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On the behaviour of $\left(1+\frac{1}{n+1}\right)^{n+1}-\left(1+\frac{1}{n}\right)^n$ I have to find the limit : (let $k\in \mathbb{R}$)
$$\lim_{n\to \infty}n^k \left(\Big(1+\frac{1}{n+1}\Big)^{n+1}-\Big(1+\frac{1}{n}\Big)^n \right)=?$$
My Try :
$$\lim_{n\to \infty}\frac{n^k}{\Big(1+\frac{1}{n}\Big)^n} \left(\frac{\Big(1+\frac{1}{n+1}\Big)^{n+1}}{\Big(1+\frac{1}{n}\Big)^n}-1\right)$$
we know that :
$$\frac{\Big(1+\frac{1}{n+1}\Big)^{n+1}}{\Big(1+\frac{1}{n}\Big)^n}>1$$
now what do i do ?
|
$$\lim_{n\to \infty}n^k \left((1+\frac{1}{n+1})^{n+1}-(1+\frac{1}{n})^n \right)=
\lim_{n\to \infty}n^k \left(\frac{e}{2n^2}+O((\frac{1}{n^3})) \right)$$
for n<2 limit is 0, for n=2 limit is e/2, for n>2 limit is infinity
|
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|
Area of projected parallelogram onto a plane. Say you have a parallelogram which is defined by the to vectors: $\vec u$, $\vec v$. Prove that the area of its projection on a plane with a perpendicular vector $\vec n$ (where $|\vec n|=1$) is: $E=|(\vec u \times \vec v)\ \vec n|$. Now I know that the area of the original parallelogram is: $|\vec u \times \vec v|$, but i can't relate this with the other area, or find it from scratch.
|
The geometric intuition is that the projected area is equal to the original area multiplied by $\,\cos \theta\,$ where $\,\theta\,$ is the angle between the planes. But $\,\vec u \times \vec v\,$ is a vector along the normal to the plane spanned by$\,(\vec u, \vec v)\,$, so the angle between $\,\vec u \times \vec v\,$ and $\,\vec n\,$ is precisely the angle between the two planes. The dot product of $\,\vec u \times \vec v\,$ with unit vector $\,\vec n\,$ then introduces the projection factor of $\,\cos \theta\,$.
Outline of an algebraic proof (where $\,\vec \cdot \,$ arrows are omitted, and $\, a \cdot b\,$ is the dot product):
*
*the projection of $ u$ onto the normal $\, n\,$ is $\,( u \cdot n)\, n\,$, so the projection onto the given plane orthogonal to $\, n\,$ is $\, u - ( u \cdot n)\, n\,$, and the same goes for $\, v\,$
*the projected parallelogram is the parallelogram formed by the projections of the two original vectors, so its area is the magnitude of $\,\left( u - ( u \cdot n)\, n\right) \times \left( v - ( v \cdot n)\, n\right)\,$
*the latter simplifies, using the triple product identity $\, a \times ( b \times c)=(a \cdot c)b - (a\cdot b)c\,$, to:
$$\require{cancel}
\begin{align}
\left( u - ( u \cdot n)\, n\right) \times \left( v - ( v \cdot n)\, n\right) &= u \times v- ( v \cdot n) u \times n - ( u \cdot n) n \times v + \cancel{( u \cdot n)( v \cdot n) n \times n} \\[5px]
&= u \times v - \big(( v \cdot n) u - ( u \cdot n) v\big) \times n \\[5px]
&= u \times v - \big( n \times \left( u \times v\right)\big) \times n \\[5px]
&= \cancel{ u \times v} + \left( n \cdot ( u \times v)\right) n - \cancel{\left( n \cdot n\right) u \times v} \\[5px]
&= \left(( u \times v) \cdot n\right) n
\end{align}
$$
|
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Suppose $\lim _{x\to \infty} f(x) = \infty$. Calculate $\lim_{x\rightarrow \infty} \left(\frac{f(x)}{f(x)+1}\right)^{f(x)}$ Suppose $\lim _{x\to \infty} f(x) = \infty$. Calculate: $\lim_{x\rightarrow
\infty} \left(\dfrac{f(x)}{f(x)+1}\right)^{f(x)}$
I figured the limit is $\dfrac{1}{e}$, but I have to prove it using the definition of limit, not sure how. Thanks
|
Notice that $$\left(\frac{f(x)}{f(x)+1} \right)^{f(x)} = \frac{1}{\left(1+\frac{1}{f(x)} \right)^{f(x)}}$$
|
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|
Dedekind domain and converse of Krull's principal ideal theorem It is clear that Dedekind domain's all primes other than $(0)$ are maximal. Its dimension must be $\leq 1$. Suppose this domain is not a field. So $\dim=1$.
So all maximal must be minimal over a principal element.
Here is Krull's principal ideal theorem converse.
Thm: Any prime of height $c$ is minimal over an ideal generated by $c$ elements.
Consider $\mathbb Z[\sqrt{-5}]$. Then $m=(3,2+\sqrt{-5})$ is a maximal ideal. One can check this easily by lifting to $\mathbb Z[x]/(x^2+5)$.
Q1. This maximal ideal $m$ is generated by 2 elements. Why this maximal ideal is not minimal over itself? $m$'s primary decomposition is itself. So it is minimal over itself.
The reason to ask this question is that $m$ should be minimal over principal ideal rather than $(3,2+\sqrt{-5})$. If $m$ minimal over itself, this says $ht(m)\leq 2$ from Krull PIT. Now converse says $m$ minimal over principal by $m$ having $ht(m)=1$.
Q2. $m$ cannot be minimal over itself and a principal ideal simultaneously.
|
The ideal $m$ is minimal over itself, but this does not contradict the theorem: it simply says that $m$ is minimal over a principal ideal. And indeed, since $(3) = (3, 1 + \sqrt{-5})(3, 2 + \sqrt{-5})$, $m$ is minimal over the principal ideal $(3)$.
|
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Is the mapping $T : x \mapsto \int_{0}^{\bullet} \tau^{-1/2}x(\tau)\,d\tau$ uniformly continuous? Consider $X = C([0,1])$ with its natural metric induced by $\| \cdot \|_{\sup}$ and $Y = C([0,1])$ with the metric $d_1(x,y) = \int^1_0 |x(t)-y(t)| \, dt$. Let
$$T: X\to Y : x(t) \mapsto y(t) = \int_0^t \frac{1}{\sqrt \tau} \ x(\tau) \, d\tau$$
Is the mapping T uniformly continuous?
Definition. $T:(X,d_x) \to (Y,d_y)$ is uniformly continuous if $\forall \epsilon$, $\exists \delta = \delta(\epsilon)$ :
$$ \quad T(B(a,\delta)) \subset B(T_a,\epsilon), \qquad \forall a \in X$$
Usage of $\delta$ in the definition confuses me how can I prove this?
|
I believe this is uniformly continuous. We can estimate
$$d(T(x(t)), \ T(y(t))) = \int_0^1\bigg|\int_0^t \frac{1}{\sqrt \tau}\big[x(\tau) - y(\tau) \big] d\tau \bigg | dt \leq \int_0^1\int_0^t \bigg|\frac{1}{\sqrt \tau} \bigg| \ \delta \ d\tau \ dt$$
Where we assume that the distance between $x(t)$ and $y(t)$ is $\delta$. We will determine exactly what $\delta$ should be in terms of $\epsilon$, and this will complete our proof. Since we're working in the sup norm in $X$. Now it just remains to figure out how to choose $\delta$ to make this quantity smaller than $\epsilon$. Since $\frac{1}{\sqrt\tau}$ is positive on $(0,1)$, we drop the bars.
$$d(T(x(t)), \ T(y(t))) \leq \delta \int_0^1\int_0^t \frac{1}{\sqrt \tau} d \tau \ dt$$
And this is easily evalauted to be $4\delta/3$. So taking $\delta = 3\epsilon/4$, we're done.
Since $x(t)$ and $y(t)$ were arbitrary, this proves uniform continuity.
|
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|
Morphisms to categorical product completely determined by composition with canonical projections? Consider the diagram:
$$\begin{array}{} &&&& A_1 \\ &&& \overset{\pi_1}\nearrow \\ X & \overset f{\underset g \rightrightarrows} & A_1 \times A_2 \\ &&& \underset{\pi_2}\searrow \\ &&&& A_2 \end{array}$$
where $X$ is any arbitrary object, $(A_1 \times A_2, \pi_1, \pi_2)$ is the category product of $A_1$ and $A_2$, and $f$ and $g$ are any maps from $X$ to $A_1 \times A_2$.
My question is whether $\pi_1 \circ f = \pi_1 \circ g$ and $\pi_2 \circ f = \pi_2 \circ g$ imply $f=g$, i.e. whether a morphism from any object to a categorical product is uniquely determined by the composition with the two projection morphisms.
|
Yes, this is exactly the content of the universal property of the categorical product.
*
*For every pair of maps $f_1,f_2\colon X\to A_1,A_2$ there is a map $f\colon X\to A_1\times A_2$ satisfying $\pi_1\circ f=f_1$ and $\pi_2\circ f=f_2$
*If there are two maps $f,g\colon X\to A_1\times A_2$ satisfying $\pi_1\circ f=f_1$ and $\pi_2\circ f=f_2,$ then they are equal.
Together, the two statements say "for any pair, there is one and only one arrow to the product." So yes, $f_1$ and $f_2$ determine $f$ by statement 1, and do so uniquely (any two such arrows are equal) by statement 2.
|
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Determining if points lie on a vertical plane This is slightly related to a recent post of mine. If I am dealing with three-dimensional Cartesian coordinates and have at least three points, how can I easily tell if the best-fit plane to the data points is vertical (or near-vertical) in the $z$ dimension?
My concern is that I have many sets of points and would like to fit a plane to each. In many cases, this is no issue. However, when the points lie on a plane that is vertical, I cannot fit a plane of the form $ax+by+c=z$ (just as I wouldn't be able to fit vertical points in the $xy$-coordinate system with an equation $y=mx+b$). I would like to automatically identify these instances but am not sure the best way to do so.
The very naive solution is to say that if two points have the same $x$ and $y$ coordinates, then the points lie on a vertical plane, but this is a very bad assumption because that one point could be an outlier in an otherwise horizontal plane (for instance). Note that I do not have the equation of the plane in advance, just the data points.
|
Consider three points with coordinates $\vec v_i=(x_i,y_i,z_i)$ with $i=1,2,3$. The obvious step for your problem would be to find the normal to this plane. You can do that by using the cross product $(\vec v_1 -\vec v_0)\times (\vec v_1 -\vec v_0)$. You can now look at the $z$ component of this vector product. It the magnitude (absolute value) of the $z$ component is small compared to magnitude of the cross product, it means that the normal is almost in the horizontal plane, so the plane is vertical. You can repeat this procedure to as many points as you want, just replace $\vec v_2$ with $\vec v_3$ and so on.
However, you can easily fit a plane to a set of points, just don't use the formula where the coefficient of $z$ is one. For vertical plane you will get $|a|=|b|=|c|=\infty$. For example, look at this question.
|
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Almost surely convergence related with independent suppose that $ x_1 , x_2 $ are independent observation of a random variable x
prove that $x_1 +x_2$ has the same distribution as $x \iff x=0 $ a.s
please please help me for this problem
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If $X_{1}+X_{2}$ has the same distribution as $X_{1}$, then
\begin{align*}
Var[X_1]
&= Var[X_1+X_2] \\
&= Var[X_1]+Var[X_2]
& \text{Independence}
\end{align*}
Conclude that $Var[X_2]=0$. Since the variables are identically distributed,
$X_1=X_2=k$. Since $X_{1}+X_{2}$ has the same distribution as $X_{1}$, $2k=k$, that is $k=0$.
|
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Evaluating $\big(\cot \frac{\pi}{18}-3\cot \frac{\pi}{6}\big)\cdot \big(\csc \frac{\pi}{9}+2\cot \frac{\pi}{9}\big)$
Finding value of $\displaystyle \bigg(\cot \frac{\pi}{18}-3\cot \frac{\pi}{6}\bigg)\cdot \bigg(\csc \frac{\pi}{9}+2\cot \frac{\pi}{9}\bigg)$
Try: $$\cot \frac{\pi}{18}\csc \frac{\pi}{9}-3\sqrt{3}\csc \frac{\pi}{9}+2\cot \frac{\pi}{18}\cot\frac{\pi}{9}-6\sqrt{3}\cot \frac{\pi}{9}$$
could some help me how can i simplify it,thanks
|
$$\cot x-3\cot3x=\dfrac{\cos x\sin3x-3(\cos3x\sin x)}{\sin x\sin3x}$$
Again,
\begin{align}
2\cos x\sin3x-3(2\cos3x\sin x)
&=\sin4x+\sin2x-3(\sin4x-\sin2x)\\[4px]
&=4\sin2x-2\sin4x \\[4px]
&=4\sin2x(1-\cos2x)\\[4px]
&=4\sin2x(2\sin^2x)
\end{align}
Finally,
\begin{align}
\csc2x+2\cot2x
&=\dfrac{1+2\cos2x}{\sin2x}\\[4px]
&=\dfrac{1+2(1-2\sin^2x)}{\sin2x}\\[4px]
&=\dfrac{\sin3x}{\sin x\sin2x}
\end{align}
for $\sin x\ne0$ using $\sin3x$ formula.
Can you identify $x$ here?
|
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Problem about convergence of operator norm and compactness of an operator. Problem-$1$ Given the sequence of continuous linear operators $T_n : l^2 \to l^2$ defined by $$T_n(x) = (0, 0, \ldots, x_{n+1}, x_{n+2}, \ldots)$$ for every $x \in l^2$. Then for every $x \neq 0$ in $l^2$ i want to check whether $\|T_n\|$ and $\|T_nx\|$ converge to $0$ or not?
Problem-$2$ Let the continuous operator $T: l^2 \to l^2$ defined by $$T_n(x) = (0, x_1, 0, x_3, \ldots,)$$ for every $x = (x_1, x_2, \ldots) \in l^2$. To find whether $T$ is compact or not?
According to me for $(1)$, $$\|T_nx\|^2 = \sum_{k = n+1}^{\infty} \|x_k\|^2$$ Now as $x \in l^2$, so there exist $N \in \mathbb{N}$ such that $$\sum_{k = N+1}^{\infty} \|x_k\|^2 < \epsilon$$ we conclude that $\|T_n(x)\| \to 0$. What about the others? am i correct?
|
What you did for problem 1 is correct (maybe it would be better to say that for any positive $\varepsilon$, there exists an $N$ such that...). Your computation shows that $\left\lVert T_n\right\rVert\leqslant 1$. Considering $x$ as the vector whose $(n+1)$th coordinate is $1$ and all the others are zero, you can show the reverse inequality.
For problem 2, the range of the unit ball contains the collections of elementary vectors $e_{2i+1}$, where $e_j$ is the element of $\ell^2$ whose $j$-th coordinate is $1$ and all the others are zero. Is $\left\{e_{2i+1},i\in\mathbb N\right\}$ compact?
|
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|
Relation between integral, gamma function, elliptic integral, and AGM The integral $\displaystyle\int\limits_0^{\infty}\frac {\mathrm dx}{\sqrt{1+x^4}}$ is equal to $\displaystyle \frac{\Gamma \left(\frac{1}{4}\right)^2}{4 \sqrt{\pi }}$.
It is calculated or verified with a computer algebra system that $\displaystyle \frac{\Gamma \left(\frac{1}{4}\right)^2}{4 \sqrt{\pi }} = K\left(\frac{1}{2}\right)$ , where $K(m)$ is the complete elliptic integral of the first kind. This is in relation to what is called the elliptic integral singular value.
It is also known or verified that
$\displaystyle K\left(\frac{1}{2}\right) =\displaystyle \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt{1-\frac{\sin ^2(t)}{2}}} \, dt= \frac{1}{2} \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt{\sin (t) \cos (t)}} \, dt$.
Can one prove directly or analytically that
$\displaystyle\int\limits_0^{\infty}\frac {\mathrm dx}{\sqrt{1+x^4}} =\frac{1}{2} \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt{\sin (t) \cos (t)}} \, dt =\displaystyle \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt{1-\frac{\sin ^2(t)}{2}}} \, dt = K\left(\frac{1}{2}\right) $ ?
|
Let $t=\frac{1}{{1+x^4}}$ and then
$$ dx=-\frac14(1-t)^{-3/4}t^{5/4} $$
So
\begin{eqnarray}
&&\int\limits_0^{\infty}\frac {\mathrm dx}{\sqrt{1+x^4}}\\
&=&\frac14\int\limits_0^{1}(1-t)^{-3/4}t^{1/4}\mathrm dx\\
&=&\frac14B(\frac14,\frac54)\\
&=&\sqrt{\frac{\pi}{2}}\frac{\Gamma(\frac54)}{\Gamma(\frac34)}\\
&=&\frac{\Gamma \left(\frac{1}{4}\right)^2}{4 \sqrt{\pi}}.
\end{eqnarray}
|
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Pascal's rule : I don't understand why we can do this... So I understand that they are equal, but I can't get my head around the wording. Why can we choose "1 special element, and $k − 1$ from the remaining $n − 1$, or choose all $k$ from the $n − 1$"?
To choose $k$ elements from $n$ we can either choose 1 special element, and $k-1$ from the remaining $n-1$, or choose all $k$ from the $n-1$. This [sic] we have Pascal's Rule $$C_k^n = C_{k-1}^{n-1} + C_k^{n-1}. $$ This leads on to the so-called Pascal's Triangle.
Sorry if this basic but sometimes I have an issue getting around wording I'm hope there is a simpler wording for it!
|
To see the correspondence, note that the number $C_k^n = \binom{n}{k}$ is the number of a teams we can construct of size $k$ with $n$ players. Now assign one of the players as "special". This player can be either included or not included in the team, so if it's included there are $C_{k-1}^{n-1} = \binom{n-1}{k-1}$ ways to choose the rest players. If he's not we can choose the team in $C_k^{n-1} = \binom{n-1}{k}$ ways.
The punchline is that the special player must be included OR not included in the team and they are two disjoint cases, so we can find the number of combinations in both cases and add them to obtain the total sum.
|
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Determine $\lim\limits_{x\to 0, x\neq 0}\frac{e^{\sin(x)}-1}{\sin(2x)}=\frac{1}{2}$ without using L'Hospital How to prove that
$$\lim\limits_{x\to 0, x\neq 0}\frac{e^{\sin(x)}-1}{\sin(2x)}=\frac{1}{2}$$
without using L'Hospital?
Using L'Hospital, it's quite easy. But without, I don't get this. I tried different approaches, for example writing $$e^{\sin(x)}=\sum\limits_{k=0}^\infty\frac{\sin(x)^k}{k!}$$
and
$$\sin(2x)=2\sin(x)\cos(x)$$
and get
$$\frac{e^{\sin(x)}-1}{\sin(2x)}=\frac{\sin(x)+\sum\limits_{k=2}^\infty\frac{\sin(x)^k}{k!} }{2\sin(x)\cos(x)}$$
but it seems to be unrewarding. How can I calculate the limit instead?
Any advice will be appreciated.
|
Hint Let $f(x)=e^{sin(x)}-1$. Then
$$\lim_{x \to 0} \frac{f(x)-f(0)}{x-0}=f'(0)$$
Also,
$$\lim_{x \to 0} \frac{x}{\sin(2x)}$$ can be easily be deduced from the fundamental trigonometric limit.
Alternately canceling $\sin(x)$ you get
$$\frac{e^{\sin(x)}-1}{\sin(2x)}=\frac{\sin(x)+\sum\limits_{k=2}^\infty\frac{\sin(x)^k}{k!} }{2\sin(x)\cos(x)}=\frac{1+\sum\limits_{k=2}^\infty\frac{\sin(x)^{k-1}}{k!} }{2\cos(x)}$$
|
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Convergence/absolute convergence of $\sum_{n=1}^\infty \left(\sin \frac{1}{2n} - \sin \frac{1}{2n+1}\right)$
Does the following sum converge? Does it converge absolutely?
$$\sum_{n=1}^\infty \left(\sin \frac{1}{2n} - \sin
\frac{1}{2n+1}\right)$$
I promise this is the last one for today:
Using Simpson's rules:
$$\sum_{n=1}^\infty \left(\sin \frac{1}{2n} - \sin
\frac{1}{2n+1}\right) = \sum_{n=1}^\infty 2\cos\frac{4n+1}{4n² + 2n}\sin\frac{1}{8n² + 4n}$$
Now, $$\left|2\cos\frac{4n+1}{4n² + 2n}\sin\frac{1}{8n² + 4n}\right| \leq \frac{2}{8n² + 4n}$$
hence by the comparison test, the series converges absolutely, and hence it also converges. Is this correct?
|
$$|\sin(\frac{1}{2n})-\sin(\frac{1}{2n+1})| = |cos(\xi)(\frac{1}{2n}-\frac{1}{2n+1})|$$
The above is followed by mean value theorem, then
$$|\sin(\frac{1}{2n})-\sin(\frac{1}{2n+1})| \leq \frac{1}{2n}-\frac{1}{2n+1}$$
So the series is absolutely convergent.
|
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$z ≤ x + y$ implies $z/(1 + z) ≤ x/(1 + x) + y/(1 + y)$
Suppose $x,y,z $ be nonnegative reals. Show that $z ≤ x + y\implies z/(1 + z) ≤ x/(1 + x) + y/(1 + y)$.
My Proof: If $z=0$, then we are done. So, suppose $z>0$. Since $z ≤ x + y$, and $x$ and $y$ nonnegative, $z ≤ x + y+2xy+xyz$, which leads to $z(1+x+y+xy)\le (x+y+2xy)(1+z)$ and since $z\ne 0$, we get $z/(1 + z) ≤x/(1 + x) + y/(1 + y)$.
Is my proof correct?
|
Yes your proof is correct. I have provided another proof as follows. Notice that $f(t) = \dfrac{t}{1+t}$ is an increasing function on its domain. Therefore if $x$, $y$, and $z$ are non-negative real numbers and $z$ is less than or equal to $x+y$ then f(z) is less than or equal to f(x+y). That is $$\dfrac{z}{1+z} <= \dfrac{x+y}{1+x+y} <= \dfrac{x}{1+x} + \dfrac{y}{1+y}$$
|
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Isometries of $\mathbb R^2$ with maximum norm. I have been asked to prove that all isometries of $\mathbb R^2$ with suprenum norm
$$ \|(x,y)\|_{\infty}=\max \{ |x|,|y| \} $$
Are $T(x,y)=(ax+by, cx+dy)$ where
$b=c=0$ and $a, d \in \{1, -1\}$ or
$ a=d=0$ and $b, c \in \{1, -1\}$.
It is obvious that these are isometries. Why are these all of them.
It seems that by computation we can show these are all of these are isometries. But can we use classic theorems to answer the question?
|
The concept of extreme points helps here, even though it does not involve the norm itself. The definition implies that if $p$ is an extreme point of a set $K$, and $T$ is an invertible linear map, then $T(p)$ is an extreme point of $T(K)$ .
Consider the closed unit ball of this space, which is a square $Q$. Its extreme points are $\{(\pm 1, \pm 1)\}$. Any isometry maps $Q$ onto itself, and therefore maps the extreme points onto extreme points. This leaves only a few choices:
*
*$(1, 1)$ can be mapped to $(1, 1)$, $(1, -1)$, $(-1, 1)$, or $(-1, -1)$.
*$(1, -1)$ has to be mapped to one of those points too, but it can't be the same as the image of $(1, 1)$; it can't be the negative of it, either.
Since $(1, 1)$ and $(1, -1)$ form a basis of the space, an isometry is determined by their images.
This leaves us with at most $4\times 2$ isometries, and you already found as many.
|
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Integrals of form $\int^{∞}_{-∞} x^{2k} \exp(-b^2 (x+x_0)^2)dx$, $k=1,2$. Please can anybody help me to solve these integrals:
$$\int^{∞}_{-∞} x^2 \exp(-b^2 (x+x_0)^2)dx\,,\,\,\int^{+∞}_{-∞} x^4 \exp(-b^2 (x+x_0)^2)dx\;\;?$$
|
One may start with the gaussian integral
$$
\int^{∞}_{-∞} \exp(-b^2 u^2)du=\frac{\sqrt{\pi}}{b},\qquad b>0,
$$ getting, by differentiation with respect to the parameter $b$,
$$
\int^{∞}_{-∞} u^2\exp(-b^2 u^2)du=\frac{\sqrt{\pi}}{2b^3},
$$$$
\int^{∞}_{-∞} u^4\exp(-b^2 u^2)dx=\frac{\sqrt{\pi}}{4b^4}.
$$ Then, by the change of variable
$$
u=x+x_0,\qquad du=dx,
$$ one obtains
$$\int^{∞}_{-∞} x^2 \exp(-b^2 (x+x_0)^2)dx=\int^{∞}_{-∞} (u-x_0)^2\exp(-b^2 u^2)du
$$$$\int^{∞}_{-∞} x^4 \exp(-b^2 (x+x_0)^2)dx=\int^{∞}_{-∞} (u-x_0)^4\exp(-b^2 u^2)du
$$ and one may conclude by expanding the integrand, using the parity and the above results.
|
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What does homomorphism mean in the GLOVE paper? It is mentioned in the GloVe: Global Vectors for Word Representation. It says:
where $w_i$, $w_j$ and $\tilde{w}_k$ are all word vectors and $F$ is just an unknown function. The author then assumes $F$ as $exp$. What is homomorphism exactly? Why equation (4) can make $F((w_i - w_j)^T\tilde{w}_k)$ homomorphism?
|
copying verbatim from Wikipedia
A homomorphism is a map between two algebraic structures of the same
type (that is of the same name), that preserves the operations of the
structures. This means a map ${\displaystyle f:A\to B}$ between two
sets $A$, $B$ equipped with the same structure such that, if $∗$ is an
operation of the structure (supposed here, for simplification, to be a
binary operation), then $${\displaystyle f(x*y)=f(x)*f(y)}$$ for every
pair $x, y$ of elements of $A$
|
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$\lim_\limits{x \to 0}(x\sec x)=0$? $$\lim_{x \to 0}(x\sec x)$$
So putting in $x=0$ you get the answer $0$.
$$\lim_{x \to 0}(x\sec x)=0$$
My question is is this a correct way to solve?
edit : So from the answers below, I've understood that if a function is continuous, then $\lim_{x \to a}f(x)=f(a)$
But how do you figure out if a function is continuous?
from the graph? But what if it's a function that I don't know the graph of?
|
Yes. For continuous functions, you can just plug in the value, because you can switch a limit and a continuous function.
I.e., if $g$ is continuous, then $$\lim_{x \to a} g(f(x)) = g \left(\lim_{x\to a}f(x)\right)$$
provided the limit on the right exists and gives a value on which $g$ is defined.
|
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Need a help in finding the inverse of an operator . The question and part of its answer is given as follows:
13. Let $K$ be an operator of a finite rank on a Hilbert space $H$. For $\varphi \in H$,
$$ K\varphi = \sum_{j=1}^{n} \langle \varphi, \varphi_j\rangle\psi_j. $$
Suppose $\psi_j \in \operatorname{sp}\{ \varphi_1, \cdots, \varphi_n\}^{\perp}$ for $j = 1, \cdots, n$. Prove that $\mathrm{I} + \alpha K$ is invertible for any $\alpha$ and find its inverse.
Solution. Let $\alpha \in \mathbb{C}$, $K'\varphi = \sum_{j=1}^{n} \langle \varphi, \varphi_j\rangle (-\alpha \psi_j)$. By Theorem 7.1,
\begin{align*}
\text{$\mathrm{I} + \alpha K$ is invertible}
&\quad \Leftrightarrow \quad
\text{$\mathrm{I}-K'$ is invertible} \\
&\quad \Leftrightarrow \quad
\det(\delta_{ij}-\langle(-\alpha\psi_j),\varphi_i\rangle)_{i,j=1}^{n}\neq 0.
\end{align*}
But
$$\det(\delta_{ij}-\langle(-\alpha\psi_j),\varphi_i\rangle)_{i,j=1}^{n}
= \det(\delta_{ij}+\alpha\langle\psi_j,\varphi_i\rangle)_{i,j=1}^{n}
= 1
\neq 0$$
(because $\psi_j \in \operatorname{sp}\{ \varphi_1, \cdots, \varphi_n\}^{\perp}$ for $j = 1, \cdots, n$). Thus $\mathrm{I}+\alpha K$ is invertible.
A part of the theorem is given as follows:
7.1. Theorem. Suppose $K \in L(H)$ is of finite rank, say
$$ Kx = \sum_{j=1}^{n} \langle x, \varphi_j\rangle\psi_j. $$
The operator $I-K$ is invertible if and only if
$$\det(\delta_{ij}-\langle\psi_j,\varphi_i\rangle)_{i,j=1}^{n} \neq 0. $$
In this case, for every $y \in H$,
$$ (I-K)^{-1}y = y - \frac{1}{\det(a_{ij})}\det\begin{pmatrix}
a_{11} & a_{12} & \cdots & a_{1n} & \langle y, \varphi_1\rangle \\
a_{21} & a_{22} & \cdots & a_{2n} & \langle y, \varphi_2\rangle \\
\vdots & \vdots & & \vdots & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{nn} & \langle y, \varphi_n\rangle \\
\psi_1 & \psi_2 & \cdots & \psi_n & 0
\end{pmatrix} $$
Then the action of the inverse of the operator on $y$ is $y-I$, but how can this tell me what is the inverse of the operator $K$, could anyone clarify this for me please?
Thanks!
|
Your hypotheses on $\{\psi_1,\dotsc,\psi_n\}$ and $\{\varphi_1,\dotsc,\phi_n\}$ make an appeal to that theorem completely and utterly unnecessary. Since $\{\psi_1,\dotsc,\psi_n\} \subset \{\varphi_1,\dotsc,\phi_n\}^\perp$, it immediately follows that $K^2 = 0$. What, then, is $I^2 - \alpha^2 K^2$, and why should this suggest to you an explicit inverse for $I + \alpha K$?
|
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Limit of recurrence sequence I have to find a limit (or prove it doesn't exist) for the following recurrence sequence.
$a_1 = 2; a_{n+1} = \frac{1}{2}(a_n + \frac{2}{a_n})$
Now I know, in order to find the limit, I first need to prove that the sequence is monotonic and bounded. I've made a partial table of values and concluded that the sequence is decreasing, thus to prove monotonicity, I've written down:
$ a_{n+1} < a_n \rightarrow a_n > \sqrt{2} $
And that's all I could think of. I don't think the inequality above proves anything so I don't know how to continue. I tried to calculate limit of the sequence by using limits of elements as follows:
$ \lim a_{n+1} = \frac{1}{2}(\lim a_n + \lim \frac{2}{a_n}) = a\Rightarrow a = \sqrt{2}$
But without proving monotonicity and bounding, there's no proof the limit exists at all.
Thank you for any help in advance.
|
one prove easy that $$a_n>0$$ for all $n$ then we have by $AM-GM$
$$a_{n+1}=\frac{1}{2}\left(a_n+\frac{2}{a_n}\right)\geq \sqrt{a_n\cdot \frac{2}{a_n}}=\sqrt{2}$$
|
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Equivalent formulations of Baire's Lemma / Theorem I use the following notation $int_Z(A)$ for the interior of $A \subset Z$ in a metric space $Z$.
Now I have the following problem understanding a proof:
Let $(M,d)$ be complete. Then we have that statement i) implies ii).
i) If $M= \bigcup_{j=0}^{\infty}A_j$ with $A_j$ all closed, then there exists a $j_0$ with $int_M(A_{j_0}) \neq \emptyset$.
ii) If $int_M(\bigcup_{j=0}^{\infty}A_j) \neq \emptyset$ with closed $A_j$'s, then there exist a $j_0$ with $int_M(A_{j_0}) \neq \emptyset$
I came as close to understanding the proof as this:
Define the metric space $(X,d)=(\overline{B_{r}(x)},d)$ where $B_{2r}(x) \subset int(\bigcup_{j=0}^{\infty}A_j) \subset \bigcup_{j=0}^{\infty}A_j$ (since it has non-empty interior). Then by using (i) on the sets $X\cap A_j$ we get a $j_0$ with $int_{X}(X\cap A_{j_0})\neq \emptyset$.
Now how do I show that I have $int_M(A_{j_0}) \neq \emptyset$, if I only know that $int_{X}(X\cap A_{j_0})\neq \emptyset$?
Thanks a lot in advance!
|
Let $p\in int_X(X\cap A_{j_0})=S\cap X$ where $S$ is open in $M$. Let $b>0$ such that $B_b(p)\subset S.$
Since $p\in X=\overline {B_r(x)}$ we may take some $q\in B_b(p)\cap B_r(x).$
Let $c>0$ be small enough that $B_c(q)\subset B_b(p)$ and $B_c(q)\subset B_r(x).$
Then $B_c(q)\subset B_b(p)\subset S$ and $B_c(q)\subset B_r(x)\subset X.$ Therefore $B_c(q)\subset S\cap X=int_X(X\cap A_{j_0}).$
Addendum : More simply: In any space $M,$ if $S$ is open in $M$ and $X=\overline Y$, then $S\cap X\ne \phi \implies S\cap Y\ne \phi.$..... So in the Q, with $Y=B_r(X)$ we have $\phi\ne S\cap Y\subset S\cap X=int_X(X\cap A_{j_0}),$ and $S\cap Y$ is open in $M.$
|
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Find: $\lim_{x\to\infty} \frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}.$
Find: $\displaystyle\lim_{x\to\infty} \dfrac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}.$
Question from a book on preparation for math contests. All the tricks I know to solve this limit are not working. Wolfram Alpha struggled to find $1$ as the solution, but the solution process presented is not understandable. The answer is $1$.
Hints and solutions are appreciated. Sorry if this is a duplicate.
|
A fun overkill: it is well known (at least among Ramanujan supporters) that for any $x>1$ we have
$$ \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}} = \tfrac{1}{2}+\sqrt{x+\tfrac{1}{4}} $$
hence $\frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}$ is bounded between $1$ and $\frac{\sqrt{x}}{\sqrt{x+\frac{1}{4}}+\frac{1}{2}}$, whose limit as $x\to +\infty$ is also $1$.
The claim hence follows by squeezing.
|
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Antisymmetry and Totality implies Reflexivity One can check that antisymmetry and totality imply re
exivity. Thus,
a totally ordered set is equivalent to a partially ordered set in which the binary
relation is total.
I am reading Khovanov's notes for Representation theory of finite groups, and confused by this statement. Can someone help me understand this. Thank you very much!
|
Let $X$ be a set and $R \subseteq X \times X$ be a binary relation on $X$. We say that:
*
*$R$ is reflexive if $x \, R \, x$ for all $x \in X$;
*$R$ is antisymmetric if, for all $x, y \in X$, if $x \, R \, y$ and $y \, R \, x$ then $x = y$;
*$R$ is total if, for all $x, y, \in X$, either $x \, R \, y$ or $y \, R \, x$.
If $R$ is total then $R$ is reflexive (notice that antisymmetry is not needed). Indeed, given $x \in X$, by totality (take $y = x$) either $x \, R \, x$ or $x \, R \, x$, i.e. $x \, R \, x$.
|
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Constant of integration change So, sometimes the constant of integration changes, and it confuses me a bit when and why it does. So for example, we have a simple antiderivative such as $$\int \frac{1}{x} dx $$ and we know that the result is $$\log|x| + C$$ and the domain is $$x\in\mathbb R \backslash \{0\} $$ If we want to show all the solutions, we need to do something like $$\begin{cases}
\log x+C_1 & x>0\\
\log(-x)+C_2 & x<0
\end{cases}$$
Do we need to do change the constant every time there is a gap in the domain or is it just when the expression changes? For example, $$ \int \frac {x^5} {x^2-1} dx$$ which is $$ \frac {1} {2} \log |x^2-1| + \frac {x^4} {4} + \frac {x^2}{2} + C$$ and the domain is $$x \in \mathbb R \backslash \{-1,1\}$$ When we want to write all the solutions, is it something like $$ \begin{cases} \ \frac {1} {2} \log (x^2-1) + \frac {x^4} {4} + \frac {x^2}{2} + C_1 & x>1\\ \frac {1} {2}\log (-x^2+1) + \frac {x^4} {4} + \frac {x^2}{2} + C_2 & -1<x<1\\ \frac {1} {2} \log (x^2-1) + \frac {x^4} {4} + \frac {x^2}{2} + C_3 & x<-1 \end{cases}$$ or is it that since the the first and last expression are the same they only have one constant associated? Meaning, the solutions are actually $$ \begin{cases} \ \frac {1} {2} \log (x^2-1) + \frac {x^4} {4} + \frac {x^2}{2} + C_1 & x \in \mathbb R \backslash [-1, 1]\\ \frac {1} {2}\log (-x^2+1) + \frac {x^4} {4} + \frac {x^2}{2} + C_2 & -1<x<1\\ \end{cases}$$
|
Here is a 1-line compactification of Βασίλης Μάρκος' answer $\ddot\smile$
It can be shown that on every sub-interval of the domain of the original function, any two anti-derivatives differ by an additive constant, but constants for different sub-intervals can very well be different.
|
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The identification $(\pi^{-1}\mathcal{G})_p \to \mathcal{G}_{\pi(p)} $ induces a continuous map between the "space of sections" Let $\pi:X\to Y$ be a continuous map and $\mathcal{G}$ be a sheaf on $Y$, then there is a natural isomorphism $f_p:(\pi^{-1}\mathcal{G})_p \to \mathcal{G}_{\pi(p)} $ (by using the adjunction of inverse image and pushforward or whatever means). $f_p's$ will induce a map $f: \bigsqcup_{p\in X}(\pi^{-1}\mathcal{G})_p \to \bigsqcup_{q\in Y}\mathcal{G}_{q}$.
I wonder if this map $f$ is also a continuous map with respect to the "space of section" (espace etale) topology, whose base are sets $\{(x,s_x):x\in U\}$, where $U\subset X$ is open and $s\in \mathcal G(U)$. The main difficulty for me is to find the explicit formula for $f_p$'s.
I ask this because I need this result to prove:
The pullback of the "space of sections" is the same as the "space of sections" of the pullback
|
Yes it is. Moreover, if we let $\mathcal E=\coprod_{q\in Y}\mathcal{G}_q$ be the etale space of $\mathcal G$ and $\mathcal{E}'
=\coprod_{p\in X}\mathcal{G}_{\pi (p)}$ be the etale space of $\pi^{-1}\mathcal G$
then we have a commutative square
$\require{AMScd}$
\begin{CD}
\mathcal{E'} @>f>> \mathcal{E}\\
@V V V @VV V\\
X @>>\pi> Y
\end{CD}
which is a pullback (fibre product) in the category of topological spaces. Indeed we can define the inverse image of sheaf maps by insisting
that $\mathcal{E}'=\mathcal{E}\times_Y X$ be the fibre product.
|
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A question about AP.
How is the encircled step justifiable? According to my knowledge I can substitute m=any variable but how can I substitute m=2m-1, isn't it the same as assuming m=1?
|
The question itself is confusing. What it means is that the condition is true for any $m,n$ you care to choose, rather than a specific $m,n$.
Given that it is true for any $m,n$ it is somewhat confusing to use the same symbols in the suggested answer, as has been done.
Maybe clearer to say that if it is true for any $m,n$ then we can pick the specific values $m=2r-1, n=2t-1$ and go from there.
It doesn't matter what you call your variables - you can give them what names you like. So having worked out an expression in $r,t$, which is true for any $r$ and $t$, it is true in particular for $r=m, t=n$.
The essential thing here is that the expression in $r$ and $t$ has been shown to be true for any choice of $r$ and $t$ (or any choice of interest) and does not depend on the original relation with $m$ and $n$.
|
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The value of $f'(\sqrt{\pi})+g'(\sqrt{\pi}).$ Let $f(x)=(\int_{0}^{x}e^{-t^{2}}dt)^{2}$ and $g(x)=\int_{0}^{1}\frac{e^{-x^{2(1+t^{2})}}}{1+t^{2}}dt.$ Then what is the value of $f'(\sqrt{\pi})+g'(\sqrt{\pi})?$ According to me $g'(\sqrt{\pi})$ is equal to zero. But i dont't know how to find $f'(\sqrt{\pi})$. I am thinking it by fundamental theorem of calculus. Please suggest me to find its value. Thanks.
|
I dont't know how to find $f'(\sqrt{\pi}).$
By applying the chain rule to
$$
f(x)=\left(\int_{0}^{x}e^{-t^{2}}dt\right)^{2}
$$ one gets
$$
f'(x)=2e^{-x^{2}}\int_{0}^{x}e^{-t^{2}}dt
$$ putting $x=\sqrt{\pi}$ gives a result in terms of the error function.
|
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Circular permutation on seating There are 21 people ,15 boys and 6 girls.how many ways are there to seat at least 2 boys between any two adjacent girls in a round table?.
I get my answer 708480.i m wrong ,i think.please help me.
|
Distinguishing only by gender, with the usual assumption of unnumbered seating since it is not explicitly specified otherwise, the problem simplifies to placing $15$ green marbles and $6$ red marbles on an unnumbered circle with the given stipulations.
The $6$ red marbles in a circle will have $6$ gaps or "compartments", in which we can initially place $2$ green marbles each, with the remaining $3$ green marbles placements just enumerated to be $10$:
$|\; 3\; |\; 0\; |\; 0\; |\; 0\; |\; 0\; |\; 0 \; |\; (3)\quad\quad|\; 2\; |\; 1\; |\; 0\; |\; 0\; |\; 0\; |\; 0 \; |\; (2)\quad\quad|\; 2\; |\; 0\; |\; 1\; |\; 0\; |\; 0\; |\; 0 \; |\; (2)$
$|\; 2\; |\; 0\; |\; 0\; |\; 1\; |\; 0\; |\; 0 \; |\; (2)\quad\quad|\; 2\; |\; 0\; |\; 0\; |\; 0\; |\; 1\; |\; 0 \; |\; (2)\quad\quad|\; 2\; |\; 0\; |\; 0\; |\; 0\; |\; 0\; |\; 1 \; |\; (2)$
$|\; 1\; |\; 1\; |\; 1\; |\; 0\; |\; 0\; |\; 0 \; |\; (1)\quad\quad|\; 1\; |\; 1\; |\; 0\; |\; 1\; |\; 0\; |\; 0 \; |\; (1)\quad\quad|\; 1\; |\; 1\; |\; 0\; |\; 0\; |\; 1\; |\; 0 \; |\; (1)$
and the last one,$\;|\; 1\; |\; 0\; |\; 1\; |\; 0\; |\; 1\; |\; 0 \; |\; (1)$
|
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Circles intersecting at two points orthogonally. I am finding the following much harder than it probably is!
If a circle $A$ intersects the circle $B$ at two points orthogonally, then why can't $A$ pass through the centre of B?
|
Let $X $ and $Y $ be the centres of $A $ and $ B $, respectively, and let $Z $ be one of the two points where those circles intersect each other. The triangle $\triangle XZY$ is a right-angled triangle with hypotenuse $XY $, thus $XY $ is the longest side of the triangle: $XY\gt XZ $ and, since $XZ $ is the radius of $A $, the point $Y $ lies outside of $A $
|
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|
find the limit of this function? Evaluate $$\lim_{n\to \infty}\frac{1}{(n!)^{1/n}}={?}$$
My try:
$$\lim_{n\to \infty}\frac{1}{(n!)^{1/n}}=\exp{\lim_{n\to \infty}\frac{-\ln(n!)}{n}}$$
$$=\exp{\lim_{n\to \infty}\frac{-\ln(1\times2\times3\ldots(n-1)\times n)}{n}}$$
$$=\exp{\lim_{n\to \infty}\frac{-(1+2+3+\ldots+(n-1)+n)}{n}}$$
$$=\exp{(-\lim_{n\to \infty}\left(\frac1n+\frac2n+\ldots+\frac{n-1}{n}+\frac nn\right)})$$
$$=\exp(-1)=\frac1e$$
I don't know if my answer is correct please correct me if i am wrong. thanks
|
You were almost there! Remember that: $$1 + 2 + \cdots + n = \frac{n(n+1)}{2}.$$ If you use this, then you will see that the expression inside $\operatorname{exp}$ actually goes to $- \infty$.
|
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Prove the geodesic on 2-sphere is the great circle I want to use the Killing vector fields to prove the geodesic on the sphere is the great circle. First of all, the given metric is $$ds^2=d\theta^{2}+\sin^2\theta d\phi^{2},$$ where I set the radius to be $1$. By Killing equation, $$\nabla_\mu K_\nu+\nabla_\nu K_\mu =0,$$ and some computation, I have the following three Killing vectors:
\begin{align*}
K_1 &= \partial_{\phi} \\
K_2 &= \cos\phi \, \partial_{\theta} - \cot\theta \sin\phi \partial_{\phi} \\
K_3 &= -\sin\phi \partial_{\theta} - \cot\theta \cos\phi \partial_{\phi}
\end{align*}
I want to use the fact that
$$\frac{d}{d\lambda}\left\{K_\mu \frac{dx^{\mu}}{d\lambda}\right\}=0,$$
where $x$ is the curve and $\frac{dx^{\mu}}{d\lambda}$ is the tangent vector. However, as everyone knows, the geodesic is the great circle. Therefore, I thought the final equation would be of the form,
$$aX+bY=0,$$
where $a$ and $b$ are some constant, while $X$ and $Y$ are positions on the sphere. The equation passes the origin. But I can't see how to get the desired results, please give me some help.
|
This should probably be a comment, but I do not yet have the reputation to make a comment. However, I believe your final form of the solution is incorrect. A great circle does not pass through the origin of a sphere: the center of the circle does, but the center of the circle does not belong to the curve. If $X$ and $Y$ are positions on the sphere, then you should have the final form
$$X^2+Y^2=R^2$$
where $R$ is the radius of the sphere. However, I do not think this will tell you much because $X$ and $Y$ can be arbitrary, so long as they belong to the sphere: it is not necessarily true that three points on the sphere can be connected by a great circle. I suggest working in spherical coordinates, and from there, you expect that at the equator of the sphere, a geodesic follows the equator.
|
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|
Plotting $f(x) = x\lfloor 1/x \rfloor$ I want to plot $f(x) = x\lfloor 1/x \rfloor$ near the point zero for finding its limit but I can't choose proper intervals and plot it .
|
I switched to $1/x$ for large positive $x,$ picture gives better idea this way. I would have preferred using a different letter, say $t = 1/x,$ but I do not yet know how to do that. So, think of the graphs as indicating that the limit as $t \rightarrow +\infty \; \; \; \lfloor t \rfloor /t = 1,$ and
$t \rightarrow -\infty \; \; \; \lfloor t \rfloor /t = 1.$ That can then be applied to $x = 1/t$ at $0^+$ and $0^-$
Apparently the intended question is to apply the floor function again. In the graphs below, one can see easily what happens in the positive case and the negative case, when $t$ is an integer and when $t$ is not an integer. Just look at the graphs, think of what the floor function does if applied that one more time. Note: I did try writing an extra floor function in the graph, the computer did not do it correctly.
Next for negative $x$
|
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|
If $G$ is a group of order $250,000 = 2^4 5^6$, show that $G$ is not simple. If $G$ is a group of order $250,000 = 2^4 5^6$, show that $G$ is not simple.
By the Sylow theorem, we have that the number of $2$-sylow subgroups of $G$ $n_2$ satisfy that
$$
n_2 \equiv1\mod2\mbox{ and } n_2\mid5^6
$$
Similarly for $n_5$ we have,
$$
n_5 \equiv1\mod5\mbox{ and } n_5\mid2^4
$$
Hence,
$$
n_2 \in \{1,5,5^2,5^3,5^4,5^5,5^6\}, \mbox{ and } n_5 \in \{1,16\}
$$
But assuming that none of the $n_p'$s are one and using sylow theorems, I can't surpass the order of $G$ as the professor show us in class with one example. Now I am pretty sure I will need another approach but nothing comes to my mind. Any help would be appreciated.
|
As you state, $n_5$ is either $1$ or $16$. If it's the former, we are done. If it is $16$, then there is a homomorphism from $G\to S_{16}$ given by the fact that $G$ acts on the $5$-sylow subgroups by conjugation. But notice that the prime factorization of $|S_{16}|=16!$ contains exactly three copies of $5$ (coming from $5$, $10$ and $15$), but $|G|=2^45^6$ contains six copies of $5$. So this homomorphism cannot be injective so that it's kernel is a normal subgroup of $G$. So $G$ is not simple.
This expounds the well known aphorism that "Groups, as men, will be known by their actions".
|
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|
Will assuming the existence of a solution ever lead to a contradiction? I'm reading Manfredo Do Carmo's differential geometry book. In section 1-7, he discusses the "Isoperimetric Inequality" which is related to the question of what 2-dimensional shape maximizes the enclosed area for a closed curve of constant length. He mentions that
A satisfactory proof of the fact that the circle is a solution to the isoperimetric problem took, however, a long time to appear. The main reason seems to be that the earliest proofs assumed that a solution should exist. It was only in 1870 that K. Weierstrass pointed out that many similar questions did not have solutions.
This line of reasoning would suggest that assuming the existence of a solution might lead to a contradiction (such as an apparent solution that is not in fact valid). Is this actually a problem?
Are there any problems that produce invalid solutions under the (flawed) assumption that a solution exists at all? If so, what is an example and how does it differ from the statement of the isoperimetric problem?
|
A famous example is the existence of a solution for $$p^2 = 2q^2, p,q \in \mathbb Z$$
and $p,q$ share no common prime factors.
It is a rephrasing of a classical proof that $\sqrt{2}$ is irrational by assuming it can be written $\sqrt{2} = p/q$ for two such integers.
One comes to the conclusion that a square integer must have just 1 ( an odd number ) of the prime 2 in it's prime expansion, which is of course impossible, since a square must have each prime occurring an even number of times in it's prime factorization.
|
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|
Intuition behind logarithm change of base I try to understand the actual intuition behind the logarithm properties and came across a post on this site that explains the multiplication and thereby also the division properties very nicely:
Suppose you have a table of powers of 2, which looks like this: (after revision)
$$\begin{array}{rrrrrrrrrr}
0&1&2&3&4&5&6&7&8&9&10\\
1&2&4&8&16&32&64&128&256&512&1024
\end{array}$$
Each column says how many twos you have to multiply to get the number in that column. For example, if you multiply 5 twos, you get $2\cdot2\cdot2\cdot2\cdot2=32$, which is the number in column 5.
Now suppose you want to multiply two numbers from the bottom row, say $16\cdot 64$. Well, the $16$ is the product of 4 twos, and the $64$ is the product of 6 twos, so when you multiply them together you get a product of 10 twos, which is $1024$.
I found that very helpful to understand the actual proofs for this property.
I still struggle to get the idea behind the change of base rule. I'm familiar with the proof that goes like:
$$\log_a x = y \implies a^y = x$$
$$\log_b a^y = \log_b x$$
$$y \cdot \log_b a = \log_b x$$
$$y = \frac{\log_b x}{\log_b a}$$
But can somehow provide a explanation in the style of the quoted answer why this actually works?
|
Intuition is always tricky to get across, but I can try.
$\log_bx$, as you noted, tells you how many $b$s you need to multiply together to get $x$. Now if you need $\log_ba$ number of $b$s to multiply to get $a$, and you need $\log_ax$ number of $a$s to multiply to get $x$, we can "expand" each of those $a$s into a number of $b$s. There will be $\log_ba$ number of $b$s for each $a$, so the total number of $b$s will be $\log_ax \log_ba$. These $b$s multiply to $x$, so $\log_ax \log_ba = \log_bx$.
For example, take $b=2, a=8, x=64$. We start with $\log_ax = 2$, which tell us we need two $8$s to get $64$:
$$
8 \cdot 8 = 64
$$
We use $\log_ba = 3$, i.e., $2 \cdot 2 \cdot 2 = 8$, to expand each $8$:
$$
(2 \cdot 2 \cdot 2) \cdot (2 \cdot 2\cdot 2) = 64
$$
Now the total number of $2$s we are multiplying is $2 \cdot 3 = 6$, so $log_2 64 = 6$
|
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|
Change of variable in Elliptic Curve using Maple I used Maple to get the change of variable for the quartic
v^2 = p^4 - 2p^3 + 5p^2 + 8p + 4
This is the output :
In other words, from the output I obtained from Maple:
x^3-(121/3)x-1690/27+y^2
x=-(1/3)*(5*p^2+24*p-12*v+24)/p^2
y=-(4*(p^3-5*p^2+2*p*v-12*p+4*v-8))/p^3
p=(-72*x-264+36*y)/(9*x^2+30*x-119)
v=(-162*x^4+540*x^3-648*x^2*y+13176*x^2-4752*x*y+62340*x-16488*y+153994)/(81*x^4+540*x^3-1242*x^2-7140*x+14161)
The problem arises with this output is when I rearrange the elliptic curve to become the standard form :
$$y^2 = x^3-(121/3)x-1690/27$$
I noticed that the change of variable x,y,p,v changes as well because I tried substituting them back in the elliptic curve (that I rearranged) and they no longer satisfy the curve. Is there a way to fix this?
Also, I actually did lots of thinking and calculation on this since yesterday and found out (using Sage)
p,x,v,y= var('p x v y')
x=(1/3)*(5*p^2+24*p-12*v+24)/p^2
y= -(4*(p^3-5*p^2+2*p*v-12*p+4*v-8))/p^3
eq1=expand(y^2 - x^3 + (121/3)*x - 1690/27) #Elliptic curve E4
eq=eq1.subs({v: sqrt(p^4 - 2*p^3 + 5*p^2 + 8*p + 4)})
eq.simplify_full() = 0
Explanation : I just changed the negative sign on x to be positive and wrote my elliptic curve as y^2 = x^3 - (121/3)*x + 1690/27.
Next problem : No idea what to do with the change of variable for p, v to satisfy this curve too. Is there a better way to deal with this?
|
I don't understand why you say that the the substitutions produced by the Weierstrassform command do not satisfy the elliptic curve.
The entry k[1] is,
x^3 - (121/3)*x - 1690/27 + y^2
not,
x^3 - (121/3)*x - 1690/27 - y^2
Let's do some substitutions, using those results given by the Weierstrassform command.
restart;
f := v^2 - ( p^4 - 2*p^3 + 5*p^2 + 8*p + 4 ):
with(algcurves):
k := Weierstrassform(f,p,v,x,y):
lprint( k[1] );
x^3-(121/3)*x-1690/27+y^2
map( lprint, [ x=k[2], y=k[3], p=k[4], v=k[5] ] ):
x = -(1/3)*(5*p^2+24*p-12*v+24)/p^2
y = -4*(p^3-5*p^2+2*p*v-12*p+4*v-8)/p^3
p = (-72*x-264+36*y)/(9*x^2+30*x-119)
v = (-162*x^4+540*x^3-648*x^2*y+13176*x^2-4752*x*y+62340*x-16488*y+153994)/(81*x^4+540*x^3-1242*x^2-7140*x+14161)
lprint( solve(f, v) );
(p^4-2*p^3+5*p^2+8*p+4)^(1/2), -(p^4-2*p^3+5*p^2+8*p+4)^(1/2)
# Substitute for x and y in k[1], then substitute using
# v=solve(f, v)[1], and then simplify.
simplify( eval(eval(k[1], [x=k[2], y=k[3]]), [v=solve(f, v)[1]]) );
0
# Substitute for x and y in k[1], then substitute using
# v=solve(f, v)[2], and then simplify.
simplify( eval(eval(k[1], [x=k[2], y=k[3]]), [v=solve(f, v)[2]]) );
0
# Substitute for x and y in k[1], then simplify using
# f=0 as a side-relation.
simplify( eval(k[1], [x=k[2], y=k[3]]), {f} );
0
# Substitute for p and v in f, then simplify using
# k[1]=0 as a side-relation.
simplify( simplify( eval(f, [p=k[4], v=k[5]]) ), {k[1]} );
0
|
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|
Proving that the function $f(x) = \int_0^\infty \cos (w^3/3 - x w ) d w$ satisfies the equation $f'' + x f = 0$ It is assumed that $x$ is real.
Formally, we have
$$ f'' = \int_0^\infty -\cos (w^3/3 - x w ) w^2 d w , $$
and hence
$$f'' + x f = \int_0^\infty \cos (w^3/3 - x w ) (-w^2 + x ) d w \\ = -\int_0^\infty \cos (w^3/3 - x w ) d(w^3/3 - x w ) \\ = - \sin(w^3/3- x w ) |_0^\infty . $$
The problem is that $\sin(w^3/3- x w ) $ does not converges as $w\rightarrow \infty$.
Apparently, the problem is rooted in the fact that the expression for $f''$ is not well defined---it does not converge.
So, could anyone give a simple, elementary proof?
|
Your derivation is fine and leads to an ambiguous form, as you state. To disambiguate this, consider the following.
Your integral is
$$
-\int_0^\infty \cos (w^3/3 - x w ) d(w^3/3 - x w )
$$
If you make the transformation $t = w^3/3 - x w$ you have, for any given (finite) $x$,
$$
-\int_0^\infty \cos (t) d t
$$
which is a special case of
$$
I(\nu) = -\int_0^\infty \cos (\nu \, t) d t
$$
for $\nu = 1$.
By symmetry, this is
$$
I(\nu) = -\frac12 \int_{-\infty}^\infty \cos (\nu \, t) d t = -\frac12 \Re {\Large(} \int_{-\infty}^\infty e^{j \nu \, t} d t \Large)
$$
, i.e. the real part of a complex integral.
To disambiguate your expression you can now interpret the last integral in a distribution sense. Indeed, we have the Fourier transformation of the delta distribution which can be (symbolically) stated as
$$
\delta (\nu ) = \int_{-\infty}^\infty e^{j \nu \, t} d t
$$
Hence
$$
I(\nu) = -\frac12 \Re {\Large(} \delta (\nu ) \Large)
$$
and in particular, the desired
$$
I(\nu = 1) = 0
$$
Indeed, many treatments of the Airy function (we have the integral representatin of the Airy function here) go via the Fourier Transform.
Happy New Year 2018!
|
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|
Show that $f(x) = x^7 + x^5 + x^3 + x$ is bijective I want to show that the real polynomial function $f: \mathbb R \to \mathbb R, f(x) = x^7 + x^5 + x^3 + x$ is bijective. I want to show this without using the inverse or the derivative.
I'm struggling to prove injectivity, because I see no easy way to arrive at $x = y$. What I have so far:
Surjective:
Because the degree is odd, we have $\lim_{x \to +\infty}(f(x)) = +\infty$ and $\lim_{x \to -\infty}(f(x)) = -\infty$. Because a polynom is continuous, we can apply the IVT to the interval $I = (-\infty,+\infty) = \mathbb R$ so that for every $y \in I$ there is a $x$ such that $f(x) = y$.
Injective:
Let $f(x) = f(y)$. Then $x^7 + x^5 + x^3 + x = y^7 + y^5 + y^3 + y$. Then ??? , so $x = y$.
|
Hint: This is an odd function.
|
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find $a_1+a_3+a_5+\cdots+a_{37}+a_{39}$
Let $(1+x-2x^2)^{20} = \sum_{r=0}^{40} a_r x^r.\;\;$ Find
$$a_1+a_3+a_5+\cdots+a_{37}+a_{39}$$
I mean, at first, I don't have any idea how this can be solved. I tried factorizing the LHS: $((2x-1)(1-x))^{20}$ and then using binomial theorem, I get something like:
$$\Biggl(\binom{20}0(2x)^{20}-\binom{20}1(2x)^{19}+\binom{20}2(2x)^{18}-\binom{20}3(2x)^{17}+\cdots-\binom{20}{19}(2x)+1\Biggr)\Biggl(\binom{20}0(x)^{20}-\binom{20}1(x)^{19}+\binom{20}2(x)^{18}-\binom{20}3(x)^{17}+\cdots-\binom{20}{19}(2x)+1\Biggr)$$
Expanding the RHS:
$$a_0+a_1x^1+a_2x^2+\cdots+a_{39}x^{39}+a_{40}x^{40}$$
But I don't get what to do after this. Taking $x$ common doesn't help, nor in the former equation($2x$).
So how do we do it? A hint would be nice.
|
thanks guys, i have solved it too.
$$(1+x-2x^2)^{20} = \sum_{r=0}^{40} a_r x^r$$
putting $x = 1$
$$\sum_{r=0}^{40} a_r=0\;\;\;\;\;\;\;\;\;\;\;\;\;-(1)$$
putting $x = -1$
$$-a_1+a_2-a_3+a_4-a_5+\cdots+a_{40}=2^{20}\;\;\;\;\;\;\;\;\;\;\;\;\;-(2)$$
calculating $(1)-(2)$
$$a_1+a_3+a_5+a_7+\cdots+a_{39}=-2^{19}$$
|
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|
linearly independent elements of W I want to ask about the number of linearly independant elements of
$W=(u_{1},u_{2},...,u_{m},u_{1}-u_{2},u_{2}-u_{3},...,u_{m-1}-u_{m},u_{m}-u_{1})$.
I say that it is $m$ and i need someone to confirm or not that.
Thanks
|
The answer depends on the number of linearly independent vectors among $u_1, \ldots, u_m$.
Notice that $$\operatorname{span}\{u_1, u_2, \ldots, u_m, u_1 - u_2, u_2 -
u_3, \ldots, u_{m-1} - u_m, u_m - u_1\} = \operatorname{span}\{u_1, u_2, \ldots, u_m\}$$
because all $u_i - u_{i+1}$ are linear combinations of $u_1, u_2, \ldots, u_m$.
Thus, the number of linearly independent vectors in $W$ is equal to the number of linearly independent vectors in $\{u_1, u_2, \ldots, u_m\}$.
In particular, assuming $u_1, u_2, \ldots, u_m$ are linearly independent, the number or linearly independent vectors in $W$ is $m$.
|
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|
Spectral radius is not matrix norm. I have seen an example of matrix
$$A = \begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix}$$
whose spectral radius is zero therefore the spectral radius is not matrix norm. Why the spectral radius is not matrix norm in this case Is it possible that $\|A\|=\epsilon$?
|
From this Wikipedia page, the spectral norm of a matrix $A\in\mathbb{C}^{n\times n}$ is defined as
$$\rho\left(A\right)=\max_{1\leq i\leq n}\left\{\left|\lambda_{i}\right|\right\}$$
where the $\lambda_{i}$'s are the eigenvalues of the matrix. In your case
$$A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$$
is triangular, so the diagonal entries are the eigenvalues. Thus you have $\lambda_{1}=\lambda_{2}=0$, and it follows immediately that
$$\rho\left(A\right)=0$$
This is not a norm since $A\neq 0$ and a norm $\left\|\cdot\right\|$ must satisfy
$$\left\|A\right\|=0\:\Longleftrightarrow\: A=0$$
by definition.
|
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Given $x_1 := \sqrt{2}$ and $x_{n+1} :=\sqrt{2x_n} $, prove $\sqrt{2} ≤ x_n ≤ 2$
Let $x_1 := \sqrt{2}$ and $x_{n+1} :=\sqrt{2x_n} $ for all $n \in \mathbb{N}$.
By using proof by induction:
(i) Prove that $\sqrt{2} ≤ x_n ≤ 2$ for all $n \in \mathbb{N}$.
(ii) Prove that $x_n ≤ x_{n+1}$ for all $n \in \mathbb{N}$.
For (i)
Let's prove the case base case, for $n=1$ we have that $\sqrt{2} \leq x_1\leq 2$ which is clearly true by $x_1=\sqrt{2}$
Now we assume true for $n=k$ $\Longrightarrow$ $\sqrt{2}\leq x_k\leq2$ and we are required to prove $\sqrt{2}\leq x_{k+1} \leq 2$.
From our hypothesis : $\sqrt{2}\leq x_k\leq2$ however we know that $x_{k+1} =\sqrt{2x_k}\Longrightarrow x_k = \frac{(x_{k+1})^2}{2}$ so we have $\sqrt{2}\leq \frac{(x_{k+1})^2}{2}\leq2 \Leftrightarrow \sqrt{2\sqrt{2}}\leq x_{k+1} \leq 2$.
How do I proceed?
|
From
$$\sqrt2<x_n<2$$ you can deduce
$$\sqrt{2\sqrt2}<\sqrt{2x_n}<2$$
and obviously
$$\sqrt{2}<\sqrt{2x_n}<2,$$ which is nothing but
$$\sqrt{2}<x_{n+1}<2.$$
|
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|
Is there formulas for $\sum_{g\in G}|\operatorname{Fix}(g)|^n$? I saw a problem in Show $r(x)|G|=\sum_{g\in G}|Fix(g)|^2$ where $(G,X)$ is a transitive group and $r(x)$=#{different orbits of X under $Stab(x)$}
Let $(G,X)$ be a transitive group action, that is, for any $x,y\in X$,there exists an element $g\in G$ s.t. $g*x=y$. Fix $x\in X$. Let $r(x)$ be the number of different orbits of $X$ under $\operatorname{Stab}(x)$. Show that $r(x)|G|=\sum_{g\in G}|\operatorname{Fix}(g)|^2$.
Can this be generalized such that if the assumptions stays on the same and $n\in \mathbb{N}$ then $\sum_{g\in G}|\operatorname{Fix}(g)|^n$ can be represented in some closed form with respect to the variables $r(x)$ and $|G|$? If there is not known general solution, what do we know about some fixed powers of $n$?
|
$|\text{Fix}(g)|^n$ is the number of fixed points of $G$ acting diagonally on $X^n$. By Burnside's lemma, it follows that
$$\frac{1}{|G|} \sum_{g \in G} |\text{Fix}(g)|^n$$
is the number of orbits of $G$ acting on $X^n$.
If $G$ acts transitively on $X$ then we can count the number of orbits of $X^2$ as follows. First, $X^2$ splits into the diagonal elements $(x, x)$, which consist of one orbit, and then the off-diagonal elements $(x, y), y \neq x$. Since $G$ acts transitively we may pick representatives of each orbit of the form $(x_0, y)$ for any fixed $x_0 \in X$. By orbit-stabilizer the number of such orbits is then the number of orbits of $y$ under the action of $\text{Stab}(x_0)$.
You can do a similar but more complicated analysis for $X^n$ by breaking its orbits up according to which elements are equal to which other elements (equivalently, breaking it up according to the orbits of the action of $\text{Sym}(X)$), which shows in particular that the above sum is at least the number of partitions of $\{ 1, 2, \dots n \}$ into at most $|X|$ nonempty subsets (this is a sum of Stirling numbers of the second kind $S(n, k), k \le |X|$), with equality iff the action of $G$ is $k$-transitive for $k \le n$.
|
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|
Why is this combination of nearest-integer functions --- surprisingly --- continuous? Alright, I didn't know the best way to formulate my question. Basically, whilst doing some physics research, I naturally came upon the function
$$ f(x) = 2x[x] - [x]^2 $$
where I use $[x]$ as notation for the `nearest-integer function' (i.e. rounding off). Usually this function has to have a caveat of how we exactly define the value for $x \in \frac{1}{2} \mathbb Z$, but interestingly for this function it does not matter, since it turns out to be continuous! In fact, it turns out $f(x)$ is exactly given by the glued function of taking all the tangent lines of $x^2$ at integer values of $x$:
(Note: due to properties of $x^2$, the tangent lines exactly intersect at half-integer values of $x$.)
So my question is not literally `why is it continuous?', but rather: considering it is continuous, and considering that that is not a generic property of functions which are defined in terms of nearest-integer functions, is there a better (i.e. more insightful) way of expressing $f(x)$? Relatedly, is there some part of mathematics where functions similar to these naturally arise?
|
Let $g(x,y)$ be any continuous function such that $g(x,-\frac12)=g(x,\frac12)$. Then $f(x)=g(x,x-[x])$ is continuous.
In particular, your function is given by $g(x,y)=x^2-y^2$. Consequently, we can write $f(x)=x^2 - (x-[x])^2$.
|
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|
How to estimate a limit value of ordinary differential equations $$\frac{dx}{dt}=(x-y)(1-x^2-y^2)\\
\frac{dy}{dt}=(x+y)(1-x^2-y^2)$$
Initial condition $x(0),y(0)$ are nonzero real numbers.
How to estimate the solution $x(t)$ as $t \to \infty$?
|
Note that if $(x(0),y(0))= (0,0)$ or $x^2(0)+y^2(0) = 1$, then
$(x(t),y(t))=(x(0),y(0))$ for all $t$.
Let $s(t) = x^2(t)+y^2(t)$, then
$\dot{s} = f(s)=2(1-s)s$.
Note that $f(0)=f(1) = 0$, if $s \in [0,1]$ then $f(s) \ge0$ and
if $s \ge 1$ then $f(s) \le 0$.
In particular, if $s(0) >0$ then $\lim_{t \to \infty} s(t) = 1$. In fact, since this is a one dimensional ODE, we know that if $s(0) \in (0,1]$ then $\lim_{t \uparrow \infty} s(t) = 1$ and if
$s(0) \ge 1$ then $\lim_{t \downarrow \infty} s(t) = 1$.
We can find an explicit estimate for the convergence of $s(t)$. Note that for $s\ge 1$ we have $f(s) \le 2(1-s)$ and so $\dot{(s-1}) \le -2 (s-1)$ which gives $0 \le s(t)-1 \le (s(t_0)-1) e^{-2(t-t_0)}$ for $s(t_0) \ge 1$.
Note that for any $\alpha <2$ we can find some $\delta>0$ such that if $s \in [1-\delta,1]$ then $f(s) \ge \alpha (1-s)$. Hence if $s(t_0) \in [1-\delta,1]$, we have $0 \le 1-s(t) \le (1-s(t_0)) e^{-\alpha(t-t_0)}$.
Hence for sufficiently large $t_0$ we have the estimate
$|1-s(t)| \le |1-s(t_0)| e^{-\alpha(t-t_0)}$.
If we let $w=(x,y)$ then we can write the ODE as
$\dot{w} = (1-\|w\|^2) Aw$ for some matrix $A$ and we are given $w(0) \neq 0$. Let $s(t)= \|w(t)\|^2$, then we note that $s(t) \to 1$ hence
the solution is bounded, and so there is some $K$ such that
$\|\dot{w}\| \le K \|A\| |1-s(t)|$, in particular $t \mapsto \|\dot{w}(t)\|$ is integrable and hence $w$ converges.
|
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Proper subset of the set of irrationals such that it is countable and dense in $\Bbb R$ We know that $\Bbb R$ is separable i.e. it contains a dense subset which is countable. We have $\Bbb Q$ and ${\Bbb R} - {\Bbb Q}$ to be dense subsets respectively countable and uncountable. I was looking for a countable dense subset of $\Bbb R$ which is a proper subset of either (i) $\Bbb Q$ or (ii) ${\Bbb R} - {\Bbb Q}$ .
For (i), by considering the set of dyadic rationals i.e. $\{\frac{a}{2^b} | a \in \Bbb Z , b \in \Bbb N \}$ or more generally for any fixed prime $p \in \Bbb N$, consider, $\{\frac{a}{p^b} | a \in \Bbb Z , b \in \Bbb N \}$ .
It is a countable proper subset of $\Bbb Q$ which is dense.
But I could not come up with any example for (ii) . Thanks in advance for help.
|
You may also pick a countable dense set whose elements are all trascendental (and in such a way that this set is not the set of all rational multiples of some fixed transcendental, which would be the right boring answer). As algebraic numbers are countable, the set $T$ of all transcendentals is dense in $\mathbb{R}$. "Having a countable base" is a hereditary topological property, so the space $T$ of transcendentals has a countable base. Spaces with a countable base are separable, that is, have a countable dense subset (just pick one point in each basic open set). This means that $T$ contains a countable dense subset $D$. As $D$ is dense in $T$ and $T$ is dense in $\mathbb{R}$, $D$ is dense in $\mathbb{R}$. So, $D$ is a countable dense subset of the real line whose elements are all transcendental. From time to time I pose such question to my students...
|
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|
Finding the limit of the sequence $a_n\cdot a_{n+1}=n,\,n=1,2,3,\cdots.$ Let $(a_n)_{n>=1}$ be a sequence of real numbers defined by the below recurrence relation:
$$a_n\cdot a_{n+1}=n,\quad n=1,2,3,\cdots.$$
Prove that $\lim_{n\to \infty}a_n=+\infty.$
Edit: $a_1>0$
|
Since $a_n\cdot a_{n+1}=n$ and $a_{n+1}\cdot a_{n+2}=n+1$,
$$
\frac{a_{n+2}}{a_n}=\frac{n+1}n\ge\sqrt{\frac{n+2}n}
$$
Therefore, for even $n$
$$
\begin{align}
a_n
&\ge a_2\sqrt{\frac n2}\\
&=\frac1{a_1}\sqrt{\frac n2}
\end{align}
$$
and for odd $n$
$$
a_n\ge a_1\sqrt{n}
$$
Thus,
$$
\bbox[5px,border:2px solid #C0A000]{a_n\ge\sqrt{n}\min\!\left(a_1,\frac1{a_1\sqrt2}\right)}
$$
|
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|
Why Do Mersenne Primes Only Occur at Terms of Prime Index? A Mersenne prime is a prime of the form $2^n-1$.
Only when $n$ is a prime itself is there a chance that $2^n-1$ is a Mersenne primes. The largest primes discovered are almost always Mersenne primes. Some of the more known Mersenne primes are $3, 7, 31, 127$, e.t.c.
Now on to the question.
Why do non-prime values of $n$ never yield a prime?
I have always heard from my teachers that Mersenne primes occur at only the prime values of $n$ but no one ever explained it to me. Is there any way of proving this? Or are there any exceptions for $n>1$?
P.S. As you may have guessed from my writing "teachers" instead of "professors", I am only in grade $10$ and not that skilled so I would prefer if you could give me simple explanations. Thanks in advance!
|
Because if $m=kl$, with $k,l>1$, then\begin{align}2^n-1&=2^{kl}-1\\&=(2^k)^l-1^l\\&=(2^k-1)\bigl((2^k)^{l-1}+(2^k)^{l-2}+\cdots+1\bigr).\end{align}
|
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|
Discuss the monotonicity of $\sqrt[n]{n!}$. It seems that $\sqrt[n]{n!}$ is increasing because it turns to $+\infty$ as $n\to+\infty$. How to prove it?
|
It is enough to show that $\frac{1}{n}\log(n!)=\frac{1}{n}\sum_{k=1}^{n}\log(k)$ is increasing.
Since $\log(x)$ is concave on $\mathbb{R}^+$, this is a simple consequence of Karamata's inequality:
$$ \tfrac{1}{n}\cdot\log(1)+\tfrac{1}{n}\cdot\log(2)+\ldots+\tfrac{1}{n}\cdot\log(n)+0\cdot \log(n+1)\\ \leq \tfrac{1}{n+1}\cdot\log(1)+\tfrac{1}{n+1}\cdot\log(2)+\ldots+\tfrac{1}{n+1}\log(n+1).$$
As an alternative approach, $n!^{n+1}\leq (n+1)!^n$ is equivalent to $n!\leq (n+1)^n$ which is trivial.
|
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|
Simplify this equation if $x$ is negative I am trying to simplify $7 \cdot\sqrt{5x}\cdot\sqrt{180x^5}$, given that $x$ is negative. The answer is $-210x^3$, but I am getting $210x^3$. Below is my reasoning:
For a $ k > 0 $, let $ k = - x $. Then, $7\cdot\sqrt{5x}\cdot\sqrt{180x^5}$ = $7\cdot5\cdot6\cdot i\sqrt{k} \cdot i\sqrt{k^5}$ = $210\cdot(-1)\cdot k^3$ = $ 210 \cdot (-1) \cdot k^3 = 210\cdot (-1) \cdot (-x)^3 = 210x^3$.
What am I doing wrong?
Any help is greatly appreciated
|
$$\text{$7 \cdot\sqrt{5x}\cdot\sqrt{180x^5} \ $ where $ \ x<0$}$$
Lets let $x=-1$ to see what we should expect.
\begin{align}
\left. \left( 7 \cdot\sqrt{5x}\cdot\sqrt{180x^5}\right)\right|_{x=-1}
&= 7 \cdot\sqrt{-5}\cdot\sqrt{-180} \\
&= 7 \cdot i \cdot\sqrt{5}\cdot i \cdot \sqrt{180} \\
&= -7 \cdot \sqrt{900} \\
&= -210
\end{align}
Now Lets do it with $x=-n$ where $n > 0$.
\begin{align}
\left. \left( 7 \cdot\sqrt{5x}\cdot\sqrt{180x^5}\right)\right|_{x=-n}
&= 7 \cdot\sqrt{-5n}\cdot\sqrt{-180n^5} \\
&= 7 \cdot i \cdot\sqrt{5n}\cdot i \cdot \sqrt{180n^5} \\
&= -7 \cdot \sqrt{900n^6} \\
&= -210n^3 \\
&= -210(-x)^3 \\
&= 210x^3
\end{align}
|
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|
Sum of exactly n perfect square divisors I'm doing a Number Theory question, and if someone could offer a hint, that would be greatly appreciated;
The question is: Find the sum of the perfect square divisors of the smallest integer with exactly 6 perfect square divisors.
My reasonings:
*
*My method so far has been simply listing out pairs of possible exponents that are multiples of 2, and listing them out, or using combinations to calculate the total number of combinations. Obviously, this brute-force method is not working out well, and if anyone could provide a hint, that would be greatly appreciated.
*I know how to find the product of all divisors, but not their sum. Furthermore, how would one "target" only perfect square divisors? Perhaps complementary counting?
|
Conversation in the comments is correct, but I would suggest the following approach. First we find the lowest number with exactly $6$ divisors, and then we square that number. This squared number will then have exactly $6$ square divisors.
To find the number, we use the number of divisors formula
$$d\left(\prod p_i^{a_i}\right) = \prod(a_i +1)$$
Now the divisors of $6$ are $6,1$ and $2,3$, so now we choose the lowest primes and attach these exponents to them as follows.
$$2^5=32 \quad 2^2*3=12$$
Now we have the lowest number with exactly $6$ divisors, $12$. So the lowest number with 6 square divisors is $144$. The square divisors are the squares of the divisors of $12$, so now we add those up to get our answer.
$$1^2+2^2+3^2+4^2+6^2+12^2=210$$
|
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What is the density of a brownian bridge $X_t=B_t-tB_1$? The original question is:
For $t\in [0,1]$, we define $X_t=B_t-tB_1$, where $\{B_t:t\geq 0\}$ is a standard Brownian motion. Find the density of $X_t$ .
After reading several resources, I think $X_t$ is a normal distribution. However, since most of the books concerning on the process itself other than the distribution of $X_t$, can someone confirm this?
|
Yes, it is normal. Brownian motion is a Guassian process, which means that its finite dimensional distributions are all multivariate normal, and in particular any linear combination of $B_{t_1},\ldots,B_{t_n}$ is normal for any indices $t_1,\ldots,t_n$.
To find the density of $X_t$, we need only calculate its mean and variance. Its mean can be found as
$$E(X_t)=E(B_t)-tE(B_1)=0$$
and its variance can be found by calculating its second moment using $E(B_sB_t)=s\wedge t$:
$$\operatorname{Var}(X_t)=E(X_t^2)=E(B_t^2)-2tE(B_tB_1)+t^2E(B_1^2)=t-2t^2+t^2=t(1-t).$$
Hence for $t\in(0,1)$, $X_t$ is $\mathcal N(0,t(1-t))$ and so has density
$$f(x)=\frac1{\sqrt{2\pi t(1-t)}}e^{-x^2/2t(1-t)}.$$
|
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$f : \mathbb{R} \to \mathbb{R}$ continuous and positive. Show $\lim\limits_{t\to 0^+} \frac{1}{t^3} \int_t^{t+t^2} xf(x)dx =f(0)$
$f : \mathbb{R} \to \mathbb{R}$ continuous and positive. Show $$\lim_{t\to 0^+} \frac{1}{t^3} \int_t^{t+t^2} xf(x)dx =f(0)$$
$f$ is continuous so $\forall \epsilon >0\,, \exists \delta > 0, |x| < \delta \implies |f(x) -f(0)| < \epsilon$
And, $f$ is continuous on $[t,t+t^2]$ so $\exists M>0, |f(x)| < M, x \in [t,t+t^2] $
As $f$ is positive, I can't wlog take $f(0)=0$.
I've tried two approaches:
1) $\bigg | \frac{1}{t^3} \int_t^{t+t^2} xf(x) - f(0) dx \bigg|$
2) $\bigg | \frac{1}{t^3} \int_t^{t+t^2} xf(x) dx \bigg| = \bigg | \frac{1}{t^3} \int_t^{t+t^2} xf(x)dx - \frac{1}{t^3} \int_t^{t+t^2} xf(0)dx + \frac{1}{t^3} \int_t^{t+t^2} xf(0)dx \bigg|$
The first seems to fail because I can't get to $x(f(x)-f(0))$ nicely and the second seems to fail because it doesn't seem to get me to a situation where I have $f(0) + \epsilon$ remaining.
|
Hint: $$\frac1{t^3}\int_t^{t+t^2}x\,dx=\frac1{2t^3}((t+t^2)^2-t^2)
=1+\frac t2,$$so $$f(0)-\frac1{t^3}\int_t^{t+t^2}xf(x)\,dx
=-t\frac{f(0)}{2}+\frac1{t^3}\int_{t}^{t+t^2}x(f(0)-f(x))\,dx.$$
|
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|
What is the sideways tension on a hanging chain under gravity? I'm sure you are familiar with the question of deducing the shape of a free hanging chain but if not, here's my solution so far summarised:
By taking $\tan(\theta)$ to be $f'(x)$ and considering the forces on a
general particle, one can show that the shape of a free hanging chain
obeys $f''^2(x) = (1+f'^2(x))k^2$ whose solution is
$\frac{1}{k}\cosh(kx)$ and $k = \frac{g\rho}{T_0}$ where g is
gravitational acceleration, $\rho$ is length-density $(pL = m)$ and $T_0$ is
sideways tension (which must remain constant otherwise particles in
the chain would accelerate sideways). The general solution to the diff
equation is also works for sideways and vertical shifts of this
function but these can be set to zero.
Now my problem is finding $T_0$ given initial conditions. Note that the differential equation is not affected by changing the distance between the two endpoints (call it $2x_0$) so that is also an initial condition.
I tried looking at the endpoint of the chain where $2T_0 = \frac{mg}{\sinh(kx_0)}$ but this leads to an equation which gives a nonsensical answer when $kx_0$ approaches zero. If $kx_0$ approaches zero then $2T_0 = \frac{mgx_0}{gm/(LT_0)}$ so $x_0 = L/2$, which seems nonsensical. Maybe this is because $T_0$ becomes small with $x_0$, preventing the $kx_0$ approaching zero? A problem I'm having is that there are no real numerical solutions when I plug this function into wolfram alpha. Please help me
|
The answer on Physics SE, linked by cgiovanardi in a comment, considers a non-symmetric configuration, where the two ends are at different heights. If you are only interested in the symmetric case, it is somewhat simpler than that, though you can still apply the same approach, and in the end the equation is transcendental anyway.
The usual method is to consider the length of the chain, which is also an initial condition. From the length, together with distance between endpoints, you can get the $k$ - that's just the geometry of the curve. Then $k$ tells you what $T_0$ is.
Your approach doesn't do that explicitly, but still gets the same result because $m = \rho L$, so you end up with the same equation (though you have a subtle mistake in that the coordinates of the endpoints are $\pm x_0/2$ with your definition of $x_0$):
$$ \sinh\frac{kx_0}2 = \frac{kL}2 \;\;\;\Leftrightarrow\;\;\; \frac{\sinh(kx_0/2)}{kx_0/2} = \frac{L}{x_0} $$
Notice that $x_0 \rightarrow L \Rightarrow kx_0 \rightarrow 0$ but it still makes perfect sense - that is a too-tightly-stretched chain with $k \rightarrow 0 \Rightarrow T_0 \rightarrow \infty$; while as $x_0 \rightarrow 0$, you just get a vertically hanging doubled-up chain, which will not have a horizontal component of tension: $kx_0 \rightarrow \infty \Rightarrow k \rightarrow \infty \Rightarrow T_0 \rightarrow 0$.
|
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|
Proving particular solution in differential equation
Given the second-order ordinary differential equation:
$$
{y}''+y=f(x)
$$
prove that:
$$
y_p(x)=\int_{0}^{x}f(u)\sin(x-u)du
$$
is the particular solution of the equation.
I know this is homework but I've been trying to solve it for the past few days and I can't. I even asked my teacher for help but he doesn't answer.
Thanks.
|
Hint. We assume our $f$ is nice enough to be allowed to use the Leibniz rule,
$$
\frac{d}{dx} \left (\int_{0}^{b(x)}f(x,u)\,du \right) = f\big(x,b(x)\big)\cdot \frac{d}{dx} b(x) + \int_{0}^{b(x)}\frac{\partial}{\partial x} f(x,u) \,du
$$ giving here
$$
\begin{align}
y'_P(x)&=\frac{d}{dx} \left (\int_{0}^{x}f(u)\sin(x-u)\,du \right)
\\\\&= 0+ \int_{0}^{x}f(u) \cos(x-u) \,dt
\end{align}
$$ then differentiate once more using the same tool and see what happens.
|
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|
How many tries on average before I see the same value $N$ times in a row? If I repeatedly roll a fair, $X$-sided die, on average how many rolls should I expect to make before I happen to roll the same value $N$ times in a row?
I've found questions on here with answers when $X=2$, or where $N=2$, or where you're looking for a specific result $N$ times in a row; I'm interested in the situation where I don't care what the specific value is, I just want it to be $N$ times in a row. The closest I can find is Suppose we roll a fair $6$ sided die repeatedly. Find the expected number of rolls required to see $3$ of the same number in succession., but I can't quite figure out how to generalize it beyond $N=3$.
|
$N_1$ - No of tosses for 1st Heads
$N_2$ - No of tosses for 2 consecutive Heads
$N_3$ - No of tosses for 3 consecutive Heads
$\mathbb E[N_1] = 2$
$\mathbb E[N_2] = (1+\mathbb E[N_1])2 = 6$
$\mathbb E[N_3] = (1+\mathbb E[N_2])2 = 14$
A general formula can be created
$\mathbb E(1) = X$
$\mathbb E(2) = (X+1)X$
$\mathbb E(3) = \big((X+1)X+1\big)X$
and so on.
$$\mathbb E(N) = \sum_{i=1}^N X^i = \frac{X(X^N-1)}{X-1} $$
I have made a video explanation and made a simple formula at the end
https://youtu.be/d72rcsBblRE
|
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|
If $x_n\to y $ and $d(x_n, x)<\epsilon$, $\forall x_n$. Is it true $d(y, x)<\epsilon$? Let the sequence $\{x_n\}_{n=0}^\infty$ be given and for all $x_n$, we have $d(x_n, x)<\epsilon$ and $x_n\to y$.
Is it true that $d(y, x)<\epsilon$? ( It is important for me to know that $d(y, x)\neq \epsilon$)
|
Try $x_n=1/n\in\Bbb R$, $y=0$ and $x=\epsilon$. (Start the sequence with $n$ sufficiently large.) ... So, no, you may well have $d(y,x) = \epsilon$.
|
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|
Determining: $\lim_{n\rightarrow \infty}e^{-n}\sum_{k=n}^\infty n^k/k!$ What is the value of $\lim_{n\rightarrow \infty}e^{-n}\sum_{k=n}^\infty n^k/k!$ ?
I have tried initially but could not proceed any further. What I have tried is:
$$\lim_{n\rightarrow \infty}e^{-n}\sum_{k=n}^\infty{n^k \over k!}\\
=\lim_{n\rightarrow \infty}e^{-n}\left[e^n-\sum_{k=0}^{n-1}{n^k \over k!}\right]$$
I got no clue after this. I am not sure if how that limit can be be determined.
Any kind of help will be welcome.
|
we have $$e^{-n}\sum_{k=n}^\infty \frac{n^k}{k!}
=e^{-n}\left[e^n-\sum_{k=0}^{n-1} \frac{n^k}{k!}\right]= \left[1+ \frac{e^{-n}n^n}{n!}-e^{-n}\sum_{k=0}^{n} \frac{n^k}{k!}\right] $$
By Stirling formula $$ \frac{e^{-n}n^n}{n!}\sim \frac{1}{\sqrt{2n\pi}}\to0$$
from and from here:Evaluating $\lim\limits_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$ we have
$$\lim_{n\rightarrow \infty}\left[e^{-n}\sum_{k=0}^{n} \frac{n^k}{k!}\right] = \frac12$$
thus $$\lim_{n\rightarrow \infty}e^{-n}\sum_{k=n}^\infty \frac{n^k}{k!}=1-\frac12$$
|
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Prove that $\int_0^\infty\frac1{x^x}\, dx<2$
Prove that $$\int_0^\infty\frac1{x^x}\, dx<2.$$
Note: This inequality is rather tight. The integral approximates to $1.9955$.
Integration by parts is out of the question. If we let $f(x)=\dfrac1{x^x}$ and $g'(x)=1$ then $f'(x)=-x^{-x}(\ln x + 1)$ by implicit differentiation and $g(x)=x$. The integral $\int f'(x)g(x)\, dx$ looks even harder to evaluate.
Expressing the left-hand side as a Frullani integral $$\int_0^\infty\frac{f(ax)-f(bx)}x\, dx=(f(0)-f(\infty))\ln\frac ba$$ means that $f(ax)-f(bx)=x^{1-x}$. However, I can't seem to find a continuous function $f$ that satisfies the functional equation. Is there such a function?
(For context, user371838's post below proves sophomore's dream which I also asked about originally.)
|
$$\int_{0}^{+\infty}e^{-x\log x}\,dx = \underbrace{\int_{0}^{1}e^{-x\log x}\,dx}_{I_1}+\underbrace{\int_{1}^{+\infty}e^{-x\log x}\,dx}_{I_2} $$
$$ I_1=\sum_{n\geq 0}\frac{(-1)^n}{n!}\int_{0}^{1}x^n\left(\log x\right)^n\,dx = \sum_{n\geq 0}\frac{1}{(n+1)^{n+1}}=\sum_{n\geq 1}\frac{1}{n^n}\tag{A}$$
$$ I_2 = \int_{0}^{+\infty}e^{-(x+1)\log(x+1)}\,dx=\int_{0}^{+\infty}\frac{e^{-x}}{W(x)+1}\,dx\tag{B} $$
where $(A)$ gives $I_1\leq 1.292$ and high-order Padé approximants give $I_2\leq 0.705$.
It is a very tight inequality: I wonder if it can be proved in a more elementary way, maybe by writing the whole integral as $\int_{1}^{+\infty}e^{-x}g(W(x))\,dx$.
|
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Calculate the determinant of $A-5I$ Question
Let $ A =
\begin{bmatrix}
1 & 2 & 3 & 4 & 5 \\
6 & 7 & 8 & 9 & 10 \\
11 & 12 & 13 & 14 & 15 \\
16 & 17 & 18 & 19 & 20 \\
21& 22 & 23 & 24 & 25
\end{bmatrix}
$.
Calculate the determinant of $A-5I$.
My approach
the nullity of $A$ is $3$, so the algebraic multiplicity of $\lambda = 0$ is $3$, i.e. $\lambda_1 = \lambda_2 = \lambda_3 = 0.$
Now trace($A$) = $\lambda_4 + \lambda_5 = 1+6+13+19+25 = 65$
Then det($A-5I$) = $(\lambda_1-5)(\lambda_2-5)(\lambda_3-5)(\lambda_4-5)(\lambda_5-5)=(-5)^3(\lambda_4\lambda_5 - 5 \times 65 + 25)$
We need to calculate the value of $\lambda_4 \lambda_5$, which includes sum of lots of determinant of $2 \times 2$ matrices.
Is there any fast way to calculate the determinant?
|
Let the standard basis of $\mathbb R^5$ be denoted $\{e_1,e_2,\dotsc,e_5\}$. Let $\vec l$ denote the leftmost column of $A$, and let $\vec 1$ be the column vector of all $1$s. We have that for all $i \in \{1,\dotsc,5\}$: $$\begin{align}&Ae_i = \vec l+(i-1)\vec1 \\&\vec l = \sum_{i=1}^5 ie_i \\&\vec 1 = \sum_{i=1}^5 e_i\end{align}$$
Since every vector in the image (or range, if you call it that) of $A$ is a linear combination of $\vec 1$ and $\vec l$, it follows that the rank of $A$ is $2$. From which it follows that $3$ of $A$'s eigenvalues are $0$.
We now proceed to find the remaining eigenvalues. We focus on the subspace of $\mathbb R^5$ spanned by $\vec 1$ and $\vec l$. We see that $$A \vec 1 = 10\vec 1 + 5\vec l \\ A \vec l= 160\vec 1 + 55\vec l$$
Restrict $A$ to the subspace $\operatorname{span}\{\vec 1, \vec l\}$, and change to the basis $\{\vec 1, \vec l \}$. You get:$$B=\begin{bmatrix}10 & 160 \\ 5 & 55 \end{bmatrix}$$
Any eigenvalue of $B$ is an eigenvalue of $A$. The eigenvalues of $B$ are $λ_{12} = \frac52 (13 \pm \sqrt{209})
$.
We've now found all of $A$'s eigenvalues.
Finally, the determinant of $A$ is $\prod_{i=1}^5(\lambda_i - 5) = \frac52(11 + \sqrt{209})\frac52(11 - \sqrt{209})(-5)(-5)(-5)=68750$.
|
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|
Taylor approximation of inverse square root Given the function $f(x)=\sqrt{1+mx+\mathcal{O}(x^2)}$ I am reading that $g(x) = \frac{1}{f(x)}$, the inverse square root, can be computed with first order Taylor approximation and take $g(x) = 1 - \frac{m}{2}x + \mathcal{O}(x^2)$. So given, $f'(x) = \frac{m+\mathcal{O}(x^2)}{2\sqrt{1+mx+\mathcal{O}(x^2)}}$, I started with the first order Taylor approximation of $f(x)\approx f(a) + f'(a)(x-a)+\mathcal{O}(x^2)$ in point $a$ but I do not know how to continue. Could you please give some help?
|
It is easier to write $g(x)=\frac{1}{f(x)}=(1 + m x + O(x^2))^{-1/2}$. Doing the derivative on g(x) (not f(x)) gives you $g'(x)= -m/2 (1 + m x +O(x^2))^{-3/2} +O(x^2)$, so the Taylor expansion (you have a sign error and should read $h(x)=h(a)+h'(a)(x-a)+O(x^2))$ at $a=0$ is:
$g(x)=g(0) + g'(x) (x-0) + O(x^2)= 1 - \frac{m}{2} x + O(x^2)$
|
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|
What is undefined times zero? Einstein's energy equation (after substituting the equation of relativistic momentum) takes this form:
$$E = \frac{1}{{\sqrt {1 - {v^2}/{c^2}} }}{m_0}{c^2}
% $$
Now if you apply this form to a photon (I know this is controversial, in fact I would not do it, but I just want to understand the consequences), you get the following:
$$E = \frac{1}{0}0{c^2}% $$
On another note, I understand that after dividing by zero:
*
*If the numerator is any number other than zero, you get an "undefined" = no solution, because you are breaching mathematical rules.
*If the numerator is zero, you get an "indeterminate" number = any value.
Here it seems we would have an "indeterminate" [if (1/0) times 0 equals 0/0], although I would prefer to have an "undefined" (because I think that applying this form to a photon breaches physical/logical rules, so I would like the outcome to breach mathematical rules as well...) and to support this I have read that if a subexpression is undefined (which would be the case here with gamma = 1/0), the whole expression becomes undefined (is this right and if so does it apply here?).
So what is the answer in strict mathematical terms: undefined or indeterminate?
|
Undefined is not a number. There is no such number as undefined, for which you could define the multiplication operation.
You could extend a set of a numbers (the set of the real numbers or the set of the complex numbers) with a new element, what you call "undefined", and then define a multiplication on this set as usual. It might be possible (although there are major problems solve).
However, from this moment, you have a defined value what you still call "undefined". Well, paper can hold everything, but it does not really sound as useful mathematics.
|
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|
validity of a formula with one existential quantifier and one variable Let $\sigma$ a dictionary without equlity symbol that contains at least one constant symbol.
Let $\varphi$ a formula over $\sigma$ without quantifiers such that $FV(\varphi)=\{x\}$.
Prove that: $\exists x\varphi$ is valid $\iff$
There exist $s_1,\dots,s_n$ ground terms over $\sigma$, such that $\varphi[s_1/x]\lor\dots\lor\varphi[s_n/x]$ is valid.
My approach:
Suppose that there exist $s_1,\dots,s_n$ ground terms over $\sigma$, such that $\varphi[s_1/x]\lor\dots\lor\varphi[s_n/x]$ is valid.
Let $\mathcal{M}$ a model and $v$ an interpretation in it.
Then $\mathcal{M},v\vDash\varphi[s_1/x],\dots,\varphi[s_n/x]$, so $\mathcal{M},v\vDash\varphi[s_i/x]$ for some $i$ $\implies \mathcal{M},v[\bar{v}(s_i)/x]\vDash\varphi$.
So there exists some $d\in D^\mathcal{M}$ such that $\mathcal{M},v[d/x]\vDash\varphi\implies\mathcal{M},v\vDash\exists x\varphi$.
For the other direction I only managed to observe that:
since $\varphi$ has no quantifiers and $FV(\varphi)=\{x\}$ we have that $x$ is the only variable that appers in $\varphi$.
|
I assume that in your notation, the statement "$\psi_1,\dots,\psi_k$ is valid" is equivalent to "$\bigvee_{i=1}^k \psi_i$ is valid". Is that right?
For the converse, suppose $\exists x\,\varphi(x)$ is valid. Consider the set of sentences $T = \{\lnot \varphi(t)\mid t\text{ is a ground term}\}$. Suppose for contradiction that $T$ is consistent. Then there is a model $\mathcal{M}\models T$. Let $\mathcal{N}$ be the substructure of $\mathcal{M}$ generated by the constants (with domain $\{t^{\mathcal{M}}\mid t\text{ is a ground term}\}$). For every ground term $t$, since $\varphi(t)$ is quantifier-free and $\mathcal{N}$ is a substructure of $\mathcal{M}$, $\mathcal{M}\models \lnot \varphi(t)$ implies $\mathcal{N}\models \lnot \varphi(t)$, so $\mathcal{N}\models \lnot \exists x\, \varphi(x)$, contradiction.
So we conclude that $T$ is inconsistent. By the compactness theorem, a finite subset $T'\subseteq T$ is inconsistent, say $T' = \{\lnot \varphi(s_1),\dots,\lnot \varphi(s_n)\}$, where $s_1,\dots,s_n$ are ground terms. To say that $T'$ is inconsistent is to say that every $\sigma$-structure satisfies $\varphi(s_i)$ for some $1\leq i\leq n$, which is what we wanted to prove.
|
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Why does Hausdorff measure go to zero as diameter power increases? For example, why does the Hausdorff measure of a flat disc go to zero when the power that the diameter is raised to (in the definition of the Hausdorff measure) reaches 3?
|
A flat disc $D$ of radius $1$ has area $\pi.$ For $n\in \Bbb N$ let $S_n$ be a set of open discs, each of diameter $1/n$ or less, with $\cup S_n \supset D$ and $$\sum_{t\in S_n}A(t)<2\pi,$$ where $A(t)$ is the area of $t.$
For $t\in S_n$ let $d(t)$ be the diameter of $t.$ Then $A(t)=\pi d(t)^2/4.$
Let $r>0.$ Then $$\sum_{t\in S_n} \pi d(t)^{2+r}/4=\sum_{t\in S_n}A(t)d(t)^r\leq \sum_{t\in S_n}A(t)(1/n)^r<2\pi(1/n)^r.$$ As $n\to \infty$ we have $2\pi (1/n)^r\to 0.$
|
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Question on surds rule Through the textbook, I've been taught the rule $\frac{\sqrt a}{\sqrt b} = \sqrt\frac{a}{b}$, however I realized that if all numbers are assumed to be real, and $a<0 ,b<0$, then the rule is not true as $\frac{\sqrt{-a}}{\sqrt{-b}} = \sqrt\frac{a}{b}$, whereas in $\frac{\sqrt{-a}}{\sqrt{-b}}$, the process makes it invalid as square rooting negative numbers is impossible. So does this mean that the surds rule (multiplication of surds included) is invalid for negative numbers? Or does it mean complex numbers have to be introduced in all cases?
|
For all nonnegative real numbers $a$ and $b$ (with $b > 0$), we have
$$ \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}. $$
After that, all bets are off. The basic problem is that the function $x \mapsto x^2$ is not injective (one-to-one) over the real numbers. Thus to define the principal square root, which is what the surd represents, we must restrict the domain of this function to the nonnegative real numbers. Thus if we are working with real variables, then, for example, $\sqrt{a}$ is nonsense for $a < 0$.
Okay, so real variables cause problems. What if we work with complex variables, instead? Unfortunately, even if we are careful, we generally cannot get the "surd rule" to work as you want. For example, if we could extend it, then we might have
$$ 1 = \sqrt{1} = \sqrt{(-1)^2} = \sqrt{-1}^2 = i^2 = -1, $$
which is clear nonsense. The problem again comes down to how we define the square root function, and on what domain that definition holds. There are generally two possible choices, and these choices are more or less arbitrary.
|
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Evaluation of limiting value of given series For a given sequence, $a_1=1$ and $a_n=n(1+a_{n-1})$ $\forall n\geq 2$, then value of given limit:
$$\lim_{n\to \infty} \bigg(1+\frac{1}{a_1}\bigg)\bigg(1+\frac{1}{a_2}\bigg)\cdots\bigg(1+\frac{1}{a_n}\bigg)$$
Usually such type of questions are solved by squeeze theorem or by converting them into definite integral but don't see neither working here. Could someone give me little help to proceed
|
Hint: $\displaystyle \frac{a_n}{n!} = \frac{a_{n - 1}}{(n - 1)!} + \frac{1}{(n - 1)!} \ (n \geqslant 2)$ and$$
\prod_{k = 1}^n \left(1 + \frac{1}{a_k}\right) = \left. \prod_{k = 1}^n (1 + a_k) \middle/ \prod_{k = 1}^n a_k\right. = \left.\prod_{k = 1}^n \frac{a_{k + 1}}{k + 1} \middle/ \prod_{k = 1}^n a_k\right. = \frac{1}{a_1} \frac{a_{n + 1}}{(n + 1)!}.
$$
|
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Prime divisors of sequences of integers I have the following problem and I need your support for it
Problem. Let P(x) be a polynomial with integer coefficients, such that $\deg P>0$ and $\lim_{x\to+\infty}P(x)=+\infty$. Prove that there exist infinitely many prime numbers $p$ such that for some natural number $n$
$$p\mid \left\lfloor \log\left(2017^{P(n)}+1\right)\right\rfloor.$$
|
As I pointed out in my comments, what matters here is the growth. First we prove :
Proposition : Given $f : \mathbb{N} \mapsto \mathbb{N}$ strictly increasing, and $k \ge 0$ such that $f(n) = O_{n \to +\infty} \big( n^k \big)$, the set $\{ p \mbox{ prime } | \ \exists n:\ p|f(n) \}$ is infinite.
Proof : ad absurdum, assume we can write this set $p_1, ..., p_r$. Let us denote $A = \{p_1^{k_1} \cdot\cdot\cdot p_r^{k_r}\ |\ k_i \ge 0\}$, $B = f(\mathbb{N})$, and for all $n$, $A_n = A \cap [\![1, n]\!]$ and $B_n = B \cap [\![1,n]\!]$. As $B_n \subset A_n$, $|B_n| \le |A_n|$.
However, as all $p_i$ are $\ge 2$, $|A_n| \le (\mbox{log}_2(n)+1)^r$.
And as $f$ is strictly increasing, $|B_n| = \max(\{m | f(m)\le n\})$. It is easy to conclude that $n^{1/k} = O_{n \to +\infty} (|B_n|)$.
Conclusion : $n^{1/k} = O_{n \to +\infty}\big( (\mbox{log}_2(n)+1)^r\big)$, which is absurd.
Regarding your problem, it is easy to check that $f :n \mapsto \left \lfloor \mbox{log}\Big( 2017^{P(n)}+1 \Big)\right \rfloor$ is strictly increasing (the quantity between the floor brackets increases by more than $\mbox{log}(2017)>1$ between $n$ and $n+1$).
Moreover, $f(n)\le P(n)*\mbox{log}(2017)+\mbox{log}(2017)+1$. We can apply the proposition.
|
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Given a set $A \subseteq \{10,11,12,13...98,99\}$ such that $|A|=10$. Prove using Pigeonhole Principle there are 2 disjoint subsets with the same Sum. Given a set $A \subseteq \{10,11,12,...98,99\}$ such that $|A|=10$. Prove using Pigeonhole Principle there are 2 disjoint non-empty subsets of $A$ with the same Sum.
Direction or Hint would be appreciated.
|
The range of sums of a subset of $|A| \le 10$ is $\le 21 (10+11)$ and $945\ge (90+91+...+99)$, therefore $925$ different sums you can get from a subset of $\{10,11,...,98,99\}$ if $|A|\le10.$
In a set $A, |A| = 10$, there are $2^{|A|} - 1$ non-empty subsets = $1023$.
$1023 \gt 945$, therefore considering $945$ "pigeonholes", atleast $2$ disjoint sets will share the same sum. We can force our sets $A,B\, s.t \,\Sigma A = \Sigma B$ to be disjoint easily by -
$$ A = A \backslash \{A\cap B\} $$
$$ B = B \backslash \{B\cap A\} $$
As required.
|
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How many $3$-digit numbers can be formed using the digits $ 2,3,4,5,6,8 $ such that the number contains the digits $5$ and repetitions are allowed? The solution I have is by counting the complement which gives an answer of $91$. But I think that it should be solved as follows-
Let the $3$-digit number be denoted by $3$ boxes. We can put the digit $5$ in one of the three boxes and we can put fill remaining two boxes with $6$-digits. Hence answer is $6\cdot 6\cdot 3= 108$. Help!
|
Take all the possible numbers with $5$
then substract all the numbers without $ 5$
Result: $6^3-5^3=91$
|
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Number of integer triangles.
Number of integer isosceles or equilateral triangle none of whose
sides exceed $2c$ is?
I substituted $c =3$ checked, and got $27= 3 \times (3^2)$ i.e. $3c^2$ triplets. It gives the right answer($3c^2$) but how do I write a proper formal proof of this?
I tried it this way: We have three vacant places for three integers. First place can be filled in $2c$ ways, 2nd one in $2c$ ways, 3rd one in $2c$ way but it involves over-counting, is it possible to eliminate the extras?
|
Let $b$ be the length of the base and $a$ be the length of the legs of such a triangle. Given $c\geq1$ we want
$$1\leq b\leq 2c, \quad {b\over2}<a\leq 2c\ .\tag{1}$$
If $b=2k-1$ with $1\leq k\leq c$ is odd the condition $(1)$ enforces $k\leq a\leq 2c$ and allows of $2c-(k-1)$ different integer values for $a$.
If $b=2k$ with $1\leq k\leq c$ is even then $(1)$ enforces $k+1\leq a\leq 2c$ and allows of $2c-k$ different integer values for $a$.
The total number $N$ of isosceles triangles having sides $\leq2c$ therefore is given by
$$N=\sum_{k=1}^c\bigl(2c-(k-1)\bigr)+\sum_{k=1}^c(2c-k)=4c^2-\sum_{k=1}^c(2k-1)=3c^2\ .$$
|
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Find possible values of angle C In an acute triangle $ABC$, $O$ is the circumcenter, $H$ is the orthocenter and $G$ is the
centroid. Let $OD$ be perpendicular to $BC$ and $HE$ be perpendicular to $CA$, with $D$ on $BC$ and $E$
on $CA$. Let $F$ be the midpoint of $AB$. Suppose the areas of triangles $ODC, HEA$ and $GFB$ are
equal. Find all the possible values of angle $C$.
My approach :
Let $R$ be the circumradius of $△ABC$ and $∆$ its area. We have $OD = R \cos A$ and
$DC =\frac{a}{2}$, so
$$[ODC] = \frac{1}{2}· OD · DC$$ $$=\frac{1}{2}· R \cos A · R \sin A $$ $$=\frac{1}{2}. R^2.\sin A \cos A .$$
Again $HE = 2R \cos C \cos A$ and $EA = c \cos A$. Hence
$$[HEA] = \frac{1}{2}· HE · EA$$
$$=\frac{1}{2}· 2R \cos C \cos A · c \cos A$$
$$=2R^2.\sin C .\cos C.{\cos}^2A .$$
Further
$$[GFB] = \frac{∆}{6}=\frac{1}{6}· 2R^2.\sin A.\sin B.\sin C$$ $$=\frac{1}{3}.R^2.\sin A.\sin B.\sin C$$
What to do next? Any help would be greatly appreciated.
|
From $[ODC]=[HEA]$ one gets
$$
\tag{1}
\sin C\cos C={1\over4}{\sin A\over \cos A}.
$$
From $[ODC]=[GFB]$, taking into account that $\sin B=\sin(A+C)=\sin A\cos C+\cos A\sin C$, one gets
$$
\sin A \sin C\cos C+\cos A \sin^2C={3\over2}\cos A,
$$
that is, using $(1)$:
$$
{1\over4}{\sin^2 A\over \cos A}+\cos A \sin^2C={3\over2}\cos A,
$$
whence:
$$
\tag{2}
\sin^2 C={7\cos^2 A-1\over4\cos^2A}.
$$
Squaring $(1)$ and inserting there $(2)$ one finally gets
$$
20\cos^4A-9\cos A+1=0,
\quad\hbox{that is:}\quad
\cos^2A={1\over4}
\ \ \hbox{or}\ \
\cos^2A={1\over5}.
$$
From $(2)$ one then obtains
$$
\sin^2 C={3\over4}\ \hbox{and}\ C=60°,
\ \ \hbox{or}\ \
\sin^2C={1\over2}\ \hbox{and}\ C=45°.
$$
The first case obviously leads to an equilateral triangle, while the second case is not trivial.
|
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Solving $\cos x + \cos 2x - \cos 3x = 1$ with the substitution $z = \cos x + i \sin x$ I need to solve
$$\cos x+\cos 2x-\cos 3x=1$$
using the substitution$$z= \cos x + i \sin x $$
I fiddled around with the first equation using the double angle formula and addition formula to get
$$\cos^2 x+4 \sin^2x\cos x-\sin^2 x=1$$
which gets me pretty close to something into which I can substitute $z$, because $$z^2= \cos^2 x-\sin^2 x+2i\sin x\cos x$$
I have no idea where to go from there.
|
Using the hint,
$$\Re(z+z^2-z^3)=1,$$ or
$$z+z^2-z^3=1+iw.$$
This can be factored as
$$-(z+1)(z-1)^2=iw$$ but I see no easy way to exploit it.
Direct resolution of the cubic equation looks terrible.
|
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System of homogeneous linear equations with coefficients in a field In the book I started to read, at almost the very beginning it is stated (without proof):
If $m<n$ and we have a system $$a_{11}x_1+...+a_{1n}x_n=0$$
$$\vdots$$
$$a_{m1}x_1+...+a_{mn}x_n=0$$
of homogeneous linear equations, with coefficients in a field, then there is a non-trivial solution in the field.
So, I have some questions about this:
Where a proof can be found? Does this also holds for finite fields (or it is implicitly assumed that we talk about infinite fields)? Is this just an application of matrix algebra (that is, this matrix of coefficients will always have an inverse if $m<n$?
|
To start, your last sentence is not correct in that if a matrix has an inverse it will only have the trivial solution to the homogenous system. Also, an invertable matrix must be square.
The result itself follows from the fact that the system is underdetermined and will have multiple solutions (proof via Gaussian algorithm). You could also likely view it from a linear dependence argument.
Any introductory linear algebra text should show this result (you could just replace the real numbers with an arbitrary field). Lay's Linear Algebra is a good one.
|
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surface area of implicit surface Ok, so I read ( on Thomas' book) that the surface area of an implicit surface is the double integral of the magnitude of the function's gradient divided by the magnitude of the dot product between such gradient and the unit normal vector p over the projection of this surface onto a plane. The p vector would be represented by the i, j or k vector, depending on which plane we projected the surface. The problem is... I don't understand this at all. Like, I get that for an implicit surface the gradient is normal to it, but I don´t get why we would integrate the gradient, nor why there is a dot product in the middle of all of this, between the gradient and p. Could someone explain it to me, or give me some intuition? Thanks
|
If you project a piece $S$ of a planar region in the plane with unit normal vector $\vec n$ onto a region $R$ in the $xy$-plane (with normal vector $\vec k$), then it's a little bit of geometry to see that
$$\text{area}(S) = \frac1{|\cos\gamma|}\text{area}(R) = \frac1{|\vec n\cdot\vec k|}\text{area}(R).$$
(Here $\gamma$ is the angle between the two planes, which coincides with the angle between the two unit normal vectors.)
At a point of your implicit surface $f(x,y,z)=\text{constant}$, the unit normal vector is $\vec{\nabla f}/\|\vec{\nabla f}\|$, and so the same formula works for a little bit of surface area, approximating the little bit of the surface by a little region in the tangent plane. Note that
$$\frac1{|\vec n\cdot \vec k|} = \frac1{\left|\frac{\vec{\nabla f}}{\|\vec{\nabla f}\|}\cdot\vec k\right|} = \frac{\|\vec{\nabla f}\|}{|\vec{\nabla f}\cdot\vec k|}.$$
|
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Find minimum value of $\sum \frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}$ If $a,b,c$ are sides of triangle Find Minimum value of
$$S=\sum \frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}$$
My Try:
Let $$P=\sqrt{a}+\sqrt{b}+\sqrt{c}$$
we have $$S=\sum \frac{1}{\frac{\sqrt{b}}{\sqrt{a}}+\frac{\sqrt{c}}{\sqrt{a}}-1}$$
$$S=\sum \frac{1}{\frac{P}{\sqrt{a}}-2}$$
Let $x=\frac{P}{\sqrt{a}}$, $y=\frac{P}{\sqrt{b}}$,$z=\frac{P}{\sqrt{c}}$
Then we have $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$$
By $AM \ge HM$
$$\frac{x+y+z}{3} \ge \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$$
Hence
$$x+y+z \ge 9$$
Any way to proceed further?
|
When $a = b = c$, $S = 3$. Next it will be proved that $S \geqslant 3$ for all possible $a, b, c$.
Denote $\displaystyle u = \frac{\sqrt{a}}{\sqrt{a} + \sqrt{b} + \sqrt{c}}$, $\displaystyle v = \frac{\sqrt{b}}{\sqrt{a} + \sqrt{b} + \sqrt{c}}$, $\displaystyle w = \frac{\sqrt{c}}{\sqrt{a} + \sqrt{b} + \sqrt{c}}$, then $\sum u = 1$. Since$$
\sqrt{a} < \sqrt{b + c} < \sqrt{b} + \sqrt{c},
$$
then $\displaystyle 0 < u < \frac{1}{2}$. Analogously, $\displaystyle 0 < v, w < \frac{1}{2}$. It suffices to prove$$
S = \sum \frac{u}{v + w - u} = \sum \frac{u}{1 - 2u} \geqslant 3.
$$
Define $\displaystyle f(x) = \frac{x}{1 - 2x} \ (0 < x < \frac{1}{2})$. Because $\displaystyle f''(x) = \frac{4}{(1 - 2x)^3} > 0$, by Jensen's inequality,$$
S = \sum f(u) \geqslant 3 f\left(\frac{1}{3} \sum u\right) = 3f\left(\frac{1}{3}\right) = 3.
$$
Therefore the minimum of $S$ is $3$.
|
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Perpendicular from incenter of a triangle to any side is equal to the radius of the incircle Given a triangle $ABC$ with incenter $I$, it is said that the perpendicular line segment from $I$ to any of the sides $AB$, $AC$, or $BC$ is equal to the radius of the incircle. (See the second picture on this page: http://mathworld.wolfram.com/RightTriangle.html )
I tried to prove it without any success. Can someone please give me a hint?
|
Well the definition of an incenter is the center of the largest circle that fits into the triangle. So the circle is externally tangent to each side of the triangle. A well-known circle theorem is that the radius at the point where a tangent touches the circle is perpendicular to the tangent.
|
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|
Is there an algebraic proof for $\sum_{m=k}^{n-k} \binom{m}{k}\binom{n-m}{k} = \binom{n+1}{2k+1}, n\ge2k\ge0$ $\sum_{m=k}^{n-k} \binom{m}{k}\binom{n-m}{k} = \binom{n+1}{2k+1}, n\ge2k\ge0$
An combinatorial proof of the identity above states as follow:
(1)Number of ways of picking (2k+1) numbers from 1 to (n+1) should be $\binom{n+1}{2k+1}$
(2)We pick (2k+1) numbers from 1 to (n+1) with median value (m+1). Then, k numbers must be selected from 1~m, and the other k numbers must be chosen from (m+2)~(n+1). Thus there are $\binom{m}{k}\binom{n-m}{k}$ ways for picking (2k+1) numbers with median value (m+1). Since $n-k\ge m\ge k$, there are total $\sum_{m=k}^{n-k} \binom{m}{k}\binom{n-m}{k}$ ways.
Since (1)=(2), the statement is true.
But is it possible to sketch an algebraic proof that doesn't require building combinatorial models?
|
Here is an algebraic proof based upon generating functions. It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. This way we can write for instance
\begin{align*}
[z^k](1+z)^n=\binom{n}{k}
\end{align*}
We obtain
\begin{align*}
\color{blue}{\sum_{m=k}^{n-k}}&\color{blue}{\binom{m}{k}\binom{n-m}{k}}\\
&=\sum_{m=0}^{n-2k}\binom{m+k}{m}\binom{n-m-k}{k}\tag{1}\\
&=\sum_{m=0}^{n-2k}\binom{-k-1}{m}(-1)^m\binom{n-m-k}{k}\tag{2}\\
&=\sum_{m=0}^\infty[z^m](1-z)^{-k-1}[u^k](1+u)^{n-m-k}\tag{3}\\
&=[u^k](1+u)^{n-k}\sum_{m=0}^\infty\left(\frac{1}{1+u}\right)^{m}[z^m](1-z)^{-k-1}\tag{4}\\
&=[u^k](1+u)^{n-k}\left(1-\frac{1}{1+u}\right)^{-k-1}\tag{5}\\
&=[u^k](1+u)^{n-k}u^{-k-1}(1+u)^{k+1}\tag{6}\\
&=[u^{2k+1}](1+u)^{n+1}\tag{7}\\
&\color{blue}{=\binom{n+1}{2k+1}}\tag{8}
\end{align*}
and the claim follows.
Comment:
*
*In (1) we shift the index to start with $m=0$ and use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.
*In (2) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
*In (3) we apply the coefficient of operator twice and we set the upper bound of the series to $\infty$ without changing anything since we are adding zeros only.
*In (4) we use the linearity of the coefficient of operator and we apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.
*In (5) we apply the substitution rule of the coefficient of operator with $z=\frac{1}{1+u}$
\begin{align*}
A(u)=\sum_{m=0}^\infty a_m u^m=\sum_{m=0}^\infty u^m [z^m]A(z)
\end{align*}
*In (6) we do some simplifications.
*In (7) we do some more simplifications and apply the same rule as we did in (4).
*In (8) we select the coefficient of $u^{2k+1}$.
|
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Group of order $3$ acting on the tetrahedron It is well-known that the group of (orientation-preserving) symmetries of the tetrahedron is isomorphic to $A_4$. Since $\mathbb{Z}/3$ is a quotient of $A_4$, $\mathbb{Z}/3$ also acts on the tetrahedron.
Is there a way to see this actions geometrically? i.e., are there $3$ natural parts of the tetrahedron that $\mathbb{Z}/3$ acts on?
For example, we can ask the same question about the copy of $S_3$ that is a quotient of $S_4$, the group of symmetries of the cube. In this case, this copy of $S_3$ acts on the lines connecting the centers of opposite faces. My question is: is there a similar geometric description for the tetrahedron? Where does "$3$" come from in the tetrahedron?
|
The tetrahedron has three (unordered) pairs of opposite edges. The stabiliser of each is the fours-group whose quotient is cyclic of order $3$. I think this is what you are interested in.
|
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Existence of a special continuous function I have stuck at a problem. " does there exist any continuous function that takes every real value exactly twice?"
Intuitively, I think such function cannot exist, as $x^2$, $|x|$, although they take every possitive value exactly twice, but they take $0$ only once. I tried to apply LMVT, but cannot make a concrete logic. Please help
|
You're right that no such function exists. If $f$ were one then there would be two values $a < b$ such that $f(a)=f(b)=1$. Then on the closed interval $[a,b]$ the function $f$ would have a maximum value $M$. We can assume $M > 1$ (if not, then use the minimum value).
If the maximum value $M$ appeared twice, at $c < d$ then find an $x$ near $c$ such that $1 < f(x) < M$. (You can do that using continuity). Then by the intermediate value theorem, $f$ takes on the value $f(x)$ between $a$ and $c$ and between $c$ and $b$. That's three times.
|
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Show $\int_{0}^{\infty} x^tf(x)dx$ is uniformly convergent for $t\in[a,b]$. Assume $\int_{0}^{\infty} x^tf(x)dx$ converges at $t=a$ and $t=b$. Show
$$\int_{0}^{\infty} x^tf(x)dx$$
is uniformly convergent for $t\in[a,b]$.
It seems that the common tests cannot work. May I consider Cauchy's convergence test？
|
Absolute convegence case:
Here we consider the case where $\int^\infty_0 x^t\,|f(x)|\,dx<\infty$ for $t\in\{a,b\}$.
Since $\left|\int_A x^t\,f(x)\,dx\right|\leq \int_A x^t|f(x)|\,dx$ for all $A\subset [0,\infty)$, it suffices to assume $f\geq0$.
Given $\varepsilon>0$, there exists $C>0$ such that
$$
\begin{align}
\int^v_u(x^a + x^b)\,f(x)\,dx<\varepsilon\tag{0}\label{zero}
\end{align}
$$
Edit: One approach as suggested by "kimchi lover" above, is to use the obvious inequality
$$x^y f(x)\leq (x^a+x^b)f(x),\qquad x\geq0, \,a\leq y\leq b$$
This implies that
$$
\int^v_u x^y\, f(x)\,dx \leq \int^v_u (x^a+x^b)\,f(x)\,dx<\varepsilon
$$
for all $u,v>C$ and $a\leq y\leq b$, which is a uniform bound in $y$.
Another approach (my original solution in fact) relies on well known result in integration, namely Holder's inequality.
measure space $(X,\mathscr{F},\mu)$, if $\frac1p+\frac1q=1$, and $f\in L_p(\mu)$ and $g\in L_q(\mu)$, then
$$\int|f\,g |\,d\mu\leq \Big(\int_X|f|^p\,d\mu\Big)^{1/p}\Big(\int_X|g|^q\,d\mu\Big)^{1/q}$$
Remark: It is possible to accurate this inequality for the setting of proper Rieman integrals, but I will leave details out. [Here][1] is one posting that deals with this. With a little effort, that posting maybe adapted to the present setting: replace $[0,1]$ by $(0,\infty)$ and $dx$ by $f(x)\,dx$
For the OP, consider the measure space $([0,\infty),\mathscr{B}([0,\infty)),f(x)\,dx)$.
By assumption, the function $\phi(x)=x$ is in $L_a$ and $L_b$. Then, for any $a<t<b$, there is $0<\lambda<1$ such that $t=\lambda a+(1-\lambda)b$. The function $g(x)=x^{a\lambda}\in L_{1/\alpha}$ and $h(x)=x^{b(1-\lambda)}\in L_{1/(1-\lambda)}$.
Applying Holder's inequality and $\eqref{zero}$
$$\int^v_u x^t\,f(x)\,dx =\int^v_u x^{a\lambda}x^{b(1-\lambda)}f(x)\,dx\leq\Big(\int^u_v x^a f(x)\,dx\Big)^{\lambda}\Big(\int^u_v x^b f(x)\,dx\Big)^{1-\lambda}<\varepsilon$$
for all $v>v>C$.
Conditional convergence case:
The case where $\int^\infty_0 x^t\,f(x)\,dx$, $t\in\{a,b\}$, converges conditionally ($\int^\infty_0x^t|f(x)|\,dx=\infty$), I have no definitive answer. I suspect that uniform convegence will depend on the function $f$.
The following classical example preserves uniform convergence:
$ \phi(y):=\int^\infty_0 \frac{\sin}{t^y}\,dx$ where $0<y\leq 1$.
Integration by parts gives
$$\int^v_u \frac{\sin x}{x^y}\,dx= -\frac{\cos x}{x^y}\Big|^v_u - y\int^v_u\frac{\cos x}{x^{1+y}}\,dx$$
Hence, for $1\leq u<v$
$$
\Big|\int^v_u\frac{\sin x}{x^y}\,dx\Big|\leq \frac{1}{u^y}+\frac{1}{v^y} + y\int^v_u\frac{1}{x^{1+y}}\,dx = \frac{2}{u^y}
$$
So, for $0<a\leq y\leq b\leq 1$, we have uniform convergence since $\frac{1}{u^y}\leq \frac{1}{u^a}$.
At the moment, I have no a counterexample (an example of $f$ rather) where uniform convergence does not hold.
[1]: Proof of Hölder's inequality for improper integrals
|
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Find the radius of curvature at the origin
Find the radius of curvature at the origin for the curve $3x^2+4y^2=2x$.
Below is my attempt:
We know that the radius of curvature $\rho=\dfrac{{(1+y_1^2)}^{\frac{3}{2}}}{y_2}$ .
Computing $y_1$ at $(0,0)$ : Differentiating we get $3x+4y\frac{dy}{dx}=1\implies \dfrac{dy}{dx}=\dfrac{1-3x}{4y}.$
How can I evaluate $\dfrac{dy}{dx}$ at the point $(0,0)$?
Please help.
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If we consider the differential $(6x-2)dx+8ydy$, then at $(0,0)$, we see $(6x-2) \neq 0$. By the implicit function theorem, we may write $x$ as a function of $y$.
Complete the square:
$$3x^2 -2x +4y^2=0$$
$$3(x-\frac{1}{3})^2+4y^2=\frac{1}{3}$$
From here we get:
$$x(y)=\frac{1}{3}+\sqrt{\frac{1}{9}-\frac{4}{3}y^2}$$
Let $r:(-\epsilon,\epsilon) \rightarrow \mathbb{R}^{2}$ be given by
$$r(y) =(x(y),y)$$
The curvature is given by the general formula:
$$\kappa=\frac{|x'y''-x''y'|}{(x'^{2}+y'^{2})^{\frac{3}{2}}}$$
Using our parameterization, we have:
$$\kappa(y)=\frac{|\frac{d^2x}{dy^2}|}{(1+(\frac{dx}{dy})^2)^\frac{3}{2}}$$
Now $\frac{dx}{dy}=0$ when $y=0$. If we compute $\frac{d^2
x}{dy^2}$ we have:
$$\frac{d^2
x}{dy^2}=y(junk)+\frac{4}{3}\bigg(\frac{1}{9}-(\frac{4}{3}y^2)\bigg)^\frac{-1}{2}$$
Evaluating at $y=0$ we find,
$$\kappa(0)=4$$.
Thus $\rho= \frac{1}{\kappa}=\frac{1}{4}$
|
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Examples of when a "new way of thinking" led to a solution I was reading William Thurston's On Proof and Progress in Mathematics in which he discusses the value of the different ways people think about the same mathematical structure. He claims that many mathematical solutions are the result of different ways of thinking about the same underlying mathematics.
What are some examples of problems that seem very difficult, but yield to a new way of thinking about the relevant structures?
Answers should include a description of the problem, the “way of thinking”, and the rigorous solution.
|
The probabilistic method has been successfully used for proving the existence of mathematical objects non constructively, by proving that the probability of choosing an object in that class is not zero.
|
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Product of a exponential and discrete Distribution Let $U \sim e(1)$ and $V$ a discrete random variable independent of $U$ such that
$p_V(v)=1/2$ if $v \in \{-1,1\}$ and $p_V(v)=0$ otherwise.
Problem:
Let $W=UV$. Find the distribution function of $W$ $\forall w\in \mathbb{R}$.
My try:
\begin{align}
P(W\leq w)&=P(UV\leq w)=P(U\leq w \mid V=1)P(V=1)+P(-U\leq w \mid V=-1)P(V=-1)\\
&= P(U\leq w)\frac{1}{2}+P(U\geq -w)\frac{1}{2}\\
&=P(U\leq w)\frac{1}{2}+(1-P(U\leq-w)\frac{1}{2}\\
&=\left(\int_0^we^{-u} \, du\right)\frac{1}{2}+\left(1-\int_0^{-w} e^{-u} \, du \right) \frac{1}{2}\\
&=(1-e^{-w})\frac{1}{2}+\frac{1}{2}e^w = F_W(w)
\end{align}
The correct answer in my book is the following
$$F_W(w)=\frac{1}{2}(1-e^{-w})+\frac{1}{2}, w>0$$ and
$$F_W(w)=\frac{1}{2}e^{w}, w\leq 0$$
Can anyone tell me what i am doing wrong?
|
You wrote:
\begin{align}
& P(W\leq w) = P(UV\leq w) \\[10pt]
= {} &P(U\leq w \mid V=1)P(V=1)+P(-U\leq w \mid V=-1)P(V=-1).
\end{align}
What is $P(U\le w\mid V=1)$?
It is $\begin{cases} 1-e^{-w} & \text{if } w\ge0, \\ 0 & \text{if } w<0. \end{cases}$
What is $P(-U\le w\mid V=-1)$?
It is $\begin{cases} e^w & \text{if } w\le0, \\ 1 & \text{if } w>0. \end{cases}$
You need to consider the cases $w\ge0$ and $w<0$ separately.
If you find the density function, you will be able to write it more simply by using the notation $|w|.$
|
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Compute $A^n$ where $A^2+bA+cI=0$ Let $A$ be a complex matrix such that $$A^2+bA+cI=0,$$
where $I$ is the identity matrix and $b,c\in \mathbb{C}$. I am interested in finding a formula for $A^n$ in terms of $A$ and $I$. The binomial formula is not giving an answer I think. Maybe using $A^2=-bA-cI$, then $A^3=-bA^2-cA$, etc. But the computation becomes so complicated to find a formula by induction.
|
You have $$ A^{n+2} = b_nA + c_nI $$
where $b_0 = -b, c_0 = -c$
Multiplying both sides by $A$ gives
$$ A^{n+3} = b_nA^2 + c_n A = b_n(-bA-cI)+c_nA = (c_n-bb_n)A - cb_nI $$
therefore
$$ b_{n+1} = -bb_n + c_n, \ c_{n+1} = -cb_n $$
You may find an implicit relation by solving the linear system
$$ \left(\matrix{b_n\\c_n}\right)=M^n\left(\matrix{-b\\-c}\right) $$
where
$$ M = \left(\matrix{-b&c\\-c&0}\right) $$
then we have the characteristic
$$ \lambda^2 + b\lambda + c^2 = 0 $$
The two eigenvalues are $\lambda_1,\lambda_2$, where $b = -(\lambda_1+\lambda_2), c^2 = \lambda_1\lambda_2$
The eigenvectors are
$$ \vec{v_1} = \left(\matrix{-\lambda_1\\c}\right), \
\vec{v_2} = \left(\matrix{-\lambda_2\\c}\right) $$
Thus
$$ M^n\left(\matrix{-b\\-c}\right) = -M^n(\vec{v_1}+\vec{v_2}) = -{\lambda_1}^n\ \vec{v_1} - {\lambda_2}^n\ \vec{v_2} $$
If $\lambda_1 = \lambda_2$ then we have the generalized eigenvectors
$$ \vec{v_1} = \lambda \left(\matrix{-1\\1}\right), \ \vec{v_2} = \left(\matrix{-2\\1}\right) $$
Then
$$ M^n\left(\matrix{2\lambda\\-\lambda}\right) = -M^n \lambda \vec{v_2} = \lambda^{n+1}\vec{v_2} + n\lambda^n \vec{v_1} $$
|
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When is the symmetric product of a matrix and a diagonal matrix negative definite on $\sum_i x_i=0$? Let $A$ be a matrix and $b$ be a strictly positive vector. Denote by $B=\text{diag}(b)$, a matrix with $b$ on its diagonal and zeros elsewhere. I am interested in the symmetric matrix
$$
C = AB+(AB)'.
$$
Now, consider the subspace $S:\sum_i x_i=0$. I want to find sufficient conditions on $A$ such that $C$ is negative definite on $S$, i.e. for all $x\in S$ we have $x'Cx\leq -d \left\Vert x\right\Vert $ for some $d>0$.
I can show that this is true if $A$ is full of ones except for the diagonal which is full of zeros. In this case, we can write $A=O-I$, where $O$ denotes a matrix full of 1 and $I$ the identity, and
$$\begin{align}
x'Cx &= x'((O-I)B+((O-I)B)')x \\
&= -2x'Bx \\
&< 0.
\end{align}$$
for $x\neq 0$ and where I used the fact that $x'OBx=0$.
I can find other hollow matrices $A$ (zeros on the diagonal) such that $C$ seems negative definite but there are also counterexamples, mostly when $A$ has many zeros. I am hoping that a more general result exists. I'm mostly interested in matrices $A$ that are binary, so if a result exists for these matrices it would be great.
|
With your definition of negative definiteness, $C$ is never negative definite when $n>2$. Note that your example is not correct: you do have $x'Cx=-2x'Bx<0$ for all nonzero $x\in S$, but you cannot find $d>0$ such that $-2x'Bx\le-d\|x\|^2$ on $S$ for every positive diagonal matrix $B$. In fact, when $x'=(1,-1,0)$ and $B=\operatorname{diag}(\epsilon,\epsilon,1)$, we have $-2x'Bx=-4\epsilon>-2d=-d\|x\|^2$ when $\epsilon$ is sufficiently small.
We can obtain, however, a necessary and sufficient condition for $x'Cx$ to be $\le0$ on $S$, regardless of $B$. Let $\{e_1,\ldots,e_n\}$ be the standard basis of $\mathbb R^n$. If $C$ is negative semidefinite on $S$, then when we pass $B$ to the limit $\operatorname{diag}(e_j)$, we have
$$
x'A\operatorname{diag}(e_j)\,x=\sum_ia_{ij}x_ix_j\le0\text{ on }S.
$$
We claim that $a_{ij}=a_{kj}$ for every $i,k\ne j$. Suppose the contrary that $a_{ij}>a_{kj}$. Then $x=me_i+(-m-1)e_k+e_j\in S$ and
$$
\lim_{m\to+\infty}\sum_ia_{ij}x_ix_j=\lim_{m\to+\infty}\left(ma_{ij}+(-m-1)a_{kj}+a_{jj}\right)=+\infty,
$$
which is a contradiction to the negative semidefiniteness of $C$ on $S$.
Now, suppose all off-diagonal entries among each column are equal to each other. Then
$$
\sum_ia_{ij}x_ix_j= a_{1j}\left(\sum_{i\ne j}x_i\right)x_j + a_{jj}x_j^2 =-(a_{1j}-a_{jj})x_j^2.
$$
So, for this quantity to be $\le0$ on $S$, we must have $a_{jj}\le a_{1j}$. Conversely, suppose $a_{jj}\le a_{ij}=a_{kj}$ whenever $i,k\ne j$. Then
$$
x'ABx=\sum_jb_{jj}x'A\operatorname{diag}(e_j)\,x=\sum_j-(a_{1j}-a_{jj})b_{jj}x_j^2
\le0.
$$
|
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|
How can I solve $a_{n+2} + 4a_n = n \cdot 2^n$ for non homogeneous solution?
Solve $a_{n+2} + 4a_n = n \cdot 2^n\;$ for non homogeneous $a_0 =1$ and $a_1 =0$
I was trying but I think it's wrong. I solved a question with $2^n$ but that had a form with $A \cdot 2^n$ but I think this requires a different solution than this.
|
We have here a linear recurrence relation.
Solve first the homogeneous equation : $$h_{n+2}+4h_n=0$$
It has characteristic equation $x^2+4=0$ thus $\pm 2i$ are roots.
The solutions of homogeneous equation are then given by $h_n=2^n(A\cos(\frac{n\pi}2)+B\sin(\frac{n\pi}2))$
Now we have to find a particular solution of the full equation: $$p_{n+2}+4p_n=n\times2^n$$
When the RHS is of the form $P(n)\times r^n$ with $P$ a polynomial, we have to look whether $r$ is a root of the characteristic equation or not.
Here $r=2$ is not a root, so we will search for a solution $Q(n)\times 2^n$ with $\deg Q=\deg P$.
[In case $r$ is a root, we search for $\deg Q=\deg P+1$, or even $+2$ is $r$ is a double root, and so on]
Thus with $p_n=(Cn+D)2^n$ we get $b_{n+2}+4b_n=2^n(8Cn+8C+8D)=n2^n\iff\begin{cases}C=\frac 18\\ D=-\frac 18\end{cases}$
Finally we apply the initial conditions:
The general solution is : $a_n=h_n+p_n=\left(\frac{n-1}8+A\cos(\frac{n\pi}2)+B\sin(\frac{n\pi}2)\right)2^n$
$a_0=-\frac 18+A=1\iff A=\frac 98$
$a_1=2B=0\iff B=0$
$a_n=\left(n-1+9\cos(\frac{n\pi}2)\right)2^{n-3}$
|
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|
Conflicting definitions of isogeneous and the relevance of separability The following questions seem to be related.
Firstly, let $E_1$ and $E_2$ denote elliptic curves. Silverman defines that $E_1$ and $E_2$ are isogeneous if and only if there exists a basepoint preserving regular map $E_1 \rightarrow E_2$ that is non-constant.
Galbraith et. al. further require that the map be separable.
Question 1. Are these conditions equivalent?
Secondly, Galbraith et. al. (on the same page) state that a separable isogeny can be identified with its kernel. However, it seems to me that any non-constant isogeny can be identified with its kernel (up to isomorphism, if we view that isogeny as an object in the relevant coslice category.) This is because non-constant isogenies are surjective.
Question 2. Is separability really necessary for Galbraith et. al.'s statement to be correct?
I'm not sure if these are secretly the same question or not. If they're not related, feel free to comment and I'll happily break this into two different questions.
|
*
*I don't think they're equivalent, since you can use google to see that authors use the phrase "separably isogenous." I don't know how to construct a counterexample but Frobenius twists might work.
*I think Galbraith is taking the kernel on $\overline{\mathbb{F}_p}$-points, not the kernel as a group scheme. The kernel in this sense doesn't identify the isogeny unless the kernel group scheme is étale.
|
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|
Calculating probability using combinations We have 50 chairs. 2 out of them are broken. What is probability that out of 5 taken chairs 4 or 5 are not broken?
We can calculate it by using $P(4 not broken) + P( 5 not broken)$
$P( 5 not broken ) = \frac{48}{50} * \frac{47}{49} * \frac{46}{48} * \frac{45}{47}* \frac{44}{46}$
But what about $P(4 not broken)$? If i take one chair probability that another chair will be broken/ not be broken is reduced. In this case can i use
$P(4 not broken) = C(5,4)* (\frac{48}{50} * \frac{47}{49} * \frac{46}{48} * \frac{45}{47})^{4} * \frac{2}{46}$ ?
Do i need to calcualte it using combinations?
Thanks for answers.
|
Note that the broken chair can be chosen either:
*
*first with probability: $$P_1 = \frac {2}{50}\times \frac {48}{49}\times \frac {47}{48}\times \frac {46}{47}\times \frac {45}{46} $$
*second with probability: $$P_2 = \frac {48}{50}\times \frac {2}{49}\times \frac {47}{48}\times \frac {46}{47}\times \frac {45}{46}$$
Surely you can complete the other cases.
|
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|
Probability of a letter coming from a city I got this question:
A letter is known to have come either from $\text{TATANAGAR}$ or from $\text{CALCUTTA}$. On the envelope just two consecutive Letters $\text{TA}$ are visible. What is the probability that the letters came from $\text{TATANAGAR}$?
My attempt:
Total number of cases $=3$, as there are three such pairs.
Probability $={{\text{Favourable}}\over{\text{Total}}}=\frac23$
However, the answer is given to be $7\over11$.
How is this possible? Please help.
|
$$T\equiv \text{comes from Tatanagar}\quad \text{and}\quad C\equiv \text{Comes from Calcutta}\rightarrow \left\{ \begin{array}{lcc}
p(T)=\dfrac{1}{2} \\
\\ p(C)=\dfrac{1}{2}
\end{array}
\right.$$
$$\text{Possible choices of two consecutive letters:}\left\{ \begin{array}{lcc}
\text{In TATANAGAR}: 8 \\
\\ \text{In CALCUTTA}: 7
\end{array}
\right.$$
$$\text{Event} D\equiv \text{The chosen couple of letters is TA}\rightarrow \left\{ \begin{array}{lcc}
p(D|T)=\dfrac{2}{8} \\
\\ p(D|C)=\dfrac{1}{7}
\end{array}
\right.$$
$$p(T|D)=\dfrac{p(T\cap D)}{p(D)}=\dfrac{\frac{1}{2}\cdot \frac{2}{8}}{\frac{1}{2}\cdot \frac{2}{8}+\frac{1}{2}\cdot \frac{1}{7}}=\dfrac{\frac{1}{8}}{\frac{7+4}{56}}=\dfrac{7}{11}$$
|
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|
Evaluate $\int (2x+3) \sqrt {3x+1} dx$ Evaluate $\int (2x+3) \sqrt {3x+1} dx$
My Attempt:
Let $u=\sqrt {3x+1}$
$$\dfrac {du}{dx}= \dfrac {d(3x+1)^\dfrac {1}{2}}{dx}$$
$$\dfrac {du}{dx}=\dfrac {3}{2\sqrt {3x+1}}$$
$$du=\dfrac {3}{2\sqrt {3x+1}} dx$$
|
Let us just split $2x+3 = \frac23 (3x+1)+\frac73$ and simplify to give us: $$I = \int (2x+3)\sqrt{3x+1}\, dx = \frac23 \int (3x+1)^{\frac32}\, dx+ \frac73 \int \sqrt {3x+1}\, dx$$ which can be easily solved.
|
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|
Directional derivative doubt I have two functions.
$e(t,z)=\cos(\omega t)\cos(\beta z)$, $h(t,z)=\frac{1}{\eta}\sin(\omega t)\sin(\beta z)$
Let's set the direction $\hat{u} = \hat{z} - c\hat{t}$, with $\dfrac{\epsilon_0}{c}=\dfrac{1}{\eta}$ and $\dfrac{\omega}{c}=\beta$. I can calculate $\dfrac{\partial( h-\frac{1}{\eta} e)}{\partial u}$ and I find it to be 0. So the quantity $h-\dfrac{1}{\eta} e$ is constant along that direction and since I know that for some $(t,z)$ points both $e$ and $h$ are $0$, that constant is $0$. So I'd like to say that if I change $z$ and $t$ according the $u$ direction, the ratio $\dfrac{e}{h}$ stays constant. It seems to me quite wrong but I'd like to get some comments about this misunderstanding.
|
There seems to be some confusion regarding directional derivatives vs. partial derivates.
*
*Your function $f(t,z)=h-\frac{1}{\eta} e= -\frac{1}{\eta} \cos(\beta(z+ct))$ is constant in the direction $\hat{u}=(-\frac{1}{c},1)$ in the $(t,z)$-plane, since:
$$ -\frac{1}{c} \partial_t f + \partial_z f = 0$$
*If you make a change of variables, say $u=z-ct$, $v=z+ct$, your function becomes $\hat{f}(u,v)=-\frac{1}{\eta} \cos (\beta v)$ which is independent of $u$, whence the partial derivative w.r.t. u is zero of this function in the coordiantes $(u,v)$. It seems that you are mixing the two notions.
If you happen to know that $f(t_0,z_0)=0$ then along the $\hat{u}$ direction,
$f(t_0-\frac{1}{c}s, z_0+s)=0$ for all $s\in {\Bbb R}$ implying that $h=\frac{1}{\eta} e$ along that line, so the ratio $e/h=\eta$ whenever defined (note that sometimes both vanish).
Explicitly, we have that $\beta(z+ct)=\beta z + w t$ is constant equal to $\pi/2+2\pi k$ for some integer $k$ and since $\;\cos(\pi/2-x)=\sin(x)\;$ we get $e(t,z)= \cos(wt) \sin(wt) = \eta h(t,z)$ along that line.
|
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|
Show that $RU$ intersects $AB$ at the midpoint (of $AB$). $PA, PB$ are tangents. $PU$ is a reflection of $PS$ over $PO$ where $O$ is the center of the circle. Show that $RU$ intersects $AB$ at the midpoint (of $AB$).
It looks so obvious. And I can see a lot of similar triangles. But I can't really solve it.
Please avoid solutions that use inversive/projective geometry.
|
Note that $TS$ and $RU$ meets on $AB$ because $P$ is the pole of the line $AB$.
On the other hand $TS$ and $RU$ meets on the simmetry axis $PO$ because they corresponds under the reflection.
Thus $TS\cap RU=AB\cap PO$ and $AB\cap PO$ is the mid point of $AB$ because $OP$ is axis of $AB$.
|
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|
Entropy solution - Burgers' equation Why doesn't the following problem have a solution for $t\ge1$?
$u_{t}+uu_{x}=0\\
u(0,x)=-x$.
The characteristics don't intersect and they cover the whole space above t=1.
|
All characteristics intersect at $(t,x) = (1, 0)$.
Indeed, the characteristic starting from $(0, x_0)$ is $x(t) = x_0 - x_0 \, t = (1-t) x_0$.
|
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|
Finding solutions to a non homogeneous differential equation knowing the solutions of its homogenous counterpart
Let $f(x)$ and $xf(x)$ be the particular solutions of a differential
equation $$y''+R(x)y'+S(x)y=0$$ Find the solution of the differential
equation $$y''+R(x)y'+S(x)y=f(x)$$ in terms of $f(x)$.
I was trying to find a solution if the form $(ax^2+bx+c)f(x)$. Now as $(bx+c)f(x)$ is a solution to the homogeneous equation, I just need to assume that $ax^2f(x)$ is a particular solution of the non homogeneous equation, and find the suitable value of $a$. But I am getting stuck. Please help.
|
Once you found two linearly independent solution of the homogeneous equation say $y_1$ and $y_2$, the particular solution is $$ y_p= u_1y_1+u_2 y_2$$Substitute in your inhomogeneous equation and solve for $u_1$ and $u_2.$ You may study the method of varion of constant for specific examples.
|
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|
Any example of a connected space that is not locally connected? A topological space $X$ is said to be locally connected at a point $x \in X$ if, for every neighborhood $U$ of $x$ (i.e. open set $U$ such that $x \in U$), there exists a connected neighborhood $V$ of $x$ such that $V \subset U$. If $X$ is locally connected at each of its points, then it is said to be locally connected.
Now is there any easy enough example of a connected space that fails to be locally connected at some point?
One example adduced by Munkres is the so-called topologist's sine curve, but I'm not sure why it is not locally connected.
Any other examples, please?
|
Another standard example that is even path connected but not locally connected is the comb space
$$ [0,1]\times\{0\} ~\cup~ \{0\}\times[0,1] ~\cup~ \bigcup_{k\ge 1} \{\tfrac1k\}\times[0,1] $$
Every neighborhood of the point $(0,1)$ contains the top end of infinitely many of the teeth, but if the neighborhood is so small that it doesn't contain the base of the comb, it will be disconnected -- and making it smaller cannot possibly make it connected.
|
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|
Continuous $f$ such that $f(x)=f(x^2)$ is constant? Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function such that $f(x)=f(x^2)$ for all $x\in\mathbb{R}$. I've proven that also $f\left(y^{(2^{-n})}\right)=f(y)$ for all $n\in\mathbb{N}, y\geq0$. How can I deduce that $f$ is constant?
|
If $f$ is continous, for all sequence $(x_n)_{n \in \mathbb{N}}$ such as $x_n \underset{n \rightarrow +\infty}{\rightarrow}x$ then
$$
f\left(x_n\right) \underset{n \rightarrow +\infty}{\rightarrow}f\left(x\right)
$$
It should help you conclude.
|
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|
Counterexample to $M$ a $R$-module or rank $n$ then if $N\leq M$, $N$ is a $R$-module of rank $\leq n$ I've seen the following theorem
Let $R$ be a PID (principal ideal domain), If now $M$ is a free $R$-module of rank $n$ and $N\leq M$ a submodule then $N$ is a free $R$-module of rank $\leq n$.
I would like to see the dependence on the PID through a counterexample.
Can I build some counterexample with $(2,x) \trianglelefteq \mathbb{Z}[x]$?
Could anything of the following be used?
Let $R=\mathbb{Z}[x]$ and $M=\mathbb{Z}[x]$ be a $R$-module then $\{1\}$ is a basis of $M$. Because, choose a $f\in M$, then it is possible to create it through choosing $\phi = f \in R$ and considering $\phi \cdot 1 = f$. Now consider $N=(2,x)$ which is submodule of $M$ (right?)
I'm somewhat confused of the rank of $N$ now. $\{1\}$ still seems to be a generator for the module? But this generates to much?
|
Take any non-PID $R$ and a projective $R$-module that isn't free. One definition of projective module is that $P$ is projective if there exists $N$ such that $P \oplus N$ is a free $R$-module. In this case $P$ is a submodule of the free module $P \oplus N$ but it's not free.
If $R$ is local or a PID then a projective $R$-module is free.
For example, you can take $R = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ and then the module $M = \mathbb{Z}/2\mathbb{Z} \times 0$. This is clearly a direct summand of $R$ itself, but it's not free.
|
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|
$P(X \geq 2Y) = ?$ where $P(n) = 2^{-n} , n =1,2,3,...$. Let $X,Y$ are independent identically distributed random variables.Then $P(X \geq 2Y) = ?$ where $P(n) = 2^{-n} , n =1,2,3,...$.
So it is the discrete case, So in order to calculate this -
$P(X \geq 2Y) = \sum_{y=1}^{\infty} \sum_{x = 2y}^{\infty} 2^{-x-y} = \sum_{y=1}^{\infty}2^{-y} (\frac{2^{-2y}}{1 - 2^{-2}}) = \sum_{y=1}^{\infty} \frac{4}{3}. 2^{-3y} =\frac{4}{3}. \frac{2^{-3}}{1 - 2^{-3}} = \frac{4}{21}.$
Is the above correct?
|
Above Answer is not Correct. It should be
$P(X \geq 2Y) = \sum_{y=1}^{\infty} \sum_{x = 2y}^{\infty} 2^{-x-y}
= \sum_{y=1}^{\infty}2^{-y} (\frac{2^{-2y}}{1 - 2^{-1}})
= \sum_{y=1}^{\infty} \frac{2}{1}. 2^{-3y} =\frac{2}{1}. \frac{2^{-3}}{1 - 2^{-3}}
= \frac{2}{7} $
|
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|
Prove that $a(u,v)=\int_\Omega A\nabla u\cdot \nabla v$ is continuous if $A$ uniformly elliptic. Let $\Omega \subset \mathbb R^d$ smooth, bounded and connected domain. Let $A\in \mathbb R^{d\times d}$ symetric and uniformly elliptic, i.e. there is $C>0$ such that $$C^{-1}\|x\|^2\leq Ax\cdot x\leq C\|x\|^2.$$
How can I prove that $$a(u,v)=\int_\Omega A \nabla u\cdot \nabla v,$$
continuous ? I know that $$|a(u,v)|\leq \int_\Omega \|A\nabla u\|\|\nabla v\|.$$
I suppose that $|A\nabla u|\leq C|\nabla u|$, but I can't prove it since $|A\nabla u|$ is not $A\nabla u\cdot \nabla u$. I also tried as $$|A\nabla u|^2=A^2\nabla u\cdot \nabla u,$$
but is also $A^2$ uniformly elliptic ? If yes how can I prove it ? If no, how can I conclude ?
|
If $A$ is a symmetric matrix, then $Ax \cdot x \le C \|x\|^2$ implies that all eigenvalues of $A$ are bounded from above by $C$. Hence, $\|A x\| \le C \|x\|$.
|
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