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system ψ(x,y) = (x,y,0)? Answer: In order to compute the Christoffel symbols, first we need to compute the partials of E, Fand G: E ρ = 0 E θ = 0 F ρ = 0 F θ = 0 G ρ = 2ρ G θ = 0. Polar coordinates are in the form r, , where is the independent variable. We want to nd another way to get to the point (x;y). The Polar Coordinate System is a different way to express points in a plane. The polar form of (a,b) is illustrated in Figure 1. Number Planes. Polar coordinates and applications Let's suppose that either the integrand or the region of integration comes out simpler in polar coordinates (x= rcos and y= rsin ). Plane polar coordinates pdf Plane polar coordinates pdf Plane polar coordinates pdf DOWNLOAD! DIRECT DOWNLOAD! Plane polar coordinates pdf Polar Coordinates r, θ in the plane are described by r distance from the origin and θ 0, 2π is the counter-clockwise angle. It is sometimes convenient to refer to a point by name, especially when this point occurs in multiple \draw commands. generally start to learn to think in terms of polar coordinates. Consider the top which is bounded above by z= p 4 x2 y2 and bounded below by z= p x2 + y2, as shown below. 2 We can describe a point, P, in three different ways. Extensions Graphing the polar equations will help the students to make the connection when they are learning to change polar coordinates to rectangular coordinates and back The students can visually see the points on the polar axis and compare the point on the rectangular axis. Double Integrals in Polar Coordinates 1. Finally, the Coriolis acceleration 2r Ö. 5355 0 -10] x = 1×4 5. Stirling's Web Site. Compass Labels on Polar Axes. | {
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1. Dec 12, 2013
### Appleton
1. The problem statement, all variables and given/known data
$\int \frac{1}{\sqrt{1-x^{2}}} dx$
I seem to be getting two incompatible answers when I substitute x with $sin \theta$ and $cos \theta$. Could someone please help me with where I'm going wrong.
2. Relevant equations
3. The attempt at a solution
first attempt (the correct answer I believe):
$let\ x = sin \theta$
$dx = cos \theta\ d\theta$
$\int \frac{cos \theta}{\sqrt{1-sin^{2}\theta}} d\theta$
$= \int \frac{cos \theta}{\sqrt{cos^{2}\theta}} d\theta$
$= \int \frac{cos \theta}{{cos\theta}} d\theta$
$= \int 1\ d\theta$
$= \theta$
$x = sin \theta$
$\theta = arcsin x$
second attempt (the wrong answer I believe)
$let\ x = cos \theta$
$dx = -sin \theta\ d\theta$
$\int \frac{-sin \theta}{\sqrt{1-cos^{2}\theta}} d\theta$
$= \int \frac{-sin \theta}{\sqrt{sin^{2}\theta}} d\theta$
$= \int \frac{-sin \theta}{{sin\theta}} d\theta$
$= \int -1\ d\theta$
$= -\theta$
$x = cos \theta$
$\theta = -arccos x$
2. Dec 12, 2013
### sandyp
Its simple...we know that d/dx arc sinx = (1-x^2)^-1/2
d/dx arc cos x = -(1-x^2)^-1/2 (notice the minus sign)
as integral is antiderivative of a function you are getting the same values
yes integral arc sinx=(minus)integral arc cos x..Hope your doubt is clarified
3. Dec 12, 2013
### ShayanJ
In fact $\int \frac{1}{\sqrt{1-x^2}}dx=\sin^{-1}{x}+C_1 \ and \ \int \frac{1}{\sqrt{1-x^2}}dx=-\cos^{-1}{x}+C_2$
If you take a look at the plots of the functions $\sin^{-1}{x}$ and $\cos^{-1}{x}$,you can see that you can get one of them by multiplying the other by a minus sign and then adding a constant(or vice versa!)...and you can see that you have the negative sign and also the constant in the results of the integral!
4. Dec 12, 2013
### Appleton
Thanks for clearing that up for me.
5. Dec 12, 2013
### Staff: Mentor
No one has mentioned this. There's a trig identity that can shed some light here. | {
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### Staff: Mentor
No one has mentioned this. There's a trig identity that can shed some light here.
$sin^{-1}(x) + cos^{-1}(x) = \pi/2$
The two functions differ by a constant. If an indefinite integral produces two different antiderivatives, those functions can differ by at most a constant. | {
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Integral - substitution method problem
Yankel
Active member
Hello all
I am working on this integral
$\int \frac{x^{2}+1}{x^{4}+1}dx$
Now, I have tried this way:
$u=x^{2}+1$
after I did:
$\int \frac{x^{2}+1}{\left ( x^{2}+1 \right )\left ( x^{2}-1 \right )}dx$
But I got stuck, I got:
$\frac{1}{2}\cdot \int \frac{1}{u\sqrt{u-1}}dx$
I thought of making another substitution, but I tried and failed. Help required
Prove It
Well-known member
MHB Math Helper
Hello all
I am working on this integral
$\int \frac{x^{2}+1}{x^{4}+1}dx$
Now, I have tried this way:
$u=x^{2}+1$
after I did:
$\int \frac{x^{2}+1}{\left ( x^{2}+1 \right )\left ( x^{2}-1 \right )}dx$
But I got stuck, I got:
$\frac{1}{2}\cdot \int \frac{1}{u\sqrt{u-1}}dx$
I thought of making another substitution, but I tried and failed. Help required
Well first of all, \displaystyle \displaystyle \begin{align*} x^4 + 1 \end{align*} is NOT equal to \displaystyle \displaystyle \begin{align*} \left( x^2 + 1 \right) \left( x^2 - 1 \right) \end{align*}.
I would try
\displaystyle \displaystyle \begin{align*} x^4 + 1 &= x^4 + 2x^2 + 1 - 2x^2 \\ &= \left( x^2 + 1 \right) ^2 - \left( \sqrt{2} \, x \right) ^2 \\ &= \left( x^2 - \sqrt{2}\, x + 1 \right) \left( x^2 + \sqrt{2}\,x + 1 \right) \end{align*}
and so | {
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and so
\displaystyle \displaystyle \begin{align*} \int{ \frac{x^2 + 1}{x^4 + 1}\,dx} &= \int{ \frac{x^2 - \sqrt{2}\,x + 1 + \sqrt{2}\,x}{ \left( x^2 - \sqrt{2}\,x + 1 \right) \left( x^2 + \sqrt{2}\,x + 1 \right) } \, dx} \\ &= \int{ \frac{1}{x^2 + \sqrt{2}\,x + 1} \,dx} + \sqrt{2} \int{ \frac{x}{ \left( x^2 - \sqrt{2}\,x + 1 \right) \left( x^2 + \sqrt{2}\,x + 1 \right) } \, dx} \\ &= \int{ \frac{1}{x^2 + \sqrt{2}\,x + \left( \frac{\sqrt{2}}{2} \right) ^2 - \left( \frac{\sqrt{2}}{2} \right) ^2 + 1 } \, dx} + \sqrt{2} \int{ \frac{x}{x^4 + 1} \, dx} \\ &= \int{ \frac{1}{ \left( x + \frac{\sqrt{2}}{2} \right) ^2 + \frac{1}{2} } \, dx} + \frac{\sqrt{2}}{2} \int{ \frac{2x}{ \left( x^2 \right) ^2 + 1 }\, dx} \end{align*}
Now in the first integral, let \displaystyle \displaystyle \begin{align*} x + \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}\tan{(\theta)} \implies dx = \frac{\sqrt{2}}{2}\sec^2{(\theta)}\,d\theta \end{align*} and in the second integral let \displaystyle \displaystyle \begin{align*} x^2 = \tan{(\phi)} \implies 2x\,dx = \sec^2{(\phi)}\,d\phi \end{align*} and the integrals become | {
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\displaystyle \displaystyle \begin{align*} \int{ \frac{1}{\left( x + \frac{\sqrt{2}}{2} \right) ^2 + \frac{1}{2} } \, dx} + \frac{\sqrt{2}}{2} \int{ \frac{2x}{ \left( x^2 \right) ^2 + 1 } \, dx} &= \int{ \frac{1}{ \left[ \frac{\sqrt{2}}{2} \tan{(\theta)} \right] ^2 + \frac{1}{2} } \cdot \frac{\sqrt{2}}{2}\sec^2{(\theta)}\,d\theta } + \frac{\sqrt{2}}{2} \int{ \frac{\sec^2{(\phi)}}{ \tan^2{(\phi)} + 1 } \, d\phi } \\ &= \frac{\sqrt{2}}{2} \int{ \frac{\sec^2{(\theta)}}{ \frac{1}{2}\tan^2{(\theta)} + \frac{1}{2} } \, d\theta} + \frac{\sqrt{2}}{2} \int{ \frac{\sec^2{(\phi)}}{\sec^2{(\phi)}}\,d\phi} \\ &= \sqrt{2} \int{ \frac{\sec^2{(\theta)}}{ \tan^2{(\theta)} + 1 } \, d\theta} + \frac{\sqrt{2}}{2} \int{ 1 \, d\phi } \\ &= \sqrt{2} \int{ \frac{\sec^2{(\theta)}}{\sec^2{(\theta)}}\,d\theta} + \frac{\sqrt{2}}{2} \phi + C_1 \\ &= \sqrt{2} \int{ 1\,d\theta } + \frac{\sqrt{2}}{2} \arctan{ \left( x^2 \right) } + C_1 \\ &= \sqrt{2} \, \theta + C_2 + \frac{\sqrt{2}}{2}\arctan{ \left( x^2 \right) } + C_1 \\ &= \sqrt{2} \arctan{ \left( \sqrt{2}\, x + 1 \right) } + \frac{\sqrt{2}}{2} \arctan{ \left( x^2 \right) } + C \textrm{ where } C = C_1 + C_2 \end{align*}
Yankel
Active member
sorry, I thought it was minus when I tried, of course x^4+1 can't go the way I wrote it.
I have a basic question, how did you got to trigonometry ? did you move to polar system or something ? I don't get it. I mean, how did you know you need trigonometry in the first place ?
Thanks !
paulmdrdo
Active member
I would try this method but this is open for any correction
$\displaystyle \int\frac{x^2+1}{x^4+1}dx$
by letting
$\displaystyle u\,=\,x^2$
$\displaystyle du\,=\,2xdx$
$\displaystyle dx\,=\,\frac{du}{2x}$
$\displaystyle x\,=\,u^{\frac{1}{2}}$
can you continue..?
ZaidAlyafey
Well-known member
MHB Math Helper
For practice try
$$\displaystyle \int \frac{1}{1+x^4}\,dx$$
$$\displaystyle \int \frac{1}{\sqrt[4]{1+x^4}}\, dx$$
$$\displaystyle \int \frac{x^2}{1+x^4}\, dx$$ | {
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$$\displaystyle \int \frac{1}{\sqrt[4]{1+x^4}}\, dx$$
$$\displaystyle \int \frac{x^2}{1+x^4}\, dx$$
They are a little bit tougher .
[*] By the way I have a way to solve the integral using complex numbers , the answer is nasty .
MarkFL
Staff member
Using partial fractions, we may write:
$$\displaystyle \int\frac{x^2+1}{x^4+1}\,dx=\frac{1}{2}\int\frac{1}{x^2+\sqrt{2}x+1}+\frac{1}{x^2-\sqrt{2}x+1}\,dx$$
Completing the squares...
$$\displaystyle \int\frac{x^2+1}{x^4+1}\,dx=\frac{1}{2}\int\frac{1}{\left(x+\frac{1}{\sqrt{2}} \right)^2+\frac{1}{2}}+\frac{1}{\left(x-\frac{1}{\sqrt{2}} \right)^2+\frac{1}{2}}\,dx$$
Now, we may simply apply the formula:
$$\displaystyle \int\frac{du}{u^2+a^2}=\frac{1}{a}\tan^{-1}\left(\frac{u}{a} \right)+C$$
to get (after simplification):
$$\displaystyle \int\frac{x^2+1}{x^4+1}\,dx=\frac{1}{\sqrt{2}} \left(\tan^{-1}\left(\sqrt{2}x+1 \right)+ \tan^{-1}\left(\sqrt{2}x-1 \right) \right)+C$$
Prove It
Well-known member
MHB Math Helper
sorry, I thought it was minus when I tried, of course x^4+1 can't go the way I wrote it.
I have a basic question, how did you got to trigonometry ? did you move to polar system or something ? I don't get it. I mean, how did you know you need trigonometry in the first place ?
Thanks !
It's a method known as trigonometric substitution. I suggest you read what Mark has written in post #8 of this thread. | {
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# Convex set with empty interior is nowhere dense?
Suppose $C\subseteq\mathbb R^n$ is a convex set and $C^o=\varnothing$. Is it necessarily true that $(\overline C)^o=\varnothing$? In general, is this true if $\mathbb R^n$ is replaced by a topological vector space $X$? Or can a counterexample be found?
I know that if $C^o\neq\varnothing$, then $C^o=(\overline C)^o$, so my question is whether this result can be generalized to the case when $C^o=\varnothing$.
Update: It's not true in general topological vector spaces. Indeed, if $X$ is a topological vector space and $Y$ is a proper dense subspace (such examples can be constructed), then $Y$ is convex and has empty interior, but $(\overline Y)^o=X^o=X$ is clearly not empty. Yet, I'm still not sure if the result holds for finite-dimensional vector spaces (in which every proper subspace is closed and thus no proper subspace is dense, so the preceding counterexample doesn't work).
• Seems obvious that a convex subset of $\mathbb R^n$ with empty interior is contained in a hyperplane, which is a nowhere dense closed set. What am I missing? – bof Dec 30 '14 at 20:54
This is true in $\mathbb{R}^n$. Say $n = 3$. If $\overline{C}$ has non empty interior that $C$ is dense in some ball hence it contains the vertices of a tetrahedron (just take a tetrahedron inside the ball and move its vertices slightly to land on a point in $C$). But now by convexity, $C$ must contain the whole solid tetrahedron which contradicts the fact that $C$ has empty interior. In higher dimensions, you can start with a generalized cube and move its vertices slightly to enter $C$. | {
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• I see that the intuitive argument works, but I still have trouble operationalizing it rigorously. I will keep trying. Thank you for the answer! – triple_sec Dec 31 '14 at 16:03
• Maybe I'm missing something, but it seems to me that you don't need to resort to "move its vertices slightly". If $C$ is dense in some ball, then $C$ must contain $4$ non-coplanar points (because no subset of a plane can be dense in a ball). Use these points as the vertices of a tetrahedron. – Dave L. Renfro Jan 8 '15 at 17:07
• That works too. – user203787 Jan 8 '15 at 18:00
Here is a rigorous—and, accordingly, a bit lengthy—operationalization of the intuition offered by the answer by @OohAah.
$$\textbf{Proposition}\phantom{---}$$ Let $$C\subseteq \mathbb R^n$$ ($$n\in\mathbb N$$) be a convex set. If $$C$$ has no interior, then nor does $$\overline C$$.
$$\textit{Proof}\phantom{---}$$ Suppose that $$(\overline C)^o\neq\varnothing$$. I will show that $$C^o$$ is not empty, either. Let $$\mathbf x_0\in (\overline C)^o$$. Then, there exists some $$\varepsilon>0$$ such that $$\mathbf x_0\in B(2\varepsilon,\mathbf x_0)\subseteq\overline C$$, where $$B(2\varepsilon,\mathbf x_0)$$ denotes the open ball of radius $$2\varepsilon$$ around $$\mathbf x_0$$ with respect to the Euclidean norm. Let $$\{\mathbf e_i\}_{i=1}^n$$ denote the standard basis of $$\mathbb R^n$$ and define $$\mathbf x_i\equiv \mathbf x_0+\varepsilon \mathbf e_i$$ for each $$i\in\{1,\ldots,n\}$$. Clearly, $$\{\mathbf x_i\}_{i=0}^n\subseteq B(2\varepsilon,\mathbf x_0)\subseteq \overline C.$$ That is, $$\{\mathbf x_i\}_{i=0}^n$$ is included in the closure of $$C$$, so that, for each $$i\in\{0,1,\ldots,n\}$$, there exists some $$\mathbf y_i$$ such that
• $$\mathbf y_i\in C$$; and
• $$\mathbf y_i\in B((2\sqrt n)^{-1}\varepsilon,\mathbf x_i)$$. | {
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• $$\mathbf y_i\in C$$; and
• $$\mathbf y_i\in B((2\sqrt n)^{-1}\varepsilon,\mathbf x_i)$$.
For each $$i\in\{1,\ldots,n\}$$ define $$\mathbf b_i\equiv \mathbf y_i-\mathbf y_0$$. I claim that the vectors $$\{\mathbf b_i\}_{i=1}^n$$ are linearly independent. Indeed, suppose that $$\lambda_1,\ldots,\lambda_n\in\mathbb R$$ satisfy $$\sum_{i=1}^n\lambda_i\mathbf b_i=0.$$ Suppose, for the sake of contradiction, that not all of $$\{\lambda_i\}_{i=1}^n$$ are zero. Then, $$\sum_{i=1}^n|\lambda_i|>0\tag{1}.$$ In addition, \begin{align*} 0=&\,\sum_{i=1}^n\lambda_i\mathbf b_i=\sum_{i=1}^n\lambda_i(\mathbf y_i-\mathbf y_0)=\sum_{i=1}^n\lambda_i(\mathbf y_i-\mathbf x_i+\mathbf x_i-\mathbf x_0+\mathbf x_0-\mathbf y_0)\\=&\,\sum_{i=1}^n\lambda_i(\mathbf y_i-\mathbf x_i+\varepsilon\mathbf e_i+\mathbf x_0-\mathbf y_0), \end{align*} or \begin{align*} -\varepsilon\sum_{i=1}^n\lambda_i\mathbf e_i=\sum_{i=1}^n\lambda_i(\mathbf y_i-\mathbf x_i+\mathbf x_0-\mathbf y_0). \end{align*} Clearly, the Euclidean norm of the left-hand side is $$\varepsilon\sqrt{\sum_{i=1}^n\lambda_i^2}$$, so the following chain of inequalities holds true: \begin{align*} &\varepsilon\sqrt{\sum_{i=1}^n\lambda_i^2}=\left\|\sum_{i=1}^n\lambda_i(\mathbf y_i-\mathbf x_i+\mathbf x_0-\mathbf y_0)\right\|\leq\sum_{i=1}^n|\lambda_i|\left(\|\mathbf y_i-\mathbf x_i\|+\|\mathbf x_0-\mathbf y_0\|\right)\\ \underset{\text{see (1)}}{<}&\,\sum_{i=1}^n|\lambda_i|\left(\frac{\varepsilon}{2\sqrt{n}}+\frac{\varepsilon}{2\sqrt{n}}\right)=\sum_{i=1}^n|\lambda_i|\frac{\varepsilon}{\sqrt{n}}\leq\sqrt{\sum_{i=1}^n|\lambda_i|^2}\sqrt{\sum_{i=1}^n\frac{\varepsilon^2}{n}}=\varepsilon\sqrt{\sum_{i=1}^n\lambda_i^2}, \end{align*} where I used the Cauchy–Schwarz inequality. This is a contradiction, which implies that $$\lambda_1=\ldots=\lambda_n=0$$. Hence, the vectors $$\{\mathbf b_i\}_{i=1}^n$$ are linearly independent, and since $$\dim\mathbb R^n=n$$, it follows also that they constitute a basis. | {
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If $$\mathbf z\in\mathbb R^n$$, there exists a unique $$(\mu_i)_{i=1}^n\in\mathbb R^n$$ such that $$\mathbf z=\sum_{i=1}^n\mu_i\mathbf b_i$$. Define $$\|\mathbf z\|_b\equiv\sum_{i=1}^n|\mu_i|$$. Then, $$\|\cdot\|_b$$ is a norm and since $$\mathbb R^n$$ is finite-dimensional, it must be equivalent to the Euclidean norm. Therefore, there exists some $$\xi_b>0$$ such that $$\|\cdot\|_b\leq \xi_b\|\cdot\|$$.
Let $$D\equiv\left\{\sum_{i=0}^n\alpha_i\mathbf y_i\,\Bigg|\,\alpha_i\geq0\text{ for all i\in\{0,1,\ldots,n\} and }\sum_{i=0}^n\alpha_i=1\right\}.$$ Since $$\{\mathbf y_i\}_{i=0}^n\subseteq C$$ and $$C$$ is convex, it follows that $$D\subseteq C$$. Define $$\mathbf w\equiv\sum_{i=0}^n\frac{1}{n+1}\mathbf y_i.$$ Clearly, $$\mathbf w\in D$$ and $$\mathbf w-\mathbf y_0=\sum_{i=1}^{n}\frac{1}{n+1}(\mathbf y_i-\mathbf y_0)=\sum_{i=1}^{n}\frac{1}{n+1}\mathbf b_i.$$ | {
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Let $$\delta\equiv\frac{1}{n(n+1)\xi_b}>0.$$ I will show that $$B(\delta,\mathbf w)\subseteq D$$. To this end, pick any $$\mathbf z\in B(\delta,\mathbf w)$$, so that $$\|\mathbf z-\mathbf w\|<\delta.$$ Since $$\{\mathbf b_i\}_{i=1}^n$$ is a basis, there exists a unique $$(\mu_i)_{i=1}^n\in\mathbb R^n$$ such that $$\mathbf z-\mathbf y_0=\sum_{i=1}^n\mu_i\mathbf b_i$$. Then, $$\mathbf z-\mathbf w=(\mathbf z-\mathbf y_0)-(\mathbf w-\mathbf y_0)=\sum_{i=1}^n\left(\mu_i-\frac{1}{n+1}\right)\mathbf b_i.$$ Consequently, for any $$i\in\{1,\ldots,n\}$$, it follows that \begin{align*} \left|\mu_i-\frac{1}{n+1}\right|\leq\sum_{j=1}^n\left|\mu_j-\frac{1}{n+1}\right|=\|\mathbf z-\mathbf w\|_b\leq \xi_b\|\mathbf z-\mathbf w\|<\xi_b\delta=\frac{1}{n(n+1)}. \end{align*} Hence, $$0\leq\frac{(n-1)}{n(n+1)}=\frac{1}{n+1}-\frac{1}{n(n+1)}<\mu_i<\frac{1}{n+1}+\frac{1}{n(n+1)}=\frac{1}{n},$$ for each $$i\in\{1,\ldots,n\}$$ so that $$\sum_{i=1}^n\mu_i<1$$. It follows that \begin{align*} \mathbf z=\mathbf y_0+\sum_{i=1}^n\mu_i\mathbf b_i=\mathbf y_0+\sum_{i=1}^n\mu_i(\mathbf y_i-\mathbf y_0)=\left(1-\sum_{i=1}^n\mu_i\right)\mathbf y_0+\sum_{i=1}^n\mu_i\mathbf y_i, \end{align*} so $$\mathbf z\in D$$.
Conclusion:
• $$\mathbf w\in B(\delta,\mathbf w)\subseteq D\subseteq C$$; and
• $$B(\delta,\mathbf w)$$ is a non-empty open set; so that
• $$\mathbf w\in C^o$$.
In particular, $$C^o$$ is not empty. $$\blacksquare$$
Intuitively, the points $$\{\mathbf x_i\}_{i=0}^n$$ in $$\overline C$$ span an $$n$$-dimensional simplex, whose vertices are perturbed slightly so that the new vertices do not lie in the same hyperplane and are all in $$C$$. The convex hull of these new vertices gives rise to a “distorted simplex” fully contained in $$C$$. Then, the centroid of this distorted simplex can be surrounded by a small ball still included in the distorted simplex, and hence in $$C$$. | {
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Suppose that ##A## is a diagonalizable ## n \times n ## matrix. Then it is similar to a diagonal matrix ##B##. My question is, is ##B## the only diagonal matrix to which ##A## is similar?
I have thought about this, but am unsure if my answer is correct. My claim is that ##B## is the only diagonal matrix to which ##A## is similar, up to a rearrangement (or permutation) of the eigenvectors of ##A## in an ordered eigenbasis for ##ℝ^{n}##. Thus, if you rearrange the vectors in your eigenbasis, you will obtain a different diagonal matrix for ##[L_{A}]_{β}## but this will be merely a rearrangement of the diagonal entries of B.
Is this true? If so, how might I go about proving this conjecture of mine?
BiP
Related Linear and Abstract Algebra News on Phys.org
Yes, it's true.
A hint for the proof: the diagonal entries are the eigenvalues of ##A##.
CompuChip
Homework Helper
If ##A = X^{-1}BX = Y^{-1}B'Y## then ##B = (Y X^{-1})^{-1} B' (Y X^{-1})##.
So if A is similar to two diagonal matrices, then these matrices are similar to each other.
I guess a starting point would be to check which similarity transformations keep diagonal matrices diagonal, and see what you can say about X and Y that way.
HallsofIvy
Homework Helper
Yes, it's true.
A hint for the proof: the diagonal entries are the eigenvalues of ##A##.
Not quite. A matrix can be similar to two different diagonal matrices if they have the same numbers on the diagonal in different orders.
Not quite. A matrix can be similar to two different diagonal matrices if they have the same numbers on the diagonal in different orders.
That's exactly what the OP said:
Thus, if you rearrange the vectors in your eigenbasis, you will obtain a different diagonal matrix for ##[L_A]_\beta## but this will be merely a rearrangement of the diagonal entries of B.
mathwonk | {
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# Fourier Transform Of Triangular Pulse | {
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To do so, you just have to divide the pulse by its norm, i. Signal Transmission Through LTIC Systems. ∆(t), from Problem 3. Fourier Series approach and do another type of spectral decomposition of a signal called a Fourier Transform. The Fourier transform of a gate pulse is: a. For any parameter, ≠. otherwise. I am attaching a Vi in which first i convoluted a Square pulse with triangular pulse. Therefore, the departure of the roll-off from that of the sinc function can be ascribed to aliasing in the frequency domain , due to sampling in the time domain. Re: Help me get a Fourier transform of sinc It show the Fourier Transform of sinc(x) function --> a rectangular pulse. See under Discrete Fourier transform data compression use of, 494-495 decomposition. Fourier transforms of deltas and sinusoids Fourier transform of periodic signals Preface Not Just One 1 2 3 0. 1: The Fourier transform of a triangular pulse. Then the Fourier series expansion of the output function y(t) literally gives the spectrum of the output! B. periodic relationship, 222 discrete time Fourier series (DTFT),. 8 Boxcar Pulse Bc(t) Bc(t) = u(t a)u (t b) Fourier Transform of Boxcar pulse can be treated as an application of Fourier Transform property u(t) !F A˝sinc! ˝ 2 Thus u(t j!˝˝) !F e A˝sinc! ˝ 2 Therefore u(t a)u (t b) !F ej!a A˝sinc! ˝ 2. 2 The Fourier Transform for Periodic Signals 4. In Fourier Transform Nuclear Magnetic Resonance spectroscopy (FTNMR), excitation of the sample by an intense, short pulse of radio frequency energy produces a free induction decay signal that is the Fourier transform of the resonance spectrum. 4 Find The Fourier Transform Of The Triangular Pulse (Fig. 5 Signals & Linear Systems Lecture 10 Slide 12 Fourier Transform of a unit impulse train XConsider an impulse train. Plot sine wave. The LCFBG in the system performs three functions: temporally stretching the input ultrashort pulse, shaping the pulse spectrum, and temporally compressing the spectrum-shaped pulse. | {
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ultrashort pulse, shaping the pulse spectrum, and temporally compressing the spectrum-shaped pulse. Amplitude-modulated optical pulse trains undergo the discrete Fourier transform (DFT) realized by the temporal Talbot effect in a dispersive fiber. Fourier series: Given a 2ˇ-periodic complex-valued function f (think of it as a function on an interval T of length 2ˇ), its Fourier (series) transform is the sequence of its Fourier coecients: (f) = fb= fck: k 2 Zg, which can be thought of as a complex-valued function of the discrete frequency variable k. The series produced is then called a half range Fourier series. A square wave is approximated by the sum of harmonics. The S-transform can be effectively utilized for the analysis of nonstationary data. Output kernel Figure 5. The Gaussian shape is often called a bell shape. I'm at a computer without MATLAB at the moment. d R pp T P, where R pp is the auto-correlation of p(t) d. 1 FOURIER SERIES FOR PERIODIC FUNCTIONS This section explains three Fourier series: sines, cosines, and exponentials eikx. Still, many problems that could have been tackled by using Fourier transforms may have gone unsolved because they require integration that is difficult and tedious. T T 0 elsewhen. There are similar convolution theorems for inverse Fourier transforms. 2 p693 PYKC 10-Feb-08 E2. The Fourier Transform 1. That is: since the is an odd function, its contribution is zero. Tips If a , b , and c are variables or expressions with variables, triangularPulse assumes that a <= b <= c. DSP First and it's accompanying digital assets are the result of more than 20 years of work that originated from, and was guided by, the premise that signal processing is the best starting point for the study of electrical and computer engineering. time signal. School of Health Information Sciences. Definition of Fourier Transform The forward and inverse Fourier Transform are defined for aperiodic signal as: Already covered in Year 1 Communication course | {
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Transform are defined for aperiodic signal as: Already covered in Year 1 Communication course (Lecture 5). Signal Distortion over a Communication Channel. The Fourier series of this general pulse train is:. Equation of the Day #11: The Fourier Transform. Except now we're going to build a composite wave form that is a triangle wave. (We shall use this in the assignment). In order for F(t) to be real, F (- t) = F* ( t) must hold, Example 9. Plotting a triangular signal and finding its Fourier transformation in MATLAB. A “Brief” Introduction to the Fourier Transform This document is an introduction to the Fourier transform. What is the FT of a triangle function? - To be able to do a continuous Fourier transform on a signal before and after repeated signals in the Fourier domain. time signal. This is a good point to illustrate a property of transform pairs. This pulse can be used to represent an electrostatic discharge, an electromagnetic pulse, or a lightning event. The LCFBG in the. Applying the time-convolution property to y(t)=x(t) * h(t), we get: 2(ω) e-jωτ. For1secondofdatasampledat40,000. Which are the only waves that correspond/ support the measurement of phase angle in the line spectra? a. t/, use the derivative property to find the Fourier transform of x. Let's look at the triangular pulse and its Fourier transform,. Fourier transform of a triangular pulse is sinc 2, i. Note: This is why the Fourier transform of a rectangular pulse is shaped like a sinc function, and the Fourier transform of a sinc function is shaped like a rectangular pulse 3. I am attaching a Vi in which first i convoluted a Square pulse with triangular pulse. t2 exp( 2 ) 2. Inverse Fourier Transform maps the series of frequencies (their amplitudes and phases) back into the corresponding time series. Discrete Fourier Transform (DFT) Fast Fourier Transform (FFT) The Fourier Integral (FI) is a mathematical technique of transforming an ideal mathematical expression in the time domain into a | {
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a mathematical technique of transforming an ideal mathematical expression in the time domain into a description in the frequency domain. Realisability of One Port Network – Consider the network function H(s) which is the ratio of Laplace transform of the output R(s) to the Laplace transform of the excitation E(s). The line spectrum, obtained from the Fourier series coefficients, indicates how the power of the signal is distributed to harmonic frequency components in the series. "Fourier Series--Triangle Wave. The Gaussian shape is often called a bell shape. Click the button "Real Transform" in the Fourier graph. Inverse Fourier Transform maps the series of frequencies (their amplitudes and phases) back into the corresponding time series. We then generalise that discussion to consider the Fourier transform. (a) below is X(Ω) = sin Ω 2 Ω 2!2. The Fourier series can also be written in its more convenient but somewhat less intuitive form: x(t) = X1 n=1 c ne jn2ˇf 0t (2) The transform of the Fourier series can be found to be: X(f) = X1 n=1 c n (f nf 0) (3) X(f) is a summation of impulses. A group of algorithms is presented generalizing the fast Fourier transform to the case of noninteger frequencies and nonequispaced nodes on the interval $[ - \pi ,\pi ]$. School of Health Information Sciences. Consequently, the square wave has a wider bandwidth. The Fourier transform of a Gaussian signal in time domain is also Gaussian signal in the frequency domain −𝝅 ↔ −𝝅 Option (c) 10. Sometimes there is a big spike at zero so try taking the log of it before plotting. Figures (c) & (d) show that a triangular pulse in the time domain coincides with a sinc function squared (plus aliasing) in the frequency domain. Then the Fourier series expansion of the output function y(t) literally gives the spectrum of the output! B. Signals & Systems - Triangular Signal Watch more videos at https://www. 2) Here 0 is the fundamental frequency of the signal and n the index of the harmonic such. 2 The | {
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2) Here 0 is the fundamental frequency of the signal and n the index of the harmonic such. 2 The Fourier Transform Having now presented the idea of the frequency response H(s)js=j! = H(j!) of linear systems, it is appropriate to recognise that in the special case of this Laplace transform value H(s)being evaluated at the purely imaginary value s = j!, it is known as the ‘Fourier transform’. x(t)= γ (t)-γ (t-T 0) a) by using a table of Fourier Transforms and Properties (this is just a rectangular pulse function of width T 0 that is not centered on the origin). , sin(x)/x] in the frequency domain. Consequently, the square wave has a wider bandwidth. Improving images by “ deconvolution ” of. In this chapter much of the emphasis is on Fourier Series because an understanding of the Fourier Series decomposition of a signal is important if you wish to go on and study other spectral techniques. FFT onlyneeds Nlog 2 (N). Conceptually, this occurs because the triangle wave looks much more like the 1st harmonic, so the contributions of the higher harmonics are less. The FT of a rectangle function is a sinc. tutorialspoint. The rectangular function is an idealized low-pass filter, and the sinc function is the non-causal impulse response of such a filter. The Fourier Transform of the Box Function. I am attaching a Vi in which first i convoluted a Square pulse with triangular pulse. Triangle Pulse 0 0. periodic relationship, 222 discrete time Fourier series (DTFT),. Exams are approaching, and I'm working through some old assignments. The transform of a triangular pulse is a sinc2 function. Try taking the real part of it with real(). 5 Signals & Linear Systems Lecture 10 Slide 12 Fourier Transform of a unit impulse train XConsider an impulse train. If a function is defined over half the range, say 0 to L, instead of the full range from -L to L, it may be expanded in a series of sine terms only or of cosine terms only. Tips If a , b , and c are variables or expressions with | {
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of sine terms only or of cosine terms only. Tips If a , b , and c are variables or expressions with variables, triangularPulse assumes that a <= b <= c. In this particular SPICE simulation, I’ve summed the 1st, 3rd, 5th, 7th, and 9th harmonic voltage sources in series for a total of five AC voltage sources. Relation of Z transform to Laplace transform. 2005-03-15. We say that f(t) lives in the time domain, and F(ω) lives in the frequency domain. We begin by discussing Fourier series. On the other hand, as Fourier transform can be considered as a special case of Laplace transform when the real part of the complex argument is zero:. Baskakov, O I; Civis, S; Kawaguchi, K. After simplification the sinc squared function is obtained as the Fourier transform of a triangular pulse with unit area. 30) yields 2 2 ( ) ˆ p p T T. FFTs of Functions. '' Figure 8 shows an example for. Discrete Fourier Transform of Windowing Functions. )2 Solutions to Optional Problems S9. More Advanced Topics Up: Fourier Series-What, How, and Why Previous: The Fast Fourier Transform Using the Fourier Transform. Fourier Transform of Useful Functions. Since f(t) is even then g(w) is real. For any parameter, ≠. It is reversible, being able to transform from either domain to the other. You can buy my book 'ECE for GATE' here https://goo. We then generalise that discussion to consider the Fourier transform. SAMPLING RATE CRITERIA A rule-of-thumb states that the sampling rate for the input time history should be at least ten times greater than the highest shock response spectrum calculation frequency. The Fourier transform of the triangular pulse shown in Fig. The curve should be symmetrical with respect to the origin in 1024 points. Ultra-high-resolution Fourier transform ion cyclotron resonance mass spectrometry (FT-ICR MS) analysis enables the identification of thousands of masses in a single measurement. Recall that normalized Fourier transform of triangular pulse is $sinc^{2}(f)$$. The transform of | {
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Recall that normalized Fourier transform of triangular pulse is $sinc^{2}(f)$$. The transform of a triangular pulse is a sinc 2 function. Compare the result with pan (b). Find the Fourier Transform of a rectangular pulse that is zero everywhere except between t=0 and t=T 0 where it has a height of one:. Fast Fourier Transform(FFT) • The Fast Fourier Transform does not refer to a new or different type of Fourier transform. However, at 45 degrees, the radon projection clearly is NOT a rectangular pulse (but rather a triangle) and its FFT is clearly NOT a sinc (note that the oscillations never run negative). These impulses may only occur at integer multiples (harmonics) of the fundamental frequency f 0. Fourier Transform of the Gaussian Konstantinos G. The reason why Fourier analysis is so important in physics is that many (although certainly. Homework Statement What is the Fourier transform of the function graphed below? According to some textbooks the Fourier transform for this function Fourier Transform (Triangular Pulse) | Physics Forums. Cvetkovic, IntechOpen, DOI: 10. Basic Triangle Preparation for DFT(Discrete Fourier Transform) (Please send message to MathFreeOn Facebook page manager. In this tutorial numerical methods are used for finding the Fourier transform of continuous time signals with MATLAB are presented. The fast Fourier transform (FFT) is a fast algorithm for calculating the Discrete Fourier Transform (DFT). A major challenge in the data analysis process of NOM using the FT-ICR MS technique is the need to sort the entire data set and to present it in an accessible mode. Often we are confronted with the need to generate simple, standard signals ( sine, cosine , Gaussian pulse , squarewave , isolated rectangular pulse , exponential decay, chirp signal ) for. 1 The Fourier transform and series of basic signals Triangular pulse 1. The graph at the bottom of the screen is the Fourier transform of this pulse. An extension of the time-frequency relationship | {
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the screen is the Fourier transform of this pulse. An extension of the time-frequency relationship to a non-periodic signal s(t) requires the introduction of the Fourier Integral. (a) below is X(Ω) = sin Ω 2 Ω 2!2. Definition of Fourier Transform The forward and inverse Fourier Transform are defined for aperiodic signal as: Already covered in Year 1 Communication course (Lecture 5). The series does not seem very useful, but we are saved by the fact that it converges rather rapidly. Fourier Transform Fourier Transform maps a time series (eg audio samples) into the series of frequencies (their amplitudes and phases) that composed the time series. This is the principle on which a pulse Fourier transform spectrometer operates. Discrete-Time Fourier Transform (DTFT) Chapter Intended Learning Outcomes: (i) Understanding the characteristics and properties of DTFT (ii) Ability to perform discrete-time signal conversion between the time and frequency domains using DTFT and inverse DTFT. The fast Fourier transform (FFT) is a computationally efficient method of generating a Fourier transform. Figures (c) & (d) show that a triangular pulse in the time domain coincides with a sinc function squared (plus aliasing) in the frequency domain. Sometimes there is a big spike at zero so try taking the log of it before plotting. In the synthesis, we are going to obtain a network from the given network function which may be admittance function or impedance function. Fourier Transforms For additional information, see the classic book The Fourier Transform and its Applications by Ronald N. Sometimes fft gives a complex result. rectangular pulse triangular pulse periodic time function unit impulse train (model of regular sampling). The example in this figure pertains to an output sampling rate which is times that of the input signal. Recall that normalized Fourier transform of triangular pulse is [math]sinc^{2}(f)$[math]. The Fourier transform is an integral transform widely used in physics | {
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is [math]sinc^{2}(f)$[math]. The Fourier transform is an integral transform widely used in physics and engineering. In this article, we will discuss the fact that choice of different window functions involves a trade-off between the main lobe width and the peak sidelobe (PSL). Improving images by “ deconvolution ” of. How to get a triangular pulse? Discrete Fourier Transform using scipy. 4 Reconstruction of a triangular pulse. What do we hope to achieve with the Fourier Transform? We desire a measure of the frequencies present in a wave. Can you explain this answer? is done on EduRev Study Group by Electrical Engineering (EE) Students. In some cases, as in this one, the property simplifies things. Triangle Pulse Sinc Pulse. What is the FT of a triangle function? – To be able to do a continuous Fourier transform on a signal before and after repeated signals in the Fourier domain. DAS RISIN and MAIZAN MUHAMAD. Fourier Series approach and do another type of spectral decomposition of a signal called a Fourier Transform. A Tables of Fourier Series and Transform Properties 321 Table B. periodic relationship, 222 discrete time Fourier series (DTFT),. The Fourier transform of this convolution is the product w 1 w 2 of the two transforms, each one a series of parallel walls, and differs from zero only when both factors are different from zero. We demonstrate both amplitude and phase tailoring by generating a picosecond squarelike pulse as well as trains of femtosecond pulses with a terahertz-range repetition rate from either a. The LCFBG in the. Digital signal processing (DSP) vs. For the bottom panel, we expanded the period to T=5, keeping the pulse's duration fixed at 0. Relation of Z transform to Laplace transform. Ideal and Practical Filters. Sibbett A Wollaston prism is used in the design of a polarizing Fourier. The impulse response for the triangle in Fig. The graph at the bottom of the screen is the Fourier transform of this pulse. † The Fourier series is then f(t) = | {
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the bottom of the screen is the Fourier transform of this pulse. † The Fourier series is then f(t) = A 2 ¡ 4A …2 X1 n=1 1 (2n¡1)2 cos 2(2n¡1)…t T: Note that the upper limit of the series is 1. Signals & Systems - Triangular Signal Watch more videos at https://www. Exams are approaching, and I'm working through some old assignments. Signal Distortion over a Communication Channel. Fourier transform theorem table 4 1 jpg bax blog fourier transform table rh baxtyfraze blo com Pdf Fourier Transforms And Their Application To Pulse Amplitude. I One-Dimensional Fourier Transform. The Fourier transform of a Gaussian signal in time domain is also Gaussian signal in the frequency domain −𝝅 ↔ −𝝅 Option (c) 10. The amplitudes of the harmonics for this example drop off much more rapidly (in this case they go as 1/n 2 (which is faster than the 1/n decay seen in the pulse function Fourier Series (above)). Slow Fourier Transforms Consider a general 1D Fourier transform relating two vectors of length : contains the values in real-space contains the frequency components The Slow Fourier Transform : Suppose that we precompute and store the coefficients c(j,k) = exp(i 2pi jk/n). I am attaching a Vi in which first i convoluted a Square pulse with triangular pulse. I'm at a computer without MATLAB at the moment. This Demonstration illustrates the relationship between a rectangular pulse signal and its Fourier transform. Signal Energy and Energy Spectral Density. Minimalist Tall Rectangular Planter On 4 Or 6 Zebra Onion Grass. F(ω) is just another way of looking at a function or wave. A LABORATORY DEMONSTRATION OF HIGH-RESOLUTION HARD X-RAY AND GAMMA-RAY IMAGING USING FOURIER-TRANSFORM TECHNIQUES David Palmer and Thomas A. A two parts tutorial on Fourier series. For voltage signals, the power per unit frequency is proportional to |H(f)|2 and is called the power spectrum or spectral power density of h(t). I One-Dimensional Fourier Transform. Sampling theorem. 1 Fourier Series Representation | {
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of h(t). I One-Dimensional Fourier Transform. Sampling theorem. 1 Fourier Series Representation of Periodic Signals 3. 900 950 1000 1050 1100 1150 0. It is reversible, being able to transform from either domain to the other. Then using the relation: one can show that. PDF | Fourier transform ultrashort optical pulse shaping using a single linearly chirped fiber Bragg grating (LCFBG) is proposed and experimentally demonstrated in this letter. Solution: g(t) is a triangular pulse of height A, width W, and is centered at t 0. Therefore the FT of a triangle function is the product of two identical sincs, or a sinc^2. 4*Pi); # The base function is f0 = f restricted to [0,T] f0:=x->x; # Then f(x)=f0(trw(x,T))=trw(x,T) # INTEGRATION STEPS FOR FOURIER COEFFICIENTS # Possible also by hand using an integral table. The fast Fourier transform (FFT) is a computationally efficient method of generating a Fourier transform. Note that as long as the definition of the pulse function is only motivated by the time-domain experience of it, there is no reason to believe that the oscillatory interpretation (i. Cosine waves c. Plotting a triangular signal and finding its Fourier transformation in MATLAB. The Fourier Transform In this appendix, the most important facts pertaining to the Fourier trans form are surveyed without proof. NEW! Updated labs, visual demos, an update to the existing chapters, and hundreds of new homework problems and solutions. Basic Triangle Preparation for DFT(Discrete Fourier Transform) (Please send message to MathFreeOn Facebook page manager. Properties of the Fourier Transform • Linearity: • Let and • then • Time Scaling: • Let • then Compression in the time domain results in expansion in the frequency domain Internet channel A can transmit 100k pulse/sec and channel B can transmit 200k pulse/sec. Signals & Systems - Triangular Signal Watch more videos at https://www. Variable position. Since the pulse is getting narrower and narrower in the limit, the | {
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https://www. Variable position. Since the pulse is getting narrower and narrower in the limit, the multiplication with this other function has less and less effect, until it has no effect once the. This is the example given above. The Fourier transform (English pronunciation: / ˈ f ʊr i eɪ /), named after Joseph Fourier, is a mathematical transformation employed to transform signals between time (or spatial) domain and frequency domain, which has many applications in physics and engineering. Signals & Systems - Reference Tables 1 Table of Fourier Transform Pairs Function, f(t) Fourier Transform, F( ) Definition of Inverse Fourier Transform. Fourier transform unitary, angular frequency Fourier transform unitary, ordinary frequency Remarks 10 The rectangular pulse and the normalized sinc function 11 Dual of rule 10. t2 exp( 2 ) 2. As is an even function, its Fourier transform is Alternatively, as the triangle function is the convolution of two square functions ( ), its Fourier transform can be more conveniently obtained according to the convolution theorem as:. Periodicity of the Fourier transform; Fourier transform as additive synthesis. In the first part an example is used to show how Fourier coefficients are calculated and in a second part you may use an applet to further explore Fourier series of the same function. The LCFBG in the system performs three functions: temporally stretching the input ultrashort pulse, shaping the pulse spectrum, and temporally compressing the. Note that as long as the definition of the pulse function is only motivated by its behavior in the time-domain experience, there is no reason to believe that the oscillatory interpretation (i. Another representation of signals that has been found very useful is frequency domain representation. We can sample a function and then take the FFT to see the function in the frequency domain. The triangle peak is at the integral of the signal or sum of the sequence squared. ), the frequency response of | {
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peak is at the integral of the signal or sum of the sequence squared. ), the frequency response of the interpolation is given by the Fourier transform, which yields a sinc function. Single-pulse, Fourier-transform spectrometer having no moving parts M. Fourier transform: F(s) = - f(x)e-'2n"dr Here, two sidebands have been introduced (in addition to the carrier frequency), and the signal of frequency s has been shifted into the region of the carrier frequency, so. Today I want to start getting "discrete" by introducing the discrete-time Fourier transform (DTFT). Now, if we're given the wave function when t=0, φ(x,0) and the velocity of each sine wave as a function of its wave number, v(k), then we can compute φ(x,t) for any t by taking the inverse Fourier transform of φ(x,0) conducting a phase shift, and then taking the Fourier transform. profile closer to Gaussian. DFT needs N2 multiplications. Of course, we must sample often enough to avoid losing content. P d P2 f , where is Fourier Transform of p(t) c. 5 The multiplication Property 4. 1, is a triangular pulse of height 1, width 2, and is centered at 0. In the first row is the graph of the unit pulse function and its Fourier transform , a function of frequency. For example, a rectangular pulse in the time domain coincides with a sinc function [i. The Fourier series can also be written in its more convenient but somewhat less intuitive form: x(t) = X1 n=1 c ne jn2ˇf 0t (2) The transform of the Fourier series can be found to be: X(f) = X1 n=1 c n (f nf 0) (3) X(f) is a summation of impulses. The actual Fourier transform are only the impulses. As in the case of ideal sampling, the spectrum contains uniformly spaced (scaled) copies of 𝑋(𝑓), with a spacing of 1⁄𝑇 Hz. † The Fourier series is then f(t) = A 2 ¡ 4A …2 X1 n=1 1 (2n¡1)2 cos 2(2n¡1)…t T: Note that the upper limit of the series is 1. 8 Boxcar Pulse Bc(t) Bc(t) = u(t a)u (t b) Fourier Transform of Boxcar pulse can be treated as an application of Fourier | {
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Bc(t) = u(t a)u (t b) Fourier Transform of Boxcar pulse can be treated as an application of Fourier Transform property u(t) !F A˝sinc! ˝ 2 Thus u(t j!˝˝) !F e A˝sinc! ˝ 2 Therefore u(t a)u (t b) !F ej!a A˝sinc! ˝ 2. Ultra-high-resolution Fourier transform ion cyclotron resonance mass spectrometry (FT-ICR MS) analysis enables the identification of thousands of masses in a single measurement. Cosine waves c. Moreover, the amplitude of cosine waves of wavenumber in this superposition is the cosine Fourier transform of the pulse shape, evaluated at wavenumber. Mapping s plane semiellipse to Z plane. The corresponding intensity is proportional to this transform squared, i. To do so, you just have to divide the pulse by its norm, i. 3 Properties of the Continuous-Time Fourier Transform 4. Prince California Institute of Technology, Pasadena, CA 91125 Abstract A laboratory imaging system has been developed to study the use of Fourier-transform techniques in high-resolution hard x-ray andγ. The dotted-line is a sinc function that doesn't apply to this question, but gives the notion that this transform has something to do with the transform of a square pulse (i. Ideal and Practical Filters. F(ω) is called the Fourier Transform of f(t). Plot sine wave. 4 The Convolution Property 4. Today I want to follow up by discussing one of the ways in which reality confounds our expectations and causes confusion. Fn = 6 shows that as T/t increases the lines get closer together and the spectrum begins to look like that of the Fourier Transform. They are widely used in signal analysis and are well-equipped to solve certain partial. Signal Distortion over a Communication Channel. profile closer to Gaussian. periodic relationship, 222 discrete time Fourier series (DTFT),. Fourier series and square wave approximation Fourier series is one of the most intriguing series I have met so far in mathematics. The actual Fourier transform are only the impulses. The primary reason for it’s success is | {
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The actual Fourier transform are only the impulses. The primary reason for it’s success is that it plots any course of constant bearing (angle w. I would guess it would be a triangle because I think the transform of a triangular pulse centered at zero is something like T sinc 2 (pi * f * T) I would restate your original question as follows: What is the transform of the following periodic rectangular function: per-rect(t) which has period T=T1+T2; is 1 during each T1 duration of its. 5 Signals & Linear Systems Lecture 10 Slide 3 Connection between Fourier Transform and Laplace. 1 The Fourier transform and series of basic signals Triangular pulse 1. 6: Roll-off of the rectangular-window Fourier transform. We describe a pulse-shaping technique that uses second-harmonic generation with Fourier synthetic quasi-phase-matching gratings. time signal. In the case of natural sampling, however, the spectral copies have different scaling factors: 𝑃(𝑘⁄𝑇)/𝑇. For above triangular wave: The square wave has much sharper transition than the triangular wave. These impulses may only occur at integer multiples (harmonics) of the fundamental frequency f 0. PSfrag replacements PROBLEM: Let x. 1 FOURIER SERIES FOR PERIODIC FUNCTIONS This section explains three Fourier series: sines, cosines, and exponentials eikx. Inverse Z transform using inversion integral. Click the button "Real Transform" in the Fourier graph. 1, is a triangular pulse of height 1, width 2, and is centered at 0. In the first part an example is used to show how Fourier coefficients are calculated and in a second part you may use an applet to further explore Fourier series of the same function. Square waves (1 or 0 or −1) are great examples, with delta functions in the derivative. An extension of the time-frequency relationship to a non-periodic signal s(t) requires the introduction of the Fourier Integral. Fourier Transforming the Triangular Pulse. Fourier Slice Theorem or central Slice Theorem. Translation (that is, | {
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the Triangular Pulse. Fourier Slice Theorem or central Slice Theorem. Translation (that is, delay) in the time domain goes over to complex phase shifts in the frequency domain. , use ^ M(n)/k^ M kinstead of ^ M(n). Then i deconvoluted the convoluted signal using triangular pulse with the deconvolution tool given in labview. 1-1) Since u(t) = 0 for t < 0, eq. Signal Transmission Through LTIC Systems. 2to create successive recon-structions of the pulse. Fourier transforms are used widely, and are of particular value in the analysis of single functions and combinations of functions found in radar and signal processing. SAMPLING RATE CRITERIA A rule-of-thumb states that the sampling rate for the input time history should be at least ten times greater than the highest shock response spectrum calculation frequency. Z transforms and difference equations. Time-harmonic impulse response calculations—The time-harmonic pressure generated by these triangular source geometries is proportional to the Fourier transform of the impulse response. Derpanis October 20, 2005 In this note we consider the Fourier transform1 of the Gaussian. A "Brief" Introduction to the Fourier Transform This document is an introduction to the Fourier transform. tutorialspoint. It is reversible, being able to transform from either domain to the other. Inverse Z transform using inversion integral. See Discrete Fourier Transform discrete vs. ∆(t), from Problem 3. The first example has a duty cycle of 0. The Fourier Transform: Examples, Properties, Common Pairs Square Pulse Spatial Domain Frequency Domain f(t) F (u ) 1 if a=2 t a=2 0 otherwise sinc (a u ) = sin (a u ) a u The Fourier Transform: Examples, Properties, Common Pairs Square Pulse The Fourier Transform: Examples, Properties, Common Pairs Triangle Spatial Domain Frequency Domain f(t. The Fourier transform (English pronunciation: / ˈ f ʊr i eɪ /), named after Joseph Fourier, is a mathematical transformation employed to transform signals between time | {
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after Joseph Fourier, is a mathematical transformation employed to transform signals between time (or spatial) domain and frequency domain, which has many applications in physics and engineering. Sometimes fft gives a complex result. Padgett, A. But what we're going to do in this case is we're going to add them. Fourier Series approach and do another type of spectral decomposition of a signal called a Fourier Transform. Applying the time-convolution property to y(t)=x(t) * h(t), we get: That is: the Fourier Transform of the system impulse response is the system Frequency Response L7. property, we can compute the Fourier transform of the phasor: 0 ( )0 F e f fj tω = −δ This allows us to compute the Fourier transform of periodic signals. Triangular waves d. This is the example given above. (Browser settings for best viewing results) (How to cite this work). We then generalise that discussion to consider the Fourier transform. Fourier Transforms and Theorems. A) shows the original pulse and the real part of its Fourier transform. This transform pair isn't as important as the reason it is true. " From MathWorld--A Wolfram Web Resource. t/ D X1 nD1. In mathematics, the continuous Fourier transform is one of the specific forms of Fourier analysis. 003 EECS MIT discrete operator transforms. 6 depicts a resistor and capacitor in series. Using the above, we obtain: [ ] 2 0 ( ) ( )0 j nf t n. the Fourier transform function) should be intuitive, or directly understood by humans. One imagines a delta function to be a square pulse of unit area in the limit as the base of the pulse becomes narrower and narrower and higher and higher. 082 Spring 2007 Fourier Series and Fourier Transform, Slide 22 Summary • The Fourier Series can be formulated in terms of complex exponentials – Allows convenient mathematical form – Introduces concept of positive and negative frequencies • The Fourier Series coefficients can be expressed in terms of magnitude and phase. However, at 45 degrees, the | {
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Series coefficients can be expressed in terms of magnitude and phase. However, at 45 degrees, the radon projection clearly is NOT a rectangular pulse (but rather a triangle) and its FFT is clearly NOT a sinc (note that the oscillations never run negative). Discrete Fourier Transform (DFT) Fast Fourier Transform (FFT) The Fourier Integral (FI) is a mathematical technique of transforming an ideal mathematical expression in the time domain into a description in the frequency domain. Exams are approaching, and I'm working through some old assignments. The magnitude of the Fourier transform of a rectangular pulse equals the absolute value of a sinc. Tips If a , b , and c are variables or expressions with variables, triangularPulse assumes that a <= b <= c. Existence of the Fourier Transform. 104 Chapter 5. jaj<1 (n+ 1)anu[n] 1 (1 ae j. In the second row is shown , a delayed unit pulse, beside the real and imaginary parts of the Fourier transform. The amplitudes of the harmonics for this example drop off much more rapidly (in this case they go as 1/n 2 (which is faster than the 1/n decay seen in the pulse function Fourier Series (above)). The fundamental frequency is 50 Hz and each harmonic is, of course, an integer multiple of that frequency. Four chapters on analog signal processing systems, plus many updates and enhancements. This is a good point to illustrate a property of transform pairs. In other words, the input signal is upsampled by a factor of using linear interpolation. | {
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Question
# (Diet problem): A dietician wishes to mix two types of foods in such a way that vitamin contents of the mixture contain atleast $$8$$ units of vitamin A and $$10$$ units of vitamin C. Food I contains $$2$$ units/kg of vitamin A and $$1$$ unit/kg of vitamin C. Food II contains $$1$$ unit/kg of vitamin A and $$2$$ units/kg of vitamin C. It costs $$Rs 50$$ per kg to purchase Food I and $$Rs. 70$$ per kg to purchase Food II. Formulate this problem as a linear programming problem to minimise the cost of such a mixture.
Solution | {
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Solution
## Let the mixture contain $$x$$ kg of Food I and $$y$$ kg of Food II. Clearly, $$x \geq 0,y \geq 0$$. We make the following table from the given data:ResourcesFoodI$$(x)$$FoodII$$(y)$$RequirementVitamins A (units/ kg)$$2$$$$1$$$$8$$Vitaminc C (units/ kg)$$1$$$$2$$$$10$$Cost (Rs/ kg)$$50$$$$70$$Since the mixture must contain at least $$8$$ units of vitamin A and $$10$$ units of vitamin C, we have the constraints:$$2x + y \geq 8$$$$x + 2y \geq 10$$Total cost $$Z$$ of purchasing $$x$$ kg of food I and $$y$$ kg of Food II is$$Z = 50x 70y$$Hence, the mathematical formulation of the problem is:Minimise $$Z = 50x + 70y ... (1)$$subject to the constraints :$$2x + y \geq 8 ... (2)$$$$x + 2y \geq 10 ... (3)$$$$x, y \geq 0 ... (4)$$Let us the graph the inequalities $$(2)$$ to $$(4)$$. The feasible region determined by the system is shown in the Fig. Here again, observe that the feasible region is unbounded.Let us evaluate $$Z$$ at the corner points $$A(0,8), B(2,4)$$ and $$C(10,0)$$.Corner Point$$Z = 50x + 70y$$$$(0, 8)$$$$560$$$$(2, 4)$$$$380\leftarrow Minimum$$$$(10, 0)$$$$500$$In the table, we find that smallest value of $$Z$$ is $$380$$ at the point $$(2,4)$$. Can we say that the minimum value of $$Z$$ is $$380$$? Remember that the feasible region is unbounded. Therefore, we have to draw the graph of the inequality$$50x + 70y < 380$$ i.e., $$5x + 7y < 38$$to check whether the resulting open half plane has any point common with the feasible region. From the Fig, we see that it has no points in common.Thus, the minimum value of $$Z$$ is $$380$$ attained at the point $$(2, 4)$$. Hence, the optimal mixing strategy for the dietician would be to mix $$2$$ kg of Food I and $$4\ kg$$ of Food II,and with this strategy, the minimum cost of the mixture will be $$Rs. 380$$.Mathematics
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# Given relatively prime positive integers $a,b>1$, how many positive integers are there which are not non-negative integer combination sum of $a,b$?
Let $a,b>1$ be relatively prime positive integers. I know that $ab-a-b$ is the largest positive integer which can not be written as a sum of non-negative integer combination of $a,b$. My question is : can we explicitly determine how many positive integers are there which can not be written as a non-negative integer combination of $a,b$ ?
• @B.Goddard $a, b$ are assumed to be relatively prime. Also, the linear combinations here are assumed to have non-negative coefficients. That changes the problem quite a bit from the one you describe. – Arthur Jun 20 '18 at 13:11
• The most explicit I can get is a summation. If $a\lt b$ then there are: $$\sum_{k=1}^{a-1}\left\lfloor\frac{bk}{a}\right\rfloor$$ such numbers. Although you should definitely double-check this result. – N. Shales Jun 20 '18 at 16:19
• Having thought a bit more I further conjecture that the result simplifies to $$\frac{1}{2}(a-1)(b-1)\, .$$ – N. Shales Jun 20 '18 at 17:16
• @N.Shales your conjecture is correct: \begin{align*} R &= \sum_{k=1}^{a-1} \left\lfloor \frac{bk}{a} \right \rfloor\\ &= \sum_{k=1}^{a-1} \left\lfloor \frac{(a-k)b}{a} \right\rfloor \\ &= \sum_{k=1}^{a-1} b+\left\lfloor \frac{-kb}{a} \right\rfloor \\ &= \sum_{k=1}^{a-1} b-1-\left\lfloor \frac{kb}{a} \right\rfloor \\ &= (a-1)(b-1) - \sum_{k=1}^{a-1} \left\lfloor \frac{kb}{a} \right\rfloor \\ &= (a-1)(b-1) - R \end{align*} So that $2R = (a-1)(b-1)$. The key is the equality $\lfloor (-kb)/a\rfloor = -\lfloor kb/a \rfloor - 1$ which holds when $a$ does not divide $kb$. – Yong Hao Ng Jun 28 '18 at 8:20
• @N.Shales I saw it a few days ago but it took me a while before I found a not very involved solution to the floor-sum. – Yong Hao Ng Jun 29 '18 at 13:11 | {
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Let $N = ab$, we count the number of integers $0\leq t \leq N$ that can be written as $$t = ma+nb$$ such that $m,n\geq 0$. Let there be $R$ such numbers. Since the largest non-representable (positive)-integer is $ab-a-b$, all non-representable numbers are in $[0,N]$ and hence we know there are $N+1 - R$ of them. | {
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If $n\geq a$, then we replace it with $$ma+nb = (m+b)a + (n-a)b = m'a+n'b$$ So with repeated application we may assume that $0\leq n < a$. We require $$0 \leq ma+nb \leq N \implies 0 \leq ma \leq N-nb = (a-n)b$$ Hence we get the bounds of $m$ as $$0 \leq m \leq \left \lfloor \frac{(a-n)b}{a}\right \rfloor$$ This means the number of integers we can represent for a given $n$ is $\left \lfloor \frac{(a-n)b}{a} \right \rfloor + 1$. Now summing this over all possibilities $n=0,1,\dots a-1$, the number of representable integers $R$ is \begin{align*} R = \sum_{k=0}^{a-1} \left\lfloor \frac{(a-k)b}{a} \right\rfloor +1 &= \left\lfloor \frac{(a)b}{a} \right\rfloor + \left\lfloor \frac{(a-1)b}{a} \right\rfloor + \cdots + \left\lfloor \frac{(2)b}{a} \right\rfloor + \left\lfloor \frac{(1)b}{a} \right\rfloor + a\\ R &= a + b + \sum_{k=1}^{a-1} \left\lfloor \frac{kb}{a} \right\rfloor \\ \end{align*} Next we observe that if $a$ does not divide $kb$, then for some integer $t$ $$t < \frac{kb}{a} < t+1 \Longleftrightarrow -t-1 < \frac{-kb}{a} < -t$$ In particular this applies for $1\leq k\leq a-1$, since $\gcd(a,b)=1$ and $a>1$. We use the inequalities to obtain $$\left \lfloor \frac{-kb}{a}\right \rfloor = -t-1 = - \left \lfloor \frac{kb}{a}\right \rfloor -1,$$ This lets us get another equation of $R$ as \begin{align*} R = \sum_{k=0}^{a-1} \left\lfloor \frac{(a-k)b}{a} \right\rfloor +1 &= \sum_{k=0}^{a-1} \left\lfloor \frac{-kb}{a} \right\rfloor + b + 1 \\ &= a(b+1) + \sum_{k=0}^{a-1} \left\lfloor \frac{-kb}{a} \right\rfloor\\ &= a(b+1) + \sum_{k=1}^{a-1}\left\lfloor \frac{-kb}{a} \right\rfloor \\ &= a(b+1) + \sum_{k=1}^{a-1}-\left\lfloor \frac{kb}{a} \right\rfloor -1\\ R &= ab + 1 - \sum_{k=1}^{a-1}\left\lfloor \frac{kb}{a} \right\rfloor \end{align*} Therefore adding both equations of $R$, we have $$2R = ab + a + b + 1 = (a+1)(b+1)$$ giving us $R=(ab+a+b+1)/2$. Notice that $R$ is not an integer if and only if $a,b$ are both even, which is not possible since $\gcd(a,b)=1$. | {
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Finally, the number of integers considered is $ab+1$, so the number of integers not representable is $$ab+1 - \frac{ab+a+b+1}{2} = \frac{ab -a - b+1}{2} = \frac{(a-1)(b-1)}{2}$$
Proposition. Let $a,b$ be relatively prime integers greater than $1,$ and let $m=ab-a-b.$
If $x\in\{0,1,2,\dots,m\},$ then exactly one of the two numbers $x$ and $m-x$ can be expressed as a linear combination of $a$ and $b$ with nonnegative integer coefficients.
It follows that exactly half the elements of $\{0,1,2,\dots,m\}$ can be expressed as nonnegative integral linear combinations of $a$ and $b;$ that is, there are exactly $\frac{m+1}2=\frac{(a-1)(b-1)}2$ numbers which cannot be so expressed. | {
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# How many answers to $|3^x-2^y|=5$?
How many answers are there to the equation $|3^x-2^y|=5$ such that $x$ and $y$ are positive integers? Are there infinite? I've found $(2,2)$, $(3,5)$, and $(1,3)$. It seems to explode with larger values, but it's not a steady increase and there seems to be many dips. Do we KNOW that there are no large values for $x$ and $y$ where a power of 3 comes close to a power of 2?
• Its a putnam problem . – DeepSea Feb 5 '16 at 23:12
• Kf-Sansoo, had to google "putnam problem." Do you mean literally one, or one that fits the general qualifications? Do I need to add a tag? If it is one that's been mentioned before, could you give a link? – Elem-Teach-w-Bach-n-Math-Ed Feb 5 '16 at 23:18
• – Lucian Feb 6 '16 at 0:01
• @Kf-Sansoo: so what? Does that mean the question shouldn't be discussed on MSE? Does it mean that there is helpful information about the question to be found elsewhere? Or what? – Rob Arthan Feb 6 '16 at 0:01
• @Lucian: So, as it's conjectured that $Ax^n-By^m=C$ has a finite # of solutions for $(x,y,n,m)$ then $3^n-2^m=5$ can be conjectured to also have a finite # or solutions for $(n,m)$? In other words, it's unproven in a more general sense, so with the greater constraints it should be even less likely? If there are finite answers, my question could then be rephrased, "Is there a 4th answer?" – Elem-Teach-w-Bach-n-Math-Ed Feb 6 '16 at 0:18 | {
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There are only a finite number of solutions. It was proved by Pillai that $a^x - b^y = k$ where $a,b,k$ are fixed positive integers, $a > 1, b > 1, k \neq 0,$ with positive integer variables $x,y,$ has finitely many solutions. This is from page 51 in Shorey and Tijdeman, Exponential Diophantine Equations. The two papers by Pillai are 1931 and 1936. Both are in the Journal of the Indian Mathematical Society. A detail: if $k$ is larger than some bound that depends on $a,b,$ there is only one solution. Since we have more than one solution for $k = -5,$ it appears Pillai's bound is not tight enough to finish this problem. We just know one solution for $k=5.$
• I'll probably end up going with this as best answer. I've got an addendum though for your perusal, if I may. Since $|3^x-2^y|=5$ is proven to have finite solutions, I'd like to carry this knowledge to similar problems: i.e. $|(3^x)(5^y)-2^z|=7$, $|(2^x)(3^y)-5^z|=7$, $|(2^x)(3^y)-5^z|=7$. Namely working my way up to prime solutions such as $|(2^r)(7^s)(11^t)(13^u)(23^v)-(3^w)(5^x)(17^y)(19^z)|=29$ (or similarly creating primes less than the square of the largest prime used; in this case 23). Any thoughts? Probably throw this out there as a new ?. – Elem-Teach-w-Bach-n-Math-Ed Feb 8 '16 at 22:39
Here's an elementary self-contained argument that there is no solution with $y>5$.
A power of $3$ is congruent to either $1$ or $3 \bmod 8$, so once $y \geq 3$ we must have $3^x - 2^y = -5$.
Once $y \geq 6$, we then have $3^x \equiv -5 \bmod 2^6$, and thus $x \equiv 11 \bmod 16$.
But then $3^x + 5 \equiv 12 \bmod 17$, and no power of $2$ is congruent to $12 \bmod 17$ (the powers of $2 \bmod 17$ are $2,4,8,-1,-2,-4,-8,1,2,4,8,-1$ etc.), QED. | {
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There is a large literature on such Diophantine questions. One key phrase is "$S$-unit equations". In general it has been known for some time that there are finitely many solutions, and indeed for equations of the form $$\prod_i p_i^{n_i} - \prod_j q_j^{m_j} = r$$ this already follows from Thue's theorem (1909); and by now we even have effective algorithms known to find all solutions. There's still no elementary technique known in general, but in your case (where only the primes 2,3,5) appear an elementary solution is contained in a 1976 paper
This is not a complete answer, but I'd just like to note a connection with the concept of irrationality measure. First recall that $\alpha \in \mathbf{R}$ has irrationality measure $\mu(\alpha)$ if, for all $r>\mu(\alpha)$, there are only finitely many pairs of integers $p,q$ such that $$\left| \alpha - \frac{p}{q} \right| < \frac{1}{q^r}.$$ Note that $\mu(\alpha)=\infty$ is allowed (but we have $\mu(\alpha)=2$ for all but a measure-zero set of reals $\alpha$).
With this definition out of the way, note that $$\left| 3^x - 2^y \right| = 5$$ implies $$3^x = 2^y \pm 5.$$ Take logs: $$\log(3^x) = \log(2^y) + \log(1 \pm 5 \cdot 2^{-y}).$$ So by the Taylor expansion of $\log(1+t)$, $$\left| \log(3^x) - \log(2^y) \right| < 5 \cdot 2^{-y}+\frac{5^2 \cdot 2^{-2y}}{2}+\cdots < 6 \cdot 2^{-y}$$ for $y$ sufficiently large. Hence, using $\log(3^x) = x \log 3$ and $\log(2^y) = y \log 2$ $$\left| \frac{\log 3}{\log 2} - \frac{y}{x} \right| < \frac{6}{x \log 2} \cdot 2^{-y}.$$ Since the RHS decreases much faster than anything polynomial in $x$ (at least for $y/x$ within some fixed small interval around $\log(3)/\log(2)$), the existence of infinitely many solutions to your equation would imply that $\mu(\log(3)/\log(2))$ were infinite. It is probably known that this latter statement is false, probably by methods similar to the proof in the Shorey/Tijdeman book mentioned by Will Jagy. | {
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# istriu
Determine if matrix is upper triangular
## Syntax
• `tf = istriu(A)` example
## Description
example
````tf = istriu(A)` returns logical `1` (`true`) if `A` is an upper triangular matrix; otherwise, it returns logical `0` (`false`).```
## Examples
collapse all
### Test Upper Triangular Matrix
Create a 5-by-5 matrix.
`A = triu(magic(5))`
```A = 17 24 1 8 15 0 5 7 14 16 0 0 13 20 22 0 0 0 21 3 0 0 0 0 9```
Test `A` to see if it is upper triangular.
`istriu(A)`
```ans = 1 ```
The result is logical `1` (`true`) because all elements below the main diagonal are zero.
### Test Matrix of Zeros
Create a 5-by-5 matrix of zeros.
`Z = zeros(5);`
Test `Z` to see if it is upper triangular.
`istriu(Z)`
```ans = 1 ```
The result is logical `1` (`true`) because an upper triangular matrix can have any number of zeros on the main diagonal.
## Input Arguments
collapse all
### `A` — Input arraynumeric array
Input array, specified as a numeric array. `istriu` returns logical `0` (`false`) if `A` has more than two dimensions.
Data Types: `single` | `double`
Complex Number Support: Yes
collapse all
### Upper Triangular Matrix
A matrix is upper triangular if all elements below the main diagonal are zero. Any number of the elements on the main diagonal can also be zero.
For example, the matrix
$A=\left(\begin{array}{cccc}1& -1& -1& -1\\ 0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& -2& -2\\ 0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& -3\\ 0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\end{array}\right)$
is upper triangular. A diagonal matrix is both upper and lower triangular.
### Tips | {
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is upper triangular. A diagonal matrix is both upper and lower triangular.
### Tips
• Use the `triu` function to produce upper triangular matrices for which `istriu` returns logical `1` (`true`).
• The functions `isdiag`, `istriu`, and `istril` are special cases of the function `isbanded`, which can perform all of the same tests with suitably defined upper and lower bandwidths. For example, ```istriu(A) == isbanded(A,0,size(A,2))```. | {
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Mathematics
# If $\displaystyle\int x^5e^{-x^2}dx=g(x)\cdot e^{-x^2}+C$ then the value of $g(-1)$ is?
$-\dfrac{5}{2}$
##### SOLUTION
Put $x^2=t$
$2xdx=dt$
$\displaystyle\int t^2e^{-t}\dfrac{dt}{2}$
$=\dfrac{1}{2}[-t^2\cdot e^{-t}+2\displaystyle\int te^{-1}dt]+c$
$=\dfrac{1}{2}[-t^2\cdot e^{-t}-2te^{-t}+\displaystyle\int 2e^{-t}dt]+c$
$=\dfrac{1}{2}(-t^2e^{-t}-2(te^{-1}+e^{-t}))+c$
$=\dfrac{-(x^4+2x^2+2)e^{-x^2}}{2}+c$
$g(x)=\dfrac{-(x^4+2x^2+2)}{2}$
$g(-1)=-\dfrac{5}{2}$.
Its FREE, you're just one step away
Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86
#### Realted Questions
Q1 Subjective Hard
Evaluate the given integral.
$\displaystyle\int { { e }^{ x } } \left( \cfrac { \sin { x } \cos { x } -1 }{ \sin ^{ 2 }{ x } } \right) dx$
1 Verified Answer | Published on 17th 09, 2020
Q2 Subjective Medium
Evaluate the given integral.
$\int { x\sec ^{ 2 }{ x } } dx\quad \quad$
1 Verified Answer | Published on 17th 09, 2020
Q3 Single Correct Hard
The value of the definite integral $\int_\limits{ 1}^e( (x+1)e^x\ln x) dx$ is-
• A. $e$
• B. $e^e(e-1)$
• C. $e^{x+1}$
• D. $e^{e+1}+e$
1 Verified Answer | Published on 17th 09, 2020
Q4 Single Correct Hard
The integral $\displaystyle \int (1+x-\displaystyle \frac{1}{x})e^{x+\frac{1}{x}} dx$ is equal to
• A. $(x-1)e^{x+ \frac{1}{x}} +c$
• B. $(x+1)e^{x+\frac{1}{x}} +c$
• C. $-xe^{x+\frac{1}{x}} +c$
• D. $xe^{x+\frac{1}{x}} +c$
The average value of a function f(x) over the interval, [a,b] is the number $\displaystyle \mu =\frac{1}{b-a}\int_{a}^{b}f\left ( x \right )dx$
The square root $\displaystyle \left \{ \frac{1}{b-a}\int_{a}^{b}\left [ f\left ( x \right ) \right ]^{2}dx \right \}^{1/2}$ is called the root mean square of f on [a, b]. The average value of $\displaystyle \mu$ is attained id f is continuous on [a, b]. | {
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# Simple integral - but a question of methodology
ianb
simple integral -- but a question of methodology
$$\int \frac{(x-1)}{2} dx$$
First, try substitution. Let $$u = x - 1$$, and $$du = dx$$
$$\int \frac{u}{2} du$$
$$\frac{1}{2} \int u \ du$$
$$\frac{1}{2} \times \frac{u^{2}}{2}$$
$$\frac{1}{2} \times \frac{(x-1)^2}{2}$$
$$\frac{(x-1)^{2}}{4} + C$$
Or we could do this:
$$\int \frac{(x-1)}{2} dx$$
$$\int \frac{1}{2}(x-1)dx$$
$$\frac{1}{2} \int(x-1)dx$$
$$\frac{1}{2} (\frac{x^2}{2} - x) + C$$
$$\frac{1}{4} (x^2 - 2x) + C$$
Why should one of them be right?
Last edited:
ZioX
They're both right. Antiderivatives differ by a constant. (Differentiating eliminates the constant term). Show that those two answers differ by a constant.
Homework Helper
...
$$\frac{1}{4} (x^2 - \frac{x}{2}) + C$$
Why should one of them be right?
The last line is wrong. You've pulled out the factor 1 / 4 incorrectly.
$$\frac{1}{4} \left( x ^ 2 - 2x \right)$$
Both are correct, the two results differ from each other by a constant 1/4.
You should note that if F(x) is one anti-derivative of f(x), then F(x) + C is also f(x)'s anti-derivative, where C is any constant.
Proof: Differentiate the LHS with respect to x, we have:
(F(x) + C)' = F'(x) + C' = F'(x) + 0 (the derivative of a constant is just 0)
= F'(x) = f(x)
Can you get this? :)
ianb
Well I can certainly show that, just expand it. But the issue arises when you evaluate the area, say from 0 to 1.7, and using either ways give different answers. In fact, the answers differ by 0.25.
Thoughts?
ianb
Ah, my apologies. I was rushing through latex'ing my equations (still a newbie)--should have double checked it.
Homework Helper
Well I can certainly show that, just expand it. But the issue arises when you evaluate the area, say from 0 to 1.7, and using either ways give different answers. In fact, the answers differ by 0.25. | {
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Thoughts?
Nope, they are identically the same, let F(x) + C1, and F(x) + C2 be two anti-derivatives of f(x), we have:
$$\int_a^b f(x) dx = (F(b) + C_1) - (F(a) + C_1) = F(b) - F(a)$$
$$\int_a^b f(x) dx = (F(b) + C_2) - (F(a) + C_2) = F(b) - F(a)$$, they are still the same.
The constant will automatically vanish when you subtract the two terms. :)
ianb
Hm. What if the lower bound is zero, I don't think it does. Look at the substitution method, and evaluate it from 0 to 1.7. You will still get a difference of 0.25 than if you compare it with the other method. Of course, you might say that the constants do indeed cancel even with 0 set as the lower bound, but that will require me to expand--why? I mean, why would the answers be the same after expansion, and different without it? Does integration by substitution require expansion of factors?
Last edited:
ianb
I'm sorry, I should have been clearer. Message edited accordingly.
Homework Helper
Hm. What if the lower bound is zero, I don't think it does. Look at the substitution method, and evaluate it from 0 to 1. You will still get a difference of 0.25 than if you compare it with the other method. Of course, you might say that the constants do indeed cancel even with 0 set as the lower bound, but that will require me to expand--why? I mean, why would the answers be the same after expansion, and different without it? Does integration by substitution require expansion of factors?
Uhm, well, no, they are not different.
Well, here we go:
$$\int_0 ^ 1 \frac{x - 1}{2} dx = \left. \left( \frac{(x - 1) ^ 2}{4} \right) \right|_0 ^ 1 = \frac{(1 - 1) ^ 2}{4} - \frac{(0 - 1) ^ 2}{4} = 0 - \frac{1}{4} = -\frac{1}{4}$$
$$\int_0 ^ 1 \frac{x - 1}{2} dx = \left. \frac{1}{4} \left( x ^ 2 - 2x \right) \right|_0 ^ 1 = \frac{1}{4} \left( (1 ^ 2 - 2 \times 1) - (0 ^ 2 - 0) \right) = \frac{1}{4} (-1 - 0) = -\frac{1}{4}$$
They are the same. :)
ianb
Wow. I'm embarrassed to say the least. You, sir, rock my world. | {
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# Fewest steps to reach $200$ from $1$ using only $+1$ and $×2$
This is a problem from the AMC 8 (math contest):
A certain calculator has only two keys $$[+1]$$ and $$[\times 2]$$. When you press one of the keys, the calculator automatically displays the result. For instance, if the calculator originally displayed “$$9$$” and you pressed $$[+1]$$, it would display “$$10$$”. If you then pressed $$[\times 2]$$, it would display “$$20$$”. Starting with the display “$$1$$”, what is the fewest number of keystrokes you would need to reach “$$200$$”?
Intuitively I worked back from $$200$$, dividing by $$2$$ until I reached an odd number, subtracting $$1$$ when I did, etc..to reach the correct answer of $$9$$ steps.
However, I can't figure out how to convince myself beyond any doubt that it is the optimal solution. In other words, I can't prove it mathematically. The best I can come up with is that beyond the first step from $$1$$ to $$2$$, multiplication by $$2$$ is always going to yield a larger step than addition by $$1$$ and therefore I should take as many $$[\times 2]$$ steps as I can. This doesn't feel rigorous enough, though.
EDIT: Just to be clear, I am asking for a proof or at least more rigorous explanation of why this is the optimal solution.
• Do you want the fewest steps to get to exactly $200$ or at least $200$? – John Douma Mar 17 at 19:37
• @JohnDouma: Exactly $200$, obviously. Otherwise eight steps would suffice. – TonyK Mar 17 at 19:43
• @TonyK That is not obvious. In fact, the last paragraph before the EDIT implies otherwise. – John Douma Mar 17 at 19:44
• @JohnDouma It's exactly $200$, but I disagree with your comment that the last paragraph implies otherwise. I can try to take as many X2 steps as I can and still intend to not get past $200$. – jeremy radcliff Mar 17 at 19:48
Look at what the operations $$+$$ and $$\times$$ do to the binary expansion of a number: | {
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Look at what the operations $$+$$ and $$\times$$ do to the binary expansion of a number:
• $$\times$$ appends a $$0$$, and increases the length by one, leaving the total number of $$1$$'s unchanged;
• if the final digit is $$0$$, then $$+$$ increases the number of $$1$$'s by one, but doesn't change the length;
• if the final digit is $$1$$, then $$+$$ doesn't increase the total number of $$1$$'s (it may in fact decrease it), and doesn't increase the total length by more than $$1$$.
Therefore, with a single key press:
• you can increase the length by one, but this won't increase the number of $$1$$'s;
• you can increase the number of $$1$$'s by one, but this won't increase the length.
The binary expansion of $$200$$ is $$200_{10}=11001000_2$$. This has three $$1$$'s, and a length of eight. Starting from $$1$$, we must increase the length by seven, and increase the number of $$1$$'s by two. So this requires at least nine steps.
• Beautiful, this allows you to determine the optimal solution and path for an arbitrary number from it's binary representation. – Jared Goguen Mar 18 at 1:34
• There actually exist two ways of getting $200$ using nine steps. $$1+1+1\times2\times2\times2+1\times2\times2\times2=200\\ 1\times2+1\times2\times2\times2+1\times2\times2\times2=200$$ – Kay K. Mar 18 at 2:13
• @KayK.: Yes, but only because there are two ways of getting to 2. The rest is uniquely determined. – TonyK Apr 4 at 16:05
You can proceed by induction on $$n$$, showing that the quickest way to reach any even number $$2n$$ involves doubling on the last step, which is clearly true for the base case $$n=1$$ (where doubling and adding $$1$$ have a tomato-tomahto relationship). | {
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Now if the last step to reach $$2n+2$$ isn't doubling, it can only be adding $$1$$ from $$2n+1$$. But $$2n+1$$ can only be reached by adding $$1$$ from $$2n$$, at which point the inductive hypothesis says the next previous number was $$n$$. But you can get from $$n$$ to $$2n+2$$ more quickly in two steps: add $$1$$, then double. So the last step in the quickest route to $$2n+2$$ is doubling from $$n+1$$.
• This is nice as it formalizes the OPs intuition, whereas the binary representation answer (which is super slick) might feel a little out of the blue. – jacob1729 Mar 17 at 22:55
Setting the display to binary base, $$[\times2]$$ inserts a $$0$$ to the right and $$[+1]$$ increments; if the rightmost digit is a zero, it just turns it to a $$1$$.
Using these rules you build a number of $$o$$ ones and $$z$$ zeroes in $$o-1+z$$ keystrokes, starting from $$1$$. This seems close to optimal.
• you just exactly reproduces the binary 200, why should we think it is not optimal? – dEmigOd Mar 17 at 19:54
• @dEmigOd: I didn't prove that inserting the bits one by one with $\times2$ or $\times2+1$ is optimal. – Yves Daoust Mar 17 at 19:57
I'll make a try
Since $$200=2^7+2^6+2^3$$ you will need at least $$8$$ steps to reach $$200$$ (since we start from $$1$$ and we get a number of the form $$2^a+...+2^l$$) so it remains to show that $$8$$ steps are not enough.
Maybe you could try to show that if there was a solution with $$8$$ steps then it would contain only one $$+1$$ which contradicts the fact that in $$200=2^7+2^6+2^3$$ we have two $$+$$
Working backwards, use the tree diagram and stop lagging branches (e.g. if $$\color{red}{99}$$ in step $$3$$ leads to $$1$$, then it takes one more operation than $$\color{red}{99}$$ in step $$2$$):
$$\hspace{3cm}$$ | {
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There are three formulas for calculating the area of a sector. The formula for the area of a sector is: A = r² * θ / 2. Find the central angle of a sector whose radius is 56 cm and area, is 144 cm2. Geometric skills. How to use the calculator Enter the radius and central angle in DEGREES, RADIANS or both as positive real numbers and press "calculate". Area of a sector when the central angle is given in degrees If the angle of the sector is given in degrees, then the formula for the area of a sector is given by, Area of a sector = (θ/360) πr2 A = (θ/360) πr2 Find the radius of a semi – circle with the area of 24 inches squared. Math A level Syllabus, 2016. The area of a sector of a circle with a radius of 5 cm, with an angle of 2 radians, is 25 cm 2. Then, the area of a sector of circle formula is calculated using the unitary method. 40pie units squared. Show that 2θ-3sinθ=0. The area of the minor segment as shown in Fig 5 . Area of a cyclic quadrilateral. Area of sector of circle = (1/2)r²Ө , Ө must be in radians. Example 1 Find the arc length and area of a sector of a circle of radius … Step 1: Find the area of the circle. Katelyn is making a semi-circular design to put on one of her quilts. Thus, if you are not sure content located Radians, Arc Length and Sector Area Radians Radians are units for measuring angles. They can be used instead of degrees. Show that 2θ-3sinθ=0. If the radius of the circle is , what is the area of the semi-circular design? With the help of the community we can continue to Each of these formula is applied depending on the type of information given about the sector. Yes, both the semicircle and the quadrant are special types of the sector of a circle with angles $$180^{\circ}$$ and $$90^{\circ}$$ respectively. Find the area of a sector if the radius of the circle is 4, and the angle of the sector is radians. In simple words, the area of a sector is a fraction of the area of the circle. We can find the area of a sector of a circle in a | {
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a sector is a fraction of the area of the circle. We can find the area of a sector of a circle in a similar manner. Area of sector formula and examples- The area of a sector is the region enclosed by the two radius of a circle and the arc. Jul 2009 448 89. Please follow these steps to file a notice: A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; Displaying top 8 worksheets found for - Area Of A Sector In Radians. Both can be calculated using the angle at the centre and the diameter or radius. So in the below … The area of the circle is equal to the radius square times . as shown in Fig 4, we consider the sector as a fraction of the circle hence: Area of a segment. Introducing Radians; 9. If the diameter of a circle is 6, find the area of a sector with a sector angle of . Varsity Tutors LLC , where is the angle measure of the sector in radians, and is the radius of the circle. or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing The full angle is 2π in radians, or 360° in degrees, the latter of which is the more common angle unit. The measure of central angle XYZ is 1.25 radians. The formulas to find the area of a sector in Degrees (D°) or Radians (R°) are shown below: Area (Degrees) = πr2 x θ/360 Area (Radians) = ½r2θ r, D° r, R° r, s r, A D°, s Use 3.14 for . Track your scores, create tests, and take your learning to the next level! Area of a sector given the arc length. The non-shaded area of the circle shown below is called a SECTOR. Area of an ellipse. ChillingEffects.org. The angle AOB is in radians. Area of sector. You can find it by using proportions, all you need to remember is circle area formula (and we bet you do! What is the area of Fiona's circle? An arc of a circle when bounded by two radii with the centre of the circle then we get a slice of a circle. Find the radius of a sector whose area is 47 meters squared and central angle is 0.63 | {
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of a circle. Find the radius of a sector whose area is 47 meters squared and central angle is 0.63 radians. Use … Send your complaint to our designated agent at: Charles Cohn Circle sector area calculator - step by step calculation, formulas & solved example problem to find the area of circle sector given input values of corcle radius & the sector angle in degrees in different measurement units between inches (in), feet (ft), meters (m), centimeters (cm) & millimeters (mm). Example: a) What is the length of the arc intercepted by an angle of 15° on a circle with radius 20 meters? Use prior knowledge on length of circumference and area of circle to deduce formulae to calculate arc length and sector area. Given, the length of the arc, the area of a sector is given by. You can find it by using proportions, all you need to remember is circle area formula (and we bet you do! The number inside the sector is the area. degree radian; area S . If r is in "m", the area will be in "m" 2. Now, this looks messy, but we can simplify it to get: Next, use your calculator to find a decimal answer, and then round to get our final answer. Find the area of a sector whose arc is 8 inches and radius, is 5 inches. a And then we just can solve for area of a sector by multiplying both sides by 81 pi. Find the area of the sector with radius 7\ "cm" and central angle 2.5 radians. There are two types of sectors, minor and major sector. I remember this formula as it is quite easy to remember. SECTORS . The portion of the circle's circumference bounded by the radii, the arc, is part of the sector. Perimeter of sector = r + 2r = r( + 2) Where is in radians If angle is in degrees, = Angle × π/(180°) Let us take some examples: Find perimeter of sector whose radius is 2 cm and angle is of 90° First, We need to convert angle in radians = Angle in degree × π/(180°) = 90° × π/(180° ) = π/4 Convert degrees into radians and viceversa. What is the area of the shaded sector? Fiona draws a circle with a | {
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into radians and viceversa. What is the area of the shaded sector? Fiona draws a circle with a diameter of 14 meters. A Terminal side Vertex B Initial Side C B, ABC, CBA, and are all notations for this angle. Questionnaire. which specific portion of the question – an image, a link, the text, etc – your complaint refers to; (see diagrams below) There is a lengthy reason, but the result is a slight modification of the Sector formula: I have managed to get: 3=½r²θ and 2=½r²sinθ Therefore: ½r²θ-3=0 and ½r²sinθ-2=0 But I'm unsure where to go from there. Area of a circle. Find the area of a sector if the radius is 1 and the angle of sector is radians. Find the area of a sector with the radius of 1 and angle of . In order to calculate the area of a sector, you need to know the following two parameters: With the above two parameters, finding the area of a circle is as easy as ABCD. Infringement Notice, it will make a good faith attempt to contact the party that made such content available by 3. The arc length formula is used to find the length of an arc of a circle; ℓ = rθ ℓ = r θ, where θ θ is in radian. Arc Length Formula - Example 1 Discuss the formula for arc length and use it in a couple of examples. is obtained by the arc AB of the centre O is given by: where is in radians. Finding a Missing Angle With the Sine Rule; 13. Section 4.2 – Radians, Arc Length, and the Area of a Sector 1 Section 4.2 Radians, Arc Length, and Area of a Sector An angle is formed by two rays that have a common endpoint (vertex).One ray is the initial side and the other is the terminal side.We typically will draw angles in the coordinate plane with the Express the answer in terms of . Side of polygon given area. r O 1 radian is the size of the angle formed at the centre of a circle by 2 radii which join the ends of an arc equal in length to the radius. Some of the worksheets for this concept are Arc length and sector area, Area of a sector 1, L 2r, Find the area of the shaded sector in the | {
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are Arc length and sector area, Area of a sector 1, L 2r, Find the area of the shaded sector in the following, Radians arc length and area of a sector, Radians, Mcr3ui radian work, Area and arc length of a sector. Exercise worksheet on 'Find the area of a sector of a circle when the angle is given in radians.' Figure 6. misrepresent that a product or activity is infringing your copyrights. 222 r Ar r Note, to use this formula, the measure of the central angle must be given in radians. Calculate the area of the sector shown below. Calculate the area of a sector with a radius of 10 yards and an angle of 90 degrees. A-Level Maths Edexcel C2 June 2008 Q7b ExamSolutions Northeastern University, Bachelor of Science, Industrial Engineering. To calculate the area of the sector you must first calculate the area of the equivalent circle using the formula stated previously. The non-shaded area of the circle shown below is called a SECTOR. If the angle of the sector is given in degrees, then the formula for the area of a sector is given by. Find the area of a sector with a radius and angle of . © 2007-2020 All Rights Reserved, Computer Science Tutors in Dallas Fort Worth, ISEE Courses & Classes in San Francisco-Bay Area, SSAT Courses & Classes in San Francisco-Bay Area. A sector is created by the central angle formed with two radii, and it includes the area inside the circle from that center point to the circle itself. In this example the sector subtends a right-angle (900) at the centre of the circle. Area of a sector of a circle. A-level : area and arc length of a sector tutorial In this tutorial you are shown how to find the area of a sector and arc length when the angle is in degrees or radians. Problem Solving With the Cosine Rule; 15. Find the angle of a sector whose arc length is 22 cm and area, is 44 cm2. When the angle at the centre is 360°, area of the sector, i.e., the complete circle = πr² When the angle at the center is 1°, area of the sector = Thus, when the angle | {
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complete circle = πr² When the angle at the center is 1°, area of the sector = Thus, when the angle is θ, area of sector, OPAQ = Write the formula for the area of a sector in radians. Using this formula, and approximating , the area of the circle is . Sep 2, 2009 #2 For #1. Recognize parts of a circle and use appropriate terminology. Your Infringement Notice may be forwarded to the party that made the content available or to third parties such The circular face of a watch has an area that measures between 800 and 900 square millimeters. We know that the area of the whole circle is equal to πr². The ratio of the area of the sector to the area of the full circle will be the same as the ratio of the angle θ to the angle in a full circle. When angle of the sector is 360°, area of the sector i.e. Area of a regular polygon. Find the area of the sector. How to find the area of a sector whose central angle is in radian: formula, 1 example, and its solution. Section 4.2 – Radians, Arc Length, and the Area of a Sector 1 Section 4.2 Radians, Arc Length, and Area of a Sector An angle is formed by two rays that have a common endpoint (vertex).One ray is the initial side and the other is the terminal side.We typically will draw angles in the coordinate plane with the | {
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1. ## Piecewise Function problem
Let f(x) =
(0) if x< 0
(x) if 0 <= x <= 1
(2 - x) if 1 < x <= 2
(0) if x > 2
and g(x) = [integrate] f(t)dt on [0, x]
i) Find an expression for g(x) similar to the one for f(x).
ii) Where is f differentiable? Where is g differentiable?
I'm pretty much stuck, and don't know how to start this question.
I know first of all that by the fundamental theorem of Calculus, g'(x) = f(x).
But from that, how do I figure out an expression for g(x) from the piecewise function?
Any and all help is appreciated here.
2. Originally Posted by Tulki
Let f(x) =
(0) if x< 0
(x) if 0 <= x <= 1
(2 - x) if 1 < x <= 2
(0) if x > 2
and g(x) = [integrate] f(t)dt on [0, x]
i) Find an expression for g(x) similar to the one for f(x).
ii) Where is f differentiable? Where is g differentiable?
I'm pretty much stuck, and don't know how to start this question.
I know first of all that by the fundamental theorem of Calculus, g'(x) = f(x).
But from that, how do I figure out an expression for g(x) from the piecewise function?
Any and all help is appreciated here.
i)
for $\displaystyle x < 0$, $\displaystyle f(x) = 0$ and $\displaystyle g(x) = \int f(t) ~dt = 0$
for $\displaystyle 0 \leq x \leq 1$, $\displaystyle f(x) = x$ and $\displaystyle g(x) = \int_{0}^{x} f(t) ~dt = \int_{0}^{x} t ~dt = \frac{1}{2}x^2$
for $\displaystyle 1 < x \leq 2$, $\displaystyle f(x) = 2 - x$ and $\displaystyle g(x) = \int_{0}^{x} f(t) ~dt = \int_{0}^{1} f(t) ~dt + \int_{1}^{x} f(t) ~dt=$$\displaystyle \int_{0}^{1} t ~dt + \int_{1}^{x} (2-t) ~dt= \frac{1}{2} + (2x - 2) - (\frac{1}{2}x^2 - \frac{1}{2})$
for $\displaystyle x > 2$
3. Hello, Tulki!
Here's part (a).
Did you make a sketch?
$\displaystyle \text{Let }\:f(x) \;=\;\left\{\begin{array}{cccc}0 && x < 0 \\ \\[-4mm] x && 0 \leq x \leq 1 \\ \\[-4mm] 2 - x && 1 < x \leq 2 \\ \\[-4mm] 0 && x > 2 \end{array}\right\}$
and: .$\displaystyle g(x) \:=\: \int_0^x f(t)\,dt$ | {
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and: .$\displaystyle g(x) \:=\: \int_0^x f(t)\,dt$
a) Find an expression for $\displaystyle g(x)$ similar to the one for $\displaystyle f(x).$
$\displaystyle g(x)$ gives the area under the graph from $\displaystyle 0\text{ to }x$
The graph looks like this:
Code:
|
|
1+ *
| *:::*
| *:::::::*
| *:::::::::| *
- * * * - - - * - + - * * * - -
0 1 x 2
Then: .$\displaystyle \displaystyle g(x) \;=\;\left\{ \begin{array}{cccc} 0 && x < 0 \\ \\[-3mm] \int^x_0t\,dt && 0 \leq x \leq 1 \\ \\[-2mm] \frac{1}{2} + \int^x_1 (2-t)\,dt && 1 < x \leq 2 \\ \\[-3mm] 1 && x > 2 \end{array} \right\}$
4. Because all of the graphs are straight lines, you don't really need to integrate, just find the areas of the polygons involved.
For x< 0, f(x)= 0 so its integral is 0 (and its graph is just the x-axis so there is 0 area).
For x between 0 and 1, y= x so the "area under the curve" is the area of a right triangle with base x and height x: its area is $\displaystyle \frac{1}{2}x^2$.
For x between 1 and 2, y= 2- x and we can break the "area under the currve" from 0 to x into two parts: The area of the right triangle with base 1 and height 1 is $\displaystyle \frac{1}{2}(1)(1)= \frac{1}{2}$. The area from 1 to x is a trapezoid with bases 1 and 2-x and height x. It's area is $\displaystyle \frac{2-x+ 1}{2}x= \frac{(3-x)x)}{2}= \frac{3x- x^2}{2}$.
The total area is $\displaystyle \frac{1}{2}+ \frac{3x- x^2}{2}= \frac{1+ 3x- x^2}{2}$. AT x= 2, the integral is the entire area of a triangle with base 2 and height 1 which is 1.
For x> 2, the graph is again the x-axis so no new area is added. For all x> 2, the function is the constant 1.
5. Thank you very much, all of you!
I suppose it was only the piecewise function that scared me. Graphing it DOES indeed help greatly. The pieces of the function are pretty darn simple, in retrospect. | {
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# Dense subset of $S^{n-1}$ with antipodal complement
Let $$n \in \mathbb{N}$$, and let $$S^{n} \subseteq \mathbb{R}^{n+1}$$ be the unit $$n$$-sphere.
Does there exist a dense subset $$D$$ of $$S^n$$ such that $$\textbf{x} \in D \implies -\textbf{x} \notin D$$?
This is true for $$n=1$$. Note that $$S^1 \cong A := [0,1]/\mathbb{Z}$$, where the equivalent question is: does there exists a dense subset $$D'$$ of $$A$$ such that $$x \in D \implies x \pm \frac{1}{2} \notin D'$$?
Consider the function $$f : [0,1) \to \{0,1\}$$,
$$f(x)= \begin{cases} 0 & x< \frac{1}{2},\text{ any decimal expansion of } x \text{ has finitely many 3s} \\ 1 & x < \frac{1}{2}, \;x \text{ has a decimal expansion with } \infty \text{ many 3s} \\ 1 & x = \frac{1}{2} \\ 1 & x > \frac{1}{2}, \text{ any decimal expansion of } x \text{ has finitely many 3s}\\ 0 & x > \frac{1}{2}, \;x \text{ has a decimal expansion with } \infty \text{ many 3s} \end{cases}$$
Then $$D' = f^{-1}(\{0\})$$ is a suitable dense set.
However, I am not sure how to approach the cases $$n >1$$, since $$S^n$$ is no longer homeomorphic to a very simple object and my topological toolkit is quite unsophisticated. If the question can be answered with elementary methods, I welcome any hints. I am not even sure whether or not I expect such a set $$D$$ to exist for $$n> 1$$. | {
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• It is also equivalent to ask the analogue of the question in $\Bbb R^n$, if you’ve figured out what the antipodes look like after stereographic projection Aug 7 at 14:42
• I don't have time for a full answer, but you could consider the stereographic projection $\mathbb{S}^n\setminus\{(0,\ldots,0,1)\}\rightarrow\mathbb{R}^n$, then look at where the intersection $H_\pm^{n+1}=\{x\in\mathbb{R}^{n+1}|\operatorname{sign}(x_0)=\pm 1\}$ with $\mathbb{S}^n$ lands, and then consider the primages of their intersection with $\mathbb{Q}^n$ and $\mathbb{R}^n\setminus\mathbb{Q}^n$ respectivly. Aug 7 at 14:58
• @SamuelAdrianAntz I almost understand your comment, but it's not quite clear. I am a little suspicious of your last step because note that $D$ quite naturally bijects with its complement, so both are uncountable. I understand the stereographic projection to be a bijection, so the preimage of $\mathbb{Q}^n$ would be countable? Or have I misunderstood? Aug 7 at 15:19
• @FShrike That's certainly a possible approach. But even if I could figure out where they antipodes go, I am not certain that this simplifies matters much (or enough for me). After all, $\mathbb{R}$ is still much simpler than $\mathbb{R}^n$, $n>1$. Aug 7 at 15:28
This answer is an expansion of my comment, but I simplified it a bit without using the stereographic projection.
First of all, the obvious way of taking the intersection of $$\mathbb{Q}^{n+1}$$ with $$\mathbb{S}^n\subset\mathbb{R}^{n+1}$$ won't work, as it can even be empty as explained in this answer here. Consider the half-spaces $$\mathbb{H}_\pm^{n+1}=\{x\in\mathbb{R}^{n+1}|\operatorname{sign}(x_0)=\pm 1\}$$ and the (surjective) projections: $$\operatorname{pr}_\pm\colon \mathbb{H}_\pm^{n+1}\cap\mathbb{S}^n \twoheadrightarrow\mathbb{D}^n, (x_0,x_1,\ldots,x_n)\mapsto(x_1,\ldots,x_n)$$ flattening down both half-spheres. | {
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The subsets $$A=\operatorname{pr}_-^{-1}(\mathbb{D}^n\cap\mathbb{Q}^n)$$ and $$B=\operatorname{pr}_+^{-1}(\mathbb{D}^n\cap(\mathbb{R}^n\setminus\mathbb{Q}^n))$$ are both dense (in the closure of their respective half-sphere) as preimages of dense sets under continuous maps. $$A\cup B$$ is therefore dense in $$\mathbb{S}^n$$ as the closure operator commutes with finite unions (See here).
For $$x\in A$$, we have $$x_0<0$$, therefore $$-x\notin A$$, as well as $$x_1,\ldots,x_n\in\mathbb{Q}$$, therefore $$-x\notin B$$.
For $$y\in B$$, we have $$y_0>0$$, therefore $$-y\notin B$$, as well as $$y_1,\ldots,y_n\notin\mathbb{Q}$$, therefore $$-y\notin A$$.
• So one can use effectively the same idea as for $n=1$. Fair enough. I convinced myself it would be much more difficult and didn't try it. Also, your profile description is fairly entertaining. Aug 7 at 21:26
• As a minor correction, $A$ and $B$ are images, not preimages. Aug 7 at 21:33
• Thanks a lot! :-) The mistake with $A$ and $B$ is also corrected now. Aug 7 at 21:40
• No worries, and thanks for the answer. I'm sorry, but I think I have just observed another typo. I think the projection maps surject onto $D_1(0)$ in $\mathbb{R}^{n-1}$, the interior of the $(n-1)$-sphere. Aug 7 at 21:45
• Oh, yes! Honestly, I'm really not that good in writing down my thoughts in the correct notation. Aug 7 at 21:47 | {
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# Math Help - Difficult Trigonometric Integral
1. ## Difficult Trigonometric Integral
I was working through my homework and stumbled across this problem. I have tried many methods, but still have no luck. Rationalizing the denominator didn't seem to get me anywhere. WolframAlpha wasn't much either help in showing me the steps.
$\int {sin^2 x \over \sqrt {1+cosx}}$
Thanks for the help!
2. Originally Posted by Tylerzzz
I was working through my homework and stumbled across this problem. I have tried many methods, but still have no luck. Rationalizing the denominator didn't seem to get me anywhere. WolframAlpha wasn't much either help in showing me the steps.
$\int {sin^2 x \over \sqrt {1+cosx}}$
Substitute $u = \cos x$. Then $du = -\sin x\,dx$ and the integral becomes
$-\int \frac{\sin x}{\sqrt {1+cosx}}(-\sin x)dx = -\int \frac{\sqrt{1-u^2}}{\sqrt{1+u}}du = -\int\sqrt{1-u}\,du = \frac23(1-u)^{3/2} = \frac23(1-\cos x)^{3/2}$ (plus a constant).
3. Wow! I never really thought about solving it that way. Thanks a lot!
4. EDIT: Apologies to Opalg, I mistook it for [LaTeX ERROR: Convert failed]
5. Originally Posted by TheCoffeeMachine
EDIT: Apologies to Opalg, I mistook it for [LaTeX ERROR: Convert failed]
You had me worried for a moment.
6. Hello, Tylerzzz!
Another approach . . .
$\displaystyle \int \frac{\sin^2\!x}{\sqrt{1+\cos x}}\,dx$
We have: . $\dfrac{\sin^2\!x}{\sqrt{1+\cos x}}$
Multiply by $\frac{\sqrt{1-\cos x}}{\sqrt{1-\cos x}}\!:\;\;\dfrac{\sin^2\!x}{\sqrt{1+\cos x}} \cdot\dfrac{\sqrt{1-\cos x}}{\sqrt{1-\cos x}} \;=\;\dfrac{\sin^2\!x\sqrt{1-\cos x}}{\sqrt{1-\cos^2\!x}}$
. . . $\;=\;\dfrac{\sin^2\!x\sqrt{1-\cos x}}{\sqrt{\sin^2\!x}} \;=\;\dfrac{\sin^2\!x\sqrt{1-\cos x}}{\sin x} \;=\;\sin x\sqrt{1-\cos x}$
The integral becomes: . $\displaystyle \int(1-\cos x)^{\frac{1}{2}}(\sin x\,dx)$
Let: . $u \:=\:1-\cos x \quad\Rightarrow\quad du \:=\:\sin x\,dx$
Substitute: . $\displaystyle \int u^{\frac{1}{2}}\,du \;=\;\tfrac{2}{3}u^{\frac{3}{2}} + C$ | {
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Substitute: . $\displaystyle \int u^{\frac{1}{2}}\,du \;=\;\tfrac{2}{3}u^{\frac{3}{2}} + C$
Back-substitute: . $\frac{2}{3}(1-\cos x)^{\frac{3}{2}} + C$
7. Originally Posted by Opalg
You had me worried for a moment.
Haha! In my defence, I blame the UCAS for the plight of my eyesight today. | {
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# Thread: Need Help for Half Life Problems using Logs
1. ## Need Help for Half Life Problems using Logs
Hi, I need help with solving half life problems using natural log. Or normal log. I learned this before, but totally forgot how to do it right now before the Final Exam, please help.
The equation that I got was:
N = No (1/2) ^ t
N is final amount, No is the initial ammount and t is time.
I was given the data:
Time (min) 10 20 30 40
Amount (g) .83 .68 .56 .46
and I need to find the regression equation, write an equation in terms of base e, and use the equation to predict the half life.
How do I do this? I have no clue...
2. Do you have Excel?. You can do all sorts of regressions with it. Also, many calculators do regressions. Do you have one?. If so, what is it. Maybe I can step you through it.
Anyway, using Excel I got an equation of
$y=1.0091e^{-.0196x}$
This has an R^2 of 1, so it is as good as it gets.
Enter in x=10, do you get .83?.
3. hi thanks for replying, I have a TI-83+.
4. I would have to re-familiarize myself. I suppose the 83 does exponential regressions. I am not sure.
5. it does, i did it before, I think.
I just forgot how to do it.
I think it's storing the data into 2 list and then compare them or something...
but then, how do I set the equations up?
I mean that N = No (1/2) ^ t thing , hate math final, lol.
Maybe you can show me how you did it in excel?
6. You can find the half life by using the equation $k=\frac{-1}{T}ln(2)$
We know k=-.0196
So, $-.0196=\frac{-1}{T}ln(2)$
$t=35.36$
We could also do it using the equation.
Half of 1.0091 is .50455. Because 1.0091 is the amount at t=0
So, $.50455=1.0091e^{-.0196t}$
$t=35.36$
Same as before.
To find a regression on Excel, I make the graph, then add a trendline.
Click on Insert, then Chart, then Scatter
Highlight your data which is entered in columns A and B and proceed.
It is pretty much self-explanatory. Then onece you have the graph, click on | {
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It is pretty much self-explanatory. Then onece you have the graph, click on
Chart, Add Trendline, Exponential, Click on Options and check the box that says to display the equation on the chart.
There you have it.
7. Originally Posted by Billwaa
Hi, I need help with solving half life problems using natural log. Or normal log. I learned this before, but totally forgot how to do it right now before the Final Exam, please help.
The equation that I got was:
N = No (1/2) ^ t
N is final amount, No is the initial ammount and t is time.
I was given the data:
Time (min) 10 20 30 40
Amount (g) .83 .68 .56 .46
and I need to find the regression equation, write an equation in terms of base e, and use the equation to predict the half life.
How do I do this? I have no clue...
$N = N_0 \left ( \frac{1}{2} \right ) ^{kt}$
$ln(N) = ln(N_0) - [k~ln(2)]t$
Put t on the horizontal axis, ln(N) on the vertical axis, and do a linear regression. The slope will be $k~ln(2)$ which you can then solve for k, and the intercept will be $ln(N_0)$.
-Dan
8. thanks guys, this clear it up a little bit. ^_^
9. Originally Posted by Billwaa
hi thanks for replying, I have a TI-83+.
To do this on the 83+, input the data into two lists:
L1: {10,20,30,40}
L2: {.86,.68,.56,.46}
Now, hit [STAT] and scroll over to [CALC] Select option [0]:ExpReg.
This is an exponential regression. The solution will have the form of $y=a\times b^x$.
At the home screen, you should now see:
ExpReg
Now hit [2nd][1][,][2nd][2]
You should now have the following on the screen:
$\text{ExpReg } L_1\text{,}L_2$
Hit [ENTER]
You'll get the regression equation:
This is what the screen should say:
ExpReg
\begin{aligned}
y&=a*b^x \\
a&=1.00914514 \\
b&=.9805442185
\end{aligned}
Note that TheEmtpySet's answer was $y=1.0091e^{-.0196x}$
We can rewrite the answer that the calculator gave us to get his solution.
Note that $b^x= e^{x\ln(b)}$.
Thus, $.9805442185^x=e^{x\ln(.9805442185)}=e^{-.0196475365x}$ | {
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Note that $b^x= e^{x\ln(b)}$.
Thus, $.9805442185^x=e^{x\ln(.9805442185)}=e^{-.0196475365x}$
Rounding the regression equation, we get:
$\color{red}\boxed{y=1.0091e^{-.0196x}}$
Hope this makes sense! | {
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# Is the limit of this infinite step construction an equilateral triangle?
Just for fun (inspired by sub-problem described and answered here):
Let's pick three points on a circle, say $A,B,C$. Move one point ($A$ for example) until the triangle becomes isosceles ($A'BC$) with all angles acute:
Now we have triangle with sides $AB$ and $AC$ equal. Pick any of the two, say $AC$ and move $B$ until the triangle becomes isosceles again, with all angles acute:
Now we have a triangle with sides $AB$ and $BC$ equal. Pick any of the two, say $BC$ and move $A$ until the triangle becomes isosceles again, with all angles acute:
Repeate the same process infinite number of times.
Can we prove that the end result is always an equilateral triangle? It looks so but I might be wrong. I have checked several initial configurations and always ended up with something looking like an equilateral triangle.
• Regarding the title question, here's a more complicated procedure: 1. Prove whether P=NP or not. 2. Draw an equilateral triangle with compass and straightedge (or any other method). – user253751 Sep 18 '18 at 1:32
• @immibis Very funny :) – Oldboy Sep 18 '18 at 5:08
• @user202729 Title changed as suggested. Thanks! – Oldboy Sep 18 '18 at 5:09
• @immibis On second thought, your procedure is not more complicated. Assuming P=NP is true or not the proof will always take finite time. And my “construction” takes infinite number of steps, each step taking constant time. Just kidding :) – Oldboy Sep 18 '18 at 21:39
• However the steps are all the same, which make it pretty simple :) – user253751 Sep 18 '18 at 22:57
Think about what happens to the maximum difference between angles over time.
For simplicity, let's start with an isoceles triangle with angles $x,y,y$. This triangle has "maximum angle difference" $\vert y-x\vert$. Then when we move one of the $y$-angled points, our new triangle will have angles
$$y, {x+y\over 2}, {x+y\over 2}$$ | {
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$$y, {x+y\over 2}, {x+y\over 2}$$
since the angle of the point being moved doesn't change.The maximum difference of angles in this new triangle is
$$\left\vert {y\over 2}-{x\over 2}\right\vert={1\over 2}\vert y-x\vert.$$
So each time we perform this transformation, the maximum angle difference goes down by a factor of two. Whatever the initial value $\vert y-x\vert$ was, this means that the maximum angle difference goes to zero,$^*$ which in turn means that in the limit the angles are equal.
$^*$This is because it's a geometric sequence with ratio in $(-1,1)$ (namely, ${1\over 2}$): if $r\in(-1,1)$ then for any $a$ we have
$$\lim_{n\rightarrow\infty}ar^n=0.$$
Note that it would not have been enough to simply know that the maximum angle difference decreases, since not every decreasing sequence goes to zero!
• I have also noticed that the difference between angles decreases. But having a decreasing sequence does not mean that the sequence converges to zero. You have proved the key point! – Oldboy Sep 18 '18 at 5:18
• Angles $x$ and $y$? Aaah mine eyes! – user332714 Sep 18 '18 at 7:08
By the Inscribed Angle Theorem, moving a vertex around the circle preserves the angle at that vertex. Now, suppose that, at stage $i$, the apex angle is $\theta_i$, so that the base angles are $\frac12(\pi - \theta_i)$. But this apex angle was the base angle of the previous step, giving this recurrence $\theta_{i} = \frac12(\pi-\theta_{i-1})$. Thus, \begin{align}\theta_n &= -\frac12\theta_{n-1} + \frac12\pi \\[6pt] &=\frac12\left(-\frac12(\pi-\theta_{n-2})+\pi\right) = \frac14\theta_{n-2}+\frac12\pi-\frac14\pi \\[6pt] &= \cdots \\[6pt] &= \left(-\frac12\right)^{n}\theta_0 \;-\; \sum_{i=1}^n\left(-\frac12\right)^{n}\pi \\[6pt] \lim_{n\to\infty}\theta_n &= 0\cdot\theta_0 \;-\; \frac{(-1/2)}{1-(-1/2)}\pi \\ &=\frac{\pi}{3} \end{align}
Thus, in the limit, the triangle becomes equilateral. $\square$ | {
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Thus, in the limit, the triangle becomes equilateral. $\square$
Assume WLOG that the initial triangle is isoceles. Let $\alpha$ be the apical angle, and let $\beta$ be a remaining angle. Then the transformation in question sends
$$\begin{bmatrix}\alpha \\ \beta \end{bmatrix}\mapsto \begin{bmatrix}0 & 1 \\ \tfrac{1}{2} & \tfrac{1}{2} \end{bmatrix} \begin{bmatrix}\alpha \\ \beta \end{bmatrix}\text{.}$$ Let $\mathsf{X}$ be the $2\times 2$ transformation matrix on the rhs. $\mathsf{X}$ has characteristic polynomial $x^2-\tfrac{1}{2}x-\tfrac{1}{2}=0.$ By the Cayley–Hamilton theorem, $$\mathsf{X}^2=\tfrac{1}{2}\mathsf{X}+\tfrac{1}{2}\text{.}$$ Therefore we have a Sylvester formula $$f(\mathsf{X})=f(1)\left(\frac{1+2\mathsf{X}}{3}\right)+f(-\tfrac{1}{2})\left(\frac{2-2\mathsf{X}}{3}\right)$$ for any polynomial $f$ (where the matrices in brackets are the Frobenius covariants). Thus, $$\mathsf{X}^n=\frac{1+2\mathsf{X}}{3}+(-\tfrac{1}{2})^n\left(\frac{2-2\mathsf{X}}{3}\right)\text{.}$$ The second term converges to zero, so $$\begin{split} \lim_{n\to\infty}\mathsf{X}^n&=\frac{1+2\mathsf{X}}{3}\\ &=\frac{1}{3}\begin{bmatrix}1 & 2 \\ 1 & 2\end{bmatrix}\\ &=\frac{1}{3}\begin{bmatrix} 1\\ 1\end{bmatrix}\begin{bmatrix}1&2\end{bmatrix}\text{,} \end{split}$$ $$\lim_{n\to\infty} \begin{bmatrix}0 & 1 \\ \tfrac{1}{2} & \tfrac{1}{2} \end{bmatrix}^n \begin{bmatrix}\alpha \\ \beta \end{bmatrix}=\begin{bmatrix}\tfrac{\alpha+2\beta}{3}\\ \tfrac{\alpha+2\beta}{3}\end{bmatrix}\text{.}$$ i.e., the apical and side angles approach equality as the operation is repeated. | {
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# Homework Help: Definite integral of step function
Tags:
1. Jun 18, 2015
### Byeonggon Lee
I need to prove whether this expression is true or false:
$\sum\limits_{k=1}^{n}\int_{k-1}^{k}[x]dx = \frac{n(n-1)}{2}$
I'm so confused because as I know, definite integral is possible only when the target function is continuous in closed interval.
In this case, function $[x]$ should be continuous in interval $[k-1,k]$
but actually [x] is not continuous in $[k,k-1]$
$[x]=k-1$ when $k-1\leq x<k$, $[x]=k$ when $x=k$
So I thought that this expression is false,
but in my book's answer, it is true:
"""
Because$[x]=k-1$ when $k-1\leq x<k$, $[x]=k$ when $x=k$
$\sum\limits_{k=1}^{n}\int_{k-1}^{k}[x]dx$
$=\int_{0}^{1}[x]dx+\int_{1}^{2}[x]dx+\int_{2}^{3}[x]dx+...+\int_{n-1}^{n}[x]dx$
$=\int_{0}^{1}0dx+\int_{1}^{2}1dx+\int_{2}^{3}2dx+...+\int_{n-1}^{n}(n-1)dx$
$=0+\bigg[x\bigg]_{1}^{2}+\bigg[2x\bigg]_{2}^{3}+...+\bigg[(n-1)x\bigg]_{n-1}^{n}$
$=0+1+2+...+(n-1)$
$=\frac{n(n-1)}{2}$
"""
The book also says that I need to think interval as $k\leq x<k+1$
and I don't understand how is it possible to omit equal sign at $x<k+1$
2. Jun 18, 2015
### DEvens
An integral is an area. The troublesome points at the end of the interval mean you are uncertain about the area at a point. What is the difference in the area under a curve if you move one single point up or down by 1?
Another way to approach it is this. Think of the integral as a limit of a sum of areas. It happens that in this case you can very easily construct the limit because the function you are looking at is piece-wise constant.
3. Jun 18, 2015
### Ray Vickson | {
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3. Jun 18, 2015
### Ray Vickson
A definite integral certainly can exist for a discontinuous function. Jump discontinuities in f(x) at an interval's endpoint a or b does not affect the area under the curve y = f(x) from x = a to x = b. In other words,
$$\int_{[a,b]} f(x) \, dx = \int_{(a,b]} f(x) \, dx = \int_{[a,b)} f(x) \, dx = \int_{(a,b)} f(x) \, dx$$
We can denote all four of these by the common symbol $\int_a^b f(x) \, dx$.
4. Jun 18, 2015
### RUber
You can read a little more on this idea at https://en.wikipedia.org/?title=Riemann_integral.
For example, to your point on continuity:
"A function on a compact interval [a, b] is Riemann integrable if and only if it is bounded and continuous almost everywhere (the set of its points of discontinuity has measure zero, in the sense of Lebesgue measure)."
Set of measure zero means a finite (or countably infinite) set of distinct points. In the case of the step function is discontinuous at the integers which for a maximum n is finite, and without a maximum is countably infinite. Both sets are measure zero, so you are okay to integrate.
5. Jun 18, 2015
### WWGD
Just to nitpick, or to add a bit: while in this context of step functions discontinuities are finite, you may have in other cases uncountably-infinite sets of measure zero, e.g., the Cantor set.
6. Jun 18, 2015
### SammyS
Staff Emeritus
I believe your function is also called the floor function, $\displaystyle \ \lfloor x\rfloor \,,\$ and the greatest integer function, $\displaystyle \ [\![ x]\!]\ .\$
On the interval $\displaystyle \ [k-1\,,\ k)\,,\$ we have $\displaystyle \ [x]=k-1\ .$
Take the limit: $\displaystyle \ \lim_{a\to k^-} \int_{k-1}^a [x]\,dx \ . \$ | {
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# $f: \mathbb{R} \to \mathbb{R}$ satisfies $(x-2)f(x)-(x+1)f(x-1) = 3$. Evaluate $f(2013)$, given that $f(2)=5$
The function $f : \mathbb{R} \to \mathbb{R}$ satisfies $(x-2)f(x)-(x+1)f(x-1) = 3$. Evaluate $f(2013)$, given that $f(2) = 5$.
• What math contest is this from? Apr 20, 2013 at 17:38
• @Potato: KöMaL, Problem B. 4538. Apr 28, 2013 at 14:44
• Another example of a question that should have been closed before anybody answered it! Apr 30, 2013 at 11:03
Hint 1: $3=(x+1)-(x-2)$
$(x-2)f(x)-(x+1)f(x-1)=3\Leftrightarrow (x-2)f(x)-(x+1)f(x-1)=(x+1)-(x-2)\Leftrightarrow (x-2)(f(x)+1)-(x+1)(f(x-1)+1)=0$
Hint 2: Is there a function closely related to $f$ that would verify a simpler equation?
$g(x)=f(x)+1$
$(x-2)g(x) - (x+1)g(x-1)=0 \Leftrightarrow (x-2)g(x)=(x+1)g(x-1) \Leftrightarrow g(x)=\cfrac{x+1}{x-2}g(x-1)$
$g(x)=\cfrac{x+1}{x-2}g(x-1)=\cfrac{x+1}{x-2}\cfrac{x+1-1}{x-2-1}g(x-1-1) = \left(\prod\limits_{k=0}^{n-1} \cfrac{x+1-k}{x-2-k}\right) g(x-n)$
Now give the good value to $n$, simplify the product and you'll have the expression of $g$ you need.
• Yeah right. Some terms cancel each other >_< Apr 20, 2013 at 20:42
• What is "the good value"? Apr 28, 2013 at 14:31
• The value sot hat you can evaluate $g(x-n)$ Apr 28, 2013 at 14:39
• Big fan of answers with this format +1 May 1, 2013 at 3:26
• Same here, format+1. May 5, 2013 at 1:11
Let $x-1=y$, then, $$(y-1)f(y+1)-(y+2)f(y)=3$$ $$\implies f(y+1)=\frac{3+(y+2)f(y)}{y-1} \;\;\;\;\;(1)$$
Lemma: $\forall \; n \geq 2 \in \mathbb{N}$, $f(n)=n(n-1)(n+1)-1$.
Base Case: If $n=2$ then $f(n)=1 \cdot 2 \cdot 3 -1=5$ which is true by information provided in the question.
Inductive Step: Assume for $n=k$ that $f(k)=k(k-1)(k+1)-1$.
Then by equation $(1)$, $$f(k+1)=\frac{3+(k+2)(k(k-1)(k+1)-1)}{k-1}$$ $$=\frac{k^4+2k^3-k^2-3k+1}{k-1}$$ $$=\frac{(k-1)(k^3+3k^2+2k-1)}{k-1}$$ $$=k^3+3k^2+2k-1$$ $$=k(k^2+3k+2)-1$$ $$=k(k+1)(k+2)-1$$
Thus completing the induction. | {
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Thus completing the induction.
Hence our Lemma is true and $f(n)=n(n-1)(n+1)-1 \; \forall \; n \geq 2 \in \mathbb{N}$
In particular, if $n=2013$, $f(2013)=2013 \cdot 2012 \cdot 2014-1=8157014183$ (I confess I used a calculator).
P.S I found the lemma by trying small cases.
The conditions allow you to calculate $f(x+1)$ if you know $f(x)$. Try calculating $f(3), f(4), f(5), f(6)$, and looking for a pattern.
Running the following Mathematica program:
f = DifferenceRoot[Function[{f, x}, {(x - 2) f[x] - (x + 1) f[x - 1] == 3, f[2] == 5}]];
f[2013]
We get the answer = 8157014183. | {
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Which of the following complex numbers is equivalent to $\frac{3-5i}{8+2i}$?
I encountered this question in one of my SAT practice tests.
I know the answer is option C, however the only way I got the answer was by trial and error of trying multiple ways to simplify the equation and I ended up rationalising it to get answer choice C.
Is there any other, perhaps easier or more direct method, that I can use to solve these types of questions?
• Mutiplying top and bottom by the conjugate of the denominator is the standard approach. No need for trial and error -- the standard method will always work. – quasi Mar 29 '17 at 6:36
• I ended up rationalising it Don't know that there is anything more direct than that. You don't even need to carry out all calculations, just figure out the sign of the imaginary part to decide between C and D. (That's assuming you discarded choices A, B upfront, as expected) – dxiv Mar 29 '17 at 6:39
• As a multiple choice question, there will be a minus sign, but A is likely to be wrong – Henry Mar 29 '17 at 6:45
• As a multiple choice question ,you can compare $|z|=|\dfrac{3-5i}{8+2i}|=\dfrac{|3-5i|}{|8+2i|}=\dfrac{\sqrt{2(17)}}{\sqrt{4(17)}}=\dfrac{\sqrt2}{2}$ – Khosrotash Mar 29 '17 at 6:56
• @Khosrotash: interesting proposal, but unfortunately it leads to harder computation than the standard way (because of $(7^2+23^2)/34^2$) :-( – Yves Daoust Mar 29 '17 at 7:10
Two ways to approach this problem. First: As quasi suggests in the comments, multiply by the conjugate of the denominator.
\begin{align} \frac{3-5i}{8+2i} & = \frac{3-5i}{8+2i} \times \frac{8-2i}{8-2i} \\ & = \frac{24-40i-6i-10}{8^2+2^2} \\ & = \frac{14-46i}{68} = \frac{7-23i}{34} \end{align}
The second approach, given that it's a multiple choice problem, is to multiply each of the answers by $8+2i$ and see if you obtain $3-5i$. I think the first approach is simpler, but they'll both work. | {
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Multiplying the top and bottom by the complex conjugate is how you handle this. That gives:
$$\frac{3-5i}{8+2i}\cdot\frac{8-2i}{8-2i}=\frac{(3-5i)(8-2i)}{8^2+2^2}=\frac{7-23i}{34}$$
Notice that at the second step we are guaranteed to have a real denominator because $(a+bi)(a-bi)=a^2+b^2$ is always real.
I think the method you are using easier one.
$$\frac{3-5i}{8+2i}$$
Multiply and divide the fraction by conjugate of denominator,
$$\frac{3-5i}{8+2i} \cdot \frac{8-2i}{8-2i}$$
$$\frac{24-6i-40i+10i^2}{64-4i^2}$$
$$\frac{24-6i-40i-10}{64+4}$$
$$\frac{14-46i}{68}$$
$$\frac{7-23i}{34}$$
Alternatives:
Let the answer be $a+ib$. Rewrite the initial equation as $$(a+ib)(8+i2)=3-i5.$$
Expanding, you get the $2\times2$ system $$\begin{cases} 8a-2b=3,\\ 2a+8b=-5.\end{cases}$$
Then by Cramer or simply adding four times the first equation and the second
$$34a=7$$
and subtracting the first from four times the second,
$$34b=-23.$$
Get rid of the denominators and try the products
$$(3\pm i20)(8+i2)=24\mp40+i(6\pm160)\to64-i154,$$ $$(7\pm i23)(8+i2)=56\mp46+i(14\pm144)\to102-i170.$$
(We select the signs that match those of $3-i5$.)
Then we obtain the identity
$$(7-i23)(8+i2)=34(3-i5).$$
The real way:
Use the division formula
$$\frac{a+ib}{c+id}=\frac{ac+bd}{c^2+d^2}+i\frac{bc-ad}{c^2+d^2},$$
giving
$$\frac{14}{68}-i\frac{46}{68}.$$
Then compare to the proposed answers. As C seems to match but D is similar, double check the signs or try the product, for safety. | {
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# Regular average calculated accumulatively
is it possible to calculate the regular average of a sequence of numbers when i dont know everything of the sequence, but just everytime i get a new number i know the total count of numbers and the average for the numbers - 1.
for example: 2 3 10 the average is of course: 5
but in the last step to calculate i only have access to the previous average of 2 and 3: 2.5 the next number: 10 and the count of numbers: 3
if this is possible, how?
Yes, and you can derive it from the expression for the average. Let the average of the first $n$ numbers be $\mu_n$. The formula for it is
$$\mu_n = \frac{1}{n} \sum_{i=1}^n x_i$$
Then you can derive
$$n \mu_n = \sum_{i=1}^nx_i = x_n + \sum_{i=1}^{n-1} x_i = x_n + (n-1)\mu_{n-1}$$
and hence, dividing by $n$,
$$\mu_n = \frac{(n-1) \mu_{n-1} + x_n}{n}$$
i.e. to calculate the new average after then $n$th number, you multiply the old average by $n-1$, add the new number, and divide the total by $n$.
In your example, you have the old average of 2.5 and the third number is 10. So you multiply 2.5 by 2 (to get 5), add 10 (to get 15) and divide by 3 (to get 5, which is the correct average).
Note that this is functionally equivalent to keeping a running sum of all the numbers you've seen so far, and dividing by $n$ to get the average whenever you want it (although, from an implementation point of view, it may be better to compute the average as you go using the formula I gave above. For example, if the running sum ever gets larger than $10^{308}$ish then it may be too large to represent as a standard floating point number, even though the average can be represented). | {
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• As you're computing the previous sum $(n-1)\mu_{n-1}$ as part of the formula, I think this wouldn't help is the sum gets too large. Jun 23, 2016 at 1:12
• @danijar Completely true - a better approach is to keep running sums of the $x_n$ using a method that is robust to rounding error (e.g. Kahan summation) and a separate running count of $n$, and divide whenever you need the mean. That way, your error is bounded by the accuracy of your floating point type. Jun 23, 2016 at 7:44
• If you are likely to end up with numbers larger than $10^{308}$ then you either need to scale down your inputs, or use a more capacious floating point type. Jun 23, 2016 at 7:46
A very simple thought process results in the same formula for running average. If you have $N$ previous measures (of course the measures could all be different) the average you calculate is exactly the same as if all measures were the same as the computed average value. Then, computing the running average of the $N+1$ is equal to $N$ times the previously computed average plus the $N+1$ measure all divided by $N+1$. I know that this is the same as the formula posted in the other answer but no derivation with sums is needed or more obscure mathematical thinking is needed (OK, maybe not really obscure).
What you are asking for is commonly called sequential estimation. A general approach is described in [Robbins, H. and S. Monro (1951). A stochastic approximation method. Annals of Mathematical Statistics 22, 400–407.]
To add to the derivation of Chris Taylor, I personally like this rewriting as it goes quite intuitively (easy to remember). $$\mu_n = \mu_{n-1} + \frac{1}{n}(x_n - \mu_{n-1})$$
# algorithmically: sequential average computation
avg += (x_n - avg)/n | {
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## Monday, July 27, 2009
### The Broken Stick Experiment
I found several interesting video lectures at the BLOSSOMS Video Library. One in particular that I enjoyed is called The Broken Stick Experiment [Flash]. In it, Professor Richard Larson asks the seemingly simple question,
If you break a stick randomly in two places, what is the probability that you can form a triangle from the three pieces?
If one of the pieces is too long, then the other two can't meet to form the triangle.
The lecture is only 33 minutes, and Professor Larson builds up a very pretty geometric proof, so I won't spoil it by posting the answer here. I will provide code for a simulation, though (yes, this is still a programming blog). See if you can guess the answer before running the simulation or watching the video.
import java.util.Arrays;import java.util.Random;public class BrokenStick { public static void main(String[] args) { Random random = new Random(); int trials = 10000; int triangles = 0; double[] breaks = new double[2]; double[] sides = new double[3]; for( int i=0; i < trials; ++i ) { breaks[0] = random.nextDouble(); breaks[1] = random.nextDouble(); Arrays.sort(breaks); sides[0] = breaks[0]; sides[1] = breaks[1] - breaks[0]; sides[2] = 1.0 - breaks[1]; Arrays.sort(sides); if( sides[2] < (sides[0] + sides[1]) ) { triangles++; } } System.out.println("Triangles: " + triangles); System.out.println("Trials: " + trials); double p = (double)triangles / trials; System.out.println("Proportion: " + p); }} | {
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The simulation breaks 10,000 sticks each into three random pieces using Java's Random class to come up with random lengths. If the longest piece of broken stick is longer than the two remaining pieces, a triangle can't be formed. Ten-thousand trials should be enough to accurately estimate the proportion of triangles made from the broken sticks to two decimal places. How close was your guess?
Related:
The Broken Stick Revisited
rbonvall said...
Thanks, I had a fun time solving it :)
At least to me, this problem lends itself nicely to a geometric solution. In the a-b-c space, the possible breaks must satisfy a+b+c=1, which is a plane. Then, it reduces to find the fraction of the "positive part" of the plane that satisfies the constraints. And I always have fun drawing intersecting planes :P
Bill the Lizard said...
rbonvall,
"And I always have fun drawing intersecting planes :P"
Your comment reminds me of the first lecture (or maybe the second, I can't quite remember) in MIT's Linear Algebra video series. Gilbert Strang shows several methods for solving systems of equations, and makes a complete mess of drawing intersecting planes. He uses that as an illustration of why solving systems of equations using matrices is far superior. :)
John | Retro Programming said...
Thanks, it's an interesting problem. I think I've got the solution, I just need to run a program to double check my result :-)
Alexey Busygin said...
I think it should be 33,3%.
Bill the Lizard said...
Alexey,
It's not 33%, but your initial guess is closer than mine was.
Trevel said...
I estimate at just under 25% triangle. Trying to figure if there's a flaw in my logic, but I can't pick it out...
tg said...
I agree with Trevel. Independent breaks which both must occur on the same half of the stick. <25%
tg said...
Got the right answer, but not sure if the logic is right: | {
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tg said...
Got the right answer, but not sure if the logic is right:
If two breaks occur on the same half of the stick, then on piece is longer than the sum of the other two. These are independent events. The probability of one break on 50% of the stick is 1/2, the probability of two (independent) breaks on the same half is 1/2*1/2 = 0.25.
Bill the Lizard said...
tg,
Wouldn't that indicate that there's a 25% chance of not forming a triangle?
tg said...
back to the drawing board....
Bill the Lizard said...
tg,
Don't feel bad, you almost had me convinced. It took me ten minutes to spot the flaw, and I've watched the video twice! :)
tg said...
How about one more try using the same concept (except using it correctly to represent all three lengths, as I didn't do the first go round):
Probability of NOT forming P(NF) =
P(a > b+c)*P(b>a+c)*P(c>b+a)
As mentioned earlier P(a>b+c) = probability of two breaks on 1/2 of the stick = 0.25.
P(NF) = 0.25*0.25*0.25 = 0.75
Thus P( NOT NF) = 0.25
Bill the Lizard said...
tg,
0.25 * 0.25 * 0.25 = 0.015625
Back to your original statement, though, you said "If two breaks occur on the same half of the stick, then on piece is longer than the sum of the other two." This is correct, but it's only one way that you could get one piece that's longer than the other two. You could also break the stick on opposite ends of the stick and have the middle piece be too long.
tg said...
B the L, you are a patient individual.
One final attempt to redeem myself. The "*" was supposed to be "+". The risks of writing at work.
We are looking for the Probability of (a>b+c) OR (b>a+c) OR (c>a+b).
a + b + c = L
P(a>b+c) = P(a>L/2)
For a>L/2, two breaks must occur on the right half:
P(of a break occurring on the right half) = 0.5
P(of two independent breaks occurring on the right)=
P(a>L/2) = 0.5*0.5 = 0.25 | {
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For b>L/2, two independent breaks must occur on either the left quarter, the right quarter, or both breaks on the left quarter, or both on the right quarter. The probability of ONE break landing on that 50% of the stick = 0.5. The probability of another break independently landing on the same 50%of the stick = 0.5*0.5 = 0.25.
c is the same as a, so
P(c>b+a) = P(a>b+c) = 0.25
For any one length > L/2 = (a>b+c) OR (b>a+c) OR (c>a+b) = 0.25+0.25+0.25.
If this is wrong, spare me the indignity and just delete my comments.
Bill the Lizard said...
tg,
Now you have it! I like the approach you used because you're showing the exact opposite of what the problem asks for, the probability of NOT forming a triangle. This kind of thinking is a very powerful tool in mathematics, and is often overlooked.
Sam152 said...
Very interesting post, my nifty c++ program calculated about 18.9% successes after 10,000 tries. How close is this?
Bill the Lizard said...
Sam152,
That's a little bit low; lower than rounding error would explain. Do you want to share your code? You could either post it here, or as a Stack Overflow question and I'd be happy to take a look.
fiberfiend6891 said...
Just a heads up - the link directly to the video has http doubled and doesn't work when clicked directly (I had to copy/paste and get rid of the double):
http://http//blossoms.mit.edu/video/larson-watch.html
This is my first time seeing your blog, and I absolutely LOVE it!
Bill the Lizard said...
fiberfiend6891,
Tom Raywood said...
Bill the Lizard,
Hi. Great question.
There is one thing I'm not clear on, though, from the description. Do obtuse triangles count here, or is the solution limited to acute?
Bill the Lizard said...
Tom,
You could end up with an acute, obtuse, or (rarely) equilateral triangle. You've got me curious now, so I'll have to modify the code later and run another simulation to see if acute or obtuse triangles are more common. Good question!
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Tom Raywood said...
Bill the Lizard,
Good. If you limit outcomes to acute triangles, (including equilaterals), what probability do you get?
Oh, and another question [intended to confirm an assumption on my part]: Is this strictly an ordered set or can any of the three pieces form the base?
Bill the Lizard said...
Tom,
Any of the three sides could form the base (I'm assuming by base you mean the longest side). In the code I gave you can see how I work around this by calculating the length of the three sides, then immediately sorting them (on line 24) so that I always know where the longest side ends up.
I'm working on an entire post to answer your other question. :)
Tom Raywood said...
Bill the Lizard,
Gee, no I hadn't imagined that the base of the triangle had to be the largest of the three pieces. All of these unstated parameters make it very difficult to approach the problem in a systematic way.
So far, yes, the triangle can be of any sort, the pieces can be moved around at will and, finally, (if I understand your last answer), the base of the triangle is expected to be whichever piece [of the stick] that's the longest.
Is this correct? Is there anything else in the way of allowances and/or limitations?
Bill the Lizard said...
Tom,
Disregard my last answer, as I had misunderstood your question. There are no constraints on what kind of triangle you can get (acute, obtuse, right, or no triangle at all) or on the order of the pieces. Any of the pieces can be the longest when you randomly select the two break-points. The only reason I sort the pieces in my code is because I need to know which piece is longest, so that I can tell if the pieces form a triangle or not. This has no impact on the final outcome.
Tom Raywood said...
Bill the Lizard,
Whoa, hey, no, it can't be as simple as that can it? | {
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Tom Raywood said...
Bill the Lizard,
Whoa, hey, no, it can't be as simple as that can it?
All it takes for the three pieces to form a triangle is that none of the pieces have a length greater than or equal to 1/2 the length of the stick. And that's a full 50% of the time.
What gives? I thought certainly the question would pose a greater challenge than that.
Bill the Lizard said...
Tom,
Close. "All it takes for the three pieces to form a triangle is that none of the pieces have a length greater than or equal to 1/2 the length of the stick." This much is true, but that doesn't happen a full 50% of the time (which was also my initial guess before I watched the full lecture).
Tom Raywood said...
Bill the Lizard,
So true. (I'm feeling better already.) After 100 simulations with n=1,000,000 I'm seeing slightly less than 50%. More specifically I'm seeing 49.994067% and am definitely curious as to what explains this slight deviation. I think I'd rather contemplate it for a while though, that is, to see if I can figure it out myself.
Tom Raywood said...
Bill the Lizard,
To follow up here, beginnings of a guess are as follows: For some small range [in the form of a difference], as two legs of the triangle both approach 1/2, the small difference between them cannot be bridged by the infinitesimally small remaining leg. Be nice to make a formal statement.
Bill the Lizard said...
Tom,
Check your simulation code against mine. Your answer is off by quite a bit more than just rounding error. You can post your code here, or on Stack Overflow if you'd like me to take a look at it (provided it's in a language I know reasonably well, Java, C, C++, Python, PHP, or any dialect of Basic).
Tom Raywood said...
Bill the Lizard,
Definitely courting tedium at this point.
n R1 R2 R3 T P1 P2 P3 C1
These represent column headings in Excel.
R1, R2 and R3 denote random numbers using the RANDBETWEEN function; I chose for each number the range from 1 to 10,001. | {
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T denotes the sum of R1, R2 and R3 for each row.
P1, P2 and P3 denote which percentage of T is entailed of R1, R2 and R3 respectively.
C1 (for condition 1) represents a check to see whether P1, P2 and P3 all three entail percentages less than 0.5, posting a 1 if they do and a 0 if they don't.
n denotes the number of iterations which, as I said, I took the time to set at 1 million.
Totaling the C1 column provides a sum which, divided into 1 million, quite closely approximates the probability that no 'leg of the triangle' will be greater than or equal to one half.
Additionally I generated a macro which recalculates the worksheet 100 times, posts each result to a table and then finds the average of that set.
You can imagine my surprise to hear that the actual probability varies significantly from what this arrangement points up.
I'll take a look at that lecture this weekend.
Bill the Lizard said...
Tom,
From your explanation it looks like you're generating three random lengths to represent the three pieces of stick. The problem with this approach is that there's nothing to make sure the three pieces all add up to the original length of the stick.
Try it by generating only 2 random numbers (x and y) that represent the break points. If 10,000 is the length of the original stick, and you pick 2 random numbers between 0 and 10,000 then the lengths of the three pieces will be x, y-x, and 10,000-y when x < y, or y, x-y, and 10,000-x when x > y. (It's easier if you just select two random numbers then sort them so x < y, then just always use x, y-x, and 10,000-y for the three lengths, which is essentially what I did in my code.)
I'll try to get my next post up this weekend. I have the answer to your question about acute and obtuse triangles, but proving why I'm getting the number I'm getting is proving to be a little bit complicated.
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Tom Raywood said...
Bill the Lizard,
I'll work this up and see what gives. It's definitely counterintuitive though that always having the same stick length would even matter, that is, any more than it would matter that triangle of type Q presents as q1, q2, ...qn for n of any size. Be nice to hear the why behind your assertion. Maybe the lecture will help in this regard.
Bill the Lizard said...
Tom,
The particular value of the stick length doesn't matter as much as making sure the three pieces add up to whatever length you decide on. I used a stick length of 1.0 in my similation, but it doesn't hurt anything if you choose a length of 10,000.
What really makes a difference is how you choose your random lengths. If you choose three random lengths then add them together, you don't really know the length of the stick you started with. If you start with a specific length, then make two random breaks within the boundaries of the stick, you're guaranteed to end up with three random lengths that add up to your original stick length.
It's possible that I just didn't correctly understand the explanation of your spreadsheet. Without seeing the formulas you used I can't be sure that this part of your simulation isn't correct. The errors you're getting could be from another part of the calculation, but this is the most logical place to start looking.
Tom Raywood said...
Bill the Lizard,
I suspect most people would anticipate that having a consistent or particular stick length shouldn't matter here. What's at issue is where the two breaks appear, that is, relative to overall length. A million sticks all of different lengths and all randomly broken in two places should demonstrate the same probabilistic characteristics of a million sticks all of the same length because, again, what's at issue is relative in any event. | {
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This principle can be demonstrated as follows. In my simulation each stick length is represented by T, that is, the sum of R1, R2 and R3. If I wanted/needed to ensure that each and every occurrence of T always comes out to equal a certain value, a few more columns could be inserted which accomplish this simply by adjusting the sizes of R1, R2 and R3 by some definitive amount. For example, if for one row T presented as equal to 16,000 [compared to, say, the value of 21,000 I'd predetermined as a uniform stick length], obviously all I'd have to do is increase each of the original random numbers by 21/16.
So even though now all million stick lengths are the same, the actual percentage of that length occupied by each R1, R2 and R3 hasn't changed at all. The outcome would still match the 50% probability that my simulation currently produces.
In short, Woe to The Foe, heh heh heh.
Here are the formulas I employ for...
...{R1,R2,R3}
................ =RANDBETWEEN(1,10001)
..........{T}
................ =SUM(E2:G2), because R1, R2 and R3 occupy columns E, F and G.
...{P1,P2,P3}
................ =E2/H2; =F2/H2; and =G2/H2, because T occupies column H.
.........{C1}
................ =IF((J2<0.5)*(K2<0.5)*(L2<0.5),1,0), because P1, P2 and P3 occupy columns J, K and L.
And fun was had by all.
Spartacus said...
1. Probability that both cuts are on opposite sides of the midpoint: 1/2
2. Probability that two cuts on opposite sides of the midpoint are less than half the stick length apart: 1/2
Therefore the probability of forming a triangle is 1/2*1/2=1/4
Here is a more difficult problem with an even more surprising result: You break the stick into two pieces, then randomly choose one of the two pieces and break it into two pieces. What is the probability that the resulting three pieces can form a triangle?
You will be amazed by the solution!
(Hint: the answer is not just a simple fraction like the original problem; however there is a very nice closed form) | {
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Bill the Lizard said...
Spartacus,
I've read about half-a-dozen variations on this problem (mostly linked from the notes on the BLOSSOMS page) but I haven't seen that one. This might take a while. :)
Anonymous said...
Sam152's solution is correct; it is the answer the Spartacus's variant.
Bill's original problem statement was under-specified, admitting two interpretations that yield different answers. (Do you choose two cutpoints independently and uniformly on the original stick; or do you make one uniform random cut, and then make another uniform random cut on one of the pieces?)
See http://www.cut-the-knot.org/Curriculum/Probability/TriProbability.shtml
Anonymous said...
This issue might be very old and might already been resolved. but just for clarification your value for second side is wrong and should be taken as follows:
breaks[1] = random.nextDouble()*(1-breaks[0]);
With this correction, the probability comes to 0.193 which is the correct solution.
Bill the Lizard said...
Anonymous,
No, that's incorrect. The second break can take place at any point along the length of the stick. Your change restricts it to only the length of the first piece broken off.
Anonymous said...
Ok yes you are correct, this imposes a restriction and thus reduces the probability. the correct answer is 0.25 only and not 0.193.
thanks.
asha rani said...
Is it ok for each triangle n What is n?? | {
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# Modular arithmetic - efficiently calculating the remainders of factorials
When working on this question regarding the divisibility of the sum of factorials, I decided to write some code to test "small values" of the problem using the following code.
f[p_] := Total[Mod[#!, p] & /@ Range[p - 1]];
Table[Mod[f@Prime@i, Prime@i], {i, 1, 500}]
Basically, what the code does is sum up all the factorials $$1!+2!+3!+\dots+(p-1)!$$
and find the remainder modulo $p$, for prime $p$.
Unfortunately, my code as written takes a very long time to run. Checking the first 500 primes takes 88.280966 seconds on my computer, but checking the first 2000 primes took me about 4 hours.
Is there any way to improve the code, or is it already the best we can do?
As for optimizations not involving the code, I used Wilson's Theorem, which states that for all primes $p$,
$$(p-1)!\equiv-1 \bmod p$$
Using the above theorem, we can modify the code as follows.
h[p_] := Total@Flatten[{Mod[#!, p], PowerMod[(# - 1)!*(-1)^(#), -1, p]} & /@ Range[(p - 1)/2]];
Table[Mod[h@Prime@i, Prime@i], {i, 1, 500}]
This is considerably faster than the previous code, since checking the first 500 primes takes only 25.896166 seconds. However, checking the first 2000 primes still takes an inordinately long time.
This is bit faster:
toPrime = 500;
sums = Accumulate@FoldList[Times, 1, Range[2, Prime@toPrime - 1]];
primes = Prime[Range[toPrime]];
Mod[sums[[primes - 1]], primes]
Precompute factorial sums and primes. Mod is fast on lists. | {
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Precompute factorial sums and primes. Mod is fast on lists.
• you are way too humble! calculating your sum for even the first 2000 primes takes less than a second. However, is there a way to get around storing large numbers in sums in memory? It keeps crashing my computer when I try toPrime=5000. – Vincent Tjeng Mar 26 '13 at 6:03
• +1 (that's freaking fast!) Can you explain why Accumulate@FoldList[#1 #2 &, 1, Range[n] is so much quicker to Accumulate@Array[#! &, n] + 1? I really don't get it. – gpap Mar 26 '13 at 11:23
• @gpap Calculating factorial so many times costs a lot. Since we know we want all the factorials up to Prime[toPrime]-1, we ultimately gain a lot keeping the intermediate results with FoldList. – Michael E2 Mar 26 '13 at 12:04
• @MichaelE2 Yes, worked it out myself in the meantime - you multiply the previous result and don't calculate a factorial at every step. Thanks – gpap Mar 26 '13 at 12:11
• Is there any reason why you used #1 #2 & instead of Times? – J. M. is away Jun 15 '15 at 12:53
Let $x \equiv r_1 \bmod p$ and $y \equiv r_2 \bmod p$. Then, $x y \equiv r_1 r_2 \bmod p$. So, we can compute the sum of the factorials mod p using:
f[p_] := Mod[ Total @ FoldList[ Mod[Times[##], p]&, Range[p-1]], p]
Let's compare this to the naive implementation:
t[p_] := Mod[Sum[k!, {k, p-1}], p]
For example:
f[Prime[500]] //AbsoluteTiming
t[Prime[500]] //AbsoluteTiming
{0.00058, 1813}
{0.085628, 1813}
The nice thing about using Mod[Times[##], p] as the FoldList function is that the output should be a machine number unless you are working with very large primes. That means that f can be compiled:
fc = Compile[{{p, _Integer}},
Mod[ Total @ FoldList[ Mod[Times[##], p]&, Range[p-1]], p],
RuntimeAttributes->{Listable}
];
Let's compare for a large prime:
f[Prime[10^6]] //AbsoluteTiming
fc[Prime[10^6]] //AbsoluteTiming
{1.83041, 9308538}
{1.05449, 9308538} | {
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{1.83041, 9308538}
{1.05449, 9308538}
Faster, but the real difference is that fc is Listable. Hence, comparing timings on a list shows a much larger difference. For example:
r1 = f /@ Prime[Range[5000, 5500]]; //AbsoluteTiming
r2 = fc[Prime[Range[5000, 5500]]]; //AbsoluteTiming
r1 === r2
{2.06691, Null}
{0.395402, Null}
True
Finally, since the list of all factorials is not stored anywhere, the memory used is much more manageable. Here is the memory and timing for the first 5000 primes:
r1 = fc[Prime[Range[5000]]]; //MaxMemoryUsed //AbsoluteTiming
{3.03463, 462848}
This is quite a bit smaller than @MichaelE2's answer:
MaxMemoryUsed[
toPrime = 5000;
sums = Accumulate@FoldList[Times, 1, Range[2, Prime@toPrime - 1]];
primes = Prime[Range[toPrime]];
r2 = Mod[sums[[primes - 1]], primes]
] //AbsoluteTiming
r1 === r2
{2.40441, 4064607008}
True
The compiled answer uses .462KB while the approach where all the factorials are precomputed takes 4GB, and the timing is not too different. | {
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# While plotting polar coordinate system graphs, should we include the negative of angle $\theta$?
While plotting polar coordinate system graphs, should we include the negative of angle $\theta$?
For example, while plotting $r=\theta$ for negative values the spiral comes out to be reverse of the graph plotted using positive values. Upon searching in google I find only one spiral. So are both spirals part of the curve?
• There's no reason to omit negative values of $\theta$, other than that the pictures are sometimes less pretty. – Aaron Montgomery Oct 30 '17 at 17:46
• Negative values of $\theta$ are perfectly fine. Often times when doing integrals in polar coordinates, it is convenient to use a parameterization with negative values rather than positive ones. – superckl Oct 30 '17 at 17:49
It's not really about $\theta$, which of course can be both positive or negative. Rather it's about the values of $r$ — should we allow negative values of $r$ or not? And this is not really a mathematical question, but rather a matter of convention. Most textbooks I've seen allow both positive and negative values of $r$. However, some require that $r$ must be non-negative ($r\ge0$). | {
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# How many points uniquely define a square?
I was wondering how many points uniquely define a square. Now it is clear to me that if we have $$n$$ random points on a square then this does not necessarily uniquely define the square (for example they may all lie on the same edge).
My question was more the following: supposing I was able to choose exactly which points of my square to use, what is the minimum number of points I could pick to uniquely define the square, and which points should I pick?
Edit: the points need not lie on the vertices.
• Isn’t it 3? Pick three corner points. Jun 27 at 18:17
• @EDX Which four points are enough? If you pick the four corners, then I can for example draw a different square for which those four points lie on the midpoints of the sides, so that's not enough. Jun 27 at 18:23
• @EDX I think it depends on how we understand the question. If we take a square and pick the four corner points, there will actually be infinitely many squares containing those points, although they won't be in the corners. Jun 27 at 18:24
• @subrosar yes the points need not lie on the corners
– JLB
Jun 27 at 18:24
• I can not think of a way to have four points uniquely define a square, but five points should suffice. so long as exactly one set of the three points are colinear. From the three that are colinear we can tell the orientation of the square as any other edges must come at 90 degree angles to it. From there you should be able to draw lines which the edges will fall upon. You should be able to then find the distance between opposite lines which will be the edge length and you should be able to complete the square from there. Jun 27 at 18:37
Four points can be enough to unambiguously define a square.
Take two of the corners, an additional point from the edge between those corners, and an additional point from the middle of the edge opposite the edge the first three points lied on. | {
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A second observer coming to look at these points having been told they all lie on the same square will learn from the three colinear points the orientation of the square and that these lie on the same edge and additional edges will either be perpendicular or parallel to this.
They will then note that since if drawing a perpendicular from the fourth point to this edge it lands in the middle of the edge formed by our colinear points that this must be on the opposite edge. Next, since they know this opposite edge must be parallel to the original edge we now know the lengths of the edges must be equal to this minimum distance between our fourth point and the edge formed by our three colinear points.
Finally, having noted that this length is equal to the length between our colinear points that these two must have been corners of our square.
• That's easier than my construction; well done! Jun 27 at 18:58
Four points are enough, if chosen correctly.
In particular, I claim that the only square which passes through the four points $$A(0,0)$$, $$B(0,0.3)$$, $$C(0.4,0)$$, and $$D(1,1)$$ is the unit square with corners $$(0,0), (0,1), (1,0), (1,1)$$. The solution lets you find many more similar examples.
To prove this claim, here are some inequalities about the distances between points on a square with side length $$s$$:
• Two points on the same side are at a distance between $$0$$ and $$s$$;
• Two points on adjacent sides are at a distance between $$0$$ and $$\sqrt2 s$$;
• Two points on opposite sides are at a distance between $$s$$ and $$\sqrt 2 s$$.
We can check that if $$A,B,C,D$$ lie on a square, then they can't all lie on two adjacent sides of the square: there's no two lines that contain all four points and intersect at right angles. (It's enough to check that $$AB$$ is not perpendicular to $$CD$$, $$AC$$ is not perpendicular to $$BD$$, and $$AD$$ is not perpendicular to $$BC$$.) So there must be some pair of points on opposite sides. | {
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The six distances between the points are $$AB=0.3$$, $$AC=0.4$$, $$BC=0.5$$, $$CD \approx 1.16$$, $$BD \approx 1.22$$, and $$AD \approx 1.41$$. So we see that if any two of $$A,B,C$$ were on opposite sides, the side length $$s$$ would be at most $$0.5$$, but $$D$$ is more than $$0.5 \sqrt2 \approx 0.71$$ away from all the rest. Therefore $$A,B,C$$ are all on two adjacent sides of the squares.
In particular, one side of the square must contain two of $$A,B,C$$. Can a side of the square contain $$B$$ and $$C$$? No: line $$BC$$ separates $$A$$ from $$D$$, which a side of the square can't do. If we assume $$A,B$$ are on one side of the square and $$C$$ is on an adjacent side, or that $$A,C$$ are on one side of the square and $$B$$ is on an adjacent side, we get the same conclusion: the square has a corner at $$A$$ with one side containing segment $$AB$$ and one side containing segment $$AC$$.
Since $$D$$ does not lie on either line, it must be on one of the other sides of the square: a side parallel to $$AB$$, and a side parallel to $$AC$$. We get the same square and try both.
• If we draw line $$AB$$, line $$AC$$, and a line through $$D$$ parallel to $$AB$$, those lines must contain three sides of the square. The only such square is the unit square.
• If we draw line $$AB$$, line $$AC$$, and a line through $$D$$ parallel to $$AC$$, those lines must contain three sides of the square. The only such square is the unit square.
Three points are never enough.
Take three points $$A,B,C$$. If they form a right triangle, we can draw ever-bigger squares with that right angle as a corner, containing all three. If they form an obtuse triangle, that's even better; suppose without loss of generality that the altitude from $$C$$ lands on line $$AB$$ at a point $$H$$ outside segment $$AB$$. Then we can draw ever-bigger squares with one corner at $$H$$ containing all three points. | {
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For an acute triangle, begin by rotating the points arbitrarily. Let $$x_{\min}, x_{\max}, y_{\min}, y_{\max}$$ be the lowest and highest $$x$$- and $$y$$-coordinates among the three points; exclude the finitely many orientations where there are ties. Draw the four lines $$x = x_{\min}, x=x_{\max}, y = y_{\min}, y = y_{\max}$$; each of them passes through one point and together they contain all three (otherwise we'd get an obtuse angle). One of the points lies on two of the lines.
Without loss of generality, $$A$$ lies on $$x = x_{\max}$$ and $$y = y_{\max}$$ and also $$x_{\max} - x_{\min} \ge y_{\max} - y_{\min}$$. Then erase the line through $$y = y_{\max}$$ and instead draw the line through $$y = y_{\min} + x_{\max} - x_{\min}$$. These four lines define a square containing all three points. Since we get this for all but finitely many ways to rotate $$A,B,C$$ (equivalently, for all but finitely many choices of a horizontal and vertical direction) we get infinitely many squares.
• Ahhhhh now I see why three doesn’t suffice. Jun 27 at 20:19
If you specify what points you chose, then two are sufficient: choose opposite vertices. A diagonal can then be drawn between them, and a perpendicular bisector of the same length can be drawn, with the ends being the other two vertices. Then the sides can be drawn between them. | {
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• Sometimes the simple answer is the correct answer… “supposing I was able to choose exactly which points of my square to use, what is the minimum number of points I could pick”
– Matt
Jun 28 at 7:27
• @Matt There is still some ambiguity. If Alice wants to communicate to Bob what square she has, two points being sufficient requires not only that Alice can choose which points to tell Bob, but that Bob knows that Alice chose those points. If all that Bob knows is that these two points are on the square, but doesn't know that they are opposite vertices, then Bob will not know which square they're from. Jun 29 at 0:55
• I understand your point, but also Bob knows it is a square and not some other completely irregular shape (otherwise the problem would be impossible). So it is reasonable Bob could know the two points are opposite vertices
– Matt
Jun 29 at 6:42
• Acccumulation : you require some communication between two persons must succeed and Matt : you assume such persons are perfectly reasonable in optimizing. But OP only is about one person figuring it out on its own. If that person knows about his own invented conventions (one could claim that to be cheating), two points would be enough. Jul 3 at 22:30
Two points are enough if they are ordered - a corner A, and the corner B clockwise from A. This is the minimum, as a square has 4 degrees of feeedom - center, size, and orientation.
A slighly different question is: whats the minimum number of points such that one and only one square will lie on them. 5 is enough - four corners and somewhere on an edge. Four might be enough. | {
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## anonymous one year ago Does the series $\sum_{n\ge1}\frac{1}{\text{lcm}\{n,n+1\}}$converge?
1. anonymous
Using the fact that $$\text{lcm}\{n,n+1\}=\dfrac{n(n+1)}{\text{gcd}\{n,n+1\}}$$, I'm almost tempted to say that $\sum_{n\ge1}\frac{1}{\text{lcm}\{n,n+1\}}\sim\sum_{n\ge1}\frac{1}{n^2}$ but I don't think I can jump to that conclusion just yet.
2. mathmate
We agree that lcm(n,n+1) is actually n(n+1). (The gcd is 1, by Euclid's algorithm) Since S=$$\sum_{n\ge1}\frac{1}{\text{lcm}\{n,n+1\}}=\sum_{n\ge1}\frac{1}{n(n+1)}<\sum_{n\ge1}\frac{1}{n^2}$$ The strict inequality holds for each and every term, and since $$\sum_{n\ge1}\frac{1}{n^2}$$ is a geometric series that is known to converge (=pi^2/6 prove it), so by comparison, S<pi^2/6 so S converges.
3. mathmate
* series, not geometric
4. anonymous
Right, since $$n$$ and $$n+1$$ are consecutive, that makes their lcm very easy to determine. And we can find the value of the series while we're at it quite easily: $\sum_{n\ge1}\frac{1}{n(n+1)}=\sum_{n\ge1}\left(\frac{1}{n}-\frac{1}{n+1}\right)=1$Neat!
5. ganeshie8
Nice! any two consecutive integers are coprime, gcd(n, n+1) = gcd(n, 1) = 1 so lcm(n, n+1) = n(n+1)
6. anonymous
consider that the least common multiple of $$n,n+1$$ is obviously $$n(n+1)$$ since $$n+1-n=1$$ and thus they must be coprime. this reduces to $$\sum_{n=1}^\infty\frac1{n(n+1)}=\sum_{n=1}^\infty\frac{n+1-n}{n(n+1)}=\sum_{n=1}^\infty\left(\frac{n+1}{n(n+1)}-\frac{n}{n(n+1)}\right)\\\quad =\sum_{n=1}^\infty\left(\frac1n-\frac1{n+1}\right)$$ which is obviously telescoping and since $$1/(n+1)\to0$$ it converges
7. imqwerty | {
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7. imqwerty
well i read that a series that converges has got some bounds and that it approaches a specific value. so in this case its going like 1/2 , 1/6, 1/12 , 1/20..... so u mark a line of length 1 on a number line and u plot the point 1/2 u knw 1/2 is still left between 1/2 and 1 and then u add 1/6 still 1/3 is left so u have bounds and u knw that it can never go > 1 so yea it converges... https://en.wikipedia.org/wiki/Convergent_series
8. anonymous
Could we make a more general claim here? Numerically, it would seem that $\sum_{n\ge1}\frac{1}{\text{lcm}\{n,n+k\}}$also converges to $$1$$ for $$k\in\mathbb{N}$$, though much more slowly.
9. anonymous
for prime $$k$$ the problem is barely any harder -- you merely have to be careful about the multiples of $$k$$: $$\sum_{n=1}^\infty\frac1{[n,n+k]}=\sum_{m=1}^\infty\sum_{n=mk+1}^{(m+1)k}\frac1{[n,n+k]}\\\quad=\sum_{m=1}^\infty\left(\sum_{n=mk+1}^{mk+k-1}\frac1{n(n+k)}+\frac1{(m+1)(m+2)k}\right)$$
10. ParthKohli
$\gcd (n,n+k) \le k$$\Rightarrow {\rm lcm }(n,n+k) = \frac{n(n+k)}{\gcd(n,n+k)} \ge \frac{n(n+k)}{k}$$\Rightarrow \frac{1}{{\rm lcm} (n,n+k) }\le \frac{k}{n(n+k)}$We can sum the right-side telescopically so I guess it is true that the sum is convergent?
11. anonymous
yeah, showing its convergence along those lines is exactly what I figured
12. anonymous
the sum of the terms for $$n$$ coprime to $$k$$ is accomplished with relatively little effort $$\sum_{n=mk+1}^{(m+1)k-1}\frac1{n(n+k)}=\frac1k\left(\frac1{mk+1}-\frac1{(m+2)k-1}\right)$$ so we get: $$\frac1k\sum_{m=0}^\infty\left(\frac1{mk+1}-\frac1{(m+2)k-1}+\frac1{(m+1)(m+2)}\right)$$now split it up and telescope, I suppose? $$\sum\left(\frac1{mk+1}-\frac1{(m+2)k-1}\right)=1+\frac1{k+1}\\\sum\left(\frac1{m+1}-\frac1{m+2}\right)=1$$ so we get $$\frac1k\left(2+\frac1{k+1}\right)=\frac2k+\frac1{k(k+1)}=\frac3k-\frac1{k+1}$$ for prime $$k$$
13. anonymous | {
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13. anonymous
oops, ignore that -- I read $$(m+2)k+1$$ rather than $$(m+2)k-1$$, so that series does not quite telescope so cleanly
14. anonymous
our problem term for prime $$k$$ is this: $$\frac1{mk+1}-\frac1{(m+2)k-1}$$
15. anonymous
Would that I could hand out more medals...
16. thomas5267
How do you show $$\gcd (n,n+k) \leq k$$ without using Bézout's Identity?
17. anonymous
i mean, if (positive) $$d$$ divides (positive) $$n$$ and $$n+k$$ then it also divides their difference, so $$d$$ must divide $$k$$. it follows that the greatest common divisor is at most $$k$$
18. anonymous
you don't necessarily need the full strength of Bezout's identity, just the fact that divisibility is preserved by sums which is very elementary
19. ganeshie8
$$\gcd(n,n+k)=\gcd(n.k)\le k$$
20. thomas5267
\begin{align*} \sum_{n=1}^\infty\frac1{[n,n+k]}&=\sum_{m=0}^\infty\sum_{n=mk+1}^{(m+1)k}\frac1{[n,n+k]}\\ &=\sum_{m=0}^\infty\left(\sum_{n=mk+1}^{mk+k-1}\frac1{n(n+k)}+\frac1{(m+1)(m+2)k}\right)\\ \end{align*} \begin{align*} &\phantom{{}={}}\sum_{m=0}^\infty\sum_{n=mk+1}^{mk+k-1}\frac1{n(n+k)}\\ &=\sum_{m=0}^\infty\left(\frac{1}{mk+1}-\frac{1}{(m+1)k-1}\right)\\ &=\sum_{m=0}^\infty\left(\frac{1}{mk+1}-\frac{1}{(m+1)k+1}+\frac{1}{(m+1)k+1}-\frac{1}{(m+1)k-1}\right)\\ &=\sum_{m=0}^\infty\left(\frac{1}{mk+1}-\frac{1}{(m+1)k+1}-\frac{2}{(m+1)^2k^2-1}\right)\\ &=1-\sum_{m=0}^\infty\frac{2}{(m+1)^2k^2-1} \end{align*} I think the sum converges but I couldn't prove it.
21. anonymous
Comparing to the series $$\sum\frac{1}{n^2}$$ would suffice for convergence. It has a somewhat daunting closed form: $\sum_{m=0}^\infty \frac{2}{(m+1)^2k^2-1}=\frac{k-\pi\cot\dfrac{\pi}{k}}{2k}$
22. anonymous
(closed form courtesy of Mathematica)
23. thomas5267
I thought $$\dfrac{2}{(m+1)k^2-1}\geq\dfrac{1}{n^2}$$ for some reason! $\min(k)=2\\ \frac{2}{4n^2-1}-\frac{1}{n^2}=\frac{2n^2-4n^2+1}{n^2(4n^2-1)}=\frac{-2n^2+1}{n^2(4n^2-1)}\leq0$
24. anonymous | {
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24. anonymous
$\sum_{m\ge0}\frac{1}{(m+1)^2k^2-1}=\sum_{m\ge1}\frac{1}{m^2k^2-1}\le\sum_{m\ge1}\frac{1}{m^2}$ since $1\le k^2-1~~\implies~~ m^2\le m^2k^2-1~~\implies~~\frac{1}{m^2k^2-1}\le\frac{1}{m^2}$(The first inequality is true since $$k\ge2$$.) | {
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# Ten people are seated at a rectangular table - Permutations homework
I got the following question for homework.
Ten people are to be seated at a rectangular table for dinner. Tanya will sit at the head of the table. Henry must not sit beside either Wilson or Nancy. In how many ways can the people be seated for dinner?
The approach I have used is to take $9!$ and subtract it from the seats where Henry, Wilson or Nancy can sit. I end up getting the answer $211680$ but for some reason, the back of the book says $201 600$. What am I doing incorrectly?
-
This looks like an error in the book: your answer appears to be correct.
We can count the allowed seatings directly. Suppose that Henry sits next to Tanya. Then there is one seat forbidden to Wilson and Nancy, so there are $\binom72$ ways to choose seats for Wilson and Nancy. There are $2$ ways to seat Wilson and Nancy in those $2$ seats and $6!$ ways to seat the unnamed people. Finally, Henry can be on either side of Tanya, so there are altogether $2\cdot2\cdot6!\cdot\binom72$ acceptable seatings with Henry next to Tanya.
Now suppose that Henry is not seated next to Tanya. Then there are $2$ seats forbidden to Wilson and Nancy, so there are $\binom62$ ways to choose their seats. As before there are $2$ ways to seat them in those two seats and $6!$ ways to seat the unnamed people. Finally, there are $7$ possible choices for Henry’s seat, so there are altogether $7\cdot2\cdot6!\cdot\binom62$ acceptable seatings with Henry not next to Tanya. Altogether, then, there are
$$2\cdot6!\left(2\binom72+7\binom62\right)=1440(42+105)=211,680$$
acceptable seatings.
-
Using inclusion-exclusion, it's possible to see where the error in the back of the book may have come from. | {
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