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Distance of the Focus of an Hyperbola to the X-Axis For a National Board Exam Review:
How far from the $x$-axis is the focus of the hyperbola $x^2 -2y^2 + 4x + 4y + 4$?
Answer is $2.73$
Simplify into Standard Form:
$$ \frac{ (y-1)^2 }{} - \frac{ (x+2)^2 }{-2} = 1$$
$$ a^2 = 1 $$
$$ b^2 = 2 $$
$$ c^2 = 5 $$
Hyperbola is Vertical:
$$ C(-2,1) ; y = 1 $$
$$ y = 1 + \sqrt5 = 3.24 $$
$$ y = 1 - \sqrt5 = 1.24 $$
Both answers don't match; What am I doing wrong?
|
$\frac{(x + 2)^2}{2} - \frac{(y - 1)^2}{1}; C = (-2, 1)$
$C = \sqrt{2 + 1} = \sqrt{3}$
Answer: $1 + \sqrt(3) = 2.73$
|
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Is there any example of a Category without generators? I was looking of an example of a category (a non trivial example) without generators (seperators). Is there any nice example? Or every category that we are using has generators?
|
It depends on what you mean by "generators." There is a notion of a collection $S$ of objects being a "family of generators" of a category $C$, which means that the functors $\text{Hom}(s, -) : C \to \text{Set}, s \in S$ are jointly faithful. Any category has a (possibly large) family of generators given by taking every object, so the interesting question is when a category has a small (set-sized) family of generators.
An example of a category, even an abelian category, without this property is the category of ordinal-graded vector spaces (that is, collections $V_{\alpha}$ of vector spaces, one for each ordinal) with bounded support (that is, for all objects $V_{\bullet}$ there is some $\alpha$ such that $V_{\beta} = 0$ for all $\beta > \alpha$). Any set-sized collection of objects $S$ collectively has bounded support (namely the supremum of the supports of the objects in $S$) and so cannot detect $V_{\bullet}$ with support above that bound.
The property of admitting a small family of generators implies and is closely related to the property that a category is concretizable.
|
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What is a residue? I've heard of residues in complex analysis, contour integration, etc. but all I really know it to be is the $c_{-1}$ term in the Laurent series for a function. Is there some sort of intuition on what a "residue" actually is? The terminology makes it seem like something left over, or something like that.
|
The holomorphic functions have an extraordinary property: if you compute an integral along a path, the value of the integral does not depend on the path !
More precisely, if the function is holomorphic everywhere inside a closed path, the integral is just zero. But if the function has poles (zeroes at the denominator, $c_{-k}$ terms in the Laurent series), every pole brings a nonzero contribution called its residue. You can shrink the path as much as you want, even turning it to infinitesimal circles around every pole, provided you keep the poles in.
So the residues are what is left (as regards integration) after you removed all the holomorphic parts of the domain.
|
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Find a mistake type of math problems I am interested in the problems where the formulation of the problem has some kind of mistake in it and as a consequence gives unexpected answer.
Can't explain it better than this example:
For example: $4^2 = 4 \cdot 4 = 4+ 4 + 4+4$ (sum $4$ times)
Similarly: $$\frac{d}{dx}x^2=\frac{d}{dx}(x\cdot x)=\frac{d}{dx}(x+x+...+x) = 1 + 1 + ... + 1 = x$$
Since the $1$'s are summed $x$ times. I hope you see what went wrong :) If you have encountered problems of this type, please share.
EDIT: I am aware that my proposed example has a mathematically inconsistent step, this type of expansion is only allowed for natural numbers making the function non-differentiable. However this is the kind of inconsistencies I find amusing.
|
Your equation does not make sense for many possible values of $x$. The equation
\begin{align}
x \cdot x &= \underbrace {x + x + \dots + x}_{x \text{ times}},
\end{align}
only holds when $x$ is a nonnegative integer. But $x$ is an indeterminant which can take on real values as well. So you can't write $x^2$ as the sum of $x$ copies of $x$ when $x$ is not an integer. It simply does not make sense.
Moreover, you cannot differentiate the identity, as it only holds for the integers. You at least need the identity to hold about a neighborhood of a point $x$ in the real numbers, since the derivative is a local property.
|
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Finding the distance from the origin to the surface $xy^2 z^4 = 32$ using the method of Lagrange Multipliers Problem: Find the distance from the origin to the surface $xy^2z^4 = 32$.
Attempt: The Lagrange equation for this problem is $L(x,y,z, \lambda) = x^2 + y^2 + z^2 + \lambda (xy^2 z^4 - 32)$. Setting the first partials to zero we have \begin{align*} \frac{\partial L}{\partial x} &= 2x + \lambda y^2 z^4 = 0 \qquad (1) \\ \frac{\partial L}{\partial y} &= 2y + 2 \lambda x y z^4 = 0 \qquad (2) \\ \frac{\partial L}{\partial z} &= 2z + 4 \lambda x y^2 z^3 = 0 \qquad (3) \\ \frac{\partial L}{\partial \lambda} &= xy^2 z^4 - 32 = 0 \qquad (4)
\end{align*}
Now I'm having a hard time solving this system for $x,y$ and $z$. Here is what I did so far. From $(1)$ and $(2)$ we get \begin{align*} \frac{2x}{y^2 z^4} = - \lambda \qquad \text{and} \qquad \frac{1}{xz^4} = - \lambda \end{align*} Thus $\frac{2x}{y^2 z^4} = \frac{1}{xz^4} $ or $y^2 = 2x^2$ after simplification. Also, from $(2)$ and $(3)$ we can deduce that \begin{align*} \frac{1}{xz^4} = - \lambda = \frac{2z}{4xy^2 z^3} \end{align*} so that $2y^2 = z^2$ after simplification. Now I used all this and substituted it into $(4)$. This gave me \begin{align*} x(2x^2) (4y^4) - 32 = 0 \end{align*} or (since $y^4 = 4x^4)$ \begin{align*} 8x^3 (4x^4) - 32 = 0 \end{align*}
This means that $32x^7 - 32 = 0$, so that $x = 1$. Then $y^2 = 2$, so that $y = \pm \sqrt{2}$. Then $z^2 = 4$, so that $z = \pm 2$. So I found the points $(x,y,z) = (1, \sqrt{2}, 2)$ and $(1, - \sqrt{2}, -2)$. They both give me the distance $\sqrt{x^2 + y^2 + z^2} = \sqrt{7}$, so I'm guessing they are equal? Is my reasoning correct?
|
A hint:
Multiply $(1)$ by $x$, $(2)$ by $y$, and $(3)$ by $z$ and look at the three equations you got.
|
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Cutting a paper with the smallest number of cuts You want to cut a piece of paper of length $N$ to $N$ pieces of length 1.
It is not allowed to fold the paper, but if two or more previously-cut pieces of paper have the same length, it is allowed to put them one over the other.
If $N=2$ you need 1 cut.
If $N=3$ you need 2 cuts.
If $N=4$ you can use 3 cuts, but there is a better way: use one cut to create two pieces of length 2, then put the two pieces one over the other and use one cut to create four pieces. The total number of cuts is 2.
In general, if $N$ is a power of 2 then the total number of cuts required is $\log_2{N}$.
What is the smallest number of cuts required when $N$ is not a power of 2?
I thought of the following recursive algorithm:
Cut(N)
If $N=2m+1$ then cut a piece of length 1 and use Cut(2m) on the remainder.
If $N=2m$ then cut two pieces of length $m$,
put one over the other and use Cut(m) on the remainder.
For every $N$, after at most 2 cuts the remaining length will be at most $N/2$. Hence, the number of required cuts in the above algorithm is at most:
$$2 \lfloor\log_2(N)\rfloor$$
The worst case is when $N=2^k-1$; then, $2(k-1)$ cuts are required.
Is there an algorithm that requires a smaller number of cuts?
NOTE: The present question is simpler than Fold, Gather, Cut because here it is not allowed to fold the paper.
|
Your problem is equivalent to that of finding the shortest addition chain for $N$. You can order the operations by descending lengths of the cut pieces, so you never need to cut the same length twice, so what you're doing is to prescribe for each length that occurs how to compose it from two smaller lengths, which is exactly what an addition chain does.
The lengths of the shortest addition chains are in OEIS sequence A003313. Some interesting results and references are given, but no closed formula.
|
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Pretty easy equations of elements in a group Problem
$G$ is a group generated by $a,b\in G$ such that $a^5=e$, $aba^{-1}=b^2$ and $b\ne e$. I want to find the order of $b$.
Attempt
I tried to multiply the second equation from right by $a^{4}$: $$ba^{-1}=a^{4}b^{2}$$ Then $$ba^{-1}=(ba^{2})^{2}=ba^{2}ba^{2}$$
Then by multiplying by $a$ from right and $b^{-1}$ from left we get
$$a^{2}ba^{3}=e$$ Then by multiplying by $a^{-2}$ from the left and $a^{-3}$ from right we get $$b=a^{-5}=(a^{5})^{-1}=e$$ but $b\ne e$, so is it right to say that the order is $\infty$? Or did I do something wrong?
|
Note that for every $m$ we have
$$ab^ma^{-1} = (aba^{-1})^m = b^{2m}.$$
Thus by induction we have
$$a^kba^{-k} = b^{2^k}\tag{$\ast$}$$
for all $k\in\mathbb{N}$. On the other hand, we have $a^5ba^{-5} = b$ since $a^5 = e$. Putting that and $(\ast)$ together reveals the order of $b$.
|
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Prove that $(a+b)(b+c)(c+a) \ge8$
Given that $a,b,c \in \mathbb{R}^{+}$ and $abc(a+b+c)=3$,
Prove that $(a+b)(b+c)(c+a)\ge8$.
My attempt: By AM-GM inequality, we have
$$\frac{a+b}{2}\ge\sqrt {ab} \tag{1}$$
and similarly
$$\frac{b+c}{2}\ge\sqrt {bc} \tag{2},$$
$$\frac{c+d}{2}\ge\sqrt {cd} \tag{3}.$$
Multiplying $(1)$, $(2)$ and $(3)$ together we reach
$$(a+b)(b+c)(c+a) \ge8abc.$$
Now, I need to show that $abc = 1$.
Again by AM-GM inequality we have
$$\frac{a+b+c}{3}\ge\sqrt[3]{abc}
\implies \frac{\frac{3}{abc}}{3}\ge\sqrt[3]{abc}
\implies abc\le1$$
Now if we show somehow that $abc\ge1$, we are done and that's where I am stuck. Can someone please explain how to show that?
Other solutions to the above question are also welcomed.
|
From $abc(a+b+c)=3$ we have $$a^2+a(b+c)=\frac {3}{bc}.$$ Therefore we have $$(a+b)(b+c)(c+a)\geq 8\iff$$ $$(a+b)(a+c)\geq \frac {8}{b+c}\iff$$ $$a^2+a(b+c)\geq -bc+ \frac {8}{b+c)}\iff$$ $$(\bullet ) \quad \frac {3}{bc}\geq -bc+\frac {8}{b+c}.$$ Now $b+c\geq 2\sqrt {bc},$ so $$\frac {8}{b+c}\leq \frac {4}{\sqrt {bc}}.$$ So with $x=\sqrt {bc} $ we see that ($\bullet $ ) is satisfied if $\frac {3}{x^2}\geq -x^2+\frac {4}{x} $, or, equivalently, $\frac {3} {x^2}+x^2-\frac {4}{x}\geq 0 ,$ for all $x>0.$
We have $\frac {3}{x^2}+x^2-\frac {4}{x}=x^{-2}(x-1)(x^3+x^2+x-3). $ The terms $(x-1)$ and $(x^3+x^2+x-3)$ always have the same sign when $x\geq 0.$ (Case 1:$\;x\geq 1.$ Case 2: $\;0<x<1.$)
Inelegant compared to the answer by Booldy, but it works.
|
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Tangent planes to $2+x^2+y^2$ and that contains the $x$ axis I need to find the tangent planes to $f(x,y) = 2+x^2+y^2$ and that contains the $x$ axis, so that's what I did:
$$z = z_0 + \frac{\partial f(x_0,y_0)}{\partial x}(x-x_0)+\frac{\partial f(x_0,y_0)}{\partial y}(y-y_0) \implies \\ z = 2 + x_0^2 + y_0^2 + 2x_0(x-x_0) + 2y_0(y-y_0) \implies \\ 2xx_0 + 2yy_0-z-x_0^2-y_0^2+2=0$$
So since the plane must contain the $x$ axis, its normal vector must have the form $(0,y,z)$. The normal vector fot the plane I found is:
$$(2x_0, 2y_0, -1)$$
so $x_0 = 0, y_0 = y_0$
our plane has the form:
$$2y_0y -z -y_0^2+2 = 0$$
but when I plot this graph for some values of $y_0$ I only get 1 tangent plane at $y_0\approx 1.5$
|
Taking advantage of the fact that the surface is a quadric, here’s a way to solve this without calculus. Working in homogeneous coordinates, all of the planes that contain the $x$-axis are of the form $[0:\lambda:\mu:0]$, for $\lambda$ and $\mu$ not both zero. These planes are tangent to the given surface iff they satisfy the dual conic equation $$\begin{align} \begin{bmatrix}0&\lambda&\mu&0\end{bmatrix}\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&0&-\frac12\\0&0&-\frac12&2\end{bmatrix}^{-1}\begin{bmatrix}0 \\ \lambda \\ \mu \\ 0\end{bmatrix} &= \begin{bmatrix}0&\lambda&\mu&0\end{bmatrix}\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&-8&-2\\0&0&-2&0\end{bmatrix}^{-1}\begin{bmatrix}0 \\ \lambda \\ \mu \\ 0\end{bmatrix} \\ &= \lambda^2-8\mu^2 = 0\end{align}$$ (the $4\times4$ matrix on the left-hand side is obtained by writing the equation $x^2+y^2-z+2=0$ in the form $\mathbf x^TC\mathbf x=0$.) therefore $\lambda=\pm2\sqrt2\mu$. Since these are homogeneous vectors, we can choose to set $\mu=1$, producing the two equations $z=\pm2\sqrt2 y$ for the tangent planes that contain the $x$-axis.
|
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Integrating unit impulse function Given that,
$$ \delta(t) = \begin{cases} \infty & \text{if } t = 0 \\
0 & \text{if } t \ne 0\\
\end{cases}$$
How is it that,
(A)
$$
\int_{-\infty}^\infty \delta(t) dt = 1
$$
(B)
$$
\int_{-\infty}^\infty f(t) \delta(t) dt = f(0)
$$
considering $f$ continuos at $t=0$
Thanks in advance
|
Can I assume you're a physicist and avoid the mathematical complexities a bit?
The integral finds the area under functions. $\delta(x)$ is a really skinny and really tall rectangle. In fact, it's width is $\epsilon$ and its height is $\omega$ so the area is given by $ \epsilon \cdot \omega$. The variable $\epsilon={1 \over {\omega}}$, so $ \epsilon \cdot \omega=1$. You'd just let $\omega \to \infty$ to get the integral of $\delta(x)$. Using this, I encourage you to figure out the answer to your other question.
More mathematical here. The delta 'function' is not a function in any typical sense. It's not continuous, differentiable, or integrable in the Riemann sense.
However, if you define it as a measure, you can look at it in a more rigorous way. A measure, is basically a way to assign mass, or weight, to subsets of the x axis. For instance, the way density can be integrated is a good example of a measure. There is some similarity with distributions as well.
Define the measure $\delta(dx)$ to be $1$ if $dx$ includes the value $0$ and $0$ otherwise. Using this definition, you can integrate with respect to the measure. Doing this, you get,
$$ \int_{-\infty}^{\infty} f(x) \ \delta(dx)=f(0)$$
Here's more information about the Dirac delta function, skip to the "As a Measure" section if you want.
|
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Fibonacci proof question $\sum_{i=1}^nF_i = F_{n+2} - 1$ The sequence of numbers $F_n$ for $n \in N$ defined below are called the Fibonnaci numbers.
$F_1 = F_2 = 1$, and for $n \geq 2$, $F_{n+1} = F_n + F_{n-1}$.
Prove the following facts about the Fibonnaci numbers
For each of the following, $n \in N$
a) $\displaystyle \sum_{i=1}^nF_i = F_{n+2} - 1$
I started this proof by using strong induction
To prove something by strong induction, you have to prove that if all natural numbers strictly less than $N$ have the property, then $N$ has the property.
$n \geq 2$, $F_{n+1} = F_n + F_{n-1}$
Check basis step $n=2$:
$F_1 + F_2 = 1 + 1 = 2 = 3-1 = F_4 - 1$, therefore TRUE
I'm unsure how to go any further, when it comes to strong induction I'm lost on how to set up my IH, and proceed for the rest of the steps
|
Combinatorial Proof:
For each $n\in\mathbb{N}_0$, the number of ways to tile an $1$-by-$n$ array with $1$-by-$1$ squares and $1$-by-$2$ dominos is $F_{n+1}$. Hence, the number of ways to tile an $1$-by-$(n+1)$ array in such a manner with at least one domino is given by $F_{n+2}-1$.
Now, let $k$ be the earliest position of the dominos. For each $k$, we are left with the $1$-by-$(n-k)$ array to tile, which can be done in $F_{n-k+1}$ ways. Since $k$ can be anything from $1$ to $n$, we have
$$F_{n+2}-1=\sum_{k=1}^n\,F_{n-k+1}=\sum_{i=1}^n\,F_i\,.$$
|
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Is this linearly independent? What is the dimension span?
Consider the vectors in $\mathbb{R}^4$ defined by
$v_1 = (1,2,10,5), v_2 = (0,1,1,1), v_3 = (1,4,12,7)$.
Find the reduced row echelon form of the matrix which has these as its rows. What is its rank? Is $\{v_1 , v_2 , v_3 \}$ linearly independent? What is the dimension of $ \text{span}(v_1, v_2, v_3)$?
I'm not sure if i'm right.
I was able to take the system $av_1+av_2+av_3=0$
$$\begin{bmatrix} 1&0&1\\2&1&4\\10&1&12\\5&1&7 \end{bmatrix} = \begin{bmatrix} 0\\0\\0\\0 \end{bmatrix}$$
And reduce it down to
$$\begin{bmatrix} 1&0&1\\0&1&2\\0&0&0\\0&0&0\end{bmatrix}= \begin{bmatrix} 0\\0\\0\\0 \end{bmatrix}$$
which means rank is $2$ because we have $2$ leading ones.
Also i think it's not linearly independent as row 1 $= a+c=0, \text{row} \hspace{.1cm} 2 = b+2c=0, a=-c$ and $b=-2c$
hence $a$,$b$ and $c$ are not equal to zero.
If it was linearly independent this would mean that the basis are vectors 1 and 2 but not 3. Meaning $v_1$ and $v_2$ spans $\mathbb{R}^2$ and hence the dimension span is $\mathbb{R}^4$
There is also a second part which i don't understand. But it's ok i'm mainly interested in the previous part might post this on a separate question.
Find a homogeneous linear system for which the space of solutions is exactly the subspace of $\mathbb{R}^4$ spanned by
$v_1, v_2,$ and $v_3$.
|
The vectors are linearly dependent since $v_3=v_1+2v_2$. As you noted, you have the two leading ones, and it is clear that the vectors span a space with dimension two, the say way that $Span\{\vec i,\vec j\}$ spans two dimensions in $\Bbb R^3$
|
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|
Euler phi function and powers of two For small values of n$\ge$1 it appears that one has the inequality
$\phi(2^n)$ $\le$ $\phi(2^n + 1)$.
However, it seems unlikely that this would hold for all n .
Question: Are there any explicit values of n known for
which the inequality is violated ?
Thanks
|
For each prime factor $p|2^n+1$ you get a reduction of $\frac {p-1}P$ in the $\varphi$ function. The $p$ you are looking for are ones for which $2^n\equiv -1 \bmod p$. Then for odd $k$, $2^{kn}\equiv -1 \bmod p$. So if we work with odd exponents we can collect primes - the question is can we collect enough small primes.
Now we have $2^1\equiv -1\bmod 3, 2^5\equiv -1 \bmod 11, 2^7\equiv -1 \bmod 43$ and the list of powers and primes continues $(1:3), (5:11), (7:43), (9:19), (11:683), (13:2731), (15:331)$ - for these the relevant factor is approximately $0.5581$ and we need it below $0.5$
No doubt someone can search more efficiently than I can to provide a definitive conclusion.
|
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Convergence and Limit of a Sequence I am looking at the sequence and I'm trying to see what happens as $n\to\infty$
$$
a_n = \frac{\prod_{1}^{n}(2n-1)}{(2n)^n} = \frac{1\cdot3\cdot5\cdot...\cdot(2n-1)}{(2n)^n}
$$
By inspection, I can see that the denominator increases faster than numerator which indicates that the sequence converges, but I'm not sure how to show that mathematically.
I tried using the ratio test to test for convergence but I got stuck on the following last step
$$
\begin{align*}
\lim_{n\to\infty}\left| \frac{a_{n+1}}{a_n}\right| &= \lim_{n\to\infty}\left|\frac{1\cdot3\cdot5\,\cdot\,...\,\cdot\,(2n-1)(2n+1)}{(2(n+1))^n}\cdot \frac{(2n)^n}{1\cdot3\cdot5\,\cdot\,...\,\cdot\,(2n-1)}\right| \\ &= \lim_{n\to\infty}\frac{n^n(2n+1)}{(n+1)^n} \\ &=\exp\left[\lim_{n\to\infty}\ln\left(\frac{n^n(2n+1)}{(n+1)^n}\right)\right]\\ &=\exp\left[\lim_{n\to\infty}n\ln \left(\frac{n}{n+1}\right) + \ln(2n+1)\right]
\end{align*}
$$
In short, how do I prove that this sequence is convergent, and how can I evaluate $\lim_\limits{n\to\infty}a_n$ ?
|
We have
$$
\frac{\prod_{1}^{n}(2k-1)}{(2n)^{n}} < \frac{2n-1}{(2n)^{2}} \to 0
$$
as $n$ grows.
|
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Show that if $|G| = 30$, then $G$ has normal 3-Sylow and 5-Sylow subgroups.
Show that if $|G| = 30$, then $G$ has normal $3$-Sylow and $5$-Sylow subgroups.
Let $n_3$ denote the number of 3-Sylow subgroups and $n_5$ the number of $5$-Sylow subgroups. Then, by the third Sylow theorem, $n_3$ divides $10$ and $n_3 \equiv 1 \mod 3$. From these two, we see that $n_3 = 1$. This implies that $G$ has a normal 3-Sylow subgroup.
Similarly, let $n_5$ denote the number of $5$-Sylow subgroups. Then, by the third Sylow theorem, $n_5$ divides $6$, and $n_5 \equiv 1 \mod 5$. So, we can infer that either $n_5 = 1$ or $n_5 = 6$. I don't know how to proceed from here. Can someone please help me? (This is not homework, only self study)
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This is more delicate and difficult than most problems about Sylow subgroups, and it must be tackled by relating the 5-sylow subgroups and the 3-sylow subgroups. In general such a problem in group theory at the undergraduate level is either going to be trivially easy, or require this technique.
As you established $n_3 = 1$ is one possibility but it was noted in the comments that $n_3 = 10$ is also possible. Suppose $n_5 = 6$ so that the $5$-sylows are not normal. Then they must be their own normalizers because the number of them is equal to the index of the normalizer of any one, and they are the unique subgroups of order $5$. However the fact that only $5$ divides $30$ and not $25$ allows us to know that all of the elements of order $5$ in this group are contained in a $5$-sylow, and that these sylows have trivial intersection. But that means there are $(5-1)*6 = 24$ elements of order $5$ in this group. But this means that there cannot possibly be $10$ subgroups of order $3$ otherwise there would be $20$ elements of order $3$ and that would give $47$ elements in the group which would be absurd.
So we know that the $3$-sylows would have to be normal in this situation. Then we have that if $H_3$ is a $3$-sylow and $H_5$ is a $5$-sylow then $H_5H_3$ is a subgroup because $H_3$ is normal. But then $H_5H_3/H_3 \cong H_5/(H_5 \cap H_3)$ and counting orders we know that $H_5H_3$ has order $15$. But there is a unique group of order $15$ (there are many ways to show this, but in our particular situation the fact that $H_5H_3 \cong \mathbb{Z}_{15}$ comes down to the fact that $H_5$ is normal by a sylow argument on this subgroup) and thus this group is abelian and this contradicts that $H_5$ is its own normalizer. Thus $H_5$ is normal.
Finally we can now count again. Since $H_5$ is normal we can take any one of these $H_3$ and do the same argument as the above to realize that $H_3$ has a normalizer of order at least $15$ and thus $H_3$ must be normal because it is supposed to have normalizer or index either $1$ or $10$.
Hope this helps.
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Prove that $(2n+1)k_{n+1}=(2n+1)k_{n}+\cos^{2n+1} (x)$ Given that $$k_n=\int \frac{\cos^{2n} (x)}{\sin (x)} dx$$ Prove that $$(2n+1)k_{n+1}=(2n+1)k_{n}+\cos^{2n+1} (x)$$
I have tried to prove this is true by differentiating both sides with product rule:
$$2k_{n+1}+\frac{\cos^{2n+1} (x)}{\sin (x)}(2n+1)=2k_n+\frac{\cos^{2n} (x)}{\sin (x)}(2n+1)+(2n+1)\cos^{2n} (x) \sin (x)$$
I am stuck here as I met a dead end upon grouping and expanding. Please help. Thank you in advance!!
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Verifying by differentiation is a good idea. We wish to prove that
$$k_{n+1}=k_n+\frac{1}{2n+1}\cos^{2n+1}(x).$$
Recall that indefinite integrals are only determined up to a constant. So we really need to show that the derivative of the left-hand side is equal to the derivative of the right-hand side. By the Fundamental Theorem of Calculus, we need to show therefore that
$$\frac{\cos^{2n+2}(x)}{\sin x}=\frac{\cos^{2n}(x)}{\sin x}-\sin x\cos^{2n}(x).$$
Equivalently, we want to show that
$$\sin x\cos^{2n}(x)=\frac{\cos^{2n}(x)}{\sin x}-\frac{\cos^{2n+2}(x)}{\sin x}.\tag{1}$$
The right-hand side of (1) is $\frac{\cos^{2n}(x)}{\sin x}(1-\cos^2 x)$. Replace $1-\cos^2 x$ by $\sin^2 x$, cancel a $\sin x$ from top and bottom, and we obtain the left-hand side of (1).
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Of the three lines $x+\sqrt3y=0,x+y=1$ and $x-\sqrt3y=0$,two are equations of two altitudes of an equilateral triangle Of the three lines $x+\sqrt3y=0,x+y=1$ and $x-\sqrt3y=0$,two are equations of two altitudes of an equilateral triangle.The centroid of the equilateral triangle is
$(A)(0,0)\hspace{1cm}(B)\left(\frac{\sqrt3}{\sqrt3-1},\frac{-1}{\sqrt3-1}\right)\hspace{1cm}(C)\left(\frac{\sqrt3}{\sqrt3+1},\frac{1}{\sqrt3+1}\right)\hspace{1cm}(D)$none of these
All options (A),(B),(C) appears to be the answers.But correct answer is given (A).It is not given which two are altitude equations and which are not.Please guide me.
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Notice, we have the following equations of the lines $$x+\sqrt 3y=0\iff y=\frac{-1}{\sqrt 3}x\tag 1$$
$$x+y=1\iff y=-x+1\tag 2$$
$$x-\sqrt 3y=0\iff y=\frac{1}{\sqrt 3}x\tag 3$$
Let the slopes of the above lines be denoted by $m_1=-\frac{1}{\sqrt 3}$, $m_2=-1$ & $m_3=\frac{1}{\sqrt 3}$ then the angles between them are calculated as follows
*
*The angle between lines (1) & (2) $$\theta_{12}=\tan^{-1}\left|\frac{m_1-m_2}{1+m_1m_2}\right|$$
$$=\tan^{-1}\left|\frac{-\frac{1}{\sqrt 3}-(-1)}{1-\frac{1}{\sqrt 3}(-1)}\right|=\tan^{-1}\left|2-\sqrt 3\right|=15^\circ$$
*The angle between lines (2) & (3) $$\theta_{23}=\tan^{-1}\left|\frac{-1-\frac{1}{\sqrt 3}}{1+(-1)\frac{1}{\sqrt 3}}\right|=\tan^{-1}\left|2+\sqrt 3\right|=75^\circ$$
*The angle between lines (1) & (3) $$\theta_{13}=\tan^{-1}\left|\frac{-\frac{1}{\sqrt 3}-\frac{1}{\sqrt 3}}{1-\frac{1}{\sqrt 3}\frac{1}{\sqrt 3}}\right|=\tan^{-1}\left|\sqrt 3\right|=60^\circ$$
We know that the angle between any two altitudes of an equilateral triangle is $60^\circ$ or $120^\circ$ Hence, the lines (1) & (3) are the altitudes of an equilateral triangle hence the centroid of the triangle is the intersection point of (1) & (3) i.e. origin $(0, 0)$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{Centriod of equilateral triangle}\equiv\color{blue}{(0, 0)}}}$$
Hence, option (A) $(0, 0)$ is correct.
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Finding range of $m$ in $x^2+mx+6$. Find the range of values of $m$ in the quadratic equation $x^2+mx+6=0$ such that both the roots of the equation $\alpha,\beta<1$.
My attempt - it is given that
$\alpha<1$ and $\beta<1$
$\rightarrow \alpha+\beta<2$
But $\alpha+\beta=-m$
Thus $m>-2$.
But this solution doesn't involve the coefficient term i.e. $6$.
Any solution to the above question is appreciated.
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You made a logical fallacy as old as time.
What you have proven is:
If $\alpha, \beta<1$, then $m>-2$
What you HAVE NOT PROVEN is:
If $m>-2$, then $\alpha, \beta < 1$.
You can easily see that the second statement is false, since if $m=0>-2$, the equation has no real solutions.
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Decreasing sequence of sets: Power of natural numbers Let $P(N)$ be the set of all the possible subsets of natural numbers (power set of $N$). Suppose that we have a decreasing sequence of sets $S_n$, ie $S_{n+1} \subseteq S_n\;,\in P(N)$ such that they are all finite and a set $M$ such that $\#M \leq \#S_n$, for all $n$. It is possible to say that $$\#M \leq \#\cap_{n=0}^\infty S_n$$?
My intuition says it does, but I couldnt prove it. It seems that the intersection must be one of the $S_n$. Any hint of what should I do?
Thanks!
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Let $S$ denote the intersection of the $S_n$. Start with a fixed $m$.
The set $S_m-S$ is finite and for every element $s\in S_m-S$ some $n_s$ exists with $s\notin S_{n_s}$.
Then $S_n=S$ if $$n\geq\max\{n_s|s\in S_m-S\}\in\mathbb N$$
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Prove $(1+ \tan{A}\tan{2A})\sin{2A} = \tan{2A}$. Prove the following statement
$$ (1+ \tan{A}\tan{2A})\sin{2A} = \tan{2A}. $$
On the left hand side I have put the value of $\tan{2A}$ and have then taken the LCM.
I got $\sin{2A}\cos{2A}$. How do I proceed?
Thanks
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Noting that
$$\cot \theta\tan\theta=1$$
We have
$$(1+\tan A\tan 2A)\sin 2A = (1+\tan A\tan 2A)\sin 2A\cot 2A\tan 2A$$
Now just show that
$$(1+\tan A\tan 2A)\sin 2A\cot 2A = 1$$
:)
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Multiple choice: $S = {x | 0 ≤ x < 280 ∧ x ≡ 3 (mod 7) ∧ x ≡ 4 (mod 8)}$ The question is:
Consider the following set of integers:
$$ S = \left\{x \left| 0 \le x < 280
∧ x \equiv 3 \mod 7
∧ x \equiv 4 \mod 8 \right. \right\}. $$
How many integers are there in S?
$0$? $1$? $2$? $5$? $10$? $280$?
*
*It's a multiple choice answer with the correct answer of 5. I just can't seem to wrap around the idea.
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$x \equiv 3 \pmod{7}$ is equivalent to $x+4 \equiv 0 \pmod{7}$.
$x \equiv 4 \pmod{8}$ is equivalent to $x+4 \equiv 0 \pmod{8}$.
Thus we want to know which numbers $x$ are such that $x+4$ is divisible by both $7$ and $8$. Since $7$ and $8$ are coprime, it would be exactly those numbers such that $x+4$ is divisible by $7 \times 8 = 56$. You can count easily now.
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Completeness of Normed spaces. I want to prove the following proposition
If $(X,||\cdot||)$ and $(X,||\cdot||')$ are homeomorphic, then $(X,||\cdot||)$ is complete if and only if $(X,||\cdot||')$ is complete.
So, I only know the definition of homeomorphic, and can't figure out how only with that I can prove that proposition. Can someone help me to prove this please?
Thanks a lot in advance :)
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Take a Cauchy sequence in $(X,||\cdot||)$ and show that is also a Cauchy sequence in $(X,||\cdot||')$.
This proves the assertion since convergence is a topological property which is preserved by homeomorphisms.
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probabilities of arrival time John and Mary arrive under the clock tower independently. Let X be John's arrival time and let Y be Mary's arrival time. If John arrives first and Mary is not there then he will leave. If Mary arrives first then she will wait up to one hour before leaving. John's arrival time, X, is exponentially distributed with mean of 1. Mary's arrival time has density $f(y) = {2y\over 9}$, $0≤y≤3$. Calculate the probability they will meet.
What I've tried.
I figured this problem should be the summation of two probabilities. The probability that Mary arrives before John + The probability that John arrives within an hour of Mary.
the joint distribution is $f(x,y) = {2y\over 9}e^{-x}$, therefore the first probability could be written as $P[X>Y]$ of the joint distribution.
$$\int_0^3 \int_0^y f(x,y) dxdy$$
this gives the answer of .822, which is well above the answer given in the book of .1125, without the second probability even being found.
I'm not even sure how to set up the limits for the second probability.
Any help would be appreciated.
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The probability that Mary arrives first, and within an hour of John can be calculated as follows.
$$P[Y<X<Y+1] = \int_{0}^{3}\int_{0}^{y} \frac{2ye^{-x}}{9} = .822$$
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limit with greatest integer function How can I evaluate this limit?
$$\lim_{x\to-7} \frac{[x]^2+15[x]+56}{\sin(x+7)\sin(x+8)}$$ where $[x]$ denotes the greatest integer less than or equal to $x$
I could easily factor the numerator. But I cannot apply any of the standard limits due to $[x]$.
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The way the question is phrased makes me wonder if you've understood the question, either the nature of limits or the nature of the greatest-integer function. Your final sentence makes me suspect the latter.
If $x<-7$ then $\lfloor x\rfloor=-8$.
If $x>-7$ then $\lfloor x\rfloor= -7$.
It you need that explained to you you then you haven't understood the greatest-integer function and that would be what you need to look at.
In one case the numerator is $(-8)^2+15(-8)+56=0$ and in the other case it is $(-7)^2 + 15(-7)+56 = 0$. Either way the fraction is equal to $0$ in small neighboorhoods about $-7$ unless the denominator is $0$. And in small neighborhoods about $-7$, the denominator is $0$ only if $x=-7$, so essentially your asking for $\lim\limits_{x\to-7} 0$.
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Is the following solvable for x? I have the following equation and I was wondering if I can solve for x given that it appears both as an exponent and a base:
$[\frac{1}{\sqrt {2\pi}.S}.e^{-\frac{(x-M)^2}{2S^2}}-0.5\frac{1}{\sqrt {2\pi}.S}.e^{-\frac{(x-M)^2}{2S^2}}.\frac{4.(\frac{x-N}{\sqrt {2}.D})}{\sqrt {\pi}.e^{-\frac{(x-N)^2}{2D^2}}+\sqrt {\pi.e^{-2\frac{(x-N)^2}{2D^2}}+16.\frac{(x-N)^2}{2D^2}}} ]+[ \frac{1}{\sqrt {2\pi}.D}.e^{-\frac{(x-N)^2}{2D^2}}-0.5\frac{1}{\sqrt {2\pi}.D}.e^{-\frac{(x-N)^2}{2D^2}}.\frac{4.(\frac{x-M}{\sqrt {2}.S})}{\sqrt {\pi}.e^{-\frac{(x-M)^2}{2S^2}}+\sqrt {\pi.e^{-2\frac{(x-M)^2}{2S^2}}+16.\frac{(x-M)^2}{2S^2}}}] = \frac{1}{v}$
In other words, can I find an expression for x as a function of all the other variables?
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Let's simplify the equation a bit so we can better see its form:
$$C_{1}e^{X_{1}} + C_{2}e^{X_{1}}\cdot \frac{X_{2}}{C_{3}e^{X_{3}} \sqrt{C_{4}e^{X_{3}} + C_{5}X_{3}}} + C_{6}e^{X_{4}} + C_{7}e^{X_{4}}\cdot \frac{X_{5}}{C_{8}e^{X_{6}} \sqrt{C_{9}e^{X_{6}} + C_{10}X_{6}}} = C_{11}$$
With constants as $C_{n}$ above and expressions involving x as $X_{n}$ above.
Even if all of the constants were 1, we would get:
$$e^{X_1} + e^{X_4} + X_{2}\frac{e^{X_1}e^{-X_{3}}}{\sqrt{e^{X_{3}} + X_{3}}} + X_{5}\frac{e^{X_4}e^{-X_{6}}}{\sqrt{e^{X_{6}} + X_{6}}} = 1$$
Which could, with some further magical simplifications (letting all of the expressions in x be the same), be written as
$$e^{2X} + \frac{2X}{\sqrt{e^{X} + X}} = 1$$
Heck, even if the denominator was 1, we would get (letting $Y = 2X$):
$$e^{Y} + Y = 1$$
Which in and of itself is not solvable for Y.
All of this was just to show that even with certain simplifications your equation reduces to a form which is very much unsolvable, so it is extremely unlikely that your original equation is solvable for x.
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finding all solutions of an equation using counting lets say I have
$$ x_1 + x_2 + x_3 + x_4 = 17 $$
What are all the solutions for $$ x_i \ge 0? \quad \text{ where } i=1,2,3,4$$
How about if $$ x_i \ge 0 \text{ ?} $$ And if $$ x_i \gt 1 \quad\text{ where } i =1,2,3,4$$
I tried following the example of the book. For the last one I got $$ C(4+10-1, 4-1) $$
No clue if it's correct. And confused on how to tackle the first two. Thanks.
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Well this is a classic stars and bars problem. If $x_i \ge 0$ then you have:
$$\binom{17 + 4 - 1}{4-1} = \binom{20}{3} = 1140 \text{ solutions}$$
For the second problem write $x_i = 2 + k_i$, then the equation is reduced to $k_1 + k_2 + k_3 + k_4 = 9$, which can be easily solved simularly to the first problem.
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Angle of elevation and distance to an object A person walking a straight slope sees (from ground level) an Object across the valley at an angle of $45^{\circ}$, after another 50 meters walking the angle is $60^{\circ}$, How far away is the Object at that moment?
With this much given info I do not know how to approach solving this problem.
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Hint...you need to draw a picture and use trigonometry
If $h$ is the height of the object and $x$ is the distance you want, you have$$\frac hx=\tan60$$ and $$\frac{h}{x+50}=\tan45$$
Eliminate $h$
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Area of quadrilateral: $S \leq \frac{(a+b)(c+d)}4 $ I got stuck on this problem:
Given a convex quadrilateral of area $S$ and sides $a$, $b$, $c$ and $d$, prove that: $$S \leq \frac{(a+b)(c+d)}4$$
What I've done so far was to proof that $$S \leq \frac{(a+c)(b+d)}4$$ using the relation that for a given triangle $ABC$, $A_{ABC} \leq \frac12 {AB}\cdot{AC}$ or $A_{ABC} \leq \frac12 {AB}\cdot{BC}$ etc.
This is derived from $A_{ABC} = \frac12 AB \cdot AC \cdot \sin (BAC)$ and $0<\sin (BAC)\leq1$.
Anyway, this isn't enough to solve the problem. Some suggestions would be appreciate.
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Let $s:=\frac{a+b+c+d}{2}$. By Bretschneider's Formula,
$$S=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\left(\frac{\alpha+\gamma}{2}\right)}\,,$$
where $\alpha$ and $\gamma$ are two opposite angles of the given convex quadrilateral. Hence,
$$S\leq \sqrt{(s-a)(s-b)(s-c)(s-d)}\;\leq \left(\frac{(s-a)+(s-b)}{2}\right)\left(\frac{(s-c)+(s-d)}{2}\right)\,,$$
where we have used the AM-GM Inequality. Thus,
$$S\leq \frac{(a+b)(c+d)}{4}\,$$
as required. The equality holds iff the quadrilateral is cyclic with two pairs of equal adjacent sides: $a=b$ and $c=d$ (or equivalently, the quadrilateral is a kite with two opposite right angles).
You can also modify this proof to show that $S\leq \frac{(a+c)(b+d)}{4}$. The equality holds iff the quadrilateral is a rectangle.
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Decide if each is a basis for $P_2$. (a) $(x^2 + x - 1, 2x + 1, 2x - 1)$ I'm using Linear Algebra by Jim Hefferon (freely available, links below with solution).
I'm having trouble understanding Exercise 1.18 on page 117.
1.18 Decide if each is a basis for $P_2$.
(a) $(x^2 + x - 1, 2x + 1, 2x - 1)$
First, I try to prove that it spans $P_2$ $(ax^2 + bx + c)$. However, I do not understand how to set up the matrix. I usually do not have any trouble when there are column vectors given to me and I simply have to row-reduce using Gauss' Method, however whenever given equations with variables I have trouble.
Can someone walk through this step by step? That would be really helpful. I'm trying to teach myself Linear Algebra so there may be many missing gaps of knowledge.
Book: http://joshua.smcvt.edu/linearalgebra/book.pdf
Answer Key: http://joshua.smcvt.edu/linearalgebra/jhanswer.pdf
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First, set up the matrix so that every row is a vector. eg,
$$
\begin{bmatrix}x^2 & x & -1 \\0 & 2x & 1 \\ 0 & 2x & -1 \end{bmatrix}
$$
if you want to, you can simply represent the variables in a "normalized" basis by simply deciding that $\bar b_1 = x^2$. You can now use a "change of basis" from $\bar b_n$ to $\bar e_n$.
Then, row reduce and determine the linear independence.
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$\int_{0}^{\pi}|\sqrt2\sin x+2\cos x|dx$ $\int_{0}^{\pi}|\sqrt2\sin x+2\cos x|dx$
MyAttempt
$\int_{0}^{\pi}|\sqrt2\sin x+2\cos x|=\int_{0}^{\pi/2}\sqrt2\sin x+2\cos x dx+\int_{\pi/2}^{\pi}|\sqrt2\sin x+2\cos x| dx$
I could solve first integral but in second one,i could not judge the mod will take plus sign or minus sign or how to break it in further intervals?
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$\displaystyle A \sin \ x + B \cos \ x = \sqrt{A^2 + B^2}\sin( x + \phi )$, where $\displaystyle \cos \phi = \frac{A}{\sqrt{A^2 + B^2}}$.
In our case $\displaystyle \cos\phi=\frac{1}{\sqrt{3}} \Rightarrow \phi\in\left(0;\frac{\pi}{2}\right)$.
Thus: $\displaystyle \begin{aligned}\int\limits_0^\pi\left|\sqrt{2}\sin x+2\cos x\right|dx&=\sqrt{6}\int\limits_0^\pi\left|\sin(x+\phi)\right|dx=\sqrt{6}\int\limits_0^{\pi-\phi}\sin(x+\phi)dx+\sqrt{6}\int\limits_{\pi-\phi}^\pi(-\sin(x+\phi))dx=\\&=\left.-\sqrt{6}\cos(x+\phi)\right|_0^{\pi-\phi}+\left.\sqrt{6}\cos(x+\phi)\right|_{\pi-\phi}^\pi=-\sqrt{6}\left(\cos\pi-\cos\phi\right)+\sqrt{6}\left(\cos(\pi+\phi)-\cos\pi\right )=\\&=\sqrt{6}+\sqrt{2}-\sqrt{2}+\sqrt{6}=2\sqrt{6}\end{aligned}$
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$\int_{0}^{1}\frac{\sin^{-1}\sqrt x}{x^2-x+1}dx$ $\int_{0}^{1}\frac{\sin^{-1}\sqrt x}{x^2-x+1}dx$
Put $x=\sin^2\theta,dx=2\sin \theta \cos \theta d\theta$
$\int_{0}^{\pi/2}\frac{\theta.2\sin \theta \cos \theta d\theta}{\sin^4\theta-\sin^2\theta+1}$ but this seems not integrable.Is this a wrong way?Is there a different way to solve it?
|
Let, $$I=\int_{0}^{1}\frac{\sin^{-1}(\sqrt x)}{x^2-x+1}dx\tag 1$$
Now, using property of definite integral $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$ we get
$$I=\int_{0}^{1}\frac{\sin^{-1}(\sqrt {1-x})}{(1-x)^2-(1-x)+1}dx$$
$$I=\int_{0}^{1}\frac{\sin^{-1}(\sqrt {1-x})}{x^2-x+1}dx\tag 2$$ Now, adding (1) & (2), we get $$I+I=\int_{0}^{1}\frac{\sin^{-1}(\sqrt {x})}{x^2-x+1}dx+\int_{0}^{1}\frac{\sin^{-1}(\sqrt {1-x})}{x^2-x+1}dx$$
$$2I=\int_{0}^{1}\frac{\sin^{-1}(\sqrt {x})+\sin^{-1}(\sqrt {1-x})}{x^2-x+1}dx$$
$$I=\frac{1}{2}\int_{0}^{1}\frac{\sin^{-1}(\sqrt {x}\sqrt{1-1+x}+\sqrt{1-x}\sqrt{1-x})}{x^2-x+1}dx$$
$$=\frac{1}{2}\int_{0}^{1}\frac{\sin^{-1}(x+(1-x)}{x^2-x+1}dx$$
$$=\frac{1}{2}\int_{0}^{1}\frac{\sin^{-1}(1)}{x^2-x+1}dx$$
$$=\frac{1}{2}\int_{0}^{1}\frac{\frac{\pi}{2}}{x^2-x+1}dx$$
$$=\frac{\pi}{4}\int_{0}^{1}\frac{dx}{x^2-x+1}$$
$$=\frac{\pi}{4}\int_{0}^{1}\frac{dx}{\left(x-\frac{1}{2}\right)^2+1-\frac{1}{4}}$$
$$=\frac{\pi}{4}\int_{0}^{1}\frac{dx}{\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt 3}{2}\right)^2}$$
$$=\frac{\pi}{4}\left[\frac{2}{\sqrt 3}\tan^{-1}\left(\frac{\left(x-\frac{1}{2}\right)}{\frac{\sqrt 3}{2}}\right)\right]_{0}^{1}$$
$$=\frac{\pi}{2\sqrt 3}\left[\frac{\pi}{6}+\frac{\pi}{6}\right]$$
$$=\frac{\pi^2}{6\sqrt 3}$$
|
{
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"url": "https://math.stackexchange.com/questions/1403619",
"timestamp": "2023-03-29T00:00:00",
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|
$\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt{x+x^2+x^3}}$ $$\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt{x+x^2+x^3}}$$
My Attempt:
$$\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt x\sqrt{1+x(1+x)}}$$
Replacing $x$ by $1-x$,we get
$$\int_{0}^{1}\frac{x}{2-x}.\frac{dx}{\sqrt{1-x}\sqrt{1+(1-x)(2-x)}}$$
Then I got stuck. Please help.
|
The integral can also be found using a self-similar substitution of $u = \dfrac{1 - x}{1 + x}$.
Here we see that $x = \dfrac{1 - u}{1 + u}$ such that $dx = -\dfrac{2}{(1 + u)^2} \, du$.
Writing the integral as
$$I = -\int \frac{1 - x}{(1 + x) x \sqrt{x + 1 + \frac{1}{x}}} \, dx,$$
if we observe that
$$x + 1 + \frac{1}{x} = \frac{u^2 + 3}{1 - u^2},$$
then under the self-similar substitution the integral becomes
$$I = \int \frac{2u}{\sqrt{(1 - u^2)(3 + u^2)}} \, du,$$
or
$$I = \int \frac{dt}{\sqrt{(1 - t)(3 + t)}},$$
after setting $t = u^2$. In the denominator, if we complete the square then integrate one has
$$I = \int \frac{dt}{\sqrt{2^2 - (t + 1)^2}} = \sin^{-1} \left (\frac{t + 1}{2} \right ) + C,$$
or
$$\int \frac{x - 1}{(x + 1) \sqrt{x^3 + x^2 + x}} \, dx = \sin^{-1} \left [\frac{1}{2} \left (\frac{1 - x}{1 + x} \right )^2 + \frac{1}{2} \right ] + C.$$
|
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|
is this inner product positive-definite? $$\left \langle u, v \right \rangle = pu_{1}v_{1}+qu_{1}v_{2}+qu_{2}v_{1}+pu_{2}v_{2}\\\text{ for }\\ \text{p >0} \text{ and } p^{2}\geq q^{2}$$
The solution breaks down $$\left \langle u, u \right \rangle = pu_{1}u_{1}+qu_{1}u_{2}+qu_{2}u_{1}+pu_{2}u_{2} $$ into
$$p(u_{1}+\frac{q}{p}u_{2})^{2}+(p-\frac{q^{2}}{p})u_{2}^{2}$$
In this instance, Mathematica will not usually give me the desired answer to solve the question. I can't imagine having to 'guess' under a time constraint during examinations. Is there a procedural manner to obtain this solution?
This is a trivial question. The tedious part is the algebra manipulation involved which is a fairly 'unfair' question should it pops out during any examination. But just in the case it does, are there any tricks around getting to the solution?
|
$p(u_1+\frac{q}{p} u_2)^2$ is nonnegative because it is the product of $p$ (which is strictly positive) with a square.
$(p-\frac{q^2}{p}) u_2^2$ is also nonnegative, being $\frac{1}{p} (p^2 - q^2) u_2^2$, and we're given that $p>0$ and $p^2 \geq q^2$.
Hence the sum is nonnegative.
When is it zero? Consider the second term, which must be zero (since the sum of nonnegative quantities being zero means that the summands are all zero). Either $p^2 = q^2$, or $u_2 = 0$. In the latter case, now consider the first term; obtain that $u_1 = 0$. In the former case: have $\frac{q}{p} = \pm 1$, so from the first term we need $u_1 \pm u_2 = 0$. That may not always be positive, so if $p^2 = q^2$ then the inner product is not positive definite.
|
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|
Prove there exists a unique local inverse. I've been set this problem recently and I'm having a lot of trouble with it. Any help would be much appreciated!
Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a function with continuous derivatives of all orders and suppose that, for some $x\in \mathbb{R}$ the derivative $f'(x)$ is non-zero. Thus there exists an interval D containing x such that $f'(y)\neq 0$ for all $y\in D$. We can also show that f is Lipschitz with Lipschitz constant less than 1.
Show there is a unique function $g:f(D)\rightarrow D$ such that $f\circ g$ and $g\circ f$ are the identity maps on $D$ and $f(D)$ respectively. Also show that g is continuous and differentiable.
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$f'$ is nonzero on $D$ so $f$ is strictly monotonic. Thus it is one-to-one. This means that $f:D\to f[D]$ has an inverse $g:f[D]\to D$. $g$ is diffentiable with $g'(x)=\frac{1}{f'(g(x))}$.
|
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|
$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$ $$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$$
I tried to solve it.
$$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx=\int\frac{1+x^2}{(x^2+1)^2+3x^3+x^2-3x+1}dx=\int\frac{1+x^2}{(x^2+1)^2+3x(x^2-1)+x^2+1}dx$$
But this does not seem to be solving.Please help.
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Please note that you have: $\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$, which on dividing numerator and denominator by $x^2$ becomes:
$\int\frac{1+1/x^2}{(x^2+1/x^2)+3+3(x-1/x)}dx$=$\int\frac{1+1/x^2}{(x-1/x)^2+5+3(x-1/x)}dx$
Now put x-1/x =t so that (1+1/$x^2$)dx=dt and thus you get:
$\int\frac{1}{t^2+5+3t}dt$;which can be rearranged to get:$\int\frac{1}{t^2+5+3t}dt$=$\int\frac{1}{(t+3/2)^2+11/4}dt$=(2/$\sqrt11$)arctan[(t+3/2)2/$\sqrt11$]+c=(2/$\sqrt11$)arctan[(x-1/x+3/2)2/$\sqrt11$]+c;where c is integration constant.
PS:x>$0$ or x<$0$,only for these x , above solution is valid.At,x=$0$,arctan[(x-1/x+3/2)2/$\sqrt11$]becomes discontinuous so Fundamental theorem doesn't hold.
|
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|
Norm of a Block Matrix Let $X\in M_{m,n}(R)$ and $l=m+n$. Now consider the block matrix
$$
Y=\left[
\begin{array}{cc}
0 &X \\ \
X^T &0
\end{array}\right]
$$
where $Y\in M_l(R)$.
I want to show that $||X||=||Y||$ where for any $A\in M_{m,n}(R)$,
we define $||A||=max\{||Ax||: x\in R^n,||x||=1\}$ and $R^n$ has the standard Euclidean inner product.
I have got upto $$
||Y||^2=||Y^TY||=|| \left[
\begin{array}{cc}
XX^T &0 \\ \
0 &X^TX
\end{array}\right]||
$$
Thanks for any help.
|
Let $u\in\mathbb R^m$, $v\in\mathbb R^n$, and $w = (u,v)$. Then, we have
\begin{align}
\|Yw\| &= \left\| \begin{bmatrix} 0 & X \\ X^T & 0 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} \right\| \\
&= \left\| \begin{bmatrix} Xv \\ X^T u \end{bmatrix} \right\| \\
&= \sqrt{ \|Xv\|^2 + \|X^T u\|^2 } \\
&\le \|X\| \sqrt{ \|v\|^2 + \|u\|^2 } \\
&= \| X \| \|w\|.
\end{align}
That is, $\|Y\|$ is bounded from above by $\| X \|$.
Now, for $u\ne 0$ with $\|X^T u\| = \|X^T\|\|u\|$ and $v=0$, we additionally have
$$ \| Yw \| = \| X^T u \| = \|X^T \|\|u\| = \|X \| \|w\|. $$
Thus, the upper bound $\|X\|$ is attained and we obtain $\|Y\| = \|X\|$.
|
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|
Why does Abel's identity imply either $W = 0$ or $W \neq 0$ everywhere? Let $y_1$ and $y_2$ be solutions to the linear differential equation $A(x)y'' + B(x)y' + C(x)y = 0 $ and let $W = W(y_1, y_2)$ be the Wronskian of the solutions. Why does Abel's identity
$\displaystyle W(y_1, y_2)(x) = W(y_1, y_2)(x_0)\cdot exp\left(-\int{\frac{B(x)}{A(x)}dx}\right) $
imply the Wronskian is either zero everywhere or nonzero everywhere?
I know that $W \neq 0$ if the solutions are linearly independent and $W=0$ if the solutions are linearly dependent. But how does Abel's identity show this for all $x$ in the domain?
Thanks for the help.
|
Suppose that $W$ were ever zero. Since the $\exp$ term is never zero, that means the $W(y_1, y_2)(x_0)$ term must be zero. But that means that $W$ is zero everywhere.
|
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|
Anti-symmetric if $AB= 1$ and $BA=0$ but every vertex has loops? I'm creating a directed graph from an adjacency list. The $0$ present that there is no relation while the $1$ represent that there is.
So i have a quick question regarding this.
Lets assume that $AB = 1$ that is that it has a connection.
$BA = 0$ which means that is does not have a connection. This continues throughout the graph. So we assume that it is anti symmetric. However the graph has loops. So $BB = 1$ $CC= 1$ etc.
It still considered anti symmetric?
|
Yes. A relation $\mathrel{R}$ is antisymmetric if
$$x\mathrel{R}y\quad\text{and}\quad y\mathrel{R}x\quad\text{implies}\quad x=y\;.$$
This means that if $x\ne y$, you can’t have both $x\mathrel{R}y$ and $y\mathrel{R}x$: you can have at most one of them. It says nothing at all about what you can (or must) have when $x=y$.
In terms of the associated graph, if you never have both an edge $A\to B$ and an edge $B\to A$ when $A\ne B$, then the relation is antisymmetric; you can have as many or as few loops as you like. (If you have a loop at every vertex, the relation is reflexive.)
|
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|
Are there reasons not to use product of vectors as dot product? The dot product of two vectors is defined as following:
$$
\langle \vec v, \vec u \rangle =
\left<
\begin{pmatrix}
v_1 \\
v_2 \\
\dots \\
v_n
\end{pmatrix},
\begin{pmatrix}
u_1 \\
u_2 \\
\dots \\
u_n
\end{pmatrix}
\right>
= v_1 \cdot u_1 + v_2 \cdot u_2 + \dots + v_n \cdot u_n
$$
Still the multiplication of transposition of $\vec v$ and u gives:
$$
\vec v^T \cdot \vec u = (v_1, v_2, \dots, v_n) \cdot
\begin{pmatrix}
u_1 \\
u_2 \\
\dots \\
u_n
\end{pmatrix}
= v_1 \cdot u_1 + v_2 \cdot u_2 + \dots + v_n \cdot u_n
$$
so the result is the same!
It may be just a silly observation but I'm just surprised because I have never seen using it.
Are these two notations the same thing or is there something important in their definitions that don't allow interchanging them?
|
Strictly speaking, a row vector represents a linear form, i.e. a lineap map from $\mathbf R^n\to\mathbf R$. So they're different in essence.
However, to the vector $\vec v$, you can associate the linear map
\begin{align*}\varphi_{\vec v}\colon\mathbf R^n&\longrightarrow\mathbf R\\\vec u&\longmapsto \langle\vec v,\vec u\rangle \end{align*}
and in this association, the column vector that represents $\vec v$, becomes its transpose. So it is quite normal the results are the same.
|
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|
Show that $\displaystyle{\int_{0}^{\infty}\!\frac{x^{a}}{x(x+1)}~\mathrm{d}x=\frac{\pi}{\sin(\pi a)}}$ Show that for $0<a<1$
$$\int_{0}^{\infty}\frac{x^{a}}{x(x+1)}~\mathrm{d}x=\frac{\pi}{\sin(\pi a)}$$
I want to solve this question by using complex analysis tools but I even don't know how to start. Any help would be great.
|
I know it's not technically a complex analysis route, but where would we be without the obligatory beta function route?
Rewrite
$$\begin{align}\int_0^{\infty} \frac{x^a}{x (x+1)} &= \int_0^{\infty} \frac{x^{a-1}}{x+1}\\ &= \int_1^{\infty} dx \frac{(x-1)^{a-1}}{x} \\ &= \int_0^1 \frac{dy}{y} \left (\frac1{y}-1 \right )^{a-1} \\ &= \int_0^1 dv \, v^{-a} (1-v)^{-(1-a)}\\ &= \frac{\Gamma(1-a) \Gamma(a)}{\Gamma(1)} \\ &= \frac{\pi}{\sin{\pi a}}\end{align}$$
|
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|
Show that if $\{X_n\}$ is a Markov Chain
Show that, if $\{X_n\}$ is a Markov Chain then
$$P(X_n=j\mid X_k=l,X_m=i)=P(X_n=j\mid X_m=i),0\leq k<m<n$$
What I did is
$$P(X_n=j\mid X_k=l,X_m=i)=\frac{P(X_n=j,X_k=l,X_m=i)}{P(X_k=l,X_m=i)}=\frac{\sum_{i_{n-1},\dots,i_0}P(X_n=j,X_{n-1}=i_{n-1},\dots,X_{m+1}=i_{m+1},X_m=i,\dots,X_k=l,\dots,X_0=i_0)}{P(X_k=l,X_m=i)}$$
$$=\frac{\sum_{i_{n-1},\dots,i_0}P(X_n=j,X_{n-1}=i_{n-1}, \dots,X_{m+1}=i_{m+1}\mid X_m=i,\dots,X_k=l,\dots,X_0=i_0)P(X_m=i,\dots,X_k=l,\dots,X_0=i_0)}{P(X_k=l,X_m=i)}$$
$$=P(X_n=j\mid X_m=i)*\frac{\sum_{i_{n-1}, \dots,i_0}P(X_m=i,\dots,X_k=l,\dots,X_0=i_0)}{P(X_k=l,X_m=i)}=P(X_n=j\mid X_m=i)$$
Is it right? Is there an easier way?
|
Yes. That's it.
A Markov chain is a sequence of random variables $\{X_k\}_{k\in\{0..n\}}$ representing $n+1$ subsequent states of a system, such that for all supported values $\{i_k\}_{k\in\{0..b\}}$, and $i_c$, where $0\leq a< b< c\leq n$ we have:
$$\mathsf P(X_c=i_c\mid \bigcap_{k\in\{0..b\}} X_k=i_k)= \mathsf P(X_c=i_c\mid X_b=i_b)$$
So if we let: $K =\{0..a-1\}\cup\{a+1..b-1\}$, then:
$$
\begin{align}
& \mathsf P(X_c=i_c\mid X_b=i_b, X_a=i_a)
\\[2ex] = & \sum_{\{i_k\}_{k\in K}} {\mathsf P(X_c=i_c\mid X_b=i_b, X_a=i_a,\bigcap_{k\in K}X_k=i_k)\;\mathsf P(\bigcap_{k\in K}X_k=i_k)}
\\[2ex] = & \mathsf P(X_c=i_c\mid X_b=i_b)\sum_{\{i_k\}_{k\in K}} \mathsf P(\bigcap_{k\in K}X_k=i_k)
\\[2ex] = & \mathsf P(X_c=i_c\mid X_b=i_b)
\end{align}
$$
Which is what you said; but perhaps a little more compactly.
|
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|
How many subgroups are there in an elementary-$p$ group $C_p\times C_p$ has $1+1+(p-1)=p+1$ subgroups of order $p$ (all of which $\cong C_p$), so it has $1+(1+p)+1=3+p$ subgroups totally.
But how many subgroups in $C_p\times C_p \times C_p\times C_p$?
(its seems hard to count subgroups of order $p^2$ and order $p^3$)
Furthermore, could we easily point out how many subgroups which isomorphic to $C_p\times C_{p^2}$ in group $C_{p^2}\times C_{p^2}\times C_{p^2}$?
I only figure out how to find all order-p subgroups in the elementary-$p$ group:
For example in $C_p\times C_p \times C_p$: $p^3=x(p-1)+1\Leftrightarrow x=(p^2+p+1)$, so it has $p^2+p+1=(p-1)^2+C_3^1(p-1)+C_3^1$ order-$p$ subgroups in total. So there are 3 types order-p subgroups:
$\{(x,1,1)| x\in C_p\}$, $\{(x,x^{\alpha},1)| x\in C_p\}$, $\{(x,x^{\alpha},x^{\beta})| x\in C_p\}$, where $\alpha$ and $\beta$ $\in Aut {C_p}$
|
Subgroups of elementary abelian groups are also elementary abelian, so the problem in that context reduces to one of vector spaces: given a vector space $V$ of dimension $n$ over a finite field $\Bbb F_q$ (for us our $q$ is prime, but the question's difficulty is unchanged by generalizing), how many subspaces of dimension $d$ are there? This is a textbook problem - the answer is $q$-binomial coefficients. See the discussion in this question for instance. (Note this hints at finite sets being vector spaces over the field with one element, since $q$-binomial coeffs when $q=1$ are just binomial coeffs.)
|
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|
Prove that the length of the common chord is $\frac{2ab\sin \theta}{\sqrt{a^2+b^2+2ab\cos \theta}}$ Two circles ,of radii $a$ and $b$,cut each other at an angle $\theta.$Prove that the length of the common chord is $\frac{2ab\sin \theta}{\sqrt{a^2+b^2+2ab\cos \theta}}$
Let the center of two circles be $O$ and $O'$ and the points where they intersect be $P$ and $Q$.Then angle $OPO'=\theta$
$\cos \theta=\frac{a^2+b^2-OO'^2}{2ab}$
$OO'^2=a^2+b^2-2ab\cos\theta$
In triangle $PO'Q$,angle $PO'Q=\pi-\theta$
$\cos(\pi-\theta)=\frac{b^2+b^2-l^2}{2b^2}$ Then i am stuck.Please help me to reach upto proof.
|
Let $A$ & $B$ be the centers of the circles with radii $a$ & $b$ respectively such that they have a common chord $MN=2x$ & intersecting each other at an angle $\theta$.
Let $O$ be the mid-point of common chord $MN$. Point $O$ lies on the line AB joining the centers of circle then we have $$MO=ON=x$$
In right $\triangle AOM$ $$AO=\sqrt{AM^2-MO^2}=\sqrt{a^2-x^2}$$
Similarly, in right $\triangle BOM$ $$OB=\sqrt{BM^2-MO^2}=\sqrt{b^2-x^2}$$
$$AB=AO+OB=\sqrt{a^2-x^2}+\sqrt{b^2-x^2}$$
Now, applying cosine rule in $\triangle AMB$ as follows $$AB^2=AM^2+BM^2-2(AM)(BM)\cos (\angle AMB)$$
Setting the corresponding values, we get
$$(\sqrt{a^2-x^2}+\sqrt{b^2-x^2})^2=a^2+b^2-2(a)(b)\cos (180^\circ-\theta)$$
$$a^2-x^2+b^2-x^2+2\sqrt{(a^2-x^2)(b^2-x^2)}=a^2+b^2+2ab\cos\theta$$
$$\sqrt{(a^2-x^2)(b^2-x^2)}=x^2+ab\cos\theta$$
$$(a^2-x^2)(b^2-x^2)=(x^2+ab\cos\theta)^2$$
$$a^2b^2-b^2x^2-a^2x^2+x^4=x^4+a^2b^2\cos^2\theta+2abx^2\cos\theta$$
$$(a^2+b^2+2ab\cos\theta)x^2=a^2b^2-a^2b^2\cos^2\theta$$ $$x^2=\frac{a^2b^2\sin^2\theta}{a^2+b^2+2ab\cos\theta}$$
$$x=\frac{ab\sin\theta}{\sqrt{a^2+b^2+2ab\cos\theta}}$$
Hence, the length of the common chord MN,
$$MN=2x=\frac{2ab\sin\theta}{\sqrt{a^2+b^2+2ab\cos\theta}}$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{Length of common chord}=\color{blue}{\frac{2ab\sin\theta}{\sqrt{a^2+b^2+2ab\cos\theta}}}}}$$
|
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|
A sequence of numbers prime to $4n$ with $n$ being odd Let $n$ be odd. If I consider the sequence of $n$ numbers in the form $4k-1$ with $k$ running from $1$ to $n$ and take those with greatest common divisor with $4n$ being $1$ ( means those being prime to $4n$ ) I get exactly $\phi(n)$ of them.
I miss now a proof/argument why this true.
I tested the case $n=15$.
|
The function $f_4: \mathbb{Z}_{n}\rightarrow \mathbb{Z}_{n},\;f_4(k) \equiv 4k-1 \pmod n$ is a bijection because $\gcd(4,n)=1$. The same applies if you substitute $4$ with any number $m$ with $\gcd(n,m)=1,\,$ and the additive constant $1$ can be replaced be any number.
Check this e.g. for $n=15, m=7, f(k) = 7k+4$
|
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|
How to find the MacLaurin series of $\frac{1}{1+e^x}$ Mathcad software gives me the answer as: $$ \frac{1}{1+e^x} = \frac{1}{2} -\frac{x}{4} +\frac{x^3}{48} -\frac{x^5}{480} +\cdots$$
I have no idea how it found that and i don't understand. What i did is expand $(1+e^x)^{-1} $ as in the binomial MacLaurin expansion. I want until the $x^4$ term. So what i found is the following:
$$ (1+e^x)^{-1} = 1 - e^x + e^{2x} + e^{3x} + e^{4x}+\cdots\tag1 $$
Then by expanding each $e^{nx}$ term individually:
$$ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24}+\cdots\tag2 $$
$$ e^{2x} = 1 + 2x + 2x^2 + \frac{3x^3}{2} + \frac{2x^4}{3}+\cdots\tag3 $$
$$ e^{3x} = 1 + 3x + \frac{9x^2}{2} + \frac{9x^3}{2} + \frac{27x^4}{8}+\cdots\tag4 $$
$$ e^{4x} = 1 + 4x + 8x^2 + \frac{32x^3}{3} + \frac{32x^4}{3}+\cdots\tag5 $$
So substituting $(2),(3),(4),(5)$ into $(1)$ i get:
$$ (1+e^x)^{-1} = 1 +2x+5x^2+ \frac{22x^3}{3} + \frac{95x^4}{12} +\cdots$$
Which isn't the correct result. Am i not allowed to expand this series binomially? I've seen on this site that this is an asymptotic expansion. However, i don't know about these and i haven't been able to find much information on this matter to solve this. If someone could help me understand how to solve it and why my approach isn't correct, i would be VERY grateful. Thanks in advance.
|
$\begin{array}\\
f(x)
&=\frac{1}{1+e^x}\\
&=\frac{1}{2+(e^x-1)}\\
&=\frac12\frac{1}{1+(e^x-1)/2}\\
&=\frac12\sum_{n=0}^{\infty}(-1)^n\frac{(e^x-1)^n}{2^n}\\
&=\frac12\sum_{n=0}^{\infty}(-1)^n\frac{(x+x^2/2+x^3/6+...)^n}{2^n}\\
&=\frac12\sum_{n=0}^{\infty}(-1)^n(x/2)^n(1+x/2+x^2/6+...)^n\\
&=\frac12\sum_{n=0}^{\infty}(-1)^n(x/2)^n(1+nx/2+...)\\
&=\frac12\left(1-\frac{x}{2}(1+\frac{x}{2}+...)+\frac{x^2}{4}(1+...)+...\right)\\
&=\frac12\left(1-\frac{x}{2}-\frac{x^2}{4} +\frac{x^2}{4}(1+...)+...\right)\\
&=\frac12\left(1-\frac{x}{2}+O(x^3)\right)\\
&=\frac12-\frac{x}{4}+O(x^3)\\
\end{array}
$
|
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|
A questions on the groups by a copy of $\Bbb Z$ Let $G$ be an abelian group and $H$ a subgroup of $G$ such that $G/H$ contains a copy of $\Bbb Z$. Is this true that $G$ contains a copy of $\Bbb Z$? ($\Bbb Z$ is the group of integer numbers)
|
Another viewpoint using a bit simpler terms. Containing a copy of $\mathbb Z$ means that it has an element of infinite order, so we have $o(g+H)=\infty$ for some $g\in G$. Now, under the natural projection map, $o(\phi (g))|o(g)$so $\infty |o(g)$, hence $o(g)=\infty$
|
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|
Distribution of sum of 2 circular uniform random variables I hope you can help me resolve the following problem:
Let $\Phi_1$ and $\Phi_2$ circular uniform random variables such that $0\leq\Phi_i\leq 2\pi$ (with $i=1,2$). Then the probability density function (pdf) and the cumulative distribution function (cdf) are given by
\begin{align}
f_{\Phi_i}(\phi_i) &= \frac{1}{2\pi}
\\
F_{\Phi_i}(\phi_i) &= \frac{\phi_i}{2\pi}.
\end{align}
I'd like to find the distribution of sum $\Phi=\Phi_1+\Phi_2$ so that I've calculated the cdf of $\Phi$ as follows:
\begin{align}
F_{\Phi}(\phi) = Pr\{\Phi_1+\Phi_2\leq\phi\}
&= \int_0^{2\pi} Pr\{\Phi_1\leq\phi-\phi_2\bigl|\Phi_2=\phi_2\bigr.\} f_{\Phi_2}(\phi_2) d\phi_2
\\
&= \int_0^{2\pi} F_{\Phi_1}\left(\Phi_1\leq\phi-\phi_2\right) f_{\Phi_2}(\phi_2) d\phi_2
\\
&= \int_0^{2\pi} \frac{(\phi-\phi_2)}{2\pi} \frac{1}{2\pi} d\phi_2
\\
&= \frac{\phi}{2\pi} - \frac{1}{2}.
\end{align}
Due to $\Phi=\Phi_1+\Phi_2$, then $0\leq\Phi\leq 4\pi$. Unfortunately, we have
\begin{align}
F_{\Phi}(4\pi) = \frac{4\pi}{2\pi} - \frac{1}{2} = \frac{3}{2} > 1.
\end{align}
This is really unreasonable! Can anyone help me find the exact distribution of $\Phi$, please?
|
The immediate problem I see is the expression of the conditional probability in the integrand in the first step. To help you understand why it is not correct, consider the case where $\Phi_1 = 3\pi/2$, $\Phi_2 = \pi$. Then their sum is $$\Phi = \Phi_1 + \Phi_2 = 5\pi/2 \equiv \pi/2$$ where in the last step we have to reduce the angle modulo $2\pi$. So you cannot write the conditional probability this way: $$\Pr[\Phi \le \phi \mid \Phi_2 = \phi_2] \ne \Pr[\Phi_1 \le \phi - \phi_2].$$ You have to take into consideration the modulo operation.
|
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|
$\Gamma$ function is continuous on $]0,\infty [$ I would like to show that $\Gamma$ is continuous on $]0,\infty [$.
Let $x>0$,
$$|\Gamma(x+h)-\Gamma(x)|\leq\int_0^\infty t^{x-1}e^{-t}|t^h-1|dt=\underbrace{\int_0^1 t^{x-1}e^{-t}\underbrace{|t^h-1|}_{\leq |h|}dt}_{\underset{h\to 0}{\longrightarrow0 }}+\int_1^\infty t^{x-1}e^{-t}|t^h-1|dt$$
If $t> 1$, there is a $\delta>0$ such that $|t^h-1|\sim \ln(t)h\leq th$,
therefore, if $|h|<\delta$,
$$\int_1^\infty t^{x-1}e^{-t}|t^h-1|dt\leq h\int_1^\infty t^{x}e^{-t}dt\leq |h|\Gamma(x+1)\underset{h\to 0}{\longrightarrow }0 .$$
We conclude that $$|\Gamma(x+h)-\Gamma(x)| \underset{h\to 0}{\longrightarrow }0$$
and thus $\Gamma$ is continuous on $]0,\infty [$.
Questions
1) Is this proof correct ?
2) I can't use theorem of dominated convergence. But If I were authorized to use it, would it be possible to use it here ? And if yes, how ?
Thanks,
|
1: Yes, your proof is correct:
2: For dominated convergence, you need a dominating function. Consider the function
$$
g(x) = \max\{t^{x + \delta}e^t,t^{x - \delta}e^t\}
$$
Where we can choose any $\delta \in (0, x)$. Once we know that this function is integrable, dominated convergence applies.
|
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|
Integrating $\int \sqrt{x+\sqrt{x^2+1}}\,\mathrm{d}x$ Integrating $$\int \sqrt{x+\sqrt{x^2+1}}\,\mathrm{d}x$$
Using substitution of $x=\tan \theta$, I got the required answer. But is there a more elegant solution to the problem?
|
HINT
$$
\sqrt{ x + \sqrt{ x^2 + 1 } }
= \sqrt{ \frac{ x + i }{ 2 } } + \sqrt{ \frac{ x - i }{ 2 } }
$$
That would be enough simple to solve the integral...
We get
$$
\begin{eqnarray}
\int \sqrt{ x + \sqrt{ x^2 + 1 } } dx
&=& \int \left\{ \sqrt{ \frac{ x + i }{ 2 } } + \sqrt{ \frac{ x - i }{ 2 } }
\right\} d x\\\\
&=& \frac{4}{3} \left\{ \sqrt{ \frac{ x + i }{ 2 } }^3
+ \sqrt{ \frac{ x - i }{ 2 } }^3 \right\} \quad \textrm{(*)}\\\\
&=& \bbox[16px,border:2px solid #800000] {\frac{4}{3} \sqrt{ x + \sqrt{ x^2 + 1 } } \left\{
x - \frac{1}{2} \sqrt{x^2+1}\right\}}
\end{eqnarray}
$$
(*) Where we have used
$$
a^3 + b^3 = \Big( a + b \Big) \Big( a^2 + b^2 - a b \Big)
$$
|
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|
Question of maps in Mayer-Vietoris sequence We obtain MV-seq. from short exact sequence
$$
0\to C_n(A\cap B) \to C_n(A)\oplus C_n(B)\to C_n(A+B)\to 0
$$
So map i wonder that map $H_n(A\cap B)\to H_n(A)\oplus H_n(B)$ maps $[a]$ to $([a],[-a])$. But for example in this case it's not quite clear why $1 \mapsto (2,-2)$. Why $1 \not\mapsto (1,-1)$? How to make argument 'wraps twice' rigorous?
And what about other maps in this sequence?
|
Sam Nead has the correct suggestion here. In most instances of Mayer Vietoris you actually need to compute what the inclusion map induces or even, god forbid, what the snake homomorphism gives you. Here the boundary of a mobius strip includes into each mobius strip in such a way that it retracts onto the circle going around of the center of the mobius band. Because the mobius band is a fiber bundle, this is very trivial to show.
But the homology of the boundary of the mobius band is isomorphic to the homology of a circle, so it is generated by a cycle given by triangulating the circle. However when you use the retraction of the mobius band onto its equator, you will find that this simplex includes as twice the generator of the homology group given by the meridian. This is pretty hard to see explicitly with simplicial homology, it is easier in singular cohomology. However this is very easily seen via the Hurewicz theorem, which gives a functorial isomorphism $\pi_1(S^1) \to H_1(S^1, \mathbb{Z})$.
|
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|
Solving a pair of ODEs I'm trying to solve a pair of ODEs for which I've obtained a solution. However, my problem is that my answer is slightly different from mathematica's answer.
$$ \frac{dA}{dt} = \theta - (\mu + \gamma)A, \ \ A(0) = G$$
$$ \frac{dT}{dt} = 2 \mu A - (\mu + \gamma)T, \ \ T(0) = B$$
Using an integrating factor of $e^{(\mu + \gamma)t}$, I got the following solution to the first ODE:
$$ A(t) = \frac{\theta}{\mu + \gamma} + \left(G -\frac{\theta}{\mu + \gamma}\right)e^{-(\mu + \gamma)t}$$
For simplicity, let $\mu + \gamma = \alpha$ such that:
$$ A(t) = \frac{\theta}{\alpha} + \left(G -\frac{\theta}{\alpha}\right)e^{-\alpha t}$$
For the second ODE (again using an integrating factor of $e^{(\mu + \gamma)t}= e^{\alpha t}$ ):
$$ e^{\alpha t}\frac{dT}{dt} + e^{\alpha t}\alpha T = 2 \mu A e^{\alpha t} $$
$$ T(t)e^{\alpha t} = \int 2 \mu e^{\alpha t}\left(\frac{\theta}{\alpha} + \left(G -\frac{\theta}{\alpha}\right)e^{-\alpha t}\right)dt $$
$$ T(t) = \frac{2 \mu \theta}{\alpha^2} + \left(B - \frac{2 \mu \theta}{\alpha^2}\right)e^{-\alpha t} $$
However, when I computed these two ODEs in mathematica, it gave back the following solution:
$$ T(t) = Be^{-\alpha t} + 2G \mu te^{-\alpha t}+ \frac{2 \mu \theta}{\alpha^2} - \frac{2 \mu \theta e^{-\alpha t}}{\alpha^2} - \frac{2 \mu \theta t e^{-\alpha t}}{\alpha} $$
I've tried solving my equation over and over again but I can't seem to understand why my solution is different from mathematica's. The only I thought about was possibly in the substitution of arbitrary constant. Am I missing something obvious here?
|
You forgot the integration constant. Following on from your integral for $T(t)$, we find
\begin{align}
T(t)e^{\alpha t} &= \int 2 \mu e^{\alpha t}\left(\frac{\theta}{\alpha} + \left(G -\frac{\theta}{\alpha}\right)e^{-\alpha t}\right)dt \\
&= \int 2 \mu e^{\alpha t}\left(\frac{\theta}{\alpha}\right) + 2 \mu \left(G -\frac{\theta}{\alpha}\right) dt \\
&= 2 \mu e^{\alpha t}\left(\frac{\theta}{\alpha^{2}}\right) + 2 \mu \left(G -\frac{\theta}{\alpha}\right)t + C \\
\implies T(t) &= 2 \mu \left(\frac{\theta}{\alpha^{2}}\right) + 2 \mu \left(G -\frac{\theta}{\alpha}\right)t e^{-\alpha t} + Ce^{-\alpha t} \\
T(0) &= B \\
&= 2 \mu \left(\frac{\theta}{\alpha^{2}}\right) + C \\
\implies C &= B - 2 \mu \left(\frac{\theta}{\alpha^{2}}\right) \\
\implies T(t) &= 2 \mu \left(\frac{\theta}{\alpha^{2}}\right) + 2 \mu \left(G -\frac{\theta}{\alpha}\right)t e^{-\alpha t} + \bigg(B - 2 \mu \left(\frac{\theta}{\alpha^{2}}\right) \bigg)e^{-\alpha t} \\
&= \frac{2 \mu \theta}{\alpha^{2}} + 2 \mu G t e^{-\alpha t} - \frac{2 \mu G \theta}{\alpha}t e^{-\alpha t} + B e^{-\alpha t} - \frac{2 \mu \theta}{\alpha^{2}} e^{-\alpha t} \\
&= B e^{-\alpha t} + 2 \mu G t e^{-\alpha t} + \frac{2 \mu \theta}{\alpha^{2}} - \frac{2 \mu G \theta}{\alpha}t e^{-\alpha t} - \frac{2 \mu \theta}{\alpha^{2}} e^{-\alpha t}
\end{align}
|
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|
Check Differentiability chech whether the function is differentiable at $x=0$
$$f(x)=\left\lbrace \begin{array}{cl}
\arctan\frac{1}{\left | x \right |}, & x\neq 0 \\
\frac{\pi}{2}, & x=0\\
\end{array}\right.$$
I feel that this is differentable at given point but I am unable to proceed with it.
|
Simply use the identity
$$\arctan (1/x)=\frac{\pi}{2}\text{sgn}(x)-\arctan(x)$$
Then, the limit of the difference quotient is
$$\lim_{h\to0}\frac{\arctan(1/|h|)-\frac{\pi}{2}}{h}=-\lim_{h\to0}\frac{\arctan(|h|)}{h} \tag 1$$
We see that the limit in $(1)$ does not exist since the limit from the right side does not equal the limit from the left side. In fact, a quick check using say L'Hospital's Rule shows that the limit from the right side is $-1$, while the limit from the left side is $1$.
We conclude, therefore, that $\arctan(1/|x|)$ is not differentiable at $x=0$.
|
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|
Differentiability of a two variable function $f(x,y)=\dfrac{1}{1+x-y}$ We're given the following function :
$$f(x,y)=\dfrac{1}{1+x-y}$$
Now , how to prove that the given function is differentiable at $(0,0)$ ?
I found out the partial derivatives as $f_x(0,0)=(-1)$ and $f_y(0,0)=1$ ,
Clearly the partial derivatives are continuous , but that doesn't guarantee differentiability , does it ?
Is there any other way to prove the same ?
|
If all partial derivatives of a function (over all possible variables) are continuous at some point, then the function is differentiable at that point.
|
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|
If $r:X\to A$ is a Retraction, Then $H_n(X)\cong H_n(A)\oplus H_n(X,A)$ $\DeclareMathOperator{\im}{Im}$
Let $A$ be a subspace of a topological space $X$ such that there is a retraction $r:X\to A$ of $X$ onto $A$.
Then $H_n(X)=H_n(A)\oplus H_n(X, A)$ for all $n$.
What I tried:
Let $i:A\to X$ be the inclusion map. Then we have $r\circ i=id_A$.
We have induced homomorphisms $i_*:H_n(A)\to H_n(X)$ and $r_*:H_n(X)\to H_n(A)$. Since $r_*\circ i_*=(r\circ i)_*=(id_A)_*$ is an isomorphism, we deduce that $i_*$ is split-injective.
Therefore, $H_n(X)=i_*(H_n(A))\oplus \ker r_*$.
So is $r_*$ were kind enough to satisfy $\ker r_*\cong H_n(X, A)$ then we are done.
I am unable to show this. For suppose $\sigma+\im \partial_{n+1}^X\in \ker r_*$ for some $\sigma\in \ker \partial_n^X$, then we have $r\circ \sigma+\im \partial_{n+1}^A=0$. This means that $r\circ \sigma\in \im \partial_{n+1}^A$.
I don't see where to go from here.
|
For $A\subset X$ you have long exact sequence
$$
\dots\to H_n(A)\to H_n(X)\to H_n(X,A)\to\dots
$$
The composition $r_*\circ i_*:H_n(A)\to H_n(X)\to H_n(A)$ is identity, so we see that $i_*:H_n(A)\to H_n(X)$ is inclusion for all $n$. Thus, we can write
$$
0\to H_n(A)\to H_n(X)\to H_n(X,A)\to0,
$$
and $r_*$ gives us splitting of this short sequence.
|
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|
Solving $x^{2n} = \frac{1}{2^n}$ for $x$ What is the principle behind solving for a variable that is raised to another variable? I came across this problem doing infinite sums: I had to solve the equation
$$x^{2n} = \frac{1}{2^n}$$
for $x$. I posed the question in the online forum and the TA said the answer is
$$x = \frac{1}{\sqrt{2}}.$$
I don't see how he got there. If someone could explain how to get from one to the other, I would appreciate it!
|
$x^{2n}=(x^2)^n=\frac{1}{2^n}$. So if $n\neq 0$, for $x\in \mathbb{R}$ you have $x^2=\frac{1}{2}$ then $x=\frac{1}{\sqrt{2}}$ or $x=-\frac{1}{\sqrt{2}}$.
|
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|
Is an algorithm to find all primes up to $n$ that runs in $O(n)$ time fast? I kindly ask you if it is useful or fast for a prime number generator to run in $O(n/3)$ time?
I believe I have a way to generate all $P$ primes up to $n$, quickly and neatly, in $P$ comparisons and $n/3$ calculations.
|
Calculating the primes up to $n$ in $O(n)$ time isn't particularly fast, and nor is it particularly slow. A naive Sieve of Eratosthenes works in time $O(n \log n \log \log n)$ and is very easy to implement. It can be sped up to run faster than $O(n)$ using some wheel techniques. I believe one can reduce the time to $O(n/\log n)$. See Paul Pritchard, A sublinear additive sieve for finding prime numbers, Communications of the ACM 24 (1981) for more, or look up Pritchard's Wheel.
Another common algorithm is the Sieve of Atkin, which naively runs in $O(n)$, but which can also be sped up to sublinear time.
So no, it's not particularly interesting or special to come up with another algorithm to find the primes up to $n$ in time $O(n)$.
As an aside, you might read what allowed others to speed up the Sieve of Eratosthenes and the Sieve of Atkin to see if it would allow you to speed up your algorithm.
|
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|
Does the integral $\int_0^{\pi} \frac{dx}{\sin(2x)+\cos(3x)}$ exist? This link
http://www.wolframalpha.com/input/?i=integral%28x%3D0%2Cpi%2C1%2F%28sin2x%2Bcos3x%29%29
shows the visual representation of the integral
$$\int_0^{\pi} \frac{dx}{\sin(2x)+\cos(3x)}$$
Looking at the picture, I tend to believe that the integral does not exist.
The integrand contains three poles, but from this I cannot conclude that the
integral does not exist, or can I ?
*
*How can I check, if the integral exists or not ?
|
Note at $x = {\pi \over 2}$, the denominator $\sin 2x + \cos 3x$ is zero. If $f(x) = \sin 2x + \cos 3x$, then $f'(x) = 2 \cos 2x - 3 \sin 3x$, so that $f'({\pi \over 2}) = 1$. Hence the function $f(x)$ behaves as ${1 \over x - {\pi \over 2}}$ near $x = {\pi \over 2}$; in particular, since $|{1 \over x - {\pi \over 2}}|$ has infinite integral in any interval centered at ${\pi \over 2}$, so does $|{1 \over \sin 2x + \cos 3x}|$. Thus in the usual sense your integral does not exist.
There are more general notions of integral where you remove intervals of the form $({\pi \over 2} - \epsilon, {\pi \over 2} + \epsilon)$ (and analogous things for the other poles), and then take limits as $\epsilon \rightarrow 0$. If you do this, sometimes you can get a finite number as your limit for integrals such as yours.
|
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|
Showing $\bigcup_{k}\{(a_{1},...,a_{k}):a_{j}\in [0,1]\cap \mathbb{Q} ,\sum^{k} a_{j}=1\}$ is countable Showing $M=\bigcup_{k}\{(a_{1},...,a_{k}):a_{j}\in [0,1]\cap \mathbb{Q} ,\sum^{k} a_{j}=1\}$ is countable.
I find this hard to believe because say for fixed k and any $b,c\in [0,1]\cap \mathbb{Q}$, we have $(\frac{1}{k-2},...,\frac{1}{k-2},b,c)\in M$. So I can embed in M any element of $ P([0,1]\cap \mathbb{Q})$, which is uncountable.
thanks
|
Fix $k$. The set $\{(a_1,...,a_k); a_j\in [0,1]\cap \mathbb{Q}, \sum_ja_j=1\}\subset \mathbb{Q}^k$. Therefore it is countable. Countable union of countable sets in countable.
|
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|
multiple sets of complex roots of a number? I am not sure if this question was asked before but I couldn't find the right keywords to choose for searching.
So today I discovered a weird problem:
If we take this equation: $$x^2=1=e^{(0i)}$$
this equation has 2 solutions : $\{1, -1\}$
Until now everything is normal.
but if we wroted like this: $$x^2=x^{\frac{4}{2}}=\sqrt{x^4}=1$$
what happens is that this becomes: $x^4=1^2=1$
and now it has another set of roots: $\{1,-1,i,-i\}$
And we can continue as much as we want getting an infinite number of sets of roots
My question is how is this happening?
Did I make an illegal operation?
Did I miss something?
P.S.: I am not a mathematician, so I could have missed something here (that I probably didn't learn)
UPDATE
My most important reason for asking is how to deal with equations like this: $$x^{\frac{a}{b}}=c$$ where a and b are 2 integers and you can simplify $\frac{a}{b}$
UPDATE
Obviously i and -i are not solution for $x^2=1$
But if we had from the beginning this equation: $x^\frac{4}{2}=1$ (which is exactly the problem I am waiting for its solution as I mentioned in my previous update above)
we have 2 options simplify to $x^2=1$=>we have only two roots
or do not simplify=>we have four roots and they are all valid: $i^\frac{4}{2}=\sqrt{1^4}=1$ (valid)
so 2 sets of roots
tl;dr: to simplify or not to simplify?
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Certain exponent laws and properties go out the door when you start dealing with complex numbers. That's why you're getting to different solutions when going about this in two different way.
In general, if $n$ is an integer, then the solution set to the problem $z^n = 1$ is the set of roots
$\{1=e^{2\pi i * (0/n)},e^{2\pi i * (1/n)},e^{2\pi i * (2/n)},\ldots,e^{2\pi i * ((n-1)/n)}\}$
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If $y= |\sin x| + |\cos x|$, then $dy/dx$ at $x=2\pi/3$ is? If $$y= |\sin x| + |\cos x|,$$
then $dy/dx$ at $x = 2\pi/3$ is?
Ans: $(\sqrt{3} - 1)/2$
How can we differentiate modulus functions? Can anyone explain?
|
you only need to worry about the behaviour in the vicinity of $x=2 \pi /3$ where $\sin x $ is positive and $ \cos x $ is negative so
$$y= \sin x - \cos x $$
$$y'= \cos x + \sin x $$
$$y'(2 \pi /3)= \cos (2 \pi /3) + \sin(2 \pi /3) = -\frac 12 + \frac{\sqrt 3}{2} = \frac{\sqrt 3 -1}{2}$$
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Write the co-ordinates of E such that the parallelogram ABCE is a rhombus. I'm unsure how to do this and it's always in my exams. (The original shape was a triangle and E was originally not a point)
A:(1,0) B:(0,8) C(7,4)
Gradient of AC:2/3
AC equation:2x - 3y - 2 = 0
Coordinates of the midpoint D of AC: 4,2
*
*AC is perpendicular to BD
*ABC is an isosceles
Area of triangle = 26 u
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The diagonals of a rhombus bisect each other at right angles (Proof)
If $O(p,q)$ is the midpoint of $AC, B E$
If $E(h,k)$ $$\dfrac{1+7}2=p=\dfrac{0+h}2$$ and $$\dfrac{0+4}2=q=\dfrac{8+k}2$$
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Proof that transpose of Hadamard Matrix is also a Hadamard matrix The question is self explaining from the title, but let me elaborate it. In most of the articles/books I've read, fact that the transpose of Hadamard matrix is also a Hadamard matrix is used, but I was not able to find or deduce a proof for it. I can basically state that transposing a matrix will (probably?) not affect its orthogonality, but I would prefer a correct proof.
Thank you,
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What you need is the fact, from elementary linear algebra, that, if a square matrix $A$ has a right inverse $B$ (i.e., $AB=I$) then $B$ is also a left inverse for $A$ (i.e., $BA=I$). In the case of an $n\times n$ Hadamard matrix, the definition in terms of orthogonal rows gives $\frac1n AA^\top=I$. So the linear algebra fact gives $\frac1n A^\top A=I$, which is the characterization by orthogonal columns.
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What's theoretical maximum information compression rate? Let's say I've got a random bit sequence s and a reversible function f(s), for which the following statement f'(f(s)) = s is true. What is the theoretical maximum average compression rate of such function?
IIRC, most if not all compression algorithms of today tend to identify particular patterns and map them with standard where it's possible. This makes the maximum and minimum compression rate 1/[s] and 1 correspondingly.
NB: Higher rate — weaker compression (worse)
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The question is answered by Shannon's source coding theorem.
For i.i.d. input, the theorem establishes that the minimum compressed bit rate is given by the entropy of the source.
For statistically dependent input bits the same result applies, but entropy has to be defined in a more general manner to take dependence into account. See for example John G. Proakis, "Digital Communications", 2nd edition, section 2.3.
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existence of a sequence of continuous functions with two conditions
*
*$\displaystyle \int_0^1 \lim_{n\to\infty} f_n(x)\,dx = \lim_{n\to\infty}\int_0^1f_n(x)\,dx $
*There is no function $\,g:\left[0,1\right]\to \mathbb R\,$ lebesgue integrable such that $\,\left\lvert f_n (x)\right\rvert\le g(x)\,$ for $\,0\le x\le 1\, $ and $\, n\ge 1 \,$
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Take the sequence:
$$f_n(x) = \begin{cases} n\qquad \mbox{if }\; \frac{1}{n+1}\leq x< \frac{1}{n} \\ 0 \qquad \mbox{otherwise}\end{cases}$$
We have $\int_{0}^1 f_n(x) dx = \frac{1}{n+1}$. So $\int_0^1 f_n(x) dx \to 0$.
What happens to $g(x)=\max\{f_n(x): n\in \mathbb{N}\}$?
Now you need to smoothen $f_n$ a little bit, to make them continuous...
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How to solve an exponential and logarithmic system of equations? $$
\left\{\begin{array}{c}
e^{2x} + e^y = 800
\\ 3\ln(x) + \ln(y) = 5
\end{array}\right.$$
I understand how to solve system of equations, logarithmic rules, and the fact that $\ln(e^x) = e^{\ln(x)} = x$. However, any direction I seem to go with this problem causes me difficulties. For example, when solving for y, I can get $y = \ln\left(800 - e^{2x}\right)$ or $y = \frac{e^5}{x^3}$. Both directions seem to lead me into a dead end. Although system of equations seems to be a frequent question on here, I haven't found anything helpful towards this problem.
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So here is another solution for your problem, again using Newton's method we get by using as initial value $x_0=3.2$ (it is very sensitive!) the following line of iterates
$$x_0=3.2$$
$$x_1=3.330187$$
$$x_2=3.303383$$
$$x_3=3.302208$$
$$x_4=3.302206$$
which gives us as an approximate solution
$$
f(3.302206)\approx-5.277731 \cdot10^{-5}
$$
Of course it still possible that there are other solutions outside just waiting for us to be discovered, a graph analysis might give some clearance about that.
For the graph analysis please check the comment by Mark just below the answer! tl;dr: there are at most two solutions + there are at least two solutions $\to$ there are exactly two solutions
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Which is greater, $98^{99} $ or $ 99^{98}$? Which is greater, $98^{99} $ or $ 99^{98}$?
What is the easiest method to do this which can be explained to someone in junior school i.e. without using log tables.
I don't think there is an elementary way to do this. The best I could find was on Quora, in an answer by Michal Forišek on a similar question here, which is to consider $\frac{98^{99}}{99^{98}}=98.(\frac{98}{99})^{98}=98.(1-\frac{1}{99})^{98} \approx\ \frac{98}{e} >1$ and hence $98^{99} > 99^{98}$.
But the approx sign step does use definition of $e$ in terms of limits and thus cannot be considered elementary. Any other way?
Edit- I was hoping something that does not involve calculus, that is why I tagged it in number theory, but as it seems it is almost impossible to avoid calculus when exponentials are involved.
All the answers are fine, and can be explained to students in classes above seventh or eigth. My aim for this qustion was to check with you all, if I have missed some elementary trick or not,I guess I did not. I was looking for the easiest solution someone can come up with. Thanks!
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The only "elementary" way I can think of is to write $99^{98} = (98 + 1)^{98}$ and then expand using the binomial expansion formula, and then show you get a sum of $99$ terms where each term is less than or equal to $98^{98}$, and the sum of the last two terms $98 + 1$ is strictly less than $98^{98}$. Then your sum is strictly less than $98 \cdot 98^{98} = 98^{99}$.
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Proof for parallelogram law of vector addition The Statement of Parallelogram law of vector addition is,
If two vectors are considered to be the adjacent sides of a parallelogram, then the resultant of two vectors is given by the vector that is a diagonal passing through the point of contact of two vectors.
But how do we justify that the resultant is along the diagonal? Is it based on experimental evidence, or is it something that can be proved?
I know, the question might sound pretty obvious, but I'm new to this stuff. :)
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Addition of Vectors basically found their origin from the Triangular Law of Vector Addition
The triangle law of vectors basically is a process that allows one to take two vectors, draw them proportional to each other, connect them head to tail, then draw the resultant vector as a result of the third side that is missing.
The Parallelogram law is just a furthermore explanation of Triangular law,
If two vectors are considered to be the adjacent sides of a Parallelogram, then the resultant of two vectors is given by the vector which is a diagonal passing through the point of contact of two vectors.
The proof for the resultant vector in Parallelogram addition is as follows,
Consider a parallelogram $OABC$ as shown in the figure,
Let $P \;\&\; Q$ be two adjacent sides of parallelogram, and $R$ be the resultant vector obtained by addition of vectors $P \;\&\; Q$,
Now, drop a perpendicular from $C$ on $OA$ so that they meet at $A$.
From right angled triangle $\Delta OCD \;,\; OC^2=OD^2+DC^2$
$$[OD=OA+AD]$$
$$R^2=(OA+AD)^2+DC^2$$
$$R^2=OA^2+AD^2+2OA.AD+DC^2$$
From $\Delta$ADC, $AC^2=AD^2+DC^2$
and also $\cos\theta= \frac{AD}{AC}$
And hence $R^2= OA^2+AC^2+2 OA.AC \cos\theta$
And substituting A and B $$R^2=A^2+B^2+2 A.B \cos\theta$$
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Prove an improper double integral is convergent I need to prove the following integral is convergent and find an upper bound
$$\int_{0}^{\infty} \int_{0}^{\infty} \frac{1}{1+x^2+y^4} dx dy$$
I've tried integrating $\frac{1}{1+x^2+y^2} \lt \frac{1}{1+x^2+y^4}$ but it doesn't converge
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Finding the exact value of $\int_0^\infty\frac{dx}{a^2+x^2}$ is just a calc I exercise. Let $a=\sqrt{1+y^4}$ and see what happens...
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Show That There Is One and (Essentially) Only One Field With 3 Elements My thought process is the following: By default, 0 and 1 have to be in this set of 3 elements since they are the neutral additive and neutral multiplicative elements, respectfully. So the set of the 3 elements that make up this field must be {0,1,x}. But for any element "a" in the set, there must be some "-a" element such that a + (-a) = 0. Therefore there is only one option for x; x=-1. My set must then be {-1,0,1}. Is this correct? It seems too simple, so I wanted a second opinion.
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Your answer is correct, your reasoning is not. I realize you're new to abstract algebra, so I'll just say this: when we write down things like $1$ and $0$, we mean only that these things are the multiplicative and additive identities, respectively. We do NOT think of them as subsets of the real line. For example, we can define a perfectly good field with only two elements $\mathbb{Z}_2=\{0,1\}$ where multiplication is defined as usual, and addition is defined as $1+1=0$.
Now, we want to show that there's only one field with three elements. We know $0$ and $1$ must be in this field, and that $0\neq 1$. So if we call the remaining element $x$, we have to find if there is more than one possible thing for $x$ to be.
We know, by closure, that $x+1$ has to be in our field, so either $x+1=0$, $x+1=1$, or $x+1=x$. In the first case, we get $x=-1$, where by $-1$ I do NOT mean "negative one", but rather simply "the additive inverse of one". In the second case we get $x=0$, which is wrong because we want $x$ to be a NEW element of our set. In the third case, we get $1=0$, which is also bad. So we know $x=-1$!
Now, we still haven't fully determined our field. Why? Well, we know there's only one possible third element, but we still have to fully define what addition does on the elements. In particular, we want to show that there's only ONE way to define addition on the elements consistent with the field axioms. We already know what happens if you take $1+0$, $1+-1$, and $-1+0$. We just need to figure out what happens when you take $1+1$ and $-1+-1$. We know $1+1$ has to be in our set, so $1+1=1,0,-1$. If $1+1=1$ then $1=0$, if $1+1=0$ then $-1=1$ and there's only two elements in your field. So $1+1=-1$. Similarly, $-1+-1=1$. Then you just have to check there's only one way to define multiplication (there is), and you're done.
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How can one compare these two 4-manifolds We would like to compare the following two real 4 dimensional manifolds:
1)$M$=The tangent bundle of $S^{2}$
2)$N$= The total space of the canonical line bundle over $\mathbb{C}P^{1}\simeq S^{2}$
Are these two manifolds homeomorphic? Are they diffeomorphic?
Some remarks: They have the same homotopy type. So there is no an obvious obstruction for $M$ and $N$ to be homeomorphic. On the other hand $M$ is parallelizable. So it is natural to ask: Is $N$ parallelizable, too?
Is the later equivalent to ask " Is $TS^{2}\oplus \ell$ a trivial real 4 dimensional bundle. Where $\ell$ is the realification of the canonical line bundle?"?
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Here's another (probably unnecessarily elaborate) proof that they are not diffeomorphic using Eliashberg's amazing theorem about which open $2$-handlebodies admit Stein structures and the adjunction inequality for Stein surfaces. I am sorry if this is not really accessible for the asker.
$M$ is the 4-manifold given by gluing an open $2$-handle along the unknot with framing coefficient $2$, and $N$ is the $4$-manifold given by gluing an open $2$-handle to the unknot with framing coefficient $1$. Therefore $\bar M$ ($M$ with the opposite orientation) is the $4$-manifold given by gluing a $2$-handle with framing $-2$ to the mirror of the unknot (which is still the unknot), and $\bar N$ is given by gluing a $2$-handle with framing $-1$ to the (mirror of the) unknot.
Now the unknot $K$ has a Legendrian embedding with $tb(K)=-1$ (with the Legendrian projection the simple one with only two cusps), hence by Eliashberg's construction of Stein manifolds $\bar M$ admits a Stein structure.
Now we need to show $N$ and $\bar N$ cannot admit a Stein structure. If we take a disk bounded by the attaching unknot in $\Bbb R^3$ union the core of the $2$-handle we get an embedded $2$-sphere $\Sigma$ with $[\Sigma]^2= \pm1$. By the adjunction inequality for Stein surfaces, we have $$[\Sigma]^2 + |\langle c_1,[\Sigma]\rangle |\leq2g(\Sigma)-2$$
In our case this is just $\pm1+|\langle c_1,[\Sigma]\rangle | \leq -2$ which is a contradiction. So neither $N$ nor $\bar N$ admit Stein structures.
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Is there any element of order $51$ in the group $U(103)$ Does there exist an element of order $51$ in the multiplicative group $U(103)$ ?
Now if the element exist say $x$ then it satisfies the equation $$x^{51}\equiv 1\pmod {103}$$ . Now $103$ being a prime it is clear that there is an element $y$ in $U(n)$ satisfying $$y^{102}\equiv 1\pmod {103}$$
So exactly an element of half order is required to be found.
Is there any result in number theory that might imply that $$a^{p-1}\equiv 1\pmod p$$ ensures the existence of some $b$ such that $$b^{{p-1}\over {2}}\equiv {1}\pmod p$$ where $p$ is a prime not dividing $a$
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I suppose that $U(103)$ denotes the multiplicative group of the ring $\mathbb{Z}_{103}$.
Note that $\mathbb{Z}_{103}$ is a field; the multiplicative group of a finite field is cyclic; and a cyclic group of order $n$ contains elements of order $d$ whenever $d$ divides $n$ (take the $\frac{n}{d}$-th power of a generator).
You can replace $103$ by some non-primes $n$, and still $U(n)$ remains cyclic.
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$\lim_{y \rightarrow b} \lim_{x \rightarrow a} f \neq \lim_{(x,y)\rightarrow (a,b)} f \neq \lim_{x \rightarrow a} \lim_{y \rightarrow b} f$ Can someone give me an example to show that in general
$\lim_{y \rightarrow b} \lim_{x \rightarrow a} f(x,y) \neq \lim_{(x,y)\rightarrow (a,b)} f(x,y) \neq \lim_{x \rightarrow a} \lim_{y \rightarrow b} f(x,y) $
I have been able to construct examples when the first and last limit exist but the middle one does not, but I can't find one where the middle limit exists but is not equal to the other two (both of which exist).
Does the existence of the middle limit as well as the other two imply that they must be equal to the middle one?
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Now since $\lim_{x\rightarrow a} \lim_{y\rightarrow b} f(x,y)$ and $\lim_{y\rightarrow b} \lim_{x\rightarrow a} f(x,y)$ both exist it is obvious that $\lim_{x\rightarrow a} f(x,y)$ and $\lim_{y\rightarrow b} f(x,y)$ have to exist. For if not then the former two limits cannot be be calculated.
Lemma - If $\lim_{(x,y)\rightarrow (a,b)} f(x,y) = l$ and if both $\lim_{x\rightarrow a} f(x,y)$ and $\lim_{y\rightarrow b} f(x,y)$ exist then $\lim_{x\rightarrow a} \lim_{y\rightarrow b} f(x,y) = l = \lim_{y\rightarrow b} \lim_{x\rightarrow a} f(x,y)$.
Proof -
Define the functions $g_1(x) = \lim_{y\rightarrow b} f(x,y)$ and $g_2(y) = \lim_{x\rightarrow a} f(x,y)$.
Given an, $\epsilon > 0$
*
*There exists a $\delta_1 >0$ such that $|f(x,y) -l| < \epsilon /2$ whenever $0 < \sqrt{(x-a)^2 + (y-b)^2} < \delta_1$
*There exists a $\delta_2 > 0$ such that $|f(x,y) - g_1(x)| < \epsilon /2$ whenever $0 < |y - b| < \delta_2$
Let $\delta = \min\{\delta_1,\delta_2\}$
When both $0<|y-b|<\delta /\sqrt{2}$ and $0<|x-a|<\delta /\sqrt {2}$ we have $0 < \sqrt{(x-a)^2 + (y-b)^2} < \delta$. So,
$$|g_1(x) - l| = |g_1(x) - f(x,y) + f(x,y) - l| \leq |f(x,y) - g_1(x)| + |f(x,y) - l| < \epsilon /2 + \epsilon /2 = \epsilon$$
Thus $\lim_{x\rightarrow a} g_1(x) = l$.
Similarly for $g_2(y)$. $\square$
Thus if all three limits in the question exist then they have to be equal.
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Prove that $\left|(|x|-|y|)\right|\leq|x-y|$
Prove that $\left|(|x|-|y|)\right|\leq|x-y|$
Proof:
$$\begin{align} \left|(|x|-|y|)\right| &\leq|x-y| \\
{\left|\sqrt{x^2}-\sqrt{y^2}\right|}&\leq \sqrt{(x-y)^2} &\text{($\sqrt{a^2}=|a|)$}\\
\sqrt{\left(\sqrt{x^2}-\sqrt{y^2}\right)^2}&\leq \sqrt{(x-y)^2} &\text{($\sqrt{a^2}=|a|)$} \\
\left(\sqrt{x^2}-\sqrt{y^2}\right)^2 &\leq (x-y)^2&\text{$(0\leq a\leq b\implies a^2\leq b^2)$}\\
\left(\sqrt{x^2}\right)^2-2\sqrt{x^2}\sqrt{y^2}+\left(\sqrt{y^2}\right)^2&\leq x^2-2xy+y^2 &\text{(Distribution)} \\
|x|^2-2|x||y|+|y|^2&\leq x^2-2xy+y^2 &\text{($\sqrt{a^2}=|a|)$}\end{align} \\ $$
From here, we can see that $|x|^2+|y|^2=x^2+y^2$, and it suffices to show that the following holds: $$-2|x||y|\leq -2xy.$$
When $x>0,y>0$ or $x<0,y<0$, the product $xy$ is positive, and so
$-2|x||y|=-2xy$ holds.
When $x<0,y>0$ or $x>0,y<0$, the product $xy$ is negative, and so $-2xy>0$, and it follows that:
$$-2|x||y|<0<-2xy. $$
$\square$
Is this proof valid? I am unsure if I overlooked anything, also I have some specific questions about the proof (if it is valid up to this point). From the 3rd to the 4th inequality, it feels like I am making a leap in knowing that $a\leq b$ prior to proving what's to be proved by using the result $a^2\leq b^2$. Is this a problem?
What are some alternative ways of showing this inequality?
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Because of the triangle inequality, there is: $\left| x+y \right| \le \left| x \right| +\left| y \right| $. Using this fact:
\begin{align}
\left| x \right| &= \left| (x-y)+y \right| \\
&\le \left| x-y \right| +\left| y \right| \\
\left| x \right| -\left| y \right| &\le \left| x-y \right| \tag{1}
\end{align}
Proceeding similarly:
\begin{align}
\left| y \right| &= \left| (y-x)+x \right| \\
\left| y \right| &\le \left| y-x \right| +\left| x \right| \\
\left| y \right| -\left| x \right| &\le \left| y-x \right| \tag{2} \\
\end{align}
And finally combining $(1)$ and $(2)$ :$$\left|(|x|-|y|)\right|\leq|x-y|$$
|
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Prove the trigonometric identity $\cos(x) + \sin(x)\tan(\frac{x}{2}) = 1$ While solving an equation i came up with the identity $\cos(x) + \sin(x)\tan(\frac{x}{2}) = 1$.
Prove whether this is really true or not. I can add that $$\tan\left(\frac{x}{2}\right) = \sqrt{\frac{1-\cos x}{1+\cos x}}$$
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An easy way to do that is:
sinx = sqrt((1+cosx)(1-cosx))
so LHS becomes:
cosx + sqrt((1+cosx)(1-sinx))sqrt((1-cosx)/1+cosx))
cosx + 1 - cosx = 1 = RHS
|
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Does this game make you arbitrarily rich with probability one? We toss a coin. If it's heads we win $\$ 1$, otherwise we lose $ \$ 1$. Fix some large sum. Will we be winning this amount with probability one at some point? We assume that we have infinitely many throws and an infinite amount of money.
More formally (please correct me if if there are errors). Consider the following Markov process: $$X_{i+1} = X_{i}+d_i,$$
where each $d_i$ is a random variable taking values 1 and -1 with an equal probability (if $i \neq j$ $d_i$ and $d_j$ are independent). Let $M \in \mathbb{Z}_+$ and define a Markov time $$\tau_M = \min \left\{ i \: \middle| \: X_i = M \right\}.$$
Is it true that $$ P(\tau_M < \infty ) = 1$$
for all $M$?
I'd guess it's true and that there is a simple proof. Are there some canonical ways of dealing with these type of questions? Not a homework, something I just thought about.
|
Let we say that a sequence of $2n$ throws is balanced if at any point we are not winning any dollar, but with the last throw we are losing zero. It is well-known that the number of balanced sequences of length $2n$ is given by the Catalan number:
$$ C_n = \frac{1}{n+1}\binom{2n}{n}. $$
If we say that a sequence of $2n$ throws is returning if at any point we are losing a positive amount of dollars, but with the last throw we are losing zero, then the number of returning sequences of length $2n$ is given by:
$$ R_n = \frac{1}{n}\binom{2n-2}{n-1} \geq \frac{2^{2n}}{4\sqrt{\pi} \left(n-\frac{1}{4}\right)^{3/2}},$$
so the probability to be winning $k$ dollars at some time is greater than:
$$ \frac{1}{2^k}\sum_{n=1}^{+\infty}\frac{1}{4\sqrt{\pi}\left(n-\frac{1}{4}\right)^{3/2}}\geq \frac{13}{28\cdot 2^k}.$$
Since $$\sum_{n\geq 1}R_n z^n = \frac{1-\sqrt{1-4z}}{2}$$
by evaluating the RHS at $z=\frac{1}{4}$ we have that any losing streak ends with probability $1$, as well as any winning streak. That gives that we go back to the original configuration with probability $1$, so the "zero money" state is a recurring state for our random walk, as well as any other state, since the probability that we may go on playing forever without winning or losing more than $k$ dollars is clearly zero.
|
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A complex matrix with real eigenvalues Let $A$ be a $10\times 10$ matrix with complex entries and all eigenvalues non-negative real numbers and at least one eigenvalue strictly positive . Then there exist a matrix $B$ such that
$A$. $AB-BA=B$
$B$. $AB-BA=A$
$C$. $AB+BA=A$
$D$. $AB+BA=B$
Complex matrix with real eigenvalues , is that Hermitian then?
Some lead please I am clueless.
|
Of course A and D must be true: we can take $B = 0$.
Note that $\operatorname{trace}(AB - BA) = 0$ for any $A,B$ (why)? So, B can only be true if $A$ has a trace of $0$, which is necessarily not the case from the premise of the question.
The answer to C is yes. Try $B = \frac 12 I$.
|
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How to prove by Mathematical Induction. I want to know how to prove this inequality by mathematical induction:
$a_k's$ are nonnegative numbers. Prove that$$a_1a_2\cdots a_n\leq \left(\frac{a_1+a_2+\cdots+a_n}{n}\right)^n.$$
In the inductive step, I tried using the inequality $ab\leq \frac{k}{k+1}a^\frac{k+1}{k}+\frac{1}{k+1}b^{k+1}$ but I got over estimate of what I was looking for.
Is there a better way of proving this by induction?
|
Maybe this is one of those cases in which an induction proof is easier if one makes the statement to be proved stronger, because then one has a stronger induction hypothesis to use. The inequality in the question is
$$
a_1^{1/n}\cdots a_n^{1/n} \le \frac 1 n \left(a_1 + \cdots+ a_n\right).
$$
These are a geometric mean and an arithmetic mean of $n$ numbers with equal weights $1/n$. More generally one has weights $p_1,\ldots,p_n\ge 0$, $p_1+\cdots+p_n=1$. The inequality is then
$$
a_1^{p_1} \cdots a_n^{p_n} \le p_1 a_1 + \cdots + p_n a_n.
$$
Now let's do the induction step:
\begin{align}
a_1^{1/(n+1)} \cdots a_{n+1}^{1/(n+1)} & = \Big(a_1^{1/n} \cdots a_n^{1/n}\Big)^{n/(n+1)} \Big( a_{n+1} \Big)^{1/(n+1)} & \text{(weights $\frac n {n+1}$ and $\frac 1 {n+1}$)} \\[10pt]
& \le \frac n {n+1} \Big(a_1^{1/n} \cdots a_n^{1/n}\Big) + \frac 1 {n+1} a_{n+1} \\
& {}\qquad \text{(by the induction hypothesis in case 2)} \\[12pt]
& \le \frac n {n+1} \Big( \frac 1 n \left( a_1+\cdots+a_n \right) \Big) + \frac 1 {n+1} a_{n+1} \\
& {} \qquad \text{(by the induction hypothesis in case $n$)} \\[10pt]
& = \frac 1 {n+1} \left(a_1 + \cdots + a_{n+1} \right).
\end{align}
|
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If points in a convex set $C$ escape to infinity roughly in direction $v$ then an infinite ray in that direction exists Let $C\subseteq \mathbb R^n$ a convex set.
Assume there is a sequence $\{c_k\}_{k\in\mathbb N}$ with $c_k\in C$, $|c_k|\to\infty$ such that $v:=\lim \frac 1{|c_k|}c_k$ exists. Does this imply that there exists $a\in C$ with $a+[0,\infty)\cdot v\subseteq C$? Is the set of such $a$ dense in $C$?
This looks intuitively clear, e.g., in the case $n=2$. I was thinking of projecting the $c_k$ on a hyperplane $\perp v$ through some inner point of $C$ and use these to find a suitable $a$. But these projections don't seem to be well-behaved ...
|
You probably need that $C$ is closed.
Assume $0\in C$ without loss of generality. Otherwise, if $a\in C$, then $|c_k-a|\to \infty$ and
$$ \frac{c_k-a}{|c_k-a|} = \frac{c_k}{|c_k|}\frac{|c_k|}{|c_k-a|} - \frac{a}{|c_k-a|} \to v.$$
Let $r\in[0,\infty)$. As $|c_k|\to\infty$, for $k$ sufficiently large, we have $r < |c_k|$ and
$\frac r{|c_k|} c_k\in C$ by convexity. If $C$ is closed than, we have $\frac r{|c_k|} c_k \to r v \in C$.
Note:
If $C$ is closed, we have in fact $C+[0,\infty)v \subseteq C$.
|
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What is the pre-requisite knowledge for generating my own integer sequence? I've recently come across the On-Line Encyclopedia of Integer Sequences and I'm completely fascinated by it; something about how easy integers are to grasp and yet how complex the sequences are. I find it intriguing and I'd love to be able to contribute to it. Naturally the sequences that are easier to come by have already been submitted and having only studied mathematics up to pre-calculus thus far I feel it is beyond my current abilities to discover something original.
And so my questions are:
*
*What would I require asides ingenuity to create/discover an original sequence?
*What topics would be helpful to study? Algebraic number theory?
*What topic should I study to build upon sequences & series that I've encountered in pre-calculus?
*Would I need to be able to programme? Or have a working knowledge of mathematica or another programming language?
Essentially, what knowledge would I require that would allow me to investigate this topic further and create my own sequence? Any books or other resources would also be appreciated.
Thank you.
|
You can come up with any sequence you like. If you look around OEIS, you'll find from the ubiquitous Fibonacci numbers to the Look and say sequence.
But consider that they prominently state that they have a huge backlog of proposed sequences, so you should make sure (a) it isn't a variant of something already in there, and (b) is is a really interesting sequence. For both, you might want to ask for opinions here...
|
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Complementary textbook algebraic geometry I don't know where to ask this or if it is allowed to do it, so please let me know any details for further questions of this kind.
I am taking an algebraic geometry class and am using the textbook "Ideals, Varieties, and algorithms" by David Cox et. al. I was wondering if any of you had any suggested textbooks to use as complementary textbooks for this book or books that help get a deeper understanding of the basic algebraic geometry material.
|
You might like to take a look at Joe Harris: Algebraic Geometry.
|
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Where did I go wrong in my evaluation of the integral of cosine squared? $$\int{\cos^2(x)}dx$$
Where did I go wrong in my evaluation of this integral?
$$=x\cos^2x - \int-2x\sin(x)\cos(x)\,dx$$
$$=x\cos^2x + \int x\sin(2x)\,dx$$
$$=x\cos^2x + \left(\frac {-x\cos(2x)}2 -\int \frac{-\cos(2x)}2\,dx\right)$$
$$=x\cos^2x + \left(\frac {-x\cos(2x)}2 + \frac 12\cdot\frac{\sin(2x)}2\right)$$
$$x\cos^2x-\frac{x\cos(2x)}2+\frac{\sin(2x)}4 + C$$
And this is clearly wrong, but I don't know where I messed up in my calculations. Would anyone mind correcting me somewhere?
|
Using the formula $\displaystyle \bullet \; 2\cos^2 x = 1+\cos 2x$
So $$\displaystyle I = \frac{1}{2}\int 2\cos^2 xdx =\frac{1}{2}\int \left[1+\cos 2x\right]$$
So $$\displaystyle I = \frac{1}{2}\int 1 dx + \frac{1}{2}\int \cos 2x dx = \frac{1}{2}x+\frac{1}{4}\sin 2x+\mathcal{C}$$
|
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Calculate correct dose for pet if I have a solution that contains 750 mg of curcumin in 10 ml of water and I want to give a dose that is equal to 25 mg curcumin How many ml do I give?
Need to medicate my cat and I need to be sure to give correct dose. Thanks
|
We apply the related ratios
$$\frac{750\,\,\text{mg}}{10\,\,\text{m}\ell}=\frac{25\,\,\text{mg}}{x}$$
whereupon solving for x gives
$$\bbox[5px,border:2px solid #C0A000]{x=\frac{10\,\,\text{m}\ell\times 25\,\,\text{mg}}{750\,\,\text{mg}}=\frac13\,\,\text{m}\ell}$$
|
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Change of basis and inner product in non-orthogonal basis I have a vector, originally expressed in the standard coordinates system, and want to perform a change of basis and find coordinates in another basis, this basis being non-orthogonal.
*
*Let $B = \{e_1, e_2\}$ be the standard basis for $\Bbb R^2$.
*Let $B' = \{e_1', e_2'\}$ be a non-orthogonal basis for $\Bbb R^2$.
*Let $v$ be some vector in $\Bbb R^2$.
The standard inner product is
$\langle a, b \rangle = \sum_{i=0}^n a_i b_i.$
I want to define an inner product in the non-orthogonal basis $B'$ so that $\langle e_1', e_2' \rangle_{B'} = 0$ since $\sum_{i=0}^n e_{1i}' e_{2i}' \neq 0$.
Basically, I want to use this new inner product to get the component/coordinates of the vector $v$ on the basis $B'$.
|
What you want to do is change the basis of the vectors you are working with, then take the inner product on that. since $e_1', e_2'$ are a basis for $\mathbb{R}^2$, every vector $v$ = $v_1 e_1' + v_2 e_2'$ for unique values $v_1, v_2$. Then you set $\langle v, w \rangle = v_1 w_1 + v_2 w_2$
To find an explicit method of converting a vector into this new coordinate system, look into change-of-basis matrices. if $e_1' = a_1 e_1 + a_2 e_2, e_2' = b_1 e_1 + b_2 e_2$, take the matrix with
$$ \left( \begin{array}{cc}
a_1 & b_1 \\
a_2 & b_2 \end{array} \right)
$$
and take its inverse. Then if $\langle v, w \rangle'$ is your new inner product, you have $\langle v, w \rangle'= \langle M^{-1}v, M^{-1}w \rangle$.
|
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Prove that $\frac{2^{x+1}+(x+1)^2}{2^x+x^2}\rightarrow 2$ as $x \rightarrow \infty$ Could someone please show me the proof that $\frac{2^{x+1}+(x+1)^2}{2^x+x^2}\rightarrow 2$ as $x \rightarrow \infty$
I have no idea where to begin with this one.
Thanks.
|
Divide numerator and denominator by $2^n$. It's not hard to show that $\frac{x^2}{2^n}\to 0$ (use L'Hôpital's rule, for example) to see $$\lim_{x\to\infty}\frac{2^{x+1}+(x+1)^2}{2^x+x^2}=\lim_{x\to\infty}\frac{2+(x+1)^2/2^x}{1+x^2/2^x}=2$$
|
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How is the Radius of Convergence of a Series determined? Consider $$\sum_{n=0}^{\infty}\frac{(-1)^nx^n}{(n+1)^2}$$ which by the ratio test the ratio of two consecutive terms converges to $|x|$ as $n\rightarrow \infty$ and has a radius of convergence equal to $1$.
Now consider $$\sum_{n=0}^{\infty}(-1)^n(2^n+n^2)x^n$$ which by the ratio test the ratio of two consecutive terms converges to $2|x|$ as $n\rightarrow \infty$ and has a radius of convergence equal to $\frac{1}{2}$.
My question is why the radius of convergence takes these values ($1$ for the former and $\frac{1}{2}$ for the latter)?
You have my sympathy if this is blatantly obvious to you, but it is not clear to me. So could someone please explain in simple English why the radius of convergence takes those values above?
Or, put in another way, from the second summation you know by the ratio test that the ratio of two of its consecutive terms converges to $2|x|$ as $n\rightarrow \infty$ how then do you proceed to determine the radius of convergence?
Many thanks.
|
An interesting aspect is to evaluate the series of the question.
Consider
\begin{align}
S_{1}(x) &= \sum_{n=0}^{\infty} \frac{(-1)^{n} \, x^{n}}{(n+1)^{2}} \\
S_{2}(x) &= \sum_{n=0}^{\infty} (-1)^{n} \, (2^{n} + n^{2}) \, x^{n}.
\end{align}
The first series:
\begin{align}
\partial_{x} \left(x \, S_{1}(x) \right) &= \sum_{n \geq 0} \frac{(-1)^{n} \, x^{n}}{n+1} = \sum_{n \geq 1} \frac{(-1)^{n-1} \, x^{n-1}}{n} = \frac{\ln(1+x)}{x}.
\end{align}
or
$$S_{1}(x) = - \frac{1}{x} \, Li_{2}(-x),$$
where $Li_{2}(x)$ is the dilogarithm function. The second series follows from
$$\sum_{n=0}^{\infty} n^{2} \, x^{n} = \frac{x(1+x)}{(1-x)^{3}}$$
and is seen to be
$$S_{2}(x) = \frac{1}{1+2x} - \frac{x(1-x)}{(1+x)^{3}}.$$
|
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How to prove that the following sequence will never contains number greater than 3 You may now the following sequence:
1
11
21
1211
111221
312211
13112221
Explanation of the sequence (I've put an hint just for the ones who want to search a bit ;) )
Where, in each iteration you count the number of occurrence of a digit then put the count and the digit in the new child. (First you've 1, then you've got one 1 (11), then you've got two 1(21)...)
I've the feeling that the numbers of the sequence will never contains a digit >3 but my mathematics knowledge are way to poor to prove it. So I've made a little program that compute the terms for me to found out if there is a number greater than 3 but up to the 42th term it doesn't find any.
As the length of the string is growing very fast, I can't compute a lot more terms and, even if I was able to do so, this would not be a proof.
So here's the question: How can I prove my feeling?
|
This is a typical proof by induction. The base case is just that the first term ($1$) does not contain any digit greater than $3$ nor more than $3$ consecutive, equal digits.
If some term had a digit greater than $3$, it would have had to come from more than $3$ (that is, $4$ or more) consecutive, equal digits in the previous term. But the first four of such digits (say that the digit is $k$) would had have to come from $k$ consecutive $k$s followed by $k$ consecutive $k$s, which makes no sense, because we would have written $2k$ and after $k$.
Example: if a term contains $41$ then the previous term contains $1111$ and then, the previous one contains "one $1$ and one $1$", that is, $11$, which should be transformed to $21$ and not to $1111$.
|
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Calculate $\lim_{n\to\infty} (n - \sqrt {{n^2} - n} )$ Calculate limit:
$$\lim_{n\to\infty} (n - \sqrt {{n^2} - n})$$
My try:
$$\lim_{n\to\infty} (n - \sqrt {{n^2} - n} ) = \lim_{n\to\infty} \left(n - \sqrt {{n^2}(1 - \frac{1}{n}} )\right) = \lim_{n\to\infty} \left(n - n\sqrt {(1 - \frac{1}{n}})\right)$$
$$\sqrt {(1 - \frac{1}{n}} ) \to 1$$
$$\lim_{n\to\infty} \left(n - n\sqrt {(1 - \frac{1}{n}})\right) = \lim_{n\to\infty} (n - n) = 0$$
Ok, it's not correct. In articles i found method:
$$\lim_{n\to\infty} \frac{{(n - \sqrt {{n^2} - n} )(n + \sqrt {{n^2} - n} )}}{{n + \sqrt {{n^2} - n} }} = \lim_{n\to\infty} \frac{n}{{n + \sqrt {{n^2} - n} }} = \lim_{n\to\infty} \frac{1}{{1 + \sqrt {1 - \frac{1}{n}}}} = \frac{1}{2}$$
Why the second method is valid, but the first one not? Whats difference?
|
The reason is that inside a limit, you can't substitute an expression with the limit it is approaching. The expresion $\sqrt{1-1/n}$ does go to $1$, but it does so at a slow pace, while the expression $n$ goes to infinity at a certain pace as well, so $n\sqrt{1-1/n}$ does not necessarily grow arbitrarily close to $n$.
|
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Finding the rank of an non-invertible matrix I have a $3\times3$ matrix with three different eigenvalues $0,1, 2$.
The question is: what is the rank of this matrix? If the matrix was invertible, I could say that the rank was equal to $n=3$. But as zero is an eigenvalue of this matrix, this matrix does not satisfy the Invertible Matrix Theorem.
How should I determine the rank? Thanks in advance.
|
*
*All eigenvalues are different, then the matrix is diagonalizable.
*The corresponding diagonal matrix has the eigenvalues on the diagonal, i.e.
$$
S^{-1}AS=D=\left[\matrix{2 & 0 & 0\\0 & 1 & 0\\0 & 0 & 0}\right].
$$
*The matrices $A$ and $D$ have the same rank.
|
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The number of ordered pairs of positive integers $(a,b)$ such that LCM of a and b is $2^{3}5^{7}11^{13}$ I started by taking two numbers such as $2^{2}5^{7}11^{13}$ and $2^{3}5^{7}11^{13}$. The LCM of those two numbers is $2^{3}5^{7}11^{13}$. Similarly, If I take two numbers like $2^{3-x}5^{7-y}11^{13-z}$ and $2^{3}5^{7}11^{13}$, the LCM is $2^{3}5^{7}11^{13}$. Here $0\le x\le3$, $0\le y\le7$, $0\le z\le13$
The number of ways of choosing 3 numbers $x,y,z$ is $^4C_1\cdot^8C_1\cdot^{14}C_1$
The above value is only for $b=2^{3}5^{7}11^{13}$.But now I have to consider another value of $b$ and the the solution becomes lengthy. Is this the correct approach?
|
Let $a = 2^{x_{1}}\cdot 5^{y_{1}}\cdot 11^{z_{1}}$ and $b = 2^{x_{2}}\cdot 5^{y_{2}}\cdot 11^{z_{2}}\;,$ Then Given $\bf{LCM(a,b)} = 2^{3}\cdot 5^{7}\cdot 11^{13}$
So Here $0\leq x_{1},x_{2}\leq 3$ and $0\leq y_{1},y_{2}\leq 7$ and $0\leq z_{1},z_{2}\leq 13.$
Here ordered pairs of $(x_{1},x_{2}) = \left\{(0,3),(1,3),(2,3),(3,3),((3,0),(3,1),(3,2)\right\}$
So Total $7$ ordered pairs
Similarly ordered pairs for $(y_{1},y_{2}),$ We get $15$ ordered pairs..
Similarly ordered pairs for $(z_{1},z_{2}),$ We get $27$ ordered pairs..
So Total ordered pairs of $(a,b) = 7\times 15 \times 27$
|
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Confused about rules for solving systems of linear equations Why are we allowed to add equations together/eliminate variables when solving systems of linear equations? I get that it works to find the solution but I don't understand why it works.
Also, why can't we do the same things with nonlinear systems of equations?
For example, the system
$x^2 -y = 1$
$x+y = 5$
If you add them together and solve for $x$ you get the correct $x$ coordinates for the two points of intersection.
EDIT: I know that this system is nonlinear and I know how to solve it for the points of intersection. My main question is why are the rules of adding equations and eliminating variables valid for systems of linear equations but not nonlinear equations in general?
|
If (x,y) is a solution to the first equation then the left side is 1, and if (x,y) is a solution to the second equation, then the left side is 5. When you add the two equations, the result on the left side must equal 6 only if (x,y) satisfies both equations i.e. the point of intersection. The same idea applies to a pair linear equations. The solution is usually the (single) point of intersection between the lines represented by the two linear equations (in 2D).
|
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|
parallel resistors Consider the set $E_b = \left\{1, 1.2, 1.5, 1.8, 2.2, 2.7, 3.3, 3.9, 4.7, 5.6, 6.8, 8.2\right\}$. This is our base set. Let's define the set $E$ as follows:
$$ E = \left\{ 10^k e \mid k=0,1,2,\ldots, \text{for every} e \in E_b \right\}$$
and $\Omega$ as the class of all subsets of $E$. We are intereted in defining a mapping $f$ from $\mathbb{N} \setminus \left\{1 \right\}$ to $\Omega$ such that $f(n)$ is mapped to a set $E_k \in \Omega$ according to the following rule:
$$ \frac{1}{n} = \sum_{e_k \in E_k}{\frac{1}{e_k}}$$
Here are some examples:
*
* $f(4) = \left\{10, 12, 15\right\}$ as $\frac{1}{4} = \frac{1}{10} + \frac{1}{12} + \frac{1}{15}$
* $f(12) = \left\{12\right\}$
* $f(19) = $ I haven't found a set that could be mapped to $19$, it could be an empty set
* $f(20) = \left\{22, 220 \right\}$ as $\frac{1}{20} = \frac{1}{22} + \frac{1}{220}$
This question/puzzle was brought to my attention by a colleague (an electrical engineer), the idea is to synthesize a desired resistor (an integer-valued in this case) with a given set of standard resistors (set $E$) in parallel (two parallel resistors result in $\frac1{\frac{1}{R_1}+\frac{1}{R_2}}$).
Now I'm interested to know if we can synthesize a resistor with a minimum numbers of standard resistors, in other words a lower bound on the cardinality of $f(n)$. More interestingly, I'd like to know if someone can come up with an algorithm to construct such a mapping for every $n \in \mathbb{N} \setminus \left\{1 \right\}$, not necessarily with minimum cardinality. Also, for what numbers $f(n)$ is an empty set?
I hope I've articulated the question in mathematical terms clear enough.
|
This is a partial answer, identifying some $n$ that cannot be synthesized with a finite subset of values from $E$.
The LCM of $E_{\text{b}}$ is $N=3541509972=2^2\cdot 3^3\cdot 7\cdot 11\cdot 13\cdot 17\cdot 41\cdot 47$. Thus $\frac{N}{e}$ is an integer for $e\in E_{\text{b}}$. For general $e\in E$, $\frac{N}{e}$ therefore has fractional part with finite decimal length.
Consequently, if $f(n)$ is a finite set, then the sum for $\frac{N}{n}$ necessarily has fractional part with finite decimal length.
This implies that if $n$ has prime divisors other than those of $5N$, or prime divisors other than $2$ and $5$ that occur in greater multiplicity than in $N$, no finite set $f(n)$ can be given because $\frac{N}{n}$ would have infinite decimal expansion then. For example $19$ and $81$ cannot be synthesized.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1409596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
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|
Locally flat manifold from Frobenius, differential forms approach It is a known fact that a Manifold $M$ is locally flat if and only if the Riemann curvature tensor (Levi-Civita connection) vanishes. To show the if part is easy, but I am trying to follow the details of the proof given here but I am at a loss in reaching a full proof. I am interested only in a proof of this statement using the Frobenius theorem.
Here is what I have tried: We start with a coframe $\Phi$, it obeys the structure equations ($R=0,~T=0$)
$$
d\Phi + \omega \wedge \Phi =0, \qquad d\omega + \omega \wedge \omega=0
$$
from the second equation it can be seen (Frobenius thm) that the first equation is integrable. Now, since $\Phi$ is a coframe I can ask whether a local rotation matrix $R(x) \in SO(n)$ will be capable of rotating the coframe $\Phi$ to a closed coframe $\Theta$, $d\Theta=0$ (If we manage to do that then Poincaré lemma implies that $\Theta = d \Psi$ locally, and such $\Psi$ are the sought euclidean coordinate functions, we are done). Calculation gives:
$$
d(R \Phi) = d \Theta = (dR - R \omega)\wedge \Phi
$$
and this will vanish if the equation $R^{-1}dR = \omega $ is solved by some $R$. I can see that the structural equation implies the integrability of this equation, so a solution $R$ exists, but how can we be sure that such solution will be a matrix in $SO(n)$?
Here is my question: How do we know that there is an $R \in SO(n)$ that solves $dR = R \omega$?
|
Following Élie Cartan, you want to think of your differential system $dR - R\omega = 0$ on $M\times SO(n)$. This $\mathfrak{so}(n)$-valued $1$-form is integrable, as you said, because of vanishing curvature. The integral manifolds of this differential system will locally be the graphs of functions $R\colon U\to SO(n)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1409665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
What do you call a space whose only compact sets are finite? What do you call a topological space where a subset is compact iff it's finite? Is there a technical name?
For example, take the discrete topology, or the countable complement topology.
|
As others have noted, the term that you want is anticompact. This is an example of the kind of property studied by Paul Bankston in The total negation of a topological property, Illinois J. of Math., Vol. 23, Nr. 2 (1979), 241-252. I quote:
Let $K$ be a topological class. The spectrum $\operatorname{Spec}(K)$ of $K$ is the class of cardinal numbers $\kappa$ such that any topology on a set of power $\kappa$ lies in $K$. For example, any topology on a finite set must be compact; and any infinite set supports noncompact topologies. Thus $\operatorname{Spec}(\{\text{compact spaces}\})=\omega$. Other spectra can be computed quite readily, such as $\operatorname{Spec}(\{\text{connected spaces}\})=2(=\{0,1\})$, and $\operatorname{Spec}(\{\text{perfect spaces}\})=1$.
Now let $K$ be a class of spaces and define $\operatorname{Anti}(K)$ to be the class of spaces $X$ such that whenever $Y\subset X$, $Y\in K$ iff $|Y|\in\operatorname{Spec}(K)$. Thus $X\in\operatorname{Anti}(K)$ iff the only subspaces of $X$ which are in $K$ are those which “have to be” on account of their cardinalities. Clearly $\operatorname{Anti}(K)$ is always hereditary.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1409825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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|
Rearrangement and Cauchy Let $a_1, \ldots, a_n$ be distinct positive integers. I want to prove that
$$\frac{a_1}{1^2} + \frac{a_2}{2^2} + \cdots + \frac{a_n}{n^2} \geq \frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{n}.$$
I've been considering using Rearrangement and Cauchy Schwarz, but cannot make any progress
|
First we quote (part of) the Rearrangement Inequality, in the notation used by Wikipedia:
$$x_ny_1 + \cdots + x_1y_n \le x_{\sigma (1)}y_1 + \cdots + x_{\sigma (n)}y_n\tag{1} $$
for every choice of real numbers
$$ x_1\le\cdots\le x_n\quad\text{and}\quad y_1\le\cdots\le y_n,$$
and every permutation
$$ x_{\sigma(1)},\dots,x_{\sigma(n)}\,$$
We can assume that the $a_i$ are a permutation of $\{1,2,\dots,n\}$. For if that is not the case, and we replace the smallest of the $a_i$ by $1$, the second smallest by $2$, and so on, we weaken the inequality.
We want to prove that
$$\frac{a_n}{n^2}+\frac{a_{n-1}}{(n-1)^2}+\cdots +\frac{a_1}{1^2}\ge \frac{1}{n}+\frac{1}{n-1}+\cdot +\frac{1}{1}.\tag{2}$$
Let $y_i=\frac{1}{(n+1-i)^2}$ and let $x_i=i$. Then the Rearrangement Inequality (1) directly gives (2).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1409922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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|
What can we say about open unit balls of sup-norm and integral-norm
Consider the normed linear spaces $X_1=(C[0,1], ||.||_1)$ and $X_{\infty}=(C[0,1],||.||_{\infty})$ , where $C[0,1]$ denotes the vector space of all continuous real valued functions on $[0,1]$ and $$||f||_1=\int_0^1|f(t)|\,dt.$$and $$||f||_{\infty}=\sup\{|f(t)|:t\in [0,1]\}.$$Let, $U_1$ and $U_{\infty}$ be the open unit balls in $X_1$ and $X_{\infty}$ respectively. Then
*
*$U_{\infty}$ is a subset of $U_1$.
*$U_1$ is a subset of $U_{\infty}$
*$U_{\infty}=U_1$
*Neither $U_{\infty}$ is a subset of $U_1$ nor $U_1$ is a subset of $U_{\infty}$.
I really don't know from where I start to solve and conclude about open balls..Please help..
|
Since it seems that you want to solve it yourself, I'll just give you a hint. If you want a more accurate answer, just leave a comment.
HINT: In a normed space the unit open balls is the set of all elements which have a norm strictly less than 1. Now, what does it mean that $\left\Vert f \right\Vert_1 < 1$? What does it mean that $\left\Vert f \right\Vert_{\infty} < 1$?
Edit: We have
$$\left\Vert f\right\Vert_1 = \int_0^1 |f(t)| ~ dt \leq \int_0^1 \left\Vert f \right\Vert_{\infty} ~ dx = \left\Vert f \right\Vert_{\infty} \cdot \int_0^1 1 ~ dx = \left\Vert f \right\Vert_{\infty}$$
for every $f \in C[0,1]$. It follows that if $\left\Vert f \right\Vert_{\infty} < 1$, then $\left\Vert f \right\Vert_1 < 1$ also. Hence, $U_{\infty} \subseteq U_1$, i.e. 1. is true.
This already shows us that 4. is false.
Now to see that 2. and 3. (which are now equivalent) are false: Consider the function
$$f(x) = \begin{cases} 2 - 8x & , x \leq \frac{1}{4} \\ 0 &, \text{ otherwise} \end{cases}$$
Then, $f \in C[0,1], \left\Vert f \right\Vert_{\infty} = 2$ but
$$\left\Vert f \right\Vert_1 = \int_0^{1/4} 2 - 8x ~ dx = [2x - 4x^2]_{x=0}^{1/4} = \frac{1}{4} < 1$$
Hence, $f \in U_1$ but $f \notin U_{\infty}$. This shows that neither 2. nor 3. can be true.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1410019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Does $\lim\limits_{n \to +∞} \sum_{k=1}^n \frac{n\cdot \ln (k)}{n^2+k^2}$ diverge? Does the limit of this summation diverge?
$$\lim\limits_{n \to +∞} \sum_{k=1}^n \frac{n\cdot \ln (k)}{n^2+k^2}$$
Thanks!
|
$$\sum\limits_{k = 1}^n {\frac{{n\ln k}}{{{n^2} + {k^2}}}} = \frac{1}{n}\sum\limits_{k = 1}^n {\frac{{\ln k}}{{1 + {{(\frac{k}{n})}^2}}}} = \frac{1}{n}\sum\limits_{k = 1}^n {\frac{{\ln (\frac{k}{n})}}{{1 + {{(\frac{k}{n})}^2}}}} + \frac{1}{n}\sum\limits_{k = 1}^n {\frac{{\ln n}}{{1 + {{(\frac{k}{n})}^2}}}} $$
When $n \to \infty $, the first term is the definite integral $$\int_0^1 {\frac{{\ln x}}{{1 + {x^2}}}dx} $$
which obviously converge.
To investigate the second term, we note that
$$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{n}\sum\limits_{k = 1}^n {\frac{1}{{1 + {{(\frac{k}{n})}^2}}}} } \right] = \int_0^1 {\frac{1}{{1 + {x^2}}}dx} $$
Hence your sequence diverges.
PS:
This method can be used to obtain estimate on your sequence, using
$$\frac{\pi }{4} - \frac{1}{n}\sum\limits_{k = 1}^n {\frac{1}{{1 + {{(\frac{k}{n})}^2}}}} = \frac{1}{{4n}} + o(\frac{1}{n})$$
we have
$$\sum\limits_{k = 1}^n {\frac{{n\ln k}}{{{n^2} + {k^2}}}} = - G + \frac{\pi }{4}\ln n - \frac{{\ln n}}{{4n}} + o(\frac{{\ln n}}{n})$$
with $G=-\int_0^1 {\frac{{\ln x}}{{1 + {x^2}}}dx}$ is the Catalan constant.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1410108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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|
Is $\sqrt[3]{-1}=-1$? I observe that if we claim that $\sqrt[3]{-1}=-1$, we reach a contradiction.
Let's, indeed, suppose that $\sqrt[3]{-1}=-1$. Then, since the properties of powers are preserved, we have: $$\sqrt[3]{-1}=(-1)^{\frac{1}{3}}=(-1)^{\frac{2}{6}}=\sqrt[6]{(-1)^2}=\sqrt[6]{1}=1$$ which is a clear contradiction to what we assumed...
|
The notation $\sqrt[3]{-1}$ is a little bit ambiguous, since there are exactly three third roots of $-1$ over the complex numbers (in general, there are exactly $n$ $n-$roots of any complex number $z$ so the notation $\sqrt[n]{z}$ is ambiguous too).
Since
$$(-1)^3 = -1$$
$-1$ is one of those roots, but there are other two, namely, the roots of the equation
$$x^2-x+1$$
Which arise from the factorization
$$x^3+1=(x+1)(x^2-x+1)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1410185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
}
|
Calculating $\int_0^{\infty } \left(\text{Li}_2\left(-\frac{1}{x^2}\right)\right)^2 \, dx$ Do you see any fast way of calculating this one? $$\int_0^{\infty } \left(\text{Li}_2\left(-\frac{1}{x^2}\right)\right)^2 \, dx$$
Numerically, it's about
$$\approx 111.024457130115028409990464833072173251135063166330638343951498907293$$
or in a predicted closed form
$$\frac{4 }{3}\pi ^3+32 \pi \log (2).$$
Ideas, suggestions, opinions are welcome, and the solutions are optionals.
Supplementary question for the integrals lovers: calculate in closed form
$$\int_0^{\infty } \left(\text{Li}_2\left(-\frac{1}{x^2}\right)\right)^3 \, dx.$$
As a note, it would be remarkable to be able to find a solution for the generalization below
$$\int_0^{\infty } \left(\text{Li}_2\left(-\frac{1}{x^2}\right)\right)^n \, dx.$$
|
$$I=\int_0^\infty\operatorname{Li}_2^2\left(-\frac1{x^2}\right)\ dx\overset{IBP}{=}-4\int_0^\infty\operatorname{Li}_2\left(-\frac1{x^2}\right)\ln\left(1+\frac1{x^2}\right)\ dx$$
By using $$\int_0^1\frac{x\ln^n(u)}{1-xu}\ du=(-1)^n n!\operatorname{Li}_{n+1}(x)$$ setting $n=1$ and replacing $x$ with $-\frac1{x^2}$ we can write
$$\operatorname{Li}_2\left(-\frac1{x^2}\right)=\int_0^1\frac{\ln u}{u+x^2}\ du$$
Then
\begin{align}
I&=-4\int_0^1\ln u\left(\int_0^\infty\frac{\ln\left(1+
\frac1{x^2}\right)}{u+x^2}\ dx\right)\ du\\
&=-4\int_0^1\ln u\left(\frac{\pi}{\sqrt{u}}\ln\left(\frac{1+\sqrt{u}}{\sqrt{u}}\right)\right)\ du, \quad \color{red}{\sqrt{u}=x}\\
&=-16\pi\int_0^1\ln x\left(\ln(1+x)-\ln x\right)\ dx\\
&=-16\pi\left(2-\frac{\pi^2}{12}-2\ln2-2\right)\\
&=\boxed{\frac43\pi^3+32\pi\ln2}
\end{align}
Proof of $\ \displaystyle\int_0^\infty\frac{\ln\left(1+
\frac1{x^2}\right)}{u+x^2}\ dx=\frac{\pi}{\sqrt{u}}\ln\left(\frac{1+\sqrt{u}}{\sqrt{u}}\right)$
Let $$J=\int_0^\infty\frac{\ln\left(1+\frac1{x^2}\right)}{u+x^2}\ dx\overset{x\mapsto\ 1/x}{=}\int_0^\infty\frac{\ln(1+x^2)}{1+ux^2}\ dx$$
and $$J(a)=\int_0^\infty\frac{\ln(1+ax^2)}{1+ux^2}\ dx, \quad J(0)=0,\quad J(1)=J$$
\begin{align}
J'(a)&=\int_0^\infty\frac{x^2}{(1+ux^2)(1+ax^2)}\ dx\\
&=\frac1{a-u}\int_0^\infty\left(\frac1{1+ux^2}-\frac1{1+ax^2}\right)\ dx\\
&=\frac1{a-u}*\frac{\pi}{2}\left(\frac1{\sqrt{u}}-\frac1{\sqrt{a}}\right)\\
&=\frac{\pi}{2}\frac{1}{\sqrt{u}a+u\sqrt{a}}
\end{align}
Then $$J=\int_0^1 J'(a)\ da=\frac{\pi}{2}\int_0^1\frac{da}{\sqrt{u}a+u\sqrt{a}}\\=\frac{\pi}{2}\left(\frac{2}{\sqrt{u}}\ln\left(\frac{1+\sqrt{u}}{\sqrt{u}}\right)\right)=\frac{\pi}{\sqrt{u}}\ln\left(\frac{1+\sqrt{u}}{\sqrt{u}}\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1410293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 2
}
|
Geodesics of Sasaki metric I would like to ask the community for a reference on the following question:
Let $(M,g)$ be a Riemannian manifold and $(T^1M,g_S)$ be the unit tangent bundle with the Sasaki metric. Is it true that the orbits of the geodesic flow $\varphi:T^1M\longrightarrow T^1M$ are geodesics of $(T^1M,g_S)$?
Any help/reference would be really appreciated!
|
(1) Consider a projection $\pi : (T_1M,G) \rightarrow (M,g) $ We
want to define a metric $G$. If $\widetilde{c_1} (t),\
\widetilde{c_2}(t) $ are curves starting at $(p,v)$ then we write
$\widetilde{c_i}(t)=(c_i(t),v_i(t)),\ c_i(0)=p,\ v_i(0)=v$ so that
let
$$ G(\widetilde{c_1}'(0),\widetilde{c_2}'(0)) =g(c_1'(0),c_2'(0)) +
g(\nabla_{c_1'} v_1,\nabla_{c_2'} v_2)\ \ast $$
And if $c_1(t)=p$, then we use $\frac{d}{dt}\bigg|_{t=0}\ v_1(t)$
instead of $\nabla_{c_1'} v_1$.
(2) Now we will interpret this definition. Fix a curve $c$ at
$c(0)=p\in M$. If $v(t)$ is a parallel vector field along $c(t)$
with $v(0)=v$, then define $\widetilde{c}(t)=(c(t),v(t))$ So we have
${\rm length}\ c ={\rm length}\ \widetilde{c}$. That is any curve in
$M$ can be lifted into some curve of same length to any
point.
(3) If $c$ is a unit speed geodesic in $M$, then
$\widetilde{c}(t)=(c(t),c'(t))$, lift of $c$, is a geodesic in
$T_1M$ : Let $\widetilde{p}=\widetilde{c}(0),\
\widetilde{q}=\widetilde{c}(\epsilon )$ Assume that
$\widetilde{\gamma} =(\gamma (t),v(t))$ is a geodesic in $T_1M$
between two points s.t. $\gamma$ has unit speed. Hence since
$${\rm length}\ \gamma\geq {\rm length}\ c,$$ then
$\widetilde{\gamma}$ is $\widetilde{c}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1410371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
}
|
bend measurement and calculating $\int_4^8 \sqrt{1+{\left(\frac{{x^2-4}}{4x}\right)^2}} $ How can i get the measure of this bend : $y=\left(\frac{x^2}{8}\right)-\ln(x)$ between $4\le x \le 8$. i solved that a bit according to the formula $\int_a^b \sqrt{1+{{f'}^2}} $:$$\int_4^8 \sqrt{1+{\left(\frac{x^2-4}{4x}\right)^2}} $$ $$= \cdots$$I don't know how I calculate this integration.the answer is: 6+ln2
|
You just need to expand the square:
$$
\int_{4}^{8}{\sqrt{1+\left(\frac{x^2-4}{4x}\right)^2}dx}=\int_{4}^{8}{\sqrt{1+\frac{x^4-8x^2+16}{16x^2}}dx}=\int_{4}^{8}{\sqrt{\frac{x^4+8x^2+16}{16x^2}}dx}=\int_{4}^{8}{\sqrt{\left(\frac{x^2+4}{4x}\right)^2}dx}=\int_{4}^{8}{\frac{x^2+4}{4x}dx}=\int_{4}^{8}{\frac{x}{4}+\frac{1}{x}dx}=\frac{8^2}{8}-\frac{4^2}{8}+\ln(8)-\ln(4)=6+\ln(2)
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1410466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
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