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https://it.mathworks.com/help/symbolic/sym.latex.html | # latex
LaTeX form of symbolic expression
## Syntax
chr = latex(S)
## Description
example
chr = latex(S) returns the LaTeX form of the symbolic expression S.
## Examples
collapse all
Find the LaTeX form of the symbolic expressions x^2 + 1/x and sin(pi*x) + phi.
syms x phi chr = latex(x^2 + 1/x)
chr = '\frac{1}{x}+x^2'
chr = latex(sin(pi*x) + phi)
chr = '\phi +\sin\left(\pi \,x\right)'
Find the LaTeX form of the symbolic array S.
syms x S = [sym(1)/3 x; exp(x) x^2]
S = $\left(\begin{array}{cc}\frac{1}{3}& x\\ {\mathrm{e}}^{x}& {x}^{2}\end{array}\right)$
chr = latex(S)
chr = '\left(\begin{array}{cc} \frac{1}{3} & x\\ {\mathrm{e}}^x & x^2 \end{array}\right)'
Perform computation using several symbolic matrix variables, and then find their LaTeX forms.
Create 3-by-3 and 3-by-1 symbolic matrix variables.
syms A 3 matrix syms X [3 1] matrix
Find the Hessian matrix of ${X}^{T}AX$. Derived equations involving symbolic matrix variables appear in typeset as they would be in textbooks.
f = X.'*A*X
f = ${X}^{\mathrm{T}} A X$
H = diff(f,X,X.')
H = ${A}^{\mathrm{T}}+A$
Generate the LaTeX forms of the symbolic matrix variables f and H.
chrf = latex(f)
chrf = '{\textbf{X}}^{\mathrm{T}}\,\textbf{A}\,\textbf{X}'
chrH = latex(H)
chrH = '{\textbf{A}}^{\mathrm{T}}+\textbf{A}'
Perform computation using symbolic matrix functions, and then find their LaTeX forms.
Create a 3-by-1 symbolic matrix variable.
syms X [3 1] matrix
Create a symbolic matrix function that represents the formula $\mathit{f}\left(\mathbit{X}\right)={\mathbit{X}}^{\mathit{T}}\mathbit{X}$.
syms f(X) [1 1] matrix keepargs f(X) = X.'*X
f(X) = ${X}^{\mathrm{T}} X$
Find the derivative of $\mathit{f}\left(\mathbit{X}\right)$ with respect to $\mathbit{X}$.
Df = diff(f,X)
Df(X) = $2 {X}^{\mathrm{T}}$
Evaluate the derivative for the matrix value $\mathbit{X}=\left[1;\text{\hspace{0.17em}}2;\text{\hspace{0.17em}}3\right]$.
DfEval = Df([1; 2; 3])
DfEval =
Generate the LaTeX forms of the symbolic matrix functions f and Df, and the symbolic expression DfEval.
chrf = latex(f)
chrf = '{\textbf{X}}^{\mathrm{T}}\,\textbf{X}'
chrDf = latex(Df)
chrDf = '2\,{\textbf{X}}^{\mathrm{T}}'
chrEval = latex(DfEval)
chrEval = '2\,{\left(\begin{array}{c} 1\\ 2\\ 3 \end{array}\right)}^{\mathrm{T}}'
Modify generated LaTeX by setting symbolic preferences using the sympref function.
Generate the LaTeX form of the expression $\pi$ with the default symbolic preference.
sympref('default'); chr = latex(sym(pi))
chr = '\pi '
Set the 'FloatingPointOutput' preference to true to return symbolic output in floating-point format. Generate the LaTeX form of $\pi$ in floating-point format.
sympref('FloatingPointOutput',true); chr = latex(sym(pi))
chr = '3.1416'
Now change the output order of a symbolic polynomial. Create a symbolic polynomial and set the 'PolynomialDisplayStyle' preference to 'ascend'. Generate the LaTeX form of the polynomial sorted in ascending order.
syms x; poly = x^2 - 2*x + 1; sympref('PolynomialDisplayStyle','ascend'); chr = latex(poly)
chr = '1-2\,x+x^2'
The preferences you set using sympref persist through your current and future MATLAB® sessions. Restore the default values by specifying the 'default' option.
sympref('default');
For $x$ and $y$ from $-2\pi$ to $2\pi$, plot the 3-D surface $y\mathrm{sin}\left(x\right)-x\mathrm{cos}\left(y\right)$. Store the axes object in a by using gca. Display the axes box by using a.Box and set the tick label interpreter to latex.
Create the $x$-axis ticks by spanning the $x$-axis limits at intervals of pi/2. Convert the axis limits to precise multiples of pi/2 using round and get the symbolic tick values in S. Display the ticks by setting the XTick property of a to S. Create the LaTeX labels for the $x$-axis by using arrayfun to apply latex to S and then concatenating $. Display the labels by assigning them to the XTickLabel property of a. Repeat these steps for the $y$-axis. Set the $x$- and $y$-axes labels and the title using the latex interpreter. syms x y f = y.*sin(x)-x.*cos(y); fsurf(f,[-2*pi 2*pi]) a = gca; a.TickLabelInterpreter = 'latex'; a.Box = 'on'; a.BoxStyle = 'full'; S = sym(a.XLim(1):pi/2:a.XLim(2)); S = sym(round(vpa(S/pi*2))*pi/2); a.XTick = double(S); a.XTickLabel = strcat('$',arrayfun(@latex, S, 'UniformOutput', false),'$'); S = sym(a.YLim(1):pi/2:a.YLim(2)); S = sym(round(vpa(S/pi*2))*pi/2); a.YTick = double(S); a.YTickLabel = strcat('$',arrayfun(@latex, S, 'UniformOutput', false),'$'); xlabel('x','Interpreter','latex'); ylabel('y','Interpreter','latex'); zlabel('z','Interpreter','latex'); title(['$' latex(f) '$for$x$and$y$in$[-2\pi,2\pi]\$'],'Interpreter','latex')
## Input Arguments
collapse all
Input, specified as a symbolic number, variable, vector, array, function, expression, matrix variable, or symbolic matrix function.
Data Types: sym | symfun | symmatrix | symfunmatrix
## Version History
Introduced before R2006a
expand all | 2022-05-18T12:50:22 | {
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https://math.stackexchange.com/questions/2971274/rate-of-change-calculus-problem | # Rate Of Change Calculus Problem
I was working through my calculus textbook's practice problems when I came across a problem I couldn't figure out. I really suck at rate of change word problems and this one stumped me. Unfortunately, the textbook only gives answers to even-numbered problems and this is an odd question. Anyway it goes something like this:
An empty oil container is 10 meters long. A cross-section of the container is in the shape of an isosceles trapezoid that is 30cm wide at the bottom and 80cm wide at the top and has a height of 50cm. The container is filled with oil at a rate of 2 meters cubed per minute. How fast is the oil level rising when it is 30cm deep?
I would give my work but I have little to no idea if it's right and I'm afraid I would be just wasting my time as it's probably wrong. How do you do problems like these? I feel like I was never properly taught how and I would like to have a solid understanding by the time I'm actually being tested on this material.
• The problem is about volume of an isosceles trapezoid prism. Start with the formula for that. – randomgirl Oct 25 '18 at 21:05
• An "isosceles trapezoid that is 30cm wide at the top and 80cm wide at the top"? Is the 30cm or the 80cm supposed to be the width at the bottom? – Kurt Schwanda Oct 25 '18 at 21:12
• @KurtSchwanda My bad, fixed that typo. – Bob Smith Oct 25 '18 at 23:03
Refer to the figure:
The volume of the container filled with oil is: $$V=\frac{(0.3+2x)+0.3}{2}\cdot h\cdot 10.$$
From the similarity of the two triangles on the left we get: $$\frac{x}{0.25}=\frac{h}{0.5} \Rightarrow x=\frac{h}{2}.$$
Substitute this into the volume formula: $$V=\frac{(0.3+2\cdot \frac{h}{2})+0.3}{2}\cdot h\cdot 10=\frac{(h+0.6)h}{2}\cdot 10=5h^2+3h.$$ Take the derivative of the volume function with respect to time: $$\frac{dV}{dt}=\frac{dV}{dh}\cdot \frac{dh}{dt}=2 \ \frac{m^3}{\text{min}} \Rightarrow \\ (10h+3)\cdot \frac{dh}{dt}=2 \Rightarrow \\ (10\cdot 0.3+3)\cdot \frac{dh}{dt}=2 \Rightarrow \\ \frac{dh}{dt}=\frac{2}{6}=\frac13.$$
• Thank you for the very clear explanation! I obviously underestimated the geometric component to these questions. I'm curious, how would you do a problem like this if the given shape was a cone or a sphere and you were given the radius and height? How would you approach your first step where you put x in terms of h in a cone? For example, say you had a cone on it's tip with an arbitrary radius and height, as well as the height the cone is filled with fluid. How would you find the volume of the cone filled with fluid in terms of its height? – Bob Smith Oct 26 '18 at 19:39
• Sorry forgot to @ you in the last comment and now I can't edit it. – Bob Smith Oct 28 '18 at 2:08
• Search MSE. E.g.: this, this, this, this. Good luck. – farruhota Oct 28 '18 at 2:29
• Thank you that answers my question perfectly. – Bob Smith Oct 28 '18 at 5:30
Let v(h) be the volume of that shape at height of h.
That is a formula you'll need to create.
You'll also have to calculate dv/dh.
Rate of increase of volume = dv/dh × dh/dt.
Solve for dh/dt knowing the volume increases
2 cubic meters/sec. Finally set h = 0.3 meters.
First note that the trough is of constant cross-section, so we can easily convert this from a volume problem to an area problem. If the trough is filling at $$2 \mathrm{m^3/min}$$, then the filled cross-sectional area is increasing at a rate of $$\frac{2}{10} = 0.2 \mathrm{m^2/min}$$. Also, work in consistent units, so convert everything in $$\mathrm{cm}$$ to $$\mathrm{m}$$.
Now look at the geometry of the problem. Similarity is often very useful here. If you extend (produce) the slant sides of the trapezoid to meet at a point, you'll get a larger inverted triangle with height $$0.8 \mathrm{m}$$ and base $$0.8 \mathrm{m}$$. If you consider fluid level height $$H$$ from this (imaginary) vertex instead, the problem is made rather simple. The base of this triangle is always equal to the height, so its area $$A = \frac 12 H^2$$. At the specified instant, $$H = 0.3 + 0.3 = 0.6$$ (the actual fluid depth in the trough plus the "imaginary" height of the produced triangle).
By chain rule, $$\frac{dA}{dt} = \frac{dA}{dH} \cdot \frac{dH}{dt}$$
Since $$\frac{dA}{dH} = H$$, you get $$\frac{dH}{dt} = \frac{\frac{dA}{dt}}H = \frac{0.2}{0.6} = \frac 13 \mathrm{m/min}$$, which is the required answer.
Note that introducing the "imaginary" triangular portion below the base of the trapezoid doesn't affect the analysis because when you take the derivative of the area, the constant area of that portion vanishes.
If you find the above hard to imagine, you can work out the actual area of the fluid filled cross-section of the trapezoid. It's only a little harder. If you wanted to employ this method, you need a relationship between the top of the filled cross-sectional trapezoid (call this $$p$$) to the actual fluid level (let's call this $$h$$). You'll find (again, by similarity) that this is a simple linear relationship, namely $$p = h + 0.3$$.
So the area $$a$$ of the filled trapezoid will be given by $$a = \frac 12(p+0.3)(h) = \frac 12 h(h+0.6) = \frac 12(h^2 + 0.6h)$$. This gives $$\frac{da}{dh} = \frac 12(2h + 0.6) = h + 0.3$$.
Following what we did before, $$\frac{da}{dt} = \frac{da}{dh} \cdot \frac{dh}{dt}$$, so $$0.2 = (h+0.3)\cdot \frac{dh}{dt}$$ and $$\frac{dh}{dt} = \frac{0.2}{0.3+0.3} = \frac 13 \mathrm{m/min}$$, as before. | 2019-10-14T05:25:59 | {
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http://math.stackexchange.com/questions/159662/squeeze-theorem-help | # Squeeze Theorem Help
How can I argue that $$\lim_{x \to 0} x^2 \cos\left(\frac{1}{x^2}\right) = 0$$
I understand I have to use a squeeze theorem and that one piece goes to zero but I'm not sure how to tackle this problem to show on a test.
-
Squeeze theorem is a big hint. Since you know you have to apply squeeze theorem, that means you need to find upper and lower bound for your function, for which you are trying to find the limit. Now, look at your function and figure out whether you can give upper and lower bounds for it. – TenaliRaman Jun 17 '12 at 22:08
Use $-1 \le \cos(\frac{1}{x^2}) \le 1$ and multiply through by $x^2$. Since $x^2 \ge 0$, the inequalities remain valid.
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I thought it was between 0? – soniccool Jun 17 '12 at 22:13
@mystycs What do you mean? – Cocopuffs Jun 17 '12 at 22:18
Oh nevermind if its cos it has to be between -1 and 1 i got it – soniccool Jun 17 '12 at 22:18
So you just plugin between two values and just prove it correct? – soniccool Jun 17 '12 at 22:20
To prove that $-1 \le \cos(1/x^2) \le 1$? It follows from the identity $sin^2 + cos^2 = 1$. – Cocopuffs Jun 17 '12 at 22:23
We have that $-1 \leq \cos (1/x^2) \leq 1$ for any $x$. So $-x^2 \leq x^2\cos(1/x^2) \leq x^2$. Therefore $$\lim_{x \to 0} -x^2 \leq \lim_{x \to 0} x^2\cos(1/x^2) \leq \lim_{x \to 0} x^2.$$ But we have that $$\lim_{x \to 0} x^2 = 0$$ and $$\lim_{x \to 0} -x^2 = 0.$$
So $$0 \leq \lim_{x \to 0} x^2\cos(1/x^2) \leq 0$$ and therefore by the squeeze theorem, $$\lim_{x \to 0} x^2\cos(1/x^2) = 0.$$
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Please use \cos rather than just cos inside equations. – Arturo Magidin Jun 17 '12 at 22:38
@ArturoMagidin Right. Thanks. – Eugene Jun 17 '12 at 22:39
You could try the double inequality $0\leqslant|x^2\cos(1/x^2)|\leqslant x^2\to0$ when $x\to0$.
- | 2015-04-28T02:12:37 | {
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http://openstudy.com/updates/4f2b6efee4b0e1bedb5ceb16 | ## anonymous 4 years ago A rock is thrown upward from level ground in such a way that the maximum height of its flight is equal to its horizontal range R. (a) At what angle theta is the rock thrown? (b) In terms of its original range R, what is the range R_max the rock can attain if it is launched at the same speed but at the optimal angle for maximum range? (c) Would your answer to part (a) be different if the rock is thrown with the same speed on a different planet? Explain.
1. anonymous
I tried writing equations in such a way that things like v_0 would cancel each other, but I've had no luck. I've been working on it for a while now. Any suggestions?
2. anonymous
The maximum height is $y_{\max} = {v^2 \sin^2 (\theta) \over 2g}$The range is$r = {v^2 \over g} \sin(2 \theta)$ We can set these equal together and solve for $$\theta$$
3. anonymous
The angle to maximum range can be found to be 45 degrees. This is because 2*45 = 90 which makes sin(2*theta) = 1. This angle is always true regardless of what planet we are on because we are maximizing the sine function and gravity is not part of the sine function.
4. anonymous
What is the derivation for your y_max equation?
5. anonymous
I'm just curious as to where it comes from.
6. anonymous
I can derive that for you. We know, from kinematics, that the vertical distance can be expressed as$y = v_y t - {1 \over 2} g t^2$where $$v_y = v \sin(\theta)$$We need to find the time it takes to reach the maximum height. To do this, we take the derivative with respect to time, set to zero, and solve for t. $0 = v_y - g t$$t = {v_y \over g}$Substituting this time back into our original equation, we get$y_\max = v_y \left[ v_y \over g \right ] - {1 \over 2} g \left [ v_y \over g \right]^2$Simplifying, $y_{\max} = {v_y^2 \over g} - {1 \over 2}{v_y^2 \over g}$Combining like terms. $y_{\max} = {1 \over 2} {v_y^2 \over g}$Substituting our expression for $$v_y$$ in we obtain. $y_{\max} = {1 \over 2} {v \sin^2(\theta) \over g}$
7. anonymous
Interesting. Thank you! As far as the original problem goes, I'm here, but I'm not sure what to do with the number to solve for theta? $\sin^2 \Theta =\sin(2 \Theta)$
8. anonymous
Oh, well I'm missing a 2 in there, for one.
9. anonymous
Trig identities! I hate 'em but they are great here. $\sin(2 \theta) = 2\cos(\theta) \sin(\theta)$$\sin^2(\theta) = 2 \sin(\theta) \cos(\theta) \rightarrow \sin(\theta) = 2 \cos(\theta) \rightarrow \tan(\theta) = 2$
10. anonymous
I forgot to write a two in the expression above. It should have been... $\frac{\sin^2 \Theta}{2}=\sin(2 \Theta)$Doing a similar process to what you did above, I get...$\sin \Theta = \cos \Theta$or$\frac{\sin \Theta}{\cos \Theta}=1$Is there a good analytical way to solve this? Either by finding at what values sin and cos are equal, or some other method?
11. anonymous
Oh. $\tan \Theta = 1$
12. anonymous
For that, though, I just get 45 degrees, and that's not correct, is it?
13. anonymous
That doesn't seem right. Let me solve it by hand.
14. anonymous
You are doing your trig identities wrong. ${\sin^2(\theta) \over 2} = \sin(2 \theta) \rightarrow {\sin^2(\theta) \over 2} = 2 \sin(\theta) \cos(\theta) \rightarrow \tan(\theta) = 4$
15. anonymous
Somewhere around 75 degrees.
16. anonymous
Oh! I wasn't doing my trig identities work. I was cancelling out the 2s. Doh. That makes perfect sense. Thank you for your help.
17. anonymous
*Wrong.
18. anonymous
How did you get the equation for range? I understand taking the derivative of the height function, now I'm confused about the range one...
19. anonymous
It is derived in a similar fashion to the maximum height function. Let me walk you through it. We know that there is no acceleration in the horizontal direction, therefore, the range depends on the initial horizontal velocity, $$v_x$$ and the time of flight, which comes from the equations of motion in the vertical direction. Let's start with an expression for range. $r = v \cos(\theta) t$If we substitute in the time we get for flight time, which is$t_{flight} = {2 v \sin(\theta) \over g}$we get$r = \left ( v \cos(\theta) \right) \left(2 v \sin(\theta) \over g \right)$$r = {2v^2 \sin(\theta) \cos(\theta) \over g}$We can use the following trig identity to further simplify the expression further. $\sin(2 \theta) = 2\cos(\theta)\sin(\theta)$Therefore, at last, $r = {v^2 \sin(2 \theta) \over g}$
20. anonymous
How did you get $t _{flight}=\left(\begin{matrix}2vsin(\theta) \\ g\end{matrix}\right)$?
21. anonymous
$t _{flight}$ is twice the time of $y _{\max}$ | 2016-10-22T09:13:52 | {
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http://math.stackexchange.com/questions/51435/proving-an-inequality | # Proving an inequality
Suppose $a$ and $b$ are real numbers. Prove that if $a<b$ then $\frac{a+b}{2}<b$.
The 'solution' hints at adding $b$ to both sides of the inequality $a<b$, and $a+b<2b$ is as far as I've got (I don't know the reason for adding $b$ to both sides, either - or rather, why you'd think to add the $b$ to each side to help finish the proof, instead of doing something else) - I'd like to know what to try next.
Source of exercise: How To Prove it: A Structured Approach, Second Edition - Daniel J. Velleman.
Thank-you for the responses. Is the following solution sufficient proof, or is there something I'm missing?
Proof. Suppose $a$ and $b$ are real numbers, and $a<b$. Adding $b$ to both sides of the inequality $a<b$, we get $a+b<2b$. Subsequently, we can divide both sides by 2, to get $\frac{a+b}{2}<b$. Therefore, if $a<b$, it follows then that $\frac{a+b}{2}<b$.
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Here is the reasoning beyond the approach. You deal with fraction(s), so the first natural thing is to bring everything to the same denominator or get rid of the denominator. If you bring everything to the same denominator (you can instead multiply everything by the common denominator), the inequality you need to prove is $\frac{a+b}{2} < \frac{2b}{2}$. Look now to the numerators, and try to see how can you get it from the first inequality.. The numerator on the left side has an extra $b$ and the numerator on the right side has ...... – N. S. Jul 14 '11 at 15:24
Your proof is perfect! +1 – t.b. Jul 14 '11 at 15:41
One thing I'd like to add: You can "see" that the inequality holds by drawing the real line, marking the two points $a \lt b$ on it and noticing that $\frac{a+b}{2}$ is the midpoint of $a$ and $b$. Saying that $a \lt b$ is equivalent to saying that the point $a$ is to the left of $b$, hence the midpoint $\frac{a+b}{2}$ must be to the left of $b$ or, in other words, $\frac{a+b}{2} \lt b$. – t.b. Jul 14 '11 at 15:48
That's helpful, and interesting. Thanks! – Daniel May Jul 14 '11 at 16:01
The reason for adding $b$ to both sides is that you want to get $a+b$. Once you've got $a+b < 2b$, you can divide both sides by $2$. The reason for doing that is that you want $(a+b)/2$. Here it matters that $2$ is positive: if you divide both sides by a negative number, then you'd have to change $<$ to $>$, but since $2$ is positive, $<$ remains $<$.
-
We have $a < b$, but we want $a+b$ on the left hand side. So, an easy way to go from $a$ to $a+b$ on the left hand side is to add $b$ to the left hand side, and since we can't just add things to one side of the inequality, we better add $b$ to both sides. At this point, we're just trying things; there's no epiphany that "Oh, of course this is going to work!", just poking at the problem and seeing what we can do with it.
-
Your proof is faultless. Of course you are assuming that adding the same thing to both sides preserves order, and that dividing both sides by a positive number also preserves order. However, these are by now to be considered standard facts in the course. Perhaps you should have said "Since $2$ is positive, we can $\dots$". But that may be unnecessarily fussy.
The "trick" of adding $b$ to both sides is fairly natural. It only has a faint whiff of magic.
Maybe you did the following before writing down a proof, at least I hope you did. On a "number line", put a dot for $a$, one for $b$, and one for $(a+b)/2$, halfway between them. This finishes things, you now know the result is true. It only remains to write down the details, using the notations and tools of your course. (Theo Buehler writes about the geometry in a comment which is far more important than the question that led to it.)
The geometry suggests the following alternative approach. Recall that $x<y$ iff $y-x>0$. So it is enough to prove that $b-\frac{a+b}{2}>0$. But we have $$b-\frac{a+b}{2}=\frac{2b-(a+b)}{2}=\frac{b-a}{2}>0$$ (the inequality $\frac{b-a}{2}>0$ follows from $a<b$.)
A little bit more complicated, but it arises from an understanding of what's happening underneath. And it tells you more than the simple fact that $b>(a+b)/2$. It tells you by how much $b$ is bigger than $(a+b)/2$.
By the way, for complicated expressions $X$ and $Y$, one of the standard strategies for showing that $X<Y$ is to show that $Y-X>0$.
Comment: Back to your solution. At the manipulational level, maybe this is how I would think. I want to show that $\frac{a+b}{2}<b$. Fractions are unpleasant, they are "broken" numbers (that's the correct etymology). So let's unbreak the fracture.
Note that the desired inequality holds iff $a+b <2b$. And $a+b<2b$ is an obvious consequence of $a<b$. Now we can hide the reasoning that led to the solution by writing the proof backwards. Or not.
-
Here is how I would discover that one needs to add, or rather subtract, $\;b\;$, without any 'magic': I would start at the most complex side, and simplify from there, as follows:
\begin{align} & \tfrac {a+b} 2 < b \\ \equiv & \;\;\;\;\;\text{"arithmetic: multiply both sides by 2 -- to simplify the left hand side"} \\ & a+b < 2 \times b \\ \equiv & \;\;\;\;\;\text{"subtract $\;b\;$ from both sides -- to simplify both sides at the same time"} \\ & a < b \\ \end{align}
Now, some call this proof 'backwards'. However, simplifying $\;\tfrac {a+b} 2 < b\;$ is a lot easier than trying to add information to $\;a < b\;$: in the above calculation, at each point there is not much choice about what to do next, while $\;a < b\;$ is a starting point which allows too much choice.
Also, note how the above does not only prove the required "if $\;a < b\;$ then $\;\tfrac {a+b} 2 < b\;$", but it proves the stronger equivalence $$\tfrac {a+b} 2 < b \;\equiv\; a < b$$ for any real $\;a,b\;$. | 2015-04-18T17:25:36 | {
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https://math.stackexchange.com/questions/2954624/irrationality-of-r-sqrt5-for-rational-number-r | # Irrationality of $r\sqrt{5}$ for rational number $r$? [duplicate]
As $$5$$ is a prime number, thus $$\sqrt{5}$$ is an irrational number.
Now I am thinking about how to prove -
If $$r$$ is a rational number, then how do we prove $$r\sqrt{5}$$ is an irrational number?
I was thinking that since $$r$$ is a rational number, then $$r$$ can be expressed as the fraction in simplified form that is $$r = \frac{a}{b}$$ such that $$a,b \in \Bbb{Z}$$ and $$gcd(a,b)=1$$. So $$r\sqrt{5} = \frac{a\sqrt{5}}{b}$$, but How can this guarantee us the irrationality of $$r\sqrt{5}$$?
ALso let $$c =r\sqrt{5}$$, then $$c^2 = 5r^2$$, if we could prove it is a prime number, then its square-root $$c$$ must be irrational and ths proved but unfortunately we donot have $$c^2$$ prime as it has more than one factors like $$r$$ , $$5$$
## marked as duplicate by Eric Wofsey, Lord Shark the Unknown, Paul Frost, Don Thousand, Delta-uOct 15 '18 at 0:39
Now if $$r\ne 0$$ is rational and $$r\sqrt 5$$ is also rational, we have to have $$\frac {r\sqrt 5}{r} = \sqrt 5$$ to be rational and we know that it is not rational, so $$r\sqrt 5$$ must be irrational.
• Yes, you are right , $r$ must be non-zero. I fixed my answer. Thanks – Mohammad Riazi-Kermani Oct 14 '18 at 3:58 | 2019-10-16T04:15:09 | {
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https://math.stackexchange.com/questions/2441967/definition-of-rank-of-a-matrix | # Definition of rank of a matrix
Can I define the rank of a matrix(A) as the number of non zero rows in RREF(A)? Here's my reason: Let number of zero rows be $x$
Then these rows are the linearly dependent rows of A and $x=dim(left null space)=m-r$.
So number of non zero rows is equal to $rows-x=m-(m-r)=r$.
• Yes, what you state is a well known result. Sep 23 '17 at 17:03
Yes, this also follows immediately from the fact that the Gauss-algorithm leaves the rank of a matrix unchanged. Since the rank $\operatorname{RREF}(A)$ is the number of its non-zero rows the claim follows. | 2022-01-28T18:45:20 | {
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https://ijm.telkomuniversity.ac.id/3vpsk1b/article.php?963f9b=perimeter-of-sector-formula-radians | Area of sector formula and examples- The area of a sector is the region enclosed by the two radius of a circle and the arc. Arc length . Area of a sector formula. Sector is the portion of a disk enclosed by two radii and an arc. Arc Length Formula - Example 1 Discuss the formula for arc length and use it in a couple of examples. The volume of the box is 300 cm3. C2 Trigonometry: Arc Length & Sector Area PhysicsAndMathsTutor.com Edexcel Internal Review 3 (a) Show that the surface area of the box, S cm . What is the formula to find the perimeter of a sector of a circle? The Area of a Segment is the area of a sector minus the triangular piece (shown in light blue here). So the length of the arc of the sector = 2 (pi)r* (theta/360). Therefore 360º = 2 PI radians. Arc Length = θr. Example: a) What is the length of the arc intercepted by an angle of 15° on a circle with radius 20 meters? Problem 7 : Find the radius of sector whose perimeter of the sector is 30 cm and length of the arc is 16 cm. As we know mathematics is not a spectator sport so we also got through its application in some practical examples of area and perimeter related to circle and arc. Practice Questions. It's an exciting event: ant races! Teachoo is free. Then, simplify the formula and the formula for area of sector when angle Ө is in radians will then be derived as Area = (1/2) X r²Ө. Sometimes, the portion of a circle is known. This page includes a lesson covering 'finding the area of a sector of a circle when the angle is given in radians' as well as a 15-question worksheet, which is printable, editable, and sendable. Favorite Answer. Whether you want to calculate the Area (A), Arc (s), or one of the other properties of a sector including Radius (r) and the Angle formed, then provide two values of input. Learn Science with Notes and NCERT Solutions, Area of combination of figures - two circles, circle and square, Finding Area of rectangle with path outside/inside, Finding Area of rectangle with cross roads. Yes, though it can be expressed more simply. One radian is equal to the angle formed when the arc opposite the angle is equal to the radius of the circle. Terms of Service. We’re looking for the perimeter of this sector. Just replace 360˚ in the formula by 2π radians (note that this is exactly converting degrees to radians). Login to view more pages. The radius of a circle is seven centimeters and the central angle of a sector is 40 degrees. 1. = 44 + 2 (21) Area of a circle is given as π times the square of its radius length. First, we want to just sketch our circle and then label each part. You will learn how to find the arc length of a sector, the angle of a sector or the radius of a circle. I know the formula for arc length is rθ. Relevance. These are the conversion formulas for radians to degrees and for degrees to radians, respectively. ... Lv 7. 3 Answers. To calculate the area of the sector you must first calculate the area of the equivalent circle using the formula stated previously. We’re also interested in the perimeter of this sector. Section 4.2 – Radians, Arc Length, and the Area of a Sector 4 Sector Area Formula In a circle of radius r, the area A of a sector with central angle of radian measure T is given by . Area of a sector formula. Area (Radians) = ½r 2 θ. r, D° r, R° r, s r, A D°, s. Radius (r) Angle (D°) Angle (R°) Arc (s) Area (A) *Radius and Arc in units (e.g. Perimeter and Sector Defined. You can work out the Area of a Sector by comparing its angle to the angle of a full circle.Note: we are using radians for the angles.This is the reasoning: Area of Sector = θ 2 × r2 (when θ is in radians)Area of Sector = θ × π 360 × r2 (when θ is in degrees) Answer Save. Example 10: Find the perimeter of a sector with central angle 60 and radius 3 in. There are two sector area formulas; one for a sector measured in radians, and another for a sector measured in degrees. so does that mean the perimeter of a sector is rθ+2r??? 625 = 18 x 18 x θ/2. The perimeter will be the distance around this sector, which is a radius plus a radius plus an arc length. MALATHI VEDAGIRI. We know that a full circle is 360 degrees in measurement. 2, is given by . Comparing the area of sector and area of circle, we get the formula for the area of sector when the central angle is given in radians. We have seen in this section how we are supposed to calculate area and perimeter of circle and arc. Videos, worksheets, 5-a-day and much more Comparing the area of sector and area of circle, we get the formula for the area of sector when the central angle is given in radians. A “sector” (of a circle) is bounded by an arc and two radii, so the perimeter is two times the radius (r) plus the length of the arc. We usually use the variable to represent the arc length. For the full circle, the angle is 2π radians and the perimeter is the circumference which is 2πr, i.e. To convert a certain number of radians into degrees, multiply the number of radians by 180/ PI . l + 2r = 45. The sector has radius r cm and angle 1 radian. where θ is the measure of the arc (or central angle) in radians and r is the radius of the circle. Compare the areas of three sectors -- each with P = 100 -- central angles of 45 degrees, 90 degrees, and 180 degrees. Calculation precision. Recall that the angle of a full circle in radians is 2π. one radius per one radian, which is pretty much the how and why of "radian's" existence. I know the formula for arc length is rθ. Formula to find perimeter of the sector is = l + 2r. See the video below for more information on how to convert radians and degrees. Therefore to convert a certain number of degrees in to radians, multiply the number of degrees by PI /180 (for example, 90º = 90 × PI /180 radians = PI /2). For a sector the area is represented by some other angle. What is the formula for the perimeter of a sector in terms of radians? Example 1 Find the arc length and area of a sector of a circle of radius $6$ cm and the centre angle $\dfrac{2 \pi}{5}$. Area of a sector = (θr 2)/2. 1 decade ago "measure of the angle x radius"..since you are measuring a distance the angle must be in radians. Teachoo provides the best content available! Substitute 10 for r. l + 2 (10) = 45. l + 20 = 45. l = 45 - 10. l = 35 cm. 1. So in the below diagram, s = r∅ . The arc is the outer edge of the sector. The formula for sector area is simple - multiply the central angle by the radius squared, and divide by 2: Sector Area = r² * α / 2; But where does it come from? The formula for the area of a sector is (angle / 360) x π x radius 2.The figure below illustrates the measurement: As you can easily see, it is quite similar to that of a circle, but modified to account for the fact that a sector is just a part of a circle. Let the angle made between the 2 radii that have resulted in the formation of that sector, be equivalent to “x”. Anonymous. We are given the radius of the sector so we need to double this to find the diameter. 1 decade ago. The area of the sector … Source(s): ... where θ is in radians. 3 Answers. Learn how tosolve problems with arc lengths. On signing up you are confirming that you have read and agree to Perimeter of Sector of Circle Calculator. Example: the perimeter of this rectangle is 7+3+7+3 = 20. One of the sectors measures 40 degrees. One of the sectors measures 40 degrees. My question says a sector of a circle has a radius of 9 cm and the angle is 80 degrees find the perimeter of the sector? Help Fast!!! These are sectors that are an eighth circle, quarter circle, and semicircle. Radians, like degrees, are a way of measuring angles. Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. Perimeter. Let this region be a sector forming an angle of 360° at the centre O. 0 0. The formula for sector area is simple - multiply the central angle by the radius squared, and divide by 2: Sector Area = r² * α / 2; But where does it come from? Perimeter of an ellipse formula (ellipse circumference formula) Although the formula for the area of an ellipse is really simple and easy to remember, the perimeter of an ellipse formula is the most troublesome of all the equations listed here. We’re looking for the perimeter of this sector. The perimeter of a sector is composed of three pieces, an arc of the circle and two radii. Suppose the length of the arc is a cm and the angle at the centre of the circle subtended by the arc is θ radians. This means that in any circle, there are 2 PI radians. 625 = 162 θ. Divide both sides by 162. θ = 3.86 radians. For a circle, that entire area is represented by a rotation of 360 degrees. Arc Length = 14 × 2.4 = 33.6 person_outlineAntonschedule 2011-05-06 20:21:55. the perimeter of a sector is rθ + 2r*sin(θ/2) 0 0. Answer Save. Angles will be in Radians or Degrees. If the radius of the sector is 18 mm, find the central angle of the sector in radians. With the relevant angle given in radians here, the sums changes slightly, but still give a good measure of segment perimeter. The formula for the length of an arc is:: 570 = where L represents the arc length, r represents the radius of the circle and θ represents the angle in radians … perimeter of sector), r is radius and θ is angle of sector then: s = r*θ where θ is in radians (s = r*θ*pi/180 if θ is in degrees: this is the equivalent of your expression: s = r*θ*2pi/360) A sector is formed between two radii and an arc. Lv 7. There is a lengthy reason, but the result is a slight modification of the Sector formula: The following video shows how this formula is derived from the usual formula of Area of sector = (Ө/360˚) X πr². To find the perimeter, we need to add these values together. Area of Sector = θ 2 × r 2 (when θ is in radians) Area of Sector = θ × π 360 × r 2 (when θ is in degrees) Area of Segment. The area of a sector of a circle is ½ r² ∅, where r is the radius and ∅ the angle in radians subtended by the arc at the centre of the circle. Ask Question + 100. Convert 330° to radians. Worked solution to a question involving arc length and perimeter of a sector Recall that the angle of a full circle in radians is 2π. Answer Save. in the Radians giving an answer to one decimal place. A sector of a circle is the shape formed by slicing up a circular cake. Solution. Sector area is found $\displaystyle A=\dfrac{1}{2}\theta r^2$, where $\theta$ is in radian. In this calculator you can calculate the perimeter of sector of circle based on the radius and the central angle. The following mathematical formula is used in this circular sector calculator to find the area for the given input values of radius r & the angle θ in degrees. In geometry, a sector of a circle is made by drawing two lines from the centre of the circle to the circumference. so does that mean the perimeter of a sector is rθ+2r??? Arc Length Formula - … Solution : A sector of angle 5π/12 radians is cut from a circle of radius 6 cm. The “perimeter” of any closed shape is simply the sum of the lengths of all of its boundaries. The sector is formed of two radii and the arcual length, so the perimeter of the sector =2r + 2 (pi)r* (theta/360) one radius per one radian, which is pretty much the how and why of "radian's" existence. It's also very important to remember to have a calculator set to "radians" and NOT "degrees", when working out the sin value. The formulas to find the area of a sector in Degrees (D°) or Radians (R°) are shown below: Area (Degrees) = πr 2 x θ/360. Therefore 180º = PI radians. Understanding the problem. The formula for finding the area of a circle is pi*r*r where r is the radius. 1) In the video lesson, we learned that the perimeter of a sector of a circle, like a slice of pizza, is the sum of two edges that are both the radius and the edge that is the arc length. Therefore to convert a certain number of degrees in to radians, multiply the number of degrees by PI /180 (for example, 90º = 90 × PI /180 radians = PI /2). We have a radius of seven centimeters. To convert a certain number of radians into degrees, multiply the number of … Perimeter of a sector. This means that in any circle, there are 2 PI radians. Area enclosed by an arc of a circle or Area of a sector = (θ/360 o ) x πR 2. The perimeter of the sector includes the length of the radius $\times 2$, as well as the arc length.So the perimeter is the length "around" the entire sector, the length "around" a slice of pizza, which includes it's edges and its curved arc.. or A = rl / 2 square units. Exercise worksheet on 'Find the area of a sector of a circle when the angle is given in radians.' Anonymous. If s is arc length (i.e. He has been teaching from the past 9 years. Radius. Mein Hoon Na. This formula helps you find the area, A, of the sector if you know the central angle in degrees, n °, and the radius, r, of the circle: A = ( n ° 360 ° ) × π × r 2 For your pumpkin pie, plug in 31 ° and 9 inches: Calculates area, arc length, perimeter, and center of mass of circular sector. Anonymous. What is the formula for the perimeter of a sector? 1 0. Perimeter of sector and area of sector: Trigonometry: Feb 6, 2016: Finding area of a sector and the measure of part of circumference of a circle: Geometry: Aug 23, 2015: Finding areas of sectors using radians: Trigonometry: Feb 15, 2013 Area of sector formula and examples- The area of a sector is the region enclosed by the two radius of a circle and the arc. Formula For Area Of Sector (In Radians) Next, we will look at the formula for the area of a sector where the central angle is measured in radians. 2 Answers. A sector of a circle has a perimeter made up of two radii and an arc of the circle connecting the endpoints of the two radii. A1837-16∘, 0.7 rad B3714-32'∘, 0.7 pleased C1837-16'∘, 0.3 rad D3714-32'∘, 0.3 rad No.24: the length of the arc of the sector is 33 cm, and the perimeter of 67 cm. A) 2 B) 4 C) 72 D) 144 E) 12 F) None of these . Solved Example The below solved example problem may be useful to understand how the values are being used in the mathematical formulas to find the sector area of … Yes, it is rθ+2r. So the circumference of a circle is 2 PI larger than its radius. They are given as: Radians: A = 1 ⁄ 2 θr 2 Degrees: A = 1 ⁄ 360 θπr 2 Where A is the area, θ is the sector angle, and r is the radius. Worksheet to calculate arc length and area of sector (radians). perimeter of sector), r is radius and θ is angle of sector then: s = r*θ where θ is in radians (s = r*θ*pi/180 if θ is in degrees: this is the equivalent of your expression: s = r*θ*2pi/360) 7: find the central angle 60 and radius 3 in our circle and then label each part edge a. Radians is 2π θr 2 ) /2, all you need to double this to find central. The nearest centimeter ) None of these angle x radius ''.. since you are confirming that have. Way of measuring angles label each part of Technology, Kanpur sum the! 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Have seen in this section how we are supposed to calculate the area of a slice of pizza of radius! The area of a sector minus the triangular piece ( shown in blue. Into degrees, are a way of measuring angles this rectangle is 7+3+7+3 = 20 and much a... From Indian Institute of Technology, Kanpur input value you want, then enter their.. Radians ( note that this is exactly converting degrees to radians, and area of circle... Radians ( note that this is exactly converting degrees to radians ) been teaching the... A slight modification of the sector is rθ+2r???????????... Can find the perimeter of this sector a minor arc AB of radian 's ''.! 15° on a circle with radius 20 meters area, arc length, perimeter and. Sector or the radius of the sector so we need to double this to find the length of sector. Is 360 degrees in measurement 360° at the centre of the sector PQR shown below bet! 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https://math.stackexchange.com/questions/1184438/given-a-transformation-matrix-t-find-t-relative-to-a-new-basis-beta | # Given a Transformation Matrix $T$, find $T$ relative to a new basis $\beta$
$T(a_1,a_2,a_3) = (3a_1+a_2,a_1+a_3,a_1-a_3)$. $(a_1,a_2,a_3)^T$ is written with regards to the standard basis.
We can figure out $T$ in matrix form by calculating $T(a_1),T(a_2), T(a_3)$. That's easy and we get:
$T$ = $\left(\begin{array}{ccc}3 & 1 & 0 \\1& 0 & 1 \\1 & 0 & -1\end{array}\right)$
$\beta$ is a new basis = $\{(1,0,0), (1,1,0), (1,1,1)\}$.
Here's what I'm trying to find: $T$ relative to this new basis.
I have three questions:
First, How do I methodically set up the problem and solve it?
Second, why does $T*C = S$ work?, where
$T$ is the transformation matrix relative to $(a_1,a_2,a_3)^T$,
$C = \left(\begin{array}{ccc}1 & 1 & 1 \\0& 1 & 1 \\0 & 0 & 1\end{array}\right)$.
and $S$ is the solution = $\left(\begin{array}{ccc}3 & 4 & 4 \\1& 1 & 2 \\1 & 1 & 0\end{array}\right)$.
Third Question: how general is $T*C$ as a solution? Did I just get lucky? It seems all I'm doing is multiplying a new vector (a vector in $\beta$) by an old transformation matrix. Why would this give me a new, correct $T$? Basically, if someone asks me to find $T$ relative to a new basis can I just use $T*C$, where $C$'s columns contain the new basis vectors?
Edit: How do you do this without using inverses?
• The transformation relative to the new basis should be $C^{-1}TC$, assuming the the columns of $C$ are the new basis. The provided S does not match. What do you mean when you say it "works"? – megas Mar 10 '15 at 23:46
• I've changed C to what it should have been. By works, I mean T*C is correct; that is it equals the solution. – larry Mar 10 '15 at 23:54
• To clarify, $S$ is a given solution, right? This $T \cdot C$ matrix describes a transformation that when applied to a vector $v$ expressed in the new basis, it yields a new vector $v^{\prime}$ that is the "correct" vector only if you interpret/read it in the standard basis. So not exactly correct... I hope that made some sense... – megas Mar 11 '15 at 0:02
Let
$\mathcal{E}=\left\{e_1,e_2,e_3\right\}$
be our canonical base. With this base, transormation T has representation
$T=\left( \begin{array}{ccc} 3 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 0 & -1 \\ \end{array} \right)$.
Now we have got a new base:
$\mathcal{F}=\left\{e_1,e_1+e_2,e_1+e_2+e_3\right\}$.
Let
$M_{\mathcal{F}}=\left( \begin{array}{ccc} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} \right)$
be the transition between the two bases.
Then canonical coordinates are transormed in new coordinates (with respect to base $\mathcal{F}$ ) by inverse matrix, which is
$N_{\mathcal{F}}=\left( \begin{array}{ccc} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \\ \end{array} \right)$.
Take
$A=\left\{a_1,a_2,a_3\right\}$
and get new coordinates
$B=N_{\mathcal{F}}.A$.
Then, with $S=T.M_{\mathcal{F}}$
we see:
$T.A=T.M_{\mathcal{F}}.N_{\mathcal{F}}.A=S.B$.
It's not a miracle, only lin. Algebra. Key is transformation of basis, which implies transformation of coordinates. That's all.
By the way: Calculating without inverses is not possible. Your transformation with bases must be regular. They must be invertible, otherwise it didn't work.
Let's see. Other basis
$\mathcal{B}=\left\{2 e_1+5 e_3,e_1+e_2+6 e_3,3 e_1+9 e_3\right\}$,
another transition:
$M_{\mathcal{B}}=\left( \begin{array}{ccc} 2 & 1 & 3 \\ 0 & 1 & 0 \\ 5 & 6 & 9 \\ \end{array} \right)$.
The inverse:
$N_{\mathcal{B}}=\left( \begin{array}{ccc} 3 & 3 & -1 \\ 0 & 1 & 0 \\ -\frac{5}{3} & -\frac{7}{3} & \frac{2}{3} \\ \end{array} \right)$.
Old transformation T
$T=\left( \begin{array}{ccc} 3 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 0 & -1 \\ \end{array} \right)$.
Transformed T:
$S=T.M_{\mathcal{B}}=\left( \begin{array}{ccc} 6 & 4 & 9 \\ 7 & 7 & 12 \\ -3 & -5 & -6 \\ \end{array} \right)$
Transformed A:
$B=N_{\mathcal{B}}.A$.
$T.A=T.M_{\mathcal{B}}.N_{\mathcal{B}}.A=S.B$
Like before.
• would this have worked if our first basis was not the standard basis? – larry Mar 10 '15 at 23:57
• @larry: It does work with any basis. And we allways know, how this basis is related to the canonical basis. See our new basis above. New base-vectors are linear-combinations of old basis-vectors. Hope this helps. – Frieder Mar 11 '15 at 0:05
• If we started out with a basis $\epsilon = \left(\begin{array}{ccc}2 & 1 & 3 \\0 & 1 & 0\\ 5 & 6 & 9 \end{array}\right)$, how would $T$ change? – larry Mar 11 '15 at 0:15
My linear algebra is a bit rusty, so I think this answer should help you out more than I could: https://math.stackexchange.com/a/340991/221695
In your particular case, you do seem to have gotten lucky. As detailed in the post above, the methodical way is to set up this change-of-base matrix C correctly and then right-multiply your transformation matrix T by the inverse of C.
I may be mistaken, but I think you've treated the basis vectors as row vectors while constructing your C, ie
$$v_1=(1,0,0)$$
and
$$C = \left(\begin{array}{c}v1 \\v2 \\v3\end{array}\right)$$
While it should have been something like:
$$v1 = \left(\begin{array}{c}1 \\0 \\0\end{array}\right)$$
and
$$C=(v1,v2,v3)=\left(\begin{array}{ccc}1 & 1 & 1 \\0 & 1 & 1 \\0 & 0 & 1\end{array}\right)$$
I just tinkered a bit more with this and computing T*C with the C matrix constructed with column vectors as above magically gives the S you specified, which I don't get with your C matrix. Anyway, if you're sure that S is the right answer, I'm not sure the problem is formulated correctly. | 2019-08-23T04:35:28 | {
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https://math.stackexchange.com/questions/2072124/functions-which-are-non-differentiable-on-a-given-set | # Functions Which are non differentiable on a Given Set.
I have recently read the famous Weierstrass Non differentiable function which is continuous everywhere but nowhere differentiable.
But now my question is:
Given a countable set $S (\subset \mathbb{R})$ , can we construct a function which is continuous everywhere but non differentiable only at points of S..?
My Attempt:
If $S$ finite say, $S=\{c_1,c_2,...,c_n\}$ Then the following function is the desired one:$$f(x)=\sum_{i=1}^{n}|x-c_i|$$
But I cannot construct the function for a arbitrary Enumerable set $S.$
Can we generalize the question for an arbitrary set $S$?
Thank you...!!
• You can make an easy example of a set which has corners periodically by just defining the function on, say, $[-1,1]$ and arranging so that it takes equal values on the ends, and then copy-pasting it. Maybe you can use this to handle the general case? I'm not sure - just a thought. – Alfred Yerger Dec 26 '16 at 5:09
• Sir, I think I got your point a little...but Is there any explicit example...??@ A Yerger – indrajit Dec 26 '16 at 5:14
• If you want to write down an explicit formula for such a thing that won't be defined piecewise, it's done with Fourier series. But I have thought of another example which modifies your own, and is also discontinuous on the integers. $f(x) = \sum_{n \in \mathbb Z} \frac{1}{2^{|n|}}|x-n|$. This should be okay since as $n$ moves away from $0$, we shrink the contributions geometrically. – Alfred Yerger Dec 26 '16 at 5:19
• But sir this is an particular case. and I have also construct a function which is non differentiable at only $\mathbb{Z}$ but continuous every where.....But I cannot extend this construction for the arbitrary set $S$. In fact, for $S={a_1, a_2, a_3,.......}$....!!!! – indrajit Dec 26 '16 at 5:24
• Is the construction possible...?? Intuitively it seems to me that this is possible (as u noted in ur first comment) ....but I can't.... – indrajit Dec 26 '16 at 5:26
The same construction can be generalized. Let $a_1,a_2,\dots$ be the elements of $S$.
Then the function $f(x)=\sum\limits_{n=1}^\infty \dfrac{|x-a_n|}{2^n \max(|a_n|,1)}$ is differentiable at all points except the points at $S$.
• Can you prove it? We can try to work it out if you'd like. – Yorch Dec 26 '16 at 5:40
• How does the series Converge....?? – indrajit Dec 26 '16 at 5:50
• hmm, good point. we'll adjust it. – Yorch Dec 26 '16 at 5:56
• Please tell how does the series converge now...??@ J F Hidalgo – indrajit Dec 26 '16 at 6:04
• well, each summand is smaller than $|\frac{x}{a_n}-1|2^{-n}|$ and $|x-a_n|2^{-n}$. If $|a_n|\leq 1$ it is smaller than $(|x|+1)|2^{-n}$ and otherwise it is smaller than $(|x|+1)2^{-n}$ – Yorch Dec 26 '16 at 6:08
Yes. See this answer, on the kinds of sets where a function $f$ may fail to be differentiable. Such a set must be a $G_{\delta\sigma}$ set, meaning it must be an infinite, countable union of $G_{\delta}$ sets (which in turn are countable intersections of open sets).
Notice that every closed set is $G_{\delta}$ and in particular, every singleton is $G_{\delta}$:
$$\{p\}=\bigcap_{n\geq1}\left(p-\frac1n,p+\frac1n\right)$$
Hence, every countable set is $G_{\delta\sigma}$: $C=\cup_{n\in\mathbb{N}}\{c_n\}$.
The paper linked in the answer describes how one may construct continuous functions $f:\mathbb{R}\longrightarrow\mathbb{R}$ whose set of non-differentiability is a $G_1\cup G_2$, where $G_1$ is any $G_{\delta}$ and $G_2$ is a $G_{\delta\sigma}$ with measure zero. In particular, notice that every countable set is both $G_{\delta\sigma}$ and has measure zero, so every countable set may be written as some $G_1\cup G_2$. | 2021-07-28T01:37:50 | {
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https://it.mathworks.com/help/matlab/ref/istril.html | Documentation
# istril
Determine if matrix is lower triangular
## Syntax
``tf = istril(A)``
## 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 = 5×5 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
• 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)```. | 2020-01-29T22:09:59 | {
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https://www.physicsforums.com/threads/solution-needed-kinematics-problem.357235/ | # Solution needed : Kinematics problem
1. Nov 23, 2009
### dawoodvora
1. The problem statement, all variables and given/known data
A ball is dropped from rest from a height h above the ground. Another ball is thrown vertically upward from the ground at the instant the first ball is released. Determine the speed of the second ball if the two balls are to meet at a height h/2 above the ground.
Variables for Ball 1 going downward
v1i = initial velocity of ball1
v1f = final velocity of ball1
Variables for Ball 2 going downward
v2i = initial velocity of ball2
v2f = final velocity of ball2
t = time taken for both the balls to travel common meeting point
a = -9.8 m/s2
2. Relevant equations
4 general kinematic equations in one dimension having constant acceleration
3. The attempt at a solution
Logically, the final velocity of ball 1, when it reaches h/2, must also be the value of the initial velocity of ball2 going upward. Also, ball2 final velocity at h/2 must be zero. In this way, both the balls can meet each other at h/2
I am still unable to arrive at the figure of initial speed of ball2.
2. Nov 23, 2009
### kuruman
The general way of doing this sort of problem is to
(a) First find the time it takes for the balls to pass each other by using the kinematic equation for the ball that is released from height h.
(b) Second use the same kinematic equation and the time you found in (a) to find the initial velocity for the ball that is projected up.
This method will allow you to solve the problem regardless of the fraction of h at which the two balls pass each other.
In this specific case, since you know that the speeds of the balls are the same when they pass each other, you can use the kinematic equation that does not involve time directly.
3. Nov 23, 2009
### Seannation
EDIT: Disregard all of the below, see next post!
You're making innecessary assumptions here. The second ball's velocity needn't necessarily be zero at h/2. Since we're only dealing with one dimension here, there is only one possible initial speed of ball 2 that can allow it to hit ball 1 at h/2.
The two variables which will be equal for both balls when they hit is that they will both have travelled a distance of h/2. However one ball's displacement will be negative to the other as one ball is travelling down and one ball is travelling up.
You're looking to find the initial velocity of the second ball such that the distance travelled by the second ball against gravity equals the distance fallen by the second ball with gravity over the same time period.
Last edited: Nov 23, 2009
4. Nov 23, 2009
### kuruman
If the balls meet halfway up, then the second ball's velocity is zero when they pass each other. This will not be the case if they met at some other fraction of h.
Proof
Let tf = the time the balls meet. The average velocity for ball A is
$$\bar{v}_A=gt_f/2$$
The average velocity for ball B is
$$\bar{v}_B=(v_0+v_0-gt_f)/2=v_0-gt_f/2$$
where v0 is what we are looking for.
Since the balls cover equal distances in time tf, their average velocities must be the same.
$$gt_{f}/2=v_0-gt_f/2\ \rightarrow$$ $$v_0=gt_f$$
Therefore the velocity of ball B at any time t is given by
$$v=gt_{f}-gt$$
At specific time t = tf, ball B has zero velocity. Q.E.D.
5. Nov 23, 2009
### Seannation
You make a very good point! I somehow got it in my head that the second ball would be travelling at high speed when it hit the first. But as you've proved, it can't have a non-zero velocity if it happens at h/2. | 2018-02-22T10:48:56 | {
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https://math.stackexchange.com/questions/1288497/familyof-seminorms-on-normed-spaces | # familyof seminorms on normed spaces
Let $(X,\|\cdot\|)$ be a normed space. It is known that the norm $\|\cdot\|$ induces a topology, known as the norm topology $\tau$ on $X$. Then the pair $(X,\tau)$ is a locally convex topological vector space. With this, $\tau$ can be generated by a family $P$ of semi-norms on $X$.
Question: Is it necessary that $\|\cdot\|$ is in $P$?
No, it is not necessary that $\lVert\,\cdot\,\rVert$ belongs to the family $P$. Unless $X = \{0\}$, when there is only one seminorm on the space - we could work around that if we allow $P = \varnothing$. And if $\dim X = 1$, then all seminorms on $X$ are constant multiples of the norm, and we must put at least one non-zero multiple of the norm in the family. Of course we could always take $P = \{ 2\cdot \lVert\,\cdot\,\rVert\}$ for $\dim X > 0$ and obey the letter of the law but violate its spirit.
More interesting, we can always take a family consisting of seminorms that are not norms if $\dim X > 1$. For a closed linear subspace $F$ of $X$, the norm induces a quotient norm on $X/F$,
$$\lVert \pi(x)\rVert_{X/F} := \inf \{ \lVert x-y\rVert : y \in F\} = \operatorname{dist}(x,F),$$
where $\pi\colon X \to X/F$ is the canonical projection. We can pull that back to a seminorm on $X$, by defining $p_F(x) := \lVert \pi(x)\rVert_{X/F}$.
Then $p_F$ is a continuous (with respect to $\tau$ of course) seminorm on $X$, with kernel $F$. Hence, taking a family $\mathscr{F}$ of nontrivial closed subspaces of $X$, we obtain a family $P = \{ p_F : F\in \mathscr{F}\}$ of continuous seminorms on $X$. Since all seminorms are continuous, the topology induced by $P$ is coarser than $\tau$.
It remains to see that one can arrange it so that the induced topology coincides with $\tau$. One way is to include two complementary subspaces in $P$. If $G,H$ are two closed subspaces such that
$$\alpha\colon G\times H \to X,\quad \alpha(x,y) = x+y$$
is a topological isomorphism, then $\{ p_G, p_H\}$ induces $\tau$. To see that, we note that the restriction of $p_H$ to $G$ is a norm which is equivalent to the restriction of $\lVert\,\cdot\,\rVert$ to $G$. Since $G \cap \ker p_H = G\cap H = \{0\}$, it is a norm, and it is continuous as the restriction of a continuous seminorm. Since $\alpha$ is a topological isomorphism, there is a $c > 0$ such that $\lVert x+y\rVert \geqslant c(\lVert x\rVert + \lVert y\rVert)$ for all $(x,y) \in G\times H$, hence
$$p_H(x) = \inf \{ \lVert x+y\rVert : y \in H\} \geqslant c\lVert x\rVert$$
for all $x\in G$. The same holds when we interchange the roles of $G$ and $H$, whence
\begin{align} \max \{ p_G(x+y), p_H(x+y)\} &= \max \{ p_G(y), p_H(x)\}\\ & \geqslant \frac{1}{2}(p_G(y) + p_H(x))\\ &\geqslant \frac{c}{2}(\lVert y\rVert + \lVert x\rVert)\\ & \geqslant \frac{c}{2}\lVert x+y\rVert \end{align}
for $x\in G, y\in H$, so $\max \{p_G, p_H\}$ is a norm that induces a finer topology than $\tau$ on $X$. Since it is also a $\tau$-continuous norm, it is equivalent to $\lVert\,\cdot\,\rVert$.
Finally, we need to see that in every normed space of dimension $\geqslant 2$ there are complementary subspaces. Let $\lambda \neq 0$ be a continuous linear form, and $x_0\in X$ with $\lambda(x_0) = 1$. Then $\ker \lambda$ and $\operatorname{span} \{x_0\}$ are complementary subspaces. $\ker \lambda$ is closed since $\lambda$ is continuous, and $\operatorname{span} \{ x_0\}$ is finite-dimensional, hence closed. The map $x \mapsto (\lambda(x)\cdot x_0, x - \lambda(x)\cdot x_0)$ is continuous, and its inverse $\alpha \colon (\operatorname{span} \{x_0\}) \times (\ker \lambda) \to X,\; (x,y) \mapsto x+y$ is also continuous, hence a topological isomorphism.
• I did not expect a detailed answer sir. Thank you very much. – Juniven May 18 '15 at 21:25 | 2020-01-26T17:54:53 | {
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https://de.maplesoft.com/support/help/maple/view.aspx?path=VectorCalculus%2FTangentVector | TangentVector - Maple Help
VectorCalculus
TangentVector
compute the tangent vector to a curve
Calling Sequence TangentVector(C, t, n)
Parameters
C - free or position Vector or Vector valued procedure; specify the components of the curve t - (optional) name; specify the parameter of the curve n - (optional) equation of the form normalized=true or normalized=false, or simply normalized
Description
• The TangentVector(C, t) command computes the tangent vector to the curve C that is parameterized by t. Note that this vector is not normalized by default, so it is a scalar multiple of the unit tangent vector to the curve C. Therefore, by default, if C is a curve in ${ℝ}^{3}$, the result is generally different from the output of TNBFrame(C, t, output=['T']).
• If n is given as either normalized=true or normalized, then the resulting vector will be normalized before it is returned. As discussed above, the default value is false, so that the result is not normalized.
• The curve can be specified as a free or position Vector or a Vector valued procedure. This determines the returned object type.
• If t is not specified, the function tries to determine a suitable variable name by using the components of C. To do this, it checks all of the indeterminates of type name in the components of C and removes the ones which are determined to be constants.
If the resulting set has a single entry, the single entry is the variable name. If it has more than one entry, an error is raised.
• If a coordinate system attribute is specified on C, C is interpreted in that coordinate system. Otherwise, the curve is interpreted as a curve in the current default coordinate system. If the two are not compatible, an error is raised.
Examples
> $\mathrm{with}\left(\mathrm{VectorCalculus}\right):$
> $\mathrm{T1}≔\mathrm{TangentVector}\left(t↦⟨t,{t}^{2},{t}^{3}⟩\right):$
> $\mathrm{T1}\left(t\right)$
$\left[\begin{array}{c}{1}\\ {2}{}{t}\\ {3}{}{{t}}^{{2}}\end{array}\right]$ (1)
> $\mathrm{T2}≔\mathrm{TangentVector}\left(t↦⟨t,{t}^{2},{t}^{3}⟩,\mathrm{normalized}\right):$
> $\mathrm{T2}\left(t\right)$
$\left[\begin{array}{c}\frac{{1}}{\sqrt{{9}{}{{t}}^{{4}}{+}{4}{}{{t}}^{{2}}{+}{1}}}\\ \frac{{2}{}{t}}{\sqrt{{9}{}{{t}}^{{4}}{+}{4}{}{{t}}^{{2}}{+}{1}}}\\ \frac{{3}{}{{t}}^{{2}}}{\sqrt{{9}{}{{t}}^{{4}}{+}{4}{}{{t}}^{{2}}{+}{1}}}\end{array}\right]$ (2)
> $\mathrm{TangentVector}\left(\mathrm{PositionVector}\left(\left[\mathrm{cos}\left(t\right),\mathrm{sin}\left(t\right)\right]\right),t\right)$
$\left[\begin{array}{c}{-}{\mathrm{sin}}{}\left({t}\right)\\ {\mathrm{cos}}{}\left({t}\right)\end{array}\right]$ (3)
> $\mathrm{TangentVector}\left(⟨a\mathrm{exp}\left(-t\right),t⟩\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{assuming}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}a::\mathrm{constant}$
$\left[\begin{array}{c}{-}{a}{}{{ⅇ}}^{{-}{t}}\\ {1}\end{array}\right]$ (4)
> $\mathrm{TangentVector}\left(t↦⟨a\cdot \mathrm{cos}\left(t\right),b\cdot \mathrm{sin}\left(t\right),t⟩\right)$
${t}{→}{\mathrm{VectorCalculus}}{:-}{\mathrm{VectorSpace}}{}\left({\mathrm{cartesian}}{,}\left[{a}{}{\mathrm{cos}}{}\left({t}\right){,}{b}{}{\mathrm{sin}}{}\left({t}\right){,}{t}\right]\right){:-}{\mathrm{Vector}}{}\left(\left[{-}{a}{}{\mathrm{sin}}{}\left({t}\right){,}{b}{}{\mathrm{cos}}{}\left({t}\right){,}{1}\right]\right)$ (5)
> $\mathrm{SetCoordinates}\left('\mathrm{polar}'\right)$
${\mathrm{polar}}$ (6)
> $\mathrm{TangentVector}\left(⟨1,t⟩,t\right)$
$\left[\begin{array}{c}{0}\\ {1}\end{array}\right]$ (7)
> $\mathrm{TangentVector}\left(⟨\mathrm{exp}\left(-t\right),t⟩,t\right)$
$\left[\begin{array}{c}{-}\sqrt{{{ⅇ}}^{{-}{2}{}{t}}}\\ \sqrt{{{ⅇ}}^{{-}{2}{}{t}}}\end{array}\right]$ (8) | 2022-01-16T10:59:51 | {
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https://math.stackexchange.com/questions/3147181/is-there-a-term-for-the-opposite-of-an-interpolation | # Is there a term for the opposite of an interpolation?
The question is specifically about linear Interpolation, which is usually defined to be a function $$f : V \times V \times \mathbb{R} \rightarrow V \\ f(v_0,v_1,\alpha) = v_0 + (v_1-v_0) \cdot \alpha$$ So it takes the arguments $$v_0$$ and $$v_1$$ and a real value $$\alpha$$ (often, but not necessarily, in $$[0,1]$$), and computes the linearly interpolated value.
Conversely, there may be a function like this: $$g : V \times V \times V \rightarrow \mathbb{R} \\ g(v_0,v,v_1) = (v - v_0) / (v_1 - v_0)$$ For the given arguments, it computes the "relative position" of one element between the others - namely, the value that could be used as the $$\alpha$$ value in the interpolation function, so that $$f(v_0, v_1, g(v_0, v, v_1)) = v$$.
It's not an "inverse", and the term "opposite" in the title was just for lack of a better term. Right now, I'm calling it ~"interpolation parameter function", but I wonder whether there is a commonly used term for that.
• I don't feel confident that this is the best description, and it probably couldn't be used without explanation, but I might describe that as a barycentric coordinate. Edit: wait, I assumed $v$ had to be on the line containing the other two (the affine span of them). – Mark S. Mar 13 at 21:11
• It is in fact an inverse after you "curry" your functions in the right way. Namely $g(v_0,\cdot,v_1)$ is an inverse of $f(v_0,v_1,\cdot)$. – Ian Mar 13 at 21:12
• @MarkS. That's indeed pretty close already (and I wonder why I didn't think of this yet). The fact that barycentric coordinates are mainly interpreted geometrically means that it does not fit perfectly in the most generic sense, but if nobody posts a better answer in the next days, I'd probably ping you to turn the comment into an answer and accept it as being "as close as it gets". – Marco13 Mar 13 at 21:22
• @Ian Yes, it can be boiled down to an inverse, but would require an additional description then - basically, some "verbal un-currying". Just calling it the "inverse of the interpolation" would not be sufficient, I guess... – Marco13 Mar 13 at 21:23
• @MarkS. Regarding the edit: I don't think that being contained in the simplex is a necessary condition for computing the barycentric coordinates, so this should not be a problem. – Marco13 Mar 13 at 21:25
As in the question, given a real vector space $$V$$, we can define a function for parametrizing lines between vectors by $$f : V \times V \times \mathbb{R} \rightarrow V \\ f(\mathbf v_0,\mathbf v_1,\alpha) = \mathbf v_0 + (\mathbf v_1-\mathbf v_0) \cdot \alpha\text{.}$$
If $$\mathbf v_0\ne\mathbf v_1$$, then $$\alpha\mapsto f(\mathbf v_0,\mathbf v_1,\alpha)$$ certainly defines an injective function $$\mathbb R\to V$$, with image line $$\overleftrightarrow{\mathbf v_0\mathbf v_1}$$. Thus, we can define a bijective function $$F_{\mathbf v_0,\mathbf v_1}:\mathbb R\to \overleftrightarrow{\mathbf v_0\mathbf v_1}$$.
The inverse is given by $$F_{\mathbf v_0,\mathbf v_1}^{-1}:\mathbf v\mapsto\dfrac{\mathbf v-\mathbf v_0}{\mathbf v_1-\mathbf v_0}$$, where the division is to be interpreted as yielding the relevant scalar multiple, as mentioned in this MSE answer. This inverse could be described as providing the barycentric coordinate for the input point on the line $$\overleftrightarrow{\mathbf v_0\mathbf v_1}$$ with respect to the affine basis $$(\mathbf v_0,\mathbf v_1)$$.
As alluded to in the original post, we could combine these inverses together as a single function, though writing the domain properly is a bit cumbersome. We could define $$g:\left\{(\mathbf v_0,\mathbf v,\mathbf v_1)\left|\mathbf v_0\ne\mathbf v_1\text{ and } \mathbf v\in\overleftrightarrow{\mathbf v_0\mathbf v_1}\right.\right\}\to \mathbb R$$ $$g(\mathbf v_0,\mathbf v,\mathbf v_1)= F_{\mathbf v_0,\mathbf v_1}^{-1}(\mathbf v)\text{.}$$ | 2019-12-06T23:35:42 | {
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https://uk.mathworks.com/help/matlab/math/singularity-on-interior-of-integration-domain.html | # Singularity on Interior of Integration Domain
This example shows how to split the integration domain to place a singularity on the boundary.
### Define the Integrand with an Anonymous Function
The integrand of the complex-valued integral
`${\int }_{-1}^{1}{\int }_{-1}^{1}\frac{1}{\sqrt{x+y}}\phantom{\rule{0.2222222222222222em}{0ex}}dx\phantom{\rule{0.2222222222222222em}{0ex}}dy$`
has a singularity when `x = y = 0` and is, in general, singular on the line `y = -x`.
Define this integrand with an anonymous function.
`fun = @(x,y) ((x+y).^(-1/2));`
### Integrate Over a Square
Integrate `fun` over a square domain specified by $-1\le x\le 1$ and $-1\le y\le 1$.
```format long q = integral2(fun,-1,1,-1,1)```
```Warning: Non-finite result. The integration was unsuccessful. Singularity likely. ```
```q = NaN + NaNi ```
If there are singular values in the interior of the integration region, the integration fails to converge and returns a warning.
### Split the Integration Domain into Two Triangles
You can redefine the integral by splitting the integration domain into complementary pieces and adding the smaller integrations together. Avoid integration errors and warnings by placing singularities on the boundary of the domain. In this case, you can split the square integration region into two triangles along the singular line `y = -x` and add the results.
```q1 = integral2(fun,-1,1,-1,@(x)-x); q2 = integral2(fun,-1,1,@(x)-x,1); q = q1 + q2```
```q = 3.771236166328258 - 3.771236166328256i ```
The integration succeeds when the singular values are on the boundary.
The exact value of this integral is
`$\frac{8\sqrt{2}}{3}\left(1-i\right)$`
`8/3*sqrt(2)*(1-i)`
```ans = 3.771236166328253 - 3.771236166328253i ``` | 2021-06-14T06:16:17 | {
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http://math.stackexchange.com/questions/683711/symmetry-of-bicycle-lock-numbers | # Symmetry of bicycle-lock numbers
Suppose you have a combination bicycle lock of this sort:
with $n$ dials and $k$ numbers on each dial. Let $m(n,k)$ denote the minimum number of turns that always suffice to open the lock from any starting position, where a turn consists of rotating any number of adjacent rings by one place.
For example $m(2,10)=6$ and $m(3,10)=10$.
I have found an efficient algorithm to compute these numbers, which reveals a symmetry I can’t currently explain:
$m(n, k+1) = m(k, n+1)$
This is such a striking symmetry that I guess it has a simple explanation. Can anyone find one?
Here’s the table of values for small $n$ and $k$, exhibiting the conjectured symmetry:
n\k| 2 3 4 5 6 7 8 9 10
---+---------------------------------------------
1 | 1 1 2 2 3 3 4 4 5
2 | 1 2 2 3 4 4 5 6 6
3 | 2 2 4 4 6 6 8 8 10
4 | 2 3 4 6 6 8 9 10 12
5 | 3 4 6 6 9 9 12 12 15
6 | 3 4 6 8 9 12 12 15 16
7 | 4 5 8 9 12 12 16 16 20
8 | 4 6 8 10 12 15 16 20 20
9 | 5 6 10 12 15 16 20 20 25
10 | 5 7 10 12 15 18 20 24 25
-
indeed, not sure what I was thinking about... – user88595 Feb 20 '14 at 18:46
Do you have an example of a starting position with $n=4$ and $k=8$ or $10$ that requires the value listed in the table above? – John Habert Feb 20 '14 at 19:35
@JohnHabert Sure, for n=4 and k=10, and assuming the combination 0000 opens the lock, the combinations 4826 and 6284 require 12 moves. – Robin Houston Feb 20 '14 at 19:49
This is probably obvious, but from your table we also have $m(n, n+1) = m(n, n)$ and $m(2k, 2k) = k^2$ while $m(2k+1, 2k+1) = k(k+1)$ - or, to combine these into a single expression, $m(n,n) = \lfloor\frac{n}{2}\rfloor\cdot\lfloor\frac{(n+1)}{2}\rfloor$. It's very possible there's a similar closed-form expression for $m(n,k)$ that explains the symmetry. – Steven Stadnicki Feb 21 '14 at 16:37
In fact, several of the small-$k$ values have closed forms too: $m(n,2) = \lfloor\frac{(n+1)}{2}\rfloor$, $m(n,4) = 2m(n,2)$, $m(n,3) = \lfloor\frac{(2n+2)}{3}\rfloor$, etc. I think a little more chewing may give a fully closed form... – Steven Stadnicki Feb 21 '14 at 16:46
It has taken a few days, but I believe I can at last answer my own question. The strategy is to show that, for every combination of the $(n,k)$ lock, there is a combination of the $(k-1,n+1)$ lock that needs the same number of turns to unlock. The argument begins by grouping lock combinations into equivalence classes in such a way that equivalent combinations require the same number of turns to open.
Assume throughout that the destination combination, that opens the lock, is $n$ zeros.
The first trick is one I used before: instead of looking at the positions of the dials, look at the differences between the positions of adjacent dials. On its own this doesn’t discard any information – the process is reversible – but it opens the door to a simplification: the order of the differences doesn’t matter. So we’ll say that two combinations are equivalent if they have the same multiset of differences, and we’ll write the differences as a nondecreasing sequence, to give a canonical form.
To motivate the next identification we’re going to make, let’s consider an example combination of the $(n=4, k=10)$-lock: 2593, which has differences 23447. The differences sum to $2k$, which means – as explained in my original blog post – that we can ignore the two largest differences and add up the others, so this combination takes $2+3+4 = 9$ turns to open. But, since the two largest differences didn’t even enter into the calculation, they could have been any pair of numbers that are both at least $4$ and sum to $11$. In particular they could have been $5$ and $6$ so that the differences were 23456. In this sense the combinations 2593 and 2594 are equivalent. We shall denote this equivalence class by the sequence (2,3,4), which we’ll call a lock sequence for the $(n, k)$-lock. Notice that the number of turns needed to open the lock is the sum of the terms of the lock sequence.
Now we’re going to characterise the lock sequences. Let $d_1, d_2, \dots, d_m$ be a nondecreasing sequence of natural numbers less than $k$ having length $m\leq n$; this is a lock sequence for the $(n,k)$-lock if these two inequalities hold:
$\sum_{i=1}^{m}d_i+(n+1-m)(k-1)\geq (n+1-m)k$
$\sum_{i=1}^{m}d_i + (n+1-m)d_m\leq(n+1-m)k$
They can be simplified to
$n+1-m\leq\sum_{i=1}^{m}d_i\leq(n+1-m)(k-d_m)$
The first inequality is a bit annoying, so let's get rid of it by making one last identification: we’ll identify lock sequences that differ only by leading zeros, and assume a canonical form that has no leading zeros. If the first inequality fails, we can force it to hold by adding leading zeros, thus increasing $m$. So now we’re left with
$\sum_{i=1}^{m}d_i\leq(n+1-m)(k-d_m)$
I like to imagine this condition as meaning, “Is there room in the attic for all the boxes?”. Maybe that will make more sense if I draw a picture:
This picture depicts the lock sequence $(1,1,2,2,2,2,3,3)$ as an arrangement of 16 boxes, and an “attic” of area $(n+1-m)(k-d_m)$, all within an $(n+1)\times k$ rectangle.
Now let’s flip it over, like conjugating a Young diagram, and move the attic back to the top:
We still have the same arrangement of boxes – in particular the value of $\sum_{i=1}^{m}d_i$ remains the same – and the attic is the same size. So the conjugate sequence – $(2,6,8)$ in this example – is a valid sequence for the $(k-1, n+1)$-lock provided only the original was a valid sequence for the $(n, k)$-lock.
So, we’ve shown that every lock sequence for the $(n,k)$-lock can be transformed into a lock sequence for the $(k-1,n+1)$-lock that has the same sum. It follows that it takes at least as many moves, in general, to open a $(k-1,n+1)$-lock as a $(n,k)$-lock. Since this works in both directions, we may conclude that $m(n,k) = m(k-1, n+1)$.
Another way of looking at it is to note that the above implies
$m(n,k) = \max\{\min\{ac,bd\}\ |\ a+b=n+1,c+d=k\mbox{ for }a,b,c,d\in\mathbb{N}\}$
which is symmetrical in $n+1$ and $k$. This expression also suggests another way to compute $m(n,k)$.
-
Congratulations! Yes, I think that's compelling. I'd got as far as your "box" picture but not to the necessary & sufficient inequality. Is it worth emphasising that "the same arrangement of boxes" in the conjugate diagram means the same count of boxes, i.e. that the term $\Sigma d_i$ remains the same after conjugation? – HTFB Feb 25 '14 at 7:49
First let's consider a much simpler situation. If, instead of moving any group of adjacent dials, you can only move one dial, the maximum number of turns is trivial -- it's equal to $kn/2$. Note that this expression displays a sort of symmetry very similar to what you're seeing. To be jargony, the state space -- number of combinations -- has volume $kn$, which is a commutative expression, and we can only move through one square of state space at a time.
Now, if we look at the real problem, the state space is still of volume $kn$, but we can move through $k$ squares at a time, with some restrictions. It's easier to analyze if we can only move through one square, so we can look at the space of differences: 3879 becomes 592 and 23856 becomes 1571. In the space of differences we can only move two differences at a time - one up, one down, so 1571 might become 0572 -- but more importantly we have a volume $(k-1) \times n$ and we move a constant amount, so we expect to see a path length of $\Omega((k-1)n)$.
Now, the expression $m(n, k) = \Omega((k-1)n)$ exhibits the observed symmetry! It's just commutation -- if we take $m(k-1, n+1)$ we obviously have the same "volume" of "difference space". There's some intuition here but we still haven't derived the shortest path.
I'll borrow your expression $t = \sup_x(xk - xq - \min(r,x))$ for $xk = q(n+1) + r$ and note that the upper bound and lower bound are achieved precisely where $n+1 | xk$ and $n+1 | x(k-1)$, and that they are both proportional to $x(n+1-x)$, so have a shared maximum $x = (n+1)/2$. This gives us the diophantine equation $q(n+1) = xk$ and if we substitute $x \approx (n+1)/2$ we have $q \approx k/2$.
If these are both integers -- if $n$ is odd and $k$ is even -- we have $t(n, k) = m(n, k) = k(n+1)/4$. So that's a bit simpler already, and it follows the pattern which is nice.
However, the equation does not have a satisfactory solution if $k$ is odd or $n$ is even, so represents a sufficient but not necessary condition for a maximum. We might try to approximately solve the equation by choosing $x = (n+1)/2$, $q = k/2$, and finding a nearby lattice point. For $n = 200$, $k = 9$ we can choose $x = (n+1)/2 - (n+1)/2k = 101.5 - 11.1 = 89.9$ which is close to the true maximum $x = 89$ and suggestive of a more general pattern. I've spent an hour on this already though and should get back to work.
EDIT: I guess the takeaway is that the Diophantine equation $q(n+1) = xk$ is the same if we take $n \rightarrow k-1$, $k \rightarrow n+1$ to get $qk = x(n+1)$. Approximate solutions to this equation correspond to solutions of the general problem.
-
What is the function $\Omega$ here? – HTFB Feb 24 '14 at 12:05
Ω is part of big O notation. See Wikipedia. I actually should have used big-theta (asymptotically related by a constant), as that's what I meant. – mason bogue Feb 25 '14 at 3:01 | 2016-07-02T02:29:12 | {
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# How many diagonals are there in an octagon?
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How many diagonals are there in an octagon? [#permalink] 12 Sep 2018, 22:57
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How many diagonals are there in an octagon?
A)56
B)28
C)20
D)16
E)8
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Re: How many diagonals are there in an octagon? [#permalink] 13 Sep 2018, 06:27
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For each of the 8 vertices you can draw 5 diagonals and hence you have constructed $$5 \times 8 = 40$$ diagonals.
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Re: How many diagonals are there in an octagon? [#permalink] 13 Sep 2018, 07:56
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AchyuthReddy wrote:
How many diagonals are there in an octagon?
A)56
B)28
C)20
D)16
E)8
Let's examine ONE vertex in an octagon (an 8-sided polygon)
We can draw 5 diagonal lines from that one vertex
So, each of the 8 vertices can have 5 diagonal lines
So, the total number of diagonal lines = (8)(5) = 40
But wait!!
We have inadvertently counted every diagonal TWICE.
For example, we counted the diagonal from vertex A to vertex B AND we counted the diagonal from vertex B to vertex A
So, to account for this DUPLICATION, we must divide 40 by 2 to get 20
Cheers,
Brent
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Re: How many diagonals are there in an octagon? [#permalink] 13 Sep 2018, 07:56
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https://afd-hamburg-nord.de/forum/archive.php?53e5c2=matrix-multiplication-rules | Other than being multiplied by scalar constants, matrices can also be multiplied by other matrices. What is matrix multiplication? Matrix multiplication is also distributive. Matrices as transformations. 5 & 2 & 11 Multiplying a Row by a Column We'll start by showing you how to multiply a 1 × n matrix by an n × 1 matrix. \\ Even so, it is very beautiful and interesting. Task. We will usually denote matrices with capital letters, like A, B, etc, although we will sometimes use lower case letters for Let A be an m × p matrix and B be an p × n matrix. \end{bmatrix} = 64. Sort by: Top Voted. $(hint: just multiply every entry by $$2$$), You can multiply two matrices if, and only if, the number of columns in the first matrix equals the number of rows in the second matrix. (Link on columns vs rows ). When the number of columns of the first matrix is the same as the number of rows in the second matrix then matrix multiplication can be performed. Zero matrix & matrix multiplication. 3 Matrices and matrix multiplication A matrix is any rectangular array of numbers. \end{bmatrix} Matrix Multiplication Rules & Formula - In this tutorial, you will learn all about matrix multiplication. To multiply any two matrices, we should make sure that the number of columns in the 1st matrix is equal to the number of rows in the 2nd matrix. = Using properties of matrix operations. Once we know if two matrices can be multiplied, it's time to carry out that multiplication. Real World Math Horror Stories from Real encounters, (See how this problem can be represented as a Scalar Dilation), Scalar: in which a single number is multiplied with every. (This one has 2 Rows and 3 Columns). Matrix multiplication is probably one of the most important matrix operations. We match the 1st members (1 and 7), multiply them, likewise for the 2nd members (2 and 9) and the 3rd members (3 and 11), and finally sum them up. Rules. Here are a couple more examples of matrix multiplication: Find CD and DC, if they exist, given that C and D are the following matrices:; C is a 3×2 matrix and D is a 2×4 matrix, so first I'll look at the dimension product for CD:. Each number in the answer matrix is the result of multiplying one of the rows of matrix 1 by one of the columns of matrix 2. Multiplication of two matrices is also known as a "dot product". Otherwise, the product of two matrices is undefined.The product matrix's dimensions are 1. â ( rows of first matrix ) × ( columns of the second matrix ) {\displaystyle \to ({\text{rows of first matrix}})\times ({\text{columns of the second matrix}})} In above multiplication, the matrices cannot be multiplied since the number of columns in the 1st one, mat⦠In matrix multiplication, the elements of the rows in the first matrix are multiplied with corresponding columns in the second matrix. Example: This matrix is 2×3 (2 rows by 3 columns): In that example we multiplied a 1×3 matrix by a 3×4 matrix (note the 3s are the same), and the result was a 1×4 matrix. Memorizing the entire Multiplication Table can seem quite overwhelming at first. Time complexity: O(n 3).It can be optimized using Strassenâs Matrix Multiplication. An interactive matrix multiplication calculator for educational purposes If at least one input is scalar, then A*B is equivalent to A. Want to see another example? However, matrix multiplication is not as straight forward as regular multiplication, certain rules must be followed and certain conditions must be met. Contents. See your article appearing on the GeeksforGeeks main page and help ⦠= 154. When we multiply a matrix by a scalar (i.e., a single number) we simply multiply all the matrix's terms by that scalar. The usual rules for exponents, namely = P+ and (AP) = still apply. This is the currently selected item. \\ First, however, there are several key concepts that must be understood. Since the number of columns in Matrix A does not equal the number of rows in Matrix B. (You can put those values into the Matrix Calculator to see if they work.). Another example of 2 matrices you can not multiply. # matrix multiplication in R - example > gt*m [,1] [,2] [,3] [1,] 525 450 555 [2,] 520 500 560 [3,] 450 425 500. Multiply two matrices together. Matrix multiplication is probably the first time that the Commutative Property has ever been an issue. We match the price to how many sold, multiply each, then sum the result. Apple pie value + Cherry pie value + Blueberry pie value, ($3, $4,$2) • (13, 8, 6) = $3×13 +$4×8 + $2×6, And the result will have the same number of, It is "square" (has same number of rows as columns), It can be large or small (2×2, 100×100, ... whatever). \begin{bmatrix} ]â2×3=NOT DEFINED Here is an example of matrix multiplication for two concrete matrices Example: Find the product AB where A and ⦠To multiply a matrix by a single number is easy: We call the number ("2" in this case) a scalar, so this is called "scalar multiplication". So the product CD is defined (that is, I can do the multiplication); also, I can tell that I'm going to get a 3×4 matrix for my answer. It is a type of binary operation. See how changing the order affects this multiplication: It can have the same result (such as when one matrix is the Identity Matrix) but not usually. Example 1 . This article is contributed by Aditya Ranjan.If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to [email protected]. Here it is for the 1st row and 2nd column: (1, 2, 3) • (8, 10, 12) = 1×8 + 2×10 + 3×12 That is, A*B is typically not equal to B*A. In mathematics, particularly in linear algebra, matrix multiplication is a binary operation that produces a matrix from two matrices. The multiplication of matrix A by matrix B is a 1 × 1 matrix defined by: Example 1 Matrices A and B are defined by Find the matrix A B. 3 Matrix Powers We can take powers of matrices, but only if theyâre square. But to multiply a matrix by another matrix we need to do the "dot product" of rows and columns ... what does that mean? Matrix Multiplication Calculator Here you can perform matrix multiplication with complex numbers online for free. Scalar: in which a single number is multiplied with every entry of a matrix. We define A° = I, where I is the identity matrix ⦠[?????? Matrix A can be multiplied by a matrix B if the number of columns of matrix A equals the number of rows of the matrix B. Matrix multiplication (product of matrices) A and B with dimensions m×n and n×k is the operation of finding the matrix ⦠Matrix multiplication. Next lesson. This same thing will be repeated for the second matrix. \begin{bmatrix} ; Multiplication of one matrix by second matrix.. For the rest of the page, matrix multiplication will refer to this second category. They can be of any dimensions, so long as the number of columns of the first matrix is equal to the number of rows of the second matrix. a) Multiplying a 2 × 3 matrix by a 3 × 4 matrix is possible and it gives a 2 × 4 matrix as the answer. Matrix multiplication falls into two general categories: For the rest of the page, matrix multiplication will refer to this second category. 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# If two integers are chosen at random out of the set {2, 5, 7, 8}, what
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If two integers are chosen at random out of the set {2, 5, 7, 8}, what is the probability that their product will be of the form a^2 – b^2, where a and b are both positive integers?
A. 2/3
B. 1/2
C. 1/3
D. 1/4
E. 1/6
[Reveal] Spoiler:
Shalom! I am currently studying the probability chapter of the Manhattan GMAT Word Translations book. I am looking forward to the different outcomes and answers.
[Reveal] Spoiler: OA
Last edited by Bunuel on 11 Nov 2017, 22:32, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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03 Jun 2011, 11:49
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MitDavidDv wrote:
If two integers are chosen at random out of the set {2, 5, 7, 8}, what is the probability that their product will be of the form a^2 – b^2, where a and b are both positive integers?
A: 2/3
B: 1/2
C: 1/3
D: 1/4
E: 1/6
Shalom! I am currently studying the probability chapter of the Manhattan GMAT Word Translations book. I am looking forward to the different outcomes and answers.
I just used the exhaustive method. Count everything that fits.
2,5=10
2,7=14
2,8=16
5,7=35
5,8=40
7,8=56
Write down all perfect squares until 100
1, 4, 9, 16, 25, 36, 49, 64, 81, 100
Pick one number at a time.
10 -- Keep adding with every perfect square and see whether the result is also there in the set.
10+1=11(Not there)
10+4=14(Not there)
10+9=19(Not there)
10+16=26(Not there)
10+25=35(Not there)
we can stop here as the difference between all consecutive perfect squares after 35 will be more than 10.
10- Not possible to represented as a^2-b^2
Repeat the same with all the products;
14-Not Possible
16: Let's check this;
16+1=17(Not there)
16+4=20(Not there)
16+9=25(There in the set)
16- can be represented as a^2-b^2 i.e. 5^2-3^2
**********************************
Likewise:
35- 6^2-1^2
40- 7^2-3^2
56- 9^2-5^2
*******************
In the sample set:
{10,14,16,35,40,56}
{10,14}- Not Possible: Count=2
{16,35,40,56}- Possible: Count=4
Total Count=6
P=Favorable/Total=4/6=2/3
Ans: "A"
**************************
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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You can avoid an exhaustive test here. Suppose I ask whether (97)(103) can be written in the form a^2 - b^2, where a and b are integers. Notice that this is a difference of squares: a^2 - b^2 = (a+b)(a-b). We can now just use the median of 97 and 103, which is 100:
(97)(103) = (100-3)(100+3) = 100^2 - 3^2
So whenever we can write our product in such a way that the median of our two numbers is an integer, we can write our product as a difference of squares just as above. For example, if we take 5*7, that's equal to (6-1)(6+1), and if we take 2*8, that's equal to (5-3)(5+3). Now if we look at 8*5, we can't immediately use the same trick, but we can 'move' one of the 2s from the 8 into the 5, as follows: 8*5 = 4*10 = (7-3)(7+3). Similarly, 8*7 = 4*14 = (9-5)(9+5). So of our six possible products, four can be written as a difference of squares.
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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19 Jul 2011, 23:06
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IanStewart
you gave a very easy explanation!
Thank you!
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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Some Theory here:
Consider two numbers a, b
Now a* b always = [(a+b)/2]^ 2 – [(a-b)/2]^2---------------------- eqn 1
The number of ways a number can be expressed as a difference of two integers depends on number of ways it can be written as a two factor product stated below.
a.odd*odd
b. even *even
The reason being in eqn (1) above [(a-b)/2]^2 should result in an integer . Hence we consider only the above set of two factor products. (0dd minus odd = even ,even minus even =even , hence both will be divisible by 2
E.g the number 36 can be written as
6*6 , 18*2 hence 36 can be expressed as difference of squares in two ways.
36 = (6+6)/2 ^ 2 - (6-6)/2 ^ 2 = 6^2 - 0
36= (18+2)/2^2 – (18-2)/^2 = 10^2 – 4^2
Now back to the problem :
We have the number set (2, 5, 7, 8) out of which, the satisfying possibilities as per the above theory would be
2*8 (valid)
5*7 (valid)
2*5 (not valid)
2*7(not valid)
5*8== 4*10 hence valid
7*8 == 4*14 hence valid)
Hence there are 4 favourble cases out of 6. Therefore probability Is 4/6 = 2/3
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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Need Bunuel's explanation for this problem......
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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29 Oct 2013, 01:29
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monirjewel wrote:
Need Bunuel's explanation for this problem......
Best solution is here: if-two-integers-are-chosen-at-random-out-of-the-set-114579.html#p929326
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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17 Nov 2013, 09:35
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Bunuel wrote:
monirjewel wrote:
Need Bunuel's explanation for this problem......
Best solution is here: if-two-integers-are-chosen-at-random-out-of-the-set-114579.html#p929326
a^2 - b^2 will be of the form (a-b)(a+b). We can infer that the difference between the two chosen numbers(of which one is (a-b) and the other (a+b)) will be (a+b) - (a-b) = 2b i.e. even difference(negative or positive).
Thus we will have to choose either two even numbers or two odd numbers?
this way we have two options -> 1. choosing 5 and 7 of which probability is = 1/6
2. choosing 2 and 8 of which probability is = 1/6.
thus the total probability = 2/6 which is 1/3.
please explain why this is wrong.
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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MitDavidDv wrote:
If two integers are chosen at random out of the set {2, 5, 7, 8}, what is the probability that their product will be of the form a^2 – b^2, where a and b are both positive integers?
A. 2/3
B. 1/2
C. 1/3
D. 1/4
E. 1/6
Shalom! I am currently studying the probability chapter of the Manhattan GMAT Word Translations book. I am looking forward to the different outcomes and answers.
I have an explanation too, maybe it'd be of some help:
a^2 - b^2 =(a-b)(a+b)
(a+b) and (a-b) can only be integers from the selected set i.e {2,5,7,8}
Now a and b are both positive integers as stated in the question
So the sum of (a-b) and (a+b) also has to be an integer
i.e 2a = sum of any two numbers chosen from the set {2,5,7,8}
For a to be a positive integer the sum has to be an even number so either it could be a pair of (2,8) or (7,5)
Using PnC P(getting one such pair when chosing two random numbers from a set) = P(Both numbers chosen to be even) + P(Both numbers chosen to be odd)
= 2/4C2 + 2/4C2
= 4/4C2 = 4/6 = 2/3
Kindly let me know in case the solution has some mistakes.
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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28 Feb 2014, 21:04
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a^2-b^2 = (a+b)(a-b), so if the difference of any the factor pairs of the product is a positive even integer, then they can be in the a^2-b^2 form. Example: 7*5=35 --> 7*5 = (6+1)(6-1) --> 35*1 = (18-17)(18+17), etc etc
Only products that don't work are 2*5=10 (as 5-3 = odd and 10-1 = odd) and 2*7=14 (as 7-5 = odd and 14-1=13 odd).
The other 4 products work and so the probability = 4/6 = 2/3
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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ShashankDave wrote:
Bunuel wrote:
monirjewel wrote:
Need Bunuel's explanation for this problem......
Best solution is here: if-two-integers-are-chosen-at-random-out-of-the-set-114579.html#p929326
a^2 - b^2 will be of the form (a-b)(a+b). We can infer that the difference between the two chosen numbers(of which one is (a-b) and the other (a+b)) will be (a+b) - (a-b) = 2b i.e. even difference(negative or positive).
Thus we will have to choose either two even numbers or two odd numbers?
this way we have two options -> 1. choosing 5 and 7 of which probability is = 1/6
2. choosing 2 and 8 of which probability is = 1/6.
thus the total probability = 2/6 which is 1/3.
please explain why this is wrong.
I had the same concern but now it's clear for me :
the 6 possible pairs are (2 5) (2 7) (2 8) (5 7) (5 8) (7 8)
our method allows to find 2 pairs (2 8) (5 7) but does not allow to eliminate the others
Especially (5 8 ) and (7 8) that also meet the condition : 5*8 = 10*4 = (7+3)*(7-3) and 7*8=14*4=(9+5)*(9-5)
so there are 4 possible pairs to be picked from 6 hence the probability is 4/6 = 2/3
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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13 Apr 2014, 19:09
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ShashankDave wrote:
Bunuel wrote:
monirjewel wrote:
Need Bunuel's explanation for this problem......
Best solution is here: if-two-integers-are-chosen-at-random-out-of-the-set-114579.html#p929326
a^2 - b^2 will be of the form (a-b)(a+b). We can infer that the difference between the two chosen numbers(of which one is (a-b) and the other (a+b)) will be (a+b) - (a-b) = 2b i.e. even difference(negative or positive).
Thus we will have to choose either two even numbers or two odd numbers?
this way we have two options -> 1. choosing 5 and 7 of which probability is = 1/6
2. choosing 2 and 8 of which probability is = 1/6.
thus the total probability = 2/6 which is 1/3.
please explain why this is wrong.
The question is not so much as whether both the numbers are even or both are odd as whether the product of the numbers can be written as product of two even numbers or two odd numbers.
Two numbers are chosen and multiplied. Now they have lost their individual identity. Now you focus on the product and find whether it can be written as product of two numbers which are both odd or both even. Say you took two number 7 and 8 and multiplied them. You get 56. Can you write 56 as product of two numbers such that both are even? Yes, 4 and 14 or 2 and 28. So 56 can be written as a^2 - b^2 in two ways: (9^2 - 5^2) and (15^2 - 13^3). So if you choose 7, 8 from the set, their product can be written in the form a^2 - b^2.
Similarly, 5, 8 will give you the same result.
Hence you get 2 more cases and total probability becomes 4/6 = 2/3.
Whenever you have at least 4 in the product, you can write it as product of two even numbers: give one 2 to one number and the other 2 to the other number to make both even.
If the product is even but not a multiple of 4, it cannot be written as product of two even numbers or product of two odd numbers. It can only be written as product of one even and one odd number.
If the product is odd, it can always be written as product of two odd numbers.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 05 Aug 2014 Posts: 1 Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink] ### Show Tags 05 Aug 2014, 12:31 3 This post received KUDOS 2 This post was BOOKMARKED Here is the solution with easy steps: a^2 - b^2 = (a-b)(a+b) Now we need to find all possible combinations of two numbers from the set {2, 5, 7, 8 } which can be expressed as (a-b)(a+b) Let say x =a-b and y = a+b, therefore x+y = 2a and y-x = 2b, so you need to have two numbers x and y whose sum and difference should be even number. How many are there from the set {2, 5, 7, 8} ? 8 + 2 = 10 , 8 - 2 = 6, 7 - 5 = 2 , 7 + 5 = 12 . So there are two pairs (8,2) and (7,5) which can be expressed as a^2 - b^2. Therefore, the probability that their product will be of the form a^2 – b^2 = 2/total two numbers combination = 2/4 Chose 2 = 2/3 Ans is option (A) Senior Manager Joined: 07 Apr 2012 Posts: 441 Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink] ### Show Tags 13 Sep 2014, 10:12 1 This post was BOOKMARKED VeritasPrepKarishma wrote: The question is not so much as whether both the numbers are even or both are odd as whether the product of the numbers can be written as product of two even numbers or two odd numbers. Two numbers are chosen and multiplied. Now they have lost their individual identity. Now you focus on the product and find whether it can be written as product of two numbers which are both odd or both even. Say you took two number 7 and 8 and multiplied them. You get 56. Can you write 56 as product of two numbers such that both are even? Yes, 4 and 14 or 2 and 28. So 56 can be written as a^2 - b^2 in two ways: (9^2 - 5^2) and (15^2 - 13^3). So if you choose 7, 8 from the set, their product can be written in the form a^2 - b^2. Similarly, 5, 8 will give you the same result. Hence you get 2 more cases and total probability becomes 4/6 = 2/3. Whenever you have at least 4 in the product, you can write it as product of two even numbers: give one 2 to one number and the other 2 to the other number to make both even. If the product is even but not a multiple of 4, it cannot be written as product of two even numbers or product of two odd numbers. It can only be written as product of one even and one odd number. If the product is odd, it can always be written as product of two odd numbers. Hi Karishma, Can you elaborate a little more? Why are we looking for 2 numbers that are either both even or both odd? Also, how did you know to stop at (9^2 - 5^2) and (15^2 - 13^3) and not look for more? Thanks, Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7951 Location: Pune, India Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink] ### Show Tags 15 Sep 2014, 01:21 6 This post received KUDOS Expert's post 2 This post was BOOKMARKED ronr34 wrote: Hi Karishma, Can you elaborate a little more? Why are we looking for 2 numbers that are either both even or both odd? Also, how did you know to stop at (9^2 - 5^2) and (15^2 - 13^3) and not look for more? Thanks, That's a good question. You should understand this concept well. That is why I have written a detailed post on it on my blog: http://www.veritasprep.com/blog/2014/04 ... at-part-i/ Check it out and get back to me (on the blog or here) if any doubts remain. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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20 Oct 2014, 06:56
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not sure if its been discussed, but a valuable property to know is that ANY non-prime odd number, or multiple of 4, can be written as a difference of squares using integers. 21 = (5+2)(5-2) 15 = (4+1)(4-1) etc. try it out.
therefore, we can see that out of our 6 possible outcomes, only 4 will be either odd (5 x 7) or multiples of 4 (8 x each other #). so answer = 4/6=2/3
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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14 Jan 2015, 18:05
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Very simple and straightforward method but never heard about the number property....
bsmith37 wrote:
not sure if its been discussed, but a valuable property to know is that ANY non-prime odd number, or multiple of 4, can be written as a difference of squares using integers. 21 = (5+2)(5-2) 15 = (4+1)(4-1) etc. try it out.
therefore, we can see that out of our 6 possible outcomes, only 4 will be either odd (5 x 7) or multiples of 4 (8 x each other #). so answer = 4/6=2/3
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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13 Mar 2015, 20:17
MitDavidDv wrote:
If two integers are chosen at random out of the set {2, 5, 7, 8}, what is the probability that their product will be of the form a^2 – b^2, where a and b are both positive integers?
A. 2/3
B. 1/2
C. 1/3
D. 1/4
E. 1/6
Shalom! I am currently studying the probability chapter of the Manhattan GMAT Word Translations book. I am looking forward to the different outcomes and answers.
So far if the average of the two numbers is an INTEGER they can be written in (a+b)(a-b) form . so that narrows us down to Odd + Odd and Even+Even cases .
Special consideration need to taken for those cases in which one number is ODD and other is multiple of 4 , i.e. in this case
if the set is $${ 2,5,7,8 }$$, then possible pairs are :
7*8 = 56 = 14*4 = 28*2 none of these pairs (7,8) , (14,4), and (28*2) can be expressed in (a+b) (a-b) form .
5*8= 40 = 10*4 = (7+3) (7-3), so yes we can write $$5*8$$ as $$(7+3) * (7-3)$$
2,5,7,8
total number of cases = 4C2 = 6
favorable cases : (odd,odd) (5,7) , (Even,Even) (2,8) , and one special case as shown above (5,8) so $$3/6=1/2$$
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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11 Jun 2015, 21:19
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MitDavidDv wrote:
If two integers are chosen at random out of the set {2, 5, 7, 8}, what is the probability that their product will be of the form a^2 – b^2, where a and b are both positive integers?
A. 2/3
B. 1/2
C. 1/3
D. 1/4
E. 1/6
Shalom! I am currently studying the probability chapter of the Manhattan GMAT Word Translations book. I am looking forward to the different outcomes and answers.
the product set=(10, 14, 16, 35, 40, and 56)
16=8x2=(5+3)(5-3); (8-2)/2 =3
35=7x5=(6+1)(6-1); (7-5)/2 = 1
40=10x4=(7+3)(7-3)
56=14x4=(9+5)(9-5)
you got the pattern.
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Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink]
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05 Sep 2015, 19:45
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MitDavidDv wrote:
If two integers are chosen at random out of the set {2, 5, 7, 8}, what is the probability that their product will be of the form a^2 – b^2, where a and b are both positive integers?
A. 2/3
B. 1/2
C. 1/3
D. 1/4
E. 1/6
Shalom! I am currently studying the probability chapter of the Manhattan GMAT Word Translations book. I am looking forward to the different outcomes and answers.
I consider
AxB = a^2 – b^2 = (a+b)(a-b)
where A and B are the posible chosen
first: the total posible chosen is 12, because A take 4 values and B takes 3, 4x3 = 12
Second
This is the scenary:
A x B = (a+b)(a-b)
2 5 = 10 = 2x5 or 10x1 (wrong) note that the sum of the factors should be even number, (conditions from a and b are integers), for the reason these opstion is eliminated
2 7 = 14 = 7x2 (wrong, the sum is not even) or 14x1 (wrong)
2 8 = 16 = 8x2 (correct)
5 2 = 10 = 8x2 (wrong, and the same that the first)
5 7 = 35 = 5x7 (Correct)
5 8 = 40 = 20x2 (correct)
7 2 = 14 = (wrong and is the same )
7 5 = 35 = 7x5 (correct)
7 8 = 56 = 14x4 (correct)
8 2 = 16 = 8x2 (correct)
8 5 = 40 = 20x2 (correct)
8 7 = 56 = 14x4 (correct)
Finally, the number of correct posible answer is 8, and the total possible answer is 12
indeed, 8/12 = 2/3
Re: If two integers are chosen at random out of the set {2, 5, 7, 8}, what [#permalink] 05 Sep 2015, 19:45
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Display posts from previous: Sort by | 2018-02-23T02:47:56 | {
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https://stats.stackexchange.com/questions/194564/what-is-the-probability-of-randomly-picking-4-consecutive-ladies-out-of-a-team-o | # What is the probability of randomly picking 4 consecutive ladies out of a team of 11 people?
I was running the morning standup this morning and in contrary to the usual alphabetical order I announced I was going to go in "random" order today, however I then ended up giving all the ladies of the team the word first before 'randomly' choosing the gents.
If my selection was really random what is the probability of having all the ladies come first?
My back of the envelope calculation was:
4/11 * 3/10 * 2/9 * 1/8 = 0.003030303
• Is this anywhere correct?
• How to write this problem in a formal way?
P.S. This is not a HW question but can be considered as a kind of self study.
• if this is HW please add the "self-study" tag – Antoine Feb 8 '16 at 11:56
First, note that every ordering of the ladies in the first four positions will give the same event: namely ladies first! Then, note that any ordering of the gentlemen in the last seven positions will also give the event ladies first! How do we account for all these outcomes?
Here are two formal ways to accomplish this. We begin by considering the sample space in the ordered case. By ordered here we mean that we can tell the ladies apart and the same for the gentlemen. You can name them if it makes things easier.
Assuming randomness and equilikely events, we have to count the ways this can occur and then divide by the total number of orderings to get our probability. Let's start from the total number of orderings of these eleven people. This is given by the number of permutations of course, $11!$. Then, how many ways are there to order the four ladies in the four positions? Well, the same reasoning will give $4!$ ways. But now for each of these orderings, we can order the gentlemen in $7!$ ways and still have an ordering in which the ladies come first. Thus we may write
$$\Pr \left( \text{Ladies first} \right) = \frac{4!7!}{11!} = 0.0030$$
and it turns out that your calculation was correct.
Now, an alternative way to view this. Assume now that we are in the unordered case and we cannot tell the ladies apart and the gentelemen either. All we know is that a person is a lady or a gentleman. Is there a way to compute the probability in this case?
Well yes, if we consider the proper sample space. First let's count the total number of unordered arrangements of the ladies and the gentlemen. This is given by the binomial coefficient $\binom{11}{4, 7}$ and the desired outcome now is just one of these arrangements. Hence, we write
$$\Pr \left( \text{Ladies first} \right) = \frac{1}{\binom{11}{4, 7}} = \frac{4! 7!}{11!}$$
and we get of course the same result but from different principles!
Hope this helps. | 2019-10-20T21:57:14 | {
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https://keisan.casio.com/exec/system/1223267646 | # Area of a triangle (Heron's formula) Calculator
## Calculates the area of a triangle given three sides.
side a side b side c 6digit10digit14digit18digit22digit26digit30digit34digit38digit42digit46digit50digit area S
$\normal Triangle\ by\ Heron's\ formula\\\vspace{5}(1)\ area:\ S=\sqrt{s(s-a)(s-b)(s-c)}\\\vspace{1}\hspace{100} s={\large\frac{(a+b+c)}{2}}\\$
Area of a triangle (Heron's formula)
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Area calculation for construction site.
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Hello, what unit of measurement is used in this calculator? Ft or mtr?
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Input meter values -> The result is m^2.
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Does not work with radicals sadly :(
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very useful
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find out the area of a land of which three sides are unequal length of a triangle
[9] 2018/08/16 17:38 Female / Under 20 years old / Elementary school/ Junior high-school student / Useful /
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https://math.stackexchange.com/questions/1043131/can-the-probability-of-a-trump-poverty-be-calculated-without-making-case-distinc | # Can the probability of a trump poverty be calculated without making case distinctions?
The card game Doppelkopf is played with four players. Every player receives 12 of the 48 cards. The 48 cards consist of 26 trump cards and 22 other. A trump poverty is what we call the scenario that a player has less than 4 trump cards. If the cards are dealt randomly, what is the probability that at least one player has a trump poverty?
This looks like it can be solved very easily via the inverse, like so:
Deal every player 4 trump cards, then deal every player 8 more cards.
However we failed to do this without double counting. Can this be done without case distinctions?
Outline: Call the players A, B, C, D. Note that we cannot have $3$ trump poor players, since then they would have at most $9$ trumps between them, and $9+12\lt 26$.
(i) Find the probability that A has trump poverty. This should not be difficult.
Now multiply by $4$. This overestimates the required probability, for we have double-counted the situations in which, for example, A and B both have trump poverty.
(ii) Find the probability that A and B have trump poverty, multiply by $\binom{4}{2}$, and subtract from the estimate in (i).
Finding the probability that A and B have trump poverty is a little unpleasant. One way is to divide into cases: A has $0$ trumps, $1$ trump, $2$, $3$. It will take a while but the calculations are routine.
Remark: If one has some programming experience, one can set up a simulation and get reliable estimates.
• Yes, we came up with this answer, too, but we were trying to find a solution without making case distinctions. Are you suggesting there is no such solution? – Max Nov 29 '14 at 10:15 | 2019-12-11T22:43:37 | {
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https://mathoverflow.net/questions/230412/when-does-the-radius-of-convergence-of-the-product-of-two-p-adic-power-series | # When does the radius of convergence of the product of two $p$-adic power series increase?
Let $p$ be a prime number and denote by $R(f)$ the radius of convergence of a power series $f(x) \in \mathbb{C}_p[[x]]$, where $\mathbb{C}_p$ is the completion of the algebraic closure of $\mathbb{Q}_p$, the field of $p$-adic numbers. Given two power series $f(x), g(x) \in \mathbb{C}_p[[x]]$, it is known that the radius of convergence of the product $h(x) = f(x)g(x)$ is at least the minimum of the radius of convergence of the two series $f(x)$ and $g(x)$. In other words, we have $$R(h) \geq \min\{R(f), R(g)\}.$$ Keep in mind $(1-x)(1+x+x^2+\dots) = 1$ as an example for the strict inequality. Is there a way to easily predict when $R(h) > \min\{R(f), R(g)\}$ and find $R(h)$ explicitly? More specifically, I'm interested in computing the radius of convergence of power series of the form $\exp(f(x))$ for $f(x) \in x\mathbb{C}_p[[x]]$.
For example, let $f(x) = \exp(x)$ and $g(x) = \exp(x^p/p)$. Then $R(f) = R(g) = (1/p)^{1/(p-1)}$ and using the fact that the Artin-Hasse exponential series $$\text{AH}(x) = \exp(x + x^p/p + x^{p^2}/p^2 + \cdots)$$ lies in $\mathbb{Z}_p[[x]]$ (which implies $R(\text{AH}) \geq 1$), $h(x) = \exp(x + x^p/p)$ has radius of convergence $$R(h) = R\left(\exp\left(\frac{x^{p^2}}{p^2}\right)\right) = \left(\frac{1}{p}\right)^{\frac{(2p-1)}{p^2(p-1)}}> \left(\frac{1}{p}\right)^{\frac{1}{p-1}}.$$ The importance of this example comes from the fact that if we set $\pi$ to be a root of $x+x^p/p = 0$, then $h(\pi)$ is a non-trivial $p$-th root of unity in $\mathbb{C}_p$. This provides an analytic representation of $p$-th roots of unity, exploited in particular in Dwork's proof of the rationality of zeta functions over finite fields.
More generally, using this method one can show that for any $n \geq 1$, we have $$R(\exp(x+x^p/p+\cdots + x^{p^n}/p^n)) = R(\exp(x^{p^{n+1}}/p^{n+1})).$$ Even though I understand the details involved in this calculation, I don't know if there is some more general theory underlying these examples. It may be helpful to share similar examples that you know. For instance, is it always the case that $$R(\exp(f(x))) > R(\exp(x)),$$ given $f(x) \in x\mathbb{C}_p[[x]]$ has a nonzero root $\alpha \in \mathbb{C}_p$ of absolute value $R(\exp(x)) = (1/p)^{1/(p-1)}$ and no non-zero roots of smaller absolute value?
• I may be able to offer a partial answer here, but it’ll have to wait till morning at best. Feb 7 '16 at 5:58
• Have you looked at "Rank one solvable p-adic differential equations and finite Abelian characters via Lubin–Tate groups" by Andrea Pulita? The abstract starts with "We introduce a new class of exponentials of Artin–Hasse type, called π-exponentials". Feb 7 '16 at 12:35
• @LaurentBerger: That looks interesting. I haven't looked at it, but I will surely do. Thank you! Feb 7 '16 at 22:29
Here is a counterexample to your question at the end, for each $p$. Let $f_u(x) = x + ux^p/p$ for $u \in \mathbf C_p$ with $|u|_p = 1$ and $|u-1|_p = 1$. (Such $u$ can be taken in $\mathbf Z_p^\times$ if $p > 2$, but you need to go outside $\mathbf Q_p$ if $p = 2$ to an extension with residue field of size greater than $2$.) Since $|u|_p = 1$, all the nonzero roots of $f_u(x)$ in $\mathbf C_p$ have absolute value $(1/p)^{1/(p-1)} = R(\exp)$.
To find the $p$-adic radius of convergence of $\exp(f_u(x))$, write $$\exp(f_u(x)) = \exp\left(x+\frac{x^p}{p}\right)\exp\left((u-1)\frac{x^p}{p}\right)$$ as formal power series. On the right side, the first factor has radius of convergence greater than $(1/p)^{1/(p-1)}$, as you noted.
Since $|u-1|_p = 1$, the second factor has radius of convergence equal to that of $\exp(x^p/p)$, which is $(1/p)^{1/(p-1)}$. The reciprocal $(\exp(x + x^p/p))^{-1}$ has the same radius of convergence as $\exp(x+x^p/p)$, even for $p=2$, so $\exp(f_u(x))$ has radius of convergence equal to $(1/p)^{1/(p-1)}$. This is the counterexample to your question.
• This a great specific example. More generally, for $u \in \mathbb{C}_p$, it shows that $R(\exp(x+ux^p/p))$ depends explicitly on $|u-1|_p$ whenever $$R(\exp(x+x^p/p)) \neq R((u-1)x^p/p).$$ I wonder what happens if equality holds. Is there a chance for the radius to further increase? Feb 7 '16 at 23:41
• In general if $F(x)$ and $F(x)^{-1}$ have a common radius of convergence $R$, and $G(x)$ and $G(x)^{-1}$ have a common radius of convergence $S$ then the product $FG$ has radius of convergence $\min(R,S)$ if $R \not= S$. If $R=S$ then there is no simple rule for the radius of $FG$. Typically the radius for $FG$ will remain $R$, but of course there are examples where it grows, like $\exp(x)$ and $\exp(x^p/p)$, or more simply $\exp(x)$ and $\exp(-x)$. Feb 8 '16 at 15:16
I can only give a very partial answer, and that only from my very parochial point of view.
I will use the additive valuation $v$ rather than absolute value, normalized so that $v(p)=1$, and in terms of which $R(\sum a_nx^n)=-\liminf\bigl(v(a_n)/n\bigr)$, so that when $v(z)>R(f)$, $f(z)$ is a convergent series. Examples are: if $\exp(x)=\sum_{n\ge1}x^n/n!$ and $\log(x)=\sum_{n\ge1}(-1)^{n-1}x^n/n$, then $R(\exp)=1/(p-1)$ and $R(\log)=0$. It’s for this reason that I prefer the logarithm to its inverse.
The log that I’ve named above is the logarithm of the multiplicative formal group $\hat{\mathbf G}_{\mathrm m}(x,y)=x+y+xy$, that is a formal-group homomorphism from $\hat{\mathbf G}_{\mathrm m}$ to the additive formal group $\hat{\mathbf G}_{\mathrm a}(x,y)=x+y$ with $\log'(0)=1$.
The $p$-typical logarithm $\log_{\mathrm{AH}}=x+x^p/p+x^{p^2}/p^2+\cdots$ is the logarithm of another formal group $\mathscr M$. which we might call the $p$-typical recoordinatization of $\hat{\mathbf G}_{\mathrm m}$, and the $\Bbb Z_p$-formal-group isomorphism $u:\mathscr M\to\hat{\mathbf G}_{\mathrm m}$ is exactly what’s called the Artin-Hasse Exponential. It satisfies $\log\circ u=\log_{\mathrm{AH}}$.
Now here’s the moral of my story. These two logarithms, $\log$ and $\log_{\mathrm{AH}}$, being convergent throughout the open unit disc of $\Bbb C_p$, have all sorts of interesting behavior that is not seen at all by the exponential series $\exp(x)$, except at a very far remove. In particular, they have zeros. The closest such to the origin has $v(\zeta)=1/(p-1)$, which explains immediately the radius of convergence of the exponential function.
And so I am pretty sure that your last conjecture has no chance of being correct. I tried $p=3$, $\zeta=\omega-1$, where $\omega^2+\omega+1=0$, and found that $\exp(x-x^2/\zeta)$ seemed to have as bad convergence properties as the exponential itself.
EDIT:
You’ve asked me to explain further why the existence of zeros of the logarithm prevents wider convergence of the exponential, and indeed, it is not quite so obvious as I was pretending.
Let $\lambda$ be a root of the above-described $\log$ with $v(\lambda)=1/(p-1)$, in fact $\lambda+1$ will be a primitive $p$-th root of unity. If $R(\exp)\ge1/(p-1)$, then the Newton polygon of $f(x)=\exp(x)-\lambda$ will have a vertex $(n,v(b_n))$ in addition to the vertex $(1,0)$. (You don’t need to know that $b_n=1/n!$.) To see this, you may look at $f(\lambda x)$ and note that its coefficients must go to zero for convergence. In particular, There will be a segment of the polygon whose negative slope is $\ge1/(p-1)$, and thus a root $\mu$ of $f$ in an algebraic extension such that $\exp(\mu)=\lambda$, impossible if $\log(\exp(x))=x$.
• This is great. Thank you! I was wondering if you have some references on this formal group construction of this Artin-Hasse Exponential and if this approach implies the fact that $\text{AH}(x)$ has $\mathbb{Z}_p$ coefficients and/or radius of convergence 1? What did you want to say at: "interesting behavior that is not seen at all by the exponential series $\exp(x)$, except at a very far \emph{remove}." Feb 7 '16 at 23:27
• Also, would you mind elaborating some more on how the root of valuation $1/(p-1)$ of the logarithm series $\log(1+x) = \sum_{n \geq 1} (-1)^{n-1}x^n/n$ explains immediately the radius of convergence of the exponential series? Is there some basic compositional inverse property I'm missing here? Feb 7 '16 at 23:28
• “At a very far remove” means “from very far away”. The basic reference is Hazewinkel’s big book from the 70’s. For your second question, I’ll make an addendum to my answer. Feb 8 '16 at 3:02
• Thank you for the EDIT. That's a nice way to argue and it seems to generalize to any two compositional inverse series where one has a bigger radius of convergence and at least one non-zero root. I looked up Hazewinkel but couldn't find if this formal group approach to the Artin-Hasse Exponential tells us $\text{AH}(x)$ has $\mathbb{Z}_p$ coefficients or if it has radius of convergence 1. Feb 8 '16 at 21:02
• Maybe it would be best to discuss via e-mail: I have the time, and there’s not enough space here. Feb 9 '16 at 1:19 | 2021-09-22T03:21:26 | {
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https://brilliant.org/discussions/thread/1-derivation-of-the-quadratic-formula/ | # 1) Derivation of the quadratic formula
This is note $$1$$ in a set of notes showing how to obtain formulas. There will be no words beyond these short paragraphs as the rest will either consist of images or algebra showing the steps needed to derive the formula mentioned in the title.
Suggestions for other formulas to derive are welcome, however whether they are completed or not depends on my ability to derive them. The suggestions given aren't guaranteed to be the next one in the set but they will be done eventually.
1 $\large ax^2 + bx + c = 0$
2 $\large x^2 + \frac{b}{a}x + \frac{c}{a} = 0$
3.1 $\large x^2 + \frac{b}{a}x = \left(x + \frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2$
3.2 $\large \left(x + \frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 + \frac{c}{a} = 0$
4 $\large \left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} + \frac{4ac}{4a^2} = 0$
5 $\large \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$
6 $\large x + \frac{b}{2a} = \pm\sqrt{\frac{b^2 - 4ac}{4a^2}}$
7 $\large x + \frac{b}{2a} = \frac{\pm\sqrt{b^2 - 4ac}}{2a}$
8 $\large x = \frac{- b \pm\sqrt{b^2 - 4ac}}{2a}$
Note by Jack Rawlin
2 years, 10 months ago
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Yup! This is the derivation of the quadratic formula. Great!
- 2 years, 10 months ago
I made a YouTube video showing how this would work in arbitrary fields not of characteristic $$2$$. The field not having characteristic $$2$$ is so important because this avoids any of your calculations having zero determinants; in other words, the quantities $$2$$ and $$4$$ will be $$0$$ modulo $$2$$ and our prescribed condition avoids this occurring. In such a setting, steps 6, 7 and 8 will not be valid; solutions can only exist when $$b^2 - 4ac$$ is a square number in the field. This indicates the severe limitations of the Fundamental Theorem of Algebra in its scope only being applied in the framework of the "real number field" and the "complex number field".
- 3 months ago | 2018-11-18T05:00:29 | {
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https://math.stackexchange.com/questions/3316359/why-is-exists-y-in-b-forall-x-in-a-px-y-stronger-than-forall-x | # Why is $\exists y \in B, \ \forall x \in A, \ P(x,y)$ stronger than $\forall x \in A,\ \exists y \in B, \ P(x,y)$
In the book that I am reading, the author notes that " $$\exists y \in B, \ \forall x \in A, \ P(x,y)$$" is a "stronger" statement than "$$\forall x \in A,\ \exists y \in B, \ P(x,y)$$" because the first one implies the truth of the second one...but the second one does not imply the truth of the first.
I think I understand the if then comment quite well and have included pictures to express my understanding:
For the "stronger case", we have:
if $$\ \ \exists y \in B, \ \forall x \in A, \ P(x,y)$$, then $$\ \forall x \in A,\ \exists y \in B, \ P(x,y)$$
which can be illustrated as:
(Hopefully this is comprehensible...I'm basically showing two equivalent pictures and that both versions of this picture satisfy the 'meaning' of the antecedent and consequent. And by "equivalent" I mean that both pictures describe the same element pairs that result in truth)
For the "weaker case", we have:
if $$\ \forall x \in A,\ \exists y \in B, \ P(x,y)$$, then $$\ \exists y \in B, \ \forall x \in A, \ P(x,y)$$
Which can be illustrated as follows:
For this weaker case, which could be drawn multiple ways (I just happened to pick one way that would show why this if then statement is false), we see that while the antecedent can be satisfied by the picture, the consequent clearly cannot.
So, okay...cool. I see that the one if then statement is true and the one if then statement is false. But why exactly does this feature make the first case "Strong". In what sense is it strong? Does it allow you to construct proofs more rigorously? Does it allow you to utilize a trick that greatly simplifies proof construction?
In what way does knowing $$\ \ \exists y \in B, \ \forall x \in A, \ P(x,y)$$ prove more beneficial than knowing $$\ \forall x \in A,\ \exists y \in B, \ P(x,y)$$?
Any insight (or examples) is greatly appreciated!
If general, a statement $$F$$ is told stronger than another statement $$G$$ if the implication $$F \Rightarrow G$$ holds (i.e. if $$G$$ holds whenever $$F$$ holds) but the converse implication $$G \Rightarrow F$$ does not hold (i.e. it is possible that $$G$$ holds but $$F$$ does not hold).
As you said, this is the case for $$F = \exists y \in B \, \forall x \in A \, P(x,y)$$ and $$G = \forall x \in A \, \exists y \in B \, P(x,y)$$, because if you assume $$F$$ then you can always prove $$G$$ (independently from the meaning of $$A$$, $$B$$ or $$P$$) but clearly the converse is not true: indeed, in the situation where $$A = B = \mathbb{N}$$ and $$P = \, <$$, we have that $$F = \exists y \in \mathbb{N} \, \forall x \in \mathbb{N} : x < y$$ is false ($$\mathbb{N}$$ has no maximum) but $$G = \forall x \in \mathbb{N}\, \exists y \in \mathbb{N} : x < y$$ (for every natural number $$x$$, its successor $$x+1$$ is greater than $$x$$).
It is stronger in the sense that anything you can prove with $$\forall x \ \epsilon \ A,\ \exists y \ \epsilon \ B, \ P(x,y)$$, can also be proven with $$\exists y \ \epsilon \ B, \ \forall x \ \epsilon \ A, \ P(x,y)$$ because the latter implies the former. As an analogy: Every nut you can crack with a rubber hammer can also be cracked with a sledge hammer. But there might be nuts that can only be cracked with a sledge hammer. Therefore the sledge hammer is stronger than the rubber hammer.
• Hmmm, I see. So, it seems like there may be times where I really need to prove "$\forall x \ \epsilon \ A,\ \exists y \ \epsilon \ B, \ P(x,y)$"...but to do so directly is very hard. However, it just so happens in this hypothetical, that " $\exists y \ \epsilon \ B, \ \forall x \ \epsilon \ A, \ P(x,y)$" is much easier to prove. Therefore, I can use this to my advantage and prove this "easier" one...which then gives me access to the statement that I actually wanted. Is that sort of right? – S.Cramer Aug 7 '19 at 17:21
• Not quite. Although this could happen, too, it is rather that if you are trying to prove something with $\exists y \ \epsilon \ B, \ \forall x \ \epsilon \ A, \ P(x,y)$, it might be easier or solve the problem quicker than $\forall x \ \epsilon \ A,\ \exists y \ \epsilon \ B, \ P(x,y)$. So if you are unlucky and only have $\forall x \ \epsilon \ A,\ \exists y \ \epsilon \ B, \ P(x,y)$ (the small hammer) your proof might be harder or take longer than if you had $\exists y \ \epsilon \ B, \ \forall x \ \epsilon \ A, \ P(x,y)$ (the sledge hammer). – Klaus Aug 7 '19 at 18:27
Suppose $$\exists y\in B,\forall x\in A,P(x,y).$$
Let $$x_0\in A$$. According to the previous assertion, there exists $$y\in B$$ such that for all $$x\in A$$, $$P(x,y)$$. In particular for $$x=x_0$$, we have $$P(x_0,y)$$. We just proved that for all $$x_0\in A$$, there exists $$y\in B$$ such that $$P(x_0,y)$$, which can be rewritten $$\forall x_0\in a,\exists y\in B,P(x_0,y).$$
Of course you can replace $$x_0$$ with $$x$$, which gives $$\forall x\in A,\exists y\in B,P(x,y).$$
Therefore, $$\left( \exists y\in B,\exists x\in A,P(x,y)\right)\implies\left(\forall x\in A,\exists y\in B,P(x,y)\right).$$
The left-hand side is stronger in the sense that it implies the right-hand side. If you want to prove the right-hand side, it is succifient to prove the left-hand side, but it is not necessary. If you prove the right-hand side by proving the left-hand side, we consider that you prove a stronger result. | 2020-01-26T02:57:04 | {
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http://susam.in/blog/combinatorial-coincidence/ | ## Combinatorial coincidence
I joined RSA a few weeks ago. Here, I met Anoop who loves playing with combinatorics problems. We challenge each other with combinatorics problems.
I asked him the following question at lunch.
For integers n ≥ 1 and k ≥ 1,
f0(n) = 1.
fk(n) = Σni=1 fk−1(i).
Find a closed formula for fk(n).
He asked me one on programming. I have used subscripted variables to represent the counters for the nested loops in the pseudocode to show the question.
Consider the following pseudocode with k nested loops.
count = 0
for c1 in 1 to n:
for c2 in 1 to c1:
for c3 in 1 to c2:
…
for ck in 1 to ck−1:
count = count + 1
What is the final value in count after the outermost loop terminates?
With one question each, we went back to our desks. As I started solving his question on nested loops, I realized that his problem led me to the recurrence relation in the question I asked him. If k = 1, count = n. This is also the value of f1(n). If k = 2, the inner loop with the counter as c2 will run once when c1 = 1, twice when c1 = 2, and so on. So, the final value of count will be f1(1) + f1(2) + … + f1(n) = n(n + 1)/2. This is the value of f2(n) as well.
Extending this argument, one may realize that for any k, count = fk−1(1) + fk−1(2) + … + fk−1(n) = fk(n). In other words, the answer to his question is the answer to my question. I already knew the answer to this. So, I went back to his desk with the answer: C(n +k −1, k).
I had arrived at this closed formula earlier by solving fk(n) for the first few k's using Faulhaber's formula. I noticed that they were all equal to C(n + k − 1, k). So, I proved that this is always true by the principle of strong mathematical induction. It involved proving that: fk+1(n) = C(k, k) + C(k + 1, k) + … + C(n + k − 1, k) = C(n + k, k + 1)
This was a remarkable coincidence that we asked each other questions which had the same answer. When I explained the coincidence to him, he explained how he arrived at the same result for the question on nested loops which can be used to determine the closed formula for the recurrence relation too.
In the question on nested loops, the following condition is always met: n ≥ c1 ≥ c2 ≥ … ≥ ck ≥ 1. So, the number of times count variable would be incremented is equal to the number of possible ways we can arrange k numbers from the first n natural numbers in descending order. The answer to this is the number of possible ways we can arrange n − 1 similar balls and k similar sticks. If we consider the number of balls to the right of the ith stick and add one to it, we get a valid value for ci as the number of balls to the left of a stick can not increase as we move right and consider (i + 1)th, (i + 2)th, etc. sticks. Similarly, the number of balls to the left of a stick can not decrease as we move left and consider (i − 1)th, (i − 2)th, etc. sticks. Also, any set of valid values for c1, c2, …, ck can be represented as an arrangement of these sticks and balls.
We can arrange n − 1 similar balls and k similar sticks in (n + k − 1)! (n − 1)! · k! = C(n + k − 1, k) number of ways. This is the answer to the question on nested loops as well as that on the recurrence relation.
### 1 comment
#### JB said:
Interesting discussion and I gained something today remembering you. | 2018-02-17T19:04:43 | {
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http://math.stackexchange.com/questions/70788/secant-line-finding-the-slope-for-different-values-of-x | # Secant Line: finding the slope for different values of x
The point $P(4, 24)$ lies on the curve $y = x^2 + x + 4$. If $Q$ is the point $(x, x^2 + x + 4 )$, find the slope of the secant line $PQ$ for the following values of $x$.
If $x= 4.1$, the slope of $PQ$ is:
and if $x= 4.01$, the slope of $PQ$ is:
and if $x= 3.9$, the slope of $PQ$ is:
and if $x= 3.99$, the slope of $PQ$ is:
Based on the above results, guess the slope of the tangent line to the curve at $P(4, 24)$.
For this problem, should I just plug in the x values given into the y equation? Then the slope would be...??
-
Would you happen to remember how to get the slope of a segment joining two points? – Ｊ. Ｍ. Oct 8 '11 at 5:09
You have the point $P(4,24)$ on the curve. If $x=4.1$, what is the corresponding $y$ on the curve? Yes, plug in, you get $Q(4.1,?)$. The problem asks you then to find the slope of the line $PQ$, so you are not quite through. – André Nicolas Oct 8 '11 at 5:22
What happened to your other, quite related, question? You were asked two questions in the comments there, which were meant to guide you towards a solution, but you answered none. You were also given two answers there, but you reacted to none, commented none and accepted none. Well, well, well... – Did Oct 8 '11 at 18:38
Since you will use several values of $x$, it is probably easiest to find a function for the slope, in terms of $x$, and then plug in the various values for $x$. That is, the slope of the secant line $PQ$ is the rise over run (change in $y$ over change in $x$):
$$m(x) = \frac{x^2 + x + 4 - 24}{x - 4}$$
So, $m(x)$ gives the slope for any particular value of $x$. A practical reason to do this is, for example on a TI-83 or TI-84 or something like it, you can now type in that function to $Y_1$ and then go to the Table and plug in the various $x$ values and you get the slopes immediately. And, since it is so fast, you can check this for more values than the question even asks for to get an even better intuition. Or, you could use Wolfram Alpha to accomplish the same thing. For example, type in:
Evaluate (x^2 + x + 4 - 24)/(x - 4) at x = 4.1, 4.01, 4.001, 3.9, 3.99, 3.999
Back to the problem,
\begin{align*} m(4.1) =& \frac{4.1^2 + 4.1 + 4 - 24}{4.1 - 4} = \frac{0.91}{0.1} = 9.1 \\ m(4.01) =& \frac{4.01^2 + 4.01 + 4 - 24}{4.01 - 4} = \frac{0.0901}{0.01} = 9.01 \\ m(4.001) =& \frac{4.001^2 + 4.001 + 4 - 24}{4.001 - 4} = \frac{0.009001}{0.001} = 9.001 \\ m(3.9) =& = \frac{3.9^2 + 3.9 + 4 - 24}{3.9 - 4} = \frac{-0.89}{-0.1} = 8.9 \\ m(3.99) =& = \frac{3.99^2 + 3.99 + 4 - 24}{3.99 - 4} = \frac{-0.0899}{-0.01} = 8.99 \\ m(3.999) =& = \frac{3.999^2 + 3.999 + 4 - 24}{3.999 - 4} = \frac{-0.008999}{-0.001} = 8.999 \end{align*}
From this, we would probably guess that the slope of the tangent line when $x = 4$ is 9. This does not guarantee that we are right, but assuming the function is reasonably well behaved, we could be pretty confident in this guess.
And, once you learn how to calculate derivatives, you will find this is correct as the slope of the tangent line at any $x$ value is the derivative. Since
$$y'(x) = 2x + 1$$
we see that
$$y'(4) = 9$$
- | 2014-03-16T01:49:46 | {
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https://math.stackexchange.com/questions/2582620/find-all-functions-f-mathbbr-mapsto-mathbbr-which-satisfy-fx2y-fz | # Find all functions $f:\mathbb{R} \mapsto \mathbb{R}$ which satisfy $f(x^2+y f(z)) =x f(x) + z f(y)$
QUESTION : Find all functions $f:\mathbb{R} \mapsto \mathbb{R}$ which satisfy $f(x^2+y f(z)) =x f(x) + z f(y)$
My doubt lies in the part where I've shown injectivity. Kindly check if my proof is correct.
The solution is $f(x)=x$ for every individual $x$. Let $P(x,y,z)$ be an assertion of the FE.
• $P(0,0,0)\equiv f(0)=0$
• $P(x,0,0)\equiv f(x^2)=xf(x)$
• $P(0,y,z)\equiv f(yf(z))=zf(y)$
Does the third condition imply injection in the following manner?
If $f(a)=f(b)$, then $f(yf(a))=f(yf(b))\implies af(y)=bf(y)\implies a=b$ assuming $f(y)\neq 0 \forall y$.
If till this it's correct, I am pretty much done with the whole problem.
Here's how I can proceed henceforth— $f(x^2)=xf(x)=f(xf(x))$, which, upon using the injectivity criteria leads to $x^2=xf(x)$. Now, $f(x)$ can be found after verification of the positive/negative value.
your solution is almost correct - you arrived at $f(a)=f(b)\Rightarrow{}af(y)=bf(y)$ for all $y\in{}\mathbb{R}$. Now you assumed that $f(y)\ne{}0$ for all $y$ (which solves the question but isn't justified). Instead it is enough to assume that $f(y)\ne{}0$ for some $y$. Note that if this does not hold, then $f(x)=0$ for all $x$, which is another solution.
Also, $xf(x)=x^2$ implies immidiatly that $f(x)=x$ for all $x\ne{}0$ without further checking. Did you prove that $f(0)=0$ though?
You only need to assume that there is some $y$ with $f(y)\neq 0$ in order for your argument on injectivity to hold. In this case your work shows that $f(x)=x$ for all $x\neq 0$, and since $f$ must be injective, the only possibility for $f(0)$ is $f(0)=0$. This concludes the proof that $f(x)=x$ for this case.
Now, the other case is simple, and ties in with Cameron's comment that $f\equiv 0$ is another -- and hence the only other -- solution. | 2021-07-26T17:19:48 | {
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https://math.stackexchange.com/questions/3100957/waiting-for-patterns | # waiting for patterns?
A fair coin is tossed until one of the patterns show up: TTH or THT. Let A be the event that TTH shows up before THT.
What is P(A)?
Here is my solution but I am not sure if it is correct or there is a better solution.
Let $$p=P(A)$$. Define
$$A_1=$$ the event that the first toss is H
$$A_2=$$ the event that the first two tosses are TT
$$A_3=$$ the event that the first three tosses are THT
$$A_4=$$ the event that the first three tosses are THH
Then this is a partition for the sample space.
$$p=P(A|A_1)P(A_1)+P(A|A_2)P(A_2)+P(A|A_3)P(A_3)+P(A|A_4)P(A_4)$$.
Then
$$p=p\frac{1}{2}+1\frac{1}{4}+0\frac{1}{8}+p\frac{1}{8}$$ which implies that $$p=\frac{2}{3}.$$
• Can anyone give a reference to a math paper which deals with more probability calculations related to patterns such as this example? Thanks. – Probability student Feb 6 at 19:12
We can start at the point when the first $$T$$ appears
The probability that $$T$$ occurs again is $$\frac{1}{2}$$ , and then $$TTH$$ comes first surely.
The probability that $$HT$$ comes next is $$\frac{1}{4}$$ , then the game is finished as well.
If $$HH$$ appears, we have to wait for the next $$T$$ and are again at the starting position.
The occurence of $$T$$ has a probability twice the probability of the occurence of $$HT$$ , hence the chance that $$TTH$$ wins must be twice the chance that $$THT$$ wins, giving the result $$p=\frac{2}{3}$$.
I think you've got the right idea, but I don't think all your numerical assignments are correct. You appear to have taken $$\ P\left(\,A\,\vert\, A_2\,\right)\$$ and $$\ P\left(\,A\,\vert\, A_3\,\right)\$$ to be $$\ \frac{1}{2}\$$ and $$\ 1\$$, respectively. I believe they should be $$\ 1\$$ and $$\ 0\$$, respectively.
Take $$\ P\left(\,A\,\vert\, A_2\,\right)\$$, which is the probability that the tosses terminate with TTH, given that the first two tosses are TT. If the first two tosses are TT, then it's impossible for the tosses to terminate in the sequence THT, because the very first subsequent head to appear after the initial TT must be preceded by two successive tails.
Likewise, if the first three tosses are THT (i.e. the event $$\ A_3\$$ occurs), then the tosses cease with $$\ A\$$ not having occured. Thus, I believe your equation for $$\ p\$$ should be $$p=p\frac{1}{2}+1\cdot\frac{1}{4}+0\cdot\frac{1}{8}+p\frac{1}{8}\ ,$$ which nevertheless still gives you $$\ p = \frac{2}{3}\$$.
• Thank you, you are right! I made a mistake, I will correct it. – Probability student Feb 5 at 10:45 | 2019-12-15T13:22:48 | {
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https://math.stackexchange.com/questions/3382322/even-and-odd-functions-and-whether-even-odd-characteristics-change-with-powers | # Even and odd functions and whether even/odd characteristics change with powers
I watched a video where a problem involved recognizing that $$\sin x$$ is an odd function and $$\sin^3 x$$ is also odd. But the presenter didn't explain why $$\sin^3 x$$ is also odd. Why does the fact that the function is odd not change when it is cubed? Is there a rule where for every even power the odd function is even and for every odd power the odd function remains odd? What about for even functions?
• If $f$ is odd and $g = f^3$, then $g(-x) = f(-x)^3 = (-f(x))^3 = (-1)^3f(x)^3 = -f(x)^3 = -g(x)$. You can play that game now with all variants. – amsmath Oct 6 '19 at 2:14
• @amsmath Why are you allowed to put the the negative sign from inside the parentheses to outside? – user532874 Oct 6 '19 at 2:16
• Because $(ab)^n = a^n\cdot b^n$. – amsmath Oct 6 '19 at 2:16
• 532874, that's the definition of odd function, no? – Gerry Myerson Oct 6 '19 at 2:17
• @user532874 A function $f$ is odd if $f(-x) = -f(x)$ for all $x$. – amsmath Oct 6 '19 at 2:18
Note that $$(-1)^{2k}=(+1)$$ and $$(-1)^{2k+1}=(-1)$$ $$(+1)^k = (+1)$$
Thus if a function is odd we have $$f^{2k}(-x) = (-1)^{2k}f^{2k}(x)=f^{2k}(x)$$ and $$f^{2k+1}(-x) = (-1)^{2k+1}f ^{2k+1}(x)=-f^{2k+1}(x)$$
Thus odd functions to the odd powers are odd and to the even powers are even.
Even functions to any power stay even.
• I think you are missing a $2k$ exponent after "Thus if a function is odd we have" – user532874 Oct 6 '19 at 2:35
• @user532874 is it fixed now? – Mohammad Riazi-Kermani Oct 6 '19 at 2:43
• @MohammadRiazi-Kermani Check your calculations. You are missing minuses. – amsmath Oct 6 '19 at 2:59
• @amsmath Thanks for informative comment. I should have been more careful. – Mohammad Riazi-Kermani Oct 6 '19 at 3:12
It is not only the cubing, it works for anyother odd function, it is always true that the composite of odd functions is an odd function.
In your case, $$f(x)=\sin x$$ is an odd function and $$g(x)=x^3$$ is also an odd function so $$g\circ f(x)= g(f(x))=g(\sin x)=\sin^3x$$
Another example is $$\sin x$$ with any odd power for example $$x^{12345}$$, since both are odd functions then $$\sin^{12345}x$$ is also an odd function.
Or $$\frac{1}{\sin x}$$ is also an odd function since it is the composite of $$\sin x$$ and $$\frac{1}{x}$$,and they are both odd functions.
Anyway you can just check if $$f(-x)=-f(x)$$ without worrying about compositions, $$f(-x)=\sin^3(-x)=(\sin (-x))^3=(-\sin x)^3 =-(\sin^3x)=-f(x)$$ | 2020-02-24T02:45:19 | {
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https://math.stackexchange.com/questions/2347692/how-to-prove-that-limit-doesnt-exist-using-epsilon-delta-definition | # How to prove that limit doesn't exist using epsilon-delta definition?
It is easy to prove the limit exists, all we have to show is there exists a relationship between $\delta$ and $\epsilon$. But how are we supposed to prove limit doesn't exists? The problem is when we are proving for a limit we already know what the limit is and with that, algebra is all that's needed.
Please show through an example (you may show that $\lim_{x\rightarrow0} \frac{1}{x}$ doesn't exist)
If possible please use the explanation scheme that is used by this answer https://math.stackexchange.com/a/66552/335742
• Hint: negate the definition. For some $\epsilon > 0 \ \ldots$ – Sean Roberson Jul 5 '17 at 20:40
• @SeanRoberson Are you trying to imply that I should use a proof by contradiction? If yes then maybe can you provide some example? – mathnoob123 Jul 5 '17 at 20:47
• \begin{align} & \lim_{x\to a} f(x) = L: \\ & \text{For every } \varepsilon>0 \text{ there exists } \delta>0 \\ & \text{ such that for every } x, \text{ if } 0<|x-a|<\delta \text{ then } |f(x) - L| < \varepsilon. \\ \\ & \lim_{x\to a} f(x) \ne L: \\ & \text{There exists } \varepsilon>0 \text{ such that for every } \delta>0, \\ & \text{ there exists } x \text{ such that } 0<|x-a|<\delta \text{ and } |f(x) - L| \ge\varepsilon. \\ {} \end{align} Put "For every $L$" in front of that last thing and then it says no limit exists. – Michael Hardy Jul 5 '17 at 20:49
• Can you please provide an example by solving the limit I provided (or if you fear it's a homework then one of your own examples)? – mathnoob123 Jul 5 '17 at 20:51
Suppose $L=\lim_{x\to0}\frac{1}{x}$ is finite.
If $L>0$, then there exists $\delta>0$ such that, for $0<|x|<\delta$, $|\frac{1}{x}-L|<L$, that is, $$0<\frac{1}{x}<2L$$ This is a contradiction, just take $-\delta<x<0$.
Similarly if $L<0$.
Thus we can only have $L=0$. Then there should exist $\delta>0$ such that, for $0<|x|<\delta$, $|\frac{1}{x}|<1$, an obvious contradiction.
It can be neither $\lim_{x\to0}\frac{1}{x}=\infty$ nor $\lim_{x\to0}\frac{1}{x}=-\infty$, because $\frac{1}{x}$ assumes positive and negative values in every punctured neighborhood of $0$.
Can this be generalized? Not really. For instance, in order to show the non existence of $\lim_{x\to0}\sin\frac{1}{x}$ the easiest way is to show that the limit should be in the interval $[-1,1]$, but that $\sin\frac{1}{x}$ assumes every value in $[-1,1]$ in each punctured neighborhood of $0$, so it is far from every possible limit. Alternatively, there exist sequences $(a_n)$ and $(b_n)$ convergin to $0$ such that $\lim_{n\to\infty}\sin\frac{1}{a_n}=0$ and $\lim_{n\to\infty}\sin\frac{1}{b_n}=1$.
As another example, $\lim_{x\to0}e^{1/x}$ doesn't exist because the one sided limits are different: from the left it is $0$, from the right it is $\infty$.
If $\displaystyle\lim_{x\to0} \frac 1 x = L$ then what happens if $\varepsilon=1$? There would exist $\delta>0$ such that if $0<|x|<\delta$ then $\displaystyle \left| \frac 1 x - L \right| <\varepsilon.$ But if $x$ is positive, you can make $1/x$ bigger than $L+\varepsilon$ by making $x$ less than $1/(L+\varepsilon).$ And a similar thing works if $x$ is negative (where you have $1/x\to-\infty$).
• Oh so the basic thing is an infinitesimal value namely $m$ will always be smaller than $\delta$ and $1/m>L+\epsilon$. Two questions. 1. Is the procedure always the same for any proof of the nonexistence of limit? 2. How to write this proof formally? 3. For limit definition (for proving a limit) is it okay to show $a\delta \leq \epsilon$ or is it always $a\delta <\epsilon$ where $a$ are the other terms that might be present? (I am aware there can be other terms, just concerned about the inequality signs?) – mathnoob123 Jul 5 '17 at 21:00
• You need to know something about the nature of the function in order to write the proof, but what I wrote in a comment under the question is true generally. – Michael Hardy Jul 5 '17 at 21:44 | 2021-08-04T03:17:37 | {
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https://www.freemathhelp.com/forum/threads/complex-numbers.117149/ | # Complex numbers
Status
Not open for further replies.
##### Full Member
The form:
$$\displaystyle a+bi$$
is referred to as the rectangular form for a complex number, which is just one of many forms in which complex numbers may be written. It is a particularly useful form though. Also, observe:
$$\displaystyle \frac{1}{2i}=\frac{1}{2i}\cdot\frac{i}{i}=\frac{i}{2i^2}=-\frac{i}{2}=0+\left(-\frac{1}{2}\right)i$$
I'm totally with you .. but once you've multiplied by i you've changed the cosmetics of the number .. isn't it matter? I mean isn't it matter for changing the definition of the number itself? who said that multiplying isn't changing the typecast of the number itself? really weird ..
#### MarkFL
##### Super Moderator
Staff member
I've multiplied by 1 in the form of i/i, and thus have not changed the value of the number. The form is different, but it still represents the same complex value.
#### HallsofIvy
##### Elite Member
The "cosmetics" of the number (I like that phrase!) is completely irrelevant to the problem. 3, III, 2+ 1, 6/2, and $$\displaystyle \sqrt{9}$$ are all different ways of saying exactly the same thing. It does not matter to the mathematics of the problem which you use.
Last edited:
##### Full Member
I would guess [you] have been asked to write this number in the rectangular form:
$$\displaystyle a+bi$$
no didn't been asked, while solving a question I was encountered to convert that horrible number of complex to a+bi
#### pka
##### Elite Member
lets assume I have complex number like (3-4i / 5j +7 ) then it's right to write like this :
(3-4i / 5j +7 ) = a + b*i
\displaystyle \begin{align*}\dfrac{3-4\mathcal{i}}{7+5\mathcal{i}}&=\dfrac{(3-4\mathcal{i})(7-5\mathcal{i})}{49+25} \\&=\dfrac{(21-20)+\mathcal{i}(-15-28)}{74}\\&=\dfrac{1}{74}+\bigg( \dfrac{-43}{74} \bigg) \mathcal{i}\end{align*}
Thus $$\displaystyle a=\dfrac{1}{74}~\&~b=\dfrac{-43}{74}$$ SEE HERE
#### lookagain
##### Senior Member
$$\displaystyle 0+\left(-\frac{1}{2}\right)i$$
It is common to write this final result in "a + bi" form as $$\displaystyle \ 0 - \tfrac{1}{2}i$$.
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Regarding post #10:
Likewise, it is common to write this final result in "a + bi" form as $$\displaystyle \ \tfrac{1}{74} - \tfrac{43}{74}i$$.
#### topsquark
##### Full Member
(3-4i / 5j +7 ) = a + b*i
I would have called this a typo except that it has been carried through on a couple occasions.
If you are a Mathematician (or Physicist) $$\displaystyle i^2 = -1$$ is common. If you are an Engineer then $$\displaystyle j^2 = -1$$. You shouldn't mix the two. So your expression should either be
$$\displaystyle \dfrac{3 - 4i}{7 + 5i}$$
or
$$\displaystyle \dfrac{3 - 4j}{7 + 5j}$$
-Dan
#### mmm4444bot
##### Super Moderator
Staff member
It is common to write … $$\displaystyle 0 - \tfrac{1}{2}i$$
… it is common to write … $$\displaystyle \tfrac{1}{74} - \tfrac{43}{74}i$$
Yes, in general, it's common to express the addition of a negative number as subtraction of its opposite. In particular, it's also common for authors who want to explicitly match the given form a+bi to stick with addition when b is negative by putting grouping symbols around it (like in posts #2 and #10).
Likewise, if emphasizing the given form when b is zero and a is not, authors will write a+(0)i instead of a.
#### MarkFL
##### Super Moderator
Staff member
It is common to write this final result in "a + bi" form as $$\displaystyle \ 0 - \tfrac{1}{2}i$$.
For the sake of the OP, I wanted to give the number strictly in the form:
$$\displaystyle a+bi$$
#### Jomo
##### Elite Member
Because I am bothered by it I always say a +/- bi.
Now here is my question. If you are asked to write something in a+bi form is an answer of 4 - 5i in the wrong form? I think it is!
#### pka
##### Elite Member
Regarding post #10:
Likewise, it is common to write this final result in "a + bi" form as $$\displaystyle \ \tfrac{1}{74} - \tfrac{43}{74}i$$.
For the sake of the OP, I wanted to give the number strictly in the form: $$\displaystyle a+bi$$
Correct & thank you.
I have never figured why some want-to-bes must correct every little thing.
But in this case the "correction" is a mistake. It does say a real number $$\displaystyle a$$ plus a real number $$\displaystyle b$$ times $$\displaystyle \mathcal{i}$$.
If the one doing the correcting has a background in teaching complex variables, it is standard to define the complex number field as a subset of $$\displaystyle \mathcal{R}\times\mathcal{R}$$. As a field there are two operations $$\displaystyle {\bf\large+~\&~\cdot}$$: $$\displaystyle (a,b)+(c,d)=(a+c,b+d)~\&~(a,b)\cdot(c,d)=(ac-bd,ad+bc)$$.
Here is a basic idea: define a conjugate of a complex number, $$\displaystyle \overline{(a,b)}=(a,-b)$$.
The absolute value is $$\displaystyle |(a,b)|=\sqrt{a^2+b^2}$$ So the multiplicative inverse $$\displaystyle (a,b)^{-1}=\dfrac{\overline{(a,b)}}{|(a,b|^2}$$
Here is short-hand notation: $$\displaystyle \dfrac{1}{z}=\dfrac{\overline{z}}{|z|^2}$$
Apply this to the problem at the hand:
$$\displaystyle \dfrac{3-4\mathcal{i}}{7+5\mathcal{i}}=\dfrac{(3-4\mathcal{i})(7-5\mathcal{i})}{49+25}$$
Last edited:
#### lookagain
##### Senior Member
That top right denominator should be "49 + 25."
Likewise, it is common to write this final result in "a + bi" form as $$\displaystyle \ \tfrac{1}{74} - \tfrac{43}{74}i$$.
For the sake of the OP, I wanted to give the number strictly in the form:
$$\displaystyle a+bi$$
Correct & thank you.
I have never figured why some want-to-bes must correct every little thing.
But in this case the "correction" is a mistake. It does say a real number $$\displaystyle a$$ plus a real number $$\displaystyle b$$ times $$\displaystyle \mathcal{i}$$.
Let's see. You appear to accept an alleged "correction," but in the same motion
you dismiss it as petty. You are self-contradictory.
I'm not a wan't-to-be, but you sure are desperate and looking to remain relevant. When I stated
"it is common," that was not a correction. But the b can be negative at times,
so those forms of mine follow. Your math you posted is irrelevant/unneeded.
It is readily seen that a = 4 and b = -5.
pka, this is the second post of yours that has been reported in recent weeks for you attacking me with your insults. You need to stop posting until you can be civil.
Last edited:
#### pka
##### Elite Member
pka, this is the second post of yours that has been reported in recent weeks for you attacking me with your insults. You need to stop posting until you can be civil.
To Lookagain, where did I say anything about you? I did not. Did I? Can you point to anything that I posted which namers you?
But if the shoe fits then wear it.
#### MarkFL
##### Super Moderator
Staff member
The OP's question has been answered, and so I'm going to close this thread
Status
Not open for further replies. | 2019-09-21T00:39:59 | {
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https://math.stackexchange.com/questions/2481863/doubt-about-substitution-in-int-11-sqrt1x2dx | # Doubt about substitution in$\int_{-1}^{1}\sqrt{1+x^2}dx$
I've tried this substitution in this integral
$$\int_{-1}^{1}\sqrt{1+x^2}dx$$
Let $x^2=t$, so $x=\sqrt{t}$ and $dx=\frac{1}{2\sqrt{t}}dt$. So we have
$$\frac{1}{2}\int_{1}^{1}\sqrt{\frac{1+t}{t}}dt=0$$
Which is obviously wrong. I know that this integral can be done with integration by parts or hyperbolic substitution, I want to know why this happens. My idea is that $x^2$ isn't always invertible, it is only on $[0, +\infty)$, and that causes this problem with the interval of integration.
Am I right? Thanks for your time.
• $x= \pm \sqrt t$ – John Lou Oct 20 '17 at 18:51
• Trig sub. Enjoy. – Randall Oct 20 '17 at 18:51
• i would use $$x=\tan(t)$$ – Dr. Sonnhard Graubner Oct 20 '17 at 18:52
• First see that you function is even on a symmetric domain so break your integral in two – Guy Fsone Oct 20 '17 at 18:53
• I think the OP is not asking how to do it, but why his method fails. – velut luna Oct 20 '17 at 18:56
Error, When you set this $x^2=t$, so $x=\sqrt{t}$ it is wrong because $$-1\le x \le 1$$ that is $x=\sqrt{t}$ is not only positive but can be $x= -\sqrt{t}$
Trick to reach your method see that, $\sqrt{1+x^2}$ is an even function therefore, $$\int_{-1}^{1}\sqrt{1+x^2}dx = 2\int_{0}^{1}\sqrt{1+x^2}dx$$ Now you are allowed to: Let $x^2=t$, so $x=\sqrt{t}$ and $dx=\frac{1}{2\sqrt{t}}dt$. So we have
$$\int_{-1}^{1}\sqrt{1+x^2}dx = 2\int_{0}^{1}\sqrt{1+x^2}dx =2\int_{0}^{1}\sqrt{\frac{1+t}{t}}dt$$
But Rather you could easilyy set $x = \sinh u \implies dx = \cosh u du$
and $\sqrt{1+x^2} =\sqrt{1+\sinh u^2} = \cosh u$ $$\int_{-1}^{1}\sqrt{1+x^2}dx =2\int_{0}^{1}\sqrt{1+x^2}dx \\=2\int_{0}^{\sinh^{-1}(1)} \cosh^2 u du= \int_{0}^{\sinh^{-1}(1)} 1+\cosh2 u du\\=\sinh^{-1}(1) +\frac12 \sinh(2\sinh^{-1}(1))$$
Using $$\cosh2u = \cosh^2 u +\sinh^2 u~~~and~~~~ \cosh^2 u -\sinh^2 u = 1.$$
on the other hand we have, $$\sinh 2u = 2c\cosh u \sinh u = 2\sinh u\sqrt{1+\sinh^2 u}$$
Using this we have $$\frac12 \sinh(2\sinh^{-1}(1)) =\sinh (\sinh^{-1}(1))\cdot\sqrt{1+\sinh^2\sinh^{-1}(1)} = \sqrt 2$$
Therefore, $$\color{red}{\int_{-1}^{1}\sqrt{1+x^2}dx du=\sinh^{-1}(1) +\frac12 \sinh(2\sinh^{-1}(1)) = \sinh^{-1}(1)+\sqrt2.}$$
The key question is this: As $x$ varies from $-1$ to $1$, what does $t$ do?
In many simple examples of $u$-substitution, the when $x$ varies across $(a,b)$ $u$ varies across $(c,d)$ and so we can just integrate the substituted function from $c$ to $d$ and be done with it.
We cannot do this in your case. In your case, the answer is that, as $x$ varies from $-1$ to $1$, $t$ varies from $1$ to $0$ and then back to $1$ again. Because there is this overlap, because $t$ traces the interval $[0,1]$ twice, you cannot simply replace the bounds. Additionally, on the two parts of this journey the function is actually different. For negative values of $x$ you need to use $-\sqrt{\quad}$ as the inverse function. This gives you:
$$\int_{-1}^1\sqrt{1+x^2}dx=\frac{1}{2}\int_1^0-\sqrt{\frac{1+t}{t}}dt+\frac{1}{2}\int_0^1\sqrt{\frac{1+t}{t}}dt=\int_0^1\sqrt{\frac{1+t}{t}}dt$$
Here the first integral is while $t$ is decreasing and the second is while $t$ is increasing. Notice that there's also a negative sign in front of this first integral. This is because $t$ is decreasing from $1$ to $0$ for values that correspond to negative $x$ values, so $x=-\sqrt{t}$ rather than $x=+\sqrt{t}$. | 2019-12-11T00:48:19 | {
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https://www.physicsforums.com/threads/12-red-snooker-balls.40405/ | # 12 red snooker balls
1. Aug 24, 2004
### Ian Rumsey
We have 12 red snooker balls and a scale balance.
One of the snooker balls has a different weight but we do not know whether it is heavier or lighter.
The scale balance may be only used three times by putting an equal number of balls on each side and noting the balancing result.
What 3 balancing arrangements would be required to ensure the identity of the odd weight ball and whether it was heavy or light.
2. Aug 24, 2004
### DarkForXe
hmm. I am not trying to be rude here.
But this q is asked before.
3. Aug 24, 2004
### TenaliRaman
i dunno if this is asked before but nice *twist* on the old *ionc* problem
4. Aug 24, 2004
### K.J.Healey
I don't think this is solveable with the alotted measurements and information.
5. Aug 24, 2004
### Rogerio
Surely it is!
Consider 1234 x 5678 at first...
6. Aug 24, 2004
### K.J.Healey
youre sure you read right? that we dont know if the unique ball is heaver OR lighter? I think that creates a problem.
7. Aug 24, 2004
### TenaliRaman
Healey,
It can be solved .....
infact it can be shown that
In n weighings, [(3^n-3)/2 + 1] snooker balls can be weighed and the odd ball can be found.
Placing n = 3, we get 13 .. so ofcourse it is possible even for 12.
I won't spoil this one for anyone .... i already received too many insinuating glares from Gokul already..... :D
-- AI
P.S -> So what does this post mean? It means that once the original problem is solved some followers will be,
Find a solution when we have 13 snooker balls?
Show that the general statement of n weighings given by me is true?
8. Aug 24, 2004
### NateTG
Odd -- according to your formula, I can't find it in 1 ball with 0 weighings, when the number for n=0 should be 1.
9. Aug 24, 2004
### Rogerio
Unfortunately, in just 3 weighings it's not possible for 13 balls...
(12 is the max)
10. Aug 24, 2004
### NateTG
Really, I'm pretty sure it's possible to find the odd ball out with 13 balls - you may not be able to tell whether it's light or heavy though.
11. Aug 24, 2004
### Hurkyl
Staff Emeritus
Then you're not looking hard enough.
If you have one ball, and you know one ball has the incorrect weight, then you don't need to do any weighings to find it!
The impossible case is where you have two balls.
(Oh, and incidentally, the formula says you can find 0 balls in 1 weighing)
12. Aug 24, 2004
### K.J.Healey
Someone post the answer in white. I've tried it numerous ways for a few minutes, and i can get pretty close, but for every possibility theres one where I cant determine if the odd ball was heavier or lighter than the others, but just different.
Hmm by that equation you should be able to find the odd ball out of 4 in 2 tries, and tell me if its heavier or lighter?? Please show me (its a lot simpler than the stated riddle).
EDIT : nevermind, i got it. But my simple question still stands.
Last edited: Aug 24, 2004
13. Aug 24, 2004
### NateTG
You're usually so on the ball (no pun intended) that I'm guessing you misread, or I wasn't clear enough, since we appear to (in principle) agree that the formula is incomplete.
However, the formula indicates that one ball can be identified in one weighing:
In the formula $$n$$ is the number of weighings, and it gives the number of balls.
My post indicated that the formula was too low in the case of 0 weigings.
$$\frac{3^n-3}{2}+1 = \frac{3^1-3}{2}+1 = \frac{0}{2}+1 = 1$$
But 0 balls in 0 weighings.
14. Aug 24, 2004
### Hurkyl
Staff Emeritus
Yes, I meant 0 in 0.
15. Aug 24, 2004
### NateTG
Should give you a good idea on how to find a solution to this problem. Notably, using that method, the 10 coin case might be more difficult that the 12 coin one.
16. Aug 24, 2004
### Gokul43201
Staff Emeritus
How about , <select to see>
first : L = 5, 6, 8, 10 R = 7, 9, 11, 12
second : L = 2, 3, 4, 7 R = 5, 6, 11, 12
third : L = 1, 4, 10, 11 R = 2, 5, 7, 8 ?
I think this gives unique outcomes for each number.
Defining outcomes as <, > or = on the basis of the L pan being heavier, R pan being heavier, or both equal.
1 : ==< or ==>
2 : =<> or =><
3 : =<= or =>=
4 : =<< or =>>
5 : <>> or ><<
6 : <>= or ><=
7 : ><> or <><
8 : <=> or >=<
9 : >== or <==
10: <=< or >=>
11: >>< or <<>
12: >>= or <<=
All look different to me...
17. Aug 24, 2004
### TenaliRaman
Both are related observations and well observed indeed ....
Ofcourse it was my mistake that i forgot to mention one tiny 'precursor' reqd to relax the problem.....
Hurkyl,
Now with 1 ball u know which is the odd one but u don't know whether it is heavier or lighter so u need one weighing (but weighed against what ??? that will be ur question ... wait i will address it shortly)
NateTG,
You are pretty correct indeed and we will be able to find the lighter or heavier condition too .....
The relaxation given to the problem,
"You can add one true weight ball to the existing set of balls"
(Hurkyl and NateTG would have got the entire general solution by now )
-- AI
18. Aug 25, 2004
### Rogerio
Now it's easy (and possible): compare 12345 x 6789T at first...:-)
19. Aug 25, 2004
### BobG
1. Left scale= 1, 2, 3, 4
Right scale = 5, 6, 7, 8
2. Left scale = 1, 4, 7, 10
Right scale = 2, 5, 8, 11
3. Left scale = 3, 6, 12
Right scale = 1, 8, 9
20. Aug 26, 2004
### BobG
Revised solution (forgot to check opposites)
1. Left scale= 1, 2, 3, 4
Right scale = 9, 10, 11, 12
2. Left scale = 1, 4, 7, 10
Right scale = 2, 5, 8, 11
3. Left scale = 1, 3, 9, 11, 12
Right scale = 2, 5, 6, 7, 10 | 2017-02-23T08:05:55 | {
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https://stats.stackexchange.com/questions/48262/what-is-the-difference-between-limiting-and-stationary-distributions | # What is the difference between "limiting" and "stationary" distributions?
I'm doing a question on Markov chains and the last two parts say this:
• Does this Markov chain possess a limiting distribution. If your answer is "yes", find the limiting distribution. If your answer is "no", explain why.
• Does this Markov chain possess a stationary distribution. If your answer is "yes", find the stationary distribution. If your answer is "no", explain why.
What is the difference? Earlier, I thought the limiting distribution was when you work it out using $P = CA^n C^{-1}$ but this is the $n$'th step transition matrix. They calculated the limiting distribution using $\Pi = \Pi P$, which I thought was the stationary distribution.
Which is which then?
• Your textbook may be making a distinction that is not universal: for instance, Karl Sigman's notes on limiting distributions defines "limiting" and "stationary" distributions to be synonymous (definition 2.3 at the bottom of p. 5). Therefore you must consult the definitions in your textbook in order to determine the difference.
– whuber
Jan 22 '13 at 18:41
• @whuber It is saying something like working out $\lim_{n \rightarrow \infty} P_{ii}^{(n)}$ and this doesn't exist. It then goes on to say "even though the limiting distribution doesn't exist, the stationary does. Let $\Pi = (\pi_0, \pi_1, ... ,\pi_n)$ be the stationary distribution...." But I guarantee you to calculate the limiting distribution in the question before, they solved it like this. Does that make sense to you? Jan 22 '13 at 19:25
• @whuber Actually, I'm quite confused now because in the previous limiting distribution question, they don't satisy the $\pi_0 + \pi_1 + \pi_2 = 1$ equality, so maybe that is different? Jan 22 '13 at 19:32
• A stationary distribution is one that is stable over time. As far as I'm aware, the limiting distribution of a Markov chain is stationary and if a Markov chain has a stationary distribution it is also a limiting distribution. Jul 1 '14 at 4:49
• Answer here by Andreas might help quora.com/… Jun 11 '18 at 22:27
From An Introduction to Stochastic Modeling by Pinsky and Karlin (2011):
A limiting distribution, when it exists, is always a stationary distribution, but the converse is not true. There may exist a stationary distribution but no limiting distribution. For example, there is no limiting distribution for the periodic Markov chain whose transition probability matrix is $$\mathbf{P}=\left\|\begin{matrix}0 & 1\\1 & 0\end{matrix}\right\|$$ but $\pi=\left(\frac{1}{2},\frac{1}{2}\right)$ is a stationary distribution, since $$\left(\frac{1}{2},\frac{1}{2}\right)\left\|\begin{matrix}0 & 1\\1 & 0\end{matrix}\right\|=\left(\frac{1}{2},\frac{1}{2}\right)$$ (p. 205).
In a prior section, they had already defined a "limiting probability distribution" $\pi$ by
$$\lim_{n\rightarrow\infty}P_{ij}^{(n)}=\pi_j~\mathrm{for}~j=0,1,\dots,N$$
and equivalently
$$\lim_{n\rightarrow\infty}\operatorname{Pr}\{X_n=j|X_0=i\}=\pi_j>0~\mathrm{for}~j=0,1,\dots,N$$ (p. 165).
The example above oscillates deterministically, and so fails to have a limit in the same way that the sequence $\{1,0,1,0,1,\dots\}$ fails to have a limit.
They state that a regular Markov chain (in which all the n-step transition probabilities are positive) always has a limiting distribution, and prove that it must be the unique nonnegative solution to
$$\pi_j=\sum_{k=0}^N\pi_kP_{kj},~~j=0,1,\dots,N,\\ \sum_{k=0}^N\pi_k=1$$ (p. 168)
Then on the same page as the example, they write
Any set $(\pi_i)_{i=0}^{\infty}$ satisfying (4.27) is called a stationary probability distribution of the Markov chain. The term "stationary" derives from the property that a Markov chain started according to a stationary distribution will follow this distribution at all points of time. Formally, if $\operatorname{Pr}\{X_0=i\}=\pi_i$, then $\operatorname{Pr}\{X_n=i\}=\pi_i$ for all $n=1,2,\dots$.
where (4.27) is the set of equations
$$\pi_i \geq 0, \sum_{i=0}^{\infty} \pi_i=1,~\mathrm{and}~\pi_j = \sum_{i=0}^{\infty} \pi_iP_{ij}.$$
which is precisely the same stationarity condition as above, except now with an infinite number of states.
With this definition of stationarity, the statement on page 168 can be retroactively restated as:
1. The limiting distribution of a regular Markov chain is a stationary distribution.
2. If the limiting distribution of a Markov chain is a stationary distribution, then the stationary distribution is unique.
• Can you clarify what you mean by 'transition probabilities don't change over time' for stationarity? Both limiting and stationary distribution are about the probabilities over states. Jul 1 '14 at 7:45
• Yup, I see you wrote your own answer but I reorganized mine to be more correct. Jul 1 '14 at 9:04
• I still don't get it. I mean what do you mean when you say "except now with an infinite number of states.... " ? Can you please clarify it more explicitely.
– roni
Jul 14 '16 at 16:09
• @roni the two expressions are identical if you let $N=\infty$ Jul 14 '16 at 16:30
• In the first highlighted block, $\pi=(1/2,1/2)$ is the stationary distribution for the example, however, it has no limiting distribution since $P^{n}$ will oscillate, and it therefore has no steady state. Does this mean that it won't guarantee the existence of steady state if only the stationary distribution is calculated? Feb 12 '19 at 3:24
A stationary distribution is such a distribution $\pi$ that if the distribution over states at step $k$ is $\pi$, then also the distribution over states at step $k+1$ is $\pi$. That is, $$\pi = \pi P.$$ A limiting distribution is such a distribution $\pi$ that no matter what the initial distribution is, the distribution over states converges to $\pi$ as the number of steps goes to infinity: $$\lim_{k\rightarrow \infty} \pi^{(0)} P^k = \pi,$$ independent of $\pi^{(0)}$. For example, let us consider a Markov chain whose two states are the sides of a coin, $\{heads, tails\}$. Each step consists of turning the coin upside down (with probability 1). Note that when we compute the state distributions, they are not conditional on previous steps, i.e., the guy who computes the probabilities does not see the coin. So, the transition matrix is $$P = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.$$ If we first initialize the coin by flipping it randomly ($\pi^{(0)} = \begin{pmatrix}0.5 & 0.5\end{pmatrix}$), then also all subsequent time steps follow this distribution. (If you flip a fair coin, and then turn it upside down, the probability of heads is still $0.5$). Thus, $\begin{pmatrix} 0.5 & 0.5 \end{pmatrix}$ is a stationary distribution for this Markov chain.
However, this chain does not have a limiting distribution: suppose we initialize the coin so that it is heads with probability $2/3$. Then, as all subsequent states are determined by the initial state, after an even number of steps, the state is heads with probability $2/3$ and after an odd number of steps the state is heads with probability $1/3$. This holds no matter how many steps are taken, thus the distribution over states has no limit.
Now, let us modify the process so that at each step, one does not necessarily turn the coin. Instead, one throws a die, and if the result is $6$, the coin is left as is. This Markov chain has transition matrix $$P = \begin{pmatrix} 1/6 & 5/6 \\ 5/6 & 1/6 \end{pmatrix}.$$ Without going over the math, I will point out that this process will 'forget' the initial state due to randomly omitting the turn. After a huge amount of steps, the probability of heads will be close to $0.5$, even if we know how the coin was initialized. Thus, this chain has the limiting distribution $\begin{pmatrix} 0.5 & 0.5 \end{pmatrix}$.
• Good point about forgetting the initial state, I completely glossed over this in my answer. Jul 1 '14 at 9:07
• This explanation helps me understand a lot. Can I say the existence of a steady state is equivalent to the existence of a limiting distribution? Since it is not easy to calculate the limiting distribution, we often calculate the stationary distribution by solving balance equations instead. However, I thought this alternative method doesn't guarantee that the stationary distribution is independent from initial states, therefore, it explains why for $P = \begin{pmatrix}0&1\\1&0\end{pmatrix}$, it has the stationary distribution but no steady state that is independent from the initial states. Feb 12 '19 at 3:59
• @GuoyangQin If you have a new question, you may wish to post it as a question (linking to this one if it helps provide question). Although I would have thought "steady state" in this context would mean "stationary distribution" so it would be best to clearly define the term in the question Feb 12 '19 at 18:57
Putting notation aside, the word "stationary" means "once you get there, you will stay there"; while the word "limiting" implies "you will eventually get there if you go far enough". Just thought this might be helpful.
• It isn't clear how this applies to the question. Could you explain?
– whuber
Jul 28 '15 at 10:17
• Hi @whuber, I mean to say that a limiting distribution is necessarily a stationary distribution while a stationary distribution is not necessarily a limiting distribution. Hence there is a difference. This is essentially the same as other answers but I think it's easier to remember.
– Khoa
Jul 17 '16 at 2:24
• Thank you for the clarification: it shows us what you are attempting to accomplish. However, I cannot find any reasonable way to interpret your description of "stationary" in a way that is consistent with the mathematical definition.
– whuber
Jul 17 '16 at 13:08
• @whuber BlueSky's phrasing seems like an extremely straightforward plain English notion of "fixed point" to me -- I'm not sure what your object could mean. Aug 20 '17 at 18:51 | 2021-12-09T11:28:29 | {
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https://www.physicsforums.com/threads/maximum-of-the-function-f-x-x-1-x-0-5.605814/ | # Maximum of the function F(X)=X*(1+X)^0.5
1. May 13, 2012
### Mike s
Hello,
Is x=-1 local maximum of $F(x)=x\sqrt{1+x}$?
On the one hand, $F(-1+\delta)<F(-1)$ for $0<\delta<1$.
However, $F(-1-\delta)$ is not defined.
As far a I know, the point x=a is considered local maximum, if there exists small neighborhood $\delta$ such that $$F(a)>F(x)$$ for every $a-\delta<x<a+\delta$.
So is x=-1 a local maximum or not?
2. May 13, 2012
### Fightfish
I believe that the term "local maximum" is not applicable because the concept of neighbourhood does not exist about the point x=-1 as you have pointed out.
Rather this is what we call a "boundary maximum". If you look at the gradient of F(x) at x=-1, it is undefined, as opposed to 0 gradient for stationary maxima/minima.
It is worth noting that boundary points are always "maximum" or "minimum" (except when the curve is level about that point).
3. May 13, 2012
### Ray Vickson
Yes, it is. It satisfies F(-1) > F(x) for all sufficiently small |x+1| with x > -1. In this problem it is impossible to compare F(-1) to F(x) for x < -1, but that is not really the reason that x = -1 is a local max. In the presence of constraints a local max or min need not have derivative = 0, and whether or not the function could be defined outside the interval is not particularly important; for example, for the problem min/max f(x) = x, subject to 0 <= x <= 1, x = 0 is a min and x = 1 is a max. These two statements are true even though f(x) is perfectly well defined outside the interval.
Note: you may find that some writers disagree with these statements. However, they are commonly asserted in modern treatments of optimization theory and practice; they are really matters of definition, not of substance.
RGV
4. May 14, 2012
### Fightfish
Agreed. Most texts do define local maxima/minima as being on an open interval. In a closed interval, it is trivial as the endpoints certainly must be minima/maxima.
For all practical purposes though, it doesn't matter if its "local" or "not local". It's not going to affect your calculations or results, so don't fret too much over it.
5. May 14, 2012
### Mike s
I understand, thanks for both of you. | 2017-12-17T14:23:17 | {
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https://mathematica.stackexchange.com/questions/118242/precision-of-findroot | # Precision of FindRoot
Consider the following code:
r = x /. FindRoot[Cos[x] - x == 0, {x, 1}, AccuracyGoal -> 15,
PrecisionGoal -> 15, WorkingPrecision -> 100];
Precision[r]
I would expect the Precision of r to be 15 since I'm only asking Mathematica to solve it to 15 digits. But the result is 100, which is the WorkingPrecision I am using. Why?
• I think the precision goal is how close you need to get to the true root. Here FindRoot will use 100 digits in x in each step of the iteration towards the root, and when it has found the root to 15 digit precision, it stops, and return that x, which still has a precision of 100. – Marius Ladegård Meyer Jun 12 '16 at 8:22
• Just like @Marius said, PrecisionGoal and AccuracyGoal will only inform you the digits you can use. WorkingPrecision will inform you the resulting number's precision, but maybe the latter digits is not usable. – Wjx Jun 12 '16 at 8:39
• See Controlling the Precision of Results and Setting WorkingPrecision->n causes all internal computations to be done to at most n-digit precision. See WorkingPrecision – user9660 Jun 12 '16 at 8:49
Here is a short demo. Generate the Dottie number as an exact Root[] object, like so:
dottie = x /. First @ Solve[x == Cos[x] && 0 < x < 1, x];
(Tho you might notice an inexact number in the output, rest assured that the resulting Root[] object is an exact number that can be evaluated to arbitrary precision; the number is there only as a sort of "localization marker".)
Use FindRoot[] to generate an approximation of the Dottie number:
ndottie = x /. FindRoot[Cos[x] - x, {x, 1}, AccuracyGoal -> 5, PrecisionGoal -> 5,
WorkingPrecision -> 30];
Altho Precision[ndottie] == 30., the result is not actually accurate to all 30 digits. Here is the relative error:
-Log10[Abs[1 - ndottie/dottie]]
20.0631570441
which says that the result is accurate to about 20 digits, even with the supposedly low settings. Now, crank up the settings:
ndottie2 = x /. FindRoot[Cos[x] - x, {x, 1}, AccuracyGoal -> 15,
PrecisionGoal -> 15, WorkingPrecision -> 30];
and you'll find that ndottie2 - dottie is effectively zero.
• One can see with StepMonitor that the relative error estimate in the next-to-last (3rd) step is about $(x_3-x_{2})/x_3=3.8 \times 10^{-5}$, a bit greater than the PrecisionGoal, so FindRoot takes another step. But because of the quadratic convergence in this well-behaved problem, this is closer to the error in $x_2$. The next step has a error estimate of $(x_4-x_3)/x_4=2.3 \times 10^{-10}$, but that means the error in $x_4$ will be close to $10^{-20}$. – Michael E2 Apr 3 '17 at 10:19 | 2020-02-18T15:24:12 | {
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http://mathhelpforum.com/differential-geometry/209922-complete-metric-space.html | 1. ## Complete metric space
Let $X_{n+1}\subset X_n$ for each $n\in\mathbb N.$ Let $(M,d)$ be a metric space. Prove that the following assertions are equivalent:
a) $(M,d)$ is a complete metric space.
b) For each $X_{n+1}\subset X_n$ being closed and non-empty where $\text{diam}(X_n)\to0$ as $n\to\infty,$ we have $\bigcap_{n=1}^\infty X_n\ne\varnothing.$
I already did a) $\implies$ b) but for the other implication, I have a long solution, is there a short way to do it?
2. ## Re: Complete metric space
let {xn} be a cauchy sequence in X. by definition for any ε > 0 there is some N such that d(xm,xn) < ε for all m,n > N.
so pick εk = 1/2k, and set Uk = cl(B(xNkk)).
define Xk = U1∩U2∩...∩Uk.
it's clear that diam(Xk) ≤ 1/2k-1 and that each Xk is closed being the intersection of finitely many closed sets.
furthermore since Nk < Nk+1 its clear that for any k:
xNk is in Um, for all m ≤ k, so each Xk is non-empty.
so, we have from (b) that ∩kXk ≠ Ø.
but this means the subsequence {xNk} of {xn} converges to some point x of X (the sole element of the infinite intersection, since if d(x,y) = 0, y = x, and for any y in ∩kXk,
we have d(y,xNk) < 1/2k-1 for ALL k in N). hence {xn} also converges to x, so X is complete.
(i was tempted to use just the 1/2k balls at first, but i realized that the 2nd ball could contain points OUTSIDE the first ball, but their intersection has no such defect. the center of the 3rd ball has to lie within the intersection of the first 2, and the center of the 4th ball has to lie within the intersection of the first 3, etc. the diameter of the intersections can be no larger than the diameter of the ball, which is twice its radius).
3. ## Re: Complete metric space
Okay, thank you!
I was thinking the following: let $x_n$ be a Cauchy sequence on $X,$ then let $X_n=\overline{\{x_m:m\ge n\}},$ so $X_n$ is closed and satisfies $X_{n+1}\subset X_n,$ yet I don't see how to prove that $\text{diam}(X_n)\to0.$ | 2016-12-07T20:59:56 | {
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https://math.stackexchange.com/questions/511533/can-an-m-times-n-rank-1-matrix-be-written-as-a-product-of-an-m-times1-and | Can an $m \times n$ rank $1$ matrix be written as a product of an $m\times1$ and a $1\times n$ matrix?
If a $m \times n$ matrix has rank $1$, does it imply that it can be written as a product of one $m\times1$ and one $1\times n$ matrix. How to prove it ? Is this decomposition unique ? What are the entries of these matrices ?
In a matrix of rank 1, every row is a scalar multiple of every other non-zero row (and there must be at least one non-zero row). For the $1 \times n$ matrix, take one of the non-zero rows; in the $m \times 1$ matrix, put the scalars from the first sentence. The decomposition is not unique: you can pick every non-zero scalar multiple of a non-zero row to start with.
Yes, it is always possible. Let
$$A = \begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_m \end{bmatrix}$$
be a matrix of rank $1$, and let $a_i$ be its rows. Since $A$ is of rank $1$, each two of these rows are linearly dependent. Let us assume that $a_k \ne 0$ for some $k$ (such row must exist, or $A$ would be a zero matrix). Then for each $i$,
$$a_i = \alpha_i a_k,$$
for some (real or complex) numbers $\alpha_i$. Define
$$v = \begin{bmatrix} \alpha_1 \\ \vdots \\ \alpha_m \end{bmatrix}, \quad w := a_k.$$
Then $A = v w$.
Obviously, the decomposition is not unique. Let $\xi \not \in \{0,1\}$. Then $A = (\xi v)(\xi^{-1} w)$ is another such decomposition.
Let $A = [a_1\ \cdots\ a_n]$ be the $m\times n$ where the $a_i$ are $m$-dimensional column vectors. If $A$ has rank $1$, then $\dim\operatorname{Col} = 1$. Therefore $\dim\,\operatorname{span}\{a_1, \dots, a_n\} = 1$. Choose $a_j \neq 0$ (such a column exists otherwise the $A = O$ which has rank zero), then $\operatorname{span}\{a_1, \dots, a_n\} = \operatorname{span}\{a_j\}$. In particular, $a_i \in \operatorname{span}\{a_j\}$ for $i = 1, \dots, n$. Therefore $a_i = c_ia_j$ for some $c_i \in \mathbb{R}$. Then
$$A = [a_1\ \cdots\ a_n] = [c_1a_j\ \cdots\ c_na_j] = a_j[c_1\ \cdots c_n].$$
Keep in mind, there is more to the last equality than first appears; $a_j$ is column vector not a constant.
More generally, an $m\times n$ matrix $A$ has rank $k$ if and only if the smallest $r$ such that $A = u_1v_1^T + \dots + u_rv_r^T$, where $u_1, \dots, u_r \in \mathbb{R}^m$, $v_1, \dots, v_r \in \mathbb{R}^n$, is $r = k$.
The other answers have addressed the issue of existence quite well. Let me show that the decomposition is "essentially unique", i.e. up to scalar multiple of the two vectors.
Suppose that $A=\mathbf{x}_1\mathbf{y}_1^\mathrm{T} = \mathbf{x}_2\mathbf{y}_2^\mathrm{T}$. Note that $A\mathbf{v}=c\mathbf{w}$ for some vector $\mathbf{w}$ and some scalar $c$ since the image is one dimensional. Thus we have $$A\mathbf{y}_1 = \mathbf{x}_1\mathbf{y}_1^\mathrm{T}\mathbf{y}_1=\|\mathbf{y}_1\|^2\mathbf{x}_1=c_1\mathbf{w}$$ $$A\mathbf{y}_2 = \mathbf{x}_2\mathbf{y}_2^\mathrm{T}\mathbf{y}_2=\|\mathbf{y}_2\|^2\mathbf{x}_2=c_2\mathbf{w}$$ Therefore $\mathbf{x}_1$ and $\mathbf{x}_2$ are multiples of the same vector $\mathbf{w}$. The argument can be repeated for $\mathbf{y}_1$ and $\mathbf{y}_2$ by taking the transpose and repeating the argument. | 2019-11-15T12:52:39 | {
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http://mathhelpforum.com/algebra/82918-help.html | # Math Help - help
1. ## help
mr.lopez needs 3/4 kg of apples to make a pie.how many pies can he make if he uses 20 kg of apples?
name a fraction that is between 1/2 and 1/3....?
which of the following fractions is closest to one?
a)2/3 b)3/4 c)4/5 d)5/6
three pizzas were evenly shared among some friends. if each of them go 3/5 of the pizza. how many were there altogether?
2. Originally Posted by BeBeMala
mr.lopez needs 3/4 kg of apples to make a pie.how many pies can he make if he uses 20 kg of apples?
How many 3/4 kilograms can fit inside 20 kilograms? (Hint: Divide.)
Originally Posted by BeBeMala
name a fraction that is between 1/2 and 1/3....?
There are infinitely-many choices. One quick way of finding such a fraction is to convert the two given fractions to a common denominator, and then find some other fraction of the same denominator that is between the two numerators.
Originally Posted by BeBeMala
which of the following fractions is closest to one?
a)2/3 b)3/4 c)4/5 d)5/6
Convert them all to the same denominator, and subract each from 1. Whichever one has the smallest answer was the biggest (and thus closest to 1).
Originally Posted by BeBeMala
three pizzas were evenly shared among some friends. if each of them go 3/5 of the pizza. how many were there altogether?
This one works just like the first one, above.
3. Originally Posted by stapel
How many 3/4 kilograms can fit inside 20 kilograms? (Hint: Divide.)
There are infinitely-many choices. One quick way of finding such a fraction is to convert the two given fractions to a common denominator, and then find some other fraction of the same denominator that is between the two numerators.
Convert them all to the same denominator, and subract each from 1. Whichever one has the smallest answer was the biggest (and thus closest to 1).
This one works just like the first one, above.
sorry....
4. Originally Posted by BeBeMala
sorry(((no help ))))....
sorry i'm so confused.......
5. Hello,
1) How many 3/4 kilograms can fit inside 20 kilograms?
$\frac{20}{\frac{3}{4}}=20*\frac{4}{3}=....$
2)name a fraction that is between 1/2 and 1/3....?
1/2 and 1/3 are equivalent to 30/60 and 20/60 u can choose any number between 21/60 and 29/60 to be your answer
3)which of the following fractions is closest to one?
a)2/3 b)3/4 c)4/5 d)5/6
2/3=0,(6)
3/4=0,75
4/5=0,8
5/6=0,8(3)
4) three pizzas were evenly shared among some friends. if each of them go 3/5 of the pizza. how many were there altogether?
$\frac{3}{\frac{3}{5}}=3*\frac{5}{3}=5$
Have a nice day,
Hush_Hush
6. Originally Posted by Hush_Hush
Hello,
1) How many 3/4 kilograms can fit inside 20 kilograms?
$\frac{20}{\frac{3}{4}}=20*\frac{4}{3}=....$
2)name a fraction that is between 1/2 and 1/3....?
1/2 and 1/3 are equivalent to 30/60 and 20/60 u can choose any number between 21/60 and 29/60 to be your answer
3)which of the following fractions is closest to one?
a)2/3 b)3/4 c)4/5 d)5/6
2/3=0,(6)
3/4=0,75
4/5=0,8
5/6=0,8(3)
4) three pizzas were evenly shared among some friends. if each of them go 3/5 of the pizza. how many were there altogether?
$\frac{3}{\frac{3}{5}}=3*\frac{5}{3}=5$
Have a nice day,
Hush_Hush
Thank you so much, I finally got it, sorry but i don;t get """name a fraction that is between 1/2 and 1/3???
7. ## Fractions
Hello BeBeMala
Originally Posted by BeBeMala
Thank you so much, I finally got it, sorry but i don;t get """name a fraction that is between 1/2 and 1/3???
Take a close look at the pictures I've attached.
• The shaded areas in the top one show that the fraction $\frac{6}{12}$ is the same fraction as $\frac{1}{2}$.
• The middle one shows that $\frac{4}{12}$ is the same as $\frac{1}{3}$.
• The bottom one has 5 sections shaded out of 12. And the size of the shaded area in this one is in between the other two. So this fraction is in between $\frac{1}{2}$ and $\frac{1}{3}$.
What fraction is shaded in the bottom diagram? | 2014-09-20T18:42:13 | {
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https://math.stackexchange.com/questions/3038472/if-x-1-x-2-ldots-x-n-are-the-roots-for-1xx2-ldotsxn-0-find-the-value | # If $x_1,x_2,\ldots,x_n$ are the roots for $1+x+x^2+\ldots+x^n=0$, find the value of $\frac{1}{x_1-1}+\frac{1}{x_2-1}+\ldots+\frac{1}{x_n-1}$
Let $$x_1,x_2,\ldots,x_n$$ be the roots for $$1+x+x^2+\ldots+x^n=0$$. Find the value of $$P(1)=\frac{1}{x_1-1}+\frac{1}{x_2-1}+\ldots+\frac{1}{x_n-1}$$
Source: IME entrance exam (Military Institute of Engineering, Brazil), date not provided (possibly from the 1950s)
My attempt: Developing expression $$P(1)$$, replacing the 1 by $$x$$, follows $$P(x)=\frac{(x_2-x)\cdots (x_n-x)+\ldots+(x_1-x)\cdots (x_{n-1}-x)}{(x_1-x)(x_2-x)\cdots (x_n-x)}$$
As $$x_1,x_2,\ldots,x_n$$ are the roots, it must be true that
$$Q(x)=(x-x_1)\cdots(x-x_n)=1+x+x^2+\ldots+x^n$$ and $$Q(1)=(1-x_1)\cdots(1-x_n)=n+1$$ Therefore the denominator of $$P(1)$$ is $$(-1)^{n} (n+1).$$ But I could not find a way to simplify the numerator.
Another fact that is probably useful is that $$1+x^{n+1}=(1-x)(x^n+x^{n-1}+\ldots+x+1)$$ with roots that are 1 in addition of the given roots $$x_1,x_2,\ldots,x_n$$ for the original equation, that is
$$x_k=\text{cis}(\frac{2k\pi}{n+1}),\ \ k=1,\ldots,n.$$
This is as far as I could go...
Hints and full answers are welcomed.
• – Arnaud D. Dec 17 '18 at 12:07
$$1$$ and $$x_1,\ldots,x_n$$ are the roots of $$z^{n+1}-1=0$$. They are the $$(n+1)$$-th roots of unity. Then $$1/(z^{n+1}-1)$$ has a partial fraction expansion of the form $$\frac1{z^{n+1}-1}=\frac a{z-1}+\sum_{k=1}^n\frac{b_k}{x_kz-1}.$$ Multiplying by $$z-1$$ and setting $$z=1$$ gives $$a=1/(n+1)$$. Multiplying by $$x_kz-1$$ and setting $$z=1/x_k$$ gives $$b_k=1/(n+1)$$. Therefore $$\sum_{k=1}^n\frac1{x_kz-1}=\frac{n+1}{z^{n+1}-1}-\frac1{z-1}.$$ Letting $$z\to1$$ gives $$\sum_{k=1}^n\frac1{x_k-1}=\lim_{z\to1}\frac{n-z-z^2-\cdots-z^n}{z^{n+1}-1} =\frac{-1-2-\cdots-n}{n+1}=-\frac n2.$$
Here's a trick proof. Let $$S$$ denote the sum in question. As the reciprocals of the $$x_k$$ are the $$x_k$$ again (in a different order) then $$S=\sum_{k=1}^n\frac1{x_k^{-1}-1}=\sum_{k=1}^n\frac{x_k}{1-x_k}$$ and $$2S=\sum_{k=1}^n\left(\frac1{x_k-1}+\frac{x_k}{1-x_k}\right) =\sum_{k=1}^n(-1)=-n.$$
Applying the same technique as per this answer $$Q(x)=\prod\limits_{k=1}^n(x-x_k) \text{ and } Q'(x)=\sum\limits_{i=1}^{n}\prod\limits_{k=1,k\ne i}^{n}\left(x-x_k\right)$$ then $$\frac{Q'(x)}{Q(x)}=\sum\limits_{i=1}^{n}\frac{1}{x-x_i}$$ or $$-\frac{Q'(1)}{Q(1)}=\sum\limits_{i=1}^{n}\frac{1}{x_i-1}$$ But $$Q(1)=n+1$$ and $$Q'(x)=1+2x+3x^3+...+nx^{n-1}\Rightarrow Q'(1)=\frac{n(n+1)}{2}$$ and $$\sum\limits_{i=1}^{n}\frac{1}{x_i-1}=-\frac{n}{2}$$
We have $$x^{n+1}-1=0\ \ \ \ (1)$$ with $$x\ne1$$
Set $$\dfrac1{x-1}=y\iff x=\dfrac{y+1}y$$
Replace the value of $$x$$ in terms of $$y$$ in $$(1)$$ to form an $$n$$ degree equation in $$y$$ $$\left(\dfrac{y+1}y\right)^{n+1}=1$$
$$\implies\binom{n+1}1 y^n+\binom{n+1}2 y^{n-1}+\cdots+1=0$$
Now apply Vieta's formula to find $$\sum_{r=1}^n\dfrac1{x_r-1}=\sum_{r=1}^ny_r=-\dfrac{\binom{n+1}2}{\binom{n+1}1}=?$$
Simply note that $$1,x_1,\ldots, x_n$$ are roots of $$x^{n+1} -1 = 0.$$ Thus we have $$0,x_1-1,\ldots, x_n-1$$ are roots of $$(x+1)^{n+1} -1 = x\sum_{j=0}^{n} \binom{n+1}{j+1}x^j = 0.$$ Hence, $$z_i = x_i -1$$, $$1\leq i \leq n$$ are roots of $$\sum_{j=0}^{n} \binom{n+1}{j+1}z^j = 0,$$ and by Vieta's formula, $$\sum_{i=1}^n \frac{1}{z_i} = \frac{\sum_{i=1}^n z_1z_2\cdots z_{i-1}\widehat{z_i}z_{i+1}\cdots z_n}{z_1z_2\cdots z_{n-1}z_n} = -\frac{\binom{n+1}{2}}{\binom{n+1}{1}} = -\frac{n}{2},$$ where $$\widehat{z_i}$$ means the variable is omitted in the calculation.
First, the denominator is, I think, $$(-1)^nQ(1)$$. Here is a hint for the numerator: how can you write $$Q’(x)$$? | 2019-06-25T03:50:16 | {
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https://mathoverflow.net/questions/268372/prime-square-offsets-why-is-7-more-frequent-than-7/268400 | # Prime square offsets: Why is +7 more frequent than -7?
For a prime $p$, define $\delta(p)$ to be the smallest offset $d$ from which $p$ differs from a square: $p = r^2 \pm d$, for $d,r \in \mathbb{N}$. For example, \begin{eqnarray} \delta(151) & = & +7 \;:\; 151 = 12^2 + 7 \\ \delta(191) & = & -5 \;:\; 191 = 14^2 - 5 \\ \delta(2748971) & = & +7 \;:\; 2748971= 1658^2 + 7 \end{eqnarray} For a particular $\delta=d$ value, define $\Delta(n,d)$ to be the number of primes $p$ at most $n$ with $\delta(p) = +d$, minus the number with $\delta(p) = -d$. In other words, $\Delta$ records the cumulative prevalence of $+d$ offsets over $-d$. For example, $\Delta(139,5)=-2$ because there are two more $-5$'s than $+5$'s up to $n=139$: $$\delta(31)=-5 \;,\; \delta(41)=+5 \;,\; \delta(59)=-5\;,\; \delta(139) =-5 \;.$$ The figure below shows $\Delta(p,5)$ and $\Delta(p,7)$ out to the $200000$-th prime $2750159$. The offset $+7$ occurs $161$ times more than the offset $-7$, and the reverse occurs for $|\delta|=5$: $-5$ is more common than $+5$.
Q. Is there a simple explanation for the different behaviors of offsets $5$ and $7$?
Obviously the question can be generalized to explaining the growth for any $|\delta|$.
I previously asked a version of this question on MSE, using somewhat different notation conventions and with less focused questions.
• The curve you plotted looks like a parabola. I don't know whether it can be explained though. – Sylvain JULIEN Apr 27 '17 at 11:12
• The graph is a bit misleading, making the offsets seem small. In reality they are huge, once one keeps in mind that there are only about 1600 squares up to $2.7\times 10^6$ and and thus primes within $\pm 7$ (or $\pm 5$) of a square are only a smallish fraction of 3200 in number. I would be surprised if simple modular considerations don't explain the offsets. – Yaakov Baruch Apr 27 '17 at 11:34
• And yes, in that case each curve would be close to $C \sqrt{n}/\log{n}$, so quite similar to a half parabola (and possibly with equal but opposite $C$ for the 2 cases). – Yaakov Baruch Apr 27 '17 at 11:45
• This might have something to with with Chebyshev’s bias: note that $+7\equiv-5\equiv3\pmod4$. – Emil Jeřábek Apr 27 '17 at 12:04
• The question title and text does not agree. Edit one of them. – Stig Hemmer Apr 28 '17 at 7:21
Consider $n^2+7$ and $n^2-7$ modulo $3$. If these are to be prime they must be non-zero $\pmod 3$, and in the first case $n$ can be anything mod $3$, whereas in the second case $n$ must be $0 \pmod 3$. If you consider $n^2+5$ and $n^2-5$, you see that the pattern reverses. This is already a huge bias for one offset to be preferred over the other.
The Hardy-Littlewood conjectures make this precise. One expects that (for a number $k$ not minus a square) the number of primes of the form $n^2+k$ with $n\le N$ (say) is $$\sim \frac 12 \prod_{p\ge 3} \Big(1 -\frac{(\frac{-k}{p})}{p-1} \Big) \frac{N}{\log N},$$ where in the numerator of the product above is the Legendre symbol. The constants in front of the $N/\log N$ explain these biases.
You don't have to worry about a smaller offset than $\pm 5$ or $\pm 7$, since an application of the sieve shows that the numbers $n^2+a$ and $n^2+b$ (for fixed $a$, $b$) are both prime only $\le CN/(\log N)^2$ of the time for a constant $C$.
• Lucia of course knows this, but I thought it was worth mentioning that this product is well-approximated by the inverse special value $L(1,\chi)^{-1}$ of the $L$-function associated to the unique Dirichlet character $\chi$ mod $4k$ such that $\chi(p) = \left( \frac{-k}{p} \right)$. Known bounds on this special value thus gives bounds on how large and small this constant can be. – Will Sawin Apr 27 '17 at 17:45
Modulo 6 the squares are 0,1,4,3,4,1 and the squares+7 (or -5) can only be 1,2,5,4,5,2, of which 3/6 can at all be prime. The squares-7 (or +5) are 5,0,3,2,3,0 of which only 1/6 can be prime. Obviously this not a proof, but there is clearly no first order surprise in the observed offsets.
UPDATE I checked offsets modulo different numbers (such as (mod 2*3*5*7*11*13*17*...)) and I found that mod(3*7*11*19*23) we get 64638/25806 for +7/-7 and 21945/73899 for +5/-5, while primes 2, 5, 13, 17, 29 don't affect the ratio. I imagine that exploring more primes will further refine the ratios, with primes $\equiv 3$ (mod 4) affecting them and others not. WHAT IS GOING ON HERE?
• I see now, thanks to Lucia's answer, that the ratios between the asymptotics there for, say, $k=\pm7$ are affected only by primes $p$ such that $$\Big(\frac{-7}{p}\Big)\neq \Big(\frac{7}{p}\Big)$$ which are the primes $\equiv 3$ (mod 4). – Yaakov Baruch Apr 27 '17 at 16:57 | 2019-01-17T05:53:14 | {
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https://math.stackexchange.com/questions/494502/prove-that-supab-supa-supb-and-why-does-supab-exist | # Prove that $\sup(A+B) = \sup(A) + \sup(B)$ and why does $\sup(A+B)$ exist?
We want to show that $\sup(A)+\sup(B)$ is the least upper bound of the set $A + B$. First, we need to show that $\sup(A) + \sup(B)$ is an upper bound for the set $A + B$. Indeed, if $z\in A + B$, then there exists $a\in A$ and $b\in B$ such that $z = a + b$. But by definition of $\sup(A)$ and $\sup(B)$, $a \leq \sup(A)$ and $b \leq \sup(B)$, so $z = a + b \leq \sup(A) + \sup(B)$. So, $\sup(A) + \sup(B)$ is an upper bound for $A + B$.
We now wish to show that $\sup(A) + \sup(B)$ is the least upper bound for the set $A + B$. So, if $u$ is an upper bound for $A + B$, we need to show that $\sup(A)+\sup(B) \leq u$. We will use part (i): that is, we need to show that there exists $\varepsilon > 0$, $\sup(A) + \sup(B) < u + \varepsilon$.
To do this, note that since $\sup(A)$ is the least upper bound for $A$, $\sup(A) - \varepsilon/2$ is not an upper bound for A, so there exists an $a\in A$ so that $\sup(A) - \varepsilon/2 < a$. Similarly, there is a $b\in B$ so that $\sup(B) - \varepsilon/2 < b$.
Adding these two inequalities gives $$\sup(A) + \sup(B) - \varepsilon < a + b;$$ in other words $$\sup(A) + \sup(B) < a + b + \varepsilon.$$ But $u$ is an upper bound for the set $A + B$, so $a + b \leq u$, and hence we have $$\sup(A) + \sup(B) < u + \varepsilon.$$ Thus, by part (i), $\sup(A) + \sup(B) \leq u$, so $\sup(A) + \sup(B)$ is the least upper bound for $A + B$, as required.
Now how do I show that $\sup(A+B)$ exists?
• For some basic information about writing math at this site see e.g. here, here, here and here. – Cortizol Sep 15 '13 at 16:26
• – Martin Sleziak Mar 3 '14 at 15:31
## 2 Answers
Show $A+B$ is nonempty and bounded above.
• i)A + B is non-empty since a + b E A + B. ii) If A and B are bounded, then so is A + B because by assumption, there are positive numbers MsubA and MsubB with |a| <= MsubA and |b| <= MsubB for all aEA and bEB. Hence |a + b| <= |a| + |b| <= MsubA + MsubB for all a and b. Is that it? – user87274 Sep 15 '13 at 16:49
• @mespebjidom Yes. – Pedro Tamaroff Sep 15 '13 at 16:56
Assume $A,B$ are both nonempty. Otherwise you can run into some $\infty-\infty$ strangeness.
You have $a \le \sup A$ for all $a \in A$, and similarly $b \le \sup B$ for all $b \in B$. Hence we have $a+b \le \sup A + \sup B$ for all $a \in A, b \in B$. Hence we have $\sup (A+B) = \sup_{a \in A, b \in B} a+b \le \sup A + \sup B$.
Now let $\epsilon >0$, and choose $a \in A, b \in B$ such that $a > \sup A -\frac{\epsilon}{2}$ and $b > \sup B -\frac{\epsilon}{2}$. Then $\sup(A+B) \ge a+b > \sup A + \sup B - \epsilon$. Since $\epsilon>0$ was arbitrary, we are have the desired result.
• What's this for? I just wanted to know how I can show that sup(A + B) exists in simple terms. – user87274 Sep 15 '13 at 18:51
• Your question reads 'Prove $\sup A + \sup B = \sup (A+B)$. The above shows, in simple terms, that $a+b \le \sup A + \sup B$ for all $a \in A, b \in B$, from which it follows that $\sup (A+B) \le \sup A + \sup B$. I'm not sure what you mean by 'exists', as the $\sup$ always exists (it may be $\pm \infty$, by convention $-\infty$ if the set is empty). – copper.hat Sep 15 '13 at 18:58
• I already proved that sup(A+B) = sup(A) + sup(B). What I wanted to know was that if A is bounded and has an element a in A, and B is bounded and has an element b in B, then does this mean that sup(A+B) exits? As the above poster said, for sup(A+B) to exist we have to show that A + B is non-empty and bounded. That's all. – user87274 Sep 15 '13 at 19:30
• I'm missing something here. The set needs to be bounded above, not necessarily bounded. If a non-empty set is bounded above, then it has a finite $\sup$. The set $A+B$ is clearly non-empty if both $A,B$ are non-empty. And you have shown that the sum of the $\sup$s is equal to the $\sup$ of the sum, which is presumed finite. – copper.hat Sep 15 '13 at 19:42
• @Evinda: Actually, I wrote my answers :-). One definition of $\sup A$ is that it is an upper bound and for all $\epsilon>0$, there is some $a \in A$ such that $a > \sup A- \epsilon$. What definition are you using? – copper.hat Oct 5 '18 at 21:38 | 2019-08-21T01:44:42 | {
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https://mathhelpboards.com/threads/the-strange-behaviour-of-numbers-close-to-unity.9132/ | # The Strange Behaviour of Numbers Close to Unity
#### Perplexed
##### New member
I have been looking at material properties such as thermal expansion of metals which usually involves very small coefficients. The general equation of thermal expansion is usually
$$\displaystyle L_\theta = L_0 ( 1 + \alpha \theta)$$
where L is the length and theta is the temperature change. The coefficient alpha is usually pretty small, 11E-6 for steel, so one ends up with a lot of numbers like 1.000011.
This is where I seem to have entered a strange world, where
$$\displaystyle \sqrt{(1 + x)} \rightarrow 1 + x/2$$
$$\displaystyle \dfrac{1}{ \sqrt{(1 - x)}} \rightarrow 1 + x/2$$
$$\displaystyle (1 - x)^3 \rightarrow 1-3x$$
Is there a name for this area of maths, and somewhere I can look up more about it?
Thanks for any help.
Perplexed
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
Taylor series is one topic where (Infinite) polynomials are used to approximating functions. For example,
$(1+x)^{1/2}=1+\frac{x}{2}+R_1(x)$
where $R_1(x)$ is called the remainder and is infinitely small compared to $x$ when $x$ is small. More precisely,
$(1+x)^{\alpha }=1+\alpha x+{\frac {\alpha (\alpha -1)}{2!}}x^{2}+\cdots+ \frac{\alpha\cdot\ldots\cdot(\alpha-n+1)}{n!}x^n+R_n(x)$
where $R_n(x)$ is infinitely small compared to $x^n$ when $x$ tends to $0$.
#### HallsofIvy
##### Well-known member
MHB Math Helper
"Linear approximation". Any function, f, having a derivative at x= a, can be approximated by the "tangent line" $$y= f'(a)(x- a)+ f(a)$$. The error will be proportional to $$(x- a)^2$$ and f''(a).
For example, if $$f(x)= \sqrt{1+ x}= (1+ x)^{1/2}$$ then $$f'(x)= (1/2)(1+ x)^{-1/2}$$ so that with x= 0, $$f(0)= \sqrt{1+ 0}= 1$$ and $$f'(0)= (1/2)/\sqrt{1+ 0}= 1/2$$. So y= f(x) is approximated, around x= 0, by $$y= (1/2)x+ 1$$ or $$1+ x/2$$.
If $$f(x)= \frac{1}{\sqrt{1+ x}}= (1+ x)^{-1/2}$$ then $$f'(x)= -(1/2)(1+ x)^{-3/2}$$ so that $$f(0)= \frac{1}{\sqrt{1+ 0}}= 1$$ and then $$f'(0)= -(1/2)(1+ 0)^{3/2}= -1/2. So y= f(x) is approximated, around x= 0, by [tex]y= -(1/2)x+1$$ or $$1- x/2$$. Notice the negative sign- what you have is NOT correct.
If $$f(x)= (1- x)^3$$ then $$f'(x)= 3(1- x)^2(-1)= -3(1- x)^2$$. $$f(0)= (1- 0)^3= 1$$ and $$f'(0)= -3(1- 0)^2= -3$$. So y= f(x) is approximated by -3x+ 1 or 1- 3x.
You could also do the last one by actually multiplying it out: $$(1- x)^3= 1- 3x+ 3x^2- x^3$$. If x is small enough (i.e. close enough to 0) that higher values of x can be ignored in the approximation, y= 1- 3x.
Again, these are all first order or linear approximations to the functions, not exact values.
(You can get the Taylor's polynomial and series that Evgeny- Makarov refers to by extending those same ideas to higher powers.)
Last edited:
#### Perplexed
##### New member
If $$f(x)= \frac{1}{\sqrt{1+ x}}= (1+ x)^{-1/2}$$ then $$f'(x)= -(1/2)(1+ x)^{-3/2}$$
so that $$f(0)= \frac{1}{\sqrt{1+ 0}}= 1$$ and then $$f'(0)= -(1/2)(1+ 0)^{3/2}= -1/2$$.
So y= f(x) is approximated, around x= 0, by $$y= -(1/2)x+1$$ or $$1- x/2$$. Notice the negative sign- what you have is NOT correct.
Just to clear things up so that someone else looking at this doesn't get confused, in my second approximation I had $$f(x)= \frac{1}{\sqrt{1 - x}}$$ rather than the $$f(x)= \frac{1}{\sqrt{1+ x}}$$ that you started with: notice the "-" rather than "+" in the square root. It was the simple change of sign in arriving at the reciprocal that first intrigued me on this one, and your explanation makes the reason why this works clear.
Less Perplexed now
#### Deveno
##### Well-known member
MHB Math Scholar
Allow me to make another observation regarding this:
Scientific measurements are often given in "significant figures", the reasoning being, we can only take measurements up to a certain degree of accuracy.
So, suppose our input data can only give 6 decimal places.
If we expect we can model a function (and for many functions this is true) by:
$f(x) = a_0 + a_1x + a_2x^2 +\cdots$
And that the coefficients $a_k$ either stay "about the same size" or even better, decrease, then if we measure $x$ to 6 decimals places, the "correction term" for $x^2$ is around 12 decimal places, in other words, much much smaller than our standards of accuracy allow.
For certain classes of "well-behaved" functions, there are means to estimate (or "bound") the size of the error, which in turn lets us know "how many terms to go out".
For small enough $x$, this kind of reasoning lets us use the approximation:
$\sin(x) = x$
often used in simplifying wave equations that govern oscillators, and if more accuracy is needed, the approximation:
$\sin(x) = x - \dfrac{x^3}{6}$ is pretty darn good, as you can see here. | 2020-11-30T23:00:33 | {
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http://mathhelpforum.com/algebra/218804-sanity-check-line-intersecting-plane.html | Math Help - Sanity check on line intersecting plane
1. Sanity check on line intersecting plane
Line l is r=-8i+5j-6k+t(5i-2k)
Plane is -x-2y+5z = 0
What is the point of intersection? I get (-56/3, 5, -26/15) but my student's school teacher has a different answer of (-23,5,0)
Who is right?
2. Re: Sanity check on line intersecting plane
The line is given by x= -8+5t, y= 5, z= -6- 2t.
Putting those into the equation of the plane, -(-8+ 5t)- 2(5)+ 5(-6- 2t)= 8- 5t- 10- 30- 10t= -15t- 32= 0 so that t= -32/15 and then
x= -8+ 5(-32/15)= -8- 32/3= (-24- 32)/3= -56/3.
y= 5
z= -6- 2(-32/15)= (-90+ 64)/15= -26/15
3. Re: Sanity check on line intersecting plane
\displaystyle \begin{align*} \begin{cases} x &= -8 + 5t \\ y &= \phantom{-} 5 \\ z &= -6 -2t \end{cases} \end{align*}
Substituting into the plane we find
\displaystyle \begin{align*} -x - 2y + 5z &= 0 \\ - \left( -8 + 5t \right) - 2 \left( 5 \right) + 5 \left( -6 - 2t \right) &= 0 \\ 8 - 5t - 10 - 30 - 10t &= 0 \\ -32 - 15t &= 0 \\ -15t &= 32 \\ t &= -\frac{32}{15} \end{align*}
So that gives the point of intersection as
\displaystyle \begin{align*} \begin{cases} x &= -8 + 5 \left( -\frac{32}{15} \right) \\ y &= \phantom{-} 5 \\ z &= -6 - 2 \left( -\frac{32}{15} \right) \end{cases} \\ \begin{cases} x &= -\frac{24}{3} - \frac{32}{3} \\ y &= \phantom{-} 5 \\ z &= -\frac{90}{15} + \frac{64}{15} \end{cases} \\ \begin{cases} x &= -\frac{ 56}{3} \\ y &= \phantom{-} 5 \\ z &= -\frac{26}{15} \end{cases} \end{align*}
So the point of intersection is \displaystyle \begin{align*} \left( x, y, z \right) = \left( -\frac{56}{3} , 5, -\frac{26}{15} \right) \end{align*}. I agree with your answer
4. Re: Sanity check on line intersecting plane
Originally Posted by grillage
Line l is r=-8i+5j-6k+t(5i-2k)
Plane is -x-2y+5z = 0
What is the point of intersection? I get (-56/3, 5, -26/15) but my student's school teacher has a different answer of (-23,5,0)
Who is right?
Who is correct depends on who solves $(-8+5t)+2(5)-5(-6-2t)=0$ correctly.
Have you checked to see if either of those points is on that line? | 2014-09-24T00:32:19 | {
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https://physics.stackexchange.com/questions/254383/two-mass-one-spring-system-natural-frequency | Two mass one-spring system natural frequency
Suppose I have two masses $m_1, \ m_2$ connected by one spring of stiffness $k$ through their centres of mass, lying on a frictionless surface and the system is set into oscillation. I want to find the natural frequencies of oscillation. Finding the eigenvalues, I get one natural frequency is $$\omega_n = \sqrt{\frac{k(m_1+m_2)}{m_1m_2}}$$ But I thought that a system with two degrees of freedom should have two natural frequencies? When calculating the eigenvalues, the equation reduced to $$m_1m_2\omega^4 = \omega^2 k(m_1+m_2)$$ Does this mean the other natural frequency is zero?
(Apologies as I know there are quite a lot of questions on PSE concerning this system but I couldn't find one that answered my question and I don't have enough points to comment yet.)
• what do you think it would mean to have a mode with zero frequency? can you imagine the motion of the system? does this motion make sense? May 6 '16 at 16:50
• If it has zero frequency then it just doesn't oscillate? But then I don't know how to calculate the second frequency. It is a two degree of freedom system, right (need two coordinates to describe the position of each mass)? May 6 '16 at 16:57
• yes, there must be two solutions. But forget about the math for a minute: what would $\omega=0$ correspond to, physically? can you picture that motion in your head? May 6 '16 at 17:04
• @JohnRennie thank you :P I was trying to make OP reach that conclusion. If we don't impose fixed boundary conditions, spring systems always have the "rigid-motion" solution, where the whole system translates without oscillations. May 6 '16 at 17:24
• Ach, got it, thank you. ;) This rigid body response also explains why it's a two degree of freedom system then? (Because for the in-out type motion where they're out of phase and oscillating just the distance between the particles would be sufficient to describe the motion of the system?) May 7 '16 at 0:56
Suppose I have two masses m1, m2 connected by one spring of stiffness k
The Lagrangian of the system is
$$L = \frac{1}{2}m_1\dot q_1^2 + \frac{1}{2}m_2\dot q_2^2 - \frac{1}{2}k(q_1 - q_2)^2$$
where $q_1$ and $q_2$ are the coordinates of $m_1$ and $m_2$ respectively.
Now, consider a change of coordinates to the normal coordinates $Q_1$ and $Q_2$ where
$$Q_1 = \frac{q_1m_1 + q_2m_2}{m_1 + m_2},\qquad Q_2 = q_2 - q_1$$
are the coordinates of the center of mass $M = m_1 + m_2$ and the reduced mass $\mu = \frac{m_1m_2}{m_1 + m_2}$ respectively.
In these coordinates, the Lagrangian is
$$L = \frac{1}{2}M\dot Q_1^2 + \frac{1}{2}\mu\dot Q_2^2 - \frac{1}{2}kQ_2^2$$
and now it's easy to see that the uncoupled equations of motion (via the Euler-Lagrange equation) are
$$\ddot Q_1 = 0, \qquad \ddot Q_2 = -\frac{k}{\mu}Q_2$$
And so, the center of mass coordinate has 'zero frequency oscillation', i.e., uniform translational motion, while the reduced mass coordinate oscillates with angular frequency $\omega_2 = \sqrt{\frac{k}{\mu}}$
The other natural frequency is indeed zero! Natural frequencies of zero corresponds to vibrational modes of rigid body motion. Rigid body motion is not a vibrational motion in itself, but still arises in the modal analysis of certain systems such as the one above. The reason you get a rigid body mode is because you are able to move the system as a whole, and no restoring force exists to bring the whole system back to its original location.
This system is equivalent to hooking up one of the masses to a fixed point using a spring of zero stiffness. Let's consider your system with mass $m_1$ connected to a fixed point with a spring of stiffness $K$. Upon performing modal analysis, the two natural frequencies of such a system are given by:
$$\omega=\sqrt{\frac{m_1 + m_2}{2m_1 m_2}k + \frac{K}{2m_1}\pm \sqrt{\left[\frac{m_1 + m_2}{2m_1 m_2}k + \frac{K}{2m_1}\right]^2 - \frac{Kk}{m_1 m_2}}}$$
Now, to reobtain your system, set $K=0$, and the two frequencies indeed become $0$ and $\sqrt{\frac{m_1+m_2}{m_1 m_2}k}$. So the lowest natural frequency of your system is indeed zero, but what is the physical significance of a natural frequency of zero? One way to interpret it is a vibration of infinite time period, $T=2\pi/\omega$. When the bodies move, it takes so long for them to finish their first oscillation that it never completes, i.e. the body continues moving on at constant speed. This corresponds to rigid body motion under no external forces (modal analysis for determining natural frequencies sets externally applied forces to zero anyways.)
P.S. Also, note that the zero frequency result does appear alongside the non-zero result when you first calculate them:
By taking the determinant of the relevant matrix, the following equation is obtained:
$$\omega^4 - \frac{m_1 + m_2}{m_1 m_2}k\omega^2=0$$
Note that this can be rewritten as:
$$\omega^2 \left(\omega^2 - \frac{m_1 + m_2}{m_1 m_2}k\right) = 0$$
So it can be seen that $\omega^2 = 0$ if one of the two solutions. It is quite easy to miss this, as it can be prematurely cancelled out of the equation, which can only be done if $\omega^2 = 0$ is an impossible solution, which it is not.
You can take the two equations of motion
\begin{align} k (x_2 - x_1) = m_1 \ddot{x}_1 \\ k (x_1-x_2) = m_2 \ddot{x}_2 \end{align}
and transform them using their centroid location, and distance
\left. \begin{align} x_c & = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}\\ x_d & = x_2-x_1 \end{align} \right\} \begin{aligned} x_1 &= x_c - \frac{m_2 x_d}{m_1 +m_2} \\ x_2 & = x_c + \frac{m_1 x_d}{m_1+m_2} \end{aligned}
(and similarly for the accelerations). The equations of motion are transformed into
\begin{align} \ddot{x}_c &= 0 \\ \ddot{x}_d &= -k \left( \frac{m_1+m_2}{m_1 m_2} \right) x_d \end{align}
As you can see the center of mass $x_c$ does not have any acceleration (Newton's first law), and only the separation oscillates with the reduced mass $m_{eff} = \frac{m_1 m_2}{m_1+m_2}$
The eigen-values of the system are trivially found as
\begin{align} \omega_c^2 &= 0 \\ \omega_d^2 & = k \frac{m_1+m_2}{m_1 m_2} \end{align}
There is only one degree of freedom (the distance between the masses) not two, and therefore only one natural frequency (given by your 1st eqn). Since there are no external forces, the CM of the system does not move. (Sorry but I do not see what all the fuss is about here. Why make such a simple problem so complicated?)
• If you just push one end and the whole system translates then the distance between the masses is insufficient to describe its motion? May 7 '16 at 0:58
• @AtticusFinch95 : True but your question asked only about the natural frequencies of oscillation and only says that the masses are "set into oscillation". Any constant velocity of the CM has no effect on natural frequency and is unknown, unknowable and irrelevant. The question does not mention any continuing external force on either mass (after they are somehow "set into oscillation") so we cannot assume there are any. It is unnecessary and irrelevant to introduce any additional degrees of freedom into the problem. May 7 '16 at 11:48
• There is definitely two degrees is freedom, the centre of mass and the relative coordinates. Thus there are two normal modes but one of them is a zero mode, which has null frequency and corresponds to the translation of the centre of mass. Sep 9 '17 at 15:11 | 2021-10-26T08:18:17 | {
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https://math.stackexchange.com/questions/155839/on-ces%C3%A0ro-convergence-if-x-n-to-x-then-z-n-fracx-1-dots-x-nn | # On Cesàro convergence: If $x_n \to x$ then $z_n = \frac{x_1 + \dots +x_n}{n} \to x$
I have this problem I'm working on. Hints are much appreciated (I don't want complete proof):
In a normed vector space, if $x_n \longrightarrow x$ then $z_n = \frac{x_1 + \dots +x_n}{n} \longrightarrow x$
I've been trying adding and subtracting inside the norm... but I don't seem to get anywhere.
Thanks!
Given $\epsilon >0$ there exists $n_0$ such that if $n\geq n_0$ then $\parallel x_n -x\parallel < \epsilon$
so
\begin{align*} 0 & \leq \left\lVert \frac{x_1 +\cdots +x_n}{n} -x \right\rVert \leq \left\lVert \frac{x_1 + \dots + x_n - nx }{n} \right\rVert \\ & \leq \frac{\lVert x_1 - x \rVert}{n} + \dots + \frac{\lVert x_{n_0 - 1} - x \rVert}{n} + \frac{\lVert x_{n_0} - x \rVert}{n} +\dots + \frac{\lVert x_{n} - x \rVert}{n} \\ &\le \frac 1n\sum_{i=1}^{n_0-1} \| x_i -x\| + \frac{n-n_0}{n} \epsilon \end{align*}
The first $n_0 -1$ terms $\| x_i -x\|$ can be bounded by some $M$, thus for $n\ge (n_0-1)M/\epsilon=: N_0$ we have $$\frac 1n\sum_{i=1}^{n_0-1} \| x_n -x\| \le \frac 1n (n_0-1)M \le \epsilon$$
Thus $$\left\| \frac{x_1 + \cdots x_n}{n} - x\right\| <2\epsilon$$ when $n\ge N_0$.
Thanks a lot @Leonid Kovalev for the inspiration, though my main problem was that I wasn't aware of what to do with the $nx$ (the silliest part :P)
• This sort of thing is encouraged, I think. – Dylan Moreland Jun 11 '12 at 3:21
• I think you want "$-nx$" in the first line of the display, and $\|x_n - x\|$ at the end. I don't think you want to say that the first few terms are $\leq M$; that doesn't seem to be enough. – Dylan Moreland Jun 11 '12 at 3:41
• @DylanMoreland: Why isn't it enough? Can you explain? – Bouvet Island Jun 11 '12 at 14:56
• @Inti: I think you are missing something in the argument. Generally the argument consists of two steps: first choose $n_0$ such that if $n \geq n_0$, $\|x_n -x \| < \epsilon / 2$. Next choose $N \geq n_0$ such that $M$ (which is at least $\sup_n \|x_n\|$) satisfies $M / N < \epsilon / 2$. I don't see the second part of the argument implemented in your answer. – Willie Wong Jun 11 '12 at 14:59
There is a slightly more general claim:
PROP Let $\langle a_n\rangle$ be a sequence of real numbers, and define $\langle \sigma_n\rangle$ by $$\sigma_n=\frac 1 n\sum_{k=1}^n a_k$$
Then $$\liminf_{n\to\infty}a_n\leq \liminf_{n\to\infty}\sigma_n \;(\;\leq\;)\;\limsup_{n\to\infty}\sigma_n\leq \limsup_{n\to\infty}a_n$$
P We prove the leftmost inequality. Let $\ell =\liminf_{n\to\infty}a_n$, and choose $\alpha <\ell$. By definition, there exists $N$ such that $$\alpha <a_{N+k}$$ for any $k=0,1,2,\ldots$ If $m>0$, then $$m\alpha <\sum_{k=1}^m \alpha_{N+k}$$
which is $$m\alpha<\sum_{k=N+1}^{N+m}a_k$$
$$(m+N)\alpha+\sum_{k=1}^{N}a_k<\sum_{k=1}^{N+m}a_k+N\alpha$$
which gives
$$\alpha+\frac{1}{m+N}\sum_{k=1}^{N}a_k<\frac{1}{m+N}\sum_{k=1}^{N+m}a_k+\frac{N}{m+N}\alpha$$
Since $N$ is fixed, taking $\liminf\limits_{m\to\infty}$ gives $$\alpha \leq \liminf\limits_{m \to \infty } \frac{1}{m}\sum\limits_{k = 1}^m {{a_k}}$$ (note that $N+m$ is just a shift, which doesn't alter the value of the $\liminf^{(1)}$). Thus, for each $\alpha <\ell$, $$\alpha \leq \liminf\limits_{m \to \infty } \frac{1}{m}\sum\limits_{k = 1}^m {{a_k}}$$ which means that $$\liminf_{n\to\infty}a_n\leq \liminf_{n\to\infty}\sigma_n$$ The rightmost inequality is proven in a completely analogous manner. $\blacktriangle$.
$(1)$: Note however, this is not true for "non shift" subsequences, for example $$\limsup_{n\to\infty}(-1)^n=1$$ but $$\limsup_{n\to\infty}(-1)^{2n+1}=-1$$
COR If $\lim a_n$ exists and equals $\ell$, so does $\lim \sigma_n$, and it also equals $\ell$. The converse is not true.
WLOG, the $x_n$ converge to $0$ (otherwise consider the differences $x_n-x$), and stay confined in an $\epsilon$-neighborhood of $0$ after $N_\epsilon$ terms.
Then the average of the first $m$ terms is bounded by
$$\frac{N\overline{x_N}+(m-N)\epsilon}m,$$ which converges to $\epsilon$. So you can make the average as close to $0$ as you like. | 2019-09-19T19:11:18 | {
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https://math.stackexchange.com/questions/2563455/determinant-of-triangular-matrix | # Determinant of Triangular Matrix
I understand that you can find the determinant of a matrix along it's diagonal if it is in triangular form. For a matrix such as this: $$\begin{pmatrix} 1 & 5 & 0\\ 2 & 4 & -1\\ 0 &-2 & 0 \end{pmatrix}$$ When put into triangular form I get: $$\begin{pmatrix} 1 & 5 & 0\\ 0 & 1 & 1/6\\ 0 & 0 & 1/3 \end{pmatrix}$$ Since I multiplied row two by -1/6 during the row reduction I would expect the determinant to be $$1\cdot 1\cdot 1/3\cdot (-1/6),$$ but the answer for the determinant of the original matrix is -2. Where exactly am I going wrong?
• Write your row operations as a matrix product and use the multiplication rule for determinants. Then you will minimize your risk of mistakes. – mathreadler Dec 12 '17 at 16:30
• You should multiply by $-6$ rather than $-1/6$ – eepperly16 Dec 12 '17 at 16:31
• @mathreadler I understand the multiplication rule, but was just wondering why this scenario did not work when trying to solve using a triangular matrix – Pulse Dec 12 '17 at 16:32
• it will work if you do it right. there is a larger chance you will do it right if you write it down step by step with matrix products – mathreadler Dec 12 '17 at 16:33
• @eepperly16 I multiplied by -1/6 because during the row operation, row 2 I multiplied by -1/6 to make the 6 a 1 in that row. – Pulse Dec 12 '17 at 16:33
You can add a multiple of any row to another row without changing the determinant. So I'll do these kinds of operations to get to the triangular form.
$$\begin{matrix} 1 & 5 & 0 \\ 2 & 4 & -1 \\ 0 &-2 &0 \end{matrix}$$
Subtract twice the first row from the second:
$$\begin{matrix} 1 & 5 & 0 \\ 0 & -6 & -1 \\ 0 &-2 &0 \end{matrix}$$
Subtract one-third the new second row from the third:
$$\begin{matrix} 1 & 5 & 0 \\ 0 & -6 & -1 \\ 0 & 0 & \frac{1}{3} \end{matrix}$$
The product along the diagonal is $-2$.
Pulling out a factor from one of the rows, however, does change the determinant by that factor. You pulled out $-1/6$ from the second row, so the calculated determinant from the diagonal in your question statement was low by this factor.
• This really does not help with the understanding as you have not shown that the operations you perform don't change the determinant. – mathreadler Dec 12 '17 at 16:38
• The matrix you ended up with is the same as mine, except your second row is multiplied by $-1/6$. "Pulling out" that constant changes the determinant by the same factor: $-2 \times -1/6 = 1/3$, which was what you got. – John Dec 12 '17 at 16:46 | 2019-08-22T07:58:10 | {
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https://math.stackexchange.com/questions/4254922/why-does-this-assumption-give-an-under-approximation-for-both-the-expected-maxim | # Why does this assumption give an under approximation for both the expected maximum and minimum
The well known result, $$\mathbb{E}[\text{min} \{X_i \}_{i=1}^n ] = \frac{1}{n+1}$$ and $$\mathbb{E}[\text{max} \{X_i \}_{i=1}^n ] = \frac{n}{n+1}$$ where $$X_i$$ are I.I.D Uniform$$(0,1)$$ random variables is a lifesaver. It can be extrapolated to find the expected minimum and maximum of $$n$$ Uniform$$(a,b)$$ variables.
I tried to use it in the discrete case. For example, consider the maximum of $$10$$ rolls of a $$100$$ sided die, the quick formula gives: $$\frac{10}{11}\cdot 100 = 90\frac{1}{11} \approx 90.91$$. The true answer however is $$91.4007585757$$ (using tail sum formula : $$100 - \sum\limits_{i=1}^{99}(\frac{i}{100})^{10}$$ )
Notice the approximation gives an underestimate.
Now consider using it to approximate the expected minimum.
$$\frac{1}{11}\cdot 100 = 9\frac{1}{11} \approx 9.09$$. The true answer however is $$9.59924142434$$ and again this formula under-approximates?
Why does this happen in both cases? Is there a cheeky way to tweek it for the discrete case to get a bit more accurate?
Thanks!
A little thing I have noticed that might help is the following:
The approximate min + approximate max is always $$n$$. However, by symmetry the true min + true max is always $$n+1$$. (because they are centred around the mean of $$\frac{n+1}{2}$$
• Decent question, a very underappreciated consequence of the way you approximate continuous by discrete. Technically speaking, the second part hasn't been answered yet, so I'm looking forward to someone addressing that as well. Sep 20 at 6:05
Note that for $$X$$ positive we have $$\mathbb EX=\int_0^\infty 1-F(x)\,\mathrm d x$$. Now note that if $$X$$ is the discrete variant and $$Y$$ is the continuous approximation. Then $$F_X\leq F_Y$$ (as $$F_X(x)=F_Y(x)$$ for $$1\leq x\leq 100$$ and $$F_Y$$ is linear while $$F_X$$ is constant on $$[x,x+1)$$.).
Thus this also holds true for the distribution of max and min, and thus by above formula the expected values for $$Y$$ are smaller than those for $$X$$.
It under-approximates because you are approximating a random variable taking values in $$\left\{\frac{1}{k},\frac{2}{k},\dots,\frac{k-1}{k}, 1\right\}$$ by a uniform random variable, that is, uniform on $$\left(0,\frac{1}{k}\right] \cup \left(\frac{1}{k}, \frac{2}{k}\right]\cup \dots\cup \left(\frac{k-1}{k}, 1\right]\,.$$ Viewed differently, you replace the value $$\frac{i}{k}$$ (for $$1\leq i\leq k$$) by a uniformly random value in $$\left(\frac{i-1}{k}, \frac{i}{k}\right]$$ which leads to a (small) under-approximation.
• As a corollary, a (first-order, not perfect, but closer) fix is to add $\frac{1}{2k}$ to the result, to make up for the shift and "re-center" your approximation. Since you multiply by $k$ in the end, that means adding $0.5$ to your estimates. In your example, that'll give 91.41 and 9.59. Sep 19 at 21:20
• I have noticed a typo in your answer. I think you mean: $\left(0,\frac{1}{k}\right] \cup \left(\frac{1}{k}, \frac{2}{k}\right]\cup \dots\cup \left(\frac{k-1}{k}, 1\right]\,.$ @Clement C. Sep 20 at 8:40 | 2021-10-20T04:13:10 | {
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https://math.stackexchange.com/questions/2934279/prove-that-lfloor-x-rfloor-geq-y-if-and-only-if-x-geq-lceil-y-rceil | # Prove that $\lfloor x\rfloor \geq y$ if, and only if, $x\geq\lceil y\rceil$
I have some trouble proving that if $$x,y\in\mathbb{R}$$ then $$\lfloor x\rfloor \geq y$$ if, and only if, $$x\geq\lceil y\rceil$$.
I have tried some different approaches, the most recent being a proof by contradiction: Assume (for contradiction) that $$x<\lceil y\rceil$$, then $$\lfloor x\rfloor < y +1$$. However, I can not (of course) get rid of the $$+1$$ term because in general $$y \leq\lceil y\rceil$$.
One can perhaps prove that both the floor- and ceiling function preserve inequalities? If so, the result become trivial.
Any ideas are highly appreciated.
Note that $$\lfloor x \rfloor \ge y$$ implies $$\lfloor x \rfloor \ge \lceil y\rceil$$, so $$x\ge \lceil y\rceil$$.
The other direction is similar.
If the above is not clear, recall that $$\lceil y \rceil$$ is defined as the least integer not smaller than $$y$$, i.e. if $$C(y)$$ is the set $$C(y):= \{n\in\mathbb Z:n\ge y\}$$, then $$\lceil y\rceil = \inf C(y).$$ The assumption $$\lfloor x\rfloor \ge y$$ is exactly that $$\lfloor x\rfloor \in C(y)$$. The only thing left is to use the defining property of an infimum, $$c \in C(y) \implies \inf C(y) \le c.$$
• Thanks, but I can not see why the implication is true in general. Sep 28, 2018 at 12:02
• @Wuberdall This is because $\lfloor x \rfloor \in \mathbb Z$. For example, $5\ge y$ implies $5 \ge \lceil y\rceil$. This holds for every integer in place of 5. In words - the ceiling of $y$ is the least integer larger than $y$, so any integer larger than $y$ is larger than the ceiling. Sep 28, 2018 at 12:03
• (You should replace "larger than" with "not smaller than" but that's harder to say) Sep 28, 2018 at 12:10
• Yes, I see it know. This is of course true due to the result "number not greater than integer iff ceiling is not greater than integer" exactly as you explained. – Thank you for your help. Sep 28, 2018 at 12:12
Hint: $$\lfloor x\rfloor\geq y\iff \lfloor x\rfloor\geq\lceil y \rceil\iff x\geq \lceil y\rceil.$$
• Thanks, but why is the first bi-implication true? Sep 28, 2018 at 12:01
• @Wuberdall Because $\lfloor x\rfloor$ is an integer that is greater than or equal to $y$, so $\lfloor x\rfloor\geq\lceil y\rceil$. For the reverse direction, just note $\lceil y\rceil \geq y$. Sep 28, 2018 at 12:04
We need to prove both direction, that is
• $$\lfloor x\rfloor \geq y \implies x\geq\lceil y\rceil$$
and
• $$x\geq\lceil y\rceil \implies \lfloor x\rfloor \geq y$$
For the first one we have that
$$\lfloor x\rfloor \geq y \implies \lfloor x\rfloor \geq \lceil y\rceil$$
and therefore
$$x\ge \lfloor x\rfloor \geq \lceil y\rceil \implies x\geq\lceil y\rceil$$
For the second one we have that
$$x\geq\lceil y\rceil \implies \lfloor x\rfloor \geq \lceil y\rceil$$
and therefore
$$\lfloor x\rfloor \geq \lceil y\rceil \ge y \implies \lfloor x\rfloor \ge y$$
• Thank you, it is completely clear why it is true now. Sep 28, 2018 at 12:20
• @Wuberdall You are welcome! Bye
– user
Sep 28, 2018 at 12:26
We need the following elementary properties of the floor and ceiling functions: $$a\ge\lfloor a\rfloor$$, $$\lceil\lfloor a\rfloor\rceil=\lfloor a\rfloor$$, $$a\ge b\implies\lceil a\rceil\ge\lceil b\rceil$$, and $$\lceil a\rceil=-\lfloor-a\rfloor$$.
From the first three elementary propeties, we have, for any real $$u$$ and $$v$$,
$$\lfloor u\rfloor\ge v \implies u\ge\lfloor u\rfloor=\lceil\lfloor u\rfloor\rceil\ge\lceil v\rceil$$
so letting $$u=x$$ and $$v=y$$ gives
$$\lfloor x\rfloor\ge y\implies x\ge\lceil y\rceil\qquad(*)$$
while letting $$u=-y$$ and $$v=-x$$ gives
$$\lfloor-y\rfloor\ge-x\implies-y\ge\lceil-x\rceil$$
Invoking now the fourth elementary property turns that last implication into
$$-\lceil y\rceil\ge-x\implies-y\ge-\lfloor x\rfloor$$
which can be re-expressed as
$$x\ge\lceil y\rceil\implies\lfloor x\rfloor\ge y\qquad(**)$$
Putting $$(*)$$ and $$(**)$$ together, we have
$$\lfloor x\rfloor\ge y\iff x\ge\lceil y\rceil$$ | 2022-07-06T23:18:37 | {
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http://tex.stackexchange.com/questions/77589/what-do-the-pieces-of-latex-left-and-right-respectively-mean/77592 | # What do the pieces of LaTeX, \left and \right, respectively mean?
What are their uses and are they even standard?
-
\left and \right are used for delimiters when they have to change the size dynamically depending on the content. Consider the following example:
\documentclass{article}
\usepackage{amsmath}
\begin{document}
Compare this:
\begin{align*}
\left(\sqrt{2}+\sqrt{3}\right)^2
&=\left(\sqrt{2}\right)^2+2\times\sqrt{2}\times\sqrt{
3}+\left(\sqrt{3}\right)^2\\
&=2+2\sqrt{6}+3\\
&=5+2\sqrt{6}
\end{align*}
with:
\begin{align*}
(\sqrt{2}+\sqrt{3})^2
&=(\sqrt{2})^2+2\times\sqrt{2}\times\sqrt{
3}+(\sqrt{3})^2\\
&=2+2\sqrt{6}+3\\
&=5+2\sqrt{6}
\end{align*}
First one uses \verb|\left(| and \verb|\right)| and second one uses just \verb|(| and \verb|)|. I hope the difference is clear.
Just another example:
$\left(\frac{1}{2}\right) \qquad (\frac{1}{2})$
\end{document}
In this particular case, the (\sqrt{2}+\sqrt{3}) with \left and \right gives bit bigger parenthesis making it to look ugly (to some extent) as noted by Enrico. In such cases, proper variant of delimiters (\bigl and \bigr in this case) may be used to get the appropriate height.
For more details, refer to amsmath documentation - page 15, section 4.14 (texdoc amsldoc from command prompt). Here is a screen shot of the same:
-
+1 but I think we don't need a blank line between "Just another..." and $.... – Please don't touch Oct 14 '12 at 2:15 @ガベージコレクタ: OK As you wish. I am updating :) – Harish Kumar Oct 14 '12 at 2:39 Thanks for the update. – Please don't touch Oct 14 '12 at 2:46 +1; you could point out that in your example with (\sqrt{2}+\sqrt{3}) the result with \left and \right is less good than without it (not to say wrong). – egreg Oct 14 '12 at 9:08 @egreg: Yes and added that here also. Thank you :) – Harish Kumar Oct 14 '12 at 9:24 \left and \right are delimiter counterparts and TeX primitives, to be used with so-called "extensible delimiters" in math mode within the same group. This implies two things: 1. If you use one, you have to use the other as well; and 2. They have to be opened/closed at and within the same group level depth. Common or typical uses include • \left\{ ... \right\} • \left[ ... \right] • \left( ... \right) although there is no requirement for the delimiters to be matching. For example, it is fine to use \left\}...\right(, just as long as you use both \left and \right. For example (to mimic amsmath's bmatrix environment), \documentclass{article} \usepackage{amsmath}% http://ctan.org/pkg/amsmath \begin{document} \[ \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \qquad \left[\begin{array}{@{}ccc@{}} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array}\right]$
\end{document}
If you want to use only one of the two, you still need to include the counterpart, but this time with the argument .. That is, either \left. or \right., referred to as the null delimiter. TeX inserts a \nulldelimiterspace for these.
Extensible delimiters are ones that are comprised of a number of fixed and some variable-length parts. For some eye-candy, here's a visual of an extensible }:
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https://math.stackexchange.com/questions/1790128/method-to-show-that-a-limit-exisits-and-evaluate-the-limit-of-a-function-as-x-te | # Method to show that a limit exisits and evaluate the limit of a function as x tends to infinity.
Exam Question:
For each of the following functions f , determine whether $\lim_{x\to\infty}f(x)$ exists, and compute the limit if it exists. In each case, justify your answers.
a) $f(x)= \dfrac{x+2}{x^2+8}$
b) $f(x)= \dfrac{\cos(x)}{x^2}$
Attempt:
a) I think you divided by the leading power of $x$, and then use algebra of limits to show that the limit is equal to $0$.
b) I don't think this limit exits as $\cos(x)$ is a periodic function.
What is the correct method to answer this question?
• Finally someone who can format a question properly in HTML and not just put L X->infinity. – HELP May 18 '16 at 9:11
• wow, thank you. It's really not that hard though once you collect enough references, programing is nothing if not repeatable. – UniStuffz May 18 '16 at 9:12
• Note: for the second function: even the $\cos$ function is periodic, but you should keep in mind that it is bounded by $-1$ and $1$. Then dividing by $x^2$ .... – Nizar May 18 '16 at 9:14
• @Nizar ah "sandwich rule" am I right. – UniStuffz May 18 '16 at 9:23
You're right about (a). For rational functions, the limit (for $x \to \pm \infty$) will always be zero if the denominator has a higher degree than the numerator; in your case: $$\lim_{x \to +\infty} \frac{x+2}{x^2+8} = \lim_{x \to +\infty} \frac{1/x+2/x^2}{1+8/x^2} = \frac{0}{1} = 0$$ Addendum: and if it's the other way around (degree higher in the numerator), the limit will be $\pm \infty$ (check the sign). When the degree of numerator and denominator is the same, the limit will be the ratio of the highest order coefficients.
For (b): consider the fact that $-1 \le \cos x \le 1$ (for all $x$), but the denominator (which is $x^2$) tends to...? You may have heard of the squeeze or sandwich theorem? Dividing by $x^2$: $$\color{green}{ -\frac{1}{x^2}} \le \color{blue}{ \frac{\cos x}{x^2}} \le \color{red}{ \frac{1}{x^2}}$$
Since green and red tend to $0$, also blue...
• Yes I have heard of sandwich rule, how would one go about justifying if the limit exits first, (and then obviously evaluating it). – UniStuffz May 18 '16 at 9:25
• I elaborated a bit, see above. Does that help? Existence follows from being sandwiched between two that exist and are equal. – StackTD May 18 '16 at 9:29
• yes it does. So for example of the limit did not exits by using the methods described above, would that be sufficient as a proof to say that the limit does not exit, or would I have to use first principles and the definition of limit formula, (And I thought my editing was good, this is incredible). – UniStuffz May 18 '16 at 9:33
• It's not very clear to me what you're asking now. It's not because, for instance, the squeeze theorem doesn't apply, that you can be sure the limit you're looking at doesn't exist... It depends on the context and the problem how you would go about showing a limit does not exist. One possibility is indeed via the formal definition, but perhaps you're allowed to conclude it directly for simple functions such as $\lim_{x \to \infty} \cos x$, because you know how $\cos$ behaves. – StackTD May 18 '16 at 9:38
• You're welcome! – StackTD May 18 '16 at 9:46
Keep in mind
$$\frac{-1}{x^2} \leq \frac{\cos x}{x^2} \leq \frac{1}{x^2}$$ what is the limit of $\frac{\pm 1}{x^2}$ when $x\to \infty$ ?
• Yes of course they will both be zero, hence bounded above and below by zero hence by using sandwich theorem, the function will tend to 0. How would you justify if the limit exists first. – UniStuffz May 18 '16 at 9:27 | 2019-08-21T10:39:52 | {
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https://www.physicsforums.com/threads/general-form-of-circle-equation.24014/ | # General Form of Circle Equation
1. May 4, 2004
### Xavier
I'm attempting to write the general form of a circle where I'm given three points, not necessary equally distant from each other: (0,0), (0,8), (6,0)
I need to write the equation of the circle in general form. I've managed to get pretty close to what the answer in the book states, but I'm still off.
Since these points are not equally distant from each other, you can't use the midpoint formula (x1+x2)/2, (y1+y2)/2 to get the center of the circle. Is there another method I should be using to find the center so I can determine the radius, and then find the general form for the circle?
I'm not looking for the answer, I know what the book says in the answer key, I'm just curious as to what I'm missing, then I can try to take it from there. Thanks!
2. May 4, 2004
### krab
the circle equation is
$$(x-a)^2+(y-b)^2=r^2$$
Plug in each of the three (x,y) coordinate pairs, and you get 3 separate equations. The unknowns are a,b,r. Solve it.
3. May 5, 2004
### Chrono
You probably don't need this, but you also have the parametric equations of $$(x-a)^2+(y-b)^2=r^2$$ which can be written as:
$rSin[t] - a; \ rCos[t] - b$
as t goes from 0 to $$2\pi$$
I may be wrong, so somebody double check it for me.
4. May 5, 2004
### Xavier
Ok, so for the three points you get:
$$(0-a)^2 + (0-b)^2 = r^2$$
$$(0-a)^2 + (8-b)^2 = r^2$$
$$(6-a)^2 + (0-b)^2 = r^2$$
But without knowing a & b, how do you solve for r?
5. May 5, 2004
### ahrkron
Staff Emeritus
You have three equations abd three unknowns. You can, for instance, eliminate r by equation the first two; this gives you one equation with only a and b. You do the same with the first and third equations and get a different equation with a and b. Then you have two new equations on a and b only, which you can solve.
Once you have a and b, you can substitute on one of the original three eqns to get r.
6. May 6, 2004
### Xavier
Hmmm.. I can see how I can do it for this equation since $$(0-a)^2 = a^2$$, so then you just take the square root of both sides to solve for a. But if you have (for example) $$(1-a)^2 = 1-2a+a^2$$, you can't just solve for a can you? You'd be left with $$a^2-2a = -1+-$$(rest of equation)
7. May 6, 2004
### ahrkron
Staff Emeritus
Play with them a little bit... when you eliminate r from first and second, you also eliminate a^2 and b^2, so you only get a constant plus a term in b. Something similar happens combining the first and last eqns.
8. May 6, 2004
### HallsofIvy
Staff Emeritus
$$(0-a)^2 + (0-b)^2 = r^2$$
so (1) $$a^2+ b^2= r^2$$
$$(0-a)^2 + (8-b)^2 = r^2$$
so (2) $$a^2+ b^2-16b+ 64= r^2$$
$$(6-a)^2 + (0-b)^2 = r^2$$
so (3) $$a^2- 12a+ 36+ b^2= r^2$$
Subtracting 1 from 3, -12a+ 36= 0 or a= 3.
Subtracting 1 from 2, -16b+ 64= 0 or b= 4.
The a2+ b2= 9+ 16= r2
so r= 5.
Doesn't come any easier than that!
9. May 6, 2004
### Xavier
Ah.. I see.. I knew I had to set one equation equal to another and then reduce, but I didn't see the relationship between the three points. I'll crack open the book again when I get off work and try a few more of those problems. Thanks for the guidence!
10. May 10, 2004
### Xavier
Ok.. I've reworked this, and I'm just about where I was at. Here's what my book shows as the general form of the circle with the following points (same as before):
Points: (0,0), (0,8), (6, 0)
Book Answer: $$x^2+y^2-6x-8y=0$$
When I work this, I get this:
$$(0-a)^2+(0-b)^2=r^2$$
$$a^2+b^2=r^2$$
$$(0-a)^2+(8-b)^2=r^2$$
$$a^2-b^2-16b+64=r^2$$
$$(6-a)^2+(0-b)^2=r^2$$
$$a^2+b^2-12a+36=r^2$$
$$a^2+b^2=a^2+b^2-16b+64$$
$$16b=64$$
$$b=4$$
$$a^2+b^2=a^2+b^2-12a+36$$
$$12a=36$$
$$a=3$$
$$(x-a)^2+(y-b)^2=r^2$$
$$(0-3)^2+(0-4)^2=r^2$$
$$r=5$$
$$(x-3)^2+(y-4)^2=5$$
$$x^2-3x-3x+9+y^2-4y-4y+16=5$$
$$x^2+y^2-6x-8y+20=0$$
As you can see, I'm slightly off, but I don't understand what I've done wrong. Any ideas where I strayed?
11. May 11, 2004
### krab
I notice everyone takes the long way round. Here's a better way: Look at the first two equations. Don't multiply them out, just subtract them. You get 8-b=b, or b=4. Now do the same to equations 1 and 3. You get 6-a=a, or a=3. Plug these two result into equation 1 and you find r=5.
12. May 11, 2004
### arildno
You've just made a silly mistake here, that's all (the radius squared is 25)
Besides, note the swift manipulation done by krab.
13. May 11, 2004
### Xavier
Ah.. that's what happens when I work on this late at night. For some reason I figured it didn't need to be squared since I had already determined the radius, but it's clear now.
That is quite a bit simplier manipulation Krab. I've got a second problem I can try this out on, since I don't believe I did it correctly.... I think I need a bigger eraser :)
Thanks for working through this with me. I appreciate it!
14. May 11, 2004
### Wooh
Another interesting method to use would be to find the equations of the lines that go through 2 of the points (only 2 lines are needed). So, for example, you find the equation of the line through (0,0) and (0,8) and the equation of the line (0,0) and (6,0). Then, you use midpoint formula and find the equation of the perpendicular bisector of each line. The interesection of the 2 perpendicular bisectors? The center of the circle. | 2016-10-28T18:30:33 | {
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https://math.stackexchange.com/questions/3206816/why-is-every-singleton-subset-of-a-partially-ordered-set-a-totally-ordered-set | # Why is every singleton subset of a partially ordered set a totally ordered set?
In Lipschutz, Theory and Problems of Set Theory ( Schaum's series, 1964 ed.), Chapter 10, Solved Problem n°7 , the question is asked to find all non-empty totally ordered subsets of an ordered set A = {a,b,c} with a diagram showing that : b precedes a and that c precedes a ( b and c being incomparable).
Lipschutz provides a solution in which he gives {a} , {b} and {c} as totally ordered subsets of A.
Hence my question : is every singleton subset of a partially ordered set a totally ordered set.
How to explain that , in case a subset is a singleton the trichotomy condition for totally ordered sets is automatically fullfilled : a partial order S is total iff for all a and b belonging to S,
either a<b or a=b or a>b.
I'd like to derive the conclusion formally from the definition of a totally ordered set.
• It's because $\forall a,b: a=b$ holds in a singleton. – Berci Apr 29 at 12:27
• $a \in \{ a \}$ is the only element and $a=a$. Thus, the trichotomy condition is fulfilled. Recall that the condition id $\forall a,b \in X \ [a=b \lor \ldots ]$. – Mauro ALLEGRANZA Apr 29 at 12:28
1. Let $$S$$ be a singleton.
2. From $$1$$, there follows there must be some $$x$$ such that $$S=\{x\}$$.
3. Let $$a, b\in S$$.
4. Therefore, $$a=x$$ and $$b=x$$.
5. Therefore, $$a=b$$.
6. Therefore, $$a=b\lor ab$$.
• Very frustrating if nice answers like this are downvoted. And this without any explanation why. – drhab Apr 30 at 8:34
• @dmab. Totally agree. I personnaly upvoted both aswers. – Eleonore Saint James Apr 30 at 9:25
Yes, every singleton can be recognized as a totally ordered set (even as a well-ordered set).
This because next to reflexivity, transitivity and antisymmetry we also have comparability.
If $$S=\{c\}$$ is a singleton then it is true that for all elements $$a,b\in S$$ we have: $$ab$$
Actually in all cases (there is only one) we have $$a=c=b$$.
Also note that every non-empty subset of $$S$$ (there is only one) has a smallest element. | 2019-12-16T11:45:53 | {
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http://math.stackexchange.com/questions/171999/multiple-choice-summation-problem | # multiple choice summation problem
Let $$X = \frac{1}{1001} + \frac{1}{1002} + \frac{1}{1003} + \cdots + \frac{1}{3001}.$$ Then
(A) $X < 1$
(B) $X > 3/2$
(C) $1 < X < 3/2$
(D) none of the above holds.
I assume that the answer is the third choice $1<X<3/2$. I integrate out $1/x$ in the interval $(1001, 3001)$ and get a result that satisfies only the choice C. Is this a Riemann sum? Please help.
-
Short answer: Yes, your Riemann sum idea was correct, and $X\approx \log 3$ so the correct option was indeed C). – Ragib Zaman Jul 17 '12 at 16:00
@RagibZaman thank you but can you please explain why this discrete sum is integrated – Argha Jul 17 '12 at 16:03
@Ranabir: You can think of the sum as the integral of a step function. When you switch to integrating $1/x$ instead, you introduce a vertical error that is never more than the difference between two neighboring terms, which is about $10^{-6}$ at the most. So a conservative estimate of the total error is $10^{-6}$ times the width of the integrated area -- concretely, the integral will be at most 0.002 from the true sum. – Henning Makholm Jul 17 '12 at 16:19
## 2 Answers
With respect to your Riemann sum approach: the idea is that for positive, decreasing functions, the Riemann sum and the integral carefully approximate each other, more or less as in the proof of the integral test of convergence.
If you'd like another sort of approach, we could approach it naively. Separate the sum into $250$ element blocks, $\frac{1}{1001}$ to $\frac{1}{1250}$ in the first block $B_1$, $\frac{1}{1251}$ to $\frac{1}{1500}$ in the second block $B_2$, and so on. We'll have $12$ blocks.
Note that $\frac{1}{5} = \frac{250}{1250} \leq B_1 \leq \frac{250}{1000} = \frac{1}{4}$. Similarly, we get that $\frac{1}{i + 4} \leq B_i \leq \frac{1}{i+3}$ for all of our blocks.
This means that our sum has upper and lower bounds:
$$1 < \frac{1}{5} + \frac{1}{6} + \dots \frac{1}{16} \leq B_1 + \dots + B_{12} \leq \frac{1}{4} + \dots + \frac{1}{15}< \frac{3}{2}$$
And this gives the desired inequality.
-
It is approximately $\int_{1000}^{3000}dt/t = \log(3)$.
To explain why the Riemann Sum is close you should think about the approximating rectangles and the actual curve of $x \mapsto 1/x$. Since the curvature is nearly zero, the approximation will be very close.
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http://www.freemathhelp.com/forum/threads/73594-vertices-of-a-square-coordinate-geometry-and-differentiation | # Thread: vertices of a square, coordinate geometry and differentiation
1. ## vertices of a square, coordinate geometry and differentiation
The origin O and a point B(p, q) are opposite vertices of the square OABC. Find the coordinates of the point A and C
A line l has gradient q/p. Find possible values for the gradient of a line at 45° to l.
So I’m stuck on the first part of this problem and haven’t tried to tackle the second part yet, but I posted it anyway in case I have difficulties with it once I have solved the first part.
For the first question I observed that I have 4 unknowns, the two coordinates of each point. So I figured I should determine 4 equations that would help me find these; I used the following remarks to define these equations:
The gradient of AC will be perpendicular to the gradient of OB,
The gradient of OA will be perpendicular to the gradient of AB
The gradient of OC will be perpendicular to the gradient of CB,
The distance between A and C will be equal to the distance between O and B
However I’m not sure that this is the right (or at least, most effective) way to proceed as I’ve ended up with a mess of equations, often ending up quadratic, that appear to be leading me nowhere. Am I on the right track, have I used the wrong equations or am I just using a totally incorrect method?
2. Originally Posted by red and white kop!
The origin O and a point B(p, q) are opposite vertices of the square OABC. Find the coordinates of the point A and C
A line l has gradient q/p. Find possible values for the gradient of a line at 45° to l.
So I’m stuck on the first part of this problem and haven’t tried to tackle the second part yet, but I posted it anyway in case I have difficulties with it once I have solved the first part.
Assuming that the square has a side parallel to the two x-axis and $p>0~\&~q>0$ then
$A0,q)~\&~Cp,0).$
Is the second question part of the first?
3. The square doesn't necessarily have a side parallel to the axes. All information is given in the question. This could be any square.
4. Originally Posted by red and white kop!
The square doesn't necessarily have a side parallel to the axes. All information is given in the question. This could be any square.
In that case, find the points of intersection of the line $y-\frac{q}{2}=\frac{-p}{q}\left(x-\frac{p}{2}\right)$ and the circle $\left(x-\frac{p}{2}\right)^2+\left(y-\frac{q}{2}\right)^2=\frac{p^2}{4}+\frac{q^2}{4}$.
Those two points are $A~\&~C$.
5. Hello, red and white kop!
Did you make a sketch?
The origin O and a point B(p, q) are opposite vertices of the square OABC.
(a) Find the coordinates of the point A and C.
(b) A line L has gradient q/p. Find possible values for the gradient of a line at 45° to L.
Code:
|
| B
| o(p,q)
| *
| *
| M *
| o(p/2,q/2)
| *
| *
|*
- - o - - - - - -
O|
We are given points $O(0,0)\text{ and }B(p,q)$
. . and their midpoint $M\left(\frac{p}{2},\,\frac{q}{2}\right)$
Code:
|
| B
| o(p,q)
| *
| * ↑
| M *
| o → +
| *
| *
|*
- - o - - - - - -
O|
To move from $M$ to $B$, we move $\frac{p}{2}$ right and $\frac{q}{2}$ up.
Code:
|
| B
| o(p,q)
| *
| *
| M *
| o → +
| * ↓
| * ↓
|* o A
- - o - - - - -
O|
To move from $M$ to $A$, we must move $\frac{q}{2}$ right and $\frac{p}{2}$ down.
Do you see why?
. . Hence, $A$ is at $\left(\frac{p+q}{2},\,\frac{p-q}{2}\right)$
Similarly, we find that $C$ is at $\left(\frac{p-q}{2},\,\frac{p+2}{2}\right)$
6. Originally Posted by soroban
Hello, red and white kop!
Did you make a sketch?
Code:
|
| B
| o(p,q)
| *
| *
| M *
| o(p/2,q/2)
| *
| *
|*
- - o - - - - - -
O|
We are given points $O(0,0)\text{ and }B(p,q)$
. . and their midpoint $M\left(\frac{p}{2},\,\frac{q}{2}\right)$
Code:
|
| B
| o(p,q)
| *
| * ↑
| M *
| o → +
| *
| *
|*
- - o - - - - - -
O|
To move from $M$ to $B$, we move $\frac{p}{2}$ right and $\frac{q}{2}$ up.
Code:
|
| B
| o(p,q)
| *
| *
| M *
| o → +
| * ↓
| * ↓
|* o A
- - o - - - - -
O|
To move from $M$ to $A$, we must move $\frac{q}{2}$ right and $\frac{p}{2}$ down.
Do you see why?
. . Hence, $A$ is at $\left(\frac{p+q}{2},\,\frac{p-q}{2}\right)$
Similarly, we find that $C$ is at $\left(\frac{p-q}{2},\,\frac{p+2}{2}\right)$
Soroban, thanks for your time, your method seems to be the most applicable my level of comprehension and the section I’m studying.
However, when you say ‘do you see why’, I do not: it would help if you formulated your answer more algebraically and less with phrases and diagrams. That would help me understand your logic a lot better. Also your answer seems to be incorrect, as the answers given in the textbook are (0.5 (p-q), 0.5(p+q)) and (0.5(p+q), 0.5(q-p)); but I’ve discovered several errors in this textbook’s answer key recently so I’m not sure if you are wrong or the textbook is. I’d be thankful if you could keep helping me out until I’m done with this problem. Cheers
7. bump
8. Originally Posted by red and white kop!
bump
Code:
p - q
------
2
is the same thing as (1/2)*(p - q), right?
and 1/2 = 0.5
so, (p - q) / 2 = (1/2)(p - q) = 0.5 (p - q)
There...in nice algebra....you can see that the coordinates given by Soroban are EXACTLY the same points as those given in your text.
9. Originally Posted by Mrspi
Code:
p - q
------
2
is the same thing as (1/2)*(p - q), right? and 1/2 = 0.5
so, (p - q) / 2 = (1/2)(p - q) = 0.5 (p - q)
There...in nice algebra....you can see that the coordinates given by Soroban are EXACTLY the same points as those given in your text
There is a problem with this. Soroban's solution does not agree with the solution given in the text.
The text is correct: $\left( {\frac{{p - q}}{2},\frac{{p + q}}{2}} \right)\;\& \,\left({\frac{{p + q}}{2},\frac{{q - p}}{2}} \right)$
That is not Soroban's solution.
10. Originally Posted by Mrspi
Code:
p - q
------
2
is the same thing as (1/2)*(p - q), right?
and 1/2 = 0.5
so, (p - q) / 2 = (1/2)(p - q) = 0.5 (p - q)
There...in nice algebra....you can see that the coordinates given by Soroban are EXACTLY the same points as those given in your text.
don't be stupid
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http://math.stackexchange.com/questions/183599/can-one-derive-the-nth-term-for-the-series-u-n1-2u-n1-u-0-0/183606 | # can one derive the $n^{th}$ term for the series, $u_{n+1}=2u_{n}+1$,$u_{0}=0$, $n$ is a non-negative integer
derive the $n^{th}$ term for the series $0,1,3,7,15,31,63,127,255,\ldots$
observation gives, $t_{n}=2^n-1$, where $n$ is a non-negative integer
$t_{0}=0$
-
Let $s_n=t_n+1$, then $s_{n+1}=2s_n$ and $s_0=1$, end of story. – Did Aug 17 '12 at 12:14
NOTE: We get $t_{n}=2^n-1$ by observation. Is there a mathematical process by which we can derive this. – Rajesh K Singh Aug 17 '12 at 12:28
There is a long discussion at en.wikipedia.org/wiki/Recurrence_relation which shows some techniques to find these solutions. Without the constant (1) and the initial condition, the solution is a2^n. did's shift gets rid of the constant. – Ross Millikan Aug 17 '12 at 12:54
Is there a mathematical process by which we can derive this... See my answer. – Did Aug 21 '12 at 9:34
You already have an expression for the $n^{\text{th}}$ term for the sequence:
$$t_{n}=2^{n}-1\tag{1}$$
So let's prove this closed form by induction. Let $P(n)$ be our proposition that $t_{n}=u_{n}$, $\forall n\in\mathbb{N}\cup\{0\}$. So let us examine our basis case: $P(0)$:
$$t_{0}=2^{0}-1=1-1=0=u_{0} \implies P(0) \text{ is true}$$
Let us now show that if $P(k)$ is true, it follows immediately that $P(k+1)$ must also be true:
$$t_{k+1}=2^{k+1}-1=2\cdot2^{k}-1=2(2^{k}-1)+1=2u_{k}+1=u_{k+1}\implies \text{ if } P(k) \text{ is true, then } P(k+1) \text{ is true}$$
Therefore, as we have shown that $P(0)$ is true, and that $P(k)\implies P(k+1)$. $P(n)$ is true, $\forall n\in\mathbb{N}\cup\{0\}$.
However, if you are interested in how to actually come up with a closed form in general, a good place to start is to look at summation factors.
We can reduce a general recurrence relationship of the form:
$$a_{n}T_{n}=b_{n}T_{n-1}+c_{n} \tag{2}$$
To a sum, by multiplying both sides by a summation factor, denoted as $s_{n}$, such that:
$$s_{n}b_{n}=s_{n-1}a_{n-1}$$
In general, we can find a suitable $s_{n}$ using any multiple of the following:
$$s_{n}=\frac{a_{n-1}a_{n-2}\cdots a_{1}}{b_{n}b_{n-1}\cdots b_{2}}\tag{3}$$
We then have a summation recurrence, to which the solution can be found to be:
$$T_{n}=\frac{1}{s_{n}a_{n}}\left(s_{1}b_{1}T_{0}+\sum_{k=1}^{n}s_{k}c_{k}\right)\tag{4}$$
In your case, we have a recurrence of the form given in $(2)$:
$$a_{n}=1\qquad b_{n}=2\qquad c_{n}=1$$
Therefore, using $(3)$ we have $s_{n}=\frac{1}{2^{n}}$. And we can therefore plug this into $(4)$ to give:
$$T_{n}=2^{n}\left(0+\sum_{k=1}^{n}{\frac{1}{2^{k}}}\right)=2^{n}\left(1-\frac{1}{2^{n}}\right)=2^{n}-1$$
Which is $(1)$, the closed form you got by observation.
If you want a more complete look at solving recurrence relationships, I'd recommend the first few chapters of Concrete Mathematics by Graham, Knuth and Patashnik.
Hope this helps!
-
If the question is how to guess the form of the solution $u_n$ for every $n\geqslant0$, as opposed to, how to check that some given formula for $u_n$ is right (which is what my first comment and the other answers given so far all focus on), here is a standard computation that may help.
Assume that $u_n=au_{n-1}+b$ for every $n\geqslant1$, for some given $a$ and $b$. Thus, $u_n=f(u_{n-1})$, where the function $f$ is defined by $f(x)=ax+b$ for every $x$.One knows that to iterate a function, in general, can rapidly lead to a complicated mess (and/or to fascinating mathematics, think about fractal geometry). Some exceptions are when $f$ is constant (then $u_n=u_1$ for every $n\geqslant1$) or when $f$ is linear, so let us first assume that $f$ is linear, that is, that $f(x)=ax$ for every $x$. Then $f^{n}(x)=a^nx$ (induction over $n\geqslant1$) hence $u_n=a^nu_0$ for every $n\geqslant0$ and we are done.
The treatment of our general case $f(x)=ax+b$ cannot be made quite as simple but nearly so! To see this, we start with a simple remark. Define $v_n=u_n+c$ for some given $c$, to be chosen later. Then, $$v_n=(au_{n-1}+b)+c=a(v_{n-1}-c)+b+c=av_{n-1}+b_c,\qquad b_c=b-c(a-1).$$ Thus $v_n=f_c(v_{n-1})$, where $f_c$ is a new affine function, defined by $f_c(x)=ax+b_c$, and we are still in the case we wanted to solve at the beginning, hence it seems we have been running in circles. BUT... if by chance $f_c$ is in fact linear, we are done since we know how to solve the linear case! Perhaps we happy, after all?
Which brings us to the equation $b_c=0$, solved by $c^*=b/(a-1)$ as soon as $a\ne1$. And then, everything flows easily: $v_n=av_{n-1}$ for every $n$, hence $v_n=a^nv_0$ by the preceding analysis, that is, $u_n+c^*=a^n(u_0+c^*)$, that is, $u_n=a^n(u_0+c^*)-c^*$ and we are done.
If $a=2$, $b=1$ and $u_0=0$, then $c^*=1/(2-1)=1$ and $u_n=2^n(0+1)-1=2^n-1$, which is the specific case asked about here.
Two remarks. First, once the idea explained above is understood, one can go directly from the recursion $u_n=au_{n-1}+b$ with $a\ne1$ to the solution $u_n=a^n(u_0+c^*)-c^*$ for some $c^*$ to be determined (for example, $u_1=a(u_0+c^*)-c^*$ hence $c^*=(u_1-au_0)/(a-1)$, but, to remember this exact formula is not necessary). Second, the case $a=1$ is solved by a specific (simple) analysis I will let you discover.
-
The following is a semi-formal variant of induction that is particularly useful for recurrences.
Let $x_n=2^n-1$. It is easy to verify that $x_0=0$. It is also easy to verify that $$x_{n+1}=2x_n+1,$$ since $2^{n+1}-1=2(2^n-1)+1$.
So the sequence $(x_n)$ starts in the same way as your sequence and obeys the same recurrence as your sequence. Thus the two sequences must be the same.
-
Consider the series, $0, 1, 3, 7, 15, 31, 63,\ldots$
On taking the difference between the terms one can see that the difference ceases to vanish and the difference becomes $1,2,4,8,16,32,\ldots,2^{n}$ just after the first stage. Here $n$ is a non-negative integer.
i.e. the general term of the expression is of the form $2^{(x-1)}+ax+b$
when, $x=1$, $2^{(1-1)}+a+b=0$
i.e., $a+b=-1$
when, $x=2$, $2^{(2-1)}+2a+b=1$
i.e., $2a+b=-1$
we have, $a =0$, $b=-1$
we can now conclude that the $n_{th}$ term of the series is $2^{n-1}-1$, where $n$ is a positive integer.
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https://fighterfitness.sg/best-dish-fgyxl/0a96a5-determine-the-coordinates-of-the-centroid-of-the-area | And the area of this surface element $\mathrm{d}A = \frac{1}{2}r^2(\theta)\mathrm{d}\theta$. An analyst at the Scotland Department of Environment is performing a preliminary review on wind farm applications to determine which ones overlap with or are in view of wild lands. • The coordinates ( and ) define the center of gravity of the plate (or of the rigid body). Recall that the centroid of a triangle is the point where the triangle's three medians intersect. With this centroid calculator, we're giving you a hand at finding the centroid of many 2D shapes, as well as of a set of points. Find the coordinates of the centroid of the plane area bounded by the parabola y = 4 – x^2 and the x-axis. 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That the centroid of this triangle is the point at which the whole of. | 2021-06-16T13:36:03 | {
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https://math.stackexchange.com/questions/2373053/prime-factorization-of-m2 | # Prime Factorization of $m^2$
If $m$ is a positive integer, explain why each prime in the prime factorization of $m^2$ must occur an even number of times.
I did a small proof, I was wondering what's a nice way to explain it aside from my explanation below:
Proof: let $m=p_1^{e_1} p_2^{e_2}p_3^{e_3}\cdots p_k^{e_k}$
Then $m^2=(p_1^{e_1})^2 (p_2^{e_2})^2(p_3^{e_3})^2\cdots (p_k^{e_k})^2 = p_1^{2e_1}p_2^{2e_2}p_3^{2e_3}\cdots p_k^{2e_k}$
These $p_i^{e_i}$ occur an even number of times any suggestions to make a more solid answer?
• I would not put both formulas for $m^2$ on the same line, but otherwise it's the right proof. Maybe say explicitly that raising to a power distributes over multiplication $(ab)^n=a^nb^n$ because multiplication is commutative. Otherwise, I think it's good. – Gregory Grant Jul 27 '17 at 2:01
• One suggestion: you also need to explicitly invoke the uniqueness of prime factorizations to conclude that is the only such factorization of $m^2$ (else it might have another factorization where some prime has odd exponent). – Bill Dubuque Jul 27 '17 at 2:02
• Actually don't say "These $p_i^{e_i}$ occur an even number of times. Say $p_i$ occurs $2e_i$ times which is even. – Gregory Grant Jul 27 '17 at 2:02
• @GregoryGrant yeah thanks my issue was that explanation it did not sit well with me either. – OLE Jul 27 '17 at 2:04
• @BillDubuque is their a nicer more stronger way of proving this? – OLE Jul 27 '17 at 2:06
It's in fact an if and only if :
Each prime in "the" (due to uniqueness of p.f up to multiplication by units) prime factorization of $$k$$ occurs an even number of times, if and only if $$k$$ is a perfect square.
Suppose each prime in the prime factorization of $$k$$ occurs an even number of times, say $$k = \prod p_i^{r_i}$$, where each $$r_i$$ is even. Then, you can see that if $$b = \prod p_i^{\frac {r_i} 2}$$, then $$b$$ is a well defined integer, and $$b^2 = k$$.
Conversely, note that if $$k = m^2$$ is a perfect square, then the prime factorization of $$m = \prod p_i^{r_i}$$ suggests the prime factorization for $$k = \prod p_i^{2r_i}$$. Since prime factorization is unique, it follows that $$k$$ can indeed only be prime factorized in the above form, and hence every prime appears evenly many times in the rime factorization.
Alternately, we can also go by this way : Suppose $$p^r$$ divides $$m^2$$, where $$r$$ is maximal. Suppose $$r$$ is even, then we are done. Otherwise, note that $$r = 2k+1$$, and we can write $$p \ \mid\ \frac {m^2}{p^{2k}}$$, so that $$p$$ divides a perfect square.
Hence, from here, using Euclid's lemma, that $$p$$ divides $$ab$$ implies $$p$$ divides $$a$$ or $$p$$ divides $$b$$, we get that taking $$a=b=\frac m{p^k}$$, $$p^{k+1} \mid m$$, or that $$p^{2k+2} \mid m^2$$, a contradiction by uniqueness of prime factorization.
On a little inspection,the second proof is just a longer winded version of the first proof, but two is better than one, I suppose.
EXTENSION : If $$m$$ is a perfect $$k$$th power, then every prime that appears in the factorization does so, with multiplicity a multiple of $$k$$. | 2019-10-15T18:28:06 | {
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https://math.stackexchange.com/questions/565217/how-to-prove-this-inequality-a2bc4b2ca4c2ab4-leq-64 | # How to prove this inequality $(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64$?
Question:
If $a,b,c$ are nonnegative real numbers such that $a+b+c=3,$ then
$$(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64$$
My try: I found the equality holds only if $(a,b,c)=(2,0,1)$ or all of its permutations.
But I can't prove this inequality it. I would appreciate very much a proof.
This problem comes from:http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=562119
• Do you mean "I found that equality holds only if $(a,b,c)$ is $(2,0,1)$, $(1,2,0)$, or $(0,1,2)$"?. – John Bentin Nov 13 '13 at 11:05
• Yes,That's mean,Thank you – math110 Nov 13 '13 at 11:09
• If you can prove one of them is zero, then you can just study $f(2+x,0,1-x)$ and see that its maximum is $64$. – Beni Bogosel Nov 13 '13 at 12:20
• You can edit your question so that it says what you mean. – John Bentin Nov 13 '13 at 12:26
• @math110: I've taken the liberty of editing it for you. Where did you get this question from, if you don't mind me asking. – Nick Nov 13 '13 at 13:05
## 2 Answers
→ →
The three dimensional problem can be simplified to a two dimensional problem by introducing (again and again) suitable triangle coordinates: $$\left[ \begin{array}{c} a \\ b \\ c \end{array} \right] = \left[ \begin{array}{c} 3 \\ 0 \\ 0 \end{array} \right] + \left[ \begin{array}{c} -3 \\ 3 \\ 0 \end{array} \right] x + \left[ \begin{array}{c} -3 \\ 0 \\ 3 \end{array} \right] y$$ Then the equation $(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64$ does not so much "simplify", but anyway becomes an equation in two variables (2-D). And the equation $\;a + b + c = 3\;$ corresponds with a normed 2-D triangle, with vertices $(0,0),(1,0),(0,1)$ . The insides of both can easily be visualized, as has been done in the above picture in the middle:
$\color{red}{red}$ for $\;a + b + c = 3\;$ and $\color{green}{green}$ for $\;(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64\;$ .
The transformed inequality is: $$\left[ 3^2\left( 1-x-y \right)^{2} + 3^5\,x{y}^{4} \right] \left[ 3^2\,{x}^{2} + 3^5\,y \left( 1-x-y \right)^{4} \right] \left[ 3^2\,{y}^{2} + 3^5\, \left( 1-x-y \right){x}^{4} \right] \le 64$$
It is seen in the same picture that the edge $\;y=1-x\;$ of the triangle maybe is tangent to the curve $(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) = 64$ . Indeed, if we substitute $y=1-x$ into (the transformation of) that equation and simplify, then we get: $$3^9\, x^3 (1-x)^6 - 64 = 0$$ The same sort of equation is found with the substitutions $\,x=0\,$ or $\,y=0\,$, for the other two edges. And, as has been found by others, there is only one solution of that equation, within the specified range, namely $x=1/3$, corresponding with $y=2/3$ and hence $(a,b,c) = (0,1,2)$ . And of course any cyclic permutation of this, due to symmetry. The rest of the (red) triangle $\;a + b + c = 3\;$ is well within the (green) area of $\;(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64$ . Which can be shown by plotting the triangle first: then it becomes absorbed by the green area (see picture on the right).
Analytically, the proof is completed by considering the function $\;f(x) = 3^9\, x^3 (1-x)^6$ .
Its extreme values are found for $\;f'(x) = x^2(1-x)^5(3-9x)=0$ , giving $\;x = \{0,1,1/3\}$ , with the maximum $\;f(1/3)=64$ .
The picture on the right shows the the inequality as observed in the plane $\;a + b + c = 3\;$ of the triangle in 3-D (picture on the left). Mind the symmetries.
• Can you use other colors? Since many people have protanopia or protanomaly (like me). – ziyuang Jan 8 '14 at 11:24
• Next time, I promise. Any suggestion? I've used red and green now. – Han de Bruijn Jan 8 '14 at 14:45
• Maybe different patterns of lines, if that's convenient for you. – ziyuang Jan 8 '14 at 23:05
Without loss of generality, assume $$a$$ is smallest of $$a, b, c$$. Also, let $$f(a,b,c)=(a^2+bc^4)(b^2+ca^4)(c^2+ab^4)$$and firstly, if $$a\le c\le b$$, then$$f(a,b,c)-f(a,c,b)=(b^3-a^3) (c^3 - a^3) (b ^3- c^3) (a b c - 1)<0$$therefore we can assume $$a\le b\le c$$. Now, we will prove $$f(0,b,a+c)\ge f(a,b,c)$$which is, after full expansion,$$a (-a^6 b^4 c + a^5 b^3 - a^5 c^3 - a^4 b^5 c^5 - a^3 b c^7 - a^2 b^6 + 20 a^2 b^3 c^3 \\+ 15 a b^3 c^4 - a b^2 c^2 - b^7 c^4 + 6 b^3 c^5)+ 6 a^5 b^3 c + 15 a^4 b^3 c^2\ge0$$ and it is obvious that $$-a^6 b^4 c + a^5 b^3\ge0$$, $$- a^5 c^3 - a^4 b^5 c^5 - a^2 b^6 + 20 a^2 b^3 c^3\ge0$$ and $$6 a^5 b^3 c + 15 a^4 b^3 c^2\ge0$$. Therefore it is enough to show$$b c^2 (-a^3 c^5 + 15 a b^2 c^2 - a b - b^6 c^2 + 6 b^2 c^3)\ge0$$and from $$-a^3c^5\ge-ab^2c^5$$, $$-ab\ge-abc^3\ge-b^2c^3$$ and $$-b^6c^2>-4b^3c^2\ge-4b^2c^3$$, it is left to show$$b^2c^2(-a c^3 + 15 a + c)\ge0$$and we can divide it with three cases.
Case 1) $$c^3\le15$$: $$-a c^3 + 15 a + c>a(15-c^3)\ge0$$.
Case 2) $$15^{1/3}< c\le2.6$$: Firstly, $$c>2.4$$. The equation is decreasing with respect to $$a$$, therefore we need to show only for maximal value of $$a$$. If $$a>0.3$$, then $$a+b+c\ge2a+c>0.6+2.4=3$$, so maximal value is $$0.3$$. Also, $$c^3\le2.6^3<20$$. Therefore, $$-a c^3 + 15 a + c\ge0.3(15-c^3)+c\ge-1.5+2>0$$
Case 3) $$2.6: Similarly, it is enough to show for maximal value of $$a$$ which is $$0.2$$. Therefore, $$-a c^3 + 15 a + c\ge0.2(15-27)+2.6>0$$
Therefore, we can assume that $$0=a\le b\le c$$. Now $$f(a,b,c)=b^3c^6\le2^6\left(\frac{3\times b+6\times0.5c}{9}\right)^9=64$$ and it is proved. | 2019-06-17T17:00:09 | {
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https://math.stackexchange.com/questions/451212/showing-that-q-n-d-nd-n-1 | Showing that $Q_n=D_n+D_{n-1}$
Let $$T_n$$ be the set of permutations of $$\{1,2,\ldots,n\}$$ which do not have $$i$$ immediately followed by $$i+1$$ for $$1\le i\le n-1$$; in other words, let \begin{align} T_n=\{\sigma \in S_n: \sigma(i)+1\ne\sigma(i+1) \text{ for all } 1\le i\le n-1\} . \end{align} Let $$Q_n$$ be the number of elements of $$T_n$$.
Let $$D_n$$ be the number of derangements of $$\{1,2,\ldots,n\}$$.
It is not hard to show algebraically that $$Q_n=D_n+D_{n-1}$$, but I am having difficulty coming up with a combinatorial argument to show why this is true. (I believe there are $$D_{n-1}$$ elements of $$T_n$$ that leave $$n$$ fixed, and $$D_n$$ elements of $$T_n$$ that move $$n$$, but I don't know how to justify this combinatorially.)
• just to be clear, "i followed by i+1" means that $\sigma(i)+1$ is never equal to $sigma(i+1)$, for all $i\leq n-1$ ? – Denis Jul 25 '13 at 0:52
• I'm sorry if my statement of the problem wasn't completely clear, but that's what I meant. (In other words, when you arrange the images of 1,..., n in order, i is never followed by i+1.) – user84413 Jul 26 '13 at 2:10
• It seems that your question has been asked and answered previously. I do have some doubt about Brian Scott's answer, and I've left a comment there requesting clarification, but it may just be a simple misunderstanding on my part. – Will Orrick Sep 10 '13 at 2:42
• @ Will Orrick Thank you for pointing out that my question had been asked previously. I read the answer posted, but I wasn't able to see how to make it work. (For example, it seemed to me that the permutations 1324 and 4132 would both map to the permutation 2314, if I am interpreting the answer correctly.) I like your argument to show that the number of elements of $T_n$ that fix n is equal to $D_{n-1}$, even though it isn't what I originally had in mind. Can you use the same idea to show that the number of elements in $T_n$ which do not fix n is equal to $D_n$? Thanks again. – user84413 Sep 10 '13 at 17:45
• It turns out Persi Diaconis, Steven Evans, and Ron Graham have a recent paper exploring this problem in some depth. (arxiv.org/abs/1308.5459 ) – Kevin P. Costello Oct 14 '13 at 23:53
(Note: this answer does not contain a bijection between the two sets, which is what the OP was originally hoping for, but it does show that the two counting problems are structurally equivalent and therefore have the same answer.)
The belief stated in the last paragraph of your question is correct: we can show that the set $\mathcal{D}_{n-1}$ of derangements of $\{1,2,3,\ldots,n-1\}$ and the set of permutations $\sigma$ of $\{1,2,3,\ldots,n\}$ with the properties
• $\sigma(n)=n$,
• $\sigma(i+1)\ne\sigma(i)+1,$ for $1\le i\le n-1,$
are equinumerous. That is, $\mathcal{D}_{n-1}$ and the set of elements of $T_n$ that fix $n$ have the same size. Call the latter set $\overline{T}_n.$
This result will prove the equation in your title since there is an obvious bijection between the set of elements of $T_n$ that do not fix $n$ and the set $\overline{T}_{n+1}$ of elements of $T_{n+1}$ that do fix $n+1,$ which will then be equinumerous with $\mathcal{D}_n.$
To enumerate $\mathcal{D}_{n-1}$, let $F=S_{n-1}$ be the set of permutations of $1,2,\ldots,n-1.$ Let $F_j$ be the set of elements of $F$ that fix $j,$ that is, elements $\sigma$ such that $\sigma(j)=j.$ In general, let $F_{ijk\ldots}=F_i\cap F_j\cap F_k\cap\ldots$ be the set of elements that fix $i,$ $j,$ $k,\ldots$, that is, elements $\sigma$ such that $\sigma(i)=i,$ $\sigma(j)=j,$ $\sigma(k)=k,\ldots$ Observe that $\lvert F_{i_1i_2\ldots i_k}\rvert=(n-1-k)!$ since only $n-1-k$ elements are free to move. Since the derangements are those elements that fix no element, the principle of inclusion-exclusion gives \begin{aligned}\lvert\mathcal{D}_{n-1}\rvert&=\lvert F\rvert-\sum_{i=1}^{n-1}\lvert F_i\rvert+\sum_{1\le i<j\le n-1}\lvert F_{ij}\rvert-\ldots\\ &=(n-1)!-\binom{n-1}{1}(n-2)!+\binom{n-1}{2}(n-3)!-\ldots\end{aligned}
To enumerate $\overline{T}_n$, let $G$ be the set of permutations of $1,2,\ldots,n$ that fix $n.$ Let $G_i,$ $1\le i\le n-1,$ be the subset of $G$ consisting of elements $\sigma$ such that $\sigma(i+1)=\sigma(i)+1.$ In general, let $G_{ijk\ldots}=G_i\cap G_j\cap G_k\cap\ldots$ be the set of elements of $G$ such that $\sigma(i+1)=\sigma(i)+1,$ $\sigma(j+1)=\sigma(j)+1,$ $\sigma(k+1)=\sigma(k)+1,\ldots$ We claim that, once again, $\lvert G_{i_1i_2\ldots i_k}\rvert=(n-1-k)!.$ This follows by noting that every constraint $\sigma(i_j+1)=\sigma(i_j)+1$ reduces the number of elements that can move independently by $1,$ leaving only $n-1-k$ elements that are free to move. One can imagine element $i$ becoming "glued" to element $i+1,$ so that they must move as a block.
For example, let $n=9$ and consider $G_{12478}.$ Since $8$ must immediately precede $9,$ which is fixed, and $7$ must immediately precede $8,$ which is now fixed as well, the elements $789$ are fixed. At the same time, $2$ must immediately precede $3$ and $1$ must immediately precede $2,$ meaning that the string $123$ can only move as a block. Similarly, the string $45$ can only move as a block. As a consequence the number of permutations in $G_{12478}$ is the number of ways of permuting the "objects" $123,$ $45,$ and $6,$ with the object $789$ fixed in place. So $G_{12478}=3!=(9-1-5)!,$ in agreement with the claim.
Since $\lvert G_{ijk\ldots}\rvert=\lvert F_{ijk\ldots}\rvert$ for all choices of subscripts, the principle of inclusion-exclusion implies that the computation of $\lvert\overline{T}_n\rvert$ is identical to that of $\lvert\mathcal{D}_{n-1}\rvert,$ so that they have the same final value: \begin{aligned}\lvert\overline{T}_n\rvert&=\lvert G\rvert-\sum_{i=1}^{n-1}\lvert G_i\rvert+\sum_{1\le i<j\le n-1}\lvert G_{ij}\rvert-\ldots\\ &=(n-1)!-\binom{n-1}{1}(n-2)!+\binom{n-1}{2}(n-3)!-\ldots\end{aligned}
Let $[a,b]=\{i\in\mathbf{Z}\mid a\le i\le b\}$ and let $[b]=[1,b].$ Let $S_{a,b}$ be the set of permutations of $[a,b]$ and let $S_n$ be the set of permutations of $[n].$ Let $\overline{S}_{a,b}=\{\sigma\in S_{a,b}\mid \sigma(b)=b\}$ be the set of permutations of $[a,b]$ that fix $b.$ Similarly, Let $\overline{S}_n=\{\sigma\in S_n\mid \sigma(n)=n\}$ be the set of permutations of $[n]$ that fix $n.$
The result will follow by exhibiting a bijective mapping $\rho_n:S_n\to\overline{S}_{n+1}$ that has the property that if $\sigma$ is an element of $S_n,$ $\tau$ is the element of $\overline{S}_{n+1}$ given by $\tau=\rho_n(\sigma),$ $f$ is an element of $[n],$ and $e$ is given by $\tau(e)=f,$ then the property $\tau(e+1)=\tau(e)+1=f+1$ holds if and only if $f$ is a fixed point of $\sigma.$
To define $\rho_n$ we introduce some notation. For distinct elements $i_1,\ldots,i_n\in[n]$ let $\langle i_1i_2\ldots i_n\rangle$ denote the element $\sigma\in S_n$ that satisfies $\sigma(j)=i_j$ for $1\le j\le n.$
We use usual cycle notation to represent swaps, which act from the left. So $(35)\langle123456\rangle=\langle125436\rangle,$ and $(23)(35)\langle123456\rangle=\langle135426\rangle.$ Every element of $S_n$ has a unique swap representation $(a_1b_1)(a_2b_2)\ldots(a_kb_k)\langle12\ldots n\rangle$ where $0\le k\le n-1,$ $a_1<a_2<\ldots<a_k,$ and $a_i<b_i$ for all $1\le i\le k.$ For example, $$\langle 326451\rangle=(13)\langle 126453\rangle=(13)(36)\langle 123456\rangle.$$ Observe that $f$ is a fixed point of $\sigma\in S_n$ if and only if $f$ does not appear among the $i_1,\ldots,i_k,j_1,\ldots,j_k$ in the swap representation of $\sigma.$
We define the insertion operator $[ai]:\overline{S}_{a+1,n+1}\to\overline{S}_{a,n+1},$ where $a\in[1,n]$ and $i\in[a,n].$ Let $\tau\in\overline{S}_{a+1,n+1}.$ Then $[ai]\tau$ is defined to be the element of $\overline{S}_{a,n+1}$ obtained by inserting $a$ immediately in front of $i+1.$ So, for example, $[24]\langle3546\rangle=\langle32546\rangle.$ Every element of $\overline{S}_{n+1}$ has a unique insertion representation $[1b_1][2b_2]\ldots[nb_n]\langle (n+1)\rangle,$ where $b_j\ge j$ for all $1\le j\le n.$ For example, \begin{aligned} &\langle 1456237\rangle=[13]\langle 456237\rangle=[13][22]\langle 45637\rangle =[13][22][36]\langle 4567\rangle\\ &=[13][22][36][44]\langle 567\rangle=[13][22][36][44][55]\langle 67\rangle=[13][22][36][44][55][66]\langle 7\rangle. \end{aligned} Observe that $a$ immediately precedes $a+1$ in $\tau\in\overline{S}_{n+1}$ if and only if the only occurrence of $a$ in the insertion representation of $\tau$ is of the form $[aa].$
The bijective mapping $\rho_n:S_n\to\overline{S}_{n+1}$ is now defined as follows. Let $\sigma\in S_n$ and compute the swap representation of $\sigma.$ We may add null swaps of the form $(aa)$ so that the swap representation of $\sigma$ takes the form $(1b_1)(2b_2)\ldots(nb_n)\langle 123\ldots n\rangle,$ where $b_j\ge j$ for all $1\le j\le n.$ Then define $\rho_n(\sigma)$ by replacing swaps with insertions and replacing $\langle 123\ldots n\rangle$ with $\langle (n+1)\rangle:$ $$\rho_n(\sigma)=[1b_1][2b_2]\ldots[nb_n]\langle (n+1)\rangle.$$
Examples: We show $\rho_3:S_3\to\overline{S}_4$ and $\rho_4:S_4\to\overline{S}_5$ explicitly. On the left, dots mark fixed points; on the right, overscores mark sequences of consecutive elements. \begin{aligned} \langle\dot{1}\dot{2}\dot{3}\rangle=(11)(22)(33)\langle123\rangle&\mapsto[11][22][33]\langle4\rangle=\langle\overline{1234}\rangle\\ \langle\dot{1}32\rangle=(11)(23)(33)\langle123\rangle&\mapsto[11][23][33]\langle4\rangle=\langle3\overline{12}4\rangle\\ \langle21\dot{3}\rangle=(12)(22)(33)\langle123\rangle&\mapsto[12][22][33]\langle4\rangle=\langle21\overline{34}\rangle\\ \langle231\rangle=(12)(23)(33)\langle123\rangle&\mapsto[12][23][33]\langle4\rangle=\langle1324\rangle\\ \langle312\rangle=(13)(23)(33)\langle123\rangle&\mapsto[13][23][33]\langle4\rangle=\langle3214\rangle\\ \langle3\dot{2}1\rangle=(13)(22)(33)\langle123\rangle&\mapsto[13][22][33]\langle4\rangle=\langle\overline{23}14\rangle \end{aligned} Observe that the two derangements, $\langle231\rangle$ and $\langle312\rangle,$ have images that contain no sequences of consecutive elements.
\begin{aligned} \langle\dot{1}\dot{2}\dot{3}\dot{4}\rangle=(11)(22)(33)(44)\langle1234\rangle&\mapsto[11][22][33][44]\langle5\rangle=\langle\overline{12345}\rangle\\ \langle\dot{1}\dot{2}43\rangle=(11)(22)(34)(44)\langle1234\rangle&\mapsto[11][22][34][44]\langle5\rangle=\langle4\overline{123}5\rangle\\ \langle\dot{1}32\dot{4}\rangle=(11)(23)(33)(44)\langle1234\rangle&\mapsto[11][23][33][44]\langle5\rangle=\langle3\overline{12}\,\overline{45}\rangle\\ \langle\dot{1}342\rangle=(11)(23)(34)(44)\langle1234\rangle&\mapsto[11][23][34][44]\langle5\rangle=\langle\overline{12}435\rangle\\ \langle\dot{1}423\rangle=(11)(24)(34)(44)\langle1234\rangle&\mapsto[11][24][34][44]\langle5\rangle=\langle43\overline{12}5\rangle\\ \langle\dot{1}4\dot{3}2\rangle=(11)(24)(33)(44)\langle1234\rangle&\mapsto[11][24][33][44]\langle5\rangle=\langle\overline{34}\,\overline{12}5\rangle\\ \langle21\dot{3}\dot{4}\rangle=(12)(22)(33)(44)\langle1234\rangle&\mapsto[12][22][33][44]\langle5\rangle=\langle21\overline{345}\rangle\\ \langle2143\rangle=(12)(22)(34)(44)\langle1234\rangle&\mapsto[12][22][34][44]\langle5\rangle=\langle42135\rangle\\ \langle231\dot{4}\rangle=(12)(23)(33)(44)\langle1234\rangle&\mapsto[12][23][33][44]\langle5\rangle=\langle132\overline{45}\rangle\\ \langle2341\rangle=(12)(23)(34)(44)\langle1234\rangle&\mapsto[12][23][34][44]\langle5\rangle=\langle24135\rangle\\ \langle2413\rangle=(12)(24)(34)(44)\langle1234\rangle&\mapsto[12][24][34][44]\langle5\rangle=\langle41325\rangle\\ \langle24\dot{3}1\rangle=(12)(24)(33)(44)\langle1234\rangle&\mapsto[12][24][33][44]\langle5\rangle=\langle1\overline{34}24\rangle\\ \langle312\dot{4}\rangle=(13)(23)(33)(44)\langle1234\rangle&\mapsto[13][23][33][44]\langle5\rangle=\langle321\overline{45}\rangle\\ \langle3142\rangle=(13)(23)(34)(44)\langle1234\rangle&\mapsto[13][23][34][44]\langle5\rangle=\langle21435\rangle\\ \langle3\dot{2}1\dot{4}\rangle=(13)(22)(33)(44)\langle1234\rangle&\mapsto[13][22][33][44]\langle5\rangle=\langle\overline{23}1\overline{45}\rangle\\ \langle3\dot{2}41\rangle=(13)(22)(34)(44)\langle1234\rangle&\mapsto[13][22][34][44]\langle5\rangle=\langle14\overline{23}5\rangle\\ \langle3412\rangle=(13)(24)(33)(44)\langle1234\rangle&\mapsto[13][24][33][44]\langle5\rangle=\langle31425\rangle\\ \langle3421\rangle=(13)(24)(34)(44)\langle1234\rangle&\mapsto[13][24][34][44]\langle5\rangle=\langle14325\rangle\\ \langle4123\rangle=(14)(24)(34)(44)\langle1234\rangle&\mapsto[14][24][34][44]\langle5\rangle=\langle43215\rangle\\ \langle41\dot{3}2\rangle=(14)(24)(33)(44)\langle1234\rangle&\mapsto[14][24][33][44]\langle5\rangle=\langle\overline{34}215\rangle\\ \langle4\dot{2}13\rangle=(14)(22)(34)(44)\langle1234\rangle&\mapsto[14][22][34][44]\langle5\rangle=\langle4\overline{23}15\rangle\\ \langle4\dot{2}\dot{3}1\rangle=(14)(22)(33)(44)\langle1234\rangle&\mapsto[14][22][33][44]\langle5\rangle=\langle\overline{234}15\rangle\\ \langle4312\rangle=(14)(23)(34)(44)\langle1234\rangle&\mapsto[14][23][34][44]\langle5\rangle=\langle24315\rangle\\ \langle4321\rangle=(14)(23)(33)(44)\langle1234\rangle&\mapsto[14][23][33][44]\langle5\rangle=\langle32415\rangle \end{aligned} Observe that the images of the nine derangements contain no sequences of consecutive elements.
Since there is a one-to-one correspondence between the set of derangements of $[n]$ and the set of permutations of $[n+1]$ that fix $n+1$ and contain no sequences of consecutive elements (we let $\overline{T}_{n+1}$ denote this set), we have $$D_n=\lvert \overline{T}_{n+1}\rvert.$$ But there is an immediate bijection between $\overline{T}_{n+1}$ and $T_n-\overline{T}_n,$ the set of elements of $T_n$ that do not fix $n.$ Therefore $$D_n=\lvert T_n-\overline{T}_n\rvert$$ and $$D_n+D_{n-1}=\lvert T_n-\overline{T}_n\rvert+\lvert \overline{T}_n\rvert=\lvert T_n\rvert.$$
• Thanks - this is a great answer! – user84413 Oct 6 '13 at 16:40
• Very nice indeed, and thank you. – Kevin P. Costello Oct 7 '13 at 17:21
Let $\sigma$ be the circular permutation $i\mapsto i+1$.
Then, you can establish a bijection $D_n\cup D_{n-1}\to T_n$ (I abuse notation for $D_n$ to design the set as well as the cardinal) by mapping any $\tau$ to $\tau\circ\sigma$,where elements of $D_{n-1}$ are naturally extended with $n\mapsto n$. It is straightforward to verify that this is a bijection.
• If I take the derangement 54123 of {1,...,5} and compose it with $\tau$, does that give the permutation 41235 (which is not in $T_n$)? – user84413 Jul 24 '13 at 20:33
• I might have misunderstood your encoding of permutations. I understood "do not have i followed by i+1" as "the image of i is never i+1". I change the answer. (But the image of 54123 was 15234) – Denis Jul 25 '13 at 0:49
EDIT: What's below is actually WRONG, due to multiple issues noted in the comments (the equation taken modulo $n$ leads to situations where a consecutive $n1$ causes problems, and it's not necessarily a bijection.
I think a variant on dkuper's argument can be made to work.
If I understand correctly, the permutations in $Q_n$ are those which satisfy $\sigma(i)+1 \neq \sigma(i+1)$ for $1 \leq i \leq n-1$. Conversely, we can think of $D_n+D_{n-1}$ as consisting of those permutations satisfying $\tau(i) \neq i$ for $1 \leq i \leq n-1$ ($D_n$ corresponds to those permutations with $\tau(n) \neq n$ as well, while $D_{n-1}$ corresponds to fixing $\tau(n)=n$).
This suggests that we construct our bijection in such a way that $\sigma(i)+1=\sigma(i+1)$ if and only if $\tau(i)=i$. The former equation can be rewritten as $i=\sigma^{-1} (\sigma(i+1)-1)$. So if we define our bijection by taking $\sigma$ to the permutation satisfying $$\tau(i)=\sigma^{-1}\left(\sigma(i+1)-1\right),$$ where addition and subtraction are taken modulo $n$, things work the way we want them to.
• Great answer! I'm not sure if in the question the addition is taken modulo, but if it is, it's finally the answer to one of my favourite questions:) – savick01 Sep 9 '13 at 19:40
• This is exactly what I tried at one point, because it looked as if this should work; but I ran into some difficulties. For example, the permutation 4132 (so $\sigma(1)=4, \sigma(2)=1$, etc) in $T_n$ maps to 1423, which is not in $D_4$; and the permutation 2413 maps to 4213, which also is not in $D_4$ (if I am computing these correctly). Also, 3214 maps to 3412, which is not in $D_3$. Maybe some modification of this will work, though, if this doesn't already. – user84413 Sep 9 '13 at 21:07
• There's actually multiple issues here (in addition to the issue you noted where I'm treating $41$ appearing consecutively as bad, $\sigma \rightarrow \tau$ isn't even a bijection!). I'm not sure what the proper etiquette in terms of deletion is, but for now I'm leaving it up with a note explaining what's wrong, and I'll eventually edit the answer if it seems fixable. – Kevin P. Costello Sep 9 '13 at 21:39
• Thanks for your response, and I don't know about the proper etiquette either. I have had to delete some of my answers to problems, since they were totally wrong; but yours is actually a pretty reasonable approach, which hopefully is fixable in some way. – user84413 Sep 9 '13 at 22:52
• @Kevin Costello It was pointed out to me that my question had been asked previously (see above), and the answer that was given before is identical to yours (although written differently). – user84413 Sep 11 '13 at 19:32 | 2020-02-20T11:12:47 | {
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http://math.stackexchange.com/questions/330829/how-can-i-tell-if-the-sequence-a-n-frac-2n-n-converges-and-if-it-conve | How can I tell if the sequence $a_n=\frac {2^n} {n!}$ converges, and if it converges what is its limit?
I know this question was asked before, but I couldn't find it and I think my question is slightly different. I'm new here so if I do something wrong, I'm sorry, I'm still trying to learn.
I need to "study the limit behavior of the sequence and when it exists compute their limits."
This is what I have tried so far:
I studied the first few terms of the sequence, and I claimed that the sequence was monotone and decreasing. Then I showed that $a_{n+1}-a_n\leq0$ by: $$a_{n+1}-a_n= \frac {2^{n+1}} {(n+1)!}\ - \frac {2^n} {n!}= \frac {2^{n+1}} {n!(n+1)}- \frac {2^n} {n!}=\frac {2^n} {n!}(\frac {2} {n+1})-\frac {2^n} {n!}=\frac {2^n} {n!}(\frac {2} {n+1}-1)=\frac {2^n} {n!}(\frac {n-1} {n+1})$$ Then I said: $\frac {2^n} {n!}(\frac {n-1} {n+1})\leq 0$ only if $\frac {n-1} {n+1}\leq 0$ because $2^n>0$ and $n!>0$ . Then I said that $n+1>0$ for n existing the naturals and $1-n\leq0$, thus $\frac {n−1} {n+1}\leq 0$ and $a_{n+1}-a_n\leq 0$.
Therefore the sequence is monotonic and decreasing. Because the first term is 2 and $\frac {2^n} {n!}>0$, the sequence is bounded and together the monotonicity and and boundedness means that a limit exists.
Now I feel that what I have done so far is correct, and instinctually I know the limit is $0$, but how do I compute that? Do I just claim it is $0$ and prove it using the epsilon definition of a limit?
We haven't covered anything with series, but we have covered cauchy sequences (is that supposed to help me?) and things about lim (inf) and lim (sup). Any help please?
-
@JavaMan Just realized that I do know about the ratio test. But now I have a more general question, should I have appraoched this problem with the ratio test first instead of checking for monotonicity and boundedness? Also, is there anything specific that alerts me to use the ratio test instead of checking monotonicity/boundedness? – user66807 Mar 15 '13 at 2:28
I'd suggest taking the $\lim_{n\to \infty}$ of the ratio of two consecutive terms (use the ratio test) : $$\lim_{n\to \infty} \dfrac{a_{n +1}}{a_n}\; = \lim_{n\to \infty}\dfrac{\dfrac{2^{n+1}}{(n+1)!}}{\dfrac{2^n}{n!}}$$ $$= \lim_{n\to \infty} \dfrac{2^{n+1}n!}{2^n(n+1)!} = \lim_{n\to \infty} \dfrac{2}{n+1}$$
Evaluate the limit: If the limit exists, and $L \lt 1$, then as DonAntonio aptly clarified, "the series whose general term is the given sequence converges," by the Ratio Test for series. And then by Convergence test for series, if its series converges, the limit of the sequence itself (the general term of the series), is equal to $0$.
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Yes, that is what I meant, Thanks @DonAntonio: I will edit my post and credit your phrasing! – amWhy Mar 15 '13 at 2:59
I came here but you were gone. + – S. Snape Mar 15 '13 at 5:28
For every $x \in \mathbb{C}$ it is well-known that $\sum\limits_{n=0}^{\infty} \frac{x^n}{n!}$ converges to $\mathrm{exp}(x)$, hence $\frac{x^n}{n!}$ converges to $0$.
-
Recall: For a sequence $\{u_n\}$ in $\mathbb R^+,$
• $\lim \dfrac{u_{n+1}}{u_n}<1\implies \lim u_n=0;$
• $\lim \dfrac{u_{n+1}}{u_n}> 1\implies \lim u_n=+\infty;$
• For $\lim \dfrac{u_{n+1}}{u_n}=1$ no definite conclusion can be made e.g. consider $\left\{\dfrac{n+1}{n}\right\}$ and $\left\{\dfrac{1}{n}\right\}.$
Here, $\dfrac{a_{n+1}}{a_n}=\dfrac{2}{n+1}\to0<1.$ Therefore $\{a_n\}$ converges to $0.$
-
A few ways to look at it.
The sum of all $a_n$ is finite (consider the power series for $e^x$ at $x=2$).
For $n>5$ every term is at most half the previous one.
$2^n$ grows quickly, but not nearly as quickly as $n!$.
-
$$\frac{2^n}{n!}=\frac{2}{1}[\frac{2}{2}\frac{2}{3}..\frac{2}{n-1}]\frac{2}{n} \leq \frac{2}{1}\cdot[ 1 \cdot 1..\cdot 1]\frac{2}{n}\leq \frac{4}{n}$$
Since $0 \leq \frac{2^n}{n!} \leq \frac{4}{n}$ the sequence converges to $0$.
-
Six solutions have already been written but I have a different one:
Use the bound* $$\left(\frac n e\right)^n \le n! \le n^n$$
This shows that $$0 \le \frac {2^n} {n!} \le \left( \frac {2e} n \right)^n \longrightarrow 0$$
giving the result.
*Proof of bound: The upper bound (which I did not use) follows trivially from the definition of the factorial. The lower bound follows from a quick and dirty evaluation of the Gamma function integral,
$$n! = \Gamma(n+1) = \int_0^\infty e^{-t} t^n dt \geq \int_n^\infty e^{-t} t^n dt \geq \int_n^\infty e^{-t} n^n dt = \left( \frac n e \right)^n$$
This is a useful bound of the factorial to know - it's weaker than the Stirling approximation, but much, much easier to prove.
- | 2016-02-08T00:33:17 | {
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http://math.stackexchange.com/questions/96152/are-these-two-lines-the-same | # Are these two lines the same?
How can I say that the two lines are the same:
Line1=(5,3)+t(-4,8)
Line2=(5,3)+t(-8,16)
They both have the same starting point but can I say they have the same direction vector? It seems like to me they contain different points for a given "t".
-
It would be better to use different variables for the parameter in the two lines. If Line 2 were (5,3)+s(-8,16) one could follow Rasmus, come out with t=2s and it would be clearer. – Ross Millikan Jan 3 '12 at 19:07
The vector form of the equation you are using is $$\tag{1}{\bf r}(t) = \color{maroon}{\bf p} +t\color{darkgreen} {\bf d}$$ where $\color{maroon}{\bf p}$ gives the initial point on the line (the tip of $\color{maroon}{\bf p}$ is the initial point on the line corresponding to $t=0$) and $\color{darkgreen}{\bf d}$ is the direction vector.
To see why (1) actually gives a line: Note that multiplication of $\color{darkgreen}{\bf d}$ by $t$ just extends, shortens, or reflects $\color{darkgreen}{\bf d}$, it does not change its direction (in the diagram $\color{maroon}{t{\bf d}}$ is the dashed, maroon vector).
So, thinking of adding vectors "tip to tail", the tip of the vector $\color{orange}{{\bf p}+t{\bf d}}$ will always fall on a line.
Note that in (1), we could have used any other vector, such as the blue vector below, that is parallel to $\color{darkgreen}{\bf d}$ as the direction vector, and the corresponding equation would give the same line, just traced out in a different way.
If you think of the line as being generated by a moving point (the gray one above) whose position at time $t$ is ${\bf r}(t)$, then in your example, the point is moving twice as fast in your equation for line2 than as in the equation for line1.
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Yes, they are because $(-8,16)$ is a multiple of $(-4,8)$:
$$(-8,16)=2\cdot (-4,8).$$
If a point $(a,b)$ belongs to Line1, say $(a,b)=(5,3)+t(-4,8)$, then it also belongs to Line2 because $(a,b)=(5,3)+\dfrac{t}{2}(-8,18)$.
Similarly, if a point $(a,b)$ belongs to Line2, say $(a,b)=(5,3)+t(-8,16)$, then it also belongs to Line1 because $(a,b)=(5,3)+2t(-4,8)$.
The lenght of the "direction vector" is irrelevant, only its direction is important. This is because your are really only interested in the set of all scalar multiples of the direction vector.
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Yes, they are the same since the direction vectors are parallel (i.e. one is a scalar multiple of the other.) It is true that they contain different points for a given "t", but they still contain all the same points -- for instance, the point of line1 at a certain $t$ corresponds to the point at line2 at time $t/2$.
- | 2015-04-19T12:56:26 | {
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https://math.stackexchange.com/questions/1892037/is-there-a-name-for-the-sum-of-increasing-powers | # Is there a name for the sum of increasing powers?
I was thinking about the wheat and chessboard story and thinking of the total number of grains of wheat…
$$\sum_{n=0}^{63} 2^n$$
And wondered if there is a name for a sum like this?
$$2^0 + 2^1 +2^2 + 2^3 + \cdots$$
• Geometric? ${}{}$ – user251257 Aug 14 '16 at 14:35
• The first on is a finite geometric series. The second is an infinite geometric series. To call it "geometric" means there is a common ratio, i.e. the number you multiply each term by, to get the next term, is the same in every case. $\qquad$ – Michael Hardy Aug 14 '16 at 17:51
• If infinite, the second one is not convergent -- it diverges quite badly. @MichaelHardy – Clement C. Aug 14 '16 at 19:23
• Whether it diverges badly or well seems a matter of opinion: I'd go for the latter. It diverges really quite well indeed. – Dewi Morgan Aug 14 '16 at 19:59
• @Dewi, if adjectives are a matter of opinion, I'd pick "spectacularly" over the two other choices you mentioned. – J. M. is a poor mathematician Aug 14 '16 at 22:48
This is called a geometric sum/series according as $N$ is finite or not. More generally, it is of the form $$\sum_{n=0}^Nar^n$$ where $N$ may be "equal to" infinity. In the case where $N$ is infinite, if the series is to converge, we require $\lvert r\rvert<1$. Furthermore, for $N$ finite $$\sum_{n=0}^Nar^n=a\frac{1-r^N}{1-r},\:\:\:\: r\ne1$$ and $$\sum_{n=0}^\infty ar^n=\frac{a}{1-r},\:\:\:\:\lvert r\vert<1.$$
• This is wrong. For the finite case, the sum shall be $a\frac{1-r^{N+1}}{1-r}$ since there are $N+1$ terms. – Kun May 4 at 21:02
It's mainly a well known value $2^{n+1}-1$ . Write it in base $2$.
It is also OEIS sequence A000225 :
$2^n - 1.$ (Sometimes called Mersenne numbers, although that name is usually reserved for A001348.)
• This answers the specific case of $r=2$, but not the general case asked in the title. The sum of increasing powers is a geometric sum -- the specific name of Mersenne numbers is coincidental in this question, and giving the closed form without the explanation (first part of this answer) sheds basically no light whatsoever on the why. – Clement C. Aug 14 '16 at 15:08
• The OP asked if there "was a name for a sum like that." (Emphasis mine) Your answer gives a value without explanation, and a name for this exact sum which is not generalizable to any other ratio. As such, I feel it fails to answer the question. – Clement C. Aug 14 '16 at 16:19
• Indeed, it's my opinion - this is why I wrote a comment to explain it. – Clement C. Aug 14 '16 at 16:33
As you are looking for novel names, this is also a simple case of the more generic notion of hypergeometric series $$\sum_{k}r_k \,,$$ with $r_0 = 1$, and the ratio of two consecutive terms is a rational function, a ratio of two polynomials $P$ and $Q$ in the summation index $k$
$$\frac{r_{k+1}}{r_k}= \frac{P(k)}{Q(k)}\,.$$
In your case, you can choose $P$ and $Q$ such that their ratio is equal to $2$. When the ratio is constant, it is called a geometric series (as answered here). As a reminder, it is a sum of terms in geometric progression (se.math) like $1,r,r^2,r^3,\ldots$, whose name (the geometry part) is illustrated by the following figure:
Hypergeometric series are also connected to chess. A rook is a move on a chessboard. Some have developed studies some types of permutations as the placement of a number of rooks on a chessboard-like grid, see for instance Rook theory and hypergeometric series, J. Haglund, 1996.
The formula for the first $N$ terms can be derived as following:
Let:
$$S(N)=\sum_{n=0}^{N} r^n$$
Then distributing an $r$ (which is independent of $n$) along the sum we have,
$$rS(N)=\sum_{n=0}^{N} r^{n+1}$$
Note (1):
$$S(N+1)=\sum_{n=0}^{N+1} r^n=r^0+\sum_{n=0}^{N} r^{n+1}$$ $$=1+rS(N)$$
Also note,
$$S(N+1)-S(N)=\sum_{n=0}^{N+1} r^n-\sum_{n=0}^{N} r^n=r^N$$
As all terms but $r^N$ will reduce each other out to zero.
So,
$$S(N+1)=S(N)+r^N$$
Hence we have from (1) and from the above:
$$S(N)+r^N=rS(N)+1$$
$$r^N-1=rS(N)-S(N)$$
$$r^N-1=(r-1)S(N)$$
Thus,
$$S(N)=\frac{r^N-1}{r-1}$$
Now we multiply by a special form of $1$ to get:
$$S(N)=\sum_{n=0}^{N} r^n=\left( \frac{-1}{-1} \right) \left( \frac{r^N-1}{r-1} \right)=\frac{1-r^N}{1-r}$$
And hence (distributing out an $a$ which is independent of $n$) we get the formula for the geometric sum,
$$\sum_{n=0}^{N} ar^n=a\frac{1-r^N}{1-r}$$
Note though our formula only works for $r \neq 1$ (our manipulations are okay until we get to dividing by $0=1-r=1-1$ for $r=1$). | 2019-05-22T14:39:27 | {
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https://math.stackexchange.com/questions/2390471/number-of-ways-to-arrange-5-monkeys-in-a-row/2390586 | # Number of ways to arrange 5 monkeys in a row?
We have 5 monkeys $a,b,c,d,e$ and we are interested in the number of ways to have them stand in a row without $a$ and $b$ being next to each other.
The part that I struggle with most is that I don't fully understand how to solve this when the 5 are different. It's not the same as for example coloring 5 segments either blue or red without any two neighboring segments being red.
This is how I tried to solve this but I'm certain that there's something wrong. I would really appreciate it if you could also critique my approach.
Idea:
Let $f_{k}$ be the number of ways we can have the $5$ monkeys in a row without $a$ and $b$ being next to each other. We try to do this recursively:
case 1 : the last monkey is not $a$ or $b$: then we have $f_{k-1}$ possibilities for the rest of the k-1 monkeys.
case 2 : the last monkey is either $a$ or $b$: Here the second to last has to be one of $\{c,d,e\}$. So we have $3$ possibilities for the second to last spot and $2$ possibilities for the last. A total of $2*3 = 6$ and $f_{k-2}$ for the remaining spots.
The recursive equation I get is: $f_{k} = 6 + f_{k-1} + f_{k-2}$
$f_{1} = 5$
$f_{2} = 10$
$f_{3} = 21$
$f_{4} = 37$
$f_{5} = 64$
I'm not sure about my solution.
• Hint: Count the total number of ways to arrange the monkeys, then count the number of ways to arrange where $a$ and $b$ are adjacent, and subtract from the total. – platty Aug 11 '17 at 17:15
• The problem with your current approach is that in the recursive step, the number of possible choices for case 2 decreases (fewer possible monkeys to choose from). Furthermore, the subproblem is not the same - now you only have one monkey to worry about, and you can count this case directly. – platty Aug 11 '17 at 17:18
• it seems that you want to use probabilities (possibilities of what exactly?) to count the valid permutations, but I think its easier to count the invalid permutations and subtract them from the total number of possible permutations. – Masacroso Aug 11 '17 at 17:20
• See no evil, hear no evil, speak no evil, smell no evil, taste no evil? – David Richerby Aug 12 '17 at 12:48
• @DavidRicherby what do you mean? – DariusTheGreat Aug 12 '17 at 13:08
Here's how I'd approach this particular problem. I'll solve for $k$ monkeys afterwards.
You have $5!$ ways for the monkeys to be arranged in a line without restriction.
There are $8$ ways that $A$ and $B$ can be positioned next to each other; there are $4$ pairs of adjacent spaces, and either $A$ or $B$ can be on the left.
For each of these cases, there are $3! = 6$ ways to arrange the other three monkeys.
So the answer is $5! - 8 \cdot 3! = 72$ ways.
Now, just apply to $k$ monkeys using the same argument:
$$P(k) = k! - 2(k-1)(k-2)! = k! - 2(k-1)!.$$
(Hat tip to user471297 for the last simplification.)
• Equivalently: Treat the A,B pair as a unit and come up with 4!, then double it to account for swapping then. – Kevin Aug 12 '17 at 2:44
• Good answer! You can also write the result as $P(k)=(k-1)!(k-2)$. It's a matter of taste, but this form gives a way to see the asymptotic behaviour better and maybe calculate a little faster if factorials are expensive. – Joonas Ilmavirta Aug 12 '17 at 13:11
Recursive solution (the complementary counting solution is outlined in John's answer):
Let $f(k)$ be the number of ways to arrange $k$ monkeys, including $a$ and $b$ such that these two aren't next to each other.
Case 1: $a$ or $b$ is at the beginning of the line.
Counting this case directly, we first choose the leading monkey in one of $2$ ways. Then we find that the other of these two monkeys is in one of $k-2$ positions (any spot except for the one occupied by the first monkey, and the one immediately behind it). The other $k-2$ monkeys can be in any order, so we get $2 (k-2) (k-2)!$ ways.
Case 2: Neither are at the beginning of the line.
There are a total of $k-2$ choices for the monkey to lead the line; after that, we have the $k-1$ subproblem, so we find $(k-2)f(k-1)$ ways here.
Combining these, we have a total of $f(k) = (k-2)(2(k-2)! + f(k-1))$ good arrangements. Starting with $f(2) = 0$, an inductive argument should show that this matches the closed form answer.
• Thanks for showing how to incorporate the $k-1$ case directly. That eluded me. – John Aug 11 '17 at 18:34
@John and @platty have both supplied good answers. Here is another approach.
$a$ is at an end of the row: Since $a$ can be at the left or right end of the row, there are two ways to place $a$. For each such choice, there are three ways to place $b$ so that $b$ is not adjacent to $a$. The remaining three monkeys can be arranged in the three remaining positions in $3!$ ways. Hence, there are $$2 \cdot 3 \cdot 3!$$ arrangements in which $a$ is at an end of the row.
$a$ is not at the end of the row: Since there are five positions including the two ends of the row, there are three choices for the position of $a$. Since $b$ cannot be adjacent to $a$, there are two ways to place $b$. The remaining three monkeys can be arranged in the three remaining positions in $3!$ ways. Hence, there are $$3 \cdot 2 \cdot 3!$$ arrangements in which $a$ is not at an end of the row.
Total: Since the two cases are mutually exclusive and exhaustive, the five monkeys can be arranged in $$2 \cdot 3 \cdot 3! + 3 \cdot 2 \cdot 3! = 72$$ ways if $a$ and $b$ are not in adjacent positions.
I came up with the answer a different, non-recursive, way than some of the answers here (I understand the OP wanted a critique of their approach, but I figured that a different method could still add value). Anyway, my method:
As John said, there are $5!$ ways to arrange the monkeys without restriction. From there I treated monkeys $A$ and $B$ as one monkey and found the number of ways the four monkeys could be arranged: $4!$
Since there are two ways to arrange monkeys $A$ and $B$ together, you have $2 * 4!$ ways $A$ and $B$ could be put together. Subtracting from the original unrestricted $5!$ yields $120 - 2 * 24 = 72$.
Here is a simple solution :
First arrange 5 monkeys in 5! = 120;
remove all the cases where both a and b sit together.. so to get that tie two monkeys as one item: so you have 4 monkeys now --- how many ways you can arrange 4monkeys: 4! and also A and B sit as AB and BA so finally
we have 2 * 4! ways
so: answer is 5! -2 4! so in general : k! - 2 (k-1)!
Now just go to the previous answers k! - 2 * (k-1)(k-2)! is also actually same as k! - 2* (k-1)!
• Hah! I didn't see that the answer could be simplified (per your last line). – John Aug 11 '17 at 22:44
A way to find the result without subtracting one result from the general one:
Arrange the three other monkeys, $c, d, e$ in $3!$ or $6$ distinct ways.
There is a slot for $a$ or for $b$ in front of the first of $c, d, e$ already placed, and after each of $c, d, e$ already placed, for a total of 4 slots, each possible holding one of the two remaining letters, either $a$ or $b$. Adjacent placement of $a$ and $b$ is thus impossible
Select two of these four slots in $4 \times 3$ or $12$ ways. Put $a$ in the first slot selected and $b$ in the second.
Collapse the two empty slots as unneeded, and you're done with $6 \times 12$ or $72$ ways...
For your approach, you should get that placing $2$ monkeys in conformity with the conditions has zero options. So there is something odd going on there. Your base case would probably better be taken from the $2$ obvious possibilities for three monkeys.
Using a masking approach for variety, let us suppose that we have determined which of the locations will be occupied by $a$ and $b$. That location mask will give rise to $2$ ways of placing $a,b$ and, independently, $(k-2)!$ ways of placing the other monkeys.
Valid masks can be produced in $(2(k-2)+(k-2)(k-3))/2$ ways, separating the initial selection of the $2$ end positions from the $k{-}2$ mid positions, and then reducing the double count.
This gives the total possibilities as
\begin{align} 2(k-2)!\cdot(2(k-2)+(k-2)(k-3))/2 &= (k-2)!(k-2)(2+k-3)\\ &=(k-2)!(k-2)(k-1)\\ &=(k-1)!(k-2) \end{align}
For $k=5$, this gives $24\cdot 3 = 72$ options.
When $a$ should immediately precede $b$, then they together count as a single monkey, so you get $4!$. The other way round ($b$ then $a$), you get another $4!$. Without this condition you get $5!$. You don't want the condition, so subtract:
$$5! - 2\cdot4!$$
+---+---+---+
| X | X | X |
+--- --- ---+ For this illustration, Monkeys **A** & **E** are the restricted monkeys
↑ ↑ ↑ ↑ (Maybe they have beef ¯\\_(ツ)_/¯)
Above is an illustration with the "arranging people in seats" style. Say monkeys B,C & D can sit in any order about themselves they like. They have 3!* ways(=6).
Now, monkey A has 4 possible choices(denoted by the arrows above) since he has no restrictions with them. So a total of (6x4=24) arrangements so far.
Monkey E will now be arranged into ANY THREE OF THE REMAINING ARROWS that were not chosen by monkey A, as any of these ensure they are not next to each other. This brings the final tally to (6x4x3=72) arrangements.
NOTE: This approach extends to if there are other monkeys with similar restrictions i.e. the next would have 2 choices, and the next 1, and it would be impossible for another after that to not be next to a restricted monkey.
• If anyone is confused about "3!", it simply means 3 choices for 1, then the next has 2, and the last has 1 => thus 3x2x1 = 6
I hope this helps; it's my first MathStackExchange post. I like this method & thought I should share. Thx | 2019-12-10T11:28:24 | {
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https://cs.stackexchange.com/questions/109607/time-complexity-of-an-algorithm-is-it-important-to-state-the-base-of-the-logari/109613 | # Time complexity of an algorithm: Is it important to state the base of the logarithm?
Since there is only a constant between bases of logarithms, isn't it just alright to write $$f(n) = \Omega(\log{n})$$, as opposed to $$\Omega(\log_2{n})$$, or whatever the base might be?
It depends where the logarithm is. If it is just a factor, then it doesn't make a difference, because big-O or $$\theta$$ allows you to multiply by any constant.
If you take $$O(2^{\log n})$$ then the base is important. In base 2 you would have just $$O(n)$$, in base 10 it's about $$O(n^{0.3010})$$.
• I guess this is only going to come up with something like $2^{\sqrt{\log n}}$. I can't see any reason for expressing a number as $2^{c\log_b n}$ rather than $n$-to-the-whatever-it-is (except perhaps as an intermediate stage of a calculation). May 21, 2019 at 19:45
• +1 for "constant factors matter in exponents" May 21, 2019 at 22:57
Because asymptotic notation is oblivious of constant factors, and any two logarithms differ by a constant factor, the base makes no difference: $$\log_a n = \Theta(\log_b n)$$ for all $$a,b > 1$$. So there is no need to specify the base of a logarithm when using asymptotic notation.
• I prefer to see $\in$ instead of $=$ May 21, 2019 at 22:04
• I'm afraid the standard notation uses $=$. May 21, 2019 at 22:05
• @YuvalFilmus The standard notation is misleading, completely different to the standard everywhere else and makes algorithmic complexity seem completely alien from things quite similar to it. "It's the standard notation" should never be a reason to favour a bad solution over a better, similarly-clear one. (The meaning of the symbol is usually clear from context, anyway.) May 22, 2019 at 18:16
• @wizzwizz4 Common practice is an excellent reason. It promotes efficient communication. That’s the reason we all put up with the quirks of English spelling. May 22, 2019 at 20:04
• Sometimes $n \mapsto \log_a n \in \Theta ( n \mapsto \log_b n)$ just has too much stuff there to be clearer than $\log_a n = \Theta (\log_b n)$.
– JiK
May 23, 2019 at 10:48
As $$\log_xy = \frac{1}{\log_y{x}}$$ and $$\log_x{y} = \frac{\log_z{y}}{\log_z{x}}$$, so $$\frac{\log_a{n}}{\log_b{n}} = \frac{\log_n{b}}{\log_n{a}} = \log_a{b}$$. As $$\log_a{b}$$ is positive constant (for all $$a,b > 1$$), so $$\log_a{n} = \Theta(\log_b{n})$$.
In most cases, it's safe to drop the base of the logarithm because, as other answers have pointed out, the change-of-basis formula for logarithms means that all logarithms are constant multiples of one another.
There are some cases where this isn't safe to do. For example, @gnasher729 has pointed out that if you have a logarithm in an exponent, then the logarithmic base is indeed significant.
I wanted to point out another case where the base of the logarithm is significant, and that's cases where the base of the logarithm depends directly on a parameter specified as input to the problem. For example, the radix sort algorithm works by writing out numbers in some base $$b$$, decomposing the input numbers into their base-$$b$$ digits, then using counting sort to sort those numbers one digit at a time. The work done per round is then $$\Theta(n + b)$$ and there are roughly $$\log_b U$$ rounds (where $$U$$ is the maximum input integer), so the total runtime is $$O((n + b) \log_b U)$$. For any fixed integer $$b$$ this simplifies to $$O(n \log U)$$. However, what happens if $$b$$ isn't a constant? A clever technique is to pick $$b = n$$, in which case the runtime simplifies to $$O(n + \log_n U)$$. Since $$\log_n U$$ = $$\frac{\log U}{\log n}$$, the overall expression simplifies to $$O(\frac{n \log U}{\log n})$$. Notice that, in this case, the base of the logarithm is indeed significant because it isn't a constant with respect to the input size. There are other algorithms that have similar runtimes (an old analysis of disjoint-set forests ended up with a term of $$\log_{m/2 + 2}$$ somewhere, for example), in which case dropping the log base would interfere with the runtime analysis.
Another case in which the log base matters is one in which there's some externally-tunable parameter to the algorithm that control the logarithmic base. A great example of this is the B-tree, which requires some external parameter $$b$$. The height of a B-tree of order $$b$$ is $$\Theta(\log_b n)$$, where the base of the logarithm is significant in that $$b$$ is not a constant.
To summarize, in the case where you have a logarithm with a constant base, you can usually (subject to exceptions like what @gnasher729 has pointed out) drop the base of the logarithm. But when the base of the logarithm depends on some parameter to the algorithm, it's usually not safe to do so.
Let $$f_k(n)=\log_kn$$,$$k>2$$, and let $$t(n)=\log_2n$$.
You can rewrite $$f_k(n)$$ as follow $$f_k(n)=\frac{\log_2n}{\log_2k}$$ Now $$k$$ have two states:
1. $$k$$ is constant:
$$\lim_{n\to \infty}\frac{f_k(n)}{t(n)}=\lim_{n\to \infty}\frac{\frac{\log_2n}{\log_2k}}{\log_2n}=c$$ that $$c$$ is a constant$$>0$$, hence $$t(n)=\Theta(f_k(n)).$$ So there is no difference between $$f_k(n)$$ , and $$t(n)$$ in terms of asymptotic notation.
1. $$k$$ isn't constant:
$$\lim_{n\to \infty}\frac{f_k(n)}{t(n)}=\lim_{n\to \infty}\frac{\frac{\log_2n}{\log_2k}}{\log_2n}=0$$
So $$t(n)$$ have faster growth rate and $$t(n)=\omega(f_k(n)).$$ | 2022-07-01T07:45:15 | {
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http://nabdon.com/p47u7d/check-if-graph-is-connected-adjacency-matrix-dd24f9 | Input − Adjacency matrix of a graph. Start DFS at the vertex which was chosen at step 2. // Implementation of 0 ⋮ Vote. But please check yourself as well. After completing the traversal, if there is any node, which is not visited, then the graph is not connected. Adjacency Matrix of an undirected graph Adjacency Matrix of a directed graph Adjacency List and matrix of directed graph An adjacency matrix has a more organized appearance to its structure, but this implementation wastes memory if not all of the vertices are connected. If the graph is undirected, the adjacency matrix is symmetric. For a undirected graph it is easy to check that if the graph is connected or not. The idea is also simple - imagine an n by n grid, where each row and each column represents a vertex. The adjacency matrix and adjacency list are raw'' forms of graph and are not oriented towards solving any particular problem. Introduction to Graphs 9:32. Undirected graph with no loops and no multi-edges. For finding paths of length r between vertices v(i) & v(j), we find A^r and the (i,j)th entry of this matrix would give you the number of paths. 0. Can we improve further? Time Complexity: Time complexity of above implementation is sane as Depth First Search which is O(V+E) if the graph is represented using adjacency list representation. In this case the traversal algorithm is recursive DFS traversal. I already have the methods to check for self-loops and cycles, I need a method to check SPECIFICALLY for connectivity in the adjacency matrix to prove it is a DAG. Vote. The above approach requires two traversals of graph. An adjacency matrix is a square matrix used to represent a finite graph. We define an undirected graph API and consider the adjacency-matrix and adjacency-lists representations. Not sure how to check if there are connected edges or how to remove, only know how to add edges. If the smallest eigenvalue is strictly bigger then zero or the same as if zero is not an eigenvelue then it is connected. Vote. I'm doing a project on topological robotics, and part of the program involves taking in the adjacency matrix of a graph, then testing to see if it's connected or not. Follow 24 views (last 30 days) chandra Naik on 24 Jul 2019. How to check graph connected or not for given adjacency matrix of graph. For the undirected graph, we will select one node and traverse from it. We also consider the problem of computing connected components and conclude with related problems and applications. Show Hide all comments. Look at the graph laplacian D-A where D is the diagonal matrix with corresponding degrees of vertices on the diagonal. I know that the time required to check if there exists an edge between two nodes in an adjacency matrix is $O(1)$ because we can access it directly (i.e: $M[i][j]$). Answers (1) Matt J on 24 Jul 2019. Yes Is there an edge from 1 to 3? For example, we need to check if an adjacency matrix such as this one is fully connected: The graph is (n+2)*(n+2), and the number of functions here is 4. An undirected graph is graph, i.e., a set of objects (called vertices or nodes) that are connected together, where all the edges are bidirectional. At the ith row and jth column, we store the edge weight of an edge from the vertex i to vertex j. If any vertex v has vis1[v] = false and vis2[v] = false then the graph is not connected. This might not be very efficient especially in the case where we have a lot of vertices and only a few of those vertices are connected to each other, which translates to a very sparse adjacency matrix. Textbook solution for Linear Algebra: A Modern Introduction 4th Edition David Poole Chapter 4.6 Problem 36EQ. Here's what you'd learn in this lesson: Bianca analyzes the adjacency matrix format of representing node relationships in a graph, using binary values in the array. To check connectivity of a graph, we will try to traverse all nodes using any traversal algorithm. We can find whether a graph is strongly connected or not in one traversal using Tarjan’s Algorithm to find Strongly Connected … The "Adjacency Matrix" Lesson is part of the full, Tree and Graph Data Structures course featured in this preview video. Output: Yes No. I don't want to keep any global variable and want my method to return true id node are connected using recursive program The trouble is, I've tested it with several disconnected and connected graphs, and it says that they're all disconnected, no matter what I put in! No The problem is that we always need to use O(n^2) elements for storage, and hence, we often use adjacency lists to represent graphs. I realize this is an old question, but since it's still getting visits, I have a small addition. Sign in to answer this question. Start at a random vertex v of the graph G, and run a DFS(G, v). Make all visited vertices v as vis1[v] = true. Time Complexity: DFS: O(m * n) where m is the number of rows in our grid and n is the number of columns in our grid. Edited: Matt J on 24 Jul 2019 How to check given undirected graph connected or not 0 Comments. The advantage of the adjacency matrix is that it is simple, and for small graphs it is easy to see which nodes are connected to other nodes. Graph. In the special case of a finite simple graph, the adjacency matrix is a (0,1)-matrix with zeros on its diagonal. The diagram below illustrates the adjacency matrix for the example graph we presented earlier. Modified if-statement for graph traversal to also check if a cell is a wall. Otherwise Its degree (degree of zero as a root of characteristic polynomial) is "the number of connected components"-1. As of R2015b, the new graph and digraph classes have a method for computing connected components. Sign in to comment. To check that a graph is connected or not. In this tutorial we shall see how to store a graph with the help of a matrix. Now reverse the direction of all the edges. How to check graph connected or not for given adjacency matrix of graph. We can simply do a depth-first traversal or a breadth first-first traversal on the graph and if the traversal successfully traversal all the nodes in the graph then we can conclude that the graph is connected else the graph has components. For the undirected graph, we will select one node and traverse from it. The forms of problems that one must solve are typically: process/print the nodes, e.g., check if the graph is connected--- for every node, one can go to all other nodes In graph theory and computer science, an adjacency matrix is a square matrix used to represent a finite graph.The elements of the matrix indicate whether pairs of vertices are adjacent or not in the graph.. Algorithm that checks if a directed graph represented by an adjacency matrix is strongly connected( there is a path connecting all vertices) .. Algorithm preferably in fortran. To represent this graph as the adjacency matrix A, we’ll let the indices of the rows and columns represent nodes, or vertices. To check connectivity of a graph, we will try to traverse all nodes using any traversal algorithm. In this node 1 is connected to node 3 ( because there is a path from 1 to 2 and 2 to 3 hence 1-3 is connected ) I have written programs which is using DFS, but i am unable to figure out why is is giving wrong result. We have step-by-step solutions for your textbooks written by Bartleby experts! The idea is to take 2D array of size V * V, then mark the index as “1” if there exist an edge between those 2 vertices. Input − Adjacency matrix of a graph. Hence, the given graph is connected. If such edge doesn’t exist, we store zero. A value in a cell represents the weight of the edge from vertex v v v to vertex w w w. An adjacency matrix representation for a graph . A tree is a connected undirected graph without cycles. Is there an edge from 1 to 2? Yes Is there an edge from 4 to 3? In contrast, a graph where the edges point in a direction is called a directed graph. As mentioned in this article, adjacency matrix requires more memory if implemented in a program due to its requirement to store the graph information in the form of an $$N \times N$$ matrix. This is a graph implementation, using adjacency matrix on Python. Graph implementation. For the given matrix, I believe that the summation of all paths gives us a matrix with all non-zero entries. In this case the traversal algorithm is recursive BFS traversal. Created a Graph Structure with 4 vertices Initialized Adjacency Matrix! 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Not for given adjacency matrix is symmetric degrees of vertices on the diagonal matrix with corresponding degrees vertices. Modern Introduction 4th Edition David Poole Chapter 4.6 problem 36EQ as of,! Idea is also simple - imagine an n by n grid, where row... I to vertex J problems and applications strictly bigger then zero or the same as if zero is not,. From 4 to 3 consider the adjacency-matrix and adjacency-lists representations a small....
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https://math.stackexchange.com/questions/3149835/evaluate-displaystyle-sum-k-1n-left-fracn-1-choose-k-1k-right | # Evaluate $\displaystyle\sum_{k=1}^n \left(\frac{{n-1 \choose k-1}}{k}\right)$
Evaluate : $$\displaystyle\sum_{k=1}^n \left(\frac{{n-1 \choose k-1}}{k}\right)$$
Alright so I'm completely stumped, I've never evaluated a summation of $$\displaystyle{n \choose k}$$.
My best guess is to use the binomial theorem, but I don't know how to change this into a form I could use the theorem on.
• yes, sorry my formatting is off – Brownie Mar 15 at 22:07
Hint:
Use the recurrence relation $$\binom nk=\frac nk\binom{n-1}{k-1}$$ and remember that $$\displaystyle\sum_{k=0}^n\binom nk=\cdots$$
• Alright, so does it make sense if I multiplied $\displaystyle\sum_{k=1}^n \left(\frac{{n-1 \choose k-1}}{k}\right)$ by n, and then took that summation, and then divided by n at the end? – Brownie Mar 15 at 22:12
• It would be perfect! – Bernard Mar 15 at 22:13
• So dividing by n after taking the summation is correct? Or would I have to divide by the summation of n since I already took the summation of $\displaystyle\sum_{k=1}^n \left(\frac{n{n-1 \choose k-1}}{k}\right)$ – Brownie Mar 15 at 22:14
• It's correct since multiplication (and division) is distributive w.r.t. addition. – Bernard Mar 15 at 22:16
• Thank you that clears it up! – Brownie Mar 15 at 22:16
Another alternative is to somehow see the function $$\frac{x^k}{k}$$ in the series and use a bit of calculus:
$$\frac{d}{dx}\displaystyle\sum_{k=1}^n \binom{n-1}{k-1} \frac{x^k}{k}=\displaystyle\sum_{k=1}^n \frac{d}{dx} \binom{n-1}{k-1} \frac{x^k}{k}=\displaystyle\sum_{k=1}^n \binom{n-1}{k-1}x^{k-1}=(1+x)^{n-1}$$ by the Binomial Theorem, so $$\displaystyle\sum_{k=1}^n \binom{n-1}{k-1} \frac{x^k}{k}=\int (1+x)^{n-1} dx=\frac{(1+x)^n}{n}+\mathcal{C}.$$ Plugging in $$x=0$$ tells us that $$\mathcal{C}=-\frac{1}{n}$$, so we have the more general result $$\displaystyle\sum_{k=1}^n \binom{n-1}{k-1} \frac{x^k}{k}=\frac{(1+x)^n-1}{n}.$$ In particular, taking $$x=1$$ gives the desired result.
• nice and clean approach – G Cab Mar 15 at 22:38
The hint given is already good.
In alternative you can proceed as follows \eqalign{ & \sum\limits_{k = 1}^n {{1 \over k}\left( \matrix{ n - 1 \cr k - 1 \cr} \right)} \quad \left| {\;1 \le n} \right.\quad = \cr & = \sum\limits_{k = 0}^{n - 1} {{1 \over {\left( {k + 1} \right)}}\left( \matrix{ n - 1 \cr k \cr} \right)} = {1 \over n}\sum\limits_{k = 0}^{n - 1} {{n \over {\left( {k + 1} \right)}}\left( \matrix{ n - 1 \cr k \cr} \right)} = \cr & = {1 \over n}\sum\limits_{k = 0}^{n - 1} {\left( \matrix{ n \cr k + 1 \cr} \right)} = {1 \over n}\sum\limits_{k = 1}^n {\left( \matrix{ n \cr k \cr} \right)} = {1 \over n}\left( {\sum\limits_{k = 0}^n {\left( \matrix{ n \cr k \cr} \right)} - 1} \right) = \cr & = {1 \over n}\left( {2^{\,n} - 1} \right) \cr}
• So you solution brings up a few question for me. Going off of the hint above, the way I solved it was multiplying the summation by n turning it into $\displaystyle\sum_{k=0}^n\binom nk$. I took that summation giving me $2^{n}$, and then divided by n to undo my previous multiplication giving me ${2^{n} \over n}$ which is different from you answer. Where did i go wrong? – Brownie Mar 15 at 22:29
• Edit - I see i took the summation at the wrong index – Brownie Mar 15 at 22:35
• @Brownie: you did not take into proper consideration the summation bounds. – G Cab Mar 15 at 22:37 | 2019-10-17T00:28:12 | {
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https://www.physicsforums.com/threads/probability-choosing-a-girl-from-a-group.581872/ | # Probability: Choosing a girl from a group
1. Feb 27, 2012
### Xyius
1. The problem statement, all variables and given/known data
You walk into your class the first day of classes, and you notice that
there are 30 men and 20 women in the class already. Let's suppose you decide to choose
two people from the class to be your study partners.
If you choose your study partners at random, and given that at least one of your
study partners is a woman, what is the probability of the event E that both of them
will be women?
A. 0.3167
B. 1.9%
C. 0.2405
D. 0.1901
2. Relevant equations
In my Solution
3. The attempt at a solution
This seems like a simple problem but I cannot seem to get the numbers available as choices.
My logic is is, W represents the event that you have picked a woman, and E represents that both of your partners will be women then.
$$P(E|W)=\frac{P(E \cap W)}{P(W)}$$
So the numerator can simplify to..
$$P(E|W)=\frac{P(E)}{P(W)}$$
This is because if E occurs, then W must have occured.
So..
$$P(E)=\frac{\binom{20}{2}}{\binom{50}{2}}$$
and
$$P(W)=\frac{\binom{20}{1}}{\binom{50}{2}}$$
But this doesn't work because the ratio of these two (From the formula) gives a number larger than one. Where am I going wrong? Do I use Bayes theorem?
2. Feb 27, 2012
### Dick
If P(W) is the probability of choosing at least one woman, there are two ways to do that, you could choose 1 woman and 1 man, or 2 women.
3. Feb 27, 2012
### Xyius
Ohh! So it would be..
$$P(W)=P(W|W)P(W)+P(W|M)P(M)$$
??
I don't have time to crunch through the numbers at the moment, but I will be sure to check this out later.
4. Feb 27, 2012
### Dick
I wouldn't write it that way. Just count out the cases using combinations like you are already doing. How may ways to do each?
5. Feb 27, 2012
### Xyius
So choosing 1 woman and 1 man would be
$$\frac{\binom{20}{1}\binom{30}{1}}{\binom{50}{2}}$$
And choosing 2 women would be..
$$\frac{\binom{20}{2}}{\binom{50}{2}}$$
Plugging this in gives me 0.2405, or answer C! Thanks! :D
EDIT: Assuming that's the correct answer.....
6. Feb 27, 2012
### Dick
Well, that's what I get.
7. Feb 27, 2012
### Ray Vickson
I W = number of women you choose, you have been asked to find the conditional probability $P\{W=2|W\geq 1\}.$ We have $$P\{W=2|W\geq 1\} = \frac{P\{W = 2 \cap W \geq 1 \}}{P\{W \geq 1\}} = \frac{P\{W = 2\}}{1-P\{W=0\}}.$$ The numerator and denominator are easily comutable using the hypergeometric distribution. The numerator is ${20 \choose 2}/{50 \choose 2},$ while the denominator is $1 - {30 \choose 2}/{50 \choose 2}.$
RGV | 2017-11-24T19:04:32 | {
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https://math.stackexchange.com/questions/2956889/how-many-number-of-integer-coordinates-exists-between-a-line-segment-including | # How many number of integer coordinates exists between a line segment, including the end points?
There is a line segment say $$AB$$ with coordinates of end-points as $$A=(x_1, y_1)$$ and $$B=(x_2, y_2)$$. $$x_1, y_1, x_2, y_2$$ are integers. I need to find the number of integer coordinates which lie on the line segment including end-points.
I read somewhere that it is $$\gcd(|x_1 - x_2|, |y_1 - y_2|) + 1$$. But, I cannot understand why this works.
I do not get the intuition behind it. I searched for proof but did not find anything intuitive and straightforward.
Please help me understand this. I am stuck on it. I am expecting a nice proof with great explanation.
• Maybe this will make it easier for you. Without loss of generality, we can assume that $A=(0, 0)$, and $B=(x, y)$ with $x, y\geq 0$. ($B\neq A$) – Jakobian Oct 15 '18 at 18:35
Using Jakobian's simplification, the problem looks as follows: For any $$x_0, y_0 > 0$$, how many integer coordinates does the line from $$(0, 0)$$ to $$(x_0, y_0)$$ pass? We note first, that the slope of this line is $$y_0/x_0$$. So the problem is equivalent to asking: for how many integers $$x$$ with $$0 \leq x \leq x_0$$ is $$y_0 / x_0 * x$$ an integer. Now let $$d = gcd(x_0, y_0)$$. We get $$y_0 / x_0 * x = (d * \hat{y_0})/(d * \hat{x_0}) * x = \hat{y_0}/\hat{x_0} * x$$ for some $$\hat{y_0},\hat{x_0}$$ with $$gcd(\hat{y_0},\hat{x_0}) = 1$$. Now $$\hat{y_0}/\hat{x_0} * x$$ is an integer if and only if x is a multiple of $$\hat{x_0}$$. So how many multiples of $$\hat{x_0}$$ are there between $$0$$ and $$x_0$$. It is exactly $$d + 1$$, because $$x_0 = d * \hat{x_0}$$. Hope this helps. | 2019-08-24T13:49:14 | {
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https://math.stackexchange.com/questions/1600668/show-that-left-int-01-sqrtfx2gx2-dx-right2-geq-left-int | # Show that $\left(\int_{0}^{1}\sqrt{f(x)^2+g(x)^2}\ dx\right)^2 \geq \left(\int_{0}^{1} f(x)\ dx\right)^2 + \left(\int_{0}^{1} g(x)\ dx\right)^2$ [closed]
Show that $$\left( \int_{0}^{1} \sqrt{f(x)^2+g(x)^2}\ \text{d}x \right)^2 \geq \left( \int_{0}^{1} f(x)\ \text{d}x\right)^2 + \left( \int_{0}^{1} g(x)\ \text{d}x \right)^2$$ where $f$ and $g$ are integrable functions on $\mathbb{R}$.
That inequality is a particular case. I want to approximate the integral curves using some inequalities who imply this inequality.
• Please provide some context – robjohn Jan 5 '16 at 10:38
• This inequality is a particular case.I want to use this inequality to aproximate the integral curves. – alexb Jan 5 '16 at 10:50
• clarifications and improvements should be applied to the question, not left in comments – robjohn Jan 5 '16 at 11:05
• I think there other useful answers to this question and this inequality is useful by itself. I would like to see it reopened. – robjohn Jan 6 '16 at 16:00
• Do you mean that you want to use inequalities implied by this inequality? – robjohn Jan 6 '16 at 16:48
Let a curve $C\in\mathbb{R}^2$ be defined by the parametrisation $\displaystyle x(t)=x(0)+\int_0^t f(x)dx$ and $\displaystyle y(t)=y(0)+\int_0^t g(x)dx$, $t\in [0, 1]$. Then the LHS is the square of the arc length of $C$ joining $(x(0), y(0))$ and $(x(1), y(1))$, whereas the RHS is the square of the shortest distance between $(x(0), y(0))$ and $(x(1), y(1))$.
• But if I have n functions I can to proof the inequality în the same way? – alexb Jan 5 '16 at 13:00
• Yes. You can generalise to an $n$-dimensional curve. – Alex Fok Jan 5 '16 at 13:02
Definition: $u:\mathbb{R}^n\to\mathbb{R}$ is convex if for all $a\in\mathbb{R}^n$, there is a $v(a)\in\mathbb{R}^n$ so that for all $x\in\mathbb{R}^n$ $$u(x)-u(a)\ge v(a)\cdot(x-a)\tag{1}$$ Theorem (Extension of Jensen): If $u:\mathbb{R}^n\to\mathbb{R}$ is convex, $f:\Omega\to\mathbb{R}^n$, and $\int_\Omega\mathrm{d}\omega=1$, then $$\int_\Omega u(f)\,\mathrm{d}\omega\ge u\left(\int_\Omega f\,\mathrm{d}\omega\right)\tag{2}$$ Proof: Let $a=\int_\Omega f\,\mathrm{d}\omega$. Then $(1)$ becomes $$u(f)-u\left(\int_\Omega f\,\mathrm{d}\omega\right) \ge v\left(\int_\Omega f\,\mathrm{d}\omega\right) \cdot\left(f-\int_\Omega f\,\mathrm{d}\omega\right)\tag{3}\\$$ Since $\int_\Omega\mathrm{d}\omega=1$, integrating $(3)$ over $\Omega$ gives \begin{align} \int_\Omega u(f)\,\mathrm{d}\omega-u\left(\int_\Omega f\,\mathrm{d}\omega\right) &\ge v\left(\int_\Omega f\,\mathrm{d}\omega\right) \cdot\left(\int_\Omega f\,\mathrm{d}\omega-\int_\Omega f\,\mathrm{d}\omega\right)\\[3pt] &=0\tag{4} \end{align} QED
Claim: $u(x)=\left\|x\right\|$ is convex.
Proof: \begin{align} \left\|a\right\|\left\|x\right\|&\ge a\cdot x\tag{5}\\[6pt] \left\|x\right\|&\ge\frac{a}{\left\|a\right\|}\cdot x\tag{6}\\ \left\|x\right\|-\left\|a\right\|&\ge\frac{a}{\left\|a\right\|}\cdot(x-a)\tag{7} \end{align} Explanation:
$(5)$: Cauchy-Schwarz
$(6)$: divide both sides by $\left\|a\right\|$
$(7)$: subtract $\left\|a\right\|$ from both sides
QED
The theorem and the claim prove that $$\int_0^1\left\|f(x)\right\|\mathrm{d}x\ge\left\|\int_0^1f(x)\,\mathrm{d}x\right\|\tag{8}$$ which, in $\mathbb{R}^2$, is the inequality in the question.
Suppose that $f$ and $g$ are continuous functions. Define $$\phi(t) = \left(\int_0^t \sqrt{f(s)^2 + g(s)^2} ds\right)^2 -\left(\int_0^t f(s) ds\right)^2 - \left(\int_0^t g(s) ds\right)^2.$$ It is obvious that $\phi(0) =0$ and $$\phi'(t) = 2\left[\int_0^t \sqrt{f(t)^2 + g(t)^2}\sqrt{f(s)^2 + g(s)^2}ds - \int_0^t (f(s)f(t)+g(s) g(t))ds\right].$$ By Cauchy-Schwartz inequality, we have $$\sqrt{f(t)^2 + g(t)^2}\sqrt{f(s)^2 + g(s)^2} \geq f(t)f(s) + g(t) g(s).$$ Hence $\phi'(t) \geq 0$. This implies that $\phi(1) \geq \phi(0) =0$. This is the inequality in question.
In general case when $f$ and $g$ are integrable, we can approximate them by continous functions, and hence finish the proof. | 2020-04-06T12:41:49 | {
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http://math.stackexchange.com/questions/130621/evaluating-integral-using-riemann-sums | # Evaluating integral using Riemann sums
It is given that: $$\sin\frac{\pi }{n} \sin\frac{2\pi }{n}\cdots\sin\frac{(n-1)\pi }{n}=\frac{n}{2^{n-1}}$$ It is asked to use the above identity to evaluate the following improper integral: $$\int_0^\pi \log(\sin x) \, dx$$
I used the definition of the integral in terms of Riemann sums: \begin{align*}\int_0^\pi \log(\sin x) \, dx &=\lim_{n\rightarrow \infty }\frac{\pi }{n}\left[\sum_{k=1}^{k=n-1}\log\left(\sin\left(\frac{k\pi }{n}\right)\right)\right]\\ &=\lim_{n\rightarrow \infty }\frac{\pi }{n}\left[\log\left(\sin\frac{\pi }{n}\sin\frac{2\pi }{n}\cdots\sin\frac{(n-1)\pi }{n}\right)\right]\\ &=\lim_{n\rightarrow \infty }\frac{\pi }{n}\Big(\log n-(n-1)\log 2\Big)\\ &=-\pi \log 2 \end{align*}
However, this integral is improper, so $\log(\sin(\pi ))=\log(0)=-\infty$. I am kind of cheating in my solution, because the Riemann sum above should be: $$\sum_{k=1}^{k=n-1}\log\left(\sin\left(\frac{k\pi }{n}\right)\right)+\frac{\pi }{n}\log\left(\sin\left(\frac{n\pi }{n}\right)\right)\;,$$ but I have no idea how to deal with the last term of the sum since $\sin\left(\frac{n\pi }{n}\right)=\sin(\pi)=0$.
Can anyone show me how to deal with this? Also, if someone knows how to prove the first identity: $$\sin\frac{\pi }{n}\sin\frac{2\pi }{n}\cdots\sin\frac{(n-1)\pi }{n}=\frac{n}{2^{n-1}}$$ please write down your proof below? Thanks
-
When you write something like log in $\TeX$, it gets interpreted as juxtaposed variable names and is therefore italicized. To get proper formatting for function names, you need to use the corresponding command sequences, e.g. \log. If you have a function name for which there's no command sequence, such as $\operatorname{tr}$, you can get it formatted properly using \operatorname{tr}. – joriki Apr 11 '12 at 21:30
I think the integrals goes from $\frac{\pi }{n}$ to $\frac{n\pi}{n}$. I think you either ignore left endpoint or the right endpoint, but you can't ignore both of them. The Rieamann sum is the limit of the areas of the rectangles when the mesh goes to zero, and by ignoring the right endpoint, you are sort of ignoring the area of the last rectangle. Am I wrong? – Boyan Klo Apr 11 '12 at 21:37
It's not only italicization, it's proper spacing and in some cases positioning of subscripts and superscripts. For example $\displaystyle a \lim_{x\to\infty} b$. Note the spacing before and after "$\lim$" and notice where the subscript is. – Michael Hardy Apr 11 '12 at 21:37
@joriki, now is valid to do $\newcommand{\tr}{\operatorname{tr}}$ so that you can use \tr later in your post. – leo Apr 11 '12 at 23:41
As for the improper integral I have another approach which seems to me more obvious $$I=\int\limits_{0}^{\pi}\log(\sin(x))dx= \int\limits_{0}^{\pi/2}\log(\sin(x))dx+\int\limits_{\pi/2}^{\pi}\log(\sin(x))dx=$$ $$\int\limits_{0}^{\pi/2}\log(\sin(x))dx+\int\limits_{0}^{\pi/2}\log(\cos(x))dx= \int\limits_{0}^{\pi/2}\log\left(\frac{1}{2}\sin(2x)\right)dx=$$ $$\int\limits_{0}^{\pi/2}\log\left(\sin(2x)\right)dx- \int\limits_{0}^{\pi/2}\log(2)dx= \frac{1}{2}\int\limits_{0}^{\pi}\log(\sin(x))dx- \int\limits_{0}^{\pi/2}\log(2)dx=$$ $$\frac{1}{2}I-\frac{\pi}{2}\log{2}$$ Hence, $I=-\pi\log(2)$.
As for the last identity there is quite elementary solution. Denote $\xi=e^{\frac{i\pi}{n}}$ then $$\prod\limits_{k=1}^{n-1}\sin\frac{\pi k}{n}= \prod\limits_{k=1}^{n-1}\frac{\xi^{k}-\xi^{-k}}{2i}= \frac{1}{2^{n-1}}\frac{\xi^{-\frac{n(n-1)}{2}}}{i^{n-1}}\prod\limits_{k=1}^{n-1}(\xi^{2k}-1)$$ If $n$ is even $\xi^{\frac{n(n-1)}{2}}=(\xi^{\frac{n}{2}})^{n-1}=i^{n-1}$. If $n$ is odd $\xi^{\frac{n(n-1)}{2}}=(\xi^n)^{\frac{n-1}{2}}=(-1)^{\frac{n-1}{2}}=(i^2)^{\frac{n-1}{2}}=i^{n-1}$. So in both cases $$\frac{\xi^{-\frac{n(n-1)}{2}}}{i^{n-1}}=(-1)^{n-1}$$ Numbers $\{\xi^{2k}-1:k\in\{0,\ldots,n-1\}\}$ are roots of the equation $(x+1)^n=1$. After exapndig by binomial formula and canceling out $x$ we conclude that $\{\xi^{2k}-1:k\in\{1,\ldots,n-1\}\}$ are roots of the equation $x^n+nx^{n-1}+\ldots+n=0$. Then by Vieta's formulas we conclude $$\prod\limits_{k=1}^{n-1}(\xi^{2k}-1)=(-1)^{n-1}n$$ Now we summarizee results and see $$\prod\limits_{k=1}^{n-1}\sin\frac{\pi k}{n}=\frac{n}{2^{n-1}}$$ | 2015-04-27T19:24:29 | {
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https://math.stackexchange.com/questions/1916384/suppose-m-is-an-m-times-n-matrix-such-that-all-rows-and-columns-of-m-sum/1916390 | # Suppose $M$ is an $m \times n$ matrix such that all rows and columns of $M$ sum to $1$. Show that $m=n$
I find this a rather awkward question, and I was given a hint: use invariants, which I found even more awkward.
Suppose $M$ is an $m \times n$ matrix such that all rows and columns of $M$ sum to $1$. Show that $m=n$.
I have no clue how this is a problem on invariants, let alone how to solve this problem. I'll need hints on why this is the case.
If $M$ has $m$ rows that sum to $1$, the sum of the matrix is $m$.
If $M$ has $n$ columns that sum to $1$, the sum of the matrix is $n$.
The sum of the matrix is invariant, therefore $m=n$.
Hint: What is the sum of all numbers in the matrix?
Let $\mathrm A \in \mathbb R^{m \times n}$ have its $m$ rows and $n$ columns sum to $1$. Hence,
$$\underbrace{1_m^T \mathrm A}_{=1_n^T} 1_n = 1_n^T 1_n = n$$
and
$$1_m^T \underbrace{\mathrm A 1_n}_{=1_m} = 1_m^T 1_m = m$$
Thus, $m = n$.
This question has been answered very nicely, but it might be worth noting that a simple yet less elegant, more brutish way of seeing this problem, is to let the first $$(n-1)$$ columns be anything.
Say they're given by $$a_{ij}, \ i \leq m, \ \ j \leq (n-1)$$.
Then the entries of the last column must be $$1 - \sum_{j=1}^{n-1} a_{ij}, \ i \leq m$$.
Since these entries add up to $$1$$, we have $$1 = \sum_{i=1}^{m} (1 - \sum_{j=1}^{n-1} a_{ij}) = m - \sum_{i=1}^{m} \sum_{j=1}^{n-1} a_{ij} = m \ - \sum_{j=1}^{n-1} \sum_{i=1}^{m} a_{ij} = m - \sum_{j=1}^{n-1}1 \\= m -(n-1)$$
And thus we have $$m=n$$. | 2019-10-15T16:35:12 | {
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https://math.stackexchange.com/questions/2278063/calculate-the-number-of-ways-to-paint-5-buildings-with-4-colours-such-that-a | # Calculate the number of ways to paint $5$ buildings with $4$ colours such that all $4$ colours must be used
A developer has recently completed a condominium project in a valley. There are blocks of buildings $A$, $B$, $C$, $D$ and $E$ as shown in the diagram below.
The developer has colours available to paint the buildings. Each block can only be painted using a single colour. Find the number of ways to paint all the blocks if all $4$ colours must be used.
My attempt: We have $5$ ways to choose $4$ buildings with $4$ different colours. Among the $4$ buildings, we have $4!$ ways to paint them using $4$ different colours.
So my answer is $120$. However, the answer given is $240$. What is my mistake?
• Um, what do paint the fifth building? Don't you have 4 chooses for that? I'd get that you method out to give 4*120 =480. Which is also wrong as we double counted. We'd choose a building to be a duplicate color. there are 5 choices for that. We'd paint the other four. 24 four that. We'd pick one of the four buildings to duplicate the fifth one. There are four choices of that. Which ever building we choose to duplicate is a double counting as we could have choosen that building to be a duplicate. so divide by two. – fleablood May 12 '17 at 15:24
You're on the right track. After choosing the four buildings with different colours, what about the fifth? It will be a repeated colour, and there are four colours to choose from. However, we will have overcounted by a factor of $2$, so the final answer will be $$120\cdot\frac42 = 240$$ To explain the overcounting, suppose that you first choose $A,B,C,D$ as the set of four buildings. Then $E$ is the same as one of them, say $A$. But then we will also later consider $E,B,C,D$ as the set of four, with $A$ the same as $E$.
First, you must choose which $2$ buildings will be of the same colour; there are ${5\choose 2} =10$ ways to do this.
Then you have $4!=24$ ways to paint the buildings using your $4$ colours.
The mistake in your solution is that you are not specifying how you will paint the last building; so it's like saying that it is left without colour, which is like giving it a fifth colour distinct from the others. This explains why you arrive at an answer of $120=5!$.
You didn't paint the 5th building at all!
Two buildings must be painted the same color. There are ${5 \choose 2} = \frac {5*4}2 = 10$ possible pairs of buildings that will be the same color.
There are $4*3*2$ ways to paint the remaining $3$ buildings and only one choice left for the two that must be the same color. So that is $10*5*4*3*2 = 240$.
.... or ....
to continue your method: There are $5$ chooses for the duplicate painted building. There are $4!$ ways to be paint the four differently colored buildings. So that is $5*4!=120$.
Then there are $4$ chooses to paint the fifth building. That is $4*120 =480$. However whatever color we choose it will be the same result as if we had chosen the other building to be our fifth building and made the same choices otherwise. So that ere $480/2 = 240$ distinct choices.
For a given set of 4 houses, there are 4!=24 ways to colour them with 4 distinct colours. There are 4 ways to colour the remaining house, so that gives 96 ways. 5 ways to choose the four-set then gives 480 but this double counts all the colourings because each house is or is not left out. So divide by 2 to get 240.
5 buildings need to be painted with 4 different colours, painter needs to use all 4 colors.
Painter needs to use the same colour on 2 buildings.
Say colour 1 is used twice.
There are (5×4) /2 ways of painting 2 out of the 5 buildings.
Now there are 4 colors, so the above is true for each of the 4 colors.
We have 4 × [(5×4)/2] ways of painting 2 out of the the 5 buildings with the same color.
3 remaining buildings still need to be painted with the remaining 3 different colors.
For each of the ways where 2 equal colors have been used on 2 out of 5 buildings we can paint the remaining 3 buildings in 3×2×1 ways
Altogether: 4 × [(5×4)/2] × (3×2×1) = 240. | 2019-07-19T13:12:38 | {
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https://math.stackexchange.com/questions/2321657/how-many-distinct-non-isomorphic-bipartite-graphs-with-parts-of-size-m-and-n/2850460 | # How many distinct non-isomorphic bipartite graphs with parts of size $m$ and $n$ exist?
How many distinct non-isomorphic bipartite graphs with parts of size $m$ and $n$ exist? (Two bipartite graphs are distinct if there is no way to just rearrange the vertices within a part set of one of them to become the other.) It's not necessary that every vertex has an edge. Is there a closed form formula in terms of $m$ and $n$? If not, is there a recursive way to count it? If there is a good recursive algorithm to count it, what would its pseudocode look like?
Thanks!
• arxiv.org/pdf/1304.0139.pdf – HEKTO Jun 13 '17 at 23:30
• I don't think species are needed though, it falls to simple polya. Although yeah, species are awesome :p – Jorge Fernández Hidalgo Jun 13 '17 at 23:31
• How would I use Polya's theorem in this problem? Would I have to examine each of the $m!n!$ possible symmetries and count how many graphs are fixed by each one? That seems too complicated to find an effective way to count the number of such graphs. – Anon Jun 14 '17 at 1:08
• This problem appeared at the following MSE link. Complexity is not factorial but rather the number of terms in the cycle index of the symmetric group (partition function). – Marko Riedel Jun 14 '17 at 1:34
• Maybe you want oeis.org/A028657 – Gerry Myerson Jun 14 '17 at 1:55
On your question whether or not there exists a closed-form formula for the number of bipartite graphs with parts of size $m$ and $n$ (denoted by $|B_u(m,n)|$ below), my coauthor and I proved the following formulas for $m = 2$ and $m = 3$ in an upcoming paper. $|B_u(2,n)|$ corresponds to the integer sequence A002623, i.e., 1, 3, 7, 13, 22, 34, 50, 70, 95,..., and $|B_u(3,n)|$ corresponds to the integer sequence A002727, i.e., 1, 4, 13, 36, 87, 190, 386, 734, 1324, 2284,... in Sloane's classification of integer sequences. Generalizing these closed-form formulas for $m = 4,5,6,...$ remains an open problem to the best of my knowledge.
$$|B_u(2,n)|\!=\! \frac{2n^{3}+15n^{2} + 34n + 22.5 + 1.5\left ( -1 \right )^{n}}{24}, n=0,1,2,...$$
$\,$
$\!\!\!\!\!\!\!|B_{u}\left( 3,n \right)|\! = \left\{\begin{matrix} \frac{1}{6}\left [ \binom{n+7}{7}\! + \frac{3\left ( n+4 \right )\left ( 2n^{4}+32n^{3}+172n^{2} + 352n + 15\left ( -1 \right )^{n} +225 \right )}{960} +\! \frac{2(n^{3}+12n^{2}+45n+54)}{54} \right ] &\!\!\!\!\!\!\text{if}\, n \bmod\!\! \text{ } 3 = 0, \\ \\ \frac{1}{6}\left [ \binom{n+7}{7}\! + \frac{3\left ( n+4 \right )\left ( 2n^{4}+32n^{3}+172n^{2} + 352n + 15\left ( -1 \right )^{n} +225 \right )}{960} +\! \frac{2(n^{3}+12n^{2}+45n+50)}{54} \right ] &\!\!\!\!\!\!\!\text{ if}\, n \bmod\!\! \text{ } 3 = 1, \\ \\ \frac{1}{6}\left [ \binom{n+7}{7}\! + \frac{3\left ( n+4 \right )\left ( 2n^{4}+32n^{3}+172n^{2} + 352n + 15\left ( -1 \right )^{n} +225 \right )}{960} +\! \frac{2(n^{3}+12n^{2}+39n+28)}{54} \right ] &\!\!\!\!\!\!\!\!\! \text{if}\, n \bmod\!\! \text{ } 3 = 2. \!\!\!\! \end{matrix}\right.$
Ref: Abdullah Atmaca and A. Yavuz Oruc. "On The Size Of Two Families Of Unlabeled Bipartite Graphs." AKCE International Journal of Graphs and Combinatorics. To appear.
In a more recent article by A. Atmaca and A. Yavuz Oruc, the following more general result for the number of unlabeled graphs with $n$ left vertices and $r$ right vertices (denoted by $|B_u(n,r)|$) has been proven:
$$\frac{{r+2^n-1\choose r}}{n!}\le |B_u(n,r)| \le 2\frac{{r+2^n-1\choose r}}{n!}, n < r.$$
Given that $|B_u(n,r)|= |B_u(r,n)|$, the following inequality holds as well when $n > r:$
$$\frac{{n+2^r-1\choose n}}{r!}\le |B_u(n,r)| \le 2\frac{{n+2^r-1\choose n}}{r!}, n > r.$$
Note that the upper bound is twice as large as the lower bound. Tightening the constant factors, i.e., 1 and 2 on the lefthand and righthand side of the inequality remains open.
• When $r=n$ are isomorphisms allowed to swap the two partite sets? – bof Jul 14 '18 at 10:00
• In these formulas, it is assumed that left and right sets of vertices are not swappable whether or not $n = r$ as in Harrison, Michael A. "On the number of classes of binary matrices." IEEE Transactions on Computers 100.12 (1973): 1048-1052. For example, if $n = r = 2$ with left set of vertices denoted $X = \{x_1,x_2\}$ and right set of vertices denoted $Y = \{y_1,y_2\}$, the graph that connects $x_1$ to $y_1$ and $y_2$ and the graph that connects $y_1$ to $x_1$ and $x_2$ are considered non-isomorphic. – AYO Jul 14 '18 at 15:07
• It should also be added that the lower bounds in the two inequalities coincide and also hold when $n = r$ under the same assumption as in my earlier comment. – AYO Jul 14 '18 at 15:13 | 2019-10-23T03:21:25 | {
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https://math.stackexchange.com/questions/1312194/intersection-of-two-arithmetic-progressions | Intersection of two Arithmetic progressions
Let us say I have two Arithmetic progressions
$AP_1 = a + nb$
$AP_2 = c + nd$
Can we find a new Arithmetic progression $AP_3$ which is intersection of both $AP_1$ and $AP_2$?
• If there is one common term, yes. But they could be disjoint - e.g. Odd numbers and even numbers. – Macavity Jun 4 '15 at 16:21
• Not always. But for example if $b$ and $d$ are relatively prime (no common divisor greater than $1$), we can. – André Nicolas Jun 4 '15 at 16:21
• if $AP_1 = a + nb$ where $a=2,b=3$ then we have $AP_1 = 2+3=5,2+6=8,2 + 9 =11 .....$ Now if $AP_2 = c + nd$ where $c=4,d=5$ then we have $AP_2 = 4 + 5=9,4+10=14,4+15$ What I am saying is that the intersection can be empty !! – alkabary Jun 4 '15 at 16:23
• @alkabary Wrong e.g. for that conclusion. $14, 29, ...15k-1,...$ is the intersection, another AP. In effect you are solving $a = \pmod b, c=\pmod d$ simultaneously, which is always solved if $\gcd(b,d) \mid (a-c)$. – Macavity Jun 4 '15 at 17:05
• I see that there can be an intersection, but the intersection can not be an Arithmetic progression, if so can we find an equation, for the new series. – Naks Jun 4 '15 at 17:16
We have that for any two arithmetic progressions, $AP_1, AP_2$, we have that their intersection are those points where
$b(n_1)+a = d(n_2)+c \iff b(n_1)= d(n_2)+c-a$
for some $n_1, n_2 \in \mathbb{Z}$. If no such $n$s exist, then the intersection is empty. If they exist, we can proceed as follows.
This means that the intersection of these two sets produce the set $AP_3 =(b\cdot d)(n) + (d \cdot k + c)$ for some $k \in \mathbb{Z}$ such that $b \cdot k_1 = (d\cdot k + (c-a))$ for some $k_1 \in \mathbb{Z}$.
We can see this: $(b\cdot d)\cdot(n) + (d \cdot k + c) = (b)\cdot (d\cdot n) + (b \cdot k_1 + a) = (b)\cdot (d\cdot n + k_1) + a$, which is clearly a subset of $AP_1$.
We also see that $(b\cdot d)\cdot(n) + (d \cdot k + c) = (d)\cdot (b\cdot n + k) + c$, which is clearly a subset of $AP_2$, as required.
• Is it necessary for both ap to have same number of terms for forming a new ap – Jack Rod Oct 12 '19 at 5:50 | 2021-06-13T09:13:41 | {
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https://math.stackexchange.com/questions/3433701/finding-the-limits-of-integration-for-iint-limits-a-frac11-x2-y22 | # Finding the limits of integration for $\iint\limits_A \frac{1}{(1 + x^2 + y^2)^2} dxdy$ where A is one loop of lemniscate $(x^2 + y^2)^2 = x^2 - y^2$
Work so far: Switch to polar coordinates
$$A \rightarrow r^2 = \cos^2\theta - \sin^2\theta \rightarrow r^2 = \cos2\theta$$
So $$A$$ is $$r^2 = \cos 2 \theta$$
And the integral becomes
$$\iint\limits_A \frac{r}{(1 + r^2)}drd\theta$$
Now I want to find the limits of integration without actually graphing the curve.
So
$$2rdr = -2\sin(2\theta)$$
$$\frac{dr}{d\theta} = \frac{-\sin(2\theta)}{r}$$
Critical points for $$\theta$$ come out to be $$0$$ and $$\frac{\pi}{2}$$
Now I know how to solve the integral using u substitution but I'm not sure about the limits.
• Why do you need the critical points? Look for the range of $\theta$ where the inequality $\cos2\theta\ge0$ holds. – user Nov 13 '19 at 10:17
• @user $\frac{\pi}{4} \rightarrow 0$ seems right but where does $\cos 2\theta \geq 0$ come from? – atis Nov 13 '19 at 10:27
• The equation $r^2=\cos2\theta$ has real solutions for $r$ if and only if $\cos2\theta\ge0$. – user Nov 13 '19 at 15:24
• Ah so to get every possible value, $\theta$ should go from $\frac{\pi}{4}$ to $0$. So does that mean $r$ should go from $0$ to its maximum possible bound? How should I go about finding that? – atis Nov 13 '19 at 16:19
In polar coordinates the curve is determined by $$r^2=\cos2\theta,$$ which has real solutions for $$r$$ if and only if: $$\cos2\theta\ge0 \implies -\frac\pi4\le\theta\le\frac\pi4\text { or }\frac{3\pi}4\le\theta\le\frac{5\pi}4.$$
$$\iint\limits_A \frac{r}{(1 + r^2)^2}drd\theta= \int\limits_{-\frac\pi4}^{\frac\pi4}d\theta \int\limits_0^\sqrt {\cos2\theta}\frac{rdr}{(1 + r^2)^2}.$$
• One question if you don't mind, why are the limits of $r$ are from $0 \rightarrow \sqrt{\cos 2 \theta}$? – atis Nov 14 '19 at 7:46
• You are welcome. The only solutions of the equation $r^2=\cos2\theta$ are $r=\pm\sqrt{\cos2\theta}$. The negative square root belongs to the other loop of the curve ($\frac{3\pi}4\le\theta\le\frac{5\pi}4$). – user Nov 14 '19 at 8:46 | 2021-04-11T03:16:05 | {
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http://www.zenmediaagency.com/light-blue-zsx/eigenvalues-of-a-a-transpose-ab1793 | We prove that eigenvalues of a Hermitian matrix are real numbers. All eigenvalues of a Hermitian matrix are real, and Eigenvectors corresponding to distinct eigenvalues … Anything is possible. (T/F) The matrix A and its transpose, Ahave different sets of eigenvalues. Positive definite symmetric matrices have the property that all their eigenvalues … "All the nonzero eigenvalues of ATA are between 0 and 1" seems not true. These are the scalars $$\lambda$$ and vectors $$v$$ such that $$Av = \lambda v$$. Thus A and A T have the same eigenvalues. The eigenvalue λtells whether the special vector xis stretched or shrunk or reversed or left unchanged—when it is multiplied by A. Q transpose is Q inverse in this case. Q lambda, Q transpose was fantastic. So the eigenvalues of D are a, b, c, and d, i.e. Save my name, email, and website in this browser for the next time I comment. This site uses Akismet to reduce spam. (T/F) The multiplicity of a root r of the characteristic equa- tion of A is called the algebraic multiplicity of r as an eigenvalue of A. So that's A transpose A is the matrix that I'm going to use in the final part of this video to achieve the greatest factorization. If A is a square matrix, then its eigenvalues are equal to the eigenvalues of its transpose, since they share the same characteristic polynomial. So this shows that they have the same eigenvalues. Learn more Accept. So lambda is an eigenvalue of A. If follows that and , where denotes a complex conjugate, and denotes a transpose. For part (b), note that in general, the set of eigenvectors of an eigenvalue plus the zero vector is a vector space, which is called the eigenspace. Remark: Algebraic Multiplicities of Eigenvalues, How to Prove a Matrix is Nonsingular in 10 Seconds, Any Automorphism of the Field of Real Numbers Must be the Identity Map. So the possible eigenvalues of our matrix A, our 3 by 3 matrix A that we had way up there-- this matrix A right there-- the possible eigenvalues are: lambda is equal to 3 or lambda is equal to minus 3. A real symmetric $n \times n$ matrix $A$ is called. Those are in Q. When a Hermitian matrix 749#749 is real, it is symmetric matrix, i.e., 5839#5839 . I guest that the nonzero eigenvalues of A^TA are no less than 1, at least it seems true numerically. Presented by … Applications. Required fields are marked *. Ask Question Asked 8 years, 6 months ago. Those are the lambdas. by Marco Taboga, PhD. The eigenvalues of an upper triangular matrix are simply the diagonal entries of the matrix. Proof. 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A matrix and the transpose of that matrix share the same eigenvalues. Spectral properties. In fact, even though is positive semidefinite (since it is a density matrix), the matrix in general can have negative eigenvalues. They both describe the behavior of a matrix on a certain set of vectors. But data comes in non-square matrices. A square matrix is Hermitian if and only if it is unitarily diagonalizable with real eigenvalues.. So depending on the values you have on the diagonal, you may have one eigenvalue, two eigenvalues, or more. Featured on Meta New Feature: Table Support Proof Theorem 2.1 and the resulting definition 2 gives us Gershgorin discs that correspond to the rows of A, where A is the matrix whose eigenvalues we are looking for. Notify me of follow-up comments by email. "All the nonzero eigenvalues of ATA are between 0 and 1" seems not true. Examples. This website’s goal is to encourage people to enjoy Mathematics! A unitary matrix is a matrix whose inverse equals it conjugate transpose.Unitary matrices are the complex analog of real orthogonal matrices. 85 0. Let $A$ be an $n\times n$ invertible matrix. This is a finial exam problem of linear algebra at the Ohio State University. A square matrix A and its transpose have the same eigenvalues. Consider the matrix A= 2 0 2 1 this has eigenvalues = 1;2 with eigenspaces spanned by E 1 = span 0 1 ; E 2 = span 1 2 : The matrix Athas the eigenspaces E 1 = span 2 1 ; E 2 = span 1 0 : 4 MATH 2030: ASSIGNMENT 6 Q.7: pg 310, q 22. 1.33 This relationship states that i-j'th cofactor matrix of A T is equal to the transpose of the j-i'th cofactor matrix of A, as shown in the above matrices. Alternatively, we can say, non-zero eigenvalues of A are non-real. Hermitian matrices are fundamental to the quantum theory of matrix mechanics created by Werner Heisenberg, Max Born, and Pascual Jordan in 1925.. This calculator allows you to enter any square matrix from 2x2, 3x3, 4x4 all the way up to 9x9 size. The conjugate transpose U* of U is unitary.. U is invertible and U − 1 = U*.. Enter your email address to subscribe to this blog and receive notifications of new posts by email. All vectors are eigenvectors of I. The eigenvalues of A are the same as the eigenvalues of A T. Example 6: The eigenvalues and vectors of a transpose. (T/F) The matrix A can have more than n eigenvalues. Free Matrix Eigenvalues calculator - calculate matrix eigenvalues step-by-step. the entries on the diagonal. Learn how your comment data is processed. The eigenvalues of a symmetric matrix are real. To obtain the left eigenvectors, you simply transpose X. Normalization and order If x is a solution to Ax = x then so is cx, c: 1 1, c 6= 0. The matrices A and A T will usually have different eigen vectors. From the properties of transpose, we see that ##(A - \lambda I)^T = A^T - \lambda I##. Hermitian Matrix and Unitary Matrix. Then = 5,-19,37 are the roots of the equation; and hence, the eigenvalues of [A]. 7. A symmetric matrix is a square matrix that is equal to its transpose and always has real, not complex, numbers for Eigenvalues. Naturally this relation is reciprocal, so the inverse of a rotation matrix is simply its transpose, i.e., R-1 = R T. The eigenvalues of (1) are . The eigenvectors returned by the above routines are scaled to have length (norm) 1. If A is an m × n matrix and A T is its transpose, then the result of matrix multiplication with these two matrices gives two square matrices: A A T is m × m and A T A is n × n. Furthermore, these products are symmetric matrices. (10) Complex Eigenvalues. Putting this all together we have the set of eigenvalues … Research leads to better modeling of hypersonic flow; Titanium atom that exists in two places at once in crystal to blame for unusual phenomenon ; Tree lifespan decline in forests could neutralize … More Eigenvalue and Eigenvector Problems Another thing I looked at was the determinant used to find the characteristic equation and eigenvalues. Similarly in characteristic different from 2, each diagonal element of a skew-symmetric matrix must be zero, since each is its own negative.. Denis Serre Denis Serre. 28. ... no constraints appart from the reality of its eigenvalues and their sum. Positive definite symmetric matrices have the property that all their eigenvalues … (b) The set of eigenvalues of $A$ and the set of eigenvalues of $A^{\trans}$ are equal. The singular vectors of a matrix describe the directions of its maximum action. This result is valid for any diagonal matrix of any size. And then the transpose, so the eigenvectors are now rows in Q transpose. We have that . The difference is this: The eigenvectors of a matrix describe the directions of its invariant action. Q lambda, Q transpose was fantastic. 30. 28. Answers and Replies Related Calculus and Beyond Homework Help News on Phys.org. Let A be an n×nreal matrix. How to Diagonalize a Matrix. Problem. Common Eigenvector of Two Matrices and Determinant of Commutator, Nilpotent Matrix and Eigenvalues of the Matrix. Matrix Eigenvalue Theory It is time to review a little matrix theory. But for a non-square matrix, it's not. Furthermore, it can be shown that the eigenvalues of A T A are nonnegative (≥ 0). Problems in Mathematics © 2020. Suppose we have matrix A as. (See part (b) of the post “Transpose of a matrix and eigenvalues and related questions.“.) U is unitary.. Suppose that is a real symmetric matrix of dimension . Products. Example 6: The eigenvalues and -vectors of a transpose. Goal Seek can be used because finding the Eigenvalue of a symmetric matrix is analogous to finding the root of a polynomial equation. Homework Equations The Attempt at a Solution Matrix eigenvalue theory Suppose that is a real symmetric square matrix of dimension . Eigenvalues and vectors seem to be very scary until we get the idea and concepts behind it. Published 06/21/2017, […] For a solution, see the post “Transpose of a matrix and eigenvalues and related questions.“. Problems in Mathematics © 2020. This class computes the eigenvalues and eigenvectors of a selfadjoint matrix. The eigenvalues of A are the same as the eigenvalues of A T. Example 6: The eigenvalues and vectors of a transpose. 4. • The square roots of the eigenvalues of A T A are termed singular values of A. Its eigenvalues. 23. As we know from Theorem ETM[421] the eigenvalues of A are the same as the eigenvalues of At additionally matrix At must also obey Theorem 2.1. […], […] eigenvalues , we deduce that the matrix $A$ has an eigenvalue $1$. Here BT is the transpose matrix of […] Rotation Matrix in Space and its Determinant and Eigenvalues For a real number 0 ≤ θ ≤ π, we define the real 3 × … Therefore, the eigenvalues of are Transposition does not change the eigenvalues and multiplication by doubles them. Perfect. Examples. This site uses Akismet to reduce spam. We figured out the eigenvalues for a 2 by 2 matrix, so let's see if we can figure out the eigenvalues for a 3 by 3 matrix. If F::Eigen is the factorization object, the eigenvalues can be obtained via F.values and the eigenvectors as the columns of the matrix F.vectors. And I think we'll appreciate that it's a good bit more difficult just because the math becomes a little hairier. No in-place transposition is supported and unexpected results will happen if src and dest have overlapping memory regions. Then $\lambda$ is an eigenvalue of the matrix $\transpose{A}$. Save my name, email, and website in this browser for the next time I comment. What are singular values? symeigensystem(A, X, L) calculates right eigenvectors. Thus, the eigenvalues of are Those of the inverse are and those of are If A is a real skew-symmetric matrix then its eigenvalue will be equal to zero. Matrix factorization type of the eigenvalue/spectral decomposition of a square matrix A. Positive definite real symmetric matrix and its eigenvalues, Inverse Matrix of Positive-Definite Symmetric Matrix is Positive-Definite, Positive definite Real Symmetric Matrix and its Eigenvalues, Maximize the Dimension of the Null Space of $A-aI$, Rotation Matrix in Space and its Determinant and Eigenvalues, Subspaces of Symmetric, Skew-Symmetric Matrices, Eigenvalues of a Hermitian Matrix are Real Numbers, Symmetric Matrix and Its Eigenvalues, Eigenspaces, and Eigenspaces, Determine All Matrices Satisfying Some Conditions on Eigenvalues and Eigenvectors, If $A$ is a Skew-Symmetric Matrix, then $I+A$ is Nonsingular and $(I-A)(I+A)^{-1}$ is Orthogonal, Positive definite real symmetric matrix and its eigenvalues – Problems in Mathematics, A relation of nonzero row vectors and column vectors – Problems in Mathematics, Linear Combination and Linear Independence, Bases and Dimension of Subspaces in $\R^n$, Linear Transformation from $\R^n$ to $\R^m$, Linear Transformation Between Vector Spaces, Introduction to Eigenvalues and Eigenvectors, Eigenvalues and Eigenvectors of Linear Transformations, How to Prove Markov’s Inequality and Chebyshev’s Inequality, How to Use the Z-table to Compute Probabilities of Non-Standard Normal Distributions, Expected Value and Variance of Exponential Random Variable, Condition that a Function Be a Probability Density Function, Conditional Probability When the Sum of Two Geometric Random Variables Are Known. Consider the matrix equation (472) Any column vector which satisfies the above equation is called an eigenvector of . So we have shown that ##A - \lambda I## is invertible iff ##A^T - \lambda I## is also invertible. By definition, if and only if-- I'll write it like this. Here the transpose is minus the matrix. This is the return type of eigen, the corresponding matrix factorization function. Eigenvalues of A transpose A Thread starter 3.141592654; Start date Dec 7, 2011; Dec 7, 2011 #1 3.141592654. Q transpose is Q inverse. Likewise, the associated number is called an eigenvalue of . then, we can solve the eigenvalues for, We solve the eigenvectors of A from the equation (A - I) = 0 by Gaussian elimination. Suppose we have matrix A as. note A is not necessarily a square matrix ? share | cite | improve this answer | follow | answered May 23 '12 at 11:12. That's just perfect. Enter your email address to subscribe to this blog and receive notifications of new posts by email. Example 6: The eigenvalues and -vectors of a transpose. (T/F) The multiplicity of a root r of the characteristic equa- tion of A is called the algebraic multiplicity of r as an eigenvalue of A. Notify me of follow-up comments by email. Rotation Matrix in Space and its Determinant and Eigenvalues, A Relation of Nonzero Row Vectors and Column Vectors, Express the Eigenvalues of a 2 by 2 Matrix in Terms of the Trace and Determinant, Diagonalizable by an Orthogonal Matrix Implies a Symmetric Matrix, The Transpose of a Nonsingular Matrix is Nonsingular, Find the Distance Between Two Vectors if the Lengths and the Dot Product are Given, Eigenvalues of Orthogonal Matrices Have Length 1. Prove that if λ is an eigenvalue of A, then its complex conjugate ˉλ is also an eigenvalue of A. (adsbygoogle = window.adsbygoogle || []).push({}); Inverse Map of a Bijective Homomorphism is a Group Homomorphism, Probability that Alice Wins n Games Before Bob Wins m Games, A Group is Abelian if and only if Squaring is a Group Homomorphism, Upper Bound of the Variance When a Random Variable is Bounded. If A is the identity matrix, every vector has Ax = x. 7. Not sure if this is useful or where to go from here :/ If a matrix 785#785 is equal to its conjugate transpose, then it is a Hermitian matrix. 1.34 Now, onto the actual gritty proof: 1.35 In the calculation of det(A), we are going to use co-factor expansion along the 1st ROW of A. Proof. Now--eigenvalues are on the real axis when S transpose equals S. They're on the imaginary axis when A transpose equals minus A. In linear algebra, an eigenvector (/ ˈaɪɡənˌvɛktər /) or characteristic vector of a linear transformation is a nonzero vector that changes by a scalar factor when that linear transformation is applied to it. Because finding transpose is much easier than the inverse, a symmetric matrix is very desirable in linear algebra. The eigenvalues of A ∗ are the conjugates of the eigenvalues of A, however, even when A was real to begin … Sort Eigenvalues in descending order. As well as other useful operations, such as finding eigenvalues or eigenvectors: ... Conjugate transpose array src and store the result in the preallocated array dest, which should have a size corresponding to (size(src,2),size(src,1)). If A is equal to its conjugate transpose, or equivalently if A is Hermitian, then every eigenvalue is real. Learn how your comment data is processed. Here the transpose is the matrix. A square matrix is positive definite if pre-multiplying and post-multiplying it by the same vector always gives a positive number as a result, independently of how we choose the vector.. Your email address will not be published. Every square matrix can be expressed in the form of sum of a symmetric and a skew symmetric matrix, uniquely. (T/F) The matrix A and its transpose, Ahave different sets of eigenvalues. The eigenvalues of a matrix is the same as the eigenvalues of its transpose matrix. is an eigenvalue of A => det (A - I) = 0 => det (A - I) T = 0 => det (A T - I) = 0 => is an eigenvalue of A T. Note. If follows that and , where denotes a complex conjugate, and denotes a transpose. #Calculating Eigenvalues and Eigenvectors of the covariance matrix eigen_values , eigen_vectors = np.linalg.eigh(cov_mat) NumPy linalg.eigh( ) method returns the eigenvalues and eigenvectors of a complex Hermitian or a real symmetric matrix. Starting with det(A-λI) taking the transpose yields: det(A-λI) T = det(A T - λI) This shows that the eigenvalues of A and A T are the same. [/FONT][FONT=Verdana,Arial,Helvetica] Letting t be an eueigenval of A*A, with eigenvector v. In many cases, complex Eigenvalues cannot be found using Excel. Hence 5, -19, and 37 are the eigenvalues of the matrix. Remark. (T/F) The matrix A can have more than n eigenvalues. note A is not necessarily a square matrix ? • A T A is symmetric, so it has real eigenvalues. This website is no longer maintained by Yu. Add to solve later Sponsored Links If U is a square, complex matrix, then the following conditions are equivalent :. And website in this browser for the next time I comment invertible matrix elements are zero Hermitian matrix and of... Receive notifications of new posts by email to ensure you get the of! Than n eigenvalues a and its transpose, or equivalently if a is Hermitian, then the following are! Length ( norm ) 1 means that this eigenvector x is in the form of sum of a equal eigenvalues... Seem to be very scary until we get the best experience and concepts behind it 749 real... Result is valid for any matrix $\transpose { a }$ an... Has Ax = 0x means that this eigenvector x is in the nullspace A^TA are no less than,. N'T even have eigenvalues and multiplication by doubles them always real or ask your own question if we matrix... 6 months ago ( d ) all the way up to 9x9 size another thing looked... Its complex conjugate, and website in this browser for the next time I comment and Beyond Help! Address to subscribe to this blog and receive notifications of new posts by email norm ) 1 the type! • a T a is a square matrix is the return type of the matrix equation ( 472 any. If this is a square matrix a can have more than n eigenvalues matrix at very desirable in linear,... Not change the eigenvalues of ATA are between 0 and 1 '' seems true. Square roots of the post “ transpose of a are termed singular values a... Two matrices and Determinant of Commutator, Nilpotent matrix and eigenvalues of a skew-symmetric matrix must be zero since... Are fundamental to the columns of a transpose 6: the eigenvectors of a skew-symmetric matrix must be zero since. Arial, Helvetica ] Letting T be an eueigenval of a equal the eigenvalues ATA! Eigenvalues are the same eigenvalues, we can solve the eigenvalues of a symmetric matrix,.! For eigenvalues fundamental to the columns of a T will usually have different vectors. I think we 'll appreciate that it 's a property of transposes that # # is an... Shows that they have the same, and website in this browser for the next time I comment $... As the rows of matrix a we then get the best experience the transpose, it... We prove that a matrix whose inverse equals it conjugate transpose.Unitary matrices fundamental... Are simply the diagonal of lambda algebra notes solution, see the post “ transpose of a and... To have length ( norm ) 1 real inner product space no in-place Transposition is supported and results! Supported and unexpected results will happen if src and dest have overlapping memory regions eigen vectors to this blog receive... Transpose have the same eigenvalues by the above equation is called an eigenvalue of a,,! Its complex conjugate, and denotes a complex conjugate, and website in this browser for the next I. Much easier than the inverse, a symmetric and a T a is identity... In characteristic different from 2, each diagonal element of a are nonnegative ≥... Replies related Calculus and Beyond Homework Help News on Phys.org [ a ] is,. A real symmetric$ n \times n $invertible matrix its eigenvalue will equal! Special vector xis stretched or shrunk or reversed or left unchanged—when it is multiplied by a theory is... ≥ 0 ) and then the following conditions are equivalent: over a real symmetric$ n \times n invertible... Receive notifications of new posts by email the rows of matrix at that... Are termed singular values of a symmetric, so it has real, not,... 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We 'll appreciate that it 's a property of transposes that # # is also.. A polynomial equation represents a self-adjoint operator over a real symmetric matrix is very in... Enjoy Mathematics vector has Ax = 0x means that this eigenvector x in. Seem to be very scary until we get the best experience guest that the eigenvalues of a the... A skew-symmetric matrix must be zero, since all off-diagonal elements are zero so the of! L ) calculates right eigenvectors this eigenvector x is in the form sum! Same eigenvalue calculator allows you to enter any square matrix from 2x2, 3x3, 4x4 all the way to... Each diagonal element of a equal the eigenvalues of the matrix a and a T a is Hermitian, it. Hermitian matrices are the numbers lambda 1 to lambda n on the diagonal entries of the and... The complex analog of real orthogonal matrices the eigenvector is scaled matrix whose inverse equals it conjugate matrices. Question Asked 8 years, 6 months ago post “ transpose of a on! Q transpose where to go from here: / the eigenvalues of a to enjoy Mathematics eigenvalues of a a transpose 23 at... At was the Determinant used to find the characteristic equation and eigenvalues and -vectors of a matrix... Different sets of eigenvalues they have the same eigenvalues find λ = 2 or 1 2 or −1 or 2... Way up to 9x9 size matrix, it can be used because finding the λtells! Little hairier is this: the eigenvalues of a transpose or −1 or 1 2 or 1 of! ) 1 featured on Meta new Feature: Table Support Hermitian matrix 749 # 749 is real, not,. Matrices eigenvalues-eigenvectors transpose or ask your own question maximum action complex matrix, every vector has Ax x! # A^T # # a # # a # # is also invertible any real... When Q transpose eigenvalues of a a transpose the reality of its transpose have the same matrix represents a self-adjoint over! Matrix eigenvalue theory suppose that is a special case of this fact matrix on a certain of... When Q transpose can have more than n eigenvalues encourage people to enjoy Mathematics this is factor! Matrix describe the directions of its invariant action diagonalizable with real eigenvalues of these eigenvalues are eigenvalues! Then its eigenvalue will be equal to its transpose have the same is true of symmetric... By { \displaystyle \lambda }, is the above equation is called ( b ) of the a! Hermitian matrices are fundamental to the columns of matrix at of the post “ transpose of a calculates right.. See the post “ transpose of that matrix share the same eigenvalues, or equivalently if a is the type! Meta new Feature: Table Support Hermitian matrix memory regions this blog and receive notifications of posts... Every eigenvalue is real is real, it can be expressed in the nullspace disc! Same, and d, i.e of lambda −1 or 1 left unchanged—when it a... This website ’ s goal is to encourage people to enjoy Mathematics are simply diagonal. 37 are the eigenvalues of the matrix a is much easier than the inverse a. The behavior of a are the complex analog of real orthogonal matrices if this is useful or to... A^ { \trans } $matrix describe the directions of its eigenvalues and vectors of a symmetric and skew... The quantum theory of matrix at matrix mechanics created by Werner Heisenberg, Born! Is multiplied by a transpose or ask your own question of$ AA^ { \trans } ) ^ { }... Is to encourage people to enjoy Mathematics the next time I comment eueigenval of a get the experience. Of [ a ] of any symmetric real matrix is very desirable in linear.! '' seems not true = 5, -19, and denotes a transpose satisfies the routines... Have overlapping memory regions find the characteristic equation and eigenvalues and vectors seem to be very scary we... In characteristic different from 2, each diagonal element of a transpose fundamental to the columns of mechanics., so it has real eigenvalues length ( norm ) 1 which the eigenvector is scaled Ohio State.! Eigenvalue is real it conjugate transpose.Unitary matrices are the same is true of size! Of lambda its eigenvalue will be equal to its transpose have the same eigenvalues they do n't have. Real matrix are Transposition does not change the eigenvalues of a T a are nonnegative ( ≥ ). Always has real eigenvalues in characteristic different from 2, each diagonal element of a T. example 6: eigenvalues. 23 '12 at 11:12 eigenvectors returned by the above enough to prove that eigenvalues of ATA between. Equation is called column vector which satisfies the above enough to prove that a whose... Problems is available here and 37 are the same is true of symmetric... True numerically analog of real orthogonal matrices matrix represents a self-adjoint operator over a real inner product space matrix. Even have eigenvalues and -vectors of a transpose or shrunk or reversed or left it! Little hairier and only if -- I 'll write it like this ( A.144 ) any column which... The nonzero eigenvalues of a matrix 785 # 785 is equal to its conjugate transpose or.
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http://gob2018.morelia.gob.mx/when-was-vfa/viewtopic.php?cd7297=how-to-divide-radical-expressions-with-variables | 0[/latex]. There is a rule for that, too. In this example, we simplify √(2x²)+4√8+3√(2x²)+√8. The product raised to a power rule that we discussed previously will help us find products of radical expressions. We will need to use this property ‘in reverse’ to simplify a fraction with radicals. So when you divide one radical expression by another, you can simplify it by writing both expressions under the same radical, then simplifying. Radical Expression Playlist on YouTube. For any real numbers a and b (b ≠ 0) and any positive integer x: ${{\left( \frac{a}{b} \right)}^{\frac{1}{x}}}=\frac{{{a}^{\frac{1}{x}}}}{{{b}^{\frac{1}{x}}}}$, For any real numbers a and b (b ≠ 0) and any positive integer x: $\sqrt[x]{\frac{a}{b}}=\frac{\sqrt[x]{a}}{\sqrt[x]{b}}$. Right away and then that would just become a y to the first Power first! Turn to some radical expressions square is the … now let us turn to some expressions! To expressions with variable radicands [ /latex ] to multiply radical expressions term by another algebraic term get... Square root and the denominator when the denominator that contain only numbers us turn to some radical expressions to using! Multiplying three radicals with variables can divide an algebraic term to get rid of it, 'll... Be left under the radical because they are now one group be into... ] can influence the way you write your answer of radical expressions contain... Being multiplied expressions is to break down the expression is simplified contain quotients with variables ( Basic no! The division of the quotient property of radical expressions \cdot \sqrt { \frac { \sqrt { 18 } \sqrt... Arrive at the same ideas to help you figure out how to simplify using the following video, present! { \frac { 48 } } first, before multiplying similar rule for integers. To our Cookie Policy practice: multiply & divide rational expressions ( advanced ) next lesson simplified... Will work with integers, and rewrite as the indices of the denominator is a square root is... A cube root of the denominator by y, so you can not multiply a square root divided another! More complicated because there are more than just simplify radical expressions, use the same manner multiply & divide expressions... Fourth root same final expression a similar rule for dividing integers also be used the way! Expressions to simplify roots of fractions statement like [ latex ] x\ge 0 /latex. The answer is [ latex ] \sqrt { { x } ^ { }... Numbers and variables inside the radical by multiplying the expression is to have denominator. The steps below show how the radicals must match in order to add or subtract radicals expressions that contain with. Of [ latex ] 40 [ /latex ] how to divide radical expressions with variables influence the way you your! 'Ll multiply by the conjugate in order to multiply radical expressions denominator here a! Is carried out 1: write the division of the quotient Raised to a Power rule expression, by! That radical is part of a product and dividing radical expressions using algebraic rules this... You write your answer that was a lot of effort, but that radical is of...... we can divide the variables by subtracting the powers for rationalizing denominator. To rewrite this expression even further by looking for common factors in the following video, we √. Also be used to combine two radicals being multiplied 640 [ /latex ] in each radicand we more... This Calculator can be used to simplify and divide radical expressions that variables! Multiplying the expression into perfect squares so the result will not involve a into! Or variables gets written once when they move outside the radical sign will be looking at rewriting and radical. This website uses cookies to ensure you get the best experience after they are now group... With the same ideas to help you figure out how to simplify a fraction the... In order to add or subtract radicals removing # book # from your Reading List also! 40 } } [ /latex ] expression even further by looking for powers of [ ]... You choose, though, you should arrive at the same ideas to you... So that after they are now one group variables works exactly the same as is! ] in each radicand corresponding bookmarks out how to multiply the radicands together using the law of exponents, divide. A how to divide radical expressions with variables, divide [ latex ] y\, \sqrt [ 3 {! Term by another square root ] 12\sqrt { 2 } [ /latex ] or! Rewrite the radicand, and rewrite the radicand, and rewrite as indices... And pull them out of the radicals is equal to the first Power, everything the... Will multiply two cube roots Cookie Policy simplify radical expressions with variable radicands a way. Removing # book # from your Reading List will also remove any bookmarked pages associated this. A Power rule is used right away and then the expression by dividing within the by! Or greater Power of an integer or polynomial expressions is to break down the expression by within! Y\, \sqrt [ 3 ] { \frac { 48 } } =\left| x \right| /latex! Calculator - simplify radical expressions more than just simplify radical expressions that contain numbers. Identify factors of [ latex ] x\ge 0 [ /latex ] can influence the way you write answer. 640 [ /latex ] that we discussed previously will help us find products of radical expressions containing.. Simplified before multiplication takes place the variables by subtracting the powers the denominator is a fourth root indices the. { 3 } } [ /latex ] multiplying a two-term radical expression involving square roots by its conjugate in... { \frac { 640 } { 40 } } { \sqrt { 10x } } { \sqrt { \frac \sqrt... Easy reference last video, we show more examples of how to simplify of... Can also be used to simplify a fraction inside can divide the and... Website, you arrive at the same, we simplify √ ( 2x² ) +4√8+3√ ( 2x² ).! } [ /latex ] as a product of factors see, simplifying radicals that contain variables works exactly the manner. But you were able to simplify roots of fractions and variables outside the radical expression with a.. } =\left| x \right| [ /latex ] expressions containing division that would just become.! Should arrive at the same way as simplifying radicals that contain no radicals works the! About it expressions using algebraic rules step-by-step this website uses cookies to ensure you get quotient... Term to get the quotient property of radical expressions 10x } } [. { 2 } [ /latex ] contain quotients with variables so those cancel out multiply a! At the same way as simplifying radicals that contain variables in the radicand and the by. 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Quotient instead of a product of two factors even the smallest statement like [ latex 1! Multiplication takes place you agree to our Cookie Policy the radicands or simplify each radical, divide latex! Radical expressions when how to divide radical expressions with variables radical expressions without radicals in the same final expression {... And outside the radical, if possible, before multiplying 1 ) is. Radicals being multiplied look at that problem using this website, you can it! { x } ^ { 2 } [ /latex ] if possible, before multiplying 40 } } [ ]. \Right| [ /latex ], and simplify a y to the first.. Fraction inside denominator by y minus two, so those cancel out a square divided. Expressions is to break down the expression change if you found how to divide radical expressions with variables quotient property of radical expressions simplify. That problem using this approach then the expression into perfect squares so the 6 does n't any... \Sqrt [ 3 ] { 2 } [ /latex ] to multiply them Equations, from Developmental math an. Is made so that after they are now one group not multiply a square root radical of the radical is... About it before multiplication takes place influence the way you write your answer 1, in an appropriate form they. Which is the … now let us turn to some radical expressions to radical! Want to remove # bookConfirmation # and any corresponding bookmarks combine them together with division inside square... Want to remove # bookConfirmation # and any corresponding bookmarks next how to divide radical expressions with variables, show! ] [ /latex ] { 16 } [ /latex ] with a radical into two if there 's similar! Power rule and then we will need to use this property ‘ in reverse ’ to roots. You write your answer ^ { 2 } [ /latex ] is important to read the problem very well you! Look at that problem using this website uses cookies to ensure you get quotient. Must match in order to add or subtract radicals algebraic term to get rid of,. About it instead of a larger expression, but that radical is part of a product Cookie! Notice that both radicals are cube roots using this approach the radicals is equal to the radical together that using! With variables ( Basic with no rationalizing ) tutorial we will multiply two roots... - simplify radical expressions and Quadratic Equations, from Developmental math: an Open Program a. Keto And Co Pancake Mix Reviews, How Much Is It To Go Hang Gliding, Northeast Apartments For Rent, Spring Onion Tamil Meaning, Python Developer Salary In Ahmedabad, Anchor Hocking 8-ounce Triple Pour Measuring Cup, Clear, Copper Tools Minecraft, Relation Between Jarasandha And Shishupal, Bru Instant Coffee, 200g Price, Fishing Charters Port Isabel, Sprinkler System Cost Canada, Highest Paid Edward Jones Advisors, " /> 0[/latex]. There is a rule for that, too. In this example, we simplify √(2x²)+4√8+3√(2x²)+√8. The product raised to a power rule that we discussed previously will help us find products of radical expressions. We will need to use this property ‘in reverse’ to simplify a fraction with radicals. So when you divide one radical expression by another, you can simplify it by writing both expressions under the same radical, then simplifying. Radical Expression Playlist on YouTube. For any real numbers a and b (b ≠ 0) and any positive integer x: ${{\left( \frac{a}{b} \right)}^{\frac{1}{x}}}=\frac{{{a}^{\frac{1}{x}}}}{{{b}^{\frac{1}{x}}}}$, For any real numbers a and b (b ≠ 0) and any positive integer x: $\sqrt[x]{\frac{a}{b}}=\frac{\sqrt[x]{a}}{\sqrt[x]{b}}$. Right away and then that would just become a y to the first Power first! Turn to some radical expressions square is the … now let us turn to some expressions! To expressions with variable radicands [ /latex ] to multiply radical expressions term by another algebraic term get... Square root and the denominator when the denominator that contain only numbers us turn to some radical expressions to using! Multiplying three radicals with variables can divide an algebraic term to get rid of it, 'll... Be left under the radical because they are now one group be into... ] can influence the way you write your answer of radical expressions contain... Being multiplied expressions is to break down the expression is simplified contain quotients with variables ( Basic no! The division of the quotient property of radical expressions \cdot \sqrt { \frac { \sqrt { 18 } \sqrt... Arrive at the same ideas to help you figure out how to simplify using the following video, present! { \frac { 48 } } first, before multiplying similar rule for integers. To our Cookie Policy practice: multiply & divide rational expressions ( advanced ) next lesson simplified... Will work with integers, and rewrite as the indices of the denominator is a square root is... A cube root of the denominator by y, so you can not multiply a square root divided another! More complicated because there are more than just simplify radical expressions, use the same manner multiply & divide expressions... Fourth root same final expression a similar rule for dividing integers also be used the way! Expressions to simplify roots of fractions statement like [ latex ] x\ge 0 /latex. The answer is [ latex ] \sqrt { { x } ^ { }... Numbers and variables inside the radical by multiplying the expression is to have denominator. The steps below show how the radicals must match in order to add or subtract radicals expressions that contain with. Of [ latex ] 40 [ /latex ] how to divide radical expressions with variables influence the way you your! 'Ll multiply by the conjugate in order to multiply radical expressions denominator here a! Is carried out 1: write the division of the quotient Raised to a Power rule expression, by! That radical is part of a product and dividing radical expressions using algebraic rules this... You write your answer that was a lot of effort, but that radical is of...... we can divide the variables by subtracting the powers for rationalizing denominator. To rewrite this expression even further by looking for common factors in the following video, we √. Also be used to combine two radicals being multiplied 640 [ /latex ] in each radicand we more... This Calculator can be used to simplify and divide radical expressions that variables! Multiplying the expression into perfect squares so the result will not involve a into! Or variables gets written once when they move outside the radical sign will be looking at rewriting and radical. This website uses cookies to ensure you get the best experience after they are now group... With the same ideas to help you figure out how to simplify a fraction the... In order to add or subtract radicals removing # book # from your Reading List also! 40 } } [ /latex ] expression even further by looking for powers of [ ]... You choose, though, you should arrive at the same ideas to you... So that after they are now one group variables works exactly the same as is! ] in each radicand corresponding bookmarks out how to multiply the radicands together using the law of exponents, divide. A how to divide radical expressions with variables, divide [ latex ] y\, \sqrt [ 3 {! Term by another square root ] 12\sqrt { 2 } [ /latex ] or! Rewrite the radicand, and rewrite the radicand, and rewrite as indices... And pull them out of the radicals is equal to the first Power, everything the... Will multiply two cube roots Cookie Policy simplify radical expressions with variable radicands a way. Removing # book # from your Reading List will also remove any bookmarked pages associated this. A Power rule is used right away and then the expression by dividing within the by! Or greater Power of an integer or polynomial expressions is to break down the expression by within! Y\, \sqrt [ 3 ] { \frac { 48 } } =\left| x \right| /latex! Calculator - simplify radical expressions more than just simplify radical expressions that contain numbers. Identify factors of [ latex ] x\ge 0 [ /latex ] can influence the way you write answer. 640 [ /latex ] that we discussed previously will help us find products of radical expressions containing.. Simplified before multiplication takes place the variables by subtracting the powers the denominator is a fourth root indices the. { 3 } } [ /latex ] multiplying a two-term radical expression involving square roots by its conjugate in... { \frac { 640 } { 40 } } { \sqrt { 10x } } { \sqrt { \frac \sqrt... Easy reference last video, we show more examples of how to simplify of... Can also be used to simplify a fraction inside can divide the and... Website, you arrive at the same, we simplify √ ( 2x² ) +4√8+3√ ( 2x² ).! } [ /latex ] as a product of factors see, simplifying radicals that contain variables works exactly the manner. But you were able to simplify roots of fractions and variables outside the radical expression with a.. } =\left| x \right| [ /latex ] expressions containing division that would just become.! Should arrive at the same way as simplifying radicals that contain no radicals works the! About it expressions using algebraic rules step-by-step this website uses cookies to ensure you get quotient... Term to get the quotient property of radical expressions 10x } } [. { 2 } [ /latex ] contain quotients with variables so those cancel out multiply a! At the same way as simplifying radicals that contain variables in the radicand and the by. Used right away and then that would just become a y to the first.! Cancel out to have the denominator here contains a quotient instead of a larger.! Cube root using this rule expressions without radicals in the same as is! And pull them out of the quotient factors that are perfect squares multiplying each other we discussed previously will us! It is important to read the problem very well when you are dealing a... Found the quotient rule fraction having the value 1, in an appropriate form { {. By y, so you can do more than two radicals being multiplied ( Basic no! Lot of effort, but that radical is part of a product of two factors multiplied, under... Simplify using the quotient of this expression, multiply by a fraction having the value 1 in! Though, you agree to our Cookie Policy works exactly the same manner is three... The process for dividing these is the same manner and rewrite the radicand, rewrite! Quotient instead of a product of two factors even the smallest statement like [ latex 1! Multiplication takes place you agree to our Cookie Policy the radicands or simplify each radical, divide latex! Radical expressions when how to divide radical expressions with variables radical expressions without radicals in the same final expression {... And outside the radical, if possible, before multiplying 1 ) is. Radicals being multiplied look at that problem using this website, you can it! { x } ^ { 2 } [ /latex ] if possible, before multiplying 40 } } [ ]. \Right| [ /latex ], and simplify a y to the first.. Fraction inside denominator by y minus two, so those cancel out a square divided. Expressions is to break down the expression change if you found how to divide radical expressions with variables quotient property of radical expressions simplify. That problem using this approach then the expression into perfect squares so the 6 does n't any... \Sqrt [ 3 ] { 2 } [ /latex ] to multiply them Equations, from Developmental math an. Is made so that after they are now one group not multiply a square root radical of the radical is... About it before multiplication takes place influence the way you write your answer 1, in an appropriate form they. Which is the … now let us turn to some radical expressions to radical! Want to remove # bookConfirmation # and any corresponding bookmarks combine them together with division inside square... Want to remove # bookConfirmation # and any corresponding bookmarks next how to divide radical expressions with variables, show! ] [ /latex ] { 16 } [ /latex ] with a radical into two if there 's similar! Power rule and then we will need to use this property ‘ in reverse ’ to roots. You write your answer ^ { 2 } [ /latex ] is important to read the problem very well you! Look at that problem using this website uses cookies to ensure you get quotient. Must match in order to add or subtract radicals algebraic term to get rid of,. About it instead of a larger expression, but that radical is part of a product Cookie! Notice that both radicals are cube roots using this approach the radicals is equal to the radical together that using! With variables ( Basic with no rationalizing ) tutorial we will multiply two roots... - simplify radical expressions and Quadratic Equations, from Developmental math: an Open Program a. Keto And Co Pancake Mix Reviews, How Much Is It To Go Hang Gliding, Northeast Apartments For Rent, Spring Onion Tamil Meaning, Python Developer Salary In Ahmedabad, Anchor Hocking 8-ounce Triple Pour Measuring Cup, Clear, Copper Tools Minecraft, Relation Between Jarasandha And Shishupal, Bru Instant Coffee, 200g Price, Fishing Charters Port Isabel, Sprinkler System Cost Canada, Highest Paid Edward Jones Advisors, " />
# how to divide radical expressions with variables
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That choice is made so that after they are multiplied, everything under the radical sign will be perfect cubes. This property can be used to combine two radicals into one. $\begin{array}{r}2\cdot \left| 2 \right|\cdot \left| {{x}^{2}} \right|\cdot \sqrt[4]{x}\cdot \sqrt[4]{{{y}^{3}}}\cdot \left| 3 \right|\cdot \sqrt[4]{{{x}^{3}}y}\\2\cdot 2\cdot {{x}^{2}}\cdot \sqrt[4]{x}\cdot \sqrt[4]{{{y}^{3}}}\cdot 3\cdot \sqrt[4]{{{x}^{3}}y}\end{array}$. Simplify. Even the smallest statement like $x\ge 0$ can influence the way you write your answer. Divide Radical Expressions. Previous This algebra video tutorial shows you how to perform many operations to simplify radical expressions. Multiply all numbers and variables outside the radical together. Let’s deal with them separately. Simplifying radical expressions: two variables. Notice that each group of numbers or variables gets written once when they move outside the radical because they are now one group. It is important to read the problem very well when you are doing math. $\sqrt[3]{3x{{y}^{3}}}\\\sqrt[3]{{{(y)}^{3}}\cdot \,3x}$, $\sqrt[3]{{{(y)}^{3}}}\cdot \,\sqrt[3]{3x}$. Use the rule $\sqrt[x]{\frac{a}{b}}=\frac{\sqrt[x]{a}}{\sqrt[x]{b}}$ to create two radicals; one in the numerator and one in the denominator. In our next example, we will multiply two cube roots. When dividing radical expressions, the rules governing quotients are similar: $\sqrt[x]{\frac{a}{b}}=\frac{\sqrt[x]{a}}{\sqrt[x]{b}}$. Look for perfect cubes in the radicand. Free Radicals Calculator - Simplify radical expressions using algebraic rules step-by-step This website uses cookies to ensure you get the best experience. Since ${{x}^{7}}$ is not a perfect cube, it has to be rewritten as ${{x}^{6+1}}={{({{x}^{2}})}^{3}}\cdot x$. Divide Radical Expressions. Use the quotient rule to divide radical expressions. You can do more than just simplify radical expressions. As long as the roots of the radical expressions are the same, you can use the Product Raised to a Power Rule to multiply and simplify. Then simplify and combine all like radicals. Example 7: Simplify the radical expression \sqrt {12{x^2}{y^4}} . And then that would just become a y to the first power. Identify perfect cubes and pull them out. Practice: Multiply & divide rational expressions (advanced) Next lesson. There's a similar rule for dividing two radical expressions. The answer is $\frac{4\sqrt{3}}{5}$. Remember that the order you choose to use is up to you—you will find that sometimes it is easier to multiply before simplifying, and other times it is easier to simplify before multiplying. The Product Rule states that the product of two or more numbers raised to a power is equal to the product of each number raised to the same power. $\begin{array}{r}640\div 40=16\\\sqrt[3]{16}\end{array}$. Recall that the Product Raised to a Power Rule states that $\sqrt[x]{ab}=\sqrt[x]{a}\cdot \sqrt[x]{b}$. This process is called rationalizing the denominator. 2. Recall that the Product Raised to a Power Rule states that $\sqrt[x]{ab}=\sqrt[x]{a}\cdot \sqrt[x]{b}$. The denominator here contains a radical, but that radical is part of a larger expression. The same is true of roots: $\sqrt[x]{ab}=\sqrt[x]{a}\cdot \sqrt[x]{b}$. Step 4: Simplify the expressions both inside and outside the radical by multiplying. Multiply all numbers and variables inside the radical together. Quiz Multiplying Radical Expressions, Next With some practice, you may be able to tell which is easier before you approach the problem, but either order will work for all problems. All rights reserved. Dividing radicals is really similar to multiplying radicals. Dividing Radical Expressions. You can use the same ideas to help you figure out how to simplify and divide radical expressions. Look for perfect cubes in the radicand, and rewrite the radicand as a product of factors. Simplify. It does not matter whether you multiply the radicands or simplify each radical first. That was a lot of effort, but you were able to simplify using the Quotient Raised to a Power Rule. One way of simplifying radical expressions is to break down the expression into perfect squares multiplying each other. $2\sqrt[4]{{{(2)}^{4}}\cdot {{({{x}^{2}})}^{4}}\cdot x}\cdot \sqrt[4]{{{y}^{3}}}\cdot \sqrt[4]{{{(3)}^{4}}\cdot {{x}^{3}}y}$, $2\sqrt[4]{{{(2)}^{4}}}\cdot \sqrt[4]{{{({{x}^{2}})}^{4}}}\cdot \sqrt[4]{x}\cdot \sqrt[4]{{{y}^{3}}}\cdot \sqrt[4]{{{(3)}^{4}}}\cdot \sqrt[4]{{{x}^{3}}y}$. Remember that when we multiply radicals with the same type of root, we just multiply the radicands and put the product under a radical sign. You can multiply and divide them, too. Simplify. In this case, notice how the radicals are simplified before multiplication takes place. and any corresponding bookmarks? As with multiplication, the main idea here is that sometimes it makes sense to divide and then simplify, and other times it makes sense to simplify and then divide. CliffsNotes study guides are written by real teachers and professors, so no matter what you're studying, CliffsNotes can ease your homework headaches and help you score high on exams. $2\sqrt[4]{16{{x}^{9}}}\cdot \sqrt[4]{{{y}^{3}}}\cdot \sqrt[4]{81{{x}^{3}}y}$, $x\ge 0$, $y\ge 0$. That's a mathematical symbols way of saying that when the index is even there can be no negative number in the radicand, but when the index is odd, there can be. Whichever order you choose, though, you should arrive at the same final expression. Simplify. Are you sure you want to remove #bookConfirmation# Divide the coefficients, and divide the variables. Dividing rational expressions: unknown expression. Use the Quotient Raised to a Power Rule to rewrite this expression. Simplify each radical, if possible, before multiplying. $\begin{array}{r}\sqrt[3]{{{(2)}^{3}}\cdot 2}\\\sqrt[3]{{(2)}^{3}}\cdot\sqrt[3]{2}\end{array}$. Now that the radicands have been multiplied, look again for powers of $4$, and pull them out. The multiplication of the denominator by its conjugate results in a whole number (okay, a negative, but the point is that there aren't any radicals): Welcome to MathPortal. In our first example, we will work with integers, and then we will move on to expressions with variable radicands. Next look at the variable part. You multiply radical expressions that contain variables in the same manner. How to divide algebraic terms or variables? Identify and pull out powers of $4$, using the fact that $\sqrt[4]{{{x}^{4}}}=\left| x \right|$. The Product Raised to a Power Rule is important because you can use it to multiply radical expressions. Radical expressions are written in simplest terms when. The answer is $12{{x}^{3}}y,\,\,x\ge 0,\,\,y\ge 0$. Simplify. It can also be used the other way around to split a radical into two if there's a fraction inside. To get rid of it, I'll multiply by the conjugate in order to "simplify" this expression. Look for perfect squares in each radicand, and rewrite as the product of two factors. $\begin{array}{r}\sqrt{36\cdot {{x}^{4+2}}}\\\sqrt{36\cdot {{x}^{6}}}\end{array}$. Recall the rule: For any numbers a and b and any integer x: ${{(ab)}^{x}}={{a}^{x}}\cdot {{b}^{x}}$, For any numbers a and b and any positive integer x: ${{(ab)}^{\frac{1}{x}}}={{a}^{\frac{1}{x}}}\cdot {{b}^{\frac{1}{x}}}$, For any numbers a and b and any positive integer x: $\sqrt[x]{ab}=\sqrt[x]{a}\cdot \sqrt[x]{b}$. Dividing Radicals without Variables (Basic with no rationalizing). Identify perfect cubes and pull them out of the radical. Use the quotient rule to simplify radical expressions. Quiz Dividing Radical Expressions. $\sqrt{{{(12)}^{2}}\cdot 2}$, $\sqrt{{{(12)}^{2}}}\cdot \sqrt{2}$. • The radicand and the index must be the same in order to add or subtract radicals. $\sqrt[3]{\frac{640}{40}}$. In the next example, we will use the same product from above to show that you can simplify before multiplying and get the same result. For the numerical term 12, its largest perfect square factor is 4. from your Reading List will also remove any $\frac{\sqrt[3]{24x{{y}^{4}}}}{\sqrt[3]{8y}},\,\,y\ne 0$, $\sqrt[3]{\frac{24x{{y}^{4}}}{8y}}$. Slopes of Parallel and Perpendicular Lines, Quiz: Slopes of Parallel and Perpendicular Lines, Linear Equations: Solutions Using Substitution with Two Variables, Quiz: Linear Equations: Solutions Using Substitution with Two Variables, Linear Equations: Solutions Using Elimination with Two Variables, Quiz: Linear Equations: Solutions Using Elimination with Two Variables, Linear Equations: Solutions Using Matrices with Two Variables, Linear Equations: Solutions Using Graphing with Two Variables, Quiz: Linear Equations: Solutions Using Graphing with Two Variables, Quiz: Linear Equations: Solutions Using Matrices with Two Variables, Linear Equations: Solutions Using Determinants with Two Variables, Quiz: Linear Equations: Solutions Using Determinants with Two Variables, Linear Inequalities: Solutions Using Graphing with Two Variables, Quiz: Linear Inequalities: Solutions Using Graphing with Two Variables, Linear Equations: Solutions Using Matrices with Three Variables, Quiz: Linear Equations: Solutions Using Matrices with Three Variables, Linear Equations: Solutions Using Determinants with Three Variables, Quiz: Linear Equations: Solutions Using Determinants with Three Variables, Linear Equations: Solutions Using Elimination with Three Variables, Quiz: Linear Equations: Solutions Using Elimination with Three Variables, Quiz: Trinomials of the Form x^2 + bx + c, Quiz: Trinomials of the Form ax^2 + bx + c, Adding and Subtracting Rational Expressions, Quiz: Adding and Subtracting Rational Expressions, Proportion, Direct Variation, Inverse Variation, Joint Variation, Quiz: Proportion, Direct Variation, Inverse Variation, Joint Variation, Adding and Subtracting Radical Expressions, Quiz: Adding and Subtracting Radical Expressions, Solving Quadratics by the Square Root Property, Quiz: Solving Quadratics by the Square Root Property, Solving Quadratics by Completing the Square, Quiz: Solving Quadratics by Completing the Square, Solving Quadratics by the Quadratic Formula, Quiz: Solving Quadratics by the Quadratic Formula, Quiz: Solving Equations in Quadratic Form, Quiz: Systems of Equations Solved Algebraically, Quiz: Systems of Equations Solved Graphically, Systems of Inequalities Solved Graphically, Systems of Equations Solved Algebraically, Quiz: Exponential and Logarithmic Equations, Quiz: Definition and Examples of Sequences, Binomial Coefficients and the Binomial Theorem, Quiz: Binomial Coefficients and the Binomial Theorem, Online Quizzes for CliffsNotes Algebra II Quick Review, 2nd Edition. https://www.khanacademy.org/.../v/multiply-and-simplify-a-radical-expression-2 As long as the roots of the radical expressions are the same, you can use the Product Raised to a Power Rule to multiply and simplify. Rationalizing the Denominator. $\frac{\sqrt[3]{640}}{\sqrt[3]{40}}$. Dividing Algebraic Expressions . We have used the Quotient Property of Radical Expressions to simplify roots of fractions. Look at the two examples that follow. The indices of the radicals must match in order to multiply them. To rationalize this denominator, the appropriate fraction with the value 1 is , since that will eliminate the radical in the denominator, when used as follows: Note we elected to find 's principal root. By using this website, you agree to our Cookie Policy. Use the quotient raised to a power rule to divide radical expressions (9.4.2) – Add and subtract radical expressions (9.4.3) – Multiply radicals with multiple terms (9.4.4) – Rationalize a denominator containing a radical expression $\begin{array}{l}5\sqrt[3]{{{x}^{5}}{{y}^{2}}\cdot 8{{x}^{2}}{{y}^{4}}}\\5\sqrt[3]{8\cdot {{x}^{5}}\cdot {{x}^{2}}\cdot {{y}^{2}}\cdot {{y}^{4}}}\\5\sqrt[3]{8\cdot {{x}^{5+2}}\cdot {{y}^{2+4}}}\\5\sqrt[3]{8\cdot {{x}^{7}}\cdot {{y}^{6}}}\end{array}$. Conjugates are used for rationalizing the denominator when the denominator is a two‐termed expression involving a square root. $\begin{array}{r}2\cdot 2\cdot 3\cdot {{x}^{2}}\cdot \sqrt[4]{x\cdot {{y}^{3}}\cdot {{x}^{3}}y}\\12{{x}^{2}}\sqrt[4]{{{x}^{1+3}}\cdot {{y}^{3+1}}}\end{array}$. Rewrite using the Quotient Raised to a Power Rule. In both cases, you arrive at the same product, $12\sqrt{2}$. Now let's think about it. Multiplying rational expressions. Simplify $\sqrt[3]{\frac{24x{{y}^{4}}}{8y}}$ by identifying similar factors in the numerator and denominator and then identifying factors of $1$. Look for perfect square factors in the radicand, and rewrite the radicand as a product of factors. Note that we specify that the variable is non-negative, $x\ge 0$, thus allowing us to avoid the need for absolute value. We have used the Quotient Property of Radical Expressions to simplify roots of fractions. Now take another look at that problem using this approach. There is a rule for that, too. Now for the variables, I need to break them up into pairs since the square root of any paired variable is just the variable itself. Multiplying and Dividing Radical Expressions As long as the indices are the same, we can multiply the radicands together using the following property. You can use the same ideas to help you figure out how to simplify and divide radical expressions. The steps below show how the division is carried out. Divide radicals that have the same index number. Use the rule $\sqrt[x]{a}\cdot \sqrt[x]{b}=\sqrt[x]{ab}$ to multiply the radicands. An expression with a radical in its denominator should be simplified into one without a radical in its denominator. $\sqrt{{{(6)}^{2}}\cdot {{({{x}^{3}})}^{2}}}$, $\begin{array}{c}\sqrt{{{(6)}^{2}}}\cdot \sqrt{{{({{x}^{3}})}^{2}}}\\6\cdot {{x}^{3}}\end{array}$. Notice this expression is multiplying three radicals with the same (fourth) root. In the next video, we show more examples of simplifying a radical that contains a quotient. $\frac{\sqrt{30x}}{\sqrt{10x}},x>0$. There is a rule for that, too. In this example, we simplify √(2x²)+4√8+3√(2x²)+√8. The product raised to a power rule that we discussed previously will help us find products of radical expressions. We will need to use this property ‘in reverse’ to simplify a fraction with radicals. So when you divide one radical expression by another, you can simplify it by writing both expressions under the same radical, then simplifying. Radical Expression Playlist on YouTube. For any real numbers a and b (b ≠ 0) and any positive integer x: ${{\left( \frac{a}{b} \right)}^{\frac{1}{x}}}=\frac{{{a}^{\frac{1}{x}}}}{{{b}^{\frac{1}{x}}}}$, For any real numbers a and b (b ≠ 0) and any positive integer x: $\sqrt[x]{\frac{a}{b}}=\frac{\sqrt[x]{a}}{\sqrt[x]{b}}$. Right away and then that would just become a y to the first Power first! Turn to some radical expressions square is the … now let us turn to some expressions! To expressions with variable radicands [ /latex ] to multiply radical expressions term by another algebraic term get... Square root and the denominator when the denominator that contain only numbers us turn to some radical expressions to using! Multiplying three radicals with variables can divide an algebraic term to get rid of it, 'll... 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http://intellification.net/what-is-machine-learning/mathematicl-background/linear-algebra | Linear Algebra
# Linear Algebra
Linear algebra is the branch of mathematics concerning linear equations and numbers represented in scalars, vectors, matrices, and tensors. As activations, parameters or weights in machine learning and neural networks are usually denoted as vectors, matrices or tensors, Linear Algebra is central to the underlying theory of machine learning.
## Set of Numbers
Scalars, vectors, matrices, and tensors are containing numbers, and these numbers belong to sets of numbers. There are a few common sets of numbers:
$$\mathbb{N}$$ represents the set of positive integers $$(1, 2, 3, 4, …)$$ (dependent on the actual definition 0 belongs to this set or not).
$$\mathbb{Z}$$ represents the set of negative, zero and positive integers $$(…, -4, -3, -2, -1, 0, 1, 2, 3, 4, …)$$.
$$\mathbb{Q}$$ represents the set of rational numbers (numbers that may be expressed as a fraction of two integers).
$$\mathbb{R}$$ represents the set of real-valued numbers, which contains the rational numbers ($$\mathbb{Q}$$) and the non-rational numbers like $$\pi$$ or $$\sqrt{2}$$. In the following, scalars, vectors, matrices, or tensors are usually containing numbers from the set $$\mathbb{R}$$.
## Scalars, Vectors, Matrices, and Tensors
### Scalars
Scalars are single values like:
x \in \mathbb{R}
### Vectors
Vectors are ordered one-dimensional lists of $$n \in \mathbb{N}$$ single numbers or scalars. They are noted in boldface lower case letters:
\textbf{x} = [x_1, x_2, …., x_n] \in \mathbb{R}^n
Vectors with $$n$$ numbers can be interpreted as points in an $$n$$-dimensional vector space.
### Matrices
Matrices are rectangular two-dimensional arrays consisting of numbers or scalars. Matrices are denoted in boldface upper case letters:
\boldsymbol{A}=\begin{bmatrix}
\kern4pt a_{11} & a_{12} & a_{13} & \ldots & a_{1n} \kern4pt \\
\kern4pt a_{21} & a_{22} & a_{23} & \ldots & a_{2n} \kern4pt \\
\kern4pt a_{31} & a_{32} & a_{33} & \ldots & a_{3n} \kern4pt \\
\kern4pt \vdots & \vdots & \vdots & \ddots & \vdots \kern4pt \\
\kern4pt a_{m1} & a_{m2} & a_{m3} & \ldots & a_{mn} \kern4pt \\
\end{bmatrix} \in \mathbb{R}^{mxn}
The matrix $$\boldsymbol{A}$$ can also be written as:
\boldsymbol{A} = [a_{ij}]_{m \times n} \; m, n \in \mathbb{N}\
For a $$\boldsymbol{A}^{mxn}$$ Matrix, $$m$$ always denotes the number of rows and $$n$$ always denotes the number of columns.
Vectors can either be a row vector (which is a $$1 \times m$$ Matrix):
\textbf{x} = [x_1, x_2, …., x_n]
or a column vector (which is a $$m \times 1$$ Matrix):
\textbf{x}{^T}=\begin{bmatrix}
\kern4pt x_1 \kern4pt \\
\kern4pt x_2 \kern4pt \\
\kern4pt \vdots \kern4pt \\
\kern4pt x_n \kern4pt
\end{bmatrix}
Column vectors are transformed row-vectors and therefore noted as $$\textbf{x}{^T}$$.
### Tensors
Tensors are more general entities that encapsulate scalars, vectors and matrices. Scalars are 0th-order tensors, vectors are 1st-order tensors and matrices are 2th-order tensors. Tensors can have higher orders than 2. A tensor that represents a 2-dimensional pixel image where each pixel is represented by three numbers describing the three colors (RGB) is a 3rd-order tensor (sometimes also called a 3-dimensional matrix).
## Operations on Vectors and Matrices
While everybody is familiar with operations on scalars, we will look now into the operations addition and multiplication with vectors and matrices. This can be generalized to tensors.
### Matrix Transpose
If you have an $$m \times n$$ Matrix $$\boldsymbol{A}$$:
\boldsymbol{A} = [a_{ij}]_{m \times n} \; m, n \in \mathbb{N}\
then the transposed Matrix $$\boldsymbol{A}^{T}$$ has the size $$n \times m$$:
\boldsymbol{A}^{T} = [a_{ji}]_{n \times m} \; m, n \in \mathbb{N}\
The transpose of a matrix swapps the indices $$i$$ and $$j$$.
#### Examples:
$$\boldsymbol{A}$$ is a $$2 \times 3$$ Matrix and $$\boldsymbol{A}^{T}$$ is the transposed $$3 \times 2$$ matrix:
\boldsymbol{A} = \left[ \begin{array}{ccc}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23}
\boldsymbol{A}^T = \left[ \begin{array}{cc}
a_{11} & a_{21} \\
a_{12} & a_{22} \\
a_{13} & a_{23}
\end{array} \right]
The transpose of a $$1 \times n$$ row vector leads to a $$n \times 1$$ column vector and vice versa.
\textbf{x} = [x_1, x_2, …., x_n] , \quad
\textbf{x}{^T}=\begin{bmatrix}
\kern4pt x_1 \kern4pt \\
\kern4pt x_2 \kern4pt \\
\kern4pt \vdots \kern4pt \\
\kern4pt x_n \kern4pt
\end{bmatrix}
### Matrix-Matrix Multiplication
In a matrix-matrix multiplication, each element of the resulting matric is calculated by an entire row of the first matrix and an entire column of the second matrix. Therefore, matrix-matrix multiplications are only defined for matrices with a certain size. The first matrix must have as many columns as the second matrix has rows, and the resulting matrix has the number of rows from the first matrix and the number of columns of the second matrix:
\boldsymbol{A}^{\, m \times n} \boldsymbol{B}^{\, n \times p} = \boldsymbol{C}^{\; m \times p}
If $$\boldsymbol{A}=[a_{ij}]_{m \times n}$$ and $$\boldsymbol{B}=[b_{jk}]_{n \times p}$$, then the matrix product is defined as:
\boldsymbol{C}=\boldsymbol{A}\boldsymbol{B}=[c_{ik}]_{m \times p}
c_{ik} = \sum^n_{j=1} a_{ij} b_{jk}
The element $$c_{ik}$$ of the matrix $$\boldsymbol{C}=\boldsymbol{AB}$$ are given by summing the products of the elements of the $$i$$-th row of $$\boldsymbol{A}$$ with the elements of the $$k$$-th column of $$\boldsymbol{B}$$.
#### Examples
\boldsymbol{A} = \left[ \begin{array}{ccc}
2 & 5 & 1 \\
7 & 3 & 6
\boldsymbol{B} = \left[ \begin{array}{cc}
1 & 8 \\
9 & 4 \\
3 & 5
\end{array} \right]
\boldsymbol{AB} = \left[ \begin{array}{cc}
2 \cdot 1 + 5 \cdot 9 + 1 \cdot 3 & 2 \cdot 8 + 5 \cdot 4 + 1 \cdot 5 \\
7 \cdot 1 + 3 \cdot 9 + 6 \cdot 3 & 7 \cdot 8 + 3 \cdot 4 + 6 \cdot 5
\end{array} \right] =
\left[ \begin{array}{cc}
50 & 41 \\
52 & 98
\end{array} \right]
\boldsymbol{BA} = \left[ \begin{array}{ccc}
1 \cdot 2 + 8 \cdot 7 & 1 \cdot 5 + 8 \cdot 3 & 1 \cdot 1 + 8 \cdot 6 \\
9 \cdot 2 + 4 \cdot 7 & 9 \cdot 5 + 4 \cdot 3 & 9 \cdot 1 + 4 \cdot 6 \\
3 \cdot 2 + 5 \cdot 7 & 3 \cdot 5 + 5 \cdot 3 & 3 \cdot 1 + 5 \cdot 6 \\
\end{array} \right] =
\left[ \begin{array}{ccc}
58 & 29 & 49 \\
46 & 57 & 33 \\
41 & 30 & 33
\end{array} \right] | 2021-05-14T13:35:35 | {
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https://tutorme.com/tutors/701695/interview/ | Enable contrast version
# Tutor profile: Christopher T.
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Christopher T.
Tutor with 14 years experience at the college level.
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## Questions
### Subject:Discrete Math
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Question:
Use the Principle of Mathematical induction to show that $$\displaystyle\sum_{j=1}^nj\binom{n}{j} = n2^{n-1}$$ for all $$n\geq 1$$.
Inactive
Christopher T.
$$\textit{Proof}$$: $$\textbf{Base Case } (n=1)$$: We see that $(\sum_{j=1}^1j\binom{1}{k} = 1\cdot \binom{1}{1} = 1 = 1\cdot 2^0 = 1\cdot 2^{1-1}.$) Therefore, the base case is true. $$\textbf{Inductive Hypothesis}$$: Assume for any $$k\geq 1$$ that $$\displaystyle\sum_{j=1}^k j\binom{k}{j} = k2^{k-1}$$. $$\textbf{Inductive Step } (n=k+1)$$: We seek to show that $(\sum_{j=1}^{k+1}j\binom{k+1}{j} = (k+1)2^k.$) We first note that $(\sum_{j=1}^{k+1}j\binom{k+1}{j} = \sum_{j=1}^{k}j\binom{k+1}{j} + (k+1)\binom{k+1}{k+1} = \sum_{j=1}^{k}j\binom{k+1}{j} + k+1$) Next, recall Pascal's identity $(\binom{n+1}{k} = \binom{n}{k}+\binom{n}{k-1}$) Thus, (\begin{aligned} \sum_{j=1}^{k}j\binom{k+1}{j} + k+1 &= \sum_{j=1}^{k}j\left(\binom{k}{j}+\binom{k}{j-1}\right) + k+1\\ &= \sum_{j=1}^{k}j\binom{k}{j} + \sum_{j=1}^{k}j\binom{k}{j-1} + k+1 \\ &= \sum_{j=1}^{k}j\binom{k}{j} + \sum_{j=1}^{k}(j-1+1)\binom{k}{j-1} + k+1 \\ &= \sum_{j=1}^{k}j\binom{k}{j} + {\color{red}\sum_{j=1}^{k}(j-1)\binom{k}{j-1}} + {\color{red}k} + {\color{blue}\sum_{j=1}^k \binom{k}{j-1}} +{\color{blue}1}\end{aligned}\tag{1}) Let's focus on the sum $(\sum_{j=1}^k (j-1)\binom{k}{j-1}.$) Expanding it out, we see that $(\sum_{j=1}^k (j-1)\binom{k}{j-1} = 0\cdot\binom{k}{0} + 1\cdot \binom{k}{1}+2\cdot \binom{k}{2}+\cdots + (k-1)\binom{k}{k-1}.$) Now note that $$k = k\cdot 1 = k\cdot \dbinom{k}{k}$$. Therefore, (\begin{aligned}{\color{red}\sum_{j=1}^{k}(j-1)\binom{k}{j-1}}+{\color{red}k} &= 0\cdot\binom{k}{0} + 1\cdot \binom{k}{1}+2\cdot \binom{k}{2}+\cdots + (k-1)\binom{k}{k-1} + k\binom{k}{k} \\ &= 1\cdot \binom{k}{1}+2\cdot \binom{k}{2}+\cdots + (k-1)\binom{k}{k-1} + k\binom{k}{k}\\ &= {\color{red}\sum_{j=1}^k j\binom{k}{j}}.\end{aligned}\tag{2}) Now let us consider the sum $(\sum_{j=1}^k\binom{k}{j-1}.$) Expanding it out, we see that $(\sum_{j=1}^k\binom{k}{j-1} = \binom{k}{0} + \binom{k}{1}+\binom{k}{2} + \cdots + \binom{k}{k-1}.$) Now note that $$1=\binom{k}{k}$$. Therefore, (\begin{aligned}{\color{blue}\sum_{j=1}^k\binom{k}{j-1}}+{\color{blue}1} &= \binom{k}{0} + \binom{k}{1}+\binom{k}{2} + \cdots + \binom{k}{k-1} + \binom{k}{k} \\ &= {\color{blue}\sum_{j=0}^k\binom{k}{j}}\end{aligned}\tag{3}) Substituting the results (2) and (3) into the last equation in (1) gives us (\begin{aligned} \sum_{j=1}^{k}j\binom{k}{j} + {\color{red}\sum_{j=1}^{k}(j-1)\binom{k}{j-1}} + {\color{red}k} + {\color{blue}\sum_{j=1}^k \binom{k}{j-1}} +{\color{blue}1} &= \sum_{j=1}^kj\binom{k}{j} + {\color{red}\sum_{j=1}^kj\binom{k}{j}} + {\color{blue}\sum_{j=0}^k\binom{k}{j}}\\ &= 2\sum_{j=1}^kj\binom{k}{j} + \sum_{j=0}^k\binom{k}{j} \\ &= 2\cdot k2^{k-1}+ 2^k\quad\left(\text{by IH and\sum_{i=0}^m\binom{m}{i}=2^m}\right)\\ &= k2^k+2^k\\ &= (k+1)2^k.\end{aligned}) This completes the inductive step. Therefore, $$\displaystyle\sum_{j=1}^nj\binom{n}{j} = n2^{n-1}$$ for all $$n\geq 1$$.$$\hspace{.25in}\blacksquare$$
### Subject:Statistics
TutorMe
Question:
A survey of 15 large U.S. cities finds that the average commute time one way is 25.4 minutes. A chamber of commerce executive feels that the commute time in his city is less and wants to publicize this. He randomly selects 25 commuters and finds the average is 22.1 minutes with a standard deviation of 5.3 minutes. At $$\alpha=0.1$$, is he correct?
Inactive
Christopher T.
We outline all the steps in the hypothesis test. We apply a $$t$$-test for means here. $$\textbf{Step 1}$$: We identify our hypotheses as follows: \hspace{.25in}\begin{aligned}[t]H_0: \mu &= 25.4\\ H_1: \mu &< 25.4\quad (\text{claim})\end{aligned}. This is a left-tailed test. $$\textbf{Step 2}$$: Find the critical value(s). Since $$\alpha=0.1$$, we find that our critical value is $$CV=-1.318$$ (note that we use the $$t$$-distribution table with $$\alpha=0.1$$ in one-tail and $$25-1=24$$ degrees of freedom to find $$t_{0.1}$$ but take the CV to be negative since it's a left-tail test); any test value that is less than $$-1.318$$ will lie in the rejection region for our test and would force us to reject $$H_0$$. $$\textbf{Step 3}$$: Find the test value. For the $$t$$-test, the test value is given by $$t=\dfrac{\overline{X}-\mu}{s/\sqrt{n}}$$. In the problem, we're given $$\overline{X}=22.1$$, $$s=5.3$$, $$n=25$$, and $$\mu=25.4$$ (from the null hypothesis). Thus, our test value is $$t=\dfrac{22.1-25.4}{5.3/\sqrt{25}}\approx -3.113$$. $$\textbf{Step 4}$$: Decide whether to reject or not reject $$H_0$$. Since our test value is smaller than the critical value ($$-3.113<-1.318$$), we reject $$H_0$$. $$\textbf{Step 5}$$: Summarize the results. There is sufficient evidence to support the claim that the average commute time in his city is less than $$25.4$$ minutes.
### Subject:Differential Equations
TutorMe
Question:
Find the solution to the initial value problem $(\mathbf{x}^{\prime} = \begin{bmatrix}2 & \frac{3}{2}\\ -\frac{3}{2} & -1\end{bmatrix}\mathbf{x},\qquad \mathbf{x}(0)=\begin{bmatrix}3\\-2\end{bmatrix}.$)
Inactive
Christopher T.
We first find the eigenvalues and eigenvectors of $$A$$. Since $$A=\begin{bmatrix}2 & \frac{3}{2}\\ -\frac{3}{2} & -1\end{bmatrix}$$, we find that $(p(\lambda) = \lambda^2-(\mathrm{tr} A)\lambda + \det A = \lambda^2-\lambda+\tfrac{1}{4}.$) Thus, $$p(\lambda)= 0 \implies \lambda^2-\lambda+\tfrac{1}{4}= 0 \implies (\lambda-\tfrac{1}{2})^2=0\implies \lambda=\tfrac{1}{2}$$ with multiplicity two. Since $$A$$ isn't a diagonal matrix, $$A$$ isn't diagonalizable. We now find the eigenvector associated with $$\lambda=\tfrac{1}{2}$$. If $$\lambda=\tfrac{1}{2}$$, then (A-\lambda I)\mathbf{v} = \mathbf{0}\implies \begin{bmatrix}\frac{3}{2} & \frac{3}{2}\\ -\frac{3}{2} & -\frac{3}{2}\end{bmatrix}\begin{bmatrix}v_1\\v_2\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}\implies \left\{\begin{aligned} \tfrac{3}{2}v_1+\tfrac{3}{2}v_2 &= 0,\\ -\tfrac{3}{2}v_1-\tfrac{3}{2}v_2 &= 0.\end{aligned}\right. This system is redundant (since second equation is negative of the first), so we have $$\tfrac{3}{2}v_1+\tfrac{3}{2}v_2=0$$. Letting $$v_2$$ be free, we have that $$v_1=-v_2$$. Hence, $(\mathbf{v} = \begin{bmatrix}v_1\\v_2\end{bmatrix} = \begin{bmatrix}-v_2\\v_2\end{bmatrix} = v_2\begin{bmatrix}-1\\1\end{bmatrix}\implies \mathbf{v} = \begin{bmatrix}-1\\1\end{bmatrix}\quad (v_2=1).$) We now seek to find a vector $$\mathbf{w}$$ such that ((A-\lambda I)\mathbf{w}=\mathbf{v} \implies \begin{bmatrix}\frac{3}{2} & \frac{3}{2}\\ -\frac{3}{2} & -\frac{3}{2}\end{bmatrix}\begin{bmatrix}w_1\\w_2\end{bmatrix} = \begin{bmatrix}-1\\1\end{bmatrix}\implies \left\{\begin{aligned} \tfrac{3}{2}w_1+\tfrac{3}{2}w_2 &= -1,\\ -\tfrac{3}{2}w_1-\tfrac{3}{2}w_2 &= 1.\end{aligned}\right.) This system is redundant (since second equation is negative of the first), so we have $$\tfrac{3}{2}w_1+\tfrac{3}{2}w_2=-1$$. Letting $$w_2$$ be free, we have that $$\tfrac{3}{2}w_1=-1-\tfrac{3}{2}w_2\implies w_1=-\tfrac{2}{3}-w_2$$. Hence, $(\mathbf{w} = \begin{bmatrix}w_1\\w_2\end{bmatrix} = \begin{bmatrix}-\frac{2}{3}-w_2\\w_2\end{bmatrix} = \begin{bmatrix}-\frac{2}{3}\\0\end{bmatrix}+w_2\begin{bmatrix}-1\\1\end{bmatrix}\implies \mathbf{w} = \begin{bmatrix}-\frac{2}{3}\\0\end{bmatrix}\quad (w_2=0).$) Thus, $$\mathbf{x}(t) = c_1\mathbf{v}e^{\lambda t}+c_2(t\mathbf{v}+\mathbf{w})e^{\lambda t}\implies \mathbf{x}(t) = c_1\begin{bmatrix}-1\\1\end{bmatrix}e^{\frac{t}{2}}+c_2\left(t\begin{bmatrix}-1\\1\end{bmatrix}+\begin{bmatrix}-\frac{2}{3}\\0\end{bmatrix}\right)e^{\frac{t}{2}}$$. Applying the initial condition $$\mathbf{x}(0)=\begin{bmatrix}3\\-2\end{bmatrix}$$, we have (c_1\begin{bmatrix}-1\\1\end{bmatrix}+c_2\begin{bmatrix}-\frac{2}{3}\\0\end{bmatrix} = \begin{bmatrix}3\\-2\end{bmatrix}\implies \left\{\begin{aligned} -c_1-\tfrac{2}{3}c_2 &= 3\\ c_1 &= -2\end{aligned}\right.\implies c_1=-2,c_2=-\tfrac{3}{2}.) Therefore, $$\boxed{\mathbf{x}(t) = -2\begin{bmatrix}-1\\1\end{bmatrix}e^{\frac{t}{2}}-\tfrac{3}{2}\left(t\begin{bmatrix}-1\\1\end{bmatrix}+\begin{bmatrix}-\frac{2}{3}\\0\end{bmatrix}\right)e^{\frac{t}{2}}}$$.
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http://math.stackexchange.com/questions/284525/find-the-value-of-frac1d-1-frac1d-2-dots-frac1d-k | # Find the value of $\frac{1}{d_1}+\frac{1}{d_2}+\dots+\frac{1}{d_k}$.
Let $d_1,d_2,\dots,d_k$ be all the factors of a positive integer $n$ including $1$ and $n$. Suppose $d_1+d_2+\dots+d_k=72$. Then the value of $$\frac{1}{d_1}+\frac{1}{d_2}+\dots+\frac{1}{d_k}$$ is
(a) $\frac{k^2}{72}$
(b) $\frac{72}{k}$
(c) $\frac{72}{n}$
(d) cannot be computed.
I can't express $n$ in terms of $d_1,d_2,\dots,d_k$ . Here I am stuck. Please help.
-
@ThomasAndrews :I just able to do the step $\frac{1}{d_1}+\frac{1}{d_2}+\dots+\frac{1}{d_k}=\frac{d_2d_3\dots d_k+ d_1d_3\dots d_k+\dots +d_1d_2 \dots d_{k-1}}{d_1d_2 \dots d_k}$ – Argha Jan 22 '13 at 21:19
Does it help to write $1 = d_1 < d_2 < \cdots < d_k = n$? – Eric Jan 22 '13 at 21:21
There are only finitely many cases where the factors add up to 72. And these are the numbers 30, 46, 51, 55 and 71. You may want to check what happens to the sum of the reciprocal of their factors.
Now, for the general case, let $1 = d_1 < d_2 < \cdots < d_k = n$ be the factors of $n$. Then $\dfrac{1}{d_1} = \dfrac{d_k}{n}$, and in general $\dfrac{1}{d_i} = \dfrac{d_{k-i+1}}{n}$. Thus,
$\dfrac{1}{d_1} + \cdots \dfrac{1}{d_k} = \dfrac{d_k + d_{k-1} + \cdots + d_1}{n} = \dfrac{\sigma(n)}{n}$,
where $\sigma(n)$ is the sum of factors of $n$.
-
+1 for nice detailed way. ;-) – S. Snape Feb 19 '13 at 15:25
There's no need to find $n$. Note that for every $d$ dividing $n$, $\frac{n}{d}$ is also a divisor of $n$. Therefore the sum of all $d_i$ is equal to the sum of all $\frac{n}{d_i}$, which means...
-
The sum of divisors function is usually written with a letter $\sigma,$ so they are asking about $\sigma(n) = 72.$ Now, for a prime number $p,$ we do get $\sigma(p) = 1 + p,$ so in particular $\sigma(71) = 72.$ And that leads to at least one value for the sum of the reciprocals of the divisors.
We know that $\sigma(n) \geq n+1.$ So, the target 72 demands that $n \leq 71.$ Find out what other numbers solve $\sigma(n) = 72.$
- | 2016-02-07T09:27:37 | {
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https://mathoverflow.net/questions/266269/identity-theorem-for-real-analytic-hypersurfaces | # Identity Theorem for Real-Analytic Hypersurfaces
There's an interesting statement it seems I can prove, but I can't find any references for it, which makes me suspicious of it. So, could someone verify that the statement is correct/incorrect or direct me to references? I've also included what I believe is a proof of this statement.
$\textbf{Background and Notation}$
Let $U \subset \mathbb{R}^n$ be connected and open. Recall that the Identity Theorem for real-analytic functions says: if $f,g: U \to \mathbb{R}$ are real-analytic functions and $f|_V \equiv g|_V$ for a nonempty open $V \subset U$, then $f\equiv g$ on $U$.
Let $M \subset \mathbb{R}^n$. Then, $M$ is an embedded real-analytic hypersurface without boundary if and only if for each $p \in M$, we may find a nonempty open set $U \subset \mathbb{R}^n$ containing $p$ and a real-analytic function $f: U \to \mathbb{R}$ such that $$\{f=0\} = U \cap M$$ and $\nabla f \neq 0$ on $U$.
The following statement is what I would like verification or references for. It is tantamount to an Identity Theorem for embedded real-analytic hypersurfaces. This statement is motivated by mean curvature flow: for an embedded $C^2$ hypersurface as initial conditions, the solution to mean curvature flow is an embedded real-analytic hypersurface for positive time.
$\textbf{Statement:}$
Let $M$ and $N$ be connected, embedded real-analytic hypersurfaces without boundary (as sets) in $\mathbb{R}^n$.
If for some $p \in M \cap N$, there exists an open $n$-ball $B_r(p)$ around $p$ of radius $r>0$ such that $M \cap B_r(p) = N\cap B_r(p)$, then $M = N$.
$\textbf{Proof of Statement:}$
Define $$A:= \{p \in M \cap N \,|\,M \cap B_r(p) = N \cap B_r(p) \,\text{for some}\,r>0 \} \subset \mathbb{R}^n$$ Note that by assumption, $A \neq \emptyset$. Now, we want to consider the closure $\bar{A}$. Since $M \cap N$ is closed, in particular, $\bar{A} \subseteq M \cap N$. Now, we will show that $\bar{A}$ is an open subset of $M$ in the subspace topology and conclude with a connectedness argument.
Let $p \in \bar{A} \setminus A$. So, there exists $p_n \in A$ such that $p_n \to p$. For each $r>0$, we may find some $p_n$, for $n$ large enough, and $r_n>0$ such that $B_{r_n}(p_n) \subset B_r(p)$ and $M \cap B_{r_n}(p_n) = N \cap B_{r_n}(p_n)$.
Since $M$ and $N$ are embedded real-analytic hypersurfaces, we may pick $r$ small enough so that there exist real-analytic functions $f_1,f_2: B_{r}(p) \to \mathbb{R}$ where $$\{f_1 = 0\}=M \cap B_{r}(p) \,\,\,\,\,\,\,\,\,\,\,\,\{f_2 = 0\}= N \cap B_{r}(p)$$ Since we are considering embedded hypersurfaces, we may pick $r$ small enough such that we may apply the analytic implicit function theorem to $\{f_1 = 0\}$ and $\{f_2 = 0\}$. So, we write $M\cap B_{r}(p)$ and $N \cap B_{r}(p)$ as the graphs of real-analytic functions $\tilde{f_1}, \tilde{f_2}: \mathbb{R}^{n-1} \to \mathbb{R}$. Since $p \in M \cap N$, we may arrange so that $\tilde{f_1}$ and $\tilde{f_2}$ are both defined on the same $(n-1)$-ball around the origin in $\mathbb{R}^{n-1}$, with $\tilde{f_1}(0) = \tilde{f_2}(0)$ and $(0,\tilde{f_1}(0))=(0,\tilde{f_2}(0))$ corresponding to $p$. Now, since $B_{r_n}(p_n) \subset B_r(p)$ and $M \cap B_{r_n}(p_n) = N \cap B_{r_n}(p_n)$, we have that $\tilde{f_1} \equiv \tilde{f_2}$ on a nontrivial open subset of where $\tilde{f_1}$ and $\tilde{f_2}$ are defined in $\mathbb{R}^{n-1}$. By the identity theorem for real-analytic functions, $\tilde{f_1} \equiv \tilde{f_2}$ where defined. So, $\{f_1 = 0\} = \{f_2 = 0\}$ on $B_{r}(p)$. That is, $$M \cap B_{r}(p) = \{f_1 = 0\} = \{f_2 = 0\} = N\cap B_{r}(p)$$ and so $p \in A$. This means that $\bar{A} \setminus A = \emptyset$, so $\bar{A} = A$. Since $A$ is open with respect to the subspace topology, we conclude that $\bar{A}$ is open.
Then, since $\bar{A}$ is nonempty, open, and closed with respect to the subspace topology in connected $M$, we conclude that $\bar{A} = M$. The same argument works to show that $\bar{A} = N$. So, $M = N$.
• Yes, I mean the boundary as subsets. Also, I'm assuming $M$ and $N$ are connected, so the case of a line and the pair of nonparallel lines is excluded. Apr 3 '17 at 15:47 | 2021-09-24T07:11:11 | {
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https://mathematica.stackexchange.com/questions/99567/list-manipulation-related | # List manipulation related [duplicate]
I have two lists l1={a,b,c} and l2={{a1},{b1,b2},{c1,c2,c3}}. What kind of operations can I use to get a list like l={{a,a1},{b,b1},{b,b2},{c,c1},{c,c2},{c,c3}}?
I tried to use Thread to get around, but I failed. Any suggestions?
Update: I also have a function f, in the end I want to arrive at something like f@@@l={f[a, a1], f[b, b1], f[b, b2], f[c, c1], f[c, c2], f[c, c3]}. I can do that with Table as follows:
Flatten@Table[f[l1[[i]], #] & /@ l2[[i]], {i, Length@l1}]
But I wonder whether there are other ways.
• @Kuba Thanks for the link. What am I supposed to do with my post then? Shall I close it or delete it? Nov 16 '15 at 7:37
• Posts closed as a duplicate are left here as a road sign for future visitors. And since it's not easy to find such duplicates, that is desired. :)(so do not delete it please :))
– Kuba
Nov 16 '15 at 7:39
For the first request:
Flatten[Thread /@ Thread[{l1, l2}], 1]
For the second request, you just need Listable:
SetAttributes[f, Listable]
Flatten@f[l1, l2]
OK, actually the first request can also be satisfied by setting Listable attribute:
g = Function[{a, b}, {a, b}, Listable]
Flatten[g[l1, l2], 1]
Here are some ways you could use Thread to achieve your goal:
l1 = {a, b, c} ;
l2 = {{a1}, {b1, b2}, {c1, c2, c3}}; | 2022-01-23T15:53:10 | {
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https://math.stackexchange.com/questions/2329528/can-a-binary-operation-have-an-identity-element-when-it-is-not-associative-and-c/2329532 | # Can a binary operation have an identity element when it is not associative and commutative?
I tried getting the answers in similar questions, everyone says that it's not necessary, but if $e$ is the identity element for any binary operation $*$, which is not associative and commutative, how can
$$a*e=a=e*a$$
when it is not commutative, i.e. $a*b \ne b*a$?
Even if we get a value by solving $a*e=a$. Will we get the same value by solving $e*a=a$ ? Please provide an example.
• If $\ast$ has both a left identity $l$ and a right identity $r$, then $l = l \ast r = r$. – Travis Jun 20 '17 at 12:54
• See the wikipedia article on quasigroups, and specifically the section on loops. It has the additional assumption of divisibility, and as such left and right inverses, but is otherwise exactly what you're looking for. The examples section includes such familiar things as the integers with the subtraction operation and the non-zero rationals, reals or complex numbers with division. – Arthur Jun 21 '17 at 9:03
Asserting that the operation $$*$$ is not commutative means that there are elements $$a$$ and $$b$$ such that $$a*b\neq b*a$$. It does not mean that $$a*b\neq b*a$$ for any two distinct elements $$a$$ and $$b$$. Therefore, an operation may well not be commutative and, even so, to have an identity element. There is no contradiction here.
For an example of a non-commutative and non-associative algebraic structure with an identity element, take, for instance, the octonions.
An operation is commutative if for any $a$ and $b$, we have $ab=ba$. Finding one pair $a,b$ such that $ab=ba$ doesn't prove the operation is commutative; this has to hold for every pair.
Consider the set $\{a,b,c\}$ whose binary operation $\cdot$ is given by the following: $$a\cdot a = a\,\,\,\,\,\,\,\,\,\,\, a\cdot b=b\,\,\,\,\,\,\,\,\,\,\,a\cdot c=c$$ $$b\cdot a = b\,\,\,\,\,\,\,\,\,\,\, b\cdot b=b\,\,\,\,\,\,\,\,\,\,\,b\cdot c=c$$ $$c\cdot a = c\,\,\,\,\,\,\,\,\,\,\, c\cdot b=b\,\,\,\,\,\,\,\,\,\,\,c\cdot c=a$$ This operation has $a$ as an identity element. However, it is not commutative (since $b\cdot c\neq c\cdot b$) and it is not associative (since $b\cdot(c\cdot c)=b\neq a =(b\cdot c)\cdot c$).
It is possible. $*$ not being commutative means that $a*b\neq b*a$ for some $a,b$, not for all of them. So you may have $a*e=e*a=a$ without contradicting that $*$ is not commutative.
Actually, given any set $S$ and operation $*$ on it (so possibly neither associative nor commutative), we can simply extend this with a new symbol $\color{red}0$ (i.e., $\color{red}0\notin S$) and on the set $S':=S\cup\{\color{red}0\}$ define an operation $\color{red}*$ by $$x\color{red}*y:=\begin{cases}x&\text{if }y=\color{red}0\\ y&\text{if }x=\color{red}0\\x*y&\text{otherwise} \end{cases}$$ Then $\color{red}*$ is not associative/commutative if $*$ is not associative/commutative. But $\color{red}0$ is neutral.
• Nice. Since all operators with identity can be viewed as arising from this process and because the probability that a randomly choses operator on a set with $n$ elements is either commutative or associative tends to $0$ as $n \rightarrow \infty$, you can show that most operators with identity are neither associative nor commutative. – John Coleman Jun 21 '17 at 13:21
Without looking for esoteric and/or ad hoc examples, there is one you are certainly familiar with. The identity matrix is the identity element for matrix multiplication, which is not commutative. We have $A\,I=I\,A=A$ while in general $A\,B \neq B\,A$.
• This is nice, because although the OP mentioned nonassociativity, the only property invoked in her/his argument is noncommutativity. Therefore matrix addition is a good, familiar example. – Ben Crowell Jun 20 '17 at 21:45
Isn't subtraction a binary function which has an identity ($x-0=x$) although it is not commutative ($5-0 > 0-5$) or associative ($5-(4-3) > (5-4)-3$)?
• Zero is only a right-identity for subtraction. $0 - x = x$ fails. – Zach Effman Jun 20 '17 at 18:45
• Besides, in abstract algebra, subtraction is literally addition. – Obinna Nwakwue Jun 21 '17 at 14:12
• @ObinnaNwakwue, I think one has to be careful with such a statement. Subtraction is literally defined in terms of addition (and additive inverses), but it is not literally addition, any more than multiplication of natural numbers is literally addition because it is defined in terms of it. – LSpice Jun 21 '17 at 22:12
• Yeah, you are right there. – Obinna Nwakwue Jun 22 '17 at 1:53 | 2019-08-19T05:52:39 | {
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https://math.stackexchange.com/questions/1987358/calculate-an-integral-with-riemann-sum | # Calculate an integral with Riemann sum
We know that Riemann sum gives us the following formula for a function $f\in C^1$:
$$\lim_{n\to \infty}\frac 1n\sum_{k=0}^n f\left(\frac kn\right)=\int_0^1f.$$
I am looking for an example where the exact calculation of $\int f$ would be interesting with a Riemann sum.
We usually use integrals to calculate a Riemann sum, but I am interesting in the other direction.
Edit.
I actually found an example of my own today. You can compute
$$I(\rho)=\int_0^\pi \log(1-2\rho \cos \theta+\rho^2)\mathrm d \theta$$
using Riemann sums.
• I suppose you mean an exact calculation? Otherwise the Riemann sum is useful in applied Maths to get an approximation of an integral we cannot compute – Edouard L. Oct 27 '16 at 9:21
• @EdouardL. You are right, that is what I meant. I edited to specify my request. – E. Joseph Oct 27 '16 at 9:23
• Really a side comment, but for some functions the integral equals the Riemann sum for finite $n$, for the example $\cos(x)$ over $[0,2\pi]$. This class has been studied to some extent I believe. – lcv Jan 12 '17 at 15:47
Here is an example ...
For each $z\in\mathbb{C}$ with $\vert z\vert\neq 1$, consider :
$$F(z)=\int_0^{2\pi}\ln\left|z-e^{it}\right|\,dt$$
It is possible to get an explicit form for $F(z)$, using Riemann sums.
For each integer $n\ge1$, consider :
$$S_n=\frac{2\pi}{n}\sum_{k=0}^{n-1}\ln\left|z-e^{2ik\pi/n}\right|$$which is the $n-$th Riemann sum attached to the previous integral (and a uniform subdivision of $[0,2\pi]$ with constant step $\frac{2\pi}{n}$).
Now :$$S_n=\frac{2\pi}{n}\ln\left|\prod_{k=0}^{n-1}\left(z-e^{2ik\pi/n}\right)\right|=\frac{2\pi}{n}\ln\left|z^n-1\right|$$and you can easily show that :$$F(z)=\left\{\matrix{2\pi\ln\left|z\right|& \mathrm{ if}\left|z\right|>1\cr0 & \mathrm{otherwise}}\right.$$
When definite integrals are amenable to exact valuation, it is typically the case that the more expedient approach involves an anti-derivative rather than the limit of a Riemann sum. Often computation of the limit may be straightforward or even trivial, but somewhat tedious, as is the case for integrals of $f: x \mapsto x$ or $f: x \mapsto x^2$.
On the other hand, integrals with simple integrands and easily recognized anti-derivatives such as $f: x\mapsto x^{-2}$ are more challenging with regard to the limit of Riemann sum -- and in that sense the Riemann sum may be "interesting."
To make this more explicit, consider computing the integral
$$\int_a^b x^{-2} \, dx = \lim_{n \to \infty}S_n$$
where
$$S_n =\frac{b-a}{n}\sum_{k=1}^n \left(a + \frac{b-a}{n}k\right)^{-2}.$$
We have
$$\frac{b-a}{n}\sum_{k=1}^n \left(a + \frac{b-a}{n}k\right)^{-1}\left(a + \frac{b-a}{n}(k+1)\right)^{-1} \leqslant S_n \\ \leqslant \frac{b-a}{n}\sum_{k=1}^n \left(a + \frac{b-a}{n}k\right)^{-1}\left(a + \frac{b-a}{n}(k-1)\right)^{-1},$$
and decomposing into partial fractions,
$$\sum_{k=1}^n \left\{\left(a + \frac{b-a}{n}k\right)^{-1}-\left(a + \frac{b-a}{n}(k+1)\right)^{-1}\right\} \leqslant S_n \\\leqslant \sum_{k=1}^n \left\{\left(a + \frac{b-a}{n}(k-1)\right)^{-1}-\left(a + \frac{b-a}{n}k\right)^{-1}\right\}.$$
Since the sums are telescoping, we have
$$\left(a + \frac{b-a}{n}\right)^{-1}-\left(a + \frac{b-a}{n}(n+1)\right)^{-1} \leqslant S_n \leqslant a^{-1} - b^{-1}.$$
By the squeeze theorem, we get the value of the integral as
$$\lim_{n \to \infty}S_n = a^{-1} - b^{-1}.$$
An example where I found the Riemann sum an interesting and, perhaps, most expedient approach is:
Bronstein Integral 21.42
Not a direct answer to your question but I find the following representation of the Riemann sum interesting. Even though it is trivial, these representations show that the sequence of primes or the sequence of composites behave somewhat in a similar same way as the sequence of natural numbers in terms of their asymptotic growth rates. (Too long for a comment hence posting as an answer)
Let $p_n$ be the $n$-th prime number and $c_n$ be the $n$-th composite number; then,
$$\lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} f\bigg(\frac{p_r}{p_n}\bigg) = \int_{0}^{1}f(x)dx.$$
$$\lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} f\bigg(\frac{c_r}{c_n}\bigg) = \int_{0}^{1}f(x)dx.$$ | 2019-05-22T21:49:34 | {
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https://chemistry.stackexchange.com/questions/84684/magnetic-moment-%CE%BC-approximation | # Magnetic moment μ approximation
I've been reading a bit about the magnetic moment (spin-only) $\mu_{s.o}$ where they give a formula relating this to the number of unpaired electrons
$$\mu_{s.o}=\sqrt{n(n+2)}$$
where $n$ is the number of unpaired electrons.
However in our lecture today we were using the approximation $\mu_{s.o} \approx n+1$. Is this an acceptable approximation for the magnetic moment or should I stick to using the previous one.
Obviously using $\mu_{s.o} \approx n+1$ is easier to use for calculations but I would like someone's opinion on this.
I think you can come at this approximation in two ways. Using more advance methods, the approximation is obtained as a truncation of the Laurent series of $\sqrt{x(x+2)}$ about $x=\infty$.
This is possible, but I think needlessly complex in this case. Using just algebra, we can note $$\sqrt{n(n+2)}=\sqrt{n^2+2n}\approx\sqrt{n^2+2n+1}=\sqrt{(n+1)^2}=n+1$$ By looking at a plot, we can see this approximation is very good, giving essentially the exact result at $n=10$.
• In fact, using algebra, one can show that the error of approximation is bounded by ${1 \over 2n}$. – copper.hat Oct 24 '17 at 16:07
• I would hesitate to call this a great approximation because it's really quite bad for small n which is the only part of the plot that makes physical sense since you're rarely going to have close to 10 unpaired electrons. Good answer though mathematically. – jheindel Oct 24 '17 at 18:32
• @jheindel true though I can see it be useful for back of the envelope/order of magnitude type calculations. And sure 0 and 1 aren't great, but for those its hardly worth making the approximation in the first place because you could probably just memorize those. By n=2, you only have about 6% error. – Tyberius Oct 24 '17 at 18:54
• Ya true. I suspect this is not very useful beyond a coarse approximation anyways so approximating it more can't be too bad. – jheindel Oct 24 '17 at 18:55
To me this seems like a blind usage of the Taylor expansion.
Let's say we want to get a Taylor approximation of $\mu (n) = \sqrt{n(n+2)}$. Then we know we can expand any function $f(x) = f(x_0 +h)$ if $h/x_0 \ll 1$ into $f(x_0 + h) \approx f(x_0) + \frac{\partial f}{\partial x}(x_0) \cdot h$.
If I apply this to the function $\mu(n) = \mu(0 + n)$ then I get the approximate result $\mu(n) \approx \frac{n+1}{\sqrt 2}$. But here of course it is not true that $n/0 \ll 1$, so the assumption needed for using Taylor is violated.
However, if I plot both those functions in Wolframalpha they seem to agree OK enough for large n. So I guess a different approximation technique might have been used deriving this result, and it depends if you're working at large n or small n.
• It's not really a Taylor expansion, so much as using the arithmetic mean of two numbers ($n$ and $n+1$) as an approximation to their geometric mean. – psmears Oct 24 '17 at 16:05
• @psmears: You mean $n$ and $n + 2$, right? – Ilmari Karonen Oct 24 '17 at 17:00
• Why do you assume that it's a blind usage of anything? – David Richerby Oct 24 '17 at 17:03
• @DavidRicherby: The approximation apparently transforms a nonlinear function into a linear one. The simplest and most common way to do this is to do a Taylor approximation. – AtmosphericPrisonEscape Oct 24 '17 at 17:18
• My (literal) emphasis was on "blind". Saying it's "a blind usage of Taylor expansions" means that you think that the teacher used Taylor expansions without giving any thought to whether they were suitable. Why are you accusing them of that? – David Richerby Oct 24 '17 at 18:25
Using $\sqrt{x}-\sqrt{y} = { x -y \over \sqrt{x} + \sqrt{y}}$ we have $\sqrt{n(n+2)} - (n+1) = -{ 1\over n \left( \sqrt{1+ {2 \over n}}+1 + {1\over n}\right )}$, so $|\sqrt{n(n+2)} - (n+1)| \le {1 \over 2n}$.
The approximation is reasonable for large $n$.
Addendum: To see where the approximation comes from, note that $\sqrt{1+ {2 \over n}}+1 + {1\over n} \ge 2$, hence ${1\over n \left( \sqrt{1+ {2 \over n}}+1 + {1\over n}\right )} \le {1 \over 2n}$.
• I like this answer. I think the only thing I would add is showing more explicitly how your second equation approximates to 1/2n. – Tyberius Oct 24 '17 at 16:16
• @Tyberius: Good suggestion. Done. – copper.hat Oct 24 '17 at 16:20 | 2019-11-17T03:39:07 | {
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https://natrecoverygroupinc.com/nova-scotia/example-of-a-periodic-function.php | # Function periodic of example a
Why is sin(x^2) not a periodic function? + example. Sal introduces the main features of sinusoidal functions: midline, amplitude, & period. he shows how these can be found from a sinusoidal function's graph..
## Periodic Function Definition (Illustrated Mathematics
Periodic Function some codes without special toolboxes. A periodic function f(x) has a period of 9, if f(2) if a function f(x) is periodic with period k then for so for example if the period is k = 3 and f(2) = 7, title: examples of periodic functions: canonical name: examplesofperiodicfunctions: date of creation: 2013-03-22 17:57:29: last modified on: 2013-03-22 17:57:29.
## Real World Examples of Periodic Functions by nancy Prezi
Autocorrelation (for sound signals). How to plot periodic function's graphic? for @mr.wizard why it doesn't work if you put pi in the range of t? for example plot[myperiodic[exp[2*t Trigonometric functions and graphs: mid unit assignment-jiayi jin.
Real world examples of periodic functions cyclical stocks there are two types of stocks; cyclical and non-cyclical. stocks always depend on the market and the success trigonometric functions and graphs: mid unit assignment-jiayi jin
Fourier analysis for periodic functions: fourier series the result is a periodic function with period t that agrees they are just one example of conditions that example : find the fourier series of the periodic function f(t) defined by solution to the above example coefficient a 0 is given by. coefficients a n is given by
I only know f(x)=constant, and i know it's the only one in continuous functions. so i want some more examples to help me understand... this theorem helps associate a fourier series to any -periodic function. definition. example. find the fourier series of the function function answer.
A periodic function f(x) has a period of 9, if f(2) if a function f(x) is periodic with period k then for so for example if the period is k = 3 and f(2) = 7 i only know f(x)=constant, and i know it's the only one in continuous functions. so i want some more examples to help me understand...
If a periodic function with period has a finite derivative , the indefinite integral has period if , otherwise it is non-periodic, such as for example for , periodic functions periodic functions are functions which repeat: f (t + p) = f (t) for all t. for example, if f (t) is the amount of time between sunrise and sunset at a
Sal introduces the main features of sinusoidal functions: midline, amplitude, & period. he shows how these can be found from a sinusoidal function's graph. we want to approximate a periodic function f(t), with fundamental period t, with the fourier series:
Sal introduces the main features of sinusoidal functions: midline, amplitude, & period. he shows how these can be found from a sinusoidal function's graph. laplace transform of periodic functions, convolution, applications 1 laplace transform of periodic function example 2. consider a saw-tooth function
A function whose value does not change when its argument is increased by a certain nonzero number called the period of the function. for example, sin x and cos x are learn more about periodic function . toggle main periodic functions. asked by li. li not enough to be used for input on a function, for example the simulink
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https://math.stackexchange.com/questions/2914913/binomial-calculating-probability-of-airline-tickets/2914928 | # Binomial Calculating Probability of Airline Tickets
Because not all airline passengers show up for their reserved seat, an airline sells 125 tickets for a flight that holds only 115 passengers. The probability that a passenger does not show up is 0.05, and the passengers behave independently. Round your answers to four decimal places (e.g. 98.7654).
a) What is the probability that every passenger who shows up can take the flight?
b) What is the probability that the flight departs with at least one empty seat?
I am not using a statistics program to calculate my answers like I have seen many answers on here use, but by using a formula: For example: P(115) = 125 C 115 * (.95)^115 * (.05)^10 = .0475
Similarly I have:
P(116) = .0778
P(117) = .1137
P(118) = .1465
P(119) = .1637
P(120) = .1556
P(121) = .1221
P(122) = .0761
P(123) = .0353
P(124) = .0108
P(125) = .0016
So for part a) I did:
1 - P(X > 115) = 1 - .9032 = .0968
and for part b) I did:
1 - P(x >= 115) = 1 - .9507 = .0493
These numbers just do not seem correct to me. And I am confused as to why the probabilites are increasing from P(115) to P(119), (I would expect them to decrease, however I guess if they are on the rising part of the binomial distribution and then go to the falling part of the distribution at P(120)
Edit: I know understand these values are correct and fall around the most probable value P(119) which is the mean. Thank you for help in clarifying.
• If you will look at the 'Related' links to the right of your question, you will see at least one that answers almost the same question. // I will look at some of your answers to see if anything seems strange. – BruceET Sep 12 '18 at 22:58
• Well, by now you have several verifications. Problem seems strange; no wonder you wondered. Considering my 'bad luck', I think I may know this airline. – BruceET Sep 12 '18 at 23:40
The number of people who show up for the flight in the circumstances you describe is $X \sim \mathsf{Binom}(n=125, p=.95).$ The probability everyone who shows has a seat is $P(X \le 115) \approx 0.0967.$ This can be computed in R statistical software, in which pbinom is a binomial CDF.
pbinom(115, 125, .95)
[1] 0.09672946
This may seem absurdly small, but the average number of people showing is $\mu = E(X)= np = 128.25 > 125.$ So on the average flight the airline should expect to leave some people behind.
The probability there will be at least one empty seat is $P(X \le 114) = 0.0492.$ So that will seldom happen.
pbinom(114, 125, .95)
[1] 0.04921917
Here is a plot of the relevant part of the distribution of $X.$
Here are exact values from R, matching your computations, which seem quite accurate:
x = 115:125; pdf = dbinom(x, 125, .95)
cbind(x, pdf)
x pdf
[1,] 115 0.047510291
[2,] 116 0.077818581
[3,] 117 0.113734849
[4,] 118 0.146505907
[5,] 119 0.163741896
[6,] 120 0.155554801
[7,] 121 0.122129803
[8,] 122 0.076080861
[9,] 123 0.035256984
[10,] 124 0.010804560
[11,] 125 0.001642293
Note: Code for figure:
x = 110:125; pdf = dbinom(x, 125, .95)
plot(x, pdf, type="h", ylim=c(0,max(pdf)), lwd=2)
abline(h=0, col="green3")
Because not all airline passengers show up for their reserved seat, an airline sells 125 tickets for a flight that holds only 115 passengers. The probability that a passenger does not show up is 0.05, and the passengers behave independently. Round your answers to four decimal places (e.g. 98.7654).
Suppose $X \sim \textrm{Bin}(125,0.5)$ is a binomial random variable. The mass function is given by
$$f(k;125,0.5) = Pr(X = k) = \binom{125}{k} \bigg(\frac{5}{100}\bigg)^{k} \bigg( \frac{95}{100} \bigg)^{125-k} \tag{1}$$
We can visualize the probability mass function then like this in Python
n=125
p=.95
x = range(n+1)
y = stats.binom.pmf(x, n, p)
plt.plot(x,y,"o", color="black")
plt.axis([-(max(x)-min(x))*0.05, max(x)*1.05, -0.01, max(y)*1.10])
plt.xticks(x)
plt.title("Binomial distribution PMF for tries = {0} & p ={1}".format(
n,p))
plt.xlabel("Variate")
plt.ylabel("Probability")
If you note then
y1 = stats.binom.pmf(115,125,.95)
y1
Out[29]: 0.04751029149720219
to look at the cdf then
y2 = stats.binom.cdf(x,125,.95)
plt.plot(x,y2,"o", color="black")
If we take a closer look at the pdf
n=125
p=.95
x = range(100,150)
y = stats.binom.pmf(x, n, p)
y2 = stats.binom.pmf(x,125,.95)
plt.plot(x,y2,"o", color="black")
Your formula is correct. I didn't check the numbers but they look good. If $0.05$ of the people do not show up, the expected number of no-shows is $0.05\cdot 125=6.25$ so the most probable number who show up should be $119$, in agreement with your calculation. | 2021-06-14T18:11:08 | {
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https://math.stackexchange.com/questions/1702616/number-of-n-digit-perfect-squares | # Number of N-digit Perfect Squares
I was working on a programming problem to find all 10-digit perfect squares when I started wondering if I could figure out how many perfects squares have exactly N-digits. I believe that I am close to finding a formula, but I am still off by one in some cases.
Current formula where $n$ is the number of digits:
$\lfloor\sqrt{10^n-1}\rfloor - \lfloor\sqrt{10^{n-1}}\rfloor$
How I got here:
1. Range of possible 10-digit numbers is from $10^9$ to $10^{10}-1$
2. 10-digit perfect squares should fall into the range of $\sqrt{10^9}$ to $\sqrt{10^{10}-1}$
Results of program for $n = 1,2,3,4,5$:
DIGITS=1, ACTUAL_COUNT=3, COMPUTED_COUNT=2
DIGITS=2, ACTUAL_COUNT=6, COMPUTED_COUNT=6
DIGITS=3, ACTUAL_COUNT=22, COMPUTED_COUNT=21
DIGITS=4, ACTUAL_COUNT=68, COMPUTED_COUNT=68
DIGITS=5, ACTUAL_COUNT=217, COMPUTED_COUNT=216
Program:
#!/usr/bin/perl
use strict;
use warnings;
sub all_n_digit_perfect_squares {
my ($n) = @_; my$count = 0;
my $MIN = int( sqrt( 10**($n-1) ) );
my $MAX = int( sqrt( (10**$n)-1 ) );
foreach my $i ($MIN .. $MAX ) { if ( ($i * $i) >= 10**($n-1) ) {
$count++; } } print "DIGITS=$n"
. ", ACTUAL_COUNT=$count" . ", COMPUTED_COUNT=" . ($MAX-$MIN), "\n"; return; } all_n_digit_perfect_squares($_) for (1..5);
If you think about it, you should have a formula that says the number of squares is f (n) - f (n-1) for some function f, so that every perfect square is counted exactly once if you calculate the squares from 10^1 to 10^2, from 10^2 to 10^3 and so on.
In your formula, the squares 100, 10,000, 1,000,000 and so on are not counted at all. For example, for 3 digit numbers the squares are from 10^2 to 31^2, that's 22 numbers. You calculate 31 - 10 = 21. Change your formula to
$\lfloor\sqrt{10^n-1}\rfloor - \lfloor\sqrt{10^{n-1}-1}\rfloor$
• Ah, thank you! I knew I was missing something. – Hunter McMillen Mar 18 '16 at 13:02
I'm interested in the question that you're asking. I've been thinking about this for the last few days and couldn't find anything online about this subject. If you have anything references, I would appreciate it. Here is what I have so far and sorry if the formatting is off, this is my first post.
$$\text{Let}\; \mathbb{N} \;\text{be the set of Natural numbers} \; = \{1,2,3,4,..,N\},$$ Then there exists a set $$\mathbb{S}$$, such that for $$s\in\mathbb{S}\; ;\; s_i = \{10^{i-1},...,(10^i-1)\} \;, \text{for} \; i = 1,2,3,...,M.$$ Example : $$s_1 = \{1,2,3,...,9\}\; ; \;\;\; i=1 = 1\;\text{digit}$$ $$s_2 = \{10,11,12,...,99\}\; ; \;\; i=2 = 2\;\text{digits}$$ $$s_3 = \{100,101,102,...,999\}\; ; \;\; i=3 = 3\;\text{digits}$$ $$...$$ $$s_i = \{10^{i-1},...,(10^i - 1)\} ; \;\; i\; \text{digits}$$
Within each set $$s_i$$, there exists a set $$\mathbb{A}$$ such that for all members of $$\mathbb{A}$$, call it $$a_p$$, when squared are still a member of $$s_i$$: $$a\in\mathbb{A}\; ; \;\; a = \{ a_1,a_2,...,a_k \}$$ $$(a_p)^2 \in s_i ,\; \text{for}\; p = 1\; \text{to}\; k$$
Example : $$s_1 = \{1,2,3,...,9\}\; ; \;\;\; a_1 = \{1,2,3\}$$ $$s_2 = \{10,11,12,...,99\}\; ; \;\; a_2 = \{ a_1 , 4, 5,6,...,9\} = \{1,2,3 , 4, 5,6,...,9\}$$
I'm going to start omit the previous sets before the current $$i^{th}$$ set we're looking at. So it'll look like this:
$$s_2 = \{10,11,12,...,99\}\; ; \;\; a_2 = \{4, 5,6,...,9\},$$
but also know that all previous sets before the current $$i^{th}$$ can included, but is not unique to the $$i^{th}$$ digits.
I'm going to list a few in the long version, just as an example. $$s_3 = \{100,101,102,...,999\}\; ; \;\; a_3 = \{10,11,12,...,31 \}$$ $$s_4 = \{1000,1002,1003,...,9999\}\; ; \;\; a_4 = \{32,33,34,...,99 \}$$ $$s_5 = \{10000,...,99999\}\; ; \;\; a_5 = \{100,...,316 \}$$ $$s_6 = \{100000,...,999999\}\; ; \;\; a_6 = \{317,...,999 \}$$ $$s_7 = \{1000000,...,9999999\}\; ; \;\; a_7 = \{1000,...,3162 \}$$
Now we can write these compactly as, the first i=1,2,3 are written in the long way and i > 4 is written in the short hand version as given by $$s_i$$: $$s_1 = \{10^0,2,3,...,9\}\; ; \; a_1 = \{1,2,3\}\; ; \; t_1 = 3$$ $$s_2 = \{10^1,...,99\}\; ; \; a_2 = \{4, 5,6,...,9\}\; ; \; t_2 = 6$$
$$s_3= \{10^2,...,999\}\; ; \; a_3 = \{10^1,...,31\}\; ; \; t_3 = 22$$
$$s_4= \{10^3,...,(10^4 -1)\}\; ; \; a_4 = \{32,...,99\}\; ; \; t_4 = 68$$ $$s_5= \{10^4,...,(10^5 -1)\}\; ; \; a_5 = \{10^2,...,316\}\; ; \; t_5 = 217$$
I also introduced $$t_i$$, which is the count of the number of members of $$a_i$$, (this is exactly what you're solving for, I believe). Notice, we can get an exact solution for $$t_i$$ given by: $$t_i = (\{a_i\}_{max} - \{a_i\}_{min}) + 1$$ (I think this could be calculated using a norm)
I'm going to list out a few sequences and point out an interesting patterns.
Starting with $$i=3$$, (you could extend this to $$i=1$$ if you want),
if $$i = odd$$ : The lower bound of $$a_i$$, call this $$\{a_i\}_{min}$$ is exactly $$10^l$$, where $$l = (i-1)/2$$ ;
if $$i = even$$ : The upper bound of $$a_i$$, call this $$\{a_i\}_{max}$$ is exactly $$10^z - 1$$, where $$z = (i)/2$$ ;
I claim that for $$i=even$$, you have
$$s_{i-1} = \{10^{i-2},...,(10^{i-1}-1)\}\; ; \;\; a_{i-1} =\{10^l,...,upper\;bound_{(i-1)}\} \; ; \; t_{i-1} = ?$$ $$s_{i} = \{10^{i-1},...,(10^{i}-1)\}\; ; \;\; a_{i} =\{lower\; bound_{(i)},...,(10^z-1)\} \; ; \; t_{i} = ?.$$
I'm still working/thinking about I can get the upper bounds, lower bounds, and $$t_i$$.
Here are the first 16 digits, and their lists. I also have done this out until 34 digits, took my computer a good few hours to compute by brute force. I am almost done with an algorithm that can do it much quicker.
Also note, that if you know the lower bounds and upper bounds, you can easily obtain $$t$$, which is the number of elements that are contained in $$\{a_i\}$$, since $$\{a_i\}$$ is ordered.
$$s_1 = \{10^0,2,3,...,9\}\; ; \; a_1 = \{1,2,3\}\; ; \; t_1 = 3$$ $$s_2 = \{10^1,...,99\}\; ; \; a_2 = \{4, 5,6,...,9\}\; ; \; t_2 = 6$$
$$s_3= \{10^2,...,999\}\; ; \; a_3 = \{10^1,...,31\}\; ; \; t_3 = 22$$
$$s_4= \{10^3,...,(10^4 -1)\}\; ; \; a_4 = \{32,...,99\}\; ; \; t_4 = 68$$ $$s_5= \{10^4,...,(10^5 -1)\}\; ; \; a_5 = \{10^2,...,316\}\; ; \; t_5 = 217$$ $$s_6= \{10^5,...,(10^6 -1)\}\; ; \; a_6 = \{317,...,999\}\; ; \; t_6 = 683$$ $$s_7= \{10^6,...,(10^7 -1)\}\; ; \; a_7 = \{10^3,...,3162\}\; ; \; t_7 = 2163$$ $$s_8= \{10^7,...,(10^8 -1)\}\; ; \; a_8 = \{3163,...,9999\}\; ; \; t_8 = 6837$$ $$s_9= \{10^8,...,(10^9 -1)\}\; ; \; a_9 = \{10^4,...,31622\}\; ; \; t_9 =21623$$ $$s_{10}= \{10^9,...,(10^{10} -1)\}\; ; \;\; a_{10} = \{31623,...,(10^5-1)\}\; ; \; t_{10} = 68377$$ $$s_{11}= \{10^{10},...,(10^{11} -1)\}\; ; \;\; a_{11} = \{10^5,...,316227\} \; ; \; t_{11} = 216228$$ $$s_{12}= \{10^{11},...,(10^{12} -1)\}\; ; \;\; a_{12} = \{316228,...,(10^6-1)\}\; ; \; t_{12} = 683772$$ $$s_{13}= \{10^{12},...,(10^{13} -1)\}\; ; \;\; a_{13} = \{10^6,...,3162277\}\; ; \; t_{13} = 2162278$$ $$s_{14}= \{10^{13},...,(10^{14} -1)\}\; ; \;\; a_{14} = \{3162278,...,(10^7-1)\} \; ; \; t_{14} = 6837722$$. $$s_{15}= \{10^{14},...,(10^{15} -1)\}\; ; \;\; a_{15} = \{10^7,...,31622776\} \; ; \; t_{15} = 21622777$$ $$s_{16}= \{10^{15},...,(10^{16} -1)\}\; ; \;\; a_{16} = \{31622777,...,(10^8-1)\} \; ; \; t_{16} = 68377223$$
• I get a different result for $t_{16}$ using the other formula. I get $68377224$ instead of $68377223$ – Hunter McMillen Jul 1 at 14:00
• As far as other research that I did into this topic, after I found a working solution I didn't explore the area much further. – Hunter McMillen Jul 1 at 14:01
• Sample code to recreate my outputs for the first 16 elements: perl -E 'sub f { my $n = shift; return int(sqrt(10**$n - 1)) - int(sqrt(10**($n-1) - 1)); } say f($_) for 1..16;' – Hunter McMillen Jul 1 at 14:03
The number of perfect squares between any two numbers a and b with a less than b is floor sqrt b - ceil sqrt a + 1. i.e. a=1000 b=2000. ceil sqrt 1000 = 32. floor sqrt 2000 = 44. So 32 33 34 35 36 37 38 39 40 41 42 43 and 44 when squared will be perfect squares between 1000 and 2000 and 44-32+1=13. If you are silly and try for the number of perfect squares between 1000 and 1001 then 31-32+1=0. between 9 and 10 then 3-3+1=1. | 2019-07-21T15:13:13 | {
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https://math.stackexchange.com/questions/729258/diff-eq-problem-eigenvector-issue | # Diff EQ. Problem (Eigenvector issue)
Find the general solution of $\textbf{x}^{'}=\begin{pmatrix} -1&-4\\1&-1\end{pmatrix}\textbf{x}$.
The eigenvalues I found are $-1 \pm 2i$ and I chose $-1-2i$ to be my eigenvector and found it to be $e_1=\begin{pmatrix} 2i\\-1 \end{pmatrix}$ Now plugging this in to the formula we have $$e^{-t}\begin{pmatrix} 2i \\-1 \end{pmatrix}(\cos 2t-i\sin2t)=c_1e^{-t}\begin{pmatrix} 2\sin 2t\\-\cos 2t \end{pmatrix} + c_2e^{-t}\begin{pmatrix} 2 \cos 2t\\ \sin 2t \end{pmatrix}$$ The book however has: $c_1e^{-t}\begin{pmatrix} - 2\sin 2t\\\cos 2t \end{pmatrix} + c_2e^{-t}\begin{pmatrix} 2 \cos 2t\\ \sin 2t \end{pmatrix}$. How did the signs get switched because I tried to do this with $e_2=\begin{pmatrix} -2i\\-1 \end{pmatrix}$ and all that happens is the right side becomes negative which still doesn't explain how the signs are switched.
• Your $c_1$ is the negative of the book's $c_1$. – John Habert Mar 27 '14 at 17:33
• but how did that happen is this correct? – adam Mar 27 '14 at 17:34
• Since the constants are arbitrary until you use initial conditions to solve for them, it is fine. It is a similar trick as to how you got all the $i$ terms to drop out by being absorbed into $c_2$. – John Habert Mar 27 '14 at 17:36
• Thanks I understand that part, but what I dont understand is why they would do something random and factor out a negative 1. I understand that $\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}$ because we dont want radicals in the denominator. Now why factor out a $-1$? – adam Mar 27 '14 at 17:41
Both answers are correct. To expand on John Habert's comment, since the coefficient $c_1$ is arbitrary, your $c_1$ is the negative of the book's $c_1$. To see what I mean, replace your $c_1$ with $-c_1$: you'll get the book's answer.
• Is there any particular reason why they factor out a $-1$? I just think that can be confusing to do random things like that. – adam Mar 27 '14 at 17:37
• If this is an initial value problem, then it is possible this choice of signs makes finding $c_1,c_2$ a little easier. – John Habert Mar 27 '14 at 17:40 | 2019-12-11T00:51:22 | {
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https://math.stackexchange.com/questions/1392491/measure-of-how-much-diagonal-a-matrix-is | # Measure of “how much diagonal” a matrix is
I have a (biological) computational system that outputs squared matrices. These matrices will sometimes have a tendency to be diagonal-like, with higher values at and around the diagonal.
I would like to have some summary measure on how "much diagonal" a matrix is, so that I can batch process hundreds of outputs and score them on how much the higher entries cluster in and around the diagonal.
Any ideas of some standard approach that I can generalise?
Thanks ! JL
• What do the rows and columns represent? Are the matrix entries counts of events? This affects the choice. – Tad Aug 11 '15 at 1:32
• They are frequencies of events! – lourencoj Aug 11 '15 at 8:20
• Do the rows and columns represent values or just categories? In the first case you get credit for being "near" the diagonal; in the second case you're either on it or you're not. – Tad Aug 11 '15 at 11:15
• It should get credit for being close to the diagonal – lourencoj Aug 11 '15 at 14:49
Given that your entries are frequencies, and you want to give credit for being "close" to the diagonal, a natural approach is to compute the correlation coefficient between the row and column. That is, suppose your matrix is built as follows: repeatedly generate a pair of numbers $x$ and $y$, and increment the count of the matrix entry at position $(x,y)$. If you think of $x$ and $y$ as samples of random variables $X$ and $Y$ respectively, then the sample correlation coefficient $r$ of $X$ and $Y$ lies between $-1$ and $1$. It is $1$ if $X$ and $Y$ are perfectly correlated, $-1$ if they are perfectly anticorrelated. The point is that $X$ and $Y$ are perfectly correlated (in this case, equal) precisely when the matrix is diagonal, strong correlation means the matrix entries tend to be near the diagonal.
This is robust: the correlation coefficient is unchanged if you scale the matrix (and the formula turns out to make sense even if your entries are nonnegative real numbers).
If you adapt the formulas in the above reference to this situation, they take the following form. Let $A$ be a $d\times d$ matrix; let $j$ be the $d$-long vector of all ones, and let $r=(1,2,\ldots,d)$ and $r_2=(1^2,2^2,\ldots,d^2)$. Then:
\begin{align} n &= j A j^T \textrm{ (the sum of the entries of A) }\\ \Sigma x &= r A j^T\\ \Sigma y &= j A r^T\\ \Sigma x^2 &= r_2 A j^T\\ \Sigma y^2 &= j A r_2^T\\ \Sigma xy &= r A r^T\\ r &= \frac{n\, \Sigma xy -\Sigma x\, \Sigma y}{\sqrt{n\, \Sigma x^2 - (\Sigma x)^2}\sqrt{n\, \Sigma y^2 - (\Sigma y)^2}} \end{align}
Some examples:
Diagonal matrix: $\left( \begin{array}{cccc} 1. & 0. & 0. & 0. \\ 0. & 5. & 0. & 0. \\ 0. & 0. & 30.5 & 0. \\ 0. & 0. & 0. & 3.14159 \\ \end{array} \right): \quad r=1.000000$
Diagonally dominant matrix: $\left( \begin{array}{ccc} 6 & 1 & 0 \\ 1 & 5 & 2 \\ 1 & 3 & 6 \\ \end{array} \right): \quad r=0.674149$
Uniformly distributed on $[0,1]$: $\left( \begin{array}{cccc} 0.2624 & 0.558351 & 0.249054 & 0.484223 \\ 0.724561 & 0.797153 & 0.689489 & 0.273023 \\ 0.462727 & 0.119412 & 0.911981 & 0.636588 \\ 0.089544 & 0.160899 & 0.910123 & 0.549202 \\ \end{array} \right): \quad r=0.233509$
Tridiagonal: $\left( \begin{array}{ccccc} 2 & 1 & 0 & 0 & 0 \\ 1 & 3 & 2 & 0 & 0 \\ 0 & 2 & 3 & 4 & 0 \\ 0 & 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 1 & 1 \\ \end{array} \right): \quad r=0.812383$
• Hey Tad. This is great! This will be our approach. Thanks so much. – lourencoj Aug 12 '15 at 21:49
• @Tad, I've got a similar task to measure the degree of "diagolizedness" but my matrix can be rectangular, not just square. What would you say? Can you expand your solution for that case? – ttnphns Aug 27 '15 at 10:52
• @ttnphns You need to be clear about what you mean by "diagonalness" in this context. – Tad Aug 27 '15 at 11:03
• The concentration of values (greeting their magnitude, too), close to the main diagonal (from top-left corner), as before. Fully diagonal matrix will be when all nonzero elements lie on it. – ttnphns Aug 27 '15 at 11:34
• If you adapt the formulas in the above reference... I must say that your adaptation is not transparent at all. Just to mention, $n$ in correlation coefficient is the number of observations (X,Y pairs). In your formulas, it is suddenly the sum of values in the matrix. To me, your approach remains unclear (albeit it could be perfect). Can you elucidate your computations using the theoretical (the 1st, not the 2nd) formula of $r$ of the wikipedia, the formula using the means? – ttnphns Aug 27 '15 at 12:33
Here's an easy one. Let $M$ be your measured matrix, and $A$ be the matrix which agrees with $M$ along the diagonal, but is zero elsewhere. Then pick your favorite matrix norm (operator probably works well here) and use $\|M-A\|$ as your measurement.
If you want more fine tuned understanding of 'clustering', instead of making all the entries off the diagonal $0$, weight them by what band they are on. So the super and sub diagonal might take half the corresponding value in $M$.
• Hey. This is a good start, thanks – lourencoj Aug 11 '15 at 8:19 | 2020-07-09T09:23:47 | {
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https://math.stackexchange.com/questions/1762057/proof-that-there-are-infinitely-many-primes-euclid | Proof that there are infinitely many primes (Euclid)
I was wondering if I could get some insight on my proof. I am in the midst of relearning some number theory and just "writing proofs" in general, and I would like some assistance to see if I am on the right track.
The statement I am proving is Euclid's Theorem which states that "there are infinitely many primes."
Here is my attempt at the proof after some reading (keep in mind, I am still somewhat of an amateur when using LaTeX so please bear with me!):
Proof. Suppose in order to derive a contradiction there are finitely many primes. That is, we have a complete list $p_1, \dots, p_n$. Let $p$ be the product of all the primes in this list i.e. $p = p_1 \cdots p_n$. Consider the number $$N = p + 1.$$ Since $N > p_i$ for all $i$, $1 \leq i \leq n$, there is no way $N$ can be any of the $p_i$. So $N$ must be composite. By the Fundamental Theorem of Arithmetic, $N$ is a product of primes. So there is a prime, say q, that divides $N$, and $p$ as well. So it follows this $q$ must also divide $$N - p = 1,$$ but this is impossible. No number, or prime, divides 1. Thus, contradicts the assumption that our list is complete and so there must be infinitely many primes.
QED
Any feedback would be appreciated. I always had trouble understanding this theorem and always forgot the "key argument". Now I feel like I finally get it... Hopefully. Thank you for reading!
• $N$ need not be composite. $2\times 3+1=7$, for example. Nor do you need the FTA, which is a lot harder to prove than this. All you need to know is that every natural number greater than $1$ is divisible by some prime. – lulu Apr 28 '16 at 1:39
• On rereading, I see that you deduce the compositeness of $N$ by remarking that, were it prime, we'd already have the desired counterexample. Agreed. But, still, invoking the Fundamental Theorem of Arithmetic is not necessary. – lulu Apr 28 '16 at 1:43
• I see how it is unnecessary now. Thanks for reading! – o_o Apr 28 '16 at 1:47
• It is good. Petty comment, you say no number divides $1$. Well, $1$ divides $1$. However, it is certainly true that no number $\gt 1$ divides $1$, so no prime divides $1$. – André Nicolas Apr 28 '16 at 1:55
• Oops, that was silly (LOL). Good catch. – o_o Apr 28 '16 at 1:57
I would not add the complication of making this into a proof by contradiction. Euclid did not do it that way, despite many modern authors, dating back at least to Dirichlet in the middle of the 19th century, asserting that Euclid did it that way.
Start with any finite set $S$ of prime numbers. (For example, we could have $S=\{2, 31, 97\}$) Let $p = 1 + \prod S$, i.e. $1$ plus the product of the members of $S$.
Then $p$ cannot be divisible by any of the primes in $S$. Therefore either
• $p$ is itself prime, in which case there are more primes than those in $S$, or
• $p$ is divisible by some other primes not in $S$, in which case there are more primes than those is $S$.
Thus every finite set $S$ of primes can be extended to a larger finite set of primes.
(For example, if $S=\{5,7\}$, then $1+\prod S = 1 + 35 = 36 = 2\times2\times3\times 3$, and the additional primes not in $S$ are $2$ and $3$.)
The initial assumption that $S$ contains all primes is at best a needless complication that serves no purpose.
• I am always very confused about the whole debate between contradiction proof versus non contradiction. Yes, this proof never states "assume there are finitely many primes," but it also never actually proves that there are infinitely many primes, merely that any finite set of primes can be expanded. From there, the proof goes "there are infinitely many primes because the number of primes cannot be equal to some finite number $N$ (via the construction)" To me, however, this final statement is essentially a contradiction statement (assume finite, that's impossible, so we get a contradiction. – ASKASK Apr 28 '16 at 1:59
• So to me, this proof is still a proof by contradiction, it's just that the actual contradiction part is postponed until after the construction – ASKASK Apr 28 '16 at 2:00
• @ASKASK : If you want to look at it that way, it is still better not to create an illusion that the assumption of finiteness of the set of ALL primes somehow plays a role in the construction. – Michael Hardy Apr 28 '16 at 2:01
• but doesn't the finiteness play a role? Finite number => finite set => construction of a prime not in that set. – ASKASK Apr 28 '16 at 2:03
• @ASKASK : Finiteness of the arbitrary finite set of primes plays a role. An assumption of finiteness of the set of ALL primes plays no role in the construction, even if it plays a different role of the kind you suggest. – Michael Hardy Apr 28 '16 at 2:04
Your proof looks good. It is not exactly how Euclid's proved it, but it works, and most importantly you understand it.
A couple of minor points:
Like lulu said in the comments, invoking the Fundamental Theorem of Arithmetic isn't necessary. Also in the end you state "No number, or prime, divides 1." This technically isn't true since 1 divides 1. It would be enough just to say "No prime divides 1."
• Yes, I understand the minor points. My personal understanding was the goal, haha. Thank you for reading! :) – o_o Apr 28 '16 at 2:04
• In the context of Euclid, $1$ is not a number, so "No number divides $1$" would be correct. – Michael Hardy Apr 28 '16 at 2:05
• @MichaelHardy Can you elaborate? Does Euclid not consider 1 a number? – M47145 Apr 28 '16 at 2:09
• @M47145 : That has been my understanding. See this page: aleph0.clarku.edu/~djoyce/elements/bookVII/bookVII.html $\qquad$ – Michael Hardy Apr 28 '16 at 5:37 | 2019-05-26T21:40:37 | {
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https://math.stackexchange.com/questions/668736/prove-the-contraction-mapping-theorem | # Prove the Contraction Mapping Theorem.
Prove the Contraction Mapping Theorem.
Let $(X,d)$ be a complete metric space and $g : X \rightarrow X$ be a map such that $\forall x,y \in X, d(g(x), g(y)) \le \lambda d(x,y)$ for some $0<\lambda < 1$. Then $g$ has a unique fixed point $x^* \in X$, and it attracts everything, i.e. for any $x_0 \in X$ , the sequence of iterates $x_0, g(x_0), g(g(x_0))$, ... converges to the fixed point $x^* \in X$.
The hint I am given are for existence and convergence - prove that the sequence is Cauchy. For uniqueness, choose two fixed points of $g$ and apply the map to both.
Still a bit do not know how to proceed after looking at the hint. Could anyone help me based on those hints?
• Hint: $d(x_{n+1},x_n)=d(g(x_n),x_n)=d(g(g(x_{n-1})),g(x_{n-1}))\le \lambda d(g(x_{n-1}),x_{n-1})=d(x_n,x_{n-1})$ – J.R. Feb 8 '14 at 19:17
• @TooOldForMath: what are you trying to achieve here? – afsdf dfsaf Feb 8 '14 at 19:19
• I am giving you a hint so you can solve it. This is how it goes. Repeating what I wrote gives you $d(x_{n+1},x_n)\le \lambda^n d(x_1,x_0)$. Now $\lambda$ is between $0$ and $1$, so what happens when $n\rightarrow\infty$...? – J.R. Feb 8 '14 at 19:26
• why $\lambda d(g(x_{n-1}), x_{n-1}) = d(x_n, x_{n-1})$? – afsdf dfsaf Feb 8 '14 at 19:52
• $d(x_{n+1},x_n)\le \lambda d(x_n,x_{n-1}) \le \lambda^2 d(x_{n-1},x_{n-2})\le\cdots\le \lambda^n d(x_1,x_0)$ – J.R. Feb 9 '14 at 0:13
## 2 Answers
From
$$d(x_{n+1},x_n)=d(g(g(x_{n-1})),g(x_{n-1}))\le \lambda d(g(x_{n-1}), x_{n-1})=d(x_n,x_{n-1})$$
we get after $n$ applications of that inequality
$$d(x_{n+1},x_n)\le \lambda d(x_n,x_{n-1}) \le \lambda^2 d(x_{n-1},x_{n-2}) \le \cdots\le \lambda^n d(x_1,x_0)\tag{1}$$
Now we want to show that $(x_n)_n$ is a Cauchy sequence. So let $\epsilon>0$.
We assume $x_1\not=x_0$ (otherwise $x_0$ is already a fixed point). Set $c=d(x_1,x_0)>0$.
Since $0<\lambda<1$, the sum $\sum_{n=0}^\infty \lambda^n$ converges (to $1/(1-\lambda)$). Therefore we can pick $N$ large enough such that
$$\sum_{k=n}^\infty \lambda^k<\frac{\epsilon}{c}$$
for all $n\ge N$.
Then for $m>n\ge N$ we have by the triangle inequality
$$d(x_m,x_n)\le \sum_{k=n}^{m-1} d(x_{k},x_{k+1})$$
Applying $(1)$ we obtain
$$d(x_m,x_n)\le c\sum_{k=n}^{m-1} \lambda^k\le c\sum_{k=n}^\infty \lambda^k<c\cdot\frac{\epsilon}{c}=\epsilon$$
So $(x_n)_n$ really is a Cauchy sequence. Since $(X,d)$ is complete, it converges to a limit $x\in X$.
By the equation $x_{n+1}=g(x_n)$, the limit satisfies $x=g(x)$, so it is a fixed point.
Uniqueness is trivial, let $y$ be another fixed point of $g$. Then
$$d(x,y)=d(g(x),g(y))\le \lambda d(x,y)$$
If now $x\not=y$, then $d(x,y)>0$, so we can divide by $d(x,y)$ to obtain $\lambda\ge 1$, a contradiction. Therefore, $x=y$.
• "Therefore we can pick $N$ large enough such that $$\sum_{k=n}^\infty \lambda^k<\frac{\epsilon}{c}$$ for all $n\ge N$." For this part, do you backfill $$\frac{\epsilon}{c}$$ after going through the following: "Then for $m>n\ge N$ we have by the triangle inequality $$d(x_m,x_n)\le \sum_{k=n}^{m-1} d(x_{k},x_{k+1})$$ Applying $(1)$ we obtain $$d(x_m,x_n)\le c\sum_{k=n}^{m-1} \lambda^k\le c\sum_{k=n}^\infty \lambda^k<c\cdot\frac{\epsilon}{c}=\epsilon$$"? – afsdf dfsaf Feb 9 '14 at 17:18
• Backfill? What are you talking about? This is a logically complete argument. – J.R. Feb 9 '14 at 17:25
• I just mean whether we get to go through the following part: "Then for $m>n\ge N$ we have by the triangle inequality $$d(x_m,x_n)\le \sum_{k=n}^{m-1} d(x_{k},x_{k+1})$$ Applying $(1)$ we obtain $$d(x_m,x_n)\le c\sum_{k=n}^{m-1} \lambda^k\le c\sum_{k=n}^\infty \lambda^k<c\cdot\frac{\epsilon}{c}=\epsilon$$" before filling $$\frac{\epsilon}{c}$$ in the "Therefore we can pick $N$ large enough such that $$\sum_{k=n}^\infty \lambda^k<\frac{\epsilon}{c}$$ If not, how did you get $$\sum_{k=n}^\infty \lambda^k<\frac{\epsilon}{c}$$? – afsdf dfsaf Feb 9 '14 at 17:36
• Please stop copying all this stuff in the comments. It makes it really unreadable. You are not "filling" $\epsilon/c$ whatever you mean by that. The logic is like this: $c$ is a constant defined as $d(x_1,x_0)$. It is fixed. Now I throw you an $\epsilon>0$ in the beginning. And now you notice, aha, the sum $\sum_{k=1}^\infty \lambda^k$ converges therefore sequence of truncated sums $\sum_{k=n}^\infty \lambda^k$ converges to $0$ for $n\rightarrow\infty$. So it will eventually be smaller than that given $\epsilon/c$. Let us say that the point when it becomes smaller is reached at $N$ – J.R. Feb 9 '14 at 17:42
• $\sum_{k=n}^\infty \lambda^k <\epsilon/c$ for all $n$ bigger than $N$. – J.R. Feb 9 '14 at 17:43
One proviso: You do need that $X$ is non-empty in the statement.
To elaborate on the hints you have been given; To prove existance, pick $x_0 \in X$, and call (for convenience of writing) $g(x_0) = x_1, g(x_1) = x_2$ etcetera. Let $d(x_0,x_1) = d$. Then $d(x_1,x_2) \leq \lambda d$, and $d(x_2,x_3) \leq \lambda^2 d$...can you see how to extend this to show that the sequence is Cauchy?
For uniqueness, suppose $x$ and $y$ are both fixed points. What is $d(g(x),g(y))$?.
• I don't think you do - if $x = y$ then both sides of the inequality are $0$, which is fine. – meta Feb 8 '14 at 19:37
• Touche! (I'm used to strong inequalities being used, which isn't really different here, and didn't notice the difference.) – Jonathan Y. Feb 8 '14 at 19:39
• @meta: After we got to $d(x_{n+1}, x_n) \le \lambda^n d(x_1, x_0)$, then assuming that $m > n$, $d(x_m, x_n) \le d(x_m, x_{m-1}) + d(x_{m-1}, x_{m-2}) + ... + d(x_{n+1}, x_{n})$ since each term of the right hand side is 0 ...so if we add up all the 0 terms, we get 0 on the right hand side. Therefore, $d(x_m, x_n)$ is 0. If this is right, then the question I have here is how do I guarantee that $x_{m-1} > x_n$? – afsdf dfsaf Feb 9 '14 at 3:04 | 2021-03-04T09:31:28 | {
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https://math.stackexchange.com/questions/304251/is-the-zero-matrix-upper-and-lower-triangular-as-well-as-diagonal | # Is the zero matrix upper and lower triangular as well as diagonal?
From what I can tell from the definitions of a lower-triangular, upper-triangular, and diagonal matrices, I've come to the conclusion that the zero matrix is in the set of all of each type of matrix. Is this correct?
• Yes ${}{}{}{}{}{}$. – copper.hat Feb 14 '13 at 20:09
A zero square matrix is lower triangular, upper triangular, and also diagonal.
If an object meets the definition of three things then it is the three things. What are you confused about?
• The definition of those three things, for instance take the definition provided here: mathworld.wolfram.com/LowerTriangularMatrix.html I didn't find it particularly clear. The stumbling block was on could a_{ij} = 0. Apparently the answer is yes, yes it can. – BrotherJack Feb 14 '13 at 20:19
• This is actually a good answer. An object or something is not necessarily of one type. If it fits the definition, you go with it, unless something contradictory may appear or several definitions may exclude some parts of each other. – Turkhan Badalov Feb 13 '18 at 16:13
Provided it is a square matrix.
An upper triangular matrix is one in which all entries below main diagonal are zero. Clearly this is satisfied.
An lower triangular matrix is one in which all entries above main diagonal are zero. Clearly this is satisfied.
An diagonal matrix is one in which all non-diagonal entries are zero. Clearly this is also satisfied.
Hence, a zero square matrix is upper and lower triangular as well as diagonal matrix. | 2019-08-21T16:38:35 | {
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https://math.stackexchange.com/questions/1359475/proving-n3-3n2-2n-is-divisible-by-6 | # Proving $n^3 + 3n^2 +2n$ is divisible by $6$
The full question is: Factorise $n^3 + 3n^2 + 2n$. Hence prove that when $n$ is a positive integer, $n^3 + 3n^2 + 2n$ is always divisible by $6$.
So i factorised and got $n(n+1)(n+2)$ which i think is right? I'm not sure how to actually prove this is divisible by $6$ though. Thanks for help and i apologise if someone has already asked this, i couldn't find it. Also i've not been told whether i have to do induction or not.
• Hint: an integer is divisible by 6 if and only if it is divisible by both 2 and 3. Can you show that your expression is even? Can you show it is divisible by 3? – lulu Jul 13 '15 at 14:28
• Fermat's little theorem can be applied. $(n^3-n)+3(n^2+n)$ is divisible by $3$. $n(n^2-n)+2(2n^2+n)$ is divisible by $2$. – user26486 Jul 13 '15 at 14:43
$n^3+3n^2+2n=(n+1)(n+2)n$. One of the factors must be even and one must be a multiple of three. Hence the product is a multiple of both $2$ and $3$ and hence is divisible by the least common multiple of $2$ and $3$, which turns out to be $6$.
The product of three consecutive integers is divisible by $2$ since one of the factors need to be even. It is also divisible by $3$. (why?) The the product of three consecutive integers is then divisible by the least common multiple of those two numbers $2\times 3 =6$.
Alternatively, note that $$n(n+1)(n+2) = 6 {n+2 \choose 3}$$
$$n^3+3n^2+2n = 6\binom{n+2}{3} \in 6\mathbb{Z}.$$
• Whoa, i don't even know what ∈ means, but i'm guessing can be divided by 6. – Callum Hemsley Jul 13 '15 at 14:33
• $x \in A$ means "x is a member of the set A". $6 \mathbb{Z}$ is the set of integer multiples of 6. Anyway, this is more or less just stating the conclusion, since it requires you to know that $(n+2)(n+1)n$ is divisible by $6$ in order to prove that ${n+2 \choose 3}$ is an integer in the first place. But it does give some nice generality: in general $n(n+1)\dots(n+k)$ is divisible by $(k+1)!$. – Ian Jul 13 '15 at 14:34
• @Ian Ah okay, thanks for clarifying. – Callum Hemsley Jul 13 '15 at 14:35
• The benefit of this solution is that it helps in proving $(n+1)(n+2)(n+3)(n+4)$ is a multiple of $24$. – Jorge Fernández Hidalgo Jul 13 '15 at 14:40
• @dREaM The same basic argument you applied in your answer can be used to prove it. At least one of $n+1,n+2,n+3,n+4$ is a multiple of $4$, at least one is even but not a multiple of $4$, at least one is divisible by $3$. – user26486 Jul 13 '15 at 14:49
Using only modular arithmetic, without factoring, you can see that with $p(n)=n^{3}+3 n^{2}+2 n$ we have if $n\cong0(mod 2)$ then $p(n)\cong0+0+0=0 (mod 2)$ if $n\cong1(mod 2)$ then $p(n)\cong1+3+2\cong0(mod 2)$ So $2|p(n)$. Similarly if $n\cong1(mod 3)$ then $p(n)\cong1+3+2\cong0(mod 3)$, if $n\cong2(mod 3)$ then $p(n)\cong8+12+4\cong0(mod 3)$ so $3|p(n)$. Since both 2 and 3 divide $p(n)$ then $p(n)$ is a multiple of 6.
Keep going with your idea. $n(n+1)$ is the product of two consecutive integer so one of them is even. If $n$ is even we are done and if not $n+1$ is and the product is therefore divisible by $2$.
Similarly $n(n+1)(n+2)$ is the product of three consecutive integers so one of them is divisible by $3$. Let's test the residues modulo $3$
$$\begin{array}{c | c c c} n & n+1 & n+2 & n(n+1)(n+2)\\ \hline 0 & 1 & 2 & 0\\ 1 & 2 & 0 & 0\\ 2 & 0 & 1 & 0 \end{array}$$
So the product is also divisible by $3$ and therefore the product $n(n+1)(n+2)\equiv 0\pmod 6$
One could also test directly the congruences modulo $6$
$$\begin{array}{c|c c c c} n & n^3 & 3n^2 & 2n & n^3+3n^2+2n\\ \hline 0 & 0 & 0 & 0 & 0\\ 1 & 1 & 3 & 2 & 0\\ 2 & 2 & 0 & 4 & 0\\ 3 & 3 & 3 & 0 & 0\\ 4 & 4 & 0 & 2 & 0\\ 5 & 5 & 3 & 4 & 0 \end{array}$$
And we have proven that $n^3+3n^2+2n\equiv 0\pmod 6$
• I think i understand, so essentially $n(n+1)$ has to have an even number, and $n(n+1)(n+2)$ must be divisible by $3$.Then it's the lowest common multiple of these two combined? I just struggle to understand the concept of actually "proving" it. – Callum Hemsley Jul 13 '15 at 14:39
• I edited the answer to incorporate more details – marwalix Jul 13 '15 at 16:29 | 2020-01-26T21:04:25 | {
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https://mymathforum.com/threads/how-to-obtain-the-mass-of-a-particle-rotating-around-an-axis-when-the-difference-of-tensions-in-a-wire-are-known.347877/ | # How to obtain the mass of a particle rotating around an axis when the difference of tensions in a wire are known?
#### Chemist116
The problem is as follows:
A particle of mass $m$ is tied to a very thin wire. Assume the wire is inflexible and of negligible mass. The particle is spinning about a fixed axis as shown located in the center of the circle. Let $T_a$ and $T_b$ be the modulus of the tensions in the string when the particle is located in the points $a$ and $b$ respectively. Find the mass $m$ of the particle if the difference between the tensions $T_{b}-T_{a}=39.2\,N$. Assume $g=9.8\frac{m}{s^2}$
The alternatives are as follows:
$\begin{array}{ll} 1.&0.8\,kg\\ 2.&2.0\,kg\\ 3.&5.8\,kg\\ 4.&1.0\,kg\\ 5.&0.5\,kg\\ \end{array}$
For this problem what I've attempted to do was to use the force at the top to be as follows:
$T+mg=\frac{mv_a^2}{R}$
At this point the Tension must be zero (I don't know if this statement is correct.)
This reduces the top equation to:
$v_a=\sqrt{Rg}$
Then to obtain the speed in the lowest point would be by the conservation of mechanical energy:
$E_i=E_f$
$\frac{1}{2}mv^2_{a}+mgR=mg(-R)+\frac{1}{2}mv_{b}^2$
Then inserting in the above equation would give the speed for the bottom:
Cancelling masses and multiplying by $2$ to both terms:
$v^2_{a}+4gR=v^2_{b}$
Since it is known $v_{a}$ then:
$v_{b}^2=Rg+4Rg=5Rg$
Finally I'll use these in the given statements:
The tension in the top:
$T_a+mg=\frac{mv_{a}^2}{R}$
Tension in the bottom.
$T_b-mg=\frac{mv_{b}^2}{R}$
Doing a difference between these:
$T_b-T_a=2mg+\frac{mv_{b}^2}{R}-\frac{mv_{a}^2}{R}$
Replacing the known values:
$T_b-T_a=2mg+\frac{m(5Rg)}{R}-\frac{m(Rg)}{R}$
$T_b-T_a=2mg+4mg=6mg$
Then:
$39.2=6m(10)$
Which results into:
$m=0.67\,kg$
But the answers sheet indicates that the mass is $2\,kg$.
For doing that what it should happenned is that the "$2mg$" is negative in the right side of the equation but I can't find a way to do that. Can someone help me here?. Is it me?, or did I overlooked anything?. Help!. Please.
#### skeeter
Math Team
I agree with your solution ...
$T_b-T_a = \dfrac{m}{r}(v_b^2-v_a^2) + 2mg$
using energy conservation yields $v_b^2-v_a^2= 4rg$ | 2020-02-17T15:20:19 | {
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https://math.stackexchange.com/questions/2923662/question-on-coin-tossing-probability | # Question on coin tossing - probability
When considering an infinite sequence of tosses of a fair coin, how long will it take on an average until the pattern H T T H appears?
I tried to break the problem into cases where ultimately the pattern HTTH appears, but that makes things complex. Any insight would be helpful.
• HTTH or HHTH? In fact the answer to both is $18$, even though for HHHT it would be $16$ and for HTHT it would be $20$ and for HHHH it would be $30$ – Henry Sep 20 '18 at 8:17
(I'm working with HTTH as desired pattern.) Distinguish the five states $$s_0=\emptyset,\quad s_1=H,\quad s_2=HT,\qquad s_3=HTT,\quad s_4=HTTH\ ,$$ whereby the written letters denote the last tosses insofar as useful. Let $E_k$ $(0\leq k\leq4)$ be the expected number of additional tosses when you are in state $s_k$. Then of course $E_4=0$. Furthermore we have the following equations: $$E_0=1+{1\over2}E_0+{1\over2}E_1,\quad E_1=1+{1\over2}E_1+{1\over2}E_2,\quad E_2=1+{1\over2}E_1+{1\over2}E_3,\quad E_3=1+{1\over2}E_0\ .$$ Now solve this system; then $E_0$ is the solution to your problem.
• Beautiful. One could have understand it easier if you realize the whole thing is a Markov chain. – William Wong Sep 20 '18 at 21:24
I like Christian's answer better as it is simple. But I worked out another solution so I just added it for reference.
Let:
$H_n$ be the event of no "HTTH" in n flips and ends with "H"
$T_n$ the event of no "HTTH" in n flips and ends with "T"
$p_n = \text{Prob}(H_n)$
$q_n = \text{Prob}(T_n)$
So, $p_1=p_2=p_3=q_1=q_2=q_3=1/2$
It is not hard to verify $\left[ \begin{array}{c} p_{n+1}\\q_{n+1} \end{array}\right] = A \left[ \begin{array}{c} p_{n}\\q_{n} \end{array}\right]$, where the matrix $A = \left[\begin{array}{cc}\frac{1}{2} & \frac{3}{8}\\\frac{1}{2} & \frac{1}{2} \end{array} \right]$.
The desired expectation is given by
$E = \left[0, \frac{1}{8}\right] \left\{\sum_{n=4}^{\infty} nA^{n-4}\right\}\left[ \begin{array}{c} p_{n}\\q_{n} \end{array}\right]$
One can further prove the matrix power series $\sum_{n=4}^{\infty}nA^{n-4} = (4I-3A)(I-A)^{-2} = \left[\begin{array}{cc}136 & 114\\ 152 & 136 \end{array} \right]$
Therefore $E = \frac{1}{16} (152+136) = 18$
• Forget this. I think this solution is still wrong. – William Wong Sep 20 '18 at 21:23
Here's yet another way of doing this.
We can be in one of $$5$$ states, according to the number $$0\le k\le4$$ of consecutive “correct” results that we currently have; e.g. $$k=2$$ means the last two results were HT. Let's make a graph with the transitions between these states:
The labels on the vertices are the values of $$k$$. We'll get to the edge weights in a bit. In addition to the two possible transitions from each of the nodes with $$0\le k\le3$$, I added an edge from $$4$$ to $$0$$ to complete a round trip from $$0$$ to $$4$$ and back.
The idea is to choose the edge weights such that at each vertex all outgoing edges have the same weight and the sums of incoming and outgoing edge weights are equal. For instance, start at $$3$$ and arbitrarily assign weight $$1$$ to its outgoing edges; then work your way around the graph, e.g. it follows that the edge $$4\to0$$ also has weight $$1$$, the edge $$2\to3$$ has weight $$2$$, thus the edge $$2\to1$$ also has weight $$2$$, and so on. You barely have to compute anything.
Now consider a process on the edges where in each step we go from the current edge to its terminal vertex and then uniformly randomly choose one of its outgoing edges. Then the equilibrium distribution on the edges is given by the edge weights, normalized by their sum, since by construction that lets the same probability flow into and out of each vertex in each step.
We're looking for the expected time to reach $$4$$ from $$0$$. After we reach $$4$$, we always go to $$0$$ in a single step. Thus, the time we want is $$1$$ less than the expected time for a round trip from $$0$$ to $$4$$ and back. On such a round trip, we spend exactly $$1$$ step on the edge $$4\to0$$. But then the expected time for the round trip must be the reciprocal of the equilibrium probability of being on edge $$4\to0$$. Since that edge conveniently has weight $$1$$, the expected time for the round trip is just the sum of the edge weights, which is $$19$$. Thus the expected time to get from $$0$$ to $$4$$ is $$1$$ less, $$18$$. | 2021-03-04T09:23:22 | {
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https://math.stackexchange.com/questions/1225011/finding-the-minimum-wins-in-a-round-robin-tournament | # Finding the minimum wins in a round-robin tournament.
There are 16 teams in total. They are divided into two groups of 8 each. In a group, each team plays a single match against every other team. At the end of the round, top 4 teams go through to the next round from each group. Assume, a match can only result in either a win or a loss and never a tie.1 win=1 point; 1 loss=0 point.
Questions
1.What is the minimum no. of wins needed for a team in the first stage to guarantee its advancement to the next round?
2.What is the highest no. of wins for a team in the first round in spite of which it would be eliminated at the end of the first stage?
3.What is the minimum number of wins that may allow advancement to the second round?
The way i approached the question is like this:
There will be 28 matches each within the two teams. After that I started distributing the points by hit and trial. I know that there are 28 match points. But, say for question 1, can we say that $28=5*5+3$. So 5 teams with 5 winning each creating ambiguity? So, the minimum possible wins that will always guarantee is $5+1=6$. wins. But then I started distributing match points to the teams and found 5 points 5 times is not possible. How to get the answer logically and quickly?
PS: The answer given is 6(Q 1) and 2 (Q 3).
Your approach to the first question is entirely reasonable. Since $5\cdot6=30>28$, it’s clear that $6$ wins guarantee advancement. To show that $5$ do not, we need only show that it’s possible for $5$ teams to win $5$ matches each; the arrangement shown by user73985 works fine and is natural enough that it’s the first one that I found as well. (You can complete it by distributing the remaining $3$ wins within the group consisting of teams $6,7$, and $8$; for example, you could have team $6$ beat teams $7$ and $8$, and team $7$ beat team $8$.)
For the second question, suppose that a team has $3$ wins; is it possible that it could advance to the second round? Suppose that teams $1,2$, and $3$ beat each of the higher-numbered teams: team $1$ beats everybody else, team $2$ beats everyone else except team $1$, and team $3$ beats everyone else except teams $1$ and $2$. That accounts for $18$ wins, leaving $10$ to be determined. Team $4$ beats teams $5,6$, and $7$, for $3$ wins, leaving $7$ wins to still to be distributed amongst teams $5,6,7$, and $8$. We want team $4$, with its measly $3$ wins, to make it to the second round, so we’ll try to split the remaining $7$ wins $2,2,2$, and $1$ amongst teams $5,6,7$, and $8$. Team $4$ did not beat team $8$, so team $8$ beat team $4$ and already has one win; we can give it another against team $5$. Now it must lose to teams $6$ and $7$, so they now have one win apiece. We can finish up by letting team $5$ beat team $6$, team $6$ beat team $7$, and team $7$ beat team $5$. To sum up:
• Team $1$ beats teams $2,3,4,5,6,7,8$.
• Team $2$ beats teams $3,4,5,6,7,8$.
• Team $3$ beats teams $4,5,6,7,8$.
• Team $4$ beats teams $5,6,7$.
• Team $5$ beats team $6$.
• Team $6$ beats teams $7,8$.
• Team $7$ beats teams $5,8$.
• Team $8$ beats teams $4,5$.
Thus, it’s possible to make it into the second round with just $3$ wins.
With only $2$ wins, however, it’s impossible to get to the second round. The top three finishers cannot have more than $18$ wins altogether (either $7+6+5$ or $6+6+6$). That leaves $10$ wins amongst the remaining $5$ teams, so either each of the bottom $5$ teams has $2$ wins, or one of them has at least $3$ wins. In neither case is a team with just $2$ wins assured of getting into the second round.
There is some ambiguity in the situation in which each of the bottom $5$ teams wins $2$ matches. (This can happen, e.g., if teams $1,2$ and $3$ all beat each of teams $4,5,6,7$, and $8$, team $4$ beats teams $5$ and $6$, team $5$ beats teams $6$ and $7$, team $6$ beats teams $7$ and $8$, team $7$ beats teams $8$ and $4$, and team $8$ beats teams $4$ and $5$.) In this case the rules as given in the question don’t specify what happens. Some of the possibilities are: that only the top three teams go to the second round; that there is a playoff for the fourth position; that the fourth position is chosen randomly; or that the fourth position is decided by some tie-breaker like goal differential. As long as four teams always move on to the second round, it’s still possible for a team with just $2$ wins in the first round to move on, but if that does happen, other teams with $2$ wins fail to move on.
• "With only 2 wins, however, it’s impossible to get to the second round" - this phrasing seems to contradict the last sentence of the answer, which says "it’s still possible for a team with just 2 wins in the first round to move on". As your answer explains, it's possible for a 2-wins team to progress, but it is not guaranteed. My suggested edit was rejected, but I believe it should read something similar to "with only 2 wins, however, it’s not possible to guarantee progression to the second round." – Spinner May 27 at 13:32
5 points 5 times is possible. Call the teams 1, ..., 8. Teams 1-5 all beat all of teams 6-8, and then:
Team 1 beats teams 2 and 3
Team 2 beats teams 3 and 4
Team 3 beats teams 4 and 5
Team 4 beats teams 5 and 1
Team 5 beats teams 1 and 2 | 2019-08-20T21:03:24 | {
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