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Proving two linear operators are equivalent iff they have the same rank From Halmos's Finite-Dimensional Vector Spaces, #7 section 51, rank and nullity. Prove that two linear operators are equivalent iff they have the same rank.
Typically when Halmos phrases exercises like that it means that they are provable, but I don't see how it's true.
What if two operators on $\mathcal{V}$ both have disjoint ranges of dimension $\rho$? They have the same rank but they are certainly not equivalent because they map any $v \in \mathcal{V}$ to different subspaces. Also, what if the two operators map to the same ranges but one is a scalar multiple of the other ($A = \alpha A$)? They aren't equivalent but they have the same rank.
Could someone explain what Halmos might mean or how this might be true?
Edit: There's no definition of equivalence in the past sections or in the exercises. The closest thing is "similarity", where $B$ and $C$ are similar if $C = A^{-1}BA$ for some linear transformation A.
|
The linear transformations $A$ and $B$ are equivalent if there exist invertible transformations $P$ and $Q$ so that (Halmos page 87, Exercise 6, (b)):
$$
A = P^{-1}BQ.
$$
If this is the case, then the rank of $A$ and $B$ are the same (Halmos, $\S50$, THM3, (10)):
$$
\rho(A) = \rho(P^{-1}BQ) = \rho(P^{-1}B)=\rho(B).
$$
To prove the converse, we do the following.
Suppose $k=\rho(A)=\rho(B)$ and that we are working in a vector space $V$ with dimension $n$.
Let $\{x_{k+1},\dots,x_n\}$ be a basis of the null space of $A$.
Add vectors $\{x_1,\dots,x_k\}$ to get a basis of $V$.
Do the same thing with $B$ to get a basis $\{y_1,\dots,y_k,y_{k+1},\dots,y_n\}$ of V, in which $\{y_{k+1},\dots,y_n\}$ is a basis of the null space of $B$.
Define the linear map $Q$ by:
$$
Qx_i = y_i,
$$
$i=1,\dots,n$.
It is clearly linear.
It is also easy to show that it is invertible (show that the null space of $Q$ is the zero subspace of $V$).
Now we will define $P$.
Set:
$$
z_i = Ax_i,
$$
and
$$
w_i = By_i,
$$
for $I=i,\dots,k$.
These sets span the ranges of $A$ and $B$, respectively.
We stop at $k$ because for $i=k+1,\dots,n$ the vectors $x_i$ and $y_i$ are mapped to zero.
We add to each set $n-k$ linear independent vectors to make two new bases of $V$, say $\{z_1,\dots,z_k,z_{k+1},\dots,z_n\}$ and $w_1,\dots,w_{k},w_{k+1},\dots,w_n$, respectively.
Make the linear, invertible map:
$$
w_i = Pz_i.
$$
We are done.
Observe that:
$$
PAx_i = Pz_i = w_i = By_i = BQx_i.
$$
|
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Probability of at least one random number out of 3 being greater than 3 other random numbers? $$\{?,?,?\mid?,?,?\}$$
There are 6 random numbers (drawn from arbitrarily large pool). What is the probability that biggest number lies in second half? Answer is $1/2$ but I tried to solve it with combination. Can you show me where I went wrong?
I thought, for one particular number in second half, the probability that it is greater than all three points in first half is $1/8$ and likewise probability that it is not greater than all three points in first half is $7/8$, so
$${3 \choose 1} \left(\frac18\right)^1 \left(\frac78\right)^2$$ is probability that only of them is bigger than first half
$${3 \choose 2} \left(\frac18\right)^2 \left(\frac78\right)^1$$ is probability that two of them are bigger than first half
$${3 \choose 3} \left(\frac18\right)^3 \left(\frac78\right)^0$$ is probability that all three of them are bigger than first half
Total is $\dfrac{169}{512}$ not $\dfrac{256}{512}$. So what is my mistake?
Edit: I think I figured out, biggest number, rank 6 must be in second half; so I can choose other two numbers
(5 choose 2) different ways * 3! configuration for right side * also 3! configuration for left side;
divided by 6! equals to 1/2. now i can generalize it to solve secretary problem, : )
|
Your problem is that the probability that the fourth number is greater than the preceding three is not $\frac 18$, it is $\frac 14$. Your calculation of $\frac 18$ assumes that the fourth number being greater than the first and the fourth number being greater than the second are independent. They are not. If the fourth number is large, it has a good chance of being greater than any of the first three.
|
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Congruences- find the number of solutions
*
*Find the number of solutions of $ a_1x_1+ ... +a_nx_n =$ b (mod m).
I was thinking of denoting $ d = (a_1,a_2,...,a_n,m)$.
Then the congruence has no solutions if $d$ doesn't divide $b$. And it has solutions when $d$ divides $b$.
I was thinking of using induction, but I didn't have a clear idea.(I think it may have $dm^{n-1}$ solutions ??)
Could someone please help me show the solutions of these questions more concretely?
1. Find the number of solutions to the congruence $ x^3-2x^2-x+2 = 0$ (mod 143) without finding the actual solutions.
|
First Question:
Theorem :
The congruence $a_1x_1+\cdots+a_nx_n\equiv b\mod m$ has exactly $gcd(a_1,\cdots,a_n,m)m^{n-1}$ when $gcd(a_1,\cdots,a_n,m)$ divides $b$, and no solution otherwise.
Proof: by induction:
*
*For $n=1$ let $d=gcd(a_1,m)$, $a=\frac{a_1}{d}$, $b'=\frac{b}{d}$ and $m'=\frac{m}{d}$ so the equation $ax \equiv b' \mod m'$ has exactly on solution $x_0$ in $\mathbb{Z_{m'}}$. which gives us $gcd(a,m)$ in $\mathbb{Z_{m}}$ ( the solutions are $x_0,x_0+m',\cdots,x_0+m'(d-1)$)
*Assume that the congruence $a_1x_1+\cdots+a_nx_n\equiv b\mod m$ has exactly $gcd(a_1,\cdots,a_n,m)m^{n-1}$ when $gcd(a_1,\cdots,a_n,m)$ divides $b$ for every integers $a_1,\cdots,a_n,m,b$. let's consider a congruence with $n+1$ variables:
$$a_1x_1+\cdots+a_nx_n+a_{n+1}x_{n+1}\equiv b\mod m \ \ \ \ (*)$$
with surely the constraint $gcd(a_1,\cdots,a_n,a_{n+1},m)$ divides $b$, let $d=gcd(a_1,\cdots,a_n,m)$ and consider that $x_{n+1}$ is fixed,the equation $$a_1x_1+\cdots+a_nx_n\equiv b-a_{n+1}x_{n+1}\mod m \ \ \ (**)$$ has exactly $dm^{n-1}$ solution for every fixed $x_{n+1}$ such that $d|b-a_{n+1}x_{n+1}$ or $a_{n+1}x_{n+1}=b \mod d$ and ther is no solution otherwise. Now let's count the number of possible values of $x_{n+1} \mod m$, every possible value of $x_{n+1}$ it's a solution of :
$$ \left(\frac{ma_{n+1}}{d} \right)x_{n+1}\equiv \frac{bm}{d} \mod m $$
and because as hypothesis, $gcd(a_{n+1},d)=gcd(a_1,\cdots,a_{n+1},m)$ divides $b$ so $k=gcd\left(\frac{ma_{n+1}}{d},m\right)=m\frac{gcd(d,a_{n+1})}{d}$ divides $\frac{bm}{d}$ hence this equation has exactly $k$ solutions, so there is exactly $k$ values possible of $x_{n+1}$ for which the equation $(**)$ has solutions and has for each value of $k$ $dm^{d-1}$ solutions . So the total number of solutions of $(*)$ is exactly:
$$kdm^{n-1}=gcd(a_1,\cdots,a_{n+1},m)m^{n} $$
which terminates the proof.
Second Question:
As pointed by @marty cohen, we have :
$$x^3-2x^2-x+2=(x-2)(x-1)(x+1) $$
this signifies that $x^3-2x^2-x+2\equiv 0 \mod 143$ if and only if $143$ divides $(x-1)(x+1)(x-2)$ so this is equivalent to $13$ divides one factor among $x-1$,$x+1$ and $x-2$ and $11$ also divides one factor among this factors. but every time we choose two factors we will deal with an equation of the form :
$$ x\equiv a_1 \mod 13\\x\equiv a_2 \mod 11$$
and because $gcd(11,13)=1$, the Chinese remainder theorem gives us the existence of an unique solution $\mod 13*11$ so the number of solutions is the number of ways of choosing two factors among $3$ which is $3^2=9$.
Verification by walframalpha.
Generalization:Given $m$ primes $p_1\cdots p_m$ and $n$ different integers $a_1,\cdots,a_n\in \mathbb{Z_{p_1\cdots p_m}}$, the equation $(x-a_1)\cdots(x-a_n)\equiv 0 \mod p_1\cdots p_m$ has exactly $n^m$ solutions.
|
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Proof by induction that $P_n(a) \neq 0$ for $n>3$. Let $a,b,c$ be 3 non-zero coprime integers and $P_n(a)=a^n+\sum_{k=1}^{n}{{n\choose{k}}a^{n-k}(c^k-b^k)}$
Show that if $P_3(a) \neq 0$ then for all $n \geq 3, P_n(a)\neq 0$
Using mathematical induction, how can I prove it?
|
Hint: $\sum_{k=1}^{n}{{n\choose{k}}a^{n-k}(c^k-b^k)} = \sum_{k=1}^{n}{{n\choose{k}}a^{n-k}c^k}-\sum_{k=1}^{n}{{n\choose{k}}a^{n-k}b^k}=(a+c)^n-(a+b)^n$
|
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Relationship between Fibonacci's secuence and $x^2 - x - 1$. On the end of Apostol's Mathematical Analysis' first chapter, one can find the following exercise (and I paraphrase):
Prove that the $n$-th term of the Fibonacci sequence is given by $$x_n = \frac{a^n - b^n}{a-b}$$ Where $a$ and $b$ are the roots of $x^2 - x - 1$.
Apostol also states that $x_{n+1} = x_n + x_{n-1}$. I've tried a proof by induction on $n$ in this formula. I assumed it hold, and I went on with the second step: I assume it holds for $k\leq n$ and tried to show it must hold for $n+1$:
\begin{array}{rcl}
x_{n+1} & = & x_n + x_{n-1} \\[0.2cm]
& = & \displaystyle \frac{a^n - b^n}{a-b} + \frac{a^{n-1} - b^{n-1}}{a-b} \\
& = & \displaystyle \frac{(a^n + a^{n-1})-(b^n + b^{n-1})}{a-b}
\end{array}
So, if I prove that $a^n + a^{n-1} = a^{n+1}$ (and the same with $b$), I should be done. So, as a lemma, I try to demonstrate this by induction, but it doesn't seem to hold.
My questions, then, are: Is there another way of proving this statement?, Is my lemma true?
|
$$x^2-x-1=0$$
$$x^2=x+1$$
Multiply that by $x$ and you get:
$$x^3=x^2+x$$
Then replace $x^2$ by $x+1$, which yields:
$$x^3=2x+1$$
Repeat this process. Using induction, you can easily prove that:
$$x^n=x_nx+ x_{n-1} $$
Since $a$ and $b$ are the solutions of $x^2-x-1=0$ then they are also the solutions of $x^n=x_nx+1$,which means that:
$$a^n=x_na+x_{n-1}$$
$$b^n=x_nb+ x_{n-1} $$
Subtracting these two equations yields:
$$a^n-b^n=x_n(a-b)$$
$$x_n=\frac{a^n-b^n}{a-b}$$
|
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Is $\mathbb Z_1$ a subgroup of $\mathbb Z_{10}$ $\mathbb Z_1 = \{0, 1\} \subseteq \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\} = \mathbb Z_{10}.$
Since $0 + 1 = 1$, $1 + 0 = 1$ and $1 + 1 = 0$, the identity in $\mathbb Z_1$ is $0$ and the inverse element is $1.$ Since $0 + 1 \in \mathbb Z_1$, the set $\mathbb Z_1$ is closed under addition. Therefore $\mathbb Z_1$ is a subgroup $\mathbb Z_{10}.$ Does that check out?
|
First of all, $\mathbb{Z}_1$ would be defined as $\{0\}$ if it were ever used. What you are referring to is $\mathbb{Z}_2$. Second, your proof is not correct. The $0$ and $1$ in the first group are completely unrelated to the $0$ and $1$ in the second group. The subset $\{0,1\}$ of $\mathbb{Z}_{10}$ is not a subgroup because for example $1+1=2$ is outside the set.
There is also a bit of abuse of notation going on here which can lead to confusion. These elements are not actually numbers, they are congruence classes of integers. To denote elements of $\mathbb{Z}_2$ it is perhaps most correct to write $[0]_2$ and $[1]_2$ and elements of $\mathbb{Z}_{10}$ could be written $[3]_{10}$ etc. to emphasize that these are not numbers but rather congruence classes modulo an integer. This way congruence classes modulo different integers would be less likely to be confused with each other.
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Open balls in euclidean space are homeomorphic to the whole space The following question is from Fred H. Croom's book "Principles of Topology"
Prove that each open ball $B(a,r), a\in \mathbb{R}^n, r>0$, considered as a subspace of $\mathbb{R}^n$, is homeomorphic to $\mathbb{R}^n$.
After much studying, I concluded the first way to approach this problem would be by showing that the unit open ball $B(\theta,1)$ with center and radius 1 is homeomorphic to $\mathbb{R}^n$. Afterwards, show how any open ball $B(a,r)$ is topologically equivalent to $B(\theta,1)$, thus ending my proof. However, I am having a hard time showing that $B(\theta,1)$ is homeomorphic to $\mathbb{R}^n$. Any suggestions?
I want to thank you for taking the time to read this question. I greatly appreciate any assistance you provide.
|
Hing: Take $B(0,1)$. The let $f(x) = \frac{x}{1-|x|}$.
|
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Is $\alpha$ algebraic or not? I am trying to find out if $\alpha = \sqrt[]{2} - i$ for $F = \mathbb{Q}$ is algebraic or not.
I found that $\alpha^2 = 1 - 2i\sqrt[]{2}$ and have been trying to construct a suitable polynomial, but I have so far been unable to. Does anyone have a nice way to do this?
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Hint: The sum of algebraic numbers is algebraic. One proof uses the resultant.
Wolfram alpha calculates the minimal polynomial of $\sqrt{2}-i$ to be $x^4-2x^2+9$. One way to come up with this on your own is to consider $1,\alpha,\alpha^2,\alpha^3,\alpha^4$ as vectors in $\mathbb{Q}[1,\sqrt{2},i]$ and find a linear combination summing to zero. We know that such a linear combination must exist since the minimal polynomials of $\sqrt{2},i$ have degree 2, and so the minimal polynomial of $\sqrt{2}-i$ has degree at most 4.
|
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For what values of $k$ is $g(x)=x^3+kx^2+x$ one-to-one? I need to find for what values of $k$ $g(x)=x^3+kx^2+x$ is one-to-one. I tried finding for what values it is strictly increasing and got the derivative to be $3x^2+2kx+1>0$, but I'm not really sure where to go from there since there are two variables.
|
Hints:
Your function $g(x)$, being a cubic polynomial, is one-to-one if and only if the derivative has at most one root. One root is allowed.
You can tell how many roots a quadratic equation $ax^2+bx+c=0$ has by examining its discriminant $b^2-4ac$. If the discriminant is positive, there are two roots; if zero, one root; if negative, no real roots.
So examine the discriminant of the derivative of $g(x)$. That discriminant will depend on $k$, so find which values of $k$ will make the discriminant non-positive.
|
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If Q is a matrix of orthonormal bases, is Q times its transpose anything special? I know that if Q is a matrix of orthonormal bases then Q transpose times Q is the identity matrix. Is Q times Q transpose anything special?
|
It is also the identity matrix. This holds in general: if $B$ is a left inverse of a square matrix $A$, i.e. $BA=I$, then $B$ is also a right-inverse of $A$, i.e. $AB=I$.
|
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Volume of a Solid, $x^2 - y^2 = a^2$ The question is
Find the volume of a solid rotated around the y axis, bounded by the given curves: $$x^2 - y^2 = a^2$$
$$x = a + h$$
I am lost by the number of variables in this question and the question does not tell me what kind of variables they are, only that they are both greater than 0.
|
This is known as a hyperboloid of revolution of one sheet.
Forget the second equation, it only means that to get both x and y real, x should be equal or grater than a by a small amount h. $y > x$ always in the first quadrant for a quarter branch of this (rectangular) hyperbola.The limits of y are $0$ and $y_1$ in:
$$ x^2 = y^2 +1 $$
$$ V = \pi \int x^2 dy = \pi \int_0^{y1} (1+y^2) dy, $$ where I have chosen $y $ as the independent variable. And you can take it from there.
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In an equilateral spherical triangle, show that $\sec A=1+\sec a$ Q. In an equilateral spherical triangle, show that $\sec A=1+\sec a$
So $A$ is the vertex or the angle of the triangle and $a$ is the side of the equilateral spherical triangle.
I started off the proof by using the law of cosines:
$$\cos(a)-\cos(a)\cos(a)=\sin(a)\sin(a)\cos(A)$$
and after simplifying it a bit, I obtained:
$$\cos(a)-\cos^2(a)=\sin^2(a)\cos(A)$$
I replaced $\sin^2(a)$ with $(1-\cos^2(a))$.
Then I obtained:
$$\cos(a)-\cos^2(a)=(1-\cos^2(a))\cos(A)$$
and I realized on the left side, I can pull a $\cos(a)$. So through factoring:
$$\cos(a)(1-\cos(a))=(1-\cos^2(a))\cos(A)$$
Either I'm not seeing it but I do not how to proceed after this. If anyone can help, I'd like that. Thanks!
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Let $\,x:=\cos(a)\,$ and $\,X:=\cos(A).\,$ Then
$$ 1+\sec(a)-\sec(A) = 1+\frac1x-\frac1X = \frac{-x+X+xX}{xX}.$$
$$ \cos(a)(1-\cos(a))-(1-\cos^2(a))\cos(A)=x(1-x)-(1-x^2)X=(x-1)(-x+X+xX).$$
Thus, if $\,x\ne 1,\,$ then
$\, \cos(a)(1-\cos(a))=(1-\cos^2(a))\cos(A)\, $ implies that
$\, 1+\sec(a)=\sec(A).$
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Finding Fixed Point If there is a continuous function from the closed unit disk to itself such that it is identity map on boundary, must it admit a fixed point in the interior of the disk?
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No. Consider the set of vertical line segments which are intersections of lines $x=c$ with the closed unit disc $x^2+y^2 \le 1.$ Each of these segments goes from a lower circle boundary point $P_c$ to an upper one $Q_c.$ Along each such segment we can choose a mapping which moves all the interior points but keeps the endpoints fixed. An example of such a map for the interval $[0,1]$ is $h(t)=t^2.$ It would seem not too difficult to, in this way, choose the collection of maps on the intervals $P_c Q_c$ in such a way that the resulting pieced together mapping is continuous, and it would fix the points on the boundary of the unit disc by construction. Also by construction there would not be interior fixed points inside the unit disc.
An explicit map: First note that for $a\le b$ one can parametize the closed interval $[a,b]$ as $(1-t)a+tb$ where $t \in [0,1].$ Here $t=0$ goes with the smaller endpoint $a$ while $t=1$ goes with endpoint $b$.
We can apply this to the unit disc $x^2+y^2 \le 1$ by letting its points be of the form $(x,y(t))$ where $$y(t)=(1-t)[-\sqrt{1-x^2}] + t [\sqrt{1-x^2}].\tag{1}$$
Note that for $x= \pm 1$ this correctly gives $y=0$ for any $t,$ so this is a continuous parametrization (in the new variables $x$ and $t$) of the disc and its interior. the points where $t=0$ correspond to the lower boundary of the disc, while points with $t=1$ go with the upper boundary.
Now to define the map in these terms, we map the point $(x,y(t))$ onto the point $(x,y(t^2)).$ In general the value of $t$ uniquely determines the point $(1-t)a+tb$ in the interior of the closed interval $[a,b]$, and also at the endpoints provided $a<b$. This assures us that our mapping has no fixed points interior to the disc, while it is clearly continuous. It fixes every point on the boundary of the disc, since $0^2=0$ and $1^2=1$ at the two extremes of $[0,1]$.
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If $ f(x \cdot f(y) + f(x)) = y \cdot f(x) + x $, then $f(x)=x$ Let $ f : \mathbb{Q} \rightarrow \mathbb{Q} $ be a function which has the following property:
$$ f(x \cdot f(y) + f(x)) = y \cdot f(x) + x \;,\; \forall \; x, y \in \mathbb{Q} $$
Prove that $ f(x) = x, \; \forall \; x, y \in \mathbb{Q} $.
So far, I've found that $f(f(x)) = x$, $f(0) = 0$ and $f(-1) = -1$.
(For $f(0)=0$, we substitute $x=0$ to arrive at $f(f(0))-yf(0)$ identically $0$ for all rational $y$; for $f(f(x))=x$, we substitute $y=0$ and use $f(0)=0$. For $f(-1) = -1$, substitute $x=y=-1$ to get $f(0)=-f(-1)-1$, and use $f(0)=0$.)
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If you have $f(f(x))=x$ it means that the function is onto
Furthermore, assume that $f(y_1) = f(y_2)$ for some $y_1 \neq y_2$
$$ f(-f(y_1)-1 ) =f(-f(y_1)-1 ) $$
$$\Rightarrow -y_1 -1 = -y_2 -1 $$
$$\Rightarrow y_1 = y_2 !! $$
Thus the function is $1$ $to$ $1$, its inverse $f^{-1}$ exists,
in addition to $f(f(x))=x$, we have $f^{-1}=f$
From above, you get $f(f(1)+1) = f(1)+1 $
$f^{-1} = f \Rightarrow f(yf(x)+x) = f^{-1}(yf(x)+x) = xf(y) +f(x) $
For some $x_0=f(1)$, $f(x_0)=f(f(1))=1$ and putting $y=1$
$x_0 f(1) + f(x_0) = f(1 f(x_0) + x_0 )$
$\Rightarrow f(1)^2 +1 = f(f(1)+1) = 2f(1) \Rightarrow f(1)=1$
$\Rightarrow f(y+1)=f(y)+1$ by putting $x=1$ into $f(yf(x)+x) = xf(y) +f(x) $
From the original definition, putting $y=1$ gives $f(x+f(x)) = f(x) +x$
Since $f(x)$ is onto, $f(x)+x$ is also onto(to be proved), therefore we can find $x$ such that $y=x+f(x)$ for all $y$
then $f(y) =y$
I cannot prove that it is onto... However, I found another way by looking at the answer of Willard Zhan. He has proven $f(1/q) = 1/q$ for integers $q$
For all $\frac{p}{q}$, it can be always written as $\frac{m+1}{q}$, where $m$ is also an integer.
putting $y=m$ and $x=\frac{1}{q}$ into $f(yf(x)+x)=xf(y)+f(x)$
$$f\left( \frac{m+1}{q}\right)=f\left( \frac{m}{q} +\frac{1}{q}\right) = \frac{f(m)+1}{q} = \frac{m+1}{q}$$
since we have proved $f(m)=m$ for integers. I think it completes the proof.
|
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|
Find generating function for sequence I am suppose to find generating function for sequence $(e_n)_0^\infty$ where $e_n$ is number of ways how to write a number $n$ as a sum of four natural odd numbers ($e_n$ is basically a number of ordered fours $(\alpha, \beta, \gamma, \delta)$ odd natural numbers that $\alpha + \beta + \gamma + \delta = n$)
I don't know how should I start or "imagine" this problem.
|
I’ll get you started. Consider the product
$$(x_1+x_1^3+x_1^5+\ldots)(x_2+x_2^3+x_2^5+\ldots)(x_3+x_3^3+x_3^5+\ldots)(x_4+x_4^3+x_4^5+\ldots)\;;$$
a typical term has the form $x_1^\alpha x_2^\beta x_3^\gamma x_4^\delta$, where $\alpha,\beta,\gamma$, and $\delta$ are odd positive integers. If you were to drop the subscripts on the indeterminates, that would be $x^{\alpha+\beta+\gamma+\delta}$. And if you were to collect like powers, the coefficient of $x^n$ would be the number of terms $x^{\alpha+\beta+\gamma+\delta}$ with $\alpha+\beta+\gamma+\delta=n$. In other words, it would be $e_n$. Thus,
$$\sum_{n\ge 0}e_nx^n=\left(x+x^3+x^5+\ldots\right)^4\;.$$
Can you take it from here to get the desired generating function?
|
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|
Probability of (1 + min(X,Y))/(1 + min(X,Z))? I have been trying to derive the probability $\Pr \left[ {\frac{{1 + \min \left( {X,Y} \right)}}{{1 + \min \left( {X,Z} \right)}} < c} \right]$, where X, Y, and Z are independent and follow exponential distribution with parameters $λ_x$, $λ_y$, and $λ_z$, respectively. c here is a constant. What I did is briefly as follows:
$
\Pr \left[ {\frac{{1 + \min \left( {X,Y} \right)}}{{1 + \min \left( {X,Z} \right)}} < c} \right] \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(1)\\
= \Pr \left[ {X < Y,X < Z} \right]\\
+ \Pr \left[ {\frac{{1 + X}}{{1 + Z}} < c} \right]\Pr \left[ {X < Y,X > Z} \right]\\
+ \Pr \left[ {\frac{{1 + Y}}{{1 + X}} < c} \right]\Pr \left[ {X > Y,X < Z} \right]\\
+ \Pr \left[ {\frac{{1 + Y}}{{1 + Z}} < c} \right]\Pr \left[ {X > Y,X > Z} \right] \;\;\;\;\;\;\;\;(2),$
from which I can obtain a closed-form expression. However, when I do simulation to verify my analytic result, I get
where the red curve is from analysis. I don't know why there is a gap between the results.
Do you think that going from (1) to (2) is problematic?
|
Let $B=(\frac{1 + \min \left( {X,Y} \right)}{1 + \min \left( {X,Z} \right)} < c)$. You have define four events:
\begin{align}
A_1&=(X < Y,X < Z)\\
A_2&=(X < Y,X > Z)\\
A_3&=(X > Y,X < Z)\\
A_3&=(X > Y,X > Z)\\
\end{align}
and then wrote \begin{align}
Pr(B)&=\sum_{i=1}^4P(B|A_i)P(A_i)\\
\end{align}
For $i=1$ you then write
\begin{align}
P(B|A_1)P(A_1)&=P(\frac{1 + \min \left( {X,Y} \right)}{1 + \min \left( {X,Z} \right)} < c|A_1)P(A_1)\\
&=P(\frac{1 + \min \left( {X,Y} \right)}{1 + \min \left( {X,Z} \right)} < c|X < Y,X < Z)P(X < Y,X < Z)\\
&=P(1 < c|X < Y,X < Z)P(X < Y,X < Z)\\
&=1\times P(X < Y,X < Z)\\
\end{align}
which is correct only if $c>1$ if $c\leq1$ then this first term is zero not one.
|
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|
Help in making an animated locus is needed
I have triangular cardboard ABC, right angled at C. As shown in the attached, initially, A and B are resting on the x- and y- axes respectively. A is then allowed to slide along the x-axis with B slides accordingly along the y-axis.
The locus of C has found to be:- $9x^2 – 16y^2 = 0$. But it needs to be shown in an animated form (in Geogebra if possible).
Considering I only have PASCAL programming experience, a detail explanation on how it is done is appreciated.
|
Here's what I tried in geogebra:
*
*Create point $A$ on X-axis.
*Make circle of radius $5$ centered at $A$.
*Intersect the circle and Y-axis at point $B$.
*Make angle of $53.13^\circ$ at $A$ and $36.87^\circ$ at $B$.
*Intersect these angles at $C$.
*Join line segments $AB$,$BC$ and $CA$ and hide everything else.
*Trace on for point $C$.
*Now just drag point $A$ on the X-axis.
Here are the results(The GIF is high quality so might take a couple seconds to load):
The locus of point $C$, as expected, is a straight line through origin with slope $\dfrac{3}{4}$. The other half of the locus is obtained when point $B$ is below the X-axis.
|
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|
Term for similarity transformation which is not a translation What's the best (i.e. most concise) term to refer to an orientation-preserving similarity transformation which is not a translation? Here are some descriptions I could think of, but all of them feel rather bulky. I hope they are as equivalent to one another as I think they are, and I hope there is something simpler equivalent to all of them.
*
*a direct similarity transformation which is not a translation (by a non-zero displacement)
*a homothety (possibly with factor $1$) followed by rotation (with possibly zero angle)
*an orientation-preserving similarity with at least one fixed point
*a direct similitude with a well-defined similitude center, or the identity
*what $z\mapsto a(z-c)+c$ describes in the complex plane, for some fixed $a,c\in\mathbb C$ (with $a\neq 0$)
As a native German, I tend to think about this using the German term “Drehstreckung” which literally translates to “rotation-dilation”. I'm somewhat surprised by the difficulty I have in finding an exact English translation for this concept.
|
I think that what you're looking for is simply called "affinity".
From Wikipedia:
Examples of affine transformations include translation, scaling, homothety, similarity transformation, reflection, rotation, shear mapping, and compositions of them in any combination and sequence.
|
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|
First order differential equation with non constant coefficients I have the following system :
$$\begin{cases}(t^2+1)x'(t)=tx+y+2t^2+1\\(t^2+1)y'(t)=-x+ty+3t\end{cases}$$
How can it be solved ?
What I have tried so far :
*
*polynomials of the first, second degree as solutions - didn't work
*One can notice that if we use $X=\begin{bmatrix}x\\y\end{bmatrix},A=\begin{bmatrix}t&1\\-1&t\end{bmatrix},B=\begin{bmatrix}2t^2-1\\3t\end{bmatrix}$ then the system becomes $(t^2+1)X'=AX+B$, and $t^2+1=\det A$. I'm pretty sure that this last result is supposed to help, but I haven't been able to find a way to use it.
|
We are given:
$$\begin{cases} (t^2+1)x'(t)=tx+y+2t^2+1\\ (t^2+1)y'(t)=-x+ty+3t\end{cases}$$
From the first equation $(t^2+1)x'(t)=tx+y+2t^2+1$, we have:
$$\tag 1 y = t^2 x' + x' - t x - 2 t^2 -1$$
Taking the derivative of $(1)$, yields:
$$\tag 2 y' = t^2 x'' + x'' + t x' -x - 4 t$$
Substituting $(1)$ and $(2)$ into the original second equation yields:
$$\tag 3 (t^2+1)((t^2+1)x'' + t x' - x - 4 t) = -x +t((t^2+1) x' - t x - 2 t^2 -1) + 3 t$$
Simplifying $(3)$, yields:
$$\tag 4 x'' = 2 \dfrac{t^3}{(t^2+1)^2} + 6 \dfrac{t}{(t^2+1)^2}$$
Integrating $(4)$ twice yields:
$$x(t) = c_1 + c_2 t + t \ln(t^2 + 1)$$
We can now use this result and $(1)$ to yield:
$$y(t) = -c_1 t + c_2 -1 + \ln(t^2+1)$$
|
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|
Proving a the square root of a function to be Riemann integrable How could I prove that the square root of a Riemann integrable function $f$ on a given interval, where $f(x) > 0$ is also Riemann integrable?
|
The most straightforward way is Reveillark's, the validity of which follows immediately from Lebesgue's criterion for Riemann-integrability.
Here is a direct proof that doesn't appeal to this result.
Let $f \geq 0$ be Riemann-integrable on $[a,b]$. Let $\varepsilon > 0$ be given. Since $f$ is integrable, there exist step functions $\varphi$, $\psi$ such that
$$0 \leq \varphi \leq f \leq \psi, \qquad \int_a^b (\psi - \varphi) \leq \varepsilon^2/(b-a).$$
Now $\sqrt{\varphi}, \sqrt{\psi}$ are step functions satisfying $\sqrt{\varphi} \leq \sqrt{f} \leq \sqrt{\psi}$ and
$$\int_a^b \left( \sqrt{\psi} - \sqrt{\varphi} \right) \leq \int_a^b \sqrt{\psi - \varphi} \leq (b-a)^{1/2} \left[\int_a^b (\psi - \phi)\right]^{1/2} \leq \varepsilon,$$
where the first inequality follows from $\sqrt{\psi} - \sqrt{\varphi} \leq \sqrt{\psi - \varphi}$ and the second follows from Cauchy-Schwarz. (Note that the argument is not circular because $\sqrt{\psi - \varphi}$ is a step function.)
Since $\varepsilon > 0$ was arbitrary, this shows that $\sqrt{f}$ is integrable.
|
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|
Problem with a calculation using homogeneous coordinates Suppose in a complex projective plane CP2, I use homogeneous coordinates
$$(x, y, z)$$
and the following transformation:
$$A =\begin{pmatrix}
\cos{\alpha} & -\sin{\alpha} & 0 \\
\sin{\alpha} & \cos{\alpha} & 0 \\
0 & 0 & 1
\end{pmatrix}
$$
The point $I=(1, i, 0)$ is transformed in $(e^{-i\alpha}, i e^{-i\alpha}, 0)$.
So $A(I) = e^{-i\alpha} I$ and $I$ is invariant.
Likewise for $J=(1,-i,0)$: $A(J) = e^{i\alpha} J$ and $J$ is invariant.
The point $P = I + J = (2,0,0)$ transforms into $P' = (2\cos{\alpha},2\sin{\alpha},0)$.
The matrix is linear and indeed:
$$A(P) = A(I+J) = A(I) + A(J) = e^{-i\alpha} I + e^{i\alpha} J = P'.$$
Now I call the projective transformation itself $B$, so:
$$B(I) = I, \quad B(J) = J \quad\mbox{and}\quad B(P) = P'.$$
I thougth the projective transformation would be linear as well. This would mean:
$$B(P) = B(I+J) = B(I) + B(J) = I + J =P.$$
Which is clearly false.
What error am I making here precisely?
|
Projective transformations take (projective) lines to (projective) lines but they don't preserve the "midpoints of lines" or "sums of points". In fact, midpoints and "sums" are not even defined in the projective geometry.
You can just consider $\mathbb{RP}^1$, homogenous coordinates and the identity. "$(1,0)+(0,1)$" is mapped to $(1,1)$ but "(1,0)+(0,2)" is not mapped to a multiple of $(1,1)$. There is no $+$ in the projective geometry.
|
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|
Global maxima of $f(x,y)=x^2y$ restricted to D Let $f(x,y) = x^2y$ and $D = \{(x,y): y\geq0 \land 2x^2+y^2 \leq a\}$ with $a>0$.
I need to find $a$ such that the global maxima of $f$ restricted to $D$ is $\frac{1}{8}$.
I found, using Lagrange multipliers, that $a$ = $\frac{3}{4}$. With that value, $f(\frac{1}{2}, \frac{1}{2}) = f(-\frac{1}{2}, \frac{1}{2}) = \frac{1}{8}$, and those points are local maxima.
Tha part where I am struggling with is to prove that those are global maxima.
The restriction now states that $2x^2+y^2 \leq \frac{3}{4}$, but from there I can't get a close enough bound for $x^2y$. What else can I try?
Thank you.
|
The function is monotonic in $y$, so that by definition, for each constant $x$, the maximum will be found on the upper half of the ellipse's boundary.
Thus, you can search for local maxima of:
$$f(x,y) = x^2\sqrt{a-2x^2}$$
Which will then be global maxima of $f(x,y)$ on $D$, and also happen to coincide with the local minima you already found.
|
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Find two linearly independent solutions of the differential equation $(3x-1)^2 y''+(9x-3)y'-9y=0 \text{ for } x> \frac{1}{3}$ I want to find two linearly independent solutions of the differential equation
$$(3x-1)^2 y''+(9x-3)y'-9y=0 \text{ for } x> \frac{1}{3}$$
Previously I have seen that the following holds for the differential equation $y''+ \frac{1}{x}y'-\frac{1}{x^2}y=0, x>0$:
*
*We are looking for solutions of the differential equation of the form $x^r$. Then the function $x^r$ is a solution of the differential equation at $(0,+\infty)$ if:
$$r(r-1)x^{r-2}+ \frac{1}{x} r x^{r-1}- \frac{1}{x^2}x^r=0 \forall x >0 \Rightarrow r=1 \text{ or } r=-1$$
*
*So, the functions $y_1(x)=x, y_2= \frac{1}{x}$ are solutions of the differential equation and it also holds that they are linearly indepedent since $W(y_1, y_2) \neq 0$
For this differential equation
$$(3x-1)^2 y''+(9x-3)y'-9y=0 \text{ for } x> \frac{1}{3}$$
I thought the following:
$(3x-1)^2 y''+(9x-3)y'-9y=0 \text{ for } x> \frac{1}{3} \Rightarrow y''+ \frac{3}{3x-1}y'-\frac{9}{(3x-1)^2}y=0$
*
*We are looking for solutions of the differential equation of the form $\left( x- \frac{1}{3}\right)^r$.
Then the function $\left( x- \frac{1}{3}\right)^r$ is a solution of the differential equation at $( \frac{1}{3},+\infty)$ if:
$$r(r-1) \left( x- \frac{1}{3}\right)^{r-2}+ \frac{1}{ \frac{3x-1}{3}} r \left( \frac{3x-1}{3}\right)^{r-1}- \frac{9}{(3x-1)^2} \left( x- \frac{1}{3}\right)=0 \Rightarrow \dots \Rightarrow r= \pm 1$$
Therefore, the functions $z_1(x)=x- \frac{1}{3}, z_2(x)=\frac{1}{x- \frac{1}{3}}$ are solutions of the differential equation at $\left( \frac{1}{3}, +\infty\right)$.
$$z_1(x) z_2'(x)-z_1'(x) z_2(x)=\frac{-2}{x- \frac{1}{3}} \neq 0$$
So, $z_1, z_2$ are linearly independent solutions of the differential equation.
Thus, the general solution of $y''+ \frac{3}{3x-1}y'-\frac{9}{(3x-1)^2}y=0$ is of the form:
$$c_1 \left( x- \frac{1}{3} \right)+ c_2 \left( \frac{1}{x- \frac{1}{3}}\right) | c_1, c_2 \in \mathbb{R}, x> \frac{1}{3}$$
EDIT:
We set $t=x-\frac{1}{3}$ and we have:
$$\frac{dy}{dt}=\frac{dy}{dx} \frac{dx}{dt}=\frac{dy}{dx}$$
$$\frac{d^2y}{dt^2}=\frac{d}{dt} \left( \frac{dy}{dt}\right)=\frac{d}{dt} \left( \frac{dy}{dx} \right)=\frac{dx}{dt} \frac{d}{dx} \left( \frac{dy}{dx} \right)=\frac{d^2y}{dx^2}$$
$$y''(x)+ \frac{1}{x-\frac{1}{3}}y'(x)-\frac{1}{\left( x-\frac{1}{3}\right)^2}y(x)=0 \\ \Rightarrow y''(t)+\frac{1}{t}y'(t)-\frac{1}{t^2}y(t)$$
Two linearly independent solutions are $y_1(t)=t$ and $y_2(t)=\frac{1}{t}, y \in (0,+\infty)$.
Thus, two linearly independent solutions of $y''+\frac{1}{x-\frac{1}{3}}y'-\frac{1}{\left( x-\frac{1}{3} \right)^2}y=0, x> \frac{1}{3}$
are $y_1(x)=x-\frac{1}{3}, y_2(x)=\frac{1}{x-\frac{1}{3}}$
Is it right or have I done something wrong?
|
Hint
I do not know how much this could help you; so, please forgive me if this is off-topic.
If you define first $y=(3x-1)u$, $$(3x-1)^2 y''+(9x-3)y'-9y=0$$ rewrite $$(3 x-1)^2 \Big((3 x-1) u''+9 u'\Big)=0$$ for which order can be reduced and integration seems simple.
|
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|
Are the Unit Ball and Any other Ball Topologically Equivalent How would I correctly show that the unit ball $B(0,1)\subset \mathbb{R}^n$ and the ball $B(a,r) \subset \mathbb{R}^n$ are Topologically Equivalent?
I know I need to find a one-to-one function $f: X\rightarrow Y$ for which $f$ and $f^{-1}$ are both continuous to prove it is Topologically Equivalent; however, I am having a hard time finding such a function.
A function that came to mind is the equation of a circle or sphere. But I wouldn't know if that's correct.
I want to thank you for taking the time to read this question. I greatly appreciate any assistance you provide
|
By proving $f:B(0,1)\rightarrow B(a,r)$ defined by $f:x\mapsto a+rx$ is a homeomorphism.
|
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|
Question on additivity of Riemann Integral . Given a Riemann integrable function , $f:[a,b] \rightarrow \mathbb{R}$ and a sequence ${b_k\to b}$ ( $b_k \in [a,b]$ ) is the following use of additivity of integrals correct ? $\int_a^bf(x)dx=\int_a^{b_1}f(x)dx +\int_{b_1}^{b_2}f(x)dx+ \int_{b_2}^{b_3}f(x)dx +...+\int_{b_k}^{b_{k+1}}f(x)dx +...$ I realized this issue when I was giving an answer here : Is $\int_a^{b}f(x) dx = \lim_{k\rightarrow \infty } \int_a^{b_k}f(x)$? Also, what happens if we talk about the same issue in terms of lebesgue integration ?
|
The definition of the Riemann integral ensures that if $f$ is Riemann integrable on the interval $[a,b]$ then it is also Riemann integrable on any subinterval. It also ensures that $f$ is bounded, so assume $|f(x)|\leq M$ for all $x\in[a,b]$. Then, for each $k$ we have
$$\int_a^b f(x)\ dx-\int_{b_k}^bf(x)\ dx = \int_a^{b_1}f(x)\ dx+\int_{b_1}^{b_2}f(x)\ dx+\cdots+\int_{b_{k-1}}^{b_k}f(x)\ dx$$
Now, we have $|\int_{b_k}^bf(x)\ dx| \leq (b-b_k)M \to 0$. Thus the sum on the right converges to $\int_a^b f(x)$, as desired.
The same result is true if $f$ is Lebesgue integrable: set $f_k=\chi_{[a,b_k]}f$. Then
$$\int_a^b f_k(x)\ dx = \int_a^{b_k} f(x)\ dx = \int_a^{b_1}f(x)\ dx+\int_{b_1}^{b_2}f(x)\ dx+\cdots+\int_{b_{k-1}}^{b_k}f(x)\ dx$$
The left side converges to $\int_a^b f(x)\ dx$ by the dominated convergence theorem, which proves the desired result.
|
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|
How do I find the antiderivative of $sin(u)e^u$? So, I have the following indefinite integral:
$$\int sin(u)e^u \ du$$
I tried solving it using integration by parts, but it just kept repeating itself. What should I do?
|
Here is a nice way to do it: by parts,
$$\int e^u\sin u\,du=e^u\sin u-\int e^u\cos u\,du\ .$$
Consider also
$$\int e^u\cos u\,du=e^u\cos u+\int e^u\sin u\,du\ .$$
Eliminating the cos integral,
$$\int e^u\sin u\,du=e^u\sin u-e^u\cos u-\int e^u\sin u\,du$$
so
$$2\int e^u\sin u\,du=e^u\sin u-e^u\cos u$$
and hence
$$\int e^u\sin u\,du=\frac12(e^u\sin u-e^u\cos u)\ \ldots$$
. . . not forgetting plus $C$ of course ;-)
|
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Finding the largest constant $C$ such that $|\ln x−\ln y| \geq C|x−y|$ for all $x, y \in (0, 1]$ Find the greatest value of C such that
$|\ln x−\ln y|≥C|x−y|$
for any $x,y∈(0,1]$. What should my approach be? I can't think of much options.
|
To find the largest $C$ so that $|\log(x)-\log(y)|\ge C|x-y|$, note that, by the Mean Value Theorem, for some $t$ strictly between $x$ and $y$
$$
\frac{\log(x)-\log(y)}{x-y}=\frac1t
$$
The minimum of $\frac1t$ on $(0,1]$ is $1$. Therefore, the largest $C$ can be is $1$.
|
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|
What are classifying spaces actually classifying? Let $G$ be a group. When we say the classifying space of $G$ we are actually meaning the classifying space of the principal $G-$bundles because the notion of classifying spaces is about classifying the principal $G-$bundles and not about classifying groups. Is my understanding correct?
|
If $G$ is a topological group, $BG$ classifies principal $G$-bundles. More precisely, for any paracompact topological space $X$, there is a natural bijection between $[X, BG]$, the set of homotopy classes of maps $X\to BG$, and $\operatorname{Prin}_G(X)$, the set of isomorphism classes of principal $G$-bundles.
There is a principal $G$-bundle $EG \to BG$ which is called the universal principal $G$-bundle (see here for more about the construction of the bundle). The natural bijection is
\begin{align*}
[X, BG] &\to \operatorname{Prin}_G(X)\\
[f] &\mapsto f^*EG.
\end{align*}
Note, the topology on $G$ matters. With the usual topology on $O(n)$, $BO(n) = \operatorname{Gr}_n(\mathbb{R}^{\infty})$ which has infinitely many non-zero homotopy groups, but if $O(n)$ is equipped with the discrete topology, the classifying space is a $K(O(n), 1)$ and therefore has only one non-zero homotopy group.
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Can a set be closed in one topology but neither open nor closed in another? Can a set be closed in one topology but neither open nor closed in another?
If we say that the complement of a open set is a closed set, i.e. if $S \subseteq X$ is open then $X \setminus S$ is closed, I argue that depending on the topology, it could actually be the case. Consider the Euclidean topology on $\mathbb{R}$. Clearly any closed interval $[a,b]$ with $a,b \in \mathbb{R}$ is a closed set in this topology.
However, given the trivial topology on $\mathbb{R}$, any interval $[a,b]$ is neither closed or open.
But according to this Wikipedia page, an equivalent definition of a closed set is that "a set is closed if and only if it contains all of its limit points". And from this point of view, the inverval $[a,b]$ is clearly closed.
So i guess that my question is: How can two (supposedly) equivalent definitios of a closed sets, have a set be closed in one and not the other?
|
The definition of a limit point depends on the topology too.
Recall the definition of a limit point:
A limit point of a set $S$ in a topological space $X$ is a point $x$ (which is in $X$, but not necessarily in $S$) that can be "approximated" by points of $S$ in the sense that every neighbourhood of $x$ with respect to the topology on $X$ also contains a point of $S$ other than $x$ itself.
With this definition, and the trivial topology on $\mathbb{R}$, any point of $\mathbb{R}$ is a limit point of $[a, b]$.
|
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Proving that $7^n(3n+1)-1$ is divisible by 9 I'm trying to prove the above result for all $n\geq1$ but after substituting in the inductive hypothesis, I end up with a result that is not quite obviously divisible by 9.
Usually with these divisibility induction problems, it falls apart nicely and we can easily factorise say a 9 if the question required us to prove that the expression is divisible by 9. However in this case, I do not end up with such a thing.
My work so far below:
Inductive Hypothesis: $7^k(3k+1)-1=9N$ where $N\in\mathbb{N}$
Inductive Step:
$7^{k+1}(3k+4)-1 \\ =7\times 7^k(3k+1+3)-1 \\ =7\times \left [ 7^k(3k+1)+3\times 7^k \right ] -1 \\ = 7 \times \left [ 9N+1 + 3 \times 7^k \right ] -1 \\ = 63N+21\times 7^k+6 \\ = 3 \left [ 21N+7^{k+1}+2 \right ]$
So now I need to somehow prove that $21N+7^{k+1}+2$ is divisible by 3, but I'm not quite sure how to proceed from here...
|
First, show that this is true for $n=1$:
$7^1\cdot(3\cdot1+1)-1=9\cdot3$
Second, assume that this is true for $n$:
$7^n\cdot(3n+1)-1=9k$
Third, prove that this is true for $n+1$:
$7^{n+1}\cdot(3(n+1)+1)-1=$
$7^{n+1}\cdot(3n+3+1)-1=$
$7^{n+1}\cdot(3n+1+3)-1=$
$7^{n+1}\cdot(3n+1)+7^{n+1}\cdot(3)-1=$
$7\cdot\color{red}{7^n\cdot(3n+1)}+3\cdot7^{n+1}-1=$
$7\cdot(\color{red}{9k+1})+3\cdot7^{n+1}-1=$
$63k+7+3\cdot7^{n+1}-1=$
$63k+3\cdot7^{n+1}+6=$
$\color{blue}{3(21k+7^{n+1}+2)}$
Now:
*
*$7\equiv1\pmod3\implies$
*$\forall{m}\in\mathbb{N}:7^m\equiv1\pmod3\implies$
*$7^{n+1}\equiv1\pmod3\implies$
*$7^{n+1}+2\equiv3\pmod3\implies$
*$7^{n+1}+2\equiv0\pmod3\implies$
*$3|7^{n+1}+2\implies$
*$\exists{p}\in\mathbb{N}:7^{n+1}+2=3p\implies$
*$3(21k+7^{n+1}+2)=3(21k+3p)=3(3(7k+p))\color{blue}{=9(7k+p)}$
Please note that the assumption is used only in the part marked red.
|
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|
Counterexamples for "every linear map on an infinite dimensional complex vector space has an eigenvalue" Every linear map on a finite dimensional complex vector space has an eigenvalue. Not so in the infinite case.
I'm interested in nice counterexamples anyone might have.
Here's one:
Consider the vector space $\mathbb C^\infty$ of sequences and the right shift map $R$ defined by
$$R(a_1, a_2, a_3, ...) = (0, a_1, a_2, a_3, ...)$$
$R$ has no eigenvalue (using the usual convention that there must be a non-trivial eigenvector).
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There is an easy generalisation of the example in the question. In the polynomial ring $\Bbb C[X]$, the $\Bbb C$-linear operator $\phi: Q\mapsto PQ$ of multiplication by a fixed non-constant polynomial $P$ cannot have any eigenvalues. This is simply so since if $Q\neq0$ were an eigenvector for$~\lambda$, then one would lave $(P-\lambda)Q=0$, contradicting that $\Bbb C[X]$, is an integral domain. The example in the question is just the case $P=X$.
You can generalise this further using other $\Bbb C$-algebras that are integral domains, of which there are many.
It might be worth noting, and I did in this answer, that though there are no eigenvalues, one can make (the image of) any nonzero vector become an eigenvector for any desired eigenvalue by taking a suitably chosen quotient of $\Bbb C[X]$ by a $\phi$-stable subspace.
|
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Could someone explain steps? I am learining about logarithm equations, and i can´t seem to understand how to solve such an equation, could someone help?
I must solve the equation/find $x$ for:
$$2^{2x} - 3\cdot2^x - 10=0$$
The final answer should be
$x=\dfrac{\log5}{\log2}$
|
Assuming you mean $2^{2x} - 3\times2^x - 10 = 0$, then since this is a quadratic in $2^x$ we can factorise as $(2^x+2)(2^x-5)=0$. Since $2^x>0$ $\forall x$, $2^x+2\neq0$, so we must have $2^x=5$. Taking logarithms and applying the power rule, $x\log2=\log5$, whence $x=\frac{\log5}{\log2}$ as required.
If you're having trouble seeing the factorisation, then try setting $y=2^x$ and solving the resulting quadratic for $y$, then using logarithms to obtain $x$.
|
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Limits and exponentials Asked to find $\lim_{n\to\infty}a_n$ where
$$a_n = \left(1+\dfrac1{n^2}\right)^n$$
I know that the limit = 1, and can get to this by saying $\ln a_n=n\ln\left(1+\dfrac1{n^2}\right)$ and going from there.
My question is: would it also be enough to simply direct substitute and say that…?
$$\left(1+\left(\frac1\infty\right)^2\right)^\infty = (1+0)^\infty = 1$$
|
In short no. If you look at the limit that yields $e$ And did the same trick what would you get?
$$
\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n = e\neq \left(1+\frac{1}{\infty}\right)^\infty = 1
$$
The last equality is more to match with your statement question than being a proper definition of 1.
|
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|
Convergence of $\sum^\infty_{n=1} \left(\frac 1 n-\ln(1+\frac 1 n)\right)$
Does the series: $\displaystyle\sum^\infty_{n=1} \left(\frac 1 n-\ln(1+\frac 1 n)\right)$ converge?
We know that $\ln (1+x) <x$ for $x>0$ so this series behaves like $\frac 1 n - \frac 1 n$ which is $0$, but that's just intuition.
I think the only test that would work is the limit comparison test, but I can't find an expression that would work. Any hints?
Note: no integral test nor integrals, nor Taylor, nor Zeta.
|
Notice that
$$
x - \ln(1+x) = \int_0^x \frac{t}{1+t}dt
$$
hence for all $x \geq 0$,
$$
0 \leq x - \ln(1+x) \leq \int_0^x t dt = \frac{x^2}{2}.
$$
Now, you can use the comparision test.
Edit: if you don't want integrals, then you can prove the same inequality by studying the variations of the functions
$$
f(x) = x - \ln(1+x),\qquad g(x) = x - \ln(1+x) - \frac{x^2}{2}.
$$
|
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Differentiable approximation $f(x) = x$ for $x>0$ and $0$ otherwise. I would like to find a twice continuously differentiable approximation of
$$f(x)= \begin{cases}
0 & x\leq 0 \\
x & x>0. \\
\end{cases}$$
Are there any approximations which are not defined piece-wise? Ideally, I'd like an approximation that will allow me to control the error given some norm.
|
For a general method of approximation, the convolution is really usefull. Take a positive function of $C^{\infty}_c([-1,1])$, as an exemple,
$$g(x) = \exp\left({-\frac{2}{(x-1)(x+1)}}\right)$$
Let's normalise it :
$$g_{norm}(x) = \frac{g(x)}{\int_{-1}^1 g(t) dt}$$
Then we define
$$g_{\epsilon}(x) = \frac{1}{\epsilon} g_{norm}(\frac{x}{\epsilon})$$
Then, for every $f \in L^1_{loc}$ you have that $g_{\epsilon} \star f \to f$ uniformly and that $g_{\epsilon} \star f \in C^{\infty}$
You also have $g_{\epsilon} \star f \to f$ in all $L^p$.
But I don't remember the error estimate...
|
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Greatest Commom Divisor In my lecture notes there is the following:
$$\newcommand{\gcd}{\operatorname{gcd}}
a, b \in \mathbb{N}, \gcd(a, b)=d \implies \exists s, t \in \mathbb{Z} : sa+tb=d \implies tb \equiv d \pmod a$$
If $\gcd(a, b)=1$ then $tb \equiv 1 \pmod a \ $ $ \ \ \ [b]_a \in \mathbb{Z}_a^{\star}$ and $[b]_a^{-1}=[t]_a$
$$r_0=a, r_1=b \\ r_0=q_1 r_1+r_2 , 0 < r_2 <r_1 \\ r_1=q_2 r_2 + r_3, 0<r_3 < r_2 \\ \dots \\ r_{l-1}=q_{l-1}r_{l-1}+r_l, 0<r_l < r_{l-1} \\ r_{l-1}=q_l r_l+0$$
$\gcd(a, b)=r_l$
$$b, b-1, b-2, \dots , 0 \\ l \leq b$$
Because there isn't any explanation what $b, b-1, b-2, \dots , 0$ means, do you have any idea about what it could mean? What does this represent?
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The point here seems to be to give a simple argument why the Euclidean algorithm terminates.
The sequences of remainders $r_i$ is strictly decreasing and $r_1 = b$. So it will stop at at most $r_b$, as the worst case would be $r_2 = b-1$, $r_3 = b-2$ and so on $r_b = b - (b-1)=1$, so then the following remainder must be $0$.
Note that there are somewhat more sophisticated arguments that give a much better bound for the number of necessary steps.
|
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Calculate $\int_0^{\pi/2}\frac{\sin(x)\log{\sin{(x)}}}{x}\,dx$ Inspired by a question I saw these days, I try to calculate in closed form
$$\int_0^{\pi/2}\frac{\sin(x)\log{\sin{(x)}}}{x}\,dx$$
So far no fruitful idea that is worth sharing. What way would you propose? Note I prefer ways suggested, not necessarily solutions, but I have nothing against any of the options you prefer.
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It's not a closed form, but I hope can be useful. Using $$\log\left(\sin\left(x\right)\right)=-\log\left(2\right)-\sum_{n\geq1}\frac{\cos\left(2nx\right)}{n}$$
we have$$\int_{0}^{\pi/2}\frac{\sin\left(x\right)\log\left(\sin\left(x\right)\right)}{x}=-\log\left(2\right)\textrm{Si}\left(\frac{\pi}{2}\right)-\sum_{n\geq1}\frac{1}{n}\int_{0}^{\pi/2}\frac{\sin\left(x\right)\cos\left(2nx\right)}{x}dx.$$
Now we use the identity $$\sin\left(x\right)\cos\left(2nx\right)=\frac{1}{2}\left(\sin\left(x-2nx\right)+\sin\left(2nx+x\right)\right)$$
to obtain $$\int_{0}^{\pi/2}\frac{\sin\left(x\right)\log\left(\sin\left(x\right)\right)}{x}=-\log\left(2\right)\textrm{Si}\left(\frac{\pi}{2}\right)+\frac{1}{2}\sum_{n\geq1}\frac{\textrm{Si}\left(\frac{\pi}{2}\left(2n+1\right)\right)-\textrm{Si}\left(\frac{\pi}{2}\left(2n-1\right)\right)}{n}.$$
Now, we have that $$\frac{\textrm{Si}\left(\frac{\pi}{2}\left(2n+1\right)\right)-\textrm{Si}\left(\frac{\pi}{2}\left(2n-1\right)\right)}{n}=O\left(\frac{1}{\pi n^{2}}\right)$$
at $n\rightarrow\infty$
so we have the approximation $$\int_{0}^{\pi/2}\frac{\sin\left(x\right)\log\left(\sin\left(x\right)\right)}{x}\simeq-\log\left(2\right)\textrm{Si}\left(\frac{\pi}{2}\right)-\frac{\zeta\left(2\right)}{2\pi}.$$ Note that numerically the integral is $-1.05585$ and my result is $-1.21193...$
|
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How to tell where parentheses go in functional notation? The professor gave us a function $f(z) = \ln r + i \theta$ (this is for a complex analysis class). He doesn't like answering students' questions and there's no assigned textbook so I don't know where to look up such a function. How can I tell where the parentheses are supposed to go for this equation?
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In addition to syntactic answers already discussed, you can think a little about which makes the most sense in context. For a complex analysis class, $f(z) = \ln(r) + i\theta$ is a common function to encounter. If $z$ is a complex number with modulus $r$ and principal argument $\theta$, then $f(z)$ is the principal branch of the complex logarithm, a function which is used all the time. On the other hand, $f(z) = \ln(r + i\theta)$ doesn't seem to be a particularly useful function.
|
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$\mathbb Z_p^*$ is a group iff $p$ is prime I'm trying to prove $\mathbb Z_p^*$ is a group if and only if $p$ is prime. I know that if $p$ is prime $\mathbb Z_p^*$ is a group, but how can I do the converse? In another words, if the equation $ax\equiv 1 (\mod m)$ has solution for every integer $a$ which is not divisible by $m$, then $m$ is prime.
I'm trying to solve this exercise:
Thanks
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The equation in $x$ : $ax\equiv 1 \bmod n$ has a solution if and only if $(a,n)=1$. If $n$ is not prime take $a$ a proper divisor of $n$. Then $a$ is not divisible by $n$ , while $ax\equiv 1$ has no solution.
Although note that $\mathbb Z_p^*$ is the group in which the elements are the congruence classes relatively prime to $p$. Not those that are not multiples of $p$.
|
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Show that if $a_n> 0$ and $\lim na_n = \ell$ with $\ell \neq 0$ then the series $\sum a_n$ diverges I'm trying to understand the solution to the following problem:
Show that if $a_n> 0$ and $\lim na_n = \ell$ with $\ell \neq 0$ then the series $\sum a_n$ diverges.
I don't understand how the sign changed in the starred equation. Can someone briefly explain the solution?
If $na_n \to 1$ and $a_n > 0$ then $l \geq 0$. Furthermore, we know that for any $\epsilon>0$ there is an $N$ for which
$n > N$ implies $|na_n-\ell| < \epsilon$. Equivalently, $|a_n-\ell/n| < \epsilon/n$. But then
$$ a_n > \frac{\ell}{n} - \frac{\epsilon}{n}. \tag{$\ast$} $$
Consider this for $\epsilon = \ell/2 > 0$. Then for $n > N$ we have $a_n > \ell/2n$. But $\ell/2 \sum 1/n$ diverges. Hence $\sum a_n$ also diverges.
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The claim $|a_n - \frac{\ell}{n}| < \frac{\epsilon}{n}$ is equivalent to
$$ -\frac{\epsilon}{n} < a_n - \frac{\ell}{n} < \frac{\epsilon}{n}. $$
In particular, the left inequality implies $(\ast)$.
Why is $|X| < Y$ equivalent to $-Y < X < Y$? By definition, $|X| = \max(X,-X)$. Now $\max(X,-X) < Y$ iff both $X < Y$ and $-X < Y$. The latter is equivalent to $X > -Y$, and put together we get $-Y < X < Y$.
|
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Prove that if $\lim\limits_{x\to 0}f\big(x\big(\frac{1}{x}-\big\lfloor\frac{1}{x}\big\rfloor\big)\big)$, then $\lim\limits_{x\to 0}f(x)=0$ Prove that if $\lim\limits_{x\to 0}f\bigg(x\bigg(\dfrac{1}{x}-\bigg\lfloor\dfrac{1}{x}\bigg\rfloor\bigg)\bigg)$, then $\lim\limits_{x\to 0}f(x)=0$
My attempt:
If I can show that $\bigg(\dfrac{1}{x}-\bigg\lfloor\dfrac{1}{x}\bigg\rfloor\bigg) \to 1$ as $x\to 0$, then we are done.
We know, $n\le \dfrac{1}{x}\le n+1$, so $\bigg\lfloor\dfrac{1}{x}\bigg\rfloor=n$.
But, I cannot do anything more to it. Please help. Thank you.
|
Use the Squeeze Theorem. First note that
$$\frac{1}{x}-1<\left\lfloor\frac{1}{x}\right\rfloor\le\frac{1}{x}$$
from which we obtain
$$ \left|x\right|\ge \left|x\right|\left( \frac{1}{x} -\left\lfloor\frac{1}{x}\right\rfloor\right)\ge 0$$
Therefore, as $x\to 0$, the Squeeze Theorem guarantees that
$$\lim_{x\to 0} x\,\left( \frac{1}{x} -\left\lfloor\frac{1}{x}\right\rfloor\right)=0 $$
which implies
$$\lim_{x\to 0}f\left(\,x\left( \frac{1}{x} -\left\lfloor\frac{1}{x}\right\rfloor\right)\right)=\lim_{x\to 0}f(x)$$
If $f$ is continuous, then
$$\lim_{x \to 0} f\left(x \left(\frac{1}{x} -\left\lfloor\frac{1}{x}\right\rfloor\right)\right)=f\left(\lim_{x \to 0} \left[x \left(\frac{1}{x} -\left\lfloor\frac{1}{x}\right\rfloor\right)\right]\right)=f(0).
$$
|
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Showing that if $xf(x)=\log x$ for $x>0$, then $f^{(n)}(1)=(-1)^{n+1}n!\left(1+\frac12+\cdots+\frac{1}{n}\right)$
Let $f(x)$ be a function satisfying $$xf(x)=\log x$$ for $x>0$. Show that $$f^{(n)}(1)=(-1)^{n+1}n!\left(1+\frac12+\frac13+\cdots+\frac{1}{n}\right),$$ where $f^{(n)}(x)$ denotes the $n$th derivative of the function $f$ evaluated at $x$.
I attempted plain differentiation of the function. But, that seemed to go the complex way. Please help. Thank you.
|
Hint: First, write out the first few derivatives of each side of $$x f(x) = \log x$$ (rather than the derivatives of $f(x) = \frac{\log x}{x}$), identify the patterns of the derivatives, and prove that they are correct (using, say, induction).
|
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Convergence and Absolute Convergence of Arithmetic Mean of a sequence Suppose $\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n |x_i|$ exists. Does $\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n x_i$ exist?
How about the converse?
My thoughts:
*
*I guess for the sequence $\{x_1,-x_1,x_2,-x_2,\ldots\}$ the converse doesn't necessarily hold.
*If I can show that the existence of $\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n x_i$ implies that $\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n x_i 1\{x_i\geq 0\}$ and $\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n |x_i| 1\{x_i\leq 0\}$ exist, then the first part would hold. But I'm not sure this is true, I need to find a counterexample.
|
Another trivial counter example for the converse could be this:
$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n x_i 1\{x_i\geq 0\}$ does not exist and $\frac{1}{n}\sum_{i=1}^n |x_i| 1\{x_i\leq 0\}=\frac{1}{n}\sum_{i=1}^n x_i 1\{x_i\geq 0\}$.
|
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Show that $F(x) = f(\|x\|)$ is differentiable on $\mathbb{R}^n$.
Let an even function $f:\mathbb{R}\to\mathbb{R}$ which is even and differentiable. We define $F:\mathbb{R}^n\to\mathbb{R}$ as $F(x) = f(\|x\|)$. Show that $F(x)$ is differentiable on $\mathbb{R}^n$.
My Work:
Let $x_0\in\mathbb{R}^n$.
*
*If $x_0\ne 0$ then the partial derivatives of the norm are well-defined and:
$$ \frac{\partial \|.\|}{\partial x_i} = \frac{x_i}{\|x\|}$$
$\frac{x_i}{\|x\|}$ continuous and since $f$ is differentiable it's partial derivatives are continuous. So we can conclude that $\frac{\partial F}{\partial x_i}$ is continuous and therefore $F$ is differentiable.
*If $x_0=0$ then the $\|x_0\| = 0$ and since $f$ is even $f(0)=0$ and it must be an extremum point. Therefore, $F'(0)=0$.
I'd like to get a critique of my work. Am I being right/rigorous?
Thanks.
|
While the case $x_0 \neq 0$ is correct and trivial, your proposed solution for the case $x_0=0$ is not a proof at all. A function can have a maximum or minimum at a point where it is not smooth (e.g. $x \mapsto |x|$ in $\mathbb{R}$).
Here you should use the fact that $f'(0)=0$ (because $f$ is even and differentiable at zero), so that
$$
f(t)=f(0)+o(t) \quad\hbox{as $t \to 0$}. \tag{1}
$$
Now plug $t=\|x\|$ into $(1)$ and deduce that
$$
F(x)=F(0)+o(\|x\|) \quad\hbox{as $x \to 0$},
$$
which shows that $DF(0)=0$.
|
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|
Quadratic formula question: Missing multiplying factor of A? I have a very simple problem which must have a simple answer and I was wondering if anyone can point out my error.
I have the following quadratic equation to factor:
$2x^2+5x+1$
Which is of the form:
$Ax^2+Bx+C$
All I want to now do is factor this into the form:
$(x+\alpha)(x+\beta)$
So using the quadratic formula
$x = \frac{-B\pm\sqrt{B^2-4AC}}{2A}$
I get
$x = \frac{-5 \pm \sqrt{17}}{4}$
So I would think then that
$(x - \frac{-5 + \sqrt{17}}{4})(x - \frac{-5 - \sqrt{17}}{4}) = 2x^2+5x+1 $
But it doesn't seem to work. Any x that I choose results in the answer being off by a factor of A (in this case A=2).
What silly thing am I missing?
|
$Ax^2+Bx+C$ factorize as : $A(x-\alpha)(x-\beta)$ wiht $\alpha + \beta= \dfrac{-B}{A}$ and $\alpha \beta=\dfrac{C}{A}$.
obviously if $A \ne 0$, and $\alpha$ and $\beta$ are roots of the the two equations:
$$
A(x-\alpha)(x-\beta)=0 \iff Ax^2+Bx+C=0
$$
and
$$
(x-\alpha)(x-\beta)=0 \iff x^2+\dfrac{B}{A}x+\dfrac{C}{A}=0
$$
|
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|
How to use Induction with Sequences? I have posted this similar question here, but with no hopes. I would just like to know:
Most of the solution I have no issue with. Look at where they say:
"Choose a representation $(n - 3^m)/2 = s_1 + ... + s_k$ in the desired form.
But to do that first, they applied induction to the set:
$$A = \{1, 2, 3, ..., n-1\}$$
You have to ensure,
$$\frac{n - 3^m}{2} \le n-1$$
How do you show that?
Lets, consider the case, $n=5$. It follows,
$$3^1 < 5 < 3^2$$, Hence, $m = 1$ which gives:
$$\frac{5 - 3}{2} \le 4 \implies 1 < 4 \checkmark$$
But the question is how to prove it?
Suppose:
$$\frac{n - 3^m}{2} \le n-1$$
I am to prove:
$$\frac{n + 1 - 3^m}{2} \le n$$
Begin with the hypothesis,
$$\frac{n - 3^m}{2} \le n-1$$
$$\frac{n - 3^m}{2} + \frac{1}{2} \le n- \frac{1}{2} < n$$
Since $n \in \mathbb{N}$
I suppose the statement is proved. Can you check it?
|
Recall that $m=\lfloor\log_3n\rfloor$. This means that $m\le\log_3n<m+1$ and hence that $3^m\le n<3^{m+1}$. It follows that $n-3^m<3^{m+1}-3^m$ and hence that
$$\frac{n-3^m}2<\frac{3^{m+1}-3^m}2=\frac{3^m(3-1)}2=3^m\le n\;.$$
That is,
$$\frac{n-3^m}2<n\;.\tag{1}$$
Since we’re assuming in this case that $n$ is odd, we know that $\frac{n-3^m}2$ is an integer, $(1)$ implies that
$$\frac{n-3^m}2\le n-1\;.$$
|
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|
Is antisymmetric the same as reflexive? Note: The following definitions from my book, Discrete Mathematics and Its Applications [7th ed, 598].
This is my book's definition for a reflexive relation
This is my book's definition for a anti symmetric relation
Is a reflexive relation just the same as a anti symmetric relation? From what I've, the only way to meet that antisymmetric requirement is to have the same ordered pair, say an element a from Set A, (a,a). If you have anything other than the same ordered pair, (1,2) and (2,1), it will not meet the antisymmetric requirement. But the overall definition of reflexive relation is that it's the same ordered pair. Are they just two ways of saying the same thing? Is it possible to have one and not the other?
|
Reflexivity requires all elements to be both way related with themselves. However, this does not prohibit non-equal elements from being both way related with each other.
That is: $\qquad\forall a\in A: (a,a)\in R$ from which we can prove: $$\forall a\in A\;\forall b\in A:\Big(a=b \;\to\; (a,b)\in R \wedge (b,a)\in R\Big)$$
Antisymetricality requires any two elements which are both way related with each other to be equal. Which means it prohibits non-equal elements from being both-way related to each other; however, this does not require all elements to be both way related with themselves.
$$\forall a\in A, \forall b\in A:\Big((a,b)\in R \wedge (b,a)\in R \;\to\; a=b\Big)$$
Or via contraposition: $\qquad\forall a\in A, \forall b\in A:\Big(a\neq b \;\to\; (a,b)\notin R \vee (b,a)\notin R\Big)$
Thus they are not the same thing.
Examples are easy to generate. From the set $A=\{0, 1\}$ The following relations are:
$$\begin{array}{c|c|c}
R & \text{Reflexive} & \text{Antisymmetric}
\\ \hline
\{(0,0), (1,1)\} & \text{Yes} & \text{Yes}
\\
\{(0,0), (0,1), (1,0), (1,1)\} & \text{Yes} & \text{No}
\\
\{(0,0), (0,1), (1,0)\} & \text{No} & \text{No}
\\
\{(0,0), (0,1)\} & \text{No} & \text{Yes}
\end{array}$$
|
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Is the number $-1$ prime? From my understanding it's not prime because it's not greater than $0$. So my followup question is why did mathematicians exclude $-1$?
The definition of prime is having only two factors.
$-1 \cdot 1 = -1$
|
In general, a definition that describes relationships between numbers without describing the domain in which such relationships are valid is incomplete.
In other words, when you say that the "definition of prime is having only two factors," you are leaving out a crucial element: for which set(s) of numbers is this relationship between primes and factors valid?
The typical definition of "prime numbers" describes a relationship between natural numbers; since $-1$ is not a natural number, any definition that only applies to natural numbers has no bearing on whether $-1$ is prime.
Thus, in order to make the question "is $-1$ prime" meaningful, you must find an extension (or generalization) of the definition of "prime" to a domain that includes $-1$.
As pointed out in @Hugh's answer, perhaps the most natural extension is that of prime elements. In this extension, $-1$ is considered a unit and is therefore by definition not prime. Another extension is
the Gaussian Primes, in which $-1 = -1 + 0i$ is non-prime because the coefficient of $i$ is $0$ and $|-1| = 1$ is not prime (according to the standard definition on the natural numbers). @BillDubuque, meanwhile, gives an example of a context in which it might be reasonable to consider $-1$ to be prime.
Now, we can perhaps make further generalizations about all possible extensions of the definition of primes within the natural numbers. Some of the other answers provide some good intuition about this; for instance, it is observed that the units are typically not considered prime, because this simplifies the definition of prime factorization. Thus, we could answer that $-1$ is not considered prime in any domain in which $-1$ is a unit (though of course this requires us to define "unit" for each such domain).
|
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If $T^2=T$ then determine whether $\ker T=\operatorname{Range}\,(T)^\perp$.
Let $T$ be linear operator on a finite dimensional inner product space $V$ such that $T^2=T$. Determine whether $\ker T=\operatorname{Range}\,(T)^\perp$.
I have proved that $\ker T=\operatorname{Range}\,(T)^\perp$ under the assumption that $T$ is Hermitian. I guessed that the answer is yes but still in trouble to make it. I also want to know whether $T^2=T$ on $V$ will imply that $T$ is Hermitian.
|
In fact, $\ker T = ({\cal R} T)^\bot$ iff $T$ is Hermitian.
Note that $\ker T = {\cal R} (I-T)$.
Suppose $\ker T = ({\cal R} T)^\bot$. Then
$\langle (I-T)x, Ty \rangle = 0$ for all $x,y$, hence $(I-T^*)T = 0$,
or $T= T^* T$, and so $T$ is Hermitian.
Now suppose $T$ is Hermitian.
If $x \in \ker T$, then $\langle x, Tz \rangle = \langle Tx,z \rangle = 0$ for all $z$ and
so $x \in ({\cal R} T)^\bot$.
If $x \in ({\cal R} T)^\bot$, then $\langle x, Tz \rangle =\langle Tx,z \rangle = 0$ for all $z$ and so $Tx = 0$, hence $x \in \ker T$.
|
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Integration: Area between curves Let $f(x)=x^2−c^2$ and $g(x)=c^2−x^2.$ Find $c>0$ such that the area of the region enclosed by the parabolas $f(x)$ and $g(x)$ is 9.
The question above is what I am having trouble with. In order to solve this problem I use the formula given as:
$\int_a^b f(x) - g(x) dx$
Here is what I have done so far:
$9 = \int x^2−c^2 - (c^2−x^2) dx$
$9 = \int 2x^2−2c^2 dx$
$9 = 2\int x^2−c^2 dx$
$\frac 9 2 = \int x^2−c^2 dx$
To find a and b:
$x^2 - c^2 = (x-c)(x+c)$ ; therefore $a = -c$ and $b = c$
Back to the original question, sub in a and b:
$\frac 9 2 = \int_{-c}^c x^2−c^2 dx$
This is as far as I have gotten. I am not sure what to do next. Thanks in advance to anyone who can help.
|
Draw a picture. I would use symmetry and say that the area is $4$ times the integral from $0$ to $c$ of $c^2-x^2$. So we want
$$\int_0^c (c^2-x^2)\,dx=\frac{9}{4}.$$
The integral is equal to $\frac{2}{3}c^3$. Now solve for $c$.
Remark: Your approach will work, except that we want $\int_{-c}^c (c^2-x^2)\,dx=\frac{9}{2}$. Note that $\int_{-c}^c (x^2-c^2)\,dx$ is negative.
|
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solution involving inverse of a rank-1 matrix I am looking for $\mathbf{y} \in \mathbb{R}^n$ that minimizes the following objective function that involves a real matrix $\mathbf{V} \in \mathbb{R}^{n\times n}$
\begin{equation}\tag{*}
\begin{array}{c}
\text{min} \hspace{4mm} \mathbf{y}^T \mathbf{V}\mathbf{y}
\end{array}
\end{equation}
\begin{align*}
\mathbf{y} &=
\begin{bmatrix}
y_{1} & y_{2} &\cdots & y_{n} \end{bmatrix}^T
\end{align*}
I have a single equality constraint i-e $y_1=1$. To accommodate this constraint, I rewrite my objective function as
\begin{equation}\tag{**}
\begin{array}{c}
\text{min} \hspace{4mm} \mathbf{y}^T \mathbf{V} \mathbf{y} - 2\lambda(\mathbf{u}^T\mathbf{y}-1)
\end{array}
\end{equation}
where $\mathbf{u}\in \mathbb{R}^n$ and is defined as
\begin{align*}
\mathbf{u} &=
\begin{bmatrix}
1 & 0 &\cdots & 0 \end{bmatrix}^T
\end{align*}
Diffrentiating (**) wrt $\mathbf{y}$ and $\lambda$ gives me the solution
\begin{align*}\tag{***}
\mathbf{y} &= \lambda \mathbf{V}^{-1}\mathbf{u} \\
\lambda &= \frac{1}{\mathbf{u}^T\mathbf{V}^{-1}\mathbf{u}}
\end{align*}
My problem is that in my setup $\mathbf{V}$ is a $rank 1$ matrix. Therefore $\mathbf{V}^{-1}$ does not make sense. Is there a way to fix this problem?
|
Of course, since $\mathbf V$ is not invertible one cannot use $\mathbf V^{-1}$. However, already the differentiation seems a problem. I've made some quick computation to get
$$
(\mathbf V^T+\mathbf V)\mathbf y=\lambda \mathbf u=[\lambda,0,\dots,0].
$$
Then $\lambda$ is the first component of the vect0r in the left:
$$
\lambda=\sum_i(V_{i1}+V_{1i})y_i.
$$
The other equations give the following homogeneous system to obtain $\mathbf y$
$$
0=\sum_i(V_{ij}+V_{ji})y_i\quad(1\le j<n).
$$
Now if your matrix has rank one, this system is very special. Any rank 1 matrix is of the form $\mathbf V=\mathbf v\mathbf w^T$ for some non-zero vectors $\mathbf v,\mathbf w\in\mathbb R^n$. Then the system simplifies to
$$
\mathbf w\langle \mathbf v,\mathbf y\rangle+\mathbf v\langle \mathbf w,\mathbf y\rangle=\lambda\mathbf u.
$$
To understand this system is better to consider the unknowns are the scalar products $\langle \mathbf v,\mathbf y\rangle$ and $\langle \mathbf w,\mathbf y\rangle$ not just the $\mathbf y$.
For instance, if both scalar products are zero, you're done with $\lambda=0$. And this is something, because in dimension $>2$ the orthogonal complemente of two vectors has dimension $\ge1$.
Another possible consideration: if $\mathbf v,\mathbf w$ are not independent, say $\mathbf w=\rho \mathbf v$, $\rho\ne0$, then you get
$$
\rho\mathbf v\langle \mathbf v,\mathbf y\rangle+\mathbf v\langle \rho\mathbf v,\mathbf y\rangle=\lambda\mathbf u,
$$
hence
$$
2\rho\mathbf v\langle \mathbf v,\mathbf y\rangle=\lambda\mathbf u.
$$
If $\langle \mathbf v,\mathbf y\rangle\ne0$, then looking at the equations $i=2,\dots,n$ we see that $2\rho v_i\langle \mathbf v,\mathbf y\rangle=0$, hence $v_i=0$. Thus $\mathbf v=v_1\mathbf u$ and the equation $i=1$ reads $\lambda=2\rho v_1^2y_1$.
Next suppose $\mathbf v,\mathbf w$ are independent. This can come from the coordinates $i=2,\dots,n$, but then the system would have the solution $\langle \mathbf v,\mathbf y\rangle=\langle \mathbf w,\mathbf y\rangle=0$, already discussed.
The only case remaining is that the vectors are independent by the first coordinate en some other $i=2,\dots,n$, not any else. Thus the system reduces to two independent equations whose unknowns are the two scalar products, and it can be solved for any $\lambda$ you choose.
Actually this could be much better and easier explained with a little bit of geometry and quadratic forms, but I don't know the real context of the question, and I've chosen a purely computational approach.
|
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A question on a problem on localization from Atiyah (3.8) I was having trouble with the following problem from Chapter $3$ of Atiyah-MacDonald
Let $S, T$ be multiplicatively closed sets in the ring $A$, such that $S\subseteq T$.
Let $\varphi : S^{−1}A \to T^{−1}A$ be the homomorphism which maps each $r/s \in S^{−1}A$ to $r/s$ viewed as an element of $T^{−1}A$. Show that the following are equivalent:
(i) $\varphi$ is a bijection;
(ii) $t/1$ is a unit in $S^{−1}A$ for all $t\in T$;
My Attempts
I am assuming $\varphi$ is bijective
I first tried to show this directly by noting that since $\varphi$ is bijective we have that $\varphi(a/s)=t/1$ .
This means we have that $a/s=t/1$ and therefore there is a $w \in T$ where $w(a-st)=0$. I am stuck at this point.
Another approach, I saw recommended was to consider to universal property associated to localization. I am confused on how to apply it, do I consider this with respect to $A \rightarrow T^{-1}A$ where $S$ is the closed mult. set being considered in which case there should be a map from $S^{-1}A$ to $T^{-1}A$. I'm not sure how that relates to $\varphi$ being bijective.
Any help would be appreciated.
|
For the forward direction: Let $\phi$ be a bijection and let $t\in T$. Since $t\in T$, $1/t\in T^{-1}A$. Since $\phi$ is a bijection, there is some $a/s\in S^{-1}A$ such that $\phi(a/s)=1/t$. Consider $\phi(at/s)=(a/s)t=(1/t)t=1$. Since $\phi$ is a bijection and $\phi(at/s)=1$, it follows that $at/s=1$.
On the other hand, suppose that for all $t\in T$, $t/1$ is a unit in $S^{-1}R$. Let $a/b,c/d\in S^{-1}T$ such that $\phi(a/b)=\phi(c/d)$. Then there is some $t\in T$ such that $t(ad-bc)=0$. Now, the statement $t(ad-bc)=0$ is true in $A$ as well as in $S^{-1}A$. Since $t\in T$, there is some $e/f\in S^{-1}A$ such that $(e/f)t=1$ in $S^{-1}A$. Therefore, $(e/f)t(ad-bc)=ad-bc=0$ in $S^{-1}A$. Now, since $(ad-bc)/1=0/1$ in $S^{-1}A$, there is some $s\in S$ such that $s(ad-bc)=0$. Hence $a/b=c/d$. Therefore, $\phi$ is injective.
Finally, to show that $\phi$ is surjective. Let $a/t\in T^{-1}A$. Since $t\in T$, $t$ is a unit in $S^{-1}A$, so there is some $b/c$ such that $(b/c)t=1$ in $S^{-1}A$. In other words, there is some $s\in S$ such that $s(bt-c)=0$. Since $S\subseteq T$, $s\in T$. Now, consider $\phi(ab/c)=ab/c$. I claim that $ab/c=a/t$ in $T^{-1}A$. This follows from $sa(bt-c)=0$.
|
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Martingale energy inequality I am reading a book on BMO martingales which uses a so-called energy inequality. I have not been able to find a solid reference for this. Can someone please give a reference to these inequalities. Hopefully someone has heard of them, apparently they are "well known."
|
The only reference I know of is P. A. Meyer's Probability and Potentials. It is not an easy read. Chapter VII Section 6 "A few Results on Energy" can resolve your questions. Let $(X_t)$ be a right-continuous potential, i.e. there is an integrable increasing process $(A_t)$ ($A_0 = 0$, increasing paths, right continuous, $E A_\infty<\infty$), and $(M_t)$ is a right continuous modification of $E[A_\infty | \mathscr{F}_t]$, and $X_t = M_t-A_t$. Note: $(X_t)$ is a positive supermartingale with $\lim_{t \to \infty} X_t = 0$ (the "usual" definition of potential). For $p>1$ an integer we define the $p$-energy
$$
e_p[(X_t)] := \frac{1}{p!}E[(A_\infty)^p].
$$
The $p$-energy inequality says that if $(X_t)$ is dominated by a constant $c$, then $e_p[(X_t)] \leq c^p$, i.e.
$$
E[(A_\infty)^p] \leq p!\,c^p.
$$
If you are reading Kazamaki's Continuous Exponential Martingales and BMO, he says for a BMO martingale $M$ "the energy inequalities" give
$$
E[\langle M\rangle_\infty^n] \leq n! ||M||^{2n}_{BMO_2}.
$$
Note that this is a straightforward application of what P.A. Meyer spends a chapter proving in his book (well, he proves for $p=2$ and suggests how to do it for $p>1$, but he does not assume $A$ is continuous, where I suspect there may a much shorter proof). Take $A = \langle M \rangle$, $p=n$ and $X_t$ the right continuous version of $E[\langle M\rangle_\infty- \langle M\rangle_t| \mathscr{F}_t]$. The assumption that the BMO norm is finite tells us exactly that $|X_t| \leq ||M||_{BMO_2}^2$. Another method of proving this fact is suggested in Revuz and Yor and uses optional projections and the fact that $A_\infty= p! \int_{0}^\infty \int_{u_1}^\infty\cdots \int_{u_{p-1}}^\infty d A_{u_{p-1}} \cdots dA_{u_2} d A_{u_1}$.
|
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|
How do we see that the shape of an equation is closed? Actually, I would like to understand :
When we have an equation , can we decide that its shape closed curve?
For instance, for the equation:
$$((x+y)^2) + 2((x^4)/4-(x^5)/5)=C$$
For which values of C makes the shape of this equation closed?
|
For any value of $C$ the graph is closed. To see this, define a function $f:\Bbb R^2\to \Bbb R$ by $f(x,y)=(x+y)^2+2\left(\dfrac{x^4}{4}-\dfrac{x^5}{5}\right)-C$. Clearly $f$ is continuous and so $f^{-1}(\{0\})=\left\{(x,y)\in\Bbb R^2:(x+y)^2+2\left(\dfrac{x^4}{4}-\dfrac{x^5}{5}\right)=C\right\}$ is a closed set in $\Bbb R^2$ as this set is the inverse image of the closed subset $\{0\}$ under the continuous function $f$.
|
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Proof that $p$-th total variation of a brownian motion is $0$ while $p>2$ The p-th total variation is defined as $$|f|_{p,TV}=\sup_{\Pi_n}\lim_{||\Pi_n||\to n}\sum^{n-1}_{i=0}|f(x_{i+1}-f(x_{i})|^p$$
And I know how to calculate the first total variation of the standard Brownian motion. But when dealing with high order TV, there are some problem.
At first we assume that p is even.
First I define $$\xi_i=|B_{\frac{i+1}{n}}-B_{\frac{i}{n}}|^p$$ then we can get that $$E[\xi_i]=\left(\frac1n\right)^{\frac p2}(p-1)!!$$ and $$E[\xi_i^2]=\left(\frac1n\right)^{p}(2p-1)!!$$
Next, we define $V_n=\sum^{n-1}_{i=0}\xi_i$
Then we have$$E[V_n]=\sum^{n-1}_{i=0}\left(\frac 1n\right)^{\frac p2}(p-1)!!$$
But there's something wrong in the following step, when calculating $E[V_n^2]$
$$\begin{align}
E[V_n^2] &= E\left[\sum^{n-1}_{i=0}\xi_i\sum^{n-1}_{j=0}\xi_j\right]\\
&=E\left[\sum^{n-1}_{i=0}\sum^{n-1}_{j=0}\xi_i\xi_j\right]\\
&=\sum^{n-1}_{i=0}\sum^{n-1}_{j=0}E\left[\xi_i\xi_j\right]\\
&=\sum^{n-1}_{i=0}E[\xi_i^2]+\sum_{i\neq j}E[\xi_i]E[\xi_j]\\
&=\left(\frac1n\right)^{p-1}(2p-1)!!+n(n-1)\left(\frac1n\right)^{p}\left[(p-1)!!\right]^2
\end{align}$$
And then the question is I have no idea that how to deal with this awesome equation. Do I need to brute it out or if there is any method more efficient to prove it?
|
Note that $2$-variation is not the same as quadratic variation. For the quadratic variation you take the limit as the partition gets finer, whereas for $p$-variation you take the supremum over all partitions.
In particular, the Brownian motion has finite quadratic variation on any finite interval but infinite $2$-variation (link).
Now coming to your question and assuming the interval $[0,1]$, we have that
$$\|B\|_{p-\text{var}} \geq |B_1|,$$
because the supremum is larger or equal to taking the partition that only contains $0$ and $1$.
So, the $p$-variation of Brownian motion for any $p\geq 1$ is clearly not converging to zero in any sense.
|
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|
Clarification of Proof involving $\sum_{p \le x} \frac{1}{p}$ For fun I've been doing problems from M. Ram Murty's text "Problems in Analytic Number Theory". I recently encountered the following problem:
If
$$\lim_{x \rightarrow \infty} \frac{\pi(x)}{x/\log x } = \alpha $$
then show that
$$\sum_{p \le x} \frac{1}{p} = \alpha \log\log x + \underline{o}(\log \log x).$$
Deduce that if the limit exists, it must be 1.
The solution to this exercise is provided in the book. It is as follows:
By partial summation,
$$\sum_{p \le x} \frac{1}{p} = \frac{\pi(x)}{x} + \int_2^x \frac{\pi(t)}{t^2} dt = \alpha \log \log x + \underline{o}(\log \log x) $$
Since $\sum_{p \le x} \frac{1}{p} = \log \log x + O(1)$, we know that $\alpha$ must be 1.
Unfortunately, I don't completely understand this solution. I understand the argument why $\alpha$ must be 1, and the first equality in the solution makes sense to me. However, I'm unclear on how the second equality follows. Can somebody please help explain why the second equality holds? Any assistance will be much appreciated. Thank you.
|
For proving $$\sum_{p\leq x}\frac{1}{p}=\log\left(\log\left(x\right)\right)+O\left(1\right)$$
we can use the Mertens second formula$$\sum_{p\leq x}\frac{\log\left(p\right)}{p}=\log\left(x\right)+O\left(1\right).$$
So by partial summation$$\sum_{p\leq x}\frac{1}{p}=\sum_{p\leq x}\frac{1}{\log\left(p\right)}\frac{\log\left(p\right)}{p}=\frac{1}{\log\left(x\right)}\sum_{p\leq x}\frac{\log\left(p\right)}{p}+\int_{2}^{N}\sum_{p\leq t}\frac{\log\left(p\right)}{p}\frac{dt}{t\left(\log\left(t\right)\right)^{2}}=$$ $$=1+O\left(\frac{1}{\log\left(x\right)}\right)+\int_{2}^{N}\frac{dt}{t\left(\log\left(t\right)\right)}+O\left(\frac{1}{\log\left(t\right)}\right)=$$ $$=\log\left(\log\left(x\right)\right)+1-\log\left(\log\left(2\right)\right)+O\left(\frac{1}{\log\left(x\right)}\right).$$
|
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|
How to prove this simple fact without using distribution theory? Suppose function $f(x) $ is normalized to unity, i.e.,
$$ \int dx |f(x)|^2 =1 . $$
Now consider the Fourier transform of $f$, i.e.,
$$ F(k) = \int d x f(x) e^{-i k x} . $$
Here we assume that $f $ is a very well-behaved function (say, a Gaussian function) so that $F(k)$ is well defined and also well-behaved.
It is well-known that
$$ \int d k |F(k)|^2 =2 \pi . $$
But how to prove it without using the identity
$$ \int dk e^{i k x } = 2 \pi \delta (x) . $$
|
Assume that $f$ is real, smooth, absolutely integrable with absolutely integrable derivative. Then $f \in L^{1}\cap L^{2}$ follows. Define $g(x)=f(-x)$. Then,
$$
g^{\wedge}(s) =\overline{f^{\wedge}(s)}.
$$
The convolution $f\star g$ has Fourier transform $\sqrt{2\pi}|f^{\wedge}(s)|^{2}$. Consequently,
$$
(f\star g)(x)=\int_{-\infty}^{\infty}e^{isx}|f^{\wedge}(s)|^{2}ds.
$$
Evaluating at $x=0$ gives
$$
\int_{-\infty}^{\infty}f(x)^{2}dx = (f\star g)(0)=\int_{-\infty}^{\infty}|f^{\wedge}(s)|^{2}ds.
$$
|
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|
Number of commuting involutions in Symplectic group Definition. $x$ is called involution if it has order $2$.
Problem. I have to count maximal number of pairwise commuting involutions in $Sp(2n, F)$, $\text{char}F \neq 2$.
Attempt. Firstly, let's remember that pairwise commuting matrices are simultaneous diagonalizable so in some basis they will be diagonal with $-1$ and $1$ on the diagonal. Unfortunately, $\Omega$ (which is used in definition of $Sp(2n, F)$) is also changed. Changing basis we can make
$$
\Omega =
\begin{pmatrix}
\Omega_1 &0\\
0 &\Omega_2
\end{pmatrix},
$$
where $\Omega_1$ and $\Omega_2$ are even dimensionality antisymmetric matrices and our involutions are still in diagonal form. Now I'm stuck and don't know how to proceed.
|
The largest elementary abelian $2$-subgroup of ${\rm Sp}(2n,F)$ has order $2^n$, so the answer is $2^n-1$.
To see this, let $E$ be an elementary abelian $2$-subgroup. Then $E$ fixes some $1$-dimensional subspace $\langle v \rangle$ of $F^{2n}$, and hence it stabilizes the subspace $\langle v \rangle^\perp$, which has dimension $2n-1$.
Now, since $E$ is completely reducible, it also fixes a $1$-dimensional subspace $\langle w \rangle$ with $w \in F^{2n} \setminus \langle v \rangle^\perp$, and in order to preserve the form on $\langle v,w \rangle$, the induced action of $E$ on this $2$-dimensional subspace has order at most $2$. Now $E$ preserves $\langle v,w \rangle^\perp$ of dimension $2n-2$, and by induction its induced action on that subspace has order at most $2^{n-1}$, so $|E| \le 2^n$.
There exist elementary abelian subgroups of order $2^n$ as subgroups of ${\rm Sp}(2,F)^n \le {\rm Sp}(2n,F)$.
|
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|
Deduce that if $A$ is a subset of $C$, then $\sup A\leq \sup C$.
Deduce that if $A$ is a subset of $C$, then $\sup A\leq \sup C$.
How do I begin with the above proof?
I can't see why $\sup A$ must either be equal of less than $\sup C$.
Is it not possible for some $a \in A$ to be greater than some $c\in C$?
|
Hint: Start with an upper bound of one of the two sets, and try to show that it is also an upper bound of the other. (You'll have to figure out which is which.)
|
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|
Find the Center of Tangent Circles One of my math tests has this question.
A circle has its center at $(6,7)$ and goes through $(1,4)$. Another circle is tangent at $(1,4)$ and has the same area.
What are the possible coordinates of the second circle. Show your work or explain how you found you answer.
Could someone solve a sample question to show me how to solve this one?
|
The center $Q=(h,k)$ of the second circle lies on the line defined by $P_1=(6,7)$ and $P_2=(1,4)$. What means
$$\frac{k-4}{h-1}=\frac{4-7}{1-6}=\frac{3}{5}...(1)$$
Since these circles have the same area it follows they have equal radius. Then $$\sqrt{(h-1)^2+(k-4)^2}=\sqrt{(1-6)^2+(4-7)^2}...(2)$$
Solving the system of equations formed by $(1)$ and $(2)$ you find the center of the second circle.
|
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|
Find all integral solutions to $a+b+c=abc$. Find all integral solutions of the equation $a+b+c=abc$.
Is $\{a,b,c\}=\{1,2,3\}$ the only solution? I've tried by taking $a,b,c=1,2,3$.
|
a+b+c=abc then first of all (a+b+c)>=3*(abc)^1/2 so (abc)^2>=9abc so abc>=9 which shows that a+b+c>=9 which is strictly impossible to satisfy for integer.
Other way is to think that first of all 0 being obvious solution and for other solution let a+c=0 then
b=-aab
Which again gives 0 as the only solution
a+b+c=abc
(1/ab)+(1/bc)+(1/ca)=1 which is only possible for ab=2,bc=3,ca=6 which again yields no solutions.I mean different solutions i.e another equal to b & so on.
|
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|
Probability of rolling all $6$ die faces I've been struggling with this for over an hour now and I still have no good results, the question is as follows:
What's the probability of getting all the numbers from $1$ to $6$ by rolling $10$ dice simultaneously?
Can you give any hints or solutions? This problem seems really simple but I feel like I'm blind to the solution.
|
The way I see this problem, I'd consider two finite sets, namely the set comprised by the $10$ dice (denoted by $\Theta$) and the set of all the possible outcomes (denoted by $\Omega$), in this particular situation, the six faces of the dice.
Therefore, the apparent ambiguity of the problem is significantly reduced by considering all possible mappings of the form $f:\Theta \mapsto \Omega$ that are surjective. Why exactly surjection ?
Recall that the definition of surjection implies that all values in the codomain must be hit at least once through the mapping of elements in the domain of the function. Under the circumstance, it's exactly what we're interested in, since we want to exclusively count all instances in which all the possible $6$ faces appear .
The number of surjective mappings can be found using the following identity :
$$m^n - \binom {m} {1}(m-1)^n +\binom {m} {2}(m-2)^n-\binom {m} {3}(m-3)^n + \cdots $$
(where $m$ denotes the number of elements in $\Omega$ and $n$ the number of elements in $\Theta$ ).
Let $A$ represent the happy event, given the facts I've presented you should be able to get: $$Pr(A)=0.27$$
|
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|
A question about the additive group of the $p$-adic integers Let $J_p$ be the additive group of the $p$-adic integers. I know that it is torsion-free. I'm not pretty confortable with $p$-adic.
Is it possible to find a direct sum of infinitely many cyclic subgroup in $J_p$? (essentially I'm asking if they have finite special rank or not).
EDIT
I find out on Wikipedia that they effectively have infinite rank. But I still cannot find that direct sum.
|
Complete rewrite:
For this, you need to look at the fraction field $\Bbb Q_p$ of the $p$-adic integer ring $\Bbb Z_p$. It’s a vector-space over $\Bbb Q$, and according to a well-known theorem depending on the Axiom of Choice, there’s a $\Bbb Q$-basis of this vector space. Since $\Bbb Q_p$ is uncountable, the basis must be uncountable as well. For each $\beta$ in the basis, there’s a rational number $\lambda_\beta$ such that $\lambda_\beta\beta\in\Bbb Z_p$. Now take the (uncountably-indexed) direct sum
$\bigoplus_\beta\lambda_\beta\beta\,\Bbb Z$. There you are.
|
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|
proving a wiki statement about uniform convergence and supremum This wiki page says that a sequence $f_n$ is uniformally convergenct to $f: S \to \mathbb{R}$ if and only if $||f_n - f||_{\sup} \to 0$. I am trying to prove this statement.
For the forward implication, we have that $f_n \to f$ uniformaly, so we have that $|f_n - f| < \epsilon$. Now, it seems as though the obvious option here is to simply take the supremum of this to get the result if we choose $N$ which gives us $|f_n - f| < \epsilon$. How would I do this step by step, I mean - it seems wrong to just take the supremum of everything as this doesn't use the fact that $f_n$ is uniformaly convergent to $f$
I have yet to try the backwards implication.
|
It's just another way to state the definition :
This is your base definition
$$\forall \epsilon > 0, \exists N, \forall n > N, \forall x, \quad |f_n(x)-f(x)|< \epsilon$$
This is what means $\|f_n-f\|_{\sup} \to 0$ :
$$\forall \epsilon > 0, \exists N, \forall n > N, \quad \|f_n-f\|_{\sup} < \epsilon$$
And now replace $\|f_n-f\|_{\sup}$ by it's definition :
$$\forall \epsilon > 0, \exists N, \forall n > N, \quad \sup_{x} |f_n(x)-f(x)| < \epsilon$$
And it's the same thing that
$$\forall \epsilon > 0, \exists N, \forall n > N, \forall x, \quad |f_n(x)-f(x)| < \epsilon$$
|
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|
Definite integral $\int_1^2 \sqrt{1+\left(-x^{-2}+x^2/4\right)^2}\,dx$ I'm having trouble solving this integral. It relates to an arc length question. I tried Wolfram|Alpha, but when it solves it doesn't give me the option to view step-by-step.
Integral:
$$
\int_{1}^{2} \sqrt{1+\left(-x^{-2}+\frac{x^2}{4}\right)^2}\,dx
$$
Original question:
$$
y=\frac{1}{x} + \frac{x^3}{12}
$$
Find the arc length of the function on [1,2].
I used the equation:
$$
L = \int_a^b \sqrt{1+(f'(x))^2}
dx$$
|
Simplifying the integrand proceed as follows.$$1+\left( -x^{-2}+\frac14x^2\right)^2=1+\left(\frac{x^4-4}{4x^2}\right)^2=\frac{(x^4+4)^2}{16x^4}$$Then after taking a square root, the integrand simplifies to $$\frac{x^4+4}{4x^2}=x^{-2}+\frac14x^2$$which is trivial to integrate.
|
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|
Can the expression $D\cos(x)+Ci\sin(x)$ be rewritten in the form $R\sin(x+\theta)$? I've been trying to use the auxiliary equations method to solve the most simple SHM DE:
$$
\frac{d^2x}{dt^2}+w^2x=0
$$
The auxiliary equation has roots $\lambda=\pm iw$
Hence $x=A\exp(iwx)+B\exp(-iwx)=(A+B)\cos(wx)+i(A-B)\sin(wx)$
Letting $A+B=D$, $A-B=C$, we have the expression in the question.
How can this be rewritten in harmonic form?
|
You are missing that $A$ and $B$ are also complex numbers. To get a real solution, you would need $B=\bar A$ to get $\overline{x(t)}=x(t)$. Then your expression resolves to
$$
(C+iD)(\cos(wt)+i\sin(wt))+(C-iD)(\cos(wt)-i\sin(wt))=2C\cos(wt)-2D\sin(wt)
$$
and this has a classical conversion to an amplidute-phase form.
|
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|
Symplectic lie algebra Can anyone explain me why, in the symplectic lie algebra, which is defined as $ sp(n)=\{X \in gl_{2n}:X^tJ+JX=0\}$ where $J=\begin{pmatrix}
0 & I \\
-I & 0 \\
\end{pmatrix} $ we can write its elements, in block form $X=\begin{pmatrix}
A & B \\
C & -A^t \\
\end{pmatrix} $ where $ A,B,C \in M_{n\times n}$ and $B=B^t,C=C^t$ .How does it proved?
|
Hint: The block decomposition used for $J$ (as well as the answer) suggest writing a generic matrix $X \in \mathfrak{gl}_{2n}$ as
$$X = \begin{pmatrix} A & B \\ C & D \end{pmatrix}.$$
To produce the block matrix description of $\mathfrak{sp}_{2n}$, simply substitute the block expressions for $X$ and $J$ in the definition of that algebra:
\begin{align}
X^t J + J X &= 0 \\
\begin{pmatrix} A & B \\ C & D \end{pmatrix}^t \begin{pmatrix} 0 & I \\ -I & 0 \end{pmatrix} + \begin{pmatrix} 0 & I \\ -I & 0 \end{pmatrix} \begin{pmatrix} A & B \\ C & D \end{pmatrix} &= 0 \textrm{.}
\end{align}
Now, simplify the l.h.s. of the equation to produce algebraic conditions that $A, B, C, D$ must satisfy.
|
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|
How large is the set of all Turing machines? How large is the set of all Turing machines? I am confident it is infinitely large, but what kind of infinitely large is its size?
|
It depends on how you define "distinct" in terms of turing machines. In general, two (one-tape) turing machines are not "distinct" if there is a component-wise bijection that goes from one's 7-tuple to the other's 7-tuple.(See http://en.wikipedia.org/wiki/Turing_machine#Formal_definition for the 7-tuple I am referring.). This leads to a countable number of turing machines. If instead if "distinct" is defined in a way that the 7-tuples have to be identical to be not "distinct", then there are uncountably many turing machines. In this case, there would not be a "set" of all turing machines, instead we would have a "class" of all turing machines.
|
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|
Defining multiplication on the tensor product of $R$-algebras. If $M$ and $N$ are $R$-algebras, then one can define a multiplication of elementary tensors as follows; $(m \otimes n) \cdot (m' \otimes n') = mm' \otimes nn'$.
My question is how can we show, using the universal property of the tensor product, that this is a well defined operation?
My understanding is that we would be need a 4-linear map out of $(M \times N)\times (M \times N)$ into $M \otimes N$, that would factor through $(M \otimes N) \otimes (M \otimes N)$ giving us the required linear map.
Am I on the right track? I can't seem to get my head around the details in the proof. Any pointers would be of great help to me.
|
This is explained in the first paragraph of Bourbaki, A.III.4.1.
|
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Let $N \unlhd G$ and let $N \leqslant H \leqslant G$. Show that $N \unlhd H$.
Let $N \unlhd G$ and let $N \leqslant H \leqslant G$. Show that $N \unlhd H$.
$N \unlhd G$ implies that there is some homomorphism $f$ on $G$ for which $N = ker(f)$. We want to show that there is a homormorphism $f'$ on H such that $N = ker(f')$.
Edit:
Proof. The function $f: G \to G/N$ is a surjective group homorphism with $ker(f) = N$. For $a,b \in G$, we have $f(ab) = abN$ and $f(a)f(b) = (aN)(bN) = abN$, demonstrating that $f$ is a homomorphism. For $a \in G$, we have $$a \in ker(f) \Longleftrightarrow aN = N \Longleftrightarrow a \in N.$$ This shows $ker(f) = N$. Finally, $f$ is surjective, since every left coset is of the form $aN = f(a)$ for some $a \in G$. Since $H \leqslant G$, we can define a map $f':H \hookrightarrow G \to G/N$, $h \mapsto aN$ for $h \in H \leqslant G$. It follows that $f'$ is also a surjective homomorphism and that $ker(f') = N$. Thus $N \unlhd H$. $\text{ } \Box$
Is this proof correct?
|
Consider $H \hookrightarrow G \to G/N$
Edit: (Concerning your edit)
Your proof looks good, just a few things to mention:
*
*you don't need surjectivity of any map (it is given but not necessary). Also if a map $G \to M$ is not surjective, it will factor through a surjective map, since the image $M'$ of the map is a subgroup. Hence we have $G\to M' \to M$
*probably a typo, but the map $f':H \hookrightarrow G \to G/N$ is $h \mapsto hN$
*hence $f'$ is not necessarily surjective! (but as mentioned above we don't need that)
*if you already go into detail you could mention why $ker(f')=N$ (which is easy)
As a last comment, let me mention that the properties of the quotient map are usually assumed as given, because they are very basic. But if you use this the first time it is good to prove everything.
|
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Homogeneous Fredholm Integral Equation I'm having problem obtaining the solution of the homogeneous Fredholm Integral Equation of the
2nd kind, with separable kernel. I always get a zero
if I use the normal method i was taught for the nonhomogeneous type. I have an example: $$y(x) = \lambda\int_{-1}^1(x + z)y(z)dz$$
|
certainly, your method of solving gives the general form $y(x)=c_1x+c_2$
Then, bringing it back into the Fredholm Integral Equation leads to the condition :
$$ y(x)=c_1x+c_2=2\lambda c_2x+\frac{2}{3}\lambda c_1$$
which implies $c_1=2\lambda c_2$ and $c_2=\frac{2}{3}\lambda c_1$
$ (\frac{4}{3}\lambda^2 -1 ) c_1=0$
So, they are two cases :
If $\lambda=\pm \frac{\sqrt3}{2}$ , then $c_2=\frac{2}{3}\lambda c_1=$any value, and the solutions are $y(x)=2 \lambda c_2 x + c_2$.
If $\lambda \neq \pm \frac{\sqrt3}{2}$ , then $c_1=c_2=0$ and the solution is $y(x)=0$
|
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|
Prove that if $\lim_{x\to\infty}f(x)=L$ exists and finite, and $\lim_{x\to\infty}\lfloor f(x)\rfloor$ doesn't exist then L is an integer
Let $f$ be a continuous function on $(0, \infty)$ s.t $\lim \limits_{x \to \infty}f(x) = L$ exists and finite, but $\lim \limits_{x\to \infty} \lfloor f(x) \rfloor$ doesn't exist. Prove that L is an integer .
I tried to use the fact that for each $\epsilon >0$ there is $x_o$ that for each $x>x_0$ : $|f(x)-L|< \epsilon$. and suppose negatively that L is not an integer and get a contradiction.
Thats by choosing such an epsilon that guarantees that $\lfloor f(x)\rfloor=\lfloor L\rfloor$ when $x>x_0$, or any other combination that tells us $\lim \limits_{x \to \infty}\lfloor f(x)\rfloor = \text{exists}$
I didn't know how to continue from here, or find such an epsilon.
NOTE: $\lfloor f(x)\rfloor$ is the floor function.
|
I am going to do a proof by contradiction. Assume the opposite, that $L$ is not an integer. Note that because $L$ is not an integer, $\lfloor L\rfloor<L<\lceil L\rceil$. Because $f(x)\to L$ as $x\to\infty$, we can choose $\varepsilon=\min\{L-\lfloor L\rfloor, \lceil L\rceil-L\}$ such that there exists an $M$ such that for all $x>M$, $|f(x)-L|<\varepsilon$.
If $L-\lfloor L\rfloor\le \lceil L\rceil-L$, then $\varepsilon=L-\lfloor L\rfloor$ and
$$|f(x)-L|<\varepsilon\\\Rightarrow\lfloor L\rfloor-L<f(x)-L<L-\lfloor L\rfloor\le\lceil L\rceil-L\\
\Rightarrow\lfloor L\rfloor<f(x)<\lceil L\rceil$$
If $L-\lfloor L\rfloor> \lceil L\rceil-L$, then $\varepsilon=\lceil L\rceil-L$ and
$$|f(x)-L|<\varepsilon\\\Rightarrow \lfloor L\rfloor-L<L-\lceil L\rceil<f(x)-L<\lceil L\rceil-L\\
\Rightarrow\lfloor L\rfloor<f(x)<\lceil L\rceil$$
And so, for all $x>M$, $f(x)\in(\lfloor L\rfloor,\lceil L\rceil)$. Then $\lfloor f(x)\rfloor = \lfloor L\rfloor$ for all $x>M$, and $\lim_x\lfloor f(x)\rfloor$ exists and is equal to $\lfloor L\rfloor$.
|
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Show that $\frac{3}{8}\le\int_0^{1/2}\sqrt{\frac{1-x}{1+x}}dx\le \frac{\sqrt{3}}{4}$ Show that $\displaystyle\frac{3}{8}\le\int_0^{1/2}\sqrt{\frac{1-x}{1+x}}dx\le \frac{\sqrt{3}}{4}$ without calculating the integral.
This is an exercise for training the use of mean value theorem for integral.
I tried to apply it in different ways, but the bounds I get are worst that those I need:
*
*A direct application does not work
*I also tried to bound first the function and then apply the thorem, but also failed.
I tried to split the interval and apply the Mean value theorem (in fact, is the same that take upper and lower Riemann sums).
Using MATHEMATICA, We need to split the interval in 3 parts (with equal length), to get (something better than) the upper bound; and we need to split the interval in 8 parts to get (something better than) the lower bound.
Now I conjecture there's some mistake/misprint in the original exercise.
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We have:
$$\frac{d^2}{dx^2}\sqrt{\frac{1-x}{1+x}}=\frac{(1-2x)}{(1-x)^{3/2}(1+x)^{5/2}}$$
so the integrand function is convex over $\left[0,\frac{1}{2}\right]$. That implies (see Hermite-Hadamard inequality):
$$\frac{f\left(\frac{1}{4}\right)}{2}\leq\int_{0}^{\frac{1}{2}}f(x)\,dx \leq \frac{f(0)+f\left(\frac{1}{2}\right)}{4}\tag{1}$$
or:
$$\color{red}{0.387}\ldots=\frac{1}{2}\sqrt{\frac{3}{5}}\leq\color{red}{ \int_{0}^{\frac{1}{2}}\sqrt{\frac{1-x}{1+x}}\,dx} \leq \frac{1}{4}\left(1+\frac{1}{\sqrt{3}}\right)=\color{red}{0.394}\ldots\tag{2}$$
that is way stronger that what was asked to prove.
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Finding Projection Vector Find the projection of the vector $t = [3, 3, 3]^T$ onto the subspace spanned by the vectors $\{x, y\}$, where $x = [6; 1; -3]$, $y = [1; 0; 2]$.
I was told to look at the orthogonal basis, project the vector to each basis element, and then add them up. So do I use the projection formula for $t$ and $x$, and $t$ and $y$, and then add those two together?
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The method you state is good and is basically what the other two answers recommend.
However, this involves calculating two projections. This is not onerous, but there is an easier way to project a 3-D vector onto a 2-D plane defined by two linearly independent vectors.
1) Find the cross-product of the two vectors that define the plane.
2) Find the projection of your vector onto that vector.
3) Subtract that projection from your original vector.
You now have the projection of the original vector onto the plane. This took only one projection calculation, one cross-product, and one subtraction. The other method takes two projection calculations and an addition. Since cross-products are easier than projections, this method is easier. Note also that this method did not even require that the two basis vectors of the plane are orthogonal to each other--just that they are linearly independent.
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Proving that the products GCDs of the coefficients of two polynomials is equal to the GCD of their product's coefficients? Assume that $p(x)=a_nx^n+\dots+a_0$ and $q(x)=b_nx^n+\dots+b_0$ where the coefficients are integers. Let $y$ be the gcd of $a_n,\dots,a_0$ and let $z$ be the gcd of $b_n,\dots,b_0$. How does one prove that the GCD of the coefficients of $p(x)q(x)$ is $yz$?
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Write $p(x)=y P(x)$ and $q(x)=zQ(x)$, where $P(x)$ and $Q(x)$ are primitive polynomials (i.e. the g.c.d. of their coefficients is $1$). Thus
$$p(x)q(x)=yz P(x)Q(x)$$
and it is enough to show $P(x)Q(x)$ is primitive. . It not, there exists a prime number $a$ that divides all its coefficients. Reduce the coefficients modulo $a$; in the ring $\mathbf Z/a\mathbf Z$, we have $\overline{\!PQ}=\overline{\mkern-1muP\vphantom Q}\,\overline{\mkern-1muQ}$. As $\mathbf Z/a\mathbf Z$ is an integral domain, this implies $\overline{\mkern-1muP}=0 $ or $\overline{\mkern-1 muQ}=0$, which means $a$ divides all coefficients of $P$ or all coefficients of $Q$, and is impossible since $P$ and $Q$ are primitive.
Note. This proof is valid for polynomials over any UFD, since irreducible elements in such rings are prime.
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Euclid's proof of infinitude of primes. http://en.wikipedia.org/wiki/Euclid's_theorem
I just read Euclid's proof for the existence of infinitely many primes (I have never used his proof earlier to prove this). It seems to me that he assumes it exist a finite number and he constructs the number $x=p_1 \cdot... \cdot p_n + 1$, then he says that $x$ is either prime or not, if it is we are done, if not it must exist a prime that divides it, and so on...(see link).
However this proof makes no sense, it relies on the fact that if $x$ is not prime then it exist a prime that divides it which is exacly the (weak) fundamental theorem of arithmetics, however this theorem immediately implies that it exist infinite number of primes, because if it didnt then any number $n > p_1 \cdot... \cdot p_n$ cant have a prime factorization.
What am I missing here? To me it seems as Euclid's proof is just stupid.
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Theorem: Any $n\in N$ with $n>1$ is divisible by a prime. Proof: Among the divisors of $n$ that are greater than $1$ (noting that $n$ itself is a divisor of $n$), there is a LEAST one, $M.$ Now $M$ is prime because $1<M, $ and if $M=A B$ with $A>1<B$ then $A$ would be a divisor of $n$ with $1<A<M$, which is absurd by the def'n of $M$.
Theorem (Euclid).If $p_1,...,p_k$ are primes then there is a prime $p$ not equal to any of $p_1,...,p_k.$ Proof: Let $n=1+\prod_{j=1}^kp_j.$ Now $n$ is $1$ more than a multiple of $p_j$ for $j=1,...,k$ so $n$ is not divisible by any of $p_1,...,p_k .$ But $n>1$ so by the previous theorem, $n$ has a prime divisor $p.$
Remark. For $1\leq j\leq k$ we have $n-1=p_j A_j$ with $A_j\in N.$ If $p_j$ divides $n$ then $n=p_j B_j$ with $B_j\in N.$ But then $1=n-(n-1)=p_j(B_j-A_j), $ implying that $p_j$ divides $1, $ which is absurd.
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Is the pushforward measure a categorical-theoretic pushout? Given two measurable spaces $(X,\mathscr{F}),(Y,\mathscr{G}),$ $f:X \to Y$ measurable and $\mu:\mathscr{F} \to [0,\infty)$ a measure, the pushforward of $f_*(\mu):\mathscr{G} \to [0, \infty)$ is defined as $f_*(\mu)(G)=\mu f^{-1}(G).$
Is this terminology coincidental, or is this an example of a universal object in some category? If so, which category, and why is it universal?
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Let me elaborate on the comment by @tomasz. Consider a category $\mathtt{Mes}$ where objects are measurable spaces (sets endowed with $\sigma$-algebras) and morhpisms are measurable maps. For each object $M$ you can construct another object $\mathcal M(M)$ whose elements are measures on $M$, and the $\sigma$-algebra is generated by the evaluation functions $\theta_A:\mathcal M(M)\to \Bbb R$ given by $\theta_A:\mu\mapsto \mu(A)$ for each measurable subset $A$ of $M$. You can regard $\mathcal M$ as an endofunctor on $\mathtt{Mes}$, and its action on morphisms is exactly given by the pushforward construction for measures.
Regarding your original question: I guess that the notion of pushforward for measures was called that way due to the natural idea that you push the measure from one space to another along some map. I have only basic skills in category theory, and I am not much familiar with the pushout and pullback constructions in its general setting, however from what little I know it seems that pushouts are often applied to formalize the idea of gluing sets together. In that case, even if there exists a category for which pushforward of measure is described as a category-theoretical pushout, I am afraid it may be a rather artificial construction without much insight. In a sense, the naming may indeed be just a coincidence, which does not yet rule out the possibility of a connection, just maybe this connection maybe an artificial and purely technical one, existing due to another coincidence.
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Subgroups of a permutation group The permutation group $S_{4}$ is defined as the group of all possible permutations of [1234].
i) Find the number of subgroups of $S_{4}$ that have order 2.
ii) A: { [1234], [2143], [3412], [4321] } and B: { [1234], [1243], [2134], [2143] }. Which of A and B are subgroups of $S_{4}$?
Trying to teach myself a Further Maths module on Groups is proving difficult when none of my teachers know the syllabus, any help would be appreciated! Thanks.
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Hints:
i) If a subgroup has order two then there are exactly one trivial and one non-trivial element. Furthermore the nontrivial element $a\in G$ must fulfill $a^2=e$. So just check with of the elements of $S_4$ fulfill this requirement.
ii) Just check the subgroup axioms for $A$ and $B$.
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How long would it take to guess an arbitrary integer or real number? Let's say two mathematicians play a game. One of them picks an arbitrary element from a countably infinite set (perhaps the integers, as per the title), and the other one guesses what it is. The second player has as many guesses as they need, and after each guess, they are simply told whether they were correct or not.
Would this game never end, or would it last an arbitrarily large but finite amount of time?
What if the first mathematician selects from an uncountably infinite set, such as the real numbers?
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In the first case, one constructs a bijection $f$ from the natural numbers into said countable set. Then you guess $f(1),f(2), \dots$ and eventually (after finitely many steps) you guess correctly.
In the second case, with probability 100% you will always guess incorrectly, so usually the game will go on forever. (You can make only countably many guesses anyway)
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Property on homomorphism between finite groups I was reading some lecture notes on homomorphism between finite groups and intuitively it appeared to me that for a $\phi$ an homomorphism $G \rightarrow H$, we should have:
$$|\,\textrm{Im}(\phi)\,| = \frac{|\,G\,|}{|\,\textrm{ker}(\phi)\,|}$$
where $| . |$ stands for order. Is this right? If yes, is there a short demonstration for?
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ker($\phi$) is a normal subgroup of G, so G/ker($\phi$) is well defined and |G/ker($\phi$)|=|G|/|ker($\phi$)|.
Thanks to "the first isomorphism theorem", there exists an isomorphism between Im($\phi$) and G/Ker($\phi$) moreover |Im($\phi$)|=|G|/|Ker($\phi$)|
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Relative primes What is the number of integers between 1 and 60 that are relatively prime to 60? I know that the answer is 16, but how do I go about finding the relative primes using a quick process?
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You can just use Euler's totient function. Given a prime $p$, $\phi(p^\alpha) = (p - 1)p^{\alpha - 1}$. The function is multiplicative, so $\phi(p^\alpha q^\beta) = \phi(p^\alpha) \phi(q^\beta)$.
Since $60 = 2^2 \times 3 \times 5$, you calculate $$\phi(60) = \phi(4) \phi(3) \phi(5) = 2 \times 2 \times 4 = 16.$$
To verify this answer, let's count up the primes from 7 to 59, and throw in there the odd composite numbers not divisible by 3 or 5, to get: 1, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 49, 53, 59; this is a list of 16 integers.
I suppose either way is a quick enough process for small numbers. But if the numbers get just a little bit bigger, Euler's totient function is much faster: try for example to count up the integers between 1 and 5168743489 that are relatively prime to 5168743489.
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Solving a non-linear, multivariable system of equations I'm researching the mathematics behind GPS, and at the moment I'm trying to get my head around how to solve the following system of equations:
$\sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}=r_1$
$\sqrt{(x-x_2)^2+(y-y_2)^2+(z-z_2)^2}=r_2$
$\sqrt{(x-x_3)^2+(y-y_3)^2+(z-z_3)^2}=r_3$
$(x,y,z)$ is the coordinates of a GPS receiver's location, with $(0,0,0)$ being the center of the earth.
$(x_1,y_1,z_1)$, $(x_2,y_2,z_2)$ and $(x_3,y_3,z_3)$ are the coordinates of each satellite. Hence these are just three distance formulas, where $r_1$, $r_2$ and $r_3$ are the distances from each satellite to the point $(x,y,z)$ on the Earth's surface. By assigning values for $r_1$, $r_2$, $r_3$ and coordinates of each satellite, I want to know how I can solve for the coordinates $(x,y,z)$.
I understand the Newton-Raphson Method for one variable, but I'm getting lost when it comes to using the Jacobian matrix for more than one. All I've been able to do it take the partial derivative for each variable, and then I'm not sure where to go.
For the purpose of this, use the following values. I've tested them on WolframAlpha, so I know that they work:
$r_1=20000$
$r_2=19000$
$r_3=19500$
$x_1=20000$
$y_1=19400$
$z_1=19740$
$x_2=18700$
$y_2=1800$
$z_2=18500$
$x_3=18900$
$y_3=17980$
$z_3=20000$
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I think that you can make the problem simpler the solution of which not requiring Newton-Raphson or any root finder method; the solution is explicit and direct (it does not require any iteration).
Start squaring the expressions to define $$f_i={(x-x_i)^2+(y-y_i)^2+(z-z_i)^2}-r_i^2=0$$ Now, develop $(f_2-f_1)$ and $(f_3-f_2)$ for example. As a result, you have two linear equations since terms $x^2,y^2,z^2$ disappear. You can solve these two equations for $y$ and $z$ and the result will write $y=\alpha_1+\alpha_2 x$,$z=\beta_1+\beta_2 x$. Report these expressions in $f_1$ which is a quadratic equation in $x$.
Using the numbers you gave, I found $$y=\frac{121160315}{7921}-\frac{4255 }{15842}x$$ $$z=-\frac{340394455}{7921}+\frac{43785 }{15842}x$$ and the solution of $f_1=0$ is just $x=16276.24775$ from which $y=10924.45372$, $z=2011.526168$ (just as Amzoti found).
You must take care that there is a second root for the quadratic $x=27858.39152$ from which $y=7813.607758$, $z=34022.89878$.
Depending on the starting point chosen for Newton-Raphson, you would arrive to one of these two solutions.
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Solving 2nd order ODE How would you this ODE:
$$x''+x+\cos^3t=0$$
The homogenous solution is $x=C\cos t$ so $C\cos t$ is term is part of the solution. But how would you go from there?
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The solution of the ode
$$ x''+x=0 $$
should be
$$ x(t) = c_1\cos(t)+c_2\sin(t). $$
Added: For the particular solution you can use variation of parameters method.
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Definition of previsible processes? Definition from my textbook:
A stochastic process $X = (X_n, n \in \mathbb{N}_0)$ is called predictable (or previsible) with respect to the filtration $\mathbb{F} = (\mathcal{F}_n, n \in \mathbb{N}_0 )$ if $X_0$ is constant and if, for every $n \in \mathbb{N}$
$$X_n \text{ is } \mathcal{F}_{n-1}\text{-measurable.}$$
Okay, but that surely cannot mean that I can really predict the next outcome $X_{n+1}$ if I know $X_0, \ldots, X_n$ (regardless if $X$ is also adapted to $\mathbb{F}$), right?
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Technically, you can always predict in terms of the best available information but you will NOT necessarily know the value of $X_{n+1}$ if you know the values of $X_0, \ldots, X_n$. This is easy to show since based on the definition of a previsible process, if you are at $(n-1)^{th}$ index of the filtration (i.e. $\mathcal{F}_{n-1}$), you already know the values of $X_0, \ldots, X_n$ but you don't get to know the value of $X_{n+1}$. You will know $X_{n+1}$ when you are at $n^{th}$ index of filtration (i.e. $\mathcal{F}_{n}$).
I think in the question there seems to be a confusion regarding what subscript $n$ means in $X_n$. It doesn't mean when you know $X_0, \ldots, X_n$ you are at index $n$ in the filtration. In fact, you know those values when you are at $\mathcal{F}_{n-1}$ which doesn't tell what $X_{n+1}$ is.
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To show that $7 \Bbb Z$ and $16 \Bbb Z$ are isomorphic as groups but not isomorphic as rings. To show that $7 \Bbb Z$ and $16 \Bbb Z$ are isomorphic as groups but not isomorphic as rings.
I have done the first part but finding difficult to show that they are not isomorphic as rings??
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$n \mathbf{Z}$ is generated by $n$, so that a group morphism $n \mathbf{Z} \to G$ is characterized by the value it takes on $n$. Therefore, the group morphism $7 \mathbf{Z} \to 16 \mathbf{Z}$ sending $6$ to $17$ is an isomorphism, and its inverse is defined by $16\mapsto 7$. Now take $f : 7 \mathbf{Z} \to 16 \mathbf{Z}$ a ring morphism. Then $f(7)k = 16$ for some $k$. But the $f(7) = 2^d$ for $0\leq d \leq 4$. Can you conclude from that ?
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Prove that odd polynomials $f(x)$ of degree $\leq 10$ with $f(-1) = 0$ form a vector space.
Let $P(X)$ be the usual vector space of polynomials in $x$ with real coefficients.
Let $U$ denote the subset of $P(X)$ consisting of those elements $f(x)$ which have degree less than or equal to $10$, satisfy $f(-x)=-f(x)$ for all $x$ and also satisfy $f(-1)=0$.
Prove $U$ is a subspace of $P(x)$.
Find a basis for $U$ and the dimension of $U$.
I have proved that $U$ is a subspace of $P(X)$ but am not sure how to find a basis and dimension of $U$
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$f(-x)=-f(x)$ means that the subspace $U$ consists solely of odd functions. The subspace $V$ of odd functions in $P(X)$ with degree less than or equal to 10 has basis $x,x^{3},x^{5},x^{7},x^{9}$. Indeed, all these functions are linearly independent, odd and have degree less than 10. Moreover, they span $V$: If $g \in P(X)$ has degree less than or equal to $10$, then $g(x)=a_{10}x^{10}+a_{9}x^{9}+...+a_{1}x+a_{0}$ for some $a_{i}\in \mathbb{R}, 0 \leq i \leq10$. Now $g(-x)=-g(x)$ says $a_{i}=-a_{i}$ if $i$ is even, so these must be $0$. Then $g$ is an $\mathbb{R}$-linear combination of $x,x^{3},x^{5},x^{7},x^{9}$. So $dim(V)=5$.
Notice that $x^3-x, x^5-x, x^7-x, x^9-x$ all lie in $V$, are linearly independent and are all attain the value $0$ at $x=1$ (hence at $-1$ as they are all odd functions). So $dim(U)\geq 4$. But the function $x$ is odd and doesn't equal $0$ at $x=1$, so $U$ can't be all of $V$. So $dim(U)=4$, with a basis given by $x^3-x, x^5-x, x^7-x, x^9-x$. $\square$
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Evaluate $\lim_{t \to \infty} \frac{(\sqrt t+ t^2)}{(4t - t^2)}$. Why is the limit as $t \to \infty = 0$? Would like some feedback on my process. Also, would like to gain a Calculus I understanding of limits.
A question many are having a problem explaining to me is why is the limit as x approaches infinity designated as zero. Thats is, when I get to the point of solving a limit expression or function, why is it as x approaches infinity the value is designated zero?
Surely the number can be a very high positive number also? Or am I missing something here? for example why is the limit as x approached infinity $1/x = o$? What if x is 0.0001?
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$$\lim_{t \to \infty} \dfrac{(\sqrt t+ t^2)}{(4t - t^2)}$$
$$=\lim_{t \to \infty} \dfrac{(1+t^{3/2})}{(4\sqrt t - t^{3/2})}$$
$$=\lim_{t \to \infty} \dfrac{(1+t^{3/2})}{\sqrt t(4 - t)}$$
$$=\lim_{t \to \infty} \dfrac{1}{\sqrt t(4 - t)} + \lim_{t \to \infty} \dfrac {t^{3/2}}{\sqrt t(4 - t)}$$
$$=0 + \lim_{t \to \infty} \dfrac {t}{(4 - t)}$$
$$=\lim_{T \to 0} \dfrac {1/T}{4-1/T},\; \text {with} \;T=1/t$$
$$=\lim_{T \to 0} \dfrac {1}{4T-1}$$
$$=-1$$
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If $p$ is prime and congruent to $1$, then show $((\frac{p-1}{2})!)^2 \equiv -1 \pmod p$ I got another one.
Quadratic residues are completely new to me...
Thanks!
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Take the congruences, $p-1 \equiv -1 \pmod p$, $\text{ }\text{ } p-2 \equiv -2 \pmod p$
and so on upto, $\frac{p+1}2 \equiv -\frac{p-1}2 \pmod p$.
Multiplying and rearranging, $$(p-1)!\equiv 1\cdot (-1)\cdot 2 \cdot (-2) ...\frac{p-1}2 \cdot (-\frac{p-1}2) \equiv (-1)^{\frac{p-1}2}[(\frac{p-1}{2})!]^2 \pmod p$$
Thus, by Wilson's theorem, $ -1 \equiv (-1)^{\frac{p-1}2}[(\frac{p-1}{2})!]^2 \pmod p$.
If by "$p$ is congruent to $1$", you meant that $p \equiv 1 \pmod 4$, we are done as $(-1)^{\frac{p-1}2} =1$.
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Product of functions in $H^1(B)$ where $B \subset \mathbb{R}^2$ I'm rather new to Sobolev spaces and finding myself rather deficient of intuition. So when given a problem like the below where I need to "prove or disprove", I'm finding myself stuck.
Suppose $B$ is a ball in $\mathbb{R}^2$ and $u,v$ are in the Sobolev space $H^1(B)$ (that is, $u,v \in L^2(B)$ and their weak derivatives are also in $L^2(B)$). I need to either prove $uv \in H^1(B)$ or find a counterexample where $uv \not \in H^1(B)$.
Here are my ideas so far.
I know that $H^1(B)=W^{1,2}(B)$ is in the "edge case" for the Sobolev-type inequalities, exactly between where we get integrability ($p<n$) and where we get regularity ($p>n$). This means that $H^1(B)$ embeds into $L^q(B)$ for each $1 \leq q < \infty$. I also know that the requirement $q < \infty$ cannot be dropped. This means the easy argument for the corresponding problem on $W^{1,1}(I)$ for an interval $I$ does not work.
So here $u,v \in L^q$ for every $1 \leq q < \infty$. So $u,v \in L^4$, hence $u^2,v^2 \in L^2$, hence we can use the Cauchy-Schwarz inequality to conclude that $uv \in L^2$.
Now the problem is with the integral of $|\nabla(uv)|^2$. As I recall the Leibniz rule goes through and we get $|u \nabla v + v \nabla u|^2$. This is at most $(|u \nabla v| + |v \nabla u|)^2 = |u \nabla v|^2 + |v \nabla u|^2 + 2|uv \nabla u \cdot \nabla v|$. And here I get stuck, not knowing where to go. Any help?
Edit: having given a little more thought, I am thinking that we can make $uv \not \in H^1$. First of all I think we can take $u=v$. Then $|\nabla(u^2)|^2=2u^2|\nabla u|^2$. Since $\nabla u \in L^2$, $|\nabla u|^2 \in L^1$. Since $u^2$ is in $L^p$ for $1 \leq p < \infty$, we need to get $|\nabla u|^2 \not \in L^p$ for all $p>1$ and we need $u \not \in L^\infty$. I think this can probably be done with polar coordinates and a radial gradient.
|
Let me add the counterexample I gave in the comments for completeness and with further explanations:
Following Alt (Funktionalanaysis, section 8.7), we have the charaterization:
$$|x|^\rho \in H^{m,p}(B_1) \iff \rho > m -\frac{n}{p} \tag{1}$$
From this characterization it is easy to obtain counterexamples. We are interested in the case $m=1$ and $p=2$.
For simplicity we consider $n=3$. Thus $$|x|^\rho \in H^{1,2}(B_1) \iff \rho >-\frac{1}{2}\tag{2}.$$
Consequently $f(x)=|x|^{-\frac{1}{4}}\in H^{1,2}$. However, $f^2=|x|^{-\frac{1}{2}}\not\in H^{1,2}(B_1)$ due to $(1)$ and $(2)$. In particular, these functions are both not in $L^\infty(B_1)$.
With the same idea, counter examples can be constructed for any $n\geq 3$
|
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|
How to show $n^5 + 29 n$ is divisible by $30$ Show that $n^5 + 29 n$ is divisible by $30$.
Attempt:
$n^4 ≡ 1 \pmod 5$ By Fermat Little Theorem
|
You can prove this by induction. Case $n = 1$ is obviously true.
Hint: Use the binomial theorem in the inductive step. Also, a number is divisible by $30$ if and only if it is divisible by its prime factors $2$, $3$, and $5$.
|
{
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|
Why is $\deg(f)$ well-defined? If $M$ and $N$ are boundaryless, compact, connected, oriented $n$-manifolds, and $f:M\to N$ is smooth, then if $\omega_0$ is a $n$-form on $N$ and $\int_N\omega_0\neq 0$, there is a number $a$ such that $\int_M f^*\omega_0=a\int_N\omega_0$.
In fact, if $\omega$ is any $n$-form on $N$, then $\int_M f^*\omega=a\int_N\omega$. How do we know it's the same $a$ for any $\omega$?
|
Suppose $M$ is oriented, compact and $\partial N=\emptyset$. All $n$-forms on $N$ are closed, and by Stokes theorem, $\int_N\partial \eta=0$ for any exact form, so that the map
$$\int_N:\Omega^n(N)\to\Bbb R$$ automatically factors through the isomorphism
$$\int_N:H^n_{dR}(N)\xrightarrow{\sim\,}\mathbb R\,.$$
The conclusion you are after follows from the fact that the linear map $f^*:H^n_{dR}(N)\to H^n_{dR}(M)$ is multiplication by some real number (actually an integer) $a$.
|
{
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|
A functional problem on page 268 in the book GTM 95 probability
I have noticed that if
$$
\lim_{c\downarrow 0}\int_{-\infty}^{\infty} e^{c|x|}P(dx) < \infty
$$
the system $\{1, x, x^2, \cdots \}$ is complete in $L^2$.
So I want to know which theorem or result was used here?
|
This seems to be more of a problem in complex analysis (+ a bit of
Fourier analysis) than in functional analysis.
Your assumption on $\mathbb{P}$ implies that $\int e^{c\left|x\right|}d\mathbb{P}\left(x\right)<\infty$
for some $c>0$. This easily implies $e^{\frac{c}{2}\left|x\right|}\in L^{2}\left(\mathbb{P}\right)$.
Now, let $f\in L^{2}\left(\mathbb{P}\right)$ be arbitrary. We first
show that the function
$$
G:\left\{ z\in\mathbb{C}\mid{\rm Re}\left(z\right)<\frac{c}{4}\right\} \to\mathbb{C},z\mapsto\int f\left(x\right)\cdot e^{zx}d\mathbb{P}\left(x\right)
$$
is well-defined and holomorphic.
The integral is well-defined because of
$$
\left|f\left(x\right)\cdot e^{zx}\right|=\left|f\left(x\right)\right|\cdot e^{{\rm Re}\left(zx\right)}\leq\left|f\left(x\right)\right|\cdot e^{{\rm Re}\left(z\right)\cdot\left|x\right|}\leq\left|f\left(x\right)\right|\cdot e^{\frac{c}{2}\left|x\right|}\in L^{1}\left(\mathbb{P}\right),
$$
by Cauchy Schwarz, since $e^{\frac{c}{2}\left|x\right|}\in L^{2}\left(\mathbb{P}\right)$.
Since the integrand is obviously complex differentiable, we can use
differentiation under the integral sign. We have
\begin{eqnarray*}
\left|\frac{\partial}{\partial z}\left[f\left(x\right)\cdot e^{zx}\right]\right| & = & \left|f\left(x\right)\right|\cdot\left|x\right|\cdot\left|e^{zx}\right|\\
& = & \left|f\left(x\right)\right|\cdot\left|x\right|\cdot e^{{\rm Re}\left(zx\right)}\\
& \leq & \frac{4}{c}\cdot\left|f\left(x\right)\right|\cdot\frac{c}{4}\left|x\right|\cdot e^{{\rm Re}\left(z\right)\cdot\left|x\right|}\\
& \leq & \frac{4}{c}\cdot\left|f\left(x\right)\right|\cdot e^{\frac{c}{4}\left|x\right|}\cdot e^{\frac{c}{4}\left|x\right|}\in L^{1}\left(\mathbb{P}\right),
\end{eqnarray*}
by Cauchy Schwarz, since $e^{\frac{c}{2}\left|x\right|}\in L^{2}\left(\mathbb{P}\right)$.
All in all, this implies that $G$ is holomorphic.
Now assume that $\left\langle f,x^{n}\right\rangle _{L^{2}\left(\mathbb{P}\right)}=0$
for all $n\in\mathbb{N}_{0}$. For $\left|z\right|<\frac{c}{4}$,
we have
\begin{eqnarray*}
\int\left|f\left(x\right)\right|\cdot\sum_{n=0}^{\infty}\left|\frac{\left(zx\right)^{n}}{n!}\right|d\mathbb{P}\left(x\right) & = & \int\left|f\left(x\right)\right|\cdot e^{\left|zx\right|}d\mathbb{P}\left(x\right)\\
& \leq & \int\left|f\left(x\right)\right|\cdot e^{\frac{c}{4}\left|x\right|}d\mathbb{P}\left(x\right)<\infty,
\end{eqnarray*}
which justifies the interchange of integration and summation in
$$
G\left(z\right)=\int f\left(x\right)\cdot\sum_{n=0}^{\infty}\frac{\left(zx\right)^{n}}{n!}d\mathbb{P}\left(x\right)=\sum_{n=0}^{\infty}\left[\frac{z^{n}}{n!}\cdot\int f\left(x\right)\cdot x^{n}d\mathbb{P}\left(x\right)\right]=0.
$$
But since $G$ is holomorphic, this implies $G\equiv0$.
In particular, we get
$$
0=G\left(-i\xi\right)=\int f\left(x\right)\cdot e^{-ix\xi}d\mathbb{P}\left(x\right)=\widehat{f\cdot d\mathbb{P}}\left(\xi\right),
$$
where $f\cdot d\mathbb{P}$ is a finite, complex measure since $f\in L^{2}\left(\mathbb{P}\right)\subset L^{1}\left(\mathbb{P}\right)$.
But this implies $f\cdot d\mathbb{P}\equiv0$ and thus $f=0$ almost
everywhere.
|
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|
Are there some techniques for checking whether a statement implies another without truth tables? Are there some techniques for checking whether a statement implies another without truth tables?
For example, I was asked whether $P\Longrightarrow P_{1}$ given the following statements:
$$P: [p \land (q \land r)]\lor \neg[p \lor (q \land r)],$$
$$P_{1}: [p \land (q \lor r)]\lor \neg[p\lor(q\lor r)].$$
What I did was to find the logical equivalences of the negations of $\neg[p \lor (q \land r)]$ and $\neg[p\lor(q\lor r)]$, then tested some values, with the negations done, it was slightly easier to see what was happening as I drew 0's and 1's under $p,q,r$. I find that for $p:0,q:1,r:0$, $P$ and $P_{1}$ does not have the same truth statement, so $P\nRightarrow P_{1}$, yet it was still a tedious process.
|
You could write a proof instead. Suppose $P$.
If $p\land q\land r$, then $p\land(q\lor r)$. Therefore $P_1$ tautologically.
Otherwise $\neg[p \lor (q \land r)]\Leftrightarrow \neg p \land \neg(q\land r)$. Then $p$ being true contradicts $P$, so $P\Rightarrow P_1$ in that case. In the case that $p$ is false, $P$ reduces to $\neg q \lor \neg r$ and $P_1$ reduces to $\neg q \land \neg r$. If $\neg q\not= \neg r$, then $P$ and not $P_1$, so $P\not\Rightarrow P_1$; otherwise $P$ and $P_1$ are both true or both false, so $P\Rightarrow P_1$.
Therefore $P\Rightarrow P_1$ exactly when $p$ is true or $q=r$, and $P\not\Rightarrow P_1$ exactly when $p$ is false and $q\not=r$.
|
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|
Finding a Matrix from Determinants I've stumbled upon this problem on my homework, and I have no clue how to do it, and haven't found any help online:
If I'm understanding this correctly, then $det(M) = ad - cb + eh - gf$ ?
What I don't get is how to find M from knowing this information.
Are there any suggestions on how to solve this?
|
It's unclear what the dimension of $M$ should be, but here is a start to a solution where $M$ is a $2\times 2$ matrix:
$$\begin{pmatrix}a&??\\1&d\end{pmatrix}$$
Try to finish it from here (i.e., find what goes in the entry labeled $??$).
|
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|
what is this sort of function called? I am doing an assignment but I do not know how to do this problem. I have the following:
$$
f(x)=
\begin{cases}
0 & \text{for $x<0$},\\
x & \text{for $x\geq 0$}.
\end{cases}
$$
We are meant to determine whether it is continuous or not. I do not know why there is like a bracket with a $0$ up top and an $x$ at the bottom, nor do I understand what is meant by for $x \geq 0$. Basically, I do not understand the whole thing. What is this sort of function called so I can do some research on it (I plan to look it up on khan academy)?
|
This is an example of a piecewise function.
The big bracket means that the function is evaluated differently depending on what domain $x$ is in.
When $x < 0$ then $f(x) = 0$.
When $x \geq 0$ then $f(x) = x$.
Continuous functions are functions that do not have any gaps.
This function is continuous.
An example of a function that is not continuous would be
$$
f(x)=
\begin{cases}
0 & x<0\\
1 & x\geq 0
\end{cases}
$$
This is not continuous because there is a gap between the line segments at $x = 0$.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Examples of Separable Spaces that are not Second-Countable In this post I give an example of a separable space that is not second-countable. What are other good examples?
|
$\pi$-Base is an online encyclopedia of topological spaces inspired by Steen and Seebach's Counterexamples in Topology. It currently lists twenty-one separable spaces that are not second countable. You can learn more about any of these spaces by viewing the search result.
Appert Space
Arens-Fort Space
Deleted Diameter Topology
Deleted Radius Topology
Finite Complement Topology on an Uncountable Space
Half-Disc Topology
Helly Space
Maximal Compact Topology
Niemytzki's Tangent Disc Topology
Novak Space
One Point Compactification of the Rationals
Pointed Rational Extension of the Reals
Rational Extension in the Plane
Rational Sequence Topology
Right Half-Open Interval Topology
Single Ultrafilter Topology
Sorgenfrey's Half-Open Square Topology
Stone-Cech Compactification of the Integers
Strong Parallel Line Topology
Strong Ultrafilter Topology
Uncountable Particular Point Topology
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the basis of $\mathbb R^5/W.$ I need help finding out the basis in the following question :
Let $~~W=\big<[1~~2~~1~~0~~1]^t~,[1~~0~~1~~1~~1]^t~,[1~~2~~1~~3~~1]^t\big >~$ be a subspace of $\mathbb R^5$ . Find a basis of $\mathbb R^5/W.$
I can't figure out the basis , kindly help with some hint on how to proceed ...
|
Row reduction of the matrix:
$$\begin{bmatrix}
1&1&1&x\\
2&0&2&y\\
1&1&1&z\\
0&1&3&t\\
1&1&1&u\end{bmatrix}
\rightsquigarrow
\begin{bmatrix}
1&1&1&x\\
0&1&3&t\\
0&0&6&y-2x+2t\\
0&0&0&z-x\\
0&0&0&u-x
\end{bmatrix}$$
shows $W$ has dimension $3$ and is defined by the equations $\,x=z=u,\,$ hence $\,\mathbf R^5/W\,$ has dimension $2$ and it is enough to find two linearly independent vectors that do not satisfy these equations, e.g.:
$$e_1={}^{\mathrm t}(1,0,0,0,0),\enspace e_2={}^{\mathrm t}(0,0,1,0,0) $$
|
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|
Congruence of square-free numbers Let $m$ be a square-free number. I want to prove that, given some $b \in \mathbb{Z}^+$, where $\gcd(b,m)>1$, that
$\qquad b^{\,c\,\phi(m) + 1} \equiv b \pmod m,\ $ for some $c \in \mathbb{Z}^+$.
I've verified this with a bunch of different values of $m$ and $b$, but I'm really stuck on how to prove it.
|
Let $b=p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k}$, where $p_1, p_2, \ldots p_k$ are prime divisors of b. We'll prove there exist $c$ (actually every natural number works), such that:
$$p_1^{a_1(c\phi(m)+1)} \equiv p_1^{a_1} \pmod {m}$$
$$p_2^{a_2(c\phi(m)+1)} \equiv p_2^{a_2} \pmod {m}$$
$$\cdots$$
$$p_k^{a_k(c\phi(m)+1)} \equiv p_k^{a_k} \pmod {m}$$
Get the first relation and divide with $p_1^{a_1}$ since $m$ is square integer, one is the highest power $p_1$ that devides it. Now we get:
$$p_1^{a_1c\phi(m)} \equiv 1 \pmod n$$
where $n=\frac m{gcd(m,p_1)}$, obviously $(n,p_1) = 1$ since m is squarefree integer. Now use the fact that the Euler Totient Function is multiplicative for relatively prime numbers and that $p_1^{\phi(n)} \equiv 1 \pmod n$:
$$p_1^{a_1c\phi(m)} \equiv p_1^{a_1c\phi(n) \phi(p_1)} \equiv (p_1^{\phi(n)})^{a_1c\phi(p_1)} \equiv 1^{a_1c\phi(p_1)} \equiv 1 \pmod n$$
Simularly we prove for each of the congruence relations.
Now just multiply all the congruence relations and we have:
$$(p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k})^{c\phi(m) + 1} \equiv p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k} \pmod m$$
$$b^{c\phi(m) + 1} \equiv b \pmod m$$
Hence the proof.
You can directly prove this by using a nice lemma that another MSE user has proved: here. Subsitute $e=c\phi(m) + 1$ in his answer and use that $p-1 = \phi(p)$ and the fact that $m$ is squarefree.
|
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|
Write down each of the terms in the expansion of $\sin(x^2)-(\sin(x))^2$. Write down each of the terms in the expansion of $\sin(x^2)-(\sin(x))^2$. Taylor's Theorem applies at the point $a=0$ and with $n=4$.
Got no idea how to proceed. My lecture notes have one example that I barely understand. I'd really appreciate a semi-detailed overview of what I should be doing. I can go to class and ask questions, but I need to have some idea of what I am looking at in order to ask intelligent questions.
|
we will keep track of upto powers of $x^4.$ $$\sin(x) = x - \frac 16 x^3 + \cdots, (\sin x)^2 = x^2 - \frac 13 x^4 + \cdots, \sin(x^2) = x^2+ 0x^4 + \cdots $$ therefore $$\sin (x^2) - (\sin x)^2 = \frac 13 x^4 + \cdots $$
|
{
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"url": "https://math.stackexchange.com/questions/1181055",
"timestamp": "2023-03-29T00:00:00",
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|
Is there a relationship between Turing's Halting theorem and Gödel Incompleteness Turing's proof that a Halting oracle is impossible and Gödel's proof that and omega-consistent first order theory of arithmetic must be incomplete are similar in that they use self-referential arguments. Is there an interesting relationship between them especially in light of the Curry-Howard Correspondence and especially the categorical version thereof i.e. the Curry-Howard-Lambek Correspondence.
|
Turing's proof that a Halting oracle is impossible and Gödel's proof that and omega-consistent first order theory of arithmetic must be incomplete are similar in that they use self-referential arguments. Is there an interesting relationship between them.
Well, Gödel's theorem is a simple consequence of Turing's proof.
Take a look at my Introduction to Gödel's Theorems, for example. §43.2 (in the numbering of the second edition) shows that the recursive unsolvability of the halting problem implies that the set of truths of the first-order language of arithmetic is not recursively enumerable. But the theorems in that language of a formalized theory $T$ are recursively enumerable. So there are truths that $T$ can't prove, and if $T$ is sound, can't disprove either. So it is incomplete.
§43.3 then strengthens the result by dropping the assumption that $T$ is sound in favour of the assumption of omega-consistency, together with the usual assumption that $T$ is (primitive) recursively axiomatized and includes a small amount of arithmetic (e.g. contains Robinson arithmetic Q -- the crucial thing is being strong enough to represent the (primitive) recursive functions). Then we can prove that $T$ is incomplete, going via the unsolvability of the Halting Problem (it's a half-page proof in detail, so forgive me for not reproducing it here)!
|
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"url": "https://math.stackexchange.com/questions/1181151",
"timestamp": "2023-03-29T00:00:00",
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