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H: recovering a representation from its character
The character of a representation of a finite group or a finite-dimensional Lie group determines the representation up to isomorphism.
Is there an algorithmic way of recovering the representation given the character and the conjugacy structure of the group?
AI: Below by "representation" I mean "finite-dimensional complex continuous representation."
This is false for Lie groups in general; for example, the character of a representation of $\mathbb{R}$ does not distinguish the $2$-dimensional trivial representation from the representation
$$r \mapsto \left[ \begin{array}{cc} 1 & r \\ 0 & 1 \end{array} \right].$$
(Technically speaking, any countable discrete group is also a Lie group as it is a $0$-dimensional manifold for any reasonable definition of manifold, but now I'm being far too picky.)
The correct statement is for compact groups (not necessarily Lie). It suffices to address the problem for irreducible representations. If $G$ is such a group, $\mu$ is normalized Haar measure, and $\chi_V$ is the character of an irreducible representation $V$, let $L_g : L^2(G) \to L^2(G)$ be the map which translates a function by $g$ (that is take $L_g(f(h)) = f(g^{-1}h)$). Then
$$\dim(V) \int_G \overline{ \chi_V(g) } L_g \, d \mu$$
is a projection $L^2(G) \to L^2(G)$ whose image is the $V$-isotypic component of $L^2(G)$ (exercise). Taking any nonzero vector in this image and applying suitably many elements of $G$ to it will give you explicit vectors spanning a subspace of $L^2(G)$ isomorphic to $V$; turn these into a basis, and then you get explicit matrices for the elements of $G$.
Since $L^2(G)$ is a bit large, an alternate method (which only works for Lie groups) is to start with a faithful representation $W$ and take a tensor power $W^{\otimes n} (W^{\ast})^{\otimes m}$ containing $V$ (this is always possible, see this MO question), then apply the projection
$$\dim(V) \int_G \overline{\chi_V(g)} g \, d \mu.$$ |
H: Show $(\log n)^{ (\log n) } = 2^{(\log n)(\log (\log n))}$
I am having difficulty understanding how this follows.
$$(\log n)^{ (\log n) } = 2^{(\log n)(\log (\log n))} = n^{\log \log n}$$
Which logarithmic identities are used to go through each equality?
e.g. how do you first go from
$$(\log n)^{ (\log n) } = 2^{(\log n)(\log (\log n))}$$
and then to
$$2^{(\log n)(\log (\log n))} = n^{\log \log n}$$
(The log base must be 2 or else this equality won't hold)
AI: I'm assuming the base is $2.$ Otherwise the equality doesn't hold.
$$2^{(\log n)(\log (\log n))} = 2^{(\log (\log n)) (\log n)} = (2^{(\log (\log n))})^{(\log n)}$$
and
$$ 2^{(\log (\log n))} = \log n $$
because
$$ 2^{\log x} = x.$$
Edit: to reflect the update in the question:
$$ 2^{(\log n)(\log (\log n))} = (\color{blue}{2^{(\log n)}})^{(\log (\log n))} = \color{blue}{n}^{{(\log (\log n))}} = n^{\log \log n} $$
because again
$$ 2^{\log n} = n.$$ |
H: When does a normal subgroup contain precisely one non-identity conjugacy class?
Every normal subgroup $N$ of a group $G$ is a union of conjugacy classes. Since every subgroup contains the identity, and the identity is in a class by itself, every normal subgroup already contains the conjugacy class of the identity.
So when is a normal subgroup comprised of exactly two conjugacy classes?
$N = \{1\} \cup \mathcal K$
Here is what I see so far:
Unless $|N|=2$ and $N \leq Z(G)$, the subgroup must have trivial intersection with the center, since each element in the center is contained in its own conjugacy class.
Since $|\mathcal K|$ is the index of the centralizer $C_G(k)$ of any $k\in\mathcal K$, and $|G:C_G(k)| = |N|-1$ divides $|G|$, we must have that G is divisible by the product $|N|(|N|-1)$ of two consecutive numbers. This also implies $|G|$ is even.
Any inner automorphism fixes $N$, but I don't know about outer automorphisms, so $N$ may not have to be a characteristic subgroup.
What is the full characterization of these types of normal subgroups? Do they have any important properties?
Edit: Ted is correct.
AI: All groups are finite. The following is an old result of Wielandt, I believe.
Proposition: Such a subgroup $N$ must be an elementary abelian $p$-group, and every elementary abelian $p$-group is such a subgroup in a certain group $G$.
Proof: If two elements of $G$ are conjugate, then they have the same order. Hence every non-identity element of $N$ has the same order. The order cannot be composite since $g^a$ has order $b$ if $g$ has order $ab$. Hence $N$ is a $p$-group. The commutator subgroup of $N$ is characteristic in $N$ and so normal in $G$. Hence it is either all of $N$ or just $1$; however, in a non-identity $p$-group the commutator subgroup is always a proper subgroup. In particular, $N$ is abelian and every element has order $p$.
Now suppose such an elementary abelian group $N$ is given. Let $G=\operatorname{AGL}(1,p^n)$ be the set of affine transformations of the one-dimensional vector space $K$ over the field $K$ of $p^n$ elements. That is $G$ consists of all $\{ f : K \to K :x \mapsto \alpha x + \beta ~\mid~ \alpha,\beta \in K, \alpha \neq 0 \}$. Then every non-identity element of $N=K_+=\{ f : K \to K : x \mapsto x + \beta ~\mid ~ \beta \in K \}$ is conjugate under $K^\times = \{ f : K \to K : x \mapsto \alpha x ~\mid~ \alpha \in K, \alpha \neq 0 \}$. Thus the proof is complete. $\square$. |
H: How to handle big powers on big numbers e.g. $n^{915937897123891}$
I'm struggling with the way to calculate an expression like $n^{915937897123891}$ where $n$ could be really any number between 1 and the power itself.
I'm trying to program (C#) this and therefor performance is key.
I tried
for(i <= 915937897123891)
value = n * value;
But that is obviously very very slow.
I wander if it is possible to achieve this using bitwise operators.
Also I would like to know what steps I could take to improve the way the calculate this.
Additionally I would also like to know how this should be solved without a calculator, just with (much?) paper and pen.
Feel free to ask for more information.
Thanks in advance,
Mixxiphoid
AI: The general method using exponentiation by squaring.
In some situations, the time analysis is fully determined by realizing that the direct algorithm uses $\Theta(n)$ multiplies, but the square-and-multiply algorithm only uses $\Theta(\log n)$ multiplies. However, numbers grow large in this calculation, and the cost of multiplication becomes significant and must be accounted for.
Let $N$ be the exponent and $M$ be the bit size of $n$. The direct algorithm has to, in its $k$-th multiply, compute an $M \times kM$ product (that is, the product of an $n$ bit number by a $kn$-bit number). Using high school multiplication, this costs $kM^2$, and the overall cost is
$$ \sum_{k=1}^{N-1} k M^2 \approx \frac{1}{2} N^2 M^2 $$
With $N$ as large as it is, that's a lot of work! Also, because $n$ is so small, the high school multiplication algorithm is essentially the only algorithm that can be used to compute the products -- the faster multiplication algorithms I mention below can't be used to speed things up.
In the repeated squaring approach, the largest squaring is $NM/2 \times NM/2$. The next largest is $NM/4 \times NM/4$, and so forth. The total cost of the squarings is
$$ \sum_{i=1}^{\log_2 N} \left(2^{-i} N M \right)^2 \approx \sum_{i=1}^{\infty} N^2 M^2 2^{-2i} = \frac{1}{3} N^2 M^2$$
Similarly, the worst cost case for the multiplications done is that it does one $M \times NM$ multiply, one $M \times NM/2$ multiply, and so forth, which add up to
$$ \sum_{i=0}^{\infty} M (M N) 2^{-i} = 2 M^2 N $$
The total cost is then in the same ballpark either way, and there are several other factors at work that would decide which is faster.
However, when both factors are large, there are better multiplication algorithms than the high school multiplication algorithm. For the best ones, multiplication can be done in essentially linear time (really, in $\Theta(n \log n \log \log n)$ time), in which case the cost of the squarings would add up to something closer to $MN$ time (really, $MN \log(MN) \log \log(MN)$), which is much, much, much faster.
I don't know if BigInteger uses the best algorithms for large numbers. But it certainly uses something at least as good as Karatsuba, in which case the cost of the squarings adds up to something in the vicinity of $(NM)^{1.585}$. |
H: Asking a linear algebra problem in Berkeley Problems in Mathematics, Spring 1986.
Let $V$ be a finite dimensional vector space and $A$ and $B$ two linear transformations on $V$ such that $A^{2}=B^{2}=0$ and $AB+BA=1$.
1) Prove that if $N_{A}$ and $N_{B}$ are respective null spaces of $A$ and $B$, then $N_{A}=AN_{B}$, $N_{B}=BN_{A}$, and $V=N_{A}\oplus N_{B}$.
2) Prove the dimension of $V$ is even.
3) Prove that if the dimension of $V$ is 2, then $V$ has a basis with respect to which $A$ and $B$ are represented by matrices $(0,1),(0,0)$ and $(0,0),(1,0)$ (sorry I do not know how to type matrices).
My main difficulty is to prove $N_{A}\oplus N_{B}=V$. It follows trivially that $N_{A}\cap N_{B}=\{0\}$, but I do not know how to prove $N_{A}\oplus N_{B}=V$, and I also do not know how to prove $n=2k$.
AI: $$AB+BA=I$$
$$A(AB+BA)=A$$
$$ABA=A$$
$$rank(ABA)=rank(A)$$
Combining with
$$rank(ABA)\leq \min(rank(AB),rank(A))$$
Gives that
$$rank(A)\leq \min(rank(AB),rank(A))$$
If $rank(AB)<rank(A)$, then $rank(A)$ is $\leq$ something strictly smaller than itself, a contradiction. So $rank(A)\leq rank(AB)$.
$$rank(A)\leq rank(AB)\leq \min(rank(A),rank(B))$$
Reusing the previous argument, we find that $rank(A)\leq rank(B)$.
We can start back up at the top with
$$AB+BA=I$$
$$B(AB+BA)=B$$
$$BAB=B$$
and carry through the whole previous argument again (swapping $B$ and $A$ essentially) to find that $rank(B)\leq rank(A)$.
Both of these together imply that $rank(A)=rank(B)$. Since the nullspaces span the space together, the space must be of even dimension. |
H: All real functions are continuous
I've heard that within the field of intuitionistic mathematics, all real functions are continuous (i.e. there are no discontinuous functions). Is there a good book where I can find a proof of this theorem?
AI: Brouwer proved (to his own satisfaction) that every function from $\mathbb{R}$ to $\mathbb{R}$ is continuous. Modern constructive systems rarely are able to prove this, but they are consistent with it - they are unable to disprove it. These system are also (almost always) consistent with classical mathematics in which there are plenty of discontinuous functions from $\mathbb{R}$ to $\mathbb{R}$. One place you can find something about this is the classic Varieties of Constructive Mathematics by Bridges and Richman.
The same phenomenon occurs in classical computable analysis, by the way. Any computable function $f$ from $\mathbb{R}$ to $\mathbb{R}$ which is well defined with respect to equality of reals (and thus is a function from $\mathbb{R}$ to $\mathbb{R}$ in the normal sense) is continuous. In particular the characteristic function of a singleton real is never computable. This would be covered in any computable analysis text, such as the one by Weihrauch.
Here is a very informal argument that has a grain of truth. It should appear naively correct that if you can "constructively" prove that something is a function from $\mathbb{R}$ to $\mathbb{R}$, then you can compute that function. So the classical fact that every computable real function is continuous suggests that anything that you can constructively prove to be a real function will also be continuous. This suggests that you cannot prove constructively that any classically discontinuous function is actually a function. The grain of truth is that there are ways of making this argument rigorous, such as the method of "realizability". |
H: Convergence of the sequence $f_{1}\left(x\right)=\sqrt{x} $ and $f_{n+1}\left(x\right)=\sqrt{x+f_{n}\left(x\right)} $
Let $\left\{ f_{n}\right\} $ denote the set of functions on
$[0,\infty) $ given by $f_{1}\left(x\right)=\sqrt{x} $ and
$f_{n+1}\left(x\right)=\sqrt{x+f_{n}\left(x\right)} $ for $n\ge1 $.
Prove that this sequence is convergent and find the limit function.
We can easily show that this sequence is is nondecreasing. Originally, I was trying to apply the fact that “every bounded monotonic sequence must converge” but then it hit me this is true for $\mathbb{R}^{n} $. Does this fact still apply on $C[0,\infty) $, the set of continuous functions on $[0,\infty) $. If yes, what bound would we use?
AI: Clearly, for each $x \ge 0$ the sequence $\{f_n(x)\}_n$ is increasing.
Let
$$
g:[0,\infty) \to \mathbb{R},\ g(x)=
\left\{
\begin{array}{cc}
0 & \text{ if } x=0\cr
\frac{1+\sqrt{1+4x}}{2} & \text{ if } x>0
\end{array}
\right..
$$
I claim that $f_n(x) \le g(x)$ for every $x \ge 0$. Indeed $f_n(0)=0=g(0)$ for every $n$, and $f_1(x) \le g(x)$ for every $x > 0$. If I suppose that $f_n(x) \le g(x)$ for $n \ge 1$ and $x > 0$, then
$$
f_{n+1}^2(x)-g^2(x)=x+f_n(x)-\frac{1+2x+\sqrt{1+4x}}{2}=f_n(x)-g(x)\le 0,
$$
i.e. $f_{n+1}(x)\le g(x)$. Hence $f_n(x)\le g(x)$ for every $n \ge 1$ and $x > 0$.
Thus for each $x \ge 0$, the sequence $\{f_n(x)\}_n$ is convergent (being increasing and bounded from above), and its (pointwise) limit $f: [0,\infty) \to \mathbb{R}, x \mapsto f(x)$ satisfies $f(x)=\sqrt{x+f(x)}$, i.e. $f(x)=g(x)$ for every $x \ge 0$. |
H: Distance Puzzles?
A man moves 1km east, 2km north, 3km west, 4km south, 5km east, 6km north, 7km west and so on until he travels total of 300km. So what will be the distance from origin?
AI: To travel 300 km, he would need to take 24 "steps".
For every 4 steps, he travels $2\sqrt{2}$ km south west. He will do this 6 times, so the answer is $12\sqrt{2}$ km |
H: 4 by 4 Matrix Puzzle
I was solving the puzzle for the Company interview exam. I found this puzzle, I cannot come up with the solution. How to solve it and what is the correct answer?
Determine the number of $4\times 4$ matrices having all entries 0 or 1 that have an odd number of $1$s in each row and each column.
AI: Fill the upper-left hand $3\times 3\,$ arbitrarily with $0$'s and/or $1$'s. This can be done in $2^9$ ways.
For any such choice of $0$'s and/or $1$'s, fill in the first three entries in the fourth row, and the first three entries in the fourth column, so that the number of $1$'s in each of the first three columns, and in each of the first three rows, is odd. This can be done in precisely one way.
Now put a $0$ or a $1$ in the lower right-hand corner, to make the number of $1$'s in the bottom row odd. It turns out that this makes the number of $1$'s in the rightmost column odd. To check this, work modulo $2$. |
H: Follow up on example computation of $\mathrm{Tor}_n$
I have a follow up question on this question of mine:
I can't reconstruct how I got $\operatorname{Im}{d_1^\ast} = 0$ from the following chain:
$$0 \to \mathbb Z \otimes_{\mathbb Z} (\mathbb Z / 2 \mathbb Z) \xrightarrow{d_1^\ast = \cdot 284 \otimes id} \mathbb Z \otimes_{\mathbb Z} (\mathbb Z / 2 \mathbb Z) \xrightarrow{d_0^\ast=0} 0$$
Now I think $\operatorname{Im}{d_1^\ast} = 284 \mathbb Z \otimes N$ and $\operatorname{Ker}{d_0^\ast} = \mathbb Z \otimes (\mathbb Z / 2 \mathbb Z)$.
And then $Tor^1 (\mathbb Z / 284 \mathbb Z, \mathbb Z / 2 \mathbb Z) = (\mathbb Z \otimes \mathbb Z / 2 \mathbb Z) / (284 \mathbb Z \otimes \mathbb Z / 2 \mathbb Z) $.
Is
$$\operatorname{Im}{d_1^\ast} = 284 \mathbb Z \otimes (\mathbb Z / 2 \mathbb Z) $$
and
$$\operatorname{Ker}{d_0^\ast} = \mathbb Z \otimes (\mathbb Z / 2 \mathbb Z) \cong \mathbb Z / 2 \mathbb Z$$ correct ?
And what does $(A \otimes B) / (C \otimes D)$ look like? Is it isomorphic to $(A/C) \otimes (B/D)$? Thanks for your help.
AI: Your description of the image is not correct.
You would do well to heed Jack Schmidt's warning in the comments: although
$248\mathbb Z$ is a submodule of $\mathbb Z$, this is no longer true once you
tensor with $\mathbb Z/2$. So your description of the image is not only incorrect, but the candidate image you have written down is not a subobject of the target.
I think you would also do well to follow my advice in the comments above, and to simplify the various tensor products in your complex , and then describe the maps in terms of the simplified objects, before you try to
compute its cohomology. |
H: minimal polynomial of normal endomorphism with given eigenvalues
What's the minimal polynomial of a normal endomorphism $\phi$ with eigenvalues $2, 2, 1+i, 1+i, 1-i, 1-i, 3$?
It is $\mu_\phi | (t-2)^2(t-1-i)^2(t-1+i)^2(t-3)$ but is there any more I can derive from the fact that $\phi \circ \phi^*=\phi^* \circ \phi$?
AI: An endomorphism $\phi$ is diagonalizable iff its minimal polynomial is a product of distinct linear factors (see http://en.wikipedia.org/wiki/Diagonalizable_matrix). A normal endomorphism is always diagonalizable over $\mathbb{C}$.
The minimal polynomialof $\phi$ must thus bu $(t-2)(t-1-i)(t-1+i)(t-3)$ because, as you pointed out, it divides $(t-2)^2(t-1-i)^2(t-1+i)^2(t-3)$ and shares that same roots as the characteristic polynomial of $\phi$. |
H: Conditional and joint probability manipulations when there are 3 variables
I'm having trouble verifying why the following is correct.
$$p(x, y \mid z)= p(x \mid y, z) p(y \mid z)$$
I tried grouping the $(x, y)$ together and split by the conditional, which gives me
$$p(x, y \mid z) = p(z\mid x, y) p(x, y)/p(z)$$
However, this did not bring me any closer. I'm uncertain about what kind of manipulations are allowed given more than 2 variables.
Say an expression like:
$$p(a, b, c)$$
Then I know from the chain rule that I can break it down to:
$$p(a, b, c)=p(a \mid b, c) p(b, c) = p(a \mid b, c) p(b \mid c) p(c)$$
Is it allowed to split by the second comma:
$$p(a, b, c) = p(a, b \mid c) p(c) ?$$
And even more complicated and expression like:
$$p(a|b,c)$$
Am I allowed to rewrite this expression by grouping (a|b) together to give me something like
$$p(a|b,c)=p((a|b)|c)p(c)$$
And does this expression even make sense?
AI: $\Pr(a,b,c)=\Pr(a,b\mid c)\Pr(c)$ is allowed.
You are simply saying $\Pr(d,c)=\Pr(d\mid c)\Pr(c)$ where $d = a \cap b$.
Combine this with $\Pr(a,b,c)=\Pr(a\mid b,c)\Pr(b,c)=\Pr(a\mid b,c)\Pr(b\mid c)\Pr(c)$ and divide through by nonzero $\Pr(c)$ to get $\Pr(a,b\mid c)=\Pr(a\mid b,c)\Pr(b\mid c)$. |
H: Formula for calculating the center of an arc
Is there a formula for calculating the point equidistant from the start point and end point of an arc given:
1) An arc is defined as: A center point $P$, a radius $r$ from the center, a starting angle $sA$ and an ending angle $eA$ in $radians$ where the arc is defined from the starting angle to the ending angle in a counter-clockwise direction.
2) The start point $sP$ is calculated as: $sP\{Px + \cos sA \times r, Py + -\sin sA \times r\}$
3) The end point $eP$ is calculated as: $eP\{Px + \cos eA \times r, Py + -\sin eA \times r\}$
Give the above restrictions, is there a way to calculate the point that is halfway between the start and end angles and exactly $r$ units away from the center?
AI: Had an epiphany just as I hit submit, would this work?
$cP \{ Px +$ $\cos (eA - sA) \over 2$ $\times r, Py +$ $-\sin (eA - sA) \over 2$ $\times r\}$
SOLVED:
Using the piece-wise function:
$
cP( Px +$ $\cos($ $sA + eA \over 2$ $ + n) \times r, Py +$ $-\sin($ $sA + eA \over 2$ $ + n) \times r) =
\begin{cases}
n = 0, & \text{if }eA - sA \text{ is } >= 0 \\
n = \pi, & \text{if }eA - sA \text{ is } < 0
\end{cases}
$
For you computer science-y types here's some pseudo-code:
double e = GetEndAngle();
double s = GetStartAngle();
double d = e - s;
double x = 0.0;
double y = 0.0;
double offset = 0.0;
if(d < 0.0) {
offset = PI;
}
x = (GetPosition().GetX() + std::cos(((s + e) / 2.0) + offset) * GetRadius());
y = (GetPosition().GetY() + -std::sin(((s + e) / 2.0) + offset) * GetRadius()); |
H: Geometry Puzzle - Largest circle on Chess
The following puzzle was asked in company interview round.I have no idea ,how to do it?
What is the diameter of the largest circle that can be drawn on a
chessboard so that its entire circumference
gets covered by the black squares and no part of the circumference
falls on any white space.
The side of the chess square is 2 cm.
How to solve this Puzzle?
Thank You!
AI: It's one that passes through $8$ black squares arranged in a diamond shape, passing through the corners of each. Any larger and it couldn't curve in one direction while passing only through corners, which is has to do to never hit a white square.
Any circular arc can be repeated on the board simply by reflecting it across each axis and diagonal. In other words, a circle can be considered fully defined by an $8^{th}$ of its circumference. Of course it could be defined by any section, but that's irrelevant since we want to use the board's eightfold symmetry. Our definition (on an $8^{th}$ circumference arc) involves centering the circle in a central black square and specifying the radius to the single vertex the arc must pass through. We can clearly specify such an arc, and due to the symmetry of the board we know that the circle will also pass through all the other vertices we need it to.
Find the actual diameter by looking at opposite vertices. This involves going $2\ cm$ down and $6\ cm$ across (or equivalent), giving a diameter of $2\sqrt{10}$. |
H: Probability distribution with binomial
Could someone explain to my dumb head why we are seeking $P(X \leq2)$? Is it because that the "majority" of "five" is 3? And we want to find three correct transmissions?
AI: There are alltogether $5$ bits and you are interested in the total number of correct decisions for at least $3$ of them to say that a correct $1$ or correct $0$ is sent. It can also be interpreted that at most there are $k=2$ errors. this is exactly what $P(X\leq 2)$ is. |
H: polynomial, power series, radius of convergnce
Let $p(x)$ be a polynomial of degree $N$ then the radius of convergence of the power series $$\sum_{n=0}^{\infty}p(n)x^n$$
depends on $N$
is $1$ for all $N$
is $0$ for all $N$
is $\infty$ for all $N$
Radius of convergence $r=\lim\limits_{n\to\infty}\frac{p(n)}{p(n+1)}\rightarrow1$ ?
AI: You (probably) know that if $p$ and $q$ are polynomials with the same degree and leading coefficients $p_0$ and $q_{\,0}$, then
$$\lim_{x\to \infty}\frac{p(x)}{q(x)}=\frac{p_0}{q_{\,0}}$$
Your line of thought is good. Since $p(n)$ and $p(n+1)$ have the same degree and the same leading coefficient, say $p_0$, then
$$\lim_{n\to \infty}\frac{p(n)}{p(n+1)}=\frac{p_0}{p_0}=1$$ |
H: When is a quasi-projective variety affine?
By an affine variety I mean a variety that is isomorphic to some irreducible algebraic set in $\mathbb A^n$ and by a quasi-projective variety I mean a locally closed subset of $\mathbb P^n$, with the usual Zariski topology and structure sheaf. (I am not quite familiar with the language of schemes.)
Assume that the underlying field is algebraically closed.
Let $X$ be a quasi-projective variety. Are there any effective ways to tell whether $X$ is affine?
Let me be more specific. If $X$ is affine, then $\mathcal O_X(X)$ is large enough so that it completely determines the variety. In particular, the Nullstellensatz holds, i.e. the map
$$X\to \mathrm {spm}(\mathcal O_X(X))$$
$$x\mapsto \{f\in\mathcal O_X(X)|f(x)=0\}$$
is a 1-1 correspondence, where $\mathrm{spm}(\mathcal O_X(X))$ denotes the maximal ideals in $\mathcal O_X(X)$.
Now I'd like to pose the question: If $X$ is quasi-projective and the above map is a bijection, is $X$ necessarily affine?
Thanks!
AI: If the map from $X$ to the maxspec of $\mathcal O(X)$ is a bijection, then $X$ is indeed affine.
Here is an argument:
By assumption $X \to $ maxspec $\mathcal O(X)$ is bijective, thus quasi-finite,
and so by (Grothendieck's form of) Zariski's main theorem, this map factors as an open embedding of $X$ into a variety that is finite over maxspec $\mathcal O(X)$.
Any variety finite over an affine variety is again affine, and hence $X$ is an open subset of an affine variety, i.e. quasi-affine. So we are reduced to considering the case when $X$ is quasi-affine, which is well-known and straightforward.
(I'm not sure that the full strength of ZMT is needed, but it is a natural tool
to exploit to get mileage out of the assumption of a morphism having finite fibres, which is what your bijectivity hypothesis gives.)
In fact, the argument shows something stronger: suppose that we just assume
that the morphism $X \to $ maxspec $\mathcal O(X)$ has finite non-empty fibres,
i.e. is quasi-finite and surjective.
Then the same argument with ZMT shows that $X$ is quasi-affine. But it is standard that the map $X \to $ maxspec $\mathcal O(X)$ is an open immersion when $X$ is quasi-affine,
and since by assumption it is surjecive, it is an isomorphism.
Note that if we omit one of the hypotheses of surjectivity or quasi-finiteness, we can find a non-affine $X$ satisfying the other hypothesis.
E.g. if $X = \mathbb A^2 \setminus \{0\}$ (the basic example of a quasi-affine,
but non-affine, variety), then maxspec $\mathcal O(X) = \mathbb A^2$, and the open immersion $X \to \mathbb A^2$ is evidently not surjective.
E.g. if $X = \mathbb A^2$ blown up at $0$, then maxspec $\mathcal O(X) =
\mathbb A^2$, and $X \to \mathbb A^2$ is surjective, but has an infinite fibre
over $0$.
Caveat/correction: I should add the following caveat, namely that it is not always true, for a variety $X$ over a field $k$, that $\mathcal O(X)$ is finitely generated over $k$, in which case maxspec may not be such a good construction to apply, and the above argument may not go through. So in order to conclude that $X$ is affine, one should first insist that $\mathcal O(X)$ is finitely generated over $k$, and then that futhermore the natural map $X \to $ maxspec $\mathcal O(X)$ is quasi-finite and surjective.
(Of course, one could work more generally with arbitrary schemes and Spec rather than
maxspec, but I haven't thought about this general setting: in particular, ZMT requires some finiteness hypotheses, and I haven't thought about what conditions might guarantee that the map $X \to $ Spec $\mathcal O(X)$ satisfies them.)
Incidentally, for an example of a quasi-projective variety with non-finitely generated ring of regular functions, see this note of Ravi Vakil's |
H: Keep getting generating function wrong (making change for a dollar)
Possible Duplicate:
Making Change for a Dollar (and other number partitioning problems)
I am working on the classic coin problem where I would like to calculate the number of ways to make change for a dollar with a given number of denominations. From here, I am also going to be working on how to partition the number $100$ with at least two positive integers below $100$.
I have read over all the same posts on here and other sites and still I am unsure of what I am doing wrong.
The answer to our problem ($293$) is the coefficient of $x^{100}$ in the
reciprocal of the following:
$$(1-x)(1-x^5)(1-x^{10})(1-x^{25})(1-x^{50})(1-x^{100}).$$
I do out this equation with $x = 100$ and get a really large return. My skills are very limited and most of the sites I have been reading over use terminology and operators/symbols I am unfamiliar with.
I look at this and think it seems very straight forward but, I get answers that way off. Are there any tips or steps that I could be overlooking?
AI: The coefficient of $x^{100}$ means the multiplicative factor that appears along with $x^{100}$ as some term in the expansion of the expression. For example, the coefficient of $x$ in $$(1-x)^2 = 1 - 2x + x^2$$ is $-2$.
In this case, you want the coefficient of $x^{100}$ in
$$\frac1{(1-x)(1-x^5)(1-x^{10})(1-x^{25})(1-x^{50})(1-x^{100})}.$$
Without going into any details on how to find the coefficient, let me just show you how to look it up: if you go to Wolfram Alpha and type in "power series [that expression]", the first output box "Series expansion at $x = 0$" says:
$$1 + x + x^2 + x^3 + x^4 + 2x^5 + O(x^6)$$
If you click on "More terms" it expands to something like
$$1+x+x^{2}+x^{3}+x^{4}+2 x^{5}+2 x^{6}+2 x^{7}+2 x^{8}+2 x^{9}+4 x^{10}+4 x^{11}+4 x^{12}+4 x^{13}+4 x^{14}+6 x^{15}+6 x^{16}+6 x^{17}+6 x^{18}+6 x^{19}+9 x^{20}+9 x^{21}+9 x^{22}+9 x^{23}+9 x^{24}+13 x^{25}+13 x^{26}+13 x^{27}+13 x^{28}+13 x^{29}+18 x^{30}+18 x^{31}+18 x^{32}+18 x^{33}+18 x^{34}+24 x^{35}+24 x^{36}+24 x^{37}+24 x^{38}+24 x^{39}+31 x^{40}+31 x^{41}+31 x^{42}+31 x^{43}+31 x^{44}+39 x^{45}+39 x^{46}+39 x^{47}+39 x^{48}+39 x^{49}+50 x^{50}+50 x^{51}+50 x^{52}+50 x^{53}+50 x^{54}+62 x^{55}+62 x^{56}+62 x^{57}+62 x^{58}+62 x^{59}+77 x^{60}+77 x^{61}+77 x^{62}+77 x^{63}+77 x^{64}+93 x^{65}+93 x^{66}+93 x^{67}+93 x^{68}+93 x^{69}+112 x^{70}+112 x^{71}+112 x^{72}+112 x^{73}+112 x^{74}+134 x^{75}+134 x^{76}+134 x^{77}+134 x^{78}+134 x^{79}+159 x^{80}+159 x^{81}+159 x^{82}+159 x^{83}+159 x^{84}+187 x^{85}+187 x^{86}+187 x^{87}+187 x^{88}+187 x^{89}+218 x^{90}+218 x^{91}+218 x^{92}+218 x^{93}+218 x^{94}+252 x^{95}+252 x^{96}+252 x^{97}+252 x^{98}+252 x^{99}+ \color{red}{293 x^{100}} +293 x^{101}+293 x^{102}+293 x^{103}+293 x^{104}+337 x^{105}+337 x^{106}+337 x^{107}+337 x^{108}+337 x^{109}+388 x^{110}+388 x^{111}+388 x^{112}+388 x^{113}+388 x^{114}+442 x^{115}+442 x^{116}+442 x^{117}+442 x^{118}+442 x^{119}+503 x^{120}+503 x^{121}+503 x^{122}+503 x^{123}+503 x^{124}+571 x^{125}+571 x^{126}+571 x^{127}+571 x^{128}+571 x^{129}+646 x^{130}+646 x^{131}+646 x^{132}+646 x^{133}+646 x^{134}+728 x^{135}+728 x^{136}+O(x^{137})$$
so you can see that the coefficient of $x^{100}$ is $293$.
That said, I doubt whether the generating function approach is any easier to compute with, than the more elementary way of writing down a recurrence relation, etc. |
H: Cantor's theorem
The theorem cardinality of set of real numbers is strictly larger than cardinality of set of natural numbers uses diagonal method in its proof.
Why can't we use the same argument for creating a natural number just like we do real number? i.e. take 1st digit of the first element, 2nd digit of the second element... and change the digits so that the new natural number differs from the rest in the set of natural numbers.
AI: Because, in order to be an integer, the constructed digit-string must end in an infinite string of zeros (reading right-to-left), and there is no way to guarantee this with a diagonal argument. However, the diagonal argument can be used to prove that there are an uncountable number of p-adic numbers. |
H: example of the vector (bounding expectation of a form in Rademacher functions)
Let $x\in R^{2m}$ such that $x_1+\ldots+x_{2m}=0$ and let $r_i, i=1, \ldots, 2m$ be Rademacher functions, i.e. $P(r_i=1)=P(r_i=-1)=1/2$.
I would like to find an example of the vector $x$ such that
$E(\sum_{i=1}^{2m}x_ir_i)^{2q}\geq C\sqrt{2q}\|x\|_2$.
AI: I suspect you need to add more constraints, $x=0$ will satisfy the inequality trivially.
Alternatively, assuming that the $r_i$ are independent, then
$$E(\sum_{i=1}^{2m}x_ir_i)^{2q} = E (\sum_{i_1}\cdots \sum_{i_{2q}}x_{i_1} \cdots x_{2q} r_{i_1} \cdots r_{2q}) =\sum_{i_1}\cdots \sum_{i_{2q}}x_{i_1} \cdots x_{2q} E(r_{i_1} \cdots r_{2q}) = \sum x_i^{2q}, $$
where we used independence and that $r_i^2 = 1$ in the last equality.
The problem then reduces to finding an $x$ that satisfies $\sum x_i = 0$ and $\sum x_i^{2q} \geq C \sqrt{2 q} \sqrt{\sum x_i^2}$.
Let $t\geq 0$ and let $x=(t,\cdots t,-t,\cdots, -t)$ ($m$ components are $t$ followed by $m$ components of $-t$). Clearly $\sum x_i = 0$. The inequality becomes $2m t^{2q} \geq C \sqrt{2 q} \sqrt{2 m} t$. This is satisfied whenever $t=0$ or whenever the inequality $t^{2q-1} \geq C \sqrt{\frac{q}{m}}$ is satisfied, or equivalently whenever $t \geq \sqrt[2q-1]{C \sqrt{\frac{q}{m}}}$. |
H: Simpler expression for a certain determinant.
A question in elementary linear algebra, while considering the Cayley-Menger Determinant:
Given an $n\times n$ matrix $M$, consider $$\tilde{M}=\begin{pmatrix} M & (1,1,\cdots, 1)^\top \\ (1,1,\cdots, 1)& 0\end{pmatrix}$$ Is it possible to express $\det(\tilde{M})$ in terms of $\det(M)$ and some simpler terms?
You may assume that $M$ is symmetric. (This problem, in its original setting in my research, has the condition that $M$ is symmetric. But it'll be more interesting to solve the general case.)
AI: One can certainly specialize the determinant formula for block matrices to the bordering case:
$$\begin{vmatrix}\mathbf M&\mathbf e\\\mathbf e^\top&0\end{vmatrix}=-(\mathbf e^\top\mathbf M^{-1}\mathbf e)\det\mathbf M$$
where $\mathbf e$ is the column vector whose entries are all $1$'s. |
H: Upper bound on smallest prime $p$ needed to tell two numbers $\leq n$ apart modulo $p$
I'm going through this paper:
E. D. Demaine, S. Eisenstat, J. Shallit, and D. A. Wilson. Remarks on separating words. ArXiv e-prints, March 2011.
And on page 2, there is the following lemma:
Lemma 1. If $0 \leq i,j \leq n$ and $i \ne j$, then there is a prime $p \leq 4.4 \log{n}$ such
that $i \not\equiv j$ (mod $p$).
They don't show where the $4.4 \log{n}$ comes from and I'm not sure where to look for it. Best I could find is this, but it does not provide the solution.
Could someone point me in the correct direction?
AI: High Level Idea
Suppose that $j > i$. Let $k = 4.4 \log n$. Then we need to prove that there is a prime number $p < k$ that doesn't divide $j -i$. Assume to the contrary that this is not the case. Then all primes between $2$ and $k$ divide $j -i$. Since all of them are coprime, the product of all primes between $2$ and $k$ divides $j-i$. In particular, we get
$$\prod_{\substack{2\leq p \leq k\\p\text{ is prime}}}p \leq j -i \leq n.$$
The product in the LHS is sometimes called “primorial” (see http://en.wikipedia.org/wiki/Primorial ). It is known that it roughly equals $e^k$ and therefore is greater than $n$. We get a contradiction.
An Estimate for the Primorial
Let me informally explain why the primorial of $k$ is approximately $e^k$. The factorial of $k$ is approximately $e^{k\ln k}$. In the factorial we multiply all numbers between 1 and $k$, but in the primorial only primes. The density of primes is $1/\ln k$. Therefore, the primorial is approximately $\left(e^{k\ln k}\right)^{1/\ln k} = e^k$.
Numerical Bounds
Note that this proof implies that for large enough $n$ there is a prime number as desired between $1$ and $(1+ o(1))\ln n$. However, for small $n$ the term $o(1)$ may be relatively large. I computed the value of the primorial numerically in PARI/GP for $k\in{2,\dots, 100000}$ and got that the primorial of $k$ is always at least $\exp(\frac{\ln 2}{2} k)$ (the equality is attained for $k=2$). If this is indeed true for all $k$, as my computation suggests, then there is a desired prime between $1$ and $2\, \log_2 n$. |
H: Inverse Hyperbolic Tangent type Series
Is there a name for this series?
$$\sum_{k=1}^{\infty}\frac{a^{2k}}{2k}.$$
I know that:
$$\tanh^{-1}(a)=\sum_{k=1}^{\infty}\frac{a^{2k-1}}{2k-1}.$$
AI: I don't believe there's a name for the series you have there, but,
$$\begin{align*}
-\log(1-z)&=\sum_{k=1}^\infty\frac{z^k}{k}\\
-\log(1-z^2)&=\sum_{k=1}^\infty\frac{z^{2k}}{k}\\
-\frac12\log(1-z^2)&=\sum_{k=1}^\infty\frac{z^{2k}}{2k}\\
\log\frac1{\sqrt{1-z^2}}&=\sum_{k=1}^\infty\frac{z^{2k}}{2k}\\
\end{align*}$$ |
H: Find (in terms of n) the shortest distance between lines L and M.
The line L has equation:$$r=(i-4k)+t(i+2j-2k)$$
The line M has equation $$r=(4i+nj+5k)+s(7i+3j-4k)$$
where n is a constant.
Find (in terms of n) the shortest distance between lines L and M.
Here's what I've tried:
The direction vector of the line perpendicular to both L and M is $(2i+10j+11k)$.
So I tried to find the vector equation of the line perpendicular to both L and M then I can find the intersection of this line with L and M. (I failed to continue)
Then I tried to express the distance between the two lines in terms of $n$ and $t$
((1+t)-(4+7t))i+((2t)-(n+3t))j+((-4-2t)-(5-4t))k
but I'm not sure whether the two 't' in the equation of L and M is the same..
AI: So you want $t,s,b$ such that $$i-4k+t(i+2j-2k)+b(2i+10j+11k)=4i+nj+5k+s(7i+3j-4k)$$ That's three linear equations in the three unknowns $t,s,b$ (and the parameter $n$), so you should be able to solve for the three unknowns (in terms 0f $n$). From the value of $b$ you can easily calculate the distance between the lines (as a function of $n$). |
H: Expected length of a game of Kings
The game of Kings is a drinking game played with a standard 52-card deck. The rules are irrelevant to the nature of this question; we only wish to calculate the expected length of a game of Kings. The game ends when all four kings have been drawn (without replacement of any cards).
I've tried using the definition of the hypergeometric distribution's expectation, but it has not gotten me any closer to a solution.
AI: Here's a relationship that's very useful for solving this kind of problem:
$$\langle n\rangle=\sum_n np(N=n)=\sum_np(N\ge n)\;.$$
Thus we don't need to calculate $p(N=n)$, only $p(N\ge n)$, which is easier. The probability that $4$ kings occur somewhere in the first $n$ out of $52$ cards is
$$p(N\le n)=\binom n4\Big/\binom{52}4\;.$$
Then using
$$\sum_{n=m}^k\binom nm=\binom{k+1}{m+1}$$
we get
$$
\begin{align}
\langle n\rangle
&=
\sum_{n=1}^{52}p(N\ge n)
\\
&=52-\sum_{n=0}^{51}p(N\le n)\\
&=52-\sum_{n=4}^{51}p(N\le n)\\
&=52-\sum_{n=4}^{51}\binom n4\Big/\binom{52}4\\
&=52-\binom{52}5\Big/\binom{52}4\\
&=52-\frac{48}5\\
&=42\frac25\;.
\end{align}
$$ |
H: Why the number e(=2.71828) was chosen as the natural base for logarithm functions?
Possible Duplicate:
What's so “natural” about the base of natural logarithms?
Why the number e(=2.71828) was chosen as the natural base for logarithm functions ? Mainly I am interested in knowing why is it called "natural " . The number "2" could instead have been chosen as the most natural base.
AI: The simplest answer is this:
If you draw the graphs of $y=a^x$ for varying values of $a$, you find that they all pass through the point$(0,1)$ on the $y$-axis. There is exactly one of these curves that passes through that point with a gradient of exactly 1, and that value is obtained by taking $a=2.718281828459 \dots$.
In more analytical terms, this means that this is the value of $a$ which makes the derivative of $a^x$ equal to $a^x$, rather than a constant multiple of $a^x$. |
H: Gradient of a function on the vertices of a graph
Let $G=(V,E)$ be a graph. The gradient of a function $f:V\longrightarrow R$ is defined on the edges of the graph, given by the discrete derivative
$$
\nabla f(e)=f(y)-f(x), \quad e=(x,y) \in E
$$
Take $V=S_{2n}$ the group of permutations $\pi$ of the set $\{1, \ldots 2n\}$ and $e=(\pi, \pi \tau) \in E$ for some transposition $\tau$.
Consider $f=\left|\sum\limits_{i=1}^na_{\pi(i)}-\sum\limits_{i=n+1}^{2n}a_{\pi(i)}\right|$.
Find $\|\nabla f\|_{\infty}$.
I wasn't sure, but it turns out that this is kind of rephrased question as here question involving Markov chain
Thank you.
AI: As already explained here, that is, on a page you know and should have mentioned,
$$
\|\nabla f\|_\infty\leqslant2\cdot\max\{|a_i-a_j|\,;\,1\leqslant i\leqslant n,\,n+1\leqslant j\leqslant 2n\}.
$$ |
H: How many elements $a \in \Bbb{Z}_N$ such that $ax \equiv y \mod N$
Consider the ring $\Bbb{Z}_N$ of arithmetic modulo $N$: $\{0,1,2, \ldots ,N-1\}.$
Given $x,y \in \Bbb{Z}_N,$ how many of the elements of $\Bbb{Z}_N$ when multiplied with $x \pmod{N}$ result in $y$? And how can we calculate what they are efficiently?
The answer is a function of $N, x$ and $y.$
For example if $N = 10,$ $x = 6$ and $y = 2,$ then there are two elements $\{2,7\}$:
$2x = 12 \pmod{10} = 2 = y.$
$7x = 42 \pmod {10} = 2 = y.$
AI: So we are asking how many $a$ are there such that, for a given $x,y$, we have $ax \equiv y \mod n$. This falls under the Linear Congruence Theorem, which states that there will be a solution only if $\gcd(x,n)|y$, and if there are any solutions, then there are exactly $\gcd(x,n)$ of them.
Further, it is very easy to find all solutions once one is known, and the Euclidean algorithm will quickly give one solution (if there is a solution).
EDIT added content
In a comment, the OP asks Once one solution is known how do I find the others?
On the one hand, if you'd clicked on the wiki link answers this in the third sentence (the first restates the problem, the second gives the gcd requirement for a solution, and the third gives all the solutions). On the other, self-contained answers are usually simply better.
First one uses the Euclidean Algorithm to find a solution. In this case, it's often called the extended Euclidean Algorithm, as one uses it to find $\gcd(x,n)$, then 'extends it' to find integers $\alpha, \beta$ s.t. $\alpha x + \beta n = \gcd(x,n)$. If $\gcd(x,n)|y$, then in particular $y = c\gcd(x,n)$ for some $c$. But then $c\alpha x + c \beta n = y$, giving that $c\alpha x \equiv y \mod n$. So $c \alpha$, which I will now call $a_0$ is a solution: $a_0 x \equiv y \mod n$.
To get all the other solutions, one looks at $\left\{ a_0 + k \dfrac{n}{\gcd(x,n)} \right\}$ across $k \in \mathbb{Z}$. Let's check:
$\left(a_0 + \dfrac{kn}{\gcd(x,n)}\right)x = a_0x + \dfrac{kx}{\gcd(x,n)}n \equiv a_0x \equiv y \mod n$, where I used that $\gcd(x,n)|x$ to pull the $n$ aside. Going through, one sees that this gives $\gcd(x,n)$ different solutions $\mod n$. |
H: A problem including the given norm of $u \in C_0^\infty (\mathbb R)$
Let $u \in C_0^\infty (\mathbb R)$, $v(x) := u(x) e^{-x^2 /2} $. And define the norm as $$ \| u \|_1^2 = \int_{\mathbb R} | u' (x) |^2 e^{-x^2} dx + \int_{\mathbb R} | u(x) |^2 e^{-x^2} dx $$
Then I want to prove that $$ \| u \|_1^2 = \int_{\mathbb R} ( | v' (x) |^2 + x^2 | v(x)|^2 ) dx $$
I think this is not trivial by just using the definition of the norm above.
$C_0^\infty $ means that $C^\infty$ functions with a compact support. And I have one more question.
If the condition $u \in C_0^\infty ( \mathbb R) $ changes to "$\| u \|_1^2 < \infty$", then does this still hold?
AI: We have
$$
|v(x)|^2+x^2\,|v'(x)|^2=(|u'(x)|^2-2\,x\,u(x)\,u'(x)+2\,x^2|u(x)|^2)\,e^{-x^2}.
$$
The desired equality is thus equivalent to
$$
\int_\mathbb{R}(-2\,x\,u(x)\,u'(x)+2\,x^2|u(x)|^2)\,e^{-x^2}\,dx=\int_\mathbb{R}|u(x)|^2\,e^{-x^2}\,dx.
$$
Integration by parts gives
$$\begin{align*}
\int_\mathbb{R}(-2\,x\,u(x)\,u'(x)\,e^{-x^2})\,dx&=-\int_\mathbb{R}(|u(x)|^2)'\,x\,e^{-x^2}\,dx\\
&=\int_\mathbb{R}|u(x)|^2(\,x\,e^{-x^2})'\,dx\\
&=\int_\mathbb{R}|u(x)|^2(1-2\,\,x^2)\,e^{-x^2}\,dx,
\end{align*}$$
and the result follows. The computation is valid as long as $\lim_{x\to\pm\infty}|u(x)|^2(\,x\,e^{-x^2})=0$. |
H: A question of the norm calculation of Hermite polynomials
Define the (physicist's) Hermite polynomial $H_n (x)$ by $$H_n (x) = (-1)^n e^{x^2} \frac{d^n}{dx^n} e^{-x^2} $$ then prove that $$ \int_{\mathbb R} |H_n (x) |^2 e^{-x^2} dx = 2^n n! \sqrt{\pi}$$.
AI: We have the recursive relation
$$H'_n(x)=(-1)^ne^{x^2}\left(2x\frac{d^n}{dx^n}e^{—x^2}+\frac{d^{n+1}}{dx^{n+1}}e^{-x^2}\right)=-H_{n+1}(x)+2xH_n(x).$$
We use this in the integral:
\begin{align}
\int_{\Bbb R}H_n(x)^2e^{-x^2}dx&=\int_{\Bbb R}(2xH_{n-1}(x)-H'_{n-1}(x))H_n(x)e^{-x^2}dx\\
&=\int_{\Bbb R}2xe^{-x^2}H_{n-1}(x)H_n(x)dx-\int_{\Bbb R}H'_{n-1}(x)H_n(x)e^{—x^2}dx\\
&=\left[-e^{-x^2}H_{n-1}(x)H_n(x)\right]_{—\infty}^{+\infty}+\int_{\Bbb R}e^{-x^2}(H'_{n-1}(x)H_n(x)+H_{n-1}(x))H'_n(x))\\
&-\int_{\Bbb R}H'_{n-1}(x)H_n(x)e^{—x^2}dx\\
&=\int_{\Bbb R}H_{n-1}H'_n(x)e^{-x^2}dx.
\end{align}
By induction, we get
$$\int_{\Bbb R}H_n(x)^2e^{-x^2}dx=\int_{\Bbb R}H_0(x)H^{(n)}_n(x)e^{-x^2}dx=\int_{\Bbb R}H^{(n)}_n(x)e^{-x^2}dx.$$
Indeed, denote $\langle P,Q\rangle:=\int_{\Bbb R}P(x)Q(x)e^{-x^2}dx$ for two polynomials $P$ and $Q$. We can show by an analogous way that $\langle H_n,H_{n+k}\rangle=\langle H_{n-1},H'_{n+k+}\rangle$ for integers $k$ and $n$, then the induction relation.
Now, to get the result, we need the following facts:
$H_n$ is a polynomial of degree $n$, whose leading terms is $2^n$ (show it by induction);
$\int_{\Bbb R}e^{-x^2}=\sqrt \pi$.
By the way, this shows that the Hermite polynomials are orthogonal. |
H: Probability exercise
A couple has $3$ kids, and $1$ of them is known to be male.
What is the probability that $1$ (only $1$) of the other $2$ kids is male?
AI: in generally,if we consider $3$ child,and we know that only $1$ is male,then probability of being any choosen child male is $1/3$,but if our space is consisted by two child then possibility of all combination is,let choose male as $M$,and female $F$,so
$(F,F)$,$(F,M)$,$(M,F)$,$(M,M)$ probability is $2/4=1/2$ |
H: Reference request: Extremes of composite functions
Let $g : X \rightarrow Y$ be a homeomorphism, and let $f : Y \rightarrow \mathbb{R}$
be a continuous function. Then the extremes of $f$ in $Y$ are mapped to the extremes
of the composition $f \circ g $ in $X$, and if $f$ has a unique minimizer in $Y$, then
$f \circ g $ has a unique minimizer in $X$. This seems plausible to me, but does anyone know of a reference where this is proved (and preferably related results)?
Kind regards
Olav
AI: This is one of the things that are so plausible that a formal proof makes them less so. Therefore I shall describe in so many words how I perceive the situation.
From the definition of "homeomorphism" we need only that $g:\ X\to Y$ is bijective. This means that we have a set $$S:=\bigl\{(x,y)\ \bigm|\ x\in X,\ y=g(x)\in Y\bigr\}$$ of "married pairs". Each point $p=(x,y)\in S$ can be adressed either by its first coordinate $x\in X$ or its second coordinate $y\in Y$, and the only difference between $f$ and $f\circ g$ is whether the function $$F:\ S\to {\mathbb R},\quad p=(x,y)\mapsto F(p):=f(y)=f\bigl(g(x)\bigr)$$
is expressed as $F(p):=f(y)$ or as $F(p):=f\bigl( g(x)\bigr)$.
Looking at the setup in this way both your statements become obvious: The extremes of $f$ resp. $f\circ g$ are the extremes of the encompassing idea $F$. |
H: General solution for 3D line intersection
I've been trying to get the intersection point of 2 3D lines (in general, so i can code an algorithm) using the following equations:
$$ x_0 + k_0 a_0 = x_1 + k_1 a_1 \tag{1}$$
$$ y_0 + k_0 b_0 = y_1 + k_1 b_1 \tag{2}$$
$$ z_0 + k_0 c_0 = z_1 + k_1 c_1 \tag{3}$$
where $x,y,z$ are point coordinates and $a,b,c$ direction vector coordinates.
I think the idea is to get $k_0$ or $k_1$ so I can get the point from the line ($l_0$ if i use $k_0$, $l_1$ otherwise) using it's parametric equation.
I've made some tests and if I use equations $(1)$ and $(2)$ I get different results that if I use $(2)$ and $(3)$, so i think i'm doing something wrong...
From $(1)$: $$ k_0 = \frac{x_1-x_0 + k_1a_1}{a_0} \tag{4}$$
from $(2)$: $$ k_1 = \frac{y_0-y_1 + k_0b_0}{b_1} \tag{5}$$
from $(4)$ & $(5)$: $$k_1 = \frac{a_0 (y_0 - y_1) + b_0 (x_1 - x_0)}{a_0 b_1 - b_0 a_1} \tag{6}$$
And from $(1)$ & $(3)$: $$k_1 = \frac{a_0 (z_0 - z_1) + c_0 (x_1 - x_0)}{a_0 c_1 - c_0 a_1} \tag{7}$$
I don't know why, but $(6)$ & $(7)$ get different results for $k_1$...
I'm not controlling divisions by zero because C# returns NaN ($0/0$), -Infinity ($-x/0$), +Infinity ($x/0$) when needed, so I assume that if there is no intersection there would be some division by zero or something like that, right?
Any idea of whats wrong?
Thanks a lot!!
AI: The problem is quite simplyy that if you take 2 lines in a 3D space they usually do not intersect (think about it!). So it's perfectly normal to get different results for $k_1$ : it simply means that your lines are not coplanar (you will only get a divide by zero error if they happen to be parallel). |
H: Stronger than ZF, weaker than ZFC
Can you please name axiom system that is strictly weaker than ZFC and strictly stronger than ZF? (such as DC, AC$_\omega$)
I searched for it but i could only find these two.
If there are statements equivalent to DC or AC$_\omega$ in ZF, please tell me or give me a link introducing those..
(I want some equivalent statements, since I don't know where these two can be used usefully.)
AI: There are infinitely many assertions which are between ZF and ZFC, these are often called choice principles.
Slightly more formally (but naively, as Carl indicates in the comments), we say that a sentence in the language of set theory $\varphi$ is a choice principle if it is provable from ZFC but not from ZF, we say that $\varphi$ is a weak choice principle if ZF+$\varphi$ does not prove AC.
Examples for choice principles:
the axiom of choice;
the axiom of choice for countable families;
countable unions of countable sets are countable;
the axiom of choice for countable families of finite sets;
the axiom of choice for countable families of pairs.
The first one is not a weak choice principle, it is in fact the axiom of choice in full. It can be shown that each choice principle implies the following, and that all those implications are strict. The list can be expanded much further, even if we require it will remain "linear" (that for every two statements one of them implies the other).
However there are other choice principles, for example:
The ultrafilter lemma: Every filter can be extended to an ultrafilter.
This is a particularly important choice principle with many equivalents and uses throughout mathematics. For example, it is equivalent to the compactness theorem in logic; or to the Tychonoff theorem for Hausdorff compact spaces. This choice principle implies that every set can be linearly ordered, as well the Hahn-Banach theorem and in turn - the Banach-Tarski paradox. From this family of choice principles also follow the fact that every field has an algebraic closure, and that the closure is unique up to isomorphism.
There are many many other choice principles, some of them are equivalent to the ultrafilter lemma; others to the axiom of choice for families of some sort (countable; etc.); and other principles are just "out there" and are not equivalent to the variants of these two above, for example KWP($n$) and SVC are two important examples.
Specifically for DC, we have an interesting equivalent:
The Baire category theorem for complete metric spaces.
As for the axiom of countable choice, it too has several equivalents and one of the interesting ones, for example:
$\sigma$-compact spaces are Lindelof.
To learn more about such principles you may find yourself interested in reading these books, all contain proofs, lists and diagrams of various choice principles and the known implications between them:
Herrlich, H. Axiom of Choice. Lecture Notes in Mathematics, Springer, 2006.
Jech, T. The Axiom of Choice. North-Holland (1973).
Howard, P. and Rubin, J.E. Consequences of the Axiom of Choice. American Mathematical Soc. (1998). Also see the online database for the book.
Moore, G. H. Zermelo's Axiom of Choice. Springer-Verlag (1982). |
H: Proving Quadratic Formula
purplemath.com explains the quadratic formula. I don't understand the third row in the "Derive the Quadratic Formula by solving $ax^2 + bx + c = 0$." section. How does $\dfrac{b}{2a}$ become $\dfrac{b^2}{4a^2}$?
AI: $b/2a$ does NOT become $b^2/4a^2$. All that happens in the third row is that $b^2/4a^2$ is added to both sides of the equation.
The bit about taking half of the $x$ term and squaring it is just a means of working out WHAT to add. This is often called "completing the square" - adding a constant term to an expression to turn it into a perfect square, so that one may later take its square root. |
H: What is a lift?
What exactly is a lift? I wanted to prove that for appropriately chosen topological groups $G$ we can show that the completion of $\widehat{G}$ is isomorphic to the inverse limit $$\lim_{\longleftarrow} G/G_n$$
I wasn't sure how to construct the map so I asked this question to which I got an answer but I don't understand it. Searching for "lift" or "inverse limit lift" has not helped and I was quite sure that if $[x] \in \widehat{G}$ I could just "project" $x_n$ to $$\lim_{\longleftarrow} G/G_n$$ using the map $\prod_{n=1}^\infty \pi_n (x) = \prod_{n=1}^\infty x_n$ in $G/G_n$. Would someone help me finish this construction and explain to me what a lift is? Thank you.
Edit
I don't know any category theory. An explanation understandable by someone not knowing any category theory would be very kind.
AI: See the Wikipedia article on lift:
In a category $C$, let be given two arrows $f,g$ with codomain $Z$. A lift (of $f$ with respect to $g$) is an arrow $F$ such that $g\circ F =f$. So it is a morphism from $f$ to $g$ in the over category $C/Z$.
Motivation: the idea is that $g:X\to Z$ indicates the vertical direction pointing down: $X$ lies above $Z$ via $g$. Think of $g$ as a fibre bundle, or covering map, or vertical projction. So a given $f:Y\to Z$ takes values in the lower space; and to lift $f$ (via $g$) means to 'let it take values in the upper space': find a 'modified' $F:Y\to X$ such that, after vertically projecting its values, we get $f$ back; in other words $p\circ F=f$. If we take $Y=[0,1]$ this is path-lifting.
In particular:
a lift of the identity $id:Z\to Z$ (via $g:X\to Z$) is an arrow $s:Z\to X$ such that $g\circ s=id$, i.e. a section of $g$.
(I believe in your case this is meant: a section of a quotient map $q:X\to X/\sim$ is a map $s:X/\sim\to X$ such that $s[x]\sim x$. One could say that $s[x]\in X$ is a lift of $[x]$.)
DUALLY: let be given two arrows $f,g$ with domain $Z$. A lift of $f$ (via $g$) is an arrow $F$ such that $F\circ f=g$. So it is a morphism from $f$ to $g$ in the under category $Z/C$.
Motivation: again the idea is that $g:Z\to X$ indicates the vertical direction, but now pointing up: $X$ lies above $Z$ via $g$. Think of $g$ as an
inclusion; then:
a lift of $f:Y\to X$ (via an inclusion $g:Z\to X$) is an arrow $F:Y\to X$ such that $F\circ i=g$, i.e. an extension of $f$.
Related concepts are injective and projective objects. Moreover note that lifts need not exist in general (see also Zhen Lin's comment). |
H: Example of a proper homotopy between smooth functions on manifolds
Let $h:S^{n-1}\to S^{n-1}$ be $C^{\infty}$ map. How to prove that a function $F: S^{n-1}\times[0,1]\to S^{n-1}$ given by
$$F(v,t)=(\cos{\pi t})v+(\sin{\pi t})h(v)$$
is proper $C^{\infty}$ map?
AI: Assuming $F(v,t)\in S^{n-1}$ there is not much to show. Suppose $A$ is compact in $S^{n-1}$, so $A$ is closed. The map $F$ is obviously continuous, so $F^{-1}(A)$ is closed, and since $S^{n-1}\times [0,1]$ is compact $F^{-1}(A)$ is compact, so $F$ is proper.
The fact that $F$ is $C^{\infty}$ is obvious, too, it's the composition of $C^{\infty}$ maps.
One may ask why $F(v,t)\in S^{n-1}$, though. Is $\langle h(v),v \rangle = 0$? |
H: Approximation of bounded measurable functions with continuous functions
This is not homework. I was reading a paper where the authors showed a result for all continuous functions and then just proceeded to write "the usual limiting Argument gives the result for all bounded functions" - so I am asking myself what this "usual limiting argument" might be. I do not know whether they mean uniform or pointwise convergence. As I see it pointwise convergence should suffice :D
Thus I was am wondering whether there is a theorem having or leading to the following statement:
Let $K\subset\mathbb{R}^2$ be compact. Any bounded measurable function $f:K\to\mathbb{R} $ can be approximated by a sequence of continuous functions $(g_m)$ on $K$.
Nate Eldredge suggested that I post some excerpt from the original to provide more context for the problem. Here I go:
The goal is to proof the existence of a weak limit for a tight sequence of probability measures on $\mathcal{C}^0([0,1]^2,\mathbb{R})$ associated with reflecting Brownian Motions on the compact the set $[0,1]^2$ which is a Lipshitz Domain. Thus we already, know that some weak limit must exist and it remains to show that two limit-Points agree. Weak-Convergence is generally defined via bounded measurable functions. Now let $P'$ and $P''$ be two subsequential limit points. The authors show that $f \in \mathcal{C}^0([0,1]^2,\mathbb{R})$ the following holds
(here $X_s$ the canonical process)
$E'f(X_s)=E''f(X_s)$
And now comes the actual source of my question:
"The usual limiting argument gives the result for bounded $f$ and hence the one-dimensional distributions agree." (the second part I understand only the "standard limiting argument thing" is somewhat confusing)
Any Help is much appreciated and Thanks in Advance :D
AI: It is not true that every bounded measurable function is the pointwise, or uniform, limit of continuous functions. See this MSE question.
It depends on what result the author wanted to prove, but extending results from continuous functions to bounded measurable functions often uses the Monotone Class Theorem.
Added: You have two probability measures
$\mu^\prime:=P^\prime\circ X_s^{-1}$ and $\mu^{\prime\prime}:=P^{\prime\prime}\circ X_s^{-1}$ on $[0,1]^2$ so that, for every continuous function $f$, $$\int f\,d\mu^\prime=\int f\,d\mu^{\prime\prime}.\tag1$$
Let $\cal H$ be the space of all bounded, measurable functions $f$ so that (1) holds, and let $\cal K$ be the space of continuous functions. Once you check the conditions of the functional Monotone Class Theorem, you can conclude that $\cal H$ contains all bounded functions measurable with respect to $\sigma(\cal K)$. That is, $\cal H$ includes all bounded, Borel measurable functions which means that $\mu^\prime=\mu^{\prime\prime}$. |
H: Trigonometric eigenvalue equation
In solving an eigenvalue problem, I've come to following equation ($\lambda=1$):
$$\begin{pmatrix}
\cos(\theta) & \sin(\theta) \\ \sin(\theta) & -\cos(\theta)
\end{pmatrix}\begin{pmatrix}
a \\ b
\end{pmatrix}=\begin{pmatrix}
a \\ b
\end{pmatrix}$$
Now, the solution says, "This matrix equation can be reduced to a single equation":
$$a \sin(\frac{1}{2}\theta)=b\cos(\frac{1}{2}\theta)$$
I've been rotating trigonometric formulas to get to this, but I simply can't find the way. Could you help me with this, or at least give me a hint?
AI: $$\begin{pmatrix}
\cos(\theta) & \sin(\theta) \\ \sin\theta & -\cos\theta
\end{pmatrix}\begin{pmatrix}
a \\ b
\end{pmatrix}\\=\begin{pmatrix}
a\cos\theta+b\sin\theta \\ a\sin\theta-b\cos\theta
\end{pmatrix}=\begin{pmatrix}
a \\ b
\end{pmatrix}$$
Which gives us the following system of equations:
$$a\cos\theta+b\sin\theta=a\tag{1}$$
$$a\sin\theta-b\cos\theta=b\tag{2}$$
So from $(1)$ you have:
$$b\sin\theta=a(1-\cos\theta)$$
$$2b\sin(\frac \theta{2})\cos(\frac \theta{2})=a(2\sin^2(\frac \theta{2}))$$
$$b\cos(\frac \theta{2})=a\sin(\frac \theta{2})$$ |
H: On the intersection of closed sets
In a book on beginning measure theory, the following statement is made:
"It is clear that any intersections and finite unions of closed sets are closed."
However the intersection of two disjoint closed sets is the empty set which is open by definition. Is there something wrong with my understanding, or was the statement false?
AI: You seem to believe that if a set is open then it is not closed. This is false. In particular, the empty set and the whole space are always both open and closed. |
H: Global existence of solutions
How do I see whether or not these ODE's have a global solution:
$$x'=t^2+x^2$$
$$x'=t^2+x$$
Why?
AI: Applying the weaker form of the Cauchy theorem (probably known to many as Picard–Lindelöf theorem) to $$x'=f(t,x),$$ $\partial f/\partial x$ is required to be bounded (this implies it is Lipschitz with respect to $x$, for any $t$), and this is true for the second equation where such derivatives gives $1$, but not for the first, where it gives $2x$. |
H: how is this series expansion $(\sum\limits_{i=1}^n{x_{i}})^2=(\sum\limits_{i=1}^n{x_{i}^2}+\sum\limits_{i
I'm reading my vector calculus text when I encountered below formula.
$(\sum\limits_{i=1}^n{x_{i}})^2=(\sum\limits_{i=1}^n{x_{i}^2}+\sum\limits_{i<j}{2x_{i}x_{j}})$
Is this a definition or there's a proof for above?
AI: Consider the array
$$\left[ \begin{array}{cccc} \color{red}{x_1^2} & \color{green}{x_1x_2} & \color{blue}{x_1x_3} & x_1x_4&&& \ldots &x_1x_n \\ \color{green}{x_1 x_2} & \color{red}{x_2^2} & x_2x_3 & x_2x_4
&&&\ldots & x_2x_n \\ \color{blue}{x_1x_3} & x_2x_3 & \color{red}{x_3^2} & x_3x_4 &&&\ldots & x_3x_n \\ x_1x_4 & x_2x_4 & x_3x_4 & \color{red}{x_4^2} & x_4x_5 &&\ldots & x_4x_n \\ &&&\vdots & \color{red}{\ddots} \\ &&&\vdots \\ &&&\vdots \\ x_1x_n & x_2x_n & \ldots &&&&\ldots & \color{red}{x_n^2}\end{array}\right].$$
This is the array consisting of all terms in the expansion of $(x_1+\ldots + x_n)^2$. Notice that the array is symmetric about the diagonal. So therefore we just need to notice that
$$\begin{eqnarray*} \left(\sum_{i=1}^n x_i\right)^2 = (x_1 + \ldots+ x_n)^2 &=& \text{(sum of all terms along the diagonal)} \\
&& \hspace{16mm}+ \text{($\color{red}{twice}$ the sum of all terms above the diagonal)}\\
&=& \sum_{i=1}^n x_i^2 + 2\sum_{i < j} x_ix_j.\end{eqnarray*}$$
$\hspace{6in} \square$ |
H: Find endomorphism of $\mathbb{R}^3$ such that $\operatorname{Im}(f) \subset \operatorname{Ker}(f)$
I have a problem:
Let's find an endomorphism of $\mathbb{R}^3$ such that $\operatorname{Im}(f) \subset \operatorname{Ker}(f)$.
How would you do it?
The endomorphism must be not null.
AI: Well, you could always take $f$ to be the null function... Not the only solution, but certainly the simplest :)
If you want $f$ to be non-null, then you just need to make sure that $f^2=0$. Either try to find a $3 \times 3$ matrix for which this holds (look up nilpotent matrices) or look at this endomorphism and see how it can be adapted : $$f : (x,y,z) \mapsto (y,z,0)$$ |
H: Cauchy+pointwise convergence $\Rightarrow$ uniform converges (for an operator in a Hilbert space)
Suppose that the sequence of operators in a Hilbert space $H$, $\left(T_{n}\right)_{n}$,
is Cauchy (with respect to the operator norm) and that there is an
operator $L$, such that $Lx=\lim_{n\rightarrow\infty}T_{n}x$, for
all $x\in H$ (i.e. the $T_{n}$ converge pointwise to $L$).
How
can I prove then, that $\left(T_{n}\right)_{n}$ converges with respect
to the operator norm to $L$ (i.e. $\left(T_{n}\right)_{n}$ converges
uniformly to $L$)?
AI: Fix $\varepsilon>0$; there is $N=N(\varepsilon)$ such that if $m,n\geq N(\varepsilon)$, $\lVert T_n-T_m\rVert_{B(H)}\leq \varepsilon$. In particular, for all $n\geq N$ and all $x\in H$, we have
$$\forall m\geq N,\lVert T_nx-T_mx\rVert\leq \varepsilon\lVert x\rVert.$$
Take the limit $m\to +\infty$ to get
$$\forall n\geq N,\forall x\in H,\lVert T_nx-Lx\rVert\leq \varepsilon\lVert x\rVert.$$
We get $\lVert T_n-L\rVert_{B(H)}\leq \varepsilon$. The fact that $L$ is continuous can be seen by taking the particular value of $\varepsilon:=1$. |
H: Prove $\frac{1-\tan(x/2)}{1+\tan(x/2)}=\frac{1-\sin x}{\cos x}$
Can anyone offer please help me solve the following trig identity.
$$\frac{1-\tan(x/2)}{1+\tan(x/2)}=\frac{1-\sin x}{\cos x}=\frac{\cos x}{1+\sin x}$$
My work thus far has been on the left most side.
I did
$$\frac{\;\;1-\dfrac{\sin(x/2)}{\cos(x/2)}\;\;}{1+\dfrac{\sin(x/2)}{\cos(x/2)}}$$
Unfortunately now I am not sure what to do I multiplied the numerator by $\cos$ on the the top and $\cos^2$ on the bottom to get $\cos x$,but I am not sure if this is correct.
AI: In order to make typing easier, I will let $t=x/2$. Replacing $\tan t$ by $\frac{\sin t}{\cos t}$, as you did, is a good general purpose strategy.
So the leftmost side is equal to
$$\frac{1-\frac{\sin t}{\cos t}}{1+\frac{\sin t}{\cos t}}.$$
The next step is semi-automatic. It seems sensible to bring top and bottom to the common denominator $\cos t$, and "cancel." We get
$$\frac{\cos t -\sin t}{\cos t+\sin t}.$$
The move after that is not obvious: Multiply top and bottom by $\cos t-\sin t$.
At the bottom we get $\cos^2 t-\sin^2 t$, which from a known double-angle formula we recognize as $\cos 2t$.
The top is $(\cos t-\sin t)^2$. Expand. We get $\cos^2 t-2\sin t\cos t+\sin^2 t$. The $\cos^2 t+\sin^2 t$ part is equal to $1$, and the $2\sin t\cos t$ part is equal to $\sin 2t$. So putting things together we find that the leftmost side is equal to
$$\frac{1-\sin 2t}{\cos 2t},$$
which is exactly what we want.
The second identity $\frac{1-\sin x}{\cos x}=\frac{\cos x}{1+\sin x}$ is much easier. In $\frac{1-\sin x}{\cos x}$, multiply top and bottom by $1+\sin x$. On top we now get $1-\sin^2 x$, which is $\cos^2 x$. At the bottom we get $\cos x(1+\sin x)$. Cancel a $\cos x$.
Remark: We can play a game of making the calculation more magic-seeming. The middle expression is equal to $\frac{1+\sin 2t}{\cos 2t}$. Replace the $1$ on top by $\cos^2 t+\sin^2 t$, and the $\sin t$ by $2\cos t\sin t$. Then on top we have $(\cos t+\sin t)^2$. Replace the $\cos 2t$ at the bottom by $\cos^2 t-\sin^2 t$, which is $(\cos t+\sin t)(\cos t -\sin t)$. Cancel the $\cos t+\sin t$, and we arrive at
$$\frac{\cos t+\sin t}{\cos t-\sin t}.$$
Now divide top and bottom by $\cos t$, and we get the desired $\frac{1+\tan t}{1-tan t}$. |
H: Finding $a$ from $\lim\limits_{x\rightarrow0}(1+a\sin x)^{\csc x} =4$
The question is to find the value of $a$ from the following equation:
$$\lim_{x\rightarrow 0}(1+a\sin x)^{\csc x} =4 $$
AI: You have
$$
\lim_{x\rightarrow 0}\left(1+a\sin x\right)^\frac{1}{\sin x}=\\
\lim_{x\rightarrow 0}\left(\left(1+a\sin x\right)^\frac{1}{a\sin x}\right)^a
$$
if you set $t=1/(a\sin x)$, and given that $t$ tends to $\infty$ when $x$ tends to $0$, you have
$$
\lim_{t\rightarrow\infty}\left(\left(1+\frac{1}{t}\right)^{t}\right)^a=e^a\\
$$ |
H: Open mapping of the unit ball into itself
Does there exist a continuous open function $f:B^n\to B^n$ which is not injective?
(Here $B^n\subseteq\mathbb{R}^n$ is the open unit ball)
AI: Hint for your problem: $re^{ix}\mapsto re^{2ix}$. |
H: finding distribution $\mathbb{Z}$ in problem
suppose $X_1,X_2,\ldots,X_n$ is random sample of $Exp(0,\sigma)$. if $\mathbb{S_n}=X_1+X_2+\ldots+X_n$
and $\mathbb{Z}=\max\{n:\mathbb{S_n}\leq s\}$ how can find distribution $\mathbb{Z}$?
AI: Note that $(S_n)_{n\geqslant1}$ is the set of events of a Poisson process $(N_t)_{t\geqslant0}$ with intensity $\sigma$ and that $Z=N_s$ counts the number of events before time $s$. Hence, the distribution of $Z$ is Poisson with parameter $s\cdot\sigma$. |
H: Unramified extension is normal if it has normal residue class extension
Let $K/F$ be an unramified extension such that $\rho_K / \rho_F$ (the corresponding extension of residue classes) is normal. Prove $K/F$ is normal.
I guess I need to do some polynomial lifting, but if we take $f(x)$ the minimal polynomial of $\alpha \in K$, we can't lift it to $\rho_F$ unless his coefficients are in $O_{F} = \{x\in F : |x| \leq 1\}$.
AI: I'm assuming $K$ and $F$ are complete, discretely valued fields and $K/F$ is finite separable. The term ``unramified" means the residue extension $\rho_K/\rho_F$ is separable and if $\pi$ is a uniformizer of $\mathscr{O}_F$, it is also a uniformizer of $\mathscr{O}_K$. So $[K:F]=[\rho_K:\rho_F]$. Let $f$ be a monic lift in $\mathscr{O}_F$ of the minimal polynomial for a generator $\bar{b}$ of $\rho_K$ over $\rho_F$ (which exists because the residue extension is finite separable). Then $f$ is irreducible because its reduction mod $\pi$ is. Because $\bar{f}$ is separable, we can uniquely lift $\bar{b}$ to a root $b$ in $\mathscr{O}_K$ of $f$ by Hensel's lemma. Thus we have an $F$-algebra injection $F[X]/(f(X))\hookrightarrow K$ sending $X$ to $b$. This must be an isomorphism by dimension counting. By Hensel's lemma, each root of $\bar{f}$ in $\rho_K$ lifts to a (unique) root of $f$ in $\mathscr{O}_K$. So, since $\bar{f}$ splits over $\rho_K$, $f$ splits over $K$. Thus $K$ is normal over $F$. |
H: Norm of the sum of projection operators
Is it true that $$|| a R+b P||\leq\max \{|a|,|b|\},$$where $a$ and $b$ are complex numbers and $P,R$ are (orthogonal) projection operators on finite-dimensional closed subspaces of an infinite-dimensional hilbert space ?
In finite dimensions this would be true, since a projection there has always norm one.
AI: It is true when $R,P$ are orthogonal to each other (there is an ambiguity in terminology, as "orthogonal" could mean that $PR=0$, or that $P=P^*=P^2$, $R=R^*=R^2$).
If $PR=0$ is not assumed, then the answer is no: take $P=R=I$, $a=b=1$, then $\|aR+bP\|=2$.
Assuming $PR=0$, then $\|aR+bP\|=\max\{|a|,|b|\}$. Indeed, for any $\xi$ in the range of $R$ with $\|\xi\|=1$, we have
$$
\|(aR+bP)\xi\|=\|aR\xi\|=\|a\xi\|=|a|;
$$
similarly, $\|(aR+bP)\eta\|=|b|$ if $\eta$ is in the range of $P$ and $\|\eta\|=1$. So $\|aR+bP\|\geq\max\{|a|,|b|\}$.
For an arbitrary vector $\nu$ with $\|\nu\|=1$, we can write $\nu=\alpha\nu_1+\beta\nu_2+\gamma\nu_3$, for three unit vectors with $\nu_1$ in the range of $R$, $\nu_2$ in the range of $P$, and $\nu_3$ orthogonal to both the ranges of $R$ and $P$, and $\alpha,\beta,\gamma\geq0$, $\alpha^2+\beta^2+\gamma^2=1$ (see edit below for an explanation). Then
$$
\|(aR+bP)\nu\|^2=\|a\alpha\nu_1+b\beta\nu_2\|^2=|a|^2\,\alpha^2 + |b|^2\,\beta^2\leq\max\{|a|^2,|b|^2\},
$$
so $\|(aR+bP)\nu\|=\sqrt{\alpha^2|a|^2+\beta^2|b|^2}\leq\max\{|a|,|b|\}$.
In conclusion, $\|aR+bP\|=\max\{|a|,|b|\}$.
Edit: below is a proof of the claim that, given three pairwise orthogonal subspaces $X$, $Y$, $Z$ of a Hilbert space $H$ that span the whole space, any unit vector $\nu\in H$ can be written as $$\nu=\alpha\nu_1+\beta\nu_2+\gamma\nu_3,\ \ \ \nu_1\in X,\ \nu_2\in Y, \ \nu_3\in Z,$$ with $\alpha,\beta,\gamma\geq0$ and $\alpha^2+\beta^2+\gamma^2=1$.
Let $\{x_j\}$, $\{y_j\}$, $\{z_j\}$ be orthonormal bases for $X$, $Y$, $Z$. Together, they form a basis for the whole $H$. So there exist coefficients such that
$$
\nu=\sum_j a_jx_j + \sum_jb_jy_j+\sum_jc_jz_j.
$$
As $\|\nu\|=1$, $\sum_j|a_j|^2+\sum_j|b_j|^2+\sum_j|c_j|^2=1$. Let
$$
\alpha=(\sum_j |a_j|^2)^{1/2},\ \beta=(\sum_j |b_j|^2)^{1/2},\ \gamma=(\sum_j |c_j|^2)^{1/2},\
$$
and
$$
\nu_1=\sum_j\frac{a_j}\alpha\,x_j,\ \nu_2=\sum_j\frac{b_j}\beta\,y_j,\
\nu_3=\sum_j\frac{c_j}\gamma\,z_j.
$$
Then $\nu_1,\nu_2\nu_3$ are unit vectors, $\alpha^2+\beta^2+\gamma^2=1$, and
$$
\nu=\alpha\nu_1+\beta\nu_2+\gamma\nu_3.
$$ |
H: Scalar product equals weighted sum of projection of the vectors onto the edges of a simplex
Given a triangle with the edges $e_1$, $e_2$, $e_3$, it seems (from numerical evidence) that there are coefficients $\alpha_i$ such that
$$
u^Hv = \sum_{i=1}^3 \alpha_i \, (u^He_i)\, (e_i^H v)
$$
holds for all vectors $u, v\in\mathbb{C}^2$.
There is of course the brute-force way of taking the edge coefficients in terms of their $x$- and $y$-coordinates to show this is true, but the simplicity of the statement makes me think there should be a more elegant way.
I looked at a bunch of geometry text books and couldn't find it. I'm thinking to maybe take off my geometry hat and put on my linear algebra hat. What's your take on the problem; where do you think are similar results to be found?
AI: What you're saying is that the $2 \times 2$ identity matrix is in the linear span of the three matrices $M_i = e_i \otimes e_i^H$, where $e_i$ are the (column) vectors corresponding to the edges of the triangle, and thus $e_1 + e_2 + e_3 = 0$. In fact the three matrices are linearly independent, and thus form a basis of the symmetric $2 \times 2$ matrices, as long as two of the vectors are linearly independent.
Note that if you replace $e_i$ by $f_i = U e_i$ where $U$ is any invertible $2 \times 2$ matrix, you replace $M_i$ by $U M_i U^H$, and these are linearly independent iff the original $M_i$ are. By taking a suitable $U$, we may assume $e_1 = \pmatrix{1\cr 0\cr}$ and $e_2 = \pmatrix{0\cr 1\cr}$, so $M_1 = \pmatrix{1 & 0\cr 0 & 0\cr}$, $M_2 = \pmatrix{0 & 0\cr 0 & 1\cr}$, $M_2 = \pmatrix{1 & 1\cr 1 & 1\cr}$, which are clearly linearly independent. |
H: Does this sequence of operators in Hilbert space stop at $\text{rank}T+1$ steps?
Let $\left(T_{n}\right)_{n}$ be a sequence of operator in a infinitdimensional
Hilberspace $H$, defined as restrictions of an operator $T:H\rightarrow H$,
on smaller and smaller subsets, by the algorithm in this question (were also additional information about $T$ is provided).
There it was shown, that if this sequence is finite, $T$ must have
finite rank.
My question is: Is the number $n$, for which the algorithm described
here stops, always $\text{rank}T+1$ ? How can we prove that ?
This guess came from the fact, that if $H$ were finite dimensional,
it is not too hard to show, that this sequence stops after exactly
$\text{rank}T+1$ steps.
AI: As the answer you link shows, if the algorithm stop after $n$ iterations, then the rank of $T$ is $\leq n-1$. Let $r$ the rank of $T$, and $n$ the number of iterations until the algorithm stops. We have $r\leq n-1$ hence $r+1\leq n$. If we have more than $r+2$ steps, we have extracted $r+1$ orthogonal non-zero vectors, associated with non-zero eigenvalues. This contradicts the fact that the rank of $T$ is $r$. |
H: Is there a simple explanation why degree 5 polynomials (and up) are unsolvable?
We can solve (get some kind of answer) equations like:
$$ ax^2 + bx + c=0$$
$$ax^3 + bx^2 + cx + d=0$$
$$ax^4 + bx^3 + cx^2 + dx + e=0$$
But why is there no formula for an equation like $$ax^5 + bx^4 + cx^3 + dx^2 + ex + f=0$$
I'm not sure if this has anything to do with the Galois theory, but is there a dumbed-down simple explanation as to why degree 5 polynomials (and up) are unsolvable?
AI: I'll try a "dumbed down" version, although @Robert Israel's answer plus comments are fine!
Solvable means solvable by radicals, and that means that, starting from the polynomial equation, you can only do 1) field arithmetic $(+,-,\times,\div)$, or 2) "extracting roots; e.g. square roots, cube roots, etc. It is the case, by Abel-Ruffini first and then by Galois, that there is no general "formula" for solving polynomials above degree 4. Naively, that suggests that the formula gets "too complicated" at some point. @paul garrett gets at this when he refers to the resolvent, which is a step that can simplify solving if the resolvent polynomial is of lower degree.
Galois found that the way to measure "too complicated" is by checking which roots of the polynomial can be "switched around", or permuted, while maintaining certain equations of the roots. For example, if you are working over the rational numbers, then you can't switch around any rational number without changing important relationships. That seems obvious. But what might seem strange is that for a polynomial like $x^2-2$, whose roots are $\sqrt2$ and $-\sqrt2$ , you can switch these around and not hurt any other arithmetic!
The way to formalize what it means to "switch around" roots is thought group theory, and there is a group that corresponds to how the roots of a polynomial can be switched around called the Galois group. Finally, if this group is "too complicated" (i.e. too many ways to permute the roots), then that group and its corresponding polynomial are not solvable by radicals. In the case of 5th degree polynomials, if it were possible to "invert" the polynomial $x^5-x-1$ (i.e. solve it directly like we can $x^5$), I believe this is all that would we needed for all 5th degree polynomials to be solvable by radicals. So as you see, it's just a "little bit" too complicated, and it gets worse as the degree increases.
I'm leaving out lots of details, but the other answers and links fill in those details. But I hope this gives you a flavor if what's going on. |
H: Ratio GRE question
Cashews cost 4.75 per pound and hazelnuts cost 4.50 per pound. What is
larger, the number of pounds of cashews in a mixture of cashews and
hazelnuts that costs $5.50 per pound, or 1.25? Alternatively, are they
equal, or is it impossible to calculate?
My answer:
I believe that 1.25 is larger. I conclude this because even if a 1.25lb mix were entirely composed of cashews, it would be more costly than $5.50/lb. Therefore, the mixture must have fewer than 1.25lbs of cashews, and 1.25 is greater.
The supposed answer:
"There is no way to calculate the number of pounds of either nut in the mixture. We can calculate the ratio of the number of pounds of cashews to the number of pounds of hazelnuts required of the mixture to cost $5.50 per pound, but without knowing how many total pounds the mixture is, we cannot calculate the number of pounds of either component."
Please let me know if I'm wrong, but if I'm not, I think I need to email them.
AI: First of all, there is no way any mixture of them can cost \$5.50 per pound: if we have a mixture of $A$% cashews and $(100-A)$% hazelnuts, then $x$ pounds of the mixture costs
$$x\text{ pounds}\cdot\left(\frac{A}{100}\cdot\frac{4.75\text{ dollars}}{\text{pound}}+\frac{100-A}{100}\cdot\frac{4.50\text{ dollars}}{\text{pound}}\right)$$
$$=4.75x\left(\frac{A}{100}\right)+4.5x\left(\frac{100-A}{100}\right)\text{ dollars}$$
$$\leq 4.75x\left(\frac{A}{100}\right)+4.75x\left(\frac{100-A}{100}\right)\text{ dollars}=4.75x\text{ dollars}$$
so the cost of the mixture is never more than \$4.75 per pound of mixture. Perhaps you, or the test writers, have made a typo in this respect.
But more importantly, there is no such thing as the
number of pounds of cashews in a mixture of cashews and hazelnuts
Suppose I tell you "spice" is a mixture consisting of 50% garlic and 50% salt (by weight). How many pounds of garlic are there in spice? The question doesn't make any sense - you can only talk about how many pounds of garlic there are in a specific pile of spice. The substance spice does not have any pounds of anything; it has ratios of its constituent ingredients.
Any information of the form "a mixture of cashews and hazelnuts costs $y$ dollars per pound" can only ever specify a ratio of ingredients. There is no number of pounds of cashews in a mixture; there is a number of pounds of cashews in a specified amount of a mixture. |
H: Why does a countably infinite dimensional space not have an uncountable chain of subspaces?
This question is motivated by Jyrki Lahtonen's comment in this question. It was an attempt at a short proof that for arbitrary field $k$, we have that $k^\mathbf N$ (as the full set-theoretic product with obvious linear structure) is not of countable dimension over $k$.
Don't know about one-liners. You could also argue that the subspaces $V_r$ defined for all real numbers $r$ as
$$V_r:=\{ s:\mathbf Q \to k \lvert s(q)=0\forall q>r\}$$
form an uncountable chain of subspaces (without repetitions): $V_{r_1}\subsetneq V_{r_2}$ whenever $r_1<r_2$.
At first glance, it seems like a really nice and elegant argument, but when you look a little closer, there's no obvious (to me) reason for it not to work in the case of a countably infinite dimensional space. Sure, individual $V_r$ may be smaller, but they still seem to form an uncountable strictly increasing chain of subspaces, contradicting countable dimension.
This obviously can't be right, so where does the problem lie?
Countability is probably immaterial here, you could make a very similar argument using some linear orders larger than rationals, dense in yet larger linear orders to obtain pretty much the same "result" for arbitrarily large cardinals.
AI: The subspace $V'$ tomasz is interested in consists of functions $s:\mathbb{Q}\rightarrow k$ with finite support. It seems to me that $V'$ also has an uncountable chain of subspaces.
However, this does not lead to an uncountable set of linearly independent vectors.
This is because, for example,
$$
V'\cap V_{\sqrt2}=\bigcup_{r<\sqrt2}(V'\cap V_r).
$$
This follows from the fact that any function on the l.h.s. has a finite support in
$\mathbb{Q}\cap (-\infty,\sqrt2]$ and hence also belongsto the r.h.s. union.
The same does not work in the case of the original space, where we don't require the functions to have a finite support. I repeat the argument from my comment to the linked question. The characteristic function $\chi_r$ of the set $\mathbb{Q}\cap(-\infty,r]$
is in the difference:
$$
\chi_r\in V_r,\qquad \chi_r\notin \bigcup_{r'<r}V_{r'}
$$
for all real numbers $r$. Therefore the set $\{\chi_r\mid r\in\mathbb{R}\}$
is an uncountable linearly independent set.
So the answer is that the existence of an uncountable chain of subspaces does not really prove that the space has uncountable dimension, unless we can, for each subspace in the chain, give a vector that does not belong to the union of the preceding subspaces. |
H: Taylor expansion on interval or at infinity
I'm trying to figure out how to get this result via taylor expansion as $x,y \rightarrow \infty$:
$f(x) = \sqrt{(x-1)y} = \sqrt{xy} - \frac{1}{2}\sqrt{\frac{y}{x}} + ...$
I've been told (on yahoo answers), that since you can't take taylor expansions at infinity, you do it on intervals (what does that mean?). Specifically, I was told:
You can write
f(x) = f(a) + (x-a) f'(a) + error term
or
f(a) = f(x) -(x-a)f'(x) + error term
Here your 2 ends are X and X - 1
I'm not sure what that means =P.
I am looking for a step-by-step taylor series calculation up to the first 2 terms. I've never approximated a function "at infinity" nor am I familiar with the generalized formula for taylor expansion.
So, any help would be great.
Thanks!
AI: In your case, you can just ignore the common factor $\sqrt{y}$ and concentrate on $\sqrt{x-1}$. To take the Taylor expansion at infinity is to replace $x$ by $1/u$ and take the Taylor expansion of at $u=0$ instead. Naïvely: $$\sqrt{x-1}=\sqrt{\frac1u-1}$$ which is of course singular at $u=0$, so you take out a troublesome factor: $$\sqrt{x-1}=\frac{\sqrt{1-u}}{\sqrt{u}}.$$ Now you expand the numerator as $\sqrt{1-u}=1-\frac12u+\cdots$, replace $u$ by $1/x$, and put the pieces together. The approximation will be good when $\lvert u\rvert$ is small, i.e., when $\lvert x\rvert$ is large. |
H: Factoring extremely large integers.
The question is about factoring extremely large integers but you can have a look at this question to see the context if it helps. Please note that I am not very familiar with mathematical notation so would appreciate a verbose description of equations.
The Problem:
The integer in question will ALWAYS be a power of N and will be known to us to calculate against. So let's assume N = 2 for example. That will give us a sequence of numbers like:
2, 4, 8, 16... up to hundreds of thousands of digits.
I need to find all possible factors (odd, even, prime, etc.) as efficiently as possible.
The Question:
What is the solution and how could I understand this from mathematical and computational perspectives?
EDIT:
Does the fact that each number to be factored is a power of 2 help in eliminating any complexity or computational time?
AI: If $N = 2$ (or any prime) and you can get $k$ as Ross indicated, then the divisors of $A$ are $\{1, N, N^2, \ldots, N^i, \ldots, N^{k} \}.$ If $N$ is a composite, then you will incur extra time complexity in computing the prime factorization of $N = \prod_{i = 1}^{\ell} p_i^{e_i}$ and the divisors of $A$ are all possible numbers of the form $\prod_{i=1}^{\ell} p_i^{r_i}$ where $\mathbf{0} \le r_i \le ke_i.$ Notice $r_i$ can be zero for some terms. There are $\prod_{i=1}^{\ell}(ke_i + 1)$ such possible divisors. |
H: spectral gap of the graph / Markov chain
Let $G=(V,E)$ be a graph. The gradient of a function $f:V\longrightarrow R$ is defined on the edges of the graph, given by the discrete derivative
$$
\nabla f(e)=f(y)-f(x), \quad e=(x,y) \in E
$$
Take $V=S_{2n}$ the group of permutations $\pi$ of the set $\{1, \ldots 2n\}$ and $e=(\pi, \pi \tau) \in E$ for any transposition $\tau$.
I am trying to find a lower bound of the spectral gap of the graph $(S_{2n}, E)$.
I guess it should be well-known.
Any literature would be very helpful.
Thank you.
AI: Papers by Persi Diaconis and co-authors in general, and in particular Comparison Techniques for Random Walk on Finite Groups, Persi Diaconis and Laurent Saloff-Coste, Ann. Probab. Volume 21, Number 4 (1993), 2131-2156. |
H: Prove that there do not exist positive integers $x$ and $y$ with $x^2 - y^2 = n$
I'm working on a homework problem that is as follows:
Suppose that $n$ is a positive even integer with $n/2$ odd. Prove that there do not exist positive integers $x$ and $y$ with $x^2 - y^2 = n$.
It looked like a good candidate for proof by contradiction. So I know that I still assume the argument "$n$ is a positive even integer with $n/2$ odd" but will try to show that there exists positive integers $x$ and $y$ with $x^2 - y^2 = n$, and if this reaches a contradiction then I have proven the original conjecture.
So I started with the $n/2$ is odd and rewrote it as $n/2 = 2k+1$, for some $k$ in the integers.
Then I knew that $n = 4k+2$, and then tried so equate that with $x^2 - y^2 = 4k+2$.
EDIT: I then recognized that $x^2 - y^2$ is equivalent to $(x + y)(x - y)$ but that doesn't seem to be very helpful, because if you divide one or the other out you get a term on the RHS in terms of k and x and y.
I'm going to keep playing with it but I don't really have any good strategies going forward. Any help is appreciated!
AI: A more or less mechanical approach is to work modulo $4$. Note that for any integer $k$, $k^2\equiv 0 \pmod{4}$ or $k^2\equiv 1\pmod{4}$.
So, modulo $4$, $x^2-y^2$ can only take on the values $0$, $1$, and $-1$.
More basic, and more useful, is to suppose that $x^2-y^2=n$. Then $(x-y)(x+y)=n$. Note that for any integers $x$ and $y$, the numbers $x-y$ and $x+y$ are both even or both odd. (If we want a proof, their difference $2y$ is even.)
In neither case is $(x-y)(x+y)$ twice an odd integer. You started along these lines. Note that you were one step from the end.
Remark: The reason the second idea is more useful is that when it comes to solving $x^2-y^2=n$, we express $n$ as a product $st$ of integers of the same parity, and solve the system $x-y=s$, $x+y=t$. The solution is $x=\frac{s+t}{2}$, $y=\frac{s-t}{2}$. If $n$ is twice an odd integer, then this process breaks down, because one of $s$ and $t$ will be odd and the other even, so we do not get integers $x$ and $y$. |
H: Right Coset and the associated lemma from Herstein: Appreciating whats going on?
I.N. Herstein in Page 34 (last line) and Page 35 of "Topics in Algebra" book goes on to explain a definition of right coset and a lemma like this:
Def: If $H$ is a subgroup of G, and $a \in G$, then $Ha = \left \{ha|h\in H \right \}$;then $Ha$ is the right coset of $H$ in $G$
Lemma: FOr all $a \in G $ $Ha = \left \{x \in G |a \equiv x mod H \right \}$
He goes on to define a set $[a]$ exactly like $Ha$ and trying to show $Ha \subseteq [a] $
My confusion:
Whats going on here?
More specifically, what the lemma trying to convey and why did the author go on to define $[a]$ exactly like $Ha$ and trying to show $Ha \subseteq [a] $ Isnt it trivial that every set is a subset of itself?
If you have the proof of the lemma with you, can you help me understand it. I am not able to understand why exactly are we dealing with $a(ha)^{-1}$ which I understand as motivated from $a = ha mod H$
Thanks for your time and patience
Soham
AI: The bracket notation is usually used in reference to equivalence relations. If $\sim$ is an equivalence relation on a set $G$, then $[a] = \{x \in G : x \sim a\}$.
So if I had to guess what is going on here, I would say that he defines $Ha = \{ha : h \in H\}$. Then he would have defined an equivalence relation $\sim$ on $G$ by $a \sim b$ if and only if $ab^{-1} \in H$.
Then it is easy (but not trivial) to show that $Ha \subset [a]$. |
H: Alternating sum of binomial coefficients
Calculate the sum:
$$ \sum_{k=0}^n (-1)^k {n+1\choose k+1} $$
I don't know if I'm so tired or what, but I can't calculate this sum. The result is supposed to be $1$ but I always get something else...
AI: Using the binomial theorem we have:
$$ (1 + (-1))^{n+1} = {{n+1} \choose 0} (-1)^0 + {{n+1} \choose 1} (-1)^1 + \ldots + {{n+1} \choose {n+1}} (-1)^{n+1}.$$
"Divide" by ${-1}$ to get:
$$ - (1 - 1)^{n+1} = -{{n+1} \choose 0} + \color{blue}{{{n+1} \choose 1} (-1)^0 + \ldots + {{n+1} \choose {n+1}} (-1)^{n}}.$$
This pretty much solves it. |
H: A recurrence relation for the Harmonic numbers of the form $H_n = \sum\limits_{k=1}^{n-1}f(k,n)H_k$
Working on Harmonic numbers, I found this very interesting recurrence relation :
$$
H_n = \frac{n+1}{n-1} \sum_{k=1}^{n-1}\left(\frac{2}{k+1}-\frac{1}{1+n-k}\right)H_k
,\quad \forall\ n\in\mathbb{N},n>1$$
My proof of this is quite long and complicated, so I was wondering if someone knows an elegant or concise one. Any idea would be appreciated.
Alternatively, if someone knows a reference that talks about this kind of relation, it would be of great interest for me.
Thanks.
AI: It helps to visualize the terms in a square of products $1/(ij)$ with $i$ and $j$ running for $1$ to $n$. The sum over the first term contains all products with $i\ne j$ exactly once, whereas the sum over the second term roughly corresponds to the upper left half of the square, but with the left-most column, which adds up to $H_n$, excluded. Thus we have
$$
\def\sub#1{{\scriptstyle{i\ne j}\atop{\scriptstyle i,j\le #1}}}
\sum_{k=1}^{n-1}\frac{2}{k+1}H_k=\sum_{\sub n}\frac1{ij}$$
and
$$-\sum_{k=1}^{n-1}\frac{1}{1+n-k}=H_n-\sum_{i+j\le n+1}\frac1{ij}\;.$$
Substituting this into your equation, multiplying through by $n-1$ and simplifying leads to
$$\sum_{i+j\le n+1}\frac1{ij}-\sum_{\sub n}\frac1{ij}=\frac2{n+1}H_n\;,$$
$$\sum_{i+j\le n+1}\frac1{ij}-\sum_{\sub n}\frac1{ij}=2\sum_i\frac1{n+1}\frac1i\;,$$
$$\sum_{i+j\le n+1}\frac1{ij}=\sum_{\sub{n+1}}\frac1{ij}\;.$$
This we can prove by induction: The equation is satisfied for $n=0$, and going from $n$ to $n+1$ adds
$$\sum_{i+j=n+1}\frac1{ij}=\sum_{i=1}^n\frac1i\frac1{n+1-i}$$
to the left-hand side and also
$$2\frac1{n+1}\sum_{i=1}^n\frac1i=\frac1{n+1}\sum_{i=1}^n\left(\frac1i+\frac1{n+1-i}\right)=\sum_{i=1}^n\frac1i\frac1{n+1-i}$$
to the right-hand side. |
H: About a linear map from $ \mathbb R^n \to \mathbb R^n$
I want to prove the linear map $L : \mathbb R^n \to \mathbb R^n$ is onto if and only if $L$ is one to one.
AI: This is false. The trivial map $L : \mathbb{R}^n \rightarrow \mathbb{R}^n$ defined by $L(v) = 0$ is linear, but not one to one or onto.
For the new phrasing of the question. Suppose $L$ is linear. If $L$ is injective, then $\text{ker}(L) = \{0\}$. By the first isomorphism theorem, $im(L) \simeq \mathbb{R}^n / \{0\}$. This is a $n$-dimensional vector space. The only $n$ dimensional vector subspace of a $n$-dimensional vector space is the entire vector space. So $L$ is surjective.
Suppose that $L$ is surjective. Then again by the first isomorphism theorem, $\mathbb{R}^n / \text{ker}(L) \simeq \mathbb{R}^n$. So $\text{ker}(L)$ is a zero dimensional vector subspace. Hence $\text{ker}(L) = \{0\}$. $L$ is injective. |
H: If $f(x)$ is irreducible in $k[x]$, why is it also irreducible in $k(t)[x]$, for $t$ an indeterminate?
I've been thinking on this a few days, but I'm stuck.
Let $k$ be a field, $f(x)$ irreducible in $k[x]$. Why is $f(x)$ also irreducible in $k(t)[x]$, for $t$ an indeterminate?
I write $f(x)=c_0+c_1x+\cdots+c_nx^n$ for $c_i\in k$. Suppose that $f(x)$ is reducible in $k(t)[x]$, so
$$
f(x)=(a_0(t)+a_1(t)x+\cdots+a_r(t)x^r)(b_0(t)+b_1(t)x+\cdots+b_s(t)x^s)
$$
for $r,s>0$. Since $c_0=a_0(t)b_0(t)$, $a_0(t)$ and $b_0(t)\in k$. Same goes for $a_r(t)$ and $b_s(t)$.
I tried something like letting $b_j(t)$ be the least index such that $b_j(t)\in k$. Then
$$
c_j=a_0b_j+a_1b_{j-1}+\cdots
\text{ and }
c_j-a_0b_j=a_1b_{j-1}+\cdots
$$
where $c_j-a_ob_j\in k$, and the right hand side is the sum of polynomials in $k(t)$ of degree at least $1$. I don't see if there is a contradiction to be had. What's the right approach? Thanks.
AI: If $k$ is infinite you do not need Gauss's lemma (in fact you hardly need any tools at all). Let us start the same way as you started: suppose $f$ admitted a nontrivial factorization in $k(t)[x]$:
$$f(x) = (a_0(t) + ... + a_n(t) x^n)(b_0(t) + ... + b_m(t) x^m).$$
Pick $c \in k$ such that the denominators of each $a_i(t)$ and $b_i(t)$ are nonzero when $t = c$ and such that the numerators of $a_n(t), b_n(t)$ are also nonzero (this is always possible iff $k$ is infinite). Then
$$f(x) = (a_0(c) + ... + a_n(c) x^n)(b_0(c) + ... + b_m(c) x^m)$$
is a nontrivial factorization of $f$ in $k[x]$.
(The takeaway lesson here is that a polynomial in $k(t)[x]$ really lies in $k[t][x]$ localized away from finitely many elements of $k[t]$.)
If $k$ is finite you can do something similar but sneakier: pass to the algebraic closure $\bar{k}$ to find $c$ as above. By choosing $c$ more carefully you can deduce that either the $a_i$ are all constant multiples of each other (resp. the $b_i$) or $f$ admits a nontrivial factorization in $\bar{k}[x]$ with an irreducible factor having a Galois orbit larger than the degree of $f$, which is a contradiction. |
H: Heine-Borel Theorem ($\mathbb{R}^k$) (in ZF)
Heine-Borel Theorem;
If $E \subset \mathbb{R}^k$, then $E$ is compact iff $E$ is closed and bounded.
I have proved 'closed and bounded⇒compact' and 'compact⇒bounded'.
(There exists $r\in \mathbb{R}$ such that for every $x\in E$, $|x|<r$)
The proof in Rudin PMA p.40 uses 'countable axiom of choice'
I have googled it and found some proofs, but they all used some weaker form of AC.
Please help me how to show that $compact⇒closed$ in ZF..
AI: We will prove that compact implies closed by contraposition.
Suppose that $E$ is not closed. Then there is some $x\notin E$ such that every neighbourhood of $x$ has a nonempty intersection with $E$. In particular, the collection $E_n:=E\setminus \overline {B(x,1/n)}$ (where $\overline {B(x,1/n)}$ is the closed ball centered at $x$ with radius $1/n$) is an infinite, nondecreasing open cover of $E$ (because for any $p\in E$ and $n>1/d(x,p)$ we have $p\in E_n$).
It is enough to show that $E_n$ does not stabilize. But if it did, then we would have for some $N<\omega$ that $E_N=E$, so $B(x,1/N)$ would be disjoint from $E$, so $x$ would not belong to the closure of $E$, so we're done.
This argument should work in an arbitrary Hausdorff space, though without countable character the cover will not be a sequence, but a directed set. |
H: Can manholes be made in other shapes than circles, that prevent the cover from being able to fall down its own hole?
Circular manholes are great because the cover can not fall down the hole. If the hole were square, the heavy metal cover could fall down the hole and kill some man working down there.
Circular manhole:
Can manholes be made in other shapes than circles, that prevent the cover from being able to fall down its own hole?
Semi rigid math formulation:
Let us say that we have an infinite matematical 2D plane in 3D space. In this plane is a hole of some shape. Furthermore we have a flat rigid 2D figure positioned on one side of the plane. This figure has the same shape as the hole in the plane, but infinitesimal larger.
Is it possible to find a shape, where there is no path twisting and turning the figure that brings the figure through the hole?
Here is one such shape (only the black is the the shape):
But if one put the restriction on the shape, that it needs to be without holes (topological equivalent to a circle in 2D), then I can not answer the question!?
Edit:
Because of the huge amount of comments and answers not about math, I fell the need to specify that:
I am not interested in designing manholes. I am interested in the math inspired by the manhole problem.
AI: Any manhole cover bounded by a curve of constant width will not fall through. The circle is the simplest such curve. |
H: A simple question of finite number of basis
Let $V$ be a vector space. Define
" A set $\beta$ is a basis of $V $" as "(1) $\beta$ is linearly independent set, and (2) $\beta$ spans $V$ "
On this definition, I want to show that "if $V$ has a basis (call it $\beta$) then $\beta$ is a finite set."
In my definition, I have no assumption of the finiteness of the set $\beta$. But Can I show this statement by using some properties of a vector space?
AI: I take it $V$ is finite-dimensional? For instance, the (real) vector space of polynomials over $\mathbb{R}$ is infinite-dimensional $-$ in this space, no basis is finite. And for finite-dimensional spaces, it's worth noting that the dimension of a vector space $V$ is defined as being the size of a basis of $V$, so if $V$ is finite-dimensional then any basis for $V$ is automatically finite, by definition! |
H: Pumping lemma - do I have to show every way to split string to have a complete answer?
In the pumping lemma, we have to split strings into $uvwxy$ (for example). Say the language was $a^n$$b^n$$a^n$$b^n$. We could it this way: $a^r$$a^s$$a^t$$a^u$$b^n$$a^n$$b^n$, with $uvwx$ all contained in the first $a^n$. We can pump the string, and we'll see the string doesn't belong in the language.
This is all good, but do I then have to show other ways to split the string? Or is just showing one example enough to be a complete answer (and enough to declare the language as not context free)?
AI: Say the language was $\{a^n b^n\}$. If your example was good enough to be a complete answer, the same argument would also suggest that $\{a^n b^n\}$ isn't context-free, even though it certainly is.
Since you cannot prove a false statement with a valid proof, this proof is incomplete. Check the conditions of the particular pumping lemma you're using to see what assumptions you made about $u,v,w,x,y$ that are unwarranted. Quite possibly it is your claim that $uvwx$ fits entirely into $a^n$, even though the pumping lemma makes no promise that $|uvwx| \le n$. |
H: Expected values of Binomial?
A group of $d$ students tries to solve $k$ problems in the following way. Each student picks only one problem at random and tries to solve it. Assume that all the attempts have probability $p$ of success and are independent of each other and of the students choice. Let $X$ denote the number of solved problems. Find $\Bbb E(X)$.
I am not sure if I just overestimating this, but would the answer simply be $E(X) = \dbinom{d}{k}p$
Because there are exactly $\dbinom{d}{k}$ independent trials and p is the success rate
Thank you
AI: Fix one of the problems. The probability that a particular student solves that problem is $\frac{p}{k}$ (the probability that the student selected the fixed problem, multiplied by the probability that the student solved the problem correctly). Since there are $d$ students, the probability that NOBODY solved the fixed problem correctly is $\left(1-\frac{p}{k}\right)^d$. It follows that the probability the fixed problem is solved by at least one person is $1-\left(1-\frac{p}{k}\right)^d$. The expected number of problems that are solved is then $k\left(1-\left(1-\frac{p}{k}\right)^d\right)$ |
H: How to find all solutions to equations like $3x+2y = 380$ using matrices/linear algebra?
I'm coming up blank on Wikipedia and other sources, though this seems elementary. I'd like to know what techniques or processes are used to find all (integer) solutions to an equation such as $3x+2y = 380$ using linear algebra.
AI: I can't imagine using linear algebra to find the integer solutions to your equation, when it's so simple to note that $x$ must be even, say, $x=2z$, so the equation becomes $3z+y=190$, and the solution is $z$ is arbitrary, $y=190-3z$.
EDIT: For linear diophantine equations in several variables, a good starting place is the Wikipedia piece on Bezout's identity, http://en.wikipedia.org/wiki/Bezout%27s_identity
MORE EDIT: There's a nice discussion by Gilbert and Pathria of systems of linear diophantine equations here; now linear algebra comes into it. |
H: Determine the number of factors for extremely large numbers.
An offshoot from a related question, is there a way to determine the number of possible factors (odd, even, prime, etc.) for extremely large integers without actually factoring them?
Even an estimation would help as long as it has some relevance to the number in question.
AI: As others have noted, there are bounds. But you can have, say, two 1000-digit numbers, differing only in their 778th digit, one of the numbers having zillions of factors, the other being prime, or a product of two primes. There is, in general, no way to get much information about the number of possible factors (or odd factors, or even factors, or prime factors, or repeated factors, etc., etc.) without factoring the number. |
H: Why Strongly Continuous Representations?
When working with not-necessarily-finite-dimensional representations, the topology on $GL(V)$ makes a difference. My experience has been that usually people require that the representation $\pi :G\rightarrow GL(V)$ be continuous with respect to the strong-operator topology on $GL(V)$. I would have guessed that instead we would equip $GL(V)$ with the norm topology, so that $GL(V)$ has the nice structure of a Banach manifold. Now that I know this is not the usual assumption, I am guessing this topology is too strong for some purposes. Could someone provide me with some examples demonstrating that the norm topology is the wrong choice of topology when it comes to representation theory?
AI: Take the regular representation
$$\mathbb{R} \ni t \mapsto (f(x) \mapsto f(x+t))$$
of $\mathbb{R}$ on $L^2(\mathbb{R})$. This is strongly continuous but not norm-continuous, the problem being that the norm of $f(x + t) - f(x)$ can be large even if $t$ is small. But this is clearly a representation we want. (More generally we want the regular representation of $G$ on $L^2(G)$ for $G$, say, a locally compact Hausdorff group.)
Note that if $H$ is a Hilbert space then a norm-continuous one-parameter family of unitary maps $U_t : H \to H$ is necessarily generated by a bounded self-adjoint operator $A$ in the sense that $U_t = \exp (iAt)$ (proof). In the above example the relevant self-adjoint operator is $-i \frac{d}{dx}$, which is not bounded. In general see Stone's theorem. |
H: $A' \cap (A' \cup B') \cap T= A' \cap T$
Let $A'$ denotes the complement of A with respect to $ \mathbb{R}$ and $A,B,T$ are subsets of $\mathbb{R}$. I am trying to prove $A' \cap (A' \cup B') \cap T= A' \cap T$, but I got some problems along the way.
$A' \cap (A' \cup B') \cap T= (A' \cap A') \cup (A' \cap B') \cap T= A' \cup (A \cup B) \cap T =(A' \cup A)\cup B \cap T= \mathbb{R} \cup B \cap T = B \cap T$ Something wrong?
AI: Well, given arbitrary sets $X,Y$ we always have $X\cap(X\cup Y) = X$ since the intersection of a set with a union of that same set and anything else is just the first set itself (I hope that's not too wordy). Try drawing all possibilites with sets to see this. So $A'\cap (A' \cup B') = A'$ and that gives you your answer.
Your fault is that in the second equality does not hold: for arbitrary sets $X,Y$ in general $(X \cap X) \cup (X\cap Y) \neq X \cup (X'\cup Y')$. Again, drawing all possibilites really helps to make this clear. Here I'm generalizing slightly from what you have, so for you $X$ is $A'$ and $Y$ is $B'$. Recall $A'' = A$. |
H: A question of an orthonormal system
Let $H$ be an inner product space, $e_n\;(n \in \Bbb N )$ be the orthonormal system of $H$. Here I want to prove that for any $f \in H$ , $\langle f, e_n\rangle_H \to 0$ as $n \to \infty$.
The bracket means the inner product.
AI: Let $a_n = \langle f, e_n \rangle$. One has that $||f||^2 = \sum_{i = 1}^\infty |a_i|^2$. Since $||f|| < \infty$, one must have that $|a_i| \rightarrow 0$. So $a_i \rightarrow 0$. |
H: a problem of trinomial distribution
$X_1$, $X_2$, $X_3$ are distributed according to the trinomial distribution with $n$($=X_1+X_2+X_3$) and $p_1$, $p_2$, $p_3$ ($p_1+p_2+p_3=1$). What is a correlation of $X_i$ and $X_j$? Is the conditional probability function $p(x_2,x_3|x_1)$ of $X_2$ and $X_3$ given that $X_1=x_1$ what special families of distributions? I think it need the probability function and moment generating function...but I am not sure..
AI: The probability distribution is
$$\mathbb{P}(X_1=k,X_2=l,X_3=m) = {{n}\choose{k,l,m}} p_1^k p_2^l p_3^m$$
with $k+l+m=n$ and $k,l,m \in \{0,\ldots,n\}$.
Multiply this by $x^k y^l z^m$ and sum over all possible values for the triple $(k,l,m)$ and you'll get the following generating function:
$$\Phi(x,y,z)=(p_1 x + p_2 y + p_3 z)^n \; .$$
By taking derivatives w.r.t the variables $x,y$ or $z$, you can easily find moments. For instance
$$\mathbb{E}[X_1 X_2] = \left.\frac{\partial^2\Phi}{\partial x \partial y}\right|_{(1,1,1)}=n(n-1)p_1p_2 \; .$$
Here's the wikipedia page for the multinomial distribution. Just put $k=3$ to find anything you need. |
H: What is a good technique for solving polynomials?
Say for example:
$6x^{3}-17x^{2}-4x+3=0$
I sort of look at it and don't know where to start, other than just guessing what the first one would be and trying to do from there. Is there a good technique for approaching such polynomials?
AI: First, to minimize the lead coefficient, we reverse the polynomial (which $\rm\color{brown}{reciprocates}$ the roots) and then we scale it to make the leading coefficient $1,\,$ i.e. we apply the AC-method, which yields
$$\begin{eqnarray}\rm f(x) \,&=&\,\rm 3\,x^3 - 4\,x^2-17\,x+ 6 \\
\rm \Rightarrow\ \ 9\,f(x)\, &=&\,\rm (3x)^3 - 4\,(3x)^2 - 51\,(3x) + 54 \\
\rm &=&\rm\ \ z^3 - \color{blue}4\, z^2 - 51\,z + \color{#0A0}{54},\ \ \ z = 3x
\end{eqnarray}$$
Note $\rm\:z=\color{#C00}1\:$ is a root. By Vieta, other roots have product $\color{#0A0}{-54/\color{#C00}1},\,$ sum $\color{blue}4\!-\!\color{#C00}1 = 3,\:$ so are $\:9,\,-6.\:$
Hence $\rm\:z\, =\, 1,\,9,\,-6\,\:$ so $\rm\:x\, =\, z/3\, =\, 1/3,\, 3,\, -2,\,$ which $\rm\color{brown}{reciprocated}$ yields $\:3,\,1/3,\, -1/2.$ |
H: Are non-circular swords possible?
I was reminded of this by our recent discussion of the old chestnut about possible shapes for utility hole covers. Perhaps this question is less familiar.
A sword can be made in any shape at all, but if you want to be able to put it into a scabbard, only certain shapes will do. For simplicity, let us neglect the width of the sword, and take it to be the image of a continuous mapping $[0,1]\to {\Bbb R}^3$. Clearly, a sword in the shape of a circular arc will fit into a similar scabbard, as long as the arc is less than about half of a full circle:
(For non-mathematical reasons, such swords are rarely more than about $\frac1{36}$ of circle.)
If we let the curvature go to zero, we get the special case of a straight segment, which can be of any length:
It seems to me that if one had a sword which was a helical segment, it would go into a matching scabbard. The circular and straight swords are special cases of this, where the helical pitch is zero. (A cursory search for helical swords or scabbards turned up nothing—unsurprisingly, since a sword which has to be worked into the enemy like a corkscrew is a really stupid idea.)
Are there any other shapes of swords which will go into scabbards?
AI: Clearly the curvature and torsion of the sword must be constant. A curve of nonzero constant curvature and torsion is a helix. |
H: How to conceptualize conditional expectation inductively?
Attempting the solve the following problem
A fair die is successively rolled. Let $X$ and $Y$ denote,
respectively, the number of rolls necessary to obtain a 6 and a 5.
Find
$E[ X|Y = 5]$
I am confused about how to conceptualize such a problem?
I know $E[X|Y= 1] = 1 + \frac{ 1}{ p} = 1 + \frac{ 1}{ 1/6} = 7$ building off the that we have a geometric random variable. But, I am unsure of how to extend this to a higher given expectation for $Y$.
I would like to solve this in an inductive mannor but
I am unable to conceptualize an inductive solution?
I am not sure that an inductive solution would even valid?
AI: I am not quite sure what kind of induction you are looking for, so I will just try giving a more general solution to this problem, i.e. $E[X|Y=y]$.
The point is to notice that random variable X distributed differently before and after the y-th roll, for tosses before the first one which yield 5 cannot have 5 as result, hence there are 1 out of 5 chances to get a 6.
The conditional probability is
$$P_{X|Y}(x,y)=\left\{ \begin{array}{1}(4/5)^{x-1}\times 1/5 ~ ~(X<Y)\\0 ~ ~(X=Y)\\(4/5)^{y-1}\times(5/6)^{x-y-1}\times1/6~~(X>Y)\end{array} \right.$$
Then follow the definition of conditional expectation we have:
$E[X|Y=y]\\=\sum\limits_{x}xP\{X=x|Y=y\}\\=\sum\limits_{x=1}^{y-1}x(\frac{4}{5})^{(x-1)}\frac{1}{5}+0+\sum\limits_{x=y+1}^\infty x(\frac{4}{5})^{y-1}(\frac{5}{6})^{(x-y-1)}\frac{1}{6}\\=\frac{1}{5}\sum\limits_{x=1}^{y-1}x(\frac{4}{5})^{(x-1)}+\frac{1}{6}(\frac{4}{5})^{y-1}\sum\limits_{x=y+1}^\infty x(\frac{5}{6})^{(x-y-1)}\\=\frac{1}{5}[5^{1-y} (-5(4^y)+5^{1+y}-4^yy)-y(\frac{4}{5})^{y-1}]+\frac{1}{6}(\frac{4}{5})^{y-1}6(y+6)\\=\frac{5}{4}(4-(\frac{4}{5})^y(y+4))+(\frac{4}{5})^{y-1}(y+6)$
To quickly verify this, plug in y=1 we will get 7 and for y=5 we have 5.8192.
Note:
The part of the equation to obtain $\sum_{x=y+1}^\infty x(\frac{5}{6})^{x-1}$ may not be obvious but the trick here is to use the series of $\sum_{k=0}^\infty(k+1)x^k=\frac{1}{(1-x)^2}$, which can be obtained by geometric series in the following way:
$$
x\sum (k+1)x^k-\sum (k+1)x^k=(x-1)\sum (k+1)x^k =1+x+x^2+...=\frac{1}{1-x}\Rightarrow\sum (k+1)x^k=\frac{1}{(1-x)^2}
$$ |
H: Find the limit without L'Hôpital's theorem
I'm trying to find $$ \lim_{ n\to\infty} { {n^2+1}\over {n^3+1}} \cdot {\frac {n} {1}}$$
I know the answer is $1$, but I can't remember how my professor found it so simply without using L'Hôpital's theorem.
Could you please show me the shortcut?
AI: First let's expand $$ \lim_{ n\to\infty} { {n^2+1}\over n^3+1} \cdot {\frac n 1\phantom{n^2}}$$ using regular algebra:
$$ \lim_{ n\to\infty} { {n^3+n}\over n^3+1}$$
There's a standard trick for dealing with rational functions (that is, with quotients of polynomials). One way of looking at it is that you can divide the top and bottom by $n^k$ where $k$ is the largest power that appears in the numerator or denominator; here that is $n^3$:
$$ \lim_{ n\to\infty} { {n^3+n}\over n^3+1} \cdot {\frac1{n^3}\over\frac1{n^3}\\
\lim_{ n\to\infty} { {1+\frac1{n^2}}\over 1+\frac1{n^3}} }$$
As $n$ gets very large, $\frac1{n^2}$ and $\frac1{n^3}$ become insignificant compared with 1, so we are left with $\lim_{ n\to\infty} \frac11 = 1$. This is what Tim Duff did in the comments above.
But the short version of the same thing is to observe that it lets us disgregard all but the largest terms of the polynomials in the numerator and denominator. If the numerator has the largest power of $n$, the limit is $\infty$; if the denominator has the largest power of $n$, the limit is 0, and if the powers are the same, as they are here, then you have $an^k\over bn^k$ and the limit is $\frac ab$. |
H: A question about an uncountable summation.
Possible Duplicate:
The sum of an uncountable number of positive numbers
Consider $\sum_{\lambda \in \Lambda} a_{\lambda}$ . Here all $a_\lambda $ is non-negative.
Then I want to prove that if $\sum_{\lambda \in \Lambda} a_{\lambda} < \infty $ then there exists at most countable set $ \Lambda_0 \subset \Lambda$ such that $\lambda \notin \Lambda_0 \Rightarrow a_{\lambda}=0.$
(This means that if the summation converges then there are only at most countable $a_i$'s such that $a_i \neq 0)$
AI: HINT
Given $\varepsilon>0$, can you measure the set $\{\lambda:a_\lambda>\varepsilon\}$?
Giving some background to the somewhat short hint.
Suppose $E\subset\mathbb{R}$ is uncountable, and let
$\lambda\mapsto a_\lambda$ be a non-negative function defined on $E$. The
usual definition of the expression $\sum_{\lambda\in E}a_\lambda$ is given by
$$\sum_{\lambda\in E}a_\lambda =\sup_{F\subset E,\,|F|<\infty}\sum_{\lambda\in F}a_\lambda \tag{1}$$
i.e. we take supremum over finite sets.
Now, let us choose $\varepsilon>0$ and consider the set
$$E_\varepsilon = \{\lambda\in E :a_\lambda>\varepsilon\}.$$ This set
must be finite in order for the sum in (1) to be finite, because
otherwise we may for each positive integer $n$ choose subsets
$F_n\subset E_\varepsilon$ such that $|F_n|=n$ and then
$$\sum_{\lambda\in E}a_\lambda\ge \sum_{\lambda\in F_n}a_\lambda>\sum_{F_n}\varepsilon =n\varepsilon.$$
Since $$\bigcup_{\varepsilon>0} E_\varepsilon =\bigcup_{n=1}^\infty E_{1/n} =\{\lambda\in E:a_\lambda>0\}$$ the conclusion follows.
Remark: If we exclude the assumption $a_\lambda\ge0$. I do not think there is a reasonable definition for conditional convergence of this kind. |
H: Finite Subgroups of $GL(n,\mathbb{C})$
In the Artin's book on Algebra, the author stated a theorem (Ch.9, Thm. 2.2):
"A finite subgroup $G$ of $GL(n,\mathbb{C})$ is conjugate to a subgroup of $U(n)$.
Here, $U(n)$ is the unitary group, i.e. if $\langle \,\,, \rangle$ is the standard Hermitian inner product on $\mathbb{C}^n$ given by
$\langle (x_1,\cdots,x_n),(y_1,\cdots,y_n)\rangle=\sum_{}{x_i\bar{y_i}}$ then $U(n)=\{A\in GL(n,\mathbb{C}) \colon \langle Av,Aw\rangle = \langle v,w\rangle,\forall v,w\in \mathbb{C}^n \}$.
The proof is: there is a $G$-invariant Hermitian inner product $\langle\,\, , \rangle_1$ on $\mathbb{C}^n$, and consider an orthonormal basis $B_1$ w.r.t this form. If $P$ is the matrix of transformation which changes standard basis to $B_1$, then $PGP^{-1}\leq U(n)$
Question: In the last statement in the proof, the author says that $PGP^{-1}$ is subgroup of unitary group; but, this unitary group is corresponding to the Hermitian inner product $\langle \,\,, \rangle_1$, i.e. it is the group of linear transformations which preserves the inner product $\langle \,\,, \rangle_1$. Why $PGP^{-1}$ should be subgroup of $U(n)$ which is the group of linear transformations which "preserves" the standard Hermitian inner product $\langle \,\,,\rangle$ on $\mathbb{C}^n$?
AI: To answer the question a slightly different way: choose an orthonormal basis wrt the $G$-invariant Hermitian form $\langle, \rangle_{1}.$ Let this be $\{ u_{i}: 1 \leq i \leq n \}.$
Since the form is $G$-invariant, for any fixed $g \in G,$
we have $\langle gu_{i},gu_{j} \rangle_{1} = \delta_{ij}$ for all $i,j.$ Hence writing the matrix of (the transformation represented by) $g$ with respect to this basis, the columns (and rows) are orthonormal, and the matrix is unitary. So now you have a group of unitary matrices, which is conjugate to the original one via a change of basis matrix.
To answer comment below, which took to long to leave as a comment: It means that if write $gu_{i} = \sum_{j=1}^{n} a_{ij}(g)u_{j},$ where the $a_{ij}(g)$ are complex numbers, then we have for any $r,s$, $$\delta_{rs} = \langle gu_{r},gu_{s} \rangle_{1} = \sum \sum_{p,q= 1}^{n} a_{rp}(g) {\overline a_{sq}(g)} \langle u_{p}, u_{q} \rangle_{1} .$$ But this is $$\sum_{p =1}^{n} a_{rp}(g) { \overline a_{sp}(g)}$$ because $\langle u_{p}, u_{q} \rangle_{1} = \delta_{pq}.$ So what is left is the usual innner product of the rows of the matrix $[a_{ij}(g)].$ |
H: In mathematics, what is meant by induction?
I was going through MIT video lectures on "Introduction to Algorithms " . In order to solve recurrences by substitution the professor says that we can solve them by induction.
What is actually the principle behind induction?
AI: The most basic principle is this:
For any proposition P, which is a function from the natural numbers (1,2,...) to a boolean, if:
P(n) $\Rightarrow$ P(n+1) for all n, and
P(1)
then P(n) for every n.
That is, if you can prove the base case P(1), and you can prove the implication P(n) $\Rightarrow$ P(n+1) for every n, then you have proved that P holds for all natural numbers. |
H: For what values for m does $\sum \limits_{k=2}^{\infty}\frac{1}{(\ln{k})^m}$ converges?
For what values for m does $$\sum \limits_{k=2}^{\infty}\frac{1}{(\ln{k})^m}$$ converge?
What about
$$\sum_{k=2}^{\infty}\frac{1}{(\ln(\ln{k}))^m}$$ or more generally
$$\sum_{k=2}^{\infty}\frac{1}{(\ln(\cdots (\ln{k}))\cdots)^m}$$ ?
AI: For your first series
Let $\alpha,\beta \in \mathbb{R}$
$$\sum_{k=2}^\infty \frac{1}{k^\alpha \ln(k)^\beta} $$
is convergent iif $(\alpha>1)$ or $(\alpha=1, \ \beta>1)$
So in your case it is not convergent.
As
$$\frac{\sqrt{k}}{\ln(k)^\beta} \rightarrow \infty$$ we can find $k_0$ such that for all $k \geq k_0$
$$ \frac{\sqrt{k}}{\ln(k)^\beta} > 1 $$
which yields
$$ \frac{1}{\ln(k)^\beta} > \frac{1}{\sqrt{k}} $$
The right-hand side behind divergent, your series is divergent. These series are called Bertrand series. See here for proofs.
As for your second series, just repeat the same argument. Take $\ln(k) > k_0$ you will get
$$ \frac{1}{\ln(\ln(k))^\beta} > \frac{1}{\sqrt{\ln(k)}} $$
The right-hand side was just proven to be divergent. Repeat the process for your generalized version.
Edit:
You have to be careful though about the integer that will start the series. For your first series it is obvisouly $k=2$ because $\ln(1)=0$ which yields division by zero. For the second series the division by zero occurs at $e$ and $\ln(\ln(k))$ is not defined for $k=1$ so start at $k=3=\lfloor e \rfloor+1$. For you generalized series just start at $k=\lfloor e^{e^{\cdots^e}} \rfloor +1$ |
H: What is the correct way to solve $|2K^3-2K^4|$ determinant?
Given -
$$K_{3\times3} = \begin{bmatrix} 1&1&1 \\ 3&2&1 \\ 1&2&1 \end{bmatrix}$$
$$|K| = 2$$
Find -
$$|2K^3-2K^4|$$
I tried this:
Since $|A+B|=|A|+|B|$ ( $\Leftarrow$ This is the main mistake ) -
$$|2K^3-2K^4|=|2K^3+(-2K^4)|=|2K^3|+|(-2K^4)|$$
Now using $|\alpha A_{n\times n}|=\alpha ^n|A|$ -
$$=2^3|K^3|+(-2^4)(K^4)|=8*8+(-16)*16=-192$$
AI: Now that you've edited $K$ into the question, we can get somewhere!
Use $2K^3-2K^4=(2)(K^3)(I-K)$, and $\det cA=c^nA$, and $\det A^r=(\det A)^r$, and then you just have to calculate $\det(I-K)$ directly. |
H: Complex and real forms of the Poisson integral formula
In my complex analysis book there is the expression
$$\frac{1 - |z|^2}{|1 - \bar z e^{it}|^2}$$
and it says that when $z = re^{it}$, we can write the above expression as
$$P_r(t) = \frac{1 - r^2}{1 - 2r\cos t + r^2} = \text{Re}\left( \frac{1 + z}{1 - z} \right)$$
I do not see where the $\cos t$ comes from though.
Isn't $\bar z = re^{-it}$, so the top is $1 - r^2$ and the bottom is $|1 - r|^2 = 1 - 2r + r^2$. I have not really figured out where the $\text{Re}(1 + z)/(1 - z)$ comes from either.
AI: Here is a link a of the book. $P_r(t)$ is defined at equation (5) as $\frac{1-r^2}{1-2r\cos t+r^2}$. If $z=re^{it}$, then
$$\frac{1+z}{1-z}=\frac{1+re^{it}}{1-re^{it}}=\frac{(1+re^{it})(1-re^{-it})}{(1-re^{it})(1-re^{-it})}=\frac{1-re^{-it}+re^{it}-r^2}{1-2r\cos t+r^2},$$
and the real part of $-re^{-it}+re^{it}$ is $0$.
It helps us to write a formula for $f(re^{it})$, involving $P_r$. |
H: "The space of all lines on a plane is an open Möbius Band."
I have come across this sentence when I was reading something about algebraic topology:
The space of all lines on a plane is an open Möbius Band.
I don't quite understand this, can anyone explain this to me? Thank You.
AI: From Wikipedia:
The open Möbius band is formed by deleting the boundary of the standard Möbius band. It is constructed from the set $S = \{ (x,y) \in \mathbb{R}^2 : 0 ≤ x ≤ 1 \text{ and } 0 < y < 1 \}$ by identifying (glueing) the points $(0,y)$ and $(1,1−y)$ for all $0 < y < 1$.
Occurrence and use in mathematics
The space of unoriented lines in the plane is diffeomorphic to the open Möbius band.[6]
To see why, notice that each line in the plane has an equation $ax + by + c = 0$ for fixed constants $a$, $b$ and $c$. We can identify the equation $ax + by + c = 0$ with the point $(a,b,c)$. However, the line given by $ax + by + c = 0$ is also given by $λ(ax + by + c) = (λa)x + (λb)y + (λc) = 0$ for all $λ ≠ 0$.
These equations, which give the same line, are identified with the points $(λa,λb,λc)$. Thus, the space of lines in the plane is a (proper) subset of the real projective plane; where the equation $ax + by + c = 0$ corresponds to the point $(a:b:c)$ in homogeneous coordinates. It is only a subset because some equations of the form $ax + by + c = 0$ do not give lines. We need to disallow $a = b = 0$ to be sure that the equation $ax + by + c = 0$ does indeed give a line. The space of unoriented lines in the plane is given by deleting the point $(0:0:c) = (0:0:1)$ from the real projective plane. This space is exactly the open Möbius band. |
H: What does the letter epsilon signify in mathematics?
This letter "$\varepsilon$" is called epsilon right ? What does it signify in mathematics ?
AI: The greek letter epsilon, written $\epsilon$ or $\varepsilon$, is just another variable, like $x$, $n$ or $T$.
Conventionally it's used to denote a small quantity, like an error, or perhaps a term which will be taken to zero in some limit.
It's possible that you are confusing it with the set membership symbol $\in$, which is something different. When you see $x\in X$ it means that $X$ is a set, and $x$ is a member of the set. For example,
$$1\in \{1,2,3\}$$
is true, but
$$4\in\{1,2,3\}$$
is false. |
H: $T^2=S$ implies $\text{rank}T=\text{rank}S$?
Is it true, that if $T$ and $S$ are selfadjoint and positive operators in a finite-dimensional Hilbert space, such that $T^2=S$, that then $\text{rank}T=\text{rank}S$ ?
AI: This is true more generally if $T$ is diagonalizable (recall that if $T$ is self-adjoint, it is diagonalizable).
If $T$ is diagonalisable, there is an invertible matrix $P$ and a diagonal matrix $D$ such that $T=PDP^{-1}$, the rank of $T$ is the number (counted with multiplicities) of the non-zero eigenvalues of $T$. Now since $T^2=PDP^{-1}PDP^{-1}=PD^2P^{-1}$, the rank of $T^2$ is the number of the non-zero eigenvalues of $D^2$, which is the same as the number of non-zero eigenvalues of $D$.
The ranks of $T$ and $T^2$ are thus equal. |
H: Using Trapezoid Rule to approximate $\int^{0}_{-2}\frac{1}{4+x^2}dx$
I need to use the trapezoid rule to approximate the following integral: $$\int^{0}_{-2}\frac{1}{4+x^2}dx$$
for a given step $n=5$.
I'm not exactly sure how to do a problem like this.
AI: http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/techniques/numerical.html, check if it helps.. |
H: A certain family of continuous functions on $[0,1]^2$ the closure of which linear span is $\tilde{\mathcal{C}}([0,1]^2,\mathbb{R}))$
First of all I must apologize for the vague title and am open to suggestions.
This is not a Homework Assignment but something I once again encountered while reading a very compactly written paper.
$\tilde{\mathcal{C}}([0,1]^2,\mathbb{R}))$ is defined as follows
for $f\in \tilde{\mathcal{C}}([0,1]^2,\mathbb{R}))$ the following holds:
$f$ is continuous
$f(x)=0 \quad \forall x\in [0,1]^2\setminus [0,1)^2$
I am looking for a countable family of continuous functions $\mathcal{F}:=(f_m)_{m\in\mathbb{N} }$ on $[0,1]^2$ so that the following requirements are satisfied
$f\in \mathcal{F}\implies f(x)=0 \quad \forall x\in [0,1]^2\setminus [0,1)^2$
the closure of the linear span of $(f_m)_{m\in\mathbb{N}}$ consists of all continuous Functions on $[0,1]^2$
According to the authors such a sequence is supposed to exist. But I do not know what the $f_m$ would like or why the family with the aforementioned properties exists. So I am looking for an explicit construction of the $f_m$ or a theorem which proofs the existence of such a sequence.
My first guess was $f_{n.m}(x_1,x_2):=x_1^nx_2^m$ for with Stone-Weierstrass it fullfills the second requirement. Unfortunately it does not meet the first one and I am somewhat at a loss.
Has anybody encountered a similar construction or knows of a theorem that might help?
As always thanks in advance :)
AI: We have that $\widetilde{\mathcal C}$ is a (closed) subspace of $\mathcal C[0,1]^2$. The latter is separable, and so will be $\widetilde{\mathcal C}([0,1]^2,\Bbb R)$. So a countable dense family $\mathcal F$ will do the job. |
H: Sum of all elements in a matrix
The trace is the sum of the elements on the diagonal of a matrix. Is there a similar operation for the sum of all the elements in a matrix?
AI: I don't know if it has a nice name or notation, but for the matrix $\mathbf A$ you could consider the quadratic form $\mathbf e^\top\mathbf A\mathbf e$, where $\mathbf e$ is the column vector whose entries are all $1$'s. |
H: Convex hull approximated from inside by only finite number of elements?
In approximating the convex hull "from inside", i.e.
$$ \text{conv}S = \{ x \in \mathbb{R}^n \mid x= \sum_{i=1}^k \lambda_i x^i, x^i \in S, \lambda_i \geq 0, \sum_{i=1}^k \lambda_i= 1 \} \text{,}$$
in the case where $S$ is infinite - why can we restrict ourselves to finite many $x^i \in S$?
AI: I assume you are really asking this: Let $S$ be an infinite subset of $\mathbb{R}^n$. Then why is the convex hull of $S$, denoted $\operatorname{conv}S$, defined as you show it, using only finite convex combinations of elements of $S$?
The answer is that the set defined this way really is convex (a convex combination of two finite convex combination is another finite convex combination), and this is the smallest convex set containing $S$ (since any convex set containing $S$ must contain any finite convex combination of members of $S$).
I guess the main point is that the definition of convexity nowhere speaks of inifinite convex combinations.
As pointed out in the comments, Carathéodory's theorem is worth mentioning: It states that you never need to take a convex combination of more than $n+1$ of the points if you work in $\mathbb{R}^n$.
There is an analogy with linear combinations: The linear span of an infinite set of vectors consists of finite linear combinations of the given vectors. |
H: Three problems with binomial coefficients
I found three difficult problems for me, involving binomial coefficients. They are extremely interesting I think, but I don't know if I have enough knowledge to manage. Seem really hard, can you help me with them?
Prove that every $z\in\mathbb{N}$ we can represent (in one and only one way) by $$ z={x+y+1\choose 2}+x $$ for some $x,y\in\mathbb{N}$.
Show that the number: $$ \sum_{k\ge 0}{n\choose k}F_{m+k} $$ is some Fibonacci number for $m,n\in\mathbb{N}$
How many sequences consisting of $0$ and $1$ are there, where we have $2n$ zeros and $2n$ ones, but before $n$-th $1$ in sequence we have at most $n$ zeros?
AI: Hints :
Make a table, where you put $x$ in the column, and $y$ in the row, and the corresponding $z$ in each cell. That should be enough to give you an idea of the proof
The proof is simple for $n=1$. Now, from a formula for $n$, show you can obtain any formula for $n+1$ by decomposing $F_{n+2}=F_{n+1}+F_n$
Find a recursive formula that counts your sequences... (The solution is catalan numbers, those words are the dyck words) |
H: Fourier transform of the derivative - insufficient hypotheses?
An exercise in Carlos ISNARD's Introdução à medida e integração:
Show that if $f$ and $f'$ $\in\mathscr{L}^1(\mathbb{R},\lambda,\mathbb{C})$ and $\lim_{x\to\pm\infty}f(x)=0$ then $\hat{(f')}(\zeta)=i\zeta\hat{f}(\zeta)$.
($\lambda$ is the Lebesgue measure on $\mathbb{R}$.)
I'm tempted to apply integration by parts on the integral from $-N$ to $N$ and then take limit as $N\to\infty$. But to obtain the result I seemingly need $f'e^{-i\zeta x}$ to be Riemann-integrable so as to use the fundamental theorem of Calculus.
What am I missing here?
Thank you.
AI: There is a measure-theoretic version of the fundamental theorem of Calculus, see here. |
H: Domain of an operator in functional analysis
I used to think that if we say $f$ is a function from a set $X$ to $Y$ then this implied that $f$ was defined on all of $X$. Because the definition of function is that it's a set $\{(x,y) \mid \text{ for every } x \in X \text{ there is exactly one } (x,y) \text{ where } y \in Y \}$.
Now I'm reading the definition of closed graph of a linear operator:
So a linear operator is somehow not a function. But then what is it? Why does it make sense to say that $T$ is a map from $X$ to $Y$ when it's not even defined on all of $X$?
In response to comment:
Edit
Perhaps my question is: why and when does it make sense to say $T:X \to Y$ is a linear operator when $T$ is not even defined on all of $X$?
AI: This is an admittedly confusing abuse of terminology which your book appears to make even more confusing by using somewhat nonstandard terminology. The following is more standard but also confusing in its own way. For what follows let $X$ and $Y$ be Banach spaces.
A bounded linear operator from $X$ to $Y$ is a function $T: X \to Y$ which is linear and norm continuous.
An unbounded linear operator from $X$ to $Y$ is a pair $(T,D_T)$ where $D_T$ is a linear subspace of $X$ and $T: D_T \to Y$ is a linear map.
Comments:
An unbounded linear operator from $X$ to $Y$ is not in general a function from $X$ to $Y$, and the definition does not claim it is. The phrase "from $X$ to $Y$" is part of what is being defined.
Because unbounded operators are not functions, extra care must be taken: two unbounded linear operators from $X$ to $Y$ cannot necessarily be added, and given unbounded linear operators from $X$ to $Y$ and from $Y$ to $Z$ they do not necessarily have a composition.
"Unbounded" really means "not necessarily bounded"; every bounded linear operator is an unbounded linear operator.
Some authors require that the domain $D_T$ of $T$ is a dense subspace of $X$. With this convention one can conceive of operators which are neither bounded nor unbounded. When this is not part of the definition, the assumption is added via the phrase "densely defined".
Adjoints can be very confusing: there are densely defined unbounded operators such that the largest possible domain of definition for the adjoint is $\{0\}$.
The only real way to sort through all of this language is to look at some examples and results. The theory was invented by Von Neumann in the 20's to make precise sense of what physicists were doing in quantum mechanics. They had stumbled onto the idea that classical notions like position and momentum can be profitably viewed as linear operators on Hilbert space which satisfy certain relations. Mathematicians noticed, however, that there are no bounded operators on Hilbert space which satisfy these relations, and in particular the physicists' use of spectral theory had no mathematical basis. So Von Neumann invented the language of unbounded operators to make sense of the operators that physicists were doing. In particular he generalized the spectral theorem to certain classes of unbounded operators.
Maybe it would help to look at a quick example. Consider the operator $D = i \frac{d}{dx}$, a linear map $C_{per}^\infty[0,1] \to C_{per}^\infty[0,1]$, the space of smooth periodic functions on the interval $[0,1]$. Notice that $D$ has a rich supply of eigenfunctions: setting $e_n(x) = e^{-2 \pi i n x}$, we find that $e_n$ is an eigenfunction of $D$ with eigenvalue $2 \pi n$. The set of functions $\{e_n\}$ forms an orthonormal basis for the Hilbert space $L^2[0,1]$, so even though $D$ isn't an operator on $L^2[0,1]$ we can in a sense diagonalize it and use Hilbert space techniques to study it.
By contrast, consider the same operator $D = i \frac{d}{dx}$, this time viewed as a linear map $C_0^\infty(0,1) \to C_0^\infty(0,1)$, the space of smooth functions on the interval $[0,1]$ which vanish at $0$ and $1$. Now the only eigenfunctions for $D$ are constant functions, so it is not clear how one would diagonalize $D$. Secretly we know that we should just enlarge the domain of $D$ to include all periodic functions, but this isn't immediately obvious. Part of Von Neumann's accomplishment was to understand abstractly why $D$ with the domain $C_{per}^\infty[0,1]$ is "good" but $D$ with the domain $C_0^\infty(0,1)$ is not. |
H: Is $\int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx$ still valid for infinite limits?
Maybe this question is a stupid one but, let me ask it here just to be sure. :) We know that under continuity of function $f$ on an interval $[a,b]$ wherein $a<b$: $$\int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx$$ Now,
Does this equality remain valid if we replace $a$ and $b$ with $-\infty$ and $+\infty$ respectively?
I just want to be sure that, if the definition of definite integral can be extended for infinity (as for upper and lower limits). Thanks
AI: Yes, but that is a definition of what it means to integrate 'backwards' along a (finite) interval. But a consequence of this definition and the definition of integration over an infinite interval is that the result does hold.
Pardon my disgusting abuse of notation:
$$\displaystyle \int_{-\infty}^{\infty} = \lim_{a \to -\infty} \lim_{b \to \infty} \int_a^b = \lim_{a \to -\infty} \lim_{b \to \infty} -\int_b^a = -\lim_{a \to -\infty} \lim_{b \to \infty} \int_b^a = -\int_{\infty}^{-\infty}$$ |
H: Dimension of a quotient vector space of meromorphic functions
Let $U$ be an open set of the Riemann sphere, $z_i$ be $n$ distinct points of $U$, and $E$ the vector space of meromorphic functions on $U$ with poles of order no more than 2.
Let $F$ be the subspace of $E$ whose elements are holomorphic in a neighborhood of the $z_i$.
Does $E/F$ have finite dimension ? If so, what is it ?
It is clear that it has dimension at least $2n$, since the $\frac{1}{(z-z_i)^k}$, $k=1,2$, form a free family. However, I couldn't determine if there was more (intuition suggests not).
AI: If $f$ is an element in $E$, you can subtract all principal parts of $f$ at $z_i$, then you'll get a function which is holomorphic near every $z_i$. Since every principal part is a linear combination of $\frac{1}{z-z_i}$ and $\frac{1}{(z-z_i)^2}$, we can deduce that $E/F$ is a complex vector space of dimension $2n$. |
H: yet another problem about trinomial
An enterprise has workers divided in 3 groups. $p_i$ is a proportion of the $i$-th group to all, $p_1+p_2+p_3=1$.
Select independently and equiprobably with replacement $n$ workers out of all, $X_i$ workers belong to $i$-th group. $n=X_1+X_2+X_3$. Use $Y_i=X_i/n$ as unknown proportion $p_i$. After find the joint probability function $p(y_1,y_2,y_3)$. If I have large size of sample, does the random variable $Y_i$ converge to the true value $p_i$? If $y_1>p_1$, which is more probable, $Y_2>p_2$ or $p_2>Y_2$?and what is a conditional mean E($Y_i/y_1$),(i=2,3)?
AI: The first part can be settled by just thinking about the situation. If $y_1>p_1$, that means that our sample has a larger than "expected" number of workers from group $1$. That means that the total number of workers from groups $2$ and $3$ is less than expected, so with probability $\gt 1/2$ we will have $Y_2\lt p_2$.
For the conditional expectation, please see one of the answers to this previous question of yours.
There the conditional distribution of $X_2$ given that $X_1=x_1$ is shown to be binomial, number of trials $n-x_1$, probability of "success" $p_2'=\frac{p_2}{p_2+p_3}=\frac{p_2}{1-p_1}$.
So you can find the conditional expectation of $X_2$ given $X_1=x_1=ny_1$ by using the ordinary formula for the expectation of a binomial. Then multiply by $1/n$ to find the conditional expectation of $Y_2$. |
H: Rational points on singular curves and their normalization
Let $X$ be a curve over a field $k$. Assume that $X$ is geometrically connected, geometrically reduced and stable.
Let $Y\to X$ be the normalization. Is $Y(k) = X(k)$?
AI: The answer is no. Consider the curve $C: y^2=x^2+x^3,$ and let $\pi:\widetilde C \to C$ be the blowup of $C$ at $0.$ Then the fibre $\pi^{-1}(0)=\{p,q\}$ consists of two points, while $\pi$ is an isomorphism everywhere else. Since $\widetilde C$ is a desinglularization of $C,$ it is isomorphic to the normalization of $C.$ |
H: Distribution of the sum of independent r.v.
Assume that $X_1$ and $X_2$ are independent random variables with given distribution $f(.)$ (say Normal distribution with $\mu_i$ and $\sigma_i$).
I am stuck with the calculation of:
$P(\{X_1 \leq a\} \; \cap\; \{X_2 \leq b\} \; \cap\; \{c \leq X_1 + X_2 \leq a+b\} )$
where $c < a+b$. For brevity, let's say $Z_1 = \{X_1 \leq a\}$, $Z_2 = \{X_2 \leq b\}$ and $Z_3 = \{c \leq X_1 + X_2 \leq a+b\}$.
Using the product rule I can write:
$P(Z_1 \cap Z_2 \cap Z_3) = P(Z_1)\cdot P(Z_2|Z_1) \cdot P(Z_3|Z_1\cap Z_2)$
and since $X_1$ and $X_2$ are independent, then $P(Z_2|Z_1) = P(Z_2) $ and $P(Z_1 \cap Z_2) = P(Z_1)\cdot P(Z_2)$.
But now I don't know how to calculate the third factor $P(Z_3|Z_1\cap Z_2)$ since, as far as I understand, $Z_3$ is not independent of $Z_1$ and $Z_2$.
Is there any other way to solve this problem?
Thanks.
AI: If your random variables each have a probability density (like in the case of a normal distribution )and we call these densities $f_1,f_2$, you can just write down your probability as an integral over the density functions on a suitably chosen domain.
\begin{equation}
\int_{A} f_1(x_1)f_2(x_2)dx_1dx_2.
\end{equation}
The domain $A$ should be given by the conditions $x_1 \le a$, $x_2 \le b$ and $c \le x_1+x_2 \le a + b$, where the last inequality is in fact redundant because $ x_1 + x_2 \le a+b$ follows from the individual conditions on $x_1,x_2$. |
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