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Circle bisecting the circumference of another circle If the circle $x^2+y^2+4x+22y+l=0$ bisects the circumference of the circle $x^2+y^2-2x+8y-m=0$,then $l+m$ is equal to
(A)$\ 60$
(B)$\ 50$
(C)$\ 46$
(D)$\ 40$
I don't know the condition when one circle intersects the circumference of other circle. So could not solve this question. Can someone help me in this question?
|
Circle $C_1: x^2+y^2+4x+22y+l=0$ has its center $(-2, -11)$ & a radius $\sqrt{(-2)^2+(-11)^2-l}=\sqrt{125-l}$
Similarly, circle $C_2: x^2+y^2-2x+8y-m=0$ has its center $(1, -4)$ & a radius $\sqrt{(1)^2+(-4)^2-(-m)}=\sqrt{m+17}$
Now, solving the equations of circles $C_1$ & $C_2$ by substituting the value of $(x^2+y^2)$ from $C_2$ into $C_1$, we get the $\color{blue}{\text{equation of common chord}}$ as follows $$(2x-8y+m)+4x+22y+l=0$$ $$\color{blue}{6x+14y+(l+m)=0}\tag 1$$
Now, since the circumference of circle $C_2$ is bisected by the circle $C_1$ hence the center $(1, -4)$ of circle $C_2$ must lie on the common chord or in other words, the common chord must pass through the center of circle $C_2$
Now, satisfying the above equation of common chord by center point $(1, -4)$ as follows
$$6(1)+14(-4)+(l+m)=0$$ $$6-56+l+m=0$$ $$\bbox[5px, border:2px solid #C0A000]{l+m=50}$$
|
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|
A basic root numbers question If $\sqrt{x^2+5} - \sqrt{x^2-3} = 2$, then what is $\sqrt{x^2+5} + \sqrt{x^2-3}$?
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Notice, we have $$\sqrt{x^2+5} - \sqrt{x^2-3} = 2\tag 1$$ let $$\sqrt{x^2+5} + \sqrt{x^2-3} = y\tag 2$$ Now, multiplying both (1) & (2), we get $$(\sqrt{x^2+5} - \sqrt{x^2-3} )(\sqrt{x^2+5} + \sqrt{x^2-3})=2y$$ $$(\sqrt{x^2+5})^2-(\sqrt{x^2-3})^2=2y$$ $$x^2+5-(x^2-3)=2y$$ $$8=2y\implies y=4$$ Hence, $$\bbox[5px, border:2px solid #C0A000]{\sqrt{x^2+5} + \sqrt{x^2-3} = 4}$$
|
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|
find the complex number $z^4$
Let $z = a + bi$ be the complex number with $|z| = 5$ and $b > 0$ such that the distance between $(1 + 2i)z^3$ and $z^5$ is maximized, and let $z^4 = c + di$.
Find $c+d$.
I got that the distance is:
$$|z^3|\cdot|z^2 - (1 + 2i)| = 125|z^2 - (1 + 2i)|$$
So I need to maximize the distance between those two points.
$|z^2| = 25$ means since: $z^2 = a^2 - b^2 + 2abi$ that:
$625 = (a^2 - b^2)^2 + 4a^2b^2 = a^4 + b^4 + 2a^2b^2$
But that doesnt help much.
|
Let $\theta$ denote the argument of $z$, then the arguments of $(1+2i)z^3$ and $z^5$ are respectively $3\theta +\arctan(2)$ and $5\theta$. Now imagine that these points lie on concentric circles, so in order to maximize the distance, the difference between the arguments must be $\pi$ (or $-\pi$ it won't matter):
$$5\theta - 3\theta -\arctan(2)=\pi$$
$$\tan(5\theta - 3\theta -\arctan(2))=0$$
$$\tan(2\theta -\arctan(2))=0=\frac{\tan(2\theta)-2}{1+2\tan(2\theta)}$$
$$\tan(2\theta)=2$$
$$\theta=\frac{1}{2}\arctan(2)\tag{A}$$
$$\tan(\theta)=\frac{1}{2}(\sqrt{5}-1)=\frac{b}{a}$$
$$b^2=\frac{1}{2}(3-\sqrt{5})a^2$$
$$a^2+b^2=25$$
$$a^2=\frac{25+5\sqrt{5}}{2}$$
$$b^2=\frac{25-5\sqrt{5}}{2}$$
ALSO, to get c and d directly, multiply both sides of $(\text{A})$ by 4:
$$4\theta=2\arctan(2)$$
$$tan(4\theta)=-\frac{4}{3}=\frac{d}{c}$$
and we have that
$$\sqrt{c^2+d^2}=5^4$$
Finally, solve for $c$ and $d$.
|
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Dedekind Construction Of Real Numbers If we define Dedekind-real numbers as Dedekind cuts, i.e. $\sqrt 2 = \{\text{rationals less than }\sqrt2\} \cup \{\text{rationals more than } \sqrt2\}$, can we define addition and multiplication of these real numbers as follows:
These real numbers $\mathbb R$ are a complete lattice in which the rational numbers are embedded. Take the open intervals of $\mathbb R$ and make them the basis for a topology on $\mathbb R$.
Define $+(\mathbb R,\mathbb R)$ and $\times(\mathbb R,\mathbb R)$ as the unique continuous (in this topology) extension of the already known arithmetic functions $+(\mathbb Q,\mathbb Q)$ and $\times(\mathbb Q,\mathbb Q)$?
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The idea of Dedikind cuts is that one can define $\mathbb{R}$ and its arithmetic operations and order in an elementary manner, directly in terms of $\mathbb{Q}$ and its the arithmetic operations and order. One doesn't need the concepts of analysis or topology to do this. Of course, after the real numbers and their operations are defined, one can go on to develop calculus and to prove that addition and multiplication are continuous operations, but it is not necessary to bring continuity or other concepts of analysis into the picture simply for the purpose of defining addition and multiplication of real numbers.
For example, given two Dedekind cuts $(A_1,B_1)$ and $(A_2,B_2)$ of $\mathbb{Q}$, their sum is then defined to be
$$(A_1,B_1) + (A_2,B_2) = (A_1+A_2,B_1+B_2)
$$
where the meaning of addition of subsets is simply
$$A_1 + A_2 = \{a_1 + a_2 \, \bigm| \, a_1 \in A_1 \, \text{and} \, a_2 \in A_2 \}
$$
Now there is a little work to be done, but it should be pretty elementary: one has to prove that $(A_1+A_2,B_1+B_2)$ is, indeed, a Dedikind cut.
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|
Sorting triangles by hypotenuse length I have some points in $xy$ space and I need to sort distances between these points. If I calculate real distance, then I need to perform $\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}$ and this is very time consuming operation. As I understand, I can easily omit square root for sorting purposes. But I want to go further. Can I use $|x_1-x_2| + |y_1 - y_2|$ instead of initial equation for sorting?
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No.
Consider $x_1-x_2=4$, $y_1-y_2=4$ then $(x_1-x_2)^2+(y_1-y_2)^2=32$ and $|x_1-x_2|+|y_1-y_2|=8$. But $x_1-x_2=6$, $y_1-y_2=1$ then $(x_1-x_2)^2+(y_1-y_2)^2=37$ and $|x_1-x_2|+|y_1-y_2|=7$. So this would give the wrong comparison.
|
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Trying to make sense of this proof in Hatcher So I'm trying to understand this proof in Hatcher's Algebraic Topology.
Lemma: The composition $\Delta_n(X)\xrightarrow{\partial_n} \Delta_{n-1}(X)\xrightarrow{\partial_{n-1}}\Delta_{n-2}(X)$ is zero
where here $\partial_n$ is a boundary homomorphism $$\partial_n(\sigma_{\alpha})=\sum_i (-1)^i\sigma_{\alpha}\:\vert\:[v_0\dots,\bar v_i,\dots,v_n]$$
where $[v_0,\dots,v_n]$ is an $n$-simplex and the bar $\bar v_i$ represents removing that vertex, giving a face of of the $n$-simplex.
So here's the proof:
We have $\partial_n(\sigma)=\sum_i (-1)^i\sigma\:\vert\:[v_0\dots,\bar v_i,\dots,v_n]$, and hence
$$\partial_{n-1}\partial_n=\sum_{j<i}(-1)^i(-1)^j\sigma\:|\:[v_o,\dots,\bar v_j,\dots,\bar v_i,\dots,v_n]+\sum_{j>i}(-1)^i(-1)^{j-1}\sigma\:|\:[v_o,\dots,\bar v_i,\dots,\bar v_j,\dots,v_n]$$
And the two summations cancel because after switching $i$ and $j$ in the second sum, it becomes the negative of the first.
What I'm confused about is, why are there two summations coming from this composition of the two boundary functions? And where does the $j-1$ come from in the second sum?
My lack of understanding of this proof may be from some lack of genuine understanding of these boundary maps, but I have a limited time to learn this in and this is really stumping me.
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When taking the second boundary map, one can not remove the $ i $th term since it was already removed by the first boundary map. So $ j \neq i $ in the sum, and so it is convenient to split up the sum as
$$ \sum_{\substack{ j \\ j \neq i } } \cdot = \sum_{\substack{ j \\ j < i } } \cdot + \sum_{\substack{ j \\ j > i } } \cdot
$$
Now for the signs, in the definition $ \partial_n(\sigma_{\alpha})=\sum_i (-1)^i\sigma_{\alpha}\:\vert\:[v_0\dots,\bar v_i,\dots,v_n] $, you should think of the $ (-1)^i $ as $ (-1)^{\text{number of $ v $ to the left of $ v_i $} } $. When $ j < i $, there are $ j $ terms to the left so it is $ (-1)^j $. When $ j > i $, since the $ i$ th term is already gone there are only $ j-1 $ terms to the left.
I think for these types of calculations (which happen often in algebraic topology) it is more enlightening to work through the pieces of the proof with a specific example, and even better, with a picture. Try computing $ \partial^2 [0,1,2,3] $.
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Are there papers or books that explain why Bernhard Riemann believed that his hypothesis is true? I would like to know what are the mathematical reasons for which Bernhard Riemann believed that his hypothesis is true, and I would like to know if those mathematical reasons were cited in his original paper.
My question Here : Are there papers or books or links that explain why Bernhard Riemann believed that his hypothesis is true?
Note: I do not want to know if the Riemann Hypothesis is true or false.
Thank you for any help
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Harold Edwards provides some informed speculation on this question in his book. I'm not home right now but will come back and summarize his thoughts in a couple days.
And by the way his book is a gem. It includes a translation of the original paper, and one of the more enjoyable and satisfying couple months of my "mathematical life" was spent working my way carefully through that paper ( it invites lots of side-trips and explorations along the way... I had a particularly long sidetrack with the gamma function... )
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What is the variance of the volumes of particles? According to Zimmels (1983), the sizes of particles used in sedimentation experiments often have a uniform distribution. In sedimentation involving mixtures of particles of various sizes, the larger particles hinder the movements of the smaller ones. Thus, it is important to study both the mean and the variance of particle sizes. Suppose that spherical particles have diameters that are uniformly distributed between $0.01$ and $0.05$ centimeters. Find the mean and variance of the volumes of these particles (Volume of the sphere is $\frac 4 3 \pi R^3$).
Book's solution:
Mean $= 0.0000065\pi$
Variance $= 0.0003525\pi^2$
My solution:
Volume $= \frac 4 3 \pi R^3 = \pi D^3$, where $D$ is diameter.
Mean $\displaystyle = E\left[\frac \pi 6 D^3\right] = \int_{0.01}^{0.05} \frac \pi 6 D^3 \cdot25$.
$= 0.0000065\pi$ (same answer with the book)
\begin{align}
\text{Variance} & = E\left[\frac{\pi^2}{36} D^6\right] - E\left[ \frac \pi 6 D^3\right]^2 \\[8pt]
& = \int_{0.01}^{0.05} \left[ \frac{\pi^2}{36} D^6\right] \cdot 25 - 4.225\times(10^{-11})\times \pi^2 \\[8pt]
& = 3.5254 \times (10^{-11})\times \pi^2 \text{ (about $10^7$ times smaller than the book solution)}
\end{align}
Is my solution wrong?
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I'm going to use units of hundredths of a centimeter, then convert back to centimeters.
The diameter of a given particle is modeled as a random variable $$D \sim \operatorname{Uniform}(1,5);$$ namely, $$f_D(x) = \begin{cases} \frac{1}{4}, & 1 \le x \le 5 \\ 0, & \text{otherwise}. \end{cases}$$ If $$V = \frac{4}{3}\pi R^3$$ where $R = D/2$, this suggests we wish to find $\operatorname{E}[D^3]$ and $\operatorname{E}[D^6]$. To this end, we write $$\operatorname{E}[D^k] = \int_{x=1}^5 x^k f_D(x) \, dx = \frac{1}{4} \left[ \frac{x^{k+1}}{k+1} \right]_{x=1}^5 = \frac{5^{k+1} - 1}{4(k+1)},$$ for any positive integer $k$. For $k = 3, 6$, we get $$\operatorname{E}[D^3] = 39, \quad \operatorname{E}[D^6] = \frac{19531}{7}.$$ Therefore, $$\operatorname{E}[V] = \operatorname{E}\left[\frac{4}{3} \pi \frac{D^3}{8} \right] = \frac{13}{2}\pi,$$ and $$\operatorname{Var}[V] = \operatorname{E}\left[\frac{\pi^2}{36} D^6 \right] - \operatorname{E}[V]^2 = \left( \frac{19531}{252} - \frac{169}{4}\right) \pi^2 = \frac{2221}{63}\pi^2.$$ Now converting these to centimeters gives $$\operatorname{E}[V] = \frac{13\pi}{2000000} \approx 2.04204 \times 10^{-5}$$ and $$\operatorname{Var}[V] = \frac{2221\pi^2}{63000000000000} \approx 3.47943 \times 10^{-10}.$$
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determination of probability Recently I have faced a problem related to probability.I have tried it by bayseian theorem but failed.Here is the problem:
A doctor knows that pneumonia causes a fever 95% of the time. She knows that if a person is selected
randomly from the pneumonia. 1 in 100 people suffer from fever. You can go to the doctor complaining
about the symptom of having a fever (evidence). What is the probability that pneumonia is the cause of
this symptom?
Can anyone give some hints or how can i solve it?
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There are many ambiguities in the question itself and the question can be interpreted in two ways.
*
*At first it says that "pneumonia causes a fever 95% of the time" and later it is being contradictory/(false in terms of stats) by saying " if a person is selected randomly from the pneumonia. 1 in 100 people suffer from fever." So, how come only 1 in 100 person are suffering from fever when pneumonia itself causes fever 95% of the time. If we go by this, then it should have been "...95 in 100 people suffer from fever".
Another way to interpret the question is,
*Since there a 'fullstop' between "if a person..." and "1 in 100...". Therefore it in this way, the line "She knows that if a person is selected randomly from the pneumonia" will make no sense at all.
Putting everything in one and somehow trying to get the answer, may return multiple different answers because different people will interpret this question as differently.
But the answer would be:
Consider that there are only 100 people in that town. Out of which it is given that 1 person suffers from "fever". We are also provided that pneumonia causes fever 95% of the times. Therefore, for that "1" person, who suffers from fever, there would be $ 95/100 * 100 = 95\% $ person chances that it is caused by pneumonia.
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Convexity under diffeomorphisms Let $K \subset \mathbb{R}^n$ be a compact convex subset with non-empty interior, and $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ a diffeomorphism. Then is it true that $f[K]$ is convex?
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I'm afraid that in general it's not true. In $\mathbb{R}^2$ you can construct a diffeomorphism $(x, y) \mapsto (x + hy^2, y), \; h > 0$ and apply it to unit square (or unit circle). The "top part" of this figure will be stretched much stronger than parts are near to the $Ox$ axis, which will lead to fail of convexity.
ADDITION: this is an illustration of what happens with rectangle under the action of this map (sorry for the quality, my usual drawing tool is on vacation :) )
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If $g$ is a permutation, then what does $g(12)$ mean? In Martin Lieback's book 'A Concise Introduction to Pure Mathematics', he posts an exercise(page 177,Q5):
Prove that exactly half of the $n!$ permutations in $S_n$ are even.
(Hint: Show that if $g$ is an even permutation, then $g(12)$ is odd. Try to use this to define a bijection from the set of odd permutations to the set of even permutations.)
I couldn't understand what is meant by $g(12)$. Does it mean the composite permutation function $g\circ (1 \ \ 2)$ or the function $g\circ (12)$? Also, what would be your thought process for this question?
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It is the composition of the permutation $g$ and the transposition $(1\ 2)$. Since $g$ is even and all transpositions are odd, it is elementary that $g (1\ 2) = g \circ (1\ 2)$ is odd, as the product of an even and odd permutation is odd.
(There is a space between $1$ and $2$; this is not the number $12$.)
Edit: To also answer your second question: the hint tells you that to each even permutation $g$ you may associate an odd one $g (1 \; 2)$. Note that this correspondence is injective, because $g_1 (1 \; 2) = g_2 (1 \; 2) \implies g_1 = g_2$ (just multiply by the inverse of $(1 \; 2)$ on the right, which happens to be $(1 \; 2)$ again). Now note that this correspondence is also invertible, simply because $(1 \; 2)$ has an inverse in $S_n$. Therefore, you have a bijection between the set of even permutations and the set of odd permutations, therefore these two sets must have the same number of elements, hence the conclusion.
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Find max of $S = x\sin^2\angle A + y\sin^2\angle B + z\sin^2\angle C$ Let $x$, $y$, $z$ are positive constants. $A$, $B$, $C$ are three angles of the triangle. Prove that $$S = x \sin^2 A + y \sin^2 B + z \sin^2 C \leq \dfrac{\left(yz+zx+xy\right)^2}{4xyz}$$
and find when it holds equality
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Update
This is an intricate problem, since not only the value of $S_{\max}$ is changing with the values of the parameters $x$, $y$, $z$, but also the global extremal situation.
Claim: If $0<2x<y\leq z$ then $S_{\max}=y+z$.
Proof: If one of the angles is $>{\pi\over2}$, e.g., $\alpha\leq\beta<{\pi\over2}<\gamma$, it is possible to enlarge $S$ by replacing $\alpha$, $\beta$,$\gamma$ with $$\alpha':=\alpha+\gamma-{\pi\over2}, \quad \beta':=\beta,\quad \gamma':={\pi\over2}\ .$$
It follows that we may assume $0\leq\alpha\leq\beta\leq\gamma\leq{\pi\over2}$, in this order. Write $\beta:={\pi\over2}-\beta'$, $\>\gamma:={\pi\over2}-\gamma'$, $\alpha:=\beta'+\gamma'$. Then $S$ appears as
$$S=\sin^2(\beta'+\gamma') x+(1-\sin^2\beta')y+(1-\sin^2\gamma')z\ .$$
Now by Schwarz' inequality
$$\sin(\beta'+\gamma')=\cos\gamma'\sin\beta'+\cos\beta'\sin\gamma'\leq\sqrt{\cos^2\gamma'+\cos^2\beta'} \sqrt{\sin^2\beta'+\sin^2\gamma'}\ ,$$
so that
$$\sin^2(\beta'+\gamma')\leq 2(\sin^2\beta'+\sin^2\gamma')\ .$$
It follows that
$$S\leq y+z-(y-2x)\sin^2\beta'-(z-2x)\sin^2\gamma'\leq y+z$$
with equality iff $\beta'=\gamma'=0$.$\quad\square$
On the other hand, if $x=y=z=1$, then $\alpha=0$, $\beta=\gamma={\pi\over2}$ leads to $S=2$, whereas
$\alpha=\beta=\gamma={\pi\over3}$ produce $S={9\over4}$. This shows that there is a "phase transition" where the morphology of the extremal situation changes.
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Sizes of Quotient Rings of DVRs with Finite Residue Field If $R$ is a discrete valuation ring (DVR) with maximal ideal $\mathfrak{m}$ such that $R/\mathfrak{m}$ is finite, then all quotient rings of $R,$ namely $R/\mathfrak{m}^n$ for $n \in \mathbb{N},$ are finite. My question is:
Can we say anything about the sizes of $R/\mathfrak{m}^n$ for $n>1,$ if
we know $\#(R/\mathfrak{m})$?
Many thanks!
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We have a short exact sequence: $$0\to\mathfrak m/\mathfrak m^2\to R/\mathfrak m^2\to R/\mathfrak m \to 0.$$ Since $\dim_{R/\mathfrak m}\mathfrak m/\mathfrak m^2=1$ we get $\#(\mathfrak m/\mathfrak m^2)=\#(R/\mathfrak{m})$. It follows that $\#(R/\mathfrak m^2)=\#(R/\mathfrak{m})^2$.
Can you continue from here?
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Showing that $\nabla\times(\nabla\times\vec{A}) = \nabla(\nabla\cdot\vec{A})-\Delta\vec{A}$ I have faced difficulties while trying to prove that
$$\nabla\times(\nabla\times\vec{A}) = \nabla(\nabla\cdot\vec{A})-\Delta\vec{A}$$
I don't have any clue how can I start to work with it. Any hint will be helpful.
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I am coming to this party about three years too late, but the answers posted here, and some here as well: enter link description here, may leave readers with the false impression that the vector Laplacian acting on a vector field (appearing on the right-hand side) always takes the form
$$\begin{align}
\nabla^2 \vec{A} = \nabla^2(A_i)\,\hat{e}_i ~~~ {\text{(summation convention in force)}} \tag 1
\end{align}$$
in an arbitrary basis and coordinate system of $\mathbb{R}^3$, where $\nabla^2$ is the Laplacian operator in the coordinates associated with the basis. This is true only in a cartesian coordinate system and its orthonormal basis. For example, in the cylindrical coordinate system $\{\rho, \phi, z\}$, with orthogonal (but not orthonormal) basis $\{\hat{e}_\rho, \hat{e}_\phi. \hat{e}_z\}$, it might be supposed from these derivations that
$$\begin{align}
\nabla^2 \vec{A} = \nabla^2(A_\rho)\,\hat{e}_\rho + \nabla^2(A_\phi)\,\hat{e}_\phi + \nabla^2(A_z)\,\hat{e}_z,
\end{align}$$
where
$$\begin{align}
\nabla^2(f) = \frac{1}{\rho}\frac{\partial}{\partial \rho}\left(\rho\frac{\partial f}{\partial \rho}\right) + \frac{1}{\rho^2}\frac{\partial^2 f}{\partial \phi^2} + \frac{\partial^2 f}{\partial z^2}
\end{align}$$
is the Laplacian operator in cylindrical coordinates acting on a scalar function $f$. This is quite false, as I first learned long ago from the textbook ``Mechanics of Deformable Bodies'', by the master mathematical physicist Arnold Sommerfeld. In the answers to Problems I.3 and I.4 of the book, he lays bare the fallacy of this assumption. The first two components of $\nabla^2{\vec{A}}$ contain extra terms besides $\nabla^2(A_\rho)$ and $\nabla^2(A_\phi)$.
In fact, Sommerfeld clearly states in Chapter I, page 23, what seems to have been overlooked, or at any rate, not emphasized, in many of the answers posted here: the formula $\nabla \times (\nabla \times \vec{A}) = \nabla(\nabla\cdot \vec{A}) - \nabla^2\vec{A}$ is an identity in the sense of (1) being true, only in cartesian coordinates. For any other coordinate system and basis, the formula serves to define $\nabla^2\vec{A}$ by
$$\begin{align}
\nabla^2\vec{A} := \nabla(\nabla\cdot \vec{A}) - \nabla \times (\nabla \times \vec{A}),
\end{align}$$
each term on the right-hand side being a well-defined operation in any coordinate system. So the upshot is that the formula can be shown to be true only in a cartesian coordinate system (as was done in the answers here); in any other coordinate system the formula instead defines the vector Laplacian acting on a vector field.
|
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Finding the work from $(0,0)\to(1,1)$ of $\vec F(x,y)=xy^2\hat i+yx^2\hat j$
I need to find the work from $(0,0)\to(1,0)\to(1,1)$ of the following vector field:$\vec F(x,y)=xy^2\hat i+yx^2\hat j$
My attempt:
$$\oint_{c}\vec F d\vec r=\int_{(0,0)\to (1,0)}\bigg(xy^2\; dx +yx^2 \; dy\bigg)+\int_{(1,1)\nwarrow(1,0)}\bigg(xy^2\; dx+yx^2\; dy\bigg)$$
Is it correct so far? if yes so how can I proceed from this point?
|
METHOD 1:
We can simplify the problem by noting that $\nabla \times \vec F=0$. Therefore, $\vec F$ is therefore conservative on any connected domain. By Stokes' Theorem, the integral of $\vec F$ over any contour $C$ that bounds a connected domain $S$ is
$$\oint_C \vec F\cdot d\vec \ell =\int_S \nabla \times \vec F\cdot \hat ndS$$
Thus, the integral of interest is equal to
$$\int_{C_1+C_2}\vec F\cdot d\vec \ell=-\int_{(1,1)\,\text{to}\,(0,0)}(xy^2dx+x^2ydy)=\int_0^1 2t^3=\frac12$$
METHOD 2:
If we choose to evaluate the line integral directly, that is without deforming the contour as in Method 1, then we proceed as follows.
$$\int_{C_1+C_2}\vec F\cdot d\vec \ell=\int_{C_1}\vec F\cdot d\vec \ell+\int_{C_2}\vec F\cdot d\vec \ell$$
where
$$\int_{C_1}\vec F\cdot d\vec \ell=\int_0^1 (x(0)^2)dx =0 \tag 1$$
and
$$\int_{C_2}\vec F\cdot d\vec \ell=\int_0^1 (1)^2y\,dy=\frac12 \tag 2$$
Putting together $(1)$ and $(2)$ we obtain
$$\int_{C_1+C_2}\vec F\cdot d\vec \ell=\frac12$$
as expected!
|
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Prove that if $\forall A \in \mathcal F (B\subseteq A)$ then $B \subseteq \bigcap \mathcal F $ This is Velleman's exercise 3.3.10. Suppose that $\mathcal F$ is a nonempty family of sets, B is a set, and $\forall A \in \mathcal F (B\subseteq A)$. Prove that $B \subseteq \bigcap \mathcal F $.
My approach so far:
Suppose that $A \in \mathcal F$, suppose that $x \in B$. Then $x$ is an element of any $A \in \mathcal F$ because $\forall A \in \mathcal F (B\subseteq A)$. Then $x$ will also be an element of $\bigcap \mathcal F$ because $\forall A \in\mathcal F(x \in A)$. Hence, any element of $B$ is an element of $\bigcap \mathcal F$. This shows that $B \subseteq \bigcap \mathcal F$.
Is this a valid proof? Thanks in advance.
|
Basically yes, but you should try to be a bit more methodical to obtain a 100% correct proof. Here are the steps to take (compare with what you've actually done).
1) Suppose $x\in B$. The aim is to show that $x\in \bigcap \mathcal F$.
2) To show $x$ is in an intersection, it suffices to show that $x\in A$ for all $A\in \mathcal F$.
3) So, let $A\in \mathcal F$.
4) Now finish the proof.
You will notice that the first three steps are mechanical. They do not require any thinking. This is the part you got a bit wrong in your proof.
|
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Area of a triangle with sides $\sqrt{x^2+y^2}$,$\sqrt{y^2+z^2}$,$\sqrt{z^2+x^2}$ Sides of a triangle ABC are $\sqrt{x^2+y^2}$,$\sqrt{y^2+z^2}$ and $\sqrt{z^2+x^2}$ where x,y,z are non-zero real numbers,then area of triangle ABC is
(A)$\frac{1}{2}\sqrt{x^2y^2+y^2z^2+z^2x^2}$
(B)$\frac{1}{2}(x^2+y^2+z^2)$
(C)$\frac{1}{2}(xy+yz+zx)$
(D)$\frac{1}{2}(x+y+z)\sqrt{x^2+y^2+z^2}$
I tried applying Heron's formula but calculations are very messy and simplification is difficult.I could not think of any other method to find this area.Can someone assist me in solving this problem.
|
hint: Let $a = \sqrt{x^2+y^2}, b = \sqrt{y^2+z^2}, c = \sqrt{z^2+x^2} \to a^2 = x^2+y^2, b^2 = y^2+z^2, c^2= z^2+x^2 $. Use this and Cosine Law to find $\cos^2 A$, then $\sin^2 A$, and use $S^2 = \dfrac{b^2c^2\sin^2 A}{4}$, to find $S^2$ and then take square-root to get back $S$.
|
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Can functions within a matrix adjust its size? I've been working my way through a proof, and without going into the full extent of the details it's come down to whether a function G() exists such that the 1 by 3 matrix:
\begin{bmatrix}G\begin{pmatrix}1\\0\\0\end{pmatrix}&G\begin{pmatrix}0\\1\\0\end{pmatrix}&G\begin{pmatrix}0\\0\\1\end{pmatrix}\end{bmatrix}
could become a n by 3 matrix, for any whole number n (for example in the case above I would want to resolve it to a 5 by 3 matrix.
Is this possible? Or is it back to the drawing board?
|
That is mostly a question of notation and convention. You can certainly choose to define that in your work, the notation
$$ A = \begin{bmatrix}X & Y & Z\end{bmatrix} $$
where $X$, $Y$ and $Z$ are column matrices of the same height, will mean that $A$ is the $3\times n$ matrix that has those three columns. As a matter of terminology that would mean that you're defining $A$ as a block matrix.
However, it seems unlikely that this is actually what your problem depends on. There's essentially no mathematical content in asking (effectively) "is it possible to form block matrices" -- the answer to that will only tell you how your notation works, but not anything about the underlying mathematical structure that your notation speaks about.
So when you say that your problem "comes down to" whether you can employ that notation or not, it sounds very likely that you have some conceptual confusion or mistake in your work before you reach that point. And you should probably ask another question where you give details of that work and ask whether your approach is legitimate.
|
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Geodesic connectivity implies geodesic convexity? Assume a subset $C$ of a Riemannian manifold $M$ is "geodesically-connected", that is: given any two points in $C$, there is a geodesic contained within $C$ that joins those two points.
Is it true that $C$ must be geodesically convex?
(given any two points in $C$, there is a minimizing geodesic contained within $C$ that joins those two points).
What happens if we only require existence of a minimizing geodesic among all the connecting geodesics contained in $C$? ("internal minimization?")
|
The answer is negative.
Take the sphere $S^2$ with the round metric, and cut out the part which is souther than some latitude line which is close to the south pole. (The threshold line also stays out).
We are left with an open Riemannian submanifold $M$.
Any two points in $M$ will be connected via a geodesic. For some pairs of points it will be the "long" part of the usual geodesic in $S^2$ which connects them. (The short minimzing part passes through an area we have deleted).
For such pairs, there will be no connecting minimizing geodesic:
Since $M$ is open, it is a totally geodesic submanifold of $S^2$. Assume such a pair has a minimizing geodesic between them. Then it must be a geodesic $S^2$, hence the long part of a great circle. However, we can clearly find such a pair for which exists a shorter path in $M$, contrdicting the minimality.
In particular, there are points in $M$ which do not have a minimizing path between them. (By the theorem which states any minimizing path is a geodesic).
|
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integrating product of PDF and CDF I am trying to show that the following integral:
$$
\int_{-\infty}^a F(x)~f(x)~dx = \frac{F(a)}{2!}
$$
Where $F$ is the cumulative distribution function of some continuous random variable X, and $f$ is the probability density function. Not quite sure how this conclusion was reached?
|
Clearly $F'(x)=f(x)$ (wherever CDF is continuous).
Substitute $F(x)=t$. Then $dt=f(x)dx$.
$$\int_{-\infty}^a F(x)~f(x)~dx \rightarrow \int_{0}^{F(a)} t~dt= \frac{(F(a))^2}{2}$$
|
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Iterative calculation of $\log x$ Suppose one is given an initial approximation of $\log x$, $y_0$, so that:
$$y_0 = \log x + \epsilon \approx \log x$$
Here, all that is known about $x$ is that $x>1$. Is there a general method of improving that estimation using only addition & multiplication, i.e. without exponentiation or logarithms?
$$y_1 = f(y_0, x)=\ ?$$
|
Instead of solving $y - \ln(x) = 0$ for $y$ you can solve $g(y) = e^y - x = 0$
Given initial approximation $y_0 \approx \ln(x)$ you can try to solve $y$ using Newton's medhod:
$$y_{n + 1} = y_n - \frac{g(y_n)}{g'(y_n)}= y_n - \frac{e^{y_n} - x}{e^{y_n}} = y_n - 1 +\frac{x}{e^{y_n}}$$
When exponentiation isn't allowed you can approximate $e^{y}$ with $\left(1 + \frac{y}{2^m} \right)^{2^m}$ using repeated multiplication: $$\begin{aligned} \left(1 + \frac{y}{2^m} \right)^{2^m} &= \left(1 + \frac{y}{2^m} \right)^{2^{m - 1}} \cdot \left(1 + \frac{y}{2^m} \right)^{2^{m - 1}} \\ \\2^{-m} &= \frac{1}{2} \cdot 2^{-m + 1}\end{aligned}$$
|
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Minimum value of reciprocal squares I am bit stuck at a question. The question is : given: $x + y = 1$, $x$ and $y$ both are positive numbers. What will be the minimum value of: $$\left(x + \frac{1}{x}\right)^2 + \left(y+\frac{1}{y}\right) ^2$$
I know placing $x = y$ will give the right solution. Is there any other solution?
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$(x+1+y/x)^2+ (y+1+x/y)^2 >= 1/2 (x+1+y/x+y+1+x/y)^2 >= 1/2(2+1+x/y+y/x)^2 >=25/2 $
Equality holds when $x/y=y/x$ or $x=y$
|
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Name for continuous maps satisfying $\operatorname{cl}(f^{-1}f(U))= \operatorname{cl}(U)$ I have recently come across particularly kind of continuous maps $f
\colon X \to Y$ between topological spaces with the property that
$$
\operatorname{cl}(f^{-1}f(U))= \operatorname{cl}(U),
$$
for all $U \in \wp(X)$.
If $X$ is $T_1$ then this condition implies that $f$ will be an injection. However, the spaces I consider are not necessarily $T_1$. Therefore, I would like to know if these kinds of maps have a name or if they have been considered before.
|
Look at the fibers. Your condition implies that $f^{-1}(y) ⊆ cl(\{x\})$ for every $x ∈ f^{-1}(y)$, i.e. each fiber is indiscrete. So, such mapping is injective even if $X$ is $T_0$. Conversely, if each fiber is indiscrete, then the closure contains the “fiber-closure”.
Every continuous map is a unique composition of a quotient map and an injective continuous map. This condition is equivalent to the condition that the $T_0$ quotient factorizes through the quotient part of the original map.
|
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Whether the set $A$ , $ A=\{y\in R : y= \lim_{n\rightarrow \infty} f(x_{n}) \text{ where } x_n \text{ diverges to } +\infty \}$ is connected Let $f: \mathbb R \rightarrow \mathbb R$ be continuous function and $A\subset \mathbb R$ be defined by $$ A=\{y\in R : y= \lim_{n\rightarrow \infty} f(x_{n}) \text{ where } x_n \text{ diverges to } +\infty \}$$
Then $A$ is
A. Connected
B. Compact
C. Singleton
D. None of the above
Now by taking $f(x)=\sin(x)$ I see C. do not hold but A. and B. do. But how to ensure that what must be correct for all continuous functions or not and what special role the divergence of that sequence $x_n$ plays there?
|
As has been sort of said, $x\sin(x)$ is a counterexample to (B) and to (C).
But (A) is true. (And this is really awful notation, using "A" for a certain set and also for a condition that set may or may not satisfy).
It's enough to show that if $a,c\in A$ and $a<b<c$ then $b\in A$. Say $x_n\to\infty$, $f(x_n)\to a$, $z_n\to\infty$ and $f(z_n)\to c$.
For each positive integer $k$ there exist $n$ and $m$ such that $x_n>k$, $z_m>k$, $f(x_n)<b$ and $f(z_m)>b$. The intermediate value theorem shows that there exists $y_k$ between $x_n$ and $z_m$ such that $f(y_k)=b$. And $y_k\to\infty$, since $x_n>k$ and $z_m>k$.
|
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Permutations of the elements of $\mathbb Z_p$ Let $p$ be prime. Describe all permutations $\sigma$ of the elements of $\mathbb Z_p$, having the property that $\{\sigma(i)-i: i\in\mathbb Z_p\}=\mathbb Z_p$
(Added by Robert Lewis in an attempt to provide background, motivation, and other context for this engaging problem)
This problem essentially asks for a method of representing permutations $\sigma$ of the finite field $\Bbb Z_p$ which respects the algebraic structure/computations inherent in such fields.
|
These are known as complete mappings of the finite field $\mathbb{F}_{p}$. I think you will find that it is quite a difficult problem to describe all such permutations.
Here is a good overview:
Niederreiter, Harald; Robinson, Karl H., Complete mappings of finite fields, J. Aust. Math. Soc., Ser. A 33, 197-212 (1982). ZBL0495.12018.
You will see that this is very dependent on $p$, for example when $p=7$
we have $\pm {(x+a)}^{4} +3x + b$
giving examples for any choice of $a,b \in \mathbb{F}_{7}$; but finding and describing such polynomials relies on ad-hoc methods.
|
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Determine $P(S_n\leq1)$ where $S_n=\sum_{k=1}^nX_k$
Suppose that $X_n$ are i.i.d. $Uniform(0,1)$ random variables. Let $S_n=\sum_{k=1}^nX_k$ with $S_0:=0$. Then, determine $P(S_n\leq1)$.
I know that maybe by using Characteristic function of $S_n$ I will be able to get the d.f. of $S_n$ using Inversion Theorem but I do not want to use that.
I tried the following method:
I want to find $P(S_n>1)$ instead. Let $A=\{S_n>1\}$. Then define $A_k=\{S_{k-1}\leq1,S_k>1\}$. Notice that $A_k$ are disjoint, and $\cup_kA_k=A$.
Hence $P(A)=\sum_{k=1}^nP(A_k)=\sum_{k=1}^nP(S_{k-1}\leq1,S_k>1)=\sum_{k=2}^nP(S_{k-1}\leq1,S_k>1)$ because $P(S_0\leq1,S_1>1)=0$.
Now $P(S_{k-1}\leq1,S_k>1)=P(S_{k-1}\leq1, S_{k-1}>1-X_k)=P(1-X_k< S_{k-1}\leq 1)$
It seems I can't proceed from here, because I would need knowledge of $P(S_{k-1}\leq1)$. Will induction work here, then?
In general is there any formula for $P(S_n\leq a)$ where $a>0$?
|
Note that the vector $(X_1,X_2,...,X_n)$ has density $f(x_1,x_2,...,x_n)=1$ on the unit hypercube. So $\mathbf{P}(S_n\leq 1)$ is just the volume of the region bounded by the axis planes and the plane $x_1+x_2+\cdots+x_n=1$ and equals $1/{n!}$.
|
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Limit of Fraction $$\lim_{x \to \infty} \frac{(1 + x)^{x/(1 + x)}\cos^{4}x}{e^{x}}$$
Attempt: I've tried evaluating the limits of the terms individually using the property of limits. Also, $y=1/x$ subsituition hasn't helped me, any help will be appreciated.
|
Hint: For positive $x$, the top is $\ge 0$ and $\le 1+x$.
Now use Squeezing.
|
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Continued Fraction Counting Problem The house of my friend is in a long street, numbered on this side one, two, three, and so on. All the numbers on one side of him added up exactly the same as all the numbers on the other side of him.
There is more than fifty houses on that side of the street, but not so many as five hundred. If we find the number of the house where my friend lives, the problem had one solution — house no. 204 in a street of 288 houses, i.e.
1+2+ ... 203 = 205 + 206+ . . . 288.
But without the 50-to-500 house constraint, there are other solutions. For example, on an eight-house street, no. 6 would be the answer:
1 + 2 + 3 + 4 + 5 = 7 + 8.
Ramanujan is reported to have discovered a continued fraction comprising of a single expression with all such correct answers. What is the continued fraction, and how can it be found?
|
If it is house number $x$ in a street of $y$ houses, we have
$$\frac{x(x-1)}{2}+x+\frac{x(x-1)}{2}=\frac{y(y+1)}{2}$$
which simplifes to
$$(2y+1)^2-8x^2=1\ .$$
This can be solved by computing the continued fraction
$$\sqrt8=2+\frac{1}{1+{}}\frac{1}{4+}\frac{1}{1+{}}\frac{1}{4+\cdots}\ .$$
The table of convergents is
$$\matrix{&&2&1&4&1&4&1&4&1&4&\cdots\cr
0&1&2&\color{red}{3}&14&\color{red}{17}&82&\color{red}{99}&478&\color{red}{577}&2786&\cdots\cr
1&0&1&\color{red}{1}&5&\color{red}{6}&29&\color{red}{35}&169&\color{red}{204}&985&\cdots\cr}$$
The pairs beneath the second-last entry in each period (that is, in this case, beneath the $1$s) are marked in red. They give the integer values of $2y+1$ and $x$ satisfying the equation. For example we have firstly $y=1$, $x=1$ (the trivial solution - house numbers on both sides of $1$ add up to $0$). Then $y=8$, $x=6$ as you observed. The fourth solution is Ramanujan's.
|
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How to show that the cycle $(2 5) = (2 3) (3 4) (4 5) (4 3) (3 2)$ I generally do not have any problem multiplying cycles, but I've seen on Wikipedia that
$$(2 5) = (2 3) (3 4) (4 5) (4 3) (3 2). $$
I started following the path of $2$ on the right:
$$2\to3\to4\to5\to \ ?$$
Where does $5$ go? I should stop here, right? Then $2\to 5$, that is, $(25)$. But what about $(23)(34)$?
|
This is another conjugation problem in disguise:
$(2\ 3)(3\ 4)(4\ 5)(4\ 3)(3\ 2) = (2\ 3)[(3\ 4)(4\ 5)(3\ 4)^{-1}](3\ 2)$
$= (2\ 3)(3\ 5)(2\ 3)^{-1}$ (since $(3\ 4)$ takes $4 \to 3$ and fixes $5$)
$= (2\ 5)$ (since $(2\ 3)$ takes $3 \to 2$ and fixes $5$).
|
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Consider the function f(x)=sin(x) in the interval x=[π/4,7π/4]. The number and location(s) of the local minima of this function are? This is MCQ of a competitive exam(GATE), Answer is (d) given by GATE , and from other sources ,explanation is (b) somewhere and (d) somewhere , I am going with (b) as minimum at $270$, I have drawn graph . but it produces two pictures , I confused , it asked Local minima of a trivial $\sin(x)$ function and explanation here, but didn't get obviously , finally I want to explanation .
Problem is :
Consider the function $f(x)=\sin(x)$ in the interval $x=[\pi /4,7 \pi /4]$. The
number and location(s) of the local minima of this function are
(A) One, at $\pi /2$
(B) One, at $3 \pi /2$
(C) Two, at $ \pi /2$ and $3 \pi /2$
(D) Two, at $\pi /4$ and $3 \pi /2$
|
The local minima is at $x=\frac{3\pi}{2}$
This is very obvious from the graph of $f(x)=\sin{x}$
On a second look at the graph below, I believe $x=\frac{\pi}{4}$ is also a local minimum. This is because it is lesser than all other values within its locality.
Thus we have two local minima: $x=\frac{\pi}{4}, \frac{3\pi}{2}$
|
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Does $A\cap B =\varnothing \Rightarrow B\subseteq \overline{A}$? How to prove $A\cap B =\varnothing \Rightarrow B\subseteq \overline{A}$?
If I going by definitions, there is no $x$ s.t $x\in A$ and $x\in B$.
But, what do we can tell about $\overline{A}$? What i'm missing?
|
Let's assume $A\cap B =\varnothing$ (start hypothesis)
Let $x \in B$
Since $A$ and $B$ are disjoint (start hypothesis), then $x \notin A$
By definition of $\overline A$, since $x \notin A$ then $x \in \overline A ~~~~(= \Omega - A)$
Therefore $B\subseteq \overline A$, because for all $x \in B$, we have $x \in \overline A$
Note that the reciprocal is also true.
|
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Given vector $\vec x = \left\{ x_i\right\}_{i=1}^n$ find an algebraic expression for $\vec y = \left\{ x^2_i\right\}_{i=1}^n$ Given vector
$$\vec x = \begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix},$$
How can we write out vector
$$\vec y = \begin{bmatrix} y_1 \\ \vdots \\ y_n \end{bmatrix} := \begin{bmatrix} x^2_1 \\ \vdots \\ x^2_n \end{bmatrix}$$
in terms of $\vec x$ using only matrix operations?
It is simple to write $\vec y$ in terms of $\vec x$ element wise, for example in the form of system of equations $y_i = x_i^2$ for $i = 1, \dots, n$.
However, I am struggling to do so using matrix notation and operations.
The best guess I could come up with was to write expressions like
$$
\begin{aligned}
\vec y &= \vec x ^T \cdot I_{n \times n} \cdot\vec x,
& I_{n \times n} -\text{ identity matrix, } &
& I_{n \times n} & =
\begin{bmatrix}
1 & 0 & \cdots & 0 \\
0 & 1 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 1 \\ \end{bmatrix}
\\
\vec y &= \left\langle \vec x, \vec x \right \rangle
= \left\| \vec x \right\|
& - \text{ inner product / norm, } &
& \left\| \vec x \right\| &= \left\langle \vec x, \vec x \right \rangle =
\sum_{i=1}^{n} x_i^2
\end{aligned}
$$
both of which are obviously flawed.
Any hint would be appreciated.
|
$\begin{pmatrix}
x_1 & 0 & 0\\
0 & x_2 & 0\\
0 & 0 & x_3\\
\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}x_1^2\\x_2²\\x_3^2\end{pmatrix}$
|
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|
Use stirlings approximation to prove inequality. I have come across this statement in a text on finite elements. I can give you the reference if that will be useful. The text mentions that the inequality follows from Stirling's formula. I can't prove it to myself but I think it is true from checking values with Mathematica.
Let $p,k\in \mathbb{N}$ with $k\le p$. Then we have
$$\frac{(p-k)!}{(p+k)!} \le \left(\frac{\theta}{p}\right)^{2k},\, \theta = \left(\frac{e}{2}\right)^{k/p}.$$
Does anyone have an idea of how to get this inequality from Stirling's approximation?
This is my attempt at a solution. To build intuition, let's look at the case $k=p$ we have
$$
\begin{align*}
\frac{(p-k)!}{(p+k)!} &= \frac{1}{(2p)!} \\
&\le \frac{1}{\sqrt{2\pi}} \frac{\exp(2p)}{(2p)^{2p+1/2} } \\
&= \frac{1}{\sqrt{2\pi}} \left(\frac{e}{2} \right)^{2p} \frac{1}{\sqrt{2}} \frac{1}{p^{2p}} \frac{1}{p^{1/2}}\\
& \le \frac{1}{\sqrt{2\pi}} \left( \frac{e}{2} \right)^{2p} \frac{1}{\sqrt{2}} \frac{1}{p^{2p}} \\
&= \frac{1}{\sqrt{2\pi}} \frac{1}{\sqrt{2}} \left( \frac{\theta}{p} \right)^{2k}\\
&\le \left( \frac{\theta}{p} \right)^{2k}
\end{align*}
$$
And this is what we wanted to show.
Now let's look at the case $k=p-1$, then we get
$$
\begin{align*}
\frac{(p-k)!}{(p+k)!} &=
\frac{1}{(2p-1)!} \\
&\le \frac{1}{\sqrt{2\pi}} \frac{\exp( 2p-1)}{ (2p-1)^{2p-1/2}} \\
&= \frac{1}{\sqrt{2\pi}} \frac{\exp(2p-1)}{2^{2p-1/2}} \frac{1} {(p-1/2)^{2p-1/2} }\\
& = \frac{1}{\sqrt{2\pi}} \frac{1}{\sqrt{2}} \left(\frac{e}{2} \right)^{2p-1} \frac{1}{(p-1/2)^{2p-1/2}}\\
&\le \frac{1}{2\sqrt{\pi}} \left( \frac{e}{2} \right)^{2p-1} \frac{1}{(p-1/2)^{2p-1}} \\
&\le ??
\end{align*}$$
This is where I am stuck. For the $k=p-1$ case, I can multiply and divide by $(\frac{e}{2})^{3-2/p}$ to get something of the correct form, but then it seems I would have to to prove that for all $p$
$$\left(\frac{e}{2}\right)^{3-2/p} \frac{1}{(p-1/2)^{2p-1}} \le 1.$$ This seems like a dead end to me. And this is still just a special case of the general result. Any ideas?
|
For
$\left(\frac{e}{2}\right)^{3-2/p} \frac{1}{(p-1/2)^{2p-1}}
\le 1
$,
$\begin{array}\\
(p-1/2)^{2p-1}
&=p^{2p-1}(1-1/(2p))^{2p-1}\\
&\approx p^{2p-1}(1/e)(1-1/(2p))^{-1}
\quad\text{since }(1-1/(2p))^{2p} \approx 1/e\\
\end{array}
$
so
$\begin{align*}\\
\left(\frac{e}{2}\right)^{3-2/p} \frac{1}{(p-1/2)^{2p-1}}
&\approx \left(\frac{e}{2}\right)^{3-2/p}\frac1{p^{2p-1}}e(1-1/(2p))\\
&< \left(\frac{e}{2}\right)^{3-2/p}\frac1{p^{2p-1}}e\\
&< \left(\frac{e}{2}\right)^{3}\frac1{p^{2p-1}}e\\
&= \frac{e^4}{8p^{2p-1}}\\
&\ll 1 \quad\text{for } p > 5
\end{align*}
$
So this inequality
is very strongly true.
|
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|
What is the solution of $a^b=a+b$ in terms of $a$?
Let $a, b$ be real numbers. Solve
$$a^b=a+b$$
for $a$.
If there isn't a solution with $a, b$ real, maybe $a, b$ should be complex.
But no matter how hard I try, this is proving to be very difficult to
do. Would anyway be kind enough to show me the solution to this? Thank
you!
|
I will use Lambert-W function defined as the following:
$$W(x)e^{W(x)}=x$$
Your equation:
$$a^b=a+b$$
Multiply it by $a^a$ to make the exponent more "friendly" and use some tricks
$$a^{a+b}=(a+b)a^a$$
$$\frac{1}{a+b}a^{a+b}=a^a$$
$$(a+b)a^{-(a+b)}=a^{-a}$$
$$-(a+b)a^{-(a+b)}=-a^{-a}$$
$$-\ln(a)(a+b)e^{-\ln(a)(a+b)}=-\ln(a)a^{-a}$$
$$-\ln(a)(a+b)= W(-\ln(a)a^{-a})$$
$$a+b= -\frac{W(-\ln(a)a^{-a})}{\ln(a)}$$
$$b= -\frac{W(-\ln(a)a^{-a})}{\ln(a)}-a$$
When solving for $b$ is easy, solving for $a$ is not and I'm not sure if even possible using known functions.
EDIT: Here is the plot of $a(b)$.
|
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|
$f$ is an analytic function in the disk $D=\{z\in\mathbb{C}\,:\,|z|\leq 2\}$ such that $\iint_D=|f(z)|^2\,dx\,dy\leq 3\pi$. Maximize $|f''(0)|$
Determine the largest possible value of $|f''(0)|$ when $f$ is an analytic function in the disk $D=\{z\in\mathbb{C}\,:\,|z|<2\}$ with the property that $\iint_{D}|f(z)|^2\,dx\,dy\leq 3\pi$.
I don't really know what to do with the assumption that $\iint_D|f(z)|^2\,dx\,dy\leq 3\pi$. I believe you could use Stokes' theorem to rewrite this as a line integral on $\partial D$, but I'm really rusty with my usage, so I'm kind of stuck.
If I could get a bound on $\int_{\partial D}|f(z)|^2\,dz$, I could probably use harnack's inequality for subharmonic functions to get a bound on $|f|$ then Cauchy's inequality, however I'm not very sure about my usage of green's theorem (if that's even the right way to go) any help is greatly appreciated. Thanks
|
$$\langle f,f \rangle=\iint_Df(z)\times\overline{f(z)}dxdy=\iint_D |f(z)|^2dxdy \leqslant 3\pi$$
Power series of $f(z)$ about $z=0$:
$$\sum_{n=0}^{\infty}a_nz^n=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}z^n$$
then you can easily prove (using polar coordinates, edit: or more easily by defining a base using $z^n$ then using the generalized form of Parseval's Identity) that $$\langle f,f \rangle=\pi\sum_{n=0}^{\infty}\frac{|a_n|^2\times 2^{2n+2}}{n+1} \leqslant 3\pi$$
$$\pi \frac{|a_2|^2\times 2^{2*2+2}}{2+1} \leqslant \pi\sum_{n=0}^{\infty}\frac{|a_n|^2\times 2^{2n+2}}{n+1} \leqslant 3\pi$$
$$|f''(0)/2!|^2\times\frac{64}{3} \leqslant3$$
$$|f''(0)| \leqslant\frac{3}{4}$$
CAVEAT : there might be more than a few mistakes, it's been some time since I've last done something like this.
|
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|
Math Subject GRE 1268 Question 55 If $a$ and $b$ are positive numbers, what is the value of $\displaystyle \int_0^\infty \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx$.
|
The integral being considered is, and is evaluated as, the following.
\begin{align}
I &= \int_{0}^{\infty} \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx \\
&= \int_{0}^{\infty} \frac{dx}{1 + e^{bx}} - \int_{0}^{\infty} \frac{dx}{1 + e^{ax}} \\
&= \left( \frac{1}{b} - \frac{1}{a} \right) \, \int_{1}^{\infty} \frac{dt}{t(1+t)} \mbox{ where $t = e^{bx}$ in the first and $t = e^{ax}$ in the second integral } \\
&= \left( \frac{1}{b} - \frac{1}{a}\right) \, \lim_{p \to \infty} \, \int_{1}^{p} \left( \frac{1}{t} - \frac{1}{1+t} \right) \, dt \\
&= \left( \frac{1}{b} - \frac{1}{a}\right) \, \lim_{p \to \infty} \, \left[ \ln(t) - \ln(1+t) \right]_{1}^{p} \\
&= \left( \frac{1}{b} - \frac{1}{a}\right) \, \lim_{p \to \infty} \left[ \ln\left( \frac{p}{1 + p}\right) + \ln 2 \right] \\
&= \left( \frac{1}{b} - \frac{1}{a}\right) \, \lim_{p \to \infty} \left[ \ln\left( \frac{1}{1 + \frac{1}{p}}\right) + \ln 2 \right] \\
&= \left( \frac{1}{b} - \frac{1}{a} \right) \, \ln 2
\end{align}
Note: Originally the statement "This is valid if $a \neq b$" was given at the end of the solution. Upon reflection it is believed that the statement should have been "This is valid for $a,b \neq 0$".
|
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|
How can the axiom of choice be called "axiom" if it is false in Cohen's model? From what I know, Cohen constructed a model that satisfies $ ZF\neg C $.
But if such a model exists, how can AC be an axiom? Wouldn't it be a contradiction to the existence of the model?
Only explanation I can think of, is that this model requires other axioms in addition to ZF, and these axioms contradict AC. Is it true?
|
The existence of a model of a statement does not mean that statement is "true" (whatever that means; see below). For example, the Poincare disk is a model of Euclid's first four postulates plus the negation of the parallel postulate; this does not mean that the parallel postulate is "false."
What having a model of a set of statements does mean, is: that set of statements is consistent. In particular, if there is a model of $T\cup\{\neg p\}$, then $p$ cannot be a consequence of $T$. Cohen's result, for instance, shows that $AC$ cannot be proved from $ZF$ alone.
As to your comments about its value of an axiom: knowing that there is a model of $ZF$ where $AC$ fails shows that, if we believe that $AC$ is "true" (again, see below), then we have a good reason to adopt it as an axiom. Conversely, if there were no model of $ZF+\neg AC$, that would mean (by Goedel's completeness theorem) that $ZF$ proves $AC$, so there would be no need to add $AC$ as an additional axiom.
OK, so now another question: what is going on when we use the word "true" in these contexts? Basically, we're presupposing the existence of a "correct" model. But the statement "$AC$ is true in the actual universe of set theory" doesn't have anything to do with the statement "there is some model in which $AC$ fails."
To repeat: all we know from the existence of a model of $ZF+\neg AC$ is that $AC$ is not a consequence of $ZF$.
|
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|
Derivative of sum of powers For fixed $n \geq 1$ and $p \in [0,1]$, is there a nice expression for the derivative of
$\sum_{k=0}^n p^k (1-p)^{n-k}$
with respect to p?
|
I believe you can use the fact that $\displaystyle \dfrac{\text{d}}{\text{d}p} \sum \left( \dots \right) = \sum \dfrac{\text{d}}{\text{d}p} \left( \dots \right) $. $$ \begin{aligned} \dfrac{\text{d}}{\text{d}p} \sum_{k=0}^{n} p^k (1-p)^{n-k} & = \sum_{k=0}^{n} \dfrac{\text{d}}{\text{d}p} \left( p^k (1-p)^{n-k} \right) \\ & = \sum_{k=0}^{n} \left( kp^{k-1}(1-p)^{n-k} - (n-k)p^k (1-p)^{n-k-1} \right) \\ & = \sum_{k=0}^{n} p^{k-1} (1-p)^{n-k-1} \left( k(1-p) - (n-k)p \right) \\ & = \sum_{k=0}^{n} p^{k-1} (1-p)^{n-k-1} \left( k - np \right) \end{aligned} $$
It might be handy knowing that if you were dealing with $\binom{n}{k}$ multiplied the summand, you'd indeed have $(p+(1-p))^n$ written out as a sum.
|
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|
$z^3=w^3 \implies z=w$? I've reached this in another problem I have to solve: $z,w \in \Bbb {C}$. $z^3=w^3 \implies z=w$?
I've scratched my head quite a bit, but I completely forgot how to do this, I don't know if this is correct:
$$
z^3=|z^3|e^{3ix}=|w^3|e^{3iy}
$$
I know the absolute values are equal, so I get $3ix=3iy \implies x=y \implies z=w$. I'm not sure I solved this correctly, I know this is pretty basic, but I haven't done this stuff in a year...
|
We assume $z \ne 0 \ne w$; it is obvious that in the other circumstance $z = w = 0$ is the only possible solution.
Now, since
$z^3 = w^3 \Leftrightarrow z^3 - w^3 = 0, \tag{1}$
then
$(z - w)(z^2 + zw + w^2) = z^3 - w^3 = 0; \tag{2}$
we see that if
$z^2 + zw + w^2 \ne 0, \tag{3}$
then
$z = w; \tag{4}$
furthermore, if (4) binds then
$z^2 + zw + w^2 = 3z^2 \ne 0; \tag{5}$
thus
$z = w \Leftrightarrow z^2 + zw + w^2 \ne 0; \tag{6}$
certainly $z = w$ is a possible solution to (1); (6) indicates that the any other prospects are to be found via the logically equivalent
$z \ne w \Leftrightarrow z^2 + zw + w^2 = 0; \tag{7}$
we therefore scrutinize
$z^2 + zw + w^2 = 0 \tag{8}$
for solutions of (1) other than $z = w$. Perhaps the most straightforward and clearest way to proceed is to exploit the assumption $w \ne 0$ and divide (8) through by $w^2$:
$(\dfrac{z}{w})^2 + \dfrac{z}{w} + 1 = 0; \tag{9}$
if we now set
$\omega = \dfrac{z}{w}, \tag{10}$
we find
$\omega^2 + \omega + 1 = 0. \tag{11}$
(11) is readily recognized as the equation for the non-real cube roots of unity; indeed we have, taking for the moment $z = \omega$ and $w = 1$ in (2),
$(\omega - 1)(\omega^2 + \omega + 1) = \omega^3 - 1 = 0, \tag{12}$
which shows that if
$\omega^3 = 1, \;\; \omega \ne 1, \tag{13}$
then $\omega$ satisfies (11); it is easily seen from the quadratic formula that the two such $\omega$ are
$\omega = -\dfrac{1}{2} \pm \dfrac{i}{2}\sqrt{3}; \tag{14}$
we recognize that
$\cos \dfrac{2\pi}{3} = \cos -\dfrac{2\pi}{3} = -\dfrac{1}{2},$
$\sin \dfrac{2\pi}{3} = \dfrac{\sqrt{3}}{2}; \tag{15}$
therefore, if $\omega \ne 1$ we have
$\omega = \cos \dfrac{2\pi}{3} \pm i \sin \dfrac{2\pi}{3}$
$= e^{\pm 2\pi i /3} = e^{2\pi i / 3}, e^{4\pi i / 3}; \tag{16}$
now since
$\dfrac{z}{w} = \omega = e^{\pm 2\pi i / 3}, \tag{17}$
it follows that, other than $z = w$, the solutions to (1) are
$z = e^{\pm 2 \pi i /3} w. \tag{18}$
(18) is easily verified:
$z^3 = (e^{\pm 2 \pi i / 3}w)^3 = w^3(e^{\pm 2\pi i / 3})^3$
$= w^3 e^{\pm 2 \pi i} = w^3, \tag{19}$
since $e^{\pm 2 \pi i} = 1$. We thus see that (1) entails precisely three possibilities:
$z = w, z = e^{2\pi i /3} w, z = e^{- 2 \pi i /3} = e^{4 \pi i /3} w; \tag{20}$
it's really all about equations (11)-(12); and of course, there are the nearly "self-evident truths" of the generalization of these results to the case $z^n = w^n$ for $n \in \Bbb Z$.
|
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|
Angle between two parabolas I'm a little confused about a problem that asks me to find the angle between the two parabolas $$y^2=2px-p^2$$ and $$y^2=p^2-2px$$ at their intersection. I used implicit differentiation to find the slopes $$y'=\frac{-p}{y}$$ and $$y'=\frac{p}{y}$$ Using the formula for the angle between two lines $\tan\alpha=|\frac{m_2-m_1}{1+m_2m_1}|$ with some substitutions I ended up with $\tan\alpha=|\frac{y}{x}|$. When I set the two equations equal to one another to find the point of intersection I ended up with $x=0$ and $y=±p$.
I know these values lead to tangent being undefined, can I infer from this fact that the angle between these parabolas is a right angle? The solution says that they intersect at right angles, but I'm confused about using something that is undefined to formulate my answer, if it turns out to be correct. Thanks for any help.
|
First, let's find the point of intersection. This occurs when
$$p^2-2px_0=2px_0-p^2\implies x_0=p/2\implies y=0$$
At the point of intersection, $y=0$ and thus $yy'=\pm p$ implies that $y'$ is undefined and therefore the tangent to both parabolas is a vertical line in the $x-y$ plane defined be the equation $x=p/2$.
Inasmuch as the tangent lines are parallel, the angle between them is $0$ and we are done!
|
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|
Existence of solutions of the equation with a limit. Let f be continuous function on [0,1] and
$$\lim_{x→0} \frac{f(x + \frac13) + f(x + \frac23)}{x}=1$$
Prove that exist $x_{0}\in[0,1]$ which satisfies equation $f(x_{0})=0$
I suppouse that the numerator should approach $0$ which would implicate that for
x near $0$ $f(x + \frac13)$ would be of opposite sign then $f(x + \frac23)$ or both be $0$.
Then by intermediate value theorem we would know that there is $x_{0}\in[\frac13,\frac23]$ which fulfill $f(x_{0})=0$
Yet, I have no idea how to prove that numerator $\rightarrow 0$
|
First, prove that the numerator should be zero. You can do this by contradiction. Prove that if $$\lim_{x\to 0} f(x+\frac13) + f(x+\frac23)\neq 0$$
then the original limit cannot exist.
Now, you can use continuity to show that $f(\frac23) + f(\frac13)=0$, and then use a well known theorem to finish your proof.
|
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|
Determining if function odd or even This exercise on the Khan Academy requires you to determine whether the following function is odd or even
f(x) = $-5x^5 - 2x - 2x^3$
To answer the question, the instructor goes through the following process
*
*what is f(-x)
*f(-x) = $-5(-x)^5 - 2(-x) - 2(-x)^3$
*f(-x) = $5(x)^5 +2(x) + 2(x)^3$
*f(-x) = $-(-5x^5 - 2x - 2x^3)$
*f(-x) = -f(x)
Therefore f(x) is odd.
Question. Why is everything converted to a negative both inside and outside the parenthesis at step 4?
At step 2, he's making x negative. At step 3, we get the result of making the x values negative. So why add the negatives again at step 4?
|
Let's follow the simplest process
Notice, $$f(x)=-5x^5-2x-2x^3$$ $$f(-x)=-5(-x)^5-2(-x)-2(-x)^3$$ $$ =5x^5+2x+2x^3$$ $$\implies f(x)+f(-x)=-5x^5-2x-2x^3+5x^5+2x+2x^3=0$$
Hence, $f(x)$ is odd.
|
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|
Can someone help me give a proof for this? I know there are theorems about integrals of odd and even functions, but i kept wondering about integrals that share symmetry around an axis $x=c$. I've been trying to give a proof for this but can't seem to get around it; could someone help me prove/disprove this?
$$
\large \int_{c-x}^{c+x}f(x)dx=2\int_{c}^{c+x}f(x)dx
$$
Hypothesis--------------------
$$
\large f(c-x)=f(c+x)
$$
$f(x)$ is symmetric around $x=c$ for all $x$
|
Two initial remarks:
*
*For clarity, you should not use the same letter in the limits of integragion and as the integration variable itself.
*I assume your hypothesis is that, for all $x$, $f(c-x)=f(c+x)$.
Let us prove that
$$
\large \int_{c-x}^{c+x}f(t)dt=2\int_{c}^{c+x}f(t)dt
$$
Proof:
First, let us consider:
$$ \int_{c-x}^{c}f(t)dt$$
Let $s=2c-t$, then we have: $t=2c-s$, $dt=-ds$ and
$$ \int_{c-x}^{c}f(t)dt=-\int_{c+x}^{c}f(2c-s)ds=\int_{c}^{c+x}f(c+(c-s))ds=\int_{c}^{c+x}f(s)ds$$
In the last step above, we used the hypothesis to concluded that $f(c+(c-s))=f(c-(c-s))=f(s)$.
Now we have
$$
\int_{c-x}^{c+x}f(t)dt=\int_{c-x}^{c}f(t)dt + \int_{c}^{c+x}f(t)dt=\int_{c}^{c+x}f(s)ds+ \int_{c}^{c+x}f(t)dt=2\int_{c}^{c+x}f(t)dt
$$
|
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|
"Mean value like" problem. Let $f:\mathbb{R} \longrightarrow \mathbb{R}$ be differentiable, take $a<a'<b<b'$. Prove that there exists $c<c'$ such that $$\frac{f(b)-f(a)}{b-a}=f'(c) \quad and \quad \frac{f(b')-f(a')}{b'-a'}=f'(c').$$ My first tries were connected with mean value because we can find such $c,c'$ but we don't konw if they satisfie required relation. We know though that they are in $(a',b)$. I ask for some hints.
|
Consider the function
$$
g(x)=f(x)-{f(b)-f(a)\over b-a}(x-b).
$$
It is easy to prove that $g(a)=g(b)$ and $g'( c)=0$. Moreover, if
$\displaystyle{g(x_2)-g(x_1)\over x_2-x_1}=g'(\xi)$ then also
$\displaystyle{f(x_2)-f(x_1)\over x_2-x_1}=f'(\xi)$. So we can prove the theorem for $g$ and the result will hold for $f$ too.
We can take as $c$ any stationary point of $g$ in $(a,b)$.
If we can choose $c\le a'$ then $c<c'$ and we are done.
If not, suppose $g(a')>g(a)$ (the case $g(a')<g(a)$ can be handled in a similar way and the case $g(a')=g(a)$ is easy because then we can choose $c\in(a,a')$) and take $c$ as the absolute maximum point for $g$ on $(a,b)$. Then we have $g(a)<g(a')<g( c)$ and in addition there exists $a'_1\in(c,b)$ such that $g(a'_1)=g(a')$. Let $r$ be the line in the $(x,y)$ plane passing through $(a',g(a'))$ and $(b',g(b'))$. There are two possibilities:
1) if $g(b')<g(a')$, then the line parallel to $r$ and passing through $(c, g( c))$ intersects the graph of $g$ at a point $(b'', g(b''))$ with $b''>c$, and there exists $c'$ as required in $(c,b'')$;
2) if $g(b')\ge g(a')$, then the line parallel to $r$ and passing through $(a'_1, g(a'_1))$ intersects the graph of $g$ at a point $(b'_1, g(b'_1))$ with $b'_1>a'_1>c$, and there exists $c'$ as required in $(a'_1,b'_1)$.
|
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|
Can a space $X$ be homeomorphic to its twofold product with itself, $X \times X$? Let $X$ be a topological space of infinite cardinality. Is it possible for any $X$ to be homeomorphic to $X\times X$ $?$
For example, $\mathbb R$ is not homeomorphic to $\mathbb R^{2}$, and $S^{1}$ is not homeomorphic to $S^{1} \times S^{1}$ . What other topological spaces might we consider$?$ What properties of a space may ensure or contradict this possibility$?$ From the little topology I have learnt yet, I have not seen this happening.
|
At this level of generality you can make $X=X \times X$ happen quite easily. Take a discrete space of any infinite cardinality, for instance. Or topologize $X=A^B$ by whatever means and compare $X \times X = A^{B \sqcup B}$; under various mild assumptions on $B$ those spaces would be homeomorphic.
|
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|
Proving an Inequality using a Different Method Is there another way to prove that: If $a,b\geq 0$ and $x,y>0$
$$\frac{a^2}{x} + \frac{b^2}{y} \ge \frac{(a+b)^2}{x+y}$$
using a different method than clearing denominators and reducing to $(ay-bx)^2 \ge 0$?
|
One method of solving any inequality is to determine points where the associated equality is true or where the functions involved are not continuous. That is because, as long as $a$ and $b$ are continuous, we can only go from "$a> b$" to "$a< b$", or vice-versa, by going through "$a= b$".
However here, the simplest way to solve the equality is by "clearing denominators" so apparently you would not consider that a "different" way.
|
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|
Sums of Fourth Powers While fooling around on my calculator I found:
$$7^4 + 8^4 + (7 + 8)^4 = 2 * 13^4$$
$$11^4 + 24^4 + (11 + 24)^4 = 2 * 31^4$$
I'm intrigued but I can't explain why these two equations are true. Are these coincidences or is there a formula/theorem explaining them?
|
You have a disguised version of triangles with integer sides and one $120^\circ$ angle. These are
$$ 3,5,7 $$
$$ 7,8,13 $$
$$ 5,16,19$$
$$ 11,24, 31, $$
which solve
$$ a^2 + ab + b^2 = c^2. $$
Square both sides and then double both sides and you get your identities.
These can be generated by a coprime pair of number $m,n$ with
$$ a = m^2 - n^2 $$
$$ b = 2mn+n^2 $$
$$ c = m^2 + mn + n^2 $$
|
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|
Use the discriminant to show that $mx−y + m^2 = 0$ touches the parabola $x^2 =−4y$, for all values of m. Use the discriminant to show that $mx−y + m^2 = 0$ touches the parabola $x^2 =−4y$, for all values of m.
I attempted to solve by letting them both equal each other, but it didn't work. How do I do this question?
Thank You in advance.
|
So the two functions are $y=-\frac{1}{4}x^2$ and $y=mx+m^2$.
Setting them equal, we have $-\frac{1}{4}x^2=mx+m^2$
which rearranges to $0=x^2+4mx+4m^2$
then you can use the quadratic formula (or simply complete the square) to get the result. I suspect that this is where you are supposed to use the discriminant, since in the case the discriminant would be $16m^2-16m^2=0$, showing that there is always one solution.
|
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|
sum of the residues of all the isolated simgualrities Prove that, for $n \geq 3$, the sum of the residues of all the isolated singularities of
$$\frac{z^n}{1+z+z^2+\cdots+z^{n-1}}$$
is 0
Can someone show me how to do this problem. Thank you.
|
Let
$$
F(z)=\frac{z^n}{1+z+z^2+\ldots+z^{n-1}}=\frac{P(z)}{Q(z)}.
$$
Since $Q(1)=n\ne 0$, then, for every $z\ne 1$ we have
$$
Q(z)=\frac{1-z^n}{1-z},
$$
and $F$ can be redefined as
$$
F(z)=\begin{cases}
\frac{(z-1)z^n}{z^n-1} &\mbox{ for } z\ne 1\\
\frac1n &\mbox{ for } z=1
\end{cases}
$$
Therefore, the set of isolated singularities of $F_n$ is given by:
$$
Q^{-1}(0)=\{z_{k,n}=z_n^k:\, 1\le k\le n-1\},\quad z_n=e^{i\frac{2\pi}{n}}
$$
We want to calculate the sum
$$
S_n=\sum_{k=1}^{n-1}\mathrm{Res}(F_n,z_n^k),
$$
where
$$
\mathrm{Res}(F,z_n^k)=\frac{(z_n^k-1)(z_n^k)^n}{n(z_n^k)^{n-1}}=\frac{(z_n^k)^2-z_n^k}{n},
$$
and we should assume that $n\ge 3$, because for $n=2$ the set $Q^{-1}(0)$ contains one element.
We get:
\begin{eqnarray}
nS_n&=&\sum_{k=1}^{n-1}\left[(z_n^k)^2-z_n^k\right]=\sum_{k=1}^{n-1}(z_n^2)^k-\sum_{k=1}^{n-1}z_n^k=\sum_{k=0}^{n-1}(z_n^2)^k-\sum_{k=0}^{n-1}z_n^k\\
&=&\frac{1-(z_n^2)^n}{1-z_n^2}-\frac{1-z_n^n}{1-z_n}=\frac{1-(z_n^n)^2}{1-z_n^2}-\frac{1-z_n^n}{1-z_n}.
\end{eqnarray}
Using the fact that $z_n^n=1$, we conclude that
$$
S_n=\frac1n\left[\frac{1-(z_n^n)^2}{1-z_n^2}-\frac{1-z_n^n}{1-z_n}\right]=0.
$$
|
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|
Calculate 3D Vector out of two angles and vector length What is the easiest way to calculate vector coordinates in 3D given 2 angles vector length?
Input:
*
*Angle between X and Y axis: $$\alpha \in [0, 360).$$
*Angle between Y and Z axis: $$\beta\in [0, 360).$$
*Scalar length.
Expected output:
X, Y, Z coordinates of the vector
|
You have presented a point in spherical coordinate system, which needs to be converted to Cartesian. The line joining the origin to the point will be the vector. Let the scalar length be $r$. If the $\beta$ angle is measured from $Y$ axis towards $Z$ axis and $\alpha$ from $X$ axis towards $Y$:
$$x = r \cos \beta sin \alpha$$
$$y = r \cos \beta cos \alpha$$
$$z = r \sin \beta$$
The vector is: $x \hat{i} + y\hat{j} + z\hat{k}$.
|
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|
does simply connectedness require connectedness? My question consists of two parts.
$1)$
suppose domain $D=\{(x,y)\in\mathbb R^2~|~xy>0\}$ is given. Now that is first quadrant and third quadrant with exclusion of $x$ and $y$ axis. We can easily see that $D$ is not connected, since there is a discontinuity at origin. But every closed curve we can construct in domain contains interior of it (or formally, they can shrunk to a point). So do we call it simply connected, or do we also need connectedness to say $D$ is simply connected?
$2)$
Now for second part, let origin also included in domain so that $D$ is connected. Let us construct closed curve which goes through origin. Now this curve also satisfies assumption given above. But it is not simple closed curve. Does this affects simply connectedness? Do we need simple closed curves for simply connectedness?
|
According to the Wikipedia, that definition of simply connectedness excludes all the non path connected spaces.
The injectivity of the closed curves is not mentioned and not required.
Your second example thus becomes simply connected.
|
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|
Find $\lim\limits_{x\to 0}\frac{\sqrt{2(2-x)}(1-\sqrt{1-x^2})}{\sqrt{1-x}(2-\sqrt{4-x^2})}$ I use L'Hospitals rule, but can't get the correct limit.
Derivative of numerator in function is
$$\frac{-3x^2+4x-\sqrt{1-x^2}+1}{\sqrt{(4-2x)(1-x^2)}}$$
and derivative of denominator is
$$\frac{-3x^2+2x-2\sqrt{4-x^2}+4}{2\sqrt{(1-x)(4-x^2)}}$$
Now, L'Hospitals rule must be applied again. Is there some easier way to compute the limit?
Limit should be $L=4$
|
using Bernoulli $$x \to 0 \\ {\color{Red}{(1+ax)^n \approx 1+anx} } \\\sqrt{1-x^2} = (1-x^2)^{\frac{1}{2}} \approx 1-\frac{1}{2}x^2 \\ \sqrt{4-x^2}=\sqrt{4(1-\frac{x^2}{4}})=2(1-\frac{x^2}{4})^{\frac{1}{2}} \approx 2(1-\frac{1}{2} \frac{x^2}{4})=2-\frac{x^2}{4} $$ so by putting them in limit :
$$\lim_{x \to 0} \frac{\sqrt{2(2-x)} (1-\sqrt{1-x^2})}{\sqrt{1-x}(2-\sqrt{4-x^2})}=\\\lim_{x \to 0} \frac{\sqrt{2(2-x)} (1-(1-\frac{1}{2}x^2))}{\sqrt{1-x}(2-(2-\frac{x^2}{4}))} =\\ \lim_{x \to 0} \frac{\sqrt{2(2-x)} (\frac{1}{2}x^2)}{\sqrt{1-x}(\frac{x^2}{4})}=\\\lim_{x \to 0} \frac{\sqrt{2(2-x)} 2}{\sqrt{1-x}}=4 $$
|
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|
last two digits of $14^{5532}$? This is a exam question, something related to network security, I have no clue how to solve this!
Last two digits of $7^4$ and $3^{20}$ is $01$, what is the last two digits of $14^{5532}$?
|
${\rm mod}\,\ \color{#c00}{25}\!:\, \ 14\equiv 8^{\large 2}\Rightarrow\, 14^{\large 10}\equiv \overbrace{8^{\large 20}\equiv 1}^{\rm\large Euler\ \phi}\,\Rightarrow\, \color{#0a0}{14^{\large 1530}}\equiv\color{#c00}{\bf 1}$
${\rm mod}\ 100\!:\,\ 14^{\large 2}\, \color{#0a0}{14^{\large 1530}} \equiv 14^{\large 2} (\color{#c00}{{\bf 1}\!+\!25k}) \equiv 14^{\large 2} \equiv\, 96$
|
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|
How many bit strings of length 10 contains... I have a problem on my home work for applied discrete math
How many bit strings of length 10 contain
A) exactly 4 1s
the answer in the book is 210
I solve it
$$C(10,4) = \frac{10!}{4!(10-4)!} = 210
$$
for the last 3 though I can't even get close. Did I even do part A right or is the answer only a coincidence?
B) at most 4 1s
the answer in the book is 386
C) at least 4 1s
the answer in the book is 848
D) an equal number of 0s and 1s
the answer in the book is 252
|
For part A it will be simply $C(10,4)=210$
For part B it will be $C(10,0)+C(10,1)+C(10,3)+C(10,4)=386$ because at most means maximum which we actually start from zero till that particular number
For part C it will be simply $C(10,4)+C(10,5)+C(10,6)+C(10,7)+C(10,8)+C(10,9)+C(10,10)=848$
For part D it will be simply $C(10,5)=252$
|
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|
Radical under Radical expression how to find the sum of $\sqrt{\frac54 + \sqrt{\frac32}} + \sqrt{\frac54 - \sqrt{\frac32}} $ ? Is there a method to solve these kind of equations ?
|
Why does squaring work here? Well it obviously deals with part of the square root. But there are two other things going on.
The first is that $$\left(\sqrt{a+\sqrt b}\right)^2+\left(\sqrt{a-\sqrt b}\right)^2=a+\sqrt b+a-\sqrt b=2a$$
And this means that the inner square roots disappear in part of the square.
The second is that $(a+\sqrt b)(a-\sqrt b)=a^2-b=c^2$. Here the cross term might have left a square root, but $c^2$ has a simple form.
Quite often in questions like this conjugates (expressions with the sign of the square root changed) come into play - the sum and product of two conjugate expressions both tend to be simple. So one technique for such questions is to look for a means to exploit that simplicity.
Note that the idea of conjugates can be generalised considerably, and is one of the ideas behind Galois Theory.
|
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|
If the product of two matrices, $A$ and $B$ is zero matrix, prove that matrices $A$ and $B$ don't have to be zero matrices I can give an example where product of two non-zero matrices is zero matrix,
$$
A=
\begin{bmatrix}
3 & 6 \\
2 & 4 \\
\end{bmatrix}
$$
$$
B=
\begin{bmatrix}
2 & -8 \\
-1 & 4 \\
\end{bmatrix}
$$
the product is zero matrix.
I don't know how to prove this. Could someone help?
|
whilst exhibiting a single value $a$ for which $P(a)$ is true is sufficient to establish the existential claim $\exists x\cdot P(x)$, i sympathize with OP's feeling that a single case doesn't give us much grasp of the why's and wherefores of the presence of zero divisors in matrix rings. there is, after all, considerably more that might be pointed out in this context. here just a couple of observations will suffice to offer a little contextualization of OP's question.
in the first place $1 \times 1$ matrices, whilst lacking the characteristics of the paradigmatic matrix, are nevertheless valid matrices. for the matrix ring $M_1(R)$ there are zero divisors if and only if the matrix elements are drawn from a ring $R$ which fails to be an integral domain. in particular if this ring is in fact a field, then there are no zero divisors. let us now focus on the (finite-dimensional) ring $M_n(F)$ of $n \times n$ matrices over a field $F$
1. all zero-divisors are singular (have no multiplicative inverse)
its proof is exceedingly simple, but this result is basic knowledge nonetheless. here is the left-hand-side case. the other case is easily dealt with in the same way (exercise!)
$$
(\exists A \cdot AB=I) \Rightarrow (BC=0 \Rightarrow ABC=0 \Rightarrow IC=0 \Rightarrow C=0)
$$
as is often the case to prove the converse offers greater difficulty.
2. all singular matrices are zero-divisors
the difference between showing this to be true and showing the converse (as above) is a good illustration of the power of 'strategic' mathematical ideas. instead of regarding a matrix merely as an array of numbers (or other field elements), we examine the connections between $M_n(F)$ and the endomorphism ring of the free module $F^n$. establishing this connection is the role of the concept of a basis for $F^n$.
any endomorphism $\alpha:V \to V$ has no inverse exactly when it fails to be both injective and surjective. in finite dimensions these two conditions are equivalent. such an endomorphism must annihilate some non-zero $v \in F^n$, which means that any matrix $A$ which represents $\alpha$ wrt some basis has zero as an eigenvalue. hence its characteristic equation must take the form $\lambda f(\lambda)=0$. now, by the Cayley-Hamilton theorem it follows that $Af(A)=0$, i.e. $A$ is a zero-divisor.
|
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Poisson Conditional Probability - Very lost! A discrete random variable is said to have a Poisson distribution if its possible values are the non-negative integers and if, for any non-negative integer $k$,
$$P(X=k)=e^{-\lambda}\frac{\lambda^k}{k!}$$
where $\lambda>0$. It turns out that $E(X)=\lambda$.
Minitab has a calculator for calculating Poisson probabilities, which is very similar to the calculator for Binomial probabilities.
The Poisson distribution model is widely used for modeling the number of "rare" events.
Suppose we have a Poisson random variable $X$ with mean (or expected value) equal to $2$ and another Poisson random variable $Y$ with mean $3$.
Suppose $X$ and $Y$ are independent random variables, in which case $W = X+Y$ will be a Poisson random variable with mean equal to $5 (= 2+3)$.
Find the conditional probability that $X = 5$ given that $W = 10$.
|
The distribution of the conditional probability is $$f_{X\mid W}(x\mid w)f_W(w)=f_{X,W}(x,w).$$
The probability that $X=5$ and $W=10$ is the same as the probability that $X=5$ and $Y=5$, so the joint probability on the right is the same as $f_{X,Y}(x,y)$, where $Y=W-X$. (Explicitly, this is a change of coordinates, but it has unit Jacobian). Since $X$ and $Y$ are independent, the joint probability is easy to understand. Dividing it by the distribution of $W$, which we also know, gives us the conditional probability.
|
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|
Distribution of sum of random variables Let $X_1, X_2, . . .$ be independent exponential random variables with mean $1/\mu$ and let $N$ be
a discrete random variable with
$P(N = k) = (1 − p)p^{k-1}$ for $k = 1, 2, . . . $
where $0 ≤ p < 1$ (i.e. $N$ is a shifted geometric random variable). Show that $S$ defined as $S =\sum_{n=1}^{N}X_n$ is again exponentially distributed with parameter $(1 − p)\mu$.
My approach:
$S = \sum_{n=1}^{N}\sum_{k=1}^{\infty}\frac{1}{\mu}e^{-\frac{n}{\mu}}(1-p)p^{k-1}$
How to solve this? Is it the right approach to solve this problem?
|
I would calculate the characteristic function :
$$E[ e^{itS} ] = E\left[ \prod_{i=1^N} e^{itX_1} \right]$$
$$= \sum_{k=1}^\infty (1-p)p^k \prod_{i=1}^k \int_{\mathbb{R}^+} e^{itx} \mu e^{-\mu x} dx$$
$$= \sum_{k=1}^\infty (1-p)p^{k-1} \prod_{i=1}^k \frac{\mu}{\mu-it}$$
$$ =\frac{(1-p)}{p}\sum_{k=1}^\infty \left( \frac{ \mu p }{\mu - it} \right)^k$$
$$ = \frac{(1-p)}{p} \frac{ 1 }{ 1- \frac{ \mu p }{\mu - it} } - (1-p)$$
$$= \frac{(1-p)}{p} \frac{\mu - it - (\mu - it - \mu p)}{ \mu - it - \mu p}$$
$$ = \frac{(1-p)\mu }{ (1-p) \mu - it}$$
And this is exactly the characteristic function pof an exponential variable with parameter $(1-p)\mu$
|
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Are there infinitely many pythagorean triples? I believe these questions are all asking different things, but:
*
*Are there infinitely many (integer) solutions to the pythagorean theorem?
*Is every positive integer part of a solution to the pythagorean theorem?
Also, is there a difference in multiplying the pythagorean triple by a constant factor, let's say $k$, on both sides and multiplying each number $a, b, c$ by a constant $k$?
|
That should be the easiest one:
Let n be a factor of every Integer in a pythagorean triple:
n* 3sq + n* 4sq = n* 5sq
There are infinitely many positive integers and so also infinitely many n´s. Now we just need to show that the pythagorean triples are all correct with the factor n:
n* 3sq + n* 4sq = n* 5sq (factoring out n)
n(3sq + 4sq) = n* 5sq (:n)
3sq + 4sq = 5sq
This shows that the factor can be "ignored" and every pythagorean triple of that sort works and that there are infinitely many of them.
|
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Obtaining a $3$-dimensional simple random walk from a $d$-dimensional simple random walk with $d>3$. Suppose $S_n$ is a $d$-dimensional random walk with $d>3$. Let $T_n=(S_n^{(1)},S_n^{(2)},S_n^{(3)})$, that is, we obtain $T_n$ by looking only at the first three coordinates of $S_n$. It is clear $T_n$ isn't always a simple random walk. Define a random variable $N$ by $N(0)=0$ and $$N(k+1)=\inf\{m>N(k):T_m\neq T_{N(k)}\}$$
The claim is that $T_{N(k)}$ is a simple three dimensional random walk. This is from Durret's "Probability: Theory and Examples", and is claimed without proof. Could someone provide one?
|
Here's the proof for a simple random walk, which can be generalized further. Hopefully it's clear that $T_{N(k)}$ as a function of $k$ changes exactly one coordinate for each $k$ and moreover $|T_{N(k+1)}-T_{N(k)}|_\infty=1$. More importantly, $T_{N(k+1)}|$ is a Markov process. So it suffices to verify that each step is uniformly random: the probability of a particular step is 1/6. This can be proven in a multitude of ways. One way is to just condition on $N(k+1)-N(k)$, and realize that the distribution of $T_{N(k+1)}$ is independent of this difference. Then show that $T_{N(k+1)}$ can be rewritten as $T_{N(k)}$ plus the argmin of the waiting times of jumps in the first 3 coordinates, each of which has the same distribution (but are not independent!). It's useful to remark that $T_{N(k)}$ like the bigger $d$-dimensional random walk is also invariant under permutations of its coordinates.
|
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|
Eigenvalues of a quasi-circulant matrix The following matrix cropped up in a model I am building of a dynamical system:
$$A=
\begin{bmatrix} 1 - \alpha & \alpha/2 & 0 & 0 &\cdots & 0 & 0 & \alpha/2\\
\alpha/2 & 1-\alpha & \alpha/2 & 0 &\cdots & 0 & 0 & 0\\
0 & \alpha/2 & 1-\alpha & \alpha/2 &\cdots & 0 & 0 & 0\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\
0 & 0 & 0 & 0 &\cdots & \alpha/2 & 1-\alpha & \alpha/2\\
\alpha/2 & 0 & 0 & 0 &\cdots & 0 & \alpha/2 & 1-\alpha\\
\end{bmatrix}$$
It is a stochastic matrix and a circulant matrix, and has equal values in the diagonal.
I am interested in the eigenvalues of this matrix, and it was easy to derive them from the properties listed here. It turns out that for size $n$,
$$
\lambda_k = 1 - \alpha \left(1 - \cos\frac{\pi k (n-2)}{n}\right), \qquad k\in\{0,1,\dots,n-1\},
$$
and in the limiting case,
$$
\lim_{n\rightarrow\infty} \lambda_k = 1 - \alpha(1 + (-1)^k) = \begin{cases}1-2\alpha & k \textrm{ even}\\ 1 & k \textrm{ odd}\end{cases}
$$
This is interesting for my study, because an eigenvalue of $1$ that is independent of $\alpha$ implies a marginally stable system that cannot be fully stabilized.
Now, I am interested in a slightly modified system, represented by the matrix below. This matrix is exactly like the one above save for the first and last rows, and is still a stochastic matrix with equal values in the diagonal.
I am wondering whether it is possible to derive the eigenvalues of this matrix, even if only for the limiting case.
$$A^\prime=
\begin{bmatrix} 1 - \alpha & \color{red} \alpha & 0 & 0 &\cdots & 0 & 0 & \color{red} 0\\
\alpha/2 & 1-\alpha & \alpha/2 & 0 &\cdots & 0 & 0 & 0\\
0 & \alpha/2 & 1-\alpha & \alpha/2 &\cdots & 0 & 0 & 0\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\
0 & 0 & 0 & 0 &\cdots & \alpha/2 & 1-\alpha & \alpha/2\\
\color{red} 0 & 0 & 0 & 0 &\cdots & 0 & \color{red} \alpha & 1-\alpha\\
\end{bmatrix}$$
|
Oddly enough, spectral graph theory has the analytic solution for this problem.
Let $L$ be the graph Laplacian for the path graph. Then $L$ is tri-diagonal with $[1,2,2,...,2,1]$ on the diagonal and $-1$'s on the super and sub diagonals. Then the matrix $A'$ is given by $$A'=I-\frac{\alpha}2 L.$$ Now since every vector in $\mathbb{C}^N$ is an eigenfunction of $I$, every eigenfunction $\chi_j$ of $L$ with eigenvalue $\lambda_j$ will be an eigenfunction of $A'$ with eigenvalue $1-\frac{\alpha}{2}\lambda_j$. Thus, to solve the problem stated hereinbefore, one needs only to solve eigenvalue problem of the path graph. The derivation of which can be found here.
To summarize the results, the eigenvalues of $L$ are $\lambda_j = 2-2\cos(\frac{\pi j}{N})$ and may be obtained by mapping the cycle graph with $2N$ vertices to the path graph with $N$ vertices.
Interestingly, the cycle graph with $N$ vertices also yields a method for finding the eigenvalues and eigenvectors of $A$ as $$A=I-\frac{\alpha}2 L_c,$$ where $L_c$ is the graph Laplacian for the cycle graph with $N$ vertices. If you let $v_j=\frac{1}{\sqrt{N}}[\omega^{0j}, \omega^{1j}, ..., \omega^{(N-1)j}]^T$ for $\omega=e^{2\pi i/N}$ (i.e. $v_j$ is a column of the DFT matrix) then $$L_cv_j = (2\omega^{0}-\omega^{j}-\omega^{-j})v_j = (2-2\cos(2\pi j/N))v_j.$$
Hence, $Av_j = (1-\alpha(1-\cos(2\pi j/N)))v_j$ (so I think the indexing in the question might be a bit off) and the DFT matrix forms an orthonormal eigenbasis that diagonalizes $A$. To get a real orthonormal eigenbasis for $A$ (which is guaranteed to exist because $A$ is real and symmetric), taking combinations $w_j=1/2(v_j+v_{N-j})$ and $w_{N-j} = i/2 (v_j-v_{N-j})$ will do the trick.
|
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|
Phase portrait of a $2 \times 2$ system of linear, autonomous differential eqns. with a zero eigenvalue Let $\mathbf{Y} = \begin{pmatrix} x(t) \\ y(t) \end{pmatrix}$ and $\mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ with $a, b, c, d \in \mathbb{R}$. Now, consider the system
$$
\frac{\operatorname d \mathbf{Y}}{\operatorname d t} = \mathbf{A Y}
$$
and let $\lambda_1, \lambda_2$ be its eigenvalues with $\lambda_1 = 0$ and $\lambda_2 \ne 0$, and $\mathbf{V}_1, \mathbf{V}_2$ the eigenvectors associated with them respectively. Then it can be seen that the general solution is
$$
\begin{equation}\tag{1}
\mathbf{Y} = k_1 \mathbf{V}_1 + k_2 e^{\lambda_2 t} \mathbf{V}_2\text{.}
\end{equation}
$$
Now, I understand that there is a line of equilibria along the $\mathbf{V}_1$ eigenvector. However, I do not understand why the rest of the phase portrait looks like this ($\lambda_2 > 0$ in this case):
That is, why are the rest of the solutions lines parallel to $\mathbf{V}_2$ in the phase plane? Can anyone please shed some light as to how I might infer this from looking at $\text{(1)}$.
|
If you let $t' = k_2 e^{\lambda_2 t}$, then $(1)$ becomes
$$Y = k_1 V_1 + t' V_2.$$ This is the equation for a line through $k_1V_1$ with direction $V_2$ ie lines parallel to $V_2$.
|
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|
The range of the function $f:\mathbb{R}\to \mathbb{R}$ given by $f(x)=\frac{3+2 \sin x}{\sqrt{1+ \cos x}+\sqrt{1- \cos x}}$ The range of the function $f:\mathbb{R}\to \mathbb{R}$ given by $f(x)=\frac{3+2 \sin x}{\sqrt{1+ \cos x}+\sqrt{1- \cos x}}$ contains $N$ integers. Find the value of $10N$.
I tried to find the minimum and maximum value of the function.First i simplified the function.
$f(x)=\frac{3+2 \sin x}{\sqrt{1+ \cos x}+\sqrt{1- \cos x}}=\frac{1+4\sin^2\left(\frac{x}{2}+\frac{\pi}{4}\right)}{2\sin \left(\frac{x}{2}+\frac{\pi}{4}\right)}$
Then i differentiated the function and equate it to zero to get the critical points.
Critical point equations are $\cos\left(\frac{x}{2}+\frac{\pi}{4}\right)=0$
$\sin\left(\frac{x}{2}+\frac{\pi}{4}\right)=\frac{1}{2},\sin\left(\frac{x}{2}+\frac{\pi}{4}\right)=\frac{-1}{2}$
When i checked plotted the function on desmos.com graphing calculator,i found minimum value to be $0.5$ and maximum value to be $2.5$.
Which i cannot get by my critical points.Where have i gone wrong?Please help me.
|
Put $\sqrt{1+\cos x}$ +$\sqrt{1-\cos x} = A$
$A^2 = 2\pm 2 \sin x ,\quad A^2 - 2 =\pm 2 \sin x$
$ -2\leq A^2 - 2\leq 2,\quad -2\leq A\leq2$
So $f(x) = \frac{5 - A^2}{A}$ or $\frac{A^2 + 1}{A}$
Find the minimum and maximum of $f(x)$ in the two conditions with $-2\leq A\leq 2$
|
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|
Difference between these two logical expression I am trying to solve the following problem:
Let S(x) be the predicate “x is a student,” F(x) the predicate “x is a
faculty member,” and A(x, y) the predicate “x has asked y a question,”
where the domain consists of all people associated with your school.
Use quantifiers to express each of these statements. f ) Some student
has asked every faculty member a question.
What is the difference between
$\forall y(F(y)\to\exists x(S(x)\land A(x,y)))$
and $\exists x (S(x) \land \forall y(F(y)\to A(x,y)))$? A'int they same?
|
The textbook answer says that the following represents:
Some student has asked every faculty member a question.
$\forall y(F(y) \to \exists x (S(x) \land A(x,y))) $
Which seems incorrect as I read this as saying for every faculty member there exists a student that has asked a question.
I'd argue that the following is more correct:
$\exists x (S(x) \land \forall y (F(y) \to A(x,y))) $
|
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|
How do you calculate $\lim_{z\to0} \frac{\bar{z}^2}{z}$? How do you calculate $\lim_{z\to0} \frac{\bar{z}^2}{z}$?
I tried $$\lim_{z\to0} \frac{\bar{z}^2}{z}=\lim_{\overset{x\to0}{y\to0}}\frac{(x-iy)^2}{x+iy}=\lim_{\overset{x\to0}{y\to0}}\frac{x^2-2xyi-y^2}{x+iy}=\lim_{\overset{x\to0}{y\to0}}\frac{x^2-2xyi-y^2}{x+iy}\cdot\frac{x-iy}{x-iy} \\ \\ =\lim_{\overset{x\to0}{y\to0}}\frac{(x^2-2xyi-y^2)(x-iy)}{x^2+y^2}$$
And that I could not get out, can anyone help me?
|
From your calculation :
$$=\lim_{\overset{x\to0}{y\to0}}\frac{(x^2-2xyi-y^2)(x-iy)}{x^2+y^2}$$
$$=\lim_{(x,y)\to (0,0)}\frac{x^3-3xy^2}{x^2+y^2}-i\lim_{(x,y)\to (0,0)}\frac{3x^2y-y^3}{x^2+y^2}$$
From here, show that both the limits are zero by changing polar form , $x=r\cos \theta$ , $y=r\sin \theta$.
For the first limit,
$$\left|\frac{r^3\cos^3\theta-3r^3\cos \theta\sin^2\theta}{r^2}\right|\le 4r<\epsilon$$whenever, $r^2<\epsilon^2/16$ i,e, whenever $|x|<\epsilon/\sqrt 8=\delta(say)$ , and $|y|<\epsilon/\sqrt 8=\delta(say)$.
Similarly the second limit will be zero and hence the given limit will be $0$.
|
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|
Non-additive upper logarithmic density: $\ell^\star(X \cup Y) \neq \ell^\star(X)+\ell^\star(Y)$ Let $\ell^\star$ be the upper logarithmic density on the set of positive integers, namely
$$
\forall X\subseteq \mathbf{N}^+, \,\, \ell^\star(X)=\limsup_n \frac{1}{\ln n}\sum_{x \in X\cap [1,n]}\frac{1}{x}.
$$
Problem: Let $X$ be the set of all even positive numbers, and let $Y$ be an arbitrary set of odd positive numbers. Is it necessarily the case that
$$
\ell^\star(X \cup Y) = \ell^\star(X)+\ell^\star(Y)?
$$
|
The answer is yes. More in general, if $\alpha \in [-1,\infty[$ and $\mathsf{d}^\ast(\mathfrak F; \alpha)$ is the upper $\alpha$-density (on $\mathbf N^+$) relative to a certain sequence $\mathfrak F = (F_n)_{n \ge 1}$ of nonempty subsets of $\mathbf N^+$ with suitable properties (can make this more precise if requested), viz. the function $$\mathcal P(\mathbf N^+) \to \mathbf R: S \mapsto \limsup_{n \to \infty} \frac{\sum_{i \in S \cap F_n} i^\alpha}{\sum_{i \in F_n} i^\alpha},$$ then $$\mathsf{d}^\ast(\mathfrak F; \alpha)(X \cup Y) = \mathsf{d}^\ast(\mathfrak F; \alpha)(X) + \mathsf{d}^\ast(\mathfrak F; \alpha)(Y)$$ for all disjoint $X,Y \subseteq \mathbf N^+$ such that
$$X \in \{S \in \mathcal P(\mathbf N^+): \mathsf{d}^\ast(\mathfrak F; \alpha)(S) + \mathsf{d}^\ast(\mathfrak F; \alpha)(S^c) = 1\}.$$
(I denote by $\mathcal P(\mathbf N^+)$ the power set of $\mathbf N^+$, and let $S^c := \mathbf N^+ \setminus S$ for every $S \subseteq \mathbf N^+$.)
The reason is essentially that if $(a_n)_{n \ge 1}$ and $(b_n)_{n \ge 1}$ are two real sequences for which the limit $\lim_n a_n$ exists and is finite, then $\limsup_n (a_n + b_n) = \lim_n a_n + \limsup_n b_n$.
The OP refers to the case $\alpha = -1$ and $F_n = [\![1,n]\!]$ for every $n$.
|
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|
Prove $\begin{pmatrix} a&b\\ 2a&2b\\ \end{pmatrix} \begin{pmatrix} x\\y\\ \end{pmatrix}=\begin{pmatrix} c\\2c \\ \end{pmatrix}$ Prove that
$$
\begin{pmatrix} a & b \\ 2a & 2b \\ \end{pmatrix}
\begin{pmatrix} x \\ y \\ \end{pmatrix}
=
\begin{pmatrix} c \\ 2c \\ \end{pmatrix}
$$
has solution
$$
\begin{pmatrix} x \\ y \\ \end{pmatrix}
=
\begin{pmatrix} \frac{c}{a} \\ 0 \\ \end{pmatrix}
+t
\begin{pmatrix} -b \\ a \\ \end{pmatrix}
$$
and $t$ is real number.
If I plug some numbers, all $y$ variables are vanish and the matrix has infinitely many solutions, but I have no idea how the solution could be into like that. Also I haven't reached Vector chapter.
Any help is appreciated!
|
The fast approach, as I see it is (in case $a \neq 0$):
you have $ax+by = c$ , just plug in $y=0+ta$, and you get
$$ax + bta = c$$
$$x= \frac ca - bt$$.
alternatively, in case you don't have the solution in advance, use same approach and you will get: $$ x = \frac ca - \frac ba y$$, so the general solution will be $$(x,y) = (\frac ca - \frac bat , t)$$
$$ (x,y) = (\frac ca, 0) + (-\frac ba,1)t$$ where $t$ is free. if you plug $t=a$ you will end up with the same result.
|
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|
Skorokhod space with uniform norm is Banach Let $D := D([0,t])$ be the Skorokhod space of right-continuous functions with left limits taking values in $\mathbb{R}^d$. Equip $D$ with the supremum norm $||f||_\infty = \sup_{s \in [0,t]}|f(s)|$. How would one go ahead and show that this is a Banach space? I have thought about that for any Cauchy sequence $(x_n)_{n \in \mathbb{N}}$, then $s \in [0,t]$ $x_n(s)$ converges pointwise to a limit $x$, but I am not sure if that is useful at all. Any hint would be greatly appreciated.
|
If $(f_n)$ is a Cauchy sequence then $f_n$ is uniformly Cauchy, so there exists $f$ such that $f_n\to f$ uniformly. You only need to show that $f\in\mathcal D$. This follows by exactly the same argument as one uses to show a uniform limit of continuous functions is continuous...
|
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|
Prove that $f$ is integrable if and only if $\sum^\infty_{n=1} \mu(\{x \in X : f(x) \ge n\}) < \infty$ Problem statement: Suppose that $\mu$ is a finite measure. Prove that a measurable, non-negative function $f$ is integrable if and only if $\sum^\infty_{n=1} \mu(\{x \in X : f(x) \ge n\}) < \infty$.
My attempt at a solution: Let $A_n = \{x \in X : f(x) \ge n\}$. To show that $\sum^\infty_{n=1} \mu (\{x \in X : f(x) \ge n\}) < \infty$ implies $f$ is integrable, the Borel Cantelli lemma tells us that almost all $x \in X$ belong to at most finitely many $A_n$. Thus, the set $\{x \in X : f(x) = \infty\}$ has measure $0$. Now, this, together with the fact that $\mu(X) < \infty$, should give us that $f$ is integrable, but I can't figure out how to prove that! It seems fairly obvious, but I can't figure out if it is then ok to say that $f$ is bounded almost everywhere? It seems like $f$ is then pointwise bounded, but I'm not sure if that means I can find some $M$ such that $f(x) \le M$ for all $x$.
For the reverse implication, I haven't come up with anything useful - I have been trying to show that the sequence of partial sums, $\sum^m_{n=1}\mu(A_n)$, is bounded, but I'm not sure how to do so.
|
Convince yourself that $$f(x)-1\leq \sum_{n=1}^\infty {\bf 1}_{(f\geq n)}(x)\leq f(x)$$
for all $x\in X$, then integrate with respect to $\mu$.
|
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|
Help with Spivak's Calculus: Chapter 1 problem 21 I've been stuck on this problem for over a day, and the answerbook simply says "see chapter 5" for problems 20,21, and 22. But I want to complete the problem without using knowledge given later in the book, so I've been banging my head against the wall trying all sorts of things, but nothing I do seems to lead me anywhere.
The problem is as follows:
Prove that if
$|x - x_0| < min (\frac{\varepsilon}{2(|y_0| + 1)}, 1)$ and $|y - y_0| < min (\frac{\varepsilon}{2(|x_0| + 1)}, 1)$
then
$|xy - x_0 y_0| < \varepsilon$.
Here are some of the things I've been thinking about, I don't know which of these are useful (if any), but they somewhat outline the logic behind my various attempts.
Since at most $|x - x_0| < 1$ and $|y - y_0| < 1$ then it follows that $(|x-x_0|)(y-y_0|) < |x-x_0|$ and $(|x-x_0|)(y-y_0|) < |y-y_0|$
Also, $(|x - x_0|)(|y_0| + 1) < \frac{\varepsilon}{2}$ and $(|y - y_0|)(|x_0| + 1) < \frac{\varepsilon}{2}$ so $(|x - x_0|)(|y_0| + 1) + (|y - y_0|)(|x_0| + 1) < \varepsilon$. and since $|a + b| \leq |a| + |b| < \varepsilon$ I've tried multiplying things out, and then adding them together to see if anything cancels, but I can't make anything meaningful come out of it.
Also since $|a - b| \leq |a| + |b|$ I've also tried subtracting one side from the other, but to no avail.
I was also thinking that since $(|x-x_0|)(y-y_0|) < |x-x_0|$, then I could try something along the lines of $(|x-x_0|)(y-y_0|)(|x_0| + 1) + (|x-x_0|)(y-y_0|)(|y_0| + 1)< \varepsilon$ and various combinations as such, but I just can't seem to get anything meaningful to come out of any of these attempts.
I have a sneaking suspicion that the road to the solution is simpler than I'm making it out to be, but I just can't see it.
|
I think I figured it out thanks to GeoffRobinson's hint, and from RJS (thanks guys!) so I figured I'd write out my own work here.
$|xy-x_0y_0| = |x(y-y_0) + y_0(x-x_0)| \leq |x(y-y_0)| + |y_0(x-x_0)|$
So in essence we're going to show that $|x(y-y_0)| + |y_0(x-x_0)| \leq \varepsilon$ which implies that $|xy-x_0y_0| \leq \varepsilon$.
So let's start. From
$|x - x_0| < \frac{\varepsilon}{2(|y_0| + 1)}$
then
$(|x - x_0|)(|y_0| + 1) = |y_0(x-x_0)| + |x - x_0| < \frac{\varepsilon}{2}$
Which in turn implies that $|y_0(x-x_0)|< \frac{\varepsilon}{2}$
We're halfway done with our inequality.
Next by looking at
$|x-x_0| < 1$, we can observe that $|x| - |x_0| \leq |x-x_0| \Rightarrow |x| < |x_0| + 1$
Then consider the given inequality:
$|y-y_0| < \frac{\varepsilon}{2(|x_0| + 1)}$
We can see that $|x||y-y_0| < (|x_0| + 1) \frac{\varepsilon}{2(|x_0| + 1)} = \frac{\varepsilon}{2}$ and so $|x(y-y_0)| < \frac{\varepsilon}{2}$
So finally we can show that:
$|y_0(x-x_0)| + |x(y-y_0)| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$
and because $|xy-x_0y_0| \leq |x(y-y_0)| + |y_0(x-x_0)|$
We can say $|xy-x_0y_0| < \varepsilon$
|
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|
Find $f(2015)$ given the values it attains at $k=0,1,2,\cdots,1007$
Let $f$ be a polynomial of degree $1007$ such that $f(k)=2^k$ for $k=0,1,2,\cdots, 1007$. Determine $f(2015)$.
Taking $f(x)=\sum_{n=0}^{1007} a_n x^n $, (whence $a_0=1$), I tried to combine $$a_1=1-\sum_{n=2}^{1007} a_n$$ and the easily derivable $$ a_n = \frac{2^n-1}{n^n} - \sum_{k=1}^{n-1} a_k n^{k-n} - \sum_{k=n+1}^{1007} a_k n^{k-n},$$ hoping to find several null coefficients, but that didn't work. I looked for the solution on Yahoo Answers, and I came to know it's about binomial coefficients, Tartaglia in particular; however the explanation is confusing to me, I'd like a better one, or even a different approach.
|
Hint: $f(x)=\sum_{r=0}^{1007}\,\binom{x}{r}$, where $\binom{x}{r}:=\frac{x(x-1)\cdots(x-r+1)}{r!}$ for every positive integer $r$ and $\binom{x}{0}:=1$, satisfies the required condition. Show that it is the only one satisfying this property. Then, verify that $f(2015)=2^{2014}$.
|
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|
Jackpot probablity There are two parts to a problem:
Part (a);
In a Lottery, three white balls are drawn (at random) from twenty balls numbered 1 through 20, and a blue SuperBall is drawn (at random) from ten balls numbered 21 through 30. When you buy a ticket, you select three numbers from 1-20 and one number from 21-30. To win the jackpot, the numbers on your ticket must match the three white balls and the SuperBall. (You don't need to match the white balls in order).
If you buy a ticket, what is your probability of winning the jackpot?
Part (b):
In a Lottery, three white balls are drawn (at random) from twenty balls numbered 1 through 20, and a blue SuperBall is drawn (at random) from ten balls numbered 21 through 30. When you buy a ticket, you select three numbers from 1-20 and one number from 21-30. To win a prize, the numbers on your ticket must match at least two of the white balls or must match the SuperBall.
If you buy a ticket, what is your probability of winning a prize?
Approach for part (a):
Here is how I am solving the problem.Since 3 number in white balls needs to match during the first pick, the probability is 3/20. For the second one it will be 2/19 and 3rd one 1/18. That combined with probability for blue ball will be 3/20 * 2/19 * 1/18 * 1/10
part (b):
3 cases (1) only blue ball matches and no match in white balls (2) no match in blue ball but 2 out of 3 white ball matches (3) Both matches
Probability for case (a):
1/10 * 17/20 * 16/19 * 15/18
case (b) :
9/10 * 3/20 * 2/19
case (c) :
1/10 * 3/20 * 2/19 * 1/18
Add case (a) + (b) + (c) because they are mutually exclusive.
Am I correct in my approach?
|
3 matching white balls are drawn.
B: 2 matching white balls are drawn.
C: The right blue ball is drawn.
Therefore $P(A\cup B \cup C )=P(A)+P(B)+P(C)-P(A \cap C)-P(B \cap C)$
$\left[\frac{3}{20} \cdot \frac{2}{19} \cdot \frac{1}{18}\right]+\left[3 \cdot \frac{3}{20} \cdot \frac{2}{19} \cdot \frac{17}{18}\right]+\left[\frac{1}{10}-\frac{1}{10} \cdot \frac{3}{20} \cdot \frac{2}{19} \cdot \frac{1}{18}\right]-\left[\frac{1}{10}\cdot3\cdot \frac{3}{20} \cdot \frac{2}{19} \cdot \frac{17}{18}\right]\approx 14.1\%$
|
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|
Extending continuous function from a dense set If $X$ is a metric space and $Y$ a complete metric space. Let $A$ be a dense subset of $X$. If there is a uniformly continuous function $f$ from $A$ to $Y$, it can be uniquely extended to a uniformly continuous function $g$ from $X$ to $Y$. I was trying to think of an example of a pointwise continuous function from set a rational numbers $Q$ to real line $R$ which cannot be extended to a continuous function $R$ to $R$. But could not get anywhere.
|
Define $f:\mathbb{Q}\to\mathbb{R}$ by $f(x)=0$ if $x<\sqrt{2}$ and $f(x)=1$ if $x>\sqrt{2}$. Then $f$ is continuous, but cannot be continuously extended to $\sqrt{2}$.
|
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|
Proving that cows have same weight without weighing it! My friend gave me this problem and I have no clue how to go about it:
A peasant has $2n + 1$ cows. When he puts aside any of his cows,
the remaining $2n$, can be divided into two sub-flocks of $n$ cows and each having
the same total weight. How can we prove that the cows all have the same weight?
How can I approach this?
|
We have $2n+1$ equations in $2n+1$ unknowns. This has a unique solution. As all the cows having the same weight is a solution, it is the only solution.
|
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|
Proving an inequality with given equality of the form $a^2+b^2=c$ $$\begin{Bmatrix}
a^2+b^2=2\\
c^2+d^2=4
\end{Bmatrix} \to ac+bd\le3 $$ prove that inequality
|
HINT
Note that $$(a-c)^2+(b-d)^2\ge0$$so that $$a^2+c^2+b^2+d^2\ge2ac+2bd$$
This deals with the easy inequality required. Other answers give a sharper result, which is possible because it is impossible for $a=c$ and $b=d$ with the given conditions.
|
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|
Prove identity: $\frac{1+\sin\alpha-\cos\alpha}{1+\sin\alpha+\cos\alpha}=\tan\frac{\alpha}{2}$ Prove identity: $$\frac{1+\sin\alpha-\cos\alpha}{1+\sin\alpha+\cos\alpha}=\tan\frac{\alpha}{2}.$$
My work this far:
we take the left side
$$\dfrac{1+\sqrt{\frac{1-\cos2\alpha}{2}}-\sqrt{\frac{1+\cos2\alpha}{2}}}{1+\sqrt{\frac{1-\cos2\alpha}{2}}+\sqrt{\frac{1+\cos2\alpha}{2}}}=$$
then after some calulations I come to this
$$=\frac{\sqrt{2} +\sqrt{1-\cos2\alpha}-\sqrt{1+\cos2\alpha}}{\sqrt{2} +\sqrt{1-\cos2\alpha}+\sqrt{1+\cos2\alpha}}$$ but now I'm stuck...
|
Notice, $$LHS=\frac{1+\sin\alpha-\cos\alpha}{1+\sin\alpha+\cos\alpha}$$
$$=\frac{1+\frac{2\tan\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}-\frac{1-\tan^2\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}}{1+\frac{2\tan\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}+\frac{1-\tan^2\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}}$$
$$=\frac{1+\tan^2\frac{\alpha}{2}+2\tan\frac{\alpha}{2}-1+\tan^2\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}+2\tan\frac{\alpha}{2}+1-\tan^2\frac{\alpha}{2}}$$
$$=\frac{2\tan\frac{\alpha}{2}+2\tan^2\frac{\alpha}{2}}{2+2\tan\frac{\alpha}{2}}$$
$$=\tan\frac{\alpha}{2}\left(\frac{1+\tan\frac{\alpha}{2}}{1+\tan\frac{\alpha}{2}}\right)$$$$=\tan\frac{\alpha}{2}=RHS$$
|
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|
Tale of a frog that jumps a fraction of what is left to cross the pond I recall from undergraduate courses in calculus and series analysis a tale of a frog that tries to jump a fraction (e.g. 1/2) of what is left for the frog to cross the pond.
In the limit, the fraction of the pond the frog travels is:
$1/2 + 1/2(1/2) + 1/2 (1/4) + ...$
Does this tale have a name? What about the series?
|
This looks like a geometric series, which is a series of the form
$$\sum_{n=0}^\infty a r^n$$
In your case $a = \frac{1}{2}$ and $r = \frac{1}{2}$.
|
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|
Integrate $\int_{-\infty}^{\infty} \frac{\cosh(\beta x)}{1+\cosh( \beta x )} e^{-x^2} x^2 \rm{d}x$ Integrate
$$ \int_{-\infty}^{\infty} \frac{\cosh(\beta x)}{1+\cosh( \beta x )} e^{-x^2} x^2 \rm{d}x, $$
with $\beta \in \mathbb{R}$ and $\beta > 0$.
Numerical integration shows that this integral exists, but I have been unable to find a closed analytical expression (using contour integration).
I have tried to use a rectangular contour $(-R,0) \to (R,0) \to (R,i\eta) \to (-R,i\eta) \to (-R,0)$. The vertical (imaginary direction) integrals at $\pm R$ vanish for $R \to \infty$. I am unable to find an $\eta$ that allows me to relate the horizontal parts of the contour, which would in turn allow me to equate the result to the sum of residues of the enclosed poles.
|
I have some idea, if you are willing to take into account something different by residues, and if you are willing to proceed by your own adopting some tricks in the same way I will do (this message refers to the part $2$ of what I will do in a while. I'll show you my ideas for part $1$ only).
Starting with writing $\cosh$ in exponential form, and arranging:
$$\int_{-\infty}^{+\infty} \frac{e^{\beta x} + e^{-\beta x}}{2 + e^{\beta x} + e^{-\beta x}}e^{-x^2} x^2\ \text{d}x$$
Now we split the integral into the one from $[0, +\infty)$ and $(-\infty, 0]$. I'll do the first one, then for the second one there will be surely similar methods. Use your fantasy.
I collect a tern $e^{\beta x}$ for the integrand, above and below:
$$\int_0^{+\infty} \frac{1 + e^{-2\beta x}}{1 + 2e^{-\beta x} + e^{-2\beta x}}e^{-x^2} x^2\ \text{d}x$$
Now I use the Geometric Series
$$\frac{1}{1 + 2e^{-\beta x} + e^{-2\beta x}} = \sum_{k = 0}^{+\infty}\left(-2e^{-\beta x} - e^{-2\beta x}\right)^k$$
Inserting and arranging a bit:
$$\sum_{k = 0}^{+\infty}(-1)^k \int_0^{+\infty}(1 + e^{-2\beta x})e^{-x^2} x^2 e^{-\beta k x}(2 + e^{-\beta x})^k\ \text{d}x$$
Binomial theorem will help now:
$$(2 + e^{-\beta x})^k = \sum_{j = 0}^{k} \binom{k}{j}2^k\ \left(e^{-\beta x}\right)^{k-j}$$
namely
$$\sum_{k = 0}^{+\infty}\sum_{j = 0}^{k}(-1)^k\binom{k}{j}2^k\int_0^{+\infty}(1 + e^{-2\beta x})e^{-x^2} x^2e^{-\beta k x}\ \text{d}x$$
Arranging a little bit
$$\sum_{k = 0}^{+\infty}\sum_{j = 0}^{k}(-1)^k 2^k \binom{k}{j}\int_0^{+\infty}x^2(1 + e^{-2\beta x})e^{-\beta x(k-j)}e^{-x^2}e^{-\beta k x}\ \text{d}x$$
From this point we can split the integral in the two terms: $A_1$ and $A_2$, and compute their values:
$$A_1 = \int_0^{+\infty}x^2 e^{-\beta x (k-j)}e^{-x^2 - \beta k x}\ \text{d}x$$
$$A_2 = \int_0^{+\infty}x^2 e^{-\beta x (k - j + 2\beta)}e^{-x^2 - \beta k x}\ \text{d}x$$
Now, proceeding with $A_1$, we can arrange better the exponentials collecting $x$ terms:
$$A_1 = \int_0^{+\infty}x^2 e^{-x^2 - x[2\beta k - \beta j]}\ \text{d}x$$
$$A_1 = \int_0^{+\infty}x^2 e^{-x^2 - Bx}\ \text{d}x$$
This is a well known integral, whose result is
$$ \int_0^{+\infty}x^2 e^{-x^2 - Bx}\ \text{d}x = \frac{1}{8}\left(-2B + (2 + B^2)e^{B^2/4}\sqrt{\pi}\text{Erfc}\left(\frac{B}{2}\right)\right)$$
Where of course $B = 2\beta k - \beta j$.
In the same way we have for $A_2$:
$$A_2 = \int_0^{+\infty}x^2 e^{-\beta x (k - j + 2\beta)}e^{-x^2 - \beta k x}\ \text{d}x$$
$$A_2 = \int_0^{+\infty}x^2 e^{-x^2 - x[2\beta k - \beta j + 2\beta^2]}\ \text{d}x$$
$$A_2 = \int_0^{+\infty}x^2 e^{-x^2 - Cx}\ \text{d}x$$
Which again is like before:
$$\int_0^{+\infty}x^2 e^{-x^2 - Cx}\ \text{d}x = \frac{1}{8}\left(-2C + (2 + C^2)e^{C^2/4}\sqrt{\pi}\text{Erfc}\left(\frac{C}{2}\right)\right)$$
Where of course $C = 2\beta k - \beta j + 2\beta^2$.
The end of (this part of) the story is:
$$\sum_k \sum_j A_1(k, j) + A_2(k, j)$$
$$\sum_{k = 0}^{+\infty}\sum_{j = 0}^{k}(-1)^k 2^k \binom{k}{j} \frac{1}{8}\left[-2B + (2 + B^2)e^{B^2/4}\sqrt{\pi}\text{Erfc}\left(\frac{B}{2}\right) -2C + (2 + C^2)e^{C^2/4}\sqrt{\pi}\text{Erfc}\left(\frac{C}{2}\right)\right]$$
You can eventually try to calculate it with some software. I tried with Mathematica but I believe my computer is not that powerful to perform such a calculation..
Then the other part of the integral can be manipulated similarly I guess. Give it a try.
In the meanwhile I'll watch a Supernatural new episode. I leave my computer trying to calculate, maybe we get some spider out of the hole.
|
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|
Compare growth of function and its derivative. Suppose $f:\mathbb R \rightarrow \mathbb R$, $f(x)<O(x)$ (i.e. has slower than linear growth), $\lim_{x\rightarrow \infty} f'(x)=0$. Is it possible to show that there exists a $\delta$ such that for $|x|>\delta$, $x f'(x) < f(x)$ for all functions $f$?
|
This is not true. For example, take $f(x)=\frac{1}{x}\sin x$. Then $f'(x)=\frac{1}{x}\cos x - \frac{1}{x^2}\sin x \to 0$ as $x\to \infty$. However, $xf'(x) = \cos x - \frac{1}{x} \sin x \not\to 0$ while $f(x) \to 0$.
|
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|
Question about Euler's formula I have a question about Euler's formula
$$e^{ix} = \cos(x)+i\sin(x)$$
I want to show
$$\sin(ax)\sin(bx) = \frac{1}{2}(\cos((a-b)x)-\cos((a+b)x))$$
and
$$ \cos(ax)\cos(bx) = \frac{1}{2}(\cos((a-b)x)+\cos((a+b)x))$$
I'm not really sure how to get started here.
Can someone help me?
|
Try using the identity
$$\sin A \cos B \equiv \tfrac{1}{2}\left(\sin(A + B) + \sin(A - B)\right)$$
|
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|
$2^{nd}$ order PDE: Solution I am trying to solve the following equation:
$$\frac{\partial F}{\partial t} = \alpha^2 \, \frac{\partial^2 F}{\partial x^2}-h \, F$$
subject to these conditions:
$$F(x,0) = 0, \hspace{5mm} F(0,t) = F(L,t)=F_{0} \, e^{-ht}.$$
I know that I am suppose to simplify the equation with:
$$F(x,t)=\phi(x,t)e^{-ht}$$
My initial guess is to divide by $$\alpha^2$$
and have this:
$$\frac{d^2F}{dx^2}-\frac{1}{\alpha^2}\frac{dF}{dt}-\frac{h}{\alpha^2} \, F=0.$$
I am having trouble with the next steps. Should I assume a solution of the exponential form?
|
The following field redefinition, similar to what the other answers have done, simplifies the problem:
$$F = (F_0 + \phi)e^{-ht} \implies \frac{d\phi}{dt} = \alpha^2\frac{d^2\phi}{dx^2}$$
with $\phi(0,t) = \phi(L,t) = 0$ and $\phi(x,0) = -F_0$. One can now proceed with your favoritte method like separation of variables or expanding $\phi$ in a Fourier series to obtain an equation for the coefficients.
The other answers have done this so I just want to focus on a problem one encounters when applying the initial condition. No matter what approach you take you will be lead to a Fourier series and to evaluate the coefficients you need to apply the initial condition $\phi(x,0) = -F_0$. This function does not have a non-trivial Fourier-series so a naive application would not give a solution. The usual trick to get around this is by extending the domain of $\phi$ from $[0,L]$ to $[-L,L]$ and then demaning $\phi(x,t)=-\phi(-x,t)$, i.e. we demand that the solution should be odd (since the odd completion do have a nice Fourier series). For the initial condition we therefore consider
$$\phi(x,0) = -F_0\left\{\matrix{-1 & x<0 \\1 & x > 0}\right.$$
This function, the so-called square-wave, has the Fourier series
$$\phi(x,0) = -\frac{4F_0}{\pi}\sum_{n=0}^\infty\frac{1}{2n+1}\sin\left(\frac{(2n+1)\pi x}{L}\right)$$
which can be found by applying the usual method.
Just for completeness, here is a quick runthough of the solution. We expand $\phi$ in a Fourier $\sin$-series (since we demand that $\phi$ is odd) $\phi(x,t) = \sum_{n=1}^\infty a_n(t) \sin\left(\frac{n\pi x}{L}\right)$ that satisfy the boundary conditions $\phi(0,t)=\phi(L,t) = 0$. By inserting this series into the PDE and equating the left and right hand side we get the equation for the coefficients
$$\frac{da_k}{dt} = -\frac{n^2\pi^2\alpha^2}{L^2} a_k \implies a_k(t) = a_k(0) e^{-\frac{n^2\pi^2\alpha^2}{L^2}t}$$
where $a_k(0)$ is simply the Fourier coefficients of the initial condition which here is just the square-wave given above. The full solution can therefore be written
$$F(x,t) = F_0e^{-ht}\left[1 - \frac{4}{\pi}\sum_{n=0}^\infty\frac{e^{-\frac{(2n+1)^2\pi^2 \alpha^2}{L^2}t}}{2n+1}\sin\left(\frac{(2n+1)\pi x}{L}\right)\right]$$
for all $t>0$ and $0\leq x\leq L$. Note that the formula above is not valid for $t=0$ at the two points $x=0$ and $x=L$ since we get $F(x,0)=F_0$ instead of $F(x,0)=0$.
|
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|
how did Cardano obtain three solutions for cubic? So, if I am not mistaken Complex numbers were discovered after Cardano's method. But from Cardano's Method on Wikipedia, it says to get the three solutions, we should use the root of unity. In that case, when Cardano's method was published it could only find one solution?
|
The method of solving a cubic is explained very clearly in Stroud's Further Engineering Mathematics. According to Stuart Hollingdale's book "Makers of Mathematics", Cardano plagiarized Tartaglia's solution of the cubic. Paul Halmos said that he solved this problem for himself as a high school student.
|
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|
Prove $\gcd(n, n + 1) = 1$ for any $n$ Let $n \in \mathbb Z$ be even. Then $n + 1$ is odd. So, $2$ doesn't divide $n + 1$. Thus there's no even number for which $\gcd(n, n+1)$ is not $1$. I am not sure how to show it for odd numbers. Is there a better way to prove the statement?
|
Suppose $gcd(n,n+1)=d >0$ $$\to \left.\begin{matrix}
d|n\\
d|n+1
\end{matrix}\right\}\Rightarrow d|(n+1)-(n)\Rightarrow d|1\\\frac{1}{d} \in \mathbb{Z}\Rightarrow d=\pm1 \overset{d>0}{\rightarrow} d=1$$
|
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|
Average of a list of numbers, read one at a time I have a dilemma
Lets take the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$
If we wanted to calculate the average of this set we would add up all the numbers in the set $(45)$ and then divide by the total number of items in the set $(9)$ and arrive at the correct average of $5$
What happens if we are not given the full set? What if we are only given one number at a time and have to calculate the correct average?
For instance if we are given $1$ we know the average is $1$, then if we are given $2$ then we add $1+2$ and get $3$, then divide $3$ by $2$ and arrive at the correct average of $1.5$.
Then what if we are given the next number $3$ we would add $1.5$ and $3$ giving $4.5$ and divide that by $2$ arriving at $2.25$ which is incorrect since $1+2+3=6$ and $6/3=2$.
Is there a way of calculating the correct total average like this?
|
You can:
*
*Keep track of the total number of values you have seen so far, call it $n$, and the average of the numbers so far, $A$.
Then when you see a new number $x$, map
\begin{align*}
n &\mapsto n+1 \\
A &\mapsto \frac{nA + x}{n+1}
\end{align*}
Or:
*
*Keep track of the total number of values you have seen so far, call it $n$, and the sum of the numbers so far, $S$.
Then when you see a new number $x$, map
\begin{align*}
n &\mapsto n+1 \\
S &\mapsto S+x
\end{align*}
at the end, compute the average $\frac{S}{n}$.
|
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|
Limits (Three Variable function). We're given :
$f(x,y,z) = \dfrac{xyz}{x^{2}+y^{2}+z^{2}}$ ,
Also , it's given that $\lim_{(x,y,z) \to (0,0,0)} f(x,y,z)$ exists.
We need to prove that $\lim_{(x,y,z) \to (0,0,0)} f(x,y,z) = 0$.
What I figured :
First I approach $(0,0,0)$ along $x$ -axis , and thus the limit becomes :
$\lim_{(x,y,z) \to (0,0,0)\dfrac{x.0.0}{x^{2}+0^{2}+0^{2}}}$ which is $0$.
Similarly approaching along $y$ and $z$ axis gives the limit to be $0$.
Now , since the limit at $(0,0,0)$ exists , it should be unique throughout , thus ,
$\lim_{(x,y,z) \to (0,0,0)} f(x,y,z) = 0$
Could anyone tell , am I right in making the above statement ?
|
Little observation: in general (if you do not know whether the limit exists), if the limit exists along one path it does NOT mean that it exists, so your proof for $(x,0,0)$ is not sufficient.
For the general case, therefore, I would either use the demonstration of answer $1$ , or majorize at the numerator $|xy|$ by $\frac{x^2+y^2}{2}$, and minorize at the denominator $x^2+y^2+z^2$ by $x^2+y^2$, in order to get that $0\leq |f(x,y,z)|\leq \frac{|z|}{2}\leq\frac{\sqrt{x^2+y^2+z^2}}{2}$, which goes to zero for $(x,y,z)$ going to $(0,0,0)$.
For the sandwich theorem, the limit of your function at the origin is zero.
|
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|
Prove the inequality - inequality involving surds Prove that for $r$ greater than or equal to 1:
$\displaystyle 2(\sqrt{r+1} - \sqrt{r}) < \frac{1}{\sqrt{r}} < 2(\sqrt{r}-\sqrt{r-1})$
Any help on this would be much appreciated.
|
Let $f(x) = \sqrt x$. By mean value theorem, there exist $q\in (r-1,r)$ and $s \in (r, r+1)$ that satisfy
$$f(r) - f(r-1) = f'(q),\quad f(r+1)-f(r) = f'(s).$$
By the concavity of $f$, since $q<r<s$,
$$\begin{array}{rcl}
f'(q) >& f'(r) &> f'(s)\\
f(r)-f(r-1) >& f'(r) &> f(r+1) - f(r)\\
\sqrt{r}-\sqrt{r-1} >& \dfrac1{2\sqrt r} &> \sqrt{r+1} - \sqrt{r}\\
2(\sqrt{r+1}-\sqrt{r}) <& \dfrac1{\sqrt r} &< 2(\sqrt{r} - \sqrt{r-1})
\end{array}$$
|
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|
How to approach to fitting curve? I'd like to approximate fitting curve some kind of curves like below.
(1, 3.5), (2, 4.3), (3, 7.2), (4, 8) which is having 4 points.
and I heard that this solver is PINV() of matlab function.
But I don't know how to use.
Would you please let me know how to use and find a approximation fitting curve equation?
|
Problem statement
Fit the given data set with a sequence of polynomial fits:
$$
y(x) = a_{0} + a_{1} x + \dots + a_{d} x^{d}
$$
Indicate where we can solve the linear system with the normal equations, and when we must rely exclusively upon the pseudoinverse.
Linear System
The linear system for the polynomial with highest order $d$ is
$$
\begin{align}
\mathbf{A} a &= y \\
% A
\left[ \begin{array}{cccc}
1 & x_{1} & \cdots & x_{1}^{d} \\
1 & x_{2} & \cdots & x_{2}^{d} \\
1 & x_{3} & \cdots & x_{3}^{d} \\
1 & x_{4} & \cdots & x_{4}^{d} \\
\end{array} \right]
% a
\left[ \begin{array}{cccc}
a_{0} \\
a_{1} \\
\vdots \\
a_{d} \\
\end{array} \right]
%
&=
%
% y
\left[ \begin{array}{c}
y_{1} \\
y_{2} \\
y_{3} \\
y_{4}
\end{array} \right]
%
\end{align}
$$
Least squares solution
The least squares solution is defined as
$$
a_{LS} = \left\{
a \in \mathbb{R}^{d+1} \colon
\lVert
\mathbf{A} a - y
\rVert_{2}^{2}
\text{ is minimzed}
\right\}
$$
The residual error vector is
$$
r = \mathbf{A} a - y.
$$
The total error, the quantity minimized, is $r^{2} = r\cdot r$.
Summary of results
The total error demonstrates typical behavior. Increasing the order of fit initially reduces the error. Then it plateaus, or may actually increase.
For these data, the cubic fit provides the best combination of total error and computational cost.
The amplitudes for each order are collected in the table below.
$$
\begin{array}{clllllll}
k & a_{0} & a_{1} & a_{2} & a_{3} & a_{4} & a_{5} & a_{6} \\
0 & 5.75 \\
1 & 1.65 & \phantom{-}1.64 \\
2 & 1.65 & \phantom{-}1.64 & \phantom{-}0 \\
\color{blue}{3} & \color{blue}{9.} & \color{blue}{-10.05} & \phantom{-}\color{blue}{5.25} & \color{blue}{-0.7} \\
4 & 4.04271 & \phantom{-}0.277692 & -1.97938 & \phantom{-}1.36554 &
-0.206554 \\
5 & 2.57594 & \phantom{-}1.38279 & -0.0545302 & -0.868031 &
\phantom{-}0.545109 & -0.0812778 \\
6 & 1.89099 & \phantom{-}1.41016 & \phantom{-}0.686033 & -0.172893 &
-0.615848 & \phantom{-}0.350573 & -0.0490168 \\
\end{array}
$$
Lower order fits: $d\le 3$
For $d=0$, the amplitude is given by the mean of the data $a_{0} = \bar{y}$.
For $d=1,2,3$, the normal equations
$$
\mathbf{A}^{T} \mathbf{A} a = \mathbf{A}^{T} y
$$
can be solved as
$$
a_{LS} = \left( \mathbf{A}^{T} \mathbf{A} \right)^{-1} \mathbf{A}^{T} y
$$
Higher order fits: $d\ge 4$
For $d\ge4$, the product matrix $\mathbf{A}^{T} \mathbf{A}$ is rank deficient and cannpt be inverted. The solution requires the pseudoinverse. The least squares minimizers are the affine space
$$
a_{LS} = \mathbf{A}^{+}y + \left( \mathbf{I}_{4} - \mathbf{A}^{+} \mathbf{A} \right) z, \quad z\in \mathbb{R}^{4}
$$
Sequence of plots
|
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|
Generic fiber of a scheme over a DVR What is (usually?) meant by the generic fiber of a scheme over a discrete valuation ring? I've seen in some talks now, could somebody give a precise definition?
Thank you very much in advance!
|
If $R$ is a discrete valuation ring, then $Y=\mathrm{Spec}(R)$ has two points, often denoted $\eta$ and $s$, the generic and the special (or closed) point, corresponding to the ideal $(0)$ and the unique maximal ideal $\mathfrak{m}_R$, respectively. The names are apt, as $\{\eta\}$ is dense in $Y$, while $\{s\}$ is closed in $Y$. Now a scheme over $Y$ is a scheme $X$ equipped with a morphism $f:X\to Y$. The generic (resp. special or closed) fibers of $X$ are the fibers over the generic (resp. closed) point of $Y$. As with any morphism of schemes, the fibers are equipped with scheme structures over the residue fields of the corresponding points, that is, $X_\eta$, the generic fiber, is a $k(\eta)=\mathrm{Frac}(R)$-scheme, while $X_s$, the special or closed fiber, is a scheme over the residue field $k(s)=R/\mathfrak{m}_R$.
|
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|
Distinct Roots of $x^2+(a-5)x+1=3|x|$ Problem:
$$x^2+(a-5)x+1=3|x|$$ Find 3 distinct solutions to the above problem.
A friend of mine at my coaching center came up with this problem which nobody was able to solve. Unfortunately, I have been unable to contact my professor and understand how to solve this problem. Despite thinking for a long time, I could not come up with anything.
The only things that struck me was that I should open up the modulus sign (first by taking $x\ge0\Rightarrow |x|=x$ and then by taking $x<0\Rightarrow |x|=-x$).
Also, the question could perhaps then be tackled by using relations between the roots of the quadratic equations (I know only that the sum of both roots of a quadratic equation $ax^2+bx+c$ is $\dfrac{-b}{a},$ and that their product is $\dfrac{c}{a}$).
Unfortunately I could not proceed any further. I would be truly grateful if somebody would kindly show me how to solve this problem. Many, many thanks in advance!
|
Case when $x \geq 0$:
$x^2+(a-8)x+1=0$
Then $x=\frac{(8-a)\pm \sqrt{a^2-16a+60}}{2}$
Case when $x < 0$:
$x^2+(a-2)x+1=0$
Then $x=\frac{(2-a)\pm \sqrt{a^2-4a}}{2}$
As we're only working with real solutions: for the determinants to be non-negative, we need $a \leq 6$ or $a \geq 10$ in the former, and $a \leq 0$ or $a \geq 4$ in the latter.
Since we want 3 roots, this means one of the determinants has to be zero while the other is positive. Only $a=6$ satisfies this: for the first case, $x=1$ (only root: the determinant is zero). In the second case, $x=-2 \pm \sqrt 3$. You must also remember to check that the proposed solutions satisfy the domain of $x$ which you have fixed; indeed $1>0$ and $-2 \pm \sqrt 3<0$.
So the answer necessitates $a=6$, from which it follows that $x=1$, $x=-2 \pm \sqrt 3$ are your three distinct solutions.
EDIT: Sorry, there is another solution in $a=4$, whereupon your roots are $x=2 \pm \sqrt 3$ and $x = -1$.
|
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|
Show that $\lim_{z\to 0}|z|^{\sqrt 2}f(z)=0$ in $D$
Let , $f$ be analytic in $\mathbb D\setminus\{0\}$ and unbounded near $z=0$. If the function $|z|^{\sqrt 2}f(z)$ is bounded at $z=0$ then show that $$\lim_{z\to 0}|z|^{\sqrt 2}f(z)=0\text{ & }\lim_{z\to 0}|z|^{\sqrt 2/2}f(z)=\infty.$$where , $\mathbb D=\{z\in \mathbb C:|z|<1\}.$
Since, the function $|z|^{\sqrt 2}f(z)$ is bounded at $z=0$ so, it has a removable singularity at $z=0$ and so, $\lim_{z\to 0}z.|z|^{\sqrt 2}f(z)=0$. From here , how I can show the required limit ?
|
Hint: $z^2f(z)$ is analytic in some $\{0<|z|<r\},$ and from the given information, $z^2f(z) \to 0$ at $0.$ In particular $z^2f(z)$ is bounded in this domain. Hence it has a removable singularity at $0.$ Thus $z^2f(z) = \sum a_nz^n $ in this domain. Think about the first few of the coefficients $a_n.$
|
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|
Does other solutions exist for $29x+30y+31z = 366$? I was asked this trick question:
If $29x + 30y + 31z = 366$ then what is $x+y+z=?$
The answer is $12$ and it is said to be so because $29$ , $30$ and $31$ are respectively the number of days of months in a leap year. Therefore $x + y + z$ must be $12$, the total number of months.
How accurate is this? Is it possible to say so with just a single equation? Are there not other solutions to the equation? If yes, how can one proceed to find other solutions?
|
With $x,y,z$ being reals, no, there's tons of answers. With $x,y,z$ be natural numbers, $12$ is the only answer. Proof: $11$ is too small, because even if all $11$ was in the biggest number, $11\cdot 31=341<366$, and $13$ is too big because $13\cdot 29=377>366$.
The "middle" case if you allow negative integers but not rationals/reals, I'm not sure about
|
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|
Algebra question from practice GRE exam The following is a question from the GRE exam GR9367:
Let $n > 1$ be an integer. Which of the following conditions guarantee that the equation $x^n = \sum_{i=0}^{n-1} a_ix^i$ has at least one root in the interval $(0,1)$?
I. $a_0 > 0$ and $\sum_{i=0}^{n-1}a_i < 1$
II. $a_0>0$ and $\sum_{i=0}^{n-1}a_i > 1$
III. $a_0<0$ and $\sum_{i=0}^{n-1}a_i > 1$
Through some guess work and luck I was able to get the right answer but I would prefer to understand the reasoning behind the problem. What is the underlying idea here? It feels like the right thing to use is Decartes' rule of signs but I couldn't find a way of using the condition on $\sum_{i=0}^{n-1}a_i$.
|
The key to this question lies in three known facts. Namely:
*
*A continuous function $f$ has a root in an open interval $]a,b[$ if $f(a)<0<f(b)$ or $f(b)<0<f(a)$
*All power series are continuous.
and
*$1>0$
So, to prove that a power series $\sum_{i=0}^\infty a_i x^i$ has a root between 0 and 1, wee need to show that either
*
*$a_0<0$ and $\sum_{i=0}^\infty a_i >0$
or
*$a_0>0$ and $\sum_{i=0}^\infty a_i <0$
If we look at condition III, we see that it satisfies $a_0<0$ and $\sum_{i=0}^\infty a_i >1>0$. So Condition III ensures a root between 0 and 1. The other two, however, don't. The "power series" $\frac{1}{2}$ and $2$ satisfy conditions I and II respectively, but neither have roots.
|
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|
Clarification on the difference between Brouwer Fixed Point Theorem and Schauder Fixed point theorem From Zeidler's Applied Functional Analysis
Brouwer
The continuous operator $A:M \to M$ has a fixed point provided $M$ is
compact, convex, nonempty set in a finite dimensional normed space
over $\mathbb{K}$
Schauder
The compact operator $A:M \to M$ has a fixed point provided $M$ is a
bounded, closed, convex, nonempty subset of a Banach space
$\mathbb{X}$ over $\mathbb{K}$
Claim: if $\dim(\mathbb{X}) < \infty$ then Schauder = Brouwer
Just to clarify:
Why is the operator $A$ assumed to be compact for Schauder but merely continuous for Brouwer? What does finite/infinite dimension have to do with this assumption?
|
Your claim is correct.
The point is that in finite dimension all continuous operators are compact, while in infinite dimension you can have continuous operators which are not compact, as the following example shows:
In $\ell_2(\mathbb{N})$ consider the operator $T(x)=(\sqrt{1 - \| x\|^2},x_1, x_2, \dots)$
defined for $\|x\| \leq 1$, where $x=(x_1, x_2, \dots)$ and $\|x\|^2= \sum_{i=1}^{\infty} |x_i|^2$. $T$ is continuous, in fact
\begin{align}
\|T(x) - T(y)\|^2 & = \left| \sqrt{1 - \|x\|^2} - \sqrt{1 - \|y\|^2}\right|^2 + \|x - y\|^2 \leq \\ & \leq \left| \|x\|^2 - \|y\|^2\right| + \|x - y\|^2 \leq \\ &\leq ( \|x\| + \|y\|)\|x-y\| + \|x-y\|^2 \leq \\ &\leq 2\|x-y\| + \|x-y\|^2.
\end{align}
Moreover $T$ maps the closed unit ball to its boundary because $\|T(x)\|^2= 1 - \|x\|^2 + \|x\|^2=1$. $T$ does not have fixed points, by contradiction, if we had $T(x)=x$, we would have $\|x\|=1$, but also $(0,x_1,x_2, \dots) = (x_1,x_2, \dots)$, that is $x_i=0$ for every $i$. In that case we would have $\|x\|=0 \neq 1$, which is a contradiction.
In the same book you refer to, you can read a characterization of compact operators as those well approximated by operators with finite range. In some sense that means compact operators are continuous operators with "almost" finite range. Hence, in finite dimension, continuous operators coincide with compact ones and that's the key point (as you can also easily deduce from the definition of compact operators).
Also, in finite dimension being compact for a set is nothing more than being bounded and closed, so to better understand the main difference, I would write Brouwer's theorem as follows:
The continuous operator $A:M \to M$ has a fixed point provided $M$ is
bounded, closed, convex, nonempty set in a finite dimensional normed space
over $\mathbb{K}$.
|
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|
$\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$.Then find the value of $n$ If $A_1A_2A_3.....A_n$ be a regular polygon and $\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$.Then find the value of $n$(number of vertices in the regular polygon).
I know that sides of a regular polygon are equal but i could not relate$A_1A_3$ and $A_1A_4$ with the side length.Can someone assist me in solving this problem?
|
This problem can be trivialized by Ptolemy's theorem.
First we take the l.c.m and simplify both sides of the given equation.
Let $A_1A_2=a$, $A_1A_3=b$, $A_1A_4=c$. Then we have
$$\frac{1}{a}=\frac{1}{b}+\frac{1}{c} \qquad\to\qquad b c = a b + a c \tag{1}$$
Now, note that regular polygons can always be inscribed in a circle.
Take the quadrilateral $A_1A_3A_4A_5$, which is a cyclic quadrilateral.
Applying Ptolemy's theorem we get,
$$A_1A_3\cdot A_4A_5 + A_3A_4\cdot A_1A_5 = A_1A_4\cdot A_3A_5 \tag{2}$$
Now see that $A_1A_3=b$, $A_4A_5=a$, $A_3A_4=a$, $A_1A_4=c$, $A_3A_5=b$,
so in $(2)$ we have,
$$b a + a\cdot A_1A_5 = c b \tag{3}$$
Comparing equation $(1)$ with $(3)$, we see that $A_1A_5 = c = A_1A_4$.
These are two diagonals of the regular polygon sharing a common vertex, so this diagonal must be a diagonal that bisects the area of the polygon. So on one side of the diagonal $A_1A_4$ there is $A_2$, $A_3$; and on the other side of the diagonal $A_1A_5$ there must be $A_6$, $A_7$. So there are 7 vertices of the polygon.
Hence $n=7$, leading to a heptagon.
|
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|
Binomial expansion. Find the coefficient of $x$ in the expansion of $\left(2-\frac{4}{x^3}\right)\left(x+\frac{2}{x^2}\right)^6$.
I've used the way that my teacher teach me.
I've stuck in somewhere else.
$\left(2-\frac{4}{x^3}\right)\left(x+\frac{2}{x^2}\right)^6=\left(2-\frac{4}{x^3}\right)\left(x^6\left(1+\frac{2}{x^8}\right)^6\right)$
Can anyone teach me? Thanks in advance.
|
Notice, we have
$$\left(x+\frac{2}{x^2}\right)^6=^6C_0x^{6}\left(\frac{2}{x^2}\right)^{0}+^6C_1x^{5}\left(\frac{2}{x^2}\right)^{1}+^6C_2x^{4}\left(\frac{2}{x^2}\right)^{2}+^6C_3x^{3}\left(\frac{2}{x^2}\right)^{3}+^6C_4x^{2}\left(\frac{2}{x^2}\right)^{4}+^6C_5x^{1}\left(\frac{2}{x^2}\right)^{5}+^6C_6x^{0}\left(\frac{2}{x^2}\right)^{6}$$ Now, we have
$$\left(2-\frac{4}{x^3}\right)\left(x+\frac{2}{x^2}\right)^6$$
$$=2\left(x+\frac{2}{x^2}\right)^6-\frac{4}{x^3}\left(x+\frac{2}{x^2}\right)^6$$
Hence, the coefficient of $x$ in the above expansion $$=\text{coefficient of x in the expansion of}\ 2\left(x+\frac{2}{x^2}\right)^6-\text{coefficient of x in the expansion of}\ \frac{4}{x^3}\left(x+\frac{2}{x^2}\right)^6$$$$=2(0)-4(2\cdot ^6C_1)$$$$=-8(^6C_1)=-8(6)=-48$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{coefficient of}\ x=-48}}$$
|
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|
What was Gauss' 2nd Factorization Method? Reading Jean-Luc Chabert's A History of Algorithms, I learned that Gauss, prompted by the poor state-of-the-art, designed two distinct methods for fast integer factorization.
Chabert's book discusses the first, the Method of Exclusions, and D.H. Lehmer gives a really nice explanation in his 1928 introduction of the Lehmer Sieve.
Those two documents plus a misleading MathWorld stub regarding solving Diophantine equations (which was evidently not its original purpose) are the only external references to Gauss' first method that I can find. I can't seem to find anything on the second.
Clarke's 1966 translation of the Disquisitiones Arithmeticae indicates that he actually considered his second method the superior of them (p.397), but I can't make heads or tails of his description. (pp.403-6)
Is there an explanation of Gauss' 2nd factorization method anywhere outside the Disquisitiones, or can anyone who understands it give a simple explanation?
|
Gauss's second factorization method is pretty much a take on the difference of squares method. If one finds $a^2 \equiv b^2 \ (\mathrm{mod} \ M)$, then either $\mathrm{gcd}(M,a+b)$ or $\mathrm{gcd}(M,a-b)$ is a non-trivial factor of M.
Gauss's genius idea was using quadratic forms to force a congruence of squares, by means of calculating the square root of the determinant (a quarter of the discriminant).
Considering D as the determinant of a quadratic form, if one has the following:
$$\frac {x^2}{y^2} \equiv \frac {z^2}{w^2} \equiv D \ (\mathrm{mod} \ M),$$ it is possible to manipulate this congruence and obtain a congruence of squares. Gauss then used the uniqueness of the reduced forms of negative determinant and how one could calculate them all easily to calculate these "square roots".
For the time, it was an outstanding method, even though it requires a lot of calculations.
|
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|
$\int\limits_{0}^{1}(\prod\limits_{r=1}^{n}(x+r))(\sum\limits_{k=1}^{n}\frac{1}{x+k})dx$ The value of $\int\limits_{0}^{1}(\prod\limits_{r=1}^{n}(x+r))(\sum\limits_{k=1}^{n}\frac{1}{x+k})dx$ is equal to
$(A)n\hspace{1cm}(B)n!\hspace{1cm}(C)(n+1)!\hspace{1cm}(D)n.n!$
I tried:$\int\limits_{0}^{1}(\prod\limits_{r=1}^{n}(x+r))(\sum\limits_{k=1}^{n}\frac{1}{x+k})dx$=$\int\limits_{0}^{1}(x+1)(x+2)...(x+n)(\frac{1}{x+1}+\frac{1}{x+2}+......+\frac{1}{x+n})dx$=$\int\limits_{0}^{1}(x+2)(x+3)...(x+n)+(x+1)(x+3)...(x+n)dx$
I cannot solve it further.Is my approach wrong,I am stuck.What is the right way to solve,Please help...
|
Differentiate $\prod_{k=1}^n(x+k)$ with respect to $x$, and you get the integrand. So the answer is $(n+1)!-n!=n\cdot n!$.
|
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|
Fraction of original velocity in series of elastic collisions I need to find the answer to this question:
Three particles A, B, and C, with masses $m$, $2m$ and $3m \, \mathrm k \mathrm g$ respectively, lie at rest in that order in a straight line on a smooth horizontal table. The particle A is then projected directly towards B with speed $u \, \mathrm m \mathrm s^{-1}$. Assuming the collisions are perfectly elastic, what fraction of $u$ is the speed of the particle C immediately after the second impact?
I tried to solve it like this:
$$P = mv$$
Momentum before collision = momentum after collision, therefore
$$um = mV_a + 2mV_b$$ where $V_a$ and $V_b$ are the velocities of the particles after impact.
$$\implies u = V_a + 2V_b$$
I got to there and then tried rearranging to make $V_b$ the subject and using that in the next collision but I couldn't get anything out of it in the end. How can I solve this?
|
The collisions are perfectly elastic. So you should write energy balance equation.
$$ K.E._{initial}= K.E._{final} $$
So for the $1^{st}$ collision
$$\frac{1}{2}mu^2=\frac{1}{2}mV_a^2+ \frac{1}{2}(2m)V_b^2 $$
Now we get 2 unknowns $ V_a \hspace{0.2cm} and \hspace{0.2cm} V_b $ and 2 equations namely momentum conservation and energy conservation .
Solve . Repeat for next collision.
|
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|
How do I convince students in high school for which this equation: $2^x=4x$ have only one solution in integers that is $x=4$? I would like to convince my student in high school level using a simple
mathematical way to solve this equation: $$2^x=4x$$ in $\mathbb{Z}$ which
have only one integer solution that is $x=4$ .
My question here :How do I convince students in high school for which this equation: $$2^x=4x$$ have only one solution that is $x=4$?
Note : I do not want to use substitution to convince them and by numerical methods can't give us exactly $x=4$
EDIT: I edited the question as it is very related to the precedent
Thank you for any help.
|
Hint:
plot the graph of $y=2^x$ and $y=4x$ and shows that the only other solution is between $0$ and $1$.
|
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|
Solving equations with logarithmic exponent I need to solve the equation :
$\ln(x+2)+\ln(5)=\lg(2x+8)$
With the change of base formula we can turn this into:
$\ln(x+2)+\ln(5)=\frac{\ln(2x+8)}{\ln(10)}$
We can also simplify the LHS with the product rule so:
$\ln(5(x+2))=\frac{\ln(2x+8)}{\ln(10)}$
Solving the fraction gives us: $\ln(10) \, \ln(5(x+2)) = \ln(2x+8)$
Simplifying the LHS even further:
$\ln(5x+10)^{\ln(10)}=\ln(2x+8)$
We can then see that $(5x+10)^{\ln(10)}=2x+8$
And this is where I get stuck, I can't seem to figure out how to expand this term. Does anyone know how to solve this?
|
You can find answer by researching of graphic of functions,I don't think you can get it by algebric method
|
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