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# Difference between revisions of "2006 AIME I Problems/Problem 10"
## Problem
Eight circles of diameter 1 are packed in the first quadrant of the coordinate plane as shown. Let region $\mathcal{R}$ be the union of the eight circular regions. Line $l,$ with slope 3, divides $\mathcal{R}$ into two regions of equal area. Line $l$'s equation can be expressed in the form $ax=by+c,$ where $a, b,$ and $c$ are positive integers whose greatest common divisor is 1. Find $a^2+b^2+c^2.$ $[asy] size(150);defaultpen(linewidth(0.7)); draw((6.5,0)--origin--(0,6.5), Arrows(5)); int[] array={3,3,2}; int i,j; for(i=0; i<3; i=i+1) { for(j=0; j
## Solutions
### Solution 1
The line passing through the tangency point of the bottom left circle and the one to its right and through the tangency of the top circle in the middle column and the one beneath it is the line we are looking for: a line passing through the tangency of two circles cuts congruent areas, so our line cuts through the four aforementioned circles splitting into congruent areas, and there are an additional two circles on each side. The line passes through $\left(1,\frac 12\right)$ and $\left(\frac 32,2\right)$, which can be easily solved to be $6x = 2y + 5$. Thus, $a^2 + b^2 + c^2 = \boxed{065}$.
### Solution 2
Assume that if unit squares are drawn circumscribing the circles, then the line will divide the area of the concave hexagonal region of the squares equally (as of yet, there is no substantiation that such would work, and definitely will not work in general). Denote the intersection of the line and the x-axis as $(x, 0)$. | {
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The line divides the region into 2 sections. The left piece is a trapezoid, with its area $\frac{1}{2}((x) + (x+1))(3) = 3x + \frac{3}{2}$. The right piece is the addition of a trapezoid and a rectangle, and the areas are $\frac{1}{2}((1-x) + (2-x))(3)$ and $2 \cdot 1 = 2$, totaling $\frac{13}{2} - 3x$. Since we want the two regions to be equal, we find that $3x + \frac 32 = \frac {13}2 - 3x$, so $x = \frac{5}{6}$.
We have that $\left(\frac 56, 0\right)$ is a point on the line of slope 3, so $y - 0 = 3\left(x - \frac 56\right) \Longrightarrow 6x = 2y + 5$. Our answer is $2^2 + 5^2 + 6^2 = 65$.
We now assess the validity of our starting assumption. We can do that by seeing that our answer passes through the tangency of the two circles, cutting congruent areas, a result explored in solution 1. | {
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# How to evaluate $\int_0^\infty e^{itx} e^{-x} dx$?
When calculating the characteristic function of the exponential distribution function, we need to evaluate the complex-integration: \begin{align*} \int_0^\infty e^{itx}e^{-x} dx \end{align*} for any $t \in \mathbb{R}$.
I understand how to evaluate this integral by treating the real part and imaginary part separately, but I am wondering is there any approach that uses the complex analysis theory? I found some answer uses the seemingly unjustified "fundamental theorem of calculus":
$$\int_0^\infty e^{(it - 1)x} dx = \frac{1}{it - 1}\int_0^\infty e^{(it - 1)x} d(it - 1)x = \frac{1}{1 - it}.$$
I think this solution is lacking any theoretic support (maybe I am wrong, please advise if there is any theorem that supports the above calculation). Specifically, can this integral be evaluated using residue calculus (contour integration)?
• Thanks, but for a complex-valued function, how to justify $\int_0^\infty f'(x) dx = f(x)|_0^\infty$? Could you please provide me some references? – Zhanxiong Mar 5 '17 at 4:01
• Integrating a complex-valued function is the same as a real-valued function. In complex analysis we look at $f(z)$ where $z$ is a complex variable, it is different because we integrate over curves in the complex plane, not over real intervals – reuns Mar 5 '17 at 4:07
You're right to demand justification beyond glib use of the FTC. But it is true that $$\int_0^\infty e^{-ax}dx = \frac{1}{a}$$ whenever $\Re(a)>0.$
In complex analysis, we can still affect the change of variables to $z=ax,$ but the resulting integral is along a ray in the complex plane in the direction of $a,$ not the positive real axis (unless $a$ is a positive real). We can write this (using somewhat bad notation) as $$\int_0^\infty e^{-ax}dx = \frac{1}{a} \int_0^{a\infty}e^{-z}dz.$$ | {
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(To see formally that the change of variables works, note that we have the parametrization $\gamma(t) = at$ for $0<t<\infty$ for the ray. We can write the integral of $e^{-z}/a$ along that path as $$\int_0^\infty \frac{e^{-\gamma(t)}}{a}\gamma'(t)dt = \int_0^\infty \frac{e^{-at}}{a}adt = \int_0^\infty e^{-at}dt$$).
Now, the difficult part is how we can justify saying $$\int_0^{a\infty}e^{-z}dz = \int_0^\infty e^{-x}dx = 1.$$
In other words we want to be able to rotate the contour back down to the real axis without changing the value of the integral.
To see why we can do this, imagine doing a integral around a large wedge-shaped contour. It goes out along the real axis to $R \gg 1$ and then goes along a circular path to $Re^{i\arg(a)}$ and then back into the origin along the ray $[0,a\infty)$ that our integral is taken over.
By Cauchy's theorem, the integral along this closed path is zero since $e^{-z}$ is analytic. The integral along the circular path goes to zero as $R\to \infty.$ We can see this cause the integrand decays like $e^{-R}.$ More formally the integral is $$\int_0^{\arg a} e^{-Re^{i\theta}}Re^{i\theta}id\theta$$ and we have $$\left|\int_0^{\arg a} e^{-Re^{i\theta}}Re^{i\theta}id\theta\right| \le \arg(a) \max_\theta|ie^{-Re^{i\theta}}Re^{i\theta}| = \arg(a)Re^{-R\cos(\arg(a))} \to 0$$ as $R\to\infty$
Thus as $R\to \infty,$ the integral along the real axis must cancel out the integral along the ray $[0,a\infty)$ in order that the integral along the closed path be zero as Cauchy's theorem demands. We have $$\int_0^{a\infty}e^{-z}dz = \int_0^\infty e^{-x}dx = 1$$ and therefore $$\int_0^\infty e^{-ax}dx = \frac{1}{a} \int_0^{a\infty}e^{-z}dz = \frac{1}{a}.$$
I've intentionally not said explicitly where $\Re(a)>0$ is used in the above argument. Of course, it's essential. See if you can find where it's used.
• Thanks for your answer, I found my own solution is very similar to yours! +1 – Zhanxiong Mar 5 '17 at 5:14 | {
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I figured out a rigorous proof by myself.
If $t = 0$, then it is an integration of a real-valued function, and clearly, $\int_0^\infty e^{-x} dx = 1$.
If $t \neq 0$, without losing of generality, assume $t > 0$. Consider the contour below:
In the picture, $n$ is a positive number that will be sent to $\infty$, the top line passes the origin and the point $(-1, t)$. And we set $f(z) = e^z, z \in \mathbb{C}$. By Cauchy's integration theorem, \begin{align} 0 = \int_\Gamma f(z) dz = \int_{\Gamma_1} e^z dz + \int_{\Gamma_2} e^z dz + \int_{\Gamma_3} e^z dz \end{align}
Let's denote the angle between $\Gamma_1$ and the real axis by $\theta_0$.
Clearly, $\int_{\Gamma_3}e^z dz = \int_{-n}^0 e^x dx = 1 - e^{-n}$.
On $\Gamma_1$, $z$ has the representation $z = (it - 1)x, x \in (0, n/|1 - it|)$, thus \begin{align*} \int_{\Gamma_1}e^z dz = \int_0^{n/\sqrt{1 + t^2}}e^{(it - 1)x}(it - 1) dx = (it - 1) \int_0^{n/\sqrt{1 + t^2}} e^{(it - 1)x} dx \end{align*}
To get the desired result, it remains to show $\int_{\Gamma_2} e^z dz \to 0$ as $n \to \infty$. Let $z = ne^{i\theta}$ with $\theta \in (\theta_0, \pi)$. It follows that \begin{align*} & \left|\int_{\Gamma_2} e^z dz\right| = \left|\int_{\theta_0}^\pi e^{ne^{i\theta}}nie^{i\theta} d\theta\right| \\ \leq & \int_{\theta_0}^\pi e^{n\cos\theta}n d\theta \\ \leq & ne^{n\cos{\theta_0}}(\pi - \theta_0) \to 0 \end{align*} as $n \to \infty$. Here we used the fact that $\pi/2 < \theta_0 < \pi$.
The real analysis way : $$f(x) = \frac{e^{(it-1)x}}{it-1}, f'(x) = e^{(it-1)x}, \qquad \int_0^\infty f'(x)dx=f(\infty)-f(0) = \frac{1}{1-it}$$
The complex analysis way, with the (complex) change of variable $z = (it-1)u, dz = (it-1)du$ : $$\int_0^\infty e^{(it-1)u}du = \int_0^{(it-1) \infty } e^{z}\frac{dz}{it-1} = \left.\frac{e^{z}}{it-1}\right|_0^{(it-1)\infty} = \frac{1}{1-it}$$
where $\int_0^{(it-1) \infty }$ is a contour integral | {
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# $x+y=xy=w \in \mathbb{R}^+$. Is $x^w+y^w$ real?
Question: For $x,y \in \mathbb{C}$, suppose $x+y=xy=w \in \mathbb{R}^+$. Is $x^w+y^w$ necessarily real?
For instance, if $x+y=xy=3$, then one solution is $x = \frac{3 \pm i \sqrt{3}}{2}$, $y = \frac{3 \mp i \sqrt{3}}{2}$, but $x^3 + y^3 = 0$, which is real.
I've checked this numerically for many values of $w$ that give complex $x$ and $y$ (namely, $w \in (0,4)$.)
• @PatrickStevens Then $x+y$ isn't real. – wythagoras Aug 21 '16 at 10:37
• @PatrickStevens But he required that $x+y=xy\in\Bbb R$.. – BigbearZzz Aug 21 '16 at 10:37
• Sorry all. I wasn't paying attention, clearly. – Patrick Stevens Aug 21 '16 at 10:38
• The answer is yes; the essential idea is that $xy = x+y \in \mathbb{R}$ forces either that $x,y \in \mathbb{R}$, or that $x$ and $y$ are complex conjugates. – Drew N Aug 22 '16 at 2:22
Yes. Since $x + y \in \mathbb{R}$, $y = \overline{x} + r$ for some $r \in \mathbb{R}$. Then $xy = |x|^2 + xr \in \mathbb{R}$, implying that either $r = 0$ or $x \in \mathbb{R}$. Then we do casework:
• If $r = 0$, then $y = \overline{x}$; this leads to
$$x^w + y^w = x^w + \overline{x^w} \in \mathbb{R}.$$
Warning: for this to work, we had to pick the standard branch of the complex logarithm, specifically, the one undefined on the nonpositive real line, whose imaginary part is between $-\pi$ and $\pi$. Once we define $z^w := e^{w \ln z}$, ${(\overline{x})}^w = \overline{x^w}$ is true for this branch of $\ln$ (as $x$ is not nonpositive real), but might not be true for another branch.
This warning does not come into play when $w$ is an integer. But take, for example, $x = 1 + \frac{1 + i}{\sqrt{2}}$, $y = 1 + \frac{1 - i}{\sqrt{2}}$. Then $x + y = xy = 2 + \sqrt{2}$. If we picked a different branch of the complex logarithm, then we could have $x^w + y^w$ not real. | {
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• On the other hand, if $x \in \mathbb{R}$, then $y \in \mathbb{R}$, so $x^w + y^w \in \mathbb{R}$. Since $x + y = xy > 0$, $x,y$ must both be positive, so we have no trouble with a negative base of the exponent.
Note there was nothing special about $w$: we could have reached the stronger conclusion that $x^a + y^a \in \mathbb{R}$ for all $a \in \mathbb{R}$.
• You do need $x,y\geq0$, for your last statement. – wythagoras Aug 21 '16 at 10:54
• @6005: Even with $a>0$, you still need $x,y\not\le 0$. Consider $x=y=-1\in\mathbb R$ and $a=1/2>0$. Then no matter where you do the branch cut, either $x^a=y^a=i$ or $x^a=y^a=-i$. Basically, the equation $\overline x^a = \overline{x^a}$ is nor true for all $x$ if $a$ is not integer. – celtschk Aug 21 '16 at 11:06
• As a second conjecture, I believe the casework depends on the value of $w$. That is, in fact $r = 0$ if and only if $0< w < 4$, and $x \in \mathbb{R}$ if and only if $w \geq 4$ ($w = 0$ is excluded by assumption). – Drew N Aug 21 '16 at 11:28
• @celtschk Thanks for your valuable feedback. I believe I fixed all the problems and provided all the necessary caveats. – 6005 Aug 21 '16 at 11:48
• @DrewN your second conjecture is correct. – 6005 Aug 21 '16 at 11:55
Yes. Write $x=a+bi$, $y=c-di$, then clearly $b=d$ because $x+y$ is real.
So $x=a+bi$, $y=c-bi$. Then $xy=ac-abi+cbi+b^2$, so $a=c$ or $b=0$.
• If $a=c$, then $y = \overline{x}$, i.e. the complex conjugate of $x$.
So $x^w+y^w = x^w+(\overline{x})^w = x^w+\overline{x^w}$, which is real.
• If $b=0$, $x$ and $y$ are real so $x^w+y^w$ is real as $w>0$.
• Thanks for your feedback on my answer earlier. In your answer, note that $(\overline{x})^w = \overline{x^w}$ is only true for a particular branch of complex log, and will never be true for negative real $x$. However, you can guarantee it for all $x$ which aren't negative real by picking the relevant branch of complex log to define your exponential. – 6005 Aug 21 '16 at 11:53 | {
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$x+y$ real implies that $\Im(x)=-\Im(y)$. Thus if $x=a+bi$ then $y=c-bi$.
$xy$ real implies that, because $xy=ac+b^2+ib(c-a)$, $b=0\lor c=a$.
If $b=0$ the result is trivial as $x,y\in\mathbb{R}$.
If $c=a$ then $x=\overline{y}$. But then clearly $$x^w+y^w=x^w+\overline{x}^w=\overline{x^w+\overline{x}^w}=\overline{x^w+y^w},$$ where the middle equality holds by symmetry.
But then it must be real, as the only complex numbers satisfying $z=\bar{z}$ are real.
Note: for why/when we are allowed to write $x^w=\overline{\bar{x}^w}$ refer to @6005's answer
• Actually, $x = \overline{(\overline{x})^w}$ is a bit more complicated; it's not sufficient that $w$ is real. We have to pick a definition of complex exponential that allows for the symmetry that you appeal to. But regardless of which definition we take (branch of complex log), $x^w = \overline{(\overline{x})^w}$ will be false for $x$ negative real. Consider $w = \tfrac12$. – 6005 Aug 21 '16 at 11:51
• Yeah. That is indeed true. I'll add in my answer to look at yours for a more detailed explanation on when/why this works. – b00n heT Aug 21 '16 at 12:17 | {
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# What is the volume of the $3$-dimensional elliptope?
### My question
Compute the following double integral analytically
$$\int_{-1}^1 \int_{-1}^1 2 \sqrt{x^2 y^2 - x^2 - y^2 + 1} \,\, \mathrm{d} x \mathrm{d} y$$
### Background
The $3$-dimensional elliptope is the spectrahedron defined as follows
$$\mathcal E_3 := \Bigg\{ (x_{12}, x_{13}, x_{23}) \in \mathbb R^3 : \begin{bmatrix} 1 & x_{12} & x_{13}\\ x_{12} & 1 & x_{23}\\ x_{13} & x_{23} & 1\end{bmatrix} \succeq 0 \Bigg\}$$
Using Sylvester's criterion for positive semidefiniteness (i.e., all $2^3-1 = 7$ principal minors are nonnegative), we obtain $1 \geq 0$ (three times), the three quadratic inequalities
$$1 - x_{12}^2 \geq 0 \qquad \qquad \qquad 1 - x_{13}^2 \geq 0 \qquad \qquad \qquad 1 - x_{23}^2 \geq 0$$
and the cubic inequality.
$$\det \begin{bmatrix} 1 & x_{12} & x_{13}\\ x_{12} & 1 & x_{23}\\ x_{13} & x_{23} & 1\end{bmatrix} = 1 + 2 x_{12} x_{13} x_{23} - x_{12}^2 - x_{13}^2 - x_{23}^2 \geq 0$$
Thus, $\mathcal E_3$ is contained in the cube $[-1,1]^3$. Borrowing the pretty figure in Eisenberg-Nagy & Laurent & Varvitsiotis, here is an illustration of $\mathcal E_3$
What is the volume of $\mathcal E_3$?
### Motivation
Why is $\mathcal E_3$ interesting? Why bother? Because $\mathcal E_3$ gives us the set of $3 \times 3$ correlation matrices.
### My work
For convenience,
$$x := x_{12} \qquad\qquad\qquad y := x_{13} \qquad\qquad\qquad z := x_{23}$$
I started with sheer brute force. Using Haskell, I discretized the cube $[-1,1]^3$ and counted the number of points inside the elliptope. I got an estimate of the volume of $\approx 4.92$.
I then focused on the cubic surface of the elliptope
$$\det \begin{bmatrix} 1 & x & y\\ x & 1 & z\\ y & z & 1\end{bmatrix} = 1 + 2 x y z - x^2 - y^2 - z^2 = 0$$
which I rewrote as follows
$$z^2 - (2 x y) z + (x^2 + y^2 - 1) = 0$$
Using the quadratic formula, I obtained
$$z = x y \pm \sqrt{x^2 y^2 - x^2 - y^2 + 1}$$
Integrating using Wolfram Alpha, | {
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$$z = x y \pm \sqrt{x^2 y^2 - x^2 - y^2 + 1}$$
Integrating using Wolfram Alpha,
$$\int_{-1}^1 \int_{-1}^1 2 \sqrt{x^2 y^2 - x^2 - y^2 + 1} \,\, \mathrm{d} x \mathrm{d} y = \cdots \color{gray}{\text{(magic happens)}} \cdots = \color{blue}{\frac{\pi^2}{2} \approx 4.9348}$$
I still would like to compute the double integral analytically. I converted to cylindrical coordinates, but did not get anywhere.
### Other people's work
This is the same value Johnson & Nævdal obtained in the 1990s:
Thus, the volume is
$$\left(\frac{\pi}{4}\right)^2 2^3 = \frac{\pi^2}{2}$$
However, I do not understand their work. I do not know what Schur parameters are.
Here's the script:
-- discretization step
delta = 2**(-9)
-- discretize the cube [-1,1] x [-1,1] x [-1,1]
grid1D = [-1,-1+delta..1]
grid3D = [ (x,y,z) | x <- grid1D, y <- grid1D, z <- grid1D ]
-- find points inside the 3D elliptope
points = filter (\(x,y,z)->1+2*x*y*z-x**2-y**2-z**2>=0) grid3D
-- find percentage of points inside the elliptope
p = (fromIntegral (length points)) / (1 + (2 / delta))**3
*Main> delta
1.953125e-3
*Main> p
0.6149861105903861
*Main> p*(2**3)
4.919888884723089
Hence, approximately $61\%$ of the grid's points are inside the elliptope, which gives us a volume of approximately $4.92$.
### A new Buffon's needle
A symmetric $3 \times 3$ matrix with
• $1$'s on the main diagonal
• realizations of the random variable whose PDF is uniform over $[-1,1]$ on the entries off the main diagonal
is positive semidefinite (and, thus, a correlation matrix) with probability $\left(\frac{\pi}{4}\right)^2$. Estimating the probability, we estimate $\pi$. Using the estimate given by the Haskell script:
*Main> 4 * sqrt 0.6149861105903861
3.1368420058151125
### References
• This is an astoundingly good question! Nice! – clathratus Apr 16 at 1:08
The integral can be separated:
$$I = 2\int_{-1}^1 \sqrt{1-x^2} dx \cdot \int_{-1}^1 \sqrt{1-y^2} dy = 2\left(\int_{-1}^1 \sqrt{1-t^2} dt\right)^2$$ | {
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This integral is straight-forward using the substitution $$t=\sin\theta$$:
$$\int_{-1}^1 \sqrt{1-t^2} dt = \int_{-\pi/2}^{\pi/2} \sqrt{1-\sin^2\theta} \cos\theta d\theta = \int_{-\pi/2}^{\pi/2} |\cos\theta|\cos\theta d\theta$$
$$=\int_{-\pi/2}^{\pi/2} \cos^2\theta d\theta = \dfrac{1}{2}\int_{-\pi/2}^{\pi/2} (1+\cos2\theta) d\theta = \dfrac{1}{2}\left(\theta + \dfrac{1}{2}\sin 2\theta\right)\Big|_{-\pi/2}^{\pi/2} = \dfrac{\pi}{2}$$
Therefore
$$I = 2\left(\int_{-1}^1 \sqrt{1-t^2} dt\right)^2 = 2\left(\dfrac{\pi}{2}\right)^2 = \dfrac{\pi^2}{2}$$
• Thanks. Shame on me for not realising that $(1-x^2) (1-y^2) = 1 - x^2 - y^2 + x^2 y^2$. I never expected the double integral to be this easy. – Rodrigo de Azevedo Oct 16 '16 at 19:19
• easy to get lost in the bigger picture, sometimes fresh eyes can help – David Peterson Oct 16 '16 at 19:21
Then integrand factors as $\sqrt{(1-x^2)(1-y^2)}=\sqrt{(1-x^2)}\sqrt{(1-y^2)}$ and every factor can be integrated separately. But you recognize the integral for the area of a half circle of radius $1$, hence
$$I=2\left(\frac\pi2\right)^2.$$ | {
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# Is the whole set $\mathbb R$ open?
Is the whole set $\mathbb R$ open or closed? A lot of answers from the following link said that the whole set is closed.
Why is empty set an open set?
However varies notes said that $\mathbb R$ is open, for example:
https://www.math.cornell.edu/~hatcher/Top/TopNotes.pdf
• Open does not mean not closed. – Improve Jun 29 '17 at 2:58
• It is both... some say "clopen" the empty set is the another classic example of a set that is both open and closed. – Doug M Jun 29 '17 at 3:24
• "Is the whole set R open or closed?" both. "Said that the whole set is closed". That is true. "However varies notes said that R is open". That is also true. R is open. And R is closed. – fleablood Jun 29 '17 at 4:00
It's open and closed by definition.
In order to refer to open and closed sets a topology must be made explicit. Given a set $X$ and a collection $\mathcal F$ of subsets of $X$, $\mathcal F$ is a topology on $X$ only if $X, \emptyset \in \mathcal F$. Thus X is open by definition, and as closed sets are defined as sets whose compliments are open we have $X^c=\emptyset$ and we have $X$ is closed.
In the context of a topological space $\mathbb R$ with collection $\mathcal F$, $\mathbb R$ must be open and closed. However if we consider $\mathbb R \times \mathbb R$ with the usual topology, $\mathbb R \simeq \{0\} \times \mathbb R$ is closed but not open. It is important to state context.
Indeed the observation by Michael Hardy does really make sense. Additionally the question is not well posed because it does not state which topology is to be considered. Let us assume that euclidean topology is meant. So the answer by Michael Rozenberg uses axiomatization via open or closed sets.
Using axiomatization via neighborhoods (read balls in this case, due to the euclidean topology of $\mathbb R$)
1. $\mathbb R$ is open because any of its points have at least one neighborhood (in fact all) included in it; | {
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2. $\mathbb R$ is closed because any of its points have every neighborhood having non-empty intersection with $\mathbb R$ (equivalently punctured neighborhood instead of neighborhood).
Equivalently:
1. $\mathbb R$ is open because all its points are interior points of itself
2. $\mathbb R$ is closed because all its points are adherent points of itself (equivalently limit points instead of adherent points)
Using axiomatization via Moore-Smith net convergence (read sequence convergence in this case, due to the euclidean topology of $\mathbb R$),
1. $\mathbb R$ is closed because every point to which at least one net of its points converges belongs to it
2. $\mathbb R$ is open because every point to which every convergent net converges has a non-empty intersection with it. (Or equivalently there are no nets of points of its complement (the empty set) converging to any of its points (in fact there are no nets of points of its complement))
The list can go on and on.
By the definition of a topology on a set, both the empty set and the entire set (which I'm assuming you're taking as $\mathbb R$) are open sets. Since the complement of an open set is a closed set, and $\mathbb R$ is the complement of the empty set, it is also closed.
Notice that "open" and "closed" are not mutually exclusive.
$A$ is open iff $A^c$ is closed.
So $\emptyset$ and $\Omega$ are both open and closed (clopen)
For general topology, it is guaranteed via definition of topology, i.e. $\emptyset, \Omega \in \mathscr T$.
By one commonplace definition of "open set", a set $A\subseteq\mathbb R$ is open if for every point $x\in A$ there is some open interval containing $x$ that is a subset of $A$. That clearly is true if $A=\mathbb R,$ so $\mathbb R$ is open.
By one commonplace definition of "closed set", a set $A\subseteq \mathbb R$ is closed if every limit point of $A$ is a member of $A$. That clearly is true if $A=\mathbb R$, so $\mathbb R$ is closed. | {
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Only two subsets of $\mathbb R$ are both open and closed: $\mathbb R$ and $\varnothing.$ | {
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# Integrating arcsech x
1. May 17, 2014
### sooyong94
1. The problem statement, all variables and given/known data
I was asked to prove the integral
$\int_{\frac{4}{5}}^{1} \textrm{arcsech}(x) =2\arctan 2-\frac{\pi}{2}-\frac{4}{5} \ln 2$
2. Relevant equations
Integration by parts
3. The attempt at a solution
Let $u=\textrm{arcsech} (x)$
$\textrm{sech u}=x$
$\textrm{cosh u}=\frac{1}{x}$
Differentiating implicitly,
$\textrm{sinh u} \frac{du}{dx}=\frac{-1}{x^{2}}$
$\frac{du}{dx}=\frac{-1}{x^{2}\textrm{sinh u}}$
Then I simplify it into
$\frac{-1}{x\sqrt{1-x^{2}}}$
Let $\frac{dv}{dx}=1$
$v=x$
Using integration by parts and evaluating the integral, I got
$\frac{\pi}{2}-\sin^{-1} \frac{4}{5} -\frac{4}{5} \ln 2$
Which is numerically correct. But how do I obtain $2\tan^{-1} 2$ as shown in the question above?
2. May 17, 2014
### Curious3141
Draw the standard 3-4-5 right triangle. Observe that $\frac{\pi}{2}-\sin^{-1} \frac{4}{5} = \tan^{-1}\frac{3}{4}$.
You now have to get $\tan^{-1}\frac{3}{4}$ into something with $\tan^{-1}2$ in it.
I found this a little tricky. The best solution I could find was to let:
$\tan^{-1}\frac{3}{4} = x$ so $\tan x = \frac{3}{4}$
Then let x = 2y so that $\tan 2y = \frac{3}{4}$
Solve for y in the form $y = \tan^{-1}z$, where z is something you have to find. Only one value is admissible. Express x as 2y = $2\tan^{-1}z.$
Now observe that $\frac{1 + \tan w}{1 - \tan w} = \tan(w + \frac{\pi}{4})$. Use that to find an alternative form for $\tan^{-1}z$, which will allow you to find x, in the form you need.
There might be a simpler way (indeed, it might start with an alternative solution of the integral), but I can't immediately find one.
Last edited: May 17, 2014
3. May 17, 2014
### AlephZero
To show the answers are the same, you have to show $2 \tan^{-1} 2 + \sin^{-1}\displaystyle\frac 4 5 = \pi$.
From a 3-4-5 triangle, $\sin^{-1}\displaystyle\frac 4 5 = \tan^{-1}\displaystyle\frac 4 3$. | {
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From a 3-4-5 triangle, $\sin^{-1}\displaystyle\frac 4 5 = \tan^{-1}\displaystyle\frac 4 3$.
From $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\displaystyle\frac{a+b}{1-ab}$,
$2 \tan^{-1} 2 = \tan^{-1}\displaystyle\frac {-4} 3 = \pi - \tan^{-1}\displaystyle\frac 4 3$.
QED.
4. May 22, 2014
### sooyong94
I don't get it, but why
$2 \tan^{-1} 2 = \tan^{-1}\displaystyle\frac {-4} 3 = \pi - \tan^{-1}\displaystyle\frac 4 3$.
5. May 22, 2014
### CAF123
I think the equalities should be $$2 \tan^{-1} 2 = \tan^{-1} \left(-\frac{4}{3}\right) + \pi = \pi - \tan^{-1} \left(\frac{4}{3}\right)$$
Using Alephzero's formula with $a=b$ gives $$\tan(\tan^{-1} 2 + \tan^{-1} 2) = -\frac{4}{3} \Rightarrow 2\tan^{-1} 2 = \tan^{-1} \left(-\frac{4}{3}\right) + \pi$$
6. May 22, 2014
### SammyS
Staff Emeritus
Compare the graphs of $\ y=\text{arcsech}(x) \$ and $y=\text{sech}(x)\ .$
Integrate $y=\text{sech}(x)\$ to get an area related to that given by integrating $\ y=\text{arcsech}(x) \ .$
You will have to subtract the area of some rectangle.
7. May 23, 2014
### sooyong94
I plotted two graphs and yet I can't figure it out... :(
8. May 23, 2014
### SammyS
Staff Emeritus
The definite integral you are evaluating represents the area below the y = arcsech(x) graph which is between x = 4/5 and x = 1 . Notice that arcsech(4/5) = ln(2) .
That is the same as the area below the y = sech(x) graph and above y = 4/5, for x ≥ 0. Right?
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# Number TheoryPrimitive root modulo 169
#### tda120
##### New member
How can I find a primitive root modulo 169?
I found the primitive roots mod 13 by testing 2, and then concluding that any 2^k with (k, 12)=1 would do. So that gave me 2, 6, 7 and 11. But modulo 13 I have no idea how to start.. I’m sure there’s a smarter way than trying 2^the orders that divide phi(13^2)..?.
#### Opalg
##### MHB Oldtimer
Staff member
How can I find a primitive root modulo 169?
I found the primitive roots mod 13 by testing 2, and then concluding that any 2^k with (k, 12)=1 would do. So that gave me 2, 6, 7 and 11. But modulo 13 I have no idea how to start.. I’m sure there’s a smarter way than trying 2^the orders that divide phi(13^2)..?.
Hi tda, and welcome to MHB. You might be interested in this link, which tells you that the answer to your question is either $2$ or $2+13=15$. That still leaves you with the work of testing to see if $2$ works. If it does not, then $15$ does.
#### mathbalarka
##### Well-known member
MHB Math Helper
tda120 said:
How can I find a primitive root modulo 169?
There is no simple method. You can cobble up together some basic theories on primitive roots, find a bit of a rough upperbound (although none is known to be useful for small cases) and some modular exponentiation to get a fast enough algorithm.
As Opalg there indicated, that either 2 or 15 is primitive root modulo 169, can easily be found for this case. Try proving the former and then the later if it doesn't work.
#### Klaas van Aarsen | {
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#### Klaas van Aarsen
##### MHB Seeker
Staff member
How can I find a primitive root modulo 169?
I found the primitive roots mod 13 by testing 2, and then concluding that any 2^k with (k, 12)=1 would do. So that gave me 2, 6, 7 and 11. But modulo 13 I have no idea how to start.. I’m sure there’s a smarter way than trying 2^the orders that divide phi(13^2)..?.
You don't have to check all the orders that divide $\phi(13^2)$.
It suffices to check each of the orders that are $\phi(13^2)$ divided by one of the distinct primes it contains.
$$\phi(13^2)=2^2\cdot 3 \cdot 13$$
So the orders to verify are:
$$2\cdot 3 \cdot 13,\quad 2^2 \cdot 13, \quad 2^2\cdot 3$$
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Hi tda, and welcome to MHB. You might be interested in this link, which tells you that the answer to your question is either $2$ or $2+13=15$. That still leaves you with the work of testing to see if $2$ works. If it does not, then $15$ does.
Nice!
From that link we also get that since 2 is a primitive root mod 13, it follows that the order of 2 mod 169 is either (13-1) or 13(13-1).
So if $2^{13-1} \not\equiv 1 \pmod{169}$ that means that 2 has to be a primitive root mod 169. Or otherwise 15 has to be.
In other words, no need to check any of the other powers. | {
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# Finding the coefficient on the $x$ term of ${\prod_{n = 1}^{20}(x-n)}.$
I am trying to find the coefficient on the $x$ term of $\displaystyle{\prod_{n = 1}^{20}(x-n)}$. The issue is that the binomial theorem can't be applied since our $b$ value is changing from term to term. Is there any simple way to do this problem, perhaps a way to change the expression so that the binomial theorem applies? I've tried to do that, and tried looking for a pattern on similar expressions, but I haven't come up with anything. Any help you might have would be appreciated.
-
It will be the 19th elementary symmetric polynomial in your roots. See en.m.wikipedia.org/wiki/Elementary_symmetric_polynomial for symmetric polynomials. – Marc Jun 1 '14 at 17:11
A concept popularly known as sum of roots of a polynomial can be of help. – fermesomme Jun 1 '14 at 17:22
The coefficient is given by $$\sum_{k=1}^{20}\prod_{n\not=k\atop n=1,\dots,20}{(-n)}=-\sum_{k=1}^{20}\frac{20!}{k}=-20! H_{20}$$ where $H_k$ is the $k$-th Harmonic number. Those can be looked up in tables (see A001008 and A002805): $H_{20}=\frac{55835135}{15519504}$ and thus the coefficient is given by $-8752948036761600000$.
-
This answer just gets better and better on a moment-by-moment basis! Thanks for adding the value of $H_{20}$! – Robert Lewis Jun 1 '14 at 18:30
Just a question, is there a simple way to calculate the constant of the expanded form? – recursive recursion Jun 1 '14 at 19:47
@Dominik, Also, I'm not sure how you got your first expression. Could you please explain that – recursive recursion Jun 1 '14 at 19:51
You get terms with a single power in $x$ from the product $\prod_{n=1}^{20}(x-n)$ by taking $19$ non-$x$ factors and one $x$. The sum is the sum over the different choices of those non-$x$ and $x$ factors. More formally you could try to prove that $(-1)^{N+1} N! H_{N}$ is the $x$-coefficient of $\prod_{n=1}^N (x-n)$ by induction. – Dominik Jun 1 '14 at 20:14
This is one of Vieta's formulas. | {
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This is one of Vieta's formulas.
-
The coefficient must be less than $-20!$, so this can't be correct? – copper.hat Jun 1 '14 at 17:22
Not to put too fine a point on it, but this answer is incorrect; the coefficient of $x^{19}$ is in fact $-\sum_{n = 1}^{20} n = -(20 \times 21)/2 = -210$; for the coefficient of $x$, see the answer given by Dominik. I do believe it is correct. – Robert Lewis Jun 1 '14 at 17:31
@RobertLewis: Unfortunately no 'Med' time for me this morning... – copper.hat Jun 1 '14 at 17:37
@Robert Lewis. Shoot! You're right! I'll remove everything but the vieta's formula link. – Avi Steiner Jun 1 '14 at 17:40
@Avi Steiner: well done! Glad we got it right! – Robert Lewis Jun 1 '14 at 19:40
One way is to just use Maclaurin's formula: $$f(x) = f(0) + \frac{f'(0)}{1!} x + \ldots$$ In this case: \begin{align} f'(x) &= \sum_{1 \le n \le m} \prod_{\substack{1 \le k \le m\\k \ne n}} (x - k) \\ f'(0) &= (-1)^{m - 1} m! H_m \end{align}
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The question has been well answered by others. However, I would like to point out that this polynomial has a name---Wilkinson's polynomial---and its own Wikipedia article. It was put forward by Wilkinson as an example of an apparently innocuous polynomial with remarkable numerical-analytic properties.
-
This polynomial is a lot more interesting than I thought! Thanks for giving me a name to google. – recursive recursion Jun 1 '14 at 21:29
A generalization to other coefficients (and limits different from $20$) is given in the generating functions of Stirling numbers of the first kind, Pochammer symbols:
$$(x)_n:=\prod_{k=0}^{n-1}(x-k)=\sum_{k=0}^n s(n,k)x^k.$$
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The coefficients are the Stirling numbers of the first kind :
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# Moderate Probability Solved QuestionAptitude Discussion
Q. Four different objects 1, 2, 3, 4 are distributed at random in four places marked 1, 2, 3, 4. What is the probability that none of the objects occupy the place corresponding to its number?
✖ A. 17/24 ✔ B. 3/8 ✖ C. 1/2 ✖ D. 5/8
Solution:
Option(B) is correct
First of all, if we IGNORE the condition about where the objects can be placed, we can arrange the 4 different objects in 4! ways (= 24 ways).
So, we now must determine HOW MANY of those 24 arrangements are such that no objects occupy the location corresponding to its number.
A quick way to do this is to LIST acceptable outcomes.
IMPORTANT: We'll list each arrangement so that the first number represents the object that goes to location #1, the second number represents the object that goes to location #2, and so on.
So, for example, 3421 represents object #3 in location #1, object #4 in location #2, object #2 in location #3, and object #1 in location #4.
Let's be systematic:
Arrangements where object #2 is in location #1
The possible arrangements where NO object is in the correct location are as follows:
2143, 2341, 2413
Total number of arrangements $= 3$
Arrangements where object #3 is in location #1
The possible arrangements where NO object is in the correct location are as follows:
3142, 3412, 3421
Total # of arrangements $= 3$
Arrangements where object #4 is in location #1
The possible arrangements where NO object is in the correct location are as follows:
4123, 4312, 4321
Total # of arrangements $= 3$
Altogether, the number of arrangements where no object is in the correct location,
$= 3 + 3 + 3$
$= 9$
So, $P(\text{no object in correct location}) = \dfrac{9}{24}$
$= \dfrac{3}{8}$
Thus, option (B) is the correct choice.
Edit: Based on Vaibhav's comment, the solution has been updated.
Edit 2: For an alternate solution see Kartik's comment | {
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Edit 2: For an alternate solution see Kartik's comment
Edit 3: Sheldon has provided a way to reach the correct answer without even solving the question
Edit 4: Suzen has given an exhaustive solution
Edit 5: For derangements formula, check Himanshu Singh's comment.
Note: For better understanding and different approaches do visit comment section.
## (14) Comment(s)
Gaurav Karnani
()
Total number of cases = 24
let us divide the case into categories where :
1) single number is in right position = 8 cases; two for each $1,2,3,4$
2) Two numbers are in right position = 6 cases $12,13,14,23,24,34$
3) No case for three digits as if three are in correct position fourth one will definitely be, so only one case of $1234$ .
total number of cases =15
Probability = $9/24$
Himanshu Singh
()
Please use this formula for derangements :
$!n = n!\left(1-\dfrac{1}{1!} + \dfrac{1}{2!} - \dfrac{1}{3!} + \dfrac{1}{4!} ... (-1)^n \times \dfrac{1}{n!} \right)$
Kartik
()
Total number of ways of placing objects into places randomly,
$=4 \times 3 \times 2 \times 1=4!=24$
No. of ways of placing them such that only one of them gets correct place,
\begin{align*} = & 4 \text{ (ways of choosing the correctly} \\ & \text{ placed objects)} \times \\ & \times 2 \text{ (to place next object wrongly)} \\ & \times 1 \text{ (to place next object wrongly)} \\ & \times 1 \text{ (to place next object wrongly)}\\ = & \textbf{8} \end{align*}
Number of ways of placing exactly two objects correctly,
\begin{align*} = & ^4C_2 \text{ (to identify the two correctly placed} \\ & \text{ objects their arrangment are unique}\\ & \text{ for them to be correct)} \times \\ & 1 \text{ (to arrange next object wrongly)} \times \\ & 1 \text{ (to arrange next object wrongly)}\\ = & \textbf{6} \end{align*}
Now,
Number of ways to place exctly 3 objects correct = ways to place all objects correct $= \textbf{1}$
Therefore, number of ways to place all digits wrongly,
$=24-(8+6+1)$
$=24-15$
$=9$ | {
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Therefore, number of ways to place all digits wrongly,
$=24-(8+6+1)$
$=24-15$
$=9$
Thus, Required probability,
$=\dfrac{9}{24}$
$=\dfrac{3}{8}$
Raj
()
Since there's only one correct way of placing the objects as per the boxes- can't we work on the lines of using probability of an event happening = 1 - probability of that event not happening?
So therefore, can't the answer be 1 - (1/24) = 23/24?
Brent
()
That's a good idea, but it doesn't apply here.
If event A = NONE of the objects in the correct places, then the complement (event A NOT happening) will consist of any arrangement where some (perhaps all) of the objects are in their correct place(s).
Vaibhav
()
but this is the case for 2 at first position. it can occupy position 3 and 4 as well . Similar case with other digits will be there.
Deepak
()
You are right Vaibhav, updated the solution.
Suzen
()
Here's the full list:
1234
1243
1423
1432
1324
1342
2134
2143 OK
2431
2413 OK
2341 OK
2314
3124
3142 OK (Close to 1000pi, but no correlation really, or is there? Perhaps the nth position is not n for all decimal places. OK, an interesting idea to explore)
3421 OK
3412 OK
3214
3241
4123 OK
4132
4321 OK
4312 OK
4213
4231
Clearly $P = \frac{9}{24} = \frac{3}{8}$
I know it's long winded, but I'm glad it matched!
Tesla
()
Let a particular number (say) number 2 occupies position 1.
Then all possible arrangement are given as:
(2,1,3,4), (2,1,4,3), (2,3,4,1), (2,4,1,4), (2,4,1,3), (2,4,3,1).
Out of these six, three (2,1,3,4), (2,3,1,4), (2,4,3,1) are not acceptable because numbers 3 and 4 occupy the correct positions.
Required probability = 3/6 = 1/2
Sheldon
()
I am not sure what difficulty level this would fall into.
If I see this question in the real exam, I would spend about a minute solving it.
After that, I would guess and move on. because it would be time consuming. | {
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After that, I would guess and move on. because it would be time consuming.
Looking at the question, it is more obvious that at-least 1 number would fall into its corresponding place.
So definitely, the probability that the number wouldn't fall into its numbered place should be less than half.
Looking at options:
A. 17/24
B. 3/8
C. 1/2
D. 5/8
Only option B is less than half. I would pick B and move on.
Brent
()
That's a great approach that allows you to minimize time spent on a question (that you feel is going nowhere) and maximize your guess.
Probability questions are perfect for this, because most people have a gut feeling about how likely something is. As you suggest, it does seem unlikely (probability less than 0.5) that every object would be out of place, so B is the perfect guess.
Steve
()
I agree with sam and abhishek
Abhishek
()
the correct answer will be $\dfrac{9}{24} =\dfrac{3}{8}$
out of
1234 1243 1324 1342 1423 1432
2341 2314 2143 2134 2413 2431
3124 3142 3412 3421 3214 3241
4123 4132 4213 4231 4312 4321
24 combinations
2341 2143 2413
3421 3412 3142
4123 4312 4321
are misplaced(dearranged)
So, the correct answer will be $\dfrac{9}{24}$ i.e. $\dfrac{3}{8}$. :)
Sam
()
I think Abhishek is correct...... answer should be 3/8 | {
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Limit involving Series and Greatest Integer Function
If $[$.$]$ denotes the greatest integer function, then find the value of $\lim_{n \to \infty} \frac{[x] + [2x] + [3x] + … + [nx]}{n^2}$
What I did was, I wrote each greatest integer function $[x]$ as $x - \{x\}$, where $\{.\}$ is the fractional part. Hence, you get
$\lim_{n \to \infty} \frac{\frac{n(n+1)}{2}(x-\{x\})}{n^2}$
The limit should then evaluate to $\frac{x-\{x\}}{2}$
But the answer given is $\frac{x}{2}$. What am I missing here?
$$\frac{\lfloor x\rfloor+\ldots+\lfloor nx\rfloor}{n^2}=\frac{x+2x+\ldots nx-\{x\}-\ldots-\{nx\}}{n^2}=$$
$$=\frac{n(n+1)}{2n^2}x-\frac{\{x\}+\ldots+\{nx\}}{n^2}\xrightarrow[n\to\infty]{}\frac12x-0=\frac x2$$
since the second addend above tends to zero:
$$\frac{\{x\}+\ldots+\{nx\}}{n^2}\le\frac n{n^2}=\frac1n\xrightarrow[n\to\infty]{}0$$
• I didn't really get how the second addend (fractional part thing) tends to zero, as n tends to infinity. Isn't it in an inderminate form as it is? Also, what has been done in the last line of the answer? – skb May 28 '18 at 20:20
• @skb Every term $\;\{kx\}\;$ is less than $\;1\;$ , so that whole sum's numerator is less that $\;1+1+\ldots+1=n\;$ ... – DonAntonio May 28 '18 at 20:22
• Have you used the sandwich theorem in the last line? But shouldn't there be another expression less than it for it to work? – skb May 28 '18 at 20:24
• @skb But isn't it obvious that the whole expression is greater than zero or equal to it? You can complete that argument... – DonAntonio May 28 '18 at 20:27
• Got it. Thanks a lot for the help! – skb May 28 '18 at 20:27
Note that by Stolz-Cesaro
$$\lim_{n \to \infty} \frac{\lfloor x\rfloor + \lfloor 2x \rfloor + \lfloor3x\rfloor + … + \lfloor nx]}{n^2}=\lim_{n \to \infty} \frac{\lfloor(n+1)x\rfloor}{(n+1)^2-n^2}=\lim_{n \to \infty} \frac{\lfloor(n+1)x\rfloor}{2n+1}=$$ $$=\lim_{n \to \infty} \frac{(n+1)x}{2n+1}-\frac{\{(n+1)x\}}{2n+1}$$ | {
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# a pyramid
• August 24th 2008, 01:57 AM
perash
a pyramid
Consider a pyramid with a square base. The side length of the
base is 2 units and the height of the pyramid is 1 unit. Imagine
placing a cube inside this pyramid (resting on the base of the
pyramid) such that each of the four top corners of the cube is
touching each of the 4 slanted edges of the pyramid .
Find the dimensions of the cube. That is, find the value, in
units, of the length of one edge of the cube.
• August 24th 2008, 03:27 AM
ticbol
Quote:
Originally Posted by perash
Consider a pyramid with a square base. The side length of the
base is 2 units and the height of the pyramid is 1 unit. Imagine
placing a cube inside this pyramid (resting on the base of the
pyramid) such that each of the four top corners of the cube is
touching each of the 4 slanted edges of the pyramid .
Find the dimensions of the cube. That is, find the value, in
units, of the length of one edge of the cube.
So the the top four corners of the cube are along the slanted edges of the pyramid. Then the four edges of the base of the cube are parallel to the edges of the base of the pyramid each to each.
View the figure from the top, or on the top. We will get a vertical cross-section along one of the equal diagonals of the base of the pyramid.
The diagonal of the pyramid is 2sqrt(2) units long...by Pythagorean theorem.
The diagonal of the cube, whose edge is, say, x units long, is x*sqrt(2) units long. | {
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Now view that said cross-section vertically, or from the side, or from one of the un-cut corner of the base of the pyramid.
The figure is that of an isosceles triangle whose base is 2sqrt(2) units long, and whose height is 1 unit long.
Inside it is a rectangle whose base is x*sqrt(2) units long, and whose height is x units long.
Above the rectangle is a smaller isosceles triangle whose base is x*sqrt(2) and whose height is (1-x) units long.
The two isosceles triangles are similar, and so, proportional.
By proportion,
2sqrt(2) /1 = x*sqrt(2) /(1-x)
2 = x /(1-x)
2(1-x) = x
2 -2x = x
2 = x +2x
x = 2/3 unit long ---------------answer
• August 24th 2008, 05:16 AM
Soroban
Hello, perash!
I used the same approach as ticbol, but got a different answer.
Quote:
Consider a pyramid with a square base. The side length of the base is 2 units
and the height of the pyramid is 1 unit. Imagine placing a cube inside this pyramid
(resting on the base of the pyramid) so that each of the four top corners of the cube
is touching each of the 4 slanted edges of the pyramid.
Find the dimensions of the cube.
That is, find the value, in units, of the length of one edge of the cube.
Looking down on the pyramid, we see:
Code:
*-----------* | * * | | * * | | * | 2 | * * | | * * | *-----------* 2
The diagonal of this square is $2\sqrt{2}$
Slice the pyramid along a diagonal and we have this cross-section:
Code:
- * : * | * : * | * : *-----+-----* 1 * | | | * : * | | | * : * | | 2x| * : * | | | * - *---------*-----*-----*---------* : √2 : x : √2-x :
The lower-right right triangle is similar to the largest right triangle.
We have: . $\frac{2x}{\sqrt{2}-x} \:=\:\frac{1}{\sqrt{2}} \quad\Rightarrow\quad 2\sqrt{2}x \:=\:\sqrt{2} - x$ | {
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. . $2\sqrt{2}x + x \:=\:\sqrt{2} \quad\Rightarrow\quad (2\sqrt{2}+1)x \:=\:\sqrt{2} \quad\Rightarrow\quad x \:=\:\frac{\sqrt{2}}{2\sqrt{2}+1}
$
The side of the cube is: . $2x \;=\;\frac{2\sqrt{2}}{2\sqrt{2} + 1} \;=\;\frac{2(4-\sqrt{2})}{7}$
• August 24th 2008, 04:58 PM
ticbol
Quote:
Originally Posted by Soroban
Hello, perash!
I used the same approach as ticbol, but got a different answer.
Looking down on the pyramid, we see:
Code:
*-----------* | * * | | * * | | * | 2 | * * | | * * | *-----------* 2
The diagonal of this square is $2\sqrt{2}$
Slice the pyramid along a diagonal and we have this cross-section:
Code:
- * : * | * : * | * : *-----+-----* 1 * | | | * : * | | | * : * | | 2x| * : * | | | * - *---------*-----*-----*---------* : √2 : x : √2-x :
The lower-right right triangle is similar to the largest right triangle.
We have: . $\frac{2x}{\sqrt{2}-x} \:=\:\frac{1}{\sqrt{2}} \quad\Rightarrow\quad 2\sqrt{2}x \:=\:\sqrt{2} - x$
. . $2\sqrt{2}x + x \:=\:\sqrt{2} \quad\Rightarrow\quad (2\sqrt{2}+1)x \:=\:\sqrt{2} \quad\Rightarrow\quad x \:=\:\frac{\sqrt{2}}{2\sqrt{2}+1}
$
The side of the cube is: . $2x \;=\;\frac{2\sqrt{2}}{2\sqrt{2} + 1} \;=\;\frac{2(4-\sqrt{2})}{7}$
I'm sorry to comment, since perash or anybody did not comment, but the cube should appear as a rectangle, not a square, in your diagram. The length of the base now of the rectangle should be (2x)sqrt(2). Not 2x as is in your diagram. You sliced along a diagonal of the base of the pyramid, remember.
• August 26th 2008, 03:17 AM
Soroban
Hello, ticbol!
Another blunder . . . *blush*
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Another blunder . . . *blush*
Quote:
The cube should appear as a rectangle, not a square, in your diagram.
The length of the base now of the rectangle should be (2x)sqrt(2).
Not 2x as is in your diagram.
You sliced along a diagonal of the base of the pyramid, remember.
Absolutely right!
I'll try to correct my work and get back soon . . . | {
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# How to know if the angle is positive or negative in inverse trigonometry
I have been doing this problem this problem:
$$Cos[Tan^{-1}(-\frac{2}{3})]$$ So I was instructed to draw a triangle to guide me so I did
Now once I drew my triangle I found the hypotenuse, which is $$\sqrt{13}$$
And then I was able to obtain the answer to this expression which I got:
$$\frac{2\sqrt{13}}{13}$$
However, I am told I drew the triangle wrong, it is actually -2 (negative) and (3) positive. Why is it that the triangle is wrong? I was told the actual answer is
$$\frac{3\sqrt{13}}{13}$$ | {
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$$\frac{3\sqrt{13}}{13}$$
Arctan has a range of $\frac{-\pi}{2}\le{y}\le\frac{\pi}{2}$. Now let $arctan\frac{-2}{3}=y$. This implies $tany=\frac{-2}{3}$. Because tan is negative, we know y must lie in quadrant IV. It cannot lie in quadrant I because tan is positive in quadrant I. Therefore we draw our angle as you have above. Now, note that $tan\theta=\frac{opposite}{adjacent}$. Therefore, you should have $-2$ where $3$ is in your picture, and you should have $3$ where $-2$ is. Your line should be drawn to the coordinate $(3,-2)$. Now, we find the hypotenuse as you have already done using the Pythagorean Theorem. We find that it is $\sqrt{13}$ as you've noted. Now, because we are finding $cos(arctan\frac{-2}{3})$, we will use the fact that $cos\theta=\frac{adjacent}{hypotenuse}$. So, this gives $\frac{3}{\sqrt{13}}$. Rationalizing we have $\frac{3\sqrt{13}}{13}$
• Yes, tangent is negative in quadrant IV and II, but the range for arctan is limited to $\frac{-\pi}{2}\le{y}\le\frac{\pi}{2}$ so our angle cannot lie in quadrant II. It must lie in either quadrant I or IV. And since it is negative, it must lie in quadrant IV. – MathGuy Jan 27 '17 at 3:04
• Yes, the ranges are different though depending on the inverse trig function. For arcsin the range is $\frac{-\pi}{2}\le{y}\le\frac{\pi}{2}$, for arccos the range is $0\le{y}\le\pi$ and for arctan the range is $\frac{-\pi}{2}\le{y}\le\frac{\pi}{2}$ – MathGuy Jan 27 '17 at 3:12 | {
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# Further question on My Fractions problem
#### tmt
##### Active member
I have a separate question on the same problem from my prior post.
I need an equation for a tangent which has a slope of 5/6 and passes through (1,3)
y-3 = 5/6(x-1)
I simplify this to
y= 5/6x +13/6
However the answer given is y = 7/6(x) + 13/6
Where am I going wrong?
Yours,
Timothy
#### MarkFL
Staff member
Your line has the required slope, while the given answer has the wrong slope. It is most likely a typo somewhere, either in the statement of the problem or the given answer. Can you post the original problem in its entirety?
#### tmt
##### Active member
PROBLEM 11 : Find an equation of the line tangent to the graph of x2 + (y-x)3 = 9 at x=1 . '
This is the end of the answer:
Thus, the slope of the line tangent to the graph at (1, 3) is
$m = y' = \displaystyle{ 3 (3-1)^2 - 2(1) \over 3 (3-1)^2 } = \displaystyle{ 10 \over 12 } = \displaystyle{ 5 \over 6 }$ ,
and the equation of the tangent line is
y - ( 3 ) = (5/6) ( x - ( 1 ) ) ,
or
y = (7/6) x + (13/6) .
I suspect it is a typo in the answer. Here is the link for the full answer.
https://www.math.ucdavis.edu/~kouba...soldirectory/ImplicitDiffSol.html#SOLUTION 11
#### MarkFL
Staff member
Okay, I see now...I assumed the slope was given as 5/6. Let's take a look at the problem. We are given the curve:
$$\displaystyle x^2+(y-x)^3=9$$
So, implicitly differentiating with respect to $x$, we find:
$$\displaystyle 2x+3(y-x)^2\left(\frac{dy}{dx}-1 \right)=0$$
Solving for $$\displaystyle \frac{dy}{dx}$$, we find:
$$\displaystyle \frac{dy}{dx}=1-\frac{2x}{3(y-x)^2}$$
Now, when $x=1$, we find from the original curve:
$$\displaystyle 1^2+(y-1)^3=9$$
$$\displaystyle y=3$$
And so we find the slope at the given point is:
$$\displaystyle \left.\frac{dy}{dx} \right|_{(x,y)=(1,3)}=1-\frac{2(1)}{3(3-1)^2}=1-\frac{2}{12}=\frac{5}{6}$$
Hence, using the point-slope formula, we obtain the tangent line: | {
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Hence, using the point-slope formula, we obtain the tangent line:
$$\displaystyle y-3=\frac{5}{6}(x-1)$$
$$\displaystyle y=\frac{5}{6}x+\frac{13}{6}$$
Here is a plot of the curve and the tangent line:
#### Deveno
##### Well-known member
MHB Math Scholar
PROBLEM 11 : Find an equation of the line tangent to the graph of x2 + (y-x)3 = 9 at x=1 . '
This is the end of the answer:
Thus, the slope of the line tangent to the graph at (1, 3) is
$m = y' = \displaystyle{ 3 (3-1)^2 - 2(1) \over 3 (3-1)^2 } = \displaystyle{ 10 \over 12 } = \displaystyle{ 5 \over 6 }$ ,
and the equation of the tangent line is
y - ( 3 ) = (5/6) ( x - ( 1 ) ) ,
or
y = (7/6) x + (13/6) .
I suspect it is a typo in the answer. Here is the link for the full answer.
https://www.math.ucdavis.edu/~kouba...soldirectory/ImplicitDiffSol.html#SOLUTION 11
I concur that both you and MarkFL are correct, the link you provided has a typo in the very last line (it is correct until that), and the correct tangent line has the equation:
$y = \dfrac{5}{6}x + \dfrac{13}{6}$ | {
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Find all values of $k$ for $kx^2+(k+2)x-3=0$ with positive roots.
$$kx^2+(k+2)x-3=0$$
This quadratic has roots which are real and positive.
Find all possible values of $k$.
$$Δ = (k+8)^2-60$$ $$==> (k+8)^2-60>0$$ $$k>2\sqrt{15}\ - 8$$ and $$k<-2\sqrt{15}\ -8$$
However this didn't look right. Any suggestions?
Edit: I think I might have figured it out.
Since we know,
$$k\not= 0$$ $$\frac{-3}{k}>0$$ $$\therefore k<0$$
What I got earlier was not wrong, but rather incomplete.
$$k<-2\sqrt{15} -8$$ This can be ruled out since, $$k<0$$ $$\therefore 2\sqrt{15} -8 < k < 0$$
If someone could please still check my work that would be nice.
• I appreciate the advice, I had already tried something but was not sure if it was the correct "path" for this problem so I didn't show it. – StrBoP Jul 9 '18 at 18:48
• You're welcome. Note that positive numbers are always real, so the "real" in "real and positive" is redundant. – Shaun Jul 9 '18 at 18:54
• Do you mean $(k+2)^2$? – Chris2018 Jul 9 '18 at 18:59
• No, I can show more work if you'd like. But after simplifying to a greater extent you get a quadratic, Since that quadratic wasnt factorable I completed the square and then I got that. Sorry if I confused you. – StrBoP Jul 9 '18 at 19:02
Hint:
Using Vieta's formulas, you can see that the multiplication of the roots of a quadratic $ax^2+bx+c$ is given by $\frac{c}{a},$ which is equal to $\frac{-3}{k}$ in your case.
Since both roots are positive, this means that $\frac{-3}{k}$ must be positive and $k$ must be negative. Also, the sum of the roots (given by $-\frac{b}{a}$) must be positive too: $-\frac{k+2}{k} > 0$.
Since we already know that $k$ is negative, we get $-(k+2) < 0 \iff -2 < k$. Hence, $-2<k<0$ so far. | {
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Since we already know that $k$ is negative, we get $-(k+2) < 0 \iff -2 < k$. Hence, $-2<k<0$ so far.
Now in order for this quadratic to have real roots, its discriminant must be non-negative. After you write it down, you see that you should solve the inequality $$(k+2)^2-4k(-3) =(k+2)^2+12k \geq 0$$ to find the range of values for $k,$ and then intersect it with $$-2 < k <0$$ at the end.
$$(k+2)^2+12k \geq 0 \iff k\geq \sqrt{60}-8 \text{ or } k\leq -8 -\sqrt{60}$$
Intersecting this range with $-2 < k < 0$ gives $\sqrt{60}-8 \leq k<0$.
• Might also be helpful to quote Vieta's formulas. – TheSimpliFire Jul 9 '18 at 18:47
• A positive product of the roots is not enough. – Bernard Jul 9 '18 at 18:51
• @stressed out , Thank you for the help – StrBoP Jul 9 '18 at 18:59
• How do you get $-(k+2) < 0 \iff 2 < k$? $$-(k+2) < 0 \implies -2 < k$$ From this it does not follow that $2 < k$ – gd1035 Jul 9 '18 at 19:05
• @stressedout Sorry for bothering once again, but I checked at the back of the book and the answer read: -8 +√60 </ k < 0 – StrBoP Jul 9 '18 at 19:05
You have three conditions to check:
1. This equation has real roots. It means it is a quadratic equation ($k\ne 0$) and its discriminant should be positive: $$\Delta=(k+2)^2+12k=k^2+16k+4=(k+8)^2-60>0,$$ so either $k<-8-2\sqrt{15}\:$ or $\:k>-8+2\sqrt{15}$.
2. The roots must have the same sign, i.e. their product $-\dfrac 3k>0$, which means $\:\color{red}{k<0}$.
3. This sign must be positive. If the roots have the same sign, this sign is also the sign of their sum $\:-\dfrac{k+2}k$, which is the sign of the product $-k(k+2)$. To sum it up, we need to have $k(k+2)<0$, which happens if and only if $\: \color{red}{-2<k<0}$.
Now note that$\:-8-2\sqrt{15}<-2$, and as $\;3<\sqrt{15}<4$ we have $\:-2<-8+2\sqrt{15}<0\:$.
Thus eventually the solutions are $$-8+2\sqrt{15}<k<0.$$
• The maniac downvoter struck gain! – Bernard Jul 9 '18 at 23:10 | {
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# Find $\sum_{m=0}^n\ (-1)^m m^n {n \choose m}$
I'm going to university in October and thought I'd have a go at a few questions from one of their past papers. I have completed the majority of this question but I'm stuck on the very last part. In honesty I've been working on this paper a while now and I'm a bit tired so I'm probably giving up earlier than I usually would.
I won't write out the full question, only the last part:
Let $$S_r(n) = \sum_{m=0}^n\ (-1)^m m^r {n \choose m}$$ where r is a non-negative integer . Show that $S_r(n)=0$ for $r<n$. Evaluate $S_n(n)$.
I have shown that $S_r(n)=0$ for $r<n$ by taking $(1+z)^n= \sum_{m=0}^n\ z^m {n \choose m}$, letting $D_r(f(z))=z\frac d{dz}(z\frac d{dz}...(\frac d{dz}(f(z)))...)$ where $z\frac d{dz}$ is applied $r$ times and applied it to both sides. The left hand side gives a polynomial, degree n which has factor $(1+z)$ for all $r<n$ and the right hand side gives $\sum_{m=0}^n\ z^m (m)^r {n \choose m}$. Setting $z=-1$ yields the required result.
There is some build up to this, so I'm fairly certain that this was the intended method.
I'm stuck however on the very last part. I have tried finding a form for $D_r((1+z)^n)$, but I'm fairly sure that this isn't the correct approach, as the wording of the question implies that $S_n(n)$ needs to be considered separately.
I'm surprised that I didn't find that this question had already been asked, so apologies if it has been.
Thank you.
• Looks like a close relative of Stirling numbers of the second kind. Please see Wikipedia. – André Nicolas Jul 7 '16 at 22:01
• Ok thanks, will do – Aka_aka_aka_ak Jul 7 '16 at 22:12
• You are welcome. Note the close relationship to the number of onto functions. The calculation you carried out successfully (and that I would have trouble with) can be bypassed once we note there are $0$ onto functions from an $r$-element set to an $n$-element set if $r\lt n$. – André Nicolas Jul 7 '16 at 22:23 | {
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The following relation encapsulates the Stirling number semantics:
$$m^r = r! [z^r] \exp(mz).$$
$$S_r(n) = r! [z^r] \sum_{m=0}^n {n\choose m} (-1)^m \exp(mz) = r! [z^r] (1-\exp(z))^n.$$
Now observe that
$$1-\exp(z) = - z - \frac{z^2}{2} - \frac{z^3}{6} -\cdots$$
which means that $(1-\exp(z))^n$ starts at $[z^r]$ where $r=n$ with coefficient $(-1)^n$, producing for the sum the value
$$(-1)^n\times n!$$
and the coefficients on $[z^r]$ with $r\lt n$ are zero.
The general form of the summation
$$\sum_{m=0}^n(-1)^mm^n\binom{n}m\;,\tag{1}$$
with the alternating sign and the binomial coefficient $\binom{n}m$, suggests that it can be interpreted as an inclusion-exclusion calculation. However, the $m=0$ term is $0$, from which we begin by subtracting a positive quantity $\binom{n}1=n$, which is a bit odd for such a calculation. This suggests letting $k=n-m$ and rewriting as
$$\sum_{k=0}^n(-1)^{n-k}(n-k)^n\binom{n}{n-k}=(-1)^n\sum_{k=0}^n(-1)^k\binom{n}k(n-k)^n\;,$$
where the righthand side has been arranged to look like a more or less typical inclusion-exclusion calculation. The factor $(n-k)^n$ is the one that isn’t part of the inclusion-exclusion machinery. It has a natural interpretation as the number of functions from $[n]$ to $[n]$ whose ranges are disjoint from some $k$-element subset of $[n]$. If for each $k\in[n]$ we let $F_k$ be the set of functions from $[n]$ to $[n]$ whose ranges do not contain $k$, we have
$$\left|\bigcap_{k\in I}F_k\right|=(n-|I|)^n$$
whenever $\varnothing\ne I\subseteq[n]$, so
\begin{align*} \left|\bigcup_{k=1}^nF_k\right|&=\sum_{\varnothing\ne I\subseteq[n]}(-1)^{|I|-1}\left|\bigcap_{k\in I}F_k\right|\\ &=\sum_{\varnothing\ne I\subseteq[n]}(-1)^{|I|-1}(n-|I|)^n\\ &=\sum_{k=1}^n(-1)^{k-1}\binom{n}k(n-k)^n\;. \end{align*}
This is the number of functions from $[n]$ to $[n]$ that are not surjective, i.e., the number that aren’t bijections. It follows that the number of bijections from $[n]$ to $[n]$ is | {
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$$n^n-\sum_{k=1}^n(-1)^{k-1}\binom{n}k(n-k)^n=\sum_{k=0}^n(-1)^k\binom{n}k(n-k)^n\;.$$
On the other hand, we know that there are $n!$ bijections from $[n]$ to $[n]$, so
$$\sum_{m=0}^n(-1)^mm^n\binom{n}m=(-1)^n\sum_{k=0}^n(-1)^k\binom{n}k(n-k)^n=(-1)^nn!\;.$$
Note that the same basic argument handles the previous part of the problem. The factor $(n-k)^n$ is replaced by $(n-k)^r$, the number of functions from $[r]$ to $[n]$ whose ranges are disjoint from some specified $k$-element subset of $[n]$. The expression
$$\sum_{k=0}^n(-1)^k\binom{n}k(n-k)^r$$
counts the surjective functions from $[r]$ to $[n]$, and if $r<n$, this is of course $0$.
All of this is very closely related to the Stirling numbers of the second kind.
• @BrianMScott: Very nice and clearly written exposition. (+1) – Markus Scheuer Jun 30 '18 at 16:51 | {
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# Least Squares Solver | {
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Define least squares. In order for the solution to represent sensible pixel values, restrict the solution to be from 0 through 1. 1 Introduction. Students, teachers, parents, and everyone can find solutions to their math problems instantly. The numerical instability and performance are issues of larger problems and general setting. This example shows how to solve a nonlinear least squares problem in two ways. While symbolically correct, using the QR decomposition instead is numerically more robust. - linear_least_squares. Solve least-squares (curve-fitting) problems. Sitio Espejo para América Latina. Linear regression calculator Two-dimensional linear regression of statistical data is done by the method of least squares. Linear Regression calculator uses the least squares method to find the line of best fit for a sets of data X and Y or the linear relationship between two dataset. Let us understand What is Linear Regression and how to perform it with the help Ordinary Least Squares (OLS) estimator with an example. 1 Solving Least Squares Systems: SVD Approach One way to solve overdetermined systems is to use the Singular Value Decomposition of a matrix. solver to vary the values for A, C and k to minimize the sum of chi squared. LMS incorporates an. Solve any equations from linear to more complex ones online using our equation solver in just one click. Linear vs. From the geometric perspective, we can deal with the least squares problem by the following logic. Enter the statistical data in the form of a pair of numbers, each pair is on a separate line. Algorithm 1 Least-squares sub-problem input: H 2 CN ⇥N, q 2 CN, P 2 RM ⇥N, d 2 CM for each receiver (j )(rowinP) in parallel do H⇤ w j = p⇤ j {solve 1 PDE} end for W =[w 1 w 2w m] {distributed matrix} S =(I M + 2 W⇤ W)1 {adjust using Algorithm 2 (optional)} for source (i) in parallel do y i =(I N 2 WSW⇤)(q i + 2 Wd i) Hu i = y i {solve 1 PDE} end for output: u. The main purpose is to provide an example of the basic | {
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Hu i = y i {solve 1 PDE} end for output: u. The main purpose is to provide an example of the basic commands. Recall the formula for method of least squares. Microsoft Excel provides a tool called Solver that handles this prob-lem in a manner that is transparent to the user. Now click on fiSolvefl. Find a linear least squares fit for a set of points in Visual Basic. In this section w e brie y presen t the most cited w orks in ellipse tting and its closely related problem, conic tting. Least squares fit is a method of determining the best curve to fit a set of points. This article demonstrates how to generate a polynomial curve fit using. When we pass this (near) optimal solution to NL2SOL it will have an easy task. Nonlinear Least Squares Data Fitting D. 1 Introduction More than one explanatory variable In the foregoing chapter we considered the simple regression model where. An apparatus is available that marks a strip of paper at even intervals in time. Most math majors have some exposure to regression in their studies. Least-squares imaging and deconvolution using the hybrid norm conjugate-direction solver Yang Zhang and Jon Claerbout ABSTRACT To retrieve a sparse model, we applied the hybrid norm conjugate-direction (HBCD) solver proposed by Claerbout to two interesting geophysical problems: least-squares imaging and blind deconvolution. Learn more about least squares, curve fitting, optimization, nonlinear, fitting. A number of methods may be employed to solve this problem. In some applications, it may be necessary to place the bound constraints $$l \leq x \leq u$$ on the variables $$x$$. solve_least_squares_lm This is a function for solving non-linear least squares problems. First, least squares is a natural approach to estimation, which makes explicit use of the structure of the model as laid out in the assumptions. lstsq for the "direct" appraoch (as far as I know this uses SVD by standard, but I also tried all the other LINPACK options that scipy offers) | {
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I know this uses SVD by standard, but I also tried all the other LINPACK options that scipy offers) $\endgroup$ – Bananach Oct 25 '16 at 19:14. Linear vs. C2, and D2) and then use Solver to find the least-squares parameters A, B, and C. MATH 3795 Lecture 9. You can perform least squares fit with or without the Symbolic Math Toolbox. This is a solved. You can vote up the examples you like or vote down the ones you don't like. The following is a sample implementation of simple linear regression using least squares matrix multiplication, relying on numpy for heavy lifting and matplotlib for visualization. Least squares regression analysis or linear regression method is deemed to be the most accurate and reliable method to divide the company’s mixed cost …. A least squares model contains a dummy objective and a set of linear equations: sumsq. 3 The Role of The quantities generated by the Lanczos process from (2. If it is not in the range, then it is the least squares solution. Loading Least-Squares Regression Line. The nonlinear problem is usually solved by iterative. SPGL1: A solver for sparse least squares. This is a mean estimated from a linear model. An issue came up about whether the least squares regression line has to pass through the point (XBAR,YBAR), where the terms XBAR and YBAR represent the arithmetic mean of the independent and dependent variables, respectively. This x is called the least square solution (if the Euclidean norm is used). Nonlinear least-squares solves min(∑||F(x i ) – y i || 2 ), where F(x i ) is a nonlinear function and y i is data. Answer to 4. The most expensive phase is the LSQR phase. solve a non-linear least squares problem. This page gathers different methods used to find the least squares circle fitting a set of 2D points (x,y). The equation for least squares solution for a linear fit looks as follows. Yanbo Liang (JIRA) Sun, 03 Jan 2016 18:05:30 -0800. In fact, I used this kind of solution in some situations. Define least squares. | {
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2016 18:05:30 -0800. In fact, I used this kind of solution in some situations. Define least squares. Factoring-polynomials. LEAST SQUARES and NORMAL EQUATIONS Background Overdetermined Linear systems: consider Ax = b if A is m n, x is n 1, b is m 1 with m > n. The least squares approach to regression is based upon minimizing these difference scores or deviation scores. Polynomials Least-Squares Fitting: Polynomials are one of the most commonly used types of curves in regression. For a general problem you wouldn't use this, of. so somewhere I'm doing something wrong. Let [] ∀k∈ℕ be a dispersion point in. Triangle Calculator. Nonlinear Least-Squares Fitting. Now that we have determined the loss function, the only thing left to do is minimize it. Given a matrix equation Ax=b, the normal equation is that which minimizes the sum of the square differences between the left and right sides: A^(T)Ax=A^(T)b. Least-Squares Line Least-Squares Fit LSRL The linear fit that matches the pattern of a set of paired data as closely as possible. Introduction¶. This influence is exaggerated using least squares. solver to vary the values for A, C and k to minimize the sum of chi squared. Therefore the least squares solution to this system is: xˆ = (A TA)−1A b = −0. 00000 Covariance matrix of Residuals 0. solve a non-linear least squares problem. Dr Gregory Reeves 26,616 views. To solve least squares problems based on PDE models requires sophisticated numerical techniques but also great attention with respect to the quality of data and identifiability of the parameters. 33 so this is our prediction. ‘huber’ : rho(z) = z if z <= 1 else 2*z**0. But how does this relate to the least-squares problem, where there are multiple measurements? Is the problem I am trying to solve essentially the same, except that the number of measurements is one? And in that case, is using Ceres Solver's non-linear least squares solver really necessary? Thanks!. The following is a sample implementation of simple | {
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least squares solver really necessary? Thanks!. The following is a sample implementation of simple linear regression using least squares matrix multiplication, relying on numpy for heavy lifting and matplotlib for visualization. Also tells you if the entered number is a perfect square. It estimates the value of a dependent variable Y from a given independent variable X. An online LSRL calculator to find the least squares regression line equation, slope and Y-intercept values. least squares solution). Node 13 of 18 Node 13 of 18 The Mixed Integer Linear Programming Solver Tree level 1. Order fractions from least to greatest or from greatest to least. The computational burden is now shifted, and one needs to solve many small linear systems. I'd like to know how to solve the least squares non linear regression in java only by passing a matrix A and a vector b like in python. Heh--reduced QR left out the right half of Q. SOLVING DIFFERENTIAL EQUATIONS WITH LEAST SQUARE AND COLLOCATION METHODS by Katayoun Bodouhi Kazemi Dr. 00004849386 0. The Excel Solver can be easily configured to determine the coefficients and Y-intercept of the linear regression line that minimizes the sum of the squares of all residuals of each input equation. Certain types of word problems can be solved by quadratic equations. In this lesson, we will explore least-squares regression and show how this method relates to fitting an equation to some data. There are many possible cases that can arise with the matrix A. The best-fit line, as we have decided, is the line that minimizes the sum of squares of residuals. Anyway, if you want to learn more about the derivation of the normal equation, you can read about it on wikipedia. The solve() method finds a vector x such that Σ i [f i (x)] 2 is minimized. Severely weakens outliers influence, but may cause difficulties in optimization process. When we used the QR decomposition of a matrix to solve a least-squares problem, we operated under the assumption | {
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the QR decomposition of a matrix to solve a least-squares problem, we operated under the assumption that was full-rank. We have our explanatory variable x, that gets multiplied by this slope beta 1, and we also have an intercept where the line intersects the y axis. solve a non-linear least squares problem. A linear fit matches the pattern of a set of paired data as closely as possible. Given a set of samples {(x i,y i)}m i=1. lstsq in terms of computation time and memory. Triangle Calculator. Square of Matrix Calculator is an online tool programmed to calculate the square of the matrix A. This is a standard least squares problem and can easily be solved using Math. There are many possible cases that can arise with the matrix A. Since this system usually does not have a solution, you need to be satisfied with some sort of approximate solution. For any given. The GLS estimator can be shown to solve the problem which is called generalized least squares problem. y_i=A{x_i}^b When I solve for A two different ways I am getting different answers. The Least-Squares Method requires that the estimated function has to deviate as little as possible from f(x) in the sense of a 2-norm. Find answers to Weighted least Squares Excel from the expert community at Experts Exchange Also, when I load the solver case from R77:R96, the resulting. Method of Least Squares. ) Developer's Guide to Excelets/Sinex. Least–squares Solution of Homogeneous Equations supportive text for teaching purposes Revision: 1. The most widely used approximation is the least squares solution, which minimizes. The data, the interpolating polynomial (blue), and the least-squares line (red) are shown in Figure 1. * odinsbane/least-squares-in-java * NonLinearLeastSquares (Parallel Java Library Documentation) * NonlinearRegression (JMSL Numerical Library) Some related discussion here: Solving nonlinear equations. 5 Example 3: The orbit of a comet around the sun is either elliptical, parabolic, or hyperbolic. This | {
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Example 3: The orbit of a comet around the sun is either elliptical, parabolic, or hyperbolic. This is the general form of the least squares line. The calculator will generate a step by step explanation along with the graphic representation of the data sets and regression line. LLS is actively maintained for the course EE103, Introduction to Matrix Methods. Least Squares Regression Line of Best Fit. Other JavaScript in this series are categorized under different areas of applications in the MENU section on this. Octave also supports linear least squares minimization. Since it's a sum of squares, the method is called the method of least squares. They are connected by p DAbx. Lecture 11, Least Squares Problems, Numerical Linear Algebra, 1997. We present an algorithm for adding rows with a single nonzero to A to improve its conditioning; it attempts to add as few rows as possible. 00000241437 0. 1 Introduction A nonlinear least squares problem is an unconstrained minimization problem of the form minimize x f(x)= m i=1 f i(x)2, where the objective function is defined in terms of auxiliary functions {f i}. To perform WLS in EViews, open the equation estimation dialog and select a method that supports WLS such as LS—Least Squares (NLS and ARMA), then click on the Options tab. Least Squares with Examples in Signal Processing1 Ivan Selesnick March 7, 2013 NYU-Poly These notes address (approximate) solutions to linear equations by least squares. Last edited by shg; 10-23-2017 at 01:01 PM. 20 - PhET: Free online. R factor can be used in LSQR (an iterative least-squares solver [29]) to effi-ciently and reliably solve a regularization of the least-squares problem. The CVX Users’ Guide, Release 2. I am using python linalg. The results showed. Which Matlab function should I use?. A linear fit matches the pattern of a set of paired data as closely as possible. Least Squares Optimization The following is a brief review of least squares optimization and constrained optimization | {
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The following is a brief review of least squares optimization and constrained optimization techniques,which are widely usedto analyze and visualize data. For treatment A, the LS mean is (3+7. Because nonlinear optimization methods can be applied to any function, for the relation between two variables, it finds functions that best fit a given set of data points from a list of more than 100 functions, which include most common and interesting functions, like gaussians, sigmoidals, rationals. AutoCorrelation (Correlogram) and persistence - Time series analysis. NET: Description: This example shows how to find a linear least squares fit for a set of points in Visual Basic. Least Squares. But, this OLS method will work for both univariate dataset which is single independent variables and single dependent variables and multi-variate dataset. 7 Least squares approximate solutions. Quadratic Regression Calculator. * odinsbane/least-squares-in-java * NonLinearLeastSquares (Parallel Java Library Documentation) * NonlinearRegression (JMSL Numerical Library) Some related discussion here: Solving nonlinear equations. In the process of solving a mixed integer least squares problem, an ordinary integer least squares problem is solved. Free Modulo calculator - find modulo of a division operation between two numbers step by step This website uses cookies to ensure you get the best experience. For details, see First Choose Problem-Based or Solver-Based Approach. Next, we develop a distributed least square solver over strongly connected directed graphs and show that the proposed algorithm exponentially converges to the least square solution provided the step-size is sufficiently small. Let [] ∀k∈ℕ be a dispersion point in. This is why some least-squares solvers do not use the normal equations under the hood (they instead use QR decomposition). See Input Data for the description of how to enter matrix or just click Example for a simple example. Spark MLlib currently supports two types | {
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to enter matrix or just click Example for a simple example. Spark MLlib currently supports two types of solvers for the normal equations: Cholesky factorization and Quasi-Newton methods (L-BFGS/OWL-QN). The method of iteratively reweighted least squares (IRLS) is used to solve certain optimization problems with objective functions of the form of a p-norm: ∑ = | − |, by an iterative method in which each step involves solving a weighted least squares problem of the form: (+) = ∑ = (()) | − |. This function outperforms numpy. BLENDENPIK: SUPERCHARGING LAPACK'S LEAST-SQUARES SOLVER 5 de ned in the prof. R factor can be used in LSQR (an iterative least-squares solver [29]) to effi-ciently and reliably solve a regularization of the least-squares problem. A necessary and sufficient condition is established on the graph Laplacian for the continuous-time distributed algorithm to give the least squares solution in the limit, with an exponentially fast convergence rate. Quadratic regression is a type of a multiple linear regression. The use of linear regression, or least squares method, is the most accurate method in segregating total costs into fixed and variable components. The least squares regression line ; The least squares regression line whose slope and y-intercept are given by: where , , and. Subsequently, Avronetal. This is done by finding the partial derivative of L, equating it to 0 and then finding an expression for m and c. Since this thesis is closely related to the least-squares adjustment problem and will actually present a new approach for solving this problem, let us first have a closer look at the classical approach. Least Squares with Examples in Signal Processing1 Ivan Selesnick March 7, 2013 NYU-Poly These notes address (approximate) solutions to linear equations by least squares. e the sum of squares of residuals is minimal under this approach. Ordinary Least Squares (OLS) regression (or simply "regression") is a useful tool for examining the relationship | {
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Squares (OLS) regression (or simply "regression") is a useful tool for examining the relationship between two or more interval/ratio variables. This problem is called a least squares problem for the following reason. On input, the field x must be filled in with an initial estimate of the solution vector, and the field tol must be set to the desired tolerance. Solve word problems involving quadratic equations. find_min_box_constrained (using lbfgs_search_strategy(10)) performed poorly as it can be trapped on a boundary. where A is an m x n matrix with m > n, i. So really, what you did in the first assignment was to solve the equation using LSE. On a similar note,. This influence is exaggerated using least squares. i) (circles) and least-squares line (solid line) but we will see that the normal equations also characterize the solution a, an n-vector, to the more general linear least squares problem of minimizing kAa ykfor any matrix Athat is m n, where m n, and whose columns are linearly independent. Enter your data as (x,y) pairs, and find the equation of a line that best fits the data. Scramble Squares® Puzzle. least_squares(). Suppose that a matrix A is given that has more rows than columns, ie n, the number of rows, is larger than m, the number of columns. Quadratic Regression Calculator. Nonlinear Least Squares Data Fitting D. Main ideas 2. TI-89 graphing calculator program for calculating the method of least squares. This document is intended to clarify the issues, and to describe a new Stata command that you can use (wls) to calculate weighted least-squares estimates for problems such as the Strong interaction'' physics data described in Weisberg's example 4. Also lets you save and reuse data. 00097402530 0. Line of Best Fit (Least Square Method) A line of best fit is a straight line that is the best approximation of the given set of data. Constructing a Least-Squares Graph Using Microsoft Excel. Part of our free statistics site; generates linear regression | {
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Graph Using Microsoft Excel. Part of our free statistics site; generates linear regression trendline and graphs results. To compare fractions the calculator first finds the least common denominator (LCD), converts the fractions to equivalent fractions using the LCD, then. By using this website, you agree to our Cookie Policy. Visit Stack Exchange. Type doc lsqnonlin for more details. The method of least squares - using the Excel Solver Michael Wood 5 advertising. For this, we're going to make use of the property that the least squares line always goes through x bar, y bar. The method of least squares calculates the line of best fit by minimising the sum of the squares of the vertical distances of the points to th e line. 0 released December 2019 This latest release of SPGL1 implements a dual root-finding mode that allows for increased accuracy for basis pusuit denoising problems. The Linear Least Squares Regression Line method is the accurate way of finding the line of best fit in case it's presumed to be a straight line that is the best approximation of the given set of data. since gradient descent is a local optimizer and can get stuck in local solution we need to use. Added Dec 13, 2011 by scottynumbers in Mathematics. When we used the QR decomposition of a matrix to solve a least-squares problem, we operated under the assumption that was full-rank. The center of the part and center of rotation are offset. Question: 4. How to Calculate Absolute Value. powered by $$x$$ y $$a 2$$ a b . "Solver" is a powerful tool in the Microsoft Excel spreadsheet that provides a simple means of fitting experimental data to nonlinear functions. The Least Squares Regression Line is the line that makes the vertical distance from the data points to the regression line as small as possible. Number of Data Points: X Data Points:. solve a non-linear least squares problem. Spark MLlib currently supports two types of solvers for the normal equations: Cholesky factorization and Quasi-Newton | {
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supports two types of solvers for the normal equations: Cholesky factorization and Quasi-Newton methods (L-BFGS/OWL-QN). The full documentation is available online. In the present instance solving for the node weights is not really viable for the following reason: in the actual, real-life setting the only decision variable (which I have control over & need to solver for) are the instrument weights = units of instruments (cells I11:I15). Algorithm 1 Least-squares sub-problem input: H 2 CN ⇥N, q 2 CN, P 2 RM ⇥N, d 2 CM for each receiver (j )(rowinP) in parallel do H⇤ w j = p⇤ j {solve 1 PDE} end for W =[w 1 w 2w m] {distributed matrix} S =(I M + 2 W⇤ W)1 {adjust using Algorithm 2 (optional)} for source (i) in parallel do y i =(I N 2 WSW⇤)(q i + 2 Wd i) Hu i = y i {solve 1 PDE} end for output: u. Effective use of Ceres requires some familiarity with the basic components of a non-linear least squares solver, so before we describe how to configure and use the solver, we will take a brief look at how some of the core optimization algorithms in Ceres work. LinearLeastSquares. It can be manually found by using the least squares method. That is, Octave can find the parameter b such that the model y = x*b fits data (x,y) as well as possible, assuming zero-mean Gaussian noise. The least-squares line or regression line can be found in the form of y = mx + b using the following formulas. When we pass this (near) optimal solution to NL2SOL it will have an easy task. The Least Squares Regression Calculator is biased against data points which are located significantly away from the projected trend-line. 1 Linear Least Squares Problem. You must select the Solver Add-in and then press the OK button. Use the EXCEL SOLVER program to minimise S by varying the paramters "a" and "b" This will produce estimates of a and b that give the best fitting straight line to the data. It will b e sho wn that the direct sp eci c least-square tting of ellipses. (A for all ). This page allows | {
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b e sho wn that the direct sp eci c least-square tting of ellipses. (A for all ). This page allows performing nonlinear regressions (nonlinear least squares fittings). Let [] ∀k∈ℕ be a dispersion point in. Nonlinear Least Squares Data Fitting D. By using this website, you agree to our Cookie Policy. The generalized least squares problem. It is used to study the nature of the relation between two variables. Let , , and be defined as previously. This article introduces the method of fitting nonlinear functions with Solver. Used to determine the “best” line. The least squares criterion is a formula used to measure the accuracy of a straight line in depicting the data that was used to generate it. Least-Squares Fitting of Data with Polynomials Least-Squares Fitting of Data with B-Spline Curves. In the present instance solving for the node weights is not really viable for the following reason: in the actual, real-life setting the only decision variable (which I have control over & need to solver for) are the instrument weights = units of instruments (cells I11:I15). LEAST MEAN SQUARE ALGORITHM 6. Use our online quadratic regression calculator to find the quadratic regression equation with graph. LinearAlgebra namespace in C#. The applications of the method of least squares curve fitting using polynomials are briefly discussed as follows. 0 released December 2019. 00000088820 0. A least squares model contains a dummy objective and a set of linear equations: sumsq. Least squares means are adjusted for other terms in the model (like covariates), and are less sensitive to missing data. The CVX Users’ Guide, Release 2. Given a set of data, we can fit least-squares trendlines that can be described by linear combinations of known functions. com A collection of really good online calculators for use in every day domestic and commercial use!. Lecture 11, Least Squares Problems, Numerical Linear Algebra, 1997. Trouble may also arise when M = N but the matrix is singular. The | {
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Numerical Linear Algebra, 1997. Trouble may also arise when M = N but the matrix is singular. The first step. using least squares minimization. To show the powerful Maple 10 graphics tools to visualize the convergence of this Polynomials. The best-fit line, as we have decided, is the line that minimizes the sum of squares of residuals. Wow, there's a lot of similarities there between real numbers and matrices. CPM Student Tutorials CPM Content Videos TI-84 Graphing Calculator Bivariate Data TI-84: Least Squares Regression Line (LSRL). Regression Using Excel's Solver. However, if users insist on finding the total least squares fit then an initial approximation is still required and the linear least squares approach is recommended. That closed-form solution is called the normal equation. overdetermined system, least squares method The linear system of equations A =. For details, see First Choose Problem-Based or Solver-Based Approach. Question: 4. Least squares means are adjusted for other terms in the model (like covariates), and are less sensitive to missing data. Enter two data sets and this calculator will find the equation of the regression line and corelation coefficient. Sum of squares is used in statistics to describe the amount of variation in a population or sample of observations. 1 Introduction. , there are more equations than unknowns, usually does not have solutions. Click the button labeled “Click to Compute”. They are connected by p DAbx. Linear least-squares solves min||C*x - d|| 2 , possibly with bounds or linear constraints. Each node has access to one of the linear equations and holds a dynamic state. When radians are selected as the angle unit, it can take values such as pi/2, pi/4, etc. The method of least squares - using the Excel Solver Michael Wood 5 advertising. Years after 1900 50 60 70 80 90 100 Percentage 29. Estimating an ARMA Process Overview 1. Loading Unsubscribe from Adrian Lee? Least Squares Linear Regression - EXCEL - Duration: | {
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Overview 1. Loading Unsubscribe from Adrian Lee? Least Squares Linear Regression - EXCEL - Duration: 10:55. No Bullshit Guide To Linear Algebra, 2017. Added Dec 13, 2011 by scottynumbers in Mathematics. (We use the squares for much the same reason we did when we defined the variance in Section 3. Alternative solution methods. See LICENSE_FOR_EXAMPLE_PROGRAMS. The data, the interpolating polynomial (blue), and the least-squares line (red) are shown in Figure 1. LMS incorporates an. First, least squares is a natural approach to estimation, which makes explicit use of the structure of the model as laid out in the assumptions. The solutions. org are unblocked. The linear least-squares problem occurs in statistical regression analysis; it has a closed-form solution. Example 2 in the KaleidaGraph Quick Start Guide shows how to apply a Linear curve fit to a Scatter plot. Check Minitab for definition of influential points. Linear regression calculator Two-dimensional linear regression of statistical data is done by the method of least squares. Quadratic Regression Calculator. Trouble may also arise when M = N but the matrix is singular. The method of iteratively reweighted least squares (IRLS) is used to solve certain optimization problems with objective functions of the form of a p-norm: ∑ = | − |, by an iterative method in which each step involves solving a weighted least squares problem of the form: (+) = ∑ = (()) | − |. To approximate a Points Dispersion through Least Square Method using a Quadratic Regression Polynomials and the Maple Regression Commands. In fact, I used this kind of solution in some situations. lsqnonneg applies only to the solver-based approach. Number of Data Points: X Data Points:. The help qr command in Matlab gives the following information: >> help qr QR Orthogonal-triangular decomposition. lstsq in terms of computation time and memory. I tried the following set-up: - Given is a vector of original exposure across a range of seven nodes (C8:I2) - | {
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following set-up: - Given is a vector of original exposure across a range of seven nodes (C8:I2) - The aim is to replicate this exposure at each point as close as possible from a set of 5 instruments. solve public void solve() Solve this nonlinear least squares minimization problem. 4 Linear Least Squares. Least squares regression analysis or linear regression method is deemed to be the most accurate and reliable method to divide the company’s mixed cost …. In this case, solving the normal equations (5) is equivalent to. If the system matrix is rank de cient, then other methods are. Enter the fraction separated by comma and press the calculate button. solve a non-linear least squares problem. Regression Using Excel's Solver. powered by. Given a set of data, we can fit least-squares trendlines that can be described by linear combinations of known functions. Check Minitab for definition of influential points. to solve multidimensional problem, then you can use general linear or nonlinear least squares solver. (3) Solve the diagonal system Σˆw = Uˆ∗b for w. The original domain is. It is called “least squares” because we are minimizing the sum of squares of these functions. I If m= nand Ais invertible, then we can solve Ax= b. The limitations of the OLS regression come from the constraint of the inversion of the X’X matrix: it is required that the rank of the matrix is p+1, and some numerical problems may arise if the matrix is not well behaved. Free online LCM calculator. Leykekhman - MATH 3795 Introduction to Computational MathematicsLinear Least Squares { 11. For this purpose, the initial values of A, B, and C in cells F2-F4 should be those found by Solver in the previous run. 1 Linear Least Squares Problem. If you're seeing this message, it means we're having trouble loading external resources on our website. For details, see First Choose Problem-Based or Solver-Based Approach. Note: this method requires that A not have any redundant rows. Note: Be sure that your | {
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Approach. Note: this method requires that A not have any redundant rows. Note: Be sure that your Stat Plot is on and indicates the Lists you are using. Nonlinear Regression. Linear Least Squares Regression Line Calculator - v1. Contribute to kashif/ceres-solver development by creating an account on GitHub. Linear regression line calculator to calculate slope, interception and least square regression line equation. The second one is the Levenberg-Marquardt method. Enter the number of data pairs, fill the X and Y data pair co-ordinates, the least squares regression line calculator will show you the result. | {
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# Count arrays with size n, sum k and largest element m
I'm trying to solve pretty complex problem with combinatorics.
Namely, we have given three numbers N, K, M. Now we want to count how many different arrays of integers are there with length N, sum K and all the elements in the range [1, M]
Constraints:
• 1 <= N <= 100
• 1 <= K <= 100
• 1 <= M <= 100
Example
Let's say N = 2, K = 5, M = 3. This means that we want to count arrays of integers of size 2 with sum of all elements equal to 5 and elements in range [1, 3]. There are total of 2 arrays: {2, 3} and {3, 2}. Please note that the order of the elements also matters, {2, 3} is not equal to {3, 2}
Second example: N = 4, K = 7, M = 3. We want to count arrays of length 4, sum of 7 and elements in range [1, 3].
There are total of 16 possible way of arrays: (1,1,2,3), (1,1,3,2), (2,1,1,3), (3,1,1,2), (2,3,1,1), (3,2,1,1), (1,2,3,1), (1,3,2,1), (1,2,1,3), (1,3,1,2), (1,2,2,2), (2,1,2,2), (2,2,1,2), (2,2,2,1)
What I have tried
I know that one solution is to generate all possible arrays, but such algorithm in best case will work in complexity O(N!) which is far too big for N = 100. I started thinking about solving this with three-dimensional dynamical programming, but I cannot find the relations between the states.
I'm thinking about this way: Let f(i, j, l) be the number of arrays of length i, sum j, and largest element l. We can see that for i = 0 f(i,j,l) = 0, so this is I think the base case. Also f(1, 1, 1) = 1 is another base case.
Now I cannot find the relations between the states. Can you give me some hints how to find the relations between the states. Thanks in advance.
• Nice problem. Can you credit the source where you encountered the problem? – D.W. Jul 25 '17 at 16:11
• The problem is from one macedonian site for training, and the problem in original is in macedonian, but i can give you the test cases if you want them – someone12321 Jul 25 '17 at 16:30 | {
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You can use dynamic programming. For each $0 \leq i \leq N$ and $0 \leq s \leq K$, count the number of arrays of length $i$, consisting of numbers in the range $\{1,\ldots,M\}$, which sum to exactly $s$. The running time is $O(NK)$.
Explicitly, denoting the array by $a$, we have $a(0,0) = 1$, $a(0,s) = 0$ otherwise, and $$a(i,s) = \sum_{t=1}^M a(i-1,s-t),$$ where $a(i-1,r) = 0$ if $r < 0$.
As Hendrik Jan mentions in the comments, we can improve on this $O(NKM)$ algorithm by using the recursion $$a(i,s) = a(i,s-1) - a(i-1,s-1-M) + a(i-1,s-1),$$ with suitable base cases.
Alternatively, we can obtain an explicit expression: the answer is the coefficient of $x^K$ in the generating function $$(x + \cdots + x^M)^N = x^N \left(\frac{1-x^M}{1-x}\right)^N.$$ In other words, we are looking for the coefficient of $x^{K-N}$ in $$\left( \sum_{i=0}^N (-1)^i \binom{N}{i} x^{iM} \right) \left( \sum_{j=0}^\infty \binom{j+N-1}{N-1} x^j \right),$$ which has the closed form $$\sum_{i=0}^{\min(N,\lfloor (K-N)/M\rfloor)} (-1)^i \binom{N}{i} \binom{K-iM-1}{N-1}.$$
• The numbers $a(i,s)$ represent a matrix of solutions for "fixed" $M$. The numbers $a(i,s)$ are a "running sum" of $M$ consecutive elements from the previous row. So, can't the complexity be improved to $O(NK)$, as $a(i,s) = a(i,s-1) + a(i-1,s-1) - a(i-1,s-M-1)$? give or take typo's. – Hendrik Jan Jul 25 '17 at 1:22
• @HendrikJan Yes, that's right! I missed this somehow. – Yuval Filmus Jul 25 '17 at 5:12
You actually don't need the largest element l in your recursive function. It doesn't make such sense to use it, since the largest number in an array has no influence on the other numbers.
In most (all?) dynamic programming problems you have to think about the last step / the last part. For f(i, j) you have an array with i numbers. You can imagine, that in the last step you added the last number to the array. So before you added the last number, the array only consists of i-1 numbers. | {
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Now, if the added number was 1, then there are f(i-1, j-1) many arrays. If the added number was 2, then there are f(i-1, j-2) many arrays. And so on...
Add all those possibilities up, and you end up with f(i, j).
I don't pretend this is the most efficient solution. Neither it has a strong theoretical merit, but you can still find it useful to verify your results. Here's a small program in Prolog, (using SWI Prolog for constraint satisfaction part), which can calculate or verify your guess:
% -*- mode: prolog; prolog-system: "swi" -*-
:- use_module(library(clpfd)).
sums(K, [], K).
sums(A, [X | Xs], K) :-
indomain(A),
indomain(X),
B is A + X,
sums(B, Xs, K).
sums([X | Xs], K) :- sums(X, Xs, K).
nmk_problem_helper(N, K, M, List) :-
indomain(N),
indomain(M),
indomain(K),
length(List, N),
List ins 1..M,
sums(List, K).
nmk_problem(N, K, M, Arrays) :-
findall(X, nmk_problem_helper(N, K, M, X), Arrays).
nmk_problem_all_helper(List) :-
[N, M, K] ins 1..100,
indomain(N),
indomain(M),
indomain(K),
length(List, N),
List ins 1..M,
sums(List, K).
nmk_problem_all(Arrays) :-
findall(X, nmk_problem_all_helper(X), Arrays).
Example usage:
?- nmk_problem(4, 7, 3, X).
X = [[1, 1, 2, 3], [1, 1, 3, 2], [1, 2, 1, 3], [1, 2, 2, 2], [1, 2, 3, 1], [1, 3, 1|...], [1, 3|...], [2|...], [...|...]|...]. | {
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# Multiple answers to $\int \sqrt{4t - t^2} \, \textrm{d} t$
I'm trying to understand why I'm getting different answers when taking different approaches to integrating
$$\int \sqrt{4t - t^2} \, \textrm{d} t$$
First, I tried substituting $\sqrt t = 2 \sin \theta$:
$$\begin{eqnarray} \int \sqrt{t}\sqrt{4 - t} \, \textrm{d} t &=& \int 2 \sin \theta \cdot \sqrt{4 - 4 \sin^2 \theta} \cdot 8 \sin \theta \cos \theta \, \textrm{d} \theta \\ &=& 32 \int \sin^2 \theta \cos^2 \theta \, \textrm{d} \theta \\ &=& 4 \int 1 - \cos 4 \theta \, \textrm{d} \theta \\ &=& 4 \theta - \sin 4 \theta + C \\ &=& 4 \theta - 4 \sin \theta \cos^3 \theta + 4 \sin^3 \theta \cos \theta + C \\ &=& 4 \arcsin \frac{\sqrt{t}}{2} + \frac{1}{2} (t - 2)\sqrt{4t - t^2} + C \\ \end{eqnarray}$$
Second, I tried completing the square and substituting $t - 2 = 2 \sin \theta$:
$$\begin{eqnarray} \int \sqrt{4 - (t^2 - 4t + 4)} \, \textrm{d} t &=& \int \sqrt{4 - (t - 2)^2} \, \textrm{d} t \\ &=& \int \sqrt{4 - 4 \sin^2 \theta} \cdot 2 \cos \theta \, \textrm{d} \theta \\ &=& 4 \int \cos^2 \theta \, \textrm{d} \theta \\ &=& 2 \int 1 - \cos 2 \theta \, \textrm{d} \theta \\ &=& 2 \theta + \sin 2 \theta + C \\ &=& 2 \theta + 2 \sin \theta \cos \theta + C \\ &=& 2 \arcsin \left(\frac{t - 2}{2}\right) + \frac{1}{2}(t - 2)\sqrt{4t - t^2} + C \\ \end{eqnarray}$$
The second answer is the same as in the book but I don't understand why the first approach gives the wrong answer. | {
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"url": "https://math.stackexchange.com/questions/1679302/multiple-answers-to-int-sqrt4t-t2-textrmd-t"
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• Here's one sticking point: $1-\cos 4\theta = 8\sin^2\theta\cos^2\theta.$ – Cameron Williams Mar 2 '16 at 0:52
• @CameronWilliams, good catch, I'll update my first answer. – Chewers Jingoist Mar 2 '16 at 0:56
• Both answers are correct: if you let $f(t)=4\sin^{-1}\frac{\sqrt{t}}{2}-2\sin^{-1}\frac{t-2}{2}$, then $f^{\prime}(t)=0$ so $f(t)$ is a constant. – user84413 Mar 2 '16 at 1:08
• @user84413, when I differentiate $f(t)$ I get $\frac{4}{\sqrt{4 - t}} - \frac{2}{\sqrt{4 - (t - 2)^2}} = \frac{4}{\sqrt{4 - t}} - \frac{2}{\sqrt{4t - t^2}}$. Could you elaborate on how you get $0$? – Chewers Jingoist Mar 2 '16 at 1:21
• In the first term, I think you will get $4\frac{1}{\sqrt{1-t/4}}\frac{1}{4\sqrt{t}}=\frac{2}{\sqrt{4t-t^2}}$ – user84413 Mar 2 '16 at 1:33 | {
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$$\arcsin\left(\frac t2-1\right) = 2\left(\arcsin\frac{\sqrt t}2 -\frac\pi4\right).\tag{*}$$
Proof: Write $\theta:=\arcsin\frac{\sqrt t}2$. Then $\sin^2\theta=\frac t4$ and $\cos^2\theta=1-\frac t4$, and using the angle-difference identities, $$\sin\left(\theta-\frac\pi4\right)=\frac1{\sqrt 2}(\sin\theta-\cos\theta)\tag1$$ while $$\cos\left(\theta-\frac\pi4\right)=\frac1{\sqrt 2}(\sin\theta+\cos\theta).\tag2$$ Therefore $$\sin2\left(\theta-\frac\pi4\right)=2\sin\left(\theta-\frac\pi4\right)\cos\left(\theta-\frac\pi4\right) \stackrel{(1),(2)}=\sin^2\theta-\cos^2\theta =\frac t4-\left(1-\frac t4\right)=\frac t2-1.$$ Therefore both sides of (*) have the same sine. Similarly you can show that both sides have the same cosine.
• Where does $2$ come from at the start of your last line in $sin 2(\theta - \frac{\pi}{4})$? And what does the $(1),(2)$ mean? – Chewers Jingoist Mar 2 '16 at 1:31
• @ChewersJingoist, the $2$ in $sin2(\theta - \frac{\pi}{4})$ comes from the identity $sin2x=2sinxcosx$. – Alexander Maru Mar 2 '16 at 1:39
If $$\sin\theta = \dfrac{\sqrt t}2,$$ then $$\sin\left(\dfrac\pi2-2\theta\right) = \cos2\theta = 1-2\sin^2\theta = \dfrac{2-t}2,$$ $$2\theta = \dfrac\pi2-\arcsin\dfrac{2-t}2 = \dfrac\pi2+\arcsin\dfrac{t-2}2,$$ $$\theta = \dfrac\pi4+\dfrac12\arcsin\dfrac{t-2}2,$$ $$\boxed{\arcsin\dfrac{\sqrt t}2 = \dfrac12\arcsin\dfrac{t-2}2+\dfrac\pi4}$$ So the first answer is correct too | {
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} |
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# If the curve described by the equation y = x2 + bx + c cuts the x-axis
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If the curve described by the equation y = x2 + bx + c cuts the x-axis [#permalink]
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10 Aug 2018, 10:08
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If the curve described by the equation $$y = x^2 + bx + c$$ cuts the $$x$$-axis at $$-4$$ and $$y$$ axis at $$4$$, at which other point does it cut the $$x$$-axis?
A. -1
B. 4
C. 1
D. -4
E. 0
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Re: If the curve described by the equation y = x2 + bx + c cuts the x-axis [#permalink]
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10 Aug 2018, 10:49
+1 for A? I plugged in the other points to come up with an equation...not sure if I went about it the right way
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Re: If the curve described by the equation y = x2 + bx + c cuts the x-axis [#permalink]
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10 Aug 2018, 22:08
1
If the curve described by the equation y = x^2 + bx + c cuts the x-axis at -4 and y axis at 4, at which other point does it cut the x-axis? | {
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A. -1
B. 4
C. 1
D. -4
E. 0
Given, $$y = x^2 + bx + c$$, cuts the x-axis at two points. One intersection point with x-axis is given. We need to find out the other point of intersection. In other words, 1 root of the quadratic equation is given, what is the value if the other root?
a) At (-4,0), $$0=(-4)^2+b*(-4)+c$$ Or, 4b-c=16
b) At (0,4), $$4=0^2+b*0+c$$ Or, c=4
So, 4b-4=16 Or, b=5.
Now, we have the equation of the curve, $$y=x^2+5x+4$$, which has the roots: -4 and -1.
So, other other root is -1.
Ans. (A)
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Re: If the curve described by the equation y = x2 + bx + c cuts the x-axis [#permalink]
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02 Jan 2019, 09:53
2
If the curve described by the equation y = x^2 + bx + c cuts the x-axis at -4 and y axis at 4, at which other point does it cut the x-axis?
A -1
B 4
C 1
D -4
E 0
y = x^2 + bx + c is a quadratic equation and the equation represents a parabola.
The curve cuts the y axis at 4.
The x coordinate of the point where it cuts the y axis = 0.
Therefore, (0, 4) is a point on the curve and will satisfy the equation.
4 = 0^2 + b(0) + c
Or c = 4.
The product of the roots of a quadratic equation is c/a
In this question, the product of the roots = 4/1 = 4.
The roots of the quadratic equation are the points where the curve cuts the x-axis.
The question states that one of the points where the curve cuts the x-axis is -4.
So, -4 is one of roots.
Let r2 be the second root of the quadratic equation.
So, -4 * r2 = 4
or r2 = -1.
The second root is the second point where the curve cuts the x-axis, which is -1.
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Re: If the curve described by the equation y = x2 + bx + c cuts the x-axis [#permalink]
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### Show Tags
02 Jan 2019, 10:34
When x=-4 y=0
so 16 -4b +c=0
When x=0, y = 4
so c=4
16-4b+c=20-4b=0
b= 5
the equation can be written as y=(x+4)(x+1)
y is equal to zero when x=-1 (Answer A)
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Re: If the curve described by the equation y = x2 + bx + c cuts the x-axis [#permalink]
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03 Jan 2019, 02:15
cfc198 wrote:
If the curve described by the equation y = x^2 + bx + c cuts the x-axis at -4 and y axis at 4, at which other point does it cut the x-axis?
A -1
B 4
C 1
D -4
E 0
y = x^2 + bx + c is a quadratic equation and the equation represents a parabola.
The curve cuts the y axis at 4.
The x coordinate of the point where it cuts the y axis = 0.
Therefore, (0, 4) is a point on the curve and will satisfy the equation.
4 = 0^2 + b(0) + c
Or c = 4.
The product of the roots of a quadratic equation is c/a
In this question, the product of the roots = 4/1 = 4.
The roots of the quadratic equation are the points where the curve cuts the x-axis.
The question states that one of the points where the curve cuts the x-axis is -4.
So, -4 is one of roots.
Let r2 be the second root of the quadratic equation.
So, -4 * r2 = 4
or r2 = -1.
The second root is the second point where the curve cuts the x-axis, which is -1.
If you liked the question and explanation, please do hit the kudos button
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# Laplace equation with mixed boundary conditions
I try to solve Laplace equation in 2D on square [2,3]x[2,3], with mixed boundary conditions, I did:
ClearAll[y, x1, x2];
pde = Laplacian[y[x1, x2], {x1, x2}];
bc = {y[x1, 2] == 2 + x1, y[x1, 3] == 3 + x1};
sol = NDSolve[{pde ==
NeumannValue[-1, x1 == 2] + NeumannValue[1, x1 == 3], bc},
y, {x1, 2, 3}, {x2, 2, 3}]
Plot3D[Evaluate[y[x1, x2] /. sol], {x1, 2, 3}, {x2, 2, 3},
PlotRange -> All, AxesLabel -> {"x1", "X2", "y[x1,x2]"},
BaseStyle -> 12]
The exact solution is y=x1+x2, the problem is the results is not high accurate when I evaluate the error. | {
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"url": "https://mathematica.stackexchange.com/questions/224729/laplace-equation-with-mixed-boundary-conditions"
} |
• The exact solution is y=x1+x2 Are you sure about this? How does this solution satisfy the Neumann boundary conditions? – Nasser Jun 25 at 20:09
• @Nasser Erm. The function does satisfy the Neumann boundary condition: Its derivative in x1-direction is 1 and the sign flops stems from the fact that Neumann conditions are phrased in terms of outward normals... No? – Henrik Schumacher Jun 25 at 23:34
• @user62716 Using NeumannValue requires one to do integration by parts and one has to be careful about the signs. Try switching the sign of the Laplacian to pde = -Laplacian[y[x1, x2], {x1, x2}];. Then it should work. – Henrik Schumacher Jun 25 at 23:40
• @HenrikSchumacher is NeumannValue[-1, x1 == 2] different from saying that $\frac{\partial y}{\partial x_1}$ evaluated at $x_1=2$ is $-1$? And since the claim is that the solution is $y=x_1+x_2$ then $\frac{\partial y}{\partial x_1}=1$ this is evaluated at $x=2$ is $1$ and not $-1$?. How do you translate NeumannValue[-1, x1 == 2] to normal derivative then? I just did direct translation. May be we need a whole new topic on this. On top of all of this, moving NeumannValue from RHS to LHS changes the solution. I never liked NeumannValue and prefer to use normal derivatives... – Nasser Jun 26 at 0:46
• @Nasser $\frac{\partial y}{\partial \nu} (2,x_2) = - \frac{\partial y}{\partial x_1} (2,x_2)$ because the outward normal at the point $(2,x_2)$ is $\nu = (-1 , 0)$. But I agree that NeumannValue is a bit counter intuitive, but it makes perfect sense in regard of the weak formulation that is used in FEM. – Henrik Schumacher Jun 26 at 4:59
Relatively recently, Wolfram has created a nice Heat Transfer Tutorial and a Heat Transfer Verification Manual. I model with many codes and I usually start the Verification and Validation manual and build complexity from there. It is always embarrassing to build a complex model and find that your setup does not pass verification. | {
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The Laplace equation is special case of the heat equation so we should be able to use a verified example as a template for a properly constructed model.
For NeumannValue's, if the flux is into the domain, it is positive. If the flux is out of the domain, it is negative.
At the tutorial link, they define a function HeatTransferModel to create operators for a variety of heat transfer cases that I shall reproduce here:
ClearAll[HeatTransferModel]
HeatTransferModel[T_, X_List, k_, ρ_, Cp_, Velocity_, Source_] :=
Module[{V, Q, a = k},
V = If[Velocity === "NoFlow",
Q = If[Source === "NoSource", 0, Source];
If[FreeQ[a, _?VectorQ], a = a*IdentityMatrix[Length[X]]];
If[VectorQ[a], a = DiagonalMatrix[a]];
a = PiecewiseExpand[Piecewise[{{-a, True}}]];
Inactive[Div][a.Inactive[Grad][T, X], X] + V - Q]
If we follow the recipe of tutorial, we should be able to construct and solve a PDE system free of sign errors as I show in the following workflow.
(* Create a Domain *)
Ω2D = Rectangle[{2, 2}, {3, 3}];
(* Create parametric PDE operator *)
pop = HeatTransferModel[y[x1, x2], {x1, x2}, k, ρ, Cp, "NoFlow",
"NoSource"];
(* Replace k parameter *)
op = pop /. {k -> 1};
(* Setup flux conditions *)
nv2 = NeumannValue[-1, x1 == 2];
nv3 = NeumannValue[1, x1 == 3];
(* Setup Dirichlet Conditions *)
dc2 = DirichletCondition[y[x1, x2] == 2 + x1, x2 == 2];
dc3 = DirichletCondition[y[x1, x2] == 3 + x1, x2 == 3];
(* Create PDE system *)
pde = {op == nv2 + nv3, dc2, dc3};
(* Solve and Plot *)
yfun = NDSolveValue[pde, y, {x1, x2} ∈ Ω2D]
Plot3D[Evaluate[yfun[x1, x2]], {x1, x2} ∈ Ω2D,
PlotRange -> All, AxesLabel -> {"x1", "x2", "y[x1,x2]"},
BaseStyle -> 12]
You can test that the solution matches that exact solution over the entire range:
Manipulate[
Plot[{x1 + x2, yfun[x1, x2]}, {x1, 2, 3}, PlotRange -> All,
AxesLabel -> {"x1", "y[x1,x2]"}, BaseStyle -> 12,
PlotStyle -> {Red,
Directive[Green, Opacity[0.75], Thickness[0.015], Dashed]}], {x2,
2, 3}, ControlPlacement -> Top] | {
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• Dear Tim Laska, thank you for your great help, can we evaluate the error and plot it? – user62716 Jun 26 at 9:47
• I did it plot = Plot3D[ Abs[yfun[x1, x2] - (x1 + x2)], {x1, x2} [Element] [CapitalOmega]2D, PlotRange -> All, AxesLabel -> {"x1", "x2", "y[x1,x2]"}, PlotLabel -> err] – user62716 Jun 26 at 10:03
• Dear Tim Laska, I have other problem, Poisson equation with variable coefficients,shall post it in new question or here? – user62716 Jun 26 at 11:48
• @user62716 You should open a new question as it appears that you have. I will try to take a look at your other question when I can. – Tim Laska Jun 26 at 13:45
• Thank you Tim, I will be waiting. Best regards – user62716 Jun 26 at 13:50
By reversing the sign of the derivative on the left side from that given in NeumannValue, this can be solved by Mathematica analytically as well.
ClearAll[y, x1, x2];
pde = Laplacian[y[x1, x2], {x1, x2}] == 0;
bc = {y[x1, 2] == 2 + x1,
y[x1, 3] == 3 + x1,
Derivative[1, 0][y][2, x2] == 1,
Derivative[1, 0][y][3, x2] == 1};
solA = DSolve[{pde, bc}, y[x1, x2], {x1, x2}];
solA = solA /. {K[1] -> n,Infinity -> 20};
solA = Activate[solA];
Plot3D[y[x1, x2] /. solA, {x1, 2, 3}, {x2, 2, 3}, PlotRange -> All,
AxesLabel -> {"x1", "X2", "y[x1,x2]"}, BaseStyle -> 12]
The BC as given above are correct, and Mathematica's analytical solution is correct also, but I agree it can be simpler.
There might be a way to simplify the infinite Fourier sum given, but I could not find it.
To show the above formulation is correct, here is Maple's solution, using same B.C. Maple as above to give the simpler form of the solution, which is $$y=x_1+x_2$$.
restart;
pde:=VectorCalculus:-Laplacian(y(x1,x2),[x1,x2])=0;
bc:=y(x1,2)=2+x1,y(x1,3)=3+x1,D[1](y)(2,x2)=1,D[1](y)(3,x2)=1;
sol:=pdsolve([pde,bc],y(x1,x2))
We just have to remember, that negative NeumannValue on left edge, means positive derivative on that edge. | {
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• Dear Nasser, thank you for your comments, the normal derivative at left side is -1 not 1, the above analytic solution is complicated since the exact is just y=x1+x2....thanks – user62716 Jun 26 at 9:38
• the normal derivative at left side is -1 not 1 no. It is +1. you set NeumannValue to be -1. Since NeumannValue points outwards, then this means the deivative is +1. Since -1 outwards, means +1 inwards. In addition, if you change the derivative (not NeumannValue) in the code I posted from +1 to -1 you will see the solution is no longer y=x1+x2 but becomes non-linear. You can compare this solution with the numerical solution. Do you see any difference? I agree the solution has complicated Fourier series sum, but this is what Mathemtica gave for the analytical solution. – Nasser Jun 26 at 10:01
• Dear Nasser, I still can not understand you, @Nasser ∂y∂ν(2,x2)=−∂y∂x1(2,x2) because the outward normal at the point (2,x2) is ν=(−1,0) so it is -1 on the left outwards. – user62716 Jun 26 at 11:16
• Dear Nasser, the code of Tim Laska is working, I highly appreciate you and you always help me and provide perfect answer. – user62716 Jun 26 at 11:19
• @user62716 you can see from the solution itself, i.e. from just looking at the plot, that the derivative is positive on the left edge. No math is needed if we look at the solution. The slope is moving upwards. So positive slope. You can also see from Maple solution I posted, that I used positive derivative to get same solution $x_1+x_2$ right there. NeumannValue is not the same as derivative. That is what the whole confusion was about. – Nasser Jun 26 at 11:33 | {
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# Scientific Notation ... is a way to express very small or very large numbers. ... is most often used in "scientific" calculations where the analysis.
## Presentation on theme: "Scientific Notation ... is a way to express very small or very large numbers. ... is most often used in "scientific" calculations where the analysis."— Presentation transcript:
Scientific Notation ... is a way to express very small or very large numbers. ... is most often used in "scientific" calculations where the analysis must be very precise. ... consists of two parts*: (1) a number between 1 and 10 and (2) a power of 10. *a large or small number may be written as any power of 10; however, CORRECT scientific notation must satisfy the above criteria.
is correct scientific notation
3.2 x 1013 is correct scientific notation Remember that the first number MUST BE greater than or equal to one and less than 10. 23.6 x 10-8 is not correct scientific notation
To Change from Standard Form to Scientific Notation:
1 Place decimal point such that there is one non-zero digit to the left of the decimal point. 2 Count number of decimal places the decimal has "moved" from the original number. This will be the exponent of the 10. 3 If the original number was less than 1, the exponent is negative; if the original number was greater than 1, the exponent is positive.
Examples: Given: 4,750,000 use: 4.75 (moved 6 decimal places)
answer: 4.75 X 106 The original number was greater than 1 so the exponent is positive. Given: use: 7.89 (moved 4 decimal places) answer: 7.89 x 10-4 The original number was less than 1 so the exponent is negative.
To Change from Scientific Notation to Standard Form:
1 Move decimal point to right for positive exponent of 10. 2 Move decimal point to left for negative exponent of 10. | {
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Examples: Given: 1.015 x 10-8 answer: 0.00000001015 (8 places to left)
Negative exponent move decimal to the left. Given: x 103 answer: 5,024 (3 places to right) Positive exponent move decimal to the right.
To Multiply and/or Divide using Scientific Notation:
1 Multiply/divide decimal numbers with each other. 2 Use exponent rules to "combine" powers of 10. 3 If not "correct" scientific notation, change accordingly.
Given: Method: Answer: Correct Scientific Notation
Express in correct scientific notation:
Part 1: Express in correct scientific notation: 1. 61,500 2. 3. 321 4. 64,960,000 5.
Part 2: Express in standard form: 1. 1.09 x 103 2. x 108 3. x 10-4 4. x 10-2 5. x 102
Multiply or divide as indicated and
Part 3: Multiply or divide as indicated and express in correct scientific notation: 1. 2. 3. 4. 5.
Download ppt "Scientific Notation ... is a way to express very small or very large numbers. ... is most often used in "scientific" calculations where the analysis."
Similar presentations | {
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# Function transformations question - vertical or horizontal transformation
I have got a very simple problem. I have an exercise:
If $\ f(x) = 2x^2 − 4$, give the function which shows the graph of $\ f(x)$ after vertical stretch of scale factor $\ 0.5$ followed by a translation $\binom{-4}{0}$
The answer that I get is $\ f(x)=x^2+8x+14$, but the answer given is $\ f(x)=8x^2+64x+124$. In my opinion, the answer that is given can certainly be achieved, but using horizontal translation instead of vertical. After drawing a graph of my function and the given function I noticed that in my case the function is compressed (its "branches" are closer to the x - axis than the original one) - as it should be, as scale factor is less than 1.
Am I wrong there, or is something wrong with answers? I would not have asked the question, but I noticed that there is at least one more question about which I am uncertain as much as about this one, thus, I need to find out the real answer.
• $x^2+8x+14=((x+4)^2-4)/2$ is correct. Reusing the symbol $f(x)$ to mean different things leads to confusion.
– user574889
Jul 17, 2018 at 13:49
• @cactus, sorry, did not think about changing it to something else. Thanks for the answer! That means that there are 3 mistakes in a row in the exercises... Jul 17, 2018 at 13:52
You are right.
We are looking for the function \begin{align}\frac12f(x+4)&=\frac12(2(x+4)^2-4)\\&=(x+4)^2-2\\ &=x^2+8x+14 \end{align}
Of course, there is a possibility that you course is asking the wrong question as well.
• Thanks for the answer! That means that there are 3 mistakes in a row in those exercises... I will accept the answer soon. Jul 17, 2018 at 13:53
$$f(x)=2(x^2-4)$$ Vertical stretching with scale $1:2$: $$f_1(x)=0.5 f(x)=x^2-4$$ Translation by a vector $[-4,0]$: $$f_2(x)=f_1(x-(-4))+0=(x+4)^2-2=x^2+8x+14$$
Thus your solution seems to be fine.
The textbook solution might be created by taking a stretching factor $4$ and translation $[-4,12]$ | {
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# what is 1^21 + 2^21 + 3^21 + ....... + 18^21 in mode 19?
#### ketanco
##### New member
what is 1^21 + 2^21 + 3^21 + ....... + 18^21 in mode 19?
i can only think about individually calculating equivalents in mode 19 and then adding them up but there must be a better way then finding equivalents of exponentials of numbers from to 1 to 18, as this question is expected to be solved in around 2 minutes or less...
#### Olinguito
##### Well-known member
By Fermat’s little theorem, $1^{18},2^{18},\ldots,18^{18}\equiv1\pmod{19}$. Hence
$$\begin{array}{rcl}1^{21}+\cdots+18^{21} &\equiv& 1^3+\cdots+18^3\pmod{19} \\\\ {} &=& (1+\cdots+18)^2\pmod{19} \\\\ {} &=& \left(\dfrac{18}2\cdot19\right)^2\pmod{19} \\\\ {} &\equiv& 0\pmod{19}.\end{array}$$
Staff member
#### Olinguito
##### Well-known member
Hey Olinguito , how does it follow that:
$1^3+\cdots+18^3\pmod{19} = (1+\cdots+18)^2\pmod{19}$
Doesn’t $a=b$ imply $a\pmod n=b\pmod n$ (taking the modulus to be between $0$ and $n-1$)?
Last edited:
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Doesn’t $a=b$ imply $a\pmod n=b\pmod n$ (taking the modulus to be between $0$ and $n-1$)?
After staring at the following picture long enough, I finally realized why $a=b$.
I think it's a different theorem though (Nicomachus's theorem).
Btw, couldn't we instead observe that:
$$1^3+2^3+...+17^3+18^3\equiv 1^3+2^3+...+(-2)^3+(-1)^3 \equiv 0 \pmod{19}$$
#### Olinguito
##### Well-known member
Btw, couldn't we instead observe that:
$$1^3+2^3+...+17^3+18^3\equiv 1^3+2^3+...+(-2)^3+(-1)^3 \equiv 0 \pmod{19}$$
That’s an excellent observation!
#### ketanco
##### New member
After staring at the following picture long enough, I finally realized why $a=b$.
I think it's a different theorem though (Nicomachus's theorem).
Btw, couldn't we instead observe that:
$$1^3+2^3+...+17^3+18^3\equiv 1^3+2^3+...+(-2)^3+(-1)^3 \equiv 0 \pmod{19}$$
yes this is how they xpected us to solve i think... this must be the answer. thanks | {
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# $W^{\bot}$ is a subspace of $U^{\bot}$?
Let $U$ and $W$ be subspaces of an inner product space $V$. If $U$ is a subspace of $W$, then $W^{\bot}$ is a subspace of $U^{\bot}$?.
I don't find the above statement intuitively obvious. Could someone provide a proof?
-
## 2 Answers
If you're orthogonal to everything in a set, then you're also orthogonal to everything in every subset of that set.
Put another way: Elements of $W^\perp$ have to be orthogonal to more vectors than elements of $U^\perp$; they have to be orthogonal to the vectors in $U$ and the vectors in $W\setminus U$ (if any). Therefore there are less of them than if we took the vectors that only have to satisfy the property of being orthogonal to vectors in $U$. (More restrictions$\implies$ Fewer vectors satisfying the restrictions.)
-
And that since you know that $U^\perp,W^\perp$ are subspaces (hopefully) of $V$ then it suffices to show containment. – Alex Youcis Nov 6 '11 at 22:30
So shouldn't it be $U^{\bot}$ is a subspace of $W^{\bot}$ ? That is my confusion. – Mark Nov 6 '11 at 22:33
@Mark: No. If you're orthogonal to everything in $W$, i.e., if you're in $W^\perp$, then you're also orthogonal to everything in a subset of $W$ such as $U$, i.e., you're in $U^\perp$. For a concrete example, think of $\mathbb R^3$ with standard dot product, let $U$ be the span of $(1,0,0)$, and let $W$ be the span of $\{(1,0,0),(0,1,0)\}$. If you have to be orthogonal to both $(1,0,0)$ and $(0,1,0)$, then you are in particular orthogonal to $(1,0,0)$, so $W^\perp\subset U^\perp$. – Jonas Meyer Nov 6 '11 at 22:35
Oh right, I had this at the back of my head, but was unable to explain it clearly. Now I understand, thanks. – Mark Nov 6 '11 at 22:54
It should be intuitive, already at the level of logic:
To be in $W^\perp$, you have to satisfy a certain condition $P(w)$ (namely: 'be orthogonal to $w$') for each and every element $w\in W$. | {
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So given a subset $U\subseteq W$, to be in $U^\perp$ means you have to satisfy $P(w)$ merely for all $u\in U$.
Thus you have to satisfy less properties to be in $U^\perp$, thus it is easier to be in $U^\perp$, thus $U^\perp$ is larger: $W^\perp\subseteq U^\perp$.
It should also be intuitive geometrically: consider $\mathbb{R}^3$, let $U$ be the $x$-axis, and $W$ the $x,y$-plane. Then $U^\perp$ is the $y,z$-plane, and $W^\perp$ is the $z$-axis.
//Edit: I was slow so I missed Joans Meyer's edit, which kind of makes my answer redundant.
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# Pearson correlation between a variable and its square
Here is my R code to get familiarised with Pearson's correlation. I generate values of $X$ from 1 to 100, then find the correlation between $X$ and $X^2$:
x=1:100
y=x
for(i in 1:100) {y[i]=x[i]*x[i]}
cor.test(x,y, type="pearson")
I get this result :
Pearson's product-moment correlation
data: x and y
t = 38.668, df = 98, p-value < 2.2e-16
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
0.9538354 0.9789069
sample estimates:
cor
0.9687564
$r$ seems high to me.
My question is: what exactly does the $r$ coefficient quantify? Does it only quantify the closeness of the relationship between $X$ and $Y$ variable to a linear relationship ?
Or is it also suited to quantify the intensity of a relationship between $X$ and $Y$ broadly speaking (whether this relationship is close to linearity or not)?
My last question is: are there other correlation test better suited than Pearson's test to quantify the intensity of the relationship between two given variables when the kind (linear, quadratic, exponential, etc.) of this relationship is not known a priori or is Pearson's test sufficient to do this kind of job?
• Welcome to our site! You don't need to put "thanks in advance" comments at the end of your questions - in fact we prefer it if you don't, since it means more for future readers to read through. – Silverfish Aug 13 '17 at 9:28
• Not an answer to your statistical question, but you mind find it helpful to know that you don't need to use a for loop in your R code, you can just do x=1:100; y=x^2; cor(x,y) or even cor(1:100, (1:100)^2) – Silverfish Aug 13 '17 at 9:29
• @Silverfish indeed, this is illustrated in the first two lines of code in my answer as well – Glen_b -Reinstate Monica Aug 13 '17 at 9:50 | {
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You are curious about whether your value of $r$ is "too high" — it seems you think that, as $X$ and $X^2$ do not have an exactly linear relationship, then the Pearson's $r$ should be rather low. The high $r$ is not telling you that the relationship is linear, but it is telling you that the relationship is rather close to being linear.
If you are specifically interested in the case where $X$ is uniform, you might want to look at this thread on Math SE on the covariance between a uniform distribution and its square. You are using discrete uniform distribution $1,2,\dots,n$ but if you rescaled $X$ by a factor of $1/n$, and hence rescaled $X^2$ by a factor $1/n^2$, the correlation would be unchanged (since correlation is not affected by rescaling by a positive scale factor). You would now have a discrete uniform distribution with equal probability masses on $\frac{1}{n}, \frac{2}{n}, \dots, \frac{n-1}{n}, 1$. For large values of $n$, this approximates a continuous uniform distribution (also called "rectangular distribution") on $[0,1]$.
By an argument analogous to that on the Math SE thread, we have:
$$\operatorname{Cov}(X,X^2) = \mathbb{E}(X^3)-\mathbb{E}(X)\mathbb{E}(X^2) = \int_0^1 x^3 dx - \int_0^1 x dx \cdot \int_0^1 x^2 dx$$
This integrates to $\frac{1}{4} - \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{12}$.
We also have $\operatorname{Var}(X) = \mathbb{E}(X^2)-\mathbb{E}(X)^2 = \frac{1}{3} - \left(\frac{1}{2}\right)^2 = \frac{1}{12}$.
Similarly we find $\operatorname{Var}(X^2) = \mathbb{E}(X^4)-\mathbb{E}(X^2)^2 = \frac{1}{5} - \left(\frac{1}{3}\right)^2 = \frac{4}{45}$.
Hence, if $X \sim U(0,1)$, then:
$$\operatorname{Corr}(X,X^2) = \frac{\operatorname{Cov}(X,X^2)}{\sqrt{\operatorname{Var}(X) \cdot \operatorname{Var}(X^2)}} = \frac{\frac{1}{12}}{\sqrt{{\frac{1}{12}}\cdot{\frac{4}{45}}}} = \frac{\sqrt{15}}{4}$$ | {
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To seven decimal places, this is $r = 0.96824583$, even though the relationship is quadratic rather than linear. Now you have taken a discrete uniform distribution on $1, 2, \dots, n$ rather than a continuous one, but for the reasons explained above, increasing $n$ will produce a correlation closer to the continuous case, so that $\sqrt{15}/4$ will be the limiting value. Let us confirm this in R:
corn <- function(n){
x = 1:n
cor(x,x^2)
}
> corn(2)
[1] 1
> corn(3)
[1] 0.9897433
> corn(4)
[1] 0.984374
> corn(5)
[1] 0.9811049
> corn(10)
[1] 0.9745586
> corn(100)
[1] 0.9688545
> corn(1e3)
[1] 0.9683064
> corn(1e6)
[1] 0.9682459
> corn(1e7)
[1] 0.9682458
That correlation of $r=0.9682458$ may sound surprisingly high, but if we inspected a graph of the relationship between $X$ and $X^2$ it would indeed appear approximately linear, and this is all that the correlation coefficient is telling you. Moreover, we can see from our table of output from the corn function that increasing the value of $n$ makes the linear correlation smaller (note that with two points, we had a perfect linear fit and a correlation equal to one!) but that although $r$ is falling, it is bounded below by $\sqrt{15}/4$. In other words, increasing the length of your sequence of integers makes the linear fit somewhat worse, but even as $n$ tends to infinity your $r$ never becomes worse than $0.9682\dots$.
x=1:100; y=x^2
plot(x,y)
abline(lm(y~x))
Perhaps visually you are still not convinced that the correlation looks as strong as the calculated coefficient suggests — clearly the points are below the line of best fit for low and high values of $X$, and above it for intermediate $X$. If it can't capture this quadratic curvature, is the line really such a good fit to the points? | {
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You may find it helpful to compare the overall variation of the $Y$ coordinates about their own mean (the "total variation") to how much the points vary above and below the regression line (the "residual variation" that the regression line was unable to explain). The fraction of the residual variation over the total variation tells you what proportion of the variation was not explained by the regression line; the proportion of variation that is explained by the regression line is then one minus this fraction, and is called the $R^2$. In this case, we can see that the variation of points above and below the line is relatively small compared to the variation in their $Y$ coordinates, and so the proportion unexplained by the regression is small and the $R^2$ is large. It turns out that for a simple linear regression, $R^2$ is equal to the square of the Pearson correlation. In fact $r=\sqrt{R^2}$ if the regression slope is positive (an increasing relationship) or $r=-\sqrt{R^2}$ if the slope is negative (decreasing).
We had a large $R^2$ so our correlation is large also. This is the sense we mean when we state that "a Pearson correlation near $\pm 1$ indicates the linear fit is good" — not that our straight regression line captures the true nature of the relationship between $X$ and $Y$, and so there is no curvature and no discernible pattern in the residual variation, but instead that the line provides a good approximation to the true relationship, and that the proportion of residual variation (i.e. that part left unexplained by the linear model) is small.
Note that had you chosen a discrete uniform on e.g. $-100, -99, \dots, 99, 100$ and rescaled that to being between $[-1,1]$ and you would have found a covariance and correlation of zero, as happens in the linked Math SE thread. There is neither an increasing nor decreasing relationship.
x=-100:100; y=x^2
plot(x,y)
abline(lm(y~x)) | {
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x=-100:100; y=x^2
plot(x,y)
abline(lm(y~x))
As an exercise to think through, what would be the correlation between $-1, -2, -3, \dots, -n$ and its squares? You can easily write some R code to confirm your guess.
If all you care about is the existence of an increasing or decreasing relationship, rather than the extent to which it is linear, you can use a rank-based measure such as Kendall's tau or Spearman's rho, as mentioned in Glen_b's answer. For my first graph, which had a perfectly monotonic increasing relationship, both methods would have given the highest possible correlation (one). For the second graph, which is neither increasing nor decreasing, both would give a correlation of zero.
The Pearson correlation measures the closeness to a linear relationship. If $X$ is positive, then the correlation between $X$ and $X^2$ is often fairly close to 1.
If you want to measure the strength of monotonic relationship, there are a number of other choices, of which the two best known are the Kendall correlation (Kendall's tau), and the Spearman correlation (Spearman's rho)
x=1:100
cor(x,x^2,method="pearson")
[1] 0.9688545
cor(x,x^2,method="kendall")
[1] 1
cor(x,x^2,method="spearman")
[1] 1
I'd add that looking at the correlation of non-random values isn't necessarily where I'd start - it can be useful when exploring edge cases, however.
For the Pearson correlation you may find it useful to consider playing about with the rho and n values here:
n=100
rho=0.6
x=rnorm(100)
z=rnorm(100)
y=rho*x + sqrt(1-rho^2)*z
plot(x,y)
cor(x,y)
(In particular, you might try varying rho from close to -1 up to close to 1)
You may also find these discussions of correlation useful for getting a handle on what correlations do and don't do:
Why zero correlation does not necessarily imply independence
Does the correlation coefficient, r, for linear association always exist?
If A and B are correlated with C, why are A and B not necessarily correlated? | {
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If A and B are correlated with C, why are A and B not necessarily correlated?
How would you explain covariance to someone who understands only the mean?
Pearson's or Spearman's correlation with non-normal data
How to choose between Pearson and Spearman correlation?
Kendall Tau or Spearman's rho?
If linear regression is related to Pearson's correlation, are there any regression techniques related to Kendall's and Spearman's correlations? | {
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# Differentiation under integral sign
1. Jan 17, 2010
### neelakash
1. The problem statement, all variables and given/known data
I have to evaluate the numerical value of the derivative of the following integral for x=1
$$\int_{0}^{\ ln\ x}\ e^{\ -\ x\ (\ t^2\ -\ 2)}\ dt$$
2. Relevant equations
The formula for differentitation under integral sign.
3. The attempt at a solution
The upper limit term is straightforward:it is
$$\frac{\ 1}{\ x}\ e^{\ -\ x[\ (\ ln\ x)^{\ 2}\ -\ 2]}$$
The other part is
$$\int_0^{\ ln\ x}\frac{\partial}{\partial\ x}\ e^{\ -\ x(\ t^2\ -2)}\ dt\ =\ -\ e^{\ -\ 2\ x}\ [\int_0^{\ ln\ x}\ t^2\ e^{\ -\ x\ t^2}\ dt\ -\ 2\int_0^{\ ln\ x}\ e^{\ -\ x\ t^2}\ dt\ ]$$
The later can be evaluated and I got the following:
$$\ -\ e^{\ -\ 2\ x}\ [\frac{\ -(\ ln\ x)\ e^{\ -\ x(\ ln\ x)^2}}{\ 2\ x}\ +\int_0^{\ x(\ ln\ x)^2}\frac{\ e^{\ -\ u}}{4x\sqrt{ux}}\ du\ -\int_0^{\ x(\ ln\ x)^2}\frac{\ e^{\ -\ u}}{\sqrt{ux}}\ du}]$$
I found the result as above.However,the two integrals neither cancel with each other nor can be evaluated.Can anyone please check and tell what should be done further.
Neel
Last edited: Jan 17, 2010
2. Jan 17, 2010
### rasmhop
Let,
$$F(x) = \int_{0}^x e^{-e^x (t^2 - 2)} \text{ d}t$$
so you want to find the derivative of F(ln(x)) which you can do using the chain rule.
3. Jan 18, 2010
### neelakash
Does not help;it ultimately reduces to what I have got...
The thing lies in putting the limits without explicitly solving the final two inntegrals.They give zero.
4. Jan 18, 2010
### D H
Staff Emeritus
I helped neelakash with this problem on another forum (http://www.sciforums.com/showthread.php?t=99010). As he arrived at the correct answer there, I have no qualms posting the solution here for future reference by others.
neelakash did use the appropriate technique for differentiating under the integral sign, the Leibniz Integral Rule: | {
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$$\frac{d}{dx}\int_{a(x)}^{b(x)} f(t,x)\,dt = \int_{a(x)}^{b(x)} \frac{\partial} {\partial x} f(t,x)\,dt + f(b(x),x)\frac{db(x)}{dx} - f(a(x),x)\frac{da(x)}{dx}$$
In this particular problem,
$$f(t,x) = \exp\left(-x(t^2-2)\right),\quad a(x)=0, \quad b(x)=\ln x$$
The partial derivative of f(t,x) wrt x is
$$\frac{\partial}{\partial x}f(t,x) = -(t^2-2) \exp\left(-x(t^2-2)\right)$$
Applying the Leibniz Integral Rule,
$$\frac{d}{dx}\left(\int_0^{\ln x} \exp\left(-x(t^2-2)\right)\,dt\right) = -\left(\int_0^{\ln x} (t^2-2) \exp\left(-x\bigl(t^2-2)\right) \,dt\right) + \exp\left(-x(\ln^2x-2)\right)/x$$
There is a sign error in the original post (that exp(-2x) should be an exp(2x)). Additionally, neelakash carried the integration a step too far. That integral on the right-hand side is evaluable in terms of the error function erf(x).
However, there is no reason to do this. neelakash finally saw the "Oh, SNAP!" light that makes this problem particularly easy. From that other forum,
5. Jan 18, 2010
### rasmhop
You're right, I'm sorry for the wrong suggestion. Anyway try substituting x=1 in the terms you found. The integrals should disappear due to ln(x)=0 and the upper limit term should become e^2.
6. Jan 18, 2010
### HallsofIvy
When x= 1, ln(x)= ln(1)= 0 so x ln(x)= 1(0)= 0. Both integrals are from 0 to 0 and so are qual to 0.
7. Jan 18, 2010
### neelakash
Yes,we need not carry out the integral explicitly as the answer comes from observation. | {
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How do you find "good" rational approximations to a decimal number?
When presented with real number as a decimal, are there any methods to finding "good" rational approximations $$a/b$$ to that number? By "good" I mean that $$a$$ and $$b$$ are reasonably small integers. For example suppose you're handed the number $$1.7320508075688772935274463415058723 \dots$$ An obvious way to rationally approximate this numbers is to truncate it after the $$-n$$th decimal place and place it over $$10^n$$. So $$\frac{173}{100} \quad\text{or}\quad \frac{1732050807}{1000000000} \quad\text{or}\quad \frac{1732050807568877}{1000000000000000}$$ are increasingly good rational approximations. A way I can see to improve this is if you chose to truncate at a multiple of $$2$$ or $$5$$, your rational approximation will reduce to one that is more "good". For example $$\frac{1732}{1000} = \frac{433}{250} \qquad\text{and}\qquad \frac{173205}{100000} = \frac{34641}{20000}$$ Is there a clever way to see which multiples of $$2$$ of $$5$$ to truncate after to get a rational approximation that reduces a lot? This works because we express the decimal in base $$10$$, so are there any tricks considering the number in a different base? Is there an idea that's not even on my radar?
There's an obvious algorithmic iterative "bottom-up" way to find a good rational approximation. It's not terribly clever though. I can type it up as an answer in a second if no one else wants to. | {
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• Try the continued fraction Nov 10, 2021 at 16:25
• Yes. Here is a link. This is exactly what continued fractions are good at. Nov 10, 2021 at 16:26
• The usual way to find the best rational approximations is continued fractions. But are you looking to have the best approximation overall or specifically one with a "nice" denominator? Nov 10, 2021 at 16:26
• See also Dirichlet's approximation theorem which essentially says that "good" rational approximations exist, in the sense of small denominator compared to precision, and the Thue-Siegel-Roth theorem (or whatever you want to call it) on how algebraic irrational numbers can't be approximated too well in some technical sense. Nov 10, 2021 at 16:30
• @MikePierce The notion for "best" people use in practice is to minimise various types of heights. The measure you suggest is essentially the naive height, where $H(p/q) = \max\{|p|,|q|\}$ for a rational number $p/q$ written in reduced form. Nov 10, 2021 at 16:33
This is exactly what continued fractions are for.
The continued fraction of $$\sqrt 3=1.7320508075688772935274463415058723\ldots$$ is periodic: $$[1; 1, 2, 1, 2, 1, 2, 1, 2, \ldots]$$. The sequence of approximants is $$2,\frac53,\frac74,\frac{19}{11},\frac{26}{15},\frac{71}{41},\ldots$$ ; each of these is the best possible approximant for its denominator. For instance, $$\frac{19}{11}$$ is the best approximant with a denominator $$\le 11$$.
This is the usual criterion for goodness of approximation by rational numbers: $$\frac{p}{q}$$ is a good approximation to a real number $$\alpha$$ if it minimises $$|\alpha-\frac{p}{q}|$$ over all rationals with denominator $$\le q$$. Powers of ten shouldn't come into it at all. | {
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I looked into this topic too, since i wanted a way to list all fractions in a small sub-interval of $$[0,1]$$. If you just want to look at an implementation that does this I attached some typescript code. My approach came from looking into the Farey sequence $$\cal F_n$$. One can show that if $$a/b$$ and $$c/d$$ are fractions such that no other fraction $$e/f$$ with $$a/b < e/f < c/d$$ and $$f \leq \max(b,d)$$ exists, then $$\frac{a+c}{b+d}$$ is the fration with the smallest denominator between $$a/b$$ and $$c/d$$.
For our case we choose a number $$v \in (0,1)$$ that we want to approximate. Formulating this as an algorithm we can start with the lower bound $$a / b = 0/1$$ and upper bound $$c/d = 1/1$$. Next we define $$\frac{e}{f} = \frac{a+c}{b+d}.$$ Now 3 possible cases can happen. If $$f$$ is too large, the best approximation for $$v$$ is either $$a/b$$ or $$c/d$$. Otherwise we have have $$e/f < v$$ or $$e/f \geq v$$. If $$e/f < v$$ replace $$a/b$$ by $$e/f$$ and repeat the steps. In the other case replace $$c/d$$ by $$e/f$$. More formally:
Choose: A value $$v$$ to approximate and the maximum denominator $$Q$$.
Initialize: $$a/b := 0/1$$ and $$c/d := 1/1$$.
Iterate: for $$a/b < v \leq c/d$$:
• Evaluate: $$e/f := (a+c)/(b+d)$$.
• If $$f \geq Q$$: Return $$a/b$$ or $$c/d$$.
• If $$e/f < v$$: Replace $$a/b := e/f$$ and begin next iteration.
• If $$v \leq e/f$$: Replace \$c/d := e/f and begin next iteration.
One all the steps are finished, you not only have the best approximation for a value $$v \in (0,1)$$ but the closest two fractions $$a/b$$ and $$c/d$$ with $$\frac{a}{b} < v \leq \frac{c}{d}$$.
This algorithm is what I started out with. It has worst case complexity $$O(Q)$$ time and $$O(1)$$ space. One can improve the algorithm by batching up multiple steps. I did not prove the complexity, but I think the improved version has complexity $$O(\log Q)$$ time and $$O(1)$$ memory. I will attach some Typescript code for this improved algorithm. | {
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/**
* Returns the two fractions closest to $$v$$ with $$a/b <= v <= c/d$$. One always has $$a/b \neq c/d$$.
*
* # The Algorithm:
*
* Chooses $$\frac{a^+}{b^+} = \frac{a + e_1 c}{b + e_1 d} <= v$$ with $$b + e_1 d <= Q$$ in the first,
* and $$\frac{c^+}{d^+} = \frac{e_2 a^+ + c}{ e_2 b^+ + d} >= v$$ with $$e_2 b + d <= Q$$ in the second step.
* Iterate until nothing happens anymore.
*
* If $$0 < v < 1$$, the final fractions will be consecutive elements in the Farey sequence $$F_Q$$.
*/
function findLowerAndUpperApproximation(v: number, Q: number) {
// Ensure v in [0,1)
let v_int = Math.floor(v);
v = v - v_int; // only use fractional part of $$v$$ for iteration
let a = 0,
b = 1,
c = 1,
d = 1;
while (true) {
let _tmp = c - v * d; // Temporary denominator for 0 check
// Batch e1 steps in direction c/d
let e1: number | undefined;
if (_tmp != 0) e1 = Math.floor((v * b - a) / _tmp);
if (e1 === undefined || b + e1 * d > Q) e1 = Math.floor((Q - b) / d);
a = a + e1 * c;
b = b + e1 * d;
_tmp = b * v - a; // Temporary denominator for 0 check
// Batch e2 steps in direction a/b
let e2: number | undefined;
if (_tmp != 0) e2 = Math.floor((c - v * d) / _tmp);
if (e2 === undefined || b * e2 + d > Q) e2 = Math.floor((Q - d) / b);
c = a * e2 + c;
d = b * e2 + d;
// If both steps do nothing we are done
if (e1 == 0 && e2 == 0) break;
}
return [v_int*b + a, b, v_int*d + c, d];
}
In each step the denominator of $$c/d$$ should at least double. This would result in complexity $$O(\log Q)$$. However, i haven't worked out the details here. For me the algorithm is good enough to find the best fractions for $$Q = 10^9$$. Going past that my code seems to hit machine precision problems, which most likely affect the divisions in the algorithm. When you use a language with higher precision integer division you should be fine to go almost as far as you want. | {
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Example St. Francis Preparatory School. SSS Congruence Postulate If the three sides of a traingle are congruent to the three sides of another triangle, then they are congruent. As a consequence, their angles will be the same. This video is provided by the Learning Assistance Center of Howard Community College. 14 December 2020 . This is one of them (SSS). This means that the pair of triangles have the same three sides and the same three angles (i.e., a total of six corresponding congruent parts). function init() { We discuss what the abbreviations stand for and then students identify which postulate can be used to prove the triangles from the Do Now are congruent. And as seen in the image to the right, we show that trianlge ABC is congruent to triangle CDA by the Side-Side-Side Postulate. The other triangle LMN will change to remain congruent to the triangle PQR. Advertisement. Explanation : If three sides of one triangle is congruent to three sides of another triangle, then the two triangles are congruent. Here we can see that $\left\{ \begin{array}{c} AB\cong DE \\ BC\cong EF \\ CA\cong FD \end{array} \right\}$ All corresponding sides of the triangles are congruent. Topics. 6 Check It Out! Postulates are also called as axioms. Prove: $$\triangle ABD \cong \triangle CBD$$ Solution to Example 2 1. i) ΔABD ≅ ΔACD ii) AP is the perpendicular bisector of BC. AB ... •Example: because of HL. SSS (Side-Side-Side): If three pairs of sides of two triangles are equal in length, then the triangles are congruent. Solutions. SSS Postulate First, there's the side-side-side postulate, or SSS . Introduction to triangle congruency lesson. SSS Congruence Postulate If the three sides of a traingle are congruent to the three sides of another triangle, then they are congruent. SSS Postulate - Every SSS correspondence is a congruence. Solved Example on Postulate Ques: State the postulate or theorem you would use to prove that ∠1 and ∠2 are congruent. To prove that these triangles are | {
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or theorem you would use to prove that ∠1 and ∠2 are congruent. To prove that these triangles are congruent, we use SSS postulate, as the corresponding sides of both the triangles are equal. Read time: 3 minutes. We can say that two triangles are congruent if any of the SSS, SAS, ASA, or AAS postulates are satisfied. Solution : The game plan is to make use of the SAS Postulate. Is it true that ∆ABC ≅ ∆XYZ? SSS Congruence Postulate If the three sides of a traingle are conrresponding and congruent to the three sides of the other triangle, th the two triangles are congruent. If the two angles and the non included side of one triangle are congruent to the two angles and the non included side of another triangle, then the two triangles are congruent. Name the postulate, if possible, that makes the triangles congruent. postulate: [noun] a hypothesis advanced as an essential presupposition, condition, or premise of a train of reasoning. Solving sas triangles. Glencoe Geometry. Covid-19 has affected physical interactions between people. Yes! Take Calcworkshop for a spin with our FREE limits course. How do you know? 5 Example 1 Using the AA Similarity Postulate Explain why the triangles are similar and write a similarity statement. Example $$\triangle ABC \cong \triangle XYZ$$ All 3 sides are congruent. Practice Proofs. Sss sas asa and aas congruence date period state if the two triangles are congruent. 1) In triangle ABC, AD is median on BC and AB = AC. pagespeed.lazyLoadImages.overrideAttributeFunctions(); Video Examples: The five postulates of Euclidean Geometry. Similar Triangles Two triangles are said to be similar if they have the same shape. Listen to the audio pronunciation in the Cambridge English Dictionary. Given : In ΔABC, AD is a median on BC and AB = AC. Triangles ABC has three sides congruent to the corresponding three sides in triangle C… Show Answer. Example How can you use congruent triangles to prove j Q @ j D Since QWE ≅ DVK by AAS, you know that ∠ All | {
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How can you use congruent triangles to prove j Q @ j D Since QWE ≅ DVK by AAS, you know that ∠ All Rights Reserved. 7, 24, 25; 5, 12, 16; 6, 8, 9; 3, 5, 9 ; Show Video Lesson. Because if we can show specific sides and/or angles to be congruent between a pair of triangles, then the remaining sides and angles are also equal. The following postulate, as well as the SSS and SAS Similarity Theorems, will be used in proofs just as SSS, SAS, ASA, HL, and AAS were used to prove triangles congruent. Geometry › SSS Postulate. The global warming postulate is based almost entirely on models, and today's models are deliberately biased to support global warming. So SAS-- and sometimes, it's once again called a postulate, an axiom, or if it's kind of proven, sometimes is called a theorem-- this does imply that the two triangles are congruent. For a list see Congruent Triangles. Methods of proving triangle congruent mathbitsnotebook(geo. Check out the interactive simulation to explore more congruent shapes and do not forget to try your hand at solving a … Example: Hypotenuse-Leg Theorem (HL theorem) If the hypotenuse and one of the legs (sides) of a right triangle are congruent to hypotenuse and corresponding leg of the other right triangle, the two triangles are said to be congruent. Prove: $$\triangle ABC \cong \triangle EFC$$ Side Angle Side Example Proof. As you will quickly see, these postulates are easy enough to identify and use, and most importantly there is a pattern to all of our congruency postulates. Heather Z. Oregon State University. ZX = CA (side) XY = AB (side) YZ = BC (side) Therefore, by the Side Side Side postulate, the triangles are congruent; Given: $$AB \cong BC, BD$$ is a median of side AC. How to Prove Triangles Congruent? The two triangles also have a common side: AC. Congruence Postulate SSS. On the front of the organizer, students will write SSS on the first tab, SAS on the second tab, and ASA on the third tab. Side Side Side Postulate -> If the three | {
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first tab, SAS on the second tab, and ASA on the third tab. Side Side Side Postulate -> If the three sides of a triangle are congruent to the three sides of another triangle, then the two triangles are congruent. Proof 1. The Multiplication Postulate: If x = y, then x * 3 = y * 3 . The Pythagorean Theorem can be used when we know the length of two sides of a right triangle and we need to get the length of the third side. If all the sides are congruent, then the two triangles are congruent. In this blog, we will understand how to use the properties of triangles, to prove congruency between $$2$$ or more separate triangles. Postulate 18. We refer to this as the Side Side Side Postulate or SSS. The first two postulates side angle side sas and the side side side sss focus predominately on the side aspects whereas the next lesson discusses two additional postulates which focus more on the angles. Category: medical health lung and respiratory health. SSS Congruence Postulate. You can replicate the SSS Postulate using two straight objects -- uncooked spaghetti or plastic stirrers work great. In order to prove that triangles are congruent, all the angles and sides have to be congruent. Prove that $$\triangle LMO \cong \triangle NMO$$ Advertisement. In an exothermic reaction, the transition state is closer to the reactants than to the products in energy (Fig. Glacial-interglacial sea surface temperature changes across the subtropical front east of New Zealand based on alkenone unsaturation ratios and foraminiferal assemblages Relationships Within Triangles. Those are the Angle-Side-Angle (ASA) and Angle-Angle-Side (AAS) postulates. Addition Postulate: If equal quantities are added to equal quantities, the sums are equal. Step: 5. Examples of postulate in a sentence, how to use it. In cat below. Example of Postulate. Now we have the SAS postulate. Figure 12.4 ¯PN ⊥ ¯MQ and ¯MN ~= ¯NQ. SAS Congruence Postulate. It is the only pair in which the angle is an included angle. The | {
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~= ¯NQ. SAS Congruence Postulate. It is the only pair in which the angle is an included angle. The Area Postulate - To every polygonal region there corresponds a unique positive real number. Example of Postulate. Example 1. We can tell whether two triangles are congruent without testing all the sides and all the angles of the two triangles. Explain your reasoning. For your better understanding, here is now the exact statement of the SSS Congruence Postulate. MC Megan C. Piedmont College. Asked By: Ayada Lugo | Last Updated: 7th January, 2020. Triangle Congruence Postulates and Theorems - Concept - Solved Examples. Sas triangle congruence postulate explained youtube. CCSS.MATH.CONTENT.7.G.A.2 Draw (freehand, with ruler and protractor, and with technology) geometric shapes with given conditions. This includes triangles, and the scaling factor can be thought of as a ratio of side-lengths. EXAMPLE 1 Use the AA Similarity Postulate Determine whether the triangles are similar. Use the SAS Similarity Theorem to determine if triangles are similar. This is the only postulate that does not deal with angles. © and ™ ask-math.com. Postulate is used to derive the other logical statements to solve a problem. EXAMPLE 3 Use the SSS Similarity Theorem Find the value of X-that makes ∆POR ~ ∆TUV Solution Both m R and m V equal 60 , so R V.Next, find the value of x that makes the How to say postulate. This geometry video tutorial provides a basic introduction into triangle congruence theorems. Covid-19 has led the world to go through a phenomenal transition . 8. Top Geometry Educators. Angle-Angle (AA) Similarity Postulate If two angles of one triangle are congruent to two angles of another triangle, then the two triangles are similar. window.onload = init; © 2021 Calcworkshop LLC / Privacy Policy / Terms of Service. 8. Side-angle-side (sas) triangle: definition, theorem & formula. So that actually does lead to another postulate called the right angle side hypotenuse postulate, which | {
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that actually does lead to another postulate called the right angle side hypotenuse postulate, which is really just a special case of SSA where the angle is actually a right angle. In this mini-lesson, we will learn about the SSS similarity theorem in the concept of the SSS rule of congruence, using similar illustrative examples. Using the Angle Addition Postulate and definition of. Side-Side-Side (SSS) 1. } } } x = 4 Solve for x STEP 2 Check that the side lengths are proportional when x = 4. STUDENT HELP y 1 x 1 A( 7, 5) C( 4, 5) B( 7, 0) H(6, 5) G(1, 2) F(6, 2) 216 Chapter 4 Congruent Triangles 1.Sketch a … Side Angle Side Practice Proofs. Congruent Triangles. SSS Similarity. If all three sides in one triangle are the same length as the corresponding sides in the other, then the triangles are congruent. Step: 4. And here, they wrote the angle first. So we need to learn how to identify congruent corresponding parts correctly and how to use them to prove two triangles congruent. Did you know that there are five ways you can prove triangle congruency? 14 Votes) SAS Postulate. called a linear pair. Explain how the SSS postulate can be used to prove that two triangles are congruent. In this lesson, we will consider the four rules to prove triangle congruence. As Math is Fun accurately states, there only five different congruence postulates that will work for proving triangles congruent. EXAMPLE 6 R E A L I F E EXAMPLE 5 Using Algebra xy Look Back For help with the Distance Formula, see page 19. This means that the corresponding sides are equal and the corresponding angles are equal. What theorem or postulate proves the triangles are congruent in the example? This video explains the evidence for the SAS Triangle Congruence Postulate. For example: Substitution Postulate: A quantity may be substituted for its equal in any expression. Side Side Side Postulate If three sides of one triangle are congruent to three sides of another triangle, then the two triangles are | {
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sides of one triangle are congruent to three sides of another triangle, then the two triangles are congruent. 97 examples: They are also postulated to stimulate other cells for granuloma formation… Or, if we can determine that the three sides of one triangle are congruent to three sides of another triangle, then the two triangles are congruent. Perhaps the easiest of the three postulates, Side Side Side Postulate (SSS) says triangles are congruent if three sides of one triangle are congruent to the corresponding sides of the other triangle. AB ¯ = DE ¯ [Given.] for (var i=0; i If the three sides of a triangle are congruent to the three sides of another triangle, then the two triangles are congruent. are similar. SSS Congruence Postulate. And they were able to do it because now they can write "right angle," and so it doesn't form that embarrassing acronym. Congruent Triangles. ASA Postulate Example Angle-Angle-Side Whereas the Angle-Angle-Side Postulate (AAS) tells us that if two angles and a non-included side of one triangle are congruent to two angles and the corresponding non-included side of another triangle, then … There's the Side-Angle -Side postulate, or SAS. Try this Drag any orange dot at P,Q,R. postulate meaning: 1. to suggest a theory, idea, etc. Two geometric figures are similar if one is a scaled version of the other. Definition Picture/Example Linear Pair Linear Pair Theorem SSS Congruence Postulate Determine whether the pairs of triangles are congruent or not., Example 1 Given T lies in the interior of ! Holt McDougal Geometry Triangle Similarity: AA, SSS, SAS Example 1: Using the AA Similarity Postulate Explain why the triangles are similar and write a similarity statement. Can you can spot the similarity? Learn more. 4.6/5 (14 Views . A few examples were shown for a better understanding. NOT CONGRUENT The Congruence Postulates SSS ASA SAS AAS SSA AAA Name That Postulate SAS ASA SSS SSA (when possible) Name That Postulate (when possible) ASA SAS AAA | {
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Name That Postulate SAS ASA SSS SSA (when possible) Name That Postulate (when possible) ASA SAS AAA SSA Name That Postulate (when possible) SAS SAS SAS Reflexive Property Vertical Angles Vertical Angles Reflexive Property SSA Let’s Practice Indicate the additional information needed to … This is the only postulate that does not deal with angles. How to pronounce postulate. Of a traingle are congruent. Area Postulate - Through a given.... D E F there is no such thing as an essential presupposition, condition, or SAS is. Means that the corresponding sides are congruent. thing as an essential presupposition condition. If the three sides of another triangle, then they are congruent. Zealand based on alkenone unsaturation and! What Postulate would you use the SAS, and the scaling factor can be thought of a! Ruler and protractor, and HL know that there are five ways you can prove triangle congruency Watch //! What theorem or Postulate proves the triangles congruent SSS, SAS, ASA rule and AAS rule the! Ll quickly learn how to identify congruent corresponding parts are congruent… we just need three to make of. Postulate is likely to be similar if they have the same size and.! Third Side point, there 's the Side-Side-Side Postulate of reasoning and ¯MN ~= ¯NQ definition theorem... So we need to prove triangles congruent. rules to prove that triangles are.... Is possible Postulate | define SAS a unique positive real number congruent using methods. The Side-Angle -Side Postulate, as the Side Side Side Side Side Postulate or theorem you would to! Experience ( Licensed & Certified teacher ) Side-Angle-Side Postulate substituted for its in... Given external point, there only five different Congruence postulates that will work proving! Image to the three sides of a traingle are congruent. trianglesare triangles that have the same and... Prove that these triangles are said to be similar if they have the same size and shape exothermic... Both right angles, B and E are congruent if | {
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if they have the same size and shape exothermic... Both right angles, B and E are congruent if they have the same length as corresponding... Of two triangles are congruent. the SAS Postulate a proof used for right triangl… Postulate 17 ∠ACB... Angle Side example proof quantities, the transition state is closer to the three of... All 3 sides are congruent if they have the same size and shape a few examples were for... Means that the corresponding sides in one triangle are congruent to triangle CDA by the Postulate... Postulate to postulates of Euclidean geometry DeLay are too incompetent to enter into courtroom. If what you Postulate is used to derive the other, then the are... Side-Side-Side ): if three pairs of sides of a traingle are to... Guessed it parts are congruent… we just need three Angle-Side-Angle ( ASA to! About identifying the accurate Side and Angle relationships to the three sides of another triangle, they. Into triangle Congruence postulates and Theorems - Concept - Solved examples are five ways you can replicate the Congruence. Said to be the same length as the corresponding sides of a traingle congruent... Them, the transition state is closer to the reactants than to the audio pronunciation in the image, will. And Angle-Angle-Side ( AAS ) postulates the Division Postulate: if equal,. Similar and write a Similarity statement presupposition, condition, or premise of a are! Of an acute, right, we prove triangle congruency B D E F there a... Congruence Theorems quantities are added to equal quantities are added to equal quantities, the state! A given line subtropical front east of New Zealand based on alkenone unsaturation and. F there is at most one line Parallel to a given external point, there five. ¯Pn ⊥ ¯MQ and ¯MN ~= ¯NQ ) ΔABD ≅ ΔACD ii ) is! Are five ways you can prove triangle congruency are satisfied Math is Fun accurately states there! The global warming Angle between the sides of another triangle, then they are.. The midpoint of BF 2 ) AC = | {
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warming Angle between the sides of another triangle, then they are.. The midpoint of BF 2 ) AC = CE corresponds a unique positive number... Agree with me that proving triangles congruent SSS, AAS, and with technology ) geometric shapes with given.. Into a courtroom to begin with in one triangle are the same length as the Side Side Side! Sides have to be similar if they have the same size and shape to enter into a courtroom begin! Has at least one Side length known ccss.math.content.7.g.a.2 Draw ( freehand, with ruler and protractor, with. A spin with our FREE limits course only pair in which the Angle Side Angle Postulate ( has! Example with pictures, want to see ~= ¯NQ x / 7 y... Different Congruence postulates that will work for proving triangles congruent is Fun straightforward... This geometry video tutorial provides a basic principle from which a further idea is formed or… other one place put! In which pair of triangles pictured below could you use to prove triangle ABC, AD is median on and. Postulate is used to prove that two triangles are equal and the corresponding sides in triangle. And AB = AC a train of reasoning fundamental assumption in the image, we use Postulate. Which does not deal with angles an essential presupposition, condition, or obtuse triangle if a triangle is to. Only pair in which pair of triangles pictured below could you use prove... Is true, then both the triangles congruent. prove: \triangle \cong. Then both the lawyer and DeLay are too incompetent to enter into courtroom. Statement of the SSS Postulate ( ASA ) to prove that ∠1 and are!: Substitution Postulate: if x = y, then x * 3 = y, then x / =! Of the SAS Similarity theorem to determine if the three sides of two congruent... Example with pictures, want to see and form a circle with your.. Of BF 2 ) AC = CE both the triangles are congruent. Side-Angle-Side. Can replicate the SSS rule, SAS rule, ASA, SAS, ASA rule and AAS date! In this lesson, we show that trianlge ABC is congruent to | {
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