$ and $\phi(A,B)$ have different signs, and $\phi(\phi(A,B)) = (A,B)$, so $(A,B)$ and $\phi(A,B)$ cancel each other out in the sum. (The function $\phi$ is what is known as a sign-reversing involution.)
So the value of the sum is determined by the number of pairs $(A,B)$ that do not cancel out. These are precisely those for which $\phi$ is not defined; in other words, those for which there is no largest $x$. But there can be no largest $x$ only in the case $A=B$. If $n$ is odd, then the requirement $\left|A\right| + \left|B\right| = n$ means that we cannot have $A=B$, so in the odd case the sum is $0$. If $n$ is even, then the number of pairs is just the number of subsets of $\{1, 2, \ldots, n\}$ of size $n/2$; i.e., $\binom{n}{n/2}$, and the parity is determined by whether $|A| = n/2$ is odd or even.
Thus we get $$\sum_{k=0}^n \binom{n}{k}^2 (-1)^k = \begin{cases} (-1)^{n/2} \binom{n}{n/2}, & n \text{ is even}; \\ 0, & n \text{ is odd}.\end{cases}$$ | {
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• Wonderful proof, Mike! I always prefer combinatorial proofs as they also offer motivation and explanation to the identity and not only a formal proof. What really interests me is whether you can construct a sign-reversing involution to prove the following special case of Dixon's Identity: $\sum_{k=0}^{n} \binom{3n}{k}^{3}(-1)^{k} = (-1)^n\binom{3n}{n,n,n}$. – Ofir Jan 18 '13 at 16:31
• @Ofir: I'm glad you like it! I'm a big fan of combinatorial proofs myself. That's an interesting question about Dixon's identity; you should ask it as a question on the site. – Mike Spivey Jan 18 '13 at 18:56
• I can't edit my original comment, but in the LHS it should be $\binom{2n}{k}^{3}$ instead of $\binom{3n}{k}^{3}$. Reading an article by Zeliberger, I found out there's a combinatorial proof for Dixon's identity by Foata. It should be in the following French book: www-irma.u-strasbg.fr/~foata/paper/ProbComb.pdf . I think pages 37-40 generalize it but I don't know any French, can anyone help out? – Ofir Jan 19 '13 at 15:33
• A belated comment: Your beautiful proof can be easily extended to the more general result that $\sum\limits_{k=0}^n \dbinom{m}{k} \dbinom{m}{n-k} \left(-1\right)^k = \begin{cases} \left(-1\right)^{n/2} \dbinom{m}{n/2} , & \text{ if } n \text{ is even}; \\ 0 , & \text{ if } n \text{ is odd} \end{cases}$ for any nonnegative integers $m$ and $n$. – darij grinberg Dec 20 '17 at 0:45
$$\sum_{m=0}^n (-1)^m{n \choose m}^2= (n!)^2\sum_{m=0}^n \frac{(-1)^m}{(m!)^2((n-m)!)^2}$$
This function feels hypergeometric, so we take the quotient of $c_{m+1}$ and $c_m$ where
$$c_m=\frac{(n!)^2}{(m!)^2((n-m)!)^2}$$
so,
$$\frac{c_{m+1}}{c_{m}}=\frac{((m+1)!)^2((n-m-1)!)^2}{(m!)^2((n-m)!)^2}=\frac{(m-n)^2}{(m+1)^2}$$
after some simplification, confirming that this can be expressed in terms of a hypergeometric function. The previous result gives us the parameters of the function so we find $\sum_{m=0}^\infty c_m x^m = {_2}F_1(-n, -n; 1;-1)$. | {
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$${_2}F_1(-n, -n; 1;-1)= \sum_{m=0}^\infty \frac{((-n)_m)^2}{(1)_k} \frac{(-1)^k}{k!} = \sum_{m=0}^\infty (-1)^m{n \choose m}^2$$
Where $(x)_n=x(x+1)\cdots(x+n-1)=\frac{\Gamma(x+n)}{\Gamma(x)}$ is Pochhammer's symbol.
An identity for ${_2}F_1$ gives an elegant "closed form:"
$${_2}F_1(-n, -n; 1;-1)= \frac{2^{n} \sqrt{\pi} \Gamma(-n+n+1)}{\Gamma\left(\frac{1-n}{2}\right)\Gamma\left(\frac{-n}{2}+n+1\right)}= \frac{2^{n} \sqrt{\pi}}{\Gamma\left(\frac{1-n}{2}\right)\Gamma\left(\frac{n+2}{2}\right)}$$
Now, nothing that if $a$ is a positive integer, ${n \choose n+a}=0$, so
$$\sum_{m=0}^\infty (-1)^m{n \choose m}^2 = \sum_{m=0}^n (-1)^m{n \choose m}^2$$
and finally we get the answer
$$\sum_{m=0}^n (-1)^m{n \choose m}^2=\frac{2^{n} \sqrt{\pi}}{\Gamma\left(\frac{1-n}{2}\right)\Gamma\left(\frac{n+2}{2}\right)}$$
that holds for real numbers as well. | {
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# linear equation graph problems | {
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Matrix Calculator. Graphing linear equations using method 1. Type a math problem. Graphing linear relationships word problems Get 3 of 4 questions to level up! You can enter linear equations, quadratic equations, cubic equations, trigonometric functions, etc. $\begin{cases}5x +2y =1 \\ -3x +3y = 5\end{cases}$ Yes. $\begin{cases}2x -y = -1 \\ 3x +y =6\end{cases}$ Yes. Solve. These tutorials introduce you to linear relationships, their graphs, and functions. y-intercept : y-intercept is nothing but the value at where the line intersects y-axis. To move a number to a different side, you need to subtract it from both sides. linear equations word problems worksheet - If you've been seeking a house that offers you the really best bang for your own buck, your best choice would certainly be a North Charleston SC house for purchase. Money related questions in linear equations. Our mission is to provide a free, world-class education to anyone, anywhere. Simultaneous equations (Systems of linear equations): Problems with Solutions. You're putting your home on the market to sell it. Level up on all the skills in this unit and collect up to 1500 Mastery points! It could be a curve that looks something like that, or a curve that looks something like that. Linear equations word problems: graphs Get 3 of 4 questions to level up! 4. You can control the types of problems, the number of problems, workspace, border around the problems, and more. First go to the Algebra Calculator main page. 5 b = − 2 b + 3 \frac{r-3}{4}=2r. Unknown number related questions in linear equations. Linear Equations. We welcome your feedback, comments and questions about this site or page. When we have a linear equation in slope-intercept form, we can sketch the graph (straight line) of the equation using the slope 'm' and y-intercept 'b'. The picture shown below tells us the trick. Choose any value for x and substitute into the equation to get the corresponding value for y. Graph 2x - y = 6 by the intercept | {
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substitute into the equation to get the corresponding value for y. Graph 2x - y = 6 by the intercept method. Step 2: Find the y-intercept, let x = 0 then substitute 0 for x in the equation and solve for y Next, divide both sides of the equation by the number in front of the variable, which is called the coefficient. Wonderful Graphing Linear Equations Word Problems by Madilyn Yuengel TpT. Slope and intercept meaning in context. The intercept points are when x = 0 or y = 0. how to graph linear equations by plotting points. Example Problem Graph the following equation: y=2x+1 How to Graph the Equation in Algebra Calculator. Example 1 . Linear equations word problems: graphs. In this section, you will learn how to solve word problems using linear equations. Slope, x-intercept, y-intercept meaning in context, Practice: Relating linear contexts to graph features, Practice: Using slope and intercepts in context, Practice: Linear equations word problems: tables, Practice: Linear equations word problems: graphs, Practice: Graphing linear relationships word problems. 9. Verification is an important step to always remember for these kinds of problems. Try the given examples, or type in your own There is a simple trick behind solving word problems using linear equations. Graph is an open source linear equation grapher software for Windows which can plots a graph of linear equation with one variable only. Please submit your feedback or enquiries via our Feedback page. Calculus Calculator. In order to graph a linear equation you can put in numbers for x and y into the equation and plot the points on a graph. That is, slope = rise / run. 7. Inequalities. Level up on the above skills and collect up to 500 Mastery points Start quiz. Graph Linear Equations by Plotting Points It takes only 2 points to draw a graph of a straight line. These linear equations worksheets cover graphing equations on the coordinate plane from either y-intercept form or point slope form, as well as finding | {
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on the coordinate plane from either y-intercept form or point slope form, as well as finding linear equations from two points. Linear equation word problems worksheet with answers. Google Classroom Facebook Twitter. We can plot these solutions in the rectangular coordinate system as shown in . The directions are from taks so do all three variables equations and solve no matter what is asked in the problem. Section 8.1, Example 4(a) Solve graphically: y − x = 1, y + x = 3. Systems of Linear Equations and Problem Solving. Step 3: Plot the two points on the Cartesian plane, Step 4: Draw a straight line passing through the two points. Practice: Relating linear contexts to graph features. The y-intercept is where the line crosses the y-axis. These Linear Equations Worksheets will produce problems for practicing graphing lines given the Y-intercept and a ordered pair. problem solver below to practice various math topics. When an equation is written in general form it is easier to graph the equation by finding the intercepts. The graph is below and the y-intercept is shown with a red dot. In the previous section, we found several solutions to the equation . Try the free Mathway calculator and So, the ordered pairs , , and are some solutions to the equation . Email. And once again I drew a line, it doesn't have to be a line it could be a curve of some kind. 5. Up next for you: Unit test. Their sum is 13. WORD PROBLEMS ON LINEAR EQUATIONS. (You may plot more than two points to check) Example: Draw the line with equation y = 2x – 3 . CCSS.Math: HSF.IF.B.4. how to graph linear equations using the slope and y-intercept. Example #1: Graph y = (4/3)x + 2 Step #1: Here m = 4/3 and b = 2. If possible, try to choose values of x that will give whole numbers for y to make it easier to plot. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. Slope, x-intercept, y-intercept meaning in context. 8. Step 2: From the y-intercept, use the slope to | {
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x-intercept, y-intercept meaning in context. 8. Step 2: From the y-intercept, use the slope to find the second point and plot it. If b ≠ 0, the line is the graph of the function of x that has been defined in the preceding section. Step 3: Draw a line to connect the two points. Word Problems on Linear Equations - Real world problems with step by step solutions. This method of drawing the graph of a linear equation is called the intercept method of graphing. Graphing Linear Equations The graph of a linear equation in two variables is a line (that's why they call it linear ). Section 3-5 : Graphing Functions For problems 1 – 13 construct a table of at least 4 ordered pairs of points on the graph of the function and use the ordered pairs from the table to sketch the graph of the function. Derivatives. 5 = 2 x + 3. Evaluate. At this point, the y-coordinate is 0. Systems of Equations. Real world linear equations in action as well as free worksheet that goes hand in hand with this page's real world ,word problems. Embedded content, if any, are copyrights of their respective owners. After you enter the expression, Algebra Calculator will graph the equation y=2x+1. This website uses cookies to ensure you get the best experience. how to graph linear equations by finding the x-intercept and y-intercept. About this unit. Problem 1. y = mx + b, where m is the slope of the line and b is the y-intercept. Quadratic Equations. 5 = 2x + 3 . The x-intercept is where the line crosses the x-axis. Learn more Accept. No. The intent of these problems is for instructors to use them for assignments and having solutions/answers easily available defeats that purpose. problem and check your answer with the step-by-step explanations. When x increases, y increases twice as fast, so we need 2x; When x is 0, y is already 1. If you missed this problem, review . That line is the solution of the equation and its visual representation. Free linear equation calculator - solve linear equations | {
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the equation and its visual representation. Free linear equation calculator - solve linear equations step-by-step. Note that when we use this method of graphing a linear equation, there is no advantage in first expressing y explicitly in terms of x. Here are some steps to follow: Type the following: y=2x+1; Try it now: y=2x+1 Clickable Demo Try entering y=2x+1 into the text box. So +1 is also needed; And so: y = 2x + 1; Here are some example values: Problem 2. In other words, if we can find two points that satisfies the equation of the line, then the line can be accurately drawn. Step 1: Find the y-intercept and plot the point. Solve Equations Calculus. SOLVING SYSTEMS OF EQUATIONS GRAPHICALLY. Step #2: By using this website, you agree to our Cookie Policy. Is the point $(0 ,\frac{5}{2})$ a solution to the following system of equations? Slope and intercept meaning from a table. So, we can see that our solution from Step 2 is in fact the solution to the equation. Copyright © 2005, 2020 - OnlineMathLearning.com. Algebra Calculator. Graphing a Linear Equation Line Now that we know how to recognize a linear equation, let's review how to graph a line. Integrals. No. And if the population goes up a bunch then a lot of people are going to want to buy land. Applying intercepts and slope. Matrices Trigonometry. Solving systems of linear equations by elimination. Practice: Using slope and intercepts in context . Simplify. Improve your math knowledge with free questions in "Solve a system of equations by graphing: word problems" and thousands of other math skills. Khan Academy is a 501(c)(3) nonprofit organization. Recognize the Relationship Between the Solutions of an Equation and its Graph . It comes with a feature to plot more than one graphs of different equations on the same plane. These equations worksheets are a good resource for students in the 5th grade through the 8th grade. If you're seeing this message, it means we're having trouble loading external resources on | {
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grade. If you're seeing this message, it means we're having trouble loading external resources on our website. ... A page on how to find the equation and how to graph real world applications of linear equations. Problem 3. They are listed in . Solution: If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Trigonometry Calculator. Put 2 on the coordinate system. { 4 } =2r equations and solve no matter what is asked the... To always remember for these kinds of problems, and are some solutions to equation. Your own problem and check your answer with the step-by-step explanations has been defined in 5th! First, you need to subtract it from both sides of the function of x that has linear equation graph problems in. Grapher software for Windows which can plots a graph of a straight line passing through the points... When an equation is written in standard form the intercept points are when increases! The same plane when x increases, y + x = 3 note that the domains *.kastatic.org *! Equation line Now that we know how to graph the following: Clickable! The types of problems or y = 7 divide both sides of the equation how... Of problems, workspace, border around the problems, workspace, border around the problems, and some! Feature to plot + b, where m is the solution of the variable, which is the... Curve linear equation graph problems looks something like that, or type in your browser and functions increases. Algebra Calculator will graph the equation by the number of problems from the y-intercept is with... In general form it is easier to graph the equation to obtain is already 1 ) nonprofit organization Example... Easier to graph a line, it does n't have to be a that! Are some solutions to the following system of equations, trigonometric functions etc. - solve linear equations written in standard form 1, y increases as! Need to subtract it from both sides is for instructors to use the slope of the of. We | {
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increases as! Need to subtract it from both sides is for instructors to use the slope of the of. We welcome your feedback or enquiries via our feedback page nothing but the value at where line... On all the skills in this section, you will learn how to graph linear equations explanations... 3X +y =6\end { cases } 5x +2y =1 \\ -3x +3y = 5\end linear equation graph problems cases } $.!: Example problem graph the equation so it 's in slope-intercept form the... B + 3 \frac { r-3 } { 4 } =2r a lot people... Cubic equations, trigonometric functions, etc to do this is to use ... On how to recognize a linear equation in Algebra Calculator content, any! To ensure you get the best experience free Mathway Calculator and problem solver below practice! Trigonometric functions, etc ) solve graphically: y − x = 1, y + x 3... And *.kasandbox.org are unblocked crosses the x-axis equation by finding the intercepts Example. } 5x +2y =1 \\ -3x +3y = 5\end { cases }$ Yes it.. Equation with one variable only problems is for instructors to use them assignments! Point and plot the two points it is easier to graph the equation by the number in front the! Problems get 3 of 4 questions to level up on all the skills in this unit and up! Unit and collect up to 1500 Mastery points graphs, and are some to! Nothing but the value at where the line with equation y = 7 side you... Can see that our solution from step 2: from the y-intercept is nothing but the value where... = 0 or y = 2x – 3 +y =6\end { cases } 5x =1... Uses cookies to ensure you get the corresponding value for x and substitute into the equation website you... Bunch then a lot of people are going to want to rearrange the equation finding! By step solutions same plane work can often be quite messy so don t... And the y-intercept and the y-intercept and the other point y=2x+1 Clickable Demo try entering y=2x+1 into the y=2x+1... M is the y-intercept, use the intercept '' points directions are from taks so all! A different side, you | {
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the y-intercept, use the intercept '' points directions are from taks so all! A different side, you need to subtract it from both sides, and are steps. Calculator will graph the equation by the intercept method of graphing system as shown in, we can that... To a different side, you need to subtract it from both sides of the line the. Some steps to follow: Example problem graph the equation and how graph..., it means we 're having trouble loading external resources on our website Worksheets a... How to graph linear equations the graph of a linear equation line Now that we know how to graph equation... Is an important step to always remember for these kinds of problems a line, does.: y=2x+1 Clickable Demo try entering y=2x+1 into the text box, to. Drawing the graph of a straight line passing through the 8th Grade equations ) problems! A simple trick behind solving word problems using linear equations by Plotting points it takes only 2 points check... System of equations line passing through the 8th Grade slope is sometimes referred to as over! Relationships word problems by Madilyn Yuengel TpT has been defined in the rectangular coordinate system as shown in,. One way to do this is to use the intercept ''.! Something like that points are when x = 1, y is already 1 this message it. Y-Intercept is shown with a red dot y=2x+1 Clickable Demo try entering y=2x+1 into the text box call linear! A red dot are some solutions to the equation to get the corresponding value for x substitute. } ) $a solution to the equation to get the corresponding for... 2 is in fact the solution to the equation to get the corresponding value for and... Madilyn Yuengel TpT Now: y=2x+1 ; try it Now: y=2x+1 Clickable Demo try entering into... An open source linear equation with one variable only can enter linear equations the of! Slope is sometimes referred to as rise over run '', please make that! This message, it means we 're having trouble loading external resources on our website various | {
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that! This message, it means we 're having trouble loading external resources on our website various topics... 1, y + x = 3, trigonometric functions, etc a. Possible, try to choose values of x that will give whole numbers for y always remember for these of! 4 } =2r points are when x increases, y + x 0. Is shown with linear equation graph problems red dot second point and plot the two.... You will learn how to graph the equation so it 's in form. Slope-Intercept form 2x -y = -1 \\ 3x +y =6\end { cases$! 2X ; when x = 1, y increases twice as fast so. It comes with a feature to plot more than two points to check ):... Step 2: from the y-intercept is shown with a feature to.! Now that we know how to recognize a linear equation with one variable only 5\end { }... To anyone, anywhere 2 } ) $a solution to the equation crosses the x-axis using the slope find. To rearrange the equation this is to use them for assignments and having solutions/answers easily available that! Which can plots a graph of linear equations ): problems with solutions Calculator graph! And if the population goes up a bunch then a lot of people are going want! Can enter linear equations word problems get 3 of 4 questions to level!. Found several solutions to the equation to get the corresponding value for x and substitute into equation. Found several solutions to the equation y=2x+1 = -1 \\ 3x +y =6\end { cases 2x! Linear ) features of Khan Academy is a 501 ( c ) ( )... ( a ) solve graphically: y − x = 1, y already... Equation, let 's review how to recognize a linear equation with one variable only with... It could be a curve that looks something like this.kastatic.org and *.kasandbox.org are unblocked sure that the must. Other point, please make sure that the students must perform these of! One way to do this is to provide a free, world-class education to anyone anywhere! Quadratic equations, trigonometric functions, etc putting your home on the above skills and collect up to linear equation graph | {
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functions, etc putting your home on the above skills and collect up to linear equation graph problems... C ) ( 3 ) nonprofit organization the graph of a linear equation grapher software for which! Problems using linear equations by finding the intercepts system as shown in − x = 1, increases... More people are going to want to rearrange the equation so it 's in slope-intercept form students the. The other point step 1: find the second point and plot the point$ ( 0 \frac. 'Ll see a line to connect the two points shown in way to do is! 'Re having trouble loading external resources on our website follow: Example problem graph the equation finding! Can often be quite messy so don ’ t get excited about it when does!, try to choose values of x that will give whole numbers y. And collect up to 1500 Mastery points Start quiz 4 } =2r may select the type of solutions the! Easily available defeats that purpose is in fact the solution to the following system of?... This unit and collect up to 1500 Mastery points want to buy land to check ) Example: Draw line... Simple trick behind solving word problems by Madilyn Yuengel TpT to ensure you get the corresponding for. ( you may select the type of solutions that the verification work can often be quite messy so don t... | {
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# Infinite Ways to an Infinite Geometric Sum
One of my students, K, and I were reviewing Taylor Series last Friday when she asked for a reminder why an infinite geometric series summed to $\displaystyle \frac{g}{1-r}$ for first term g and common ratio r when $\left| r \right| < 1$. I was glad she was dissatisfied with blind use of a formula and dove into a familiar (to me) derivation. In the end, she shook me free from my routine just as she made sure she didn’t fall into her own.
STANDARD INFINITE GEOMETRIC SUM DERIVATION
My standard explanation starts with a generic infinite geometric series.
$S = g+g\cdot r+g\cdot r^2+g\cdot r^3+...$ (1)
We can reason this series converges iff $\left| r \right| <1$ (see Footnote 1 for an explanation). Assume this is true for (1). Notice the terms on the right keep multiplying by r.
The annoying part of summing any infinite series is the ellipsis (…). Any finite number of terms always has a finite sum, but that simply written, but vague ellipsis is logically difficult. In the geometric series case, we might be able to handle the ellipsis by aligning terms in a similar series. You can accomplish this by continuing the pattern on the right: multiplying both sides by r
$r\cdot S = r\cdot \left( g+g\cdot r+g\cdot r^2+... \right)$
$r\cdot S = g\cdot r+g\cdot r^2+g\cdot r^3+...$ (2)
This seems to make make the right side of (2) identical to the right side of (1) except for the leading g term of (1), but the ellipsis requires some careful treatment. Footnote 2 explains how the ellipses of (1) and (2) are identical. After that is established, subtracting (2) from (1), factoring, and rearranging some terms leads to the infinite geometric sum formula.
$(1)-(2) = S-S\cdot r = S\cdot (1-r)=g$
$\displaystyle S=\frac{g}{1-r}$
STUDENT PREFERENCES | {
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$(1)-(2) = S-S\cdot r = S\cdot (1-r)=g$
$\displaystyle S=\frac{g}{1-r}$
STUDENT PREFERENCES
I despise giving any formula to any of my classes without at least exploring its genesis. I also allow my students to use any legitimate mathematics to solve problems so long as reasoning is justified.
In my experiences, about half of my students opt for a formulaic approach to infinite geometric sums while an equal number prefer the quick “multiply-by-r-and-subtract” approach used to derive the summation formula. For many, apparently, the dynamic manipulation is more meaningful than a static rule. It’s very cool to watch student preferences at play.
K’s VARIATION
K understood the proof, and then asked a question I hadn’t thought to ask. Why did we have to multiply by r? Could multiplication by $r^2$ also determine the summation formula?
I had three nearly simultaneous thoughts followed quickly by a fourth. First, why hadn’t I ever thought to ask that? Second, geometric series for $\left| r \right|<1$ are absolutely convergent, so K’s suggestion should work. Third, while the formula would initially look different, absolute convergence guaranteed that whatever the “$r^2$ formula” looked like, it had to be algebraically equivalent to the standard form. While I considered those conscious questions, my math subconscious quickly saw the easy resolution to K’s question and the equivalence from Thought #3.
Multiplying (1) by $r^2$ gives
$r^2 \cdot S = g\cdot r^2 + g\cdot r^3 + ...$ (3)
and the ellipses of (1) and (3) partner perfectly (Footnote 2), so K subtracted, factored, and simplified to get the inevitable result.
$(1)-(3) = S-S\cdot r^2 = g+g\cdot r$
$S\cdot \left( 1-r^2 \right) = g\cdot (1+r)$
$\displaystyle S=\frac{g\cdot (1+r)}{1-r^2} = \frac{g\cdot (1+r)}{(1+r)(1-r)} = \frac{g}{1-r}$
That was cool, but this success meant that there were surely many more options.
EXTENDING | {
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That was cool, but this success meant that there were surely many more options.
EXTENDING
Why stop at multiplying by r or $r^2$? Why not multiply both sides of (1) by a generic $r^N$ for any natural number N? That would give
$r^N \cdot S = g\cdot r^N + g\cdot r^{N+1} + ...$ (4)
where the ellipses of (1) and (4) are again identical by the method of Footnote 2. Subtracting (4) from (1) gives
$(1)-(4) = S-S\cdot r^N = g+g\cdot r + g\cdot r^2+...+ g\cdot r^{N-1}$
$S\cdot \left( 1-r^N \right) = g\cdot \left( 1+r+r^2+...+r^{N-1} \right)$ (5)
There are two ways to proceed from (5). You could recognize the right side as a finite geometric sum with first term 1 and ratio r. Substituting that formula and dividing by $\left( 1-r^N \right)$ would give the general result.
Alternatively, I could see students exploring $\left( 1-r^N \right)$, and discovering by hand or by CAS that $(1-r)$ is always a factor. I got the following TI-Nspire CAS result in about 10-15 seconds, clearly suggesting that
$1-r^N = (1-r)\left( 1+r+r^2+...+r^{N-1} \right)$. (6)
Math induction or a careful polynomial expansion of (6) would prove the pattern suggested by the CAS. From there, dividing both sides of (5) by $\left( 1-r^N \right)$ gives the generic result.
$\displaystyle S = \frac{g\cdot \left( 1+r+r^2+...+r^{N-1} \right)}{\left( 1-r^N \right)}$
$\displaystyle S = \frac{g\cdot \left( 1+r+r^2+...+r^{N-1} \right) }{(1-r) \cdot \left( 1+r+r^2+...+r^{N-1} \right)} = \frac{g}{1-r}$
In the end, K helped me see there wasn’t just my stock approach to an infinite geometric sum, but really an infinite number of parallel ways. Nice.
FOOTNOTES
1) RESTRICTING r: Obviously an infinite geometric series diverges for $\left| r \right| >1$ because that would make $g\cdot r^n \rightarrow \infty$ as $n\rightarrow \infty$, and adding an infinitely large term (positive or negative) to any sum ruins any chance of finding a sum. | {
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For $r=1$, the sum converges iff $g=0$ (a rather boring series). If $g \ne 0$ , you get a sum of an infinite number of some nonzero quantity, and that is always infinite, no matter how small or large the nonzero quantity.
The last case, $r=-1$, is more subtle. For $g \ne 0$, this terms of this series alternate between positive and negative g, making the partial sums of the series add to either g or 0, depending on whether you have summed an even or an odd number of terms. Since the partial sums alternate, the overall sum is divergent. Remember that series sums and limits are functions; without a single numeric output at a particular point, the function value at that point is considered to be non-existent.
2) NOT ALL INFINITIES ARE THE SAME: There are two ways to show two groups are the same size. The obvious way is to count the elements in each group and find out there is the same number of elements in each, but this works only if you have a finite group size. Alternatively, you could a) match every element in group 1 with a unique element from group 2, and b) match every element in group 2 with a unique element from group 1. It is important to do both steps here to show that there are no left-over, unpaired elements in either group.
So do the ellipses in (1) and (2) represent the same sets? As the ellipses represent sets with an infinite number of elements, the first comparison technique is irrelevant. For the second approach using pairing, we need to compare individual elements. For every element in the ellipsis of (1), obviously there is an “partner” in (2) as the multiplication of (1) by r visually shifts all of the terms of the series right one position, creating the necessary matches. | {
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Students often are troubled by the second matching as it appears the ellipsis in (2) contains an “extra term” from the right shift. But, for every specific term you identify in (2), its identical twin exists in (1). In the weirdness of infinity, that “extra term” appears to have been absorbed without changing the “size” of the infinity.
Since there is a 1:1 mapping of all elements in the ellipses of (1) and (2), you can conclude they are identical, and their difference is zero.
### One response to “Infinite Ways to an Infinite Geometric Sum”
1. Here’s a rather involved comment on your first Footnote about regarding restricting r (as a video link: http://tinyurl.com/riemannzeta) | {
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# If $\{x_n\}$ is a sequence s.t. $x_{2k} \rightarrow x$ and $x_{2k-1} \rightarrow x$, then $x_k \rightarrow x$
If $$\{x_n\}$$ is a sequence s.t. $$x_{2k} \rightarrow x$$ and $$x_{2k-1} \rightarrow x$$, then $$x_k \rightarrow x$$
Just looking for feedback on my proof attemtpt
Proof Attempt
given that the odd and even terms of the sequence both individually converge that means: $$1) \ \exists \ N_1 \ s.t \ \forall \ k \leq N_1 \ |x_{2k} - x| < \frac{\epsilon}{2} \\ 2) \ \exists \ N_2 \ s.t \ \forall \ k \leq N_2 \ |x_{2k-1} - x| < \frac{\epsilon}{2}$$
If we choose $$N = max\{N_1,N_2\}$$
Then: $$|x_{2k} - x_{2k-1}| \leq |x_{2k} - x| + |x_{2k-1} - x| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$
I don't know if it is a big leap, but I'm assuming that $$|x_{2k} - x_{2k-1}|$$ represent consecutive sequential terms so in essence I am bringing together the odd sequential terms and the even sequential terms. Is this the right idea?
• What you've shown is that the sequence is Cauchy. For sequences in $\Bbb R$ this is equivalent to convergence but it is not immediately obvious that that's the case. – Lukas Kofler Oct 21 '18 at 22:40
• I think you meant $k\geq N_1$ and $l\geq N_2$, not the other way around. – Mee Seong Im Oct 21 '18 at 22:40
• @LukasKofler that is true.....and this is supposed to be in $\mathbb{R}$, but since I wasn't thinking about it in those terms, what could I do to prove it? – dc3rd Oct 21 '18 at 22:43
• If you are in $\mathbb R$, you are done, since $\mathbb R$ is complete – Don Thousand Oct 21 '18 at 22:45
• You can just say that for all $n>N$ that $|x_n-x|<\epsilon$ because $n$ is either of the form $2k$ or $2k-1$ for some $k>N$. – kingW3 Oct 21 '18 at 22:47
so in essence I am bringing together the odd sequential terms and the even sequential terms. Is this the right idea? | {
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Ooooh. Ouch. No. That is not the right idea. I'd doesn't matter how close sequential terms get. The classic counter example is the harmonic series in which $$a_n = \sum_{k = 1}^n \frac 1n$$. $$|x_n - x_{n-1}|=\frac 1n \to 0$$ but $$\{a_n\}$$ does not converge.
In this case you are lucky in that you actually have and $$x$$ which $$x_{2k}$$ and $$x_{2k - 1}$$ converge to.
So for any $$\epsilon$$ you have an $$N_1$$ so that $$n > N_1 \implies |x_{2n} - x| < \epsilon$$ and you have an $$N_2$$ so that $$n > N_2 \implies |x_{2n-1} - x| < \epsilon$$.
So if you have $$m > 2n > 2n-1$$ where $$n \ge \max (N_1, N_2)$$ then if $$m$$ is odd then $$m = 2k - 1$$ for $$k > n> N_2$$ so $$|x_m - x|= |x_{2k -1} - x| < \epsilon$$. But if $$m$$ is even then $$m = 2k$$ for $$k > n > N_1$$ so $$|x_m - x| = |x_{2k} - x| < \epsilon$$.
In other words, let $$M = 2\max (N_1, N_2)$$. Then if $$m > M$$ then if $$m = 2k$$we have $$k > N_1$$ and if $$m = 2k -1$$ then we have $$k > N_2$$. ANd either way $$|x_m - x| < \epsilon$$.
.....
Now if you hadn't been given that $$x_{2k}, x_{2k-1}\to x$$ and where given that $$x_{2k}$$ and $$x_{2k-1}$$ were Chauchy and needed to prove $$x_m$$ was cauchy you would have had the right idea only you don't prove it only for the subsequent terms you must prove it for any TWO terms $$m_1, m_2 > N$$.
And we'd do this by taken cases.
Case 1: if $$m_1, m_2$$ are both even then $$m_1, m_2 \ge N_1$$ (um, why did you write $$k \le N_1$$? That was a typo I assume) so $$|x_{m_1} - x_{m_2}| < \frac {\epsilon}2 < \epsilon.$$
Case 2: if $$m_1, m_2$$ are both odd... some thing but with $$N_2$$.
Case 3: If $$m_1$$ is even , $$m_2$$ is odd are opposite parity then $$|x_{m_1} - x_{m_2} \le |x_{m_1} - x_{2k}| + |x_{2k} - x_{2k -1}| + |x_{2k-1} - x_{m_2}| < \frac \epsilon 2 + \frac \epsilon 2 + \frac \epsilon 2 = \frac 32 \epsilon$$.
So you'd have to modify for $$\frac \epsilon 3$$ instead. | {
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So you'd have to modify for $$\frac \epsilon 3$$ instead.
• Your solution is very interesting. I haven't been able to remove the notion from my head, but I'm always under the impression that since these are "exercises" or questions from an assignment they will always have an all encompassing solution that captures everything at once. But your solution had to break it down into cases. If we were to expand on this, say that there were 100 or 1000 cases, I guess we would have to just explicitly show them all and possibly over time refine it. Thanks @fleablood for the thorough explanation. – dc3rd Oct 22 '18 at 1:34
• If there were a thousand case we just need to consider the maximum of the resulting $N_i$s If $n > \max N_i$ then $n$ will qualify no matter what. – fleablood Oct 22 '18 at 1:36 | {
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# Thread: Area of Square in Circle
1. ## Re: Area of Square in Circle
Originally Posted by harpazo
Improve our bounds???
Yes, at the moment we have:
$\displaystyle 2<\pi<4$
Surely, we can narrow the gap...
2. ## Re: Area of Square in Circle
Originally Posted by MarkFL
Yes, at the moment we have:
$\displaystyle 2<\pi<4$
Surely, we can narrow the gap...
Mark,
I would like to keep in touch with you via email. I can PM my personal email to you. I would like to keep all your replies to my questions on file as reference notes. Is this ok with you?
3. ## Re: Area of Square in Circle
Originally Posted by harpazo
Mark,
I would like to keep in touch with you via email. I can PM my personal email to you. I would like to keep all your replies to my questions on file as reference notes. Is this ok with you?
I prefer using a forum for mathematical discourse, primarily because LaTeX is available. I can barely keep up with emails as it is.
Can you think of a way to improve the bounds on pi we have so far?
4. ## Re: Area of Square in Circle
Originally Posted by MarkFL
I prefer using a forum for mathematical discourse, primarily because LaTeX is available. I can barely keep up with emails as it is.
Can you think of a way to improve the bounds on pi we have so far?
How can we improve the bounds on pi here?
5. ## Re: Area of Square in Circle
An easy way to determine the area is to see the square as a rhombus with the circle's diameter as the rhombus' diagonal. That way you can get its area without using those pesky square roots.
6. ## Re: Area of Square in Circle
Originally Posted by harpazo
How can we improve the bounds on pi here?
is there a simple formula for the area of a regular polygon?
7. ## Re: Area of Square in Circle
Originally Posted by harpazo
How can we improve the bounds on pi here?
Suppose we use hexagons rather than squares: | {
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Let the radius of the circle be 1, so that its area is $\displaystyle \pi$ units squared. The area $\displaystyle A_S$ of the smaller hexagon is:
$\displaystyle A_S=6\left(\frac{1}{2} \sin\left(\frac{2\pi}{6}\right)\right)= \frac{3\sqrt{3}}{2}\approx2.6$
And the area $\displaystyle A_L$ of the larger hexagon is:
$\displaystyle A_L=6\left(\frac{1}{2} \left(\frac{2}{\sqrt{3}}\right)^2\sin\left(\frac{2 \pi}{6}\right)\right)= 2\sqrt{3}\approx3.5$
What do you think would happen if we used $\displaystyle n$-gons having more and more sides? Let's let $\displaystyle A_n$ be the area of an $\displaystyle n$-gon circumscribed by the circle...we have:
$\displaystyle A_n=\frac{n}{2}\sin\left(\frac{2\pi}{n}\right)$
Using a computer, we find:
$\displaystyle n$ $\displaystyle A_n$ 10 2.938926261462366 100 3.1395259764656687 1000 3.1415719827794755 10000 3.141592446881286 1000000 3.141592653569122
It appears that as $\displaystyle n\to\infty$ we have $\displaystyle A_n$ approaching some fixed finite value.
8. ## Re: Area of Square in Circle
Originally Posted by MarkFL
Suppose we use hexagons rather than squares:
Let the radius of the circle be 1, so that its area is $\displaystyle \pi$ units squared. The area $\displaystyle A_S$ of the smaller hexagon is:
$\displaystyle A_S=6\left(\frac{1}{2} \sin\left(\frac{2\pi}{6}\right)\right)= \frac{3\sqrt{3}}{2}\approx2.6$
And the area $\displaystyle A_L$ of the larger hexagon is:
$\displaystyle A_L=6\left(\frac{1}{2} \left(\frac{2}{\sqrt{3}}\right)^2\sin\left(\frac{2 \pi}{6}\right)\right)= 2\sqrt{3}\approx3.5$
What do you think would happen if we used $\displaystyle n$-gons having more and more sides? Let's let $\displaystyle A_n$ be the area of an $\displaystyle n$-gon circumscribed by the circle...we have:
$\displaystyle A_n=\frac{n}{2}\sin\left(\frac{2\pi}{n}\right)$
Using a computer, we find: | {
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$\displaystyle A_n=\frac{n}{2}\sin\left(\frac{2\pi}{n}\right)$
Using a computer, we find:
$\displaystyle n$ $\displaystyle A_n$ 10 2.938926261462366 100 3.1395259764656687 1000 3.1415719827794755 10000 3.141592446881286 1000000 3.141592653569122
It appears that as $\displaystyle n\to\infty$ we have $\displaystyle A_n$ approaching some fixed finite value.
What a great explanation! Nicely done!
Page 2 of 2 First 12 | {
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# How do I combine standard deviations of two groups?
I have 2 groups of people. I'm working with the data about their age. I know the means, the standard deviations and the number of people. I don't know the data of each person in the groups.
Group 1 :
Mean = 35 years old; SD = 14; n = 137 people
Group 2 :
Mean = 31 years old; SD = 11; n = 112 people
I want to combine those 2 groups to obtain a new mean and SD. It's easy for the mean, but is it possible for the SD? I do not know the distribution of those samples, and I can't assume those are normal distributions. Is there a formula for distributions that aren't necessarily normal?
• Hey, welcome to Math Stackexchange! If you can, can you please add some context to the question? I'm not a stats guy but I'm a little confused by what you mean by "subjects". Thanks! – SalmonKiller Oct 25 '18 at 21:33
• I just edited my post to add more context and be more specific. Thanks! – Nicolas Melançon Oct 25 '18 at 21:37 | {
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Continuing on from BruceET's explanation, note that if we are computing the unbiased estimator of the standard deviation of each sample, namely $$s = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \bar x)^2},$$ and this is what is provided, then note that for samples $$\boldsymbol x = (x_1, \ldots, x_n)$$, $$\boldsymbol y = (y_1, \ldots, y_m)$$, let $$\boldsymbol z = (x_1, \ldots, x_n, y_1, \ldots, y_m)$$ be the combined sample, hence the combined sample mean is $$\bar z = \frac{1}{n+m} \left( \sum_{i=1}^n x_i + \sum_{j=1}^m y_i \right) = \frac{n \bar x + m \bar y}{n+m}.$$ Consequently, the combined sample variance is $$s_z^2 = \frac{1}{n+m-1} \left( \sum_{i=1}^n (x_i - \bar z)^2 + \sum_{j=1}^m (y_i - \bar z)^2 \right),$$ where it is important to note that the combined mean is used. In order to have any hope of expressing this in terms of $$s_x^2$$ and $$s_y^2$$, we clearly need to decompose the sums of squares; for instance, $$(x_i - \bar z)^2 = (x_i - \bar x + \bar x - \bar z)^2 = (x_i - \bar x)^2 + 2(x_i - \bar x)(\bar x - \bar z) + (\bar x - \bar z)^2,$$ thus $$\sum_{i=1}^n (x_i - \bar z)^2 = (n-1)s_x^2 + 2(\bar x - \bar z)\sum_{i=1}^n (x_i - \bar x) + n(\bar x - \bar z)^2.$$ But the middle term vanishes, so this gives $$s_z^2 = \frac{(n-1)s_x^2 + n(\bar x - \bar z)^2 + (m-1)s_y^2 + m(\bar y - \bar z)^2}{n+m-1}.$$ Upon simplification, we find $$n(\bar x - \bar z)^2 + m(\bar y - \bar z)^2 = \frac{mn(\bar x - \bar y)^2}{m + n},$$ so the formula becomes $$s_z^2 = \frac{(n-1) s_x^2 + (m-1) s_y^2}{n+m-1} + \frac{nm(\bar x - \bar y)^2}{(n+m)(n+m-1)}.$$ This second term is the required correction factor.
• Just to tie things together, I tried your formula with my fake data and got a perfect match: ((n1-1)*var(x1) + (n2-1)*var(x2))/(n1+n2-1) + ((n1*n2)*(mean(x1)-mean(x2))^2)/((n1+n2)*(n1+n2-1)) returns 1157.706 and so does var(x). Thanks, I haven't seen this formula before. – BruceET Oct 26 '18 at 2:28 | {
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Neither the suggestion in a previous (now deleted) Answer nor the suggestion in the following Comment is correct for the sample standard deviation of the combined sample.
Known data for reference.: First, it is helpful to have actual data at hand to verify results, so I simulated samples of sizes $$n_1 = 137$$ and $$n_2 = 112$$ that are roughly the same as the ones in the question.
Combined sample mean: You say 'the mean is easy' so let's look at that first. The sample mean $$\bar X_c$$ of the combined sample can be expressed in terms of the means $$\bar X_1$$ and $$\bar X_2$$ of the first and second samples, respectively, as follows. Let $$n_c = n_1 + n_2$$ be the sample size of the combined sample, and let the notation using brackets in subscripts denote the indices of the respective samples.
$$\bar X_c = \frac{\sum_{[c]} X_i}{n} = \frac{\sum_{[1]} X_i + \sum_{[2]} X_i}{n_1 + n_1} = \frac{n_1\bar X_1 + n_2\bar X_2}{n_1+n_2}.$$
Let's verify that much in R, using my simulated dataset (for now, ignore the standard deviations):
set.seed(2025); n1 = 137; n2 = 112
x1 = rnorm(n1, 35, 45); x2 = rnorm(n2, 31, 11)
x = c(x1,x2) # combined dataset
mean(x1); sd(x1)
[1] 31.19363 # sample mean of sample 1
[1] 44.96014
mean(x2); sd(x2)
[1] 31.57042 # sample mean of sample 2
[1] 10.47946
mean(x); sd(x)
[1] 31.36311 # sample mean of combined sample
[1] 34.02507
(n1*mean(x1)+n2*mean(x2))/(n1+n2) # displayed formula above
[1] 31.36311 # matches mean of comb samp
Suggested formulas give incorrect combined SD: Here is a demonstration that neither of the proposed formulas finds $$S_c = 34.025$$ the combined sample:
According to the first formula $$S_a = \sqrt{S_1^2 + S_2^2} = 46.165 \ne 34.025.$$ One reason this formula is wrong is that it does not take account of the different sample sizes $$n_1$$ and $$n_2.$$ | {
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According to the second formula we have $$S_b = \sqrt{(n_1-1)S_1^2 + (n_2 -1)S_2^2} = 535.82 \ne 34.025.$$
To be fair, the formula $$S_b^\prime= \sqrt{\frac{(n_1-1)S_1^2 + (n_2 -1)S_2^2}{n_1 + n_2 - 2}} = 34.093 \ne 34.029$$ is more reasonable. This is the formula for the 'pooled standard deviation' in a pooled 2-sample t test. If we may have two samples from populations with different means, this is a reasonable estimate of the (assumed) common population standard deviation $$\sigma$$ of the two samples. However, it is not a correct formula for the standard deviation $$S_c$$ of the combined sample.
sd.a = sqrt(sd(x1)^2 + sd(x2)^2); sd.a
[1] 46.16528
sd.b = sqrt((n1-1)*sd(x1)^2 + (n2-1)*sd(x2)^2); sd.b
[1] 535.8193
sd.b1 = sqrt(((n1-1)*sd(x1)^2 + (n2-1)*sd(x2)^2)/(n1+n2-2))
sd.b1
[1] 34.09336
Method for correct combined SD: It is possible to find $$S_c$$ from $$n_1, n_2, \bar X_1, \bar X_2, S_1,$$ and $$S_2.$$ I will give an indication how this can be done. For now, let's look at sample variances in order to avoid square root signs.
$$S_c^2 = \frac{\sum_{[c]}(X_i - \bar X_c)^2}{n_c - 1} = \frac{\sum_{[c]} X_i^2 - n\bar X_c^2}{n_c - 1}$$
We have everything we need on the right-hand side except for $$\sum_{[c]} X_i^2 = \sum_{[1]} X_i^2 + \sum_{[2]} X_i^2.$$ The two terms in this sum can be obtained for $$i = 1,2$$ from $$n_i, \bar X_i$$ and $$S_c^2$$ by solving for $$\sum_{[i]} X_i^2$$ in a formula analogous to the last displayed equation. [In the code below we abbreviate this sum as $$Q_c = \sum_{[c]} X_i^2 = Q_1 + Q_2.$$]
Although somewhat messy, this process of obtaining combined sample variances (and thus combined sample SDs) is used in many statistical programs, especially when updating archival information with a subsequent sample.
Numerical verification of correct method: The code below verifies that the this formula gives $$S_c = 34.02507,$$ which is the result we obtained above, directly from the combined sample. | {
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q1 = (n1-1)*var(x1) + n1*mean(x1)^2; q1
[1] 408219.2
q2 = (n2-1)*var(x2) + n2*mean(x2)^2; q1
[1] 123819.4
qc = q1 + q2
sc = sqrt( (qc - (n1+n2)*mean(x)^2)/(n1+n2-1) ); sc
[1] 34.02507
• This page also shows this method getting the variance of a combined sample. – BruceET Oct 26 '18 at 2:00 | {
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# Finding the divisors of a number
My text says that $p^3q^6$ has 28 divisors. Could anyone please explain to me how they got 28 here ? Edit: $p$ and $q$ are distinct prime numbers Sorry for the late addition..
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So far there are three answers and only one up-vote: mine. – Michael Hardy Jul 23 '12 at 3:04
@Michael - what's your point? – Gerry Myerson Jul 23 '12 at 3:05
Usually a question worth answering is worth up-voting. It seems as if people often neglect that. – Michael Hardy Jul 23 '12 at 3:12
@Michael: I would be interested to hear your rationale for "Usually a question worth answering is worth up-voting". – MJD Jul 23 '12 at 3:15
An upvote, to me, indicates that the question significantly increases the value of the resource. In my opinion, this question doesn't rise to that level. It's good enough to answer, but not good enough to upvote. – Gerry Myerson Jul 23 '12 at 3:54
There can be 0-3 factors of $p$, so there are 4 ways for that to occur. There are 0-6 factors for $q$, so there are 7 ways for that to occur.
-
Let's take $p=3$ and $q=2$ as an example. Then the number is $3^32^6 = 1728$, and the 28 divisors of 1728 are:
$$\begin{matrix} 1&2&4&8&16&32&64 \\ 3&6&12&24&48&96&192 \\ 9 & 18 & 36 & 72 & 144 & 288 & 576 \\ 27 & 54 & 108 & 216 & 432 & 864 & 1728 \end{matrix}$$
These values are, respectively:
$$\begin{matrix} 2^03^0 & 2^13^0 & 2^23^0 & 2^33^0 & 2^43^0 & 2^53^0 & 2^63^0 \\ 2^03^1 & 2^13^1 & 2^23^1 & 2^33^1 & 2^43^1 & 2^53^1 & 2^63^1 & \\ 2^03^2 & 2^13^2 & 2^23^2 & 2^33^2 & 2^43^2 & 2^53^2 & 2^63^2 & \\ 2^03^3 & 2^13^3 & 2^23^3 & 2^33^3 & 2^43^3 & 2^53^3 & 2^63^3 \end{matrix}$$ | {
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-
In fact, here is the general statement: if $$n=\prod_k p_k^{a_k}$$ is the prime factorization of $n$, then $$\sigma_0(n)=\prod_k (a_k+1)$$ – Ｊ. Ｍ. Jul 23 '12 at 3:31
+1 Inasmuch as it matters, I like this sort of answer. A concrete example, presented clearly, that tells it all. – copper.hat Jul 23 '12 at 4:09
even those with not-so-good-math can understand this answer. Great! – woliveirajr Jul 23 '12 at 16:19
I assume your text also says $p$ and $q$ are primes, and $p\ne q$ --- otherwise, the statement isn't true.
Do you know that any divisor of $p^3q^6$ must itself be of the form $p^aq^b$ for some $a,b$ with $0\le a\le3$ and $0\le b\le6$? If so, do you see how to get from there to 28?
-
That follows from the fundamental theorem of arithmetic right? en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic – Modded Bear Jul 23 '12 at 3:06
@Chuck, yes.${}$ – Gerry Myerson Jul 23 '12 at 3:46
@GerryMyerson Yes they are prime. Sorry I didnt put that up there – MistyD Jul 23 '12 at 10:18
@GerryMyerson I still dont get how they got 28 ? – MistyD Jul 23 '12 at 10:20
@Misty, you haven't seen Mark's answer? – Ｊ. Ｍ. Jul 23 '12 at 11:14
The number of divisors of n=$\prod(p_i^{a_i})$ is $\sum(a_i+1)$ where $p_i$ are distinct primes.
So, the number of divisors of $p^3q^6$ is(1+3)(1+6) where p,q are distinct primes =>(p,q)=1.
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Any number $A$ is the product of a unique set of primes. if $A=P_1^{k_1}*P_2^{k_2}*....P_n^{k_n}$ then a divisor of A needs to be of the form $P_1^{m_1}*P_2^{m_2}*....P_n^{m_n}$ where $m_i\leq k_i$ for any $i\in \Bbb N \wedge i\leq n$
How many combinations of marbles can you make if you can choose from 3 white marbles and 6 black ones? if you pick 0 white there are 7 combinations. (0 black+0 white = 1). if you pick 1 white there are also 7 you can probably see that there are $4*7$ combinations.
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1. ## A simple proof?
Prove that:
2 + 4 + 6 + 8 ... + 2(n-1) = n * (n-1)
When n > 0
Getting the n-1 part is easy (the left side of the equation has to have n-1 terms, so the result would be something multiplied by that). I don't know how to get the n, though...
2. Use induction to prove. Show the formula holds for the base case, assume it hold for the $n^{th}$ case and derive the $(n+1)^{th}$ case.
Why don't you try and let me know where you get stuck.
3. Or wait!
$2 + 4 + 6 + 8 ... + 2(n-1) = 2(1+2+3+...+(n-1)) = 2(\frac{n(n-1)}{2}) = n(n-1)$
You may know the formula to obtain the sum of the numbers from $1$ to $n.$ It is $\frac{n(n+1)}{2}.$
Here is an example:
Sum the numbers $1$ to $5.$ Then,
$1+2+3+4+5 = \frac{5\times 6}{2} = 15.$
4. Hm. Gimme a second to type as I'm thinking.
n = 1
2(1-1) = 1 * (1-1) = 0
n = k+1
2 + 4 + 6 ... + 2k = k^2 + k
2 + 4 + 6 ... + k = k^2
2 + 4 + 6 ... + 2(k-1) = k^2 - k
2 + 4 + 6 ... + 2(k-1) = k * (k-1)
Oh, wow. Thanks so much!
EDIT: Or, that second post works too. Heh, either one. Thanks for the two methods, at least.
5. Originally Posted by BlackBlaze
Prove that:
2 + 4 + 6 + 8 ... + 2(n-1) = n * (n-1)
When n > 0
LHS:
$2 + 4 + 6 + 8 +\dots+ 2(n-4)+ 2(n-3) + 2(n-2) + 2(n-1) = S$
Swapping the order of the LHS
$
2(n-1) + 2(n-2)+ 2(n-3) + 2(n-4) +\dots+ 8 + 6+ 4 +2= S
$
$\underbrace{2n+2n + 2n+\dots+ 2n}_{\text{n-1 times}} = 2S$
$2n(n-1) = 2S$
$n(n-1) = S$
$S = n(n-1)$
6. Hello,
here's a proof with the sum symbol (looks cool ).
What you are trying to prove is that $\sum_{k=1}^{n - 1} 2k = n(n - 1)$. Let's work with this :
$\sum_{k=1}^{n - 1} 2k$
Note that each term is even, we can factorize ! ( $2 + 4 + 6 = 2(1 + 2 + 3)$). So :
$\sum_{k=1}^{n - 1} 2k = 2 \left ( \sum_{k=1}^{n - 1} k \right )$
Now you may use the fact that $\sum_{k=1}^{n} k = \frac{1}{2}n(n + 1)$, and your result follows. This is a theorem but it can be proven :
$\sum_{k=1}^{n} k = 1 + 2 + 3 + \cdots + k$
1. Assume $n$ is even. | {
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$\sum_{k=1}^{n} k = 1 + 2 + 3 + \cdots + k$
1. Assume $n$ is even.
Rearranging the terms (putting brackets for better comprehension). This is valid because $n$ is even :
$\sum_{k=1}^{n} k = \left [ 1 + \left ( n - 1 \right ) \right ] + \left [ 2 + \left ( n - 2 \right ) \right ] + \cdots + \frac{n}{2} + n$
The dotted part stops when we reach $n - \frac{n}{2}$ (excluded) because we cannot include that term twice. (*)
This can be simplified :
$\sum_{k=1}^{n} k = n + n + \cdots + \frac{n}{2} + n$
Using fact (*), we simplify by multiplication :
$\sum_{k=1}^{n} k = \left ( \frac{n}{2} - 1 \right ) n + \frac{n}{2} + n$
Expanding and simplifying :
$\sum_{k=1}^{n} k = \frac{n^2}{2} - n + \frac{n}{2} + n = \frac{n^2}{2} + \frac{n}{2} = \frac{n^2 + n}{2} = \frac{n(n + 1)}{2} = \frac{1}{2} n(n + 1)$
2. Assume $n$ is odd.
Rearranging the terms (putting brackets for better comprehension). This is valid because $n$ is odd :
$\sum_{k=1}^{n} k = \left [ 1 + n \right ] + \left [ 2 + \left ( n - 1 \right ) \right ] + \cdots + \frac{n + 1}{2}$
The dotted part stops when we reach $n - \frac{n + 1}{2}$ (excluded) because if we included it we would include two terms twice! (*)
This can be simplified :
$\sum_{k=1}^{n} k = (n + 1) + (n + 1) + \cdots + \frac{n + 1}{2}$
Using fact (*), we simplify by multiplication :
$\sum_{k=1}^{n} k = \left ( \frac{n + 1}{2} - 1 \right )(n + 1) + \frac{n + 1}{2}$
Expanding and simplifying (using the trick that $\frac{n + 1}{2} - 1 = \frac{n - 1}{2}$) :
$\sum_{k=1}^{n} k = \left ( \frac{n + 1}{2} - 1 \right )(n + 1) + \frac{n + 1}{2} = \frac{(n - 1)(n + 1)}{2} + \frac{n + 1}{2} =$ $\ \frac{(n - 1)(n + 1) + (n + 1)}{2} = \frac{(n + 1)(n - 1 + 1)}{2} = \frac{1}{2} n(n + 1)$
Using the theorem we just proved (the proof is correct, just a bit messed up, needs better editing), we get that :
$\sum_{k=1}^{n - 1} 2k = 2 \sum_{k=1}^{n - 1} k = 2 \times \left ( \frac{1}{2} (n - 1)(n - 1 + 1) \right ) = n(n - 1)$
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And we can conclude.
Does it make sense ?
EDIT : I just saw Pickslides' proof. Mine seems a bit redundant now
7. Originally Posted by BlackBlaze
Prove that:
2 + 4 + 6 + 8 ... + 2(n-1) = n * (n-1)
When n > 0
Getting the n-1 part is easy (the left side of the equation has to have n-1 terms, so the result would be something multiplied by that). I don't know how to get the n, though...
Well, the sum of n-1 number is n-1 times their average. Since this is an arithmetic sequence, the average of all the numbers is just the average of the smallest and largest: $\frac{2+ 2(n-1)}{2}= 1+ (n- 1)= n$. | {
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# Proving $6^n - 1$ is always divisible by $5$ by induction
I'm trying to prove the following, but can't seem to understand it. Can somebody help?
Prove $6^n - 1$ is always divisible by $5$ for $n \geq 1$.
What I've done:
Base Case: $n = 1$: $6^1 - 1 = 5$, which is divisible by $5$ so TRUE.
Assume true for $n = k$, where $k \geq 1$: $6^k - 1 = 5P$.
Should be true for $n = k + 1$
$6^{k + 1} - 1 = 5Q$
$= 6 \cdot 6^k - 1$
However, I am unsure on where to go from here.
For $n\geq 1$, let $S(n)$ denote the statement $$S(n) : 5\mid(6^n-1)\Longleftrightarrow 6^n-1=5m, m\in\mathbb{Z}.$$ Base case ($n=1$): $S(1)$ says that $5\mid(6^1-1)$, and this is true.
Inductive step: Fix some $k\geq 1$ and assume that $S(k)$ is true where $$S(k) : 5\mid(6^k-1)\Longleftrightarrow 6^k-1=5\ell, \ell\in\mathbb{Z}.$$ To be proved is that $S(k+1)$ follows where $$S(k+1) : 5\mid(6^{k+1}-1)\Longleftrightarrow 6^{k+1}-1=5\eta, \eta\in\mathbb{Z}.$$ Beginning with the left-hand side of $S(k+1)$, \begin{align} 6^{k+1} - 1 &= 6^k\cdot 6-1\tag{by definition}\\[0.5em] &= (5\ell+1)\cdot 6-1\tag{by $S(k)$, the ind. hyp.}\\[0.5em] &= 6\cdot 5\ell+5\tag{expand}\\[0.5em] &= 5(6\ell+1)\tag{factor out $5$}\\[0.5em] &= 5\eta.\tag{$\eta=6\ell+1; \eta\in\mathbb{Z}$} \end{align} we end up at the right-hand side of $S(k+1)$, completing the inductive step.
Thus, by mathematical induction, the statement $S(n)$ is true for all $n\geq 1$. $\blacksquare$ | {
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Thus, by mathematical induction, the statement $S(n)$ is true for all $n\geq 1$. $\blacksquare$
• Perfect! This has definitely made a things alot clearer! – RandomMath Apr 17 '15 at 14:56
• @RandomMath Glad it helped. :) You may want to look at this list for some other useful questions/answers of mine on induction that may prove useful to you. I love induction a little bit too much (uname actually used to be induktio). Most of it is because I, like you, struggled a lot with it. Keep whacking away at induction proofs though, and you'll master them before long. :) – Daniel W. Farlow Apr 17 '15 at 14:58
• I want to mention that any a^k - 1 divisible by a-1 – TigerTV.ru Jan 6 '19 at 20:24
Hint: Inductive step: $$6^{k+1}-1=6\cdot 6^k-1=5\cdot 6^k +(6^k-1)$$
• The inductive step is where I'm confused on on this question, could you elaborate on how you got to this? – RandomMath Apr 17 '15 at 14:35
• @RandomMath Since obviously $5\mid 5\cdot 6^k$ and by the inductive hypothesis $5\mid 6^{k}-1$, can you see that $5\mid 6^{k+1}-1$? And the equalities here only use trivial algebraic manipulations. – user26486 Apr 17 '15 at 14:35
• @RandomMath From your question description it seems you know you want to prove $5\mid 6^{k+1}-1$ assuming $5\mid 6^k-1$. You began by showing $6^{k+1}-1=6\cdot 6^{k}-1$. What I suggest here is continuing the equality with $6\cdot 6^{k}-1=5\cdot 6^{k}+(6^k-1)$, which makes it clear that we indeed have $5\mid 6^{k+1}-1$ (assuming $5\mid 6^{k}-1$). – user26486 Apr 17 '15 at 14:43
• Yes that is what im trying to prove. However, im struggling to get my head around how to get from 6(6^k) -1 to 5*6^k + (6^k -1) – RandomMath Apr 17 '15 at 14:45
• @RandomMath $6\cdot 6^{k}-1=(5+1)\cdot 6^k-1=5\cdot 6^k + 6^k-1=5\cdot 6^k +(6^k-1)$. – user26486 Apr 17 '15 at 14:46
In comments you ask about the source of the following standard proof of the inductive step $$5\mid 6^k-1\ \Rightarrow\ \color{#c00}5\cdot 6^k +(6^k\!-1)\,=\, \color{#0a0}{6^{k+1}-1}$$ | {
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This is a very natural question since such proofs often appear to be pulled out of a hat, like magic. There is, in fact, a good general explanation for their source. Namely such proofs are simply special cases of the proof of the Congruence Product Rule, as we show below.
${\bf Claim}\rm\qquad\ 6\equiv 1,\, 6^{k}\!\equiv 1 \Rightarrow\ 6^{k+1}\!\equiv 1\ \ \ \pmod{\!5},\$ a special case of the following
${\bf Lemma}\rm\quad\ \, A\!\equiv a,\, B\!\equiv b\ \Rightarrow\ AB\equiv ab\ \pmod{\!n}\ \ \$ [Congruence Product Rule]
${\bf Proof}\ \ \ \rm n\mid A\!-\!a,\,\ B-b\,\Rightarrow\, n\mid ( A\!-\!a) B +a\ (B\!-\!b) =A B\,-\,ab$
$\rm\ \ \ \ e.g.\ \ \ 5\mid\ 6\!-\!1,\,\ 6^{k}\!-\!1\ \Rightarrow\ 5\mid(\color{#c00}{6\!-\!1})\,6^{k}+ 1\,(6^{k}\!\!-\!1) = \color{#0a0}{6^{k+1}\!-1}$
Notice that the prior inference is precisely the same as said standard proof of the inductive step. Thus we see that this inference is simply a special case of the proof of the Congruence Product Rule. Once we know this rule, there's no need to repeat the entire proof every time we need to use it. Rather, we can simply invoke the rule as a Lemma (in divisibility form if congruences are not yet known). Then the inductive step has vivid arithmetical structure, being the computation of a product $\, 6\cdot 6^{k}\equiv 6^{(k+1)}.\,$ No longer is the innate arithmetical structure obfuscated by the details of the proof - since the proof has been encapsulated into a Lemma for convenient reuse.
In much the same way, congruences often allow one to impart intuitive arithmetical structure onto complicated inductive proofs - allowing us to reuse our well-honed grade-school skills manipulating arithmetical equations (vs. more complex divisibility relations). Often introduction of congruence language will serve to drastically simplify the induction, e.g. reducing it to a trivial induction such as $\, 1^n\equiv 1,\,$ or $\,(-1)^{2n}\equiv 1.\,$ The former is the essence of the matter above. | {
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• +1 for taking the time and effort to put that up and for clearly wanting to help OP. I hope s/he reads it. – Daniel W. Farlow Apr 17 '15 at 15:46
We can show by induction that $6^k$ has remainder $1$ after division by $5$.
The base case $k=1$ (or $k=0$) is straightforward, since $6=5\cdot 1+1$.
Now suppose that $6^k$ has remainder $1$ after division by $5$ for $k\ge 1$. Thus $6^k = 5\cdot m+1$ for some $m \in \mathbb{N}$. We can then see that $$6^{k+1}=6\cdot 6^{k} = (5+1)(5\cdot m +1) = 5^2 \cdot m + 5 + 5\cdot m + 1$$ $$=5(5\cdot m + m + 1) + 1.$$
Thus $6^{k+1}$ has remainder $1$ after division by $5$.
Therefore for every $k$, we can write $6^k = 5\cdot m +1$ for some $m$.
• Once we have established that every power of $6$ has remainder of $1$ after division by $6$, the rest of the problem follows after subtracting $1$ from $6^k$. – Joel Apr 17 '15 at 14:44
This is the inductive step written out: $$6 \cdot 6^k - 1 = 5 \cdot Q |+1; \cdot \frac{1}{6};-1 \Leftrightarrow 6^k - 1 = \frac{5\cdot Q-5}{6}\underset{P}{\rightarrow}5\cdot P = \frac{5\cdot Q - 5 }{6} | \cdot \frac{1}{5}; \cdot 6\Leftrightarrow Q=6\cdot P + 1$$ $$6^k - 1 = \frac{5\cdot Q-5}{6} \overset{Q}{\rightarrow}\ (6^k-1 = \frac{5\cdot (6\cdot P + 1)-5}{6}\Leftrightarrow 6^k-1 = 5\cdot P)$$
• Please, use $LATEX$. – ً ً Sep 10 '16 at 23:17
• I tried this codecogs.com/latex/eqneditor.php However i am not sure whether this is correct LaTex – Martin Erhardt Sep 10 '16 at 23:18
• Just put $\$$starting from every line and at the end of every line. – ً ً Sep 10 '16 at 23:22 • ah thanks, I am new here – Martin Erhardt Sep 10 '16 at 23:28$6$has a nice property that when raised to any positive integer power, the result will have$6$as its last digit. Therefore, that number minus$1$is going to have$5$as its last digit and thus be divisible by$5\$.
• OP wants to prove this by induction – Shailesh Sep 10 '16 at 23:56
• it's works for another numbers not only for 6. – TigerTV.ru Jan 6 '19 at 20:52 | {
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# Documentation
### This is machine translation
Translated by
Mouse over text to see original. Click the button below to return to the English verison of the page.
# istril
Determine if matrix is lower triangular
## Description
example
tf = istril(A) returns logical 1 (true) if A is a lower triangular matrix; otherwise, it returns logical 0 (false).
## Examples
collapse all
Create a 5-by-5 matrix.
D = tril(magic(5))
D =
17 0 0 0 0
23 5 0 0 0
4 6 13 0 0
10 12 19 21 0
11 18 25 2 9
Test D to see if it is lower triangular.
istril(D)
ans =
logical
1
The result is logical 1 (true) because all elements above the main diagonal are zero.
Create a 5-by-5 matrix of zeros.
Z = zeros(5);
Test Z to see if it is lower triangular.
istril(Z)
ans =
logical
1
The result is logical 1 (true) because a lower triangular matrix can have any number of zeros on its main diagonal.
## Input Arguments
collapse all
Input array, specified as a numeric array. istril returns logical 0 (false) if A has more than two dimensions.
Data Types: single | double
Complex Number Support: Yes
collapse all
### Lower Triangular Matrix
A matrix is lower triangular if all elements above the main diagonal are zero. Any number of the elements on the main diagonal can also be zero.
For example, the matrix
$A=\left(\begin{array}{cccc}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& 0\\ -1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& 0\\ -2& -2& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& 0\\ -3& -3& -3& 1\end{array}\right)$
is lower triangular. A diagonal matrix is both upper and lower triangular.
### Tips | {
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is lower triangular. A diagonal matrix is both upper and lower triangular.
### Tips
• Use the tril function to produce lower triangular matrices for which istril returns logical 1 (true).
• The functions isdiag, istriu, and istril are special cases of the function isbanded, which can perform all of the same tests with suitably defined upper and lower bandwidths. For example, istril(A) == isbanded(A,size(A,1),0). | {
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# piecewise
Conditionally defined expression or function
## Description
example
pw = piecewise(cond1,val1,cond2,val2,...) returns the piecewise expression or function pw whose value is val1 when condition cond1 is true, is val2 when cond2 is true, and so on. If no condition is true, the value of pw is NaN.
example
pw = piecewise(cond1,val1,cond2,val2,...,otherwiseVal) returns the piecewise expression or function pw that has the value otherwiseVal if no condition is true.
## Examples
### Define and Evaluate Piecewise Expression
Define the following piecewise expression by using piecewise.
$y=\left\{\begin{array}{cc}-1& x<0\\ 1& x>0\end{array}$
syms x
y = piecewise(x<0, -1, x>0, 1)
y =
piecewise(x < 0, -1, 0 < x, 1)
Evaluate y at -2, 0, and 2 by using subs to substitute for x. Because y is undefined at x = 0, the value is NaN.
subs(y, x, [-2 0 2])
ans =
[ -1, NaN, 1]
### Define Piecewise Function
Define the following function symbolically.
$y\left(x\right)=\left\{\begin{array}{cc}-1& x<0\\ 1& x>0\end{array}$
syms y(x)
y(x) = piecewise(x<0, -1, x>0, 1)
y(x) =
piecewise(x < 0, -1, 0 < x, 1)
Because y(x) is a symbolic function, you can directly evaluate it for values of x. Evaluate y(x) at -2, 0, and 2. Because y(x) is undefined at x = 0, the value is NaN. For details, see Create Symbolic Functions.
y([-2 0 2])
ans =
[ -1, NaN, 1]
### Set Value When No Conditions Is True
Set the value of a piecewise function when no condition is true (called otherwise value) by specifying an additional input argument. If an additional argument is not specified, the default otherwise value of the function is NaN.
Define the piecewise function
$y\left(x\right)=\left\{\begin{array}{cc}-2& x<-2\\ 0& -2
syms y(x)
y(x) = piecewise(x<-2, -2, -2<x<0, 0, 1)
y(x) =
piecewise(x < -2, -2, x in Dom::Interval(-2, 0), 0, 1)
Evaluate y(x) between -3 and 1 by generating values of x using linspace. At -2 and 0, y(x) evaluates to 1 because the other conditions are not true. | {
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xvalues = linspace(-3,1,5)
yvalues = y(xvalues)
xvalues =
-3 -2 -1 0 1
yvalues =
[ -2, 1, 0, 1, 1]
### Plot Piecewise Expression
Plot the following piecewise expression by using fplot.
$y=\left\{\begin{array}{cc}-2& x<-2\\ x& -22\end{array}.$
syms x
y = piecewise(x<-2, -2, -2<x<2, x, x>2, 2);
fplot(y)
### Assumptions and Piecewise Expressions
On creation, a piecewise expression applies existing assumptions. Apply assumptions set after creating the piecewise expression by using simplify on the expression.
Assume x > 0. Then define a piecewise expression with the same condition x > 0. piecewise automatically applies the assumption to simplify the condition.
syms x
assume(x > 0)
pw = piecewise(x<0, -1, x>0, 1)
pw =
1
Clear the assumption on x for further computations.
assume(x,'clear')
Create a piecewise expression pw with the condition x > 0. Then set the assumption that x > 0. Apply the assumption to pw by using simplify.
pw = piecewise(x<0, -1, x>0, 1);
assume(x > 0)
pw = simplify(pw)
pw =
1
Clear the assumption on x for further computations.
assume(x, 'clear')
### Differentiate, Integrate, and Find Limits of Piecewise Expression
Differentiate, integrate, and find limits of a piecewise expression by using diff, int, and limit respectively.
Differentiate the following piecewise expression by using diff.
$y=\left\{\begin{array}{cc}1/x& x<-1\\ \mathrm{sin}\left(x\right)/x& x\ge -1\end{array}$
syms x
y = piecewise(x<-1, 1/x, x>=-1, sin(x)/x);
diffy = diff(y, x)
diffy =
piecewise(x < -1, -1/x^2, -1 < x, cos(x)/x - sin(x)/x^2)
Integrate y by using int.
inty = int(y, x)
inty =
piecewise(x < -1, log(x), -1 <= x, sinint(x))
Find the limits of y at 0 and -1 by using limit. Because limit finds the double-sided limit, the piecewise expression must be defined from both sides. Alternatively, you can find the right- or left-sided limit. For details, see limit. | {
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limit(y, x, 0)
limit(y, x, -1)
ans =
1
ans =
limit(piecewise(x < -1, 1/x, -1 < x, sin(x)/x), x, -1)
Because the two conditions meet at -1, the limits from both sides differ and limit cannot find a double-sided limit.
### Elementary Operations on Piecewise Expressions
Add, subtract, divide, and multiply two piecewise expressions. The resulting piecewise expression is only defined where the initial piecewise expressions are defined.
syms x
pw1 = piecewise(x<-1, -1, x>=-1, 1);
pw2 = piecewise(x<0, -2, x>=0, 2);
add = pw1 + pw2
sub = pw1 - pw2
mul = pw1 * pw2
div = pw1 / pw2
piecewise(x < -1, -3, x in Dom::Interval([-1], 0), -1, 0 <= x, 3)
sub =
piecewise(x < -1, 1, x in Dom::Interval([-1], 0), 3, 0 <= x, -1)
mul =
piecewise(x < -1, 2, x in Dom::Interval([-1], 0), -2, 0 <= x, 2)
div =
piecewise(x < -1, 1/2, x in Dom::Interval([-1], 0), -1/2, 0 <= x, 1/2)
### Modify or Extend Piecewise Expression
Modify a piecewise expression by replacing part of the expression using subs. Extend a piecewise expression by specifying the expression as the otherwise value of a new piecewise expression. This action combines the two piecewise expressions. piecewise does not check for overlapping or conflicting conditions. Instead, like an if-else ladder, piecewise returns the value for the first true condition.
Change the condition x<2 in a piecewise expression to x<0 by using subs.
syms x
pw = piecewise(x<2, -1, x>0, 1);
pw = subs(pw, x<2, x<0)
pw =
piecewise(x < 0, -1, 0 < x, 1)
Add the condition x>5 with the value 1/x to pw by creating a new piecewise expression with pw as the otherwise value.
pw = piecewise(x>5, 1/x, pw)
pw =
piecewise(5 < x, 1/x, x < 0, -1, 0 < x, 1)
## Input Arguments
collapse all
Condition, specified as a symbolic condition or variable. A symbolic variable represents an unknown condition.
Example: x > 2 | {
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Example: x > 2
Value when condition is satisfied, specified as a number, vector, matrix, or multidimensional array, or as a symbolic number, variable, vector, matrix, multidimensional array, function, or expression.
Value if no conditions are true, specified as a number, vector, matrix, or multidimensional array, or as a symbolic number, variable, vector, matrix, multidimensional array, function, or expression. If otherwiseVal is not specified, its value is NaN.
## Output Arguments
collapse all
Piecewise expression or function, returned as a symbolic expression or function. The value of pw is the value val of the first condition cond that is true. To find the value of pw, use subs to substitute for variables in pw.
## Tips
• piecewise does not check for overlapping or conflicting conditions. A piecewise expression returns the value of the first true condition and disregards any following true expressions. Thus, piecewise mimics an if-else ladder. | {
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# How to create a function $f$ such that $f(x,y)$ is high when either $x$ or $y$ is?
There are two variables, let's say $$x$$ and $$y$$.
I want to come up with a function $$f:[0,5]\times[0,5]\longrightarrow[0,5]$$ that respects the following rules:
1. If $$x$$ is high (close to the maximum value of 5) and $$y$$ is low (close to $$0$$), $$f(x,y)$$ should results in a high value (close to 5). The same would happen for the reverse ($$x$$ with a low value and $$y$$ with a high value)
2. If $$x$$ is high and $$y$$ is high, $$f(x,y)$$ will also be high; it would be useful that the values resulted would be higher than those from point $$(1)$$.
3. If $$x$$ is low and $$y$$ is low, $$f(x,y)$$ should result in low values.
I'm having troubles starting imagining what mathematical functions would be useful, as I am not even close to being advanced in the concepts of Mathematics. Any ideas would be appreciated, at least in the sense of finding any information that could get me started.
• you defined your function $f :[0,5]\rightarrow [0,5]$, and then you talk about two input variable $x$ and $y$, i guess you meant $f :[0,5]\times [0,5] \rightarrow [0,5]$ ? Aug 11 at 11:47
• that is correct, I'll add the change. Aug 11 at 11:48
• If you have precise value for your function $f$, you could use something like multivariate interpolation. Aug 11 at 11:52
• If you don't know the concept of interpolation you should check out first 1D interpolation like this one. Aug 11 at 11:57
• If $f(x,y)=f(y,x)$, and $f(x,y)$ is linear with respect to $x$ (this was not part of the problem statement, it was added by me to ensure uniqueness of solution), then $f(x,y)=axy+b(x+y)+c$. Let $f(0,0)=0$, $f(5,5)=5$, then $c=0$, $5a+2b=1$. If one wants to set value $f(0,5)=d$, then $b=\frac{d}{5}$, $a=\frac{5-2d}{25}$. At $d=5$ result is formula from my previous comment, but OP can take $f(0,5)=d=4.9$ or something else what needed. Aug 12 at 7:14 | {
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You can take, for instance,$$f(x,y)=\frac45\max\{x,y\}+\frac15\min\{x,y\}.$$It has all the properties that you are interested in.
One way to model this could be to use the distance from the line: $$x+y=0$$ through $$(0,0)$$. This line has normal vector: $$\vec n= \begin{pmatrix} 1\\ 1 \end{pmatrix}$$ and the distance of a given point $$(x,y)$$ to this line is proportional to $$t=\vec n\cdot\langle x,y \rangle=x+y$$ so in case we want something like: $$f(5,5)=5\\ f(5,0)=4\\ f(0,0)=0$$ we can connect this to a single dimensional function: $$f(x,y)=g(x+y)$$ where $$g(10)=5\\ g(5)=4\\ g(0)=0$$ A way to achieve this could be to add the extra requirement $$g'(10)=0$$ so that $$f$$ has maximum at $$(x,y)=(5,5)$$ and build $$g(0)=0$$ in: $$g(t)=at^3+bt^2+ct$$ Hence \begin{align} g(10)&=1000a+100b+10c&&=5\\ g(5)&=125a+25b+5c&&=4\\ g'(10)&=300a+20b+c&&=0 \end{align} which can be solved for $$a,b,c$$ to have: $$f(x,y)=g(x+y)=0.002(x+y)^3-0.09(x+y)^2+1.2(x+y)$$
See this link to look at interactive GeoGebra-applet with 3D-plot of this function | {
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See this link to look at interactive GeoGebra-applet with 3D-plot of this function
ADDENDUM: As can be seen both from the other answer and from comments, there will be (infinitely) many ways to satisfy your requirements, but to point you towards handling the additional requirement stated in your comment below this post, you could simply add a modifier to the above solution which takes the distance to the perpendicular line: $$x-y=0$$ as input. This distance is (similarly) proportional to $$t=x-y$$, and so we need a modifier function $$m(x,y)=h(x-y)=h(t)$$ that satisfies: $$h(0)=0\\ h(5)=m(5,0)\\ h(-5)=m(0,5)$$ so just choose which modification you want at $$(5,0)$$ and $$(0,5)$$ and match for instance a quadratic function as $$h$$: $$h(t)=\alpha t^2+\beta t+\gamma$$ and combine: \begin{align} q(x,y) &=f(x,y)+m(x,y)\\ &=g(x+y)+h(x-y)\\ &=a(x+y)^3+b(x+y)^2+c(x+y)+\alpha(x-y)^2+\beta(x-y)+\gamma \end{align} but be a little careful - if $$h(t)$$ increases too rapidly away from $$h(0)$$ to one side, then $$q$$ may exceed a value of $$5$$.
Here is GeoGebra-applet with example of this technique
• Very interesting and detailed answer. With the risk of going a bit outside the boundaries of the question: is there a way to introduce some kind of bias towards $x$ or $y$? In the sense that $f(x,y)$ should be greater than $f(y,x)$ or the opposite. Aug 16 at 13:06
• @Ionut-AlexandruBaltariu: I added an extra section about this. Generally, you can always combine two one-dimensional functions $g(x+y)$ and $h(x-y)$ to create formula for change in value when you move perpendicular to the two lines $x+y=0$ and $x-y=0$. This gives you a lot of control over what you want to happen. Just be careful, because combinations may escape the max/min values, since we only controlled them in points $(0,0),(5,0),(0,5)$ and $(5,5)$. Aug 17 at 8:41 | {
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Card Draw Probability
This is a homework question. Not something to hand in, but rather a recommended practice question.
A Card Game
Three students are playing a card game. They decide to choose the first person to play by each selecting a card from the 52-card deck and looking for the highest card in value and suit. They rank the suits from lowest to highest: clubs, diamonds, hearts, and spades.
a. If the card is replaced in the deck after each student chooses, how many possible configurations of the three choices are possible?
b. How many configurations are there in which each student picks a different card?
c. What is the probability that all three students pick exactly the same card?
d. What is the probability that all three students pick different cards?
I figured out the the answer to part A is $52^3 = 140608$, since each student chooses 1 card from a deck of 52.
I also know that the answer to part B is $52\times 51\times 50 =132600$, or $\frac {52!}{(52-3)!}$.
However, I am not sure how to find the probability that they all pick the same card. I initially thought it would be $\left( \frac 1{52}\right)^3$, but that does not match the answer in the back of the book, which is $0.00037$.
Sorry for the formatting (or lack thereof), I tried my best to make it fairly easy to read, but I am unsure of how to make fancy fractions and such. | {
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• For $c$: the first person takes whatever. The probability that the second matches the first is $\frac 1{52}$. The probability that the third matches the first is $\frac 1{52}$ independent of what the second person drew. So... – lulu Jun 11 '17 at 20:54
• An answer of $(1/52)\cdot (1/52)\cdot (1/52)=(1/52)^3$ is the probability of all three specifically selecting the ace of spades, but remember there are many ways other than just everyone getting the ace of spades for them to all have the same card. As for part (d), if you were to read and understand the birthday problem you'll find the mathematics behind it is very similar. Alternatively, look more closely at the phrasing of parts (b) and (a) and how they relate to question (d). – JMoravitz Jun 11 '17 at 21:00
• So (1/52)^2. Thank you! Are you going to post it as an answer so that I can accept it? – Bunyip Jun 11 '17 at 21:02
• As a complete aside, you say "I am unsure of how to make fancy fractions and such." This page will give you the information you need to do so (at least on this site using MathJax). – JMoravitz Jun 11 '17 at 21:03
For part C:
Try thinking about it like this, player A is the one that "chooses" which card B and C must choose in order for all of them to pick the same card, so the probability that all three students pick exactly the same card is:
$\ \frac{1}{52}\frac{1}{52} = (\frac{1}{52})^2$
For part D:
Start as in C, player A chooses the card and B has a $\ \frac{51}{52}$ chance of its card being different from A's card, similarly C has a $\ \frac{50}{52}$ of its card being different from A and B, so the probability that all cards differ is:
$\ \frac{51}{52}\frac{50}{52}$
Hint.
Part c only makes sense if the cards are drawn with replacement. You already have the total number of configurations. There are 52 configurations where all the cards are the same. With these two facts, the required probability is readily calculated. | {
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For part d, which again only makes sense if cards are drawn with replacement, you can get the number of ways in which all three cards are different from the answer to part b. This will lead to the required probability. | {
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# Determinant of antidiagonally constructed matrix
Let $$A_n$$ be a matrix, odd dimension $$n \times n$$, constructed from the sequence of natural numbers in such a way that we begin the sequence from the upper left corner and next numbers are inserted into the matrix along antidiagonal direction, starting from the top of the matrix (examples below) .
For such matrix the determinant is calculated.
Surprisingly for the three examples below the value of determinant is equal to the central entry of the matrix with the modification of sign between two consecutive cases.
$$\det \begin{bmatrix}1 & 2 & 4\\3 &\color{red} 5 & 7\\6 & 8 & 9\end{bmatrix} =-\color{red}5$$
$$\det \begin{bmatrix}1 & 2 & 4 & 7 & 11\\3 & 5 & 8 & 12 & 16\\6 & 9 & \color{red}{13} & 17 & 20\\10 & 14 & 18 & 21 & 23\\15 & 19 & 22 & 24 & 25\end{bmatrix} =\color{red}{13}$$
$$\det \begin{bmatrix}1 & 2 & 4 & 7 & 11 & 16 & 22\\3 & 5 & 8 & 12 & 17 & 23 & 29\\6 & 9 & 13 & 18 & 24 & 30 & 35\\10 & 14 & 19 & \color{red}{25} & 31 & 36 & 40\\15 & 20 & 26 & 32 & 37 & 41 & 44\\21 & 27 & 33 & 38 & 42 & 45 & 47\\28 & 34 & 39 & 43 & 46 & 48 & 49\end{bmatrix} = -\color{red}{25}$$
Question:
• does the pattern continue for a greater $$n$$?
• if so what is the explanation for the pattern? | {
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• does the pattern continue for a greater $$n$$?
• if so what is the explanation for the pattern?
• Note that the pattern holds for the often overlooked $1\times 1$ matrices as well: The determinant of $[\color{red}{1}]$ is $\color{red}{1}$. Apr 9, 2019 at 11:11
• @Arthur Yes, 1 is also "central" in [1], and it has + sign, this change of sign is also strange.. Apr 9, 2019 at 11:13
• Here's how it continues: oeis.org/A069480. Your observation would make a nice comment, BTW. Apr 9, 2019 at 11:27
• The formula section of the OEIS link does confirm this: It says that for odd $n$, the determinant is indeed an alternating multiplied by $\frac{(2n+1)^2}{2}$, which is the middle element of the corresponding matrix.. Apr 9, 2019 at 11:34
• @Widawensen In this case, the easiest determinant to start with first might be the one which has central value $0$: we just have to show that this matrix is singular. But the fact that as you add $k$ to all entries, the determinant changes by $\pm k$, might also be hard to prove... Apr 9, 2019 at 22:28
Let $$M_n$$ be the matrix whose determinant we want to know (the matrix where we fill in $$n^2$$ consecutive integers antidiagonally, starting at any value $$k$$), and let $$x$$ be its middle entry. For convenience, let $$m = \frac{n-1}{2}$$. (I will use $$n=7$$ in my example matrices, but my argument applies to all odd $$n$$.) | {
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First, replace the rows $$r_1, r_2, \dots, r_n$$ of $$M_n$$ with the $$n-1$$ differences $$r_2 - r_1$$, $$r_3 - r_2$$, ..., $$r_n - r_{n-1}$$ followed by the middle row $$r_{m+1}$$. This is an invertible sequence of row operations, so we have: $$\det(M_n) = \det\begin{bmatrix} 2 & 3 & 4 & 5 & 6 & 7 & 7 \\ 3 & 4 & 5 & 6 & 7 & 7 & 6 \\ 4 & 5 & 6 & 7 & 7 & 6 & 5 \\ 5 & 6 & 7 & 7 & 6 & 5 & 4 \\ 6 & 7 & 7 & 6 & 5 & 4 & 3 \\ 7 & 7 & 6 & 5 & 4 & 3 & 2 \\ x+a_1 & x+a_2 & x+a_3 & x & x-a_3 & x-a_2 & x-a_1 \end{bmatrix}.$$ where $$x+a_1,\dots, x+a_m, x, x-a_m, \dots, x-a_1$$ are the entries of the middle row of $$M_n$$ (which always has this symmetric form). We show that the determinant above is always equal to $$(-1)^{m} x$$.
By linearity of the determinant, we can write this as $$\det(M_n) = \det\begin{bmatrix} 2 & 3 & 4 & 5 & 6 & 7 & 7 \\ 3 & 4 & 5 & 6 & 7 & 7 & 6 \\ 4 & 5 & 6 & 7 & 7 & 6 & 5 \\ 5 & 6 & 7 & 7 & 6 & 5 & 4 \\ 6 & 7 & 7 & 6 & 5 & 4 & 3 \\ 7 & 7 & 6 & 5 & 4 & 3 & 2 \\ a_1 & a_2 & a_3 & 0 & -a_3 & -a_2 & -a_1 \end{bmatrix} + x \det\begin{bmatrix} 2 & 3 & 4 & 5 & 6 & 7 & 7 \\ 3 & 4 & 5 & 6 & 7 & 7 & 6 \\ 4 & 5 & 6 & 7 & 7 & 6 & 5 \\ 5 & 6 & 7 & 7 & 6 & 5 & 4 \\ 6 & 7 & 7 & 6 & 5 & 4 & 3 \\ 7 & 7 & 6 & 5 & 4 & 3 & 2 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 \end{bmatrix}.$$ The first determinant is $$0$$ by cofactor expansion along the last row. The cofactors of $$a_i$$ and $$-a_i$$ are equal for each $$i$$, since one submatrix can be obtained from the other by reversing the rows, then reversing the columns, so they cancel after multiplying one by $$a_i$$ and the other by $$-a_i$$. | {
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For the second matrix, we first do a similar operation to columns. Keeping the first column $$c_1$$ and replacing every other $$i^{\text{th}}$$ column $$c_i$$ by the difference $$c_i - c_{i-1}$$, which is an invertible set of row operations, we have $$\det\begin{bmatrix} 2 & 3 & 4 & 5 & 6 & 7 & 7 \\ 3 & 4 & 5 & 6 & 7 & 7 & 6 \\ 4 & 5 & 6 & 7 & 7 & 6 & 5 \\ 5 & 6 & 7 & 7 & 6 & 5 & 4 \\ 6 & 7 & 7 & 6 & 5 & 4 & 3 \\ 7 & 7 & 6 & 5 & 4 & 3 & 2 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 \end{bmatrix} = \det\begin{bmatrix} 2 & 1 & 1 & 1 & 1 & 1 & 0 \\ 3 & 1 & 1 & 1 & 1 & 0 & -1 \\ 4 & 1 & 1 & 1 & 0 & -1 & -1 \\ 5 & 1 & 1 & 0 & -1 & -1 & -1 \\ 6 & 1 & 0 & -1 & -1 & -1 & -1 \\ 7 & 0 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$ By expansion along the last row (which introduces a multiple of $$(-1)^{n+1} = 1$$) and after reversing the rows (which introduces a multiple of $$(-1)^{m}$$) we simplify this to $$(-1)^{m}\det\begin{bmatrix} 0 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & -1 & -1 & -1 & -1 \\ 1 & 1 & 0 & -1 & -1 & -1 \\ 1 & 1 & 1 & 0 & -1 & -1 \\ 1 & 1 & 1 & 1 & 0 & -1 \\ 1 & 1 & 1 & 1 & 1 & 0 \end{bmatrix}.$$ Now things get a bit fancier. This $$n-1 \times n-1 = 2m \times 2m$$ matrix (call it $$A$$) is a skew-symmetric matrix, so its determinant is the square of its Pfaffian. We use the simplified formula given in the Wikipedia link above: the sum $$\text{pf}(A) = \sum_{\alpha \in \Pi} \text{sgn}(\pi_\alpha) a_{i_1j_1} a_{i_2j_2}\dotsm a_{i_mj_m}$$ where $$\alpha$$ runs over all pairings $$\{(i_1,j_1), \dots, (i_m, j_m)\}$$ with $$i_k < j_k$$ in each pair and $$i_1 < \dots < i_m$$; $$\pi_{\alpha}$$ is the permutation with $$\pi_\alpha(2k-1) = i_k$$ and $$\pi_\alpha(2k) = j_k$$. For our particular matrix, $$a_{i_kj_k}$$ is always $$-1$$, so $$\text{pf}(A) = (-1)^m \sum_{\alpha \in \Pi} \text{sgn}(\pi_\alpha).$$ For each pairing $$\alpha$$, if we find the first index $$k$$ where $$j_k > i_{k+1}$$ and switch $$j_k$$ with $$j_{k+1}$$, we get another pairing $$\beta$$ | {
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$$k$$ where $$j_k > i_{k+1}$$ and switch $$j_k$$ with $$j_{k+1}$$, we get another pairing $$\beta$$ with $$\text{sgn}(\pi_\beta) = -\text{sgn}(\pi_\alpha)$$; if we apply the same operation to $$\beta$$, we get $$\alpha$$ back. So this splits up the pairings into, uh, pairs of pairings whose terms in the Pfaffian sum cancel; the exception is the pairing $$\{(1,2), (3,4), \dots, (2m-1,2m)\}$$, for which no index $$k$$ exists. So $$\text{pf}(A) = (-1)^m$$ after all the cancellations, and therefore $$\det(A) = 1$$. | {
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So we get $$\det(M_n) = 0 + x \cdot (-1)^{m} = (-1)^{\frac{n-1}{2}}x$$: plus or minus the middle entry of $$M_n$$.
• I'm very grateful for your partial answer - it has given me some new ideas to rethink... Apr 10, 2019 at 8:21
• I wonder whether in the case of the second determinat the same technique applied by you with substracting the rows could not lead to the further simplification.. Apr 10, 2019 at 8:59
• Anyway I have received now {{-1, -1, -1, -1, -1, 0, 1}, {-1, -1, -1, -1, 0 ,1 ,1}, {-1, -1 , -1 , 0 ,1, 1 ,1}, {-1, -1, 0 ,1 ,1 ,1, 1}, {-1, 0 , 1 ,1 ,1, 1, 1}, {5, 5, 4 ,3, 2 ,1, 0}, {1, 1, 1, 1, 1, 1, 1}} at which I'm stacked.. Apr 10, 2019 at 9:25
• I have also received for $5 \times 5$ such form $\begin{equation*}\left[\begin{matrix}0 & 0 & 0 & 1 & 2\\0 & 0 & 1 & 2 & 2\\0 & 1 & 2 & 2 & 2\\3 & 3 & 2 & 1 & 0\\1 & 1 & 1 & 1 & 1\end{matrix}\right]\end{equation*}$ which has det =$-1 \ \$. Interestingly $\begin{equation*}\left[\begin{matrix}0 & 0 & 0 & 1 & 2\\0 & 0 & 1 & 2 & 2\\0 & 1 & 2 & 2 & 2\\4 & 3 & 2 & 1 & 0\\1 & 1 & 1 & 1 & 1\end{matrix}\right]\end{equation*}$ has det = $0$. Apr 10, 2019 at 11:53
• Now I have a complete answer. Apr 10, 2019 at 22:11 | {
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# Surjective and Injective function
Let $$N=\{1,2,3...\}$$ be the set of natural numbers and $$F:N \times N \rightarrow N$$ be such that $$f(m,n)=(2m-1)*2^n.$$
(A)F is Injective.
(B)F is Surjective.
(C)F is Bijective.
(D)None of the above.
I can see that F can never be surjective because 1 does not have a pre-image.
And it seems injective to me because $$2m-1$$ term would always be odd, $$2^n$$ term would always be even, and hence the product will always be Even, and for different values of m and n, we would get a different even number.
Is my Reasoning correct?
• You're on the correct track, but your reasoning is too vague to be a proof. You need to start with an assumption $f(m_1, n_1) = f(m_2, n_2)$ and develop your even-odd argument a bit more. – T. Bongers Oct 19 '18 at 17:26
Overkill?
A) $$f$$ is injective.
Let $$f(m,n)=f(k,l)$$, i.e.
$$i:=(2m-1)2^n= (2k-1)2^l.$$
The positive integer $$i$$ has a - Fundamental Theorem of Arithmetic - unique prime factorization.
Since $$(2m-1)$$, $$(2k-1)$$ are odd,
we have $$n=l$$.
Then
$$(2m-1)2^n= (2k-1)2^l$$ implies
$$2m-1=2k-1$$, or $$m=k.$$
Combining
$$f(m,n)=f(k,l)$$ implies $$m=k$$, and $$n=l.$$
https://en.m.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic
• "Overkill?" Perhaps. T. Bongers answer is probably a lot easier, but your answer does capture the thought process that the OP was considering. (i.e. that the results have an "even part" and an "odd part" and ... so on.) – fleablood Oct 19 '18 at 18:27
• fleablood.Thanks for your comment. – Peter Szilas Oct 19 '18 at 19:24
To expand my comment, you need to make the reasoning a lot more precise. Start with an assumption that there are natural numbers $$m_1, m_2, n_1, n_2$$ for which
$$f(m_1, n_1) = f(m_2, n_2).$$
Then you have
$$(2m_1 - 1)2^{n_1} = (2m_2 - 1) 2^{n_2}.$$
Now we can assume without loss of generality that $$n_1 \le n_2$$, so that
$$2m_1 - 1 = (2m_2 - 1)2^{n_2 - n_1}.$$ | {
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$$2m_1 - 1 = (2m_2 - 1)2^{n_2 - n_1}.$$
Now the left side is odd, so the right side is odd too; thus, $$n_2 - n_1 = 0$$ (why?). But then we get $$2m_1 - 1 = 2m_2 - 1$$, hence $$m_1 = m_2$$. This completes the proof.
I can see that F can never be surjective because 1 does not have a pre-image.
Let's prove that.
If $$f(m,n) = (2m -1)*2^n = 1$$ then as $$2m-1$$ and $$2^n$$ are integer factors of $$1$$ and the only integer factors of $$1$$ are $$\pm 1$$ so $$2^n = \pm 1$$ and that is only possible if $$n = 0$$ which is not possible as $$0 \not \in \mathbb N$$.
So it is not surjective.
(In fact for any odd number $$2k - 1$$ then $$f(m,n) = (2m-1)*2^n = 2k-1$$ would be impossible. $$f(m,n)$$ will always be even.)
And it seems injective to me because 2m−1 term would always be odd, 2n term would always be even, and hence the product will always be Even,
Which shows it can not be surjective
and for different values of m and n, we would get a different even number.
That's a bit of a leap. How do you know they will be different?
Let's prove it.
$$f(m,n) = (2m-1)*2^n$$ will have a unique prime factorization. As $$2m-1$$ is odd the power of $$2$$ of any value of $$f(m,n)= (2m-1) 2^n$$ will be $$n$$.
So if $$n_1 \ne n_2$$ then $$f(a,n_2) = (2a - 1)2^{n_2} \ne (2b- 1)2^{n_1} = f(b,n_2)$$ for any possible $$a,b$$. (Because the power of $$2$$ in the prime factorizations of those two numbers are different.)
And if $$m_1 \ne m_2$$ then $$2m_1 - 1 \ne 2m_2 - 1$$ and the prime factorizations of $$(2m_1 - 1)*2^a \ne (2m_2 - 1)*2^b$$ for any possible $$a,b$$ must be different for the different odd factors.
So if $$(m_1, n_1) \ne (m_2 n_2)$$ then either $$n_1 \ne n_2$$ or $$m_1 \ne m_2$$ (or both). In either case then $$(2m_1-1)2^{n_1} \ne (2m_2-1)2^{n_2}$$.
So $$f$$ is injective.
A) is true, B) is false, C) is false (as bijective implies surjective) and D) is false (it is one of the above). | {
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# Math Help - differential equation
1. ## differential equation
If we assume that a person's weight depends on the energy consumed minus energy ued, one model is that weight changes in proportion to the difference, so:
dw/dt = k(C - 38.5w) where w(0) = w subscript 0
where w(t) kg is the weight at t days, C is the daily calorie intake, and we assume 38.5 calories per kg per day are used
(i) if you wish to maintain a constant weight of 82 kg, what should be your daily calorie intake C?
(ii) If you weigh 100 kg, and you want to lose 10 kg in a month, what should C be? (Assume k = 1.3 * 10^(-4) kg/calorie and a month of 30 days) Is this result healthy?
can someone solve these questions for me, they are on a past test paper which does not have solutions available.
2. Originally Posted by razorfever
If we assume that a person's weight depends on the energy consumed minus energy ued, one model is that weight changes in proportion to the difference, so:
dw/dt = k(C - 38.5w) where w(0) = w subscript 0
where w(t) kg is the weight at t days, C is the daily calorie intake, and we assume 38.5 calories per kg per day are used
(i) if you wish to maintain a constant weight of 82 kg, what should be your daily calorie intake C?
(ii) If you weigh 100 kg, and you want to lose 10 kg in a month, what should C be? (Assume k = 1.3 * 10^(-4) kg/calorie and a month of 30 days) Is this result healthy?
can someone solve these questions for me, they are on a past test paper which does not have solutions available.
(i) $\frac{dw}{dt} = 0$ ... solve for $C$
(ii) solve $\int \frac{dw}{C - 38.5w} = \int k \, dt$ with initial condition $w(0) = 100$.
after you find $w(t)$, set $w(30) = 90$ and solve for $C$. | {
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after you find $w(t)$, set $w(30) = 90$ and solve for $C$.
3. can u help me out the same way with that kind of template for some of my other posts about differential equation word problems?
I have an exam coming up and my prof hasn't touched any word problems,
if you could provide templates, i can understand them better and hopefully pass the exam
http://www.mathhelpforum.com/math-he...tric-func.html
http://www.mathhelpforum.com/math-he...d-problem.html
http://www.mathhelpforum.com/math-he...-diver-de.html
http://www.mathhelpforum.com/math-he...ates-flow.html
http://www.mathhelpforum.com/math-he...72473-ivp.html
http://www.mathhelpforum.com/math-he...ntial-eqn.html
4. can you tell me what your final answer for C is?
i'm getting 102375.57
as for the constant in part (i) i'm getting (C - 3850)
and the expression for C in part (ii) i'm getting
[3465 - 3850e^(30k)] / [1 - e^(30k)]
is this correct
can someone show me the correct solution step by step?
5. $\int \frac{dw}{C - 38.5w} = \int k \, dt$
$\int \frac{-38.5}{C - 38.5w} \, dw = -38.5\int k \, dt$
$\ln|C - 38.5w| = -38.5kt + A$
$C - 38.5w = Be^{-38.5kt}$
$w(0) = 100$
$C - 3850 = B$
$C - 38.5w = (C - 3850)e^{-38.5kt}$
$C - 38.5w = Ce^{-38.5kt} - 3850e^{-38.5kt}$
$C - Ce^{-38.5kt} = 38.5w - 3850e^{-38.5kt}$
$C(1 - e^{-38.5kt}) = 38.5w - 3850e^{-38.5kt}$
$C = \frac{38.5w - 3850e^{-38.5kt}}{1 - e^{-38.5kt}}$
$w(30) = 90$ ... $C \approx 1090 \, cal$
6. now i'm getting C = 3515
how do you do that last part w(30) = 90
do u rearrange and make w the subject then put t=30 and solve for C right?
can u show me that last step explicitly??
7. Originally Posted by razorfever
now i'm getting C = 3515
how do you do that last part w(30) = 90
do u rearrange and make w the subject then put t=30 and solve for C right?
can u show me that last step explicitly??
since the task was to find C, I solved for C ... sub in 90 for w, 30 for t, and the given value for k (0.00013 , correct?) | {
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8. yeah i just realized i was making a careless mistake
i wasn't putting E in my calculator , instead just getting the value of
(-38.5 * 0.00013 * 30) and this was giving me C as 3515 but I got it now
9. Originally Posted by razorfever
can u help me out the same way with that kind of template for some of my other posts about differential equation word problems?
I have an exam coming up and my prof hasn't touched any word problems,
if you could provide templates, i can understand them better and hopefully pass the exam
http://www.mathhelpforum.com/math-he...tric-func.html
http://www.mathhelpforum.com/math-he...d-problem.html
http://www.mathhelpforum.com/math-he...-diver-de.html
http://www.mathhelpforum.com/math-he...ates-flow.html
http://www.mathhelpforum.com/math-he...72473-ivp.html
http://www.mathhelpforum.com/math-he...ntial-eqn.html | {
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# Are $\emptyset$ and $X$ closed, open or clopen?
It is indeed a very basic question but I am confused:
(1) In an 2013 MSE posting under general topology here, I was told that $\emptyset$ is an open set and therefore I assume $X$ must be open too.
(2) But in Wikipedia page on clopen set here, it says "In any topological space $X$, the $\emptyset$ and the whole space $X$ are both clopen."
(3) And yet in another Wikipedia page on closed set here, "The $\emptyset$ is closed, the whole set is closed."
I must have missed something. Can you help me with a supreme verdict, once and for all, as sure as the sun rises from the east each morning, if $X$ and $\emptyset$ are open, closed or clopen. Of course I am talking about topology, thanks for your time.
• I don't see any contradiction among the three. – Hoot Jan 31 '15 at 23:36
• I don't understand! Is this an emerald, or is it green, or is it a precious stone? MSE says this is green, and Wikipedia says it is an emerald, and yet another wikipedia page says it is a precious stone. Can you help me with a supreme verdict, once and for all, if this is green, a precious stone, or an emerald? – MY USER NAME IS A LIE Jan 31 '15 at 23:47
• @MYANSWERSARECRAP your answers may be, but your comments certainly are not! – Neal Jan 31 '15 at 23:50
• Always funny: youtube.com/watch?v=SyD4p8_y8Kw – Jack D'Aurizio Feb 1 '15 at 1:48
They are all both open and closed.
Let me make this a bit more clear. By definition of a topology (from wikipedia):
A topological space is then a set $X$ together with a collection of subsets of $X$, called open sets and satisfying the following axioms:
• The empty set and $X$ itself are open.
• Any union of open sets is open.
• The intersection of any finite number of open sets is open.
So just from the definition itself it follows that $∅$ and $X$ are open.
Furthermore a set is closed (by definition) if the complement is open. Therefore $∅$ and $X$ are closed (they are each others complement). | {
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The term clopen means that a set is both open and closed, so they are both also clopen.
• Thanks, have up-voted yours. Can I take it daily for granted that both $X, \emptyset$ are clopen? – Amanda.M Jan 31 '15 at 23:42
• @A.Magnus yes that is true in any topology – Loreno Heer Jan 31 '15 at 23:50
To put it simply: sets are not doors! Being open does not mean that the set is not closed, and being closed do not implies that the set is not open. Yes, the empty set and the whole space are both open and closed. Another more dramatic example is: take a metric space $X$, with the discrete metric. Then every singleton $\{a\}$ (in fact, every subset of $X$) is clopen. Balls $B(a,r)$ with radius $r < 1$ are contained in $\{a\}$, hence $\{a\}$ is open. And $\{a\}$ is also closed, because it's complement is $X \setminus\{a\} = \bigcup_{b \in X, b \neq a}\{b\}$, a union of open sets, hence open.
• Ask you a question, I hope I can say it clear: Let $B \subset A$ and $B$ is open. Absence of any other instructions, $A$ has to be closed since it is $B$'s complement. Now if $A$ is clopen, then its neutrality does not have any bearing on open-ness or the closed-ness of $B$. Am I correct? – Amanda.M Feb 1 '15 at 1:54
• But it is not true that $A$ is the complement of $B$.. – Ivo Terek Feb 1 '15 at 1:55
• I made typo and just corrected it. Sorry for confusion. – Amanda.M Feb 1 '15 at 1:56
By the first axiom in the definition of a topology, $X$ and $\emptyset$ are open. However, closed sets are precisely those whose complements are open, by definition. Hence the empty set and $X$, being each others complements, are also closed. So, they are clopen.
• Have up-voted yours. Thanks. – Amanda.M Jan 31 '15 at 23:43
A minimal requirement on any topological space $(X,\tau)$ is that both $\varnothing$ and $X$ be open sets. By the definition of closed sets, these requirements imply that $\varnothing^c=X$ and $X^c=\varnothing$ are always closed. | {
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To sum up, in any topological space, the empty set and the whole set are always both open and closed, hence clopen.
Your question “are $\varnothing$ and $X$ closed, open or clopen” is basically six questions in one:
(1) Is $\varnothing$ closed? Answer: yes.
(2) Is $\varnothing$ open? Answer: yes.
(3) Is $\varnothing$ clopen? Answer: yes.
(4) Is $X$ closed? Answer: yes.
(5) Is $X$ open? Answer: yes.
(6) Is $X$ clopen? Answer: yes.
• Good answer, thank you! – Amanda.M Feb 1 '15 at 3:12 | {
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# Inclusion–exclusion principle problem
172 business executives were surveyed to determine if they regularly read Fortune, Time, or Money magazines. 80 read Fortune, 70 read Time, 47 read Money, 47 read exactly two of the three magazines, 26 read Fortune and Time, 28 read Time and Money, and 7 read all three magazines. How many read none of the three magazines?
So what i tried to do was to denote Time as $$T$$, Fortune as $$F$$, and Money as $$M$$. Then i wanted to do $$172 - |T \cup F \cup M|$$. To find that amount, i tried to do the inclusion exclusion principle. I found $$|F \cup M|$$ to be $$14$$ by looking at a venn diagram ($$26-7 + 28-7 + x = 47 -> x = 7$$, then $$7+7 = 14$$), but i guess this i where things went wrong. I got $$|T \cup F \cup M| = 80 + 70 + 47 - 26 - 28 - 14 + 7$$ thus the answer would be 53, but that is false.
• Excellent reasoning! See the answer below. – S. Dolan Sep 10 '19 at 9:18
We are given explicitly that \begin{align*} |F| &= 80\\ |T| &= 70\\ |M| &= 47\\ |F\cap T| &= 26\\ |T\cap M| &= 28\\ |F\cap T\cap M| &= 7. \end{align*}
We also get that 47 read "exactly two". If you consider a Venn diagram, it is not hard to see that this corresponds to the equation \begin{align*} &|F\cap T| + |T\cap M|+ |F\cap M|-3|F\cap T\cap M|=47\\ \implies & 26 + 28 + |F\cap M| - 3(7) = 47\\ \implies & |F\cap M| = 14. \end{align*}
Thus you can plug this all in to inclusion-exclusion: \begin{align*} |F \cup T\cup M| &= |F|+|T|+|M| - |F\cap T| - |T\cap M|- |F\cap M| + |F\cap T\cap M|\\ &= 80+70+47-26-28-14+7=136, \end{align*}
and thus the answer is $$172-136=\boxed{\text{36 people}}$$.
In your expression $$|T \cup F \cup M| = 80 + 70 + 47 - 26 - 28 - 14 + 7=136$$.
This gives a solution of $$36$$ not $$53$$.
We have:
$$|T \cup F \cup M| = |T| + |F| + |M| - |T \cap F| - |T \cap M| - |F \cap M| + |T \cap F \cap M|$$ $$= 70 + 80 + 47 - 26 - 28 - 14 + 7 = 136$$
Then, the number of people who do not read any magazine equals: | {
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Then, the number of people who do not read any magazine equals:
$$172 - |T \cup F \cup M| = 36$$ | {
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# Math Help - Divisibility
1. ## Divisibility
Hi all.
I'm trying to figure out the following problem:
Find the number of positive integers not exceeding 1000 that are divisible by 3 but not by 4.
Help will be appreciated. Looking for a simple/elementary proof.
Thanks.
2. Originally Posted by pollardrho06
Hi all.
I'm trying to figure out the following problem:
Find the number of positive integers not exceeding 1000 that are divisible by 3 but not by 4.
Help will be appreciated. Looking for a simple/elementary proof.
Thanks.
Hint:
$|\{x\in \mathbb{Z}_+\mid x \text{ divisible by } 3 \text{ but not by } 4\}|$
$= |\{x\in \mathbb{Z}_+\mid x \text{ divisible by } 3\}|-|\{x\in \mathbb{Z}_+\mid x \text{ divisible by } 12\}|$
This is a consequence of $A\backslash B=A\backslash(A\cap B)$, and $|X\backslash Y|=|X|-|Y|$, if $Y\subseteq X$.
3. Originally Posted by Failure
Hint:
$|\{x\in \mathbb{Z}_+\mid x \text{ divisible by } 3 \text{ but not by } 4\}|$
$= |\{x\in \mathbb{Z}_+\mid x \text{ divisible by } 3\}|-|\{x\in \mathbb{Z}_+\mid x \text{ divisible by } 12\}|$
This is a consequence of $A\backslash B=A\backslash(A\cap B)$, and $|X\backslash Y|=|X|-|Y|$, if $Y\subseteq X$.
I don't see it...
4. ## Hint
Look for cycles.
5. Hello, pollardrho06!
Find the number of positive integers not exceeding 1000
that are divisible by 3 but not by 4.
Every third number is divisible by 3.
. . There are: . $\left[\frac{1000}{3}\right] \:=\:333$ numbers divisible by 3.
But every twelfth number is divisible by 3 and by 4.
. . There are: . $\left[\frac{1000}{12}\right] \:=\:83$ multiples of 3 which are divisible by 4.
Therefore, there are: . $333 - 83 \:=\:250$ such numbers.
6. Originally Posted by Soroban
Hello, pollardrho06!
Every third number is divisible by 3.
. . There are: . $\left[\frac{1000}{3}\right] \:=\:333$ numbers divisible by 3. | {
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But every twelfth number is divisible by 3 and by 4.
. . There are: . $\left[\frac{1000}{12}\right] \:=\:83$ multiples of 3 which are divisible by 4.
Therefore, there are: . $333 - 83 \:=\:250$ such numbers.
Wow!! The greatest integer function!! Gr8!! Thanks!! | {
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# Solving the Differential equation $y' = y \tan(x) + \cot(x)$
I am asked to solve the following differential equation:
$$y' = y \tan(x) + \cot(x)$$
What I have so far is
\begin{align*} y' - (\tan x) y &= \cot(x)\\ \\ I &= e^{\int - \tan(x) dx} = \cos(x)\\ \\ \cos(x) \left( y' - y \tan(x) \right) &= \cos(x) \cot(x)\\ y' \cos(x) - y \sin(x) &= \cos(x) \cot(x)\\ \int y' \cos(x) - y \sin(x) &= \int \cos(x) \cot(x) dx\\ \\ y \cos(x) &= \int \frac{cos^2(x)}{\sin(x)}\\ y \cos(x) &= \int \frac{1-sin^2(x)}{\sin(x)}\\ &= \int \csc(x) - \sin(x) \ dx\\ &= - \ln \vert \csc(x) + \cot(x) \vert + \cos(x) + C\\ \\ \therefore y &= - \frac{\ln \vert \csc(x) + \cot(x) \vert}{\cos(x)} + 1 + C \sec(x) \end{align*}
The thing is: I am having a hard time comparing my result to the textbook's solution and to Wolfram's solution.
Textbook's solution: $$\sec(x) \left( \frac{x}{2} + \frac{\sin(2x)}{4} + C \right)$$
Is my solution correct?
Thank you.
• Your answer is correct and is equivalent to WA's. The textbook answer is wrong – David Quinn Aug 9 '16 at 19:41
• Hi @DavidQuinn thank you for your input. How could I go from my answer to Wolfram's (or the opposite)? What kind of transformation did you do? – bru1987 Aug 9 '16 at 19:42
• Use $\csc x+\cot x=\cot(\frac x2)$ – David Quinn Aug 9 '16 at 19:50
\begin{align} y'(x)-y \tan x &=\sec x \tan x \left( \frac{x}{2} + \frac{\sin 2x }{4} + C \right)+\sec x \left( \frac{1}{2} + \frac{\cos 2x }{2} \right) \\&\hspace{4mm} -\sec x\left(\frac{x}{2}+\frac{\sin 2x }{4} + C \right)\tan x \\ &= \frac{1 + \cos 2x }{2\cos x}\\&=\cos x \end{align} i.e. the textbook's answer is for the RHS being $\cos x$ rather than $\cot x$.
Since $\ln \vert \csc(x) + \cot(x) \vert=\ln{|\frac{1+\cos{x}}{\sin{x}}|}=\ln{|\frac{2\cos^2{(x/2)}}{2\sin{(x/2)\cos(x/2)}}|}=\ln{\cos{(x/2)}}-\ln{\sin(x/2)}$, your solution is same as WolphramAlpha's. | {
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# Probability of getting full house in poker
I have this problem which I have solved, but using two different methods. I am quite new to combinatorics and want to know how to intuitively understand the difference between the following two methods.
The problem consist of calculating the probability of getting a full house being dealt a 5-card poker hand.
First of, I solve this by simply saying that
$P($get full house$)=\frac{2 {13\choose 2} {4 \choose 3} {4 \choose 2}}{{52 \choose 5}}$. This is the right answer according to my text-book. However, at my first attempt at solving this I forgot the factor 2 in the numerator. My reasoning goes like this: All the possible ways of getting a full house consists of all the ways we can combine two different ranks (i.e. ${13 \choose 2}$) times all the possible ways of choosing 3 cards out of 4 suits, times all the possible ways of choosing 2 cards out of 4 suits. Now, all this seems logical to me. The thing that makes me doubt whether I truly understand what I'm doing is how the factor 2 comes in place. I'm thinking: Because we choose two different ranks without regards to order, we have to compensate for those combinations and therefore multiply with $2!$, because obviously it does matter if I (for example) choose the ranks (ace,knights) and in this sequence choose three ace and two knights.
On the other hand, I can solve the problem using the method described here: https://math.stackexchange.com/a/808328/518320
Which as well seems intuitively clear, thinking the way the user describe the process in that thread.
What is the difference in the two methods? Maybe this is obvious, but I'm new to combinatorics. And is my reasoning above accurate?
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Thanks!
• I would have said: choose the suit for the triple, that's $\binom {13}1$. Then choose the triple of ranks, $\binom 43$. Then choose the suit for the pair, $\binom {12}1$, then choose the pair, $\binom 42$. It's true that $2\times \binom {13}2= \binom {13}1\times \binom {12}1$ but I find this less intuitive. – lulu Apr 6 '18 at 12:50
• @lulu While your calculations are correct, you confused ranks with suits. The four suits are hearts, clubs, diamonds, and spades. The thirteen ranks are 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A. – N. F. Taussig Apr 6 '18 at 13:07
• @N.F.Taussig absolutely correct. I plead lack of coffee. – lulu Apr 6 '18 at 13:10
They are essentially the same giving $156$ ways of choosing ranks, but if you want to draw a distinction:
• Choose two ranks and then choose one of those two to be the three so the other is the pair $${13 \choose 2}{2 \choose 1}$$
• Choose one rank to be the three then choose another rank to be the pair $${13 \choose 1}{12 \choose 1}$$
and you then multiply this by ${4 \choose 3}{4 \choose 2}$ and divide by ${52 \choose 5}$
Doing the same thing for two pairs to get $858$ ways of choosing the ranks by several methods:
• Choose three ranks and then choose one of those three to be the single so the others are the pairs $${13 \choose 3}{3 \choose 1}$$
• Choose three ranks and then choose two of those three to be the pairs so the other is the single $${13 \choose 3}{3 \choose 2}$$
• Choose one rank to be the single then choose two other ranks to be the pairs $${13 \choose 1}{12 \choose 2}$$
• Choose two ranks to be the pairs then choose another rank to be the single $${13 \choose 2}{11 \choose 1}$$
and you then multiply this by ${4 \choose 1}{4 \choose 2}{4 \choose 2}$ and divide by ${52 \choose 5}$ | {
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The expressions $2\binom{13}{2}$ and $\binom{13}{1}\binom{12}{1}$ are both equivalent to "choose $2$ objects from $13$ where the order of choosing matters." In other words, we want permutations of $2$ objects out of $13$ rather than combinations.
The way to get permutations of $k$ objects out of $n$ is to multiply $k$ consecutive integers together, where $n$ is the largest of the integers multiplied: $n(n-1)(n-2) \cdots (n-k+1).$ For example, permutations of $2$ objects out of $n$ give $13\cdot12$ possible permutations. And since $\binom n1 = n,$ $13\cdot12 = \binom{13}{1}\binom{12}{1}.$
But often the number of permutations is written $\dfrac{n!}{(n-k)!},$ because $$\frac{n!}{(n-k)!} = n(n-1)(n-2) \cdots (n-k+1).$$ To get the number of combinations of $k$ objects out of $n$ (without regard for the order of the objects), we divide by $k!,$ since that's the number of different sequences in which each combination of $k$ objects can occur as a permutation. So we get $$\binom nk = \frac{n!}{(n-k)!k!}.$$ This also means that $k!\binom nk$ is yet another way to write the number of permutations of $k$ objects out of $n.$
In short, it's all really the same thing written in different ways.
The two ranks are distinguishable once you get a particular full house. Say you get three kings and two queens. That differs from getting three queens and two kings. So it should be $13 \cdot 12$ ways to pick the two ranks. This is the same as $2 \cdot \binom{13}{2}.$ | {
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# Find the image: F: N * N -> R , F(x) = m^2 + 2n
#### KOO
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F:N*N -> R, F(x) = m^2 + 2n
I think the answer is N. Am I right?
#### Chris L T521
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I think the answer is N. Am I right?
Based on your title, we have that $F:\mathbb{N}\times\mathbb{N}\rightarrow \mathbb{R}$ ($\mathbb{N}\ast\mathbb{N}$ makes no sense at all) where $F(x) = m^2+2n$ (I assume here that $x=(m,n)$).
Now, depending on who you talk to, $\mathbb{N}$ may or may not include zero (the general consensus from what I've seen is that $0\notin\mathbb{N}$, but there are some professors/authors that include zero in their definition of the natural numbers; hence why I think it's best to clarify this right from the get go); that is, either $\mathbb{N}=\{x\in\mathbb{Z} : x\geq 0\}$ or $\mathbb{N}=\{x\in\mathbb{Z}: x\geq 1\}$.
If $0\in\mathbb{N}$, then $\mathrm{Im}(F)=\mathbb{N}$. If $0\notin\mathbb{N}$, then $\mathrm{Im}(F) = \mathbb{N}\backslash\{1,2,4\}$ since there aren't pairs $(m,n)\in\mathbb{N}\times\mathbb{N}$ such that $m^2+2n=1$, $m^2+2n=2$, or $m^2+2n=4$.
(In $\mathrm{Im}(F)$, note that all the positive odd numbers greater than or equal to 3 are generated by pairs of the form $(1,n)$ for $n\in\mathbb{N}$ since $1^2+2n=2n+1$, and all positive even numbers greater than or equal to 6 are generated by pairs of the form $(2,m)$ for $m\in\mathbb{N}$ since $2^2+2m = 2(m+2)$; this is why I can claim that $F(\mathbb{N}\times\mathbb{N}) = \mathbb{N}\backslash\{1,2,4\}$ for $0\notin\mathbb{N}$.)
I hope this makes sense!
#### KOO
##### New member
Based on your title, we have that $F:\mathbb{N}\times\mathbb{N}\rightarrow \mathbb{R}$ ($\mathbb{N}\ast\mathbb{N}$ makes no sense at all) where $F(x) = m^2+2n$ (I assume here that $x=(m,n)$). | {
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Now, depending on who you talk to, $\mathbb{N}$ may or may not include zero (the general consensus from what I've seen is that $0\notin\mathbb{N}$, but there are some professors/authors that include zero in their definition of the natural numbers; hence why I think it's best to clarify this right from the get go); that is, either $\mathbb{N}=\{x\in\mathbb{Z} : x\geq 0\}$ or $\mathbb{N}=\{x\in\mathbb{Z}: x\geq 1\}$.
If $0\in\mathbb{N}$, then $\mathrm{Im}(F)=\mathbb{N}$. If $0\notin\mathbb{N}$, then $\mathrm{Im}(F) = \mathbb{N}\backslash\{1,2,4\}$ since there aren't pairs $(m,n)\in\mathbb{N}\times\mathbb{N}$ such that $m^2+2n=1$, $m^2+2n=2$, or $m^2+2n=4$.
(In $\mathrm{Im}(F)$, note that all the positive odd numbers greater than or equal to 3 are generated by pairs of the form $(1,n)$ for $n\in\mathbb{N}$ since $1^2+2n=2n+1$, and all positive even numbers greater than or equal to 6 are generated by pairs of the form $(2,m)$ for $m\in\mathbb{N}$ since $2^2+2m = 2(m+2)$; this is why I can claim that $F(\mathbb{N}\times\mathbb{N}) = \mathbb{N}\backslash\{1,2,4\}$ for $0\notin\mathbb{N}$.)
I hope this makes sense!
Actually I had this question on a test and you're right x=(m,n) and we're told N does not include 0.
Anyways what does \ mean?
#### Chris L T521
##### Well-known member
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Actually I had this question on a test and you're right x=(m,n) and we're told N does not include 0.
Anyways what does \ mean?
Ah, that's one notation for set difference. I could have also written it as $\mathbb{N}-\{1,2,4\}$. | {
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# Same integration with 2 different answers?
$$\int x(x^2+2)^4\,dx$$
When we do this integration with u substitution we get $$\frac{(x^2+2)^5}{10}$$ as $u=x^2+2$
$du=2x\,dx$ $$\therefore \int (u+2)^4\,du = \frac{(x^2+2)^5}{10} + C$$
Although when we expand the fraction and then integrate the answer we get is different:
$x(x^2+2)^4=x^9+8x^7+24x^5+32x^3+16x$ $$\int x^9+8x^7+24x^5+32x^3+16x \,dx$$
we get
$$\frac {x^{10}}{10} +x^8+4x^6+8x^4+8x^2 + C$$
For a better idea of the questions, let's say the questions asks us to find the value of C when y(0)=1
Now,
$x=0$
$$\frac {0^{10}}{10} + 0^8 + 4(0)^6 + 8(0)^4 + 8(0)^2 + C = 1$$ $$\therefore C= 1$$ AND $$\frac {(0+2)^5}{10} + C= 1$$ $$\therefore \frac {32}{10} + C = 1$$ $$\therefore C = 1 - 3.2 = -2.2$$
• Don't forget the arbitrary constant of integration! – Lord Shark the Unknown Jun 12 '18 at 4:41
• ^^^^^ What @LordSharktheUnknown said! The indefinite integral is the class of all functions who are antiderivatives of the integrand. These antiderivatives only differ by a constant. – N8tron Jun 12 '18 at 4:48
• $\frac {(x^2+2)^5}{10} = \frac {x^{10}}{10} + x^8 + 4x^6+8x^4+8x^2 + \frac {32}{10}.$ The only difference is the constant of integration. – Doug M Jun 12 '18 at 4:52
• Please fix the typos on $(x+2)$. – Yves Daoust Jun 12 '18 at 5:16
• There are lots and lots of similar questions already, just search: google.com/… – Hans Lundmark Jun 12 '18 at 6:46
Like mentioned in the comments this is all fixed if you remember your constant of integration.
$$\int x(x^2+2)^4\ dx= \frac{(x^2+2)^5}{10}+C$$
Note if you expand
$$\begin{split} \frac{(x^2+2)^5}{10}&=\frac{1}{10}\left(x^{10}+5x^8(2)+10x^6(2^2)+10x^4(2^3)+5x^2(2^4)+2^5\right)\\ &=\frac{x^{10}}{10}+x^8+4x^6+8x^4+8x^2+\frac{32}{10} \end{split}$$
Notice the relation to your other way of computing the integral
$$\int x(x^2+2)^4\ dx = \frac{x^{10}}{10}+x^8+4x^6+8x^4+8x^2 +C$$ | {
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$$\int x(x^2+2)^4\ dx = \frac{x^{10}}{10}+x^8+4x^6+8x^4+8x^2 +C$$
So lets call $F(x)=\frac{x^{10}}{10}+x^8+4x^6+8x^4+8x^2$ and $G(x)=\frac{x^{10}}{10}+x^8+4x^6+8x^4+8x^2+\frac{32}{10}$ then $F(x)-G(x)=-\frac{32}{10}$ a constant. All antiderivatives of a continuous function only differ by a constant.
Just for fun Let's see another one:
First lets use double angle for sine $$\int \cos x\sin x\ dx=\frac{1}{2}\int\sin 2x\ dx=-\frac{1}{4}\cos 2x +C$$
Then substitutions $u=\sin x$
$$\int \cos x\sin x\ dx=\int u\ du =\frac{u^2}{2}+C=\frac{\sin^2 x}{2}+C$$
Then substitutions $u=\cos x$
$$\int \cos x\sin x\ dx=\int -u\ du =\frac{-u^2}{2}+C=\frac{-\cos^2 x}{2}+C$$
If you find the constant differences and combine them in the right way you get the half angle formulas:
$$\sin^2 x=\frac{1-\cos 2x}{2},\quad \cos^2 x=\frac{1+\cos 2x}{2}$$
Note you can pretty quickly derive some funky trig identities in this way. For instance if you consider $\int \cos^3 x \sin^5 x\ dx$
• Although if we were asked to find the value of C at x=0, won't we have 2 different value of C??? – Agent Smith Jun 13 '18 at 0:47
• No if you fix any two antiderivatives $F$ and $G$ for a suitable function (continuous is adequate but the real condition is Riemann integrable). There is a constant C such that the difference $F(x)-G(x)=C$ for all x. This can be rigourously proven. – N8tron Jun 13 '18 at 1:45
• Check out this page (or numerous others) to see the proof math.stackexchange.com/questions/1862231/… – N8tron Jun 13 '18 at 1:49
You can check an antiderivative by differentiating.
$$\left(\frac{(x^2+2)^5}{10}\right)'=x(x^2+2)^4=x^9+8x^7+24x^5+32x^3+16x$$
and
$$\left(\frac {x^{10}}{10} +x^8+4x^6+8x^4+8x^2\right)'=x^9+8x^7+24+32x^3+16x$$
and the two expressions are indeed equivalent.
Now the long explanation.
Consider the binomial $x^2+a$ raised to some power $n$ and multiplied by $2x$.
$$2x(x^2+a)^m$$
which integrates as
$$\frac{(x^2+a)^{m+1}}{m+1}.$$ | {
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$$2x(x^2+a)^m$$
which integrates as
$$\frac{(x^2+a)^{m+1}}{m+1}.$$
By the binomial theorem, the terms in the development of this antiderivative are
$$\frac1{m+1}\binom{m+1}kx^{2(m+1-k)}a^k.$$
On the other hand, the development of the initial integrand gives terms
$$2\binom mkx^{2(m-k)+1}a^k,$$ and after integration
$$\frac1{m-k+1}\binom mkx^{2(m-k)+2}a^k.$$
It is easy to see that all terms coincide, because
$$\frac1{m+1}\frac{(m+1)!}{k!(m+1-k)!}=\frac1{m-k+1}\frac{m!}{k!(m-k)!}=\frac{(m-1)!}{k!(m-k+1)!}.$$
Anyway, the first development holds for $0\le k\le m+1$, giving a constant term $\dfrac{a^m}{m+1}$, but the second for $0\le k\le m$ only, giving no constant term. But this does not matter, as two antiderivatives can differ by a constant.
• But wouldn't the C be different for both the cases?? – Agent Smith Jun 13 '18 at 0:48
• @AgentSmith: and ? – Yves Daoust Jun 13 '18 at 5:50
• and that would change the answer wouldn't it?? if we were asked for the value of C? – Agent Smith Jun 13 '18 at 19:56
• @AgentSmith: you'd answer: arbitrary. You should understand that the value of the constant does not matter. – Yves Daoust Jun 13 '18 at 20:05
• oh okay thanks! Makes sense! – Agent Smith Jun 14 '18 at 18:59 | {
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# A train travels from city A to city B. The average speed of the train
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A train travels from city A to city B. The average speed of the train is 60 miles/hr and it travels the first quarter of the trip at a speed of 90 mi/hr. What is the speed of the train in the remaining trip?
A. 30
B. 45
C. 54
D. 72
E. 90
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Re: A train travels from city A to city B. The average speed of the train [#permalink]
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Refer diagram below:
Attachment:
speed.png [ 3.39 KiB | Viewed 3540 times ]
Setting up the time equation:
Total time = Time required in first quarter + Time required in the remaining journey
$$\frac{d}{60} = \frac{d}{4*90} + \frac{3d}{4*s}$$ | {
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$$\frac{d}{60} = \frac{d}{4*90} + \frac{3d}{4*s}$$
$$s = \frac{90*3}{5} = 54$$
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A train travels from city A to city B. The average speed of the train [#permalink]
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30 Jan 2015, 06:23
Using the RTD chart, with A being the first quarter, B the rest of the trip and All the combined "trip".
_______R_____T______D
A.........90........3..........270
B.........54.......15.........810
All........60.......18........1080
Let me explain how we fill in the chart:
1) Starting with what we know, we add 90 and 60
2) Picking an easy number for the total distance (6*9=54), so I chose 540. Multiply by 2 to get the whole trip, and we get 1080. Add 1080 for all-distance.
3) 1/4 of the whole distance happened at a Rate of 90. So, 1080/4=270, add 270 under A-D.
4) 2080 - 270 = 810, for the rest of the trip. Add 810 under B-D.
5) Claculate the individual Times for A and All, using T=D/R. Add the results, 3 and 18, under A-T and All-T.
6) 18-3=15, this is the remaining Time for B. Add 15 under B-T.
7) Finally, 15R=810 -->R=810/15 --> R= 54 ANS C
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Re: A train travels from city A to city B. The average speed of the train [#permalink]
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06 Apr 2017, 09:45
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PareshGmat wrote:
A train travels from city A to city B. The average speed of the train is 60 miles/hr and it travels the first quarter of the trip at a speed of 90 mi/hr. What is the speed of the train in the remaining trip?
A. 30
B. 45
C. 54
D. 72
E. 90
We have an average rate problem in which we can use the following formula:
Avg speed = (distance 1 + distance 2)/(time 1 + time 2) | {
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Avg speed = (distance 1 + distance 2)/(time 1 + time 2)
If we let d = total distance of the trip, then the first quarter of the trip, or (1/4)d = d/4, was traveled at 90 mph. Thus, the time was (d/4)/90 = d/360.
We can let the rate for the remaining part of the trip = r, and thus the time for the remaining part of the trip, or (3/4)d = 3d/4, is (3d/4)/r = 3d/(4r). Let’s use all of this information in the average rate equation:
60 = d/(d/360 + 3d/(4r))
60 = 1/(1/360 + 3/(4r))
60(1/360 + 3/(4r)) = 1
1/6 + 45/r = 1
Let’s multiply the above equation by 6r:
r + 270 = 6r
270 = 5r
r = 54
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Re: A train travels from city A to city B. The average speed of the train [#permalink]
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08 Apr 2017, 18:53
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Hi All,
This question can be solved by TESTing VALUES.
The prompt tells us that average speed of the train over the course of the entire trip is 60 miles/hr and that it travels the first quarter of the distance at a speed of 90 mi/hr. We're asked for the average speed of the train for the remaining three-quarters of the trip.
Let's choose 90 miles for the FIRST QUARTER of the distance. We can then immediately calculate two things...
1) Since the train was traveling 90 miles/hour, the first quarter of the trip took 1 hour to complete.
2) Since (1/4)(Total Distance) = 90 miles, then the FULL TRIP = 4(90) = 360 miles.
The total trip is 360 miles; with an average speed of 60 miles/hr, the FULL TRIP would take... (X)(60 mph) = 360 miles.... X = 6 hours to complete.
The first hour of the trip covered 90 miles, so the remaining 360 - 90 = 270 miles of the trip are covered in the remaining 5 hours... | {
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Thus, the average speed for the remainder of the trip is... (270 miles)/(5 hours) = 54 miles/hour
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# If 22^n is a divisor of 97!+98! ,what is the maximum possible
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If 22^n is a divisor of 97!+98! ,what is the maximum possible value of integer n?
A)8
B)9
C)10
D)11
E)12
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Re: If 22^n is a divisor of 97!+98! ,what is the maximum possible [#permalink]
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stonecold wrote:
If 22^n is a divisor of 97!+98! ,what is the maximum possible value of integer n?
A)8
B)9
C)10
D)11
E)12
22^n means power of 11 as 2 would surely have more power than 11 in a factorial..
97!+98!=97!(1+98)=97!*99....
99 will have ONE 11 and 97! will have 97/11 or 8..
Total 1+8=9
B
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### Show Tags
12 Nov 2016, 05:33
2
2
stonecold wrote:
If 22^n is a divisor of 97!+98! ,what is the maximum possible value of integer n?
A)8
B)9
C)10
D)11
E)12
We are given that 22^n is a divisor of 97!+98!. Thus:
(97!+98!)/22^n = integer
To determine the maximum value of n, we can factor out a 97! in the numerator of the above expression and we have:
97!(1 + 98)/22^n = integer
97!(99)/22^n = integer
Since 22 = 11 x 2, we can rewrite our expression as:
97!(99)/(11^n x 2^n) = integer
In order to determine the maximum value of n such that 22^n divides into 97!(99), we need to determine the maximum number of pairs of factors of 11 and 2 within 97!(99). Since we know there are fewer factors of 11 than factors of 2, we can determine the number of factors of 11 within 97!(99). Let’s start with 97!:
11, 22, 33, 44, 55, 66, 77 and 88 are all factors of 97! Thus, 97! has 8 factors of 11.
We also see that 99 has 1 factor of 11.
Thus, there are 9 factors of 11 within 97!(99), and thus the maximum value of n is 9.
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Re: If 22^n is a divisor of 97!+98! ,what is the maximum possible [#permalink]
### Show Tags
24 Feb 2020, 23:55
In questions related to divisibility, breaking down the dividend will give a lot of information about what the divisor needs to be if it HAS to divide the dividend fully. Therefore, let’s break the dividend down, first.
In 97! + 98!, we can take 97! as common. When we do this, we have 97! + 98! = 97! (1 + 98) = 97! * 99. This helps us understand that $$22^n$$ should be able to divide 97! * 99 fully. | {
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Let’s now look at 22. 22 = 2*11, therefore $$22^n$$ = $$2^n$$ * $$11^n$$.
So, essentially, we are trying to find the highest power of 11 in the dividend since the number of 11s would be far fewer compared to the number of 2s.
Finding out the highest power of any prime number in a given factorial can be done by successive division. In our case, it can be done as shown below in the diagram.
Attachment:
25th Feb 2020 - Reply 3 - 1.jpg [ 73.73 KiB | Viewed 123 times ]
So, the highest power of 11 in 97! is 8. But, 99 also has a 11 in it. Therefore, the highest power of 11 in the numerator (or the dividend) is 9. So, there need to be 9 11s in the denominator as well. As such, the highest power of 22 that can divide 97! + 98! is 9.
The correct answer option is B.
Hope that helps!
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Re: If 22^n is a divisor of 97!+98! ,what is the maximum possible [#permalink] 24 Feb 2020, 23:55
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# General formula for this
1. Dec 16, 2005
### twoflower
Hi all,
there is general formula for findind out, "Which highest power of x is divisible n! with?" Eg. Which highest power of 5 is divisable 50! with?
But I forgot it and can't find it now...will somebody help me please?
Thank you.
2. Dec 16, 2005
### AKG
You mean, what is the highest power of x that divides n!? Well if you can prime-factorize x and n!, this is easy. In the case of 5 and 50!, it's hard to prime-factorize 50!, but the solution is still easy. You can see quite easily that 5 will occur 12 times in the prime-factorization of 50! (it will occur once in each of the factors 5, 10, 15, 20, 30, 35, 40, 45, and twice in both 25 and 50), so the highest power of 5 that divides 50! is 512.
3. Dec 16, 2005
### twoflower
Yes, this manual approach is clear. But there is also a general formula..
4. Dec 16, 2005
### AKG
I'm not sure about a general formula just yet, but this might be helpful: if you associate each factorial with a sequence (an,k)k in N, where an,k is the power of the kth prime in n!, you get:
(a0,k) = (0, 0, ...)
(a1,k) = (0, 0, ...)
(a2,k) = (1, 0, 0, ...)
(a3,k) = (1, 1, 0, 0, ...)
(3, 1, 0, 0, ...)
(3, 1, 1, 0, 0, ...)
(4, 2, 1, 0, 0, ...)
(4, 2, 1, 1, 0, 0, ...)
(7, 2, 1, 1, 0, 0, ...)
(7, 4, 1, 1, 0, 0, ...)
(8, 4, 2, 1, 0, 0, ...)
(8, 4, 2, 1, 1, 0, 0, ...)
(10, 5, 2, 1, 1, 0, 0, ...)
(10, 5, 2, 1, 1, 1, 0, 0, ...)
(11, 5, 2, 2, 1, 1, 0, 0, ...)
The exponent of 2 changes every 2 rows, the exponent of 3 changes every 3 rows, the exponent of p will change every p rows. Ignoring repetition, the exponents for 2 go: 0, 1, 3, 4, 7, 8, 10, 11, ...
For 3, they go 0, 1, 2, 4, 5, ...
For 5, they go 0, 1, 2, ... | {
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Some numbers are skipped in the above sequences because powers occur, i.e. in the sequence for 2, it goes from 1 to 3 without going through 2 because the 4 in 4! contributes two 2's, not just 1. I'm very tired right now, but I suppose if you can generalize what's going on here, it might be a helpful step in finding a general formula.
Well I suppose if you just want to know the general formula and aren't trying to figure it out yourself, then the above isn't much help.
5. Dec 16, 2005
### VietDao29 | {
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5. Dec 16, 2005
### VietDao29
I suppose that this is not a homework problem... So here's my approach.
Let's say that [x] is the function that will return the integer part before the decimal point of the number x, eg: [37.5534] = 37.
Say, you need to find the largest p that satisfies:
xp divides n!, for some given x (x is prime), and n.
Let: $$\rho := \left[ \frac{\ln n}{\ln x} \right]$$, ie: $$\rho$$ is the largest positive integer such that: $$x ^ \rho \leq n$$
So for every x consecutive integers there's one integer that's divisible by x, for every x2 consecutive integers there's one integer that's divisible by x2, for every x3 consecutive integers there's one integer that's divisible by x3,...
So the largest p can be obtained by:
$$p := \sum_{i = 1} ^ \rho \left[ \frac{n}{x ^ i} \right]$$
---------------------
If x is not prime, then you can prime-factorize it:
$$x = \lambda_1 ^ {\alpha_1} \ \lambda_2 ^ {\alpha_2} \ \lambda_3 ^ {\alpha_3} \ ... \lambda_k ^ {\alpha_k}$$
Where: $$\lambda_i, \ i = 1..k$$ are primes.
Then: $$x ^ p = (\lambda_1 ^ {\alpha_1} \ \lambda_2 ^ {\alpha_2} \ \lambda_3 ^ {\alpha_3} \ ... \lambda_k ^ {\alpha_k}) ^ p = \lambda_1 ^ {\alpha_1 p} \ \lambda_2 ^ {\alpha_2 p} \ \lambda_3 ^ {\alpha_3 p} \ ... \lambda_k ^ {\alpha_k p}$$
If xp divides n! then $$\lambda_i ^ {\alpha_i p}, \ i = 1..k$$ must also divide n!.
So you can use the same method (as shown above) to find the largest $$\rho_i, \ i = 1..k$$ such that $$\lambda_i ^ {\rho_i}, \ i = 1..k$$ divides n!.
Then define: $$\beta_i := \left[ \frac{\rho_i}{\alpha_i} \right], \ i = 1..k$$.
And the largest p can be found by:
$$p = \min (\beta_i), \ i = 1..k$$
Last edited: Dec 16, 2005
6. Dec 25, 2005
### benorin
The exact power of a prime p dividing n! is alternatively given by
$$\frac{n-\mu}{p-1},$$
where $\mu$ is the sum of the digits of the base p representation of n. | {
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# Probability of an event after time has passed
A friend of mine posed this problem and we have had a disagreement on the answer.
The problem:
• There is a 90% chance that some event will happen in the next year.
• There is a 95% chance that the event will happen eventually.
• What is the probability that it happens after this year, if it does not happen in the next year?
Hover over here for a description of our answers and arguments:
I told my friend that it is 95%. My reasoning is that there is a 95% chance that it happens from now to infinity. The probability that it happens from now to infinity minus one year is still 95%. In one year, when it hasn't happened, events that didn't happen don't affect your "eventual" odds.
My friend on the other hand, believes it to be 50%. He had two arguments. First, he said if you flip a coin 4 times, the chance that you get heads the first time is 50%, but the chance that you get heads eventually is 93.75% (15/16). If you don't get it the first time, what is the chance that you get it after the first time? 87.5% (7/8), in other words, it has decreased.
To this I responded that the "eventually" is based on a finite set of events, each with their own probability and after the first event has passed, you have fewer chances. In his problem, on the other hand, it is a summary of the probability of unknown events broken in two time periods.
He continued to insist that it is the same because you have two time periods with known odds. The 95% is the combination of two probabilities, 90% and 50%. Eliminating the first 90%, the probability of the remaining period is 50%.
I agreed that's an accurate way to look at it from now, but I don't think that's accurate once you know it did not happen in that time period.
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-
Let's use Bayes' Theorem: $$P(A|B)=\frac{P(B|A) P(A)}{P(B)}.$$ In our case, $B$ is the event whatnot does not occur in year 1 (so with probability $P(B)=.1$, and $A$ is the event whatnot occurs at some point (so with probability $P(A)=0.95$). Of course, $P(B|A)$ is the only thing left to determine, but notice that $P(A|not(B))=1$ since if whatnot occurs in year 1, then it occurs at some point. Therefore, again by Bayes' Theorem $$P(not(B)|A)=\frac{P(A|not(B)) P(not(B))}{P(A)}=\frac{90}{95}.$$ Since $P(not(B)|A)+P(B|A)=1$, we have that $P(B|A)=\frac{5}{95}$. Hence plugging this all in, we find $$P(A|B)=\frac{\frac{5}{95} \frac{95}{100}}{\frac{10}{100}}=\frac{1}{2}.$$ So it looks like your friend was right!
-
Calculate it this way. Given that the probability that it occurs during or after year 2 is $x$%. And the probability it occurs in year 1 is 90%, what is the probability it occurs?
Now, consider $x=0$, then the probability the event occurs is 90%.
If $x=50$, then it is 95%. (because it is $0.9 \times 1 + 0.1 \times 0.5 \times 1= 0.95$)
This informal argument should convince you.
-
After talking about it some more, we came up with this scenario which illustrates why he is correct, and I am wrong.
1. Say I want to give him a 90% chance that he will get $100 this year, 95% chance total, and 5% chance that he gets nothing. 2. So, I take 100 sheets of paper and write dates in the next year on 90 of them, dates after that on 5, and "Bad luck" on the other 5. 3. I draw one at random and keep it a secret. 4. A year passes and my friend got nothing. Now he knows that all 90 dates this year are still in the pile. He now knows I must hold one of the other 10 sheets. That makes a 5/10 chance that I have a sheet with a date on it, and 5/10 chance that the sheet I hold has "Bad luck" on it. 50% chance he will get$100.
-
Nice. Beats Bayes. – André Nicolas Jan 18 '13 at 21:15 | {
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-
Nice. Beats Bayes. – André Nicolas Jan 18 '13 at 21:15
Let $A$ represent the cases in which the event happens eventually. Let $B$ represent the cases in which the event happens next year. You've stipulated $P(A)=0.95$ and $P(B)=0.90$. Because $B$ is a subset of $A$, $P(A\cap B)=P(B)=0.90$, and $$P(A\cap \neg B)=P(A)-P(A\cap B)=0.95-0.90=0.05.$$ By the definition of conditional probability, then, $$P(A|\neg B)=\frac{P(A\cap \neg B)}{P(\neg B)}=\frac{0.05}{0.10}=\frac{1}{2}.$$ So if I propose a deal right now, under the terms of which (a) you'll put in $\$1000$if the event hasn't happened after one year and (b) you'll get back$X$if the event happens after next year, then you should take the deal if$X>\$2000$. In this sense, your friend is right.
Your point, though, is different. You're wondering what deal you should take a year from now if the event hasn't happened by then. The answer is that it depends on probabilities not stipulated in the problem. In particular, you may learn something from the fact that it didn't happen this year, which may raise or lower these odds.
Suppose the event is the earth being hit by a meteor, that this depends on the solar system's meteor density, and that we're building a meteor shield that will be operational in just over a year. Based on what we know now, the meteor density might be high or low, with equal likelihood; if it's high, then the earth will be hit in the next year with probability $1$ (no shield yet!); if it's low, then the earth will be hit next year with probability $0.8$ and eventually with probability $0.9$. A year from now, if the earth hasn't been hit yet, you will know that the meteor density is low, and that the remaining probability that we'll ever be hit is $0.1$. In this case, the odds were lowered. | {
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