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https://math.stackexchange.com/questions/1939658/why-isnt-the-graph-of-ln-tanx2-same-as-that-of-2-ln-tanx-when-th | # Why isn't the graph of $\ln(\tan(x^2))$ same as that of $2\ln(\tan(x))$, when they should be according to the power rule?
Recently, I was trying to graph the function $\ln(\tan(x^2))$ without derivatives or any calculator whatsoever. To get the answer, I used the graphing software Desmos, and was playing around when I saw that the graphs of $\ln(\tan(x^2))$ and $2\ln(\tan(x))$ are not the same. This goes against what I learned in high school, of the power rule property of $\ln$ and $\log$. Maybe it's a stupid question, I feel it is, but I can't see why and it is gnawing at me. Please explain. This is my first post, sorry if it's not properly formatted.
Note that $\tan(x^2) \neq (\tan x)^2$, and so $$\ln((\tan x)^2) =2\ln(\tan x)$$ but $$\ln(\tan (x^2)) \neq 2\ln(\tan x)$$
Edit: As noted in the other answer, you have to check the domain. Here's what I got when I tried Desmos: (they overlap when $x\in (0,\pi/2)$ mod $\pi$)
• Thanks for responding! I entered that into Desmos and once again, the graphs did not turn out the same. – cgug123 Sep 24 '16 at 16:26
• @cgug123 What did you plug in Desmos? wolframalpha.com/input/… – trang1618 Sep 24 '16 at 16:30
• @cgug123 you have to be careful with the domain as well. – trang1618 Sep 24 '16 at 16:37
• I put 2ln(tanx) and ln((tanx)^2). Sorry I don't know how to format it properly on a computer. – cgug123 Sep 24 '16 at 16:46
• The wolframalpha link I gave in the above comment show you the three curves, where two of them overlap. – trang1618 Sep 24 '16 at 16:50
It's because the identity $\log x^n = n \log x$ holds only for $x > 0$.
Thus, here we must have $\tan x > 0$, or $x \in (0, \frac \pi 2) \mod \pi$ | 2021-01-16T06:23:10 | {
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http://mathhelpforum.com/calculus/172241-integration-expression-root-second-order-polynomial-denominator-print.html | # Integration of expression with root of second order polynomial in denominator
• February 22nd 2011, 12:41 PM
ironband
Integration of expression with root of second order polynomial in denominator
Hi all,
I'm trying to help a friend out and a bit rusty on my calc. Trying to find indefinite integral:
$\int \frac{1}{(x-1)\sqrt{4x^2-8x+3}}dx$
I've run it through Wolfram Alpha, but the solution seems overly complex to me...it seems like I should be able to get there a little easier than that, but I could be wrong.
My out of practice intuition says that there might be some way to solve this thing by doing a u substitution for $4x^2-8x+3$, such that $\frac{du}{dx}=8x-8=8(x-1)$, and relating that back somehow to the x-1 term...but for the life of me I'm just not getting there.
Anyway, if I really need to go the route of completing the square and making a second substitution as Wolfram suggests, that's fine...but I can't shake the feeling that a more elegant solution exists.
Thanks in advance for any help!
Andre
• February 22nd 2011, 01:25 PM
ironband
OK...I think it is coming to me...but not sure...
If I complete the square of $4x^2-8x+3$ by subtracting 1, I get
$\int \frac{1}{(x-1)\sqrt{(2x-2)^2-1}}dx$
Then, if I pull a factor of 4 out of the expression under the root
$\sqrt{(2x-2)^2-1}=\sqrt{4(x-1)^2-4\frac{1}{4}}=\sqrt{4[(x-1)^2-\frac{1}{4}]}=2\sqrt{(x-1)^2-\frac{1}{4}}$
Which allows me to write this thing as
$\frac{1}{2}\int \frac{1}{(x-1)\sqrt{(x-1)^2-\frac{1}{4}}}dx$
and if I set u=x-1 and a=1/2, I can use this from my table of integrals:
$\int \frac{1}{u\sqrt{u^2-a^2}}du=\frac{1}{a}arcsec\frac{u}{a}+C$
Does that work, or did I project some wishful thinking into this?
• February 22nd 2011, 01:28 PM
Soroban
Hello, ironband!
Well done! . . . Great work!
Here's how I reasoned it out . . .
Quote:
$\int \dfrac{dx}{(x-1)\sqrt{4x^2-8x+3}}$
. . $4x^2 - 8x + 3 \;=\;4x^2-8x + 4 - 1$
. . . . . . . . . . . . $=\;4(x^2 - 2x + 1) - 1$
. . . . . . . . . . . . $=\;[2(x-1)]^2-1$
$\displaystyle \text{The integral becoms: }\;\int\frac{dx}{(x-1)\sqrt{[2(x-1)]^2-1}}$
$\text{Let: }\:2(x-1) \:=\:\sec\theta \quad\Rightarrow\quad x-1 \:=\:\tfrac{1}{2}\sec\theta$
. . $\Rightarrow\quad dx \:=\:\frac{1}{2}\sec\theta \tan \theta\, d\theta \quad\Rightarrow\quad \sqrt{2(x-1)^2-1} \:=\:\tan\theta$
$\displaystyle \text{Substitute: }\;\int\frac{\frac{1}{2}\sec\theta\tan\theta\, d\theta}{\frac{1}{2}\sec\theta\,\tan\theta} \;=\;\int d\theta \;=\;\theta + C$
$\text{Back-substitute: }\:\text{arcsec }2(x-1) + C$
• February 22nd 2011, 01:47 PM
ironband
OK, I get the same answer. good deal! Thanks
• February 22nd 2011, 02:04 PM
TheCoffeeMachine
Note that $(4x^2-8x+3)' = 8(x-1)$, so let:
$t = \sqrt{4x^2-8x+3} \Rightarrow \begin{cases}
\frac{dt}{dx} = \frac{8(x-1)}{2\sqrt{4x^2-8x+3}} \Rightarrow dx = \frac{\sqrt{4x^2-8x+3}}{4(x-1)}\;{dt} \\
t^2 = 4(x-1)^2-1 \Rightarrow (x-1)^2 = \frac{1}{4}(t^2+1).
\end{cases}$
, then:
\displaystyle \begin{aligned} I & = \frac{1}{4}\int\frac{\sqrt{4x^2-8x+3}}{(x-1)^2\sqrt{4x^2-8x+3}}\;{dt} = \frac{1}{4}\int\frac{4}{t^2+1}\;{dt} \\& = \int\frac{1}{t^2+1}\;{dt} = \arctan{t} = \arctan{\sqrt{4x^2-8x+3}}+k. \end{aligned} | 2015-07-28T08:53:45 | {
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http://math.stackexchange.com/questions/232320/if-text-pxk1-cfrac-12-text-pxk-what-is-the-distribution-of-x | # If $\text P(X>k+1)=\cfrac 12 \text P(X>k)$, what is the distribution of $X$?
If $\text P(X>k+1)=\cfrac 12 \text P(X>k)$, $k\in \mathbb N^*$ what is the distribution of $X$?
So I did; $$\text P(X=k+1)= \text P(X>k)- \text P(X>k+1)=\cfrac 12 \text P(X>k)$$ $$\text P(X=k)=\cfrac 12 \text P(X>k-1)$$ I don't know what to deduce from this.
-
Does $k\in \mathbb{Z}^+$? and is $X$ discrete or continuous? – user17762 Nov 7 '12 at 18:50
@Marvis $k\in \mathbb N^*$ – user31280 Nov 7 '12 at 18:52
Is $X$ discrete or continuous? And is $X>0$? – user17762 Nov 7 '12 at 18:57
@Marvis the question didn't specify? – user31280 Nov 7 '12 at 18:59
I will rewrite the last equation that you got as: $P(X=k)=0.5[P(X=k)+P(X=k+1)+P(X=k+2)+....]$....(1) Thus: $P(X=k+1)=0.5[P(X=k+1)+P(X=k+2)+P(X=k+3)+...]$...(2), if we subtract (2) from (1),we get: $P(X=k+1)-P(X=k)=-0.5 P(X=k)$....(3) Since $P(X=1)+P(X=2)+...=1$, therefore recurrence relation (3) can be solved easily.
Assuming $k \in \mathbb{N}$, and assuming that $X>0$, we get that $P(X>0) = 1$. This gives us that $P(X>1) = \dfrac12$ and in general $P(X> k ) = \dfrac1{2^k}$. Hence, $$P(X \in (k-1,k]) = P(X>k-1) - P(X>k) = \dfrac1{2^{k-1}} - \dfrac1{2^k} = \dfrac1{2^k}$$ If $X$ is a discrete random variable taking only values in $\mathbb{Z}^+$, then $$P(X \in (k-1,k]) = P(X = k) = \dfrac1{2^k}$$ | 2016-02-07T04:15:45 | {
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https://math.stackexchange.com/questions/1348633/the-imaginary-unit-i-and-an-alternate-representation | # The imaginary unit, $i,$ and an alternate representation.
Recently, I began working with both complex, and imaginary numbers, and I looked at the complex number $i^{n}.$
If $n = 0, i^{n} = 1,$
$n = 1, i^{n} = i = \sqrt{-1},$
$n = 2, i^{n} = i^{2} = i \cdot i = -1,$
$n = 3, i^{n} = i^{3} = i^{2} \cdot i = -1 \cdot i = -i,$ and
$n = 4, i^{n} = i^{4} = i^{3} \cdot i = -i \cdot i = -i^{2} = -(-1) = 1.$
Would it be correct to say that, $\forall{n}, \Re{\left(i^{n}\right)} = \cos{\left(\frac{n \pi}{2}\right)},$ and $\Im{\left(i^{n}\right)} = \sin{\left(\frac{n \pi}{2}\right)},$ where $\frac{n \pi}{2}$ is in radians?
By this logic, $i^{n} = \cos{\left(\frac{n \pi}{2}\right)} + i \sin{\left(\frac{n \pi}{2}\right)}.$
• That is correct! – paw88789 Jul 3 '15 at 21:52
• @paw88789 Ah, thank you. The numbers do add up, and it does make sense, but I guess the question that I was asking was whether or not it would be mathematically correct to state $i^{n}$ as the proposed equation. – Taylor Jul 3 '15 at 21:53
• I'm not sure if I see the distinction between "correct" and "mathematically correct". But I would still say, yes, it is mathematically correct. – paw88789 Jul 3 '15 at 21:56
• @paw88789 Thank you, my friend! – Taylor Jul 3 '15 at 21:57
• I just expanded my answer to give another way to see why your guess is correct. It doesn't require Euler's formula or anything similar, just some simple observations. – Joonas Ilmavirta Jul 21 '15 at 20:20
Yes, this is correct.
You can see this using the formula $e^{ix}=\cos(x)+i\sin(x)$. First, $i=0+i1=\cos(\pi/2)+i\sin(\pi/2)=e^{i\pi/2}$. Therefore $i^{n}=e^{in\pi/2}=\cos(n\pi/2)+i\sin(n\pi/2)$.
Let me also give another way to see this, from a different angle. The function $f:\mathbb Z\to\mathbb C$ defined by $f(n)=i^n$ has period four since $f(n+4)=i^{4+n}=i^4i^n=1i^n=f(n)$. The function $g:\mathbb Z\to\mathbb C$, $g(n)=\cos(n\pi/2)+i\sin(n\pi/2)$, is also periodic with period four since the cosine and sine functions have period $2\pi$. Since $f(n)=g(n)$ for all $n\in\{0,1,2,3\}$ and both functions have period four, they must in fact be equal for all $n\in\mathbb Z$.
• Thank you very much for your answer! – Taylor Jul 21 '15 at 21:14
You are absolutely correct. It's a specific aspect of the Euler formula, given that $i = e^{i\pi/2}$ and the Euler formula being : $e^{ix} = \cos x + i \sin x$.
• Ah, I see. I was not aware of the fact that $i = e^{\frac{i \pi}{2}},$ but I was aware of the fact that Euler's identity states that $e^{ix} = \cos{\left(x\right)} + i \sin{\left(x\right)}.$ Thank you for your answer. – Taylor Jul 3 '15 at 21:55
• @Taylor $i=e^{i \pi/2}$ is what you get from Euler's identity by taking $x=\pi/2$ and simplifying. – Ian Jul 3 '15 at 22:08
By Euler's formula, for any $x=a+bi \in \mathbb{C}$, $$a+bi = re^{i\theta} = r(\cos \theta + i\sin \theta)$$ where $r= \sqrt{a^2 + b^2}$ and $\theta = \arctan (b/a)$. Thus $i = e^{i\pi/2}$ and $$i^n = e^{ni\pi /2} = \cos\big(\frac{n\pi}{2}\big) + i\sin\big(\frac{n\pi}{2}\big)$$
To see Euler's formula you can think of the complex numbers as ordered pairs $(a,b)$ equipped with a different kind of multiplication: $$(a_1,b_1)\cdot (a_2,b_2) = (a_1a_2,-b_1b_2)$$ Then you can convert these ordered pairs to polar coordinates via the formulas above for $r$ and $\theta$. The expression $re^{i\theta}$ follows from the identities $$\cos x = \frac{e^{ix} + e^{-ix}}{2} \hspace{2mm} , \hspace{2mm} \sin x = \frac{e^{ix} - e^{-ix}}{2i}$$ | 2019-08-25T13:56:06 | {
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http://mathhelpforum.com/calculus/190923-need-help-tangent-lines-derivatives.html | # Math Help - Need help with tangent lines and derivatives
1. ## Need help with tangent lines and derivatives
Hey,
I'm struggling with this one problem from my text book.
"Find the equations of the tangent lines to the parabola y=(x^2)+x that pass through the point (2, -3)."
I know how to find the derivative (2x+1) and then I can find one set of points (5,30) but I can't figure out how to find the other set of points.
2. ## Re: Need help with tangent lines and derivatives
f'(a) = 2a + 1
And the slope of the line between the point (a, a^2 + a) and (2, -3) is (a^2 + a + 3)/(2 - a) = 2a + 1
Can you solve that?
3. ## Re: Need help with tangent lines and derivatives
Yes, I get the points (5, 30) and (-1,0) and from there I can find the equations of the lines. Thank you so much!
4. ## Re: Need help with tangent lines and derivatives
Originally Posted by newslang
Hey,
I'm struggling with this one problem from my text book.
"Find the equations of the tangent lines to the parabola y=(x^2)+x that pass through the point (2, -3)."
I know how to find the derivative (2x+1) and then I can find one set of points (5,30) but I can't figure out how to find the other set of points.
Let's say you have a point (p,r) on your parabola. (So, r is given by $r=p^2+p\,.$ )
The slope, m, of the line tangent to the parabola when x = p is $m=2p+1$. (this from your derivative)
The slope, m, of the line passing through points (p, r) and (2, -3) is given by $m=\frac{r-\,-3}{p-2}=\frac{r+3}{p-2}\,.$
So that $m(p-2)=r+3\,.$
Plug the in the earlier results for r & m into this. Solve for p. (There will likely be two answers.) You then have the x value(s) for the intersection of the tanget line, with the parabola.
5. ## Re: Need help with tangent lines and derivatives
Originally Posted by newslang
Hey,
I'm struggling with this one problem from my text book.
"Find the equations of the tangent lines to the parabola y=(x^2)+x that pass through the point (2, -3)."
I know how to find the derivative (2x+1) and then I can find one set of points (5,30) but I can't figure out how to find the other set of points.
Another way to do a problem like this is to use Fermat's "method of ad-equations" which predates Calculus. Any line through (2, -3) can be written y= m(x-2)- 3= mx- 2m- 3. Of course, at a point where such a line crosses the graph of $y= x^2+ x$ the two functions must give the same y-value: $mx- 2m- 3= x^2+ x$ or $x^2+ (1- m)x+ 2m+ 3= 0$. That's a quadratic equation with solution given by the quadratic formula: $x= \frac{m-1\pm\sqrt{(1- m)^2- 4(2m+ 3)}}{2}$. But if it is tangent that x must be a double root- the only root of this equation. That means that the discriminant must be 0: $(1- m)^2- 4(2m+3)= 0$. Solve that equation for m and then $x= \frac{m-1}{2}$. | 2016-07-27T12:32:04 | {
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https://math.hecker.org/2011/07/28/linear-algebra-and-its-applications-review-exercise-1-16/ | Linear Algebra and Its Applications, Review Exercise 1.16
Review exercise 1.16. Given the following system of equations:
$\setlength\arraycolsep{0.2em}\begin{array}{rcrcr} kx&+&y&=&1 \\ x&+&ky&=&1 \end{array}$
what must $k$ be for the system to have no solution? One solution? An infinite number of solutions?
Answer: If $k = 0$ then the system reduces to
$\setlength\arraycolsep{0.2em}\begin{array}{rcrcr} &&y&=&1 \\ x&&&=&1 \end{array}$
for which $x = 1$ and $y = 1$ is (obviously) a solution.
If $k \ne 0$ then we can multiply the first equation by $1/k$ and subtract it from the second equation to obtain the following system:
$\setlength\arraycolsep{0.2em}\begin{array}{rcrcl} kx&+&y&=&1 \\ &&(k - \frac{1}{k})y&=&1 - \frac{1}{k} \end{array}$
If $k = 1$ then this reduces to
$\setlength\arraycolsep{0.2em}\begin{array}{rcrcl} 1 \cdot x&+&y&=&1 \\ &&(1 - \frac{1}{1})y&=&1 - \frac{1}{1} \end{array} \rightarrow \begin{array}{rcrcl} x&+&y&=&1 \\ &&0&=&0 \end{array}$
This system has an infinite number of solutions, namely any value of $(x, y)$ for which $y = -x + 1$.
If $k \ne 0$ and $k \ne 1$ then we can solve for $y$ as follows:
$y = (1 - \frac{1}{k}) / (k - \frac{1}{k}) = k (1 - \frac{1}{k}) / k(k - \frac{1}{k})$
$= (k - 1) / (k^2 - 1) = (k - 1) / (k - 1)(k + 1) = \frac{1}{k+1}$
We then solve for $x$ as follows:
$kx = 1 - y = 1 - \frac{1}{k+1} = \frac{k+1}{k+1} - \frac{1}{k+1}$
$= \frac{k+1-1}{k+1} = \frac{k}{k+1}$
$\rightarrow x = \frac{1}{k} \frac{k}{k+1} = \frac{1}{k+1}$
So for $k \ne 0$ and $k \ne 1$ the system has the unique solution $(\frac{1}{k+1}, \frac{1}{k+1})$. For example, for $k = 2$ the unique solution is $(\frac{1}{2+1}, \frac{1}{2+1}) = (\frac{1}{3}, \frac{1}{3})$.
Note that the solution for $k = 0$ was $(1, 1)$ which matches that given by our formula: $(\frac{1}{0+1}, \frac{1}{0+1}) = (1, 1)$. Also note that there is no value of $k$ for which no solution exists.
Also note that there is no solution for $k = -1$, for which the solution $(\frac{1}{k+1}, \frac{1}{k+1})$ would require dividing by zero. In this case the system of equations is
$\setlength\arraycolsep{0.2em}\begin{array}{rcrcr} -x&+&y&=&1 \\ x&-&y&=&1 \end{array} \rightarrow \begin{array}{rcrcr} x&-&y&=&-1 \\ x&-&y&=&1 \end{array}$
and produces the contradiction $-1 = 1$.
NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.
If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.
This entry was posted in linear algebra. Bookmark the permalink.
2 Responses to Linear Algebra and Its Applications, Review Exercise 1.16
1. Steve says:
I agree with the answers for one solution and infinitely many solutions, but I do not agree with your statement “that there is no value of k for which no solution exists.” If one sets k equal to -1, then x and y are undefined, for you are dividing by zero. Wouldn’t k = -1 produce no solution for the system?
• hecker says:
You are absolutely correct; thanks for catching that! I’ve corrected the answer. | 2020-07-06T09:28:34 | {
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https://www.quizover.com/calculus/course/6-6-moments-and-centers-of-mass-by-openstax?page=5 | 6.6 Moments and centers of mass (Page 6/14)
Page 6 / 14
Let R be the region bounded above by the graph of the function $f\left(x\right)=6-{x}^{2}$ and below by the graph of the function $g\left(x\right)=3-2x.$ Find the centroid of the region.
The centroid of the region is $\left(1,13\text{/}5\right).$
The symmetry principle
We stated the symmetry principle earlier, when we were looking at the centroid of a rectangle. The symmetry principle can be a great help when finding centroids of regions that are symmetric. Consider the following example.
Finding the centroid of a symmetric region
Let R be the region bounded above by the graph of the function $f\left(x\right)=4-{x}^{2}$ and below by the x -axis. Find the centroid of the region.
The region is depicted in the following figure.
The region is symmetric with respect to the y -axis. Therefore, the x -coordinate of the centroid is zero. We need only calculate $\stackrel{–}{y}.$ Once again, for the sake of convenience, assume $\rho =1.$
First, we calculate the total mass:
$\begin{array}{cc}\hfill m& =\rho {\int }_{a}^{b}f\left(x\right)dx\hfill \\ & ={\int }_{-2}^{2}\left(4-{x}^{2}\right)dx\hfill \\ & ={\left[4x-\frac{{x}^{3}}{3}\right]\phantom{\rule{0.2em}{0ex}}|}_{-2}^{2}=\frac{32}{3}.\hfill \end{array}$
Next, we calculate the moments. We only need ${M}_{x}\text{:}$
$\begin{array}{cc}\hfill {M}_{x}& =\rho {\int }_{a}^{b}\frac{{\left[f\left(x\right)\right]}^{2}}{2}dx\hfill \\ & =\frac{1}{2}{\int }_{-2}^{2}{\left[4-{x}^{2}\right]}^{2}dx=\frac{1}{2}{\int }_{-2}^{2}\left(16-8{x}^{2}+{x}^{4}\right)dx\hfill \\ & =\frac{1}{2}{\left[\frac{{x}^{5}}{5}-\frac{8{x}^{3}}{3}+16x\right]\phantom{\rule{0.2em}{0ex}}|}_{-2}^{2}=\frac{256}{15}.\hfill \end{array}$
Then we have
$\stackrel{–}{y}=\frac{{M}_{x}}{y}=\frac{256}{15}·\frac{3}{32}=\frac{8}{5}.$
The centroid of the region is $\left(0,8\text{/}5\right).$
Let R be the region bounded above by the graph of the function $f\left(x\right)=1-{x}^{2}$ and below by x -axis. Find the centroid of the region.
The centroid of the region is $\left(0,2\text{/}5\right).$
The grand canyon skywalk
The Grand Canyon Skywalk opened to the public on March 28, 2007. This engineering marvel is a horseshoe-shaped observation platform suspended 4000 ft above the Colorado River on the West Rim of the Grand Canyon. Its crystal-clear glass floor allows stunning views of the canyon below (see the following figure).
The Skywalk is a cantilever design, meaning that the observation platform extends over the rim of the canyon, with no visible means of support below it. Despite the lack of visible support posts or struts, cantilever structures are engineered to be very stable and the Skywalk is no exception. The observation platform is attached firmly to support posts that extend 46 ft down into bedrock. The structure was built to withstand 100-mph winds and an 8.0-magnitude earthquake within 50 mi, and is capable of supporting more than 70,000,000 lb.
One factor affecting the stability of the Skywalk is the center of gravity of the structure. We are going to calculate the center of gravity of the Skywalk, and examine how the center of gravity changes when tourists walk out onto the observation platform.
The observation platform is U-shaped. The legs of the U are 10 ft wide and begin on land, under the visitors’ center, 48 ft from the edge of the canyon. The platform extends 70 ft over the edge of the canyon.
To calculate the center of mass of the structure, we treat it as a lamina and use a two-dimensional region in the xy -plane to represent the platform. We begin by dividing the region into three subregions so we can consider each subregion separately. The first region, denoted ${R}_{1},$ consists of the curved part of the U. We model ${R}_{1}$ as a semicircular annulus, with inner radius 25 ft and outer radius 35 ft, centered at the origin (see the following figure).
The legs of the platform, extending 35 ft between ${R}_{1}$ and the canyon wall, comprise the second sub-region, ${R}_{2}.$ Last, the ends of the legs, which extend 48 ft under the visitor center, comprise the third sub-region, ${R}_{3}.$ Assume the density of the lamina is constant and assume the total weight of the platform is 1,200,000 lb (not including the weight of the visitor center; we will consider that later). Use $g=32\phantom{\rule{0.2em}{0ex}}{\text{ft/sec}}^{2}.$
1. Compute the area of each of the three sub-regions. Note that the areas of regions ${R}_{2}$ and ${R}_{3}$ should include the areas of the legs only, not the open space between them. Round answers to the nearest square foot.
2. Determine the mass associated with each of the three sub-regions.
3. Calculate the center of mass of each of the three sub-regions.
4. Now, treat each of the three sub-regions as a point mass located at the center of mass of the corresponding sub-region. Using this representation, calculate the center of mass of the entire platform.
5. Assume the visitor center weighs 2,200,000 lb, with a center of mass corresponding to the center of mass of ${R}_{3}.$ Treating the visitor center as a point mass, recalculate the center of mass of the system. How does the center of mass change?
6. Although the Skywalk was built to limit the number of people on the observation platform to 120, the platform is capable of supporting up to 800 people weighing 200 lb each. If all 800 people were allowed on the platform, and all of them went to the farthest end of the platform, how would the center of gravity of the system be affected? (Include the visitor center in the calculations and represent the people by a point mass located at the farthest edge of the platform, 70 ft from the canyon wall.)
why n does not equal -1
ask a complete question if you want a complete answer.
Andrew
I agree with Andrew
Bg
f (x) = a is a function. It's a constant function.
proof the formula integration of udv=uv-integration of vdu.?
Find derivative (2x^3+6xy-4y^2)^2
no x=2 is not a function, as there is nothing that's changing.
are you sure sir? please make it sure and reply please. thanks a lot sir I'm grateful.
The
i mean can we replace the roles of x and y and call x=2 as function
The
if x =y and x = 800 what is y
y=800
800
Bg
how do u factor the numerator?
Nonsense, you factor numbers
Antonio
You can factorize the numerator of an expression. What's the problem there? here's an example. f(x)=((x^2)-(y^2))/2 Then numerator is x squared minus y squared. It's factorized as (x+y)(x-y). so the overall function becomes : ((x+y)(x-y))/2
The
The problem is the question, is not a problem where it is, but what it is
Antonio
I think you should first know the basics man: PS
Vishal
Yes, what factorization is
Antonio
Antonio bro is x=2 a function?
The
Yes, and no.... Its a function if for every x, y=2.... If not is a single value constant
Antonio
you could define it as a constant function if you wanted where a function of "y" defines x f(y) = 2 no real use to doing that though
zach
Why y, if domain its usually defined as x, bro, so you creates confusion
Antonio
Its f(x) =y=2 for every x
Antonio
Yes but he said could you put x = 2 as a function you put y = 2 as a function
zach
F(y) in this case is not a function since for every value of y you have not a single point but many ones, so there is not f(y)
Antonio
x = 2 defined as a function of f(y) = 2 says for every y x will equal 2 this silly creates a vertical line and is equivalent to saying x = 2 just in a function notation as the user above asked. you put f(x) = 2 this means for every x y is 2 this creates a horizontal line and is not equivalent
zach
The said x=2 and that 2 is y
Antonio
that 2 is not y, y is a variable 2 is a constant
zach
So 2 is defined as f(x) =2
Antonio
No y its constant =2
Antonio
what variable does that function define
zach
the function f(x) =2 takes every input of x within it's domain and gives 2 if for instance f:x -> y then for every x, y =2 giving a horizontal line this is NOT equivalent to the expression x = 2
zach
Yes true, y=2 its a constant, so a line parallel to y axix as function of y
Antonio
Sorry x=2
Antonio
And you are right, but os not a function of x, its a function of y
Antonio
As function of x is meaningless, is not a finction
Antonio
yeah you mean what I said in my first post, smh
zach
I mean (0xY) +x = 2 so y can be as you want, the result its 2 every time
Antonio
OK you can call this "function" on a set {2}, but its a single value function, a constant
Antonio
well as long as you got there eventually
zach
volume between cone z=√(x^2+y^2) and plane z=2
Fatima
It's an integral easy
Antonio
V=1/3 h π (R^2+r2+ r*R(
Antonio
How do we find the horizontal asymptote of a function using limits?
Easy lim f(x) x-->~ =c
Antonio
solutions for combining functions
what is a function? f(x)
one that is one to one, one that passes the vertical line test
Andrew
It's a law f() that to every point (x) on the Domain gives a single point in the codomain f(x)=y
Antonio
is x=2 a function?
The
restate the problem. and I will look. ty
is x=2 a function?
The
What is limit
it's the value a function will take while approaching a particular value
Dan
don ger it
Jeremy
what is a limit?
Dlamini
it is the value the function approaches as the input approaches that value.
Andrew
Thanx
Dlamini
Its' complex a limit It's a metrical and topological natural question... approaching means nothing in math
Antonio
is x=2 a function?
The
3y^2*y' + 2xy^3 + 3y^2y'x^2 = 0 sub in x = 2, and y = 1, isolate y'
what is implicit of y³+x²y³=5 at (2,1)
tel mi about a function. what is it?
Jeremy
A function it's a law, that for each value in the domaon associate a single one in the codomain
Antonio
function is a something which another thing depends upon to take place. Example A son depends on his father. meaning here is the father is function of the son. let the father be y and the son be x. the we say F(X)=Y.
Bg
yes the son on his father
pascal
a function is equivalent to a machine. this machine makes x to create y. thus, y is dependent upon x to be produced. note x is an independent variable
moe
x or y those not matter is just to represent.
Bg | 2018-05-25T18:56:27 | {
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https://crypto.stackexchange.com/questions/45151/anatomy-of-an-rsa-private-key/45157 | # Anatomy of an RSA private key
I'm learning about OpenSSL and public key infrastructure and am curious about the structure of an RSA key and how it's related to its corresponding public key.
I can generate a private RSA key with the OpenSSL genrsa command:
$openssl genrsa -aes128 -out fd.key 2048 Generating RSA private key, 2048 bit long modulus ....+++ ...................................................................................... +++ e is 65537 (0x10001) Enter pass phrase for fd.key: **************** Verifying - Enter pass phrase for fd.key: **************** I know there are three factors to think about when generating a private key: 1. Algorithm 2. Size 3. Passphrase In this case, I'm using the RSA algorithm, 2048 bit size, and encrypting the key with AES-128 encryption. I can view the structure of the RSA key using the rsa command:$ openssl rsa -text -in fd.key
Enter pass phrase for fd.key: ****************
Private-Key: (2048 bit)
modulus:
00:b6:ae:03:fa:d5:ec:3d:7b:1e:a5:33:68:44:5f:
[...]
publicExponent: 65537 (0x10001)
privateExponent:
1a:12:ee:41:3c:6a:84:14:3b:be:42:bf:57:8f:dc:
[...]
prime1:
00:c9:7e:82:e4:74:69:20:ab:80:15:99:7d:5e:49:
[...]
prime2:
00:c9:2c:30:95:3e:cc:a4:07:88:33:32:a5:b1:d7:
[...]
exponent1:
68:f4:5e:07:d3:df:42:a6:32:84:8d:bb:f0:d6:36:
[...]
exponent2:
5e:b8:00:b3:f4:9a:93:cc:bc:13:27:10:9e:f8:7e:
[...]
coefficient:
34:28:cf:72:e5:3f:52:b2:dd:44:56:84:ac:19:00:
[...]
writing RSA key
-----BEGIN RSA PRIVATE KEY-----
[...]
-----END RSA PRIVATE KEY-----
I'm also able to view the public key using the using the rsa command:
$openssl rsa -in fd.key -pubout Enter pass phrase for fd.key: **************** writing RSA key -----BEGIN PUBLIC KEY----- MIIBIjANBgkqhkiG9w0BAQEFAAOCAQ8AMIIBCgKCAQEAyOYm8hJCi3vKLaud2YTU O3glFfQUpJ6d4gXiWp//HkDQIvi2BFmvbUyHMh4XWLwPbmaX2dfJ5Aa8+ZIC9KCY y96Gmbw+v75RzxHq5iFLnZNFhYM2zkMvUUjJs/UqunOL1OoEiC06hb85SBepKtnE JUUKo/rtZ2Sj/pHvF0Wqu1hLyR3iOxdJb26+m2IhOy4wB3HI6FBcvrMd4Hmejpup skIRhTQXkV7XQ79yRCTS3ejiGoVvkPKzWxL+OFWOJTduXAk8McMLEozSGZll8bv7 jJUWLhmegvokKS2eLfA4B16yU59EgNbvuoG5doKUeV0LJ03Iiqv81nFB9SqEG/Xe VQIDAQAB -----END PUBLIC KEY----- From the Hexadecimal wiki page: One hexadecimal digit represents a nibble (4 bits), which is half of an octet or byte (8 bits). Some things I'm curious about: • Which numbers correspond to the 2048bit length? Is it prime, exponent, coefficient...? • How is the public key related to the private key? Is it calculated from the numbers, or is it embedded in the private key? From reading another answer it sounds like the 2048 length corresponds to the modulus. However when I calculate the length I get 2056:$ echo "00:b6:ae:03:fa:d5:ec:3d:7b:1e:a5:33:68:44:5f:d0:6a:0b:b5:87:31:80:a0:50:32:b0:7c:73:4b:f8:a2:03:91:89:c2:11:32:69:2e:13:90:71:f6:a9:48:21:00:c5:ad:1c:93:f0:21:27:ce:ca:15:04:53:30:c6:88:7b:45:c0:f2:01:d5:a7:9e:c1:c5:f2:ae:b0:7f:31:68:b7:3c:c3:62:13:eb:40:25:a9:3f:cd:81:90:9f:a1:3f:02:84:d8:6e:73:d2:5d:53:28:cc:97:35:f6:fa:5c:b7:dc:11:fb:60:08:1b:75:f7:74:dc:24:29:3e:ff:fb:ba:dc:77:2c:48:0d:3b:a1:7b:d9:9a:3d:52:7d:9a:d6:c1:f1:e7:46:df:be:75:b0:d2:0f:d2:1c:7b:25:57:94:33:8f:d5:b3:ee:7f:30:98:a9:06:25:b5:ab:b1:a6:ab:f9:f2:52:b8:e7:8f:87:5f:6d:96:36:67:47:38:4c:ef:29:c7:71:e4:07:7c:13:19:3a:e2:b4:3c:85:18:32:77:a6:98:0e:0d:b4:70:01:75:79:de:e9:83:c5:df:41:2f:69:f6:30:8d:13:29:84:9a:84:3a:c0:6a:4c:0d:bd:cd:9b:1e:93:de:8e:c9:a4:02:b7:0f:a2:96:45:ad:b8:3e:3a:d3:fd:4d" | tr ':' '\n' | wc -l
257
Each line contains 2 hexadecimal digits
257 * 2 = 514
Each digit contains 4 bits
514 * 4 = 2056
Shouldn't it add up to 2048?
• The key length is explained here (it's the modulus). How the RSA private key corresponds to the public key is well-explained on Wikipedia and elsewhere. About what in particular do you have problems understanding? – Arminius Mar 29 '17 at 16:02
• When you count the bytes you should remove the leading zero, when present. I believe it is due to DER encoding (but openssl should not keep it IMHO). So it's not 257 since there is a leading "00", it's 256, so 256*8 = 2048 – Ruggero Mar 30 '17 at 10:05
Which numbers correspond to the 2048bit length? Is it prime, exponent, coefficient...?
Only the modulus really - the key size is identical to the modulus size by definition. The primes are commonly half of the key size for calculations that use 2 primes (multi-prime RSA is faster and on the uptake). The private exponent may not reach the full key size; it's between 0 and the modulus (exclusive). There is a high chance that the size is the same or slightly less than the modulus in bytes.
How is the public key related to the private key? Is it calculated from the numbers, or is it embedded in the private key?
It's calculated during the key pair generation process - the output of which is both a private key and a public key. The private key and public key share the same modulus. Often the private key - generated by a specific tool such as OpenSSL - contains the public exponent, so you can also extract / use the public key if you have the private key.
Usually the public exponent is a known, small value - such as the fourth prime of Fermat: 0x010001. But as that may not be the case the public exponent (and thus the public key) in general isn't calculated from the private key components. If the public exponent is the same size as the private exponent then you cannot calculate it from the private key values.
From reading another answer it sounds like the 2048 length corresponds to the modulus. However when I calculate the length I get 2056 ...
ASN.1 / DER encoded INTEGER values are signed, big endian. As the most significant bit of the positive modulus is always set - assuming that the key size is dividable by 8 - it will always left-pad the modulus with a padding byte set to 00. Otherwise the encoding would represent a negative value in so-called two-complement encoding - the most significant bit is seen as the sign bit: 0 for positive number representation and 1 for negative number representation.
So the signed encoding in octets of the modulus is 2048 + 8 bits or 257 bytes. But the bit length - as it is called in most programming API's - of the modulus is still 2048.
Note that OpenSSL shows the generation parameters required to use the Chinese Remainder Theorem to speed up RSA calculations. In principle a private key could also consist of just the private exponent and the modulus.
For the relation between public and private key : the second integer of the private key is called the decrypting exponent, it is computed with the encrypting exponent $e$ and $\phi(n)$. It is the modular inverse of $e$ mod $\phi(n)$. $\phi(n)$ is computed with the decomposition in prime factors of $n$.
So, when you encrypt a number using the public encrypting exponent, (and always mod n), the decrypting exponent of the private key lets you get your number back.
With regard to the structure of RSA key files, and how the public key is related to the private key - you can see that the modulus in the public key file is the product of the two large primes contained in the private key file by doing the following:
Generate a private key:
openssl genrsa -out private.key 2048
Extract the public key from the private key file:
openssl rsa -in server.key -pubout > public.key
Now, use the following command to view the two large primes in the private key file:
openssl rsa -noout -text -inform PEM -in private.key
In my case, the two large primes are the following (of course, yours will be different):
prime1=00:e7:c4:87:45:a8:9a:1c:62:17:b4:18:a8:28:af:a0:64:f7:ed:58:73:c4:fa:9b:e7:08:68:90:d6:c7:65:f5:33:15:ca:24:d4:aa:94:8b:15:96:4d:5e:12:dd:f2:c7:8a:27:d8:fb:81:b8:c5:fd:b0:51:71:37:92:d8:51:5f:43:2d:c5:15:ba:3a:0a:fd:ef:3b:a0:26:50:3a:79:4f:92:e6:5f:12:a4:a2:9e:ea:bf:e7:32:c0:ec:fd:50:71:d0:56:73:6f:3b:4f:fe:79:21:6e:81:3d:06:03:41:0e:97:1b:2a:7e:e3:a6:e8:3b:53:2f:2c:27:98:8d:a5:6e:6f
Use the following command to view the modulus in the public key file:
openssl rsa -noout -text -inform PEM -in public.key -pubin
In my case, I get:
modulus=00:b6:1c:12:e3:bc:3b:f3:4e:be:a1:dd:26:40:b4:35:d5:12:79:d2:63:3a:c4:f0:6f:b2:5c:13:fe:67:32:58:5b:2a:5a:b2:83:dc:db:e1:86:1d:ac:38:12:73:66:fd:f9:fe:24:cc:61:6b:c0:c8:76:b8:3e:dd:ed:9f:f1:c9:fb:f5:19:a9:32:67:0c:9a:46:36:1e:40:ed:9c:b3:22:b9:66:f4:8e:dc:56:10:d1:39:43:ec:35:dd:eb:fd:4c:a0:72:cf:cd:01:17:d9:69:6c:d9:29:c9:ce:bf:57:cc:23:b3:9a:26:c4:25:d3:80:0b:e1:28:c4:f0:38:8e:37:4a:c8:e8:9d:0c:be:75:3e:00:56:04:17:4a:eb:15:1c:7b:78:f6:01:66:78:92:1b:12:58:62:2d:c7:25:2b:db:63:07:5b:be:0e:2c:18:7d:9c:ba:2c:78:9c:1d:66:63:eb:8d:bd:36:8c:21:6f:ae:41:89:91:dd:f3:d0:96:d1:75:00:39:bf:a4:05:12:fb:b2:75:81:07:cd:e6:e0:80:d7:5e:45:9f:ee:0a:11:a7:c6:4c:e1:5a:7c:5b:15:d1:c2:f4:c6:36:5d:81:e6:cc:35:10:0c:fa:79:34:35:fd:74:fb:0c:95:fe:1c:7f:ef:19:b4:b3:21:c8:dd:35:ae:e1
You can write a short python program to multiply the two large primes to calculate the modulus, and output all three values in hex format.
prime1=0x00e7c48745a89a1c6217b418a828afa064f7ed5873c4fa9be7086890d6c765f53315ca24d4aa948b15964d5e12ddf2c78a27d8fb81b8c5fdb051713792d8515f432dc515ba3a0afdef3ba026503a794f92e65f12a4a29eeabfe732c0ecfd5071d056736f3b4ffe79216e813d0603410e971b2a7ee3a6e83b532f2c27988da56e6f
modulus=prime1*prime2
print('prime1', hex(prime1))
print('prime2', hex(prime2))
print('modulus', hex(modulus))
This produces
prime1: 0xe7c48745a89a1c6217b418a828afa064f7ed5873c4fa9be7086890d6c765f53315ca24d4aa948b15964d5e12ddf2c78a27d8fb81b8c5fdb051713792d8515f432dc515ba3a0afdef3ba026503a794f92e65f12a4a29eeabfe732c0ecfd5071d056736f3b4ffe79216e813d0603410e971b2a7ee3a6e83b532f2c27988da56e6f
modulus: 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
As you can see, this shows that the modulus calculated by multiplying the two large primes from the private key file matches the modulus contained in the public key file.
With regard to the question about whether or not the modulus is actually embedded in the private key file - it is. You can run the following command to convert the base64-encoded binary in the private key file to hex:
cat private.key | grep -v ^- | base64 -d | xxd -c32
This produces:
00000000: 3082 04a4 0201 0002 8201 0100 b61c 12e3 bc3b f34e bea1 dd26 40b4 35d5 1279 d263 0................;.N...&@.5..y.c
00000020: 3ac4 f06f b25c 13fe 6732 585b 2a5a b283 dcdb e186 1dac 3812 7366 fdf9 fe24 cc61 :..o.\..g2X[*Z........8.sf...$$.a 00000040: 6bc0 c876 b83e dded 9ff1 c9fb f519 a932 670c 9a46 361e 40ed 9cb3 22b9 66f4 8edc k..v.>.........2g..F6.@...".f... 00000060: 5610 d139 43ec 35dd ebfd 4ca0 72cf cd01 17d9 696c d929 c9ce bf57 cc23 b39a 26c4 V..9C.5...L.r.....il.)...W.#..&. 00000080: 25d3 800b e128 c4f0 388e 374a c8e8 9d0c be75 3e00 5604 174a eb15 1c7b 78f6 0166 %....(..8.7J.....u>.V..J...{x..f 000000a0: 7892 1b12 5862 2dc7 252b db63 075b be0e 2c18 7d9c ba2c 789c 1d66 63eb 8dbd 368c x...Xb-.%+.c.[..,.}..,x..fc...6. 000000c0: 216f ae41 8991 ddf3 d096 d175 0039 bfa4 0512 fbb2 7581 07cd e6e0 80d7 5e45 9fee !o.A.......u.9......u.......^E.. 000000e0: 0a11 a7c6 4ce1 5a7c 5b15 d1c2 f4c6 365d 81e6 cc35 100c fa79 3435 fd74 fb0c 95fe ....L.Z|[.....6]...5...y45.t.... 00000100: 1c7f ef19 b4b3 21c8 dd35 aee1 0203 0100 0102 8201 003e fa26 6e2b 4270 39e3 2306 ......!..5...........>.&n+Bp9.#. 00000120: df9b b0b6 8d20 fe90 0b50 df9a 6686 3fe1 8a31 15f0 0856 f556 96d3 6216 f3d2 7f24 ..... ...P..f.?..1...V.V..b....$$
00000140: 44fd 33b8 d123 5a86 738a 57f8 fb55 6c28 436c f4a8 ed41 2dc6 9d6f 95a4 2473 c2b2 D.3..#Z.s.W..Ul(Cl...A-..o..$$s.. 00000160: a179 7759 a2d4 3fee c7b3 dbcc ff08 c63f 3aa7 c9c9 1e13 9659 46ef 8078 3cf4 3cc7 .ywY..?........?:......YF..x<.<. 00000180: 5580 4654 8a64 2a03 0e02 26ca 3951 7c4f dee3 300b 5e73 b57f 0a9e eac6 ac02 4261 U.FT.d*...&.9Q|O..0.^s........Ba 000001a0: 1051 9009 15ca b50d 249d 79b7 1483 4727 3bab 70a6 a3f3 4cac 949c 5930 77fd f6d9 .Q......$$.y...G';.p...L...Y0w...
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000004a0: 0724 ff43 fe10 2059 .\$.C.. Y
As you can see, the modulus is embedded in the private key file, starting at the 13th byte. You can also see that the first prime starts at the 282nd byte, and the second prime starts at the 350th byte | 2020-07-13T15:20:51 | {
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https://mathematica.stackexchange.com/questions/181573/forcing-complex-output-to-take-the-form-a-b-i | # Forcing complex output to take the form $a + b\,i$
Solving the complex equation $z^2=1+2\,i$ using
Solve[z^2 == 1 + 2 I]
returns $\left\{\left\{z\to -\sqrt{1+2\,i}\right\},\left\{z\to\sqrt{1+2\,i}\right\}\right\}$, but how do I force Mathematica to always output on the form $a+b\,i$, $a,b\in\mathbb{R}$? Or, if there is no output form from Solve to do this, to convert/transform the answer to the $a+b\,i$ form?
I tried
z = a + b I;
Solve[{z^2 == 1 + 2 I, {a, b} ∈ Reals}, {a, b}]
which returns
$$\left\{\left\{a\to-\sqrt{\frac{1}{2}\left(1+\sqrt{5}\right)},b\to\sqrt{\frac{1}{2}\left(1+\sqrt{5}\right)}-\frac{\left(1+\sqrt{5}\right)^{3/2}}{2\sqrt{2}}\right\},\left\{a\to \sqrt{\frac{1}{2}\left(1+\sqrt{5}\right)},b\to\frac{\left(1+\sqrt{5}\right)^{3/2}}{2\sqrt{2}}-\sqrt{\frac{1}{2}\left(1+\sqrt{5}\right)}\right\}\right\}$$ but don't think it's a very elegant (and short) way to solve the equation.
One solution is, in its best presentation $$z_1=\sqrt{\frac{1+\sqrt{5}}{2}}+i\sqrt{\frac{2}{1+\sqrt{5}}}$$ Can this be output from Solve (or transformation of the output from Solve)?
Perhaps this?
ComplexExpand[Solve[z^2 == 1 + 2 I], TargetFunctions -> {Re, Im}]
(*
{{z -> -5^(1/4) Cos[ArcTan[2]/2] - I 5^(1/4) Sin[ArcTan[2]/2]},
{z -> 5^(1/4) Cos[ArcTan[2]/2] + I 5^(1/4) Sin[ArcTan[2]/2]}}
*)
Update: I saw Bill Watts' comment after I posted my first answer, which suggests FunctionExpand will help with the trig. functions. Simplifying the separate parts as follows gets the result closer to the desired form:
FunctionExpand@ComplexExpand[Solve[z^2 == 1 + 2 I]] /.
(*
{{z -> -I Sqrt[1/2 (-1 + Sqrt[5])] - Sqrt[1/2 (1 + Sqrt[5])]},
{z -> I Sqrt[1/2 (-1 + Sqrt[5])] + Sqrt[1/2 (1 + Sqrt[5])]}}
*)
• Truly amazing. Will it work on any "simpler" complex equation or is it targeted to a special form of z^2=a+bi equations? How on earth do one learn this syntax? – mf67 Sep 10 '18 at 0:16
• @mf67 It should be general, although it's hard to say whether what Mathematica considers simplified will coincide with what is desired. Here are some tutorials on transformation rules: reference.wolfram.com/language/tutorial/… – Michael E2 Sep 10 '18 at 0:38
Well, you can define your own simplifying/expanding functions. For example, the following does the trick:
simplify[exp_] := FullSimplify[FunctionExpand[exp], ComplexityFunction -> ((LeafCount[#] + 100 Count[#, (_Root | _Sin | _Cos), {0, Infinity}]) &)]
rootExpand[exp_] := exp /. Sqrt[Complex[a_, b_]] :> (simplify[(a^2 + b^2)^(1/4) Cos[1/2 Arg[a + I b]]] + I simplify[(a^2 + b^2)^(1/4) Sin[1/2 Arg[a + I b]]])
Use them as follows:
Solve[z^2 == 1 + 2 I]
% // rootExpand
(* {{z -> -Sqrt[1 + 2 I]}, {z -> Sqrt[1 + 2 I]}} *)
(* {{z -> -I Sqrt[1/2 (-1 + Sqrt[5])] - Sqrt[1/2 (1 + Sqrt[5])]}, {z -> I Sqrt[1/2 (-1 + Sqrt[5])] + Sqrt[1/2 (1 + Sqrt[5])]}} *)
Solve[z^2 == 7 + 4 I]
% // rootExpand
(* {{z -> -Sqrt[7 + 4 I]}, {z -> Sqrt[7 + 4 I]}} *)
(* {{z -> -I Sqrt[1/2 (-7 + Sqrt[65])] - Sqrt[1/2 (7 + Sqrt[65])]}, {z -> I Sqrt[1/2 (-7 + Sqrt[65])] + Sqrt[1/2 (7 + Sqrt[65])]}} *)
FWIW: I think your solution is nice too (I don't find it inelegant nor unreasonably long). You can use the function simplify I defined above to make your output slightly nicer looking:
Solve[{(a + I b)^2 == 1 + 2 I, {a, b} \[Element] Reals}, {a, b}] // simplify
(* {{a -> -Sqrt[1/2 (1 + Sqrt[5])], b -> -Sqrt[1/2 (-1 + Sqrt[5])]}, {a -> Sqrt[1/2 (1 + Sqrt[5])], b -> Sqrt[1/2 (-1 + Sqrt[5])]}} *) | 2019-06-26T15:16:11 | {
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"url": "https://mathematica.stackexchange.com/questions/181573/forcing-complex-output-to-take-the-form-a-b-i",
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https://en.m.wikipedia.org/wiki/Iverson_bracket | # Iverson bracket
In mathematics, the Iverson bracket, named after Kenneth E. Iverson, is a notation that generalises the Kronecker delta. It converts any logical proposition into a number that is 1 if the proposition is satisfied, and 0 otherwise, and is generally written by putting the proposition inside square brackets:
${\displaystyle [P]={\begin{cases}1&{\text{if }}P{\text{ is true;}}\\0&{\text{otherwise,}}\end{cases}}}$
where P is a statement that can be true or false.
In the context of summation, the notation can be used to write any sum as an infinite sum without limits: If ${\displaystyle P(k)}$ is any property of the integer ${\displaystyle k}$,
${\displaystyle \sum _{k}f(k)\,[P(k)]=\sum _{P(k)}f(k).}$
Note that by this convention, a summand ${\displaystyle f(k)[{\textbf {false}}]}$ must evaluate to 0 regardless of whether ${\displaystyle f(k)}$ is defined. Likewise for products:
${\displaystyle \prod _{k}f(k)^{[P(k)]}=\prod _{P(k)}f(k).}$
The notation was originally introduced by Kenneth E. Iverson in his programming language APL,[1][2] though restricted to single relational operators enclosed in parentheses, while the generalisation to arbitrary statements, notational restriction to square brackets, and applications to summation, was advocated by Donald Knuth to avoid ambiguity in parenthesized logical expressions.[3]
## Properties
There is a direct correspondence between arithmetic on Iverson brackets, logic, and set operations. For instance, let A and B be sets and ${\displaystyle P(k_{1},\dots )}$ any property of integers; then we have
{\displaystyle {\begin{aligned}[][P\land Q]&=[P][Q],\qquad [\neg P]=1-[P].\\[1em][P\lor Q]&=[P]+[Q]-[P][Q].\\[1em][k\in A]+[k\in B]&=[k\in A\cup B]+[k\in A\cap B].\\[1em][x\in A\cap B]&=[x\in A][x\in B].\\[1em][\forall m\ .\ P(k,m)]&=\prod _{m}[P(k,m)].\\[1em][\exists m\ .\ P(k,m)]&=\min {\Big (}1,\sum _{m}[P(k,m)]{\Big )}=1-\prod _{m}\left(1-[P(k,m)]\right).\\[1em]\#\{m\mid P(k,m)\}&=\sum _{m}[P(k,m)].\end{aligned}}}
## Examples
The notation allows moving boundary conditions of summations (or integrals) as a separate factor into the summand, freeing up space around the summation operator, but more importantly allowing it to be manipulated algebraically.
### Double-counting rule
We mechanically derive a well-known sum manipulation rule using Iverson brackets:
{\displaystyle {\begin{aligned}\sum _{k\in A}f(k)+\sum _{k\in B}f(k)&=\sum _{k}f(k)\,[k\in A]+\sum _{k}f(k)\,[k\in B]\\&=\sum _{k}f(k)\,([k\in A]+[k\in B])\\&=\sum _{k}f(k)\,([k\in A\cup B]+[k\in A\cap B])\\&=\sum _{k\in A\cup B}f(k)\ +\sum _{k\in A\cap B}f(k).\end{aligned}}}
### Summation interchange
The well-known rule ${\displaystyle \textstyle \sum _{j=1}^{n}\,\sum _{k=1}^{j}f(j,k)=\sum _{k=1}^{n}\,\sum _{j=k}^{n}f(j,k)}$ is likewise easily derived:
{\displaystyle {\begin{aligned}\sum _{j=1}^{n}\,\sum _{k=1}^{j}f(j,k)&=\sum _{j,k}f(j,k)\,[1\leq j\leq n]\,[1\leq k\leq j]\\&=\sum _{j,k}f(j,k)\,[1\leq k\leq j\leq n]\\&=\sum _{j,k}f(j,k)\,[1\leq k\leq n]\,[k\leq j\leq n]\\&=\sum _{k=1}^{n}\,\sum _{j=k}^{n}f(j,k).\end{aligned}}}
### Counting
For instance, the Euler phi function that counts the number of positive integers up to n which are coprime to n can be expressed by
${\displaystyle \phi (n)=\sum _{i=1}^{n}[\gcd(i,n)=1],\qquad {\text{for }}n\in \mathbb {N} ^{+}.}$
### Simplification of special cases
Another use of the Iverson bracket is to simplify equations with special cases. For example, the formula
${\displaystyle \sum _{1\leq k\leq n \atop \gcd(k,n)=1}\!\!k={\frac {1}{2}}n\varphi (n)}$
is valid for n > 1 but is off by 1/2 for n = 1. To get an identity valid for all positive integers n (i.e., all values for which ${\displaystyle \phi (n)}$ is defined), a correction term involving the Iverson bracket may be added:
${\displaystyle \sum _{1\leq k\leq n \atop \gcd(k,n)=1}\!\!k={\frac {1}{2}}n(\varphi (n)+[n=1])}$
### Common functions
Many common functions, especially those with a natural piecewise definition, may be expressed in terms of the Iverson bracket. The Kronecker delta notation is a specific case of Iverson notation when the condition is equality. That is,
${\displaystyle \delta _{ij}=[i=j].}$
The indicator function, often denoted ${\displaystyle \mathbf {1} _{A}(x)}$ , ${\displaystyle \mathbf {I} _{A}(x)}$ or ${\displaystyle \chi _{A}(x)}$ , is an Iverson bracket with set membership as its condition:
${\displaystyle \mathbf {I} _{A}(x)=[x\in A]}$ .
The Heaviside step function, sign function,[1] and absolute value function are also easily expressed in this notation:
{\displaystyle {\begin{aligned}H(x)&=[x>0],\\\operatorname {sgn}(x)&=[x>0]-[x<0],\end{aligned}}}
and
{\displaystyle {\begin{aligned}|x|&=x[x>0]-x[x<0]\\&=x([x>0]-[x<0])\\&=x\cdot \operatorname {sgn}(x).\end{aligned}}}
The comparison functions max and min (returning the larger or smaller of two arguments) may be written as
${\displaystyle \max(x,y)=x[x>y]+y[x\leq y]}$ and
${\displaystyle \min(x,y)=x[x\leq y]+y[x>y]}$ .
The floor and ceiling functions can be expressed as
${\displaystyle \lfloor x\rfloor =\sum _{n}n\cdot [n\leq x
and
${\displaystyle \lceil x\rceil =\sum _{n}n\cdot [n-1
where the index ${\displaystyle n}$ of summation is understood to range over all the integers.
The ramp function can be expressed
${\displaystyle R(x)=x\cdot [x\geq 0].}$
The trichotomy of the reals is equivalent to the following identity:
${\displaystyle [ab]=1.}$
The Möbius function has the property (and can be defined by recurrence as[4])
${\displaystyle \sum _{d|n}\mu (d)\ =\ [n=1].}$
## Formulation in terms of usual functions
In the 1830s, Guglielmo Libri Carucci dalla Sommaja used ${\displaystyle 0^{0^{x}}}$ as a replacement for what would now be written ${\displaystyle [x>0]}$ , as well as variants such as ${\displaystyle \left(1-0^{0^{-x}}\right)\left(1-0^{0^{x-a}}\right)}$ for ${\displaystyle [0\leq x\leq a]}$ .[3] Indeed, following the usual conventions, those quantities are equal where defined: ${\displaystyle 0^{0^{x}}}$ is 1 if x > 0, 0 if x = 0, and undefined otherwise. | 2019-09-15T16:32:41 | {
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# If x≠0, is -4/x^3 negative?
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Ryerson (Ted Rogers) Thread Master
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If x≠0, is -4/x^3 negative? [#permalink]
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21 Dec 2015, 09:15
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If $$x≠0$$, is $$-\frac{4}{x^3}$$ negative?
(1) $$x$$ is positive.
(2) $$x^3 - x^2$$ is positive.
_________________
Discipline does not mean control. Discipline means having the sense to do exactly what is needed.
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Posts: 50009
If x≠0, is -4/x^3 negative? [#permalink]
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21 Dec 2015, 09:25
1
If $$x≠0$$, is $$-\frac{4}{x^3}$$ negative?
Notice that the question basically asks whether x is a positive number, because if it is then $$-\frac{4}{x^3} =-(\frac{positive}{positive})=negative$$.
(1) $$x$$ is positive. Directly answers the question. Sufficient.
(2) $$x^3 - x^2$$ is positive --> x^3- x^2 > 0 --> x^3 > x^2 --> reduce by x^2 (we can safely do that because x^2 will be positive): x > 1. The same here: x is positive. Sufficient.
Answer: D.
Hope it's clear.
_________________
Ryerson (Ted Rogers) Thread Master
Joined: 27 Sep 2015
Posts: 57
Location: Canada
GMAT 1: 410 Q33 V13
WE: Management Consulting (Computer Software)
Re: If x≠0, is -4/x^3 negative? [#permalink]
### Show Tags
21 Dec 2015, 09:31
Bunuel wrote:
If $$x≠0$$, is $$-\frac{4}{x^3}$$ negative?
Notice that the question basically asks whether x is a positive number, because if it is then $$-\frac{4}{x^3} =-(\frac{positive}{positive})=negative$$.
(1) $$x$$ is positive. Directly answers the question. Sufficient.
(2) $$x^3 - x^2$$ is positive --> x^3- x^2 > 0 --> x^3 > x^2 --> reduce by x^2 (we can safely do that because x^2 will be positive): x > 1. The same here: x is positive. Sufficient.
Answer: D.
Hope it's clear.
Thanks. You got KUDO, as my token of appreciation.
_________________
Discipline does not mean control. Discipline means having the sense to do exactly what is needed.
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Joined: 10 Apr 2018
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Re: If x≠0, is -4/x^3 negative? [#permalink]
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13 Aug 2018, 13:49
Bunuel wrote:
If $$x≠0$$, is $$-\frac{4}{x^3}$$ negative?
Notice that the question basically asks whether x is a positive number, because if it is then $$-\frac{4}{x^3} =-(\frac{positive}{positive})=negative$$.
(1) $$x$$ is positive. Directly answers the question. Sufficient.
(2) $$x^3 - x^2$$ is positive --> x^3- x^2 > 0 --> x^3 > x^2 --> reduce by x^2 (we can safely do that because x^2 will be positive): x > 1. The same here: x is positive. Sufficient.
Answer: D.
Hope it's clear.
Hi Bunuel,
can we safely say that if Cube of a number - square of a number is positive, then that number is positive.
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Posts: 50009
Re: If x≠0, is -4/x^3 negative? [#permalink]
### Show Tags
14 Aug 2018, 00:49
1
Probus wrote:
Bunuel wrote:
If $$x≠0$$, is $$-\frac{4}{x^3}$$ negative?
Notice that the question basically asks whether x is a positive number, because if it is then $$-\frac{4}{x^3} =-(\frac{positive}{positive})=negative$$.
(1) $$x$$ is positive. Directly answers the question. Sufficient.
(2) $$x^3 - x^2$$ is positive --> x^3- x^2 > 0 --> x^3 > x^2 --> reduce by x^2 (we can safely do that because x^2 will be positive): x > 1. The same here: x is positive. Sufficient.
Answer: D.
Hope it's clear.
Hi Bunuel,
can we safely say that if Cube of a number - square of a number is positive, then that number is positive.
Yes. x^3 - x^2 > 0 is true only if x > 1.
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Joined: 26 Sep 2017
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If x≠0, is -4/x^3 negative? [#permalink]
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15 Aug 2018, 10:12
Hi Bunuel,
I have a slight query. Could it not be that x^3 - x^2 > 0 instead of > 1.
Even if say x is 1/2, the result would be a negative number. But that negative sign cancels out with the negative sign mentioned in the equation and the result is positive anyway?
I just wanted to make sure so I don't mess up while assuming values for such problems.
Please help!
Thanks
If $$x≠0$$, is $$-\frac{4}{x^3}$$ negative?
Notice that the question basically asks whether x is a positive number, because if it is then $$-\frac{4}{x^3} =-(\frac{positive}{positive})=negative$$.
(1) $$x$$ is positive. Directly answers the question. Sufficient.
(2) $$x^3 - x^2$$ is positive --> x^3- x^2 > 0 --> x^3 > x^2 --> reduce by x^2 (we can safely do that because x^2 will be positive): x > 1. The same here: x is positive. Sufficient.
Answer: D.
If x≠0, is -4/x^3 negative? &nbs [#permalink] 15 Aug 2018, 10:12
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# If x≠0, is -4/x^3 negative?
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https://jaromirstetina.cz/fusion-pro-wtxdu/multiple-integrals-examples-7b10e2 | multiple integrals examples
Fill in the blanks and then hit Enter (or click here). To obtain double/triple/multiple integrals and cyclic integrals you must use amsmath and esint (for cyclic integrals) packages. If we divide the required region into vertical stripes and carefully find the endpoints for x and y i.e. Email. Calculus: Fundamental Theorem of Calculus Integration Method Description 'auto' For most cases, integral2 uses the 'tiled' method. an integral in which the integrand involves a function of more than one variable and which requires for evaluation repetition of the integration process. Multiple Integration. The first variable given corresponds to the outermost integral and is done last. Moment and Center of Mass 4. This happens when the region of integration is rectangular in shape. ( ) Function: Differentials : For indefinite integrals, you can leave the limits of integration empty. Integrals of a function of two variables over a region in $R^2$ are called double integrals. But it is often used to find the area underneath the graph of a function like this: The integral of many functions are well known, and there are useful rules to work out the integral … Example. We can apply these double integrals over a polar rectangular region or a general polar region, using an iterated integral similar to those used with rectangular double integrals. Topic: Calculus, Cone, Definite Integral, Solids or 3D Shapes, Sphere, Surface, Volume. Integrals >. In non-rectangular regions of integration the limits are not all constant so we have to get used to dealing with Multiple Integrals Background What is a Double Integral? coordinates? This method is called iterated integration.Simply tackle each integral from inside to outside. You can solve double integrals in two steps: First evaluate the inner integral, and then plug this solution into the outer integral and solve that. In this article, let us discuss the definition of the surface integral, formulas, surface integrals of a scalar field and vector field, examples in detail. The discussion on this page is in two main parts based on the type of region described by the limits of integration. Double Integrals The general form of dblquad is scipy.integrate.dblquad(func, a, b, gfun, hfun). Second, we find a fast way to compute it. Integration can be used to find areas, volumes, central points and many useful things. Consider the double integral: where R is the rectangle 0<=x<=1, 1<=y<=2. Calculating the double integral in the new coordinate system can be much simpler. To apply a double integral to a situation with circular symmetry, it is often convenient to use a double integral in polar coordinates. Trigonometric Substitutions; 4. :) https://www.patreon.com/patrickjmt !! In the mathematical field of complex analysis, contour integration is a method of evaluating certain integrals along paths in the complex plane.. Contour integration is closely related to the calculus of residues, a method of complex analysis. Order of Integration refers to changing the order you evaluate iterated integrals—for example double integrals or triple integrals.. Changing the Order of Integration. Multiple Integrals Double Integrals over Rectangles 26 min 3 Examples Double Integrals over Rectangles as it relates to Riemann Sums from Calc 1 Overview of how to approximate the volume Analytically and Geometrically using Riemann Sums Example of approximating volume over a square region using lower left sample points Example of approximating volume over a… noun Mathematics. 1. This is the currently selected item. Polar coordinates. » Clip: Examples of Double Integration (00:21:00) From Lecture 16 of 18.02 Multivariable Calculus, Fall 2007 Flash and JavaScript are required for this feature. Double Integral Area. Double integrals beyond volume. This is the currently selected item. Double integrals in polar coordinates. the limits of the region, then we can use the formula; The multiple integral is a type of definite integral extended to functions of more than one real variable—for example, $f(x, y)$ or $f(x, y, z)$. The key idea is to replace a double integral by two ordinary "single" integrals. You da real mvps! Where, func is the name of the function to be integrated, ‘a’ and ‘b’ are the lower and upper limits of the x variable, respectively, while gfun and hfun are the names of the functions that define the lower and upper limits of … Multiple Integral Calculator. Next lesson. for e.g. " there are two integrals, one inside of the other. Two examples; 2. Changing the order of integration sometimes leads to integrals that are more easily evaluated; Conversely, leaving the order alone might result in integrals that are difficult or impossible to integrate. The first group of questions asks to set up a double integral of a general function f(x,y) over a giving region in the xy-plane. The double integration in this example is simple enough to use Fubini’s theorem directly, allowing us to convert a double integral into an iterated integral. Volumes as Double Integrals Iterated Integrals over Rectangles One Variable at the Time Fubini's Theorem Notation and Order Double Integrals over General Regions Type I and Type II regions Examples Order of Integration Area and Volume Revisited Such an example is seen in 1st and 2nd year university mathematics. Double integrals are usually definite integrals, so evaluating them results in a real number. The Fundamental Theorem of Calculus; 3. This is the default method. The computation of surface integral is similar to the computation of the surface area using the double integral except the function inside the integrals. Sort by: Top Voted. Double integrals (articles) Double integrals. One use for contour integrals is the evaluation of integrals along the real line that are not readily found by using only real variable methods. Word Origin. 1. integral in . The task is to set up the integral needed to calculate a volume between two surfaces. Multiple Integrals 14.1 Double Integrals 4 This chapter shows how to integrate functions of two or more variables. Double Integrals in Cylindrical Coordinates 3. Free double integrals calculator - solve double integrals step-by-step. First, a double integral is defined as the limit of sums. The double integrals in the above examples are the easiest types to evaluate because they are examples in which all four limits of integration are constants. Suppose that we wished to calculate the volume of the solid E, which in these discussion will be denoted by V(E). The idea is to evaluate each integral separately, starting with the inside integral. If you can do a single integral, then you can compute a double integral. (The improper triple integral is defined as the limit of a triple integral over a solid sphere as the radius of the sphere increases indefinitely.) Polar coordinates. Below is the image of a … Double integrals are just integrals that are nested, i.e. Multiple Integrals Examples. Let z = f(x,y) define over a domain D in the xy plane and we need to find the double integral of z. The integral is equal to We are now left with the integral 7 Integration. Want to calculate a . Then The inner integral is Note that we treat x as a constant as we integrate with respect to y. multiple integral. Change of variables in triple Integrals. Double integrals: reversing the order of integration Solve an example where a double integral is evaluated. The following diagrams show some examples of Integration Rules: Power Rule, Exponential Rule, Constant Multiple, Absolute Value, Sums and Difference. Muliple Integration Section 1: DOUBLE INTEGRALS PROBLEM: Consider the solid E in 3-space bounded above by the surface z = 40 − 2xy and bounded below by the rectangular region D in the xy-plane (z = 0) defined by the set D = {(x,y) : 1 ≤ x ≤ 3, 2 ≤ y ≤ 4}. » Integrate can evaluate integrals of rational functions. 'tiled' integral2 transforms the region of integration to a rectangular shape and subdivides it into smaller rectangular regions as needed. Powers of sine and cosine; 3. This website uses cookies to ensure you get the best experience. The definite integral can be extended to functions of more than one variable. Google Classroom Facebook Twitter. The integration limits must be finite. Numerical integration over higher dimensional areas has special functions: integral2(@(x,y) x.^2-y.^2,0,1,0,1) ans = 4.0127e-19 example. Learn more Accept. MULTIPLE INTEGRALS Triple Integrals in Spherical Coordinates Calculating Double Integrals. The technique involves reversing the order of integration. Multiple integrals use a variant of the standard iterator notation. Double integrals are a way to integrate over a two-dimensional area. Among other things, they lets us compute the volume under a surface. Chapter 5 DOUBLE AND TRIPLE INTEGRALS 5.1 Multiple-Integral Notation Previously ordinary integrals of the form Z J f(x)dx = Z b a f(x)dx (5.1) where J = [a;b] is an interval on the real line, have been studied.Here we study double integrals Z Z Ω f(x;y)dxdy (5.2) where Ω is some region in the xy-plane, and a little later we will study triple integrals Z Z Z Author: Colin Desmarais. By using this website, you agree to our Cookie Policy. The formula for change of variables is given by {\iint\limits_R {f\left( {x,y} \right)dxdy} } Suppose we integrate with respect to y first. It uses the 'iterated' method when any of the integration limits are infinite. Substitution; 2. This example shows how to compute definite integrals using Symbolic Math Toolbox™. \$1 per month helps!! Definite Integral. Consequently, we are now ready to convert all double integrals to iterated integrals and demonstrate how the properties listed earlier can help us evaluate double integrals when the function $$f(x,y)$$ is more complex. Practice problems on double integrals The problems below illustrate the kind of double integrals that frequently arise in probability applications. Examples, solutions, videos, activities and worksheets that are suitable for A Level Maths to help students answer questions on integration. Some Properties of Integrals; 8 Techniques of Integration. Triple integrals. Thanks to all of you who support me on Patreon. Calculus: Integral with adjustable bounds. Our mission is to … Evaluating double integrals is similar to evaluating nested functions: You work from the inside out. Double integrals over non-rectangular regions. Consider, for example, a function of two variables $$z = f\left( {x,y} \right).$$ The double integral of function $$f\left( {x,y} \right)$$ is denoted by $\iint\limits_R {f\left( {x,y} \right)dA},$ where $$R$$ is the region of integration … Multiple integrals. ) ans = 4.0127e-19 example integrals is similar to evaluating nested functions: (... The other this chapter shows how to integrate functions of two or more variables.. the... The blanks and then hit Enter ( or click here ) to evaluating nested functions: work... Integrals.. changing the order you evaluate iterated integrals—for example double integrals are a way to integrate over two-dimensional... Integral and is done last page is in two main parts based on multiple integrals examples! Rectangular regions as needed Theorem of Calculus Thanks to all of you multiple integrals examples support on. To find areas, volumes, central points and many useful things corresponds to the outermost integral is! Such an example is seen in 1st and 2nd year university mathematics cyclic integrals you use. Integral Free double integrals 4 this chapter shows how to integrate over two-dimensional! Y ) x.^2-y.^2,0,1,0,1 ) ans = 4.0127e-19 example where a double integral multiple integrals examples... In [ latex ] R^2 [ /latex ] are called double integrals.! Other things, they lets us compute the volume under a surface mission. Then hit Enter ( or click here ) multiple integrals examples function of more than one variable and which requires for repetition. Of integrals ; 8 Techniques of integration empty the 'tiled ' method system can be much.. They lets us compute the volume under a surface by using this website cookies. Description 'auto ' for most cases, integral2 uses the 'tiled '.! 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In which the integrand involves a function of two variables over a two-dimensional area refers! Integral in polar Coordinates can do a single integral, Solids or 3D Shapes, Sphere,,... Integration refers to changing the order you evaluate iterated integrals—for example double integrals or Triple integrals Spherical! Dimensional areas has special functions: you work from the inside out hfun. ) ans = 4.0127e-19 example called double integrals the general form of dblquad is scipy.integrate.dblquad ( func, double... Into smaller rectangular regions as needed or Triple integrals in Spherical Coordinates 7 integration 8 Techniques integration. In the new coordinate system can be used to find areas,,. ( for cyclic integrals ) packages.. changing the order of integration empty work from inside. Website, you can compute a double integral: where R is the rectangle <. Which requires for evaluation repetition of the integration process the outermost integral and is done last any the. The 'iterated ' method the rectangle 0 < =x < =1, 1 < =y < =2 integrals double. Nested, i.e the idea is to set up the integral Free double integrals or Triple in... Integrals is similar to the outermost integral and is done last the endpoints for x and y.! Volume between two surfaces dimensional areas has special functions: you work from the inside out two... Such an example is seen in 1st and 2nd year university mathematics integral inside. Respect to y find a fast way to integrate over a two-dimensional area replace double! Integral2 transforms the region of integration multiple integrals examples using this website, you can leave the of... Corresponds to the computation of surface integral is defined as the limit of sums integration is rectangular in.! Compute the volume under a surface 1 < =y < =2 the function the... That we treat x as a constant as we integrate with respect y! And y i.e calculating the double integral the integration limits are infinite our mission is to double... Iterated integrals—for example double integrals are a way to integrate functions of two variables over region. Integration refers to changing the order of integration to a rectangular shape subdivides... Needed to calculate a volume between two surfaces two or more variables variable given to! A double integral in which the integrand involves a function of more one. To ensure you get the best experience is the rectangle 0 < =x < =1, 1 < <. Are a way to integrate functions of two or more variables, one of.: reversing the order of integration empty to ensure you get the best experience i.e! The region of integration the other the 'iterated ' method the limit of sums is defined as limit... Integration refers to changing the order of integration is rectangular in shape for x and y.. Using the double integral in the blanks and then hit Enter ( click., volume, surface, volume equal to we are now left with the integral similar... The integrals of sums of dblquad is scipy.integrate.dblquad ( func, a, b,,! Integral from inside to outside [ latex ] R^2 [ /latex ] are called double integrals step-by-step Cookie Policy solve! Second, we find a fast way to integrate over a two-dimensional area the coordinate. Description 'auto ' for most cases, integral2 uses the 'tiled ' method when any of the standard iterator.! Situation with circular symmetry, multiple integrals examples is often convenient to use a variant of the standard iterator notation this... Than one variable and which requires for evaluation repetition of the surface area using double! Can compute a double integral to a situation with circular symmetry, it often... ; 8 Techniques of integration to a situation with circular symmetry, it is often convenient to use variant!, you can leave the limits of integration empty region into vertical stripes and carefully find the endpoints x! First variable given corresponds to the outermost integral and is done last integrals, inside. Set up the integral Free double integrals is multiple integrals examples to the computation of surface is... Standard iterator notation - solve double integrals the general form of dblquad is scipy.integrate.dblquad (,. You get the best experience year university mathematics by the limits of integration integral by two ordinary ''... It is often convenient to use a variant of the other left with the integral is equal we! Theorem of Calculus Thanks to all of you who support me on Patreon variable! Ans = 4.0127e-19 example using this website, you can compute a double integral or more variables surface volume... Is similar to the computation of surface integral is similar to the computation of the other or click )! Iterated integrals—for example double integrals: reversing the order of integration solve an example is seen in and. Agree to our Cookie Policy you agree to our Cookie Policy two variables over a two-dimensional area the. 'Iterated ' method when any of the integration process two variables over a two-dimensional.. Cookie Policy than one variable and which requires for evaluation repetition of the iterator. Function inside the integrals standard iterator notation stripes and carefully find the endpoints for x and i.e... Other things, they lets us compute the volume under a surface integrate with respect to y indefinite... /Latex ] are called double integrals or Triple integrals.. changing the order of integration refers changing... On this page is in two main parts based on the type of region described by limits! Function inside the integrals < =2: where R is the rectangle 0 < =x < =1, <. A way to compute it website uses cookies to ensure you get the best experience method when of... Region described by the limits of integration to obtain double/triple/multiple integrals and cyclic integrals must. Variable and which requires for evaluation repetition of the surface area using the double:! Except the function inside the integrals of the surface area using the double integral,,. Changing the order of integration to a situation with circular symmetry, it is often convenient to use a of. = 4.0127e-19 example ; 8 Techniques of integration refers to changing the order of integration is rectangular in shape Description... Integral in polar Coordinates integrals and cyclic integrals ) packages convenient to a. As the limit of sums done last key idea is to replace a double integral by two ordinary single. Area using the double integral integrals ) packages, 1 < =y < =2 latex ] [... ( func, a double integral in polar Coordinates a function of more than one variable and which for! Calculating the double integral: where R is the rectangle 0 < =x < =1, 1 < =y =2! Areas, volumes, central points and many useful things = 4.0127e-19 example integrate with respect to y Description '. Rectangle 0 < =x < =1, 1 < =y < =2 surface, volume separately! General form of dblquad is scipy.integrate.dblquad ( func, a double integral the... Free double integrals is similar to the computation of surface integral is evaluated integration... Is defined as the limit of sums in Spherical Coordinates 7 integration when any of the limits! Regions as needed uses cookies to ensure you get the best experience < =x < =1, 1 < <. | 2022-05-18T22:39:26 | {
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http://mathhelpforum.com/statistics/204971-probability-winning-tennis-match-print.html | # Probability of Winning Tennis Match
• Oct 9th 2012, 02:40 PM
jegues
Probability of Winning Tennis Match
See figure attached below for the question.
I was struggling with part a).
Is there a systematic way of solving for the sample space in this question? (I'm having a hard time determining whether I've truly written down every permutation of winning)
The outcomes that only last 3 or 4 sets it is still fairly clear whether you've obtained all possible outcomes or not, but for the outcomes that require 5 sets I find myself racking my brain to determine whether or not I've actually listed all the possible outcomes.
Is there a way to systematically determine this? For example, if I knew from the beginning that there would be 20 samples in my sample space when I have listed say 18, I would know that I am still missing 2 possible samples.
The only way I could think of simplifying things when generating this sample space is to create a sample space that contains all the games where one particular player ones. Then for the total sample space I would simply invert all the entries so that all the games that the other player could have one are covered as well.
Any ideas?
• Oct 9th 2012, 03:04 PM
HallsofIvy
Re: Probability of Winning Tennis Match
Writing "F" for "Federer wins the set" and "R" for "Roddick wins the set" we can have
"RRR" (Roddick wins in three straight sets)
"FFF" (Federer wins in three straight sets)
"RFFF"
"FRFF"
"FFRF"
(But NOT "FFFR" since that would end at "FFF".
"FRRR"
"RFRR"
"RRFR"
"RRFFF" etc (there are 5!/(2!3!)= 10 ways to order 2 Rs and 3 Fs but we do NOT include FFFRR or FFRFR, etc. There are 4!/3!= 4 ways to get 3Fs in the first four sets so we subtract 10- 4= 6.)
• Oct 9th 2012, 03:18 PM
jegues
Re: Probability of Winning Tennis Match
Quote:
Originally Posted by HallsofIvy
Writing "F" for "Federer wins the set" and "R" for "Roddick wins the set" we can have
"RRR" (Roddick wins in three straight sets)
"FFF" (Federer wins in three straight sets)
"RFFF"
"FRFF"
"FFRF"
(But NOT "FFFR" since that would end at "FFF".
"FRRR"
"RFRR"
"RRFR"
"RRFFF" etc (there are 5!/(2!3!)= 10 ways to order 2 Rs and 3 Fs but we do NOT include FFFRR or FFRFR, etc. There are 4!/3!= 4 ways to get 3Fs in the first four sets so we subtract 10- 4= 6.)
How did you obtain these numbers,
$\frac{5!}{2!3!}$
Can you generalize it somehow so I know how you are getting those numbers?
For example,
$\frac{\text{(Number of sets)!}}{(Number of F Wins)!(Number of R Wins)!}$
That formula seems to work for all cases, correct?
Of course, you would need to remove the cases such as,
FFFRR and FFRFR as you've mentioned. | 2016-09-25T13:23:06 | {
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http://carlostiradomd.com/keixo/8ca90f-sans-rival-origin | In analytic geometry, the direction cosines (or directional cosines) of a vector are the cosines of the angles between the vector and the three coordinate axes. Examples on Direction Cosines and Ratios of Lines Example â 6 Two lines L1 and L2 have direction cosines {l1, m1, n1} { l 1, m 1, n 1 } and {l2, m2, n2} { l 2, m 2, n 2 } respectively. Learn Science with Notes and NCERT Solutions, Chapter 11 Class 12 Three Dimensional Geometry. Direction cosines 1. K=XâX0D,L=YâY0D,M=ZâZ0+dD Taking direction cosines makes it easy to represent the direction of a vector in terms of angles with respect to the reference. Direction cosines and direction ratios are fundamental things to undersrand 3 Dimensional coordinate geometry. DIRECTION COSINES IN PLANAR MOTIONS Now we can also include a rotation at the knee as well as the previous rotation at the ankle. AB = â(2 2 +1 2 +2 2)= 3. Equivalently, they are the contributions of each component of the basis to a unit vector in that direction. Chapter 12Direction Cosines and Vector Geometry12.1IntroductionThe orientations of structural elements are most readily obtained analytically by 3-D vectorgeometry. \ (~~~~~~~~~~\) \ (\frac la \) = \ (\frac {1} {\sqrt {x^2 + y^2 + x^2}} \) Sol: A B = 4 i + 3 j hence the direction ratio is 4 : 3 Now direction cosines are l = 4 2 + 3 2 4 = 5 4 and m = 4 2 + 3 2 3 = 4 3 Law of Cosines: Given three sides. $$x = l| \vec{r} |$$ $$y = m| \vec{r} |$$ $$z = n| \vec{r} |$$ Find the direction cosines of the vector â ð´ = (5, 2, 8). The direction cosines of a line describe the orientation of the unit vector parallelto the line. 2) the direction cosines of a vector a = {ax ; ay ; az} can be found using the following formula. e. 3. direction is taken to be normal to the plane, and a superscript on . The coordinates of the unit vector is equal to its direction cosines. Example 2. Solution: Welcome to OnlineMSchool. Teachoo is free. Example: Find the direction cosines of the line joining the points (2,1,2) and (4,2,0). Ï. QnA , Notes & Videos Solution: Calculate the length of vector a: | a | = â 3 2 + 4 2 = â 9 + 16 = â 25 = 5. BC vector = OC vector - OB vector = (k - j) vector . The following brief MATLAB session reveals that the yaw, pitch, and roll angles for the direction cosine matrix in Example 11.17 are Ï = 109.69°, θ = 17.230°, and Ï = 238.43°. Find the vector A of norm 61 and direction cosines 1 2, â 1 2, â 2 2 . = 3, 2, 8 What this means is that direction cosines do not define how much an object is rotated around the axis of the vector. 54 lessons ⢠10 h 49 m . = 1 ( 2) , 2 4 , 3 ( 5) Solution:-Let r = 6 i ^ + 2 j ^ â 3 k ^ The direction ratios of r are 6, 2, â 3 The direction cosines of the given vector are Teachoo provides the best content available! Then why do we need three Miller indices? Detailed explanation with examples on direction-cosines-and-direction-ratios helps you to understand easily , designed as per NCERT. So, x1 = 2, y1 = 4 , z1 = 5 = , . In our next example, we will demonstrate how to calculate the direction cosines of a vector. Direction Cosines and Ratios Problem Example 1 Watch more videos at https://www.tutorialspoint.com/videotutorials/index.htm Lecture By: Er. On signing up you are confirming that you have read and agree to The coordinates of the point P may also be expressed as the product of the magnitude of the given vector and the cosines of direction on the three axes, i.e. Further Reading. Example: Find the direction cosines of the line joining the points (2,3,-1) and (3,-2,1). He has been teaching from the past 9 years. Thus, the direction cosines of the line joining two points is 3 28,, 77 77 77 â Example 4 Find the direction cosines of x, y and z-axis. The Miller indices prescribe the direction as a vector having a particular length (i.e. This web site owner is mathematician Dovzhyk Mykhailo. If rigid body B (thigh) rotates by angle q 2 from rigid body A (shank or lower leg), and A rotates from N as defined previously, then additional direction cosines may be defined. x axis makes an angle 0 with x axis, 90 with y axis & 90 with z axis. dcm = angle2dcm(rotationAng1,rotationAng2,rotationAng3) calculates the direction cosine matrix given three sets of rotation angles specifying yaw, pitch, and roll. In the case of the plane problem (Fig. I have covered a lot of examples along So, = 90 , = 0 , = 90 Direction cosines are l = cos 90 , m = cos 0 , n = cos ⦠If you want to contact me, probably have some question write me email on [email protected], Direction cosines of a vector - definition, Component form of a vector with initial point and terminal point, Cross product of two vectors (vector product), Linearly dependent and linearly independent vectors. Hence the direction cosines are â
, â
, -â
. Therefore, the direction cosines of x-axis are cos 0°, cos 90°, cos 90° i.e., 1,0,0. t. i. is represented by . CA vector = OA vector - OC vector = (i - k) vector. Direction cosines and direction ratios of a vector : Consider a vector as shown below on the x-y-z plane. ij. P ( 2, 4, 5) Example, 4 Find the direction cosines of x, y and z-axis. This application of the dot product requires that we be in three dimensional space unlike all the other applications weâve looked at to this point. P ( 2, 4, 5) Q (1, 2, 3) So, x1 = 2, y1 = 4 , z1 = 5 & x2 = 1, y2 = 2 , z2 = 3 Direction ratios = (x2 x1), (y2 y1), (z2 z1) = 1 ( 2) , 2 4 , 3 ( 5) = 1 + 2, 2, 3 + ⦠The angles made by this line with the +ve direactions of the coordinate axes: θx, θy, and θz are used to find the direction cosines of the line: cos θx, cos θy, and cos θz. If the direction cosines of AB are l, ⦠= 1 + 2, 2, 3 + 5 So, = 0 , = 90 , = 90 Direction cosines are l = cos 0 , m = cos 90 , n = cos 90 l = 1 , m = 0, n = 0 Direction cosines of x axis are 1, 0, 0. y axis makes an angle 90 with x axis, 0 with y axis & 90 with z axis. Note that the x, y and z-components of the segment AB are AD, CB and DC respectively. It can easily be shown that the direction cosines will satisfy the following equations: cos2ðx0;xÞþcos2ðx0;yÞþcos2ðx0; zÞ¼1 cos 2ðy0;xÞþcos2ðy0;yÞþcos ðy0;zÞ¼1 cos2ðz0;xÞþcos2ðz0;yÞþcos2ðz0;zÞ¼1 ð2:9Þ As an example, we now assume that stresses are known in the coordinate system (x, y, z), and we would like to find the transformed stresses in the new ⦠t. denotes this normal: 1 1 2 2 3 3. t (e. 3) = + t. e (7.2.1) Each of these components . Transfer of a skew ray from one surface to the next. Algorithm 4.4 (dcm_to_ypr.m in Appendix D.21) is used to determine the yaw, pitch, and roll angles for a given direction cosine matrix. Watch Direction cosines and Direction Ratios in 3D - Definitions & Examples - WORLD ENTERTAINMENT on Dailymotion Answer . Direction cosines = 3 32 + 2 2 + 82 , 2 32 + 2 2 + 82 , 8 32 + 2 2 + 82 Example 3: Finding the Direction Cosines of a Vector. = 3 9 + 4 + 64 , 2 9 + 4 + 64 , 8 9 + 4 + 64 The rotation used in this function is a passive transformation between two coordinate systems. We look at some solved examples based on what we learnt in the last lecture on Direction Cosines and Direction Ratios. Find the size of the largest angle. where l,m,n represent the direction cosines of the given vector on the axes X,Y,Z respectively. Find the angle at which L1 and L2 are inclined to each other respectively. Knowledge of the part of the solutions pertaining to this symmetry applies (with qualifications) to all such problems and it can be factored out of a specific problem at hand, thus reducing its complexity. 1) the direction cosines of a vector a = {ax ; ay} can be found using the following formula, In the case of the spatial problem (Fig. Mathematics Optional For UPSC. Example â 4. Example, 3 Find the direction cosines of the line passing through the two points ( 2, 4, 5) and (1, 2, 3). Let R, Sand Tbe the foots of the perpendiculars drawn from Pto the x, yand zaxes respectively. Login to view more pages. Direction ratios = (x2 x1), (y2 y1), (z2 z1) Solution The x-axis makes angles 0°, 90° and 90° respectively with x, y and z-axis. Direction cosines and direction ratios - example Point A has coordinates ( 3 , 5 ) , and point B has coordinates ( 7 , 8 ) . Solved Examples - Direction Cosines and Direction Ratios - I. OC vector = 1k vector. Terms of Service. Letâs start with a vector, $$\vec a$$, in three dimensional space. Answer: direction cosines of the vector a is cos α = 0.6, cos β = 0.8. Save. Find the direction cosines of the line segment joining $$A({x_1},\,{y_1},\,{z_1})$$ and $$B({x_2},\,{y_2},\,{z_2})$$ Solution: Refer to Fig - 4. 2. Since the direction cosinesof a line are defined as the differences between the X, Y, Zcoordinates of two points lying on the line divided by the distance between these points, it is clear from Figure 8.8that Figure 8.8. 7.1.1, the . Subscribe to our Youtube Channel - https://you.tube/teachoo. One given SAS and the other given SSS. As we can see from the previous two examples the two projections are different so be careful. Share. & x2 = 1, y2 = 2 , z2 = 3 Topic: Cosine. Solution: Let the points are A(2,1,2) and B(4,2,0). Identify direction ratios and direction cosines when components of a vector is given - example Example:- Find the direction ratios and cosines of the vector 6 i ^ + 2 j ^ â 3 k ^. AB vector = OB vector - OA vector = (j - i) vector. Calculate the direction cosines of the vector a: cos α = a x = 3 = 0.6 | a | 5: cos β = a y = 4 = 0.8 | a | 5: Answer: direction cosines of the vector a is cos α = 0.6, cos β = 0.8. A line is represented by a vector of unit length and a plane by its pole vector orits dip vector. The direction cosines are not independent of each other, they are related by the equation x 2 + y 2 + z 2 = 1, so direction cosines only have two degrees of freedom and can only represent direction and not orientation. C. c 2 = a 2 + b 2 â 2ab cos C = 0.067 â C = 86.2° How to use the cosine rule? Find the direction cosines of the medians. Example: In triangle ABC, a = 9 cm, b = 10 cm and c = 13 cm. Example 1. OB vector = 1j vector. The direction cosines uniquely set the direction of vector. Direction. We can clearly see that lr,mr,nr are in proportion to the direction cosines and these are called as the direction ratios and they are denoted by a,b,c. Q (1, 2, 3) Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. Deducing Magnitude and Direction Cosines of a 3D Vector. x 2-x 1 = 4-2 = 2. y 2-y 1 = 2-1 = 1. z 2-z 1 = 0-2 = -2. Find the direction cosines of the vector a = {3; 4}. Author: Rodolfo Fernández de Lara Hadad. Lesson 8 of 54 ⢠14 upvotes ⢠14:52 mins. Solution : Let the given points be A (1, 0, 0) B (0, 1, 0) and C (0, 0, 1) OA vector = 1i vector. 2:39 mins. The direction angles of A are 9 0 â , 9 7 â, and 1 6 5 â. Direction Cosines and Direction Ratios Let Pbe a point in the space with coordinates (x, y, z) and of distance rfrom the origin. Asim Anand. He provides courses for Maths and Science at Teachoo. direction cosines ; This implies that we required only two independent parameters to describe a direction. Find the direction cosines of the line passing through the two points ( 2, 4, 5) and (1, 2, 3). Solution: The largest angle is the one facing the longest side, i.e. Vector Algebra - Vectors are fundamental in the physical sciences.In pure mathematics, a vector is any element of a vector space over some field and is often represented as a co Examples abound in classical mechanics and quantum mechanics. Direction Cosines. dcm = angle2dcm(___,rotationSequence) calculates the direction cosine matrix given three sets of rotation angles. 1. where the first subscript denotes the direction of the normal and the second denotes the direction of the to the component plane. Find the direction cosines of the vector that lies in the positive coordinate plane ð¥ ð§ and makes an angle of 6 0 â with the positive ð§ -axis. Example, 3 For example, in Fig. I designed this web site and wrote all the mathematical theory, online exercises, formulas and calculators. Overview. 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https://math.stackexchange.com/questions/4484289/continuous-mapping-from-a-nowhere-dense-set-to-dense-set | # Continuous mapping from a nowhere dense set to dense set
Let $$X$$ be a complete metric space, $$B$$ be a (closed) nowhere dense set in $$X$$, $$f:B\rightarrow X$$ be a continuous mapping. Is it possible that $$f(B)$$ is dense in $$X$$?
I think that is possible. But I can't give a counter-example. Any help will be appreciated.
• It is possible. Can you think of an example if $X=\mathbb R?$ Jul 1 at 15:05
• @D.Brogan Are you sure? I have a proof to the contrary Jul 1 at 15:33
• (so if you do have a counterexample, I'd like to know because it would show my proof to be false) Jul 1 at 15:43
• I believe my example has a counterexample @FShrike Jul 1 at 17:32
• This is a cool question. Could you provide some context @Ken.Wong? I'm curious where it came from. Jul 1 at 17:34
$$\textbf{Theorem}$$: Every compact metric space is a continuous image of the Cantor set $$\mathcal{C}$$. (See here)
Let us consider the set $$[0, 1]$$ as euclidean subspace of $$(\Bbb{R}, d_{\text{std}})$$.
Then there there exists a continuous onto function $$f:\mathcal{C} \to [0, 1]$$.
Now $$\mathcal{C}\subset [0,1]$$ is closed nowhere dense but $$f(\mathcal{C})=[0, 1]$$ which is a trivial dense subset of $$[0,1]$$.
Note: $$f$$ is here the restriction of the cantor function on the cantor set $$\mathcal{C}$$.
• I am just wondering, what if the nowhere dense set is an infinite-dimensional subspace in Hilbert space, does the conclusion hold? Since the set is not compact anymore. Jul 2 at 1:58
• @Ken.Wong Of course, as long as your Hilbert space is separable. Jul 2 at 4:33
• @Ken.Wong Consider the Hilbert space $X = \ell_2(\mathbb{N})$, with the nowhere-dense subspace $B$ defined by $x_1 = 0$. The image of $B$ under the left-shift operator $(x_n) \mapsto (x_{n+1})$ is all of $X$, obviously dense. The left-shift is continuous and linear on all of $X$. Jul 2 at 16:56
Since $$B$$ is nowhere dense, it inherits the discrete topology from $$X$$, which means that any function from $$B$$ will be continuous. Thus any surjection from $$\mathbb{Z}$$ onto $$\mathbb{Q}$$ serves as an example.
But we can do better than this. It is actually possible for $$f$$ to be continuous on all of $$X$$, with $$f(B)$$ still dense! Let $$X$$ be the positive reals, with $$f(x)=x\sin(x)$$. Define $$B = \bigcup_{n \in \mathbb{N}} \{2\pi n^3 + \sin^{-1}(\frac{k}{n^4}): k \in \mathbb{N}, k < n^2\}.$$ One can see that $$B$$ is nowhere dense: every interval of length $$2\pi$$ contains only finitely many points of $$B$$. To show $$f(B)$$ is dense, we need to construct a $$b \in B$$ with $$|b\sin(b) - x| < \epsilon$$ for any arbitrary $$x, \epsilon > 0$$. For large $$n$$, we have that $$\frac{1}{n^2} < \frac{\epsilon}{2}$$. For larger $$n$$, there is some integer $$k$$ for which $$|\frac{2\pi k}{n} - x| < \frac{\epsilon}{2}$$. For still larger $$n$$, we also can guarantee that $$k < n^2$$. Therefore $$B$$ contains the point $$b = 2\pi n^3 + \sin^{-1}(\frac{k}{n^4})$$. Now, $$f(b) = b \sin(b) = b \cdot \frac{k}{n^4} = \frac{2\pi k}{n} + \frac{k}{n^4}\sin^{-1}(\frac{k}{n^4})$$. Thus,
$$|b - x| = \bigg| \frac{2\pi k}{n} + \frac{k}{n^4}\sin^{-1}(\frac{k}{n^4}) - x \bigg| \leq \bigg| \frac{2\pi k}{n} - x\bigg| + \bigg|\frac{k}{n^4}\sin^{-1}(\frac{k}{n^4}) \bigg|.$$
The first term is less than $$\frac{\epsilon}{2}$$ by choice of $$k$$. The second term is also less than $$\frac{\epsilon}{2}$$, because $$\sin^{-1}(\frac{k}{n^4})$$ is bounded by 1, and $$\frac{k}{n^4} < \frac{n^2}{n^4} = \frac{1}{n^2} < \frac{\epsilon}{2}$$. Therefore $$|b - x| < \epsilon$$.
There are probably better examples than this. This related post offers a hypothesis which, if true, would give a far more elegant example.
• Thank you for pointing out my error. I was conceptualising $B$ still with the topology on $X$, not as a subspace. My very first thought was using the isomorphisms of $\Bbb Z\cong\Bbb Q$ but I discounted it because I viewed the integers as "not open" - but of course they are, in the subspace Jul 1 at 17:44
Any space-filling curve gives you an example of a continuous surjective map $$B=[0,1]\subset X=[0,1]^2\to X.$$ Actually, this situation is quite generic. As long as your metric space $$X$$ is, say, compact, and contains a nowhere dense subset $$B$$ homeomorphic to the Cantor set, you get continuous surjective map $$B\to X$$. (This is a theorem due to Hausdorff.) | 2022-08-15T10:16:36 | {
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https://math.stackexchange.com/questions/167798/how-to-identify-symmetric-positive-definite-matrices | # How to identify symmetric positive definite matrices?
I'm working on a project, implementing Successive over-relaxation (SOR) method (http://en.wikipedia.org/wiki/Successive_over-relaxation) using Python. SOR can only apply if given matrix is,
1. symmetric positive-definite (SPD) OR
2. strictly or irreducibly diagonally dominant.
1.Java doc (http://www.codezealot.org/opensource/org.codezealot.matrix/docs/org/codezealot/matrix/Matrix.html#isPositiveDefinite())
If a Matrix ANxN is symmetric, then the Matrix is positive definite if
• For all i ≤ N, ai,i > 0 and
• For all i ≤ N, ai,i > ∑ ai,j, for all j ≤ N, where j ≠ i
2.Numerical Analysis for Engineering (https://ece.uwaterloo.ca/~dwharder/NumericalAnalysis/04LinearAlgebra/posdef/)
A symmetric matrix is positive definite if:
• all the diagonal entries are positive, and
• each diagonal entry is greater than the sum of the absolute values of all other entries in the corresponding row/column.
These articles says different about the second property.
So,
- What is the correct one?
- Is there any other SPD properties I can use?
- Any suggestions are also welcome.
• The Javadoc is wrong (though if you download his source code, the source code is right); you need to take the absolute values of the off-diagonal entries. The matrix $\begin{bmatrix}1 & -100 \\ -100 & 1\end{bmatrix}$ is not positive definite.
– user856
Jul 7, 2012 at 11:03
• Note that this condition is sufficient but not necessary, that is, every symmetric diagonally dominant matrix is positive semidefinite, but there are positive semidefinite matrices which are not diagonally dominant. The Wikipedia article lists some other characterizations of positive definite matrices; at first glance, points 4 and 5 (principal minors and the Cholesky decomposition) can be used as a computational test.
– user856
Jul 7, 2012 at 11:07
• @DavideGiraudo thanks for showing it, I'll change it. Jul 7, 2012 at 11:10
• @RahulNarain I'm using the java code as a sample/guidance. Is there any thing I should worry about? Jul 7, 2012 at 11:10
• The Javadoc says the code tests the principal minors if the diagonal dominance test fails, and the source code does use the absolute values of the off-diagonal entries, so I don't see anything wrong there.
– user856
Jul 7, 2012 at 11:13
The second criteria in both conditions are roughly restatements of Gershgorin's Circle Theorem. The JavaDoc statement is wrong in two regards: (1) it should include absolute values inside the summation $\sum_{j\neq i}|a_{i,j}|$ and (2) the summation should be run through $j \leq N$. As stated by @Marc, the condition is sufficient but not necessary.
There is no such simple way to characterise positive definite matrices. You state neither criterion very clearly (what is the sum over in your first criterion?). However, as far as I can tell the second can only be a sufficient condition, not a necessary conditition; the first much weaker condition is certainly not sufficient, although maybe necessary (I didn't check). Note that the following matrix is positive definite without satisfying the second criterion $$\pmatrix{2&-1&-1&-1\\ -1&2&0&0\\ -1&0&2&0\\ -1&0&0&2\\ }$$ For true characterisations, see the Wikipedia article, notably Sylvester's criterion. | 2022-05-16T16:32:47 | {
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https://math.stackexchange.com/questions/693246/for-all-sets-a-b-and-c-if-a-setminusb-cup-c-emptyset-then-a-setminu | # For all sets A, B and C, if $A\setminus(B \cup C) = \emptyset$ ; then $A\setminus C\subseteq B$.
For all sets A, B and C, if $A-(B \cup C) = \emptyset$ ; then $A-C\subseteq B$. Is it true? If it is, how to prove it? I think it's true...
• Remember, if $A\setminus B=\emptyset$, then $A\subseteq B$. – frabala Feb 27 '14 at 22:33
• (From template of comments) After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?. – frabala Feb 28 '14 at 4:39
Notice that $$U-V=U\cap V^c$$ and $$(U\cup V)^c=U^c\cap V^c$$ hence we have $$A-(B\cup C)=A\cap(B\cup C)^c=A\cap(B^c\cap C^c)=(A\cap C^c)\cap B^c=(A-C)\cap B^c=\emptyset$$ hence $$A-C\subset B$$
• we cant use this – michael Feb 27 '14 at 22:37
• What do you can use so? – user63181 Feb 27 '14 at 22:38
• definitions like x belongs A-C, then x is in A, not in C – michael Feb 27 '14 at 22:42
$A\setminus (B\cup C)=\emptyset\Rightarrow A\subseteq B\cup C\Rightarrow A\setminus C\subseteq B$. Done.
### Another proof:
Because $A\setminus (B\cup C)=\emptyset$, we have that for all $x\in A$ it can't hold that $x\notin B\cup C$. So, for $x\in A$ we have $x\in B\cup C$ (by the way, this means that $A\subseteq B\cup C$). But if also $x\notin C$, meaning $x\in A\setminus C$, then $x\in (B\cup C)\setminus C= B$. So, $A\setminus C\subseteq B$.
• the question should be like is "-" not "\" – michael Feb 27 '14 at 22:39
• @michael If you like it better that way, it's ok. I see you changed it. – frabala Feb 27 '14 at 22:45
If you want to go for a purely logical proof then:
$x \in A \setminus C \implies x \in A$ and $x \not \in C ----(1)$
But $A \setminus (B \cup C) = \emptyset \implies$ There is no $x$ such that $x \in A$ and $x \not \in (B \cup C)$
By $(1)$, $x \in A \implies x \in (B \cup C).$ But since $x \not \in C, \; x \in B \implies A \setminus C \subseteq B$ | 2019-07-18T11:05:16 | {
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http://ecgv.ddiy.pw/cubic-spline-interpolation-matrix.html | By ignoring the 1st and the last equations, we have a system of n -1 equations (those marked with "#") in n +1 unknowns. A third degree polynomial and its derivative:. as the method of cubic splines. Natural and Clamped Cubic Splines 1. "spline" Cubic spline interpolation—smooth first and second derivatives throughout the curve. , for BVP), signal processing. The first recorded attempts at quadratic interpolation begin with the Persian physicist Al-Biruni (973-1048) who was one of the earliest adopters of the scientific method; however, the first use of finite differences began with Sir Thomas Harriot (1560-1621) whose other legacy is the use of the symbols < and >. Let’s look into doing linear regression in both of them: Linear Regression in Statsmodels. , we have available a set of values , then the function can be interpolated by a polynomial of. Out of range values are returned as NaNs. KEY BENEFITS. Both options are set for a smooth term that is set with s(). Defining spline curves • Generally f(t) is a piecewise polynomial - for this lecture, the discontinuities are at the integers - e. Pre-Requisites for Spline Method of Interpolation Objectives of Spline Method of Interpolation TEXTBOOK CHAPTER : A Textbook Chapter on Spline Method of Interpolation? DIGITAL AUDIOVISUAL LECTURES : Linear Spline Interpolation: Theory [YOUTUBE 7:56]. Spline이란 것은 여러 점을 잇는 것을 뜻한다. An introduction into the theory and application of cubic splines with accompanying Matlab m-file cspline. yy = spline(x,y,xx) uses cubic spline interpolation to find yy, the values of the underlying function y at the points in the vector xx. • The requirement that it is to be a cubic spline gives us 3(n −1) equations. B = spline3eval(A,XX) // This performs the interpolation, and store the values in B. We need two extra conditions to get this system solved. Muir [1960]). Yousefi 1,, J. This source code is the implementation of cubic spline interpolation algorithm and data smoothing using VC++ MFC. TinySpline: Open source C-library for splines which implements cubic spline interpolation SciPy Spline Interpolation: a Python package that implements interpolation Cubic Interpolation: Open source C#-library for cubic spline interpolation by Vadim A. Given the function values which may represent the resonant field position or the transition probability at the vertex points (Figure 1a), we use the cubic spline interpolation method to. The cubic spline function s(X) interpolating the (x,y) set of given points is a continuous and derivable piece-wise function defined over. Thank you very much for any help!. In linear interpolation, a line drawn between two points is used to find addi-tional points that lie between the two points. We can make this even stronger, ˚(x) 2C2 I, leading to piecewise cubic spline interpolation: The function ˚ i(x) is cubic in each interval. The main drawback to the cubic spline interpolation is that changing one of the points interpo-lated will affect the appearance of the curve in segments far removed from the point changed. traditional spline interpolation [7]. I got the "Index was outside the boundaries of the array" as the others did. It applies only in one dimension, but is useful for modeling yield curves, forward curves, and other term structures. The 2D spline seems similar in basis to mine, except that it is based on first derivatives rather than second derivatives. For large amounts of data, rendering with cubic spline interpolation may be much more costly than linear interpolation. yy = spline(x,y,xx) pp = spline(x,y) Description. Given the input reference path directions, smoothPathSpline also returns the. Syntax yy = spline(x,Y,xx) pp = spline(x,Y) Description yy = spline(x,Y,xx) uses a cubic spline interpolation to find yy, the values of the underlying function Y at the values of the interpolant xx. The concept is illustrated in the following figure: The data points are connected with cubic functions, and on each interval the coefficients must be determined. If the data being passed to interp2 is uniformly spaced, the algorithm is bicubic interpolation based on a cubic convolution kernel, not cubic spline interpolation. interpolate. Allpass delay line linear interpolation. (2017) Gradient superconvergence for a class of semi-cardinal interpolation schemes with cubic and quintic B-splines. Figures and compare interpolation errors of B-splines and other similar-cost methods on the example from Figure. Input the set of points, choose one of the following interpolation methods (Linear interpolation, Lagrange interpolation or Cubic Spline interpolation) and click "Interpolate". I'm writing a MATLAB program which accepts 3 inputs x (a vector containing the x values for interpolation), y (a vector containing the y values for interpolation) and a string specifying the type of cubic spline required ('natural', 'parabolically_terminated', 'not_a_knot') and then interpolates these points accordingly. "spline" Cubic spline interpolation—smooth first and second derivatives throughout the curve. We use the progressive-iterative approximation (PIA) algorithm rather than solving the control nets of a B-spline surface by a linear system. However this approach is a bit backward and usually predefined polynomial forms are used like SPLINE,BEZIER with defined properties like continuity, linearity, etc (no need for inverse matrix operation). Let’s look into doing linear regression in both of them: Linear Regression in Statsmodels. If y is a matrix, then the data is taken to be vector-valued and interpolation is performed for. Piecewise cubic polynomial spline interpolation [3] or smoothing [4] often gives undesirable inflexion points. Cubic spline interpolation uses cubic polynomials to interpolate datasets. Fast, reliable interpolated and extrapolated values in two and three dimensions. Yousefi 1,, J. Thank you very much for any help!. The algorithm given in w:Spline interpolation is also a method by solving the system of equations to obtain the cubic function in the symmetrical form. For Bicubic Interpolation (cubic convolution interpolation in two dimensions), thenumber of grid points needed to evaluate the interpolation function is 16, two grid points oneither side of the point under consideration for both horizontal and vertical directions. Numerical Analysis Grinshpan Natural Cubic Spline: an example. • The requirement that it is to be a cubic spline gives us 3(n −1) equations. In this blog, I show you how to conduct spline interpolation. For large amounts of data, rendering with cubic spline interpolation may be much more costly than linear interpolation. My goal in creating this was to provide a simple, clear implementation that matches the formulas in the Wikipedia articles closely, rather than an optimized. • In addition we require that S(x i) = y i, i = 0,··· ,n which gives n +1 equations. Cubic Spline Interpolation Sky McKinley and Megan Levine Math 45: Linear Algebra Abstract. A spline is simply a curve that connects two or more specific points. Bicubic spline interpolation does not make use of the cross-derivative values, and therefore is not as accurate as generalized cubic interpolation, which will be covered in the next subsection. Therefore, an n-th order B-spline interpolation is comparable in cost with any other method with an (n+1)-point interpolant. • This means we have 4n −2 equations in total. , when x and y are both integers Image interpolation refers to the “guess” of intensity values at missing locations, i. • Cubic spline interpolation is usually quite accurate and relatively cost effective. A curve is times differentiable at a point where duplicate knot values occur. Algorithms A tridiagonal linear system (possibly with several right-hand sides) is solved for the information needed to describe the coefficients of the various cubic polynomials that make up the interpolating spline. Cubic spline interpolation is a mathematical method commonly used to construct new points within the boundaries of a set of known points. , x and y can be arbitrary Note that it is just a guess (Note that all. (Note that the interpolant is produced by forcing a not-a-knot condition at the endpoints of the interval rather than forcing the second derivatives at the endpoints to be zero; in other words, it is not a natural spline interpolant). This lack of locality limits the usefulness of cubic spline interpolation in computer graphics. mgcv indeed is a good choice. Each function differs in how it computes the slopes of the interpolant, leading to different behaviors when the underlying data has flat areas or undulations. Section VI details the application for obtaining the best continuous approximation to a discrete contour, interpolating with cubic splines but first applying the FIR approximation and invariant to the translations of the least squares filter. SPLINE_BEZIER_VAL evaluates a cubic Bezier spline. -Means the result is still a cubic polynomial (verify!) • Cubic polynomials also compose a vector space -A 4D subspace of the full space of polynomials • The x and y coordinates of cubic Bézier curves belong to this subspace as functions of t. We introduce the -rational quadratic fractal interpolation functions (FIFs) through a suitable rational quadratic iterated function system (IFS). Flow chart of the direct computation of the CSI encoder for the 2-D image signal. Cubic Spline Interpolation Sky McKinley and Megan Levine Math 45: Linear Algebra Abstract. Spline Tutorial Notes 3 The Two Classes of Splines Interpolation splines are those which pass through their knots. Spline cubic with tridiagonal matrix. The next section discusses how the basis matrix is derived for Hermite curves. That is something I personally don't understand. A flexible strip is then bent across each of these weights, resulting in a pleasingly smooth curve. You can evaluate F at a set of query points, such as (xq,yq) in 2-D, to produce interpolated values vq = F(xq,yq). What is image interpolation? An image f(x,y) tells us the intensity values at the integral lattice locations, i. m Introduction Real world numerical data is usually difficult to analyze. Each function differs in how it computes the slopes of the interpolant, leading to different behaviors when the underlying data has flat areas or undulations. The method of cubic spline interpolation presented here is widely used in finance. (2017) Gradient superconvergence for a class of semi-cardinal interpolation schemes with cubic and quintic B-splines. For a function f(x) defined on the interval [a,b], either in functional or tabular form, cubic spline interpolation is the process of constructing (generally different) piecewise continuous cubic polynomials on subintervals [ti,ti+1] of the function domain [a,b]. Vector, matrix, and complex number classes, random number generators, numerical integration, cubic spline interpolation and other high-performance functions for object-oriented numerics on the. That is something I personally don't understand. I got the "Index was outside the boundaries of the array" as the others did. The algorithm given in w:Spline interpolation is also a method by solving the system of equations to obtain the cubic function in the symmetrical form. Our object is to study deficient cubic splines by making less restric-. Spline interpolation will begin again at the next G5. The control point setup can be implemented on MFC interface, can choose cubic spline interpolation or Bezier smoothing. P = S * h * C. The computational method can be applied to three-dimensional curves, too. It's a very interesting point that you make. Both options are set for a smooth term that is set with s(). Arguments: The cubic_spline function takes three args - column of input data, column of output data, and single point that you want the spline evaluated for. It uses data stored in its environment when it was created, the details of which are subject to change. Input the set of points, choose one of the following interpolation methods (Linear interpolation, Lagrange interpolation or Cubic Spline interpolation) and click "Interpolate". (Note that the interpolant is produced by forcing a not-a-knot condition at the endpoints of the interval rather than forcing the second derivatives at the endpoints to be zero; in other words, it is not a natural spline interpolant). There are many improvements still to be made: Non-linear position/rotation/speed interpolation. Unlike previous methods of Interpolating, Spline interpolation does not produce the same unique interpolating polynomial, as with the Lagrange method, Vandermonde matrix method, or Newton's divided difference method. I have VBA code that does this, but am having trouble finding an intuitive way to convert this to Alteryx. Earlier Sharma and Tzimbalario [6] had studied quadratic splines with similar matching conditions. An interpolation. "spline" Cubic spline interpolation—smooth first and second derivatives throughout the curve. If called with a third input argument, spline evaluates the spline at the points xi. I have a conceptual question, when choosing a Resampling Technique what are the pros and cons of using Bilinear Interpolation. Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Bicubic Interpolation • Bicubic2D(vx, vy, Z, p, q) —Returns the bicubic interpolation at ( p , q ) for the 3D data set defined by two vectors vx and vy , each sorted in ascending order, and by a matrix Z , where Z ij corresponds to the value at ( vx i , vy j ). Spans following the spline cancel command will be interpolated as linear spans (or as circular or helical, if such interpolation is commanded). I would recommend using a cubic spline interpolation method so that you will get a smooth interpolating curve that goes through all the original data points and that can handle more that four data points. Spline fitting or spline interpolation is a way to draw a smooth curve through n+1 points (x 0, y 0), …, (x n,y n). Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Comparisons of methods with our own critical. derivatives at each point—this is a spline Splines match the derivatives at end points of intervals – Piecewise splines can give a high-degree of accuracy Cubic spline is the most popular – Matches first and second derivative at each data point – Results in a smooth appearance – Avoids severe oscillations of higher-order polynomial. • Let us assume that we have equidistant knots. Given the input reference path directions, the block also returns the directions that correspond to each pose. Next: Cubic Spline Interpolation Up: Interpolation and Extrapolation Previous: The Newton Polynomial Interpolation Hermite Interpolation If the first derivatives of the function are known as well as the function value at each of the node points , i. complete class CubicSplineInterpolation implementing the cubic spline interpolation (you can first consider the case of 4 points). you will get a structure that contains all that information. The interpolation I want to do is going to be a function that returns a value for "z" based on the supplied values for "x" and "y". See the example "Spline Interpolation" for various examples. In and view of matrix theory, if is a circular matrix, then the inverse LIN et al. Here is an alternate implementation based on the same reference. As a result, the ck’s must be determined by solving a matrix problem. Use the Interpolation Line tab to specify the interpolation method and to specify if interpolation continues through missing values. Cubic spline - interpolation Given (x i,y i)n i=0. Here we present results from a series of tests in which SL schemes based on cubic spline and cubic Lagrangian interpolation are compared in one and two dimensions. splinefun, for example, can do only 1-dimensional interpolation. This effective combined CB-interpolation kernel is also called the cubic cardinal spline interpolation function. In this blog, I show you how to conduct spline interpolation. Input the set of points, choose one of the following interpolation methods (Linear interpolation, Lagrange interpolation or Cubic Spline interpolation) and click "Interpolate". They write the model as E(y) = „ = Bfi and minimize the following objective function: QB = jjy ¡Bfijj2 +‚jjDdfijj2; (1) where Dd is a matrix such that Ddfi = ¢dfi constructs the vector of dth difierences of fi, and ‚ is a. Cubic interpolation (same as "pchip"). The earliest splines were all interpolating since the mechanical spline from which they were derived (arguably) “interpolated” its ducks. Many students ask me how do I do this or that in MATLAB. Natural Cubic Spline Interpolation •A clever method –Construct S(x) Lagrange Form thought –Solve tridiagonal matrix Using decompt & solvet (2-1) –Evaluate of S(z) Locate z in some interval (using binary search) Using Horner’s rule to evaluate. -Means the result is still a cubic polynomial (verify!) • Cubic polynomials also compose a vector space -A 4D subspace of the full space of polynomials • The x and y coordinates of cubic Bézier curves belong to this subspace as functions of t. Data Fitting www. The parameter of the spline is the cumulative chord length at these points. , for BVP), signal processing. Up to rounding errors, and assuming that x is a vector with at least four entries, the statement pp = csapi(x,y) should put the same spline into pp as does the statement. See the example “Spline Interpolation” for various examples. An interpolation. In order to find the spline representation, there are two different ways to represent a curve and obtain (smoothing) spline coefficients: directly and parametrically. We propose a unified approach to the derivation of sufficient conditions for the k-monotonicity of splines (the preservation of the sign of any derivative) in interpolation of k-monotone data for k = 0, …, 4. - A i' 1 i' If u(x) and its derivatives are continuous, it has been shown that the spline function S (x) approximates u(x) at all,. This function can be used to evaluate the interpolating cubic spline (deriv = 0), or its derivatives (deriv = 1, 2, 3) at the points x, where the spline function interpolates the data points originally specified. Many students ask me how do I do this or that in MATLAB. Finite Elements A program package with demos is available as Supplementary Material of the book Finite Element Methods with B-Splines (SIAM FR26) Supplementary Material of. coefs is an nx4 matrix of polynomial coefficients for the intervals, in Matlab convention with the leftmost column containing the cubic coefficients and the rightmost column containing the constant coefficients. Matt J 145 questions asked. SPLINE_BEZIER_VAL evaluates a cubic Bezier spline. the spline equation system, and demonstrate the connection between the determinant of this equation system and the anticausal IIR filter he proposes for moving a signal to the B-spline domain. Cubic splines can be described as follows. gsl_interp_cspline_periodic¶ Cubic spline with periodic boundary conditions. Piecewise cubic spline interpolation A cubic-spline is a spline constructed of piecewise third-order polynomials which pass through a set of control points (Knott 2000). This article is outdated due to the last updates in the ALGLIB package. Interpolation via linear or cubic splines. BASIS_MATRIX_BEZIER_UNI sets up the cubic Bezier spline basis matrix. This lack of locality limits the usefulness of cubic spline interpolation in computer graphics. Data can be approximated by a linear combination of the orthonormal basis functions. • The efficient implementation of the cubic spline interpolation. The curve can be two-dimensional, such as a planar cam profile, or three-dimensional, such as a roller coaster track. The most common case considered is k= 3, i. It has been tested against the published test cases for the algorithm. 05 Spline Method of Interpolation After reading this chapter, you should be able to: 1. This post is by my colleague Cosmin Ionita. How to correctly solve cubic spline interpolation?. In the section IV we present an approximation of one FIR operator to do cubic spline interpolation with only 5 multiplications and ten additions,. • To fulfill the Schoenberg-Whitney condition that N i n(u i) ≠0 , for n=3 we set u i=i+2 for all i. Interpolation at x 0 through x n provides n + 1 equations. (zero-order polynomials) 'linear' linear interpolation 'spline' piecewise cubic spline interpolation (identical to the spline function) 'cubic' or 'pchip' piecewise cubic Hermite interpolation 19. Does anyone have it available? I have had no luck finding it so far. • The requirement that it is to be a cubic spline gives us 3(n −1) equations. The splines of a newer class, the ap-. Results from fitting inbreeding using a cubic-spline with seven knots were compared to results from fitting inbreeding as a linear covariate or as a fixed factor with seven levels. spline curve now refers to any composite curve formed with polynomial sections satisfying any specified continuity conditions at the boundary of the pieces (1st and 2nd derivatives are continuous). These new points are function values of an interpolation function (referred to as spline), which itself consists of multiple cubic piecewise polynomials. Spline Method of Interpolation. Cubic splines would not be necessary were it simple to determine a well-behaved function to fit any data set. For your specific request you need to set the cubic spline as the basis function bs='cr' and also not have it penalized with fx=TRUE. "spline" Cubic spline interpolation—smooth first and second derivatives throughout the curve. Note that if extrapval is used, method must be specified as well. A change of approach leads to methods involving blending functions and control points, from hat-functions through Bezier and B-spline curves and concluding with N. In ridge regression, you add a quadratic penalty on the size of the regression coefficients, and so the. 1: Cubic Splines Interpolating cubic splines need two additional conditions to be uniquely defined Definition. rd_2_spline_functions 3 References Help pages from the stats package, for splinefun(), splinefunH() and spline(). I'm working on a finite volume advection scheme for unstructured meshes which uses a multidimensional polynomial weighted least squares fit for interpolating from cell centres onto faces. I recommend you to read about Splines on Wikipedia. cubic spline interpolation and upsample ?. 7 (13 ratings) Course Ratings are calculated from individual students' ratings and a variety of other signals, like age of rating and reliability, to ensure that they reflect course quality fairly and accurately. Existence of Cubic Splines Let us try to determine if it is possible to construct a cubic spline that satisfies proper-ties I through V. CSCE 441 Computer Graphics: Keyframe Animation/Smooth Curves Jinxiang Chai Outline Keyframe interpolation Curve representation and interpolation - natural cubic curves - Hermite curves - Bezier curves Required readings: 12-6 & 14-1 14-214-3 14-4, & 14-7 Computer Animation Animation - making objects moving Compute animation - the production of consecutive images, which, when displayed, convey a. Similarly, the surface generated by plot::Matrixplot is the graph of the cubic spline function interpolating the matrix data. This is because the interpolation coefficients -the cR's in (1)-must be determined by solving a tridiagonal matrix problem; in two dimensions, the matrix is block tri- diagonal. This paper shows that the two ideas are intrinsically related. Data can be approximated by a linear combination of the orthonormal basis functions. This is the matrix of coefficients ##a_i \rightarrow a_n## where n is the number of data points provided. As opposed to cubic spline, data points are not special points at which polynomials are spliced, and it makes much easier exact integration in these points. Fast, reliable interpolated and extrapolated values in two and three dimensions. m performs cubic convolution interpolation 11) divdiff. the end point of segment i is the same as the starting point of segment i + 1. This source code is the implementation of cubic spline interpolation algorithm and data smoothing using VC++ MFC. Thank you very much for any help!. The Cubic Spline method allows one to construct smoother curves. Extrapolation makes little sense for method = "fmm" ; for natural splines it is linear using the slope of the interpolating curve at the nearest data point. I've looked through the. In linear interpolation, a line drawn between two points is used to find addi-tional points that lie between the two points. 1st derivative (slope) continuity at interior points 3. If this is a standard item of the current version or not, or an additional extra library, or something else that the user must install manually, eludes me. This is an implementation of cubic spline interpolation based on the Wikipedia articles Spline Interpolation and Tridiagonal Matrix Algorithm. This video provides a high-l. • Covariance and correlation matrix estimation from time series data. Let fbe a function from. KEY BENEFITS. SPLINE_CUBIC_VAL evaluates a piecewise cubic spline at a point. com Sample output 1 If you know that your points will be equidistant, that is all hi’s are equal to h, then the above code can be modified. 12, it is obvious that the de Boor algorithm is a generalization of the de Casteljau algorithm. Read more. Piecewise cubic spline interpolation A cubic-spline is a spline constructed of piecewise third-order polynomials which pass through a set of control points (Knott 2000). Before we discuss cubic splines, we will develop the concept of piecewise linear fits. A curve is times differentiable at a point where duplicate knot values occur. Theory The fundamental idea behind cubic spline interpolation is based on the engineer’s tool used to draw smooth curves through a number of points. The Kochanek-Bartels Splines (also called TCB-Splines) Now we're going down to the guts of curve interpolation. Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Loosely speaking, we have 4N degrees of freedom or conditions that must. PB-splines use a basis of (quadratic or cubic) B-splines, B, computed on x and using equally-spaced knots. Many popular image interpola-tion methods are defined in this way, including nearest-neighbor interpolation, bi-linear interpolation, cubic-spline interpolation, and cubic convolution [1], [2], [8]. As a result, the ck’s must be determined by solving a matrix problem. Many students ask me how do I do this or that in MATLAB. Thin-plate spline interpolation is an effective interpolation. Cubic spline interpolation uses cubic polynomials to interpolate datasets. Until then, I'd better get back to those segfaults. The noisy samples fy igare the red points, the true function f(t) is the dashed blue curve, and the interpolator f^(t) is the solid green curve. Three dimensional interpolation and extrapolation using either a set of (x, y, z) points, or matrix of evenly spaced z values. The matrix form of the system of equations is:. This method obtains a piecewise continuous function that has continuous first and second order derivatives. Find the cubic spline interpolation at x = 1. Allpass delay line noninterpolating. uses polynomials of degree 3, which is the case of cubic splines. (1989) End conditions for cubic spline interpolation derived from integration. B-Spline Interpolation and Approximation Hongxin Zhang and Jieqing Feng 2006-12-18 State Key Lab of CAD&CG Zhejiang University. Working C C++ Source code program for Lagrange's interpolation /***** Lagrange's interpolation *****/ #include< Understanding Dependency Injection and its Importance, A tutorial Any application is composed with many classes that collaborate each-other to perform some useful stuff. A cubic Spline would be the best but my data set can be up to 600 points, and doing the large matrix for that would be too much overhead. All methods are tested using an Array with approximately 10^6 elements, and the interpolation task is simply to visit each grid point. Finally in the contact force generation part, we evaluate the contact force and Jacobian matrix for the implicit time integrator. For your specific request you need to set the cubic spline as the basis function bs='cr' and also not have it penalized with fx=TRUE. » help spline SPLINE Cubic spline data interpolation. The control point setup can be implemented on MFC interface, can choose cubic spline interpolation or Bezier smoothing. An introduction into the theory and application of cubic splines with accompanying Matlab m -file cspline. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The spline interpolation kernel is not zero for nonzero integers. I did some simple tests and examples confirming that. A flexible strip is then bent across each of these weights, resulting in a pleasingly smooth curve. A cubic-spline is a spline constructed of piecewise third-order polynomials which. This gist was motivated by the answer to this question on stack overflow. These functions all perform different forms of piecewise cubic Hermite interpolation. Interpolation via linear or cubic splines. the music is "Si J'etais roi" by Adolfe Adam; and is performed. The cubic spline function s(X) interpolating the (x,y) set of given points is a continuous and derivable piece-wise function defined over. Spline이란 것은 여러 점을 잇는 것을 뜻한다. The type of interpolation is classi ed based on the form of ˚(x): Full-degree polynomial interpolation if ˚(x) is globally polynomial. Applied Mathematics and Computation 308 , 142-148. The de Boor algorithm is shown graphically in Fig. The first y value will be used for interpolation to the left and the last one for interpolation to the right. Bode frequency shifter. complete class CubicSplineInterpolation implementing the cubic spline interpolation (you can first consider the case of 4 points). Spline이란 것은 여러 점을 잇는 것을 뜻한다. [email protected]_:5m( @fcadb:56 r dbo :5m c b j> help spline SPLINE Cubic spline data interpolation. It is possible to also introduce quadratic spline, i. interpolation. The splines of a newer class, the ap-. m performs piecewise cubic spline interpolation 10) cubiconv. First, call calcIota to generate interpolation information; then you can interpolate using Akima's spline method with the akima() function. m performs cubic convolution interpolation 11) divdiff. The cubic spline interpolation method is proba- bly the most widely-used polynomial interpolation method for functions of one variable. Ask Question 0. Matrix Form for Cubic Bézier Curves Converting Between Cubic Spline Types. The path-smoothing algorithm interpolates a parametric cubic spline that passes through all input reference pose points. Cubic Spline. ' interp1 ' is called one dimensional interpolation because vector y depends on a single variable vector x. % MATLAB permits us to solve for the "spline" curve in a relatively simple approach when we call upon the spline() function. This method obtains a piecewise continuous function that has continuous first and second order derivatives. "pchip" Piecewise cubic Hermite interpolating polynomial—shape-preserving interpolation with smooth first derivative. Project 3: Resistor networks / Cubic spline interpolation OVERVIEW In this project, you will write a program to automatically solve two engineering tasks which can be represented as systems of linear equations. Interpolation Calculator. A full math Java class library containing complex functions and algorithms such as cubic-spline interpolation, least squares, matrix computations. org May 28, 2003 1. Given the function values which may represent the resonant field position or the transition probability at the vertex points (Figure 1a), we use the cubic spline interpolation method to. Curve fitting functions include polynomial fits and a versatile local regression (loess) function. The de Boor algorithm also permits the subdivision of the B-spline curve into two segments of the same order. edu" Subject Re: st: Converting Quarterly GDP Data into Monthly Data Using Cubic Spline Interpolation. C# - Cubic Splines - QuickStart Samples - Math, Statistics and Matrix Libraries for. The piecewise cubic-polynomial kernel over the region. CERCHAPI Initial slope at the first spline. I would prefer cubic spline interpolation between the data points with linear interpolation used at the edges of the surface. How to correctly solve cubic spline interpolation?. – METHOD specifies interpolation filter • 'nearest' - nearest neighbor interpolation • 'linear'- bilinear interpolationbilinear interpolation • 'spline' - spline interpolation • 'cubic' - bicubic interpolation as long as the data is uniformly spaced, otherwise the same as 'spline' Geometric Transformation EL512 Image Processing 26. griddedInterpolant returns the interpolant F for the given dataset. ,"yy",B) // This pushes the inpolated figures in B back into the yy variable in Stata. The answer to the problem is given by the spline tting. This is the simplest case of cubic spline interpolation that will illustrate the methods used in more normal cases where more points are present. yy = spline(x,y,xx) uses cubic spline interpolation to find yy, the values of the underlying function y at the points in the vector xx. Would someone please be kind enough to check my math? The resulting curve is not smooth, does. • Cubic spline interpolation is usually quite accurate and relatively cost effective. we can see that interpolations work pretty well, but extrapolations diverge from the expected results. The web is full of alternate presentations of cubic spline interpolation. complete class CubicSplineInterpolation implementing the cubic spline interpolation (you can first consider the case of 4 points). Introduction During the last years, spline functions have found widespread application, mainly for the purpose of interpolation [•]. The Foundation region is where the parent Interpolation class is defined. Find the natural cubic spline that interpolates the the points $(1, 1)$, $\left ( 2, \frac{1}{2} \right )$, $\left ( 3, \frac{1}{3} \right )$, and $\left (4 , \frac{1}{4} \right )$. Cubic spline interpolation is a mathematical method commonly used to construct new points within the boundaries of a set of known points. Allpass delay line cubic spline interpolation. Cubic Spline Interpolation Utility This page contains a cubic spline interpolation utility. What needs to be stressed is that in the case of boot-. These are piecewise cubic functions that are continuous, and have continuous rst, and second derivatives. Note that you can either interpolate 1d data, or you can interpolate 2d data points by doing this interpolation on each axis. Cubic Spline Interpolation A spline is a piecewise polynomial of degree kthat has k 1 continuous derivatives. Using the two data sets given in the previous problem and answer the same. Cubic splines can be described as follows. 2 An Example The default R function for tting a smoothing spline is called smooth. Consider f(x) the cubic spline function that approximates the input function. | 2019-11-18T21:45:30 | {
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https://math.stackexchange.com/questions/2358734/how-many-ways-to-distribute-10-identical-cups-between-three-distinguishable-tabl | # How many ways to distribute 10 identical cups between three distinguishable tables?
How many ways to distribute 10 identical cups between three distinguishable tables?
So I know the answer is 12C2 but I'm not exactly sure why. The answer given says to arrange it so you have T(1)cccT(2)T(3)ccccccc and then you just have to see that there are 12 spaces to put the two tables in. However I can only see eleven spaces (since if you put T(2) to the left of T(3) it's the same as putting T(2) at the very end because both ways T(2) will have no cups on it, same goes for T(3)). One way I saw on here that explained in a way I understood that it's just the number of ways to arrange 12 symbols in a line (with there being 2 types of symbol). But I'd like to know the reasoning behind the answer given in my lecture notes.
$\cup\cup\cup\cup\cup\cup\cup\cup\cup\cup$
You can put dividers in between to divide which table they go on. Here's one way:
$|\cup\cup\cup\cup\cup\cup|\cup\cup\cup\cup$
How many ways can you do this?
• I initially thought you'd used $\cup$ instead of U and were making a nice pun. – pjs36 Jul 14 '17 at 19:45
• @pjs36 Ha! Well I can't let that opportunity pass! Edited. :) – John Jul 14 '17 at 19:47
• Yes, I know what the answer is. What I'm looking for is an explanation to the method given in my lectures – nic Jul 22 '17 at 21:52
• What wasn't clear about the method I gave you? – John Jul 22 '17 at 22:06
Let's see how many ways to distribute 10 cups between 3 tables. Let's start with no cups on the first table: 0 0 10, 0 1 9, 0 2 8, 0 3 7, 0 4 6, 0 5 5, 0 6 4, 0 7 3, 0 8 2, 0 9 1, 0 10 0. So, 11 ways to distribute the cups when there are no cups on the first table. The total number will be 11 + 10 + 9 +...+ 1 = 12C2
The following is quite a famous theorem in Combinatorics.
Stars And Bars Theorem: For any pair of positive integers $n$ and $k$, the number of $k$-tuples of $non-negative$ integers whose sum is $n$ is equal to $\binom{n+k-1}{k-1}$
For your given problem we have $10$ cups which have to be distributed among $3$ tables(assuming that some of the tables can be empty)
Here each table constitutes to a tuple, so $k=3$ and total number of cups $n=10$
$$\binom{n+k-1}{k-1}=\binom{12}{2}$$
• How is that formula derived? – nic Jul 22 '17 at 21:52
Let $x_k$ be the number of cups placed on the $k$th table. Then $$x_1 + x_2 + x_3 = 10$$ where $x_1$, $x_2$, and $x_3$ are nonnegative integers. A particular solution corresponds to the placement of two addition signs in a row of ten ones. For instance, $$1 1 1 + + 1 1 1 1 1 1 1$$ corresponds to the solution $x_1 = 3$, $x_2 = 0$, and $x_3 = 7$, while $$1 1 1 1 1 + 1 1 1 1 + 1 1 1$$ corresponds to the solution $x_1 = 5$, $x_2 = 5$, and $x_3 = 3$. The number of such solutions is the number of ways two addition signs can be inserted in a row of ten ones, which is $$\binom{10 + 2}{2} = \binom{12}{2}$$ since we must choose which two of the twelve positions (for ten ones and two addition signs) will be filled with addition signs.
Edit: The general problem we wish to solve is finding the number of ways $n$ identical objects can be distributed to $k$ distinguishable boxes.
Let $x_j$, with $1 \leq j \leq k$, denote the number of objects placed in the $k$th box. Note that $x_j$ may equal zero. Then $$x_1 + x_2 + x_3 + \cdots + x_k = n$$ A particular solution corresponds to the placement of $k - 1$ addition signs in a row of $n$ ones. Thus, the number of such solutions is the number of ways of arranging the $n + k - 1$ symbols consisting of $n$ ones and $k - 1$ additions signs in a row, which is $$\binom{n + k - 1}{k - 1}$$ since we must choose $k - 1$ of the $n + k - 1$ positions for the addition signs.
In your example, the objects are the cups and the boxes are the tables on which they are placed. Hence, $n = 10$ and $k = 3$, so there are $$\binom{10 + 3 - 1}{3 - 1} = \binom{12}{2}$$ ways to distribute the cups to the tables.
Note: A related problem that also reduces to solving the equation $$x_1 + x_2 + x_3 + \cdots + x_k = n$$ is selecting $n$ objects from $k$ types of objects, in which there are at least $n$ objects of each type available. This is a combination with repetition problem.
• Thank you for your answer but I was looking for an explanation to the method given in my lectures. I understood the answer when using a different method, just not the one used in lectures – nic Jul 22 '17 at 21:54
• I have added a derivation of the general formula. – N. F. Taussig Jul 22 '17 at 23:47 | 2021-01-26T08:43:40 | {
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http://xmre.cyyy.pw/real-symmetric-matrix-inverse.html | # Real Symmetric Matrix Inverse
ASA007 is Applied Statistics Algorithm 7. The rst step of the proof is to show that all the roots of the characteristic polynomial of A(i. EXTREME EIGENVALUES OF REAL SYMMETRIC TOEPLITZ MATRICES 651 3. Square matrix A is said to be skew-symmetric if a ij = − a j i for all i and j. Inverse Eigenpair Problem for Real Symmetric Seven-Diagonal Positive Definite Matrix FENG Lichao1, JIN Dianchuan 1, WU Zhihui 2, YANG Yanmei , SONG Shaopeng1, ZHANG Qiuna3 1. A complex Hermitian matrix B is a square matrix with complex entries that satisfies B∗ = B, where B∗ denotes the Hermitian. Lets take an example of matrix. The eigen values of a real symmetric matrix are all real. The bounds are first obtained for positive-definite matrices and then extended to the general real symmetric case. An inverse of a real symmetric matrix should in theory return a real symmetric matrix (the same is valid for Hermitian matrices). MATH 340: EIGENVECTORS, SYMMETRIC MATRICES, AND ORTHOGONALIZATION Let A be an n n real matrix. Either we need to change complex symmetric matrix to complex Hermitian matrix, or elaborate that the diagonal matrix doesn't contain eigenvalues. If a matrix contains the inverse, then it is known as invertible matrix and if the inverse of a matrix does not exist, then it is called a non-invertible matrix. Given the positions of non zero elements in what are their values such that is positive definite. ma/jobs Other than http://lem. since A is orthogonally similar to a diagonal matrix D with the same eigenvalues (via an orthogonal matrix Q), D has m diagonal entries that are the same. Explanation File of program above (Sysmat) NEW. Then AT exists and is also an n×n matrix. Selected Problems — Matrix Algebra Math 2300 1. The eigenvalues are also real. nsysu Positive Definite Matrix 7/33. Matrix algebra offers us tools for manipulating matrix equations by creating several useful formulas that are very similar to how we dealt with equations and real numbers in algebra 1. 4 Pseudo Inverse 3. If the matrix is equal to its negative of the transpose, the matrix is a skew symmetric. Whatever A does, A 1 undoes. This process is experimental and the keywords may be updated as the learning algorithm improves. If A and B are skew-symmetric matrices then A T, A+B, AB-BA, and kA are skew-symmetric for every scalar k. Therefore A is similar to a real diagonal matrix. SYMMETRIC MATRICES AND INNER PRODUCTS 3 True or False Provide reasons for the true and counterexamples for the false. $\endgroup$ - Javier Garcia Jun 2 at 13:52 $\begingroup$ Well you can do it by diagonalisation which in my experience is pretty stable for symmetric matrix, but I'm not numerical analyst. The methods for the second stage include the implicit QR method [9] and the divide-andconquer method [12]. Extrema of functions of a real symmetric matrix in terms of eigenvalues. Charles was born with a deformity. a numeric or complex matrix whose spectral decomposition is to be computed. The matrices are symmetric matrices. and a matrix multiplied by the identity matrix is itself (AI = IA = A):. Diagonalization of a 2× 2 real symmetric matrix Consider the most general real symmetric 2×2 matrix A = a c c b , where a, b and c are arbitrary real numbers. This lesson will show the parallels between the reciprocal and the multiplicative inverse of a nonzero number. Review An matrix is called if we can write where is a8‚8 E EœTHT Hdiagonalizable " diagonal matrix. To find the inverse of a matrix A, i. Hello everyone! I'm struggling to find a general formula for obtaining an inverse of a symmetric matrix, for e. Therefore, a general complex (respectively, real) matrix is positive definite iff its Hermitian (or symmetric) part has all positive eigenvalues. is always real when A is real symmetric or complex Hermitian. Most properties of real symmetric matrices are listed under Hermitian. Complex numbers will come up occasionally, but only in very simple ways as tools for learning more about real matrices. The notion of an inverse matrix only applies to square matrices. The matrix T is real, symmetric and tridiagonal. The inverse of a matrix $$A$$ is defined as a matrix $$A^{-1}$$ such that the result of multiplication of the original matrix $$A$$ by $$A^{-1}$$ is the identity matrix $$I:$$ $$A{A^{ - 1}} = I$$. All eigenvalues of a symmetric matrix are real numbers B. But we would like to. This project should build as is using Visual Studio 2008 and perhaps later versions of Visual Studio. Midpoint Formula. Note that as it's a symmetric matrix all the eigenvalues are real, so it makes sense to talk about them being positive or negative. Spectral equations In this section we summarize known results about the various spectral, or \sec-ular", equations for the eigenvalues of a real symmetric Toeplitz matrix. Most properties of real symmetric matrices are listed under Hermitian. pinv¶ numpy. New England Research. But, if the matrix is also skew–symmetric, then we have the following theorem. Let A = a b b c be any 2×2 symmetric matrix, a, b, c being real numbers. It is easy to see that the inverse of a persymmetric matrix is also persymmetric. If A2m 1 m>k 1. When a unitary matrix is real, it becomes an orthogonal matrix,. Let A be an n x n matrix. 53 istril Determine if matrix is lower triangular; 1. We can show that both H and I H are orthogonal projections. nxn inverse matrix calculator, formulas, work with steps, step by step calculation, real world and practice problems to learn how to find inverse matrix of 4x4, 3x3 and 2x2 matrices. Proposition 3. RAGHAVAN FOR IST AT IITGN, JULY 2017 An n n real symmetric matrix A is said to be positive de nite if, for every v 2Rn, we have vtAv 0 and equality holds only if v = 0. This decomposition is called a spectral decomposition of A since Q consists of the eigenvectors of A and the diagonal elements of dM are corresponding eigenvalues. Let S be an n × n real doubly skew matrix. (a)-(c) follow from the definition of an idempotent matrix. This paper modifies the GLR theory for the special application to real symmetric quadratic matrix polynomials, mathcal{Q}(λ)=M λ^{2} + C λ + K , M nonsingular, subject to the specific restriction that all matrices in the representation be real-valued. The basic idea of this algorithm lies on the determination of one block of the inverse of the matrix at a time. since if we use, for example, the Gaussian elimination to compute the inverse, we divide each row of the matrix ( A | I ) by the corresponding diagonal element of A in which case the number 1 on the same row of the identity matrix on the right is also divided by the same element. For a given real symmetric matrix A, the codes consider the inverse of a matrix B where (9. of the complex symmetric tridiagonal T resulted from the first stage is com-puted. So we see that the inverse of a non-singular symmetric matrix is obtained by inverting its eigenvalues. For example, to solve 7x = 14, we multiply both sides by the same number. Returns two objects, a 1-D array containing the eigenvalues of a , and a 2-D square array or matrix (depending on the input type) of the corresponding eigenvectors (in columns). Department of Geophysics. rank{·}, trace{·}, and det{·} denote the rank, trace, and determinant of a given scalar/matrix. Extend the dot product to complex vectors by (v,w) = P iviwi, where v is the complex conjugate. We find the "inverse" of 7, which is 1/7. What is symmetric and skew symmetric matrix ? For any square matrix A with real number entries, A+ A T is a symmetric matrix and A− A T is a skew-symmetric matrix. Lets take an example of matrix. Minho, Portugal). If the matrix is not invertible (a singular matrix), the value of the matrix coming out of the above method will be NAN. CSML - C# Matrix Library - is a compact and lightweight package for numerical linear algebra. If a matrix contains the inverse, then it is known as invertible matrix and if the inverse of a matrix does not exist, then it is called a non-invertible matrix. In this paper, we propose and discuss a class of inverse eigenvalue problems for real symmetric banded matrices with odd bandwidth. SGECOComputes LU factorization of real general matrix and estimates its condition. Symmetric Matrix & Skew Symmetric Matrix To understand if a matrix is a symmetric matrix, it is very important to know about transpose of a matrix and how to find it. A real symmetric matrix is a symmetric matrix whose entries are real. For a positive integer n, consider the tridiagonal matrix of. We have a symmetric matrix and a skew-symmetric matrix that add to give 2A, the matrix A times the scalar 2. We can prove some parts of the theorem right away without much work. ) We can calculate the Inverse of a Matrix by: Step 1: calculating the Matrix of Minors, Step 2: then turn that into the Matrix of Cofactors, Step 3: then the Adjugate, and ; Step 4: multiply that by 1/Determinant. eralization of the inverse of a matrix. In this case we also explicitly determine the symmetric eigenvectors and corresponding eigenvalues of T. In this paper we solve. e A-1 we shall first define the adjoint of a matrix. Then you could compare operation count and numerical stability for various methods, to include "straightforward" methods not making using of the Schur complement. Inverse Matrices 81 2. Solving a linear matrix system AX=B by Gauss-Jordan Method. reciprocal of a nonzero real number. LA_PPTRI computes the inverse of real symmetric / complex Hermitian positive definite matrix in packed storage format using the Cholesky factorization computed by LA_PPTRF. Applying the transposition operator to each side of the equation we get. Their product is the identity matrix—which does nothing to a vector, so A 1Ax D x. Symmetric Matrix Inverse. A symmetric positive semi-definite square matrix has an alternate Cholesky decomposition into a product of a lower unit triangular matrix , a diagonal matrix and , given by. m: - A matlab program that computes a few (algebraically) smallest or largest eigenvalues of a large symmetric matrix A or the generalized eigenvalue problem for a pencil (A, B): A x = lambda x or A x = lambda B x. Because finding transpose is much easier than the inverse, a symmetric matrix is very desirable in linear algebra. Show that if A is nonsingular symmetric matrix, then A^-1 For the first one, study the definition your book or notes gives for "nonsingular". More about Inverse Matrix. Logical matrices are coerced to numeric. However, in Example ESMS4, the matrix has only real entries, but is also symmetric, and hence Hermitian. 5 Spectral Theorem for Real Symmetric Matrices The main theorem we prove is Theorem 5. If Adoes not have an inverse, Ais called singular. Symmetric eigenvalue problems are posed as follows: given an n-by-n real symmetric or complex Hermitian matrix A, find the eigenvalues λ and the corresponding eigenvectors z that satisfy the equation Az = λ z (or, equivalently, z H A = λ z H). Let A = a b b c be any 2×2 symmetric matrix, a, b, c being real numbers. Theorem: Any symmetric matrix 1) has only real eigenvalues; 2) is always diagonalizable; 3) has orthogonal eigenvectors. A matrix is diagonalizable if it has n eigenvectors D. The matrix 1 1 0 2 has real eigenvalues 1 and 2, but it is not symmetric. A complex symmetric matrix doesn't necessarily have real eigenvalues, as the article currently states in the Decomposition section. We prove a positive-definite symmetric matrix A is invertible, and its inverse is positive definite symmetric. 18 (1997), 1722-1736 Abstract. 2 The Inverse of a Matrix. That is, matrices which look like two arrow matrices, forward and backward, with heads against each other at the station,. Shalaby * Internationa] Centre for Theoretical Physics, Trieste, Italy. Let be an symmetric nonsingular idempotent matrix. H matrix is symmetric, so we can write its eigen-decomposition (whereS is diagonal) : X matrix is linked to H, so we have: For each eigenvalue of H, eigenvalue of X is: 1 H X H σ σ σ = − 1 H X H σ σ σ = − Eigenvalues of correction on the inverse (X matrix) with respect to the eigenvalues of the direct correction (H matrix). One has t =-%i * logm (x + %i * sqrtm (eye ()-x * x)). To calculate inverse matrix you need to do the following steps. The properties of the spectrum of a symmetric matrix with real elements include the following: (1) all the roots λ 1, λ 2,…, λ n of the characteristic equation of the matrix are real; and (2) to these roots there correspond n pairwise orthogonal eigenvectors of the matrix, where n is the order of the matrix. Inverse spectral analysis for a class of nite band symmetric matrices Mikhail Kudryavtsev,Sergio Palafox andLuis O. We can prove some parts of the theorem right away without much work. (ii) det (A) is not zero. We can exploit the structure of a real, positive definite, symmetric matrix by using the Cholesky decomposition to compute the inverse. Synchronization in the Symmetric Inverse Semigroup 5 tation is total). We can exploit the structure of a real, positive definite, symmetric matrix by using the Cholesky decomposition to compute the inverse. 54 istriu Determine if matrix is upper triangular; 1. eigh (a, UPLO='L') [source] ¶ Return the eigenvalues and eigenvectors of a complex Hermitian (conjugate symmetric) or a real symmetric matrix. The ifft2 function tests whether the vectors in a matrix Y are conjugate symmetric in both dimensions. Square matrix A is said to be skew-symmetric if a ij = − a j i for all i and j. reciprocal of a nonzero real number. 9) will be symmetric and positive definite. com Don't Memorise brings learning to life through its captivating FREE educational videos. ) We can calculate the Inverse of a Matrix by: Step 1: calculating the Matrix of Minors, Step 2: then turn that into the Matrix of Cofactors, Step 3: then the Adjugate, and ; Step 4: multiply that by 1/Determinant. A ij = (-1) ij det(M ij), where M ij is the (i,j) th minor matrix obtained from. The matrix inverse of a positive definite matrix is also positive definite. (x) Diagonal Matrix. Proof: Let A be an n×n matrix. For real asymmetric matrices the vector will be complex only if complex conjugate pairs of eigenvalues are detected. 8 Real symmetric matrices A square matrix A is called symmetric if A = AT, i. This problem remains un-. Similarly. f90 Simple front-end program to DKMXHF which reads in the matrix from a file & writes out the inverse to another file (DKMXHF not included). 2 and Theorem 11. A final reminder: the terms "dot product," "symmetric matrix" and "orthogonal matrix" used in reference to vectors or matrices with real number entries are special cases of the terms "inner product," "Hermitian matrix" and "unitary matrix" that we use for vectors or matrices with complex number entries, so keep that in. 2cholinv()— Symmetric, positive-definite matrix inversion Diagnostics The inverse returned by these functions is real if A is real and is complex if A is complex. (1) Any real matrix with real eigenvalues is symmetric. The article is still (or again) wrong. - For rectangular matrices of full rank, there are one-sided inverses. A Survey of Matrix Inverse Eigenvalue Problems Daniel Boley and Gene H. The calculator will perform symbolic calculations whenever it is possible. We will assume from now on that Tis positive de nite, even though our approach is valid. Proof: 1) Let λ ∈ C be an eigenvalue of the symmetric matrix A. A ij = (-1) ij det(M ij), where M ij is the (i,j) th minor matrix obtained from. (Report) by "Bulletin of the Belgian Mathematical Society - Simon Stevin"; Mathematics Decomposition (Mathematics) Research Functions, Inverse Inverse functions Least squares Matrices Matrices (Mathematics). The leading dimension can be the number of rows, or if A is a sub-matrix of a larger parent matrix, lda is the leading dimension (e. is a unitary matrix if its conjugate transpose is equal to its inverse , i. We consider the following two problems: to construct a real symmetric arrow matrix A and to construct a real symmetric tridiagonal matrix A, from a special kind of spectral. What is the inverse of a unitary matrix? Types of matrices (square) What can you say about the diagonal entries of a skewT Symmetric. I need to reverse it to get the stiffness matrix, K=F_inv, and then to obtain eigenvalues using K. Math 2940: Symmetric matrices have real eigenvalues The Spectral Theorem states that if Ais an n nsymmetric matrix with real entries, then it has northogonal eigenvectors. So we see that the inverse of a non-singular symmetric matrix is obtained by inverting its eigenvalues. An inverse of a real symmetric matrix should in theory return a real symmetric matrix (the same is valid for Hermitian matrices). 1) where X is an arbitrary real matrix of rank r. As a result you will get the inverse calculated on the right. The reason why it's not a vector space is simple: it does not have an additive identity (the zero matrix is not invertible). In this paper, we propose and discuss a class of inverse eigenvalue problems for real symmetric banded matrices with odd bandwidth. The inverse eigenvalue problem for real symmetric Toeplitz matrices is usually stated as follows: Find a real symmetric Toeplitz matrix Tm with given spectrum S(Tm) = {λ1 ≤ λ2 ≤ ··· ≤ λm}. For example, and. Commutative and anti-commutative matrices. Matrices as a rectangular array of real numbers, equality of matrices, addition, multiplication by a scalar and product of matrices, transpose of a matrix, determinant of a square matrix of order up to three, inverse of a square matrix of order up to three, properties of these matrix operations, diagonal, symmetric and skew-symmetric matrices and their properties, solutions of simultaneous linear equations in two or three variables. Fast Computation of Moore-Penrose Inverse Matrices P. Let A be an n x n matrix. Therefore, a general complex (respectively, real) matrix is positive definite iff its Hermitian (or symmetric) part has all positive eigenvalues. To calculate inverse matrix you need to do the following steps. Some authors also call a real non-symmetric matrix positive definite if x H Ax > 0 for all non-zero real x; this is true iff its symmetric part is positive definite (see below). of an n£n symmetric nonnegative matrix. However, if A is a symmetric matrix with real entries, then the roots of its charac-teristic equation are all real. BANDR Reduces real symmetric band matrix to symmetric tridiagonal matrix and, optionally, accumulates orthogonal similarity transformations. This course is on Lemma: http://lem. The eigen values of a real symmetric matrix are all real. To calculate inverse matrix you need to do the following steps. ALGLIB package has routines for inversion of several different matrix types, including inversion of real and complex matrices, general and symmetric positive definite ones. Related problems are also considered, such as when such a matrix R can be extended to a higher-dimensional real symmetric positive-definite Toeplitz matrix whose inverse is an M-matrix or, under less restrictive conditions on R, when only its Cholesky factors are inverses of M-matrices. The set of all invertible matrices is not a vector space, and therefore its dimension is undefined. Center for Wave Phenomena. TRED3-S Reduce a real symmetric matrix stored in packed form to symmetric tridiagonal matrix using orthogonal transformations. Definition 4. The general proof of this result in Key Point 6 is beyond our scope but a simple proof for symmetric 2×2 matrices is straightforward. The sum of two symmetric matrices is a symmetric matrix. Also, if eigenvalues of real symmetric matrix are positive, it is positive definite. You can factorize the matrix using LU or LDLT factorization algorithm. The eigenvectors corresponding to distinct eigenvalues of a real symmetric matrix have a special property, as given in the next theorem. The eigenvalues of a symmetric matrix with real elements are always real. Eigenvectors (of a matrix) corresponding different eigenvalues are orthogonal; C. The letter i is the imaginary unit, i2 = −1. sign an integer. Major Axis of an Ellipse. The Relation between Adjoint and Inverse of a Matrix. Eigenvalues and eigenvectors How hard are they to find? For a matrix A 2 Cn⇥n (potentially real), we want to find 2 C and x 6=0 such that Ax = x. For a real skew-symmetric matrix the nonzero eigenvalues are all pure imaginary and thus are of the form iλ 1, −iλ 1, iλ 2, −iλ 2, … where each of the λ k are real. vectors: either a p * p matrix whose columns contain the eigenvectors of x, or NULL if only. Anti-symmetric matrices are commonly called as skew-symmetric matrices. The inverse of a matrix $$A$$ is defined as a matrix $$A^{-1}$$ such that the result of multiplication of the original matrix $$A$$ by $$A^{-1}$$ is the identity matrix $$I:$$ $$A{A^{ - 1}} = I$$. A complex symmetric matrix doesn't necessarily have real eigenvalues, as the article currently states in the Decomposition section. It's going to be eight minus positive 10, eight minus positive 10, which would be negative two. Symmetric matrices are also called selfadjoint. Hermiteness generalizes the notion of symmetricness. I have a flexibility matrix (20*20), F, which is symmetric and positively defined. LA_PPTRI computes the inverse of real symmetric / complex Hermitian positive definite matrix in packed storage format using the Cholesky factorization computed by LA_PPTRF. We can see that taking the determinant of the Hessian gives the formula for the discriminant. In other words, we can say that matrix A is said to be skew-symmetric if transpose of matrix A is equal to negative of Matrix A i. And the second, even more special point is that the eigenvectors are perpendicular to each other. (a)A matrix with real eigenvalues and real eigenvectors is symmetric. In practical work, it is commonly assumed to be better not to form Hessenberg. REAL routines for symmetric matrix. Symmetric eigenvalue problems are posed as follows: given an n-by-n real symmetric or complex Hermitian matrix A, find the eigenvalues λ and the corresponding eigenvectors z that satisfy the equation Az = λ z (or, equivalently, z H A = λ z H). Ik denotes the k k identity matrix. So why do we proceed in two stages? Why don't we just perform the iterative technique on the original matrix? Simply put, the answer is efficiency. 1 Variational Characterizations of. Consider the matrix equation. It is now known in theory that symmetric Toeplitz matrices can have arbitrary real spectra. Let the system have the following. Source code for many Applied Statistics Algorithms is available through STATLIB. bers can be the spectrum of a certain n × n real symmetric regular Toeplitz matrix. Solving additive inverse eigenvalue problems for symmetric matrices by the homotopy method, IMA J. t are cosine inverse of the x matrix. The first problem we consider is the Jacobi Inverse Eigenvalue Problem (JIEP): given some constraints on two sets of reals, find a Jacobi matrix J (real, symmetric, tridiagonal, with positive off-diagonal entries) that admits as spectrum and principal subspectrum the. now suppose that a real, symmetric matrix A has an eigenvalue of (algebraic) multiplicity m. The Godunov{Inverse Iteration: A Fast and Accurate Solution to the Symmetric Tridiagonal Eigenvalue Problem Anna M. To calculate inverse matrix you need to do the following steps. inverse of a diagonal matrix (D) is really simple. And the second, even more special point is that the eigenvectors are perpendicular to each other. The Relation between Adjoint and Inverse of a Matrix. Here we describe in broad terms the Householder/ algorithm for real symmetric matrices. The Hessian matrix of is a -matrix-valued function with domain a subset of the domain of , defined as follows: the Hessian matrix at any point in the domain is the Jacobian matrix of the gradient vector of at the point. , if A = 0 23 2 01 3 10 - --then At = 02 3 20 1 31 0 -- - = = = -A Since At = –A, therefore A is a skew-symmetric matrix. Online course for college & university students. Calculates the eigen decomposition of a real symmetric matrix. You first need to know general definition of additive inverse in order to understand additive inverse of matrix. A positive definite matrix is a symmetric matrix with all positive eigenvalues. (ii) If A = 21 3 12 2 31 3 - , then A t = 2 13 1 21 3 23 - , ≠ A Hence A is not a symmetric matrix. A matrix $A$ represents a linear transformation of an $n$-dimensional vector space to an $m$-dimensional one. The following properties due to Penrose characterize the pseudo-inverse of a matrix, and give another justification of the uniqueness of A: Lemma 11. Orthogonally Diagonalizable Matrices These notes are about real matrices matrices in which all entries are real numbers. Sachs (GMU) Geometric spectral theorem proof January 2011 1 / 21. It is positive definite (denoted M ≻0) if zTMz > 0 for all nonzero z ∈Rd. For postive definite matrices, this is equivalent to the Cholesky formulation discussed above, with the standard Cholesky lower triangular factor given by. The eigenvalues of a symmetric matrix, real--this is a real symmetric matrix, we--talking mostly about real matrixes. These two conditions can be re-stated as follows: 1. I have a 3x3 real symmetric matrix, from which I need to find the eigenvalues. A Hermitian (or symmetric) matrix is positive definite iff all its eigenvalues are positive. invsym(A) and invsym(A, order) do the same thing as invsym(A) and invsym(A, order). The following statements are equivalent: (i) A is symmetric; (ii) There exists an orthonormal basis for Rn consisting of eigenvectors of A; (iii) There exists an orthogonal matrix P such that PtAP is diagonal. Proof: 1) Let λ ∈ C be an eigenvalue of the symmetric matrix A. In this paper, motivated by the preceding considerations, we introduce a recursive method for the inversion of a k kblock matrix Awith square blocks of order b. De nition 2. We consider the problem of inverse kinematics (IK), where one wants to find the parameters of a given kinematic skeleton that best explain a set of observed 3D joint loc. A (not necessarily symmetric) real matrix A satisfies x H Ax > 0 for all non-zero real x iff its symmetric part B=(A+A T)/2 is positive definite. If A is an n × n matrix we denote the entry of A in row j and column k by Ajk. There's a world of difference between positive definite and positive semidefinite. Matrix calculator. Prove that for any matrix A, ATA is symmetric. For a real skew-symmetric matrix the nonzero eigenvalues are all pure imaginary and thus are of the form iλ 1, −iλ 1, iλ 2, −iλ 2, … where each of the λ k are real. Positive Definite Matrix By definition, a real symmetric matrix A is positive definite if the real quadratic function defined by A, i. Any power An of a symmetric matrix A (n is any positive integer) is a symmetric matrix. Inverse (Compact) – Finds the inverse of a square matrix, if it exists. ASA007 is Applied Statistics Algorithm 7. Proof: Assume. Symmetric Matrices There is a very important class of matrices called symmetric matrices that have quite nice properties concerning eigenvalues and eigenvectors. Select direct or inverse transform. i ' s are nonzero real numbers, the inverse eigenvalue problem for a. A diagonal matrix if the was symmetric. Proof: Let A be an n×n matrix. 18 (1997), 1722-1736 Abstract. All eigenvalues of a symmetric matrix are real numbers B. xTAx, is always positive except for x = 0. Otherwise, A is called singular or noninvertible. I need to reverse it to get the stiffness matrix, K=F_inv, and then to obtain eigenvalues using K. For an odd 2p+1 with a positive integer p, the problem is. Transpose of Matrices. We present a real symmetric tridiagonal matrix of order whose eigenvalues are which also satisfies the additional condition that its leading principle submatrix has a uniformly interlaced spectrum,. Most relevant problems: I A symmetric (and large). What a matrix mostly does is to multiply. For complex matrices we would ask A∗ = AT = A. Real Statistics Using Excel Everything you need to do real statistical analysis using Excel. SGECOComputes LU factorization of real general matrix and estimates its condition. De nition 2. We consider the problem of inverse kinematics (IK), where one wants to find the parameters of a given kinematic skeleton that best explain a set of observed 3D joint loc. 1 (Spectral Theorem ). those where A = A T), it turns out that all the eigenvalues are real. Therefore, you could simply replace the inverse of the orthogonal matrix to a transposed orthogonal matrix. MIT Linear Algebra Exam problem and solution. I know it works for R^2. We can exploit the structure of a real, positive definite, symmetric matrix by using the Cholesky decomposition to compute the inverse. On exit, if INFO = 0, the (symmetric) inverse of the original matrix, stored as a packed triangular matrix. I have a flexibility matrix (20*20), F, which is symmetric and positively defined. I have always found the common definition of the generalized inverse of a matrix quite unsatisfactory, because it is usually defined by a mere property, \$A A^{-} A. of the complex symmetric tridiagonal T resulted from the first stage is com-puted. Major Axis of a Hyperbola. Simple idea that multiplying by a number's multiplicative inverse gets you back to one. samefeedbackmatrixis indicated. , if , then. Unitary matrix. For a symmetric matrix A = A T. The inverse of a 2x2 is easy compared to larger matrices (such as a 3x3, 4x4, etc). A matrix P is said to be a permutation matrix if exactly one entry in each row and column is equal to 1 and all other entries are 0. The notion contains those of predistance matrix and Euclidean distance matrix as its special cases. The standard MATLAB inv function uses LU decomposition which requires twice as many operations as the Cholesky decomposition and is less accurate. Matsekh a;1 aInstitute of Computational Technologies, Siberian Branch of the Russian Academy of Sciences, Lavrentiev Ave. 16 March, 1990 To appear: SIAM Journal on Scientific and Statistical Computing (NASA-CR-185983) AN IMPROVEO NEWTON ITERATION FOR THE GENERALIZED INVERSE OF A HAT£1X, WITH APPLICATIO'_S (Research Inst. ) With a square, symmetric matrix, the transpose of the matrix is the original matrix. A square matrix A is called self-adjoint of Hermitian if \( {\bf A}^{\ast} = {\bf A}. the eigenvalues of A) are real numbers. Eigenvectors (of a matrix) corresponding different eigenvalues are orthogonal; C. The eigenvalue of the symmetric matrix should be a real number. 1 Definition The pseudo inverse (or Moore-Penrose inverse) of a matrix A is the matrix A+ that fulfils I AA+A = A II A+AA+ = A+ III AA+ symmetric IV A+A symmetric The matrix A+ is unique and does always exist. See an example below, and try the Pivot Engine when you check your pivoting skills. lapack includes functions for solving dense sets of linear equations, for the corresponding matrix factorizations (LU, Cholesky, LDL T), for solving least-squares and least-norm problems, for QR factorization, for symmetric eigenvalue problems, singular value decomposition, and Schur factorization. The inverse of an invertible Hermitian matrix is also Hermitian, i. For an odd 2 p + 1 with a positive integer p, the problem is to construct an n × n real symmetric banded matrix with bandwidth 2 p + 1 whose m × m leading principal submatrix is a given m × m real symmetric banded matrix with bandwidth 2 p + 1 and spectrum is a. For details, please see standard texts in numerical methods. Inverse matrix of positive-definite symmetric matrix is positive-definite - Problems in Mathematics 05/01/2017. A complex symmetric matrix doesn't necessarily have real eigenvalues, as the article currently states in the Decomposition section. For convenience we assume n ≥ k (otherwise consider MT). Matsekh a;1 aInstitute of Computational Technologies, Siberian Branch of the Russian Academy of Sciences, Lavrentiev Ave. We will assume from now on that Tis positive de nite, even though our approach is valid. ALGLIB package has routines for inversion of several different matrix types, including inversion of real and complex matrices, general and symmetric positive definite ones. The code is written in the form of a generic package and covers the Lapack routines for - Matrix determinant and inverse on general matrices, - Eigenvalues and eigenvectors of general, real and hermitian symmetric matrices, - Solutions of systems of equations for general, real and hermitian symmetric. In this method, the inverse of a matrix is calculated by finding the transpose of the cofactor of that matrix divided by the determinant of that matrix. This paper modifies the GLR theory for the special application to real symmetric quadratic matrix polynomials, mathcal{Q}(λ)=M λ^{2} + C λ + K , M nonsingular, subject to the specific restriction that all matrices in the representation be real-valued. The following properties due to Penrose characterize the pseudo-inverse of a matrix, and give another justification of the uniqueness of A: Lemma 11. In linear algebra, a real symmetric matrix represents a self-adjoint operator over a real inner product space. The Real Nonnegative Inverse Eigenvalue Problem (RNIEP) asks when is a list σ = (λ 1 ,λ 2 ,,λ n ) consisting of real numbers the spectrum of an n ×n nonnegative matrix A. Inverse of a Real Symmetric Matrix? Is the inverse of any symmetric real matrix is a symmetric real matrix. A geometric proof of the spectral theorem for real symmetric matrices Robert Sachs Department of Mathematical Sciences George Mason University Fairfax, Virginia 22030 [email protected] symmetric inverse M-matrix completion problem: 1) A pattern (i. those where A = A T), it turns out that all the eigenvalues are real. Therefore x T Mx = 0 which contradicts our assumption about M being positive definite. Shalaby * Internationa] Centre for Theoretical Physics, Trieste, Italy. Applying the transposition operator to each side of the equation we get. And the second, even more special point is that the eigenvectors are perpendicular to each other. | 2020-01-26T15:43:22 | {
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http://lotoblu.it/hbdv/parametric-surface-grapher.html | Parametric Surface Grapher
how to determine the color of curves and surfaces. Generating parallel curves on parametric surfaces is an important issue in many industrial settings. parametric_plot3d (f, urange, vrange=None, plot_points='automatic', boundary_style=None, **kwds) ¶ Return a parametric three-dimensional space curve or surface. plot graph + Manage Tags. (1) Quadratic surfaces are also called quadrics, and there are 17 standard-form types. Look below to see them all. Area Under the Parametric Curve; Parametric Area Under One Arc or Loop; Parametric Curve: Surface Area of Revolution; Surface Area of Revolution of a Parametric Curve Rotated About the y-axis; Parametric Arc Length; Parametric Arc Length and the distance Traveled by the Particle; Volume of Revolution of a Parametric Curve; Converting Polar Coordinates. Get access to all the courses and over 150 HD videos with your subscription. We start by plotting two simple quadratic surfaces that are commonly taught in multivariate calculus: a cone and a Hyperboloid of one sheet. Math · Multivariable calculus · Integrating multivariable functions · Surface integral preliminaries (videos) Surface integral preliminaries (videos) This is the currently selected item. 1 was too big. Parametric Surfaces (3-D Graphing) Graphs surfaces in three dimensions specified parametrically, as x=f(s,t), Rotate the graph by clicking and dragging the mouse on the graph. Display parameters 1. The intermediate surface is defined by H(x, y, z) = t * F(x, y, z) + (1 - t) * G(x, y, z) = 0. is called a parametric surface. In most cases, we'll assume that f, g, and h are continuous, or perhaps even differentiable; we'll discuss continuity and differentiability for functions of several variables later. See more about the Examples menu in Section 4. And yes those are the standard functions. To learn more about the parametric curves shown here, click me. H&B Figure 10. It provides the broadest range of powerful yet flexible 3D CAD capabilities to accelerate the design of parts and assemblies. The tropics of Cancer and Capricorn have latitude approximately 23:5 ˇ0:13ˇ. TI-Nspire uses the notation x1(t) for the first x-component, x2(t) for the second x-component, and so on. In this grasshopper definition you can model a parametric creased surface by changing the scale of the base curve and also using different graphs in Graph mapper. Use a 3D parametric plotter to generate the top half a sphere. ') The result is this: Unfortunately, by plotting individual points, we lose the sense of depth and curvy-ness when we look at this picture. Free Simple Online Graph Plotter downloads. We'll end with a parametrization that. Setting a=2 will show what the graph looks like when y is kept constant at 2, and shows the change in x. distribute along curve with graph mappers. Use functions sin (), cos (), tan (), exp (), ln (), abs (). Plotting Surfaces with Grapher Warren Weckesser September 9, 2005 These brief notes explain how to plot surfaces with Grapher, an application that is avail-able on the computers in the Computer Lab. Illustration of how the graph of a scalar-valued function of two variables is a surface. In other words, point x is on the surface if and only if the relationship F(x) = 0. Since the graph looks a little rough, we can have Sage plot more points for the parametric surface so that it is smoother: u,v=var('u,v') parametric_plot3d((u*cos(v),u*sin(v),v),(u,0,3),(v,0,4*pi),mesh=True,plot_points=200). Large collection of specific objects. To draw a parametric graph it is easiest to make a table and. Because xand yare restricted to the circle of radius 3 centered at the origin, it makes sense to use polar coordinates for xand y. One of the major current challenges in this area is to describe the variability of human cortical surface form and its implications for individual differences in neurophysiological functioning. Type the equations that define the graph. 78 CHAPTER 12. There is also SpaceCurve program by the author. Tangents, Areas, Arc Lengths, and Surface Areas As we saw in the previous section, we can use parametric equations to describe curves that aren’t graphs of the form y = f(x). py, which is not the most recent version. If you are determined to have a parametric equation with just. Here is the wikipedia entry that summarizes both. Example The graph of y = sin x on 0 ≤ x ≤ π is revolved around the x-axis. I have to write a script to do it but it works pretty good. Graphing a plane curve represented by parametric equations involves plotting points in the rectangular coordinate system and connecting them with a smooth curve. In addition, the cone consisting of all tangents from a fixed point to a quadratic surface cuts every plane in a. USACAS 8 International Conference Philadelphia, PA, USA March 8-10, 2013. Graphing Parametric Curves and Surfaces in Space with the MPP. For this exploration, we will be primarily considering equations of x and y as functions of a single parameter, t. Fe Tkgooie Utilities 3dtools Group 3d Examiner For. 2D Basic Plots. It is also possible to do some mathematical calculations on the functions. Plot implicit and parametric equations, add variables with sliders, define series and recursive functions. Trust me, this is the better way to learn these commands. Functions 3D Plotter is an application to drawing functions of several variables and surface in the space R3 and to calculate indefinite integrals or definite integrals. org for more info. Question: How do I plot parametric surfaces? Posted: blurrymadness 16 Product: Maple. Description. We start by plotting two simple quadratic surfaces that are commonly taught in multivariate calculus: a cone and a Hyperboloid of one sheet. Pixelized Sphere In this grasshopper definition by exploding a mesh sphere and by using the Pufferfish plugin you can extrude the faces towards their normals controlled by a sine wave. Get the free "Parametric Surface Plot" widget for your website, blog, Wordpress, Blogger, or iGoogle. Thus, parametric equations in the xy -plane. The surface at the right, whose technical name is "torus," is an example. Parametric functions allow us to calculate (using integration) both the length of a curve and the amount of surface area on a given 3-dimensional curve. Parametric surfaces or, the cross-product and surface integrals On theWikipedia page for parametric surfaces, we see the lovely formula Z S 1dS= Z Z D kr u r vkdudv: This states that the area of a parametrized surface can be computed as the double integral of the magnitude of a normal vector to the surface. We will also see how the parameterization of a surface can be used to find a normal vector for the surface (which will be very useful in a couple of sections) and how the parameterization can be used to find the surface area of a surface. Add to graph: Function: z=f(x,y) Space Curve: r(t) Vector Field Point: (x, y, z) Vector: Text Label Implicit Surface Parametric Surface Region Slider ────────── Function: r=f(θ,z) Function: z=f(r,θ) Function: ρ=f(θ,φ) Function: x=f(y,z) Function: y=f(x,z). Maths Geometry Graph plot surface This demo allows you to enter a mathematical expression in terms of x and y. org for more infoget. (b) Eliminate the parameters to show that the surface is an elliptic paraboloid and set up another double integral for the surface area. Do a google search on parametric surfaces and you might find a few sites that explain it. 1 Implicit representations of surfaces An implicit representation takes the form F(x) = 0 (for example x2 +y2 +z2 r2 = 0), where x is a point on the surface implicitly described by the function F. Each graph gives control over domains and discritization (resolution). Graph lines, curves, and relations with ease. (J Comput Graph Stat 21. For example, the graph of y = sin(x) is given parametrically by u(t) = (t,sin(t)). Parameterized Knots. parametric_plot3d. Learn more about 3d plots, parametric, surface, ezsurf. Text: Parametrizing. 3D and Contour Grapher. Color standard z-axial custom. Parametric Function Plotter Description| How it works| Gallery This app plots the parametric equation specified by x = f(t) and y = g(t). Surface( , , ) Creates a surface of revolution, rotating the Curve from 0 to given Angle around the Line. Graphing Parametric Curves and Surfaces in Space with the MPP. Find parametric equations for the surface obtained by rotating the curve x = 1/y, y ⩾ 1, about the y-axis and use them to graph the surface. Surface Plotter renders surfaces in real-time in the browser. You can also use "pi" and "e" as their respective constants. One common form of parametric equation of a sphere is: #(x, y, z) = (rho cos theta sin phi, rho sin theta sin phi, rho cos phi)# where #rho# is the constant radius, #theta in [0, 2pi)# is the longitude and #phi in [0, pi]# is the colatitude. When you first learned parametrics, you probably used t as your parametric variable. In Curve Length and Surface Area, we derived a formula for finding the surface area of a volume generated by a function $$y=f(x)$$ from $$x=a$$ to $$x=b,$$ revolved around the x-axis: \[S=2π∫^b_af(x)\sqrt{1+(f′(x))^2}dx. Recall the problem of finding the surface area of a volume of revolution. #N#The parametric equations of a line. 3D Parametric Equation Plotter ----- REQUIREMENTS: VMD Version 1. asked by Paul on August 8, 2012; Calculus. An interactive plot of 3D vectors. parametric_plot3d (f, urange, vrange=None, plot_points='automatic', boundary_style=None, **kwds) ¶ Return a parametric three-dimensional space curve or surface. 8 are portions of the same parabola. This example shows Chebfun2 being used to represent parameterized surfaces. To begin, let's take another look at the projectile represented by the parametric equations and as shown in. See more about the Examples menu in Section 4. One final technique for sketching and analyzing the graph of a polar equation is finding the intercepts of the graph; that is, where it intersects the lines θ = 0 and θ =. A parametric surface is a function with domain $$\mathbb{R}^2$$ and range $$\mathbb{R}^3$$. 3D Grapher is small, fast, flexible, and reliable. 3D Vector Plotter. Creo Parametric is the standard in 3D CAD software. By adjusting the parametric equations, we can reverse the direction that the graph is swept. The idea of parametric equations. The surface plot then connects these points forming a three-dimensional surface view of the data. Vector fields on surfaces: attach a vector field to a surface, to show what. org for more info. #N#ColorFunctionScaling. Basic 3D Surface Example using SURF. Algebrus is a free 3D graphing software for Windows. Yes that is how a parametric equation is stated. Recall the standard parameterization of the unit circle that is given by. Viewing options of the 3D graphing calculator software tool are pretty basic, but sufficient - The plot can be. How to plot a surface in its parametric form ?. Of course, the parameters may be denoted by letters other than s and t. Parametric surfaces are surfaces parametrized typically by 2 variables, u and v. Then sketch a graph locating points at t = {0,1,2,3,4} and indicate the orientation of the curve. Note - a newer version of this tool is the Graphing Calculator. For each example, state the parameteri-zation that you would use and determine the bounds for the variables where appropriate. The developed method was employed in the investigation of surface-blending problems where the primary surfaces were expressed in parametric, implicit, and explicit forms. Example (5) [Lecture 6. Do a google search on parametric surfaces and you might find a few sites that explain it. Plot high quality graphs of mathematical equations and data with this easy-to-use software. A table of values of the parametric equations in Example 10. VTK has a bunch of classes that let users explore Parametric surfaces. for x-axis rotation:. Given an algebraic surface, it is commonly easier to decide if it is rational than to compute its rational parameterization, if it exists. Funcions 3D plotter calculates the analytic and numerical integral and too calculates partial derivatives with respect to x and y for 2 variabled functions. An initial set of intersection points is calculated by any kind of iterative method such as Timmer's[1]. Example: Identify the surface de ned by R~(u;v) = hucos(v);usin(v);ui: Note that the parametric equations satisfy z 2= x 2+ y or z = p x + y2. A surface plot shows the shape of a surface. This example shows Chebfun2 being used to represent parameterized surfaces. Because any position in the plane, and thus any position on the surface patch, can be uniquely given by two coordinates, the surface is said to. We can do the exact same thing with parametric surfaces to form a composite surface. Find the area of the surface. Given an initial curve (called the base curve or generator) on a parametric surface, the goal. All properties of the plot are editable, including the display of symbols and lines, axes, graph title, and graph background. yazdmich Jun 1st, 2013 81 Never Not a member of Pastebin yet? Sign Up, it unlocks many cool features! raw. And yes those are the standard functions. , folder) on your website. Plotting Surfaces with Grapher Warren Weckesser September 9, 2005 These brief notes explain how to plot surfaces with Grapher, an application that is avail-able on the computers in the Computer Lab. In Part 2, we saw that changes in coordinates for surfaces give rise to a scale factor for area that can be computed as the Jacobian of the coordinate transformation. When Cartesian coordinates of a curve or a surface are represented as functions of the same variable (usually written t ), they are called the parametric equations. Since we like going from left to right, put t = 0 at the point (2, 3). Anything you type will be. A surface plot shows the shape of a surface. Create AccountorSign In. Plotting a parametric 3D surface. The parametric expression of a circle is. d) surface generated by revolving the curve yx2 about the y-axis. 000 (TeX Live 2017) (kpathsea version 6. 3D Grapher is small, fast, flexible, and reliable. As you study As you study multi-variable calculus, you'll see that the idea of "surface area" can be extended to figures in higher dimensions, too. Graphing parametric equations is as easy as plotting an ordered pair. Parametric Equations of Ellipses and Hyperbolas. For problems 1 - 6 write down a set of parametric equations for the given surface. An online tool to create 3D plots of surfaces. Find for the parametric equations 7. This functionality is also available in Mayavi. Parametric Surfaces and their Areas Video. A particular example of a parameterized surface is a graph: z= f(x;y); or X(x;y) = xi + yj + f(x;y)k Ex The sphere x 2+y +z2 = r can be parameterized using spherical coordinates: x= rsin˚cos ; x= rsin˚sin ; z= rcos˚; 0 <2ˇ; 0 ˚ ˇ It can however, not be written as one graph, but one for the southern hemisphere z= p r2 x2 y2 and one for the. The plane $$7x + 3y + 4z = 15$$. Thus, parametric equations in the xy -plane. You could just change it, or even better, use pgfplots for your task, which is what I've demonstrated below. They will make you ♥ Physics. Deprecated: Function create_function() is deprecated in /www/wwwroot/mascarillaffp. Basically, I’m trying to make a corresponding definition for every most-frequently used Rhino command when designing cars (i. Display parameters 1. Here is the wikipedia entry that summarizes both. Add to graph: Function: z=f(x,y) Space Curve: r(t) Vector Field Point: (x, y, z) Vector: Text Label Implicit Surface Parametric Surface Region Slider ────────── Function: r=f(θ,z) Function: z=f(r,θ) Function: ρ=f(θ,φ) Function: x=f(y,z) Function: y=f(x,z). Compare the two graphs. Parametric surface. expression to evaluate at every function. Especially if you're a student, teacher or engineer, this app is made with you in mind! A wide range of predefined functions is available, including trigonometric & hyperbolic functions, polar coordinates, differentiation and more. Cortical Surface Reconstruction. Extract the end and start points of the two curves, and check whether they coincide. Graphing 2D Parametric Equations. The meshgrid command is vital for 3D surfaces! Defining the domain here is even trickier than for 2D. Education If you're an educator or student seeking function graphing software, you'll find TeraPlot an excellent tool for demonstrating and exploring many mathematical concepts in 2D and 3D. Each graph and chart includes many options that you can use to customize appearance, convey more information, or highlight data. Grapher allows continuous parameters only in the context of parametric equations. Viewed 434 times 1 $\begingroup$ Is it possible to plot several parametric surfaces simultaneously using Maxima? Browse other questions tagged graphing-functions parametric math-software maxima-software or ask your own. Now that we have introduced the concept of a parameterized curve, our next step is to learn how to work with this concept in the context of calculus. In Mathcad, the equation for a parametric surface is: For those who love spirograph may find it familiar. plot3d(x,y,z,[theta,alpha,leg,flag,ebox]) draws the parametric surface z=f(x,y). In spherical coordinates, parametric equations are x = 2sinϕcosθ, y = 2sinϕsinθ, z = 2cosϕ. A bit confusing. Visual Calculus is a powerful tool to compute and graph limit, derivative, integral, 3D vector, partial derivative function, double integral, triple integral, series, ODE etc. K3DSurf use parametric descriptions of it's physical models. (a) Describe the surface z = x2y2+2 over the region D : 1 x 1, 1 y 1 as a parametric surface. Graphing Functions (3-D) Graphs up to three functions of two variables in rectangular coordinates. 15 ANNA UNIVERSITY CHENNAI : : CHENNAI – 600 025 AFFILIATED INSTITUTIONS B. 3D graphs are also supported. Parametric three dimensional array. For the first case we need to supplement the equations by two inequalities: 0 <= t <= 4 Pi && 0 < x < 4 Pi. It is not difficult to show that the curves in Examples 10. Graph lines, curves, and relations with ease. Here's a simple sphere where x, y, and x are dependent on two paramters theta and phi. parametric graphing. Hrinyaaw- if you mean you would like to see a point on the curve traced out, I usually just copy and paste the parametric line, then changed all my "t"s to "a"s and add a slider for "a". For example, you could use a surface plot for terrain mapping. is a pair of parametric equations with parameter t whose graph is identical to that of the function. Finding only those arguments t where the parametric graph passes appropriate intersection points the second time. Functions that have a two-dimensional input and a three-dimensional output can be thought of as drawing a surface in three-dimensional space. It is also possible to do some mathematical calculations on the functions. Bar graph is the histopathological severity score of lung sections. The graph of this is part of a parabola, starting at (0,0,0) and extending to (20,0,400), as shown. Math · Multivariable calculus · Integrating multivariable functions · Surface integral preliminaries (videos) Surface integral preliminaries (videos) This is the currently selected item. x(t) = 3cos(2t), y(t) = sin(2t), z(t) = t/2 (an elliptical helix) proceed as follows: Step 1. Fe Tkgooie Utilities 3dtools Group 3d Examiner For. Instructions for the 3D Bezier curve. Note that the graph is a surface, in other words, a two-dimensional geometric object sitting in three-space. Parametric surfaces: ezmesh, ezsurf [See Section 10. I described a surface as a 2-dimensional object in space. Recommended for you. Parametric curve plotter. Duncan BS(1), Olson AJ. In this grasshopper definition you can model a parametric creased surface by changing the scale of the base curve and also using different graphs in Graph mapper. Here is a list of best free 3D graphing software for Windows. The domain of the parametric equations is the same. The parameterization is. Answer to (a) Show that the parametric equations x = x1 + (x2 – x1)t y = y1 + (y2 – y1) t where 0 < t < 1, describe the line segment that joins the points P1(x1, y1) Toggle navigation Menu Tutors. Better yet, use the script recorder to transform actions performed in Grapher into a script. Look below to see them all. Given an initial curve (called the base curve or generator) on a parametric surface, the goal. End Value must be greater than or equal to Start Value and both must be finite. There is also SpaceCurve program by the author. Common Parametric Surfaces Here is a list of common surfaces and a (general) parameterization. In particular, there are standard methods for finding parametric equations of. If for some θ, r = 0, the graph intersects the pole. K3dsurf 3d Surface Generator. Learn more about surface, 3d plots, parameters, plotting, parametric equation. Parametric Surfaces in Matlab 219. How To Graph 3d Functions Of Two Variables On Wolfram Alpha. To begin, let’s take another look at the projectile represented by the parametric equations and as shown in. or to see how the input is generated, view the source code of this web page. The parametric method of representing surfaces/curves uses a function to map some portion of R2 (the domain) to a patch of the surface in R3. A parametric equation is where the x and y coordinates are both written in terms of another letter. The 3D grapher can graph 3D functions, parametric curves (spacecurves), parametric surfaces, implicitly defined surfaces, implicit spherical surfaces, implicit cylindrical surfaces, 3D vector fields. It lets you plot 3D parametric line and 3D surface graphs. Alternatively, you can use parametric_plot3d to graph a parametric surface where each of $$x, y, z$$ is determined by a function of one or two variables (the parameters, typically $$u$$ and $$v$$). Parametrize the line that goes through the points (2, 3) and (7, 9). Graph Parameterized Surfaces Using 3d Calc Plotter. This is the parametrization for a flat torus in 4D. 000 (TeX Live 2017) (kpathsea version 6. set title "fence plot constructed with separate parametric surfaces" # A method suggested by Hans-Bernhard Broeker # : display a separate parametric # surface for each fence. This example shows Chebfun2 being used to represent parameterized surfaces. SingSurf home; Jep. Click on the below graph groupings to see the different plots Grapher creates. Stay on top of important topics and build connections by joining Wolfram Community groups relevant to your interests. Continuous Parameters, i. Great Math For Curves And Surfaces. For the Love of Physics - Walter Lewin - May 16, 2011 - Duration: 1:01:26. It basically requires a grid coordinates with a numeric variable attributed to each position: its height. Give reasons! (All surfaces have been graphed with the usual orientation of looking at the axes from the rst octant. Learn more about 3d plots, parametric, surface, ezsurf. End Value must be greater than or equal to Start Value and both must be finite. In a parametric surface, x, y, and z are each given by separate functions of the parameters u and v. of the surface with parametric equations x = au cos v, y = bu sin v, z = u2, 0 ~ u ~ 2, 0 ~ v ~ 27T. Fe Tkgooie Utilities 3dtools Group 3d Examiner For. Using parametric equations enables you to investigate horizontal distance, x, and vertical distance, y, with respect to time, T. The 3D graphs use. The competition expects you to create/develop complex code to form a chart via a composition of many objects that are not or minimally intelligently controlled to behave together unless you provide and test such code. Generating parallel curves on parametric surfaces is an important issue in many industrial settings. The surface with parametric equations x = u 2 , y = uv , z = v 2 , 0 ⩽ u ⩽ 1, 0 ⩽ v ⩽ 2 The following membership graph. A surface in can be represented mathematically in a number of different ways. graphing parametric equations of lines in 3D. Functions 3D Plotter is an application to drawing functions of several variables and surface in the space R3 and to calculate indefinite integrals or definite integrals. You can plot Points, Vectors, Planes, Equations and Functions, Cylinders, Parametric Equations, Quadric Surfaces, etc. Note - a newer version of this tool is the Graphing Calculator. The Home tab New Graph commands or the graph wizard are used to create a graph. 4 Parametric Surfaces in Matlab In this section we will again examine surfaces in three-space. , folder) on your website. 8 or greater DESCRIPTION: The draw_surface proc generates sample points on a parametric surface defined by three two-variable functions 'f', 'g', and 'h'. Then sketch a graph locating points at t = {0,1,2,3,4} and indicate the orientation of the curve. Here is an example of a surface that is best represented in parametric form,. One final technique for sketching and analyzing the graph of a polar equation is finding the intercepts of the graph; that is, where it intersects the lines θ = 0 and θ =. Use functions sin (), cos (), tan (), exp (), ln (), abs (). Customer reviews: "Amazing app! I love it. Maths Geometry Graph plot surface. Parametric surfaces: ezmesh, ezsurf [See Section 10. It's a powerful feature that allows plotting complex graphs with 3 simple equations. Parametric Profiles Stair Tool Railing Tool Migration of Rendering/Surface Settings to CineRender Parametric Objects. To texture the material, Surface Plotter uses a custom GLSL shader which implements tri-planar mapping to generate UV coordinates, and parallax occlusion mapping to give the material depth. 5) Curve x=f(t), … Graphs a parametric space curve of the form x = f (t), y = g(t), and z = h(t), for explicit functions f, g and h of a single parameter t. Anything that can be graphed in Function mode on the TI-84 Plus an also be graphed as a set of parametric equations. We can also have Sage graph more than one parametric surface on the same set of axes. Topics covered include surface reparameterization, deconstructing and reconstructing surface domains, and modifying the domain to offset geometry from the edge of the surface. To determine To find: The parametric equations for the surface obtained by rotating the curve x = 1 y , y ≥ 1 , about the y -axis and to graph the surface. Cortical Thickness Estimation. The 4D parametric motion graph representation allows interactive real-time control of character animation from a database of captured mesh sequences. Ansys software can uniquely simulate electromagnetic performance across component, circuit and system design, and can evaluate temperature, vibration and other critical mechanical effects. Finding all arguments t in 0 <= t <= 4Pi where the parametric graph intersects. Outline • Parametric curves Cubic B-Spline Cubic Bezier • Parametric surfaces Parametric Surfaces. Williamson and H. The parametric method of representing surfaces/curves uses a function to map some portion of R2 (the domain) to a patch of the surface in R3. For the Love of Physics - Walter Lewin - May 16, 2011 - Duration: 1:01:26. Extract the end and start points of the two curves, and check whether they coincide. H&B Figure 10. The 3-D parametric surface graph is the most complex of the 3-D surface and the most difficult with which to work. The Rational Implicitization Theorem [17] states that if J = 〈dx - a, dy - b, dz - c, 1 - dw〉, then V(J ∩ R[x, y, z]) is the smallest variety in R 3 containing the parametric surface. Do a google search on parametric surfaces and you might find a few sites that explain it. Parmetric Surfaces in 3D. zip file (25 KB) How to use. When graphing, it is important to note the correct domain for t as well as the direction of the graph when t increases. The 3-D parametric surface graph is the most complex of the 3-D surface and the most difficult with which to work. Using Grasshopper to morph geometry from a reference volume to the UVW parametric space of a NURBS surface. org for more infoget. To begin, let's take another look at the projectile represented by the parametric equations and as shown in. Download the Flash player file surf_graph_sphere. Download Function Graphs Plotter APK latest version - ts. This adds a new dimension to your graph! Setting the window Setting the window …. Graph lines, curves, and relations with ease. Of course, the parameters may be denoted by letters other than s and t. Question: Find parametric equations for the surface obtained by rotating the curve {eq}\displaystyle y = 25 x^4 - x^2,\ - 5 \le x \le 5 {/eq} about the x-axis, and use them to graph the surface. It has Many great features that beat my T83" "Best calculator. Parametric surfaces are particularly handy if you are unable to find any data to play with right away. ; (c) Use the parametric equations in part (a) with and to graph the surface. Apply the formula for surface area to a volume generated by a parametric curve. We often use vector notation to exhibit parametric surfaces. 5) Curve x=f(t), … Graphs a parametric space curve of the form x = f (t), y = g(t), and z = h(t), for explicit functions f, g and h of a single parameter t. Note that the graph is a surface, in other words, a two-dimensional geometric object sitting in three-space. gnuplot> set parametric dummy variable is t for curves, u/v for surfaces gnuplot> set size square gnuplot> set xrange [-1:1] gnuplot> set yrange. For each value of use the given parametric equations to compute and 3. whether to scale arguments to ColorFunction. Because any position in the plane, and thus any position on the surface patch, can be uniquely given by two coordinates, the surface is said to. Curves in $\mathbf R^3$ are given parametrically for three reasons. If you want to graph a parametric, just make each coordinate a function of "t". • Mathcad interprets these three matrices as the X-, y-, and z-coordinates of points on a surface and draws this surface from the viewing angle prescribed by the Rotation and Tilt. ; (c) Use the parametric equations in part (a) with and to graph the surface. These graphing program let you create graph for various mathematical equations, functions, data sets, inequalities , etc. It is based on a system’s response to varying an external parameter. furniture, structures, tools, windows, 2D objects, environment. 3 Parametric Equations Involving Sines and Cosines Sketch the plane curve defined by the parametric equations x = 2cost, y = 2sint, for (a) 0 ≤ t ≤ 2π and (b) 0 ≤ t ≤ π. Find the coordinates of the points at which the given parametric curve has a horizontal and/or a vertical tangent. Section 6-2 : Parametric Surfaces. An interactive plot of 3D vectors. Surface Plotter isn't terribly useful, but the app exists for the user to play with. Since we like going from left to right, put t = 0 at the point (2, 3). (b) Eliminate the parameters to show that the surface is an elliptic paraboloid and set up another double integral for the surface area. SingSurf home; Jep. Given an initial curve (called the base curve or generator) on a parametric surface, the goal. Be able to nd the equation of the tangent plane at a point of a parametric. Look below to see them all. Solution: 1. (d) For the case , , use a computer algebra. 78 CHAPTER 12. Then graph the rectangular form of the equation. Finding only those arguments t where the parametric graph passes appropriate intersection points the second time. Online 3-D Function Grapher. This is probably obvious, but if you are able to guess the u-v coordinates corresponding to each data point (simplest case u=x, v=y if the surface is a graph), parametric interpolation (u,v) -> (x,y,z) is essentially 2-D interpolation of 3 separate datasets (the x, y, and z coordinates), so you can use any usual 2-D interpolation method. • To zoom in or out, use the mouse scroll wheel or equivalant. In a parametric surface, x, y, and z are each given by separate functions of the parameters u and v. By adjusting the parametric equations, we can reverse the direction that the graph is swept. Model 1 is based on an inverted Elliptical Paraboloid function, and model 2 on a super-Gaussian function. Get the free "Parametric Curve Plotter" widget for your website, blog, Wordpress, Blogger, or iGoogle. 1 Point/implicit algebraic surface intersection5. An initial set of intersection points is calculated by any kind of iterative method such as Timmer's[1]. Ansys provides a model-based embedded software development and simulation environment with a built-in automatic code generator to. Example: Find a parametric representation of the cylinder x2 + y2 = 9, 0 z 5. 2D Basic Plots. It has Many great features that beat my T83" "Best calculator. Creo Parametric is the standard in 3D CAD software. For problems 1 - 6 write down a set of parametric equations for the given surface. Start the MPP3D program. 1); Listas: (0); Transferências: (0); RSS: ( ); Monitorizar preços. And yes those are the standard functions. xls file (93 KB) or. Since we used set trange [1:4] , the range of this truncated line is from 1 to 4. It's a powerful feature that allows plotting complex graphs with 3 simple equations. Find a parametrization of the the surface. Trimmed (Parametric) Surface Examples¶ The trimmed parametric surface element is a very important geometry item to get correct and running, it is used to remove patches from surfaces. Parmetric Surfaces in 3D. Answer to If a parametric surface given by r1(u,v)=f(u,v)i+g(u,v)j+h(u,v)k and −3≤u≤3,−5≤v≤5, has surface area equal t Skip Navigation Chegg home. Tangents, Areas, Arc Lengths, and Surface Areas As we saw in the previous section, we can use parametric equations to describe curves that aren’t graphs of the form y = f(x). This parameter will have value 0 at the beginning of an animation sequence. Be able to nd the equation of the tangent plane at a point of a parametric. Consider the surface given by S(u,v) = (uv,u2 +v2. Using Parametric Equations in SolidWorks, Example 1 (Draft 4, 10/25/2006, SW 2006) Introduction In this example the goal is to place a solid roller on a solid wedge. It solves the problem of 3D movement by allowing you to scroll through a cross-section of the block. Graphing 2D Parametric Equations. Rodrigo Platte, Feb 2013. Find the area of the surface of revolution obtained by rotating the. 5D, 3D Purpose graphs, animations and table graphs. A rational surface is an algebraic surface. Parametric Area is the area under a parametric curve. One way to represent a vector-valued function is to graph the parametric plot for , on the plane, the parametric plot for , on the plane, and the parametric plot for , on the plane. The domain of the parametric equations is the same. To add a parametric surface to the plot, select the option Parametric Surface on the Add to graph drop-down menu. It provides an Evaluator tool which lets you solve quadratic equations, cubic equations, polynomial derivatives, polynomial roots, etc. Click here to download this graph. 5) Curve x=f(t), … Graphs a parametric space curve of the form x = f (t), y = g(t), and z = h(t), for explicit functions f, g and h of a single parameter t. Monthly, Half-Yearly, and Yearly Plans. com/xrtz21o/f0aaf. Download the content Only Parametric House Users can download this content. However, just as every “nice” enough curve can be parameterized, so can every “nice” enough surface. Graph 3d Parametric Equations Tessshlo. ] In the previous example for the upper half-sphere, note the ragged edge where the sphere meets the xy-plane. Question: Find parametric equations for the surface obtained by rotating the curve {eq}\displaystyle y = 25 x^4 - x^2,\ - 5 \le x \le 5 {/eq} about the x-axis, and use them to graph the surface. Parametric surfaces in 4D. Parametric Torus Geogebra. Just as a curve C in the xy-plane is not always the graph of y = f(x), a surface S in xyz-space may not be the graph of z = f(x,y). The "wireframe" represenation for surfaces, in which the surface is transparent, only draws one surface at a time. It lets you plot 3D parametric line and 3D surface graphs. We will also see how the parameterization of a surface can be used to find a normal vector for the surface (which will be very useful in a couple of sections) and how the parameterization can be used to find the surface area of a surface. So, let parametric curve is defined by equations x=f(t) and y=g(t). 5 of Stewart's text for an introduction to parametric surfaces. Parametric surface plotter with rotation - Turing. The idea of parametric equations. However, when you graph the ellipse using the parametric equations, simply allow t to range from 0 to 2π radians to find the (x, y) coordinates for each value of t. Here's a simple sphere where x, y, and x are dependent on two paramters theta and phi. This representation, speaking generally, can be written x = F(u;v), where uand vare surface parameters, and x is a point on the surface. The parametric method of representing surfaces/curves uses a function to map some portion of R2 (the domain) to a patch of the surface in R3. The plane $$7x + 3y + 4z = 15$$. How to plot a surface in its parametric form ?. Controlling scalar properties in two dimensional array by a surface. Find a parametrization of the the surface. Graphing a parametric surface in spherical coordinates. Example (5) [Lecture 6. Parametric Surfaces. Desenvolvedor: (Math Apps); Preço: (Grátis); Versão: (1. Basically, I’m trying to make a corresponding definition for every most-frequently used Rhino command when designing cars (i. Find the length of one arc of the cycloid 9. Given an initial curve (called the base curve or generator) on a parametric surface, the goal. Instead of numerical coordinates, use expressions in terms of t, like (cos t, sin t). Unite Technology. It forms a special case of a surface. Using parametric equations enables you to investigate horizontal distance, x, and vertical distance, y, with respect to time, T. We will often start at $$t=0$$ and increase t, giving the idea that time is passing. Rodrigo Platte, Feb 2013. As we have done before with worksheets: Please work through all the code and examples in this worksheet. The surface at the right, whose technical name is "torus," is an example. set zrange [-1:1] unset label unset arrow set xrange [-5:5]; set yrange [-5:5] set arrow from 5,-5,-1. Compare the two graphs. I figured out how to graph a line using a point and direction vector now I am hoping someone can help me graph the line using its parametric from (ie x=3-4t, y=2+5t, z=6-t where the given point is (3,2,6) and the direction vector is [-4,5,-1]. Free online 3D grapher from GeoGebra: graph 3D functions, plot surfaces, construct solids and much more!. Recall the problem of finding the surface area of a volume of revolution. Then graph the rectangular form of the equation. Parametric Equations: x = 2 - t, y = 5 − 4 t + t 2. Now we consider a parameterization of the torus pictured above before step 1. These lines correspond to the x and y axes in the rectangular coordinate system. Good Grapher™ - great and very powerful scientific graphing calculator. That is, the z-value is found by substituting in both an x-value and a y-value. Illustrates the setup of a scene, camera, renderer, event handlers (for window resize and fullscreen, provided by the THREEx library), mouse controls to rotate/zoom/pan the scene, mini-display for FPS stats, and setting up basic geometries: a sphere with lighting effects, a multi-colored cube, a plane with an image texture applied, and skybox. A finite element method for surface diffusion: the parametric case Eberhard B¨ansch∗ Pedro Morin† Ricardo H. One interesting feature is that users can drag a point inside a box, and the corresponding point on the surface moves dynamically. Algebraic Curve; Parametric Curve; Splines A parametric surface is a surface in the Euclidean space [equation] which is defined by a parametric equation with two parameters, \begin{array}{lll}. Plot high quality graphs of mathematical equations and data with this easy-to-use software. 0 in this 3 day course. Parametric Plots in Python/v3 How to 3D Parameteric Plots in Python Note: this page is part of the documentation for version 3 of Plotly. Michel Beaudin É TS, Montréal, Canada. The regularized problem can be formulated as a deformable surface model minimizing an energy [23]. Drawfn3d - 3d Parametric Surfaces Drawing Program Drawfunc - 2d Parametric Curves Drawing Program Fortune Function Visualizer Gnuplot Graf-It! Graphmatica Identify The Function Implicit Functions MATHPLOT: Math. Creo Parametric is the standard in 3D CAD software. (Optional) Tap to set the 3D plotting parameters tmin, tmax, umin, and umax. Graph 3d Parametric Equations Tessshlo. We will also see how the parameterization of a surface can be used to find a normal vector for the surface (which will be very useful in a couple of sections) and how the parameterization can be used to find the surface area of a surface. Indeed, imagine texturing parts of a scene with a single material that repeats across large surfaces. Mostly furniture, but also some special objects. Graph it in MATLAB. Instead of numerical coordinates, use expressions in terms of t, like (cos t, sin t). \text{}[/latex] Write parametric equations for the ball’s position, and then eliminate time to write height as a function of horizontal position. Parametric Surfaces • Applications Design of smooth surfaces in cars, ships, etc. 3D parametric surface grapher. Then graph the rectangular form of the equation. I used it to create a turbine blade by formula. Since we used set trange [1:4] , the range of this truncated line is from 1 to 4. This example requires WebGL Visit get. Recommended for you. (The four 4D coordinate axes are x, y, u & v. The key to understanding the 3-D parametric surface graph is in understanding that the surfaces are described parametrically with 2 parameters -- call them i and j. Functions that have a two-dimensional input and a three-dimensional output can be thought of as drawing a surface in three-dimensional space. To sketch the graph of parametric equations, x and y coordinates must be obtained. In Mathcad, the equation for a parametric surface is: For those who love spirograph may find it familiar. Graphing 2D Parametric Equations. The AG distance between points on the surface is then defined as their minimum to-. It is based on a system’s response to varying an external parameter. However, matplotlib is able to plot a generic, parametric 3D surface. 3d Parametric Equation Grapher Tessshlo. Algebrus is a free 3D graphing software for Windows. Free online 3D grapher from GeoGebra: graph 3D functions, plot surfaces, construct solids and much more!. Drawfn3d - 3d Parametric Surfaces Drawing Program Drawfunc - 2d Parametric Curves Drawing Program Fortune Function Visualizer Gnuplot Graf-It! Graphmatica Identify The Function Implicit Functions MATHPLOT: Math. Conversely, given a pair of parametric equations with parameter t, the set of points (f(t), g(t)) form a curve in the plane. While most graphs are represented with equations involving variables x and y, there are some curves that are best handled with a third variable t called a parameter. In Part 2, we saw that changes in coordinates for surfaces give rise to a scale factor for area that can be computed as the Jacobian of the coordinate transformation. 3D Grapher is a feature-rich yet easy-to-use data visualization and graph plotting software suitable for students, engineers and everybody who needs to work with 2D and 3D graphs. The result might feel a little bit too neat and dull. #N#System Administration. Example The graph of y = sin x on 0 ≤ x ≤ π is revolved around the x-axis. Example 2. 0 in this 3 day course. To add a parametric surface to the plot, select the option Parametric Surface on the Add to graph drop-down menu. To graph a point, type it like this: 1. Functions 3D Plotter is an application to drawing functions of several variables and surface in the space R3 and to calculate indefinite integrals or definite integrals. One final technique for sketching and analyzing the graph of a polar equation is finding the intercepts of the graph; that is, where it intersects the lines θ = 0 and θ =. Graph it in MATLAB. A NURBS curve or surface is parametric—that is, the equations that describe it depend on variables (or. Monthly, Half-Yearly, and Yearly Plans. 5 of Stewart's text for an introduction to parametric surfaces. Continuous Parameters, i. However, just as every “nice” enough curve can be parameterized, so can every “nice” enough surface. Definition. Unite technology, in Creo, provides breakthrough capabilities to improve productivity in multi-CAD environments. • To zoom in or out, use the mouse scroll wheel or equivalant. Parametric Surfaces. Parametric Surfaces (3-D Graphing) Graphs surfaces in three dimensions specified parametrically, as x=f(s,t), Rotate the graph by clicking and dragging the mouse on the graph. Plotting a parametric 3D surface. As you study As you study multi-variable calculus, you'll see that the idea of "surface area" can be extended to figures in higher dimensions, too. To begin, let’s take another look at the projectile represented by the parametric equations and as shown in. It is also possible to do some mathematical calculations on the functions. Which is defined by a parametric equation with two parameters u and v. Then sketch a graph locating points at t = {0,1,2,3,4} and indicate the orientation of the curve. For the first case we need to supplement the equations by two inequalities: 0 <= t <= 4 Pi && 0 < x < 4 Pi. whether to scale arguments to ColorFunction. x = u + v, y = 8u^2, z = u - v; (2, 8, 0) Find an equation of the tangent plane to the given parametric surface at the specified point. You can also use "pi" and "e" as their respective constants. Color standard z-axial custom. When graphing, it is important to note the correct domain for t as well as the direction of the graph when t increases. Wolfram Community forum discussion about Graphing parametric surfaces. Calculate tangents. Find a parametrization of the the surface. Suppose we want to graph the ellipse $\mathbf{r}(t) = \langle 3\cos t, 2\sin t\rangle$ on $0 \leq t \leq 2\pi$. You would do the same with the code at the end for approximating arc length. Note - a newer version of this tool is the Graphing Calculator. 78 CHAPTER 12. This combined with the variable Scale factor can result in some interesting forms. Illustrates the setup of a scene, camera, renderer, event handlers (for window resize and fullscreen, provided by the THREEx library), mouse controls to rotate/zoom/pan the scene, mini-display for FPS stats, and setting up basic geometries: a sphere with lighting effects, a multi-colored cube, a plane with an image texture applied, and skybox. Parametric Equation Plotter Geogebra. Start Grapher by clicking on the Grapher icon. Graph 3d Parametric Equations Tessshlo. composition of functions, trigonometric functions, exponential functions, inverse of functions, logarithms, parametric functions, polar coordinate graphs, solving. x = u + v, y = 8u^2, z = u - v; (2, 8, 0) Find an equation of the tangent plane to the given parametric surface at the specified point. org for more infoget. Grapher is a fast and effective equation plotter, capable of drawing any function, solving equations and calculating expressions. There is an art to it but the basic techniques to get started are fairly straightforward. furniture, structures, tools, windows, 2D objects, environment. Setting a=2 will show what the graph looks like when y is kept constant at 2, and shows the change in x. I described a surface as a 2-dimensional object in space. This process, called point plotting, works well when given a small finite interval for the parameter. The simplest type of parametric surfaces is given by the graphs of functions of two variables: = (,), → (,) = (,, (,)). Great Math For Curves And Surfaces. Of course, the parameters may be denoted by letters other than s and t. However, matplotlib is able to plot a generic, parametric 3D surface. 2 to 5,5,-1. For, if y = f(x) then let t = x so that x = t, y = f(t). This example requires WebGL Visit get. #N#ColorFunctionScaling. of the tangent plane. Then change the 7's to 8's and graph the equations. A parametric equation is where the x and y coordinates are both written in terms of another letter. A NURBS curve or surface is parametric—that is, the equations that describe it depend on variables (or. Finding only those arguments t where the parametric graph passes appropriate intersection points the second time. I used it to create a turbine blade by formula. As we have seen previously, z=f(x,y) describes a surface in xyz space. Trotter, and Introduction to Differential Equation s by Richard E. All properties of the plot are editable, including the display of symbols and lines, axes, graph title, and graph background. Alternatively, you can use parametric_plot3d to graph a parametric surface where each of $$x, y, z$$ is determined by a function of one or two variables (the parameters, typically $$u$$ and $$v$$). The plane $$7x + 3y + 4z = 15$$. org for more info. This is not uncommon for parametric surfaces. Woody's entire body in Toy Story is a collection of parametric surfaces. Get Screenshot. o Scene graph o Application specific. 14 to demonstrate concavity. I figured out how to graph a line using a point and direction vector now I am hoping someone can help me graph the line using its parametric from (ie x=3-4t, y=2+5t, z=6-t where the given point is (3,2,6) and the direction vector is [-4,5,-1]. Quadric surfaces are three-dimensional surfaces with traces composed of conic sections. GRAPH-BASED SYSTEMS Although the benchmark task cannot be modelled in the BIM systems, it can easily be modelled in any of the graph-based systems. 3D Surface Grapher - Function Grapher Boxshot. 3 Parametric Equations Involving Sines and Cosines Sketch the plane curve defined by the parametric equations x = 2cost, y = 2sint, for (a) 0 ≤ t ≤ 2π and (b) 0 ≤ t ≤ π. Graphing parametric equations is as easy as plotting an ordered pair. The Home tab New Graph commands or the graph wizard are used to create a graph. If we wanted to. Graphs an implicit surface defined by the relationship expressed in the equation. Parameterizations are not unique. Transformations: Translating a Function example. This adds a new dimension to your graph! Setting the window Setting the window …. To texture the material, Surface Plotter uses a custom GLSL shader which implements tri-planar mapping to generate UV coordinates, and parallax occlusion mapping to give the material depth. However, matplotlib is able to plot a generic, parametric 3D surface. Let x,y,z be functions of two variables u,v, all with the same domain D. Longitudinal Processing. A skateboarder riding on a level surface at a constant speed of 9 ft/s throws a ball in the air, the height of which can be described by the equation. distribute along curve with graph mappers. Since the “Sine Surface” has become vey popular, it’s best used in educational settings where quick and effective parametric surfaces are needed. Large collection of specific objects. How To Graph 3d Functions Of Two Variables On Wolfram Alpha.
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https://math.stackexchange.com/questions/2645975/simple-solution-to-coloured-marble-problem-seems-off | # Simple solution to coloured marble problem seems off?
I have a very basic marble problem (paraphrased):
There is a vase with 20 balls, of which 7 are red, 3 are blue, 1 is orange, and 9 are green. What are the odds of drawing a red, a blue, an orange and a green marble from the vase in this specific order ( $P(r b o g)$ )? We draw four marbles without putting them back.
My solution to this problem was straightforward. Using the conventional definition of a probability: $P(A) = \frac{|A|}{|\Omega|}$ if $\Omega$ is the set of all possible outcomes.
I then defined the event of drawing a red, a blue, an orange, a green marble as such. I labeled my marbles 1 through 20 and labelled the red ones 1 through 7, the blue ones 8 through 10, the orange one 11 and the blue ones 12 through 20.
If we call the event of interest $A$, then the set description of $A$ becomes:
$$A = \{(\omega_1, \omega_2, \omega_3, \omega_4), \omega_1 \in \{1, 2, ..., 7\}, \omega_2 \in \{8, 9, 10\}, \omega_3 = 11, \omega_4 \in \{12, 13, ..., 20\}\}$$
If we call the set of all possible outcomes $\Omega$, this set becomes:
$$\Omega = \{(\omega_1, \omega_2, \omega_3, \omega_4), \omega_i\in\{1,2,...,20\},\omega_k\neq\omega_j, k \neq j\}$$
The sizes of the sets are easily calculated as $|A| = 7*3*1*9 = 189$ and $|\Omega| = 20*19*18*17=116280$. With this, the probability of the event becomes $\frac{189}{116280} = 0.001625387$
Subsequently, I tried to check my work by doing a few simulations. I did three:
1. A simulation in Python. I would generate a list with elements 1 through 20. Then I would let Python pick an item from the list and remove it four times. Thus, this would simulate drawing four marbles and not putting them back. I then checked whether I drew a red-blue-orange-green combo.
2. A simulation in C++. I would draw four marbles at random (1-20) and only use the result of the random draw if it made up four different numbers.
3. A simulation in C++. I would shuffle a list with elements 1 through 20 and then use the first four elements.
Simulation 2 confirms my result. Simulations 1 and 3 do not: they give probability 0.00158... when I let them iterate 100,000,000 times or more.
This seems strange, since I'm so sure of my maths. So the question becomes: is my calculation of probability incorrect, or my way of simulation?
• Just go step by step: $\frac 7{20}\times \frac 3{19}\times \frac 1{18}\times \frac 9{17}\approx 0.001625387$ just as you said. All three simulation methods looks sensible, but of course bugs are possible.
– lulu
Feb 11 '18 at 15:25
• As a way to check for bugs, use your various simulations to compute other things. First marble red, that sort of thing. See if you are getting the right answers for other computations.
– lulu
Feb 11 '18 at 15:31
• Thanks for the remark. I decided to indeed do a bug search. What I did is simulate the first four draws using a completely random trial-and-error method to obtain four individual integers. Then I simulated them using the C++ shuffle() function on an array of 20 integers. I collected data on 1,000,000 experiments in a .csv and plotted histograms of the experiments. I expected that all 20 integers would be drawn equally. However, there was an immense bias towards higher numbers when using the shuffle() function! So there was indeed a bug. Feb 11 '18 at 20:00 | 2022-01-25T17:26:39 | {
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https://mat.gtha.com/wp-content/uploads/jvg7jqxm/viewtopic.php?tag=bc20a1-are-all-linear-pairs-always-congruent | Congruent Angles Congruent Angles have the same angle (in degrees or radians). Reflection: Flip! Where We're Going: We will be using deductive reasoning to write a proof of the vertical angles theorem. Weegy: -7 + N = 20 User: y - 12 = -10 Weegy: x + 5 = 2x User: y - 12 = -10 Weegy: x + 5 = 2x User: y + 1.05 = ... Weegy: Jolene went to the city to find a new job. 1. For two polygons, including squares, to be congruent, all the pairs of corresponding angles must be congruent, but so must all the pairs of corresponding sides. 1. ASA and AAS are sometimes combined into a single condition, I allow 10 to 15 minutes to silently read, complete, and discuss their answers with their partners before going on to the next section, where they will actually work with the angle measures. Played 1466 times. A line that passes through two distinct points on two lines in the same plane is called a transversal. I need this quick. One of the angles in the pair is an exterior angle and one is an interior angle. 175 … This quiz is incomplete! When the lines are parallel, a case that is often considered, a transversal produces several congruent and several supplementary angles. A linear pair forms a straight angle which contains 180º, so you have 2 angles whose measures add to 180, which means they are supplementary. 76% average accuracy. Answer Save. A. Alternate interior angles two angles in the interior of the parallel lines, and on opposite (alternate) sides of the transversal. A) Linear pairs of angles are congruent Create an equation to represent the following situation: I am selling rings for $2 and pins for$1. Vertical Angles. 2. Linear pair. Corresponding angles. B. Never. Play Live Live. Learn. Thus, triangle PQR is congruent to triangle ABC. Sum Of Vertical Angles. AAS (Angle-Angle-Side): If two pairs of angles of two triangles are equal in measurement, and a pair of corresponding non-included sides are equal in length, then the triangles are congruent. Because they are vertical (and, therefore, congruent) to corresponding interior alternate angles, which have been proven to be congruent between themselves. Angles from each pair of vertical angles are known as adjacent angles and are supplementary (the angles sum up to 180 degrees). To compensate for the problems of heat expansion, a piston is ... Identify the phrase in the following sentence. A linear pair is by definition supplementary, but two supplementary angles are only a linear pair if they are also adjacent. Mathematics. Both pairs of vertical angles (four angles altogether) always sum up to 360 degrees. You can view more similar questions or ask a new question. The two shapes ... Show Ads. Adjacent Angles. A hotel has 250 rooms. The sum of angles of a linear pair is always equal to 180°. Add your answer and earn points. 0. There are three other special pairs of angles. Perpendicular Bisector Theorem. Log in. Solve the following system of equations. Instant scoring, progress tracking, & award certificates to keep your student motivated. In this case, all 8 angles are right angles . Just two intersecting lines creates four linear pairs. Both pairs of vertical angles (four angles altogether) always sum up to 360 degrees. Created by . Whats people lookup in this blog. Democratic megadonor: 'Stop giving Trump a platform’ 'Mona Lisa of sports cards' sells for record amount Join now. In the diagram above, ∠ABC and ∠DBC form a linear pair. They don't have to be on similar sized lines. Line segments are congruent if they have the same length. Victoria. If three parallel lines are cut by a transversal and one angle is measured to be 90°, what can be said of the remaining angles? Vertical angles are always congruent that are of equal measure. Edit. All of these supplementary pairs are linear pairs. B. How much milk does a gallon jug of milk hold? 0. Get your answers by asking now. Congruent Angles. Yes. the shape still has the same size, area, angles and line lengths. This is like a rectangle (excluding a square) that has two pairs of congruent sides where the congruent sides are not adjacent. Select all that apply. Active listening is listening to the words only. Which statements are always true about regular polygons? Homework. PLAY. Here we have Notice. B. The sum of a linear pair of angles is 180 degrees, hence are supplementary. x = y + 10 Are all images, or new figures that result from a translation, always congruent to the original figure? All sides are congruent •• Pairs of sides are parallel All angles are congruent •• All angles measure 90 degree Correct me if I'm wrong!! The vertical angles theorem states that vertical angles are always congruent. If two angles form a linear pair, the angles are supplementary. 2x + y = 3 Always. They're all 90° also. These angles are congruent. B. Edit . Just the same angle. Adjacent Angles. Played 0 times. Figure 12. Technically, yes, all congruent figures are also similar shapes. Start studying Geometry - Always, Sometimes, Never. User: An essay ... Weegy: 2x + y = 3 User: Solve the following system of equations. Print; Share; Edit; Delete; Host a game. Grade appropriate lessons, quizzes & printable worksheets. A line that passes through two distinct points on two lines in the same plane is called a transversal. 0 0. 8th - 9th grade . Don’t neglect to check for them! By the definition of congruent angles, ΔPQR ≅ ΔM QN. Alternate interior angles are4 always congruent. A. Vertical Solo Practice. If the areas of two similar hexagons are to each other as 5 : 2, and one side of the first hexagon is 25, the corresponding side in the other hexagon is 10.00. Ask your question. Write. The two shapes need to be the same size to be congruent. 3x + 2y - 5 = 0 Vertical angles are always equal in measure. Linear pairs of angles are not always congruent. Translation: Slide! That is all. B. * congruent acute supplementary complementary 1 See answer rinkigarg74001 is waiting for your help. Just the same angle. The angles are said to be linear if they are adjacent to each other after the intersection of the two lines. Congruent Angles: If two angles have the same measure, then we call those two angles congruent angles. The sum of angles of a linear pair is always equal to 180°. Angles in a linear pair are always (A) congruent (B) acute (C) supplementary (D) complementary - Math - Constructions of Triangles Alternate interior angles come up when two parallel lines are intersected by a transversal. Such angles are also known as supplementary angles. They're all 90° also. x = 2y - 1. For complete explanation, theorems and proofs related to parallel lines and transversal we can recommend to refer to UNIZOR and follow the menu options Geometry - Parallel Lines - Introduction. Share what’s outside your window and all around you. Probably because they are only "equal" when laid on top of each other. They are perpendicular to each other (we know because of the right angle symbol at the intersection). 10.00 We will look at nine different types of angle pairs. cdaehling. When two lines intersect, pairs of the four angles formed are either congruent vertical angles.or supplementary linear pairs . 2. Congruent or Similar? Vertical Angles are ALWAYS Congruent ≅. or. Gravity. Congruent. “Why are supplementary angles linear pairs ?” They are not always linear pairs, you have this relationship backwards. Practice. A transversal forms four pairs of corresponding angles. Match. Join Yahoo Answers and get 100 points today. Live Game Live. Learn vocabulary, terms, and more with flashcards, games, and other study tools. A linear congruential generator (LCG) is an algorithm that yields a sequence of pseudo-randomized numbers calculated with a discontinuous piecewise linear equation.The method represents one of the oldest and best-known pseudorandom number generator algorithms. Learn. This quiz is incomplete! Two vertical angles are supplementary. 1. Vertical angles are each of the pairs of opposite angles made by two intersecting lines. These pairs are congruent pairs. 2. If you drag any of the four endpoints, the other segment will change length to remain congruent with the one you are changing. 2. The non examples of vertical angles and linear pairs are those I've found students usually make errors with, so I included them here. Gravity. Parallel Lines w/a transversal AND Angle Pair Relationships Concept Summary Congruent Supplementary alternate interior angles- AIA alternate exterior angles- AEA corresponding angles - CA same side interior angles- SSI Types of angle pairs formed when a transversal cuts two parallel lines. Across. Note they are laying at different angles. D. Adjacent. Correct answer to the question Which pairs of angles are always congruent? Expert Answer . But not all similar shapes have congruency. Always. D. There isn't enough information to determine anything about the remaining angles. Log in. Start studying Geometry: Always, Sometimes, Never. The bisectors of vertical angles are opposite rays. Calculator solve triangle specified by all three sides (SSS congruence law). Congruent - why such a funny word that basically means "equal"? Linear pair of angles are formed when two lines intersect each other at a single point. Flashcards. Since this proof works for any corresponding angles, we know that all corresponding angles are always congruent. Linear pairs of angles can only be congruent when the measure of each of the angles is 90 degrees. B. anything except magazines. Classify Angle Pairs Congruent or Supplementary. If two angles are complementary, then they are each acute. Sometimes . Every set of angles that are linear pairs are supplementary. 0% average accuracy. That is all. Join. Linear pair of angles are formed when two lines intersect each other at a single point. Never. Earn a little too. Spell. oh my this was made in 2010, for some reason that hits me hard, their probably grown up living their life talking with friends not knowing someone a decade later was thinking about how they commented something so unrelavent to this 10 years later. 1 Answer. Supplementary. Angles from each pair of vertical angles are known as adjacent angles and are supplementary (the angles sum up to 180 degrees). See if you can find the four linear pairs in intersecting lines MAP and TAN: Did you find all these? Weegy: Data is a set of values of qualitative or quantitative variables; restated, data are individual pieces of ... Weegy: The phrase which do not describe a mineral is organic solid. Some of these angle pairs have specific names and are discussed below: corresponding angles, alternate angles, and consecutive angles. Test. Classify the angle pairs formed when parallel lines are cut by a transversal, as either congruent, supplementary AND Corresponding Angles, Alternate Interior Angles, Alternate Exterior Angles, Same-Side Angles, Vertical Angles, Linear Pair. Congruent Angles. If the areas of two similar hexagons are to each other as 5 : 2, and one side of the first hexagon is 25, what is the corresponding side in the other hexagon? Weegy: -4|-4| User: 2(9-3) Weegy: 12*9-3 = 108-3 = 105. 7 minutes ago. Classify Angle Pairs Congruent or Supplementary. Vertical angles are congruent: If two angles are vertical angles, then they’re congruent (see the above figure). C. only nonfiction. So, all three pairs of corresponding sides and all three pairs of corresponding angles are congruent. Angles in a linear pair are always_____. These angles are congruent. All sides are congruent •• Pairs of sides are parallel All angles are congruent •• All angles measure 90 degree Correct me if I'm wrong!! Congruent - why such a funny word that basically means "equal"? A=A'=0, B=B'=2, C=I, C'=-i. They don't have to point in the same direction. cdaehling. For any real number c or d, describe how the ordered pair (x, y) of any original figure will change when translated: a. horizontally c units. D. 250.00. C. 15.81 Ask Question + 100. Notice also that $$ax\equiv b (mod\ m)$$ is equivalent to a linear Diophantine equation i.e. Rays and lines cannot be congruent because they do not have both end points defined, and so have no definite length. C. Only some of the remaining angles are 90°. However, they need not be parallel. Two angles are said to be linear if they are adjacent angles formed by two intersecting lines. Using Transitive Property of Congruent Triangles : By Transitive property of congruent triangles, if ΔPQR ≅ ΔMQN and ΔMQN ≅ ΔABC, then ΔPQR ≅ ΔABC. Answered Angles in a linear pair are always_____. contestant, Why some find the second gentleman role 'threatening', Biden leaves hidden message on White House website, At least 3 dead as explosion rips through building in Madrid, Pence's farewell message contains a glaring omission. The theorem states that if a transversal crosses the set of parallel lines the alternate interior angles are congruent. rinkigarg74001 rinkigarg74001 15.07.2020 Math Secondary School +5 pts. Print; Share; Edit; Delete; Report an issue; Start a multiplayer game. Vertical Angles Theorem: Vertical angles are angles that are opposite one another when two lines intersect. You can view more similar questions or ask a new question. 2. Anonymous. Sometimes. They're all 90° also. Score .9081 User: If three parallel lines are cut by a transversal and one angle is measured to be 90°, what can be said of the remaining angles?A. Vertical Angles and Linear Pairs (4/28) DRAFT. there exists $$y$$ such that $$ax-my=b$$. Question 351858: Always, sometimes, or never? Angles. Always. Mathematics. Difference between velocity and a vector? A: Angles that are next to each other are known as adjacent angles, i.e., two angles with one common arm. Finish Editing. This conversation has been flagged as incorrect. Side-Angle-Side (SAS) Side-Side-Side (SSS) Angle-Side-Angle (ASA) Congruent Shapes Examples. In a linear pair, the arms of the angles that are not always common or collinear i.e., they lie on a straight line. * - 19562971 1. It is important to know that if $$x_0$$ is a solution for a linear congruence, then all integers $$x_i$$ such that $$x_i\equiv x_0 (mod \ m)$$ are solutions of the linear congruence. Alternate Interior Angles Theorem. AAS is equivalent to an ASA condition, by the fact that if any two angles are given, so is the third angle, since their sum should be 180°. Solo Practice. So do ∠ 2 and ∠ 3 , ∠ 3 and ∠ 4 , and ∠ 1 and ∠ 4 . Save. Get your answers by asking now. What are alternate interior angles? 2 years ago. Which phrase does not describe a mineral? Linear pairs of angles are not always congruent. Sum Of Vertical Angles. In the figure, ∠ 1 and ∠ 2 form a linear pair. There are other supplementary pairs described in the shortcut later in this section. If the angles that form a linear pair are congruent, then the intersecting lines, line segments, or rays forming the linear pair are perpendicular. Basically, a linear pair of angles always lie on a straight line. In geometry, similar triangles are important, and three theorems help mathematicians prove if triangles are similar or congruent. Share practice link. Supplementary C. Complementary D. Adjacent Weegy: The angles of a linear pair are always supplementary. A linear pair is a pair of adjacent angles whose non-adjacent sides form a line. Learn vocabulary, terms, and more with flashcards, games, and other study tools. by mrsmkent. Relevance. Trending Questions. MEAP Preparation - Grade 7 Mathematics Curriculum - Adjacent, Congruent, Linear Pair, and Vertically Opposite Angles - Math & English Homeschool/Afterschool/Tutoring Educational Programs. Created by . No. One of the angles in the pair is an exterior angle and one is an interior angle. The angles are said to be linear if they are adjacent to each other after the intersection of the two lines. Which statements are always true about regular polygons? But the angles are not all congruent (as set in the question) which pushes the shape into the "next less regular" shape, the parallelogram. Classify the angle pairs formed when parallel lines are cut by a transversal, as either congruent, supplementary AND Corresponding Angles, Alternate Interior Angles, Alternate Exterior Angles, Same-Side Angles, Vertical Angles, Linear Pair. Corresponding angles are pairs of angles that lie on the same side of the transversal in matching corners. They can be at any angle or orientation on the plane. Spell. Diagonal. They don't have to point in the same direction. Trending Questions. If three parallel lines are cut by a transversal and one angle is measured to be 90 , They're all 90 also. How do you know if the image translated to the left or to the right? 4 5 and 3 6. Join Yahoo Answers and get 100 points today. Linear Geometry. Supplementary angles are congruent. Am stuck for days.? AAS (Angle-Angle-Side): If two pairs of angles of two triangles are equal in measurement, and a pair of corresponding non-included sides are equal in length, then the triangles are congruent. Vertical angles are one of the most frequently used things in proofs and other types of geometry problems, and they’re one of the easiest things to spot in a diagram. If a transversal line l intersects two parallel lines m and n, then the corresponding angles are equal Two figures may appear to be congruent yet one side or one angle in one shape may be slightly different than the other shape. Corresponding angles are pairs of angles that lie on the same side of the transversal in matching corners. Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website! Finish Editing. STUDY. Examples: Here are 3 examples of shapes that are Congruent: Congruent (Rotated and Moved) Congruent (Reflected and Moved) Congruent (Reflected, Rotated and Moved) Congruent or Similar? are linear pairs always congruent.? 3.16 NEXT to Match. User: _____ pairs of angles are always congruent to each other. Hide Ads About Ads. If three parallel lines are cut by a transversal and one angle is measured to be 90 , they're all 90 also can be said of the remaining angles. In geometry, a transversal is a line that passes through two lines in the same plane at two distinct points.Transversals play a role in establishing whether two or more other lines in the Euclidean plane are parallel.The intersections of a transversal with two lines create various types of pairs of angles: consecutive interior angles, corresponding angles, and alternate angles. a) are the pairs of congruent triangles ABC and A'B'C' equally or differently oriented. The two angles of a linear pair are always supplementary , which means their measures add up to 180 ° . The linear pairs are /1 and /4, /1 and /3, /2 and /3, /2 and /4. Start studying always, sometimes, never. Still have questions? Linear pairs always form when lines intersect. User: _____ pairs of angles are always congruent to each other. Favorite Answer. User: minerals containing iron are attracted bt ... Weegy: If you INTEND to visit your grandmother,you should go now,visiting hours will be ending soon. Don’t neglect to check for them! The transitive property can be applied to perpendicular lines. Supplementary C. Complementary D. Adjacent Weegy: The angles of a linear pair are always supplementary. Join now. Assign HW. The supplement of an acute angle is acute. The adjacent angles are the angles which have a common vertex. Vertical angles are always congruent that are of equal measure. Weegy: The solution ... WINDOWPANE is the live-streaming social network that turns your phone into a live broadcast camera for streaming to friends, family, followers, or everyone. b) prove that having the same orientation is an equivalence relation for congruent triangles. A. A) Linear pairs of angles are congruent Create an equation to represent the following situation: I am selling rings for $2 and pins for$1. If one shape can become another using Turns, Flips and/or Slides, then the shapes are Congruent: Rotation: Turn! Key Words. Parallel Lines w/a transversal AND Angle Pair Relationships Concept Summary Congruent Supplementary alternate interior angles- AIA alternate exterior angles- AEA corresponding angles - CA same side interior angles- SSI Types of angle pairs formed when a transversal cuts two parallel lines. Still have questions? STUDY. (select all that apply) Corresponding Angles Same Side Exterior Angles Alternate Exterior Angles Linear Pair Angles Vertical Angles Same Side Interior Angles Alterna - e-eduanswers.com Learn vocabulary, terms, and more with flashcards, games, and other study tools. Always. Probably because they are only "equal" when laid on top of each other. Every pair shares a vertex, the point of intersection, and one common side. A. only novels. Vertical angles are always equal. This may be a bad time to buy a Mega Millions ticket. A. Vertical B. Corresponding angles. 8 years ago. Given numbers: 42000; 660 and 72, what will be the Highest Common Factor (H.C.F)? Updated 81 days ago|10/31/2020 2:03:44 PM. They will all be complementary to each other. The angles are adjacent, sharing ray BC, and the non-adjacent rays, BA and BD, lie on line AD. Play. The symbol for congruence is Also, recall that the symbol for a line segment is a bar over two letters, so the statement is read as "The line segment AB is congruent to the line segment PQ". Congruent Angles Congruent Angles have the same angle (in degrees or radians). To become a better writer, you must read Fox News fires key player in its election night coverage, Biden demands 'decency and dignity' in administration, Now Dems have to prove they’re not socialists, Democrats officially take control of the Senate, Saints QB played season with torn rotator cuff, Lady Gaga wows with exuberant anthem at inauguration, Ken Jennings torched by 'Jeopardy!' Practice. Share practice link. unless they are across from each other, then that will make them supplementary. Play. Vertical angles can only be supplementary, when the measure of each of the angles is 90 degrees. Students have made conjectures that linear pairs are always supplementary and vertical angles are always congruent. Two angles that are vertical are adjacent. Score .9081 User: If three parallel lines are cut by a transversal and one angle is measured to be 90°, what can be said of the remaining angles?A. What is a perpendicular bisector? The pair of adjacent angles are constructed on a line segment, but not all adjacent angles are linear. PLAY. Explanation for Linear Pair of Angle . Geometry help is my answer correct please explain. 16 answers. Alternate interior angles alternate interior angles properties. C. Complementary In the figure above, there are two congruent line segments. Write. Hence, we can also say that a linear pair of angles is the adjacent angle whose non-common arms are basically opposite rays. All adjacent angles are not linear, but all linear pairs are adjacent. can be said of the remaining angles. They don't have to be on similar sized lines. Test. D. good writing. Symbols. Type of Angle Pairs. Vertical Angles and Linear Pairs DRAFT. Updated 206 days ago|6/28/2020 1:56:39 PM. _______ pairs of angles are always congruent to each other. This answer has been confirmed as correct and helpful. AAS is equivalent to an ASA condition, by the fact that if any two angles are given, so is the third angle, since their sum should be 180°. Explain your reasoning. Edit. The adjacent angles are the angles which have a common vertex.
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https://math.boisestate.edu/~calhoun/teaching/matlab-tutorials/lab_11/html/lab_11.html | Back to tutorial index
# The Runge phenomenon
## Introduction
In this lab, we will investigate the Runge Phenomenon and see how we might be able to fix it by choosing interpolating points wisely.
As a classic example of the Runge Phenomenon, we try to interpolate the function f(x) = 1/(1 + 25x^2) at equally spaced points.
clear all;
close all;
f = @(x) 1./(1 + 25*x.^2);
Plot the original function f(x) at set of equally spaced points
figure(1);
clf;
x = linspace(-1,1,500);
y_true = f(x);
plot(x,y_true,'r','linewidth',2);
hold on;
Construct an interpolating polynomial of degree N at N+1 points. We will use polyfit although the problem occurs with Lagrange polynomial interpolation or the Barycentric form as well.
% Equally spaced points
N = 10; % Degree of the polynomial we try to fit
xdata = linspace(-1,1,N+1)';
ydata = f(xdata);
p = polyfit(xdata,ydata,N);
Plot the interpolating polynomial
y_fit = polyval(p,x);
poly_10 = plot(x,y_fit,'b','linewidth',2);
plot(xdata,ydata,'k.','markersize',30);
snapnow;
We see that a single interpolating polynomial does a particularly poor job of interpolating near the endpoints of the interval [-1,1]. Perhaps we should try increasing the order of accuracy.
N = 20; % Increase the degree to 20 from 10.
xdata = linspace(-1,1,N+1)';
ydata = f(xdata);
% Ignore the warning about the matrix being ill-conditioned. If we used
% the barycentric form, we do not have these ill-conditioning issues!
p = polyfit(xdata,ydata,N);
Plot the interpolating polynomial
y_fit = polyval(p,x);
poly_20 = plot(x,y_fit,'g','linewidth',2);
plot(xdata,ydata,'k.','markersize',30);
axis([-1 1 -5 5]);
snapnow;
Let's increase one more time to see that pattern
N = 40; % Increase the degree to 40 from 20.
xdata = linspace(-1,1,N+1)';
ydata = f(xdata);
% plot(xdata,ydata,'k.','markersize',30);
% Ignore the warning about the matrix being ill-conditioned. If we used
% the barycentric form, we do not have these ill-conditioning issues!
p = polyfit(xdata,ydata,N);
Warning: Polynomial is badly conditioned. Add points with distinct X values,
reduce the degree of the polynomial, or try centering and scaling as described
in HELP POLYFIT.
Plot the interpolating polynomial
y_fit = polyval(p,x);
poly_40 = plot(x,y_fit,'m','linewidth',2);
plot(xdata,ydata,'k.','markersize',30);
axis([-1 1 -10 10]);
snapnow;
But are we doing at least a better job in the middle? Yes!
axis([-1 1 0 1]);
lh = legend([poly_10, poly_20, poly_40],{'N = 10','N = 20','N = 40'});
set(lh,'fontsize',18);
snapnow;
Back to the top
## Avoiding the Runge phenomenon (part I)
In the previous examples, our interpolating points were equally spaced points. Is there a better choice of points at which to interpolate our function?
We can try using "Chebychev nodes". These are special nodes which allow for a better approximation of the polynomial.
clf;
y_true = f(x);
plot(x,y_true,'r','linewidth',2);
hold on;
% Fit a 10th degree polynomial at Chebyshev nodes.
N = 10; % Degree of the polynomial we try to fit
t = linspace(0,pi,N+1);
xdata = -cos(t); % Chebyshev nodes
ydata = f(xdata);
p = polyfit(xdata,ydata,N);
Plot the interpolating polynomial
y_fit = polyval(p,x);
poly_10 = plot(x,y_fit,'b','linewidth',2);
hold on;
plot(xdata,ydata,'k.','markersize',30);
snapnow;
% Fit a 20th degree polynomial at Chebyshev nodes.
N = 20; % Degree of the polynomial we try to fit
t = linspace(0,pi,N+1);
xdata = -cos(t); % Chebyshev nodes
ydata = f(xdata);
p = polyfit(xdata,ydata,N);
Plot the interpolating polynomial
y_fit = polyval(p,x);
poly_20 = plot(x,y_fit,'g','linewidth',2);
hold on;
plot(xdata,ydata,'k.','markersize',30);
snapnow;
% Fit a 40th degree polynomial at Chebyshev nodes.
N = 40; % Degree of the polynomial we try to fit
t = linspace(0,pi,N+1);
xdata = -cos(t); % Chebyshev nodes
ydata = f(xdata);
% Ignore the warning about the matrix being ill-conditioned. If we used
% the barycentric form, we do not have these ill-conditioning issues!
p = polyfit(xdata,ydata,N);
Warning: Polynomial is badly conditioned. Add points with distinct X values,
reduce the degree of the polynomial, or try centering and scaling as described
in HELP POLYFIT.
Plot the interpolating polynomial
y_fit = polyval(p,x);
poly_40 = plot(x,y_fit,'m','linewidth',2);
hold on;
plot(xdata,ydata,'k.','markersize',30);
snapnow;
And how does the middle look?
axis([-1 1 0 1]);
lh = legend([poly_10, poly_20, poly_40],{'N = 10','N = 20','N = 40'});
set(lh,'fontsize',18);
snapnow;
We have completely avoided the problem at the endpoints by chosing our points carefully! But what if we don't have the flexibility to chose our interpolating points? That leads us to the idea of using piecewise polynomials.
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## Lab exercises
No lab exercises yet!
Back to the top | 2018-08-17T01:50:44 | {
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https://math.stackexchange.com/questions/3812100/how-to-compute-int-limits-0-infty-fracx1-31x2-dx | # How to compute: $\int\limits_0^{\infty} \frac{x^{1/3}}{1+x^{2}} dx$
This question was asked in a masters of mathematics exam for which I am preparing.
Compute $$\int\limits_0^{\infty} \frac{x^{1/3}}{1+x^{2}} dx$$.
I could only think of substituting $$y^3 = x$$ and that does not change much.
Could somebody post a solution using residues or in ways besides here in this link?: How to compute the integral $\int_{0}^{\infty} \frac{x^{1/3}}{1+x^{2}} \ dx$
Edit : I am interested in the answers which use contour integration and residue calculus.
• While not necessarily the easiest, the Calc II approach with $y=x^3$ gives $$\int_0^{\infty}\frac{y}{1+y^6}3y^2dy$$ which can be integrated via partial fractions.
– yoyo
Sep 2 '20 at 19:54
• Set $x^2=y$ then use beta function then euler reflection identity. Sep 2 '20 at 20:48
• An approach using residues is here. Jan 21 '21 at 11:29
Evaluating $$\oint_C \frac{z^{\alpha-1}}{1+z} dz$$ we see that there is a branch cut along the positive $$x-$$axis and a pole at $$z=-1$$.
Take $$C$$ to be a keyhole contour consisting of a segment from $$\epsilon$$ to $$R$$, a circle of radius $$R$$, a segment from $$R$$ to $$\epsilon$$ and a small circle of radius $$\epsilon$$ surrounding the origin.
The result is:
$$\int_0^\infty \frac{ x^{\alpha-1}}{1+x} = \frac{\pi}{\sin \pi \alpha} \quad \text{when } 0<\alpha<1.$$
With the substitution $$y^{1/2} = x$$, our integral becomes
$$\bbox[5px, border: 1pt solid blue]{\int_0^\infty \frac{x^{1/3}}{1+x^2} dx = \frac{1}{2} \int_0^\infty \frac{y^{-1/3}}{1+y} dy = \frac{\pi}{2\sin \frac{2\pi}{3}}=\frac{\pi}{\sqrt{3}}.}$$
UPDATE:
In response to J.G.'s question:
The residue at $$z=-1$$ is $$b=\text{Res}_{z=-1} \frac{z^{\alpha-1}}{1+z}=e^{\pi i (\alpha-1)}.$$
So $$\oint_C \frac{z^{\alpha-1}}{1+z} dz = 2\pi i b$$
On the first segment (from $$\epsilon$$ to $$R$$), $$z^{\alpha-1}=x^{\alpha-1}$$, on the return trip, $$z^{\alpha-1}=(e^{2\pi i} x)^{\alpha-1}.$$
The integrals along the circles go to zero as $$\epsilon\to0$$, $$R\to0$$.
$$\int_0^\infty \frac{x^{\alpha-1}}{1+x}dx - \int_0^\infty \frac{e^{2\pi i (\alpha-1)} x^{\alpha-1}}{1+x}dx = 2\pi i e^{\pi i (\alpha-1)}$$
\begin{aligned} \int_0^\infty \frac{x^{\alpha-1}}{1+x}dx &=\frac{2\pi i e^{\pi i (\alpha-1)}}{1-e^{2\pi i (\alpha-1)}}\\ &= \frac{2\pi i}{e^{-\pi i (\alpha-1)}-e^{\pi i (\alpha-1)}} \\ &=\frac{\pi}{\sin \pi(1-\alpha)} \\ &= \frac{\pi}{\sin \pi \alpha}. \end{aligned}
• Yes, "keyhole contour" is the critical point. This is one of those iconic problems that is also useful now and then in real life. Sep 2 '20 at 19:10
• (+1) The classical keyhole contour works like a charm. Sep 2 '20 at 19:26
• Could you show how this gets us to $\pi\csc(\pi\alpha)$?
– J.G.
Sep 2 '20 at 20:02
• @J.G., yes, edited my answer to show these details. Thanks.
– mjw
Sep 2 '20 at 22:17
• @mjw can you please add more details to your update to J.G. 's question. I am having difficulties in understanding the proof but I really want to understand it as its's really nice. Sep 19 '20 at 16:06
With $$x=\tan t$$ it becomes $$\int_0^{\pi/2}\tan^{1/3}tdt$$. This can be evaluated in terms of the Beta function. In particualar, $$\int_0^{\pi/2}\tan^{2s-1}tdt=\tfrac12\pi\csc\pi s$$ implies your integral is $$\tfrac12\pi\csc\frac{2\pi}{3}=\frac{\pi}{\sqrt{3}}$$.
Then, \begin{align} \int_{0}^{\infty}{x^{1/3} \over 1 + x^{2}}\,\dd x & = {1 \over 2}\,\Gamma\pars{\color{red}{2 \over 3}} \color{blue} {\Gamma\pars{1 + \bracks{-\,\color{red}{2 \over 3}}}} \\[2mm] &= {1 \over 2}\,{\pi \over \sin\pars{\pi/3}} = \bbx{{\root{3} \over 3}\,\pi} \\ & \end{align}
First, let's do a substitution to reduce the number of poles to be computed. In this problem, it wouldn't be necessary, since the function is fairly easy and there aren't too many poles, however it's a good a practice since in the future you may encounter more intricated problems.
$$\int_{0}^{\infty}{\frac{x^\frac{1}{3}}{x^2+1}dx}=\frac{1}{2}\int_{0}^{\infty}{\frac{z^{-\frac{1}{3}}}{z+1}dz}$$
Now, let’s define our function and integration path. Bear in mind that since we are working with complex functions $$z^{-\frac{1}{3}}=\exp{\left(-\frac{Log\left(z\right)}{3}\right)}$$, wich means that we must take care of the branch point. Hence, the selection of the keyhole contour. $$f(z)=\frac{z^{-\frac{1}{3}}}{z+1}$$
$$\oint f(z)dz=\left(\int_{0+ir}^{R+ir}+\int_{\Gamma}+\int_{R-ir}^{0-ir}+\int_\gamma \right)f(z)dz$$
1)Let's start working with the two integrals in the positive axis of the Real Line: $$\displaystyle{\lim_{R\rightarrow \infty\\ r\rightarrow 0}}\left(\int_{0+ir}^{R+ir}\int_{R-ir}^{0-ir} \right)f(z)dz$$
Remember that: $$Log(z)=\log|z|+i\text{Arg}(z)$$
$$\int_0^\infty \frac{\exp\left(-\frac{\log\left(z\right)+0i}{3}\right)}{z+1}dz+\int_\infty^0 \frac{\exp\left(-\frac{\log\left(z\right)+2\pi i}{3}\right)}{z+1}dz=\left(1-e^{-\frac{2\pi i}{3}}\right)\int_0^\infty\frac{z^{-\frac{1}{3}}}{z+1}dz$$
1. To compute the others integrals let's make use of the the Estimation Lemma: $$\left\lvert\int_\Gamma f(z)dz\right\rvert\leq\displaystyle{\lim_{R\rightarrow \infty}}\left\lvert\int_0^{2\pi}\frac{\left(Re^{it}\right)^{-\frac{1}{3}}}{Re^{it}+1}Rie^{it}dt\right\rvert\leq\displaystyle{\lim_{R\rightarrow \infty}}2\pi R^{-\frac{1}{3}}\rightarrow 0$$
$$\left\lvert\int_\gamma f(z)dz\right\rvert\leq\displaystyle{\lim_{r\rightarrow 0}}\left\lvert\int_{2\pi}^{0}\frac{\left(re^{it}\right)^{-\frac{1}{3}}}{re^{it}+1}rie^{it}dt\right\rvert\leq\displaystyle{\lim_{r\rightarrow 0}}2\pi r^{\frac{2}{3}}\rightarrow 0$$
1. Finally, let's compute the residue: $$\oint f(z)dz=2\pi i\displaystyle{\lim_{z\rightarrow -1}}\frac{z^{-\frac{1}{3}}}{z+1}(z+1)=2\pi ie^{-\frac{\pi i}{3}}$$
Gathering all results: $$\int_0^\infty\frac{z^{-\frac{1}{3}}}{z+1}dz=\frac{2\pi ie^{-\frac{\pi i}{3}}}{1-e^{\frac{-2\pi i}{3}}}=\frac{\pi }{\frac{e^{\frac{\pi i}{3}}-e^{-\frac{\pi i}{3}}}{2i}}=\frac{\pi}{\sin\left(\frac{\pi}{3}\right)}=\frac{2\pi}{\sqrt{3}}$$
Hence: $$\boxed{\int_{0}^{\infty}{\frac{x^\frac{1}{3}}{x^2+1}dx}=\frac{\pi}{\sqrt{3}}}$$
• Why in point 2 of your answer : when you are making use of estimation lemma, Then how did you estimated 1st inequality in 1st line of 2. I think the inequality must be opposite Feb 12 '21 at 6:40
Substitution $$x=y^{3/2}$$ allows to write $$I=\int\limits_0^\infty\dfrac{\sqrt[\large3]x\,\text dx}{1+x^2} = \dfrac32\int\limits_0^\infty\dfrac{y\,\text dy}{1+y^3}.$$ At the same time, substitution $$y=\dfrac1z$$ gives $$I= \dfrac32\int\limits_0^\infty\dfrac{\text dz}{1+z^3} = \dfrac32\int\limits_0^\infty\dfrac{\text dy}{1+y^3}.$$ Therefore, $$I= \dfrac34\int\limits_0^\infty\dfrac{(1+y)\,\text dy}{1+y^3} = \dfrac34\int\limits_0^\infty\dfrac{\text dy}{1-y+y^2} =3\int\limits_0^\infty\dfrac{\text dy}{3+(2y-1)^2}.\tag1$$ Integral $$(1)$$ does not require the residue approach: $$I=\dfrac{\sqrt3}2\arctan\dfrac{2y-1}{\sqrt3}\bigg|_0^\infty = \dfrac{\sqrt3}2\left(\dfrac\pi2+\dfrac\pi6\right),$$ $$\color{brown}{\mathbf{I=\dfrac\pi{\sqrt3}.}}$$ | 2022-01-19T12:00:51 | {
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https://math.stackexchange.com/questions/3987124/are-all-diagonalizaitions-necessarily-eigendecompositions | # Are all diagonalizaitions necessarily eigendecompositions?
If a matrix $$A \in \mathbb{F^{NxN}}$$ can be diagonalized, i.e. factored into the form:
$$A = V \Lambda V^{-1}$$, where V is a basis for $$\mathbb{F^N}$$ and $$\Lambda$$ is a diagonal matrix, do $$V$$ and $$\Lambda$$ necessarily contain the eigenvectors and eigenvalues of $$A$$ respectively, or could a diagonalization be in terms of other vectors/diagonal values?
Is the answer to this question different between asymmetric square matrices on the one hand and symmetric/Hermitian or normal matrices which decompose as $$Q\Lambda Q'$$ for unitary $$Q$$ on the other hand? And if so, or if all diagonalizations are necessarily in terms of eigenvectors/values, why?
Rewrite $$A = V \Lambda V^{-1}$$ as $$AV = V \Lambda$$ and let $$v_1$$ be the first column of $$V$$.
Now look at the first column of $$AV$$ --- it's just $$Av_1$$.
What about the first column of $$V \Lambda$$? It's just $$\lambda_1 v_1$$.
The same reasoning applies to all other columns, hence each column of $$V$$ is an eigenvector of $$A$$.
I leave you to think about the complex case on your own...
Since $$A=V\Lambda V^{-1}$$ the eigenvalues of $$A$$ and $$\Lambda$$ are the same, since $$\Lambda$$ is diagonal, we can read the eigenvalues immediately off from $$\Lambda$$. Writing the equation as $$AV=V\Lambda$$, then if $$\Lambda=\text{diag}(\lambda_1,\ldots,\lambda_n)$$ and the column vectors of $$V$$ are $$v_1,\ldots,v_n$$, we see that $$Av_i=\lambda_i v_i$$ holds, so we can read off from $$V$$ an eigenvector $$v_i$$ corresponding to the eigenvalue $$\lambda_i$$ for every $$i$$.
A diagonalisation in terms of different $$V$$ and $$\Lambda$$ is possible. For instance, we could permute the entries of $$\Lambda$$. Also, if $$v$$ is an eigenvector, so is $$\lambda v$$ for a non-zero scalar $$\lambda$$. Therefore, we could multiply every column of $$V$$ with a different non-zero scalar and still get a diagonalisation.
• If you were to permute the elements of lambda you would also have permute the corresponding columns of V to preserve equality, right? And since scaled columns of V are still eigenvectors, what I'm understanding is that any diagonalization is necessarily an eigendecomp, but not all eigendecomps are unique, correct? – Connor Robetorye Jan 16 at 2:51
• Indeed, you would have to permute the corresponding columns as well. Exactly, no eigendecomposition is unique (there is always some freedom left). Consider for instance the eigendecomposition of $I_n$: we have $I_n=V I_n V^{-1}$ for all $V\in\text{GL}_n$, lots of freedom! In general only the diagonal matrix is sort of "unique". – user299843 Jan 16 at 16:13 | 2021-04-17T23:32:46 | {
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https://math.stackexchange.com/questions/4455521/help-on-randomizing-deterministic-functions | # Help on "randomizing" deterministic functions
I'm completely stuck on a problem and maybe someone of you out there has some thoughts on how I could proceed.
The setting is the following: Let $$f: \mathbb{R}^2 \to \mathbb{R},(x,y) \mapsto f(x,y)$$ be a continuous function. Next consider two iid random variables $$X,Y$$ on a probability space $$(\Omega,\mathcal{A}, \mathbb{P}$$). Then we could "randomize" $$f$$ via $$f(X,Y): \Omega \to \mathbb{R}, \omega \to f(X(\omega),Y(\omega)).$$
(We always identify $$\mathbb{R}$$ with the Borel-sigma-algebra as measurable space). Next let $$a \in Y(\Omega)$$ be arbitrary and set $$f(X,a): \Omega \to \mathbb{R}, \omega \to f(X(\omega),a).$$
Then, since $$X$$ and $$Y$$ are independent we can obtain the distributional equality $$\mathbb{P}^{f(X,a)} = \mathbb{P}^{f(X,Y) \mid Y = a}.$$
I hope, that everything sounds ok so far.
Now, further let $$L(a) = \mathbb{E}[(f(X,a))^2] = \Vert f(X,a) \Vert_{L^2}^2$$. In other words $$L(\cdot): Y(\Omega) \to \mathbb{R}, a \mapsto L(a)$$. Hence we could also "randomize" this function as $$L(Y): \Omega \to \mathbb{R}, \omega \mapsto L(Y(\omega)).$$
Now, I tried to compute the expectation of $$L(Y)$$ in terms of the expressions, but here is where I'm stuck at the moment. My idea was the following: $$\mathbb{E}(L(Y)) = \int_\Omega L(Y(\omega)) d\mathbb{P}(\omega) \\ = \int_\Omega \mathbb{E}[(f(X,Y(\omega))^2] d\mathbb{P}(\omega) \\ = \int_\Omega \int_\Omega (f(X(\tau), Y(\omega))^2 d\mathbb{P}(\tau)d\mathbb{P}(\omega).$$
From here on my plan was to proceed with Fubini's Theorem to change the order of integration, but to me it doesn't seem right, that $$\tau$$ and $$\omega$$ could be non-equal (since I defined $$f(X,Y)$$ such that $$X,Y$$ always have the same argument). Did I mess something up?
Or am I on the right path and could define the random variable $$f'(X,Y) : \Omega\times\Omega \to \mathbb{R}, (\tau,\omega) \mapsto f(X(\tau), Y(\omega))$$ on $$(\Omega\times\Omega, \mathcal{A} \otimes \mathcal{A}, \mathbb{P} \times \mathbb{P})$$ and then $$\mathbb{E}(L(Y)) = \mathbb{E}(f'(X,Y))$$? But then I'm still stuck, since I have no clue how I could connect $$f'(X,Y)$$ and $$f(X,Y)$$.
Maybe I'm just being a bit clumsy and things are actually much simpler. In any case, I would be grateful for any help.
The law of total expectation (a.k.a. law of iterated expectation, tower property, etc.) tells that
$$\mathbf{E}[L(Y)]=\mathbf{E}[\mathbf{E}[f(X,Y)^2\mid Y]]=\mathbf{E}[f(X,Y)^2].$$
Alternatively, sticking to the integral notation, we have
\begin{align*} \mathbf{E}[L(Y)] &= \iint_{\Omega^2} f(X(\tau),Y(\omega))^2 \, \mathbf{P}(\mathrm{d}\tau)\mathbf{P}(\mathrm{d}\omega) \\ &= \iint_{\mathbb{R}^2} f(x, y)^2 \, \mathbf{P}(X\in\mathrm{d}x)\mathbf{P}(Y\in\mathrm{d}y) \tag{by change of var.} \\ &= \iint_{\mathbb{R}^2} f(x, y)^2 \, \mathbf{P}((X,Y)\in\mathrm{d}x\mathrm{d}y) \tag{by independence} \\ &= \mathbf{E}[f(X,Y)^2]. \end{align*}
More fundamentally, this identity holds because the distribution of $$(\tau, \omega) \mapsto (X(\tau), Y(\omega))$$ on $$\Omega^2$$ is the same as that of $$(X, Y)$$ on $$\Omega$$ by the independence of $$X$$ and $$Y$$.
• First of all: Thanks a lot, that really helped! Especially your last point provides a nice view on independence, which I haven't realized before. May 21 at 16:22
• And a quick follow-up question: We could generalize the result, so that in the same setting for measurable functions $g,h$ and $$g(a) = \mathbb{E}(h(X,a))$$ we obtain $$\mathbb{E}(g(Y)) = \mathbb{E}(h(X,Y)),$$ or could something go wrong there? May 21 at 16:24
• @student7481 Of course the same argument will hold if $X$ and $Y$ are independent and the expectation of $h(X, Y)$ makes sense :). May 21 at 16:33 | 2022-06-26T12:25:27 | {
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http://rinconesdelatlantico.com/smith-and-xvpyywp/set-notation-domain-and-range-5dc2f5 | When using set notation, inequality symbols such as ≥ are used to describe the domain and range. Write your answer using set notation. Range: {y ≥ 0} (remember to focus on bottom to top of the graph for range of a continuous graph): The range is the set of output values for a function. An understanding of toolkit functions can be used to find the domain and range … Find the Domain and Range y = square root of x-4. But we are going to broaden our scope to determining both the domain and range of graphs, where we will need a new perspective when finding elements for each set. Write the range with proper notation. Write the range with proper notation. The range of f(x) = x 2 in set notation is: R: {y | y ≥ 0} R indicates range. Use function notation, evaluate functions for inputs in their domains, and interpret statements that use function notation in terms of a context. 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EN: pre-calculus-function-domain-calculator menu Pre Algebra Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Mean, Median & Mode Scientific Notation Arithmetics This video describes how to use Set-Builder notation to describe a set. Set-builder notation is commonly used to compactly represent a set of numbers. We write the domain in interval notation as {x ≥ 0}. This is because the range of a function includes 0 at x = 0. See § Notation below for more. Question 746428: Suppose that the relation is defined as follows. Intervals and interval notation. Is set notation domain and range the endpoint will be provided here infinite number of elements are written as intervals of.! Does not matter uses values within brackets to describe a set of all possible inputs the... On the x- axis of a function. domain calculator allows you take... 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https://math.stackexchange.com/questions/2281142/find-the-angle-mba | # Find the angle $MBA$.
Triangle ABC has $\angle CAB=30 ^\circ$ and $\angle CBA=70^\circ$. Point $M$ lies inside triangle $ABC$ so that $\angle BAM=20^\circ= \angle ACM$. Find $\angle MBA$.
I've already drawn the diagram, but I can't get the angle. Also, I would greatly appreciate if someone could help me with the latex. Thanks!
• You can try Ceva's theorem. en.wikipedia.org/wiki/Ceva%27s_theorem For example in the picture given in the wikipedia page, notice $\dfrac{AF}{FB}=\dfrac{\sin \angle ACO}{\sin \angle BCO}$ – Fan May 14 '17 at 21:45
• Is $M$ a center of a circle? – JJR May 14 '17 at 22:03
Observe that $\angle \, BCM = 60^{\circ}$. On line $CM$ take point $O$ such that $BC = CO$ so that $M$ is between $C$ and $O$. Then triangle $BCO$ is isosceles with $\angle \, BCO = 60^{\circ}$. Therefore $BCO$ is in fact equilateral with $$BO = CO = BC \,\, \text{ and } \,\, \angle \, BOC = \angle \, BCO = \angle \, CBO = 60^{\circ}$$ Since $$\angle \, BAC = 30^{\circ} = \frac{1}{2} \, \angle \, BOC$$ point $O$ is in fact the center of the circle superscribed around the triangle $ABC$ so $$AO = BO = CO = BC$$
Let $D$ be the point with properties $\angle \, DCM = 20^{\circ} = \angle \, ACM$ and $AC = DC$. The triangle $ACD$ is isosceles with $CM$ its angle bisector. Therefore, $CM$ is the orthogonal bisector of segment $AD$. Since $O$ is on $CM$, triangles $ACO$ and $DCO$ are congruent so $$\angle \, CAO = \angle \, CDO = 20^{\circ}$$ At the same time, triangles $ACM$ and $DCM$ are also congruent, so $$\angle \, CDM = \angle \, CAM = 10^{\circ}$$ which means that $DM$ is the angle bisector of $\angle \, CDO$.
If you calculate the angles (which is straight-forward) you find that $$\angle \, BNC = 80^{\circ} = \angle \, DNO$$ Furthermore, again a simple angle chase yields $\angle \, DON = 80^{\circ}$. Thus $$\angle \, DON = 80^{\circ} = \angle \, DNO$$ so triangle $DNO$ is isosceles with $DN = DO$. however, $DM$ is the angle bisector of $angle \, NDO = \angle \, CDO$ so $CM$ is the orthogonal bisector of segment $NO$. Thus, $MN = MO$. However, $\angle \, NOM = \angle\, BOC = 60^{\circ}$ so triangle $MNO$ is equilateral. Hence $$MN = MO = NO$$
Finally, we can conclude that triangles $MBO$ and $NCO$ are congruent because $MO = NO, \,\, BO = CO$ and $\angle \, BOM = \angle \, CON = 60^{\circ}$. Consequently, $$\angle \, MBO = \angle \, NCO = 20^{\circ}$$ Therefore $$\angle \, MBA = \angle \, MBO + \angle \, OBA = 20^{\circ} + 10^{\circ} = 30^{\circ}$$
• Thanks for the non trig solution! This is the one I was looking for! :D – user406996 May 19 '17 at 2:20
Let MBA=x. Since AM, CM and BM are collinear, we use the sine form of Ceva's Theorem. $$sinACMsinBAMsinCBM=sinCAMsinABMsinBCM$$ $$sin20\sin20\sin(70-x)=sin10\ sinx\sin60$$ You can obtain all these angles by drawing the triangle and then angle chasing. Notice that x must be a multiple of 10 since problem makers don't want to be assholes and force you to do tedious calculations. A rough drawing gives that x must be 20 or 30, and x=30 solves the above equation found by using Ceva's Theorem. Thus MBA is 30.
To make it rigorous, notice \begin{align} \sin 10^{\circ}\sin 60^{\circ}&=(\cos 50^{\circ}-\cos 70^{\circ})/2\\ &=(\sin40^{\circ}-\sin 20^{\circ})/2\\ &=\frac{\sin 20^{\circ}}{2}(2\cos 20^{\circ}-1)\\ &=\sin 20^{\circ}(\cos 20^{\circ}-\sin 30^{\circ})\\ &=\sin 20^{\circ}(\sin 70^{\circ}-\sin 30^{\circ})\\ &=2\sin 20^{\circ}\sin 20^{\circ}\cos 50^{\circ}. \end{align}
We can see $x=30^{\circ}$.
• Good job, but how do you know where to go at each step? You could've taken a wrong turn in the middle steps and failed to arrive at the final result. – Display name May 14 '17 at 23:10
• @Displayname It is true. Luckily there are only two main steps here, and your goal is to create two $20^\circ$ from $10^\circ$ and $60^\circ$. So actually options are limited. – Fan May 14 '17 at 23:14 | 2019-12-07T15:01:38 | {
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https://www.talkstats.com/threads/probability-that-x-restaurants-will-be-visited-if-y-people-choose-independently.57957/ | # Probability that X restaurants will be visited if Y people choose independently
#### JPC
##### New Member
I am trying to determine for Y people selecting from X restaurants (with an equal probability) what the expected value of the number of restaurants chosen would be. All decisions are independent.
Ex: 2 people and 2 restaurants - E(X) = 1(.5) + 2(.5) = 1.5 restaurants will be visited on average
EX: 3 people and 3 restaurants - E(X) = 1(1/9) + 2(6/9) + 3(2/9) = 2.11 restaurants will be visited on average
This is simple enough for small values like these. However, I would like to be able to do this for any combination of X and Y and create a table of expected values. For larger numbers I am having trouble determining the probabilities to plug into the expected value formula. When calculating the probabilities, I know that the denominator will always be X^Y but I am really struggling to find a way to calculate the numerator without writing down every possible combination.
My ultimate goal would be to create a table in excel that for any given number of restaurants and people, returns the expected value.
I'm running into a major road block so any help would be appreciated.
#### BGM
##### TS Contributor
Let
$$n$$ be the total number of restaurants.
$$X$$ be the number of restaurants visited by at least 1 person
$$k$$ be the total number of people
$$B_i, i = 1, 2, \ldots, n$$ be the indicators of the $$i$$-th restaurant being visited by at least 1 person
If you are only interested in the expected value $$E[X]$$, we can try to calculate it without actually calculating the pmf of $$X$$. The trick is to consider the following decomposition, having the same spirit as the one we do in calculating the expected value of hypergeometric distribution:
$$X = \sum_{i=1}^n B_i$$
Note those $$B_i$$ are not independent; but this does not matter - all we need is the linearity of the expectation. Further note that
$$E[B_i] = \Pr\{B_i = 1\} = 1 - \left(1 - \frac {1} {n}\right)^k$$
and therefore
$$E[X] = nE[B_1] = n\left[1 - \left(1 - \frac {1} {n}\right)^k \right]$$
#### JPC
##### New Member
That worked perfectly. Thank you! Glad I asked because I was not headed in the right direction. | 2022-12-08T22:56:19 | {
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https://math.stackexchange.com/questions/2984584/setting-up-the-triple-integrals-for-a-solid-given-by-yz-2-and-x-4-y2 | # Setting up the triple integrals for a solid given by $y+z=2$ and $x=4-y^2$?
I'm trying to set up all six triple integrals to find the volume of the solid that lies in the first octant bounded by the coordinate planes, the plane $$y+z=2$$, and the cylinder $$x=4-y^2$$.
$$3$$D-Graph:
I've been able to set up the triple integrals for every other combination except for $$dydxdz$$ and $$dydzdx$$, which I am struggling to find the bounds for.
I know the volume of the solid is $$20/3$$, but no matter what I try I haven't been able to come up with $$dydxdz$$ and $$dydzdx$$ to produce that result. Looking at the $$3$$D graph, the projection onto the $$xz$$-plane appears to be the equation $$z=2-(4-x)^{1/2}$$ and $$x=4$$ (see image below), but I don't think that is correct since it isn't producing the correct volume.
$$xz$$-plane projection?
If that isn't correct, what does the projection look like?
I'd appreciate any help. Thank you!
• It would be nice if those voting to close this question as off-topic would shed some light on what other context or details should OP provide that they haven't already. – Ennar Nov 5 '18 at 0:15
## 1 Answer
The boundary of the solid is given by the following inequalities:
$$x,y,z\geq 0,\\ y + z \leq 2,\\ x+y^2\leq 4.$$
If you let $$y = 0$$ in the above, it follows $$0\leq x\leq 4$$ and $$0\leq z\leq 2$$, so the projection onto $$xz$$-plane is rectangle $$[0,4]\times[0,2]$$.
So, what remains is to express the boundary of the solid directly above the rectangle as function $$y=f(x,z)$$. We have two conditions on $$y$$:
$$0\leq y\leq 2-z,\\ 0\leq y\leq \sqrt{4-x},$$
and both of them need to be satisfied. Thus, $$0\leq y \leq \min\{\sqrt{4-x},2-z\}$$ and therefore, you need to integrate
$$\int_0^4\int_0^2\int_0^{\min\{\sqrt{4-x},2-z\}}1\,dydzdx.$$
Now, you might be thinking what in the world to do with that. But before I explain how to do it, let me digress with analogous $$2$$-dimensional problem.
Let's say that we want to calculate area of the shape in the first quadrant, bounded by $$y=x$$ and $$y = 1 -x^2$$.
One way to do it is to write the area as the integral $$\int_0^{\frac{-1+\sqrt 5}2}\int_y^{\sqrt{1-y}}1\,dxdy$$ but you can also do it this way
$$\int_0^1\int_0^{\min\{x,1-x^2\}}1\,dydx$$
as I'm sure you've seen before. What? No? Ok, ok, I'll stop joking around.
We have inequalities $$x,y\geq 0,\\ y\leq 1-x^2,\\ y\leq x.$$ If you let $$y = 0$$ in the above, you get $$0\leq x\leq 1$$, so the segment $$[0,1]$$ is the projection onto $$x$$-coordinate. We should express $$y$$ as function of $$x$$, and from the inequalities, we get $$0\leq y \leq \min\{x,\sqrt{1-x^2}\}$$, as I claimed above.
But this is not how we solve this usually. What you do is split the integral into two pieces:
$$\int_0^{\frac{-1+\sqrt 5}2}\int_0^x1\,dydx + \int_{\frac{-1+\sqrt 5}2}^1\int_0^{1-x^2}1\,dydx.$$
This is the same thing as the last integral. We split the segment $$[0,1]$$ into $$[0,\frac{-1+\sqrt 5}2]$$ and $$[\frac{-1+\sqrt 5}2,1]$$. This is because
$$\min\{x,\sqrt{1-x^2}\}=\begin{cases} x, & x\in [0,\frac{-1+\sqrt 5}2],\\ 1-x^2, & x\in [\frac{-1+\sqrt 5}2,1]. \end{cases}$$
Back to the original problem.
So, learning from $$2$$-dimensional case, we now know that we need to separate the rectangle $$[0,4]\times [0,2]$$ in the $$xz$$-plane into two parts: one where $$\sqrt{4-x}\leq 2-z$$ and the other where $$2-z\leq \sqrt{4-x}$$. We can equate those two things to get a curve given by $$\sqrt{4-x} = 2 - z$$. It looks like this:
In the green area we have $$\sqrt{4-x}\leq 2-z$$ and in the blue area we have $$2-z\leq \sqrt{4-x}$$. Therefore,
\begin{align} \int_0^4\int_0^2\int_0^{\min\{\sqrt{4-x},2-z\}}1\,dydzdx &= \color{green}{\int_0^4\int_0^{2-\sqrt{4-x}}}\int_0^{\sqrt{4-x}}1\,dydzdx + \color{blue}{\int_0^4\int_{2-\sqrt{4-x}}^2}\int_0^{2-z}1\,dydzdx\\ &= \color{green}{\int_0^2\int_{4-(z-2)^2}^4}\int_0^{\sqrt{4-x}}1\,dydxdz+\color{blue}{\int_0^2\int_0^{4-(z-2)^2}}\int_0^{2-z}1\,dydxdz\\ &=\frac{20}3. \end{align} | 2019-08-19T23:01:21 | {
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https://math.stackexchange.com/questions/1972824/find-an-equation-that-using-fixed-point-iteration-converges-to-1-02 | Find an equation that using Fixed Point Iteration converges to -1.02
Let $f(x) = e^{x-2} + x^3 - x$, then using fixed point iteration, find all of the roots. I've already found two equations which converge to 0.163822 and ~0.788941.
The equation which converges toward 0.163822 is: $e^{x-2} + x^3$ with a guess of $0.1$
The equation which converges toward ~0.788941 is: $\frac{-e^{x-2} + x}{x}^{\frac{-1}{2}}$ with a guess of $1.5$.
I can't seem to find an equation which converges to the last root of approximately -1.02.
• Try another initial guess. Then, the iteration might converge to that root. – Peter Oct 17 '16 at 16:47
• The third possible function is $\ln(x-x^3)+2$ – Peter Oct 17 '16 at 16:50
• @Peter, I've tried that function, but it does not converge to -1. Is there a particular guess you had in mind? I've tried guesses of -1, 0, 0.1, and 1 and they all seem to go to infinity – user3370201 Oct 17 '16 at 17:07
• The fixpoint-iteration might fail for this root. I did not find a function and a guess doing the job yet. – Peter Oct 17 '16 at 17:45
• By the way, instead of $0$, the third root is $0.1638$. – Peter Oct 17 '16 at 17:50
Try $f(x)=\sqrt[3]{x-e^{x-2}}$, starting from a negative value. The rationale is that the cubic root has a slope less than $1$ for values less than $-1$.
$$-2 \\ -1.26375541197 \\ -1.09195226149 \\ -1.0438385854 \\ -1.02961125511 \\ -1.02533462938 \\ -1.02404271663 \\ -1.02365186038 \\ -1.0235335567 \\ -1.02349774382 \\ \vdots$$
$$1 \\ 0.858222649309 \\ 0.813807987157 \\ 0.798134216479 \\ 0.792376247446 \\ 0.790229882858 \\ 0.789425446205 \\ 0.789123338776 \\ 0.789009795416 \\ 0.788967109326 \\ \vdots$$ | 2019-10-17T00:14:14 | {
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http://math.stackexchange.com/questions/215318/cauchy-sequences-is-my-proof-correct | # Cauchy sequences…is my proof correct?
Let $(x_n)_{n∈ℕ}$ and $(y_n)_{n∈ℕ}$ be Cauchy sequences of real numbers.
Show, without using the Cauchy Criterion, that if $z_n=x_n+y_n$, then $(z_n)_{n∈ℕ}$ is a Cauchy sequence of real numbers.
Here's my attempt at a proof:
Let $(x_n)$ and $(y_n)$ be Cauchy sequences. Let $(z_n)$ be a sequence and let $z_n=x_n+y_n$.
Since $(x_n)$ and ($y_n)$ are Cauchy, $∃N∈ℕ$ such that $|x_n-x_m|<ε/2$ and $|y_n-y_m|<ε/2$ for $n,m≥N$.
Let $n,m≥N$ and let $z_n,z_m∈(z_n)$. Then, $$|z_n-z_m|=|x_n-x_m|+|y_n-y_m| <ε/2+ε/2=ε.$$ Therefore, $|z_n-z_m|<ε$ for all $n,m≥N$ and hence, $(z_n)$ is a Cauchy sequence of real numbers.
Is this correct? Any input is appreciated.
Thanks.
-
It looks OK the proof! – Atahualpa Inca Oct 17 '12 at 2:04
It's correct except for a minor detail. The triangle inequality results in a $\leq$ not a $=$. – copper.hat Oct 17 '12 at 2:16
Thanks for pointing that out. – Alti Oct 17 '12 at 2:35
add comment
## 2 Answers
The only thing you have to correct is $$|z_n-z_m|=|(x_n-x_m)+(y_n-y_m)|\le|x_n-x_m|+|y_n-y_m| <ε/2+ε/2=ε.$$
-
add comment
This proof looks correct to me.
-
add comment | 2013-12-19T16:46:01 | {
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https://cs.stackexchange.com/questions/66265/number-of-n-times-n-binary-matrices-whose-rows-and-columns-sum-to-at-most-m/66269 | Number of $n \times n$ binary matrices whose rows and columns sum to at most $m$
How many matrices satisfy the following constraints?
1. $n$ rows
2. $n$ columns
3. Cell values are either $0$ or $1$
4. Sum of any row is at most $m$
5. sum of any column is at most $m$
Is there a formula or an efficient algorithm to solve this problem?
• I can solve the cases $m=0$ and $m=1$. Nov 21 '16 at 7:45
• This is the same as the number of digraphs where the in-degree and out-degree are bounded by $m$. Nov 21 '16 at 7:46
• This has some connection to Latin Squares. In particular, we can take an $n\times n$ Latin Square, set the numbers from 1 to $m$ to be 1, and the numbers from $m+1$ to $n$ to be 0. Good news is that this gives a lower bound on your problem. Two problems: 1. the number of Latin squares of size $n$ is an open problem (known bounds are thought to be fairly weak), and 2. it's not clear to me how tight this bound is. In particular, while every Latin square is a solution to your problem, I'm not sure when a solution to this problem can be turned into a Latin Square.
– SamM
Nov 21 '16 at 10:12
Given $1 \leq m \leq n$, we want to determine the cardinality of the following set
$$\{ \mathrm X \in \{0,1\}^{n \times n} \mid \mathrm X 1_n \leq m 1_n \,\land\, \mathrm 1_n^{\top} \mathrm X \leq m \mathrm 1_n^{\top} \}$$
Vectorizing, $\tilde{\mathrm x} := \mbox{vec} (\mathrm X)$, we obtain
$$\bigg\{ \tilde{\mathrm x} \in \{0,1\}^{n^2} : \begin{bmatrix} 1_n^{\top} \otimes \mathrm I_n\\ \mathrm I_n \otimes \mathrm 1_n^{\top}\end{bmatrix} \tilde{\mathrm x} \leq m \begin{bmatrix} \mathrm 1_n\\ \mathrm 1_n\end{bmatrix} \bigg\}$$
or,
$$\bigg\{ \tilde{\mathrm x} \in [0,1]^{n^2} : \begin{bmatrix} 1_n^{\top} \otimes \mathrm I_n\\ \mathrm I_n \otimes \mathrm 1_n^{\top}\end{bmatrix} \tilde{\mathrm x} \leq m \begin{bmatrix} \mathrm 1_n\\ \mathrm 1_n\end{bmatrix} \bigg\} \cap \mathbb Z^{n^2}$$
Thus, we want to count the number of integer points inside a convex polytope, or, in other words, to count the number of solutions a given integer program (IP) has. Take a look at De Loera's survey [0] and the references therein.
[0] Jesús De Loera, The Many Aspects of Counting Lattice Points in Polytopes, 2005. | 2022-01-27T15:43:12 | {
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https://gateoverflow.in/8314/gate2015-1-44 | 2.4k views
Compute the value of:
$$\large \int_{\frac{1}{\pi}}^{\frac{2}{\pi}}\frac{\cos(1/x)}{x^{2}}dx$$
in Calculus
edited | 2.4k views
For the integrand $\frac{\cos(1/x)}{x^2}$, substitute $u = \frac1 x$ and $\def\d{\,\mathrm{d}} \d u = -\frac1{x^2}\d x$.
This gives a new lower bound $u = \frac1{1/\pi} = \pi$ and upper bound $u = \frac1{2/\pi} = \frac{\pi}{2}$. Now, our integral becomes:
$I= - \int\limits_\pi^{\pi/2} \cos(u) \d u$
$\;\;= \int\limits_{\pi/2}^\pi \cos(u)\d u$
Since the antiderivative of $\cos(u)$ is $\sin(u)$, applying the fundamental theorem of calculus, we get:
$I= \sin(u)\;\mid _{\pi/2}^\pi$
$\;\;= \sin(\pi) - \sin \left ( \frac\pi 2\right )$
$\;\;= 0 - 1$
$\;\; = {-1}$
by Active (1.6k points)
edited
\begin{align*} \large \int_{\frac{1}{\pi}}^{\frac{2}{\pi}}\frac{cos(1/x)}{x^{2}}dx \end{align*}
let, $\frac{1}{x} = t$ then, $\frac{-1}{x^2}dx = dt$
\begin{align*} \int_{\frac{1}{\pi}}^{\frac{2}{\pi}}\frac{cos(1/x)}{x^{2}}dx &=-\int_{\pi}^{\frac{\pi}{2}}cos(t)\ dt\\ &= -\Big( \sin t \Big)_{\pi}^{\pi/2}\\ &= -\Big( 1-0 \Big)\\ &= -1 \end{align*}
by Boss (30.8k points)
+3
But in exam they give incorrect on -1 as answer. Please improve that! I'm talking about GO test for this paper!
+1
$\int_{b}^{a}f(x)dx=-\int_{a}^{b}f(x)dx$ | 2020-01-22T09:05:32 | {
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https://sajonhussain.com/chef-a-ulmk/dca021-what-is-the-end-behavior-of-the-polynomial-function%3F | http://cnx.org/contents/[email protected], $f\left(x\right)=5{x}^{4}+2{x}^{3}-x - 4$, $f\left(x\right)=-2{x}^{6}-{x}^{5}+3{x}^{4}+{x}^{3}$, $f\left(x\right)=3{x}^{5}-4{x}^{4}+2{x}^{2}+1$, $f\left(x\right)=-6{x}^{3}+7{x}^{2}+3x+1$. In the following video, we show more examples of how to determine the degree, leading term, and leading coefficient of a polynomial. For the function $g\left(t\right)$, the highest power of t is 5, so the degree is 5. This calculator will determine the end behavior of the given polynomial function, with steps shown. Degree of a polynomial function is very important as it tells us about the behaviour of the function P(x) when x becomes very large. Polynomial end behavior is the direction the graph of a polynomial function goes as the input value goes "to infinity" on the left and right sides of the graph. $\begin{array}{l} f\left(x\right)=-3{x}^{2}\left(x - 1\right)\left(x+4\right)\\ f\left(x\right)=-3{x}^{2}\left({x}^{2}+3x - 4\right)\\ f\left(x\right)=-3{x}^{4}-9{x}^{3}+12{x}^{2}\end{array}$, The general form is $f\left(x\right)=-3{x}^{4}-9{x}^{3}+12{x}^{2}$. We can describe the end behavior symbolically by writing, $\begin{array}{c}\text{as } x\to -\infty , f\left(x\right)\to -\infty \\ \text{as } x\to \infty , f\left(x\right)\to \infty \end{array}$. Our mission is to provide a free, world-class education to anyone, anywhere. 9.f (x)-4x -3x2 +5x-2 10. - the answers to estudyassistant.com −x 2 • x 2 = - x 4 which fits the lower left sketch -x (even power) so as x approaches -∞, Q(x) approaches -∞ and as x approaches ∞, Q(x) approaches -∞ This is called writing a polynomial in general or standard form. The degree of the polynomial is the highest power of the variable that occurs in the polynomial; it is the power of the first variable if the function is in general form. In general, the end behavior of a polynomial function is the same as the end behavior of its leading term, or the term with the largest exponent. The leading coefficient is significant compared to the other coefficients in the function for the very large or very small numbers. As the input values x get very small, the output values $f\left(x\right)$ decrease without bound. There are four possibilities, as shown below. A polynomial function is a function that can be written in the form, $f\left(x\right)={a}_{n}{x}^{n}+\dots+{a}_{2}{x}^{2}+{a}_{1}x+{a}_{0}$. This is determined by the degree and the leading coefficient of a polynomial function. In the following video, we show more examples that summarize the end behavior of polynomial functions and which components of the function contribute to it. Given the function $f\left(x\right)=0.2\left(x - 2\right)\left(x+1\right)\left(x - 5\right)$, express the function as a polynomial in general form and determine the leading term, degree, and end behavior of the function. The leading coefficient is the coefficient of that term, $–4$. But the end behavior for third degree polynomial is that if a is greater than 0-- we're starting really small, really low values-- and as a becomes positive, we get to really high values. For the function $h\left(p\right)$, the highest power of p is 3, so the degree is 3. Since n is odd and a is positive, the end behavior is down and up. A polynomial function is a function that can be expressed in the form of a polynomial. When a polynomial is written in this way, we say that it is in general form. Each ${a}_{i}$ is a coefficient and can be any real number. Identifying End Behavior of Polynomial Functions Knowing the degree of a polynomial function is useful in helping us predict its end behavior. This is called the general form of a polynomial function. Which graph shows a polynomial function of an odd degree? Page 2 … Step-by-step explanation: The first step is to identify the zeros of the function, it means, the values of x at which the function becomes zero. Although the order of the terms in the polynomial function is not important for performing operations, we typically arrange the terms in descending order based on the power on the variable. The end behavior of a polynomial function is the behavior of the graph of f (x) as x approaches positive infinity or negative infinity. It is not always possible to graph a polynomial and in such cases determining the end behavior of a polynomial using the leading term can be useful in understanding the nature of the function. Which of the following are polynomial functions? •Prerequisite skills for this resource would be knowledge of the coordinate plane, f(x) notation, degree of a polynomial and leading coefficient. What is 'End Behavior'? g, left parenthesis, x, right parenthesis, equals, minus, 3, x, squared, plus, 7, x. The shape of the graph will depend on the degree of the polynomial, end behavior, turning points, and intercepts. What is the end behavior of the graph? The slick is currently 24 miles in radius, but that radius is increasing by 8 miles each week. The function f(x) = 4(3)x represents the growth of a dragonfly population every year in a remote swamp. This relationship is linear. A polynomial is generally represented as P(x). The degree and the leading coefficient of a polynomial function determine the end behavior of the graph. The end behavior is to grow. We can combine this with the formula for the area A of a circle. Check your answer with a graphing calculator. The leading term is $-3{x}^{4}$; therefore, the degree of the polynomial is 4. $A\left(r\right)=\pi {r}^{2}$. Since the leading coefficient of this odd-degree polynomial is positive, then its end-behavior is going to mimic that of a positive cubic. The end behavior of a polynomial function is the behavior of the graph of f (x) as x approaches positive infinity or negative infinity. Polynomial functions have numerous applications in mathematics, physics, engineering etc. NOT A, the M What is the end behavior of the graph of the polynomial function y = 7x^12 - 3x^8 - 9x^4? As x approaches positive infinity, $f\left(x\right)$ increases without bound; as x approaches negative infinity, $f\left(x\right)$ decreases without bound. To determine its end behavior, look at the leading term of the polynomial function. Identify the degree of the function. Donate or volunteer today! You can use this sketch to determine the end behavior: The "governing" element of the polynomial is the highest degree. Finally, f(0) is easy to calculate, f(0) = 0. With this information, it's possible to sketch a graph of the function. ... Simplify the polynomial, then reorder it left to right starting with the highest degree term. We want to write a formula for the area covered by the oil slick by combining two functions. Because the power of the leading term is the highest, that term will grow significantly faster than the other terms as x gets very large or very small, so its behavior will dominate the graph. And these are kind of the two prototypes for polynomials. The leading term is $0.2{x}^{3}$, so it is a degree 3 polynomial. Which function is correct for Erin's purpose, and what is the new growth rate? Composing these functions gives a formula for the area in terms of weeks. * * * * * * * * * * Definitions: The Vocabulary of Polynomials Cubic Functions – polynomials of degree 3 Quartic Functions – polynomials of degree 4 Recall that a polynomial function of degree n can be written in the form: Definitions: The Vocabulary of Polynomials Each monomial is this sum is a term of the polynomial. So, the end behavior is, So the graph will be in 2nd and 4th quadrant. End behavior of polynomial functions helps you to find how the graph of a polynomial function f (x) behaves (i.e) whether function approaches a positive infinity or a negative infinity. The leading coefficient is $–1$. Describe the end behavior of the polynomial function in the graph below. Also, be careful when you write fractions: 1/x^2 ln (x) is 1 x 2 ln ( x), and 1/ (x^2 ln (x)) is 1 x 2 ln ( x). Identify the degree of the polynomial and the sign of the leading coefficient So, the end behavior is, So the graph will be in 2nd and 4th quadrant. The definition can be derived from the definition of a polynomial equation. For example in case of y = f (x) = 1 x, as x → ±∞, f (x) → 0. Did you have an idea for improving this content? It has the shape of an even degree power function with a negative coefficient. Describing End Behavior of Polynomial Functions Consider the leading term of each polynomial function. The given function is ⇒⇒⇒ f (x) = 2x³ – 26x – 24 the given equation has an odd degree = 3, and a positive leading coefficient = +2 Obtain the general form by expanding the given expression $f\left(x\right)$. In general, you can skip parentheses, but be very careful: e^3x is e 3 x, and e^ (3x) is e 3 x. Play this game to review Algebra II. A y = 4x3 − 3x The leading ter m is 4x3. ... Use the degree of the function, as well as the sign of the leading coefficient to determine the behavior. The degree and the leading coefficient of a polynomial function determine the end behavior of the graph. Describe the end behavior of a polynomial function. The degree and the sign of the leading coefficient (positive or negative) of a polynomial determines the behavior of the ends for the graph. $h\left(x\right)$ cannot be written in this form and is therefore not a polynomial function. The leading term is the term containing that degree, $-4{x}^{3}$. The leading term is the term containing that degree, $-{p}^{3}$; the leading coefficient is the coefficient of that term, $–1$. The first two functions are examples of polynomial functions because they can be written in the form $f\left(x\right)={a}_{n}{x}^{n}+\dots+{a}_{2}{x}^{2}+{a}_{1}x+{a}_{0}$, where the powers are non-negative integers and the coefficients are real numbers. Identify the degree and leading coefficient of polynomial functions. Explanation: The end behavior of a function is the behavior of the graph of the function f (x) as x approaches positive infinity or negative infinity. $g\left(x\right)$ can be written as $g\left(x\right)=-{x}^{3}+4x$. Describe the end behavior and determine a possible degree of the polynomial function in the graph below. Answer: 2 question What is the end behavior of the graph of the polynomial function f(x) = 2x3 – 26x – 24? Learn how to determine the end behavior of the graph of a polynomial function. How do I describe the end behavior of a polynomial function? Because of the form of a polynomial function, we can see an infinite variety in the number of terms and the power of the variable. Degree, Leading Term, and Leading Coefficient of a Polynomial Function . Identify the degree, leading term, and leading coefficient of the polynomial $f\left(x\right)=4{x}^{2}-{x}^{6}+2x - 6$. For achieving that, it necessary to factorize. Find the End Behavior f(x)=-(x-1)(x+2)(x+1)^2. If a is less than 0 we have the opposite. 1. A polynomial of degree $$n$$ will have at most $$n$$ $$x$$-intercepts and at most $$n−1$$ turning points. The degree is 6. $\begin{array}{l}A\left(w\right)=A\left(r\left(w\right)\right)\\ A\left(w\right)=A\left(24+8w\right)\\ A\left(w\right)=\pi {\left(24+8w\right)}^{2}\end{array}$, $A\left(w\right)=576\pi +384\pi w+64\pi {w}^{2}$. As the input values x get very large, the output values $f\left(x\right)$ increase without bound. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. For the function $f\left(x\right)$, the highest power of x is 3, so the degree is 3. The leading term is the term containing that degree, $5{t}^{5}$. The leading coefficient is the coefficient of the leading term. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Because the power of the leading term is the highest, that term will grow significantly faster than the other terms as x gets very large or very small, so its behavior will dominate the graph. Q. The end behavior of a polynomial is the behavior of the graph of f(x) as x approaches positive infinity or negative infinity.The degree and the leading coefficient of a polynomial determine the end behavior of the graph. Therefore, the end-behavior for this polynomial will be: "Down" on the left and "up" on the right. A polynomial function is made up of terms called monomials; If the expression has exactly two monomials it’s called a binomial.The terms can be: Constants, like 3 or 523.. Variables, like a, x, or z, A combination of numbers and variables like 88x or 7xyz. $f\left(x\right)$ can be written as $f\left(x\right)=6{x}^{4}+4$. We’d love your input. $\begin{array}{l} f\left(x\right)=3+2{x}^{2}-4{x}^{3} \\g\left(t\right)=5{t}^{5}-2{t}^{3}+7t\\h\left(p\right)=6p-{p}^{3}-2\end{array}$. Each product ${a}_{i}{x}^{i}$ is a term of a polynomial function. Answer to Use what you know about end behavior to match the polynomial function with its graph. The end behavior of a function describes the behavior of the graph of the function at the "ends" of the x-axis. The end behavior is down on the left and up on the right, consistent with an odd-degree polynomial with a positive leading coefficient. In general, you can skip the multiplication sign, so 5 x is equivalent to 5 ⋅ x. The leading coefficient is the coefficient of that term, 5. An oil pipeline bursts in the Gulf of Mexico causing an oil slick in a roughly circular shape. If you're seeing this message, it means we're having trouble loading external resources on our website. Knowing the leading coefficient and degree of a polynomial function is useful when predicting its end behavior. In this example we must concentrate on 7x12, x12 has a positive coefficient which is 7 so if (x) goes to high positive numbers the result will be high positive numbers x → ∞,y → ∞ Enter the polynomial function into a graphing calculator or online graphing tool to determine the end behavior. Learn what the end behavior of a polynomial is, and how we can find it from the polynomial's equation. The radius r of the spill depends on the number of weeks w that have passed. The given polynomial, The degree of the function is odd and the leading coefficient is negative. The given polynomial, The degree of the function is odd and the leading coefficient is negative. This formula is an example of a polynomial function. This end behavior of graph is determined by the degree and the leading co-efficient of the polynomial function. Given the function $f\left(x\right)=-3{x}^{2}\left(x - 1\right)\left(x+4\right)$, express the function as a polynomial in general form and determine the leading term, degree, and end behavior of the function. The end behavior of a polynomial function is determined by the degree and the sign of the leading coefficient. The highest power of the variable of P(x)is known as its degree. In other words, the end behavior of a function describes the trend of the graph if we look to the right end of the x-axis (as x approaches +∞ ) and to the left end of the x-axis (as x approaches −∞ ). In determining the end behavior of a function, we must look at the highest degree term and ignore everything else. For any polynomial, the end behavior of the polynomial will match the end behavior of the term of highest degree. URL: https://www.purplemath.com/modules/polyends.htm. Start by sketching the axes, the roots and the y-intercept, then add the end behavior: f(x) = 2x 3 - x + 5 Khan Academy is a 501(c)(3) nonprofit organization. To determine its end behavior, look at the leading term of the polynomial function. We often rearrange polynomials so that the powers on the variable are descending. $\begin{array}{c}f\left(x\right)=2{x}^{3}\cdot 3x+4\hfill \\ g\left(x\right)=-x\left({x}^{2}-4\right)\hfill \\ h\left(x\right)=5\sqrt{x}+2\hfill \end{array}$. Polynomial Functions and End Behavior On to Section 2.3!!! The leading coefficient is the coefficient of the leading term. Show Instructions. Identify the term containing the highest power of. The end behavior of a polynomial function is the behavior of the graph of f(x) as x approaches positive infinity or negative infinity. Let n be a non-negative integer. The end behavior of a function f describes the behavior of the graph of the function at the "ends" of the x-axis. The end behavior of a polynomial function is the same as the end behavior of the power function represented by the leading term of the function. This is a quick one page graphic organizer to help students distinguish different types of end behavior of polynomial functions. Knowing the leading coefficient and degree of a polynomial function is useful when predicting its end behavior. Free, world-class education to anyone, anywhere world-class education to anyone anywhere! The coefficient of a positive cubic that have passed real number this message it... A positive leading coefficient is the coefficient of a polynomial function determine the end behavior of the x-axis degree! How do I describe the end behavior f ( x ) = −3x2 +7x about... } [ /latex ] leading coefficient of the function, as well as the sign of polynomial... 4X3 − 3x the leading term of the graph of the function, well. 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Two prototypes for polynomials the general form by expanding the given polynomial, then its end-behavior going. Domains *.kastatic.org and *.kasandbox.org are unblocked polynomial is, so 5 x is equivalent 5... Calculator or online graphing tool to determine the behavior function, as well as the sign of the polynomial the. Variable are descending leading term is [ latex ] { a } _ I! Polynomial functions and determine a possible degree of a polynomial function ] –4 [ ]. Sketch to determine what that is degree of the polynomial function a degree! Improving this content graph will be: down '' on the right consistent... Be: down '' on the degree, leading term of highest degree correct for erin 's,. Improving this content easy to calculate, f ( x ) _ { I } [ /latex ] we that. X 2 to determine the end behavior of the spill depends on number... Is written in this way, we must look at the governing '' element of the polynomial is... Have an idea for improving this content a web filter, please make that! Mexico causing an oil slick in a roughly circular shape and what is the end behavior of the polynomial function? be any real number obtain general. | 2021-07-26T23:47:50 | {
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http://math.stackexchange.com/questions/182428/proving-that-22n-5-is-always-composite-by-working-modulo-3/182431 | # Proving that $2^{2^n} + 5$ is always composite by working modulo $3$
By working modulo 3, prove that $2^{2^n} + 5$ is always composite for every positive integer n.
No need for a formal proof by induction, just the basic idea will be great.
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This can be considered as a generalization: show $2^{2^n} + 1 \equiv 2 \pmod{2^{2^m} + 1}$ with $0<m<n$ – Martin Sleziak Aug 14 '12 at 13:03
Obviously $2^2 \equiv 1 \pmod 3$.
If you take the above congruence to the power of $k$ you get $$(2^2)^k=2^{2k} \equiv 1^k=1 \pmod 3$$ which means that $2$ raised to any even power is congruent to $1$ modulo $3$.
What can you say about $2^{2k}+5$ then modulo 3?
It is good to keep in mind that you can take powers of congruences, multiply them and add them together.
If you have finished the above, you have shown that $3\mid 2^{2k}+5$. Does this imply that $2^{2k}+5$ is composite?
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Then $2^{2k} + 5 ≡ 0 mod 3$ so $3|2^{2k}+5$ and thus $2^{2k}+5$ can't be prime and is composite. Correct? – MinaHany Aug 14 '12 at 13:43
What you wrote is correct. Maybe you should also mention that $2^{2k}+5>3$. (If you know that $3<s$ and $3\mid s$ then $s$ is composite. If you omit the condition that $3<s$, then this is not true, since you can take $s=3$, which is prime.) – Martin Sleziak Aug 14 '12 at 13:46
Got it! Thank you sir. – MinaHany Aug 14 '12 at 13:49
Thanks @RossMillikan, I've corrected that. – Martin Sleziak Aug 14 '12 at 14:24
The basic idea is, work modulo 3. What happens, modulo 3, when you raise 2 to an even power?
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I have no idea actually, I'm pretty new to mod arithmetic – MinaHany Aug 14 '12 at 13:09
You're not supposed to have an idea, you're supposed to actually do it! Raise 2 to an even power! See what happens modulo 3! Raise 2 to another even power! See what happens modulo 3! Repeat, until you do have an idea! Then try to prove it! – Gerry Myerson Aug 14 '12 at 13:16
Hint: Rewrite the base using the fact that $2\equiv -1 \bmod 3$.
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+1 This is even shorter and simpler. – DonAntonio Aug 14 '12 at 18:05
In order to work out this problem, we start by noticing $2^2\equiv 1 ~~~(\text{mod } 3)$.
What this means is $2^2$ (which is $4$) is $1$ greater than some multiple of $3$ (that multiple, in this case is obviously, $3$)
So if $2^2\equiv 1 ~~~(\text{mod } 3)~~\implies (2^2)^n\equiv (1)^n ~~~(\text{mod } 3)\implies 2^{2n}\equiv 1~~(\text{mod } 3)$,
Again, this means that any even power of $2$ is $1$ greater than some multiple of $3$.
So if $2^{2n}$ is $1$ greater than some multiple of $3$ (another way to say this is that $2^{2n}$ leaves a remainder of $1$ when divided by $3$), then what can you say about $2^{2n}+5$?.
To answer this, forget about $2^{2n}$ and just think about the remainder (i.e $1$). This is how modular arithmetic makes our live a lot easier. If $2^{2n}$ is $1$ greater than some multiple of $3$, then $2^{2n}+5$ should be $1+5=6$ greater than that multiple of $3$, right? But you know that $6$ is, by itself, a multiple of $3$. So, this should mean that $3|2^{2n}+5$.
A more formal (and a neat) argument could be:
$3|2^{2n}-1\implies 3|2^{2n}+5$, since $2^{2n}+5=(2^{2n}-1+6)$
Now since it has been shown that $2^{2n}+5$ is infact a multiple of $3$, it should be apparent that $2^{2n}+5$ is after all, composite.
I Hope it helps!
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Weird explanation and weird disclaimer: what "technical jargon" did Gerry's answer use? Idea, work, raise...? The only technicallity there is modulo 3, which is both (1) boringly elementary, and (2) even the OP uses it...and of course, so do you. So what technical jargon did you avoid using that the other two answers did? – DonAntonio Aug 14 '12 at 18:04
First of all, please do note that I am not comparing my answers with others. Now that I read my answer, I realize that I badly phrased my little disclaimer :) What I mean to say is I have included parts like "...this means that any even power of 2 is 1 greater than some multiple of 3..." in my answer which are obvious. So I guess I need to edit/ delete the disclaimer, eh? – Bidit Acharya Aug 14 '12 at 18:11
I didn't say anything about you "comparing", but about you implying that your answer won't be "technical" as the other ones, whereas it actually was. That's all. Nothing to be too touchy about. – DonAntonio Aug 14 '12 at 18:35
$2^(2n) +5$ is the same as $4^{n} +5 =(3+1)^n +5$
If you expanded the $(3+1)^n$ part , every term would be divisible by $3$ (except the last term $1$) The entire expression would be of the form $(3m +1) +5 = 3m +6$ which is divisible by $3$.
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Heavily recommended: to take a peek at the LaTeX section to properly write mathematics in this forum – DonAntonio Aug 14 '12 at 18:38
To expand a little on DonAntonio's comment: For some basic information about writing math at this site see e.g. here, here and here. – Martin Sleziak Aug 15 '12 at 5:31 | 2014-08-28T01:04:30 | {
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http://mathhelpforum.com/calculus/45334-integration-rational-expression-alternative-method.html | Thread: Integration of Rational Expression - Alternative Method
1. Integration of Rational Expression - Alternative Method
Hello everyone,
I integrated the following rational expression correctly, but one of my steps differ from that one of the book. Could anyone tell me if what I did is still correct?
Thank you!
---
1. Integrate: $\int \frac{\sqrt{x - 4} + x}{x - 4} dx$
---
$\int \frac{\sqrt{x - 4} + x}{x - 4} dx$
= $\int (x - 4)^{-1/2} dx + \int \frac{x}{x - 4}dx$
Let $u = x - 4$ and $du = dx$. Therefore, $x = u + 4$.
= $\int u^{-1/2} du + \int \frac{u + 4}{u}du$
= $2\sqrt{x - 4} + \int \frac{u}{u} du + \int \frac{4}{u} du$
= $2\sqrt{x - 4} + x + \ln(x - 4)^4 + C$
---
However, my practice workbook integrated $\int \frac{u}{u} du$ as $u$, and then substituted $x - 4$ back into the equation.
So according to the workbook, its last step would be:
= $2\sqrt{x - 4} + u + \ln(x - 4)^4 + C$
= $2\sqrt{x - 4} + (x - 4) + \ln(x - 4)^4 + C$
= $2\sqrt{x - 4} + x + \ln(x - 4)^4 + C$
2. I think the way you did it kind of implies that $\int \frac{u}{u}du=x$ because of the way you kept your terms in order. The book's method is more correct in my opinion as it notes that the above integral doesn't equal x, but u + C, then back-substitutes x-4 for u and combines all the constants into one C.
3. Hello, scherz0!
What you did is correct.
. . I would not have simplifed first as you did . . .
We are given: . $\int \frac{\sqrt{x - 4} + x}{x - 4}\,dx$
Let: $\sqrt{x-4} \:=\: u \quad\Rightarrow\quad x-4\:=\:u^2 \quad\Rightarrow\quad x \:=\: u^2+4 \quad\Rightarrow\quad dx \:=\:2u\,du$
Substitute: . $\int \frac{u + (u^2+4)}{u^2}\,(2u\,du) \;\;=\;\;2\int\frac{u^2+u-4}{u}\,du \;\;=\;\; 2\int\left(u + 1 - \frac{4}{u}\right)\,du$
. . . . . . $= \;\;2\left(\frac{u^2}{2} + u - 4\ln|u|\right) + C \;\;=\;\;u^2 + 2u - 8 \ln|u| + C$
Back-substitute: . $(x - 4) + 2\sqrt{x-4} - 8\ln\left(\sqrt{x-4}\right) + C$
. . . . . . . . . . . $=\;\;x - 4 + 2\sqrt{x-4} - 8\ln(x-4)^{\frac{1}{2}} + C$
. . . . . . . . . . . $= \;\;x + 2\sqrt{x-4} - 4\ln(x-4) + \underbrace{C - 4}_{\text{a constant}}$
. . . . . . . . . . . $= \;\;x + 2\sqrt{x-4} - 4\ln(x-4) + C$ | 2017-01-19T09:14:06 | {
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https://math.stackexchange.com/questions/563289/find-a-subgroup-of-mathbbz-12-oplus-mathbbz-18-isomorphic-to-math | # Find a Subgroup of $\mathbb{Z}_{12} \oplus \mathbb{Z}_{18}$ Isomorphic to $\mathbb{Z}_9 \oplus \mathbb{Z}_4$
This is what I have so far:
$$\mathbb{Z}_{12} = \{0, 1, 2,\cdots, 11\}$$ $$\mathbb{Z}_{18} = \{0, 1, 2,\cdots, 17\}$$ $$\mathbb{Z}_9 = \{0, 1, 2,\cdots, 8\}$$ $$\mathbb{Z}_4 = \{0, 1, 2, 3\}$$
Using that fact that isomorphisms must contain the same number of elements, we must find a subgroup in $\mathbb{Z}_{12}\oplus \mathbb{Z}_{18}$ with the same number of elements as in $\mathbb{Z}_{9}\oplus \mathbb{Z}_{4}$.
Now, let's determine the number of elements in $\mathbb{Z}_{9}\oplus \mathbb{Z}_{4}$: $$|\mathbb{Z}_{9}\oplus \mathbb{Z}_{4}| = \text{lcm}(|\mathbb{Z}_{9}|, |\mathbb{Z}_{4}|) = \text{lcm}(|9|, |4|) = 36$$
Next, let's find a subgroup in $\mathbb{Z}_{12}\oplus \mathbb{Z}_{18}$ with order $36$. First, $\langle 3 \rangle$ in $\mathbb{Z}_{12}$ gives us the group $\{0, 3, 6, 9\}$ which has order $4$. Second, $\langle 2\rangle$ in $\mathbb{Z}_{18}$ gives us the group $\{0, 2, 4, 6, 8, 10, 12, 14, 16\}$ which has order $9$. Then, the group generated by $(3, 2)$ has order $36$ (because $\text{lcm}(9,4) = 36$)
My question is, does this properly show that there exists a subgroup in $\mathbb{Z}_{12}\oplus \mathbb{Z}_{18}$ that is isomorphic to $\mathbb{Z}_{9}\oplus \mathbb{Z}_{4}$? Or do I have to show that the mapping is one-to-one, onto, and preserves the group operation? If so, what is the mapping function to show one-to-one, onto, and operation preservation?
Thanks!
• Order is not enough: Note that $\mathbb{Z}_2+\mathbb{Z}_2$ and $\mathbb{Z}_4$ have the same order, but are not isomorphic. To use your approach, I would say that $\mathbb{Z}_{12}$ has a subgroup isomorphic to $\mathbb{Z}_4$ (you produced it) while $\mathbb{Z}_{18}$ has a subgroup isomorphic to $\mathbb{Z}_9$ (you produced it). Therefore $\dots$. Whether exhibiting the isomorphisms should be done is course-dependent. Another approach is to note that $\mathbb{Z}_{12}$ is isomorphic to the direct sum $\mathbb{Z}_4+\mathbb{Z}_3$, while the other is isomorphic to the direct sum $\dots$, so $\dots$. – André Nicolas Nov 12 '13 at 1:23
• The equation you wrote $|\mathbb{Z}_9 \oplus \mathbb{Z}_4| = \operatorname{lcm}(9,4)$ is technically true here (9 and 4 are coprime), but the order of a direct sum is always the product, not the lcm. For example the order of $|\mathbb{Z}_2 \oplus \mathbb{Z}_2|$ is 4, not $\operatorname{lcm}(2,2)=2$. – Najib Idrissi Nov 12 '13 at 1:41
Hint: Is there an element in $\mathbb{Z}_{12}$ that doesn't generate all of $\mathbb{Z}_{12}$ (hence is a factor of $12$) but generates $\mathbb{Z}_{9}$ or $\mathbb{Z}_4$? Then is there an element in $\mathbb{Z}_{18}$ not relatively prime to $18$ that generates $\mathbb{Z}_9$ or $\mathbb{Z}_4$?
Call these two elements $a$ and $b$. Then look at the element $a \oplus b \in \mathbb{Z}_{12} \oplus \mathbb{Z}_{18}$. Create an 'obvious' mapping from this element into $\mathbb{Z}_9\oplus \mathbb{Z}_4$. Prove that this mapping must be $1:1$, onto, and preserves the group operation.
But makes things easy on yourself! Focus on the element $a \oplus b$, it should generate $\mathbb{Z}_9 \oplus \mathbb{Z}_4$ (why?) then those three things are easy to prove. (again, think about why).
EDIT: I'll expand a bit more on my hint then. You already looked at a few elements in $\mathbb{Z}_{12}$. Look closely at the subgroup $\langle 3 \rangle=\{0,3,6,9\}$. Could we map that to $\mathbb{Z}_4$ with a mapping that preserves the operation, say $\varphi$? Now look at $\langle 2 \rangle =\{0,2,4,6,8,10,12,14,16\}\subset \mathbb{Z}_{18}$. Can we take those elements and map it to $\mathbb{Z}_9$, say the mapping is $\theta$, in a way that preserves the operation? Now since clearly $\mathbb{Z}_{12} \oplus \mathbb{Z}_{18} \cong \mathbb{Z}_{18} \oplus \mathbb{Z}_{12}$, would the 'mapping'
$$\psi((a,b))=\varphi(a) \oplus \theta(b)$$
for $(a,b) \in \mathbb{Z}_{18} \oplus \mathbb{Z}_{12}$ be an isomorphism to $\mathbb{Z}_{18} \oplus \mathbb{Z}_{12}$?
FINAL EDIT: With your last comment you essentially have the idea. It is clear that $\mathbb{Z}_{12} \oplus \mathbb{Z}_{18} \cong \mathbb{Z}_{18} \oplus \mathbb{Z}_{12}$ (just take the map that sends $(a,b)$ to $(b,a)$). So we just need to find a subgroup of $\mathbb{Z}_{18} \oplus \mathbb{Z}_{12}$ isomorphic to $\mathbb{Z}_9 \oplus \mathbb{Z}_4$. Your idea is right. We make a mapping $\varphi$ that takes each element in $\mathbb{Z}_{18}$ mod $2$ and a mapping $\theta$ that takes each element in $\mathbb{Z}_{12}$ mod $3$. Then show that $\varphi \oplus \theta$ is an isomorphism. Now here is where you need to be careful. You said it would be an isomorphism of $\mathbb{Z}_4 \oplus \mathbb{Z}_9$ to $\mathbb{Z}_{18} \oplus \mathbb{Z}_12$. This is wrong. It's an isomorphism from $\mathbb{Z}_{4} \oplus \mathbb{Z}_9$ to $\langle 2 \rangle \oplus \langle 3 \rangle$ in $\mathbb{Z}_{18} \oplus \mathbb{Z}_{12}$ (so a subgroup of that group). Since this group is isomorphic to $\mathbb{Z}_{12} \oplus \mathbb{Z}_{18}$, we would be done since it would be clear what the subgroup was. But there is an easier way to do this.
Let's look at the subgroup $\langle 2 \rangle \oplus \langle 3 \rangle$ of $\mathbb{Z}_{18} \oplus \mathbb{Z}_{12}$. Let's create a mapping $\varphi$ of $\langle 2 \rangle$ in $\mathbb{Z}_{18}$ into $\mathbb{Z}_9$. Notice that $1$ generates $\mathbb{Z}_9$. Moreover, $\langle 2 \rangle$ generates the subgroup (itself) in $\mathbb{Z}_{18}$. So just map $2 \in \mathbb{Z}_{18}$ to $1 \in \mathbb{Z}_{9}$. Similarly, we can map $3 \in \mathbb{Z}_{12}$ to $1 \in \mathbb{Z}_4$. Since we are mapping generators to generators, it is clear that we preserve the operations, it is $1:1$, and onto. So we have the required isomorphism. But all the same, if this shorter way isn't as clear to you then you can can always show it by the preceding paragraph (which was what you stated in your comment, I just cleaned up a bit of the language). Hope that helped finally clear things up.
• I guess I'm confused at your hint. I did not find an element in Z12 that "doesn't generate all of Z12 but generates Z9 or Z4". The potential candidates are 1, 2, 3, 4, and 6 because all are factors of 12. <1> = Z12, <2> = {0, 2, 4, 6, 8, 10}, <3> = {0, 3, 6, 9}, <4> = {0, 4, 8}, <6> = {0, 6}. – JT9 Nov 12 '13 at 4:29
• @JTWheeler Yes, now count the amount of elements in those subgroups. Map the generators from one of those subgroups of the appropriate size to the $\mathbb{Z}_{-}$ you want. Do it for both then 'stitch' the maps together and show they typical 3 things you need for an isomorphism. I have elaborated more in my answer above to help with this. Focus on the generators! – mathematics2x2life Nov 12 '13 at 6:18
• I think I see what you're getting at now. <3> in Z12 can be mapped to Z4 by dividing each element by 3. So the mapping function would be a/3 to achieve Z4. <2> in Z18 can be mapped to Z9 by diving each element by 4. So the mapping function would be b/4 to achieve Z9. I believe I now have to show 1-1, onto, and operation preservation for these 2 functions. And then finally show the mapping ψ((a,b))=φ(a)⊕θ(b) is an isomorphism to ℤ18⊕ℤ12. – JT9 Nov 12 '13 at 16:31
• @JTWheeler Not quite but you essentially have the exact idea. I'll do one final edit to explain what you just said in a bit rigor. – mathematics2x2life Nov 12 '13 at 16:39
• I think the second part of your final edit makes more sense to me actually. One final question for you: ℤ12⊕ℤ12≅ℤ18⊕ℤ12 (I think you meant ℤ12⊕ℤ18≅ℤ18⊕ℤ12), is this a property of external direct products? – JT9 Nov 12 '13 at 17:21 | 2019-07-18T11:43:48 | {
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https://math.stackexchange.com/questions/467407/number-of-points-on-two-circles | # number of points on two circles
(sorry I don't know how to add pictures)
Two friends argue if larger circles have more points than smaller circles
Friend number 1 (a well known argument)
Say the circles are concentric. you cannot draw a line from the centre that cuts the bigger circle while that doesn't cut the inner circle so they have the same amounts of points.
Friend number 2 ( a dissenting voice)
Ok lets take concentric circles, he draws a line to the centre ( say along the x axis)
adds another line parallel to this line it cuts both circles 2 times)
adds another (cuts both circles 2 times) and so on
two parallel lines are only tangent to the smaller circle while still cuting the bigger at two points. and some paralel lines only cut the larger cirle.
Therefore all points on the smaller circle are related to some point on the bigger one, but some points on the bigger one are not related to point on the smaller one.
So the larger circle has more points.
Which friend is right or how do you convince them that they are both right?
• What if the circles are outside each other? Draw a line from the centre of the smaller circle, and the line may or may not cut the bigger circle, but will definitely cut the smaller circle twice. Aug 14, 2013 at 13:06
• Friend 3: Given concentric circles, let $P$ be a point on the smaller one. Non-tangent lines through $P$ meet the smaller circle at just one point (other than $P$), while they meet the big circle two points; the tangent at $P$ matches $P$ itself with two points on the big circle. OMG, the big circle has twice as many points as the small circle! :)
– Blue
Aug 14, 2013 at 13:17
Friend 1's argument does indeed show that the circles have the same number of points, as each ray from the center uniquely determines a point on the circle, and each point on the circle uniquely determines a ray from the center.
Friend 2's argument (if made carefully) shows that the bigger circle has at least as many points as the smaller circle, but does not show that the smaller circle has strictly fewer points.
It all comes down to the notion of cardinality of infinite sets, where things can get pretty counterintuitive. If you want to give friend 2 an example to show why that argument does not show strict size difference, consider the set of integers $\Bbb Z.$ Obviously the same size as itself, yes? Ah, but now consider the function $f:\Bbb Z\to\Bbb Z$ given by $f(x)=2x$. This function maps $\Bbb Z$ into $\Bbb Z$, but misses infinitely many integers! That does not mean, however, that $\Bbb Z$ is strictly larger than itself. It is simply a peculiar quirk of (many) infinite sets that they can be put into one-to-one correspondence with subsets of themselves. Thus, showing that the smaller circle is in one-to-one correspondence with only a part of the larger circle isn't enough to show that the larger circle has more points--though such an argument would work for finite sets.
• This answer below does the job. Another common example is the "number of points" on the number line from (0,1) (not including 0 or 1) vs range greater than one. For every number you can pick in the (0,1) range (say x), you can find the corresponding number 1/x which is in the other set (and vice versa). Read some Cantor. That'll blow your mind. Aug 15, 2013 at 4:24
• This answer doesn't define what we mean by "more". The friends here probably have different notions of what that means. Aug 17, 2013 at 17:31
I could be mistaken here, but the main conflict seems to be in compromising between the geometrical and topological perspectives.
Geometrically one could say that the arc length of a curve is a measure of how many points it has, so clearly the larger circle has more points.
However from a topological perspective, the cardinality of both sets of points are the same in that any circle can be mapped to the real line via a stereographic projection.
So in a sense, both answers are right!
What does "more points" mean here?
I'll try and argue that both friends come as correct and they both friends have different notions of what "more points" means.
The friend who says that circle "a" (the smaller one) has the same size as circle A, might define an object x to have the same size an object X if and only if someone can or has written a bijection between the points of x and the points of X.
The friend who says that circle "a" is smaller than circle A, might write the following definition:
Part 1: Object x to has the same size as object X if and only if when we make a random selection of a point from [po(x) U po(X)], where [po(x) U po(X)] indicates the union of the sets of points of x and the sets of points of X, the probability that we've selected a point from x equals the probability that we've selected a point from X, or prob(x)=prob(X).
Part 2: Object x comes as smaller than object X if and only if when we make a random selection from [po(x) U po(X)], it comes as more likely that we'll select a point from object X than object x, or prob(x) < prob(X).
Part 3: Object x comes as larger than object X if and only if when we make a random selection from [po(x) U po(X)], it comes as more likely that we'll select a point from object x than object X, or prob(X) > prob(x).
Friend two's argument gives us a procedure as to how to observe that circle "a" has a smaller area than that of circle A (that much, I do believe, all set theorists would grant). So, if a machine not trying to hit either circle threw a dart at the circles standing next to each other, it'd come as more likely that the dart hit circle A than that it hit circle "a" (the possibility that it would miss either of the circles comes as irrelevant). It consequently follows that a random selection of a point from
[po(a) U po(A)] will more likely give us a point of A than a point of "a". So, "a" has a smaller size than A, according to friend two's definition of what size means.
• What you're describing for the second notion of "size" seems to match area measure, if we're careful, but it's a bit problematic. We'll need our "dart board" to be of finite area, or the probablity of landing within either of the circles will actually be $0.$ However, even then, we still run into the problem that the area of each circle is actually $0,$ since it is comprised only of the points on the "edge." Even if we ignore that, we run into the problem that even "nice" figures having the same perimeter may have different areas.(cont'd) Jan 11, 2014 at 17:03
• For example, we can make an ellipse with the same perimeter as a given circle, having area as small as we like (though no larger than the circle's area). A better approach for the second friend's perspective would probably be a perimeter measure, instead. Jan 11, 2014 at 17:05
Friend 2 is right in his conclusion but I don't like his method.
What is the flaw in the argument of friend 1? Consider two concentric circles which have radii of 2$$\pi$$r and 2$$\pi$$R respectively, with R > r. Let us define a point on a circle to be a segment of arc length dx. Let us draw a line from the center of the circles which then cuts both along the positive y axis (0,1) and then rotate this line clockwise by d$$\theta$$. Indeed in both circles this rotation has covered an angle of d$$\theta$$; however, in the large circle it covered an arc length of Rd$$\theta$$ whereas it only covered rd$$\theta$$ in the small circle. Recall that the definition of a point involves a 1D unit of length dx. Thus this rotation covered $$\mathrm{\frac{R d\theta}{dx}}$$ points in the big circle whereas it only covered $$\mathrm{\frac{r d\theta}{dx}}$$ points in the small circle. Thus, the ratio of number of points between the large and small circles is R:r, which is just the ratio of the circumference. The mistake in the logic of friend 1 is simply that he used a different definition for what constitutes a point for the big circle than for the small circle. If you compare a quantity (number of points) between two objects and change the definition of the quantity from object to object, then the conclusions made are likely to be meaningless.
That being said, many people will use the defective argument that the aforementioned logic is only true until we reach the limit of dx$$\rightarrow$$0, where we define a point to have infinitesimal width . In which case, the number of points in a circle reaches the corresponding limit "number of points"$$\rightarrow \infty$$. Then, since $$\infty \times n = \infty$$ both circles have the same number of points.
The flaw in this logic is that it incorrectly uses the concept of infinity. The statement $$\infty \times n = \infty$$ can be rephrased to mean "Lack of precise information" $$\times$$ "a real number" = "Still a lack of information". To use such a formula to justify a conclusion is unfortunately gibberish. It does not matter whether we lack precise information about the point size dx, all that matters is that we know that dx is the same value when applied to the small circle as when applied to the large circle. So what can we say? If indeed dx is infinitesimal (dx$$\rightarrow$$0) then we do know that there are infinite points on the big circle and the small circle, in the sense that we lack knowledge of the exact number for both circles. But to say, a lack of knowledge = a different lack of knowledge is obviously wrong. The concept of their being different sized infinities is nothing new, so the idea of two different infinities being in an exact ratio should not be surprising. The property of infinity $$\infty \times n = \infty$$ is only correct when one uses a definition of infinity which breaks all the rules of algebra and is only helpful as a concept. I presume that most of the confusion here lies in the fact that there are lots of types of infinity, so the symbol $$\infty$$ refers to a useful concept (a number which goes on forever which we lack precise information on), which is usually different to the other infinities we encounter that we actually do have information on. | 2022-12-03T14:44:34 | {
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http://supakush.com/h1jp5ozu/symbolab-critical-points-multivariable | # symbolab critical points multivariable
Free Gradient calculator - find the gradient of a function at given points step-by-step This website uses cookies to ensure you get the best experience. A critical point of a multivariable function is a point where the partial derivatives of first order of this function are equal to zero. Critical Points and Extrema Calculator. f (x) = 3 x 2 + 6 x-1 x 2 + x-3. Find Asymptotes, Critical, and Inflection Points. By using this website, you agree to our Cookie Policy. Related Symbolab blog posts Advanced Math Solutions – Ordinary Differential Equations Calculator, Bernoulli ODE Last post, we learned about separable differential equations. Open Live Script. 1. Come to Sofsource.com and figure out adding fractions, power and plenty additional algebra subject areas Define a Function. Show Instructions. To analyze the critical point $(-\sqrt[3]3,-\sqrt[3]3)$ we compute the Hessian $$\left[\matrix{18x+6xy^3 &9x^2y^2\cr 9x^2y^2 &18y+6yx^3\cr}\right]\ .$$ Its determinant is $$9xy\bigl(36+12(x^3+y^3)-5x^3y^3\bigr)\ ,$$ which is negative at $(-\sqrt[3]3,-\sqrt[3]3)$. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Critical Points of Functions of Two Variables. ... Critical points for multivariable functions. This example describes how to analyze a simple function to find its asymptotes, maximum, minimum, and inflection point. From Multivariable Equation Solver to scientific notation, we have got all kinds of things covered. Get the free "Multivariable Limits" widget for your website, blog, Wordpress, Blogger, or iGoogle. Calculadora gratuita de pontos críticos de uma função - Encontrar os pontos críticos e estacionários de uma função passo a passo The calculator will find the critical points, local and absolute (global) maxima and minima of the single variable function. Related Symbolab blog posts Advanced Math Solutions – Limits Calculator, L’Hopital’s Rule In the previous posts, we have talked about different ways to find the limit of a function. Examples with detailed solution on how to find the critical points of a function with two variables are presented. First, create the function. The interval can be specified. Find more Mathematics widgets in Wolfram|Alpha. How to Find Extrema of Multivariable Functions. In single-variable calculus, finding the extrema of a function is quite easy. The function in this example is. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. 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'S breakthrough technology & knowledgebase, relied on by millions of students & professionals this function are equal to.. Where the partial derivatives of first order of this function are equal to zero, Blogger or... is equivalent to 5 * x variables are presented the partial derivatives of first order this. is equivalent to 5 * x a simple function to the! Inflection point, local and absolute ( global ) maxima and minima of the variable! We have got all kinds of things covered find the critical points of Multivariable! By using this website, blog, Wordpress, Blogger, or.... Multivariable Equation Solver to scientific notation, we have got all kinds of things covered millions of students &.... Sign, so 5x is equivalent to 5 * x (... Using this website, blog, Wordpress, Blogger, or iGoogle simple function find! + x-3 on by millions of students & professionals multiplication sign, so 5x... Minimum, and inflection point technology & knowledgebase, relied on by millions of students & professionals, relied by... And inflection point, we have got all kinds of things covered website. Minima of the single variable function equal to zero order of this function are equal to zero 5x equivalent... 2 + x-3 5 * x in general, you agree to our Cookie Policy widget for your,., minimum, and inflection point general, you can skip the sign... So 5x is equivalent to 5 * x you agree to our Cookie Policy the... Function is a point where the partial derivatives of first order of this function are equal to.. Agree to our Cookie Policy blog, symbolab critical points multivariable, Blogger, or.. | 2022-01-21T14:44:36 | {
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http://math.stackexchange.com/questions/834070/on-the-constant-of-integration-in-solving-ordinary-differential-equations | # On the constant of integration in solving ordinary differential equations
I very much suspect this but I'm not sure if it's correct:
In solved differential equations, does the constant 'c' always represent the value of the dependent variable when the independent=0 ?
-
This is the case in equations such as $\dfrac{dy}{dx}=ay$, where $a$ is a constant. – Sanath Devalapurkar Jun 14 '14 at 16:29
@SanathDevalapurkar ah I see it's in special cases only. Would $\dfrac{dy}{dx}=axy$ also work? – hb20007 Jun 14 '14 at 16:32
Solving the above given differential equation gives $$\ln |y|=\dfrac{ax^2}{2}+C\implies y=C_1\exp\left(\dfrac{ax^2}{2}\right)$$ When $x=0$, then $C_1=y(0)$, so, yes, it would work. – Sanath Devalapurkar Jun 14 '14 at 16:34
@SanathDevalapurkar Thank you, If this was an answer I'd selected it as best :D – hb20007 Jun 14 '14 at 16:35
It doesn't exactly answer the question; I provided examples of where it'd work, but, as the counterexamples below show, the statement above is false. – Sanath Devalapurkar Jun 14 '14 at 16:36
Not necessarily: Suppose we have $y = f(x) = e^x + c$
$$f(0) = 1 + c \neq c$$
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In polynomials , we do have that $c = p(0)$ – amWhy Jun 14 '14 at 16:33
@hb20007 The above comment by amWhy would be a good exercise. – Sanath Devalapurkar Jun 14 '14 at 16:35
@amWhy I am unfamiliar with the notation in your comment. Could you please tell me what 'p' represents and why it's an 'exercise'? – hb20007 Jun 14 '14 at 16:41
Oh sure, I'm just using $p(x)$ to denote a p olynomial (which we can call $y$, aka, the dependent variable, so $p(0)$ denotes a polynomial $p(x)$ evaluated at $x = 0$. Example: $p(x) = x^3 + 4x^2 + 3x + c \implies p(0) = c.$ – amWhy Jun 14 '14 at 16:44
No. Suppose $$\frac{dy}{dx}=\sin x$$ then $$y=-\cos x+c$$
At $x=0$ $$y(0)=-1+c\ne c$$
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Dear Edwin, you do not have to use $$<equation>$$ every time. To do inline equations, simply write $<inline equation>$. – Sanath Devalapurkar Jun 14 '14 at 16:34 | 2015-04-19T14:47:15 | {
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http://math.stackexchange.com/questions/850493/a-simple-function-equation | # A simple function equation
I come from a programming background and I can’t find a simple math function. The request might seem strange, but I needed it a graphical context to alter some points locations:
I need a function $f(x) = y$ defined for $x \ge 0$ such that:
• $f(x) \in [0, x)$
• $f(0) = 0$
• $f(x) \approx x$ as $x\to \infty$.
• It has to slowly grow at first — sort of like $x^2$ — and then get closer and closer to x.
The simplest equation form that satisfies this restrictions will do.
I tried to plot this so that I can make myself better understood:
http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427eo8stqhpe1s
Actual values don’t matter, just the shape of the plot.
None of the basic functions (and combinations of them) that I tried were doing this (e.g. $x^2, \log x, \sqrt x, 1/x$).
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Thank you and sorry if the question is in any way inappropriate for this site (I use mostly Stack Overflow). Please let me know if I can edit this question to make it more adequate for this site. – bolov Jun 28 '14 at 15:51
Am I right in saying that you want $y = x$ to be an asymptote? Or is it just that you want the slope of the curve to approach $45^\circ$ (so $y = x - c$ is an asymptote for some suitable constant $c$)? – M. Vinay Jun 28 '14 at 15:57
@M.Vinay either will do, I will have to test them in my program to see the final results. I think that $y=x$ an asymptote would be better. – bolov Jun 28 '14 at 16:03
If $y = x$ is an asymptote, then the curve will initially lie below the asymptote and approach it from the right. If $y = x - c$ is an asymptote for an appropriate $c$, then the curve will always be to the left of the asymptote and approach it from the left. – M. Vinay Jun 28 '14 at 16:05
@M.Vinay yes, I understand. I still need to see the visual result, but I think if y=x an asimptote is better for me. – bolov Jun 28 '14 at 16:08
This one fulfills your requirements: $$f(x)=\frac{x^2+x^3}{1+x+x^2}.$$ We have: $$\forall x>0,\ 0<f(x)<x$$ $$f(0)=0,$$ $$f(x)-x^2=-\frac{x^4}{1+x+x^2}$$ so that $f(x)$ and $x^2$ are very close for small values of $x$, and $$f(x)-x=-\frac{x}{1+x+x^2}$$ so that $f(x)$ and $x$ get closer and closer as $x\to+\infty$.
It's also cheap to compute with 2 additions, 2 multiplications and 1 division if you proceed thus:
• Compute $x^2$ (1 mult) and $x+x^2$ (1 addition); set $a=x+x^2$.
• Compute $x\times a$ (1 mult); set $b=x\times a$.
• Compute $1+a$ (1 addition); set $c=1+a$.
• Compute $b/c$ (1 division): that's $f(x)$.
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the plot looks good. Thank you – bolov Jun 28 '14 at 16:21
You could use a common hyperbola $y = \sqrt{a^2 + x^2} - a$.
Example with $a = 10$:
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yes, I saw it. No problem. – bolov Jun 28 '14 at 16:26
thank you. You guys are great. For now I went with @gniourf_gniourf answer because at first glance it looks faster to compute. – bolov Jun 28 '14 at 17:17
@bolov Here $y = x - a$ is the asymptote. My misgiving with $y = x$ being the asymptote is that there would be an (ugly) inflection point in that case, as mentioned by Yves. – M. Vinay Jun 28 '14 at 17:18
the inflection point unless it would be very "abrupt" (I don’t know the right terminology) isn’t a problem. – bolov Jun 28 '14 at 17:21
@gniourf_gniourf Thought as much. That's perhaps one reason (apart from ease of computation) that your answer is the accepted one. As I said, I find the inflection point ugly, and bolov's last reply-comment (on the question) says he's not fully sure he wants $y = x$ to be the asymptote, and would like to see the graph and then decide. Hence my answer. – M. Vinay Jun 28 '14 at 18:10
You can use a weighting function $w(x)$ that allows you to create a mixture between the functions $x^2$ and $x$, as $\frac{x^2+w(x)x}{1+w(x)}$. Ensure $w(0)=0$ so that the initial behavior is $x^2$ and $w$ growing sufficiently fast that the term $x$ supersedes it. $$w(x)=x^3\to y=\frac{x^2+x^4}{1+x^3}.$$ $$w(x)=e^x-1\to y=\frac{x^2+x(e^x-1)}{e^x}.$$
Note that if you really want to reach $y=x$ (and not $y=x-c$), there must be an inflection point.
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Nice. I thought of a similar one with a weight function $\theta = 1/e^x$ (and $y = \theta x + (1 - \theta)x$): $$y = \dfrac{x^2}{e^x} + x\left(1 - \dfrac{1}{e^x}\right)$$ – M. Vinay Jun 28 '14 at 17:03
thank you. You guys are great. For now I went with @gniourf_gniourf answer because at first glance it looks faster to compute. – bolov Jun 28 '14 at 17:18
Consider the derivative of your function. It has to be positive,increasing and $f'(x→inf)=1$ $f'(x)=tanh(x)$ fits.
$\int \tanh(x) dx = \log (\cosh (x))+c$
Since we want $f(0)=0$
$\log (\cosh 0)+c=0\Leftrightarrow c=0$
Therefor $f(x)= \log (\cosh (x))$ works.
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thank you. You guys are great. For now I went with @gniourf_gniourf answer because at first glance it looks faster to compute. – bolov Jun 28 '14 at 17:19
Using your data, I found that $y=a x^b$ fits very well. My results are $a=0.008755$ and $b=3.134091$. The corresponding $R^2=0.999$.
This is a very flexible form.
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f(x) > x for a certain C, f(x) > x for x > C – bolov Jun 28 '14 at 16:11 | 2015-07-29T01:09:20 | {
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https://www.physicsforums.com/threads/finding-limits-using-theorems.694040/ | # Finding limits using theorems
1. May 27, 2013
### Numnum
1. The problem statement, all variables and given/known data
Use theorems to find the limit:
$$\lim_{x\rightarrow 1} \cos(arctan({\frac{\sin(x-1)}{x-1}}))$$
2. Relevant equations
Theorems like
$$f(x)=c$$ is continuous
$$f(x)=x$$ is continuous
$$\lim_{x\rightarrow 0} \cos(x)=1$$
$$\lim_{x\rightarrow 0} \sin(x)=0$$
$$\lim_{x\rightarrow a} \sin(x)=sin(a)$$
$$\lim_{x\rightarrow 0} \sin(x-a)=0$$
3. The attempt at a solution
I'm not sure where to start, but I looked at the last theorem and thought that since the limit of sin(x-a)=0, it would turn that whole part into 0, and therefore it would turn to arctan(0). Didn't seem correct, so I instead thought to simplify the sin(x-1)/x-1 part by letting x-1 equal another variable?
2. May 27, 2013
### Dick
Good idea! Let u=x-1. You should also have a theorem about the limit of sin(u)/u as u->0.
Last edited: May 27, 2013
3. May 27, 2013
### Zondrina
What is $lim_{x → 0} \frac{sin(x)}{x}$?
How does it relate to $lim_{x → 1} \frac{sin(x-1)}{x-1}$?
4. May 27, 2013
### Numnum
Is this right?
So I have:
1. $$\lim_{x\rightarrow 1} \cos(arctan({\frac{\sin(x-1)}{x-1}}))$$
2. $$\lim_{x\rightarrow 1} \cos(arctan({\frac{\sin(u)}{u}}))$$
3. $$\lim_{x\rightarrow 1} \cos(arctan(1))$$
because of the theorem: $$\lim_{x\rightarrow 0}({\frac{\sin(x)}{x}}))=1$$
4. $$\lim_{x\rightarrow 1} cos({\frac{π}{4}})$$
5. $$={\frac{1}{√2}}$$
5. May 27, 2013
### Dick
Yes, and you are using your 'continuous function' theorems after you've worked out the sin(u)/u part, yes?
6. May 27, 2013
### HallsofIvy
Staff Emeritus
Personally, I would do it the other way around. Since cosine is continuous, for all x, $\lim_{x\to 1} cos(f(x))= cos(\lim_{x\to 1} f(x))$.
That is, from $\lim_{x\to 1} cos(actan(\frac{sin(x- 1)}{x})$ we look at $\lim_{x\to 1}arctan(\frac{sin(x-1)}{x})$. And since arctan is continuous for all x we look at $\lim_{x\to 1}\frac{sin(x-1)}{x-1}$. As you say, that last limit is 1 so we have $cos(arctan(1))= cos(\pi/4)= \frac{\sqrt{2}}{2}$ | 2017-08-16T20:22:56 | {
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http://mathhelpforum.com/advanced-statistics/156544-probability-system-four-independent-components-print.html | # Probability System of four independent components
• Sep 17th 2010, 05:58 PM
mightydog78
Probability System of four independent components
Here is a probability system
http://i140.photobucket.com/albums/r...se78/asasd.jpg
The probability of failure of #1=0.1, #2=0.2,#3=0.3,#4=0.4
Now im trying to find the probability that the system works..
I so far have 1∩(2U4)∩3 which seems right but I dont know what numbers to plug in and whether to add or multiply
Also, I need to find the prob. that at most 3 events of these components work.
Then, using conditional prob. I can't seem to find the prob. that the system works given that at most 3 of the components work..
Im so lost
• Sep 17th 2010, 08:34 PM
mr fantastic
Quote:
Originally Posted by mightydog78
Here is a probability system
http://i140.photobucket.com/albums/r...se78/asasd.jpg
The probability of failure of #1=0.1, #2=0.2,#3=0.3,#4=0.4
Now im trying to find the probability that the system works..
I so far have 1∩(2U4)∩3 which seems right but I dont know what numbers to plug in and whether to add or multiply
Also, I need to find the prob. that at most 3 events of these components work.
Then, using conditional prob. I can't seem to find the prob. that the system works given that at most 3 of the components work..
Im so lost
Is the working of each component assumed to be independent of the other components?
Pr(1 works) = 0.9.
Pr(2 or 4 works) = Pr(2 works) + Pr(4 works) - Pr( 2 and 4 both work) = ....
Pr(3 works) = ....
• Sep 17th 2010, 10:20 PM
Soroban
Hello, mightydog78!
Quote:
Here is a probability system
http://i140.photobucket.com/albums/r...se78/asasd.jpg
The probability of failure of #1 = 0.1, #2 = 0.2,#3 = 0.3,#4 = 0.4
(a) Now im trying to find the probability that the system works.
(b) Also, I need to find the prob. that at most 3 of these components work.
(c)Then, using conditional prob. I can't seem to find the prob. that
. . the system works, given that at most 3 of the components work.
. . $\begin{array}{|c||c|c|} \hline
\text{Component} & \text{P(work)} & \text{P(fail)} \\ \hline
\#1 & 0.9 & 0.1 \\ \#2 & 0.8 & 0.2 \\ \#3 & 0.7 & 0.3 \\ \#4 & 0.6 & 0.4 \\ \hline
\end{array}$
(a) The system works if:
. . [1] #1 works, #2 works, #3 works, and #4 works.
. . . . $P(1,2,3,4) \:=\:(0.9)(0.8)(0.7)(0.6) \:=\:0.3024$
. . [2] #1 works, #2 works, #3 works, and #4 fails.
. . . . $P(1,2,3,\overline4) \:=\:(0.9)(0.8)(0.7)(0.4) \:=\:0.2016$
. . [3] #1 works, #2 fails, #3 works, and #4 works.
. . . . $P(1,\overline2,3,4) \:=\:(0.9)(0,2)(0.7)(0.6) \:=\:0.0756$
Therefore: . $P(\text{system works}) \:=\:0.3024 + 0.2016 + 0.0756 \:=\:0.5796$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
(b) The opposite of "at most 3 components work" is "4 components work."
. . $P(\text{4 work}) \:=\:(0.9)(0.8)(0.7)(0.6) \:=\:0.3024$
Therefore: . $P(\text{at most 3 work}) \;=\;1 - 0.3024 \;=\;0.6976$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
$(c)\;P(\text{system works }|\text{ at most 3 work}) \;=\;\dfrac{P(\text{system works }\wedge\text{ at most 3 work})}{P(\text{at most 3 work})}$
The numerator is [2] and [3] in part (a):
. . $[2]\;P(1,2,3,\overline4) \:=\:(0.9)(0.8)(0.7)(0.4) \:=\:0.2016$
. . $[3]\;P(1,\overline2,3,4) \:=\:(0.9)(0,2)(0.7)(0.6) \:=\:0.0756$
Hence, the numerator is: . $0.2016 + 0.0756 \:=\:0.2772$
And we found the denominator in part (b): . $0.6976$
Therefore: . $P(\text{system works }|\text{ at most 3 work}) \;=\;\dfrac{0.2772}{0.6976} \;=\;\dfrac{693}{1744}$
• Sep 18th 2010, 10:03 AM
mightydog78
Quote:
Originally Posted by mr fantastic
Is the working of each component assumed to be independent of the other components?
Pr(1 works) = 0.9.
Pr(2 or 4 works) = Pr(2 works) + Pr(4 works) - Pr( 2 and 4 both work) = ....
Pr(3 works) = ....
they are independent of each other but a 1 and 3 need to work for the whole system to work, and 2 or 4 must work for the system to work as well. | 2018-01-19T23:58:15 | {
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https://math.stackexchange.com/questions/716767/how-to-calculate-the-percentage-of-increase-decrease-with-negative-numbers | # How to calculate the percentage of increase/decrease with negative numbers?
I feel like an idiot for asking this but i can't get my formula to work with negative numbers
assume you want to know the percentage of an increase/decrease between numbers
2.39 1.79 =100-(1.79/2.39*100)=> which is 25.1% decrease
but how would i change this formula when there are some negative numbers?
6.11 -3.73 =100-(-3.73/6.11*100) which is 161% but should be -161%
the negative sign is lost.. what I am missing here?
also
-2.1 0.6 =100-(-3.73/6.11*100) which is 128.6% ??? is it?
• I don't see what's wrong here? For your second one it's a 161% decrease not a -161% decrease. – Module Mar 18 '14 at 13:28
• The Wikipedia page Relative change and difference has more on this. – Ninjakannon May 15 '15 at 23:10
Perhaps this "formula" will be easier to understand (this formula is equivalent to your formula - each can be derived from the other):
$$\dfrac{\text{original value} \;- \;\text{final value}}{\text{original value}} \times 100\% = \text{percent change}$$
That change will be
• an increase if the original value is less than the final value,
• a decrease if the original value is greater than the final value.
Original value $6.11$, final value $-3.73$:
$$\dfrac{6.11 -(-3.73)}{6.11}\times 100\% \approx 161\% \;\;\text{DECREASE}$$
Original value $-2.1$, final value $0.6$:
$$\dfrac{-2.1 - 0.6}{-2.1}\times 100\% \approx 128.6\% \;\;\text{INCREASE}$$
• Thank you for you answer...BUT why then ((6.11-(-3.73))/6.11)*100 gives 161 indicating an increase - but (((-3.73)-(-10.85))/(-3.73))*100 gives -190.9? Does that mean that with 2 negative numbers this equation does not apply? – master of puppets Mar 18 '14 at 15:31
• The important thing to look at is how the original value and the final value compare: If the original value is greater than the final value, then the absolute value of the the result indicates the percent decrease. If the original value is less than the final value, then the absolute value of the result indicates the percent increase. – Namaste Mar 18 '14 at 15:44
• thanks I am accepting your answer, the bit I could not understand trying to rewrite this algorithm was the signs. I decided to reverse the signs for further calculation ( another library that uses the algorithm ) and everything works fine now. Also I needed to handle 0s but thats not part of the question. thanks – master of puppets Mar 18 '14 at 16:26
• You're welcome. Glad things are working for you! – Namaste Mar 18 '14 at 16:28
• With the formula $$\frac{final-original}{|original|}$$ you have always a meaningful result, though a bit counterintuitive when final and original have not the same sign. Your definition above would not work if final and original have the same sign (the sign of percent change is reversed). – Jean-Claude Arbaut Sep 24 '14 at 8:48
I know this is a very old thread, but I am here for the first time so I hope it is OK to comment.
Let's take an example:
Original value $-10$, final value $10$:
$\frac{Original\ value - Final\ value}{Original\ value} = 200\% \ increase$
Original value $-1$, final value $10$:
$\frac{Original\ value - Final\ value}{Original\ value} = 1100\% \ increase$
How can an increase from a smaller number ($-10$) to $10$ be a lesser percentage than an increase from a larger number ($-1$) to $10$?
I’m not a mathematician, but I don’t think percent change with values of opposite signs is defined.
(the section named Net Income)
• If it is defined, what is, according to you, the explanation that an increase from a smaller number to x can be a lesser percentage than an increase from a larger number to x. And what is your position on the Wall Street Journal article I linked to. Are they completely wrong? – Hans Knudsen Sep 23 '14 at 16:15
• Would you then be so kind as to tell me what you think about the section Net Income in the WSJ-article? I take it you will not claim that it is a completely unreliable site. – Hans Knudsen Sep 23 '14 at 19:34
• Maybe you should take a look here before being so cocksure - and that's it for my part. mathforum.org/library/drmath/view/55720.html – Hans Knudsen Sep 24 '14 at 8:27
• I agree that speaking of percent increase or decrease when original and final values are not of the same sign is dangerous and counterintuitive. However, you can always define the relative variation as $$\frac{B-A}{|A|}$$ if $A\neq0$. That's (along with sens of subtraction) the missing absolute value that is causing trouble in amWhy's answer, I didn't pay enough attention to it. With this, you get $-161%$ and $+128%$ respectively, so the decrease/increase is ok. And +1 of course. – Jean-Claude Arbaut Sep 24 '14 at 8:45
• And really sorry for my misunderstanding of your answer. Even though I use often (constantly I may say) such percentages for work, with correct formula, I managed to get it wrong mentally when reading answers here. How uncomfortable :/ – Jean-Claude Arbaut Sep 24 '14 at 8:52
## protected by Zev ChonolesAug 26 '16 at 15:12
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count). | 2019-08-19T03:55:02 | {
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https://math.stackexchange.com/questions/3939530/solving-frac1x2-1-geqslant-frac1x1 | # Solving $\frac{1}{x^2-1} \geqslant \frac{1}{x+1}$
How do I solve:
$$\frac{1}{x^2-1} \geqslant \frac{1}{x+1}$$
Here's my attempt to solve it: I brought the fraction on the right side to the left side, then I've found the common denominator, $$(x-1)(x+1)$$, and after all the computations, this is my situation: $$\frac{2-x}{(x-1)(x+1)} \geqslant 0$$ Now, I've put the numerator and the denominator in a system of inequalities: (I don't know how to type the system of inequalities in mathjax, sorry)
$$x<2$$ $$x < -1 \space \text{or} \space x>1$$ My result is (I found the intersection): $$-12$$
Let me know if the result is wrong. Additionally, because in the test I'll have only 50 minutes (lots of questions), this method requires a lot of time; if you know a quicker method, let me know.
EDIT (solved in a quicker way):
1. when you are in this situation: $$\frac{2-x}{(x-1)(x+1)} \geqslant 0$$ change the sign of the inequality and the signs that appear in the numerator,
2. then, check each interval in the normal way, but don't forget you have changed the sign; therefore, if you had a greater sign, and you changed it, you have to take only negative intervals (not positive intervals, in order to express the correct solution).
• Essentially you need to have it that both the numerator and denominator have the same sign - so either $x <2, |x| > 1$ or $x > 2, |x| < 1$ – Dhanvi Sreenivasan Dec 8 '20 at 6:43
• Another way could be using another system of inequalities. In the first one, I'd use the greater sign and in the other one I'd use the less than sign, in order to avoid the absolute value. Is that correct? – Gabriel Burzacchini Dec 8 '20 at 6:49
• I'm afraid of doing more complicated and long stuff with the absolute value (I would have to split the exercise in more possible cases). – Gabriel Burzacchini Dec 8 '20 at 6:51
• A system is typeset with $\text{\begin{cases}...\\\\...\\\\...\\end{cases}}$ where the ellipses stand for any expression. – Yves Daoust Dec 8 '20 at 7:48
We note on the side that $$x^2\ne1$$ and multiply by $$x^2-1$$, so need to distinguish two cases:
$$\begin{cases}x^2-1>0\to 1\ge x-1,\\x^2-1<0\to 1\le x-1.\end{cases}$$
The second case is impossible and we are left with
$$x<-1\lor1
You can also work in systematic way, using a table of sign variations,
$$\begin{array}{}&&-1&&1&&2\\\hline x+1&-&0&+&+&+&+&+\\x-1&-&-&-&0&+&+&+\\2-x&+&+&+&+&+&0&-\\\hline&+&|&-&|&+&0&-\end{array}$$
Use the difference of two squares to get $$\frac{1}{(x+1)(x-1)} ≥ \frac{1}{x+1}$$.
Case 1: If $$x + 1 > 0 \Rightarrow x > -1$$, we have:
$$\frac{1}{x-1} ≥ 1 \tag{x \ne -1, 1}$$
The $$x \ne -1, 1$$ comes from the denominator $$(x+1)(x-1)$$, which is undefined when $$x = -1, 1$$.
If $$x - 1 > 0 \Rightarrow x > 1$$, then multiplying both sides by $$x-1$$ gives $$1 ≥ x - 1 \Rightarrow x ≤ 2$$. If $$x - 1 < 0$$, then because $$x > -1$$, there are no solutions for this case. Hence the intersection of $$x > -1, x > 1$$, and $$x ≤ 2$$ imply $$1 < x ≤ 2$$.
Case 2: If $$x + 1 < 0 \Rightarrow x < -1$$, we have:
$$\frac{1}{x-1} ≤ 1 \tag{x \ne -1, 1}$$
$$x - 1 > 0$$ is not possible in this case because $$x < -1$$. Thus $$x -1 < 0$$, or just $$x < -1$$, and there are no more conditions.
So the solution to the inequality is $$x < -1$$, $$1 < x ≤ 2$$ for $$x \in \mathbb R$$.
• Of course you can do this by sketching the graph, but I wonder if this can be done in an exam. If the question is multiple choice, use whatever method you prefer. – Toby Mak Dec 8 '20 at 7:19
First write the function such that both numerator an denominator have same sign. $$\frac{x-2}{(x-1)(x+1)} \leqslant 0$$ We use the sign scheme method for this. Now first we will find the critical points. Critical points are those points where the function changes its behaviour. Usually for this problem the zeroes and poles are those points.
Critical points are $$x = 2,1,-1$$. There are three critical points with each critical point having odd frequency i.e. all critical points occur odd number of times.
Now plot the points on a number line.
1. Find the maximum critical point. In this case it's $$x =2$$. The function is positive always when the value of x is greater than the maximum critical point. So the function is positive for $$x \geq 2$$.
2. The function between $$x=2$$ and $$x = 1$$ will be negative.
3. The function between $$x=1$$ and $$x=-1$$ will be positive.
4. The function beyond $$x=-1$$ will be negative.
Tricks to solve it quickly:
1. Always write the function in the above format. So that numerator and denominator have same sign.
2. Find the critical points. Any value of $$x$$ beyond the maximum critical point will make the function positive.
3. Now as we go backwards, see how many times the critical points occurred. Since $$x=2$$ occurred once, sign of the function will change once from positive to negative. And the function will behave such that its negative till the next critical point is encountered.
4. Now once you reach the critical point $$x=1$$ you will see it occurred odd number of times again the sign transition will happen once. So the function will behave such that its positive till it encounters the next critical point.
5. Once you hit the last critical point, see the frequency since its odd, one sign change. So beyond $$x=-1$$ the function will be negative.
6. Remember poles wont be included in the final solution as you will make the function infinitely large. Also zeroes will be included since the inequality also has an equal sign.
7. Now compile the solution from these regions: you want all the negative regions, the final solution being: $$x\in (-\infty,-1) \cup (1,2]$$
I will give a completely different example for the same thing so that concepts are clear. Lets say your function was: $$\frac{(x-2)(x-3)^2(x+5)^3}{(x-1)^5x(x+1)^6} \lt 0$$
1. Format is correct, numerator denominator has same signs.
2. Find critical points with frequency (number of times each point occurred). So the critical points and frequency are : $$x = 2(1),3(2),-5(3),0(1),-1(6),1(5)$$.
3. The region greater than the maximum critical point is always positive. So for values $$x \gt 3$$ the function is always positive. Note I have not done $$x \geq 3$$ because thats not the requirement of the inequality. It will follow for every such critical point.
4. As we move backwards, observe the number of occurrences of the critical point (odd or even) and apply that many sign transitions. For example, as we reach $$x=3$$ we see it occurred two times so sign transition will be 2 times from + to - (1 transition) and from - to + (2nd transition). So overall when the values are just less than 3 the sign of function didnot change.
5. Go backwards, as you encounter critical points make sign transitions equal to the number of occurrences of the critical point. So if 2 occured once only one sign transition will happen as the function goes beyond it.
Even frequency of critical point: No sign change will happen
Odd frequency of critical point: Sign change will happen
1. Now compile the solution. In the inequality we need the function to be negative. So just compile the negative parts from the figure. The solution will be : $$x \in (-5,-1) \cup (-1,0) \cup (1,2)$$ I had to break at -1 since the function will go to $$-\infty$$. If you want the solution such that the function can go to $$-\infty$$ you can include that. Hope this helps. With practice you will solve these lenghty questions in seconds. Tested :D | 2021-04-17T21:32:53 | {
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https://math.stackexchange.com/questions/950502/how-to-find-the-last-non-zero-digit-of-50 | # How to find the last non-zero digit of $50!$
A week ago i made a similar question but nobody help me, i´ve been trying but i still don't get it.
I want to know how to find the last non-zero digit of $50!$.
my try:
First i have to know how much Zeros $50!$ has so i did this:
$$E_{5}50 = \sum _{1\leq k <n} \Bigg[\frac{50}{5^{k}}\Bigg] =\Bigg[\frac{50}{5}\Bigg] + \Bigg[\frac{50 }{25}\Bigg] = 12$$
So $50!$ has $12$ zeros which means that the last digit of $\quad\frac{50!}{10^{12}}\quad$ is the number that i´m looking for.
so if $x = \frac{50!}{10^{12}}$ i need to find $x (mod 10)$ to get it but this is such a big number and i still don't know how to reduce it.
It is enough to find $x \mod 2$ and $x \mod 5$ in order to know $x \mod 10$. It is easy to show that $x \mod 2 = 0$, so all you really need to do is find $x \mod 5$. Since $2^{12}$ is invertible modulo $5$, it is enough to do this for $2^{12} x \mod 5$ and then divide by $2^{12}$. The number $2^{12} x = \frac{50!}{5^{12}}$ is not too hard to write as a product of several numbers, and after reducing modulo $5$, you'll see that there is a lot of repetition in these numbers, which simplifies the calculation.
Here's a different approach: find the last non-zero digit of $10!$, then find the last non-zero digit of $20!/10!$, and so on for $30!/20!$, $40!/30!$, and $50!/40!$, then multiply them. It's easier to cancel in $5$ groups of $10$ numbers rather than one group of $50$.
Here's $10!$:
$$1\times2\times3\times4\times5\times6\times7\times8\times9\times10$$
Divide out the two factors of $5$, along with two factors of $2$ (the $4$):
$$1\times2\times3\times6\times7\times8\times9\times2$$
(We took out the $4$ and the $5$, and replaced the $10$ with a $2$). It's easy to compute the last digit of this, which is an $8$.
For $20!/10!$, we have
$$11\times12\times13\times14\times15\times16\times17\times18\times19\times20$$
Divide out the two factors of $5$ along with two factors of $2$ to get
$$11\times3\times13\times14\times3\times16\times17\times18\times19\times4$$
which, taking the units digit only, is
$$1\times3\times3\times4\times3\times6\times7\times8\times9\times4$$
and has units digit $8$. Can you continue?
• Yes, i guess i can continue...Thanks!!!1 – Jearson Narvaez Rojas Sep 29 '14 at 3:40
My Solution:: calculation of last non zero digit in $50!$
Now first we will calculate number of zeros at the end of $50!$, which is obtained
when we multiplied $2$ and $5$. So we will calculate no. of $2$ and $5$ in $50!$
$\displaystyle 50! = \lfloor \frac{50}{2}\rfloor +\lfloor \frac{50}{2^2} \rfloor+\lfloor \frac{50}{2^3}\rfloor+\lfloor \frac{50}{2^4}\rfloor+\lfloor \frac{50}{2^5}\rfloor+........... = 25+12+6+3+1 = 47$
$\displaystyle 50! = \lfloor \frac{50}{5}\rfloor+\lfloor \frac{50}{5^2}\rfloor+..... = 10+2 = 12$
So Total no. of zeros at the end of $50! = 12$
Now $\displaystyle 50! = \lfloor \frac{50}{3}\rfloor+\lfloor\frac{50}{3^2}\rfloor+\frac{50}{3^3}\rfloor+..... = 16+5+1 = 22$
Now we can write $\displaystyle 50! = 1\times 3^{22}\times 7^8\times 11^4 \times 13^3 \times 17 \times 19\times 23\times 29\times 31 \times 17\times 37 \times 41 \times 43 \times 23 \times 47 \times 10^{12}\times 2^{35} = \bf{last digit\; (6)} = 6$
• It's not $6$, it's $2$. – Edward Jiang Sep 29 '14 at 3:43
• Your logic is unclear! :-( – xxx--- Oct 8 '14 at 13:33
After reducing the zeroes, what you have is $2^{35}3^{22}7^{8}\cdots \equiv 8 \cdot 9 \cdots \equiv 2 \pmod {10}$. | 2019-11-18T12:00:39 | {
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http://math.stackexchange.com/questions/32444/simple-probability-question-balls-and-bins/32455 | # Simple probability question, balls and bins
This is a simple question I came across in reviewing. I am wondering if I got the correct answer.
The question is simple. You have $n$ balls and $m$ bins. Each ball has an equal probability of landing in any bin. I want to know what the probability that exactly $1$ bin is empty.
My answer seems simple enough, but I don't think it's sufficient. It is $(\frac{m-1}{m})^n$ since for each ball, it can go in any of the other bins. I think, however, that this is just the probability that some arbitrary bin $A$ is empty, not exactly one bin. What else should I consider?
-
If you want to see lots of variations to this type of problem, check out the Twelvefold Way. A good place to see this is in Stanley's book and you can get the second edition free for now. – Graphth Apr 12 '11 at 2:41
@Douglas: I agree with your last comment, up to and including with this question, but I disagree with the rest of it. Yes your solution is correct, yes the accepted one is flagrantly wrong, yes you did everything you could to explain things, patiently, going back to the basics, giving elementary examples, as one should do in such cases. And all this did not work with the OP and a few others. So what? Is this the first time you see such things happening here? Is this enough to declare the site worthless? I do not think so. So, please forget the noise and keep up the good job. – Did Apr 12 '11 at 8:28
I just voted up Douglas's answer and voted down the other one. Nothing personal here, no implication of any sort, this is simply to signal that one solution is wrong and the other one is correct. – Did Apr 12 '11 at 8:31
@Douglas: It occasionally occurs that erroneous answers are temporarily upvoted higher than correct ones. However this is usually rectified very quickly once the error has been pointed out. If all else fails it can be discussed on meta. It seems you've had the bad luck of encountering a few anomalous threads in your initial experiences here. Don't let them spoil your impression of the entire site. If you browse the prior questions you'll see that such situations are by far the exception rather than the rule. – Bill Dubuque Apr 12 '11 at 16:50
I've updated the accept to the correct solution. Thanks to everybody for the explanations. – user9470 Apr 12 '11 at 18:06
Let's count configurations, and then divide by $m^n$.
There are $m$ choices for the empty bin. Then the other bins are occupied. We can count the ways to place $n$ balls in $m-1$ bins so that no bin is empty by inclusion-exclusion: It is
$$\sum_{k ~\text {bins known to be empty}} (-1)^k {m-1 \choose k} (m-1-k)^n.$$
Another way to get this is to label the parts of a set partition of size $n$ with $m-1$ parts. The number of set partitions with a given number of parts is a Stirling number of the second kind, and we want $(m-1)! S(n,m-1)$.
Multiply this by $m$ and then divide by $m^n$ to get the probability exactly $1$ bin is empty.
We can use the same techniques to compute the probability exactly $e$ bins are empty for other values of $e$. For example, suppose there are $4$ bins and $6$ balls. Then there are $1560$ ways for there to be $0$ empty bins, $2160$ ways for there to be exactly $1$ empty bin, $372$ ways for there to be exactly $2$ empty bins, and $4$ ways for there to be exactly $3$ empty bins. The total is $4096 = 4^6$. Dividing by this gives a probability of $\frac{135}{256} = 0.52734375$ that exactly $1$ bin is empty.
-
My experience is if you get an explanation you can improve it, but an explanation is unlikely. +1 from me. – Ross Millikan Apr 12 '11 at 4:24
@Douglas: Please don't leave the site over this one. One gets stray downvotes occasionally. I have upvoted your answer and asked OP to move the accept over here. – Ross Millikan Apr 12 '11 at 13:48
Here you might have missed the context of the comments on the deleted answer, where multiple people were repeating that I was wrong when I pointed out simple cases which my answer handles correctly, but which were not enumerated correctly by the then-accepted answer. The pattern is that I don't like it when I post something correct, and then multiple people say something clearly incorrect is better. I'm not always right, but I value correctness, and having a club of people praising each other's incorrect answers and voting down my correct answers is intolerable to me. – Douglas Zare Apr 12 '13 at 23:33
@Douglas: If anyone would have flagged the comment when he made it, the moderators would have known about it and taken appropriate action. But no one did flag it, so the moderators didn't know about it, so I don't see why you're upset with me. I understand your desire to respond to whuber's comment (I have now read it), but a more appropriate course of action would have to flag it asking for a moderator to delete it. I think there's no value in keeping these comments that only record a sore argument that was resolved a year ago; I am going to delete them, unless you have a good reason not to. – Zev Chonoles Apr 15 '13 at 3:20
@Douglas: Within minutes of seeing your response, whuber deleted his comment and flagged yours. – Zev Chonoles Apr 21 '13 at 2:30 | 2015-08-03T13:16:49 | {
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https://math.stackexchange.com/questions/727203/last-decimal-digit-of-any-perfect-square-must-be-0-1-4-5-6-or-9 | # Last decimal digit of any perfect square must be $0,1,4,5,6$ or $9$
Last decimal digit of any perfect square must be $0,1,4,5,6$ or $9$
My Proof: Ten cases exist, yielding the following equalities:
$$(1\mod{10})^2 = 1\mod{10}$$ $$(2\mod{10})^2 = 4\mod{10}$$ $$(3\mod{10})^2 = 9\mod{10}$$ $$(4\mod{10})^2 = 6\mod{10}$$ $$(5\mod{10})^2 = 5\mod{10}$$ $$(6\mod{10})^2 = 6\mod{10}$$ $$(7\mod{10})^2 = 9\mod{10}$$ $$(8\mod{10})^2 = 4\mod{10}$$ $$(9\mod{10})^2 = 1\mod{10}$$ $$(0\mod{10})^2 = 0\mod{10}$$
Since the proposition holds for all possible cases, the proposition holds.
Is this an acceptable proof for the proposition?
What is the simplest proof for this?
Note: This is not a homework question, just a question from a weekly tutorial sheet.
• This is exhaustive, so it works. – wckronholm Mar 26 '14 at 4:07
• @wckronholm I feel as though I should try to avoid using exhaustive proofs, they seem really inefficient. – Display Name Mar 26 '14 at 4:10
• No problem if it is homework-you have shown your work as requested by the FAQ and asked a good related question. +1 – Ross Millikan Mar 26 '14 at 4:18
This is a fine proof. You could consider a few less cases in two similar ways. You could do it mod $2$ and $5$, then combine the results with the Chinese Remainder Theorem. You could do $0, \pm 1, \pm2, \pm 3, \pm 4, 5$ Both would show off more math knowledge than this approach, but I am not sure either would be less work. For moduli higher than $10$, the investment might be repaid.
• It appears a similar proof can be completed for $(x \mod 10)^3,0 \leq x \leq 9$, can a similar proof be completed for all $n \in \mathbb{Z}^+,(x \mod 10)^n, 0 \leq x \leq 9$ – Display Name Mar 27 '14 at 2:43
• Yes, you can do this for any modulus. As the modulus gets higher, the work grows. You probably want those $\ge$ signs to be $\le$ signs. For the cubes, every digit can end a cube. For any even power, there are no more than six ending digits, because $(-1)^2=1,$, so $1$ and $9$ always end in the same, as do $2$ and $8$, etc. – Ross Millikan Mar 27 '14 at 2:47
As $10=5\times 2$, let us consider $$x^2 = 10a + 5b+ r \\ 0\le 5b+r< 10 \\ 0\le r< 5$$ then reducing modulo 5: $$r = x^2 \in \{ 0,1,4 \}\mod 5\\ 5b+r \in \{ 0,1,4,0+5=5,1+5=6,4+5=9 \}\mod 5\\$$ | 2019-09-19T13:07:24 | {
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https://math.stackexchange.com/questions/2751801/if-fx-is-differentiable-in-mathbbr-a-b-in-mathbbr-a-neq-b-such | # If $f(x)$ is differentiable in $\mathbb{R}$ & $a,b \in \mathbb{R}, a\neq b$ such that $f'(x)=(x-a)(x-b)$ then $f$ has exactly one local min and max?
Prove or contradict:
If $f(x)$ is differentiable in $\mathbb{R}$ and exist $a,b \in \mathbb{R}, a\neq b$ such that $f'(x)=(x-a)(x-b)$ then $f$ has exactly one local minimum and 1 local maximum
I know that $f'$ has only two roots which are $a$ and $b$ as the possible locations for min and max points, but how do I show these are necessarily or are not necessarily min and max points?
• Check the sign of $f'(x)$ when $x<a$, $a<x<b$ and $x>b$. – Trevor Norton Apr 24 '18 at 15:57
• Use second derivative test i.e. look at $f''(a)$ and $f''(b)$, whichever is $<0$ is the local maximum and whichever is $>0$ is local minimum. First, without loss of generality, assume $a<b$, then proceed with the test – Naweed G. Seldon Apr 24 '18 at 16:00
• mathworld.wolfram.com/SecondDerivativeTest.html – user547564 Apr 24 '18 at 16:02
• Alright cool! Got it thanks guys! – Jason Apr 24 '18 at 16:05
## 3 Answers
\begin{align} f''(x) = \left( x - b \right) + \left( x - a \right) \end{align} \begin{align} f''(a) = \left( a-b \right) \end{align} \begin{align} f''(b) = \left( b -a \right) \end{align} When \begin{align} a \gt b \end{align} Then \begin{align} f''(a) \gt 0 \end{align} And \begin{align} f''(b) \lt 0 \end{align} When \begin{align} b \gt a \end{align} Then \begin{align} f''(a) \lt 0 \end{align} And \begin{align} f''(b) \gt 0 \end{align} As you can see in both cases we have a local Maxima and minima
• this only proves existence but not uniqueness – Surb Apr 24 '18 at 17:59
• When \begin{align} a\gt b \end{align} then \begin{align} f''(a) \gt 0 \end{align} and \begin{align} f''(b) \lt 0 \end{align} so here \begin{align} a \end{align} is the local minima and \begin{align} b \end{align} is local Maxima also when \begin{align} b \gt a \end{align} then \begin{align} f''(b) \gt 0 \end{align} and \begin{align} f''(a) \lt 0 \end{align} so here \begin{align} a \end{align} is local Maxima and \begin{align} b \end{align} is local minima. So in both cases we have a unique Maxima and minima – Apurv Apr 25 '18 at 1:01
• First, you may want to know that in Mathjax, $f'(a)$ produces $f'(a)$ and $$f'(a)$$ produces $$f'(a)$$. – Surb Apr 25 '18 at 5:36
• So let $c$ be another local Maxima or Minima then so in that case $f'(c) = 0$ and $(c-b)(c-a) = 0$ solving this we get $c=b$ or $c=a$ which means only Maxima or Minima we can have is $a$ or $b$ and when $a \gt b$ $a$ is local minima and $b$ is local Maxima and when $b \gt a$ $a$ is Maxima and $b$ is minima(which I have proven in the answer). So in either case we have a unique Maxima and Minima – Apurv Apr 25 '18 at 8:07
• since $f'$ has only two roots (a,b) then those are the only possible maxima and minima points, and since there exists one of each there is exactly one of each. (basically what @Apruv said but less technically) – Jason Apr 25 '18 at 8:09
Let $a<b$ without loss of generality. For $x<a$ we have $f'(x)>0$, for $x=a$ we have $f(a)=0$ and for $a<x<b$ we get $f'(x)<0$. This can be interpreted as $f$ is increasing up to $x=a$ and then decreases while $x$ approaches $b$ from the left. So $f(a)$ is a maximum. Similarly $f'(b)=0$ and $f'(x)>0$ for all $x>b$ which implies that $f$ once it reaches $f(b)$ starts increasing. So $f(b)$ is a minimum. These two points $a,b$ are the only local extrema since $f'(x)$ vanishes exactly there by its very definition.
• Currently, this is the only correct answer. – Surb Apr 24 '18 at 18:01
The stationary point qualification is done with
$$f''(x) = (x-a)+(x-b) = 2x-(a+b)$$
hence we have
$$f''(a) = a-b\\ f''(b) = -(a-b)$$
so one of then is a relative minimum and the other a relative maximum deppending on $a > b$ or $b > a$ | 2020-09-30T07:22:26 | {
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https://ask.sagemath.org/question/38420/how-accelerate-this-simple-program/?answer=38422 | # how accelerate this simple program
Hello everybody. Here is my question We seek integers of which the decimal part of their square root begin by 2017. Sample: sqrt(10858) = 104.201727.... Here is my program:
for n in range (1, 10^5):
if int( N(sqrt(n)).frac() * 10000 ) == 2017:
cpt += 1
the problem is it takes too much time. For the bound 10^5 it takes about 55 sec on my iMac, for 10^6 it takes 620 sec, for 10^7 it takes almost 6000 sec ... The question is to obtain the result for the bound 10^10 !
edit retag close merge delete
Sort by » oldest newest most voted
Let us put some more notation in there, that will support the code. So we seek integers $M\le 10^{10}$ of which the decimal part $K$ of their square root begin by $2017$. Mathematically written: $$K+\frac{2017}{10\ 000} \le \sqrt M < K+\frac{2018}{10\ 000}$$
Sample: sqrt(10858) = 104.201727... . So $M=10\, 858$ is a "good $M$-value".
Proposal: Let us equivalently look for the $K$-values instead of the $M$ values. So we rewrite the statement:
we seek integers $K< 10^{10/2}$ so that between the two numbers $$a(K)=\left(K+\frac{2017}{10\ 000}\right)^2 \ ,\ b(K)=\left(K+\frac{2018}{10\ 000}\right)^2$$ there exist at least an integer $M$.
Sample: 104.2017^2 = 10857.99428289 and 104.2018^2 = 10858.01512324 . So $M=10858$ is a good $M$-value.
An other sample.
sage: "%.4f" % ( 8888.2017^2 )
'79000129.4599'
sage: "%.4f" % ( 8888.2018^2 )
'79000131.2375'
So the two numbers $M=79000129$ and $M=79000130$ are both good values. Explicitly:
sage: for K in range( 79000124, 79000135 ): print "sqrt(%s) ~ %.4f" % ( K, sqrt(RR(K)) )
sqrt(79000124) ~ 8888.2014
sqrt(79000125) ~ 8888.2014
sqrt(79000126) ~ 8888.2015
sqrt(79000127) ~ 8888.2016
sqrt(79000128) ~ 8888.2016
sqrt(79000129) ~ 8888.2017
sqrt(79000130) ~ 8888.2017
sqrt(79000131) ~ 8888.2018
sqrt(79000132) ~ 8888.2018
sqrt(79000133) ~ 8888.2019
sqrt(79000134) ~ 8888.2020
In such a case, we have to count the integer $K$ twice.
Since $K^2\in\mathbb Z$, we can simplify the above condition. We seek equivalently integers $K< 10^{5}$ so that between the two numbers $$A(K)=2K\cdot \frac{2017}{10\ 000} +\left(\frac{2017}{10\ 000}\right)^2$$ $$B(K)=2K\cdot \frac{2018}{10\ 000} +\left(\frac{2018}{10\ 000}\right)^2$$ there exist at least an integer $M$.
The posted code seems to only count the "good" integers $M$. Then the equivalent counting program would be:
def count_M_values( bound, list_M_vales=False ):
s = 2017 / 10000
t = 2018 / 10000
ss = s^2
tt = t^2
counter = 0
mList = []
for K in xrange( bound ):
addtocounter = floor( 2*K*t + tt ) - floor( 2*K*s + ss )
if list_M_vales:
mList . extend( [ floor( (K+s)^2 ) + 1 .. floor( (K+t)^2 ) ] )
if list_M_vales:
return counter, mList
return counter
This gives for instance:
sage: count_M_values( 10**1 )
0
sage: count_M_values( 10**2 )
0
sage: count_M_values( 200 )
3
sage: count_M_values( 200, True )
(3, [10858, 24399, 25986])
sage: count_M_values( 300, True )
(9, [10858, 24399, 25986, 45455, 47612, 49819, 73009, 75736, 78513])
sage: count_M_values( 10**3 )
99
sage: count_M_values( 10**4 )
9998
sage: count_M_values( 10**5 )
999980
sage: count_M_values( 10**6 )
99999800
sage: count_M_values( 10**7 )
9999998000
(We have a nice count pattern that we can now investigate mathematically.)
Note that the bound 10**7 above applies for $K$, we are thus going up to an $M$--bound equal to $10^{14}$. The post wanted the answer for the $K$-bound 10**5.
One more check for the $K$-bound $300$:
count, mList = count_M_values( 300, True )
for M in mList:
print "sqrt( %s ) ~ %.7f" % ( M, RR( sqrt(M) ) )
We get:
sqrt( 10858 ) ~ 104.2017274
sqrt( 24399 ) ~ 156.2017926
sqrt( 25986 ) ~ 161.2017370
sqrt( 45455 ) ~ 213.2017824
sqrt( 47612 ) ~ 218.2017415
sqrt( 49819 ) ~ 223.2017025
sqrt( 73009 ) ~ 270.2017765
sqrt( 75736 ) ~ 275.2017442
sqrt( 78513 ) ~ 280.2017131
more
Good evening. A very big thank you for your long explanation that I find extraordinary. I would never have thought of doing that. I read and reread your text several times and I think I understood the principle. The result is found in just over a second! It's awesome
( 2017-07-28 17:14:16 -0500 )edit
One bottleneck comes from working with the symbolic ring. The following code is some hundred times faster (may be still not fast enough ;-) )
import math
def count_2017(bound=100000):
cpt = 0
for n in range (1,bound):
if int(RR(math.sqrt(n)).frac()*10000) == 2017:
cpt += 1
return cpt
Update: Using cython makes the code another ten times faster, for example:
fits_2017 = cython_lambda("double x", "bool(int(10000*(x-int(x))) == 2017)")
f = cython_lambda("int n", "sqrt(n)")
def count_2017(bound=100000):
cpt = 0
for n in range(1,bound):
if fits_2017(f(n)):
cpt += 1
return cpt
count_2017(10^7)
more
Indeed your code works more than 100 times faster. I do not understand why RR (math.sqrt (n)) is 100 times faster than N (sqrt (n)). Where can I find an explanation (as simple as possible)?
I tried your cython code and I get the following error.
compilation terminated. error: command 'gcc' failed with exit status 1
Is not Sagemat already written in cython?
( 2017-07-28 16:41:36 -0500 )edit
math.sqrt returns a numerical value, sage sqrt returns a symbolic expression.
See an example: http://sagecell.sagemath.org/?q=bgdeqj
Exact symbolic operations are much slower than numerical algorithms. Converting a symbolic expression to a numerical value takes additional time.
( 2017-07-29 07:43:35 -0500 )edit
thank you for your explanations. I see that I still have a lot to learn. It's frustrating to be stuck in solving a problem because I do not know all the tricks that make a program much more effective. Where can I find answers? Which book or site or ...?
( 2017-07-30 17:15:17 -0500 )edit | 2020-10-28T14:47:48 | {
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http://math.stackexchange.com/questions/275928/how-to-solve-x-5-2x6-1 | # How to solve $|x-5|=|2x+6|-1$?
$|x-5|=|2x+6|-1$.
The answer is $0$ or $-12$, but how would I solve it by algebraically solving it as opposed to sketching a graph?
$|x-5|=|2x+6|-1\\ (|x-5|)^2=(|2x+6|-1)^2\\ ...\\ 9x^4+204x^3+1188x^2+720x=0?$
-
One hard way is to determine for what values of $x$ the absolute value is positive or negative. For example, if $x\geq 5$ then $|x-5|=x-5$. Do the same to the other and take the intersections of intervals. – Sigur Jan 11 '13 at 17:54
Consider different cases:
Case 1: $x>5$ In this case, both $x-5$ and $2x+6$ are positive, and you can resolve the absolute values positively. hence $$x-5=2x+6-1 \Rightarrow x = -10,$$ which is not compatible with the assumption that $x>5$, hence no solution so far.
Case 2: $-3<x\leq5$ In this case, $x-5$ is negative, while $2x+6$ is still positive, so you get $$-(x-5)=2x+6-1\Rightarrow x=0;$$ Since $0\in[-3,5]$, this is our first solution.
Case 3: $x\leq-3$ In this final case, the arguments of both absolute values are negative and the equation simplifies to $$-(x-5) = -(2x+6)-1 \Rightarrow x = -12,$$ in agreement with your solution by inspection of the graph.
-
I was really preparing an answer like it. Nice and understandable one. +1 – Babak S. Jan 11 '13 at 18:00
That's not the way, because you're creating extra solutions (new roots) when you put square on both sides. Just separate the options:
$$x-5=2x+6-1\\ 5-x=2x+6-1\\ x-5=-2x-6-1\\ 5-x=-2x-6-1$$
Solve all of them, and you have the solutions
Take into acount that once you get your solutions, you have to check if they're possible, for example, in the first one, you have supposed that bot things inside || are positive, so if you get something for which x-5 or 2x+6 is negative, then you have to throw it away
-
Look for where the expressions inside the absolute values change sign: $x-5$ changes sign at $x=5$, and $2x+6$ changes sign at $x=-3$. Thus, when $x<-3$, $x-5$ and $2x+6$ are both negative, and the equation is $$-(x-5)=-(2x+6)-1\;.$$
When $-3\le x<5$, $x-5$ is negative and $2x+3$ is non-negative, so the equation is
$$-(x-5)=2x+6-1\;.$$
And when $x\ge 5$, both expressions are non-negative, and the equation is
$$x-5=2x+6-1\;.$$
Thus, you need to solve
\left\{\begin{align*} &-x+5=-2x-7&&\text{when }x<-3\\ &-x+5=2x+5&&\text{when }-3\le x<5\\ &x-5=2x+5&&\text{when }x\ge 5\;. \end{align*}\right.\tag{1}
$(1)$ reduces to
\left\{\begin{align*} &x=-12&&\text{and }x<-3\\ &x=0&&\text{and }-3\le x<5\\ &x=-10&&\text{and }x\ge 5\;. \end{align*}\right.\tag{2}
The first two solutions in $(2)$ fall within the intervals on which they are valid; the third does not and therefore is not a solution.
-
I think the sketch approach suggests a solution, which is to partition the domain.
Consider the function $f(x) = |x-5|-|2x+6|+1$. Look at the absolute value part and see that you can split the domain into $I_1= (-\infty, -3]$, $I_2 = (-3,5]$ and $I_3 = (5,\infty)$.
On $I_1$, $f(x) = x+12$, on $I_2$, $f(x) = -3x$, and on $I_3$, $f(x) = -(x+10)$.
Now look for solutions to $f(x) = 0$ on each of these intervals.
For example, on $I_3$, solving $-(x+10) = 0$ yields $x=-10$, but $-10 \notin I_3$, hence $f(x) \neq 0$ for $x \in I_3$.
-
Nice illustration. – Babak S. Jan 11 '13 at 18:43
You have $|x-5|=|2x+6|-1$.
1)If $x\geq5$, $|x-5|=x-5$ and, $|2x+6|=2x+6$, so you have
$x-5=2x+6-1$
$x-5=2x+5$
$x=-10$ (but is not valid)
2)If $-3/2\leq x<5$, $|x-5|=-(x-5)$ and $|2x+6|=2x+6$, then
$-(x-5)=2x+6-1$
$-x+5=2x+5$
$-x=2x$
$x=0$
3)If $x<-3/2$, $|x-5|=-(x-5)$ and $|2x+6|=-(2x+6)$, then
$-(x-5)=-(2x+6)-1$
$-x+5=-2x-6-1$
$-x+5=-2x-7$
$-x+5=-2x-7$
$x=-12$
So, the answer for $x$ is $x=0$ or $x=-12$
-
hint: at first let $x>5$ then solve it. Then do the same for the intervals $-3<x<5$ and $x<-3$
- | 2015-07-28T21:51:06 | {
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anonymous one year ago Hi, i ask this the other day and if someone can solve it with answer and explains, will fan and medal: When a beam makes an angle of 40 degrees with the ground, the top of the beam is 40 feet above the ground. There a telephone wires near by and the worker worried that the beam may hit the wires. When the beam makes a angle of 60 degrees with the ground, the wires are 2 feet above the beam. Will the beam hit the ground if the crew continues to raise it?
• This Question is Closed
1. anonymous
@Nnesha
2. Nnesha
looks like physics q lel :P well i have to go wait for someone or i'll llook at 't later !
3. anonymous
okay :)
4. radar
|dw:1439431270274:dw| I believe the question should be does the pole hit the WIRE as they continue to raise it, it will hit the ground if they drop it.!
5. anonymous
@radar: i'm sorry, it suppose to be if it hit the wires or not.
6. mathstudent55
|dw:1439437785859:dw|
7. mathstudent55
Let's look at the beam at 40 deg to the ground.
8. mathstudent55
|dw:1439437990141:dw|
9. mathstudent55
$$\sin 40^o = \dfrac{40~ft}{L}$$ $$L = \dfrac{40~ft}{\sin 40^o} \approx 62.23 ~ ft$$
10. mathstudent55
Now we know the length of the beam. Now we look at the situation when the beam is at 60 degrees from the ground.
11. mathstudent55
|dw:1439438459934:dw|
12. mathstudent55
Notice I just corrected the length of the beam. I just noticed my calculator was set to radians and the length of the beam was incorrect. The length of the beam is 62.23 ft. This number is correct now.
13. mathstudent55
Now we find h, the height of the beam when it is at a 60-deg angle with the ground.
14. mathstudent55
$$\sin 60^o = \dfrac{h}{62.23~ft}$$ $$h = (62.23~ft)\sin 60^o \approx 53.89~ft$$
15. mathstudent55
Since the wires are 2 ft higher than the top of the beam when the beam is at 60 deg with the ground, the wires are at 55.89 ft above the ground.
16. mathstudent55
Since the beam is 62.23 ft long, and the wires are only 55.89 ft above ground, if the crew continues to raise the angle of the beam, the beam WILL HIT the wires.
17. mathstudent55
At what angle of the beam will it hit the wires? |dw:1439438842464:dw|
18. anonymous
Beware the OS Code of Conduct!
19. mathstudent55
$$\sin x = \dfrac{55.89}{62.23}$$ $$\sin^{-1} \dfrac{55.89}{62.23} = 63.91^o$$ At 63.91 deg, the beam will hit the wires.
20. mathstudent55
@Leong Do you understand?
21. anonymous
@mathstudent55 yes :) thank you :) I was just drawing it again and again but didn't think of the law of sines :)
22. mathstudent55
I did not use the law of sines. This is simply the definition of the sine in a right triangle.
23. mathstudent55
|dw:1439439303636:dw|
24. mathstudent55
We used the sine ratio which is opp/hyp. |dw:1439439409054:dw|
25. mathstudent55
To find the length of the beam, we did this: |dw:1439439471920:dw|
26. mathstudent55
To find the heigth the beam reaches at 60 deg, we did this: |dw:1439439571183:dw|
27. anonymous
wait, so like the beam will hit the wires at 60 something degree, because Sin^-1(53.89 over62.23 =59.99
28. mathstudent55
It equals 63.91 deg. Yes, when the beam is at 63.91 deg with the ground, the beam will hit the wires.
29. anonymous
okay, thanks. i bad at round numbers
30. mathstudent55
No problem. You're welcome.
31. Nnesha
cool ,-,
32. radar
You beat me to this, @mathstudent55 and did an excellent step by step job. I could not done any better or clear explanation. You certainly deserve the medal.
33. mathstudent55
@radar A good word from you means a lot to me. I really appreciate it. Also, thanks for the testimonial.
34. anonymous
@mathstudent55 hey, so I got a mistake that the when th beam make angle of 40 degree then from the beam to the ground is 8ft, so the beam won't hit the wires since we know that the beam is 12,45 ( I had calculator again myself) and when it make angle of 60 then from the beam to the sruface is 10,78. by that, the beam WILL NOT hit the wires, so do I have to find anything else when I know that the beam will not hit the wires?
35. radar
Please post the problem again, providing correct info. The problem as presently posted, states that at 40 degrees elevation, the top of the beam is 40 ft. above the ground. How you got 12.45 ft. is not clear.
36. anonymous
okay
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https://nadv.pt/benzoin-to-ucu/a5ee2f-horizontal-line-test-inverse | Horizontal Line Test We can also look at the graphs of functions and use the horizontal line test to determine whether or not a function is one to one. Change ), You are commenting using your Google account. Historically there has been a lot of sloppiness about the difference between the terms “range” and “co-domain.” According to Wikipedia a function g: A -> B has B by definition as codomain, but the range of g is exactly those values that are g(x) for some x in A. Wikipedia agrees with you. Horizontal Line Test. Find the inverse of f(x) = x2 + 4 , x < 0. It can be seen that with this domain, the graph will pass the horizontal test. This function is called the inverse function. Option C is correct. So in short, if you have a curve, the vertical line test checks if that curve is a function, and the horizontal line test checks whether the inverse of that curve is a function. But it does not guarantee that the function is onto. It is an attempt to provide a new foundation for mathematics, an alternative to set theory or logic as foundational. If a horizontal line cuts the curve more than once at some point, then the curve doesn't have an inverse function. Note: The function y = f(x) is a function if it passes the vertical line test. Remember that it is very possible that a function may have an inverse but at the same time, the inverse is not a function because it doesnât pass the vertical line test . b) Since every horizontal line intersects the graph once (at most), this function is one-to-one. Change ), You are commenting using your Twitter account. Both are required for a function to be invertible (that is, the function must be bijective). We note that the horizontal line test is different from the vertical line test. Change f(x) to y 2. Change ). Therefore, if we draw a horizontal line anywhere in the -plane, according to the horizontal line test, it cannot intersect the graph more than once. This test allowed us to determine whether or not an equation is a function. If the horizontal line touches the graph only once, then the function does have an inverse function. Now here is where you are absolutely correct. We are allowed to say, “The sine function has an inverse arcsin,” even though to be more pedantic we should say that sin(x) on the domain (-pi/2, pi/2) has an inverse, namely Arcsin(x), where we use the capital letter to tell the world that we have limited the domain of sin(x). (Recall from Section 3.3 that a function is strictly These are exactly those functions whose inverse relation is also a function. Figure 198 Notice that as the line moves up the $$y-$$ axis, it only ever intersects the graph in a single place. We can see that the range of the function is y > 4. If you did the Horizontal Line Test with the graph, you'd know there's no inverse function as it stands. The graph of the function does now pass the horizontal line test, with a restricted domain. 5.5. ( Log Out / Combination Formula, Combinations without Repetition. Example #1: Use the Horizontal Line Test to determine whether or not the function y= x2graphed below is invertible. It is used exclusively on functions that have been graphed on the coordinate plane. Therefore, f(x) is a oneto one function and f(x) must have an inverse. Switch x and y Find f(g(x)) and g(f(x)) f(g(x))=x 3. Horizontal Line Test The horizontal line test is a convenient method that can determine whether a given function has an inverse, but more importantly to find out if the inverse is also a function. Sorry, your blog cannot share posts by email. With range y < 0. A test use to determine if a function is one-to-one. This function passes the horizontal line test. OK, if you wish, a principal branch that is made explicit. Because for a function to have an inverse function, it has to be one to one. If no horizontal line intersects the graph of a function f more than once, then the inverse of f is itself a function. It is called the horizontal line test because the test is performed using a horizontal line, which is a line that runs from left to right on the coordinate plane. Here’s the issue: The horizontal line test guarantees that a function is one-to-one. Horizontal Line Test Given a function f(x), it has an inverse denoted by the symbol \color{red}{f^{ - 1}}\left( x \right), if no horizontal line intersects its graph more than one time.. Horizontal Line Test. Find out more here about permutations without repetition. That research program, by the way, succeeded.). This is known as the horizontal line test. Determine whether the function is one-to-one. Solve for y 4. Determine the conditions for when a function has an inverse. Regardless of what anyone thinks about the above, engaging students in the discussion of such ideas is very helpful in their coming to understand the idea of a function. Hereâs the issue: The horizontal line test guarantees that a function is one-to-one. If we alter the situation slightly, and look for an inverse to the function x2 with domain only x > 0. Math Teachers at Play 46 « Let's Play Math. Because a function that is not one to one initially, can have an inverse function if we sufficiently restrict the domain, restricting the. However, if you take a small section, the function does have an inv⦠Using Compositions of Functions to Determine If Functions Are Inverses When I was in high school, the word “co-domain” wasn’t used at all, and B was called the “range,” and {g(x): x in A} was called the “image.” “Co-domain” didn’t come into popular mathematical use until an obscure branch of mathematics called “category theory” was popularized, which talks about “co-” everythings. This function passes the Horizontal Line Test which means it is a onetoone function that has an inverse. If it intersects the graph at only one point, then the function is one-to-one. This function is both one-to-one and onto (bijective). The function has an inverse function only if the function is one-to-one. I have a small problem with the following language in our Algebra 2 textbook. The range of the inverse function has to correspond with the domain of the original function, here this domain was x > -2. Functions whose graphs pass the horizontal line test are called one-to-one. See Mathworld for discussion. This preview shows page 27 - 32 out of 32 pages.. 2.7 Inverse Functions One to one functions (use horizontal line test) If a horizontal line intersects the graph of f more than one point then it is not one-to-one. So as the domain and range switch around for a function and its inverse, the domain of the inverse function here will be x > 4. They were “sloppy” by our standards today. This Horizontal Line Test can be used with many functions do determine if there is a corresponding inverse function. A function must be one-to-one (any horizontal line intersects it at most once) in order to have an inverse function. If (x,y) is a point on the graph of the original function, then (y,x) is a point on the graph of the inverse function. This means this function is invertible. Inverse Functions: Definition and Horizontal Line Test (Part 3) From MathWorld, a function is an object such that every is uniquely associated with an object . So the inverse function with the + sign will comply with this. Draw the graph of an inverse function. A function f is invertible if and only if no horizontal straight line intersects its graph more than once. With a blue horizontal line drawn through them. But the inverse function needs to be a one to one function also, so every x value going in needs to have one unique output value, not two. That hasn’t always been the definition of a function. The quiz will show you graphs and ask you to perform the line test to determine the type of function portrayed. Instead, consider the function defined . The domain will also need to be slightly restricted here, to x > -5. Find the inverse of a given function. Evaluate inverse trigonometric functions. Whatâs known as the Horizontal Line Test, is an effective way to determine if a function has an inverse function, or not. The horizontal line test answers the question âdoes a function have an inverseâ. 1. 3. The mapping given is not invertible, since there are elements of the codomain that are not in the range of . Find the inverse of f(x) = x2 + 4x â 1 , x > -2. If you did the Horizontal Line Test with the graph, you'd know there's no inverse function as it stands. For example, at first glance sin xshould not have an inverse, because it doesnât pass the horizontal line test. The function f is injective if and only if each horizontal line intersects the graph at most once. OK, to get really, really pedantic, there should be two functions, sin(x) with domain Reals and Sin(x) with domain (-pi/2, pi/2). Yâs must be different. 4. Which gives out two possible results, +√x and -√x. Observe the graph the horizontal line intersects the above function at exactly single point. Graphically, is a horizontal line, and the inputs and are the values at the intersection of the graph and the horizontal line. Only one-to-one functions have inverses, so if your line hits the graph multiple times then donât bother to calculate an inverseâbecause you wonât find one. There is a section in Victor Katz’s History of Mathematics which discusses the historical evolution of the “function” concept. We have step-by-step solutions for your textbooks written by Bartleby experts! It’s a matter of precise language, and correct mathematical thinking. x = -2, thus passing the horizontal line test with the restricted domain x > -2. To find the inverse of a function such as this one, an effective method is to make use of the "Quadratic Formula". Also, here is both graphs on the same axis, which as expected, are reflected in the line y = x. If the horizontal line intersects the graph of a function in all places at exactly one point, then the given function should have an inverse that is also a function. Learn how to approach drawing Pie Charts, and how they are a very tidy and effective method of displaying data in Math. If no horizontal line intersects the graph of a function more than once, then its inverse is also a function. This test is called the horizontal line test. ( Log Out / Example of a graph with an inverse Use the horizontal line test to recognize when a function is one-to-one. Notice that I’m recognizing that a function is not a rule (g), but a rule, a domain, and a something. If the line intersects the graph at more than one point, the function is not one-to-one and does not have an inverse. Where as with the graph of the function f(x) = 2x - 1, the horizontal line only touches the graph once, no y value is produced by the function more than once.So f(x) = 2x - 1 is a one to one function. Old folks are allowed to begin a reply with the word “historically.”. Post was not sent - check your email addresses! I’ve harped on this before, and I’ll harp on it again. Change y to f(x)^-1 two functions are inverses if f(g(x))=x=g(f(x)) g(f(x))=x Pass How do we tell if a function has an If the horizontal line touches the graph only once, then the function does have an inverse function.If the horizontal line test shows that the line touches the graph more than once, then the function does not have an inverse function. The Quadratic Formula can put this equation into the form x =, which is what we want to obtain the inverse, solving for x . Step-by-step explanation: In order to determine if a function has an inverse, and also if the inverse of the function is also a function, the function can be tested by drawing an horizontal line the graph of the function and viewing to find the following conditions; This test states that a function has an inverse function if and only if every horizontal line intersects the graph of at most once (see Figure 5.13). What this means is that for x â â:f(x) = 2x â 1 does have an inverse function, but f(x) = x² + 1 does NOT have an inverse function. The horizontal line test is a method to determine if a function is a one-to-one function or not. Pedantic answer: I can’t tell until you tell me what its co-domain is, because a function is a triple of things and you only told me the rule and the domain. Now, what’s the inverse of (g, A, B)? Problems dealing with combinations without repetition in Math can often be solved with the combination formula. This is when you plot the graph of a function, then draw a horizontal line across the graph. Ensuring that f -1(x) produces values >-2. The image above shows the graph of the function f(x) = x2 + 4. ( Log Out / The vertical line test determines whether a graph is the graph of a function. So when I say that sin(x) on the domain of Reals has an inverse, I might mean the multi-valued function arcsin(x) whose co-domain is sets of reals, not just reals. Inverse Functions: Horizontal Line Test for Invertibility. A function has an As the horizontal line intersect with the graph of function at 1 ⦠1. Graphs that pass both the vertical line and horizontal line tests are one-to-one functions. Horizontal Line Test for Inverse Functions A function has an inverse function if and only if no horizontal line intersects the graph of at more than one point.f f One-to-One Functions A function is one-to-one if each value of the dependent variable corre-sponds to exactly one value of the independent variable. In more Mathematical terms, if we were to go about trying to find the inverse, we'd end up at Horizontal Line Test. for those that doâthe Horizontal Line Test for an inverse function. Because a function that is not one to one initially, can have an inverse function if we sufficiently restrict the domain, restricting the x values that can go into the function.Take the function f(x) = x². ( Log Out / The best part is that the horizontal line test is graphical check so there isnât even math required. But note that Mathworld also acknowledges that it is fair to refer to functions that are not bijective as having an inverse, as long as it is understood that there is some “principal branch” of the function that is understood. Inverses and the Horizontal Line Test How to find an inverse function? For each of the following functions, use the horizontal line test to determine whether it is one-to-one. Trick question: Does Sin(x) have an inverse? As such, this is NOT an inverse function with all real x values. The horizontal line test can get a little tricky for specific functions. But first, letâs talk about the test which guarantees that the inverse is a function. Math permutations are similar to combinations, but are generally a bit more involved. Because for a function to have an inverse function, it has to be one to one.Meaning, if x values are going into a function, and y values are coming out, then no y value can occur more than once. Solve for y by adding 5 to each side and then dividing each side by 2. For example: (2)² + 1 = 5 , (-2)² + 1 = 5.So f(x) = x² + 1 is NOT a one to one function. But it does not guarantee that the function is onto. f -1(x) = +âx here has a range of y > 0, corresponding with the original domain we set up for x2, which was x > 0. The function passes the horizontal line test. To obtain the domain and the range of an inverse function, we switch around the domain and range from the original function. Change ), You are commenting using your Facebook account. “Sufficient unto the day is the rigor thereof.”. If any horizontal line intersects the graph more than once, the function fails the horizontal line test and is not ⦠Therefore it must have an inverse, right? A similar test allows us to determine whether or not a function has an inverse function. There is a test called the Horizontal Line Test that will immediately tell you if a function has an inverse. The horizontal line test lets you know if a certain function has an inverse function, and if that inverse is also a function. Inverse functions and the horizontal line test. f -1(x) = +√x. Fill in your details below or click an icon to log in: You are commenting using your WordPress.com account. The graph of an inverse function is the reflection of the original function about the line y x. If the horizontal line touches the graph only once, then the function does have an inverse function. And to solve that, we allow the notion of a (complex) function to be extended to include “multi-valued” functions. You definition disagrees with Euler’s, and with just about everyone’s definition prior to Euler (Descartes, Fermat, Oresme). If no horizontal line intersects the graph of a function f more than once, then the inverse of f is itself a function. Example 5: If f(x) = 2x â 5, find the inverse. Pingback: Math Teachers at Play 46 « Let's Play Math! Any x value put into this inverse function will result in 2 different outputs. Find the inverse of a ⦠Determine the conditions for when a function has an inverse. This might seem like splitting hairs, but I think it’s appropriate to have these conversations with high school students. Solution #1: The following theorem formally states why the horizontal line test is valid. A horizontal test means, you draw a horizontal line from the y-axis. This precalculus video tutorial explains how to determine if a graph has an inverse function using the horizontal line test. Therefore it is invertible, with inverse defined . Whatâs known as the Horizontal Line Test, is an effective way to determine if a function has an inverse function, or not. Now we have the form ax2 + bx + c = 0. Stated more pedantically, if and , then . It is a one-to-one function if it passes both the vertical line test and the horizontal line test. Textbook solution for Big Ideas Math A Bridge To Success Algebra 1: Student⦠1st Edition HOUGHTON MIFFLIN HARCOURT Chapter 10.4 Problem 30E. In fact, if you put a horizontal line at any part of the graph except at , there are always 2 intersections. (Category theory looks for common elements in algebra, topology, analysis, and other branches of mathematics. At times, care has to be taken with regards to the domain of some functions. y = 2x â 5 Change f(x) to y. x = 2y â 5 Switch x and y. Notice from the graph of below the representation of the values of . Test used to determine if the inverse of a relation is a funct⦠These functions pass both the vertical line test and the horiz⦠A function that "undoes" another function. a) b) Solution: a) Since the horizontal line $$y=n$$ for any integer $$nâ¥0$$ intersects the graph more than once, this function is not one-to-one. Whatâs known as the Horizontal Line Test, is an effective way to determine if a function has an. Wrong. This new requirement can also be seen graphically when we plot functions, something we will look at below with the horizontal line test. This is when you plot the graph of a function, then draw a horizontal line across the graph. With f(x) = x² + 1, the horizontal line touches the graph more than once, there is at least one y value produced by the function that occurs more than once. Where as -âx would result in a range of y < 0, NOT corresponding with the restricted original domain, which was set at greater than or equal to zero. (You learned that in studying Complex Variables.) In this case the graph is said to pass the horizontal line test. Use the horizontal line test to recognize when a function is one-to-one. Therefore, the given function have an inverse and that is also a function. The graph of the function is a parabola, which is one to one on each side of The graphs of f(x) = x² + 1 and f(x) = 2x - 1 for x â â, are shown below.With a blue horizontal line drawn through them. Here is a sketch of the graph of this inverse function. What’s tricky in real-valued functions gets even more tricky in complex-valued functions. Example. The given function passes the horizontal line test only if any horizontal lines intersect the function at most once. Common answer: The co-domain is understood to be the image of Sin(x), namely {Sin(x): x in (-pi/2, pi/2)}, and so yes Sin(x) has an inverse. I agree with Mathworld that the function (g, A, B) has an inverse if and only if it is bijective, as you say. Consider defined . We choose +√x instead of -√x, because the range of an inverse function, the values coming out, is the same as the domain of the original function. The horizontal line test is an important tool to use when graphing algebraic functions. Do you see my problem? If the horizontal line test shows that the line touches the graph more than once, then the function does not have an inverse function. We say this function passes the horizontal line test. So there is now an inverse function, which is f -1(x) = +√x. If a horizontal line intersects a function's graph more than once, then the function is not one-to-one. More colloquially, in the graphs that ordinarily appear in secondary school, every coordinate of the graph is associated with a unique coordinate. Horizontal Line Test â The HLT says that a function is a oneto one function if there is no horizontal line that intersects the graph of the function at more than one point. Inverse trigonometric functions and their graphs Preliminary (Horizontal line test) Horizontal line test determines if the given function is one-to-one. 2. ... f(x) has to be a o⦠Let’s encourage the next Euler by affirming what we can of what she knows. To Log in: you are commenting using your Facebook account think it ’ s tricky in complex-valued.... To begin a reply with the combination formula test to determine if function... 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Need to be taken with regards to the domain and range from the vertical test... Is used exclusively on functions that have been graphed on the coordinate plane immediately tell you if a function &... Solve that, we Switch around the domain of some functions the does. + sign will comply with this is an important tool to use when graphing algebraic functions whose inverse relation also... Ensuring that & nbspf & nbsp-1 ( x ) = +√x first glance sin xshould have! For when a function has an inverse value put into this inverse with... Section in Victor Katz ’ s appropriate to have these conversations with high school.. With this domain, the function is one-to-one be taken with regards the... Test are called one-to-one note: the function is one-to-one sorry, your can! Important tool to use when graphing algebraic functions can of what she knows different... A reply with the combination formula of f is injective if and only if any horizontal test! First, letâs talk about the test which guarantees that the range of an inverse Inverses the! = x2 + 4 put a horizontal line test to recognize when a function, not. Problem with the combination formula the domain will also need to be slightly restricted here, & nbspto & x. Such, this function is both one-to-one and onto ( bijective ) foundation for mathematics, an alternative to theory... Been graphed on the coordinate plane function and f ( x ) = +√x 'd know there no! If a function they are a very tidy and effective method of data... Be extended to include “ multi-valued ” functions of function portrayed nbspand nbsp... Single point to be invertible ( that is made explicit = +√x seen graphically when we plot functions something. If it passes both the vertical line test determines whether a graph an... Function is strictly the horizontal line intersects the graph at more than once then. 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Click an icon to Log in: you are commenting using your Twitter account each side and dividing... Share posts by email are exactly those functions whose inverse relation is also a function has an inverse with! Math Teachers at Play 46 « Let 's Play Math what ’ s matter! A matter of precise language, and how they are a very tidy and effective method of data... Nbsp different outputs issue: the horizontal line test, is an way... Test which guarantees that a function is the rigor thereof. ” look at below with the following formally! Nbsp -√x every coordinate of the codomain that are not in the graphs that ordinarily in., a, b ) whatâs known as the horizontal line test guarantees that a function, which is nbsp! S the inverse function, and i ’ ve harped on this before, the. Test that will immediately tell you if a function is the reflection of function... Hairs, but are generally a bit more involved the image above shows the graph at only one,! 'S graph more than once, then the function is strictly the line!
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https://math.stackexchange.com/questions/2460174/how-many-distinct-triples-of-non-negative-lbracex-y-z-mid-x-y-z-in-mathbb-z/2460213 | How many distinct triples of non-negative $\lbrace{x,y,z\mid x,y,z\in \mathbb Z\rbrace}$ satisfy $2x+y+z=16$?
Consider $$2x+y+z=16$$ how many distinct, non-negative triples of $\lbrace{x,y,z|x,y,z\in \mathbb Z\rbrace}$ are there that satisfy the equation?
I assumed that in this question, the role of combinatorics play a vital role, so I thought of it in this order: if the triples did not have to be distinct, there would be $16$ options the first time a number is chosen, $16$ the second time a number is chosen, and $16$ the third time a number is chosen, hence there are $16^3$ permutations where repetition is allowed.
Then, when repetition is not allowed, the first time a number is chosen there $16$ options, then there are $15$ options, and then $14$, but from here not another number is chosen, so assuming there are $16!$ permutations is senseless. Therefore the remaining $13!$ permutations need to be discounted, hence there are $\displaystyle \frac{16!}{(16-3)!}$ permutations when repetition is not allowed.
So does this imply there are $3360$ possible triples? This is more or less where I encountered the flaw in my logic:
When the first number is chosen, there aren't actually $15$ remaining numbers to choose from. Because say for example you choose $y$ to be $16$, then both $x$ and $z$ must be $0$, i.e. there is actually only one choice once the first number has been chosen, and one choice once the second number has been chosen. But, what if we let $x=9$? Well, this wouldn't work at all as when $x=9$, $2x=18$, which means there are no non-negative values of $y,z$ that can satisfy the equation, hence $x$ actually has its own range; $0 \leq x \leq 8$
More or less here I kind of set aside the combinatorics approach and assumed a very basic approach: listing values. Given that $x$ has the smallest range of each of the variables, I let $x$ be the independent variable and $y,z$ be the dependant variables:
\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\\hline y & 0\leq y \leq 16 & 0 \leq y \leq 14 & 0 \leq y \leq 12 & 0 \leq y \leq 10 & 0 \leq y \leq 8 & 0 \leq y \leq 6 & 0 \leq y \leq 4 & 0 \leq y \leq 2 & y=0 \\\hline z & 0\leq z \leq 16 & 0 \leq z \leq 14 & 0 \leq z \leq 12 & 0 \leq z \leq 10 & 0 \leq z \leq 8 & 0 \leq z \leq 6 & 0 \leq z \leq 4 & 0 \leq z \leq 2 & z=0\\\hline T & 17 & 15 & 13 & 11 & 9 & 7 & 5 & 3 & 1\\\hline \end{array}
(Note; a prerequisite in each entry is that $y+z=16-2x$. Also; the $T$ in the bottom row signifies the total number of permutations provided all the conditions, including the prerequisite, are met)
From the table, the total number of distinct, non-negative triples of $\lbrace{x,y,z|x,y,z\in\mathbb Z\rbrace}$ that satisfy the equation $2x+y+z=16$ is a mere $81$.
From all of what you've just read, I have only two questions: the first of which is obviously is this answer correct, and the second of which is, arguably more obviously, how can what I've written be expressed mathematically provided it is correct?
Any responses are very appreciated, thank you.
• Just to note that the analysis from the questioner above assumes that "distinct" means $x \ne y$, $y \ne z$, $x \ne z$, but the answers below do not assume this constraint. So the answer depends on how we interpret "distinct". – gandalf61 Oct 6 '17 at 10:12
• @gandalf61 So would it be best to remove the misleading word "distinct" ? – Peter Oct 6 '17 at 10:14
• I've actually adjusted my interpretation of what is meant by "distinct" given that those answering are actually more along the right lines. It is not impossible for $x=y=z$, but otherwise this means a distinct $x,y,z$ separately, whereas for one triple to be distinct from the next, only one value actually needs to differ if the remaining two are the same. So I'll keep the word there, but I'll edit the question to display the correct interpretation. @gandalf61 – joshuaheckroodt Oct 6 '17 at 10:17
For $x=0$ , the possible values of $y$ are $0$ to $16$, so $17$ possible values.
For $x=1$ , the possible values of $y$ are $0$ to $14$, so $15$ possible values.
$\cdots$
For $x=8$ , the possible values of $y$ are $0$ to $0$, so $1$ possible value.
In total , there are $$1+3+5+7+9+11+13+15+17=\color\red{81}$$ distinct triples.
• This is something I didn't think to consider if I'm very honest, that two of the values of the triple can be similar. I have actually been assuming that by "distinct" they mean each of the values must be different, not each of the triples themselves. Interesting, thank you. – joshuaheckroodt Oct 6 '17 at 10:06
we can use generating function to find the number of solution for the problem that you have stated.
That is you will have to find the coefficient of $x^{16}$ in the following products of g.f
$(1+x^2+x^4+..)(1+x+x^2+x^3+...)^2$
The number of solutions is $\boxed{81}$
• Nice, does this also work for, lets say, $3x+4y+5z=91$ ? – Peter Oct 6 '17 at 10:21
• Yes, but the generating function is $(1+x^3+x^6+...)(1+x^4+x^8+...)(1+x^5+x^{10}+...)$ and the coefficient is $x^{91}$ – Satish Ramanathan Oct 6 '17 at 10:23
Note that $y + z$ must be even. Hence, $y$ and $z$ are either both even or both odd.
When they are both even (say $y = 2c_1$ and $z = 2c_2$), the equation reduces to $x + c_1 + c_2 = 8$. The number of non-negative integer solutions in this case is $\binom{10}{8}$. When they are both odd (say $y = 2c_1 + 1$ and $z = 2c_2 + 1$), the equation reduces to $x + c_1 + c_2 = 7$. The number of non-negative integer solutions in this case is $\binom{9}{7}$.
Thus, we get a total of $\binom{10}{8} + \binom{9}{7} = 81$ solutions.
• How have you computed to what the equation reduces to provided both $y, z$ are odd or both $y,z$ is even? – joshuaheckroodt Oct 6 '17 at 11:05
• @joshuaheckroodt, in the even case, you would get $2x + 2c_1 + 2c_2 = 16$ which is same as $x + c_1 + c_2 = 8$. The odd case is similar. – iamwhoiam Oct 6 '17 at 11:44
More generally, the equation $y+z=2n-2x$ has $2n-2x+1$ non negative solutions for all $x=0,1,\dots, n$. Therefore the number of non negative solutions is $\sum_{x=0}^n(2n+1-2x)$, that is the sum of all odd numbers in the interval $[1,2n+1]$ which is equal to $(n+1)^2$. | 2021-01-15T19:39:56 | {
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https://astronomy.stackexchange.com/tags/hubble-constant/hot | # Tag Info
18
Yes, definitely. The Hubble constant describes the expansion rate of the Universe, and the expansion may, in turn, may be decelerated by "regular" matter/energy, and accelerated by dark energy. It's more or less the norm to use the term Hubble constant $H_0$ for the value today, and Hubble parameter $H(t)$ or $H(a)$ for the value at a time $t$ or, ...
16
If you measure the gravitational waveform from an inspiralling binary, you can at any point measure the amplitude, instantaneous frequency and the rate of change of frequency. The last two give you the "chirp mass", which is related to the product and sum of the binary component masses. The amplitude of the gravitational wave then depends on the chirp mass ...
15
To make a long story short, the measurements from Planck and the Hubble Space Telescope disagree, and the reason behind this isn't known. First, let's look at the values with the uncertainties. We then have three different results that are, perhaps, not as inconsistent as they originally seemed: $70.0^{+12.0}_{-8.0}\text{ km s}^{-1}\text{ Mpc}^{-1}$ from ...
12
Cosmological parameters are measured in a variety of ways, and their values will depend on which measurements you trust the most. The paper you link to (Planck Collaboration et al. 2016) with the 2015 results from the Planck observations of the cosmic microwave background is probably the one that most people will accept, but even in that paper you will find ...
12
We can currently only detect gravitational radiation when it is extremely intense: in the last fraction of a second. For example the first gravitational wave detection lasted less 0.15 seconds. The black holes are releasing gravitational radiation with every orbit, but that radiation is too weak for us to detect. It takes a colossal amount of energy being ...
10
The Sloan Digital Sky Survey Data Release 15 contains over 4 million spectra of both galactic and extra-galactic origin from the multi-fiber spectrographs. Of these spectra, 0.7 million came from the original spectrographs during the SDSS-I/II Legacy Survey and the remainder from the upgraded spectrographs as part of the BOSS survey during SDSS-III (see SDSS ...
9
So where are these measurements of galaxies moving faster than light? They're redshift measurements. Check out the Wikipedia redshift article. It's good stuff. "we can actually observe galaxies that are moving away from us at >c" It's true. You might think it cannot be, but it can. Um, I think I missed the groundbreaking headline that said ...
8
The Hubble parameter is defined as the rate of change of the distance between two points in the universe, divided by the distance between those two points. The Hubble parameter is getting smaller because the denominator is getting bigger more quickly than the numerator. In the future, the cosmological constant, $\Lambda$ could result in an exponential ...
8
The Hubble law gives the velocity of a distant galaxy right now. A galaxy at a distance $d$ recedes at a velocity $v = H_0\,d$ right now$^\dagger$. However, the relation between $d$ and the redshift — which is the quantity that we observe — is a non-trivial function of the expansion history of the Universe, obtained by integrating the (inverse) scale factor ...
8
At a distance of $d = 87\,\mathrm{Mpc}$, with a Hubble constant of roughly $H_0 = 70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$ cosmological expansion should make the host galaxy UGC 11723 recede at $v=H_0 \,d\simeq6100 \,\mathrm{km}\,\mathrm{s}^{-1}$. However, galaxies also move through space, at typical velocities from several $100\,\mathrm{km}\,\... 7 Hubble's law is a bit more subtle than you suppose and an expansion, whether accelerating or decelerating does not invalidate it. The distance and speed that should be used are their values now. These are known as the proper distance and velocity respectively. In that form Hubble's law works just fine, providing that the cosmological principle - that the ... 7 Just a supplement to @JamesK's excellent answer. The image below (from Caltech/MIT by way of New Sciencist) shows what was detected for one collision. On the left (at the start) the blackholes orbit one another about every 0.03 seconds, but the waveform is too faint to detect. At about 0.3 seconds on the Time axis the waves start being detectable and the ... 7 The duration of a gravitational wave detection is not particularly important in detecting electromagnetic counterparts, although the fact that they are not recurrent or repeating sources is. Binary systems continually emit gravitational waves, up until the time that they merge, predominantly at twice the orbital frequency. At the same time, the power emitted ... 7 Think about it as if you were baking bread (or cake, whatever you prefer). When you bake a bread with raisins, it rises in all directions (due to the yeast). Every raisin in a rising loaf of raisin bread will see every other raisin expanding away from it. Now suppose the raisins as galaxies and the yeast as the Hubble constant. An illustration from ... 6 What the Hubble constant really depends on is how old was the universe at the time, but if you have a dynamical model of the universe, you can map that into z and come up with a function H(z). So in that sense, the answer is "yes," but be careful-- we also think of z as a measure of how far away the objects are, and H does not depend on location it depends ... 5 A finite universe is said to have a "closed geometry", or to be "positively curved", meaning that, in principle, you may travel in a straight line and eventually return back to your starting point. In the 2D analogy, the surface of Earth is positively curved, and if you travel 40,000 km straight, you're back where you started. A finite universe that does ... 5 A type 1a supernova forms when a white dwarf grows through accretion to a certain size, at which it becomes unstable. This means that the precursor object is always a white dwarf of mass 1.39 solar masses. As the precursor object is always of the same type and the same size, the supernova is thought to be the same brightness. On the other hand, type II ... 5 Am I naive to think that it is necessary to build up a nice, complete light curve with dense points in time in order to use the photometry for precision distance calculations? No, the more I read about it, the more difficult it seems to be. In principle, if you sample regularly and often, you should eventually get a light curve, unless the period is the ... 4 Without going into the technicalities of spacetime diagrams and ants, I think the quickest way to wrap your head around this is to look at it from the distant galaxy's perspective. For instance, let's take GN-z11, which actually receded from us at$v\simeq4c$when it emitted the light we see today: A photon left GN-z11 at$v=c$. Space expands, so although ... 4 No, only the Hubble law was recommended to have its name changed (I'm a member of the IAU, so hopefully I'd have known if there were more votings). However, several astronomers (including myself) found the voting a bit… weird; while acknowledging the work of George Lemaître is admirable, many more people than him and Edwin Hubble contributed to the ... 4 Hubble tension refers to the incompatibility between different measurements of the value of the Hubble constant. These measurements are incompatible up to more than$5 \sigma$. This incompatibility arises between what we measure "nearby" and what we measure further away, and indicates that there might be some physics we don't understand yet. Now there has ... 4 Here are some galaxies (in the first line), and the same galaxies a bit later (in the second line) If you are in the red galaxy (the one in the middle), galaxies that used to be one space away are now two spaces away, so they are moving away with a speed of one space per unit time. Galaxies that used to be two spaces away are now four spaces away, so they ... 3 Here is some remarks on the issue, straight from Ryden: If galaxies are currently moving away from each other, then it implies they were closer together in the past. Consider a pair of galaxies currently separated by a distance$r$, with a velocity$v = H_0r$relative to each other. If there are no forces acting to accelerate or decelerate their ... 3 While the Hubble constant describes the current expansion rate of the Universe, it should also be seen as a parameter (among others) of a given cosmological model (e.g., Lambda-CDM). All methods to measure the Hubble constant are more or less indirect in some way, and they rely on very different assumptions. For example, the emission of the Cosmic Microwave ... 3 Casertano et al. used the period-luminosity (P-L) relation of Cepheid variables as a sanity check on Gaia DR1 parallaxes. They chose Cepheid variables within the Milky Way having parallaxes in TGAS (Tycho-Gaia astrometric solution, reusing Hipparcos data for a head start). The period and apparent magnitude come from ground-based photometry (van Leeuwen et al.... 3 The Carnegie-Chicago Hubble Program. VIII. An Independent Determination of the Hubble Constant Based on the Tip of the Red Giant Branch. I hope this helps. https://arxiv.org/pdf/1907.05922.pdf 3 The simplest answer is that Milne's explanation requires a non-uniform cosmos, with a formerly crowded "starting area" that expands into empty space. Observations indicate that the cosmos is nearly uniform at very large scales. With Milne's model of a non-expanding space and uniform cosmos, galaxies that started near to us with high velocities that ended ... 3 This answer to the question “Is the Hubble constant dependent on redshift?” gives the formula (a form of the Friedmann equation) for the Hubble parameter$H(z)$as a function of redshift$z$: $$H(z)^2 = H_0^2 \left[ (1+z)^4 \Omega_r + (1+z)^3 \Omega_M + (1+z)^2 \Omega_k + \Omega_\Lambda \right]$$ where the$\Omega$terms are the fractional densities in ... 3 No. Time dilation is expected and observed as a result of universal expansion (e.g. in the light curves of type Ia supernovae and the duration of gamma ray bursts, Blondin et al. 2008; Zhang et al. 2013). There is no way of distinguishing between a Doppler shift and cosmological redshift for an individual source. 2 All of these answers start with the formula and say therefore$H_0\$ decreases, but I think I have a more intuitive explanation: If you imagine the universe as an open straight rubber band, and the expansion of the universe stretching markings (galaxies) on the band, and say one end is being held stationary (this is us), then the rate at which each galaxy ...
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https://mathhelpboards.com/threads/derivative-of-ln-sin5x.1602/ | # [SOLVED]derivative of ln(sin5x)
#### DeusAbscondus
##### Active member
Hi folks,
For some reason, the following does not make sense to me:
If 1. $$f(x)=ln(sin(5x))$$ then 2. $$f'(x)=5cot(5x)$$ but I can only get as far as
$$f(x)=ln(sin(5x))\implies\frac{1}{u}*5cosx=\frac{5cos(x)}{sin(5x)}$$
Can someone please show me how to get from here to 2.? Thanks kindly, DeusAbs
#### Sudharaka
##### Well-known member
MHB Math Helper
Hi folks,
For some reason, the following does not make sense to me:
If 1. $$f(x)=ln(sin(5x))$$ then 2. $$f'(x)=5cot(5x)$$ but I can only get as far as
$$f(x)=ln(sin(5x))\implies\frac{1}{u}*5cosx=\frac{5cos(x)}{sin(5x)}$$
Can someone please show me how to get from here to 2.? Thanks kindly, DeusAbs
Hi DeusAbscondus,
What you need here is the Chain rule of differentiation. Please refer the link given and try to understand how to use this rule so that you can easily differentiate the given function.
$f(x)=\ln(\sin(5x))$
Using the chain rule of differentiation we can write,
$\frac{d}{dx}f(x)=\frac{d}{d(\sin(5x))}f(x)\frac{d}{dx}\sin(5x)$
Try to continue from here.
Kind Regards,
Sudharaka.
#### CaptainBlack
##### Well-known member
Hi folks,
For some reason, the following does not make sense to me:
If 1. $$f(x)=ln(sin(5x))$$ then 2. $$f'(x)=5cot(5x)$$ but I can only get as far as
$$f(x)=ln(sin(5x))\implies\frac{1}{u}*5cosx=\frac{5cos(x)}{sin(5x)}$$
Can someone please show me how to get from here to 2.? Thanks kindly, DeusAbs
The derivative of $$\sin(5x)$$ is $$5 \cos(5x)$$ not $$5\cos(x)$$.
CB
#### SuperSonic4
##### Well-known member
MHB Math Helper
Hi folks,
For some reason, the following does not make sense to me:
If 1. $$f(x)=ln(sin(5x))$$ then 2. $$f'(x)=5cot(5x)$$ but I can only get as far as
$$f(x)=ln(sin(5x))\implies\frac{1}{u}*5cosx=\frac{5cos(x)}{sin(5x)}$$
Can someone please show me how to get from here to 2.? Thanks kindly, DeusAbs
I'll do this by substitution to make it easier to see. On an exam you'd be fine to just go with it.
Let
• $u(x) = 5x$
• $v(u) = \sin(u)$
• $y(v) =\ln(v)$
Their respective derivatives are
• $\dfrac{du}{dx} = 5$
• $\dfrac{dv}{du} = \cos(u)$
• $\dfrac{dy}{dv} = \dfrac{1}{v}$
The chain rule states that
$\dfrac{dy}{dx} = \dfrac{du}{dx} \cdot \dfrac{dv}{du} \cdot \dfrac{dy}{dv}$
In other words multiply your three derivatives together and simplify
$\dfrac{dy}{dx} = 5 \cdot \cos(5x) \cdot \dfrac{1}{\sin(u)}$
$\dfrac{dy}{dx} = 5 \cdot \cos(5x) \cdot \dfrac{1}{\sin(5x)}$
$\dfrac{dy}{dx} = \dfrac{5cos(5x)}{\sin(5x)}$
Use your trig identities to simplify into the desired format
#### DeusAbscondus
##### Active member
Supplementary question: Re: derivative of ln(sin5x)
Super! simply super, thx for such a full demonstration: really appreciated.
C'n Black: i'm coming to enjoy your laconic exactitude: thank you kindly.
Sudharaka: prompt, friendly, accurate and complete: thx friend.
It was obvious to me that I was missing something basic: to wit: $$sin(5x)$$calls for the chain rule, since it contains an embedded function.
Supplementary Question: does this mean that whenever the argument of a function - the $x$ bit - contains more than $x$, that one is dealing with an embedded function? (excluding of course the cases of simple derivative addition and subtraction)? Please, could one of you chaps come up with language that tidies up what I think I am right in trying to say here?
this has opened the afternoon up for me to pleasantly work away at 40 examples of a similar kind, so as to root this shared logic into my mathematically growing but immature brain.
DeusAbs
#### Sudharaka
##### Well-known member
MHB Math Helper
Re: Supplementary question: Re: derivative of ln(sin5x)
Super! simply super, thx for such a full demonstration: really appreciated.
C'n Black: i'm coming to enjoy your laconic exactitude: thank you kindly.
Sudharaka: prompt, friendly, accurate and complete: thx friend.
You are welcome.
Supplementary Question: does this mean that whenever the argument of a function - the $x$ bit - contains more than $x$, that one is dealing with an embedded function? (excluding of course the cases of simple derivative addition and subtraction)? Please, could one of you chaps come up with language that tidies up what I think I am right in trying to say here?
Yes. In mathematical language what you have is a composition of functions. Examples of using the chain rule,
$f(x)=\sin(5x)\Rightarrow\frac{d}{dx}f(x)=\frac{d}{d(5x)}\sin(5x)\frac{d}{dx}(5x)=5\cos(5x)$
$f(x)=\sin(x+10)\Rightarrow \frac{d}{d(x+10)}\sin(x+10)\frac{d}{dx}(x+10)=\cos(x+10)$
A useful video giving examples of using the chain rule can be found here.
Kind Regards,
Sudharaka.
Last edited:
#### DeusAbscondus
##### Active member
Re: Supplementary question: Re: derivative of ln(sin5x)
You are welcome.
Yes. In mathematical language what you have is a composition of functions. Examples of using the chain rule,
$f(x)=\sin(10x)\Rightarrow\frac{d}{dx}f(x)=\frac{d}{d(5x)}\sin(5x)\frac{d}{dx}(5x)=5\cos(5x)$
$f(x)=\sin(x+10)\Rightarrow \frac{d}{d(x+10)}\sin(x+10)\frac{d}{dx}(x+10)=\cos(x+10)$
A useful video giving examples of using the chain rule can be found here.
Kind Regards,
Sudharaka.
Sudharaka, thanks, but are you sure you have not made an inadvertant error here:
$$f(x)=\sin(10x)\implies f'(x)=10cos(10x)$$
doesn't it?
DeusAbs
#### Sudharaka
##### Well-known member
MHB Math Helper
Re: Supplementary question: Re: derivative of ln(sin5x)
Sudharaka, thanks, but are you sure you have not made an inadvertant error here:
$$f(x)=\sin(10x)\implies f'(x)=10cos(10x)$$
doesn't it?
DeusAbs
Yes, sorry. Corrected it.
#### DeusAbscondus
##### Active member
follow up regarding "special result" in my calc text
I'm trying to satisfy my mind as to why $$f(x)=5ln(x) \Rightarrow f'(x)=5.\frac{1}{x}=\frac{5}{x}$$
Is it valid because of the result
$$kf(x) \Rightarrow \frac{dy}{dx}=kf'(x) where 'k' is some constant$$
If so, could someone please make some brief comments illuminating this result, its proof and application (with couple of simple examples)
Thanks kindly folks, always, for the incalculable power of good you are doing me with your help.
DeusAbs
Incidentally,
i) how do I space English words in latex expressions to avoid the ugly running together of letters as in above, and
ii) is there a backslash command for If/then?, or
iii) is my use of $\Rightarrow$ appropriate for If/then expressions?
#### Jameson
Staff member
Hi DeusAbscondus,
Yes, you are correct that for constant k, $$\displaystyle \frac{d}{dx}k \cdot f(x)=k \cdot f'(x)$$ and for that reason if $$\displaystyle f(x)=5\ln(x)$$ then $$\displaystyle f'(x)=5 \cdot \frac{1}{x}=\frac{5}{x}$$.
To answer one of your other questions. You can use the Latex command \text{ } to add normal looking text.
$$\displaystyle \text{You can see this in effect right here}$$
#### DeusAbscondus
##### Active member
Hi DeusAbscondus,
Yes, you are correct that for constant k, $$\displaystyle \frac{d}{dx}k \cdot f(x)=k \cdot f'(x)$$ and for that reason if $$\displaystyle f(x)=5\ln(x)$$ then $$\displaystyle f'(x)=5 \cdot \frac{1}{x}=\frac{5}{x}$$.
To answer one of your other questions. You can use the Latex command \text{ } to add normal looking text.
$$\displaystyle \text{You can see this in effect right here}$$
Thanks kindly Jameson (btw: I thoroughly approve of the change of image, which is in keeping with the wry, comic -gently self-ironizing- tone of the former, KGB avatar; i like it!)
I love this place! The sense of support is enormous.
Yesterday I had a day during which I felt defeated by the seeming enormity of the task of learning all this stuff; after a couple of interactions here, I *always* get enlightened and/or instructed, and this, in turn, lifts my mood and encourages me to keep going.
So, now I can go $$\text{If } y=x^2\text{ then } y'=2x\text{...how cool is that}$$
Argggh, but I still have the spacing problem, even after using \text{*} (???!!!!)
Last edited by a moderator:
#### Jameson
Staff member
I'm really glad you like the site. Hearing that is why I love working here, as well as getting to interact with some great minds.
I've edited your post for you to include the spaces. If you add a space inside the \text{ } then it will appear. For example:
1) \text{The derivative of} x^2 \text{is} 2x yields $$\displaystyle \text{The derivative of} x^2 \text{is} 2x$$. Notice the problem in spacing.
2) If I add spaces inside the text tag like so: \text{The derivative of } x^2 \text{ is } 2x we get $$\displaystyle \text{The derivative of } x^2 \text{ is } 2x$$
See the difference?
#### Sudharaka
##### Well-known member
MHB Math Helper
Re: follow up regarding "special result" in my calc text
Hi DeusAbscondus,
I'm trying to satisfy my mind as to why $$f(x)=5ln(x) \Rightarrow f'(x)=5.\frac{1}{x}=\frac{5}{x}$$
Is it valid because of the result
$$kf(x) \Rightarrow \frac{dy}{dx}=kf'(x) where 'k' is some constant$$
If so, could someone please make some brief comments illuminating this result, its proof and application (with couple of simple examples)
This is a consequence of the constant factor rule in differentiation. You can find the proof on this page. Examples can be found here.
iii) is my use of $\Rightarrow$ appropriate for If/then expressions?
The symbol $$\Rightarrow$$ is used for implication. Since $$y=kf(x)$$ implies $$\frac{dy}{dx}=kf'(x)$$ it is correct to write, $$y=kf(x)\Rightarrow \frac{dy}{dx}=kf'(x)$$. Refer this.
Kind Regards,
Sudharaka.
#### CaptainBlack
##### Well-known member
Re: follow up regarding "special result" in my calc text
I'm trying to satisfy my mind as to why $$f(x)=5ln(x) \Rightarrow f'(x)=5.\frac{1}{x}=\frac{5}{x}$$
Is it valid because of the result
$$kf(x) \Rightarrow \frac{dy}{dx}=kf'(x) where 'k' is some constant$$
If so, could someone please make some brief comments illuminating this result, its proof and application (with couple of simple examples)
Thanks kindly folks, always, for the incalculable power of good you are doing me with your help.
DeusAbs
Incidentally,
i) how do I space English words in latex expressions to avoid the ugly running together of letters as in above, and
ii) is there a backslash command for If/then?, or
iii) is my use of $\Rightarrow$ appropriate for If/then expressions?
By definition:
$\frac{d}{dx}f(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$
so:
\begin{aligned} \frac{d}{dx}\left(k\; f(x)\right)&=\lim_{h \to 0}\frac{k\;f(x+h)-k\;f(x)}{h} \\ &= \lim_{h \to 0}\left(k\; \frac{f(x+h)-f(x)}{h}\right)\\ &= k\; \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ &= k\frac{d}{dx}f(x) \end{aligned}
CB
Last edited:
#### DeusAbscondus
##### Active member
Re: follow up regarding "special result" in my calc text
By definition:
$\frac{d}{dx}f(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$
so:
$\frac{d}{dx}\left(k\; f(x)\right)=\lim_{h \to 0}\frac{k\;f(x+h)-k\;f(x)}{h} = \lim_{h \to 0}\left(k\; \frac{f(x+h)-f(x)}{h}\right) = k\; \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\ = k\frac{d}{dx}f(x)$
CB
Thanks Cap'n.
Could you go one step further and instance the use of $\frac{dy}{dx}$ as opposed to $\frac{d}{dx}$ so as to make plain to me the logic and syntax of the one compared to the other?
Much appreciated,
DeusAbs
#### Jameson
Staff member
Re: follow up regarding "special result" in my calc text
Thanks Cap'n.
Could you go one step further and instance the use of $\frac{dy}{dx}$ as opposed to $\frac{d}{dx}$ so as to make plain to me the logic and syntax of the one compared to the other?
Much appreciated,
DeusAbs
Usually $$\displaystyle \frac{dy}{dx}=f'(x)=y'$$ meaning the derivative of function y with respect to x. Nothing necessarily needs to be calculated. All three of those things just represent the idea of the derivative of y with respect to x.
On the other hand, $$\displaystyle \frac{d}{dx}$$ means calculate the derivative of whatever follows with respect to x.
Example:
$$\displaystyle y=x^2$$ so $$\displaystyle \frac{d}{dx}x^2=2x=\frac{dy}{dx}$$
With $$\displaystyle \frac{d}{dx}$$ something must follow that or it doesn't make sense. $$\displaystyle \frac{dy}{dx}$$ can stand on its own.
#### DeusAbscondus
##### Active member
I'm really glad you like the site. Hearing that is why I love working here, as well as getting to interact with some great minds.
I've edited your post for you to include the spaces. If you add a space inside the \text{ } then it will appear. For example:
1) \text{The derivative of} x^2 \text{is} 2x yields $$\displaystyle \text{The derivative of} x^2 \text{is} 2x$$. Notice the problem in spacing.
2) If I add spaces inside the text tag like so: \text{The derivative of } x^2 \text{ is } 2x we get $$\displaystyle \text{The derivative of } x^2 \text{ is } 2x$$
See the difference?
Ah! Formidable, mon vieux!
This is exactly what I came on to find out about tonight! and here, you left me word to that effect days ago!
$\text{ Thanks again ... Jameson ... }$ | 2021-10-16T16:00:00 | {
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https://math.stackexchange.com/questions/1088435/definition-of-abs-function | # Definition of abs() function
Let $\text{abs}(a)$ denote the absolute value of $a$. Is it true that $\text{abs}(a)\geq{-a}$? I suppose that $\text{abs}(a)>{-a}$, but my math book says the other way. Please help me to understand is it a misprint in my book, or my misunderstanding. Thank you in advance.
• If $a\leq 0$ then $\mathrm{abs}(a)=-a$. – Thomas Andrews Jan 2 '15 at 13:22
• For non-real complex numbers $a$ one cannot even compare $a$ and $\operatorname{abs}(a)=|a|$. – Marc van Leeuwen Jan 2 '15 at 18:27
Consider the example of $a=0$. Then $\operatorname{abs}(a) = -a$.
Or consider the example of $a = -1$. Then $\operatorname{abs}(a) = -a = 1$. Similarly, $\operatorname{abs}(a) = -a$ whenever $a<0$.
• Oh poor zero always forget him ;) Thank you! – dimaastronom Jan 2 '15 at 13:07
• Don't forget zero! Always think of him first! – MJD Jan 2 '15 at 13:13
• Actually, it is for all $a\leq 0$ that $\mathrm{abs}(a)=-a$. @dimaastronom – Thomas Andrews Jan 2 '15 at 13:23
• Dont understand with -1. Abs(-1)=1 – dimaastronom Jan 2 '15 at 18:33
• If $a=-1$, both $\operatorname{abs}(a)$ and $-a$ are equal to $1$. – MJD Jan 2 '15 at 19:38
yes, it's correct - if $a\leq 0$, then $|a|=-a$, and the inequality $|a|\geq -a$ holds.
if $a>0$, then $-a<0$, and so $|a|>0>-a$.
either way, the inequality $|a|\geq -a$ holds.
We have
$$\operatorname{abs}(a)=\max(a,-a)=\left\{\begin{array}{cl}a\;&\text{if}\; a\ge0\\-a\;&\text{otherwise}\end{array}\right.$$
The $abs$ function is defined by:
$\forall{x}\in\mathbb{R},\,abs(x)=|x|=\left\{ \begin{array}{lr} x & : x\ge0\\ -x & : x <0 \end{array} \right.$
So $\forall x\in\mathbb{R},\,|x|\ge0$
Let $a\in\mathbb{R}$.
If $a\ge0$ then $|a|=a$ and so $a\le|a|$
If $a<0$ then $|a|=-a>0>a$ and so $a\le|a|$
Now, if $a\ge0$ then $-a\le0\le|a|$
If $a<0$ then $-a=|a|$ and so $-a\le|a|$. | 2019-05-23T03:37:56 | {
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http://openstudy.com/updates/4f45c0ebe4b065f388ddb7d8 | ## Dumb_as_a_Duck 3 years ago An engine can pump 30000 litres of water to a vertical height of 45 m in 10 minutes. Calculate work done by the machine and its power.
1. JamesJ
The work done is the equal to the change in gravitational potential energy of the water. Calculate that, using PE = mgh. Then Power = (Work)/(Time)
2. Dumb_as_a_Duck
given - [g = 9.8ms ^{-2} \], $density of water = 1000 kg m ^{-3}$ and $1000L = 1 m^{3}$
3. Dumb_as_a_Duck
(that was a part of the question)
4. Dumb_as_a_Duck
how do you convert the given litres of water to kg?
5. JamesJ
This is the beauty of SI units and you can deduce this from the relation you wrote down above: 1 litre of water has a mass of 1 kg.
6. Dumb_as_a_Duck
how did you get that/derive that 1 litre of water has a mas of 1 kg from the relation given?
7. eashmore
He got it from density. Note that 1000 litres are equal to 1 m^3.
8. Dumb_as_a_Duck
and then? (im not seeing the relation or the beauty here, i'm sorry.)
9. eashmore
No problem. Let's use dimensional analysis. $1000 \left [\rm kg \over m^3 \right ] \cdot \left [ \rm 1 m^3 \over 1000L \right ] = 1 \left [\rm kg \over L \right ]$
10. Dumb_as_a_Duck
WAIITTTT
11. eashmore
Waiting.
12. Dumb_as_a_Duck
i didnt get a word of that. the thing you wrote under dimensional analysis, i mean. t
13. Dumb_as_a_Duck
if you were gonna explain that, then continue.
14. Dumb_as_a_Duck
i havent learnt about dimensional analysis either.
15. eashmore
It is a fact that 1 m^3 equals 1000 L. Can you accept that? Dimensional analysis is just a method of converting units. So the above relation can be written as $\rm 1 m^3 \over 1000 L$ If we multiply this by the density that is given$1000 \left [\rm kg \over m^3 \right ] \cdot \left [\rm 1 m^3 \over 1000 L \right ]$Notice that $$m^3$$ cancel out, and that 1000/1000 = 1
16. eashmore
Therefore, we are left with $1 \left [ \rm kg \over L \right ]$
17. JamesJ
Or put another way: if 1 cubic meter of water is 1000 kg, and 1 cubic meter of water holds 1000 liters, it must be that 1 liter has a mass of 1 kg.
18. Dumb_as_a_Duck
Oh! I see :)
19. Dumb_as_a_Duck
that makes the rest of the problem obvious. thanx! :)
20. eashmore
Great! Now, we know that, as James said, $PE = mgh$and density is defined as$\rho = {m \over V}$We can write mass in terms of density as$m = \rho V$Therefore, the potential energy can be written as$PE = \rho V gh$
21. eashmore
We know that$W = \Delta PE$ in this case, and that$P = W \cdot t$Therefore, $P = \rho V ght$where $$\rho$$ is the density of water, $$V$$ is the volume of water pumped, $$g$$ is the acceleration due to gravity, $$h$$ is the height that the pump raises the water, and $$t$$ is the time that it takes to pump the given volume of water. Let me know what you get.
22. Dumb_as_a_Duck
isn't P=W/t? (Power = Work/Time)?
23. JamesJ
Yes ... eashmore made a small mistake
24. eashmore
Whoops! Sorry about that. I was just testing you. :-P
25. Dumb_as_a_Duck
oh. is my solution here right? (Given) height= 45 m time = 10 minutes = 600 seconds density of water = 1000 kg m^-3 volume of water = 30,000 litres of water = 30 m^3 (as 1000 L = 1 m^3) (solution) mass of water = density of water * volume of water = 30 * 1000 = 30,000 kg. P.E. = mgh = 30,000 * 9.8 * 45 = 1323 * 10^4 Joule Work done= P.E. = 1323 * 10^4 Joule Power = W/t = 1323 * 10^4/600 = 2205 * 10^3 watts. (or 2205 kW).
26. eashmore
Looks good. Assuming you typed it into your calculator right.
27. Dumb_as_a_Duck
I calculated it myself lol. I'll double check with a calculator.
28. Dumb_as_a_Duck
Thanks a bunch! :)
29. eashmore
Whoops! I got that it should be 22.05 kW.
30. Dumb_as_a_Duck
uhoh! did i make a conversion mistake?
31. eashmore
Nope. Everything is good up until you calculate power. Double check that guy and you should be good.
32. Dumb_as_a_Duck
I got your answer (22.05 kW) on the calculator, but that's not what i got when I quadruple-checked my calculation.
33. Dumb_as_a_Duck
><
34. eashmore
I hate when this happens. The math gods of WolframAlpha say that it is 22.05 kW. http://www.wolframalpha.com/input/?i=30000*45*9.8%2F600+W+in+kW
35. Dumb_as_a_Duck
i mean, i got 22.05 kW when i first multiplied 1323 * 10^4 and THEN divided by 600....but when i divided 1323 by 600 AND then multiplied by 10^4, its the other wrong answer. lol i suck at math too. :P
36. Dumb_as_a_Duck
well, maybe i should just listen to the math gods in silence and not be an atheist. :P
37. eashmore
You should get the same answer with both ways that you have described. Maybe you are misinterpreting the results you get.
38. Dumb_as_a_Duck
I found my mistake! I should have divided 1323 by 600, not just 6...i forgot about that :P *sheepish look*.
39. Dumb_as_a_Duck
I salute you for sticking with me til the end. :D *high five* :P
40. eashmore
No problem. That is why we are here. I salute you for doing the same. Most users simply post all their homework questions, go eat dinner, and return to copy answers. Thank you for being engaged and striving to learn. Glad I could help.
41. supercrazy92
I salute both of you.. Great work!
42. Dumb_as_a_Duck
LOL! :D | 2015-03-29T13:51:22 | {
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http://math.stackexchange.com/questions/617828/a-doubt-in-quadratic-inequality | # A doubt in quadratic inequality
While I was doing a problem I came upon this: $$(a^2-4)>0$$ $$(a-2)(a+2)>0$$
Now I thought it will be
$a>2$ or $a>-2$
but it was $a<-2$ or $a>2$
Can u explain me why it is so?
Thank you
-
An alternative to the answers given already: sketch the graph of $y=(x-2)(x+2)$. When are the $y$-values positive? – Kelvin Soh Dec 25 '13 at 7:09
$bc>0$ means that $b,c$ are not zero and have the same sign. In this case, that means that either (1) $a-2>0$ and $a+2>0$, or (2) $a-2<0$ and $a+2<0$. Case (1) means that $a>2$ and $a>-2$. Case (2) means that $a<2$ and $a<-2$. Note that case (1) is just the same as $a>2$, and case (2) is just the same as $a<-2$: If $a>2$ then clearly $a>-2$. Similarly, if $a<-2$, then also $a<2$. Now, to see that your conclusion is incorrect, note that, for example, $a=0$ satisfies $a>-2$, so it satisfies the disjunction $a>2$ or $a>-2$. However, $(0-2)(0+2)=-4<0$. – Andres Caicedo Dec 25 '13 at 7:11
Thank you all. I dont know how i missed it. I just split them into two as in equality and put the sign instaed of = – chndn Dec 25 '13 at 7:16
The most voted gets the correct answer tick :D – chndn Dec 25 '13 at 7:21
– Mhenni Benghorbal Dec 25 '13 at 23:45
Now the inequality is valid if * $(a-2)$ and $(a+2)$ are both positive or are both negative. Now $$(a-2)>0 ~\hbox{and}~ (a+2)> 0 \Rightarrow a>2$$ $$(a-2)<0 ~\hbox{and}~ (a+2)< 0 \Rightarrow a<-2$$
-
Hint
$$xy>0\iff (x>0\land y>0)\lor(x<0\land y<0)$$
-
$$(a+2)(a-2)\gt0$$ means that either "$a+2$ and $a-2$ are positive" or "$a+2$ and $a-2$ are negative", so?
-
If $x, y \in \mathbb{R}$ and $xy > 0$, either $x, y > 0$ or $x, y < 0$. As $(a-2)(a+2) > 0$, then either $a - 2 > 0$ and $a + 2 > 0$ which occurs when $a > 2$, or $a - 2 < 0$ and $a + 2 < 0$ which occurs when $a < -2$.
Alternatively, given $a^2 - 4 > 0$ we have $a^2 > 4$ so $\sqrt{a^2} > 2$ because $\sqrt{x}$ is an increasing function. As $|a| = \sqrt{a^2}$ we see that $|a| > 2$ so $a < -2$ or $a > 2$.
-
Others have already answered this question, but I'd like to give a hint. If your answer differs from the expected one, find a solution that gives different results for a quick check.
You say a > -2 or a > 2; another answer is a < -2 or a > 2. Where do these solutions differ? Most obviously, when a = 0.
Replace a with 0, and what do you get? 0*0 - 4 = -4 , therefore 0 is not a part of solution, therefore your answer is invalid. When you know that you've made a mistake, you may go on to actually searching for the error. Otherwise you may waste time trying to fix the right answer: textbooks are not immune from errors.
- | 2015-08-04T16:08:07 | {
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https://mathforums.com/threads/solving-roots-equation-neatly.345576/ | # Solving roots equation neatly
#### sigma
I am to find all real solutions for this problem.
$2\sqrt[3]{x^2}-\sqrt[3]{x}=1$
According to my textbook, the solutions are {-$\frac{1}{8}$, 1}. However, I keep getting {-$\frac{1}{2}$, 1} when I try to solve it more practically. I went on Symbolab to see how they do it and I learned if I use the exponent property $a^{n}=(\sqrt[m]{a})^{n\times m}$, I get the correct solution set. My issue with this, though, is that I'm not going to be able to remember that exponent property well and the work after that gets too tedious for this kind of problem.
Here's one of my attempts of solving this problem to give you an idea of what I've been doing.
$$2\sqrt[3]{x^2}-\sqrt[3]{x}=1$$
$$=2x^{\frac{2}{3}}-x^{\frac{1}{3}}=1$$
$$=(2x^{\frac{2}{3}}-x^{\frac{1}{3}})^3=1^3$$
$$=2x^2-x-1=0$$
$$=x(2x+1)-1(2x+1)=0$$
$$=(x-1)(2x+1)=0$$
$$x={-\frac{1}{2}, 1}$$
Of course that's a flawed solution, but can anyone help me find a more elegant approach to solving this problem? Please use TeX commands
#### romsek
Math Team
$u = \sqrt[3]{x}$
$2u^2 - u - 1 = 0$
$(2u+1)(u-1) = 0$
$u = -\dfrac 1 2,~u=1$
$x = u^3$
$x = -\dfrac 1 8,~x=1$
1 person
#### v8archie
Math Team
$$(a+b)^3 \ne a^3 + b^3$$
Specifically, $$(2x^\tfrac23 - x^\tfrac13)^3 \ne 2x^2-x$$
1 person
#### sigma
Wow. I can't believe I made that mistake. I'll try to actually expand it and see where it goes from there.
#### skipjack
Forum Staff
Letting $x = y^3\!$, so that $y = \sqrt[3]{x}$, gives $2y^2 - y - 1 = 0$. Hence $y = -1/2$ or $1$, and so $x = -1/8$ or $1$.
1 person | 2020-04-03T07:44:54 | {
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https://astronomy.stackexchange.com/tags/hubble-constant/hot | # Tag Info
18
Yes, definitely. The Hubble constant describes the expansion rate of the Universe, and the expansion may, in turn, may be decelerated by "regular" matter/energy, and accelerated by dark energy. It's more or less the norm to use the term Hubble constant $H_0$ for the value today, and Hubble parameter $H(t)$ or $H(a)$ for the value at a time $t$ or, ...
16
If you measure the gravitational waveform from an inspiralling binary, you can at any point measure the amplitude, instantaneous frequency and the rate of change of frequency. The last two give you the "chirp mass", which is related to the product and sum of the binary component masses. The amplitude of the gravitational wave then depends on the chirp mass ...
15
To make a long story short, the measurements from Planck and the Hubble Space Telescope disagree, and the reason behind this isn't known. First, let's look at the values with the uncertainties. We then have three different results that are, perhaps, not as inconsistent as they originally seemed: $70.0^{+12.0}_{-8.0}\text{ km s}^{-1}\text{ Mpc}^{-1}$ from ...
12
Cosmological parameters are measured in a variety of ways, and their values will depend on which measurements you trust the most. The paper you link to (Planck Collaboration et al. 2016) with the 2015 results from the Planck observations of the cosmic microwave background is probably the one that most people will accept, but even in that paper you will find ...
12
We can currently only detect gravitational radiation when it is extremely intense: in the last fraction of a second. For example the first gravitational wave detection lasted less 0.15 seconds. The black holes are releasing gravitational radiation with every orbit, but that radiation is too weak for us to detect. It takes a colossal amount of energy being ...
10
The Sloan Digital Sky Survey Data Release 15 contains over 4 million spectra of both galactic and extra-galactic origin from the multi-fiber spectrographs. Of these spectra, 0.7 million came from the original spectrographs during the SDSS-I/II Legacy Survey and the remainder from the upgraded spectrographs as part of the BOSS survey during SDSS-III (see SDSS ...
9
So where are these measurements of galaxies moving faster than light? They're redshift measurements. Check out the Wikipedia redshift article. It's good stuff. "we can actually observe galaxies that are moving away from us at >c" It's true. You might think it cannot be, but it can. Um, I think I missed the groundbreaking headline that said ...
8
The Hubble parameter is defined as the rate of change of the distance between two points in the universe, divided by the distance between those two points. The Hubble parameter is getting smaller because the denominator is getting bigger more quickly than the numerator. In the future, the cosmological constant, $\Lambda$ could result in an exponential ...
8
The Hubble law gives the velocity of a distant galaxy right now. A galaxy at a distance $d$ recedes at a velocity $v = H_0\,d$ right now$^\dagger$. However, the relation between $d$ and the redshift — which is the quantity that we observe — is a non-trivial function of the expansion history of the Universe, obtained by integrating the (inverse) scale factor ...
8
At a distance of $d = 87\,\mathrm{Mpc}$, with a Hubble constant of roughly $H_0 = 70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$ cosmological expansion should make the host galaxy UGC 11723 recede at $v=H_0 \,d\simeq6100 \,\mathrm{km}\,\mathrm{s}^{-1}$. However, galaxies also move through space, at typical velocities from several $100\,\mathrm{km}\,\... 7 Hubble's law is a bit more subtle than you suppose and an expansion, whether accelerating or decelerating does not invalidate it. The distance and speed that should be used are their values now. These are known as the proper distance and velocity respectively. In that form Hubble's law works just fine, providing that the cosmological principle - that the ... 7 Just a supplement to @JamesK's excellent answer. The image below (from Caltech/MIT by way of New Sciencist) shows what was detected for one collision. On the left (at the start) the blackholes orbit one another about every 0.03 seconds, but the waveform is too faint to detect. At about 0.3 seconds on the Time axis the waves start being detectable and the ... 7 The duration of a gravitational wave detection is not particularly important in detecting electromagnetic counterparts, although the fact that they are not recurrent or repeating sources is. Binary systems continually emit gravitational waves, up until the time that they merge, predominantly at twice the orbital frequency. At the same time, the power emitted ... 6 What the Hubble constant really depends on is how old was the universe at the time, but if you have a dynamical model of the universe, you can map that into z and come up with a function H(z). So in that sense, the answer is "yes," but be careful-- we also think of z as a measure of how far away the objects are, and H does not depend on location it depends ... 6 Think about it as if you were baking bread (or cake, whatever you prefer). When you bake a bread with raisins, it rises in all directions (due to the yeast). Every raisin in a rising loaf of raisin bread will see every other raisin expanding away from it. Now suppose the raisins as galaxies and the yeast as the Hubble constant. An illustration from ... 5 A finite universe is said to have a "closed geometry", or to be "positively curved", meaning that, in principle, you may travel in a straight line and eventually return back to your starting point. In the 2D analogy, the surface of Earth is positively curved, and if you travel 40,000 km straight, you're back where you started. A finite universe that does ... 5 A type 1a supernova forms when a white dwarf grows through accretion to a certain size, at which it becomes unstable. This means that the precursor object is always a white dwarf of mass 1.39 solar masses. As the precursor object is always of the same type and the same size, the supernova is thought to be the same brightness. On the other hand, type II ... 5 Am I naive to think that it is necessary to build up a nice, complete light curve with dense points in time in order to use the photometry for precision distance calculations? No, the more I read about it, the more difficult it seems to be. In principle, if you sample regularly and often, you should eventually get a light curve, unless the period is the ... 4 Without going into the technicalities of spacetime diagrams and ants, I think the quickest way to wrap your head around this is to look at it from the distant galaxy's perspective. For instance, let's take GN-z11, which actually receded from us at$v\simeq4c$when it emitted the light we see today: A photon left GN-z11 at$v=c$. Space expands, so although ... 4 No, only the Hubble law was recommended to have its name changed (I'm a member of the IAU, so hopefully I'd have known if there were more votings). However, several astronomers (including myself) found the voting a bit… weird; while acknowledging the work of George Lemaître is admirable, many more people than him and Edwin Hubble contributed to the ... 4 Hubble tension refers to the incompatibility between different measurements of the value of the Hubble constant. These measurements are incompatible up to more than$5 \sigma$. This incompatibility arises between what we measure "nearby" and what we measure further away, and indicates that there might be some physics we don't understand yet. Now there has ... 3 From the definition of the rms (e.g. here), $$\mathrm{rms}^2(x) = \langle x \rangle^2 + \sigma_x^2,$$ where$\langle x \rangle$is the mean value, and$\sigma_x$is the dispersion. For galaxies with random velocities, the mean velocity should be$\langle x \rangle = 0$, unless they drift in some direction. Hence, the velocity dispersion should be$600\,\...
3
Here is some remarks on the issue, straight from Ryden: If galaxies are currently moving away from each other, then it implies they were closer together in the past. Consider a pair of galaxies currently separated by a distance $r$, with a velocity $v = H_0r$ relative to each other. If there are no forces acting to accelerate or decelerate their ...
3
While the Hubble constant describes the current expansion rate of the Universe, it should also be seen as a parameter (among others) of a given cosmological model (e.g., Lambda-CDM). All methods to measure the Hubble constant are more or less indirect in some way, and they rely on very different assumptions. For example, the emission of the Cosmic Microwave ...
3
Casertano et al. used the period-luminosity (P-L) relation of Cepheid variables as a sanity check on Gaia DR1 parallaxes. They chose Cepheid variables within the Milky Way having parallaxes in TGAS (Tycho-Gaia astrometric solution, reusing Hipparcos data for a head start). The period and apparent magnitude come from ground-based photometry (van Leeuwen et al....
3
The Carnegie-Chicago Hubble Program. VIII. An Independent Determination of the Hubble Constant Based on the Tip of the Red Giant Branch. I hope this helps. https://arxiv.org/pdf/1907.05922.pdf
3
The simplest answer is that Milne's explanation requires a non-uniform cosmos, with a formerly crowded "starting area" that expands into empty space. Observations indicate that the cosmos is nearly uniform at very large scales. With Milne's model of a non-expanding space and uniform cosmos, galaxies that started near to us with high velocities that ended ...
3
Here are some galaxies (in the first line), and the same galaxies a bit later (in the second line) If you are in the red galaxy (the one in the middle), galaxies that used to be one space away are now two spaces away, so they are moving away with a speed of one space per unit time. Galaxies that used to be two spaces away are now four spaces away, so they ...
3
This answer to the question “Is the Hubble constant dependent on redshift?” gives the formula (a form of the Friedmann equation) for the Hubble parameter $H(z)$ as a function of redshift $z$: $$H(z)^2 = H_0^2 \left[ (1+z)^4 \Omega_r + (1+z)^3 \Omega_M + (1+z)^2 \Omega_k + \Omega_\Lambda \right]$$ where the $\Omega$ terms are the fractional densities in ...
3
No. Time dilation is expected and observed as a result of universal expansion (e.g. in the light curves of type Ia supernovae and the duration of gamma ray bursts, Blondin et al. 2008; Zhang et al. 2013). There is no way of distinguishing between a Doppler shift and cosmological redshift for an individual source.
Only top voted, non community-wiki answers of a minimum length are eligible | 2021-05-07T16:22:48 | {
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http://math.stackexchange.com/questions/101795/simple-probability-question-from-my-textbook | # Simple probability Question from my textbook
Consider a group of four people. Everybody writes down the name of one other (random) member of the group. What is the probability that there is at least one pair of people who wrote down each others name?
Answer is 17/27. I think it should be 19/27. how to calculate it ?
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Is this homework? What have you tried? – JavaMan Jan 23 '12 at 22:21
no not exactly homework. I am trying to teach myself some maths. I am stuck at some problems. my solution was to find if there was no pair so to me it seemed the solution was (3*2*2*2)/3*3*3*3 – user669083 Jan 23 '12 at 22:25
Maybe you can tell us what makes you think the answer is $19/27$? Writing the argument out explicitly might help you figure out where you are making a mistake, or persuade you even more firmly that the answer in the book is incorrect. Answers in books are incorrect in some cases. – Dilip Sarwate Jan 23 '12 at 22:28
The numerical answer is $17/27$.
Divide our set of $4$ people into groups of two.
One grouping is $\{A, B\}, \{C,D\}$. There are $2$ other groupings, $\{A, C\}, \{B,D\}$ and $\{A, D\}, \{B,C\}$.
The probability that $A$ and $B$ write each other's names is $\dfrac{1}{9}$. The same applies to $C$ and $D$. Let us compute the probability that both these things happen. It is $\dfrac{1}{81}$.
So the probability that $A$ and $B$ write each other's name, or that $C$ and $D$ do (or both), is $$\frac{1}{9}+\frac{1}{9}-\frac{1}{81}.$$
We subtract the $1/81$ to avoid "double-counting" the situations where $A$ and $B$ pick each other, and $C$ and $D$ also do. Or else we can think of it as following from the formula $$P(X\cup Y)=P(X)+P(Y)-P(X\cap Y).$$
The same calculation applies to the other two pairings. So we multiply $\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{81}$ by $3$. The result is $2/3-1/27$, which is $17/27$.
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+1! That's a much better way to think of the problem than what I was writing :-) – JavaMan Jan 23 '12 at 22:36 | 2015-04-19T14:55:52 | {
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http://mathhelpforum.com/geometry/183175-area-triangles-trapezoid-print.html | # Area of Triangles in Trapezoid
Printable View
• Jun 16th 2011, 08:25 PM
Aquameatwad
Area of Triangles in Trapezoid
How do you prove the area of Triangle A is congruent to Triangle B in this trapezoidAttachment 21679?
• Jun 16th 2011, 08:27 PM
Prove It
Re: Area of Triangles in Trapezoid
Quote:
Originally Posted by Aquameatwad
How do you prove the area of Triangle A is congruent to Triangle B in this trapezoidAttachment 21679?
Do you have the lengths of the sides of the trapezium?
• Jun 16th 2011, 08:30 PM
Aquameatwad
Re: Area of Triangles in Trapezoid
No i don't have lengths. It is supposed to apply to any trapezoid. Now i found this link Trapezoid -- from Wolfram MathWorld which gives formulas as to why the areas are equal. But i don't understand why.
• Jun 16th 2011, 10:11 PM
BAdhi
Re: Area of Triangles in Trapezoid
Quote:
Originally Posted by Aquameatwad
No i don't have lengths. It is supposed to apply to any trapezoid. Now i found this link Trapezoid -- from Wolfram MathWorld which gives formulas as to why the areas are equal. But i don't understand why.
Only having the area of the two triangles equal, does not imply that they are congruent.
• Jun 17th 2011, 04:32 AM
Aquameatwad
Re: Area of Triangles in Trapezoid
You are right badhi. So i guess the question is why do they have the same area?
• Jun 17th 2011, 05:01 AM
Soroban
Re: Area of Triangles in Trapezoid
Hello, Aquameatwad!
Quote:
How do you prove the area of Triangle A is equal to Triangle B in this trapezoid?
Code:
P Q - *-------* : * * * * : * A * * B * : * * * * h * * * * : * * C * * : * * * * : ** ** - *-------------------------------* S R
$\Delta PSR\text{ and }\Delta QSR\text{ have the same base (SR) and the same height (h).}$
$\text{Hence: }\:\text{area}(\Delta PSR) \;=\;\text{area}(\Delta QSR)$
. . . . . . . . . . $A + C \;=\;B + C$
$\text{Therefore: }\qquad\quad\; A \;=\;B$
• Jun 17th 2011, 05:18 AM
Aquameatwad
Re: Area of Triangles in Trapezoid
Wow Soroban its that simple? I was trying to prove it by somehow by showing Triangle C was congruent to the Triangle with base PQ ( i guess we can call it D). which it is. Could i have used this information to prove A=B? | 2016-10-27T12:32:17 | {
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https://mathhelpboards.com/threads/inverse-limit-of-compact-hausdorff-spaces-is-nonempty-and-compact.25915/ | # Inverse Limit of Compact/Hausdorff Spaces is Nonempty and Compact
#### joypav
##### Active member
Problem:
Suppose that $\left\{ X_n \right\}_{n=1}^{\infty}$ is a sequence of compact, Hausdorff spaces and for each $n, f_n : X_{n+1} \rightarrow X_n$ is a continuous function (not necessarily onto).
Show that:
$X = lim_{\leftarrow} \left\{ X_n, f_n \right\}_{n=1}^{\infty} \neq \emptyset$
Furthermore, show that $X$ is compact.
I have seen a proof for a general inverse limit system, with $D$ being its directed set. However, I assume the proof differs in the problem I've stated. (here I guess our $D = \Bbb{N}$).
Does anyone know of a proof for this online?
Or perhaps can give an outline for the proof?
#### Euge
##### MHB Global Moderator
Staff member
Hi joypav ,
If $X = \emptyset$, then to each $x\in X$ corresponds an index $k$ such that $x_k \neq f_k(x_{k+1})$. Since $X_k$ is Hausdorff, $x_k$ and $f_k(x_{k+1})$ are separated by some open sets $U_k \ni x_k$ and $V_k\ni f_k(x_{k+1})$; continuity of $f_k$ allows us to find an open set $W_{k+1} \ni x_{k+1}$ such that $f_k(W_{k+1}) \subset V_k$. Set $\Sigma(k) := X_1 \times X_2 \times \cdots \times X_{k-1}\times U_k \times W_{k+1}\times X_{k+2}\cdots$, for $k\in \Bbb N$. The collection $\{\Sigma(k): k\in \Bbb N\}$ is an open cover of $\prod_n X_n$. Tychonoff's theorem ensures compactness of $\prod_n X_n$ (since each $X_n$ is compact), so there are indices $k_1 < k_2 < \cdots < k_j$ such that $\prod_n X_n = \Sigma(k_1)\cup \cdots \cup \Sigma(k_j)$. If $v_k \in V_k$, then $v = (v_1,v_2,v_3,\ldots)\in \Sigma(k_m)$ for some $m$. Thus $v_{k_m}\in U_{k_m}$, so that $v_{k_m} \in V_{k_m}\cap U_{k_m}$, a contradiction.
Since $\prod_n X_n$ is compact, to show that $X$ is compact, it suffices to show $X$ is closed in $\prod_n X_n$. Take a point $x\notin X$, and let $k\in \Bbb N$ such that $x_k \notin f_k(x_{k+1})$. Using the same notation as above, consider the set $\Sigma(k)$, which is an open neighborhood of $x$. Given $y\in \Sigma(k)$, $y_k \in U_k$ and $y_{k+1}\in W_{k+1}$. Since $f(W_{k+1})\subset V_k$ and $U_k$ is disjoint from $V_k$, then $y_k \neq f_k(y_{k+1})$. Consequently, $y\notin X$. Therefore, $X$ has open complement, i.e., $X$ is closed. | 2020-06-04T03:30:56 | {
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https://plainmath.net/3638/describe-how-to-multiply-matrices | # Describe how to multiply matrices.
Question
Matrices
Describe how to multiply matrices.
2021-02-14
Step 1
Let's take an example to understand clearly
Let $$A=\begin{pmatrix}1 & 2\\3 & 4 \end{pmatrix} , B=\begin{pmatrix}4 & 6\\7 & 8 \end{pmatrix}$$
Step 2
$$A=\begin{pmatrix}1 & 2\\3 & 4 \end{pmatrix} , B=\begin{pmatrix}4 & 6\\7 & 8 \end{pmatrix}$$
$$AB=\begin{pmatrix}1 & 2\\3 & 4 \end{pmatrix}\begin{pmatrix}4 & 6\\7 & 8 \end{pmatrix}$$
Multiply the row of the first matrix by the columns of the second matrix
$$= \begin{pmatrix}1 \cdot 4 + 2 \cdot 7 &1 \cdot 6 + 2 \cdot 8 \\ 3 \cdot 4 + 4 \cdot 7 & 3 \cdot 6 + 4 \cdot 8 \end{pmatrix}$$
$$=\begin{pmatrix}18 & 22\\40 & 50 \end{pmatrix}$$
### Relevant Questions
Describe how to subtract matrices.
Multiply the given matrix. After performing the multiplication, describe what happens to the elements in the first matrix. $$\begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}$$
For each of the pairs of matrices that follow, determine whether it is possible to multiply the first matrix times the second. If it is possible, perform the multiplication.
$$\begin{bmatrix}1 & 4&3 \\0 & 1&4\\0&0&2 \end{bmatrix}\begin{bmatrix}3 & 2 \\1 & 1\\4&5 \end{bmatrix}$$
Multiply the following matrices:
$$A=\begin{bmatrix} -2 & 1 &5\\1 & 4&-5 \end{bmatrix} B=\begin{bmatrix}-1 & 7 \\2 & -2\\3&4 \end{bmatrix}$$ AB=?
Multiply the matrices:
$$\begin{bmatrix}1 & -1 &0\\2 & 1&3 \end{bmatrix} \begin{bmatrix}4 & -1 \\2 & 0\\1&1 \end{bmatrix}$$
Multiply the Following matrices:
$$\begin{bmatrix}2 & 3 \\1 & 0 \end{bmatrix} \times \begin{bmatrix}4 & 1 \\2 & 1 \end{bmatrix}$$
$$\begin{bmatrix}1 & 3 \\4 & 5\\1&2 \end{bmatrix} \times \begin{bmatrix}2 & -1&4 \\3 & 1&0 \end{bmatrix}$$
$$\begin{bmatrix}1 & 2&0 \\0 & 0&1\\0&0&0 \end{bmatrix}$$
(square roots of the identity matrix) For how many 2x2 matrices A is it true that $$A^2=I$$ ? Now answer the same question for n x n matrices where n>2 | 2021-05-11T11:30:13 | {
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http://math.stackexchange.com/questions/137375/expected-number-of-good-balls-drawn-from-an-urn | # Expected number of “good” balls drawn from an urn.
Suppose we have a $n$ balls in an urn labeled $1$ through $n$, and we draw balls without replacement. Suppose we draw a first ball , and then draw an additional $k$ balls uniformly at random without replacement. What is the expected number of balls which have a label larger than the label of the first ball?
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Let $X$ denote the number on the first ball. Then $$\mathbb{P}(X=x) = \frac{1}{n}$$ Conditioned on the number on the first ball, the remaining $n-1$ balls are split into two sets, those below $x$, there are $x-1$ of these, and those above, and there are $n-x$ of those. The number $Y|X$ of those above $x$ in the sample of size $k$ drawn without replaced follows hypergeometric distribution $\operatorname{Hyp}(k, n-x, n-1)$ with mean $$\mathbb{E}(Y|X) = \frac{k(n-x)}{n-1}$$ Thus $$\mathbb{E}(Y) = \mathbb{E}(\mathbb{E}(Y|X)) = \sum_{x=1}^n \frac{1}{n} \cdot \frac{k(n-x)}{n-1} = \frac{k}{n(n-1)} \sum_{x=0}^{n-1} x = \frac{k}{2}$$
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+1 for introducing me to hypergeometric distributions. – Austin Mohr Apr 26 '12 at 20:04
I knew there had to be a distribution for this. Bonus, the combinatorial definition of the distribution is even intuitive. en.wikipedia.org/wiki/Hypergeometric_distribution – Joe Apr 26 '12 at 20:16
Hint: what is the probability that ball #$j$ has a label larger than that of ball #$1$?
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with replacement, the answer is easy; without replacement, it seems to depend on the labels of balls $2$ through $j-1$ – Joe Apr 26 '12 at 20:08
No, it doesn't. The other balls are irrelevant. All that matters is that you're drawing two different balls (#$1$ and #$j$) and that the two possible orderings of any particular pair are equally likely. – Robert Israel Apr 26 '12 at 20:30
@Joe: I suggest to make sure you understand this answer. It happens quite often that questions such as yours about expectation values can be answered using linearity of expectation without having to worry about any of the complications caused by correlations between different draws. Note how much simpler Robert's solution is than the others. – joriki Apr 26 '12 at 21:58
@joriki ...with an indicator variable for each ball $j$, and the expected number of balls with labels greater than that of ball $1$ is the sum of the expectations of the indicator variables... – Joe Apr 26 '12 at 22:45
@Joe: Is that a question directed at me? If so, I'm afraid I don't understand it. – joriki Apr 27 '12 at 0:12
If the first ball is number $i$, the urn is left with $i-1$ balls with smaller numbers and $n-i$ balls with larger numbers. The expected number in this case is clearly $k\frac{n-i}{n-1}$, provided that $k\le n-1$. The $n$ possible values of $i$ are equally likely, so the overall expected value is $$\frac1n\sum_{i=1}^n\frac{k(n-i)}{n-1}=\frac{k}{n(n-1)}\sum_{i=1}^n(n-i)=\frac{k}{n(n-1)}\sum_{i=0}^{n-1}i=\frac{k}{n(n-1)}\cdot\frac{n(n-1)}2=\frac{k}2\;.$$
You can see this even more easily by imagining that you draw the last $k$ balls first and then draw the ‘first’ ball: by symmetry it must on average be in the middle of the $k$ balls.
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The expected number is not (at least to me) clearly $k \frac{n-i}{n-1}$, although I do see that the probability of drawing the first good ball is $\frac{n-i}{n-1}$. It's not immediately obvious to me that drawing without replacement doesn't change the expectation. – Joe Apr 26 '12 at 20:21
@Joe: You’re picking a randomly chosen $k$ of the remaining $n-1$ balls, and you might as well imagine picking all of them at once, since the draw is without replacement. On average you expect them to reflect the $i-1:n-i$ split in the urn, so you expect $\frac{i-1}{n-1}$ of them to be below $i$ and $\frac{n-i}{n-1}$ of them to be above $i$. Since you’re drawing $k$ of them, you just muliply those fractions by $k$. – Brian M. Scott Apr 26 '12 at 20:36
When randomly selecting items from a population (either with or without replacement), the distribution of the $m$'th item selected is the same as that of the first. The conditional distribution of the $m$'th item, given the first, is the same as the conditional distribution of the second item, given the first. – Robert Israel Apr 26 '12 at 21:47 | 2014-09-23T17:03:33 | {
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https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_17&diff=prev&oldid=140929 | # Difference between revisions of "2017 AMC 10B Problems/Problem 17"
The following problem is from both the 2017 AMC 12B #11 and 2017 AMC 10B #17, so both problems redirect to this page.
## Problem
Call a positive integer $monotonous$ if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3$, $23578$, and $987620$ are monotonous, but $88$, $7434$, and $23557$ are not. How many monotonous positive integers are there?
$\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048$
## Solution 1
Case 1: monotonous numbers with digits in ascending order
There are $\Sigma_{n=1}^{9} \binom{9}{n}$ ways to choose n digits from the digits 1 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in ascending order. Note that 0 is not included since it will always be a leading digit and that is not allowed. Also, $\emptyset$ (the empty set) isn't included because it doesn't generate a number. The sum is equivalent to $\Sigma_{n=0}^{9} \binom{9}{n} -\binom{9}{0} = 2^9 - 1 = 511.$
Case 2: monotonous numbers with digits in descending order
There are $\Sigma_{n=1}^{10} \binom{10}{n}$ ways to choose n digits from the digits 0 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in descending order. Note that 0 is included since we are allowed to end numbers with zeros. However, $\emptyset$ (the empty set) still isn't included because it doesn't generate a number. The sum is equivalent to $\Sigma_{n=0}^{10} \binom{10}{n} -\binom{10}{0} = 2^{10} - 1 = 1023.$ We discard the number 0 since it is not positive. Thus there are $1022$ here.
Since the 1-digit numbers 1 to 9 satisfy both case 1 and case 2, we have overcounted by 9. Thus there are $511+1022-9=\boxed{\textbf{(B)} 1524}$ monotonous numbers.
## Solution 2
Like Solution 1, divide the problem into an increasing and decreasing case:
$\bullet$ Case 1: Monotonous numbers with digits in ascending order.
Arrange the digits 1 through 9 in increasing order, and exclude 0 because a positive integer cannot begin with 0.
To get a monotonous number, we can either include or exclude each of the remaining 9 digits, and there are $2^9 = 512$ ways to do this. However, we cannot exclude every digit at once, so we subtract 1 to get $512-1=511$ monotonous numbers for this case.
$\bullet$ Case 2: Monotonous numbers with digits in descending order.
This time, we arrange all 10 digits in decreasing order and repeat the process to find $2^{10} = 1024$ ways to include or exclude each digit. We cannot exclude every digit at once, and we cannot include only 0, so we subtract 2 to get $1024-2=1022$ monotonous numbers for this case.
At this point, we have counted all of the single-digit monotonous numbers twice, so we must subtract 9 from our total.
Thus our final answer is $511+1022-9 = \boxed{\textbf{(B) } 1524}$.
## Solution 3
Unlike the first two solutions, we can do our casework based off of whether or not the number contains a $0$.
If it does, then we know the $0$ must be the last digit in the number (and hence, the number has to be decreasing). Because $0$ is not positive, $0$ is not monotonous. So, we need to pick at least $1$ number in the set $[1, 9].$ After choosing our numbers, there will be just $1$ way to arrange them so that the overall number is monotonous.
In total, each of the $9$ digits can either be in the monotonous number or not, yielding $2^9 = 512$ total solutions. However, we said earlier that $0$ cannot be by itself so we have to subtract out the case in which we picked none of the numbers $1-9$. So, this case gives us $511$.
Onto the second case, if there are no $0$s, then the number can either be arranged in ascending order or in descending order. So, for each selection of the digits $1- 9$, there are $2$ solutions. This gives $$2 \cdot (2^9 - 1) = 2 \cdot 511 = 1022$$ possibilities. Note that we subtracted out the $1$ because we cannot choose none of the numbers.
However, realize that if we pick just $1$ digit, then there is only $1$ arrangement. We cannot put a single digit in both ascending and descending order. So, we must subtract out $9$ from there (because there are $9$ possible ways to select one number and for each case, we overcounted by $1$).
All in all, that gives $511 + 1022 - 9 = \boxed{(B) 1524}$ monotonous numbers. | 2021-12-07T16:49:48 | {
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https://math.stackexchange.com/questions/2783474/ternary-strings-with-each-element-occurring-at-least-once/2783492 | # Ternary strings with each element occurring at least once
It was asked of me to count all ternary strings with size $n$, such that each element $\{a,b,c\}$ occurs at least once in the string. Exercise specific, it was when $n=5$, but I'm looking on building a recurrence relation now.
I approached the problem by counting with labeled distribution i.e. $\frac{n!}{n_1!n_2!n_3!}$, for all possible $n_k$'s in $n=5$, which are either with $2$ repetitions and $1$ single letter, or $3$ repetitions and $2$ single letters. So for $5$ elements, it's $3 \cdot \frac{5!}{3!} + 3 \cdot \frac{5!}{2!2!}$. I'm not sure if I'm correct, because I'm not sure if I'm taking all cases into consideration.
Another way I thought of would be counting all the strings with only $2$ elements and subtract them from the number of all strings which is $3^5$, for this case.
I'll now be looking into in how to build a recurrence relation, and check my previous solution. I haven't found the same question in this forum. Any insight is appreciated!
P.S. This was an exam question. I am now checking my answers.
There are three ways to fill each of the $n$ positions without restriction, so there are $3^n$ ternary strings of length $n$. From these, we must subtract those cases in which fewer than three letters are used in the string.
There are $3$ ways to exclude one of the letters from the string and $2^n$ ways to fill the $n$ positions with the remaining two letters. However, if we subtract $3 \cdot 2^n$ from $3^n$, we will have subtracted too much since we will have subtracted the three strings in which two letters are excluded twice, once for each way we could have excluded one of the two missing letters. Since we only want to subtract such strings once, we must add them back. Therefore, by the Inclusion-Exclusion Principle, the number of ternary strings of length $n$ in which each letter appears at least once is $$3^n - \binom{3}{1}2^n + \binom{3}{2}1^n = 3^n - 3 \cdot 2^n + 3$$
• I got the same result with my solution, the one with distribution. Thanks! – Kristijan Talevski May 16 '18 at 10:05
The inclusion-exclusion principle is the standard way to go. Let $A$ be the set of ternary strings containing no $a$, and similarily define $B$ and $C$. Then we're interested in the size of $A\cup B\cup C$ (or more precisely, the size of the complement of $A\cup B\cup C$).
The size of any of $A, B, C$ is $2^n$, since they're sets of binary strings (e.g. $A$ is the set of all binary strings using $b, c$).
The size of any of $A\cap B, A\cap C, B\cap C$ is $1^n = 1$, since they are sets of unary strings (e.g. $A\cap B$ contains the one string $ccc\cdots c$).
The size of $A\cap B\cap C$ is $0^n$ (which is $0$ for $n\geq 1$ and $1$ for $n = 0$), since that's the set of strings of length $n$ using no letters.
Finally, assuming $n\geq 1$, the inclusion-exclusion principle says $$|A\cup B\cup C| = |A| + |B| + |C| -|A\cap B| - |A\cap C| - |B\cap C| + |A\cap B\cap C|\\ = 3\cdot 2^n - 3$$ The complement of this union, in other words the set of ternary strings containing at least one of each letter, then has $$3^n-(3\cdot 2^n-3)$$ elements.
This is already generalised to string length $n$, and it also (relatively) easily generalises to using an alphabet of $k$ letters: you add the size of all possible intersections using an odd number of the sets, and subtract all possible intersections using an even number of the sets. There are in total $2^k-1$ different such combinations, though, so the calculation gets very long even for relatively small $k$ (for $a, b, c,\ldots,j$, there are $1023$ terms).
Another way to get it. The strings of length $n-1$ not containing $a$ will be $2^{n-1}$. Of these, one will contain only $b$ and one only $c$.
So the number of strings of length $n$ which become "eligible" by having the first $a$ in position $n$ are $2^{n-1}-2$, and $3(2^{n-1}-2)$ are the strings which become "eligible" at step $n$.
If we denote by $S(n)$ the required number of strings of length $n$ with at least one $a$,one$b$ and one $c$, then this will be given by the new-comers at step $n$, plus those obtained by adding any of the three characters to the previous $S(n-1)$ strings, thus $$\left\{ \matrix{ S(2) = 0 \hfill \cr S(3) = 3! = 6 \hfill \cr S(n) = 3\left( {2^{\,n - 1} - 2} \right) + 3S(n - 1) \hfill \cr} \right.$$
You can check that $$S(n) = 3^{\,n} - \left( {3 \cdot \left( {2^{\,n} - 2} \right) + 3} \right)$$ which means Total N. of strings minus N. of strings with two or one characters. | 2021-04-19T18:05:02 | {
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http://math.stackexchange.com/questions/41953/generating-sequences-of-numeric-partitions | # Generating sequences of numeric partitions
Definition: A tuple $\lambda = (\lambda_1, \ldots, \lambda_k)$ of natural numbers is called a numeric partition of $n$ if $1 \leq \lambda_1 \leq \cdots \leq \lambda_k$ and $\lambda_1 + \cdots + \lambda_k = n$ and is written as $\lambda \vdash n$.
Exercise: Let $p(n)$ be the amount of numeric partitions of $n$. Prove that
i) Let $f_n := \prod_{i=1}^n (1-x^i)^{-1}$. Prove that $(f_n)_{n\geq 0}$ creates a Cauchy sequence in $\mathbb{C}[[x]]$.
ii) Prove that $$\sum\limits_{n \geq 0} p(n) x^n = \lim\limits_{n \rightarrow \infty} f_n = \prod\limits_{i=1}^\infty (1-x^i)^{-1} .$$
I tried to play around with the definition of $f_n$ and tried (quite successfully as I hope) to understand what a Cauchy sequence is. But how do I prove that?
What should I do to prove the second one? Any hints?
-
expand each $(1-x_i)^{-1}$ as a geometric series. this (should) let you see that you get $p(n)$ and that the series converges in the formal power series ring. – yoyo May 29 '11 at 14:01
Do you know what a Cauchy sequence is in general? Do you know what the metric on $\mathbb{C}[[x]]$ is? – Qiaochu Yuan May 29 '11 at 17:47
@qiaochu-yuan I do know what a Cauchy sequence in general is - but I don't know the metric on $\mathbb{C}[[C]]$, could you please explain that to me? – muffel May 29 '11 at 20:54
@yoyo by expanding I get $$\left(\frac{1}{(1-x)},\frac{1}{(1-x)}+\frac{1}{(1-x^2)},\frac{1}{(1-x)}+\frac{1}{(1-x^2)}+\frac{1}{(1-x^3)},\cdots\right)$$ but I don't see the relation to neither $p(n)$ nor to $p(n)x^n$. Could you please give me another hint? – muffel May 29 '11 at 21:09
Okay, that would explain why you're confused. The metric isn't unique, but for example it can be given by $d(a, b) = 2^{-\nu(a-b)}$ where $\nu(a-b)$ is the largest power of $x$ dividing $a-b$. This metric satisfies a very strong form of the triangle inequality (see en.wikipedia.org/wiki/Ultrametric_space) and it turns out to be very easy to characterize convergence: a series converges if and only if its terms tend to $0$, and a product converges if and only if its terms are nonzero and tend to $1$. – Qiaochu Yuan May 29 '11 at 23:28
First note that $$\frac{1}{1-x^i}=\sum_{j=0}^{\infty}x^{ij}.$$ We have $f_{n+1}-f_n$ is divisible by $x^{n+1}$, so we get convergence in $\mathbb{C}[[x]]$ (see http://en.wikipedia.org/wiki/Formal_power_series or something similar, neighborhoods of zero are powers of the ideal $(x)$).
For the connection to the partition function, think about a partition of $n$. choose how many $1$'s will appear, how many $2$'s etc., and pick them out in the product $$(1+x+x^2+ \cdots)(1+x^2+(x^2)^2+\cdots)(1+x^3+(x^3)^2+\cdots) \cdots.$$ Pick the number of ones from the $(1+x)^{-1}$ (i.e., if you want three $1$'s in the partition pick $x^3$), pick the number of $2$'s from the $(1+x^2)^{-1}$ (i.e., if you want three $2$'s in the partition pick $(x^2)^3$).
Adding up every way you can do this is exactly the coefficient of $x^n$ in $\lim f_n$ (or $f_N$ for large enough $N$). (If this isnt clear just multiply out the first few coefficients or look up the partition function online and you'll see a better exposition.)
-
thank you for that great explanation! – muffel May 30 '11 at 7:31
Just one question left: What exactly do you mean by "neighborhoods of zero are powers of the ideal (x)"? – muffel May 30 '11 at 7:42
@muffel power series are "smaller" if they are more highly divisible by $x$. the wikipedia page on formal power series gives a few different equivalent ways of talking about the topology – yoyo May 30 '11 at 18:02
The function $f_n$ already has all the correct coefficients up to $p(n)$. Open the expression up (following yoyo's advice) and see if you can figure out why.
Continuing, when you move from $f_n$ to $f_{n+1}$, all the low-order coefficients remain the same, and this is the reason that the infinite product converges - since every particular coefficient is equal to some finite sum (that counts the number of integer partitions of that size).
- | 2014-08-30T12:44:09 | {
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https://math.stackexchange.com/questions/1384994/rotate-a-point-on-a-circle-with-known-radius-and-position | # Rotate a point on a circle with known radius and position
Having a circle $\circ A(x_a, y_a)$ of radius $R$ and a point on the circle $B(x_b, y_b)$, how can we rotate the point with a known angle $\alpha$ (radians or degrees, it doesn't really matter) on the circle so we will obtain a new point on the circle $C(x_c, y_c)$, like in the image below?
How to calculate the $C$ coordinates?
Here the rotation angle is $90 ^ {\circ}$. In this example, $x_b = x_a$, $y_b = R$, $\alpha = 90 ^ \circ$. From the image we see that $x_c = R$ and $y_c = y_a$.
However, I want a general solution for any $A, B, R$ and $\alpha$.
• Are you familiar with rotation matrix ? Change of origin ? – Shailesh Aug 5 '15 at 7:02
• @Shailesh I saw something on Wikipedia, but didn't really understood the things. I think I know what the origin change means. – Ionică Bizău Aug 5 '15 at 7:04
• Anyways Brent has given a complete answer. I was trying to lead you towards it. – Shailesh Aug 5 '15 at 7:41
Let's look at a simpler problem. Suppose you have the situation depicted in the figure below:
Then, given the angle $\alpha$, the coordinates of the point $C''$ are:
$$C''_x = r\cos\alpha \qquad\mbox{and}\qquad C''_y = r\sin\alpha$$
where $r$ is the radius of the circle.
Now let's look at a slightly more complicated problem, depicted below:
This is very similar to the situation above. In fact,
$$C'_x = r\cos(\alpha+\beta) \qquad\mbox{and}\qquad C'_y = r\sin(\alpha+\beta)$$
By using the trigonometric relations $\sin(\alpha+\beta) = \sin\alpha\cos\beta + \sin\beta\cos\alpha$ and $\cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$, we can write the above as follows:
$$C'_x = r\cos\alpha\cos\beta - r\sin\alpha\sin\beta \qquad\mbox{and}\qquad C'_y = r\sin\alpha\cos\beta + r\sin\beta\cos\alpha$$
But, wait... By looking at the previous situation and replacing $C''$ with $B'$ and $\alpha$ with $\beta$, we see that
$$B'_x = r\cos\beta \qquad\mbox{and}\qquad B'_y = r\sin\beta$$
Therefore, we can write
$$C'_x = B'_x\cos\alpha - B'_y\sin\alpha \qquad\mbox{and}\qquad C'_y = B'_x\sin\alpha + B'_y\cos\alpha$$
But what you want is this, instead:
Well, we can just move everything rigidly by the vector $-\vec{OA}$ so that $A$ is now the origin of the coordinate system and we get the situation just above. This amounts to subtracting $A$ from both $B$ and $C$ to get $B'$ and $C'$ in the above, and we find
$$C_x - A_x = (B_x-A_x)\cos\alpha - (B_y-A_y)\sin\alpha$$ $$C_y - A_y = (B_x-A_x)\sin\alpha + (B_y-A_y)\cos\alpha$$
Then, finally,
$$C_x = A_x + (B_x-A_x)\cos\alpha - (B_y-A_y)\sin\alpha$$ $$C_y = A_y + (B_x-A_x)\sin\alpha + (B_y-A_y)\cos\alpha$$
• This is also helpful, but I don't understand there the radius $r$ disappeared in the final solution. Isn't it supposed to be there as well? – Ionică Bizău Aug 5 '15 at 7:47
• It is! It's hidden in the coordinates of $A$ and $B$. For example, $B_x - A_x = r\cos\beta$. – wltrup Aug 5 '15 at 7:49
• Nice! Upvoted you! – Ionică Bizău Aug 5 '15 at 7:55
• Thanks. Too bad it took me too long to do the pictures. I was hoping to have my answer accepted but Brent beat me to it. lol – wltrup Aug 5 '15 at 7:57
• Oh, one last thing: my angle $\alpha$ is the negative of yours since I have the locations of $B$ and $C$ exchanged with respect to your picture. – wltrup Aug 5 '15 at 8:15
This is called an affine transformation. Basically, the idea is to temporarily shift our circle so that it's centered about the origin, apply a rotation matrix to the point as done in linear algebra, then shift it back. Using the notation you have in your problem, as well as adding $$M=\left(\begin{array}{cc} \cos(\alpha) & -\sin(\alpha)\\ \sin(\alpha) & \cos(\alpha)\\ \end{array}\right)$$ To represent the counterclockwise rotation through an angle $\alpha$ (if you want it clockwise like it appears in your picture, just swap the $-\sin(\alpha)$ with the $\sin(\alpha)$), this transformation is given by: $${C}=M(B-A)+A$$ where $A,B,C$ are the vectors representing their respective points.
• where A,B,C are the vectors representing their respective points. -- how to build these vectors? Also, how to multiply $M(B - A)$? – Ionică Bizău Aug 5 '15 at 7:19
• For example, in your picture, you have $A$ as the point $(x_a,y_a)$, so it would be the vector $\left(\begin{array}{c}x_a \\ y_a \end{array}\right)$. Multiplying $M(B-A)$ is done using matrix multiplication: en.wikipedia.org/wiki/… – Brent Aug 5 '15 at 7:34
• Ah, sounds good! Actually, it's multiplying two matrix. Thanks a lot! – Ionică Bizău Aug 5 '15 at 7:36 | 2019-10-15T16:33:39 | {
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https://cs.stackexchange.com/questions/130868/what-is-the-requirement-for-bubble-sort-to-complete-in-1-pass | # What is the requirement for bubble sort to complete in 1 pass?
I am working on a problem where you're given $$n$$ distinct numbers, and you want to find the number of permutations such that it takes bubble sort at most 1 pass to complete.
e.g., if n=3, then the following permutations would only require 1 pass
1 2 3
1 3 2
3 1 2
2 1 3
But
3 2 1
2 3 1
would require more than 1 pass. Apparently the answer is $$2^{n - 1}$$, but I am not sure how to prove this for the general $$n$$ case.
My question is, what are the general constraints for a sequence to allow for it to be sorted with 1 pass of bubble sort?
It is difficult for me to come up with a generic formula to generate the permutations for larger $$n$$.
• Hint: let the position of the maximum element to be $i$. What can you say about elements to the left of $i$ and to the right of $i$? Find the number of such permutations for all possible $i$. You'll get the recurrent formula, for which it would be easy to show the answer by induction.
– user114966
Oct 6 '20 at 2:21
• @Dmitry Doesn't both the elements to the left and right of $i$ have to be less than the element at $i$? Oct 6 '20 at 2:34
• Welcome to COMPUTER SCIENCE @SE. I think that to depend on the termination condition of the bubble sort you argue about: please specify in the question. Oct 6 '20 at 6:38
I'm assuming that a pass of bubble sort on the array $$A[1],\ldots,A[n]$$ proceeds as follows:
• If $$A[1] > A[2]$$ then swap $$A[1]$$ and $$A[2]$$.
• If $$A[2] > A[3]$$ then swap $$A[2]$$ and $$A[3]$$.
• ...
• If $$A[n-1] > A[n]$$ then swap $$A[n-1]$$ and $$A[n]$$.
Bubble sort halts after a pass in which no swaps were made.
As is evident, if bubble sort halts after one pass, then the array must have been sorted.
When does bubble sort halt after two passes? If after the first pass, the array is sorted. In this case, if we know which swaps were made, then we can recover the input permutation. Furthermore, different sets of swaps correspond to different initial permutations. It follows that there are $$2^{n-1}-1$$ permutations that cause bubble sort to halt after exactly two passes. Together with the identity permutation, it follows that there are $$2^{n-1}$$ permutations that cause bubble sort to halt within two passes.
How do these permutations look like? Suppose that the first $$k$$ swaps are made, and then the following swap isn't made. Since the first $$k$$ swaps are made but the following one isn't made, at the end of the pass the first $$k+1$$ elements of the array will be $$A[2],\ldots,A[k+1],A[1]$$, which must correspond to the permutation $$1,\ldots,k+1$$. Hence the original permutation started $$k+1,1,\ldots,k$$.
From here, it's not too difficult to describe all $$2^{n-1}$$ permutations. Starting with the identity permutation, partition it in an arbitrary way, and rotate each part one step to the right. For example, when $$n = 3$$ we get:
• $$1|2|3 \to 1|2|3$$.
• $$1|23 \to 1|32$$.
• $$12|3 \to 21|3$$.
• $$123 \to 312$$.
Each division in the partition corresponds to a step in the pass during which there was no swap.
Here is another way of describing these permutations: they avoid the patterns $$231$$ and $$321$$. This means that for any $$i, it cannot be that $$A[k] < A[i],A[j]$$. To prove this, we show that any permutation above satisfies this constraint, and then prove the converse.
Consider any of the $$2^{n-1}$$ permutations above, and let $$i. Since each element in a part is smaller than each element in a subsequent part, $$A[k] < A[i]$$ is only possible if $$i,j,k$$ are all in the same part. Moreover, $$A[k] < A[j]$$ for indices $$j in the same part only if $$j$$ is the first element in the part, which is impossible since $$j>i$$.
Suppose now that we are given a $$(231,321)$$-avoiding permutation. Let $$i$$ be any index such that $$A[i] > i$$. All elements in $$i,\ldots,A[i]-1$$ must appear before all elements in $$A[i]+1,\ldots,n$$, since otherwise there will be a copy of $$231$$. Furthermore, the former must appear in increasing order, since otherwise there will be a copy of $$321$$. Therefore the part of the array that follows $$A[i]$$ is $$A[i],i,i+1,\ldots,A[i]-1$$.
Imagine now scanning the array in the order $$A[1],A[2],\ldots$$. The first time that $$A[i] \neq i$$, necessarily $$A[i] > i$$ (since we've already seen all smaller elements), and so the following elements are $$i,\ldots,A[i]-1$$. When we continue the scan, again the first time that $$A[i] \neq i$$ we must have $$A[i] > i$$ (since we've already seen all smaller elements), and so on. Therefore the array is of the form above.
• I'm looking at your partition graphic for the $n = 3$ example. From this, I'm inducing that the number of permutation is equal to the number of possible unique partitioning of an $n$ sized array. Is this correct? For $n = 4$, we have the following $$1\ 2 \ 3\ 4 \\ 1 | 2 | 3| 4 \\ 1 \ 2 | 3| 4 \\ 1\ 2 | 3 \ 4 \\ 1 \ 2 \ 3| 4 \\ 1 | 2 \ 3 \ 4 \\ 1 | 2 | 3 \ 4 \\ 1 | 2 \ 3 | 4 \\$$ Oct 6 '20 at 13:05
• Right. It's also easy to prove combinatorially by noting that each of the internal $n-1$ positions could contain or not contain a bar. Oct 6 '20 at 14:11
• Ahhh yes, that was exactly what I was looking for that "each of the internal......could or could not contain a bar." I knew there was an easy way to see this, but I couldn't figure it out. Also btw, do you know if there's a way to see this from a recurrence approach? Specifically, I was initially viewing this problem from the perspective that we have $n$ numbers, and now we add 1 more number, how many additional valid permutations does this additional number provide us with? I couldn't figure out a good way to think about this. Oct 6 '20 at 14:22
• There might be a way to write a recurrence, but a recurrence is not the only way to count things. Oct 6 '20 at 14:23
• To prove $i < j < k$ and $A[k] < A[i], A[j]$, isn't this also clear from the fact that an element in the unsorted array can only move one space to the left max for each pass, so if you have an element in the unsorted array in which there are two elements before it that are larger, then it is not possible to sort this array in one pass? Oct 7 '20 at 22:14
Bubble sort isn’t completed when the array is sorted - it is completed when we know the array is sorted.
Take the case 1-3-2. We have x-y-z and bubblesort figures out in the first pass that x is less than y and y is greater than z. It doesn’t know how x and z are related and therefore doesn’t know whether the array is sorted after one pass or not, therefore bubblesort is not complete at this point
If you change the question to “how many passes until the array is sorted”, that’s a different question. But to answer the question: If the number of array elements is zero or one, then bubble sort completes with zero passes. Otherwise, if either the array is sorted or the array has two elements then bubble sort complets with one pass. Otherwise, two or more passes are needed. | 2022-01-26T11:47:59 | {
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https://math.stackexchange.com/questions/1110578/how-to-construct-orthogonal-complement-subspace-of-any-subspace | # How to construct orthogonal complement subspace of any subspace?
If I have one subspace $V$ belong to $\mathbb{R}^{n}$, the subspace has basis $v_{1},v_{2},\cdot\cdot\cdot,v_{k}$,where $k<n$. I want to find the orthogonal complement subspace $V^{\perp}$ of $V$. $$V^{\perp}=span(u_{1},u_2,\cdot\cdot\cdot,u_{n-k})$$ Could anyone tell me how to construct the $V^{\perp}$?
• By Gram–Schmidt process you can make change $v_1,\ldots, v_k$ with an orthogonal basis $w_1, \ldots, w_k$ of $V$. Now add to $\{ w_1, \ldots, w_k\}$ vectors $x_1, \ldots, x_{n-k}\in {\mathbb R}^n$ such that $\{ w_1, \ldots, w_k, x_1, \ldots, x_{n-k}\}$ is a basis for ${\mathbb R}$ and do the Gram–Schmidt process again. Then in the end we have basis $\{ w_1, \ldots, w_k, u_1, \ldots, u_{n-k}\}$ of ${\mathbb R}^n$, where $\{ u_1, \ldots, u_{n-k}\}$ is basis for $V^\perp}$. – Janko Bracic Jan 19 '15 at 15:17
• I do not know how to find a group vectors $x_{1},x_2,\cdot\cdot\cdot,x_{n-k}$? Do you mean the above vectors are arbitrarily selected from $\mathbb{R}^n$? How can you guarantee that vectors are linear independent? – Dajiang Lei Jan 19 '15 at 15:32
• If $\{ v_1, \ldots, v_k\}$ is a basis for $V$ and after you can find $x_1, \ldots, x_{n-k}$ such that $\{ w_1, \ldots, w_k, x_1, \ldots, x_{n-k}\}$ is a basis for ${\mathbb R}$, then $x_1, \ldots, x_{n-k}$ must be linearly independent. – Janko Bracic Jan 19 '15 at 15:36
• I think that your method is impractical. To a theory analysis, it works. But we can not find $n-k$ linearly independent vectors in $\mathbb{R}^n$ easily. – Dajiang Lei Jan 19 '15 at 15:50
What you want is the kernel (null-space) of the matrix $$\pmatrix{ -&v_1^T&-\\ -&v_2^T&-\\ &\vdots\\ -&v_k^T&- }$$ this may be found by row-reduction. If each of the $v_i$ are mutually orthogonal, however, using the Gram Schmidt process is faster.
Gram-Schmidt process: let $e_1,\dots,e_n$ denote the standard basis vectors of $\Bbb R^n$. Begin by row-reducing the matrix $$\pmatrix{ v_1 & \cdots & v_k & e_1 & \cdots & e_n }$$ there will be $n$ pivot columns once this matrix is row-reduced. $k$ of them will be in the first $k$ columns, and the rest fall in the last $n$ positions.
Let $x_1,\dots,x_{n-k}$ be the vectors $e_i$ such that $e_i$ became a pivot.
Apply the Gram Schmidt process to $\{v_1,\dots,v_k,x_1,\dots,x_{n-k}\}$.
• Do you mean the null-space of $V^{T}$ is the complement orthogonal subspace of $V$? – Dajiang Lei Jan 19 '15 at 15:53
• Calling this matrix $V^T$ would imply that $V$ is a matrix; $V$ is not a matrix, it's a subspace. However, the null-space of the matrix I describe above is indeed $V^\perp$, the orthogonal complement subspace to $V$. – Omnomnomnom Jan 19 '15 at 15:56
• In general, we have $$\operatorname{Image}(A^T) = \operatorname{Nullspace}(A)^\perp$$ for any matrix $A$. – Omnomnomnom Jan 19 '15 at 15:58
• You are right. V is not a matrix, just a subspace. I think you give the right answer. You said that "using the Gram Schmidt process is faster". Do you mean that first I should transform the basis $(v_1,\cdot\cdot\cdot,v_k)$ to orthogonal basis $(v_1^{\prime},\cdot\cdot\cdot,v_k^{\prime})$, then construct the matrix constructed by $(v_1^{\prime},\cdot\cdot\cdot,v_k^{\prime})$, finally work out the null space of the aforementioned matrix? – Dajiang Lei Jan 19 '15 at 16:04
• I see your latest edit. You are genius. You completely solve my problem. But I can not know "Let $x_1,\cdot\cdot\cdot,x_{n−k}$ be the vectors $e_i$ such that $e_i$ became a pivot". Why? – Dajiang Lei Jan 19 '15 at 16:24 | 2019-10-15T08:46:59 | {
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http://math.stackexchange.com/questions/162018/accumulation-points-cluster-points-closed-sets | # Accumulation points / Cluster points / Closed sets
In a topological space $X$, call $x\in X$ an accumulation point if $\forall$ open set $U\ni x$, $U \cap A \neq \emptyset$, and $y\in X$ a cluster point if $\forall$ open set $U\ni y$, $U\cap A\setminus \{y\} \neq \emptyset$. (These are the terminologies used by my lecturer. I'm aware that different ones exist.)
Call a set $A\subseteq X$ closed if its complement is open.
My lecturer gave us a proof that $A$ is closed iff $A$ contains all of its accumulation points (see below). However, I managed to modify it to show that $A$ is closed iff $A$ contains all of its cluster points (see below, marked with []). What went wrong here? If the latter is false in general, in what special cases is it true (I heard it's true in metric spaces)?
The proof:
($\Rightarrow$): Suppose $A$ is closed and $x_0 \in X \setminus A$. Take $U:= X\setminus A$, an open set containing $x_0$. Now $U\cap A =\emptyset$, so $x_0$ is not an accumulation point. [$x_0$ is not an accumulation point and so it is not a cluster point either.]
($\Leftarrow$): Suppose $A$ is not closed, then $X\setminus A$ is not open. $\exists x_0 \in X\setminus A$ such that no open set $U\ni x_0$ is contained in $X\setminus A$, i.e. any open set $U\ni x_0$ satisfies $U\cap A \neq \emptyset$. So $x_0$ is an accumulation point of $A$ but not in $A$. [For this $x_0$, note that $x_0 \notin U\cap A$ because $x_0 \notin A$. So any open set $U\ni x_0$ satisfies $U\cap A \setminus \{x_0\} \neq \emptyset$, i.e. $x_0$ is a cluster point of $A$ but not in $A$.]
-
In the french literature one uses the term accumulation point ("point d'accumalation" see fr.wikipedia.org/wiki/…) for what your lecturer calls cluster point, and the term limit point (or "point adherent" see fr.wikipedia.org/wiki/Point_adh%C3%A9rent) for what your lecturer calls accumulation point. The statement you are talking about is usually stated as: $A$ is closed iff $A$ contains all its limit points, meaning that $A=\bar{A}$. – Mercy Jun 23 '12 at 13:02
What your lecturer stated can be translated as: $A$ is closed iff $\partial A \subset A$ since $\bar{A}=A\cup\partial A$. – Mercy Jun 23 '12 at 13:06
Your modification isn't a new result, it's known to be equivalent to what your lecturer stated! – Mercy Jun 23 '12 at 13:12
@Mercy: The OP isn’t claiming that it’s a new result, but rather asking whether it’s correct and whether the proof given is correct. The answer to both questions is yes. – Brian M. Scott Jun 23 '12 at 13:18
Your result is correct, as is your argument. You can even prove directly that if $A$ contains all of its cluster points, then it contains all of its accumulation points. Suppose that a set $A$ contains all of its cluster points but fails to contain its accumulation point $x$. Then $x$ is not a cluster point, so $x$ has an open nbhd $U$ such that $U\cap A\subseteq\{x\}$. But $x\notin A$, so $U\cap A=\varnothing$, contradicting the assumption that $x$ was an accumulation point of $A$.
Added: Your lecturer could have proved a stronger result. Let $\operatorname{cl}A$ be the set of accumulation points of $A$; then $A$ is closed iff $A=\operatorname{cl}A$. Suppose first that $A$ is closed. You’ve already proved that $A\supseteq\operatorname{cl}A$, and it’s clear that every point of $A$ is an accumulation point of $A$, so $A=\operatorname{cl}A$. Conversely, if $A$ is not closed, you already know that it fails to contain some accumulation point, so $A\ne\operatorname{cl}A$.
This stronger result fails for cluster points. Let $X$ be any $T_1$-space with at least two points, and let $x\in X$. Then $\{x\}$ is closed, but it has no cluster points, so it can’t be equal to the set of its cluster points. | 2015-07-04T23:12:58 | {
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https://probabilityandstats.wordpress.com/tag/independent-random-variables/ | # A randomized definition of the natural logarithm constant
The number $e$ is the base of the natural logarithm. It is an important constant in mathematics, which is approximately 2.718281828. This post discusses a charming little problem involving the the number $e$. This number can be represented in many ways. In a calculus course, the number $e$ may be defined as the upper limit of the following integral:
$\displaystyle \int_1^e \frac{1}{t} \ dt=1$
Another representation is that it is the sum $\displaystyle e=\sum_{n=1}^\infty \frac{1}{n!}$. Still another is that it is the limit $\displaystyle e=\lim_{n \rightarrow \infty} \biggl( 1+\frac{1}{n} \biggr)^n$. According to the authors of a brief article from the Mathematics Magazine [1], these textbook definitions do not give immediate insight about the number $e$ (article can be found here). As a result, students come away from the course without a solid understanding of the number $e$ and may have to resort to rote memorization on what the number $e$ is. Instead, the article gives six probability oriented ways in which the number $e$ can occur. These occurrences of $e$ are more interesting and, according to the authors of [1], can potentially increase students’ appreciation of the number $e$. In two of these six examples (Example 2 and Example 5), the number $e$ is defined by drawing random numbers from the interval $(0,1)$. This post discusses Example 2.
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Random Experiment
Here’s the description of the random experiment that generates the number $e$. Randomly and successively select numbers from the interval $(0, 1)$. The experiment terminates when the sum of the random numbers exceeds 1. What is the average number of selections in this experiment? In other words, we are interested in the average length of the experiment. According to the article [1], the average length is the number $e$. The goal here is to give a proof of this result.
As illustration, the experiment is carried out 1 million times using random numbers that are generated in Excel using the Rand() function. The following table summarizes the results.
$\left[\begin{array}{rrrrrr} \text{Length of} & \text{ } & \text{ } & \text{ } & \text{Relative} \\ \text{Experiment} & \text{ } & \text{Frequency} & \text{ } & \text{Frequency} \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ 2 & \text{ } & 500777 & \text{ } & 0.500777 \\ 3 & \text{ } & 332736 & \text{ } & 0.332736 \\ 4 & \text{ } & 124875 & \text{ } & 0.124875 \\ 5 & \text{ } & 33465 & \text{ } & 0.033465 \\ 6 & \text{ } & 6827 & \text{ } & 0.006827 \\ 7 & \text{ } & 1130 & \text{ } & 0.001130 \\ 8 & \text{ } & 172 & \text{ } & 0.000172 \\ 9 & \text{ } & 14 & \text{ } & 0.000014 \\ 10 & \text{ } & 4 & \text{ } & 0.000004 \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ \text{Total} & \text{ } & 1000000 & \text{ } & \text{ } \end{array}\right]$
The average length of experiment in these 1 million experiments is 2.717001. Even though the rate of convergence to the number $e$ is fairly slow, the simulated data demonstrates that on average it takes approximately $e$ number of simulations of numbers in the unit interval to get a sum that exceeds 1.
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A proof
Let $U_1,U_2,U_3,\cdots$ be a sequence of independent and identically distributed random variables such that the common distribution is a uniform distribution on the unit interval $(0,1)$. Let $N$ be defined as follows:
$\displaystyle N=\text{min}\left\{n: U_1+U_2+\cdots+U_n>1 \right\} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$
The objective is to determine $E(N)$. On the surface, it seems that we need to describe the distribution of the independent sum $X_n=U_1+U_2+\cdots+U_n$ for all possible $n$. Doing this may be possible but the result would be messy. It turns out that we do not need to do so. We need to evaluate the probability $P(X_n \le 1)$ for all $n$. We show that
$\displaystyle F_n(x)=P(X_n \le x)=\frac{x^n}{n!} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$
for $0 \le x \le 1$ and for $n=1,2,3,\cdots$. This is accomplished by an induction proof. This is true for $n=1$ since $X_1=U_1$ is a uniform distribution. Suppose (2) holds for the integer $n-1$. This means that $\displaystyle F_{n-1}(x)=P(X_{n-1} \le x)=\frac{x^{n-1}}{(n-1)!}$ for $0 \le x \le 1$. Note that $X_n=X_{n-1}+U_n$, which is an independent sum. Let’s write out the convolution formula for this independent sum:
\displaystyle \begin{aligned} F_n(x)&=\int_{0}^x F_{n-1}(x-y) \cdot f_{U_n}(y) \ dy \\&=\int_{0}^x \frac{(x-y)^{n-1}}{(n-1)!} \cdot 1 \ dy \\&=\frac{1}{(n-1)!} \int_0^x (x-y)^{n-1} \ dy \\&=\frac{x^n}{n!} \end{aligned}
The above derivation completes the proof of the claim. We now come back to the problem of evaluating the mean of the random variable $N$ defined in (1). First, note that $N>n$ if and only if $X_n=U_1+U_2+\cdots+U_n \le 1$. So we have the probability statement $\displaystyle P(N>n)=F_n(1)=\frac{1}{n!}$.
As a result, the following is the probability function of the random variable $N$.
\displaystyle \begin{aligned} P(N=n)&=P(N>n-1)-P(N>n) \\&=F_{n-1}(1)-F_n(1) \\&=\frac{1}{(n-1)!}-\frac{1}{n!} \\&=\frac{n-1}{n!} \end{aligned}
Now evaluate the mean.
\displaystyle \begin{aligned} E(N)&=\sum \limits_{n=2}^\infty \frac{n(n-1)}{n!} \\&=\sum \limits_{n=2}^\infty \frac{1}{(n-2)!} \\&=\sum \limits_{m=0}^\infty \frac{1}{m!} \\&=e \end{aligned}
With the above derivation, the proof that $e=$ 2.718281828… is the average number of random numbers to select in order to obtain a sum that exceeds 1 is completed.
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Reference
1. Shultz H. S., Leonard B., Unexpected Occurrences of the Number e,Mathematics Magazine, October 1989, Volume 62, Number 4, pp. 269–271.
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$\copyright \ \text{2015 by Dan Ma}$
# Conditional Distributions, Part 1
We illustrate the thought process of conditional distributions with a series of examples. These examples are presented in a series of blog posts. In this post, we look at some conditional distributions derived from discrete probability distributions.
Practice problems are found in the companion blog.
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The Setting
Suppose we have a discrete random variable $X$ with $f(x)=P(X=x)$ as the probability mass function. Suppose some random experiment can be modeled by the discrete random variable $X$. The sample space $S$ for this probability experiment is the set of sample points with positive probability masses, i.e. $S$ is the set of all $x$ for which $f(x)=P(X=x)>0$. In the examples below, $S$ is either a subset of the real line $\mathbb{R}$ or a subset of the plane $\mathbb{R} \times \mathbb{R}$. Conceivably the sample space could be subset of any Euclidean space $\mathbb{R}^n$ in higher dimension.
Suppose that we are informed that some event $A$ in the random experiment has occurred ($A$ is a subset of the sample space $S$). Given this new information, all the sample points outside of the event $A$ are irrelevant. Or perhaps, in this random experiment, we are only interested in those outcomes that are elements of some subset $A$ of the sample space $S$. In either of these scenarios, we wish to make the event $A$ as a new sample space.
The probability of the event $A$, denoted by $P(A)$, is derived by summing the probabilities $f(x)=P(X=x)$ over all the sample points $x \in A$. We have:
$\displaystyle P(A)=\sum_{x \in A} P(X=x)$
The probability $P(A)$ may not be 1.0. So the probability masses $f(x)=P(X=x)$ for the sample points $x \in A$, if they are unadjusted, may not form a probability distribution. However, if we consider each such probability mass $f(x)=P(X=x)$ as a proportion of the probability $P(A)$, then the probability masses of the event $A$ will form a probability distribution. For example, say the event $A$ consists of two probability masses 0.2 and 0.3, which sum to 0.5. Then in the new sample space, the first probability mass is 0.4 (0.2 multiplied by $\displaystyle \frac{1}{0.5}$ or divided by 0.5) and the second probability mass is 0.6.
We now summarize the above paragraph. Using the event $A$ as a new sample space, the probability mass function is:
$\displaystyle f(x \lvert A)=\frac{f(x)}{P(A)}=\frac{P(X=x)}{P(A)}, \ \ \ \ \ \ \ \ \ x \in A$
The above probability distribution is called the conditional distribution of $X$ given the event $A$, denoted by $X \lvert A$. This new probability distribution incorporates new information about the results of a random experiment.
Once this new probability distribution is established, we can compute various distributional quantities (e.g. cumulative distribution function, mean, variance and other higher moments).
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Examples
Suppose that two students take a multiple choice test that has 5 questions. Let $X$ be the number of correct answers of one student and $Y$ be the number of correct answers of the other student (these can be considered as test scores for the purpose of the examples here). Assume that $X$ and $Y$ are independent. The following shows the probability functions.
$\displaystyle \begin{bmatrix} \text{Count of}&\text{ }&\text{ }&P(X=x) &\text{ }&\text{ }&P(Y=y) \\\text{Correct Answers}&\text{ }&\text{ }&\text{ } &\text{ }&\text{ }&\text{ } \\\text{ }&\text{ }&\text{ } &\text{ }&\text{ } \\ 0&\text{ }&\text{ }&0.4&\text{ }&\text{ }&0.1 \\\text{ }&\text{ }&\text{ } &\text{ }&\text{ } \\ 1&\text{ }&\text{ }&0.2&\text{ }&\text{ }&0.1 \\\text{ }&\text{ }&\text{ } &\text{ }&\text{ } \\ 2&\text{ }&\text{ }&0.1&\text{ }&\text{ }&0.2 \\\text{ }&\text{ }&\text{ } &\text{ }&\text{ } \\ 3&\text{ }&\text{ }&0.1&\text{ }&\text{ }&0.2 \\\text{ }&\text{ }&\text{ } &\text{ }&\text{ } \\ 4&\text{ }&\text{ }&0.1 &\text{ }&\text{ }&0.2 \\\text{ }&\text{ }&\text{ } &\text{ }&\text{ } \\ 5&\text{ }&\text{ }&0.1 &\text{ }&\text{ }&0.2 \end{bmatrix}$
Note that $E(X)=1.6$ and $E(Y)=2.9$. Without knowing any additional information, we can expect on average one student gets 1.6 correct answers and one student gets 2.9 correct answers. If having 3 or more correct answers is considered passing, then the student represented by $X$ has a 30% chance of passing while the student represented by $Y$ has a 60% chance of passing. The following examples show how the expectation can change as soon as new information is known.
The following examples are based on these two test scores $X$ and $Y$.
Example 1
In this example, we only consider the student whose correct answers are modeled by the random variable $X$. In addition to knowing the probability function $P(X=x)$, we also know that this student has at least one correct answer (i.e. the new information is $X>0$).
In light of the new information, the new sample space is $A=\left\{1,2,3,4,5 \right\}$. Note that $P(A)=0.6$. In this new sample space, each probability mass is the original one divided by 0.6. For example, for the sample point $X=1$, we have $\displaystyle P(X=1 \lvert X>0)=\frac{0.2}{0.6}=\frac{2}{6}$. The following is the conditional probability distribution of $X$ given $X>0$.
$\displaystyle P(X=1 \lvert X>0)=\frac{2}{6}$
$\displaystyle P(X=2 \lvert X>0)=\frac{1}{6}$
$\displaystyle P(X=3 \lvert X>0)=\frac{1}{6}$
$\displaystyle P(X=4 \lvert X>0)=\frac{1}{6}$
$\displaystyle P(X=5 \lvert X>0)=\frac{1}{6}$
The conditional mean is the mean of the conditional distribution. We have $\displaystyle E(X \lvert X>0)=\frac{16}{6}=2.67$. Given that this student is knowledgeable enough to answer some question correctly, the expectation is higher than before knowing the additional information. Also, given the new information, the student in question has a 50% chance of passing (vs. 30% before the new information is known).
Example 2
We now look at a joint distribution that has a 2-dimensional sample space. Consider the joint distribution of test scores $X$ and $Y$. If the new information is that the total number of correct answers among them is 4, how would this change our expectation of their performance?
Since $X$ and $Y$ are independent, the sample space is a square as indicated the figure below.
$\text{ }$
Figure 1 – Sample Space of Test Scores
Because the two scores are independent, the joint probability at each of these 36 sample points is the product of the individual probabilities. We have $P(X=x,Y=y)=P(X=x) \times P(Y=y)$. The following figure shows one such joint probability.
Figure 2 – Joint Probability Function
After taking the test, suppose that we have the additional information that the two students have a total of 4 correct answers. With this new information, we can focus our attention on the new sample space that is indicated in the following figure.
Figure 3 – New Sample Space
Now we wish to discuss the conditional probability distribution of $X \lvert X+Y=4$ and the conditional probability distribution of $Y \lvert X+Y=4$. In particular, given that there are 4 correct answers between the two students, what would be their expected numbers of correct answers and what would be their chances of passing?
There are 5 sample points in the new sample space (the 5 points circled above). The conditional probability distribution is obtained by making each probability mass as a fraction of the sum of the 5 probability masses. First we calculate the 5 joint probabilities.
$\displaystyle P(X=0,Y=4)=P(X=0) \times P(Y=4) =0.4 \times 0.2=0.08$
$\displaystyle P(X=1,Y=3)=P(X=1) \times P(Y=3) =0.2 \times 0.2=0.04$
$\displaystyle P(X=2,Y=2)=P(X=2) \times P(Y=2) =0.1 \times 0.2=0.02$
$\displaystyle P(X=3,Y=1)=P(X=3) \times P(Y=1) =0.1 \times 0.1=0.01$
$\displaystyle P(X=4,Y=0)=P(X=4) \times P(Y=0) =0.1 \times 0.1=0.01$
The sum of these 5 joint probabilities is $P(X+Y=4)=0.16$. Making each of these joint probabilities as a fraction of 0.16, we have the following two conditional probability distributions.
$\displaystyle P(X=0 \lvert X+Y=4)=\frac{8}{16} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P(Y=0 \lvert X+Y=4)=\frac{1}{16}$
$\displaystyle P(X=1 \lvert X+Y=4)=\frac{4}{16} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P(Y=1 \lvert X+Y=4)=\frac{1}{16}$
$\displaystyle P(X=2 \lvert X+Y=4)=\frac{2}{16} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P(Y=2 \lvert X+Y=4)=\frac{2}{16}$
$\displaystyle P(X=3 \lvert X+Y=4)=\frac{1}{16} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P(Y=3 \lvert X+Y=4)=\frac{4}{16}$
$\displaystyle P(X=4 \lvert X+Y=4)=\frac{1}{16} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P(Y=4 \lvert X+Y=4)=\frac{8}{16}$
Now the conditional means given that $X+Y=4$, comparing against the unconditional means.
$\displaystyle E(X \lvert X+Y=4)=\frac{0+4+4+3+4}{16}=\frac{15}{16}=0.9375 \ \ \ \ \ \ \ \ \text{vs} \ \ E(X)=1.6$
$\displaystyle E(Y \lvert X+Y=4)=\frac{0+1+4+12+32}{16}=\frac{49}{16}=3.0625 \ \ \ \ \ \text{vs} \ \ E(Y)=2.9$
Now compare the chances of passing.
$\displaystyle P(X \ge 3 \lvert X+Y=4)=\frac{4}{16}=0.25 \ \ \ \ \ \ \ \ \ \ \text{vs} \ \ P(X \ge 3)=0.3$
$\displaystyle P(Y \ge 3 \lvert X+Y=4)=\frac{14}{16}=0.875 \ \ \ \ \ \ \ \ \text{vs} \ \ P(Y \ge 3)=0.6$
Based on the new information of $X+Y=4$, we have a lower expectation for the student represented by $X$ and a higher expectation for the student represented by $Y$. Observe that the conditional probability at $X=0$ increases to 0.5 from 0.4, while the conditional probability at $X=4$ increases to 0.5 from 0.2.
Example 3
Now suppose the new information is that the two students do well on the test. Particularly, their combined number of correct answers is greater than or equal to 5, i.e., $X+Y \ge 5$. How would this impact the conditional distributions?
First we discuss the conditional distributions for $X \lvert X+Y \ge 5$ and $Y \lvert X+Y \ge 5$. By considering the new information, the following is the new sample space.
Figure 4 – New Sample Space
To derive the conditional distribution of $X \lvert X+Y \ge 5$, sum the joint probabilities within the new sample space for each $X=x$. The calculation is shown below.
$\displaystyle P(X=0 \cap X+Y \ge 5)=0.4 \times 0.2=0.08$
$\displaystyle P(X=1 \cap X+Y \ge 5)=0.2 \times (0.2+0.2)=0.08$
$\displaystyle P(X=2 \cap X+Y \ge 5)=0.1 \times (0.2+0.2+0.2)=0.06$
$\displaystyle P(X=3 \cap X+Y \ge 5)=0.1 \times (0.2+0.2+0.2+0.2)=0.08$
$\displaystyle P(X=4 \cap X+Y \ge 5)=0.1 \times (1-0.1)=0.09$
$\displaystyle P(X=5 \cap X+Y \ge 5)=0.1 \times (1)=0.10$
The sum of these probabilities is 0.49, which is $P(X+Y \ge 5)$. The conditional distribution of $X \lvert X+Y \ge 5$ is obtained by taking each of the above probabilities as a fraction of 0.49. We have:
$\displaystyle P(X=0 \lvert X+Y \ge 5)=\frac{8}{49}=0.163$
$\displaystyle P(X=1 \lvert X+Y \ge 5)=\frac{8}{49}=0.163$
$\displaystyle P(X=2 \lvert X+Y \ge 5)=\frac{6}{49}=0.122$
$\displaystyle P(X=3 \lvert X+Y \ge 5)=\frac{8}{49}=0.163$
$\displaystyle P(X=4 \lvert X+Y \ge 5)=\frac{9}{49}=0.184$
$\displaystyle P(X=5 \lvert X+Y \ge 5)=\frac{10}{49}=0.204$
We have the conditional mean $\displaystyle E(X \lvert X+Y \ge 5)=\frac{0+8+12+24+36+50}{49}=\frac{130}{49}=2.653$ (vs. $E(X)=1.6$). The conditional probability of passing is $\displaystyle P(X \ge 3 \lvert X+Y \ge 5)=\frac{27}{49}=0.55$ (vs. $P(X \ge 3)=0.3$).
Note that the above conditional distribution for $X \lvert X+Y \ge 5$ is not as skewed as the original one for $X$. With the information that both test takers do well, the expected score for the student represented by $X$ is much higher.
With similar calculation we have the following results for the conditional distribution of $Y \lvert X+Y \ge 5$.
$\displaystyle P(Y=0 \lvert X+Y \ge 5)=\frac{1}{49}=0.02$
$\displaystyle P(Y=1 \lvert X+Y \ge 5)=\frac{2}{49}=0.04$
$\displaystyle P(Y=2 \lvert X+Y \ge 5)=\frac{6}{49}=0.122$
$\displaystyle P(Y=3 \lvert X+Y \ge 5)=\frac{8}{49}=0.163$
$\displaystyle P(Y=4 \lvert X+Y \ge 5)=\frac{12}{49}=0.245$
$\displaystyle P(Y=5 \lvert X+Y \ge 5)=\frac{20}{49}=0.408$
We have the conditional mean $\displaystyle E(Y \lvert X+Y \ge 5)=\frac{0+2+12+24+48+100}{49}=\frac{186}{49}=3.8$ (vs. $E(Y)=2.9$). The conditional probability of passing is $\displaystyle P(Y \ge 3 \lvert X+Y \ge 5)=\frac{40}{49}=0.82$ (vs. $P(Y \ge 3)=0.6$). Indeed, with the information that both test takers do well, we can expect much higher results from each individual test taker.
Example 4
In Examples 2 and 3, the new information involve both test takers (both random variables). If the new information involves just one test taker, it may be immaterial on the exam score of the other student. For example, suppose that $Y \ge 4$. Then what is the conditional distribution for $X \lvert Y \ge 4$? Since $X$ and $Y$ are independent, the high score $Y \ge 4$ has no impact on the score $X$. However, the high joint score $X+Y \ge 5$ does have an impact on each of the individual scores (Example 3).
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Summary
We conclude with a summary of the thought process of conditional distributions.
Suppose $X$ is a discrete random variable and $f(x)=P(X=x)$ is its probability function. Further suppose that $X$ is the probability model of some random experiment. The sample space of this random experiment is $S$.
Suppose we have some new information that in this random experiment, some event $A$ has occurred. The event $A$ is a subset of the sample space $S$.
To incorporate this new information, the event $A$ is the new sample space. The random variable incorporated with the new information, denoted by $X \lvert A$, has a conditional probability distribution. The following is the probability function of the conditional distribution.
$\displaystyle f(x \lvert A)=\frac{f(x)}{P(A)}=\frac{P(X=x)}{P(A)}, \ \ \ \ \ \ \ \ \ x \in A$
where $P(A)$ = $\displaystyle \sum_{x \in A} P(X=x)$.
The thought process is that in the conditional distribution is derived from taking each original probability mass as a fraction of the total probability $P(A)$. The probability function derived in this manner reflects the new information that the event $A$ has occurred.
Once the conditional probability function is derived, it can be used just like any other probability function, e.g. computationally for finding various distributional quantities.
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Practice Problems
Practice problems are found in the companion blog.
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$\copyright \ \text{2013 by Dan Ma}$
# Picking Two Types of Binomial Trials
We motivate the discussion with the following example. The notation $W \sim \text{binom}(n,p)$ denotes the statement that $W$ has a binomial distribution with parameters $n$ and $p$. In other words, $W$ is the number of successes in a sequence of $n$ independent Bernoulli trials where $p$ is the probability of success in each trial.
Example 1
Suppose that a student took two multiple choice quizzes in a course for probability and statistics. Each quiz has 5 questions. Each question has 4 choices and only one of the choices is correct. Suppose that the student answered all the questions by pure guessing. Furthermore, the two quizzes are independent (i.e. results of one quiz will not affect the results of the other quiz). Let $X$ be the number of correct answers in the first quiz and $Y$ be the number of correct answers in the second quiz. Suppose the student was told by the instructor that she had a total of 4 correct answers in these two quizzes. What is the probability that she had 3 correct answers in the first quiz?
On the face of it, the example is all about binomial distribution. Both $X$ and $Y$ are binomial distributions (both $\sim \text{binom}(5,\frac{1}{4})$). The sum $X+Y$ is also a binomial distribution ($\sim \text{binom}(10,\frac{1}{4})$). The question that is being asked is a conditional probability, i.e., $P(X=3 \lvert X+Y=4)$. Surprisingly, this conditional probability can be computed using the hypergeometric distribution. One can always work this problem from first principle using binomial distributions. As discussed below, for a problem such as Example 1, it is always possible to replace the binomial distributions using a thought process involving the hypergeometric distribution.
Here’s how to think about the problem. This student took the two quizzes and was given the news by the instructor that she had 4 correct answers in total. She now wonders what the probability of having 3 correct answers in the first quiz is. The thought process is this. She is to pick 4 questions from 10 questions (5 of them are from Quiz 1 and 5 of them are from Quiz 2). So she is picking 4 objects from a group of two distinct types of objects. This is akin to reaching into a jar that has 5 red balls and 5 blue balls and pick 4 balls without replacement. What is the probability of picking 3 red balls and 1 blue ball? The probability just described is from a hypergeometric distribution. The following shows the calculation.
$\displaystyle (1) \ \ \ \ P(X=3 \lvert X+Y=4)=\frac{\displaystyle \binom{5}{3} \ \binom{5}{1}}{\displaystyle \binom{10}{4}}=\frac{50}{210}$
We will show below why this works. Before we do that, let’s describe the above thought process. Whenever you have two independent binomial distributions $X$ and $Y$ with the same probability of success $p$ (the number of trials does not have to be the same), the conditional distribution $X \lvert X+Y=a$ is a hypergeometric distribution. Interestingly, the probability of success $p$ has no bearing on this observation. For Example 1, we have the following calculation.
$\displaystyle (2a) \ \ \ \ P(X=0 \lvert X+Y=4)=\frac{\displaystyle \binom{5}{0} \ \binom{5}{4}}{\displaystyle \binom{10}{4}}=\frac{5}{210}$
$\displaystyle (2b) \ \ \ \ P(X=1 \lvert X+Y=4)=\frac{\displaystyle \binom{5}{1} \ \binom{5}{3}}{\displaystyle \binom{10}{4}}=\frac{50}{210}$
$\displaystyle (2c) \ \ \ \ P(X=2 \lvert X+Y=4)=\frac{\displaystyle \binom{5}{2} \ \binom{5}{2}}{\displaystyle \binom{10}{4}}=\frac{100}{210}$
$\displaystyle (2d) \ \ \ \ P(X=3 \lvert X+Y=4)=\frac{\displaystyle \binom{5}{3} \ \binom{5}{1}}{\displaystyle \binom{10}{4}}=\frac{50}{210}$
$\displaystyle (2e) \ \ \ \ P(X=4 \lvert X+Y=4)=\frac{\displaystyle \binom{5}{4} \ \binom{5}{0}}{\displaystyle \binom{10}{4}}=\frac{5}{210}$
Interestingly, the conditional mean $E(X \lvert X+Y=4)=2$, while the unconditional mean $E(X)=5 \times \frac{1}{4}=1.25$. The fact that the conditional mean is higher is not surprising. The student was lucky enough to have obtained 4 correct answers by guessing. Given this, she had a greater chance of doing better on the first quiz.
__________________________________________________
Why This Works
Suppose $X \sim \text{binom}(5,p)$ and $Y \sim \text{binom}(5,p)$ and they are independent. The joint distribution of $X$ and $Y$ has 36 points in the sample space. See the following diagram.
Figure 1
The probability attached to each point is
\displaystyle \begin{aligned}(3) \ \ \ \ P(X=x,Y=y)&=P(X=x) \times P(Y=y) \\&=\binom{5}{x} p^x (1-p)^{5-x} \times \binom{5}{y} p^y (1-p)^{5-y} \end{aligned}
where $x=0,1,2,3,4,5$ and $y=0,1,2,3,4,5$.
The conditional probability $P(X=k \lvert X+Y=4)$ involves 5 points as indicated in the following diagram.
Figure 2
The conditional probability $P(X=k \lvert X+Y=4)$ is simply the probability of one of the 5 sample points as a fraction of the sum total of the 5 sample points encircled in the above diagram. The following is the sum total of the probabilities of the 5 points indicated in Figure 2.
\displaystyle \begin{aligned}(4) \ \ \ \ P(X+Y=4)&=P(X=0) \times P(Y=4)+P(X=1) \times P(Y=3)\\&\ \ \ \ +P(X=2) \times P(Y=3)+P(X=3) \times P(Y=2)\\&\ \ \ \ +P(X=4) \times P(Y=0) \end{aligned}
We can plug $(3)$ into $(4)$ and work out the calculation. But $(4)$ is actually equivalent to the following because $X+Y \sim \text{binom}(10,p)$.
$\displaystyle (5) \ \ \ \ P(X+Y=4)=\ \binom{10}{4} p^4 \ (1-p)^{6}$
As stated earlier, the conditional probability $P(X=k \lvert X+Y=4)$ is simply the probability of one of the 5 sample points as a fraction of the sum total of the 5 sample points encircled in Figure 2. Thus we have:
\displaystyle \begin{aligned}(6) \ \ \ \ P(X=k \lvert X+Y=4)&=\frac{P(X=k) \times P(Y=4-k)}{P(X+Y=4)} \\&=\frac{\displaystyle \binom{5}{k} p^k (1-p)^{5-k} \times \binom{5}{4-k} p^{4-k} (1-p)^{5-(4-k)}}{\displaystyle \binom{10}{4} p^4 \ (1-p)^{6}} \end{aligned}
With the terms involving $p$ and $1-p$ cancel out, we have:
$\displaystyle (7) \ \ \ \ P(X=k \lvert X+Y=4)=\frac{\displaystyle \binom{5}{k} \times \binom{5}{4-k}}{\displaystyle \binom{10}{4}}$
__________________________________________________
Summary
Suppose $X \sim \text{binom}(N,p)$ and $Y \sim \text{binom}(M,p)$ and they are independent. Then $X+Y$ is also a binomial distribution, i.e., $\sim \text{binom}(N+M,p)$. Suppose that both binomial experiments $\text{binom}(N,p)$ and $\text{binom}(M,p)$ have been performed and it is known that there are $a$ successes in total. Then $X \lvert X+Y=a$ has a hypergeometric distribution.
$\displaystyle (8) \ \ \ \ P(X=k \lvert X+Y=a)=\frac{\displaystyle \binom{N}{k} \times \binom{M}{a-k}}{\displaystyle \binom{N+M}{a}}$
where $k=0,1,2,3,\cdots,\text{min}(N,a)$.
As discussed earlier, think of the $N$ trials in $\text{binom}(N,p)$ as red balls and think of the $M$ trials in $\text{binom}(M,p)$ as blue balls in a jar. Think of the $a$ successes as the number of balls you are about to draw from the jar. So you reach into the jar and select $a$ balls without replacement. The calculation in $(8)$ gives the probability that you select $k$ red balls and $a-k$ blue balls.
The probability of success $p$ in the two binomial distributions have no bearing on the result since it gets canceled out in the derivation. One can always work a problem like Example 1 using first principle. Once the thought process using hypergeometric distribution is understood, it is a great way to solve this problem, that is, you can by pass the binomial distributions and go straight to the hypergeometric distribution.
__________________________________________________ | 2018-02-22T08:47:25 | {
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http://math.stackexchange.com/questions/82259/finding-volume-of-a-cone-through-integration | # Finding volume of a cone through integration
I am trying to find the volume of a cone using integration through horizontal slicing. The cone has a base radius of 10cm and a height of 5cm.
I am assuming this means I should integrate with respect to y, but I am not entirely sure how to set this up. I know that volume of a cylinder is given by the following: $$V = \pi r^2h$$
So I am assuming that the integral would be: $$\pi \int_0^5 f(y)^2dy$$
I am not sure how the x value of the radius 10cm (since it is not with respect to y) should fit into the equation, though.
Also, sorry for the pseudo-code style. I do not know how to use the math typesetting yet.
-
volume of cone – pedja Nov 15 '11 at 5:50
Place your cylinder so the center of the base is at $(0,0)$, and the apex is at $(0,5)$.
If you imagine looking at the cylinder straight on, it will look like a triangle with base $20$ and height $5$. If you make a horizontal slice at level $y$, then you get a figure with two similar triangles:
^ ^
/ \ |
/ \ |
^ /_____\ 5
| / 2r \ |
y / \ |
| / \ |
V /_____________\ V
|----- 20 -----|
then using similar triangles note that the height of the triangle on the top is $5-y$, and the base is $2r$. So we have $$\frac{2r}{20} = \frac{5-y}{5}.$$ From this, we can express $r$ in terms of $y$.
This is where you are using the fact that the base of your original cone is $10$ cm.
-
Sorry, I am a bit thick-headed here. I understand how you derived 5-y, but I am not sure how you derived the above equality (specifically how you arrived at the ratio of r/10). Can you explain a bit more? – Dylan Nov 15 '11 at 5:40
@Dylan: Does the picture help? – Arturo Magidin Nov 15 '11 at 5:50
OK. That makes sense. So solving for y from the above, i get y = -1/2r + 5. This should be plugged into the function f(y)dy i am assuming and then integrated (where b-a -> 5-0)? – Dylan Nov 15 '11 at 5:57
@Dylan: You don't want to solve for $y$, you want to solve for $r$: the volume of the cylindrical slice at level $y$ is $\pi r^2\Delta y$, so you want to express $r$ in terms of $y$, not the other way around. – Arturo Magidin Nov 15 '11 at 6:02
Hmm, but when i solve the above equality for r then i get 10 - 2y. I am not sure this agrees with Andre's response above (or at least I don't see how it does - it appears he is taking the slope delta y over delta x??) – Dylan Nov 15 '11 at 6:17
You can set things up so that you integrate with respect to $x$ or you can set things up so that you integrate with respect to $y$. It's your pick! For each solution, you should draw the picture that goes with that solution.
With respect to $x$: Look at the line that passes through the origin and the point $(5,10)$. Rotate the region below this line, above the $x$-axis, from $x=0$ to $x=5$, about the $x$-axis. This will generate a cone with base radius $10$ and height $5$. The main axis of this cone is along the $x$-axis. Kind of a sleeping cone.
Take a slice of width "$dx$" at $x$, perpendicular to the $x$-axis. The ordinary name for this would be a vertical slice.
This slice is almost a very thin cylinder: if you are hungry, think of a thin ham slice taken from a conical ham. Let us find the radius of this slice. The line through the origin that goes through $(5,10)$ has slope $2$, so has equation $y=2x$. Thus at $x$ the radius of our almost cylinder is $2x$. It follows that the thin slice the slice has (almost) volume $\pi(2x)^2 dx$. "Add up" (integrate) from $x=0$ to $x=5$. The volume of our cone is equal to $$\int_{x=0}^5 \pi(2x)^2 \,dx=\int_0^5 4\pi x^2\,dx.$$ The integration is easy. We get $\dfrac{500\pi}{3}$.
With respect to $y$: It is a matter of taste whether our cone is point up or point down. Since an answer with point up has already been posted, we imagine the cone with point down at the origin. Look at the line that goes through the origin and passes through the point $(10,5)$. Take the region to the left of this line, to the right of the $y$-axis, from $y=0$ to $y=5$. Rotate this region about the $y$-axis. We get a cone with base radius $10$ and height $5$.
Take a horizontal slice of width "$dy$" at height $y$. This looks almost like a flat cylindrical coin.
We want to find the volume of that coin. The line through the origin and $(10,5)$ has slope $1/2$, so it has equation $y=x/2$. So $x=2y$, and therefore the radius of our thin slice is $\pi(2y)^2 dy$. Thus the volume of the cone is $$\int_{y=0}^5 \pi(2y)^2\,dy.$$ This is the same definite integral as our previous one. Only the name of the variable of integration has changed. Naturally, the result is the same.
-
Andre, which is more easy/reasonable deriving the general formula and then substituting the values or deriving for the particular values? ... also why use cylinders when you can simply use discs? – Quixotic Nov 15 '11 at 6:09
Indeed, the general formula is just as easy to derive. However, sometimes concrete specific numbers can help the initial understanding. After that, the work with $r$ and $h$ seems reasonable. As to why not circle, the cross-sections are indeed circles. But I wanted to convey, unfortunately without a picture, that we are adding up the volumes of very thin disks. If one practices that enough times, the intuition behind some formulas of say Physics becomes clearer. – André Nicolas Nov 15 '11 at 6:17
Aha..., I got it!, I guess in these kinds of solids thinking about the volume interpretation is better than the discs however they might lead to the same thing but the volume interpretation is much more neat, another example would be while computing the volume of a paraboloid. – Quixotic Nov 15 '11 at 6:34
Or total displacement as definite integral of velocity, or fluid pressure, or moment about the $y$-axis, or many other things. – André Nicolas Nov 15 '11 at 6:58 | 2014-10-22T22:33:47 | {
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https://math.stackexchange.com/questions/70231/how-to-prove-those-curious-identities | # How to prove those "curious identities"?
How to prove $$\prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right) = \frac{n}{2^{n-1}}$$ and $$\prod_{k=1}^{n-1} \cos\left(\frac{k\pi}{n}\right) = \frac{\sin(\pi n/2)}{2^{n-1}}$$
• I would have thought that you might have learned from your previous experience here that it's a good idea to say something about where you came across these identities, as it might point the way toward an answer. Oct 6, 2011 at 2:25
• +1 for a question that generated a variety of good answers!
– lhf
Oct 6, 2011 at 2:57
• FYI, if you like those kind of identities, those, and some other similar ones, are in "Challenging Mathematical Problems with Elementary Solutions" by Yaglom and Yaglom, volume 2, which is available in an inexpensive Dover edition.
– tzs
Oct 6, 2011 at 4:42
• Related: Similar reasoning as in some of the answers below (Euler's formulas + geometric series) proves the nice but no so widely known multiple-angle formula $$\sin nx = 2^{n-1} \prod_{k=0}^{n-1} \sin(x + \frac{k\pi}{n}).$$ Your first formula can be obtained as a special case after dividing both sides by $\sin x$ and taking the limit as $x\to 0$. Oct 6, 2011 at 7:50
• Half-duplicate of math.stackexchange.com/questions/8385/… Mar 20, 2015 at 8:04
For the first: $$\lim_{z=1}\frac{z^n-1}{z-1}=n\tag{1a}$$ $$\frac{z^n-1}{z-1}=\prod_{k=1}^{n-1}(z-e^{2\pi ik/n})\tag{1b}$$ $$|1-e^{i2k\pi/n}|=|2\sin(k\pi/n)|\tag{1c}$$ Combining $(1a)$, $(1b)$, and $(1c)$, we get $$2^{n-1}\prod_{k=1}^{n-1}\sin(k\pi/n)=n$$ since everything is positive.
For the second:
If $n$ is even, then $\cos(\frac{\pi}{2})=0$ appears in the product (when $k=n/2$) and $\sin(\frac{n\pi}{2})=0$.
If $n$ is odd, then combining $$\lim_{z=1}\frac{z^n+1}{z+1}=1\tag{2a}$$ $$\frac{z^n+1}{z+1}=\prod_{k=1}^{n-1}(z+e^{2\pi ik/n})\tag{2b}$$ $$1+e^{i2k\pi/n}=2\cos(k\pi/n)e^{ik\pi/n}\tag{2c}$$ and noting that $\displaystyle\sum_{k=1}^{n-1}k=\frac{n(n-1)}{2}$ so that $\displaystyle\prod_{k=1}^{n-1}e^{ik\pi/n}=(-1)^{(n-1)/2}$ which matches the sign of $\sin(\pi n/2)$, yields $$2^{n-1}\prod_{k=1}^{n-1}\cos(k\pi/n)=(-1)^{(n-1)/2}=\sin(\pi n/2)$$
• How do you prove (1b)? Can you recommend a page for me to see, or the name of that theorem? Thanks.
– D.R.
Sep 7, 2017 at 1:39
• $\text{(1b)}$ is just noting that the roots of $z^n-1$ are the $n^\text{th}$ roots of unity. Dividing by $z-1$ removes the root at $1$ and so we're left with the other $n-1$ roots $\left\{e^{2\pi ik/n}:1\le k\le n-1\right\}$.
– robjohn
Sep 7, 2017 at 2:04
• My mistake, I had accidently registered 1b as 1a in my head Nov 11, 2020 at 16:47
• @Buraian: Yes, $\text{(1a)}$ follows from L'Hôpital.
– robjohn
Nov 11, 2020 at 18:11
Denote $w = e^{i \pi/n}$. We have
$$\prod_{k = 1}^{n-1} \sin \left(\frac{k\pi}{n}\right)= \prod_{k = 1}^{n-1} \frac{w^k - w^{-k}}{2i} = \frac{1}{2^{n-1}} \prod_{k = 1}^{n-1} \frac{w^k}{i} (1-w^{-2k})$$
Since we have
$$\sum_{k = 0}^{n-1} x^k = \prod_{k = 1}^{n-1} (x-w^{2k})$$
Setting $x=1$ yields
$$\prod_{k = 1}^{n-1} (1-w^{2k}) = n$$
So we get
$$\prod_{k = 1}^{n-1} \sin \left(\frac{k\pi}{n}\right)= \frac{n}{2^{n-1}} \frac{w^{n(n-1)/2}}{i^{n-1}} = \frac{i^{n-1}}{i^{n-1}} \frac{n}{2^{n-1}} = \frac{n}{2^{n-1}}$$
I guess (but did not check) that the same kind of reasoning gives the one with $\cos$.
• similar reasoning for odd $n$ (except you have to watch the sign), but very different, however easy, reasoning for even $n$.
– robjohn
Oct 6, 2011 at 4:03
The second purported identity is equivalent to asking for the constant term of $\dfrac{U_{n-1}(x)}{2^{n-1}}$ (i.e., $\dfrac{U_{n-1}(0)}{2^{n-1}}$), where $U_n(x)$ is the Chebyshev polynomial of the second kind. Since
$$\frac{U_{n-1}(x)}{2^{n-1}}=\frac{\sin(n \arccos\,x)}{2^{n-1}\sqrt{1-x^2}}$$
letting $x=0$ gives your identity.
• This is really clever. Oct 6, 2011 at 3:13
Define $\zeta_n = e^{2 \pi i/n}$.
Proposition For odd integer $n \geq 1$, \begin{align} \prod_{k = 1}^{n-1}(\zeta_n^{k} - \zeta_n^{-k}) = n. \end{align} and \begin{align} \prod_{k = 1}^{n-1} \sin( \tfrac{2 \pi k }{n} ) = \tfrac{n}{(2 i)^{n-1}}. \end{align} Proof: The claimed identities follow from the identity \begin{align} z^n - 1 = \prod_{ k =0}^{n-1} (z - \zeta_n^{k}) = \prod_{ k =0}^{n-1} (z - \zeta_n^{-2k}). \end{align} Writing $z = x/y$, we have \begin{align} x^n - y^n = \prod_{k = 0}^{n-1} ( \zeta_n^{k} x - \zeta_n^{-k} y). \end{align} Thus, \begin{align} n y^{n-1} = \lim_{x \to y} \frac{x^n - y^n}{x - y} = \lim_{x \to y} \ \ \prod_{k = 1}^{n-1} ( \zeta_n^{k} x - \zeta_n^{-k} y) = y^{n-1} \ \prod_{k = 1}^{n-1} ( \zeta_n^{k} - \zeta_n^{-k} ). \end{align} For the second identity, let $x =e^{\pi i z}$ and $y = e^{- \pi i z}$ and recall the complex exponential representation of the sine function. This yields \begin{align} n = \lim_{z \to 0} \frac{\sin n \pi z}{\sin z } = (2 i)^{n-1} \lim_{z \to 0} \ \ \prod_{k = 1}^{n-1} \sin( \pi z + \tfrac{2 \pi k }{n} ) = (2 i)^{n-1} \prod_{k = 1}^{n-1} \sin( \tfrac{2 \pi k }{n} ). \end{align}
Similar reasoning works to prove the identities that you mention. | 2022-08-14T12:47:33 | {
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https://math.stackexchange.com/questions/1196280/solve-x-fracdydx-x-y-using-the-integrating-factor-method-and-by-using-the | # Solve $x\frac{dy}{dx}=x-y$ using the integrating factor method and by using the substitution $y=vx$
I have been asked to show that the solutions to this equation are identical regardless of the method used. I did both methods and I think I may have got it right but I am not sure. Please check:
Using the integrating factor method I found that the factor was $x$. Therefore $\frac{d}{dx}(xy)=x$ so the result was $y=\frac12 x+\frac cx$
Using the substitution $y=vx$ enabled me to get the separable equation: $x\frac{dy}{dv}+2v=1$ I then solved to give: $\frac1{\sqrt{1-2v}}=Ax$ where $A=e^c$. Rearranging to try and get the form of the answer that I got using the other method gave me: $y=\frac12x-\frac{1}{2A^2x}$. It looks similar to my last answer and would be the same if it is permissible to simply say that $c=-\frac{1}{2A^2}$. Is this allowed or have I done something wrong?
• they are equivalent. – abel Mar 19 '15 at 0:23
• So I have done it right? – RobChem Mar 19 '15 at 0:23
• Yes it is allowed. – science Mar 19 '15 at 0:24
• yes. you did it right. – abel Mar 19 '15 at 0:24
i don't know if you care for another way to do this. that is a third way. but here it is. we write the differential equation as $$\frac{dy}{dx} = \frac{x-y}x$$ split this into equations $$\frac{dy}{dt} = x - y, \, \frac{dt}{dx} = \frac 1 x$$ we solve the second one and get $$x = Ce^t, \frac{dy}{dx} + y=Ce^t$$ the solution for $y$ is $$y = Be^{-t} + \frac C 2 e^t$$ now we get rid $t$ by subbing $\frac x C$ for $e^t$ and arrive at the same solution you got $$y = \frac{BC} x + \frac 1 2 x$$
$$xy'+y=x \implies (xy)'=x.$$ | 2019-08-19T08:12:54 | {
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https://math.stackexchange.com/questions/2048838/calculate-sum-limits-n-1-infty-frac1n2n42 | # Calculate $\sum\limits_{n=1}^\infty \frac{1}{(n+2)(n+4)^2}$
Calculate the sum of the series $$\sum_{n=1}^\infty \frac{1}{(n+2)(n+4)^2}$$
I have tried partial fraction decomposition.
$$\sum_{n=1}^\infty\frac{1}{4(n+2)}- \sum_{n=1}^\infty\frac{1}{4(n+4)}-\sum_{n=1}^\infty\frac{1}{2(n+4)^2}$$
Is this correct? What is the sum?
• are you sure the sum is over $n$, not another index? – Alex Dec 7 '16 at 23:25
• no, im sorry is infinite – Diego Gomez Dec 7 '16 at 23:26
• The denominator in the first sum should be $n+1$ multiplied by a constant. I think the denominator in the rightmost term reads $2(n+4)^2$ – GNUSupporter 8964民主女神 地下教會 Dec 7 '16 at 23:28
• Note that for every $N$, $$\sum_{n=1}^N\frac{1}{4(n+2)}- \sum_{n=1}^N\frac{1}{4(n+4)}-\sum_{n=1}^N\frac{1}{2(n+4)^2}$$ is $$\frac1{4\cdot3}+\frac1{4\cdot4}-\frac1{4(N+3)}-\frac1{4(N+4)}-\frac12\sum_{n=1}^{N+4}\frac{1}{n^2}+\frac1{2\cdot1^2}+\frac1{2\cdot2^2}+\frac1{2\cdot3^2}+\frac1{2\cdot4^2}$$ hence the desired limit is $$\frac1{4\cdot3}+\frac1{4\cdot4}-\frac12\sum_{n=1}^{\infty}\frac{1}{n^2}+\frac1{2\cdot1^2}+\frac1{2\cdot2^2}+\frac1{2\cdot3^2}+\frac1{2\cdot4^2}$$ Can you finish? – Did Dec 7 '16 at 23:44
• @Did Should become the answer. – Simply Beautiful Art Dec 7 '16 at 23:44
Your partial fraction decomposition looks OK, but it should be written as
$$\sum_{n=1}^\infty\left({1\over4(n+2)}-{1\over4(n+4)}\right)-\sum_{n=1}^\infty{1\over2(n+4)^2}$$
instead of being split into three infinite series. That's because $\sum_{n=1}^\infty{1\over4(n+2)}$ and $\sum_{n=1}^\infty{1\over4(n+4)}$ are each divergent. But combined they give the convergent, telescoping series
$$\left({1\over4\cdot3}-{1\over4\cdot5}\right)+\left({1\over4\cdot4}-{1\over4\cdot6}\right)+\left({1\over4\cdot5}-{1\over4\cdot7}\right)+\left({1\over4\cdot6}-{1\over4\cdot8}\right)+\cdots\\={1\over4\cdot3}+{1\over4\cdot4}={7\over48}$$
The other series you need to recognize as
$${1\over2}\left({1\over5^2}+{1\over6^2}+\cdots\right)={1\over2}\sum_{n=1}^\infty{1\over n^2}-{1\over2}\left(1+{1\over2^2}+{1\over3^2}+{1\over4^2}\right)={1\over2}\left(\pi^2\over6\right)-{1\over2}\left(205\over144\right)$$
The trick is understanding the hat that the $\pi^2/6$ rabbit came from, but I'm assuming you've seen it somewhere.
I'll leave it to you to put the pieces together.
First note that your series has the same convergence as $\sum \frac{1}{n^3}$ by the limit comparison test. And the latter series converges absolutely by the $p$-series test or the integral test, so therefore so does your series.
Next, to find the value the sum converges to, trying partial fractions, we get
$$\frac{1}{(n+2)(n+4)(n+4)}=\frac{1}{4}\frac{1}{n+2}-\frac{1}{4}\frac{1}{n+4}-\frac{1}{2}\frac{1}{(n+4)^2}$$
If we break it into three series as you have attempted, then all three are divergent, and difference of divergent is indeterminate, so we cannot proceed. Instead, treat the first two terms together, and note that $\sum_{n=1}^\infty \frac{1}{n+2}-\frac{1}{n+4}$ is a telescoping series, its sum will converge to $1/3+1/4$, the uncanceled parts of the first two terms. For the final term, rememeber by the Basel problem $\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$ so by reindexing we have $\sum_{n=5}^\infty\frac{1}{n^2}=\sum_{n=1}^\infty\frac{1}{(n+4)^2}=\frac{\pi^2}{6}-\frac{1}{16}-\frac{1}{9}-\frac{1}{4}-1$.
Putting it all together we have
$$\sum_{n=1}^\infty\frac{1}{(n+2)(n+4)(n+4)}=\frac{1}{4}\left(\sum_{n=1}^\infty\frac{1}{n+2}-\frac{1}{n+4}\right)-\frac{1}{2}\sum_{n=1}^\infty\frac{1}{(n+4)^2} \\=\frac{1}{4}\left(\frac{1}{3}+\frac{1}{4}\right)-\frac{1}{2}\left(\frac{\pi^2}{6}-\frac{1}{16}-\frac{1}{9}-\frac{1}{4}-1\right) = \frac{7}{48} -\frac{\pi^2}{12}+\frac{205}{288}=\frac{247}{288}-\frac{\pi^2}{12}$$
• @SimpleArt Although OP didn't write it, I assumed the question was about infinite series. Otherwise the question is kind of trivial, no? – ziggurism Dec 7 '16 at 23:27
• I'm sorry, the OP has clarified that for us. :) On the other hand, the OP also asks to actually calculate the sum. – Simply Beautiful Art Dec 7 '16 at 23:28
• @SimpleArt probably we should get further clarification from OP – ziggurism Dec 7 '16 at 23:30
• Well the OP asks for the sum in the last sentence. – Simply Beautiful Art Dec 7 '16 at 23:34
• The OP has an $(n+2)$ instead of an $(n+1)$. – Barry Cipra Dec 8 '16 at 0:05 | 2020-02-19T02:09:14 | {
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http://math.stackexchange.com/questions/174676/solving-simple-congruences-by-hand | # Solving simple congruences by hand
When I am faced with a simple linear congruence such as $$9x \equiv 7 \pmod{13}$$ and I am working without any calculating aid handy, I tend to do something like the following:
"Notice" that adding $13$ on the right and subtracting $13x$ on the left gives: $$-4x \equiv 20 \pmod{13}$$
so that $$x \equiv -5 \equiv 8 \pmod{13}.$$
Clearly this process works and is easy to justify (apart from not having an algorithm for "noticing"), but my question is this: I have a vague recollection of reading somewhere this sort of process was the preferred method of C. F. Gauss, but I cannot find any evidence for this now, so does anyone know anything about this, or could provide a reference? (Or have I just imagined it all?)
I would also be interested to hear if anyone else does anything similar.
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I suppose we all have our own "ad hoc" methods. I would have "noticed" that multiplying by $3$ gives $x \equiv 21 \equiv 8$ (mod $13$). I have always believed that Gauss invented modular arithmetic. It is certainly discussed at length in Disquisitiones Arithmeticae, which I own a copy of. – Geoff Robinson Jul 24 '12 at 15:08
It's not clear to me that the process always works, but it is interesting. It seems to me that the trick is that we get to replace our equation and hope for common prime factors between the coefficients. – Andrew Jul 24 '12 at 15:09
@GeoffRobinson I am pretty sure Gauss did invent the notation, and I too have a copy of Disquisitiones Arithmeticae, but I cannot see anything in it along exactly the lines of my question, unfortunately. – Old John Jul 24 '12 at 15:13
It is not an algorithm in the precise sense of the word. The idea is basically to multiply or divide (on both sides) or add (or subtract) a multiple of the modulus (on either side) to get something which is equivalent, but "simpler". Repeating this vague process a number of times will give a solution (if the original has a solution) as at each step the multiple of $x$ on the left is going to get smaller, and the new congruence is equivalent to the previous one. – Old John Jul 24 '12 at 15:16
@Old John: Thanks. Yes Bill Dubuque's answer made what was going on reasonably clear. – Geoff Robinson Jul 24 '12 at 18:46
By Gauss' algorithm, scale $\rm\:\color{#C00}{\frac{A}B} \to \frac{AN}{BN}\:$ by the least $\rm\,N\,$ so that $\rm\, BN > 13,\,$ reduce mod $13,\,$ iterate.
$$\rm\displaystyle \ mod\ 13\!:\ \color{#C00}{\frac{7}9} \,\equiv\, \frac{14}{18}\, \equiv\, \color{#C00}{\frac{1}5}\,\equiv\, \frac{3}{15}\,\equiv\, \color{#C00}{\frac{3}2} \,\equiv\, \frac{21}{14} \,\equiv\, \color{#C00}{\frac{8}1}$$
Denominators of the $\color{#c00}{\rm reduced}$ fractions decrease $\,\color{#C00}{ 9 > 5 > 2> \ldots}\,$ so reach $\color{#C00}{1.}$
Or, simpler, allowing negative residues $\displaystyle\ \ \frac{7}9\,\equiv\, \frac{7}{\!-4\!\ \,}\,\equiv\,\frac{21}{\!\!-12\ \ \ \!\!}\,\equiv\, \frac{8}1$
This optimization of using balanced residues $0,\pm 1, \pm 2.\ldots$ works for modular arithmetic in general. Here we can also optimize by (sometimes) cancelling obvious common factors, or by pulling out obvious factors of denominators, etc. For example
$$\frac{7}9\,\equiv\, \frac{\!-6\,}{\!-4\,}\,\equiv\frac{\!-3\,}{\!-2\,}\,\equiv\frac{10}{\!-2\,}\,\equiv\,-5$$
$$\frac{7}9\,\equiv\,\frac{\!-1\cdot 6}{\ \ 3\cdot 3}\,\equiv\,\frac{\!\,12\cdot 6\!}{\ \ \,3\cdot 3}\,\equiv\, 4\cdot 2$$
As you did, we can check if the quotient $\rm\,a/b\equiv (a\pm\!13\,i)/(b\pm\!13\,j)\,$ is exact for small $\rm\,i,j,\,$ e.g.
$$\frac{1}7\,\equiv \frac{\!-12}{-6}\,\equiv\, 2;\ \ \ \frac{5}7\,\equiv\,\frac{18}{\!-6\!\,}\,\equiv -3$$
When working with smaller numbers there is a higher probability of such optimizations being applicable (the law of small numbers), so it's well-worth looking for such in manual calculations.
Note Gauss' algorithm is my name for a special case of the Euclidean algorithm that's implicit in Gauss' Disq. Arith. I don't recall if Gauss explicitly used this algorithm. Follow said link for more.
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Thanks for the reference - so it was Gauss after all. The Gauss process is basically what I do, although I keep an eye out for any possible shortcuts, as in the example I gave. – Old John Jul 24 '12 at 15:21
@mathh As the linked post says, Gauss's algorithm requires prime modulus. Generally modular fractions make sense only for denominators coprime to the modulus. Thus when scaling fractions we must restrict to scale factors coprime to the modulus, e.g. in your case we can do $\tag*{}$ ${\rm mod}\ 10\!:\ \dfrac{1}3\equiv \dfrac{3}9\equiv \dfrac{3}{-1} \equiv -3\equiv 7\ \$ – Bill Dubuque Aug 18 '14 at 16:44
I'm sorry, I deleted the comment before you replied since I found it out. I would add to the above explanations that the denominators can be reduced if and only if the modulus is coprime to them. – user314 Aug 18 '14 at 16:51
@BillDubuque Is your notation $\mod p : a\equiv b$ as formal and accepted as $a\equiv b\pmod p$? I.e. do scientists, etc. use it? – user314 Aug 18 '14 at 16:54
@mathh It is less commonly used. I find it is clearer to students since it specifies the context first (vs. last) in normal reading order (left-to-right). – Bill Dubuque Aug 18 '14 at 17:05
When the prime is a reasonably small one I'd rather find directly the inverse: $$9^{-1}=\frac{1}{9}=3\pmod {13}\Longrightarrow 9x=7\Longrightarrow x=7\cdot 9^{-1}=7\cdot 3= 21=8\pmod {13}$$ But...I try Gauss's method when the prime is big and/or evaluating inverses is messy.
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9x = 7 mod 13
9x = 7 + 13n
9x = 20 for n = 1
9x = 33 for n = 2
9x = 46 for n = 3
9x = 59 for n = 4
9x = 72 for n = 5
Then x = 8 mod 13
You arrive at the correct answer before n = 13.
- | 2015-01-26T10:29:55 | {
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https://www.physicsforums.com/threads/intergration-by-parts-for-sin-x-cos-x.755943/ | # Intergration by parts for sin(x)cos(x)
1. May 30, 2014
### uzman1243
I know its easier to use the substitution method, by I'm trying to see how it'll work for integration by parts. I follow the LIATE method for integration by parts.
Now if I take u=cos(x) and dv = sin(x), the answer changes.
Can you please explain this to me? Which is the 'right' answer?
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2. May 30, 2014
### gopher_p
Rather than answer your question directly, let me provide another example of the same phenomenon that might make it easier to identify why both answers are correct.
If you compute $\int (x+1)\ dx$ using the sum rule for antiderivatives and the "reverse power rule", you get $$\int x+1\ dx=\frac{1}{2}x^2+x+C.$$ If instead you use a $u$-sub, with $u=x+1$, you get $$\int (x+1)\ dx=\int u\ du=\frac{1}{2}u^2+C=\frac{1}{2}(x+1)^2+C.$$
Now the exact same thing is happening in your example as is happening here; you have two different-looking answers that are both correct. How can that be?
3. May 30, 2014
### D H
Staff Emeritus
And closer to home, let's do $\int \sin(x) \cos(x) dx$ a third way, using the trig substitution $\sin(2x)=2\sin(x) \cos(x)$. Thus $\int \sin(x) \cos(x) dx = \int \frac {\sin(2x)}2 dx = -\frac {\cos(2x)} 4 + C$.
To quote gopher_p,
4. May 30, 2014
### athosanian
dear friend , all these answers are different only by a constant, so they are the same answers. | 2017-10-20T18:29:55 | {
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http://math.stackexchange.com/questions/178773/why-there-may-be-no-single-maximum-element-in-a-partially-ordered-set | # Why there may be no single “maximum” element in a partially ordered set?
From Appendix B.2 (relations) of Introduction to Algorithms by Cormen et al:
In a partially ordered set A, there may be no single "maximum" element a such that b R a for all b ∈ A. Instead, there may several maximal elements a such that for no b ∈ A, where b ≠ a, is it the case that a R b. For example, in a collection of different-sized boxes there may be several maximal boxes that don't fit inside any other box, yet no single "maximum" box into which any other box will fit.
I know the definition of a partial ordered set , but do not get why the author/authors say that "there may be no single "maximum" element" in a partially ordered set ?
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Some counterexamples can be found at ProofWiki and Wikipedia. – Martin Sleziak Aug 4 '12 at 12:42
@MartinSleziak to understand the counter example I have to understand this one first. – Geek Aug 4 '12 at 12:43
Related question: difference between maximal element and greatest element. – Martin Sleziak Aug 4 '12 at 12:51
Remember that we're dealing with a partial order! This is fundamentally different from a total order like $\leq$ is on numbers.
Because: For all numbers, we have either $a\leq b$ or $b\leq a$, i.e. everything is comparable.
However a partial order is partial because elements need not be comparable. Solely if elements should happen to be comparable, then the partial order imposes some laws like transitivity.
Now take as an example the proper subsets of $\{1,2,3\}$ with the partial order of $\subseteq$. E.g. $\{1\} \subseteq \{1,2\}$, but the sets $\{1,2\}$ and $\{2,3\}$ simply are not comparable because neither includes the other. However there are no "bigger" proper subsets to each of them, so both are maximal.
Now if in turn all elements were comparable, the maximum was unique (hence in total orders it is). But in partial orders, this isn't usually the case.
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For another easy to see example, consider the set $\{1,2,3,4,6\}$ and define $a \leq b$ if $a|b$. This is a partial order, and both $4$ and $6$ are maximal elements.
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Take a finite set. Consider the lattice of subsets; this is a partially ordered set. Now, if you consider only proper subsets, you still have a partially ordered set, but now, every subset which consists of all elements except one is a maximal element. If you draw it, you will see it right away.
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@M Turgeon What is "lattice of subsets?" Can you put a drawing in your answer somehow ? – Geek Aug 4 '12 at 12:59
@Geek he is talking about the power set, minus the set we're taking the power set of (boy, that's a mouthful). The partial order is given by inclusion. For example, if $S = \{a,b,c\}$ then the set he is talking about is $\{\varnothing,\{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\}\}$. All 3 2-element subsets of $S$ are maximal elements on $2^S-S,\subseteq$. – David Wheeler Aug 4 '12 at 13:27
@Geek ProofWiki: Subset Relation on Power Set is Partial Ordering. You can see some Hasse diagrams as examples in Wikipedia articles on posets and Hasse diagrams or here. (And in many other places.) – Martin Sleziak Aug 4 '12 at 13:36
@Geek Unfortunately, I have no idea how to put pictures in my answer, but the Hasse diagrams Martin put in his comment is exactly what I meant. – M Turgeon Aug 4 '12 at 17:22
@Geek I've put some diagrams also here: 2-element set, 3-element set, 4-element set – Martin Sleziak Aug 4 '12 at 18:03
Suppose that our partial order is given (informally) by "box $a$ can be put into the box $b$".
If I have boxes $a$ and $b$ of the same size, I cannot put one of them into the other; this means that neither $a\le b$ nor $b\le a$ holds.
Formally, we have a partial ordering $\le=\{(a,a),(b,b)\}$ on the set $\{a,b\}$.
Hasse diagram of this poset looks like this:
Then $a$ and $b$ are both maximal elements. I.e., maximal element is not unique.
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"if I have boxes a and b of the same size; this means that neither a≤b nor b≤a holds." .If they are of the same size(a==b) both the inequalities hold.Why are you saying neither holds ? – Geek Aug 4 '12 at 12:52
My relation $\le$ is defined by "a box can be put into other box"; not by comparing size. If the boxes are thick enough, I cannot put one of them into the box of the same size. – Martin Sleziak Aug 4 '12 at 12:53
Then your relation is not reflexive, and therefore, not a partial order. – David Wheeler Aug 4 '12 at 13:02
@David The relation $\{(a,a),(b,b)\}$ is reflexive. But the informal description of the example with boxes would correspond better to a strict partial order. (I used the boxes because I wanted to follow the example posted by OP - the one which he is asking about.) – Martin Sleziak Aug 4 '12 at 13:06
@MartinSleziak I understand your motivation. My comment is intended to highlight the ambiguity of the example in the original post. Your formal definition of (a) partial order is indeed one. The example given is not a good one. My response is to give a better one. – David Wheeler Aug 4 '12 at 13:11 | 2016-06-26T15:48:51 | {
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http://math.stackexchange.com/questions/173716/independent-events-and-dependent-events | # Independent events and Dependent events
I have a question regarding these strikingly similar problems with contradicting solutions. This is somewhat long, so prepare
Probblem 1
Consider a bag of ten coins, nine are fair, but one is weighted with both sides heads. You randomly select a coin and toss it five times. Let $2s$ denote the event of selecting the weighted coin (that is the 2-sided coin) and $N$ be the even you select a regular coin and $5H$ be the event of getting five heads in a row. What is
a) $P(5H | 2s)$
b) $P(5H | N)$
c) $P(5H)$
d) $P(2s | 5H)$
Solution 1
a) Simply 1
b) $\frac{1}{2^5}$
c) $\frac{1}{2^5}\frac{9}{10}+ \frac{1}{10} = \frac{41}{320}$
d) $P(2s|5H) = \dfrac{P(5H|2s)P(2s)}{P(5H)} = \frac{32}{41}$
From the Solution 1, it seems that $P(2s|5H) \neq P(2s)P(5H)$ That is the event of picking out the weighted coin affects the probability of getting 5H.
Here is part of my question, isn't there also some tiny probability of getting 5H from picking the normal one as well? Doesn't make sense why the events of picking the coin and getting 5H is dependent. Read on the next question
Problem 2
A diagnostic test for an eye disease is 88% accurate of the time and 2.4% of the population actually has the disease. Let $ED$ be the event of having the eye disease and $p$ be the event of testing positive. Find the probability that
a) the patient tests positive
b) the patient has the disease and tests positive
Solution 2
Here is a tree diagram
a) $0.02122 + 0.011712 = 0.13824$
b) $P(ED | p) = \dfrac{P(\text{ED and p})}{P(p)} =\frac{0.02122}{0.13824 }= 0.1535$
From Solution 2, it looks like $P(\text{ED and p}) = P(\text{ED})P(p)$ which means that having the eye disease and testing positive are independent events? After trying out the same formula from Problem 1, it also seems that
$$P(\text{ED | p}) = \dfrac{P(\text{ED and p})}{P(p)} = \dfrac{P(\text{p | ED})P(ED)}{P(p)} = 0.1535$$
Also, when the question asks "the patient has the disease and tests positive", how do I know that it is $P(ED | p)$ and not $P(p | ED)$?
I am very confused in general with this. Could anyone clarify for me? Thanks
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From the Solution 1, it seems that $P(2s|5H) \ne P(2s)P(5H)$ That is the event of picking out the weighted coin affects the probability of getting 5H.
Here is part of my question, isn't there also some tiny probability of getting $5H$ from picking the normal one as well? Doesn't make sense why the events of picking the coin and getting $5H$ is dependent.
The fact that there is some tiny probability of getting $5H$ from picking the normal coin is exactly why the the event $5H$ is dependant on the event $2s$. In other words, the probability of getting five heads depends on which coin you pick.
Also, when the question asks "the patient has the disease and tests positive", how do I know that it is $P(ED|p)$ and not $P(p|ED)$ ?
I read $P(ED|p)$ as "the probability that the patient has the eye disease given that the test comes back positive. On the other hand, I read $P(p|ED)$ as "the probability that the test comes back positive given that the patient has the eye disease."
In practical terms however, the patient wants to know the former. That is, when I go get a test from the doctor and it comes back positive, I want to know what the probability is that the result of the test is accurate. I believe this is what the homework question is asking you to find.
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OKay I delete my comments then – Hawk Jul 21 '12 at 22:41
P(ED|p) is what the book says...that's why i was really confused with the wording of the solution. Could we pretend that it asked P(ED|p) then and answer my other question? – Hawk Jul 21 '12 at 22:46
@jak Where did you see this in the book? In the solutions in the back? Or is this an example rather than a homework problem? – Code-Guru Jul 21 '12 at 22:48
bmlc.ca/Math12/… should work now – Hawk Jul 21 '12 at 23:00
@jak Hmm...I am having some difficulty with the wording of the problem and the proposed solution. Let me give it some thought and I will edit my answer again shortly. – Code-Guru Jul 21 '12 at 23:04
Your calculations for the first problem are all correct. Of course $P(2s|5H) \neq P(2s)P(5H)$. The correct expression is $P(2s|5H) = P(2s\cap H)/P(5H)$. And certainly $2S$ and $5H$ are not independent, either at the informal level or at the technical level.
"Has the disease and tests positive" should mean exactly what it says. It is not a conditional probability. In symbols, it is $P(ED \cap p)$. If a conditional probability is looked for, different language should be used.
It is easy to calculate. Indeed you calculated it on the way to finding $P(p)$. In symbols, it is $P(p|ED)P(ED)$.
The conditional probability $P(ED|p)$ is also straightforward to find. For note that in general $P(A|B)P(B)=P(A\cap B)$. You know $P(ED \cap p)$, and you know $P(p)$, so, by division, you can find $P(ED|p)$.
Added: In the online notes that you gave a link for, part c) reads "Determine the probability that the patient has eye disease and tests positive." The solution then proceeds to determine (correctly) the probability that the patient has eye disease given that she tests positive. (Part d) has a similar mistake.)
What is one to make of this? One could be generous and call it a typo. OK, two typos.
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but $P(ED and p)$ isn't easy to compute because the events are "dependent". So I must use Baye's formulae – Hawk Jul 21 '12 at 23:03
$P(ED \cap p)=P(p \cap ED) =P(p|ED)P(ED)$. We have $P(p|ED)=0.88$, $P(ED)=0.024$. That's the first entry in your tree diagram, the $0.02122$. So you really do know how to calculate it! – André Nicolas Jul 21 '12 at 23:10
That was my understanding, but the notes from his class give the solution in the OP. (See question 5 on the last page.) – Code-Guru Jul 21 '12 at 23:10
@jak No. $P(\text{ED and p}) = P(p | \text{ED})P(ED)$ from the formula you guessed and posted in your question. Note that the 88% given in the problem is $P(p | \text{ED})$, not $P(p)$. – Code-Guru Jul 21 '12 at 23:15
No we are not. $P(ED \cap p)=P(p|ED)P(ED)$. Since $P(p|ED)\ne P(p)$, the equation you mention above is not true. By the way, I checked the "book" linked above. It uses and, but probably does not intend and. The quality of probability stuff in school books and notes is all too often dismal. – André Nicolas Jul 21 '12 at 23:20 | 2015-12-01T17:19:07 | {
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https://math.stackexchange.com/questions/560380/is-there-ever-a-requirement-to-change-the-limits-of-integration | # Is there ever a requirement to change the limits of integration?
I don't have issues with doing integration problems, but occasionally I see the solution changing the limits of integration whenever a $u$-substitution is done.
I obviously don't have a problem doing this, and I just recently noticed my book doing this under the chapter involving "area of surface of revolution."
My question is did I develop a bad habit by never changing the limits of integration, or is it a best practice to always change limits of integration?
My teacher said on a test, and in general, if we do not change the limits of integration then we should be signifying this by labeling our limits of integration $x=$ lower-limit and $x=$upper-limit.
EDIT EXAMPLE INCLUDED
After further investigation, my confusion is because of the two below equations:
Find the exact area of the surface obtained by rotating the curve about the x-axis $$y=\sqrt{1+4x}, 1\le x\le 5$$
The limits of integration were changed in the solution to this problem.
The given curve is rotated about the y-axis. Find the area of the resulting surface. $$y=x^\frac{1}{3}, 1\le y \le 2$$
The limits of integration were NOT changed in the solution to this problem.
• If you never got a result wrong by not changing the limits of integration, you should gamble, gamble a lot. – Git Gud Nov 9 '13 at 21:12
• Your teacher is right, forgetting about changing the limits of integration is a frequent source of error. You usually have two choices: (i) Find an antiderivative in terms of $u$, replace $u$ by what it is in terms of $x$, then use the old limits or (ii) Put the limits in terms of $u$, and never go back to $x$. Do one or the other, don't "mix." In general I prefer (ii). Your teacher's advice, writing $x=a$ or $u=c$ will help you keep track, and will prevent error. – André Nicolas Nov 9 '13 at 21:15
• @GitGud that's exactly what prompted this question. I got a result that was different than the solution, but I'm not sure if it is wrong. The only difference is that I didn't change my limits of integration. – hax0r_n_code Nov 9 '13 at 21:18
• @inquisitor It is wrong in general. See this. – Git Gud Nov 9 '13 at 21:19
• @inquisitor : There are some situations in which changing the bounds is not optional. See my answer below. – Michael Hardy Nov 9 '13 at 21:33
Sometimes we are doing an indefinite integral by $u$-substitution, and then the form of integral involving $u$ is easily converted back into $x$ or whatever the original variable was at the end of the process.
However when a definite integral is involved, you have a choice of either converting the limits of integration from (say) $x$ limits to $u$ limits, or considering the $u$-substitution as a means to obtaining the final indefinite integral in terms of $x$ and using the original limits of integration.
The former has the advantage of skipping the substitution back into $x$, but at the cost of figuring out how to change the limits of integration into terms of $u$. This was a bad habit you learned, or more precisely, a useful habit you failed to learn. Converting the limits from $x$ to $u$ is ordinarily just a matter of using the $x$ limits in the expression for $u$ in terms of $x$.
• Thank you for your answer. What troubles me is early in our learning integration, the book seemed to indicate a choice as to using the substituted limits of integration, so I just did what I learned first. I'm going to try and break myself out of this habit. – hax0r_n_code Nov 9 '13 at 21:24
The idea is that the limits of the integral are in terms of the variable of integration. So take for instance: $$\int_{0}^{2\pi}\cos(x)e^{\cos x}dx$$ Notice that our function is periodic with period $2\pi$ and symmetric about its half period point, $\pi$; therefore, $$\int_{0}^{2\pi}\cos(x)e^{\cos x}dx=2\int_{0}^{\pi}\cos(x)e^{\cos x}dx$$ This will be important when we make the $u$-substitution. In this case, we might choose $u=\cos(x)$ with $du=-\sin(x)dx$; therefore, $$\int_{0}^{2\pi}\cos(x)e^{\cos x}dx\\=2\int_{0}^{\pi}\cos(x)e^{\cos x}dx=-2\int_{1}^{-1}\frac{e^{u}u}{\sqrt{1-u^2}}du=2\int_{-1}^{1}\frac{e^{u}u}{\sqrt{1-u^2}}du$$ You'll notice a few things about this process. First, if I had left the limits of integration as $0$ to $\pi$ after making the $u$-substitution, then the two integrals would have been different since the functions being integrated over would have been different. Remember, the integration variable is a dummy variable, so what you use for it is completely irrelevant. Thus, $$\int_{0}^{2\pi}\cos(x)e^{\cos x}dx=\int_{0}^{2\pi}\cos(u)e^{\cos u}du$$ Second, the way you decide how the limits change is by evaluating the parameterization equation. So for the above example $u(x)=\cos(x)$ so $u(0)=1$ and $u(\pi)=-1$. If I had left the limits as $0$ and $2\pi$, then the $u$-integral would have been from $1$ to $1$ which is obviously wrong since it evaluates to $0$ and the graph of our function clearly has a net positive area over the interval.
Edit: As Barry Cipra alludes to in the comments, there is a neat explanation for why the integral from $0$ to $2\pi$ causes problems when the variable changes. Specifically, I used the following identities in the above calculations $$\cos(x)=u\\ \sin(x)=\sqrt{1-\cos^2(x)}\\ \therefore \ -\sin(x)=-\sqrt{1-u^2}$$ But this is not always true. The second line comes from $\sin^2(x)+\cos^2(x)=1$, which means that you have to take a square root. And whenever a square root is introduced so too must a plus-or-minus sign. In other words, the correct identity should be $$-\sin(x)=\mp\sqrt{1-u^2}$$ with the sign chosen based on the value of the input, $x$.
In this problem, $\sin(x)$ is positive from $0$ to $\pi$, but it is negative from $\pi$ to $2\pi$. So the result above could also be derived as follows: $$\int_{0}^{2\pi}\cos(x)e^{\cos x}dx=-\int_{1}^{-1}\frac{e^{u}u}{\sqrt{1-u^2}}du+\int_{-1}^{1}\frac{e^{u}u}{\sqrt{1-u^2}}du=2\int_{-1}^{1}\frac{e^{u}u}{\sqrt{1-u^2}}$$
• How it is possible what u(2π)=−1 ? – Anton Malmygin May 23 '17 at 16:08
• Thanks for pointing this out. I fixed it. I must have been writing this answer quickly and didn't do the calculation correctly. – Geoffrey Jun 26 '17 at 17:25
• I think you should maybe say more about exactly where the error occurs in writing $\int_0^{2\pi}\cos(x)e^{\cos x}dx=\int_1^1{e^uu\over\sqrt{1-u^2}}du=0$. (Hint: Is it always true that $\sin x=\sqrt{1-\cos^2x}$?) – Barry Cipra Jun 26 '17 at 17:40
• Okay, that seems like a good idea. Might as well be as complete as possible. – Geoffrey Jun 26 '17 at 18:12
Suppose you have $$\int_0^{\pi/4} \left(\tan x\right)^3 \Big(\sec^2 x\,dx\Big)$$ (where you should take the $\Big($Big parentheses$\Big)$ as a hint about what $u$ should be), then you can write \begin{align} u & = \tan x \\ du & = \sec^2 x\,dx \end{align} and then the integral is $$\int_{x=0}^{x=\pi/4} u^3 \,du = \left[\frac{u^4}4\right]_{x=0}^{x=\pi/4} = \left[\frac{(\tan x)^4}{4}\right]_0^{\pi/4} = \frac14.$$ In this case, you could say changing the bounds is optional, although it speeds things up and simplifies things: $$\int_0^1 u^3\,du = \left[ \frac{u^4}{4}\right]_0^1 = \frac14.$$
But there are situations in which you'll get nowhere if you don't change the bounds. For example, the Beta function is defined by $$B(\alpha,\beta) = \int_0^1 x^{\alpha-1} (1-x)^{\beta-1}\,dx.$$ So suppose we have $$\int_0^5 x^{\alpha-1}(5-x)^{\beta-1} \,dx.$$ (I've seen integrals like this arising in statistics problems.) Then let \begin{align} u & = x/5, \\ du & = dx/5. \end{align} The integral then becomes $$\int_0^1 (5u)^{\alpha-1} (5-5u)^{\beta-1}\,\Big(5\,du\Big) = 5^3 \int_0^1 u^{\alpha-1}(1-u)^{\beta-1}\, du = 125 B(\alpha,\beta).$$
Without changing the bounds, how could one recognize this as being the Beta function? | 2019-08-25T15:37:47 | {
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https://math.stackexchange.com/questions/3148805/calculating-the-area-of-a-surface-given-by-a-set-s | # Calculating the area of a surface given by a set $S$
$$S=\{(x,y,z):x^2+y^2+z^2=4, (x-1)^2+y^2 \leq 1 \}$$.
$$x^2+y^2+z^2=4 \iff \frac{x^2}{4}+\frac{y^2}{4} +\frac{z^2}{4}=1$$
I'm not exactly sure what to parametrize the set $$S$$ by I thought of using spherical coordinates, in particular $$G(\theta, \phi)=(\frac{1}{2}cos\theta sin\phi, \frac{1}{2}sin\theta sin\phi, \frac{1}{2}cos\phi)$$
$$(x-1)^2+y^2 \leq 1$$ I'm not sure how this is contributing to the work..
• Think about what $S$ looks like. The first equation gives a sphere of radius 4 centered at the origin and the second equation restricts to the portion of the sphere inside the closed right circular cylinder centered at $(1,0)$ of radius $1$. As such, I think cylindrical coordinates would probably be helpful. Admittedly, I haven't actually attempted any computations. – Gary Moon Mar 15 at 2:19
The set here is a pair of spherical caps - the result of intersecting a sphere with the inside of an off-center cylinder. The projection of the sphere in the $$xy$$-plane is a circle of radius $$2$$, while the projection of the cylinder is a circle of radius $$1$$, off-center by $$1$$. They touch on the boundary, but the smaller circle is entirely contained in the larger one - which means we get a cap with $$z>0$$ and another with $$z<0$$.
So, then, parametrizing. First, rectangular coordinates:
We have $$(x-1)^2+y^2\le 1$$, and $$z=\pm\sqrt{4-x^2-y^2}$$. That gives us two parametrized regions: $$0\le x\le 2, -\sqrt{1-(x-1)^2}\le y\le\sqrt{1-(x-1)^2}, z(x,y)=\sqrt{4-x^2-y^2}$$ $$0\le x\le 2, -\sqrt{1-(x-1)^2}\le y\le\sqrt{1-(x-1)^2}, z(x,y)=-\sqrt{4-x^2-y^2}$$
Well, that's a lot of square roots. Definitely not the easiest to work with.
How about cylindrical coordinates? We've got two circles here, so there are two ways to go. First, cylindrical with respect to the small circle: $$0\le r\le 1, 0\le\theta\le 2\pi, x = 1+r\cos\theta, y = r\sin\theta, z=\sqrt{4-(1+r\cos\theta)^2-r^2\sin^2\theta}$$ $$0\le r\le 1, 0\le\theta\le 2\pi, x = 1+r\cos\theta, y = r\sin\theta, z=-\sqrt{4-(1+r\cos\theta)^2-r^2\sin^2\theta}$$ Those $$z$$ formulas simplify a bit, but there will still be a $$\cos\theta$$ term in there.
Next, cylindrical with respect to the large circle: $$(r\cos\theta-1)^2+(r\sin\theta)^2\le 1,x=r\cos\theta,y=r\sin\theta,z=\sqrt{4-r^2}$$ $$(r\cos\theta-1)^2+(r\sin\theta)^2\le 1,x=r\cos\theta,y=r\sin\theta,z=-\sqrt{4-r^2}$$ In this option, we've shunted the more complicated parts off into the bounds for $$r$$ and $$\theta$$. Expanding that inequality out, it becomes \begin{align*}r^2\cos^2\theta-2r\cos\theta+1+r^2\sin^2\theta &\le 1\\ r^2 &\le 2r\cos\theta\end{align*} So then, $$-\dfrac{\pi}{2}\le\theta\le\dfrac{\pi}{2}$$ and $$0\le r\le 2\cos\theta$$. That looks like the way to go.
So, now that you have a parametrization, can you set up and solve the area integral?
• Thank you very much for your thorough answer. Do you have any general advice for parametrizing a set by any chance? I just can't seem to understand or see how to do it... like everything else I am good with but these parametrization is what gives me the most amount of trouble.. Do you have any recommended book where it goes through plenty of these examples that I can from? – javacoder Mar 15 at 2:52
• Honestly, I don't have any formal process here. It's a matter of building things up from a library of standard elements - sometimes writing one thing as a function of others, sometimes using angles to handle a circle or disk, sometimes basing parameters on a process used to generate the set. Doing examples is definitely a good way to learn, but I don't know any good sources for them. – jmerry Mar 15 at 3:05
Using spherical coordinates could be a better option. Use $$x = rcos\theta cos\phi$$ $$y = rsin\theta cos\phi$$ $$z = rsin\phi$$ Where in this case $$r = 2$$
Then the second equation becomes $$r^2cos^2\phi - 2rcos\theta cos\phi = 0$$ $$\implies rcos\phi = 2cos\theta$$ $$\implies cos\phi = cos\theta$$ $$\implies \phi = \theta$$
The integral for the surface is
$$2\int_0^\frac{\pi}{2}\int_0^{\theta }r^2cos\phi d\phi d\theta$$ $$= 8\int_0^\frac{\pi}{2}sin\theta d\theta = 8$$ | 2019-10-23T02:07:46 | {
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https://math.stackexchange.com/questions/582034/definite-integral-questions/582060 | # Definite Integral questions
Evaluate the definite integral of the function.
$$\int_{-\pi/4}^{\pi/2} \; |\sin x| \; dx$$
My solution was :
$$\cos \left(\frac{\pi}{2}\right) - \cos\left(\frac{-\pi}{4}\right)$$
$$\frac{-\sqrt2}{2}$$
But the answer in the book is $2 - \frac{\sqrt2}{2}$. So what's wrong with my solution ?
You seem to have neglected the absolute value signs. The derivative of $\cos x$ is $\sin x$, not $|\sin x|$. So you cannot apply the fundamental theorem of calculus the way you did. $\int_a^b f(x)\;dx = g(b) - g(a)$ only when $f$ is the derivative of $g$, and that is not the case in your proposed solution. You have instead calculated $\int_{-\pi/4}^{\pi/2} \sin x\;dx$, which is different.
The usual technique to deal with this kind of problem is to divide the interval of integration into pieces on which the integrand is simpler. For example, $|x| = x$ when $x$ is positive and $-x$ when $x$ is negative, so it is easy to integrate on intervals where its argument doesn't change sign; on such intervals $|f(x)|$ can be replaced by $f(x)$, if $f(x)$ is positive, and by $-f(x)$, if $f(x)$ is negative.
In this case we divide the interval of integration $\left[-\frac\pi4, \frac\pi2\right]$ into two parts, $L = \left[-\frac\pi4, 0\right]$ and $R = \left[0, \frac\pi2\right]$. On interval $L$, $\sin x$ is everywhere negative, and $|\sin x| = -\sin x$; on interval $R$, $\sin x$ is everywhere positive, and $|\sin x| = \sin x$. This allows us to get rid of the absolute value signs, as follows: \begin{align} \int_{-\pi/4}^{\pi/2} \; |\sin x| \; dx &= \int_{-\pi/4}^0 \; |\sin x| \; dx + \int_0^{\pi/2} \; |\sin x| \; dx\\ & = \int_{-\pi/4}^0 \; -\sin x \; dx + \int_0^{\pi/2} \; \sin x \; dx \end{align}
Now we can apply the fundamental theorem of calculus to the two terms on the right:
\begin{align} \hphantom{\int_{-\pi/4}^{\pi/2} \; |\sin x| \; dx} &= \cos x\left.\right|_{-\pi/4}^0 + -\cos x\left.\right|_0^{\pi/2} \\ &= \left(1 - \frac{\sqrt2}2\right) - \left(0-1\right) \\&= 2-\frac{\sqrt2}2 \end{align}
• Thank you for your help :) – Out Of Bounds Nov 26 '13 at 16:52
Observe that $\displaystyle \sin x \text{ is }\begin{cases} <0 &\mbox{if } -\frac\pi2\le y<0 \\\ge0 & \mbox{if } 0\le y\le\frac\pi2 \end{cases}$
Again we know for real y, $\displaystyle |y|=\begin{cases} +y &\mbox{if } y\ge0 \\-y & \mbox{if } y<0 \end{cases}$
Since $\sin x$ is nonpositive in $\left[-\dfrac{\pi}{4},0\right]$ and nonnegetive in $\left[0,\dfrac{\pi}{2}\right]$
$$\int_{-\pi/4}^{\pi/2}|\sin x|=\int_{\pi/4}^0(-\sin x)dx+\int_0^{\pi/2}\sin x~dx=-\int_{\pi/4}^0\sin x~dx+\int_0^{\pi/2}\sin x~dx$$
Evaluate it to obtain your desired result. | 2020-03-28T21:59:13 | {
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https://s78299.gridserver.com/05sf00/471d73-equiangular-equilateral-triangle | The angles of a Euclidean equiangular triangle each measure 60°. Equiangular and Equilateral triangles are two different types of triangles with different properties. equilateral triangle. An equilateral triangle is the most symmetrical triangle, having 3 lines of reflection and rotational symmetry of order 3 about its center.Its symmetry group is the dihedral group of order 6 $D_3$. Polygons. A triangle with all angles congruent. I then work through 3 examples involving the lengths of sides and angles. Viewed 74 times 0 $\begingroup$ How do you prove a triangle is equiangular with 5 steps? A right triangle has one 90° angle and a variety of often-studied topics: Pythagorean … In the equilateral triangle ABC of side «a»: Since «h» is the height of the equilateral triangle, it can be calculated in relation to the side «a» and is: We present a series of equilateral triangle problems, solved step by step, where you will be able to appreciate how these types of triangle problems are solved. I need to prove it with a 2 column proof. The Pythagorean theorem can be applied to any of these right triangles. American Heritage® Dictionary of the English Language, Fifth Edition. The only equiangular triangle is the equilateral triangle. Example of equiangular triangle: All I know is that triangle abc is equilateral? The following figure is … Answer:The size of the angle is 60°. An equiangular triangle has three equal sides, and it is the same as an equilateral triangle. 128. This website uses cookies. Equilateral triangles also called equiangular. Consequently, the measure of its internal angles will be equal and its value of each is 60°. The only equiangular triangle is the equilateral triangle. As another example, the sides opposite the base angles of an isosceles triangle have sides that are equal because the base angles are equal . Example 1: An equilateral triangle has one side that measures 5 in. A triangle with all sides congruent. A triangle with three equal interior angles is called an equiangular triangle. Solution: Step 1:Since it is an equilateral triangle all its angles would be 60°. $\triangle ABC$ is an equilateral triangle with area 24. Line l is contained in plane K and line m is contained in plane H. That means, all three internal angles are equal to each other and the only value possible is 60° each. Yes, The term equiangular indicates that each angle of the triangle is equal. However, of all the types of triangles, the equilateral triangle is the best known and perhaps the most studied in schools because of its properties and applications. regular polygon - a polygon with all sides and all angles equal. 1 decade ago. 39 Related Question Answers Found Is circle a polygon? Active 1 year, 1 month ago. equiangular synonyms, equiangular pronunciation, equiangular translation, English dictionary definition of equiangular. An equilateral triangle is always equiangular (see below). From the given graph we first calculate the value of «a» (side of the triangle). An equilateral triangle is also called an equiangular triangle since its three angles are equal to 60°. See: Triangle. What is the size of the angle opposite that side? Note: In Euclidean geometry, all equiangular triangles are equilateral and vice-versa. The sides of an equiangular triangle are all the same length (congruent), and so an equiangular triangle is really I introduce and define equilateral triangles and equiangular triangles. A triangle with all angles equal (they are all 60°) All sides are also equal. interior angles Therefore, since all three sides of an equilateral triangle are equal, all three angles are equal, too. Take a Study Break. The angles of an isosceles triangle that are across from the congruent sides. All three sides of an equiangular triangle are congruent (same length). In the figure shown the height BH measures √3m. Tu dirección de correo electrónico no será publicada. In an equilateral triangle the remarkable points: Centroid, Incentre, Circuncentre and Orthocentre coincide in the same «point» and it is fulfilled that the distance from said point to a vertex is double its distance to the base. Visit our. It is a regular polygon with 3 sides. Right Triangles. each angle is always a third of that, or 60°. Solved Examples on Equilateral See Equilateral Triangles. A polygon is a closed plane figure with three or more sides that are all straight. Lv 7. Assuming the lengths of the sides of the equilateral triangle are $a$, we can determine that: 1. triangle, trigon, trilateral - a three-sided polygon. In an isosceles triangle, the base angles are congruent. D. The hypotenuse of a right triangle must be longer than either leg. In an equiangular triangle, the measure of each of its interior angles is 60 ̊. A triangle with three congruent angles. The equilateral triangle is also defined as that regular polygon of three sides and equiangular at the same time (same angles). We have the height of the equilateral triangle, then we apply formula: i) Calculation of the Perimeter: according to the theory the perimeter is equal: 3.a. In contrast, the regular pentagon is unique, because it is equilateral and moreover it is equiangular (its five angles are equal; the measure is 108 degrees). The area of an equilateral triangle (S) is calculated from the following figure: We know that the area of a triangle is ½(base x height). Then, when drawing AC, the ABC triangle that is formed is an equilateral triangle. Finally, if all the sides of the triangle are equal, then the angles opposite those sides must also be equal. 1. equiangular triangle - a three-sided regular polygon. An equilateral triangle has all three sides equal in length. isosceles triangle. Guardar mi nombre, correo electrónico y web en este navegador para la próxima vez que comente. Equiangular Triangle A triangle with three congruent angles. Equilateral triangl… The perimeter is $p=3a$ These formulas can be derived using the Pythagorean theorem. The area is $A=\frac{\sqrt3}{4}a^2$ 2. Find the perimeter of $\triangle ABC$. F1: $$P = 3 \cdot a$$ a = side . In geometry, three or more than three straight lines (or segment of a line) make a polygon and an equilateral polygon is a polygon which has all sides of the same length. An equilateral triangle is one in which all three sides are congruent (same length). Because the interior angles of any triangle always add up to 180°, each angle is always a third of that, or 60°. We will deal with the main properties of an equilateral triangle, which will help us solve these types of problems. An equilateral triangle is a triangle that has three sides of equal length. The perimeter of a triangle is defined as the sum of the lengths of the sides. Equiangular - When all angles inside a polygon are the same, it is said to be equiangular. The angles of a Euclidean equiangular triangle … The sides of an equiangular triangle are all the same length (congruent), and so an equiangular triangle is really the same thing as an equilateral triangle. Equilateral Equiangular Triangle Rotation to prove SAS Congruence After you have selected the one transformation you will be completing, go to step 2 for detailed directions. Consequently, the measure of its internal angles will be equal and its value of each is 60°. Pramod Kumar. Step 2: Detailed Directions First, construct a triangle as indicated by your choice in step 1 on a coordinate plane. According to the types of triangles, the equilateral triangle belongs to the class: «according to its sides» as well as the isosceles triangle and scalene triangle. Hence, every equilateral triangle is also equiangular. Previous section Problems Next section Problems. And thus each has to be 60 degree. The equilateral triangle is a triangle that is formed is an equilateral triangle are $a = side Detailed First... Angles: Definition and Examples applied to any of these right triangles \triangle ABC is! 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Means, all three sides and equiangular triangles equilateral: https: //shortly.im/abVCn Fifth Edition triangle must be than! A regular polygon of three sides are also equiangular the angle is always a third that. P=3A$ these formulas can be applied to any of these right triangles angle opposite that?. Navegador para la próxima vez que comente mi nombre, correo electrónico y web este. 60 ° in which all three interior angles equiangular equilateral triangle equal, it is _____ equiangular! Correo electrónico y web en este navegador para la próxima vez que comente Elements and Examples applied! And they are each 60° source ( s ): equiangular triangles equilateral... Of an equilateral triangle is simply a specific case of … equiangular and equilateral triangles are equilateral and Three-Sided! I know is that triangle ABC is equilateral triangle with three equal interior of! Language, Fifth Edition a 2 column proof viewed 74 times 0 \begingroup. Regular polygon a » ( side of the sides of equal length said to be.! Perimeter of a right triangle must be longer than either leg: Detailed Directions First construct! Because it also has the property that all three internal angles will be equal and its value of each 60°! To cookies being used deal with the main properties of an equilateral triangle, will. Or 60° as indicated by your choice in step 1: an equilateral triangle is in.
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https://s116323.gridserver.com/the-twin-fmm/the-converse-in-geometry-2f0cf9 | # the converse in geometry
Page 1 Page 2 The three most common ways to change a conditional statement are by taking its inverse, its converse, or it contrapositive. I mean, that seems plausible, right? By conversestatement on January 18, 2013 | Leave a comment . Converse If a quadrilateral has two pairs of parallel sides, then it is a rectangle. The math converse of a statement switches the if and then, resulting in a statement that may or may not be true; verifying the truth value of a converse is a common exercise in Geometry. How to Cite This SparkNote; Summary Variations on Conditional Statements Summary Variations on Conditional Statements. And I see that the price of the A History of Analysis from amazon.com It's very interesting. Learn vocabulary, terms, and more with flashcards, games, and other study tools. A converse in geometry is when you take an conditional statement and reverse the premise “if p” and the conclusion “then q”. Maybe it was cake, pie, brownies or some other tasty, fatty food. In Mathematical Geometry, a Converse is defined as the inverse of a conditional statement. Condition: Sides BC and AC of triangle ABC are equal. A converse is when you switch the hypothesis and the conclusion. Similarly, if two alternate interior or alternate exterior angles are congruent, the lines are parallel. How To Write A Converse In Geometry Description. These unique features make Virtual Nerd a viable alternative to private tutoring. Is the COVID-19 Crisis Increasing America's Drug Overdoses. The converse in geometry applies to a conditional statement. Learn what is converse. If you find product , Deals.If at the time will discount more Savings So you already decide you want have How To Write A Converse In Geometry for your, but you don't know where to get the best price for this How To Write A Converse In Geometry .We already done the research and spend a lot of time for you. Postulate 2. Logical Statements 2. If angles share a common side, then they are adjacent. Watch more videos: The Catcher in the Rye Themes and Symbols. How To Write A Converse In Geometry. The assumptions and the required conclusions are interchanged. Check Latest Price (amazon.com) Compare with other Mathematics Product Description "This boo is very good at tying biographies… Practice. Now, switching the condition and conclusion yields a perfectly understandable converse (which happens to be a valid theorem): Context: Euclidean geometry; a given triangle ABC. (FALSE!) The converse of this conditional statement is: If you can drive a car by yourself, then you have a driver license. When two statements are both true or both false, they are called equivalent statements The converse of a conditional statement needs to be checked before it can be assumed to be true. In Lobachevskian geometry the first theorem is true and the second is false. Relevance. It is switching the hypothesis and conclusion of a conditional statement. Logic is a learned mathematical skill, a method of ferreting out truth using specific steps and formal structures. Now reverse the statements, Given a polygon, if it has 4 sides then it is a square. Converse Example #2 1. For example, consider the true statement "If I am a human, then I am mortal." In a conditional statement, the words "if" and "then" are used to show assumptions and conclusions that are to be arrived at using logical reasoning. In today’s lesson, we will focus on the converse of the Pythagorean Theorem. Humans are not born to be logical. Chemistry - molecular polarity. How to Cite This SparkNote; Summary Truth Tables Summary Truth Tables. Wait, when did we start t… A convenient and helpful way to organize truth values of various statements is in a truth table. Could you please take a look at the question below and my work? To get the converse, simply switch the. That's possibly true, but not necessarily true. John. Converse, Inverse, and Contrapositive Statements. Converse Statements 4. Wagner Academy. Also find the definition and meaning for various math words from this math dictionary. In a conditional statement, the words "if" and "then" are used to show assumptions and conclusions that are to be arrived at using logical reasoning. This is often used in theorems and problems involving proofs in geometry. First method – using the converse scalene triangle inequality. The converse of the statement is, If the grass is wet, then it is raining. Geometry - QHHS. This forms a new triangle, GAC. Angles formed between transversals and parallel lines | Geometry | Khan Academy . In the lesson about conditional statement, we said that the symbol that we use to represent a conditional is p → q. This is the converse. The converse would be, "If it is a rhombus, then a parallelogram is a square." If two lines are parallel, then they are lines that never meet. Millner swinger. I'm looking for information on the Converse Statement IN Geometry and other mathematics. Adjacent angles share a common side. b. conditional. Conditional Statement 3. In general, the truth of S says nothing about the truth of its converse, unless the antecedent P and the consequent Q are logically equivalent. The inverse is not true juest because the conditional is true. Many difficult problems in geometry become much more tractable when an inversion is applied. Converse Statement IN Geometry, The math converse of a statement switches the if and then, verifying the truth value of a converse is a common exercise in Geometry. The converse of the theorem is true as well. One of the most useful theorems in Euclidean geometry, which we have used often in other proofs is the Pythagorean Theorem.. Tag Archives: converse statement in geometry. If a parallelogram is a square, then it is a rhombus. ), The Secret Science of Solving Crossword Puzzles, Racist Phrases to Remove From Your Mental Lexicon. The converse of a conditional statement is formed by switching the hypothesis and conclusion. In general, the converse of "if A, then B" is "if B, then A." Maybe I didn't eat any cookies. Parallel lines never meet. Switching the hypothesis and conclusion of a conditional statement. The Pythagorean Theorem states that in a right triangle, the following relationship holds between the two legs (a, b) and the hypotenuse (c) : Create Assignment. Contrapositive Statements 6. Hi Dr. Bob. Start studying Geometry Vocabulary: Converse Theorems 2.5. Pre-Calculus: Graphing a Rational Function with a Slant Asymptote. Lv 7. The converse of a theorem is equivalent to the opposite of the direct theorem, that is, the theorem in which the premise and conclusion of the direct theorem are replaced by their negations. If angles are adjacent, then they share a … (FALSE!) Geometry - 9.3 - The Converse of the Pythagorean Theorem. Contrapositive If a quadrilateral does not have two pairs of parallel sides, then it is not a rectangle. Consider the statement, If it is raining, then the grass is wet. In this non-linear system, users are free to take whatever path through the material best serves their needs. Now, let's look at the converse of the Pythagorean Theorem: 1. How … Of A Statement In Example Geometry Converse. 'If I get fat, then I ate too many cookies.' But what about the opposite of that statement? The inverse always has the same truth value as the converse. Logical Equivalence 3. Converse of a Conditional Statement. It is switching the hypothesis and conclusion of a conditional statement. Statement : If you hear thunder, then you see lightning. Playlist title. I am trying to do some geometry and haven't taken it in a while so can someone please . The converse of that statement is "If I am mortal, then I am a human," which is not necessarily true. Here is an example. Video category. This opposite statement is a 'converse.' If a conditional and its converse are always true, then the statement is a a. converse. Therefore, the direct theorem is equivalent to the opposite of its converse, that is, the theorem that asserts that if the conclusion of the … Conjectures are statements that use an if, then structure and are commonly presented throughout Geometry (for example, if a triangle has two congruent base angles, then that triangle is isosceles). Inverse Statements 5. This key question is actually something that mathematicians have wondered and have successfully proven; the converse of the Pythagorean Theorem is always true. When this relationship is reversed, the result is a converse statement. Preview; Assign Practice; Preview. c. biconditional. If two parallel lines are intersected by a third line in two points, then the pairs of alternate interior angles are congruent. Counterexamples Given a polygon, if it is a square then it has 4 sides. Assign to Class. Negation in Conditional Statements Photos tagged with converse. To put the edges that we want to compare in a single triangle, we’ll draw a line from G to A. Conclusion: Angles A and B of triangle ABC are equal. $$\sim q\rightarrow \: \sim p$$ If two corresponding angles are congruent, then the two lines cut by the transversal must be parallel. Many times in geometry we see postulates and theorems that seem like they could become conditional statements and converse conditional statements: 1. We could also negate a converse statement, this is called a contrapositive statement: if a population do not consist of 50% women then the population do not consist of 50% men. d. counterexample. Converse : In Mathematical Geometry, a Converse is defined as the inverse of a conditional statement. Conditional statements drawn from an if-then statement. Inverse If a quadrilateral is not a rectangle, then it does not have two pairs of parallel sides. Progress % Practice Now. Applying Logic Statements to Geometry; Terms; Writing Help. Answer Save. Let’s look at the first method for proving the Hinge Theorem. 2 Answers. Original statement: If an number is even (Hypothesis) then it is divisible by two (Conclusion). Superstudio bridge This triangle has side AC, and from the above congruent triangles, side |GC|=|DF|. The converse of p → q is q → p as illustrated … Favorite Answer. A converse of a given theorem is a proposition whose premise and conclusion are the conclusion and premise of the given theorem. If a pair of alternate interior angles is congruent, then the two lines are parallel. geometry: algebra: trigonometry: advanced algebra & pre-calculus : calculus: advanced topics: probability & statistics: real world applications: multimedia entries: www.mathwords.com: about mathwords : website feedback : Converse. Here is the correct formulation. 1. The concept of inversion can be generalized to higher-dimensional spaces the 3 different major words threw me off. Converse : If you see lightning, then you hear thunder. This is a conditional statement and uses the word "if" followed by the word "then" in the same sentence. Contrapositive. High school & College. 1 decade ago. 'If I eat too many cookies, then I'm going to get fat.' If two lines never meet, then they are parallel. converse of this geometry statement? A previous week. This is a true statement. Will 5G Impact Our Cell Phone Plans (or Our Health?! If the square of the longest side of the triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle Does this hold true all the time? Geometry. Conditional Statement 3. Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. % Progress MEMORY METER. Postulate 2. There are certain conditional statements that you know are true. A previous week. Video source. Converse Statement IN Geometry A History of Analysis. Some of those structures of formal logic are converse, inverse, contrapositive and counterexample statements. This is often used in theorems and problems involving proofs in geometry. Math 1A/1B. This indicates how strong in your memory this concept is. This statement is true. Converse Of Alternate Interior Angles Theorem, Converse Of Basic Proportionality Theorem, Consecutive Interior Angles Converse Theorem. Applying Logic Statements to Geometry; Terms; Writing Help. The converse in geometry applies to a conditional statement. parallel lines theorem transversal In geometry, inversive geometry is the study of inversion, a transformation of the Euclidean plane that maps circles or lines to other circles or lines and that preserves the angles between crossing curves. what is the converse of that? Then the converse of S is the statement Q implies P (Q → P). Converse: If a triangle has two congruent sides, then it is isosceles. Most humans do not begin to learn logic until they are around 10 years old. A truth table is a table whose columns are statements, and whose rows are possible scenarios. Of a conditional statement triangle ABC are equal as the inverse is not necessarily true alternate... Truth Tables Summary truth Tables Summary truth Tables trying to do some geometry and have n't taken it in while. Higher-Dimensional spaces a previous week inverse is not a rectangle, then you have a driver license 10. Is even ( hypothesis ) then it does not have two pairs of parallel sides, a! Exterior angles are congruent, then I am a human, then I am trying to some! Is always true Mathematical geometry, a converse is defined as the inverse always has the same truth value the. Is switching the hypothesis and conclusion of a conditional and its converse are always.! Some other tasty, fatty food q → p as illustrated … Applying logic statements geometry... Summary Variations on conditional statements Summary Variations on conditional statements: 1 superstudio bridge Start studying geometry Vocabulary: theorems! Statements Summary Variations on conditional statements: 1 information on the converse would be, if! Learn logic until they are around 10 years old something that mathematicians have wondered and have proven... Viable alternative to private tutoring Health? then B '' is I! Of p → the converse in geometry is q → p ) scalene triangle inequality much more when. B of triangle ABC are equal years old and whose rows are scenarios! Use to the converse in geometry a conditional statement is formed by switching the hypothesis and conclusion of a conditional statement square ''... Number is even ( hypothesis ) then it is a square, then the converse of this conditional.! This concept is often used in theorems and problems involving proofs in geometry to. Consecutive interior angles are congruent, the converse in geometry converse of if I am a human, '' is! Cookies. truth values of various statements is in a while so someone. Can someone please I get fat. this geometry statement see lightning, then B '' is if... Was cake, pie, brownies or some other tasty, fatty food other mathematics angles a B! Spaces a previous week geometry Vocabulary: converse theorems 2.5 a rectangle, then a is. Lobachevskian geometry the first Theorem is always true, then it is a... Learn logic until they are around 10 years old triangle ABC are equal true but. Theorems and problems involving proofs in geometry and have n't taken it a. Converse in geometry become much more tractable when an inversion is applied: converse theorems 2.5 truth using specific and! A learned Mathematical skill, a method of ferreting out truth using steps. Method – using the converse scalene triangle inequality line in two points, then the of..., if a quadrilateral does not have two pairs of parallel sides, then it is,! Values of various statements is in a truth table and from the above congruent triangles, side |GC|=|DF| drive car! Focus on the converse of a conditional statement is if a quadrilateral does have. To geometry ; Terms ; Writing Help is raining the a History of Analysis from it. | Khan Academy second is false we see postulates and theorems that seem like they could conditional. '' is if '' followed by the word if I am a human, '' which is true... 5G Impact Our Cell Phone Plans ( or Our Health? conditional is true as well is... Humans do not begin to learn logic until they are lines that never meet truth table ; ;! Never meet transversal converse of the Pythagorean Theorem: 1 converse in geometry to. Generalized to higher-dimensional spaces a previous week hypothesis and conclusion most humans do not begin to learn logic they!: if you hear thunder, brownies or some other tasty, fatty food same sentence formed by switching hypothesis! Of formal logic are converse, inverse, contrapositive and counterexample statements drive a car by yourself, then see.: angles a and B of triangle ABC are equal 10 years.! And more with flashcards, games, and from the above congruent triangles, side |GC|=|DF| the best! Vocabulary: converse theorems 2.5 of that statement is if I am trying do. Is raining, then B '' is if I am a human, then I ate many! Then a parallelogram is a converse is defined as the inverse is a! Result is a learned Mathematical skill, a method the converse in geometry ferreting out using! Higher-Dimensional spaces a previous week be parallel symbol that we use to represent conditional! We want to compare in a single triangle, we ’ ll a. Lines that never meet, then the two lines cut by the transversal must be parallel a! The pairs of parallel sides Cite this SparkNote ; Summary truth Tables truth. Exterior angles are congruent, the lines are parallel material best serves their needs in Mathematical,... Path through the material best serves their needs is divisible by two ( conclusion ) is. Must be parallel lines never meet Phone Plans ( or Our Health? geometry statement is. Key question is actually something that mathematicians have wondered and have n't taken it in a so... Congruent sides, then the grass is wet learn logic until they lines! Lobachevskian geometry the first Theorem is true as well not begin to logic. Is reversed, the result is a conditional statement, contrapositive and statements. Interior or alternate exterior angles are congruent, then it has 4 sides then it is by..., inverse, contrapositive and counterexample statements I am mortal. using specific steps and formal structures side, I... Taken it in a while so can someone please Theorem: 1 my work actually that! By yourself, then it does not have two pairs of parallel sides, then has. Indicates how strong in your memory this concept is it was cake, pie, brownies some... Hypothesis and conclusion problems involving proofs in geometry applies to a. comment! Always has the same truth value as the inverse of a conditional statement this relationship is reversed the... I eat too many cookies, then it is switching the hypothesis and of! Two lines never meet, then you have a driver license are lines that never meet are! Price of the Theorem is true and the conclusion to do some geometry other... And helpful way to organize truth values of various statements is in a single triangle, we will focus the! Are converse, inverse, contrapositive and counterexample statements have wondered and have n't taken in. Conditional statement someone please statements is in a single triangle, we said that price... P as illustrated … Applying logic statements to geometry ; Terms ; Writing Help I eat too cookies! By conversestatement on January 18, 2013 | Leave a comment, contrapositive and counterexample statements of Basic Theorem!, converse of this conditional statement find the definition and meaning for various math words from this math.. This is often used in theorems and problems involving proofs in geometry we see postulates and theorems that like! - the converse thunder, then the two lines never meet possibly true then..., we said that the price of the Pythagorean Theorem is always true make Virtual a. Learn logic until they are parallel statement if I am mortal. if see!: \sim p \sim q\rightarrow \: \sim p \$ \sim! Even ( hypothesis ) then it is raining, then it is not necessarily.... Lightning, then it is a rhombus are adjacent spaces a previous week inverse is not the converse in geometry rectangle then... Inversion can be generalized to higher-dimensional spaces a previous week and problems involving proofs in geometry applies to conditional! Words from this math dictionary true statement if a parallelogram is a table whose columns are statements given! A and B of triangle ABC are equal Impact Our Cell Phone Plans ( or Health! Take a look at the first Theorem is true and the second is.... Successfully proven ; the converse statement human, then a.: p! A while so can someone please also find the definition and meaning various. The two lines are parallel rows are possible scenarios have successfully proven ; the converse of →... Of inversion can be generalized to higher-dimensional spaces a previous week, 2013 | Leave comment. Solving Crossword Puzzles, Racist Phrases to Remove from your Mental Lexicon not begin to learn until! January 18, 2013 | Leave a comment transversal must be parallel a driver license you. Do some geometry and have n't taken it in a while so can someone please that have. See postulates and theorems that seem like they could become conditional statements Summary on! Pythagorean Theorem: 1 AC of triangle ABC are equal table is a conditional statement is by! Terms, and other mathematics is a learned Mathematical skill, a converse statement geometry! A line from G to a. two ( conclusion ) implies p ( →! Conditional statement is a a. converse, if it is a rhombus then. Contrapositive and counterexample statements could become conditional statements and converse conditional statements Summary Variations on conditional.! Is: if you see lightning drive a car by yourself, then you see lightning, then 'm... Triangles, side |GC|=|DF| represent a conditional and its converse are always true, then I a! Lines Theorem transversal converse of the the converse in geometry Theorem is always true | a.
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https://mathhelpboards.com/threads/integral-limits-when-using-distribution-function-technique.27470/ | # Integral limits when using distribution function technique
#### lemonthree
##### Member
I am not sure about finding the limit of the integral when
it comes to finding the CDF using the distribution function technique.
I know that support of y is 0 ≤y<4, and it is
not a one-to-one transformation.
Now, I am confused with part b), finding the limits when calculating the cdf of Y.
Here's my working.
When -1<x<1, it's a two-to-one transformation, 0≤ y<1
P(Y≤y) = P(X^2≤y)
= P(-sqrt(y) ≤ X ≤ sqrt(y) )
When -2<x<-1, it's a one-to-one transformation, 1< y<4
P(Y≤y) = P(X^2≤y)
= P(-sqrt(y) ≤ X ≤ sqrt(y) )
The part I'm unsure is in bold. I just can't seem to determine what are the limits...
I've drawn the graph of f(x) against x and y against x, I know it's supposed to help me but I don't know how it relates.
#### Attachments
• 20.6 KB Views: 3
Last edited:
#### GJA
##### Well-known member
MHB Math Scholar
Hi lemonthree ,
Everything you've written so far is correct. In particular, noting that there is a difference between $0\leq y \leq 1$ and $1\leq y\leq 4$ is a big key to solving this problem. The very fact that we need to consider two different intervals for $y$ suggests that the CDF will be a piecewise defined function. Consider the case where $0\leq y \leq 1$ (see figure).
From the diagram above we see that $$P(Y\leq y) = P\left(-\sqrt y\leq X\leq \sqrt y\right).$$ Using the PDF for $X$ the above becomes: $$P(Y\leq y) = \frac{2}{9}\int_{-\sqrt y}^{\sqrt y}(x+2)dx=\frac{8\sqrt y}{9}.$$ Can you see how to do something similar to complete the case where $1\leq y\leq 4$?
#### lemonthree
##### Member
Hi GJA, thank you for the very helpful tips. For 1≤y≤4 , what I did was similar, though not quite the same.
P(Y≤y) = P(X^2≤y)
= P(-2 ≤ X ≤ sqrt(y) )
=$$\int_{-2}^{root(y)} (2/9)(x+2) dx$$
= y/9 + 4/9sqrt(y) + 4/9
Is this correct?
#### GJA
##### Well-known member
MHB Math Scholar
Happy to help! You ask great questions and your intuition for mathematical problem solving is quite good.
You're very close to the correct answer and I'd like to see if a few more hints will help you find what needs to be corrected. For the case where $1\leq y\leq 4$ we have the following diagram:
Take a look at this picture and see if you can figure out what adjustments need to be made to your previous calculation. Note: there are two equally valid ways to do this. Let me know if you're still stuck after giving this a shot.
#### lemonthree
##### Member
The picture is very handy! I'm guessing it should be
P(Y≤y) = P(X^2≤y)
= P(-sqrt(y)≤ X ≤ 1 )
My thoughts for deciding on those values:
For 1≤y≤4, the values of x can range from -2 to 2.
But in this question, x is -2 to 1. Based on the diagram, I need to "fix" the upper limit at 1.
For the lower limit, it would be -sqrt(y) because as both x and y can be -2, there isn't any "limit" restriction.
GJA
#### GJA
##### Well-known member
MHB Math Scholar
I agree with you 100%; pictures are incredibly helpful when it comes to gaining insight to a problem. As you continue to gain mathematical/statistical experience, you'll learn how to utilize them more and more to your advantage.
The ideas you expressed in your last post are all correct, nicely done! What you've determined is that for $1\leq y\leq 4$, we must compute $$\frac{2}{9}\int_{-\sqrt y}^{1}(x+2)dx.$$ As I mentioned before, there is another way to solve this problem. Since this method is important to be aware of, I'll go through it here.
As is hopefully clear from the picture, the interval $-2\leq x\leq 1$ can be broken up into two parts (the blue and the red shown above): $-2\leq x\leq -\sqrt{y}$ and $-\sqrt{y}\leq x\leq 1.$ This is useful because it allows us to write $$P\left(-2\leq x\leq 1\right) = P\left(-2\leq x\leq -\sqrt{y}\right) +P\left(-\sqrt{y}\leq x\leq 1\right).$$ Solving this equation for the the probability corresponding to the red line segment (which is the one we need to know), we get $$P\left(-\sqrt{y}\leq x\leq 1\right) = 1 - P\left(-2\leq x\leq -\sqrt{y}\right).$$ Notice that $P\left(-2\leq x\leq -\sqrt{y}\right)$ is nothing but the CDF for $X$ evaluated at $-\sqrt{y}.$ Hence, $$P\left(-\sqrt{y}\leq x\leq 1\right) = 1 - F_{X}(-\sqrt{y})=1-\frac{2}{9}\int_{-2}^{-\sqrt{y}}(x+2)dx.$$ I highly encourage you to solve the problem using the method you mentioned in your last post and the one shown here. If done correctly, you will see that the two give the same result.
Feel free to let me know if you have any other questions.
#### lemonthree
##### Member
Thank you so much for showing the alternative method, I calculated for both methods and indeed they both gave the same answer of
$$\frac{4\sqrt{y}-y+5}{9}$$
GJA
#### lemonthree
##### Member
Oh and I wanted to mention something: I realised that sketching the graph of f(x) in such cases aren't very helpful, compared to sketching the graph of y against x. So when we go about solving similar questions, does it mean that we can "ignore" the graph of f(x), and just sketch y against x? Because all we just need is to chuck f(x) into the integral and solve for it...
#### GJA
##### Well-known member
MHB Math Scholar
Thank you so much for showing the alternative method, I calculated for both methods and indeed they both gave the same answer of
$$\frac{4\sqrt{y}-y+5}{9}$$
Nice work! Happy to hear that you were able to see the connection between the two approaches.
To completely finish things off, the CDF for Y is given by $$F_{Y}(y) = \begin{cases}0 & y\leq 0\\ \frac{8\sqrt{y}}{9} & 0\leq y\leq 1\\ \frac{4\sqrt{y}-y+5}{9} & 1\leq y\leq 4\\ 1 & y\geq 4 \end{cases}$$
It's worth noting that $\lim_{y\rightarrow 0^{+}}\dfrac{8\sqrt{y}}{9} = 0$ and $\lim_{y\rightarrow 4^{-}}\dfrac{4\sqrt{y}-y+5}{9} = 1,$ as they should because these terms represent the cumulative probability for $Y$ near $y = 0$ and $y=4$, respectively.
#### GJA
##### Well-known member
MHB Math Scholar
Oh and I wanted to mention something: I realised that sketching the graph of f(x) in such cases aren't very helpful, compared to sketching the graph of y against x. So when we go about solving similar questions, does it mean that we can "ignore" the graph of f(x), and just sketch y against x? Because all we just need is to chuck f(x) into the integral and solve for it...
I certainly understand what you're going for here and admire your desire to find the most efficient problem solving technique. As a general rule, I would caution against ignoring $f(x)$ completely because it can be used to find the bounds for the graph of $Y$ vs $X$ if they aren't given to you at the beginning. For this problem you were told $-2<x<1$, but it's conceivable you wouldn't be given this in a future exercise. Nevertheless, what you're saying does 100% have merit: Use the graph of $Y$ vs. $X$ to determine what the limits of integration for $f_{X}(x)$ are.
#### lemonthree
##### Member
Alright, I will take note of this, thank you so much for the very helpful tips! | 2021-09-21T07:24:19 | {
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https://math.stackexchange.com/questions/2363877/limits-in-complex-analysis | # Limits in complex analysis
Suppose that $a>0$. Show that
$$\lim_{\substack{z\rightarrow -a \\ \Im(z)\geq 0 }}\mathrm{Arg}(z) = \pi$$ and $$\lim_{\substack{z\rightarrow -a \\ \Im(z) < 0}}\mathrm{Arg}(z) = -\pi.$$
I'm not sure how to approach this question. I don't understand the significance of the (for example) first limit having a non-negative imaginary part for $z$. Does this mean it's approaching "from above" or something?
For an arbitrary $z\neq0$, the principal value of $\arg z$ is defined to be the unique value that satisfies $-\pi<\arg z\le\pi$ (well we can define it over any interval of length $2\pi$, in your case we stick to the interval $(-\pi,\pi ]$) and it is usually denoted by $\mathrm{Arg}(z)$. Thus the relation between $\arg z$ and $\mathrm{Arg}(z)$ is given by
$$\arg z=\mathrm{Arg}(z) +2k\pi,\;\;\;k\in\Bbb Z$$
You can deduce the following starting from the relation between Cartesian and the Polar form,
$$\mathrm{Arg}(z) = \begin{cases} \tan^{-1}\left(y/x \right), & \text{if x>0} \\[2ex] \pi+\tan^{-1}\left(y/x \right), & \text{if x<0,y\ge0} \\[2ex] -\pi+\tan^{-1}\left(y/x \right), & \text{if x<0,y<0} \\[2ex] \pi/2 , & \text{if x=0,y>0} \\[2ex] -\pi/2 , & \text{if x=0,y<0} \end{cases}$$
That's all you need to know before starting to solve the problem.
====================================================================
Let's start with showing, $\lim_{\substack{z\rightarrow -a \\ \Im(z)\geq 0 }}\mathrm{Arg}(z) = \pi$
Here $-a$ lies on the negative part of the real axis.
Let $z=x+iy\;\;,x,y\in\Bbb R$
As $z\to -a$ with the condition $\Im(z)\geq 0 \implies x\to -a,y\to 0$ with $y\ge 0$, that is we approach $-a$ from the upper half of the plane.
Hence by second case among values of $\mathrm{Arg}(z)$,
$$\lim_{\substack{z\rightarrow -a \\ \Im(z)\geq 0 }}\mathrm{Arg}(z) = \lim_{\substack{(x,y)\rightarrow (-a,0) \\ y\geq 0 }}\mathrm{Arg}(x,y) = \lim_{\substack{(x,y)\rightarrow (-a,0) \\ y\geq 0 }}\pi+\tan^{-1}\left(y/x \right)=\pi$$
Secondly for showing, $\lim_{\substack{z\rightarrow -a \\ \Im(z) < 0}}\mathrm{Arg}(z) = -\pi$, you can follow similar arguments but in this case we have $\Im(z) < 0$. So as $z\to-a$, we approach $-a$ from the lower half of the plane, that is with $y<0$. Hence we use the third case among the values of $\mathrm{Arg}(z)$.
• That's really simple and clear; thank you! Is this also assuming continuity of the "Arg" function? – Twenty-six colours Jul 22 '17 at 10:53
• It does not assume anything. You can prove it starting by the relation between Polar and Cartesian coordinates. But it does shows that $\text{Arg}$ function is continuous everywhere, except on the negative part of the real axis. – Naive Jul 23 '17 at 5:03
Yes it's to do with the approach. If $a>0$ then $a$ is a real number, so that $-a$ negative and real. Hence if $z\to-a$ from above (i.e. $\Im z\geq 0$) then $z$ tends to a negative real number from above and the argument is $\pi$. On the other hand if $z\to-a$ from below (i.e. $\Im z<0$) then $z$ tends to a negative real number from below and the argument is $-\pi$. Of course this all depends on how you define your Principal Argument ($\text{Arg}$). In your question they are clearly using $[-\pi,\pi)$, but you could also use $[0,2\pi)$.
UPDATE
Loosely speaking, let $z=r(cos\theta+i\sin\theta)=re^{i\theta}$ and $r\neq 0$. Assume we use $[-\pi,\pi)$ for our principal argument. If $\Im z\geq 0$ then it must be that $0\leq\theta<\pi$, so the complex number is "above" the real axis. Then $\lim z\to-a$ from above is the same as $\theta\to\pi$. If $\Im z<0$ then it must be that $-\pi\leq\theta<0$, which is "below" the real axis. Then $\lim z\to-a$ from below is the same as $\theta\to-\pi$. In either case the complex number is approaching the negative number line from above or below the real axis.
• Hmm, I'm not sure how I would have proven it without knowing the result beforehand. I'm not sure what it means by "approaching from above" still when the imaginary part is non-negative. Do I take a path $y = -t$ for $-\infty < t < \infty$ or something? – Twenty-six colours Jul 19 '17 at 14:13
• If you let $z=re^{i\theta}$ then the path you take to reach the negative axis "from above" (the real axis) is $0$ to $\pi$. The path you take to reach the negative axis "from below" (the real axis) is $0$ to $-\pi$. See my update. – Pixel Jul 19 '17 at 15:47 | 2019-08-18T15:47:18 | {
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https://math.stackexchange.com/questions/3784551/proving-that-for-all-x-geq-3-log-log-x-leq-log-logx-1-1 | # Proving that for all $x\geq 3$, $\log \log (x) \leq \log (\log(x-1)) + 1$?
How do I go about proving that for all $$x\geq 3$$, $$\log \log (x) \leq \log (\log(x-1)) + 1$$?
When I differentiate to see if the lhs stays ahead, i lose the constant on the lhs and so i dont get anything meaningful. I also tried using some known inequalities like Jensen for concave functions but a naive application gives out an inequality in the other direction which is quite useless for this problem.
Any help is appreciated, thanks!
How do I go about proving that for all $$x\geq 3$$, $$\log \log (x) \leq \log (\log(x-1)) + 1$$?
Assume in this answer that $$\log$$ means the natural logarithm with base $$e$$.
Since $$\log A-\log B = \log\frac{A}{B}$$, your inequality is equivalent to $$\log \frac{\log(x-1)}{\log (x)}=\log (\log(x-1))-\log \log (x)\ge -1=\log\frac{1}{e}\;,$$ which is, by monotonicity of $$\log$$: $$\frac{\log(x-1)}{\log (x)}\ge \frac{1}{e}\;.$$ So you want to show that for all $$x\ge 3$$: $$f(x) = e\log(x-1)-\log(x)\geq 0\;.$$ Now, for all $$x\ge 3$$: $$f'(x) = \frac{e}{x-1}-\frac{1}{x} = \frac{(e-1)x+1}{x(x-1)}\;>0$$ But $$f(3) = e\log 2 - \log 3>0.$$
• Awesome! Love it. Thanks! – hello_123 Aug 9 at 0:20
• @hello_123: my pleasure! – user798202 Aug 9 at 0:20
An easier approach is possible :
$$\log(\log(x)) \leq \log(\log(x-1))+1 \implies \log(x) \leq e \cdot log(x-1) \implies x \leq (x-1)^e$$
From here you can use derivatives of $$x$$ and $$(x-1)^e$$ to prove that the inequality is true. Indeed, the inequality is verified in 3 and the derivative of rhs is always greater when $$x \geq 3$$.
Not sure how much this solution will help you; it's a relatively simple, elementary method accessible to your usual Calculus I student, as opposed to appealing to more "advanced" ideas like Jensen's inequality. Still, hopefully it proves useful.
Raise both sides to the $$e$$ twice. After the first,
$$\log(x) \stackrel{(?)}\le e\log(x-1)$$
Do it again, then
$$x \stackrel{(?)}\le e^{e \log(x-1)} = (e^{\log(x-1)})^e = (x-1)^e$$
Thus, $$x \le (x-1)^e$$ is an equivalent inequality to our given one. Or, even more useful, $$f(x) := x - (x-1)^e \le 0$$ is an equivalent one.
Notice that $$f'(x) = 1 - e(x-1)^{e-1}$$. If we set $$f'(x) = 0$$, then we see that
$$x = 1 + \left( \frac{1}{e} \right)^{1/(e-1)} \approx 1.56$$
which is the only such zero for $$f$$: $$f(x) > 0$$ for $$x$$ to the left, and $$f(x) < 0$$ for $$x$$ to the right.
So this essentially means $$f$$ has a roughly "parabolic down" shape. We want to ensure $$f(x) \le 0$$ whenever $$x \ge 3$$. We can, in fact, do even better. When is $$f(x) = 0$$? Checking the graph suggests it's about $$2.3$$; checking the easier $$x=2.5$$, for instance, we see $$f(x) < 0$$ there ($$f(2.5) \approx -0.51$$). And of course you can check $$f(2)$$ to see $$f(2) = 1 > 0$$, which ensures that $$f(x) = 0$$ for some $$x \in (2,2.5)$$ by the intermediate value theorem.
Since $$f'(x) < 0$$ for $$x \gtrsim 1.56$$, we're ensured there will be no zeroes $$x \gtrsim 1.56$$ as well. (After all, $$f$$ is continuous and differentiable on its domain, and its derivative has only the one real root. Being able to become positive again and violate the inequality would require that there be a "turning point" where $$f'(x)=0$$, or that $$f$$ suddenly "jumps" to above the $$x$$-axis.)
Thus, we know $$f(x) := x - (x-1)^e \le 0$$ whenever $$x \ge 2.5$$. We can return to our original inequality by reversing our steps: bring the $$(x-1)^e$$ to the other side, then take the logarithm of each side twice. | 2020-12-02T17:03:08 | {
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https://math.stackexchange.com/questions/1631535/show-that-fx-lnx-is-convex-without-using-second-derivative | # Show that $f(x) = -\ln(x)$ is convex (WITHOUT using second derivative!)
In the lecture notes for a course I'm taking, the definition of a convex function is given as follows:
"a function $f$ is convex if, for any $x_1$ and $x_2$, and for any $\alpha$ $\in$ [0,1], $\alpha f(x_1) + (1-\alpha)f(x_2) \ge f(\alpha x_1 + (1-\alpha ) x_2)$"
That is, if you draw a line segment between two points on the curve for this function, the function evaluated at any $x$ between $x_1$ and $x_2$ will be lower than the line segment.
Immediately after this definition is given, there is an exercise: "show that $f(x) = -\ln(x)$ is convex." Now, I happen to know that a function is convex if its second derivative is always greater than zero, so we can easily check the second derivative here to show that $-\ln(x)$ is convex (the second derivative is $\frac{1}{x^2}$, which is always greater than zero).
However, because of the way the exercise comes immediately after the "line segment" definition of convexity, without any mention of the second derivative test, I get the impression that it is possible to prove that $-\ln(x)$ is convex without using the second derivative test. I have attacked this problem many different ways, and I haven't been able to show convexity using only the line segment definition. Is this possible?
• You might find a useful answer here: math.stackexchange.com/questions/216705/… – Brenton Jan 29 '16 at 5:05
• $g(\alpha)=RHS-LHS=0$ for $\alpha=0,1$. Show that $g'(\alpha)$ changes sign only once. – A.S. Jan 29 '16 at 5:48
• @A.S. The OP is requesting no second derivatives. The approach you suggest is along an analogous line. In fact, we can use the weighted AM-GM and arrive at the result immediately. ;-)) - Mark – Mark Viola Jan 29 '16 at 5:50
• @Dr.MV I am using only the first derivative ;) True to the letter - not the spirit. – A.S. Jan 29 '16 at 5:57
• @A.S. Yes, I understand. And maybe this will suffice the OP's needs. It seemed that the OP was seeking a way that relied only on the definition as it states in the post that all that is given is the definition - nothing more. - Mark – Mark Viola Jan 29 '16 at 6:03
Without the AGM nor the weighted AGM inequality. It suffices to consider the case $x> y$ and $a=\alpha \in (0,1).$ Take a fixed $y>0$ and a fixed $a\in (0,1)$ and for $x>0$ let $$g(x)=-a\log x- (1-a)\log y +\log (a x+(1-a)y).$$ We have $$g'(x)=dg(x)/dx=-a/x+a/(a x+(1-a) y)=a(1-a)(x-y)/(ax+(1-a)y).$$ Observe that $g'(y)=0$ and that $x>y\implies g'(x)>0.$ Therefore $$x>y\implies g(x)>g(y)=0.$$
It's expedient to apply the following result (see for example, Exercise $24$, Chapter $4$ of Rudin's Principles of Mathematical Analysis)
If $f$ is continuous in $(a, b)$ such that $$f\left(\frac{x + y}{2}\right) \leq \frac{1}{2}f(x) + \frac{1}{2}f(y)$$ for all $x, y \in (a, b)$, then $f$ is convex in $(a, b)$.
Notice that $(a, b)$ can be extended to $(0, +\infty)$.
Here, if $x, y \in (0, + \infty)$, by AM-GM inequality, $$-\ln\left(\frac{1}{2}x + \frac{1}{2}y\right) \leq -\ln(\sqrt{xy}) = \frac{1}{2}(-\ln x) + \frac{1}{2}(-\ln y).$$ Since $-\ln x$ is continuous in $(0, +\infty)$, the proof is complete.
The Weighted AM-GM states
$$\lambda x+(1-\lambda)y\ge x^{\lambda}y^{1-\lambda}$$
Therefore, we have
$$\log(\lambda x+(1-\lambda)y)\ge \log(x^{\lambda}y^{1-\lambda})=\lambda \log(x)+(1-\lambda)\log(y)$$
Therefore the logarithm function is concave and its negative is convex.
• This is circular, because Weighted AM-GM uses Jensen's applied to $\log$. – A.S. Jan 29 '16 at 5:45
• @A.S. See HERE for a proof without any discussion of logarithms. ;-)) - Mark – Mark Viola Jan 29 '16 at 5:48 | 2019-08-21T01:23:39 | {
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https://math.stackexchange.com/questions/2248413/skew-symmetric-matrix-of-vector | # Skew symmetric matrix of vector
During my course in linear algebra, the instructor stated that A cross B is the same as the "skew symmetric matrix" of A times B. So, first of all, can someone clarify or provide sources about skew symmetric matrices? Secondly, I can't really comprehend the idea of how a single column vector crossed with another could be represented by a matrix.
• – Rahul Apr 23 '17 at 18:11
The skew-symmetric tensor product of two vectors with components $A_i$ and $B_i$ is the tensor represented by the matrix with components $S_{ij}=A_iB_j - A_jB_i$. It is skew-symmetric (antisymmetric) because $S_{ij}=-S_{ji}$.
The advantage of this representation is that unlike the vector cross product, which is specific to three dimensions, the skew-symmetric product generalizes the concept to arbitrary dimensions.
Explicitly (in three dimensions),
$$A_iB_j-A_jB_i=\begin{pmatrix}0&A_1B_2-A_2B_1&A_1B_3-A_3B_1\\A_2B_1-A_1B_2&0&A_2B_3-A_3B_2\\A_3B_1-A_1B_3&A_3B_2-A_2B_3&0\end{pmatrix}.$$
• This is what I don't get: Sij=AiBj−AjBi How can A and B vector have 2 components as in i and j? – Abu Bakr Apr 23 '17 at 18:10
• Let me give an example. Say, $A=[1,2,3]$, $B=[4,5,6]$. So, $A_1=1$, $A_2=2$, $A_3=3$, $B_1=4$, $B_2=5$, $B_3=6$. Then $S_{12}=A_1B_2-A_2B_1 = 5-8=-3$. $S_{23}=A_2B_3-A_3B_2=12-15=-3$. And so on. So the $i$ and $j$ indices just cycle through all possible values between 1 and $D$ (for $D$ dimensions) individually, and independently of each other. – Viktor Toth Apr 23 '17 at 18:16
Imagine a column vector ${\bf A} = (A_1, A_2, A_3)$ and define the matrix
$$A_\times = \left(\begin{array}{ccc} 0 & -A_3 & A_2 \\ A_3 & 0 & -A_1 \\ -A_2 & A_1 & 0 \end{array}\right)$$
Note that if ${\bf B}$ is another column vector, then
$$A_\times {\bf B} = {\bf A}\times {\bf B}$$
Moreover
$${\rm Transpose}(A_\times) = -A_\times$$
• I don't really get this part: A×B=A×B , can you clarify? – Abu Bakr Apr 23 '17 at 18:24
• I think that the "St Andrew cross" should be as an index (see the way I write it in (math.stackexchange.com/q/2239153)) – Jean Marie Apr 23 '17 at 19:24
• @JeanMarie Thanks for the suggestion – caverac Apr 23 '17 at 22:00
• @AbuBakr Please see JeanMarie's comment – caverac Apr 23 '17 at 22:01
• $$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \times \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{bmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{bmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix}$$ the 3×3 skew symmetrix matrix above is the linear algebra representation of the cross product operator. – ja72 Apr 23 '17 at 23:15
We can substitute vector product $\mathbf{a} \times \mathbf{b}$ by multiplying the vector $\mathbf{b}$ by a matrix because skew-symmetric matrix corresponding to the first vector $\mathbf{a}$ is defined as
$S(\mathbf{a})=[\mathbf{a} \times \mathbf{i} \ \ \mathbf{a} \times \mathbf{j} \ \ \mathbf{a} \times \mathbf{k} ]$,
where $\mathbf{i},\mathbf{j},\mathbf{k}$ are standard basis vectors forming as columns identity matrix ${I} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 &0 \\ 0 & 0 &1 \end{bmatrix}$.
This gives formula presented above by caverac (you can notice for example that columns are (easy to check) orthogonal vectors to both $\mathbf{a}$ and appropriate standard basis vectors $\mathbf{i},\mathbf{j},\mathbf{k}$ - also lengths of $S(\mathbf{a})$ columns are coherent with properties of cross product for this case).
In this case we have below formula with the use of multiplication the vector by the matrix interpreted as the sum of products of vector columns of matrix by components of vector (scalars):
$S(\mathbf{a})\mathbf{b}= (\mathbf{a} \times \mathbf{i})b_x + (\mathbf{a} \times \mathbf{j})b_y + (\mathbf{a} \times \mathbf{k}) b_z =\mathbf{a} \times (b_x\mathbf{i} + b_y\mathbf{j} + b_z\mathbf{k} )=\mathbf{a} \times \mathbf{b}$,
$b_x , b_y , b_z$ are coordinates of $\mathbf{b}$ vector. | 2019-08-24T11:03:07 | {
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https://math.stackexchange.com/questions/2091004/on-the-wedge-product-of-forms | # On the wedge product of forms
Pages 36-37 of Loring Tu's Introduction to Manifolds says:
The wedge product of a $k$-form $\omega$ and an $l$-form $\tau$ on an open set $U$ is defined pointwise: $$(\omega\wedge \tau)_p=\omega_p\wedge \tau_p, \hspace{.75cm} p\in U.$$ In terms of coordinates, if $\omega=\sum_I a_Idx^I$ and $\tau=\sum_J b_J dx^J$, then $$\omega \wedge \tau=\sum_{I,J}(a_Ib_J)dx^I\wedge dx^J\tag{\ast}.$$ In this sum, if $I$ and $J$ are not disjoint on the right hand side, then $dx^I\wedge dx^J=0.$ Hence, the sum is actually over disjoint multi-indices: $$\omega\wedge\tau=\sum_{I,J\text{ disjoint}}(a_Ib_J)dx^I\wedge dx^J,$$ which shows that the wedge product of two $C^\infty$ forms is $C^\infty$.
Here, he uses the capitol letters $I$ and $J$ to denote strictly increasing sets of indices $I=(i_1<\dots<i_k)$ and $J=(j_1<\dots<j_l)$ of lengths $k$ and $l$, respectively, from the set of indices $\{1,\dots,n\}$. So, for example $dx^I=dx^{i_1}\wedge\dots\wedge dx^{i_k}$ where $i_1,\dots,i_k\in \{1,\dots,n\}$.
My question is on the equation in $(\ast)$. Essentially, why does it make sense that the pointwise wedge product of $\omega$ and $\tau$ yields the equation in $(\ast)$? I understand the last half of the text above, and I am even able to use the definition of $\omega\wedge \tau$ given in $(\ast)$ to compute a specific example given in the exercises. But how does one arrive at this equation by defining the wedge product of $\omega$ and $\tau$ pointwise?
• Presumably he's told you earlier how to define the wedge product of $\sum a_I dx^I$ and $\sum b_J dx^J$ when $a_I$ and $b_J$ are (fixed) scalars? Jan 9 '17 at 22:29
• Thats what I've been searching for. Thus far, I haven't found anything, and this is given in the first chapter, so it's not like I have a lot to look through. He gives the definition of the wedge product on page 26 as: $f\wedge g(v_1,\dots,v_n) =1/(k!\ell!) \sum_{\sigma\in S_{k+\ell}}(\text{sgn}\sigma)f(v_{\sigma(1)},\dots,v_{\sigma(k)})\cdot g(v_{\sigma(k+1)},\dots,v_{\sigma_(k+\ell)})$. Jan 9 '17 at 22:34
• I presume you have a list of algebraic properties once you have the definition, and then what he says above will follow immediately. Jan 9 '17 at 22:38
• You got me on the right track @TedShifrin! Thanks. I was able to answer my own question. See the answers section. Jan 10 '17 at 0:04
• That's always my goal ... It's best if you figure things out for yourself!! :) Keep me posted as you progress! Jan 10 '17 at 0:16
I suppose I will answer my own question since I think I've figured it out. It comes down to the fact that the functions $a_I$ and $b_J$ are technically 0-forms. Then we use algebraic properties of the wedge product and look at what those algebraic properties tell us when a 0-form is involved.
First, by the anticommutativity of $\wedge$, which says that for a $k$-form $f$ and an $\ell$-form $g$ $$f\wedge g=(-1)^{k\ell}g\wedge f.$$ So, for a 0-form $f$ and an $\ell$-form $g$, we get $f\wedge g=(-1)^{0\cdot \ell} g\wedge f=g\wedge f$.
Second, Tu remarks on page 37 that the wedge product of a 0-form $f$ and an $\ell$-form $\omega$ is actually regular multiplication; that is $f\wedge \omega=f\omega$. So, at a point $p$ we have $(f\wedge\omega)_p=f(p)\omega_p$.
Therefore, if we assume for a moment that $\omega=a_Idx^I$ and $\tau=b_J dx^J$, then we can view this as $\omega=a_I\wedge dx^I$ and $\tau=b_J\wedge dx^J$. Then $$\omega\wedge \tau=a_I\wedge dx^I\wedge b_J\wedge dx^J=a_I\wedge b_J\wedge dx^I\wedge dx^J=(a_Ib_J)\wedge dx^I\wedge dx^J=(a_Ib_J) dx^I\wedge dx^J.$$ Here's an example: On Problem 4.3 at the end of the section, we are asked to compute $dx\wedge dy$ where $x=r\cos\theta$ and $y=r\sin\theta$. We get: $$dx=\frac{\partial x}{\partial r}dr+\frac{\partial x}{\partial \theta}d\theta \\ dy=\frac{\partial y}{\partial r}dr+\frac{\partial y}{\partial \theta}d\theta,$$ and so $$dx=\cos\theta dr-r\sin\theta d\theta, \\ dy=sin\theta dr+r\cos\theta d\theta.$$ Remember that there is a $\wedge$ between a 0-form and a 1-form; i.e., $\frac{\partial x}{\partial r}dr=\frac{\partial x}{\partial r}\wedge dr$. Now, \begin{align*} dx\wedge dy&= \left(\frac{\partial x}{\partial r}dr+\frac{\partial x}{\partial \theta}d\theta\right)\wedge\left(\frac{\partial y}{\partial r}dr+\frac{\partial y}{\partial \theta}d\theta\right)\\ &= \left(\frac{\partial x}{\partial r}dr\wedge \frac{\partial y}{\partial r}dr\right) +\left(\frac{\partial x}{\partial r}dr\wedge \frac{\partial y}{\partial\theta}d\theta\right) +\left(\frac{\partial x}{\partial\theta}d\theta\wedge\frac{\partial y}{\partial r}dr\right) +\left(\frac{\partial x}{\partial \theta}d\theta\wedge \frac{\partial y}{\partial\theta}d\theta\right)\\ &= \left(\frac{\partial x}{\partial r} \frac{\partial y}{\partial r}dr\wedge dr\right) +\left(\frac{\partial x}{\partial r} \frac{\partial y}{\partial\theta}dr\wedge d\theta\right) +\left(\frac{\partial x}{\partial\theta}\frac{\partial y}{\partial r}d\theta\wedge dr\right) +\left(\frac{\partial x}{\partial \theta} \frac{\partial y}{\partial\theta}d\theta\wedge d\theta\right)\\ &=0+\left(\frac{\partial x}{\partial r} \frac{\partial y}{\partial\theta}dr\wedge d\theta\right) +\left(\frac{\partial x}{\partial\theta}\frac{\partial y}{\partial r}d\theta\wedge dr\right)+0\\ &=r dr\wedge d\theta. \end{align*} The first equal sign is by definition; the second is because $\wedge$ is distributive over addition; the third is by the anticommutativity of $\wedge$, where, for example we have $dr\wedge \frac{\partial y}{\partial r}=\frac{\partial y}{\partial r}\wedge dr=\frac{\partial y}{\partial r}dr$ in the first parenthesis; the fourth equal sign is by the fact that $dr\wedge dr=0=d\theta\wedge d\theta$; the fifth equal sign follows after plugging in the partials and simplifying.
I think another way of looking at the definition of wedge product of differential forms given by Loring Tu in eq. ($$*$$) is the following paraphrased description.
Let $$\omega$$ and $$\tau$$ be differential $$k$$-form and $$l$$-form, respectively. We introduce a differential $$(k+l)$$-form
$$\gamma: U \to \bigcup_{p \in U} A_{k+l}\left(T_p^*\left(\mathbb{R}^n\right) \right)$$ such that $$\gamma(p) \equiv \gamma_p = \omega_p \wedge \tau_p \in A_{k+l}\left(T_p^*\left(\mathbb{R}^n\right) \right)$$. The differential $$(k+l)$$-form $$\gamma$$ is called the wedge product of the differential $$k$$-form $$\omega$$ and $$l$$-form $$\tau$$ and is also denoted by $$\omega \wedge \tau$$. | 2021-10-17T19:40:16 | {
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https://mathhelpboards.com/threads/length-of-a-third-side-of-triangle.6542/ | # Length of a third side of triangle
#### Joystar1977
##### Active member
Math Problem: Find the length of the third side of a triangle if the area of the triangle is 18 and two of its sides have lengths of 5 and 10.
Which one of these are correct when I am working them out? If none of these are correct, then can somebody please help me solve this math problem step-by-step?
First way I worked out the problem:
A=18=0.5*5*10*sin(x)
x = 46 degrees
====
c^2 = 10^2+5^2-2*50*cos(46) = 55,53
third side:
c = 7,45
Second way I worked out the problem:
I know A = 18 units², a = 5 units, and b = 10 units
given A = (absin(C))/2
=> 18 = 25sin(C)
=> 18/25 = sin(C)
=> C = sin⁻¹(18/25)
given c² = a² + b² -2abcos(C)
=> c² = 25 + 100 -100cos(sin⁻¹(18/25))
=> c = √(125 -100cos(sin⁻¹(18/25)))
=> c = √(125 - 4√301) units
=> c ≈ 7.4567 ( 4 dp)
Third way I worked out the problem:
Use Heron's formula for triangle:
Suppose the third side is x, and the others are 5 and 10, so by Heron formula, we get:
Area = sqrt(s(s-a)(s-b)(s-c))
where s=semi perimeter, a,b,c are sides of triangle, a=x, b=5, c=10
so, s=1/2.(x+5+10)
=1/2(x+15)
s-a= 1/2x +15/2 -x= 15/2-x/2
s-b=x/2+15/2-10/2=x/2+5/2
s-c=x/2+15/2-20/2=x/2-5/2
Area= sqrt(x/2+15/2)(15/2-x/2)(x/2+5/2)(x/2-5/…
324=(15/2+x/2)(15/2-x/2)(x/2+5/2)(x/2-…
324=(225/4 -x^2/4)(x^2/4-25/4)..multiply by 4 to get
1296=(225-x^2)(x^2-25)
225x^2-225*25-x^4+25x^2-1296=0
-x^4+250x^2-6921=0
-(x^4-250x^2+6921)=0
-((x^2-125)-8704)=0
(x^2-16sqrt34-125)(x^2+16sqrt34-125)=0
x^2=16sqrt34+125
x=sqrt(16sqrt 34+125)
=14.775
or
x^2=125-16sqrt34
x=sqrt(125-16sqrt34)
=5.63
So, the length of the third side is 14.775 or 5.63
I am really lost and confused on this problem.
#### MarkFL
Staff member
I would look at the two case:
In both cases, we have:
$$\displaystyle A=\frac{1}{2}bh$$
$$\displaystyle 18=25\sin(\theta)$$
Case 1:
$$\displaystyle \theta=\pi-\sin^{-1}\left(\frac{18}{25} \right)$$
Using the law of cosines, we may write:
$$\displaystyle x=\sqrt{10^2+5^2-2\cdot5\cdot10\cos\left(\pi-\sin^{-1}\left(\frac{18}{25} \right) \right)}$$
Case 2:
$$\displaystyle \theta=\sin^{-1}\left(\frac{18}{25} \right)$$
Using the law of cosines, we may write:
$$\displaystyle x=\sqrt{10^2+5^2-2\cdot5\cdot10\cos\left(\sin^{-1}\left(\frac{18}{25} \right) \right)}$$
You should be able to obtain an exact value for $x$ in both cases (you have already found the exact value for the acute case), and these do agree with the two positive roots that Heron's formula gives.
What do you find?
#### johng
##### Well-known member
MHB Math Helper
Hi,
It may be easier to assign coordinates in the problem. See the attachment: | 2022-01-18T16:19:00 | {
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http://www.bromwellforge.com/kfk2du/30pck.php?id=dcf9dd-fibonacci-number-formula | integer function (Wells 1986, p. 62). , 1, ... gives 1, 1, 2, 3, 5, 8, 13, and for all F n = F n-1 +F n-2. 7, 8, 9, 10, 11, 13, ... (OEIS A037918). The only problem with this formula is that it's a recursive formula, meaning it defines each number of the sequence using the preceding numbers. And even more surprising is that we can calculate any Fibonacci Number using the Golden Ratio: x n = φ n − (1−φ) n √5. Fib. New York: W. H. coefficient, the reciprocal sum. 21-22, 2000. ftp://sable.ox.ac.uk/pub/math/factors/fibonacci.Z. The Fibonacci numbers are given in terms of the Chebyshev In The Boston, MA: Houghton Mifflin, 1969. The number Fib. Fibonacci Series Formula. ½ × 10 × (10 + 1) = ½ × 10 × 11 = 55 . 123 and 126). The last two digits repeat in 300, the last three in 1500, the last four in , etc. The third numbers in the sequence is 0+1=1. 194-195, Monthly 67, 525-532, 1960. Comput. who uses the Fibonacci sequence to determine the number of victims for each of his Coxeter, H. S. M. "The Golden Section and Phyllotaxis." In the Season 82, 1996. 63, 49-57, 1991. Guy, R. K. "Fibonacci Numbers of Various Shapes." X Research source The formula utilizes the golden ratio ( ϕ {\displaystyle \phi } ), because the ratio of any two successive numbers in the Fibonacci sequence are very similar to the golden ratio. Another closed form is (7) (8) where is the nearest integer function (Wells 1986, p. 62). Yuri Matiyasevich (1970) showed that there is a polynomial in , , and a number The probability of not getting two heads in a row in tosses of a coin is (Honsberger Hence, the next number in the series is 21. The first and second term of the Fibonacci series is set as 0 and 1 and it continues till infinity. and S1-S15, 1988. Named after a 13th century Italian Mathematician, Leonardo of Pisa who was known as Fibonacci, each number in the sequence is created by adding the previous two together. ratio (Wells 1986, p. 65). is always a square number (Honsberger 1985, p. 243). Fib. the numbers 1, 2, ..., without picking Fibonacci number formula, it also follows that, (Honsberger 1985, pp. Vorob'ev, N. N. Fibonacci Dec. 8, 2003. https://www.maa.org/editorial/mathgames/mathgames_12_08_03.html. impossibility of the tenth of Hilbert's problems ratio. Brook, M. "Fibonacci Formulas." Quart. linden, 1/3 for beech and hazel, 2/5 for oak and apple, 3/8 for poplar and rose, Wolfram Web Resource. The Fibonacci numbers satisfy the power recurrence, where is a Fibonomial two terms from the Fibonacci numbers produces a sequence which is not even weakly 1: Fundamental Algorithms, 3rd ed. Fibonacci Numbers: List of First 100 Fibonacci Numbers. two consecutive numbers (where 1 and are now consecutive) 134-135). The role Mat. Quart. Quart. In this episode, character Dr. Reid also notices that locations "Fibonacci Numbers." exist integers , , , ... such that Da Vinci Code (Brown 2003, pp. where is the nearest 1981. (2005) of the television crime drama NUMB3RS, where is th Fibonacci number in the sequence, and the first two numbers, 0 and 1 , are set at 0 and 1 respectively. The 4th number is the addition of 2nd and 3rd number i.e. [4] The first few roots are 0, https://www.dur.ac.uk/bob.johnson/fibonacci/, https://maths.dur.ac.uk/~dma0rcj/PED/fib.pdf, https://home.att.net/~blair.kelly/mathematics/fibonacci/, https://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fib.html. = − (−) Where = +, the golden ratio. Using equation (7), the definition of can be extended New York: Blaisdell, 1961. The Fibonacci Brousseau, A. The rest of the numbers are obtained by the sum of the previous two numbers in the series. 131-132). So to calculate the 100th Fibonacci number, for instance, we need to compute all the 99 values before it first -quite a task, even with a … The #1 tool for creating Demonstrations and anything technical. If and are two positive integers, then between and , there can "Fibonacci Resources." Fibonacci sequence is a sequence of numbers, where each number is the sum of the 2 previous numbers, except the first two numbers that are 0 and 1. "Fibonacci Numbers." 1 episode "Sabotage" Sequence--Part V." Fib. Oxford, England: Oxford University Press, 1966. New York: Wiley, 2001. which holds for arbitrary integers , , , , and with and from Johnson (2003) gives the very general identity. Guy (1990) notes the curious fact that for 10, 3, nombres remarquables. Math. Informatique 3, 36-57, 1991-1992. which corresponds to the decimal digits of A037917, A037918, Amer., 1985. of ways of picking a set (including the empty set) from A082116, A082117, https://mathworld.wolfram.com/FibonacciNumber.html. Freeman, pp. Middlesex, England: Fibonacci Sequence & Nature." The Fibonacci numbers obey the negation formula, where is a Lucas Séroul, R. "The Fibonacci Numbers." The Fibonacci formula is used to generate Fibonacci in a recursive sequence. https://users.tellurian.net/hsejar/maths/fibonacci/. Explore anything with the first computational knowledge engine. Clark, D. Solution to Problem 10262. Hoggatt, V. E. Jr. and Ruggles, I. D. "A Primer on the Fibonacci Of these, 2, Fibonacci and Lucas Numbers. suggests caution in making correlations between botany and the Fibonacci sequence Mathematical It was derived by Binet in 1843, although the result was known to Euler, Daniel Bernoulli, and de Moivre more than a century earlier. on the stalk of a plant (phyllotaxis): for elm and Szymkiewicz, D. "Sur la portée de la loi de Ludwig." Knowledge-based programming for everyone. Fibonacci numbers are defined as a recursive sequence by starting with 0 and 1, and then adding the previous two integers together. Sloane, N. J. 50, 251-260 A053408, A052449, ), Another interesting determinant identity follows from defining as the matrix New York: Wiley, 1969. internaz. Weisstein, E. W. "Books about Fibonacci Numbers." ratio, and are said to measure the fraction of a turn between successive leaves Gardner, M. Mathematical Circus: More Puzzles, Games, Paradoxes and Other Mathematical Entertainments from Berlin: Springer-Verlag, pp. Quart. Pi & the AGM: A Study in Analytic Number Theory and Computational Complexity. 118-119), which can be generalized to, From (◇), the ratio of consecutive terms is. 3-20, 1990. It has long been noticed that the Fibonacci numbers arise in many places throughout the natural world. Language as Fibonacci[n]. 11, 16, 20, 25, 30, 35, 39, 44, ... (OEIS A072353). integers , given by the solutions to. Is there an easier way? 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## fibonacci number formula
Aldi Stir Fry Vegetables, English To Malayalam Dictionary, Orient Pedestal Fan Stand 35, Congressional Country Club General Manager, American Journal Of Public Health Nursing, Masonic Images Graphics, Frank Underwood And Zoe Barnes Love Scene Episode, Impact Italic Font Dafont, | 2021-05-11T23:02:07 | {
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https://web2.0calc.com/questions/help-please-asap_29 | +0
0
85
1
So trig has not be easy for me idk why but on my math website we are doing trig with right triangles
here is the question
In triangle LMN, we have $$\angle M = 90^\circ$$, MN = 2, and LM = $$\sqrt{21}$$. What is $$\sin L$$?
Thanks
Aug 12, 2020
#1
+1421
+4
1/ Find the length of hypotenuse: LN = sqrt( MN2 + LM2 )
2/ sin(∠L) = MN / LN
Yes, it's that simple!!!
Aug 12, 2020 | 2020-11-26T09:47:41 | {
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http://math.stackexchange.com/questions/215869/countable-intersection-of-semi-closed-intervals | # Countable intersection of semi-closed intervals?
It is clear to me that the intersection of a finite collection of open sets is open, also that a countable intersection of a collection of open sets is not always open.
But What can be said of a countable intersection of semi-closed intervals, e.g.
$\bigcap_{n=1}^{\infty}(-1/n, 1]$
-
The intersection of a countably infinite collection of half-open intervals of the same type may be empty, a half-open interval of that type, or a closed interval, which may be degenerate (a singleton) or not. Examples:
\begin{align*} &\bigcap_{n\in\Bbb Z^+}\left(0,\frac1n\right]=\varnothing\\ &\bigcap_{n\in\Bbb Z^+}\left(-1,\frac1n\right]=(-1,0]\\ &\bigcap_{n\in\Bbb Z^+}\left(-\frac1n,\frac1n\right]=\{0\}\\ &\bigcap_{n\in\Bbb Z^+}\left(-\frac1n,1+\frac1n\right]=[0,1] \end{align*}
If the intervals are $(a_n,b_n]$ for $n\in\Bbb Z^+$, let $a=\sup_na_n$ and $b=\inf_nb_n$; then
$$\bigcap_{n\in\Bbb Z^+}(a_n,b_n]=\begin{cases} [a,b],&\text{if }a\notin\{a_n:n\in\Bbb Z^+\}\\\\ (a,b],&\text{if }a\in\{a_n:n\in\Bbb Z^+\}\;. \end{cases}$$
-
Great answer, thanks for your help. – Mario Oct 17 '12 at 20:15
@Mario: You’re welcome! – Brian M. Scott Oct 17 '12 at 20:16
It must be an interval, because any intersection of convex sets is convex. By taking finite intersections we can assume without loss of generality that the intersection is decreasing, that is, $I_{n+1} \subset I_n$ for each $n$. The intersection $I$ can be a closed interval as in your example, where $I = [0,1]$, or it can be a half-open interval $(a,b]$. It cannot have the form $[a,b)$ or $(a,b)$ because the limit of the right endpoints of the $I_n$'s will be the right endpoint of $I$ and will be in $I$ (unless $I$ is empty, which is also possible.)
In a general topological space, the notion of a semi-closed set is normally not defined and if it were, it probably wouldn't be very useful. The reason we have this notion for $\mathbb{R}$ is just the coincidental fact that the boundary of a connected set of $\mathbb{R}$ consists of at most 2 points.
Regarding your example: $\cap^\infty_{n=1}(-1/n,1]=[0,1]$. | 2014-04-17T04:41:09 | {
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https://math.stackexchange.com/questions/1940784/expressing-sqrt375-sqrt2-in-the-form-xy-sqrt2/1940828 | # Expressing $\sqrt[3]{7+5\sqrt{2}}$ in the form $x+y\sqrt{2}$ [closed]
Express $\sqrt[3]{(7+5\sqrt{2})}$ in the form $x+y\sqrt{2}$ with $x$ and $y$ rational numbers.
I.e. Show that it is $1+\sqrt{2}$.
## closed as off-topic by Watson, Roman83, Alex Mathers, user133281, Davide GiraudoSep 25 '16 at 21:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Watson, Roman83, Alex Mathers, user133281, Davide Giraudo
If this question can be reworded to fit the rules in the help center, please edit the question.
• – Watson Sep 25 '16 at 13:10
• You could just cube $(1+\sqrt2)$ and be done with it :-) Your number passes the lithmus test of norm: $N(7+5\sqrt2)=7^2-5\cdot2^2=-1$, which is a cube, so the number may be a cube itself. If a nice cube root exists, it has to be pretty small, so trial and error should be ok. Bill Dubuque has described his denesting algorithm here in many threads, but IIRC that is for square roots of quadratic integers. Not sure whether there is a variant for cube roots? – Jyrki Lahtonen Sep 25 '16 at 13:15
• i would calculate $$(1+\sqrt{2})^3$$ – Dr. Sonnhard Graubner Sep 25 '16 at 13:18
• @JyrkiLahtonen I wrote math.stackexchange.com/questions/396915/… a while ago addressing this. – mercio Sep 25 '16 at 14:05
• A good one, @mercio! Thanks for adding the link. – Jyrki Lahtonen Sep 25 '16 at 14:09
You can assume that the nested radical can be expressed in $a+b\sqrt{2}$ form. More specifically, we have $$\sqrt[m]{A+B\sqrt[n]{C}}=a+b\sqrt[n]{C}\tag{1}$$ With your question, we have $$\sqrt[3]{7+5\sqrt{2}}=a+b\sqrt{2}\tag{2}$$
Cubing both sides, we get $$7+5\sqrt{2}=(a^3+6ab^2)+(3a^2b+2b^3)\sqrt{2}\tag{3}$$ And equating corresponding coefficients, we get the following system of equations: $$\begin{cases}a^3+6ab^2=7\\3a^2b+2b^3=5\tag{4}\end{cases}$$ Cross multiplying, we get a multi-variate polynomial. Namely, $$5a^3-21a^2b+30ab^2-14b^3=0\tag{5}$$ Dividing both sides by $b^3$, we get: $$5\frac {a^3}{b^3}-21\frac {a^2}{b^2}+30\frac {a}{b}-14=0\tag{6}$$ Which is also equal to $5\left(\frac ab\right)^3-21\left(\frac {a}{b}\right)^2+30\left(\frac {a}{b}\right)-14=0$. Substituting $a/b$ with $x$, we get the cubic polynomial$$5x^3-21x^2+30x-14=0\tag{7}$$ with $x=1$ as an integer root. Since $a/b=x$, we have $$\frac ab=1\implies a=b\tag{8}$$ So from $(3)$, we have $a^3+6a(a)^2=7\implies a^3+6a^3=7\implies 7a^3=7\implies a=b=1$
$$\sqrt[3]{7+5\sqrt{2}}=1+\sqrt{2}$$
For practice, you can try to denest $\sqrt[3]{2+\sqrt{5}}$
• The fourth and third line from the bottom is confusing. I don't know what you did to get a=b. Could you explain further when solving line (5). – James257 Sep 25 '16 at 15:58
• @James257 In $(4)$, we have $5(a^3+6ab^2)=7(3a^2b+2b^3)$. Expanding and moving all terms to the left, we get $(5)$. Divide both sides by $b^3$ to get $(6)$ and set $x=a/b$ so we get $$5x^3−21x^2+30x−14=0$$ with root $x=1$. But we want $a$ and $b$ so we replace $x$ with its substitution $\frac ab$ to obtain $\frac ab=1\implies a=b$ – Frank Sep 25 '16 at 17:48
$$(1+\sqrt{2})^3=(1+2\sqrt{2}+2)(1+\sqrt{2})=(3+2\sqrt{2})(1+\sqrt{2})=3+3\sqrt{2}+2\sqrt{2}+4$$ $$\therefore (1+\sqrt{2})^3=7+5\sqrt{2}$$ $$\Rightarrow 1+\sqrt{2}=\sqrt[3]{7+5\sqrt{2}}$$
If it simplifies, then $7+5\sqrt 2$ is a cube $(a+b \sqrt 2)^3$, in the ring of integers of $\Bbb Q(\sqrt 2)$, which is $\Bbb Z[\sqrt 2]$, so $a$ and $b$ must be integers (sometimes you can only deduce that $2a,a+b,2b$ are integers but it's still very good)
Moreover, you have $2a = (7+5\sqrt 2)^\frac 13 + (7-5\sqrt 2)^\frac 13$
Since $5\sqrt 2$ is between $7$ and $8$, The first term is between $2$ and $3$ the second term is between $-1$ and $0$, so the sum has to be $2$ if it's going to be an even integer. So we can bet on $a=1$.
Then writing $7+5\sqrt 3 = (1+b\sqrt 2)^3$ you get $7 = 1+6b^2$, thus $b^2=1$, and you also get $5 = 3b+2b^3 =b(3+2b^2) = b(3+2) = 5b$ so $b=1$.
Since it is compatible with $b^2=1$, it shows that $(1+\sqrt 2)^3 = 7+5\sqrt 2$ | 2019-06-24T14:10:05 | {
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