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H: Is the following True of False?
Provide a proof if true or a counterexample if false:
Let a,b be two integers (not both zero), then the gcd(a,b) divides ay+bx for all for x,y ∈ Z.
I tried with several cases such as gcd(5,10) = 5 and then multiplied by various integers and could not find a counterexamaple. I do not know how to prove this formally. (or if you see a counterexample)
AI: Among other things, $d=\gcd(a,b)$ is a common divisor of both $a$ and $b$. Hence there are integers $a',b'$ such that $a=a'd, b=b'd$. For any $x,y\in\mathbb{Z}$, we have $ay+bx=a'dy+b'dx=d(a'y+b'x)$. Hence $d$ divides $ay+bx$. |
H: Stuck at Extended Euclidean Algorithm to solve equation
I'm trying to solve the following function via the Extended Euclidean Algorithm, but I'm stuck at the last step where I need to sub in sub 2.
d * 7 = 1 (mod 180)
d = 1 / 7 (mod 180)
d = 7-1 (mod 180)
180 = 7 * 25 + 5
7 = 5 * 1 + 2
5 = 2 * 2 + 1
1 = 5 – 2 * 2
sub 2 = 7 – 5
1 = 5 – (7 – 5) * 2
How should I simplify this now? Any help would be appreciated.
AI: By Euclidean algorithm (first part):
$$\begin{align*}
180 =& 7\times25 + 5\\
7 =& 5\times1 + 2\\
5 =& 2\times2 + 1
\end{align*}$$
Then the second part of Extended Euclidean algorithm:
$$\begin{align*}
1 =& 5 - 2\times2\\
=& 5 - (7-5\times1)\times2 &&\text{Substitute }2 = 7 - 5\times1\\
=& 7\times(-2) + 5 \times3 &&\text{Group }7\text{ and }5\text{ terms}\\
=& 7\times(-2) + (180-7\times25) \times3 &&\text{Substitute }5 = 180 - 7\times25\\
=& 180\times3 - 7\times77 &&\text{Group }180\text{ and }7\text{ terms}
\end{align*}$$
Now, you should use this result to obtain $7^{-1} \pmod{180}$. |
H: Prove: If $|a_n|$ doesn't converge to $\infty$ then $a_n$ must have a finite partial limit.
Prove: If $|a_n|$ doesn't converges to $\infty$ then $a_n$ must have a finite partial limit.
My thoughts:
if $|a_n|$ doesn't converges to $\infty$ there must be two other posibilities:
$|a_n|$ converges to a finite number, $L$
$|a_n|$ doesn't have a limit at all (neither finite, nor $\infty$)
for option #1, we can infer $a_n$ is bounded by $L$ and has a converges parital limit (by BW Theorem).
EDIT:
The correct demand is to prove $a_n$ has a finite partial limit.
for option #1 I proved it using BW Theorem.
Now I need to prove $a_n$ has a partial finite limit for option #2
AI: If it doesn't converge to infinity then it must have a bounded subsequence(if it had no bounded subsequence, it would converge to infinity), this bounded subsequence then has a convergent subsequence, which itself is a subsequence of the original sequence. |
H: Just need my math checked! Integration!
Let $b \in \mathbb{R}$, and define $f:[0,2]\to\mathbb{R}$ by $$f(t)=\begin{cases}t&\text{if $0 \leq t < 1$}\\b-t^2&\text{if $1 \leq t \leq 2$}\end{cases}$$ and let $$F(x)=\int_0^xf(t)\,dt.$$ $(\text{a})$ Find a formula for $F(x)$.
$(\text{b})$ For what value(s) of $b$ is $F$ differentiable for all $x\in[0,2]$?
(a) I got $F(x)=\frac{1}{3}x^3 + bx - b - \frac{1}{6}$.
(b) I got $b = 2$, but this will only be the case if $F(x)$ is correct.
Please let me know how I did!
AI: Hints: 1) find $\int t\ dt$ and $\int (b-t^2)\ dt$ Use them to find $F(x)$. You are correct that $F(x)$ will have different expressions for $x \in [0,1]$ and $x \in (1,2]$. It will be continuous at $x=1$, but the slopes on either side will not match unless $b$ is chosen correctly. |
H: Cool simple solution to: $-A^2$ is not the identity matrix
this is not a question per se, just a simple cool solution to a potentially difficult question, that I want to share. I liked it.
The question is:
Let $A$ be a $3\times3$ matrix with real values.
Show that $A^2 \neq -I_3$
There are probably many solutions for this problem, but i really thought this one was simple.
Solution is posted down below.
AI: Solution:
Let's suppose that $A^2 = -I_3$, then $A^2+I_3=0$
That means that $A$ is a root of the polynomial $t^2+1$.
However, since we are dealing with real values only, this polynomial has no roots, and so it is irreducible in $\mathbb R$.
But it is still a polynomial that $A$ is a root of, and so, from the definition of the minimal polynomial (let's call it $m_A$):
$m_A|t^2+1$ but $t^2+1$ is irreducible and so $m_A=t^2+1$
Invoking the Cayley-Hamilton theorem, we know that the characteristic polynomial of $A$ , $p_A$ can be divided by the minimal polynomial, and it has the same roots, so we can say that:
$p_A =(m_A)^r = (t^2+1)^r$, $r \in \mathbb N$
But since $A$ is a 3-by-3 matrix, the characteristic polynomial should be in a degree of 3:
$deg(p_A)=deg((t^2+1)^r) = 2r = 3$ so 3 is an even number. Contradiction! |
H: Is every complex (smooth) manifold a scheme?
The question in the title doesn't quite make sense. I was always wondering if the scheme is the generalization of manifold. The precise statement should be like following:
If $X$ is a complex (smooth)manifold, is there a scheme $Y$ over $\mathbb{C}$ such that the corresponding analytic space $Y^{an}$ isomorphic to $X$?
I guess if $X$ is projective, then this is Chow's theorem. And I doubt the question has an affirmative answer if $X$ is non-compact (I guess that $X: y-e^x=0$ in $\mathbb{C}^2$ might be an example, but I do not know how to show this).
Edit 1: There are also example (as @hunter pointed out) of a complex torus but not an abelian variety shows that a compact manifold might not be a variety, however, is it possible to be a scheme ?(according to Hartshorne, the category of varieties is a subcategory of quasi-projective, integral, separated scheme. But could there be some non-quasiprojective scheme($Y$ has to be integral, separated in our case according to GAGA, I guess) )?
Edit 2: Another question I think of is: is (Artin) Stack a generalization of manifold? I really want to know if manifold could be think of some subcategory of some category of with "algebraic objects".
AI: This is a very interesting question and it doesn't have an easy answer. I will just give an overview over some facts.
A necessary condition for $X$ to be algebraic (i.e. it is the analytification of a scheme) is that the transcendence degree of its field of meromorphic functions $K(X)$ is equal to the dimension of $X$ (because in algebraic geometry those two numbers are always equal for nice schemes).
It is not enough for $X$ to be compact, for instance the general complex torus of dimension $\geq 2$ (all of which are compact) has no non-constant meromorphic functions at all.
On the other hand, a non-compact analytic manifold may be algebraic. This is the case e.g. for the complex line $\mathbb{C}$.
The condition $\mbox{trdeg}_{\mathbb{C}} K(X) = \dim(X)$ is also not sufficient. Compact manifolds having this property are called Moishezon manifolds but not all of them are algebraic (a famous example is called Hironaka's example). The nice thing is that all of them come from something called 'algebraic spaces' which is a generalization of a scheme.
There are also various other properties a complex manifold can have. It can be a Kähler manifold (i.e. being equipped with a special metric) and together with being Moishezon it implies being projective which is equal to being projective algebraic (this is Chow's theorem). Also there is the notion of Hodge manifold (a special kind of Kähler manifold) which is actually equivalent to being projective. This is called the Kodaira embedding theorem.
If you want to know more about all this, I recommend you have a look at the book 'Complex geometry' by Daniel Huybrechts. All the terms I used (except for algebraic spaces) are explained there. |
H: Inverses in the homotopy classes of maps into $RP^{\infty}$
One can define bilinear maps $\mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R}^{2n-1}$ by considering the elements in $\mathbb{R}^n$ as polynomials and doing multiplication. This defines an $H$-space structure on $RP^{\infty}$ since elements of this space have zeros after a finite number of entries in homogeneous coordinates. Apparently, this defines a group structure on the set $[X, RP^{\infty}]$ of homotopy classes of maps from a space $X$ into $RP^{\infty}$. I am having a hard time seeing how inverses work in this group and I can't seem to find any information in books or online. Any thoughts would be much appreciated!
AI: Not a solution, just a little long for a comment.
I think there is a theorem of Stasheff to the effect that a stricly associative (?) H-space $S$ whose set of path components $\pi_0S$ is a group (under the induced map $\mu_*$, where $\mu:S\times S\to S$ is multiplication), that such an H-space is homotopy equivalent (as an H-space) to a loop space, thus is a group object.
I don't know how this gives the inverse map here, but the hypothesis are certainly verified, as $\Bbb RP^{\infty}$ with your multiplication is a strictly associative (stricly commutative, stricly unital) path-connected H-space. Thus the homotopy sets $[X,\Bbb RP^{\infty}]$ are actually commutative groups (and not just monoids.)
It may or may not help to use the fact that $\Bbb RP^{\infty}$ is a $K(\Bbb Z/2\Bbb Z,1)$; one would have to relate the multiplicative structures.
I seem to remember this is mentioned early on in Adam's book ''Infinite Loop Spaces'', and probably in Stasheff's articles ''Homotopy Associativity for H-spaces I, II'' which are available online for free.
Maybe an answer?
It seems not too unreasonable (at first glance) that $f$ is its own homotopy inverse. For
$f^2$ is (intuitively, this is by no means a proof), at every point $x\in X$, a line in $\Bbb R^{\infty}$ generated by a vector whose greatest index with non zero coefficient is carried by a positive coefficient ($a_n^2$), and one might hope for a ''linear homotopy'' from $f^2$ to the constant map equal to $\Bbb R\in\Bbb RP^{\infty}$.
I think the following is a proof :
In a topological group $G$, multiplication of $G$ induces the same map as concatenation of loops. It seems to me this is true in a strictly unital H-space. Let $a,b$ be loops based at $1=1_{\Bbb RP^{\infty}}=\Bbb R\in\Bbb RP^{\infty}$, and let $[1]$ be the constant loop at $1$. Then $b$ is homotopic (as a loop) to $[1]\star b$, and similarly $a$ is homotopic (as a loop) to $a\star [1]$, and thus
$$\mu(a,b)\simeq\mu(a\star [1],[1]\star b)=a\star b$$
(where we have used the fact that $1$ is a strict unit) so that multiplication $\mu_*$ is equal to multiplication in $\pi_1(\Bbb RP^{\infty},1)$.
(Edit) More precisely, if $(X,e)$ is a pointed space equipped with a multiplication $\mu:X\times X\to X$ that satisfies that the two maps $X\to X$ defined by $x\mapsto\mu(x,e)$ and $x\mapsto\mu(e,x)$ are homotopic rel. $e$ to $\mathrm{id}_{X}$, then $\mu_*:\pi_1(X,e)\times\pi_1(X,e)\to\pi_1(X,e)$ is product of paths. (End of Edit)
This shows that $\mu(f,f)_*$ induces the trivial map on $\pi_1$, since $\pi_1(\Bbb RP^{\infty})=\Bbb Z/2$ (and on all other homotopy groups since they are trivial for projective space.)
When $X$ is a CW complex (or any space that is locally connected in the suitable sense), $\mu(f,f)$ lifts to a map $\widetilde{\mu(f,f)}:X\to\Bbb S^{\infty}$ (using the standard facts about covering spaces). The infinite sphere is contractible, and so $\widetilde{\mu(f,f)}$ is homotopic to a constant map, and so is its projection $\mu(f,f)$.
This shows that $f$ is its own homotopy inverse. |
H: Is the following expression a tautology?
$\forall x\,(P(x)\rightarrow Q(x))\rightarrow (\exists y\,P(y)\rightarrow\exists z\,Q(z))$
I believe the sentece is a tautoloogy. Can someone confirm?
AI: $\forall x\,(P(x)\rightarrow Q(x))\rightarrow (\exists y\,P(y)\rightarrow\exists z\,P(z))$ is trivially a logical truth, since $C = (\exists y\,P(y)\rightarrow\exists z\,P(z))$ is a trivial logical truth, and if $C$ is a logical truth, so is $A \to C$ for any $A$.
You probably meant, however, to ask whether $\forall x\,(P(x)\rightarrow Q(x))\rightarrow (\exists y\,P(y)\rightarrow\exists z\,Q(z))$ is a logical truth. But yes, this is too. If any $P$ (should there be one) is a $Q$, then if there is a $P$ then there is a $Q$. |
H: determine basis for topology on $\mathbb{R}^2$
Determine whether the collection of subsets below form a basis for a topology on $\mathbb{R}^2$.
All subsets of the form $T_{\epsilon}(x)=\lbrace (y_1,y_2) : |x_1+x_2-y_1-y_2| <\epsilon \rbrace$ for all $x \in \mathbb{R}^2$ and all $\epsilon>0$
By doing some algebra, we obtain $-\epsilon -(x_1+x_2) < y_1+y_2<\epsilon -(x_1+x_2)$, which means that the set is the region bounded by two straight lines with negative gradient and y-intercept $-\epsilon -(x_1+x_2)$ and $\epsilon -(x_1+x_2)$ respectively.
The answer given is that the collection of the subsets above does not form a basis for topology on $\mathbb{R}^2$. Why? I thought every point is contained in one of the subsets above and also intersection between any two subsets is again the region bounded by two straight lines. I don't see why it fails to be a basis.
AI: You have a sign wrong: $T_\epsilon(x)$ is the set of $y=\langle y_1,y_2\rangle$ satisfying
$$(x_1+x_2)-\epsilon<y_1+y_2<(x_1+x_2)+\epsilon\;.$$
However, you’re right: $\{T_\epsilon(x):x\in\Bbb R^2\text{ and }\epsilon>0\}$ is a base for a topology $\tau$ on $\Bbb R^2$.
In fact, let $f:\Bbb R^2\to\Bbb R:\langle x,y\rangle\mapsto x+y$; then the open sets in $\langle\Bbb R^2,\tau\rangle$ are precisely the sets $f^{-1}[U]$ such that $U$ is an open set in the usual topology on $\Bbb R$. In fact,
$$T_\epsilon(x)=f^{-1}[(f(x)-\epsilon,f(x)+\epsilon)]\;.$$ |
H: Calculus series homework with variable a
I have a series:
$$\sum_{n=1}^{\infty}(-1)^n\sin\frac{a}{n}$$
I'm supposed to check how it behaves with different $a$ values. First I would check if it converges absolutely:
$$\lim_{n\to \infty}\left| (-1)^n\sin\frac{a}{n}\right| \sim \lim_{n\to \infty} \left| (-1)^n\frac{a}{n}\right|=\lim_{n\to \infty} \frac{a}{n}$$
That limit equals $0$ if $a\in \mathbb{R}$, which means that this series always diverges absolutely.
How does it behave not absolutely?
AI: The series is clearly convergent for $a=0$.
Now let $a\ne0$ so we have by the Taylor series
$$(-1)^n\sin\left(\frac{a}{n}\right)=\underbrace{(-1)^n\frac{a}{n}}_{=u_n}+\underbrace{O\left(\frac{1}{n^3}\right)}_{=v_n}$$
The series $\sum_n u_n$ is convergent by the Leibniz theorem and the series $\sum_n v_n$ is convergent by comparison with the Riemann series. Hence the given series is convergent for all $a$.
The series isn't absolutely convergent for $a\ne0$ since
$$\left|(-1)^n\sin\left(\frac{a}{n}\right)\right|\sim_\infty \frac{|a|}{n}.$$ |
H: Give an example of relation $R$ and $S$ on $A$ such that $R$ and $S$ are nonempty, and $R \circ S$ and $S \circ R$ are empty
Let $A = \left \{a, b, c, d\right \}$, give an example of relation $R$ and $S$ on $A$ such that $R$ and $S$ are nonempty, and $R \circ S$ and $S \circ R$ are empty
I'm thinking of ways that a set could be empty, and the only thing I can think of is that if two sets are disjoint, then their intersection is empty. However, I'm not sure as to how to go about coming up with the relations. Could someone give me a hint.
AI: How about the simplest relations you can think of: $R=\{(a,a)\}$ and $S=\{(b,b)\}$
Why are $R\circ S$ and $S\circ R$ empty. Try to generalize. |
H: Several part sum of power series question
(a) Prove that $\sum_{j=0}^{\infty}x^j$ is differentiable on $(-1,1)$ and
$\frac {d}{dx}\sum_{j=0}^{\infty}x^j = \sum_{j=0}^{\infty}(j+1)x^j$.
(b) Use the fact that $\sum_{j=0}^{\infty}x^j = \frac {1}{1-x}$ on $(-1,1)$ to find a formula for $\sum_{j=0}^{\infty}(j+1)x^j$
(c) Use this to calculate $\sum_{j=0}^{\infty}\frac {j+1}{2^j}$ exactly.
work for part (a)
We can represent $\sum_{j=0}^{\infty}x^j$ with an integral, therefore it is differentiable?
Do I just need to prove that $jx^{j-1}=(j+1)(x^j)$?
In which case $j=\frac{x^2}{1-x}$..?
(b) Using $j=\frac{x^2}{1-x}$
and maybe using $\frac{d}{dx}\frac{1}{1-x} = \frac{1}{(1-x)^2}$..
Not sure how to put that all together though :(
(c) Do I use $j=\frac{x^2}{1-x}$ and then just plug and chug?
I'm really not sure how I'm supposed to go about doing this as you can tell.. :/
AI: a) The radius of convergence of the given series is $R=1$ . This can be computed by Cauchy-Hadamard formula.
Then $f(x)=\sum_{j=0}^{\infty}x^j$ is continuous on $(-1,1)$, and differentiable on $(-1,1)$ except possibly at its endpoints.
b) You're right . Taking derivative both of sides of your formula gives you the desired result.
c) Subsitute $x=\frac{1}{2}$
You can read the proof of general result here, page 9.
http://www.math.ucdavis.edu/~hunter/m125a/intro_analysis_ch6.pdf |
H: Number of equivalence relations on a set with fixed class
For A={a,b,c,d,e,f}, how much equivalence relations can we get if a,b and c are in relation?
The total is: $\sum_{k=1}^6 S(6,k)$. But since a,b and c are already in the same class, i would say the answer is $\sum_{k=1}^3 S(3,k)$.
I know that answer is not ok, but why exactly?
AI: Since $a,b$ and $c$ are together in one equivalence class, we can think of them as a single object, which I’ll call $x$; the problem then is to count the number of equivalence relations on $A'=\{x,d,e,f\}$. For instance, the equivalence relation on $A'$ whose classes are $\{x,d\}$ and $\{e,f\}$ corresponds to the equivalence relation on $A$ whose classes are $\{a,b,c,d\}$ and $\{e,f\}$. Thus, you’re simply counting the equivalence relations on a $4$-element set, and you want the Bell number
$$B_4=\sum_{k=0}^4{4\brace k}=\sum_{k=0}^4S(4,k)\;.$$ |
H: How to Prove Below Inverse Sin
How to prove below equation .
$$ \sin^{-1}\frac{3}{5}+\sin^{-1}\frac{8}{17} = \sin^{-1}\frac{77}{85}.$$
I am not able to prove the above equation how can we prove it.
As we know
$ \sin^{-1} = y$ if $\sin y = x$ where $-1\leq x\leq 1, -\pi/2 \leq y \leq \pi/2.$
AI: Remark: The question has changed from $\sin^{-1}(5/13)+\sin^{-1}(3/5)$ to
$\sin^{-1}(3/5)+\sin^{-1}(8/17)$. There is no good reason to change the answer below, since the process is basically the same.
Hint: Use the "sum" formula $\sin(a+b)=\sin a \cos b+\cos a\sin b$.
If $a$ is the angle between $0$ and $\pi/2$ whose sine is $\frac{5}{13}$, then the cosine of $a$ is $\frac{12}{13}$, since always $\cos^2 x+\sin^2 x=1$.
If $b$ is the angle between $0$ and $\pi/2$ whose sine is $\frac{3}{5}$, then the cosine of $b$ is $\frac{4}{5}$.
You will also need to show that the sum on the left is $\lt \pi/2$.
Remark: The notation $\sin^{-1} t$ is a fairly frequent source of confusion. It means the angle between $-\pi/2$ and $\pi/2$ whose sine is $t$. It does not mean $\frac{1}{\sin t}$. So for example $\sin^{-1}(1)=\pi/2$ and $\sin^{-1}(1/2)=\pi/6$. |
H: Degrees of interpolating polynomials
Given a collection of $m+1$ points $\{(x_0,y_0), (x_1,y_1), ..., (x_m,y_m)\}$, we can form the interpolating Lagrange polynomial $L(x)$: $$ L(x) = \sum_{i = 0}^{m} y_i l_i(x) \\ l_i(x) = \prod_{\substack{0 \le k \le m\\ k \ne i}} \frac{x - x_k}{x_i - x_k} $$ and it will be the unique polynomial of degree at most $m$ interpolating these points. But I'm curious:
When is this maximum degree actually obtained? When is there instead cancellation among the coefficients upon multiplying everything out?
For a rather extreme example, take any set of points on the parabola $f(x) = x^2$. This 2nd-degree polynomial trivially interpolates all of them, and if I were to use the above construction, things would always reduce down upon multiplying out, regardless of how many points (and terms) I start with.
In the other direction, can I say the following?
Any Lagrange polynomial interpolating a set of $m + 1$ points can't have degree $n < m$ if there exists a (different!) Lagrange polynomial of degree $n$ interpolating some subset of $n+1$ of those points, since this would violate uniqueness. So there seems to be some kind of recursiveness involved here: if I take any subset of $n+1$ points from the original $m+1$, and I interpolate on these instead, then the resulting Lagrange polynomial (which is of degree at most $n$) either interpolates all $m+1$ points, or these $m+1$ points require a polynomial of degree strictly higher than $n$.
If this is correct, does it get me anywhere towards answering the above question? I can't really see how, or how to approach it at all, really.
AI: The basic answer is there is almost never cancellation. Following your example of $y=x^2$, you have the data points $(0,0),(1,1),(2,4)$ which it interpolates. Now if we pick one more point, say $(3,y)$ there will only be cancellation if $y=9$. For any other value, the interpolating polynomial will be of third degree. The same is true for $n$ points. There is a unique (at most) $n-2$ degree polynomial which interpolates the first $n-1$ points. There will only be cancellation if that polynomial passes exactly through the $n^{\text{th}}$ point. |
H: Prove that for any square matrix, an invertible matrix B exists, so that BA is triangular
I'm given a matrix A, its dimensions are n x n.
I am required to prove that an invertible matrix B exists, such that the product of the matrices BA is triangular.
Any help?
AI: If $A=QR$ is a QR factorisation of $A$, then $Q^*A=R$. So the statement is true with $B=Q^*$. |
H: Remainder Question
What process do I use to show what is the remainder when 14 × 7^36 + 92 when divided by 8?
Is it the same to show the remainder of 5^2003 when divided by 7?
I tried out the problem using congruent modulo but cannot help
AI: HINT:
For the first $7^2=49\equiv1\pmod8\implies 7^{2n}\equiv1$ where $n$ is any integer $\ge0$
For the second $5^3=125\equiv-1\pmod7$ and $2003=667\cdot3+2$
$\displaystyle\implies 5^{2003}=(5^3)^{667}\cdot5^2\equiv(-1)^{667}\cdot5^2\pmod 7$ |
H: What is the hitting time distribution for white noise?
What is the distribution of the hitting time for a stochastic process
$(W_t)_{t\in [0,T]}$,
where $W_t$ are i.i.d. Gaussian random variables?
How about in cases, in which $W_t$ are i.i.d. with a common distribution other than Gaussian?
AI: If $H=\inf\{t\in\mathbb R_+\mid W_t\in B\}$ for some Borel set $B$ and if $(W_t)_{t\in\mathbb R_+}$ is i.i.d. with a distribution such that $P[W_1\in B]\ne0$, then $H=0$ almost surely.
This is because, before every time $t$, one already had infinitely many chances to hit $B$. Since $W_s\in B$ for each $s\leqslant t$ with positive probability, $H$ happens before $t$ almost surely, for every positive $t$. QED. |
H: Proving inequalities by induction
I'm having trouble understand the inductive when proving inequalities; Here's an example:
Show that $2^n \gt n^2 $ for any integer $n \gt 4 $.
Well for the basis $n=5$, it shows: $32>25$
Now, assume: $2^{n+1} \gt (n+1)^2 $ for some integer $n$.
Well, RHS:
$$(n+1)^2 = n^2 + 2n +1$$
And the problem starts there. I cannot seem to "pull" out my inductive hypothesis from here. Any input?
AI: First off, your induction hypothesis should be that $2^n>n^2$, not that $2^{n+1}>(n+1)^2$; you should be using the induction hypothesis to prove that $2^{n+1}>(n+1)^2$.
To do this, notice that $2^{n+1}=2\cdot 2^n$, so if $2^n>n^2$, as we are assuming, then $$2^{n+1}=2\cdot 2^n>2n^2\;.$$ You want to show that $2^{n+1}>(n+1)^2$; if you could show that $2n^2\ge(n+1)^2$, you’d be done, because then you’d have
$$2^{n+1}=2\cdot2^n>2n^2\ge(n+1)^2$$
and hence $2^{n+1}>(n+1)^2$. Thus, it all boils down to showing that $2n^2\ge(n+1)^2$, i.e., that $2n^2\ge n^2+2n+1$. This, of course, is equivalent to showing that $n^2\ge 2n+1$, i.e., that $n^2-2n-1\ge 0$. Now $n^2-2n-1=(n-1)^2-2$, and this is greater than or equal to $0$ provided that $n$ is ... what? |
H: equivalent functions?
I have this two functions in ($1<x<50$)
$y = -1/x$
and
$ y = \frac{x - \sqrt{x^2+4}}{2} $
why this are very similar ?
AI: The reason is that $$\sqrt{4+x^2}=x+\frac{2}{x}-\frac{2}{x^3}+O((\frac{1}{x})^5)$$
Plugging this into your equation $y=-\frac{1}{x}+\frac{1}{x^3}+O((\frac{1}{x})^5)$. |
H: $k$ colorings of the non empty subsets of $[n]$ gives the same color to two disjoint sets and their union.
This question was already asked but I didn't get enough information from the answer. Here is a link to the question.
Here is the question restated. Show that for $n$ large enough, every $k$ coloring of the non empty subsets of $[n]$ will give the same color to two disjoint sets and their union.
This is my idea so far. I want to choose $n=R_{k}(3:2)$. Defined to be the smallest $n$ such that any $k$ coloring of the edges of $K_{R_{k}(3:2)}$ yields a monochromatic triangle. Given any k-coloring $f$ on the powerset of $[n]$. Define an edge coloring $f'$ on $K_{n}$ by, $f'(ab)=f(\{a,b\})$.
This produces a monochromatic triangle. So there exists $a,b,c$ such that $\{a,b\},\{b,c\}\{a,c\}$ all have the same color. I was hoping to execute this sort of coloring on the two sets of $[n]$ then the three sets and so on until I have enough triangles to get some disjoint sets but it didn't work.
Any hint would be very nice. Thank you
AI: Define $f'(ab)=f(\{a+1,a+2,\dots,b\})$ for $0\le a\lt b\le n$. This colors the edges of a complete graph on $n+1$ vertices, so you can take $n=R_{k}(3:2)-1$. |
H: Proofs from the Book - need quick explanation
I've been recently reading this amazing book, namely the chapter on Bertrand's postulate - that for every $n\geq1$ there is a prime $p$ such that $n<p\leq2n$.
As an intermediate result, they prove that $\prod_{p\leq x}p \le 4^{x-1}$ for any real $x\geq2$, where the product is taken over all primes $p\leq x$ . While proving that, they rely on the inequality
$$
\prod_{m+1<p\le2m+1}p\leq\binom{2m+1}{m},
$$
where $m$ is some integer, $p$'s are primes.
They explain it by observing that all primes we are interested in are contained in $(2m+1)!$, but not in $m!(m+1)!$. The last part is what I don't understand.
I can understand how this principle can be applied to the bound $(2m+1)!/(m+1)! = (m + 2)\ldots(2m+1)$, but why can we safely divide this by $m!$?
Thank you!
AI: Not sure if you're analysing too much for the last part?
If we look at the inequality
$$
\prod_{m+1<p<=2m+1}p\leq\binom{2m+1}{m},
$$
we see that for any prime $p \in (m+1,2m+1]$, we have $p|(2m+1)!$ but $p\nmid m!$ and $p \nmid (m+1)!$. ["The last part" that you mention is simply because $p > m+1$.]
So $p$ is indeed a factor of the numerator of $\binom{2m+1}{m}$ but not in its denominator, which proves the inequality. |
H: Showing that $\int_0^{\pi/4} \frac{1-\cos{16x}}{\sin{2x}}\,\mathrm{d}x=\frac{176}{105}$
Wolfram Alpha tells me that $$\int_0^{\pi/4} \frac{1-\cos{16x}}{\sin{2x}}\,\mathrm{d}x=\frac{176}{105}$$
What are some quick/elegant ways of proving this?
AI: $$1-\cos(16x) = 2\sin^2(8x) = 8 \sin^2(4x) \cos^2(4x) = 32 \sin^2(2x) \cos^2(2x) \cos^2(4x)$$
Hence,
$$I = \int_0^{\pi/4}32 \sin(2x) \cos^2(2x) \cos^2(4x) dx = \int_0^{\pi/4}32 \sin(2x) \cos^2(2x) (2\cos^2(2x)-1)^2 dx$$
Now let $\cos(2x) = t$ and compute. |
H: Is there a power series which converges to $f(x) =| x|$ for all $x$?
I'm confused how to solve the following problem:
"Is there a power series which converges to $f(x)$ = $\left| x\right|$ for all $x$?"
Your help is greatly appreciated. Thanks a lot!
AI: No. Any power series defines an everywhere-differentiable function, but $|x|$ is not differentiable at $0$. |
H: Maths-Physics question, can I solve this situation for $x$?
So Let's say I have an object going at velocity $V$, initially. Each second, the current velocity $v$ is reduced by $v/x$ . After $250$ (arbitrary) seconds the velocity has been reduced to below/equal $0.01$ (arbitrary small amount). What is $x$?
The only formula I can really come up with describes only a single time-interval-step.
vnew = vold - (vold/x)
But I would like to have a formula where I can plug in:
-Initial velocity, Vinitial
-Number of time-steps, n
-Final velocity, Vfinal
...and get $x$.
(I hope this isn't too Physics-y for y'all. I'm not even sure if this can be done, but I thought that coming up with abstract formulas like this is more a mathematically-inclined thing so that's why I'm posting here.)
AI: Assuming that you mean that each step down in speed happens exactly each second which is strange.
$$V_{n+1} = V_{n}(1-\frac{1}{x})$$
Therefore:
$$V_{n+k} = V_{n}(1-\frac{1}{x})^k$$
and
$$V_{f} = V_{i}(1-\frac{1}{x})^n$$ |
H: what is so great about having an invariant measure?
I am a student who just started to learn basic concepts of ergodic theory.
It seems like that given a dynamical system, people are very excited to find various invariant measures of the system. But the books I am reading doesn't really convince me why it is good to have invariant measures.
For example, the Gauss map on the unit interval $x \mapsto \{ 1/x \}$ has the invariant measure $ 1/({1+x})$. What kind of effective results can we prove about the Gauss map using this invariant measure?
AI: One important case is when you have a probability measure, in which case if a map has an invariant measure, then it preserves probability. This is the proper setting for the Birkhoff ergodic theorem, which presumably you will soon learn. |
H: Finding a limit , dyadic pavings
I need to show that the following limit equals $\pi/4$ :
$$\lim_{k \to \infty}\sum_{n=1}^{2^k-1}\frac{ \left\lfloor\sqrt{4^k-n^2}\right\rfloor\ }
{2^{2k}}$$ I don't know if it is even possible to do so. I was trying to prove that it is possible to pave a unit disk with dyadic squares which intersect each other along their boundaries and whose total area exceed $\pi$-$\epsilon$ for any $\epsilon>0$. So I found a paving for the unit disk in the first quadrant. The total area of the paving was this sum where k shows how fine the partition is. The value of this sum when $k$ is $13$ approximates $\pi/4$ up to three decimal places.
AI: Consider the closely related sum
$$\sum_{n=0}^{2^k-1} \frac{\sqrt{4^k -n^2}}{2^{2k}}.\tag{1}$$
We can rewrite (1) as
$$\sum_{n=0}^{2^k-1} \frac{1}{2^k}\sqrt{1-\left(\frac{n}{2^k}\right)^2}.\tag{2}$$
We recognize Sum (2) as an upper Riemann sum for $\int_0^1 \sqrt{1-x^2}\,dx$, with the unit interval divided into $2^k$ equal-sized subintervals.
The given sum differs from Sum (1) in two ways: (i) The term corresponding to $n=0$ is missing and (ii) The floor function changes each term by at most $\frac{1}{2^{2k}}$.
For large (or even medium-sized) $k$, that makes little difference. The missing $n=0$ term changes things by $\frac{1}{2^k}$, and the floor function changes the sum by at most $(2^k-1)\frac{1}{2^{2k}}$.
Thus the Riemann sum (1) differs from the given sum by at most $\frac{1}{2^k}$. Since the Riemann sum has limit $\frac{\pi}{4}$ as $k\to\infty$, so does the sum of the problem. |
H: Equal balls in metric space
Let $x$ and $y$ be points in a metric space and let $B(x,r)$ and $B(y,s)$ be usual open balls. Suppose $B(x,r)=B(y,s)$. Must $x=y$? Must $s=r$?
What I got so far is that: $$r \neq s \implies x \neq y$$ but that's it.
AI: No, it’s not necessary that $x=y$ or that $s=r$. Consider the discrete metric $d$ on a set $X$:
$$d(x,y)=\begin{cases}
0,&\text{if }x=y\\
1,&\text{if }x\ne y\;.
\end{cases}$$
Then $B(x,r)=B(x,s)=\{x\}$ whenever $0<r,s\le 1$, and $B(x,r)=B(y,s)=X$ whenever $r,s>1$.
Added: If $\langle X,d\rangle$ is an ultrametric space, then $B(x,r)=B(y,r)$ whenever $d(x,y)<r$. An example is the product space $\{0,1\}^{\Bbb N}$, where $\{0,1\}$ has the discrete topology, and for $x=\langle x_n:n\in\Bbb N\rangle$ and $y=\langle y_n:n\in\Bbb N\rangle$ we define
$$d(x,y)=\begin{cases}
0,&\text{if }x=y\\
2^{-m(x,y)},&\text{if }x\ne y\;,
\end{cases}$$
where $m(x,y)=\min\{n\in\Bbb N:x_n\ne y_n\}$. For any finite sequence $\langle x_0,\ldots,x_{n-1}\rangle$ of zeroes and ones, if $y=\langle y_k:k\in\Bbb N\rangle$ and $z=\langle z_k:k\in\Bbb N\rangle$ are such that $y_k=z_k=x_k$ for $k<n$, and $2^{-(n+1)}<r,s\le 2^{-n}$, then $B(y,r)=B(z,s)$. |
H: $\int_{|z|=1} \frac{f(z) }{z-a} \, dz = 0$ for $f(z)=\sin \pi/z$.
Let $f$ be analytic for all $z$ where $0 < |z| < 2$ and $a \in \mathbb{C}$ is in this domain as well. I wish to prove that $$\int_{|z|=1} \frac{f(z) }{z-a} \, dz = 0$$ for $f(z)=\sin \pi/z$. We can rewrite the integral as a line integral
$$\int_{|z|=1} \frac{f(z) }{z-a} \, dz = i \int_0^{2\pi} \frac{\sin \left( \pi \mathrm{e}^{-it} \right) }{1-\mathrm{e}^{-it}a} \, dt.$$
To show that this is zero, we can expand the integrand into a power series using geometric series
$$\frac{\sin \left( \pi \mathrm{e}^{-it} \right)}{1-\mathrm{e}^{-it}a} = \sum_{n=0}^{\infty}a^n \mathrm{e}^{-int}\sin \left( \pi \mathrm{e}^{-it} \right).$$
So I want to show that
$$ \int_0^{2\pi} \sum_{n=0}^{\infty}a^n \mathrm{e}^{-int}\sin \left( \pi \mathrm{e}^{-it} \right) \, dt=0.$$
In order to be able to switch the summation and integration, I need to make sure the summand converges uniformly, right? So I wish to apply Weierstrass' M-test to the summand so that
$$\int_0^{2\pi} \sum_{n=0}^{\infty}a^n \mathrm{e}^{-int}\sin \left( \pi \mathrm{e}^{-it} \right) \, dt = \sum_{n=0}^{\infty}a^n \int_0^{2\pi} \mathrm{e}^{-int}\sin \left( \pi \mathrm{e}^{-it} \right) \, dt.$$ How do I do this?
And to finish the problem, I want $$\int_0^{2\pi} \mathrm{e}^{-int}\sin \left( \pi \mathrm{e}^{-it} \right) \, dt=0.$$ But I don't see how I can prove this.
Can you help?
EDIT: Added info on $f$ and $a$.
AI: Last integral can be rewritten as $$\int_0^{2\pi} \mathrm{e}^{-int}\sin \left( \pi \mathrm{e}^{-it} \right) \, dt=
i\int_0^{2\pi} \mathrm{e}^{-i(n-1)t}\sin \left( \pi \mathrm{e}^{-it} \right) \, d{(e^{-it})}.$$
Denoting $w=e^{-it},$ we have
$$\int_0^{2\pi} \mathrm{e}^{-i(n-1)t}\sin \left( \pi \mathrm{e}^{-it} \right) \, d{(e^{-it})}=\int\limits_{|w|=1}{w^{n-1}\sin{\pi w}\ dw}=0$$
for $n\geqslant{1}$ since integrand is analytic in the unit disc.
Note: The expansion into geometric series is valid only for $|a|<1.$ For $|a|>1$
$$\dfrac{1}{1-\mathrm{e}^{-it}a}=-\dfrac{1}{a\mathrm{e}^{-it}(1-\tfrac{e^{it}}{a})}=-\dfrac{1}{a\mathrm{e}^{-it}}\sum\limits_{n=0}^{\infty}{\left(\frac{e^{it}}{a}\right)^n}.$$ |
H: How to prove $\sum_{n=1}^\infty\operatorname{arccot}\frac{\sqrt[2^n]2+\cos\frac\pi{2^n}}{\sin\frac\pi{2^n}}=\operatorname{arccot}\frac{\ln2}\pi$?
How can I prove the following identity?
$$\sum_{n=1}^\infty\operatorname{arccot}\frac{\sqrt[2^n]2+\cos\frac\pi{2^n}}{\sin\frac\pi{2^n}}=\operatorname{arccot}\frac{\ln2}\pi$$
AI: Rewrite the sum as
$$\sum_{n=1}^{\infty} \arctan{\left [\frac{\sin{\left (\frac{\pi}{2^n}\right)}}{2^{2^{-n}}+\cos{\left (\frac{\pi}{2^n}\right)}}\right ]}$$
Now let
$$a_n = \arctan{\left [\frac{\sin{\left (\frac{\pi}{2^n}\right)}}{2^{2^{-n}}-\cos{\left (\frac{\pi}{2^n}\right)}}\right ]}$$
Then one may show (nontrivial!) that
$$a_n - a_{n-1} = \arctan{\left [\frac{\sin{\left (\frac{\pi}{2^n}\right)}}{2^{2^{-n}}+\cos{\left (\frac{\pi}{2^n}\right)}}\right ]}$$
so that we have a telescoping sum:
$$\sum_{n=1}^{\infty} (a_n-a_{n-1}) = a_{\infty}-a_0$$
where
$$a_{\infty} = \lim_{n\to\infty} a_n $$
Now,
$$\frac{\sin{\left (\frac{\pi}{2^n}\right)}}{2^{2^{-n}}-\cos{\left (\frac{\pi}{2^n}\right)}} \sim \frac{\pi/2^n}{1+\frac{\log{2}}{2^n}-1} = \frac{\pi}{\log{2}}$$
Note also that $a_0=0$. therefore the sum is simply
$$\arctan{\frac{\pi}{\log{2}}}$$
which is the stated result.
ADDENDUM
For completeness, I'll outline a few steps to demonstrate how to prove the difference equation above. Start with the relation
$$\arctan{p}-\arctan{q} = \arctan{\frac{p-q}{1+p q}}$$
so we need to work with the following argument of the arctangent:
$$\frac{\frac{\sin{\left (\frac{\pi}{2^n}\right)}}{2^{2^{-n}}-\cos{\left (\frac{\pi}{2^n}\right)}} - \frac{\sin{\left (\frac{2\pi}{2^n}\right)}}{2^{2^{1-n}}-\cos{\left (\frac{2\pi}{2^n}\right)}}}{1+\frac{\sin{\left (\frac{\pi}{2^n}\right)}}{2^{2^{-n}}-\cos{\left (\frac{\pi}{2^n}\right)}}\frac{\sin{\left (\frac{2\pi}{2^n}\right)}}{2^{2^{1-n}}-\cos{\left (\frac{2\pi}{2^n}\right)}}}$$
which simplifies somewhat to
$$\frac{-2^{2^{-n}} \sin \left(\pi 2^{1-n}\right)+2^{2^{1-n}} \sin \left(\pi
2^{-n}\right)+\sin \left(\pi 2^{1-n}\right) \cos \left(\pi 2^{-n}\right)-\sin
\left(\pi 2^{-n}\right) \cos \left(\pi 2^{1-n}\right)}{2^{3\ 2^{-n}}+\sin
\left(\pi 2^{1-n}\right) \sin \left(\pi 2^{-n}\right)-2^{2^{-n}} \cos
\left(\pi 2^{1-n}\right)+\cos \left(\pi 2^{-n}\right) \cos \left(\pi
2^{1-n}\right)-2^{2^{1-n}} \cos \left(\pi 2^{-n}\right)}$$
Now you may show that the numerator is equal to
$$\sin \left(\pi 2^{-n}\right) \left(2^{2^{1-n}}-2^{2^{-n}+1} \cos \left(\pi
2^{-n}\right)+1\right)$$
and the denominator is equal to
$$2^{2^{-n}} \left(2^{2^{1-n}}-2^{2^{-n}+1} \cos \left(\pi
2^{-n}\right)+1\right)+\cos \left(\pi 2^{-n}\right)
\left(2^{2^{1-n}}-2^{2^{-n}+1} \cos \left(\pi 2^{-n}\right)+1\right)$$
Cancelling the common factors in the numerator and denominator produces the desired result. |
H: Proving a Sequence's Uniform Convergence
Another homework problem that's been giving me headaches for about a week now.
Prove that the following sequence of functions $(f_n)$ converges uniformly on the interval $[1,2]$:
$$f_n(x) = \frac {nx^2 - 2}{x^4 + nx}.$$
Then, find the following integral on the same interval and justify your answer:
$$ \lim_{n\to\infty} \int_1^2 \frac {nx^2 - 2}{x^4 + nx}\mathrm dx.$$
Any help on this would be appreciated.
AI: $f_n(x) = \frac {nx^2 - 2}{x^4 + nx}$. So make a guess that $f_n \to x$. Then try to prove that for all $\epsilon \gt 0$, there exists natural $N$ such that $n \gt N, x \in [1,2] \implies |f_n(x) - x | \lt \epsilon$.
Well $\frac{nx^2 - 2}{x^4 + nx} - x = \frac{nx^2 - 2 - x^5-n x^2 }{x^4 + nx} = $ top and bottom of $x$ term multiplied by $x^4 + nx. \ $ Then $ = -\frac{x^5 + 2}{x^4 + nx}$. Clearly this can be made arbitrarily small, but let's prove it. Max of $|x^5 + 2|$ on $[1,2]$ is $M_1 = 2^5 + 2$. So $|f_n(x) - x | = |\frac{x^5 + 2}{x^4 + nx}| \leq \frac{M_1}{x^4 + nx}$ Since $[1,2]$ is an all positive interval we have that $nx \lt x^4 + nx$, so we have $RHS \lt \frac{M_1}{nx}$. And for any $n$, the minimum of the denominator is $1\cdot n$, so we have $\frac{M_1}{nx} \lt \frac{M_1}{n}$. Choose $N$ small enough so that $M_1/N \lt \epsilon$. |
H: Prove $x + y$ is divisible by $11$. Is my solution correct?
If $x$ & $y$ are natural numbers, and $56 x = 65 y$, prove that $x + y$ is divisible by $11$.
Solution)
$56$ and $65$ are relatively prime
So, $65∣x$ and $56∣y$
Let $x = 65m$ and $y = 56n$
Then,
$56x = 65y$
$56.65m = 65.56n$,
$m = n$
Thus, the solutions are of the form $x = 65k$,$y = 56k$ for integers $k$, and
$x+y = (65+56)k = 121k = 11(11k)$.
Thus, $x+y$ is even divisible by $11$
AI: Your solution is correct. Another way to prove is as follows;
We have
$$56 \equiv1\pmod{11}$$
Hence,
\begin{align}
(x+y) & \equiv 56(x+y)\pmod{11}(x+y) \equiv 56x+56y\pmod{11}\\
& \equiv 65y+56y\pmod{11} \equiv 121y \pmod{11} \equiv 0 \pmod{11}
\end{align} |
H: A sphere is a surface.
How to show a sphere is a surface.
$x^2+y^2+z^2=R^2$
Note that I need to find a homeopmhism $\phi : U \to S$ for $U$ in $\Bbb R^2$ and $S$ in $\Bbb R^3$
Let $U=[0,2\pi]\times [0,\pi]$
Then I defined a surface patch $\phi (u,v)= (R\sin u \cos u, R\sin v \sin u, R\cos v)$
After there, what do I need to do in order to show that a sphere is a surface?
AI: You need to map every open set $U$ in your chart to an open subset $V$ in the sphere (defihomorphizem from $U$ to $V$ )
.
The basic map $\phi(t,s)=(t,s,\sqrt{1-t^2-s^2}$) this will cover half of the sphere without the "Equator" to cover the all sphere you need another 5 charts, 6 charts in total |
H: Fibonacci induction stuck in adding functions together
Using Fibonacci...
I am Proving: $$f_3 + f_6 + \cdots + f_{3n} = \frac12(f_{3n+2}-1) $$
I did the assumption of $f_1$ which gave $\mathrm{LHS}=2=\mathrm{RHS}$.
For the second part where it is $n+1$ I am having problem adding the RHS:
$$f_3 + f_6 + \cdots + f_{3n}+ f_{3(n+1)} = \frac12(f_{3(n+1)+2}-1) $$
Here is the problem as I have no knowledge of how to make the function into the previous:
$$\mathrm{RHS} = \frac12(f_{3n+2}-1)+f_{3(n+1)} $$
Thanks in advance... Also, if anyone got any information on properties for functions would be greatly appreciated.
Edit: Aww I understand now because in Fibonacci we can see that F(0) + f(1) = f(2) so in that perspective you can add them like that. ^^ Thank you guys... btw, it is not letting me up vote, Mark good answer, or comment
AI: $$\begin{align}
f_3+f_6+\cdots+f_{3n}+f_{3n+3}&={1\over2}\left(f_{3n+2}-1\right)+f_{3n+3}\\
&={1\over2}\left(f_{3n+2}+f_{3n+3}+f_{3n+3}-1\right)\\
&={1\over2}\left(f_{3n+4}+f_{3n+3}-1\right)\\
&={1\over2}\left(f_{3n+5}-1\right)\\
\end{align}$$ |
H: How to find the ring of regular function on $\mathbb{P}^2\backslash\mathbb{V}(x_0^2+x_1^2+x_2^2)$
Let $X=\mathbb{P}^2$ and $U=X\backslash\mathbb{V}(x_0^2+x_1^2+x_2^2)$, could anyone show me how to find $\mathcal{O}_X(U)$?
I see examples in affine case, but have no idea how to calculate the ring in the projective case.
AI: The ring $\mathcal{O}_X(U)$ of the affine open subset $U\subset \mathbb P^2$ consists of all fractions of the form $$\frac {P(x_0,x_1,x_2)}{(x_0^2+x_1^2+x_2^2)^r}$$ with $r\geq 0$ an arbitrary integer and $P(x_0,x_1,x_2)$ a homogeneous polynomial of degree $2r$. |
H: Find length of triangle side
There is a triangle $ABC$ where $|CB|=a$, $|AC|=b$ and medians of these sides intersect at a right angle. Find |AB|.
I don't know how to use a right angle in this problem. I have a idea to link a middle of $|CB|$ with $|AC|$, let $K,L$ be a centre of these sides and we have $2|KL|=|AB|$
AI: We've the known medians theorem in geometry: the three medians to the sides of a triangle meet at one point which divides each median in a ratio of $\;1:2\;$ , with the longest segment always on the vertex side.
Thus, calling $\;M\;$ to the intersection point of the two medians, we can put
$$|AM|=2x\;,\;\;|MD|=x\;\;;\;\;|BM|=2y\;,\;\;|ME|=y$$
with $\;D=\;$ the midpoint of $\;BC\;$ , and $\;E=\;$ the midpoint of AC.
Using now Pythagoras on the triangle $\;\Delta AMB\;$ we get:
$$|AM|^2+|BM|^2=|AB|^2=4(x^2+y^2)$$
But on $\;\Delta BMD\;$ we get:
$$4y^2+x^2=\frac{a^2}4$$
and on $\;\Delta AME\;$ we get
$$4x^2+y^2=\frac{b^2}4$$
so solving the quadratic system of two unknowns above, we get:
$$|AB|=2\sqrt{x^2+y^2}=2\sqrt{\frac{4b^2-a^2}{60}+\frac{4a^2-b^2}{60}}=\ldots$$ |
H: $\frac{1}{\infty}$ - is this equal $0$?
I've seen that wolfram alpha says:
$$\frac{1}{\infty} = 0$$
Well, I'm sure that:
$$\lim_{x\to \infty}\frac{1}{x} = 0$$
But does $\frac{1}{\infty}$ only make sense when we calculate it's limit? Because for me, $1$ divided by any large amount of number will be always almost zero.
AI: The notation $\displaystyle\frac{1}{\infty}=0$ is used as a shorthand for
"As $x$ approaches infinity, the denominator blows up without bound, and hence since the numerator is constant, the value of the function approaches zero (i.e. gets arbitrarily close to zero), and hence its limit is zero."
The notation $\dfrac 1\infty$ does NOT literally mean "divide $1$ by $\infty$".
So literally, it is nonsense; taken as shorthand for the above, you'll see that notation used pretty commonly when folks evaluate limits. It's what we call "an abuse of notation." |
H: Linear algebra: Matrix multiplication problem
I need to prove something in my homework I just don't know how to approach it and need some guidance.
"Show that for a matrix $A$ ($n \times m$) and a vector $\vec{x}$ ($m \times 1$) it applies that:
$A\vec{x} = \sum_{j=1}^{j=m} x_jA_j$
s.t the multiplication of $x_jA_j$ is a multiply of vector $A_j$
and the scalar $x_j$."
AI: Well, just write that stuff down explicitly. If $\;A=(a_{ij})\;$ , then
$$\begin{pmatrix}a_{11}&a_{12}&\ldots&a_{1m}\\
a_{21}&a_{22}&\ldots&a_{2m}\\
\ldots&\ldots&\ldots&\ldots\\
a_{n1}&a_{n2}&\ldots&a_{nm}\end{pmatrix}\begin{pmatrix}x_1\\x_2\\\ldots\\x_m\end{pmatrix}=\begin{pmatrix}\sum_{k=1}^ma_{1k}x_k\\\ldots\\\sum_{k=1}^ma_{nk}x_k\end{pmatrix}=$$
$$=\sum_{k=1}^mx_k\begin{pmatrix}a_{1k}\\a_{2k}\\\ldots\\a_{nk}\end{pmatrix}$$ |
H: Substitution To Find Most General Unifier
Could someone give me some advice on how to do this problem?
For the the following pair of expressions, find the substitution that is
the most general unifier (mgu) or explain why the two expressions cannot be unified. b and c are constants, f and g are functions, and w, x, y, z are variables.
Thanks.
AI: A unifier of two terms $t$ and $t$' is a substitution function $\phi$ that maps variables to terms such that $\phi t = \phi t'$, i.e., applying $\phi$ to the variables in each term yields the same term. For example, given $t = f(x)$ and $t' = f(a)$ where $a$ is a constant and $f$ is a function name (not a variable) then $\phi\ x \mapsto a$ is a unifier since $\phi t = f(a) = \phi t'$.
The unifier $\phi$ for $t$ and $t'$ is the most general unifier if for any other unifier $\psi$ (s.t. $\psi t = \psi t'$) then their exists some other unifier $\Phi$ such that $\psi = \Phi \circ \phi$, i.e., $\psi$ factors through $\phi$. For example, given $t = f(x, a)$ and $t' = f(y, a)$ then $\psi\ x \mapsto c, \psi\ y \mapsto c$ is a unifier, but it is not the most general. The most general is $\phi\ x \mapsto y$, since $\psi$ factors though $\phi$ with $\Phi\ y \mapsto c$ (or equivalently $\phi\ y \mapsto x$ and $\Phi\ x \mapsto c$). The usual simple unification algorithm will generate an mgu; basically just pick the simplest unification (unify variables to variables, not to some other constants/ground terms). |
H: A weird idea on definition of completing a metric space
Please note the definition below, captured from Page 102, Real Analysis, Carothers, 1ed:
Completions
Completeness is a central theme in this book; it will return frequently. It may comfort you to know that every metric space can be "completed." In effect, this means that by tacking on a few "missing" limit points we can make an incomplete space complete. While the approach that we will take may not suggest anything so simple as adding a few points here and there, it is nevertheless the picture to bear in mind. In time, all will be made clear!
First, a definition. A metric space $(\hat M,\hat d)$ is called a completion for $(M,d)$ if
(i) $(\hat M,\hat d)$ is complete, and
(ii) $(M,d)$ is isometric to a dense subset of $(\hat M,\hat d)$.
Let $d(x,y)=|x-y|$ and $\hat d = d$. Then if $M=(0,1)$ and $\hat M$ is $[4,5]$, $\hat M$ is a completion of $M$ because $M$ is isometric to $(4,5)$ which is a dense set of $[4,5]$, the complete metric space. It is really interesting that intersection of $\hat M$ and $M$ is empty, say to complete a metric space, it is not necessary for us to use those old elements in $M$, which is a little ridiculous.
Is there any problem about the idea above?
AI: In my opinion, the right way of thinking about completion is not by looking at the completion $(\hat{M}, \hat{d})$ but looking at the completion and the isometry $(\hat{M}, \hat{d}, f)$.
A completion of $(M,d)$ is actually a triple $(\hat{M}, \hat{d}, f)$, where $(\hat{M}, \hat{d})$ is complete, $f(M)$ is dense in $\hat{M}$ and
$$f: (M,d) \to (f(M), \hat{d}) \, \mbox{ is an isometry} \,.$$
Any confusion will appear only if we ignore/forget $f$, but this is a part of the completion... |
H: Laguerre polynomials and least squares polynomial
A similar question from Orthogonal polynomials and Gram Schmidt says:
Use the Laguerre polynomials, i.e, $L_1(x)=x-1$,$L_2(x)=x^2-4x+2$, and $L_3(x)=x^3-9x^2+18x-6$, to compute the least squares polynomials of degree one, two, and three on the interval $(0,\infty )$ with respect to the weight function $w(x)=e^{-x}$ for $f(x)=x^3$.
I know that the answer is $6L_0(x) + 18L_1(x) + 9L_2(x) - L_3(x)$, but I'm not sure on how to obtain it either by hand or by maple.
I am working on it and meanwhile if anyone has any ideas to help me then please share it here. Thanks.
AI: I presume that by least squares polynomial, you mean determine coefficients $c_i$ in
$$
F(x):=c_0L_0(x)+c_1 L_1(x)+c_2L_2(x)+c_3 L_3(x)
$$
such that
$$\|f-F\|_w^2:=\int_0^\infty |f(x)-F(x)|^2 w(x)\,dx$$ is minimized, where $w(x)=e^{-x}$.
In that case, the orthogonality of the $L_i(x)$ (and you would need to add $L_0(x)$) together with the Best Approximation Theorem say that the coefficients should be chosen according to $$c_i={\langle f,L_i\rangle_w\over \langle L_i,L_i\rangle_w}={\int_0^\infty f(x)L_i(x)e^{-x}\,dx\over \int_0^\infty L_i^2(x)e^{-x}\,dx}, \quad i=0,1,2,3.
$$ |
H: Proving $(2n-1)^n + (2n)^n ≈ (2n+1)^n$
As I do, I was messing around and I thought to myself this simple thing:
$3^2 + 4^2 = 5^2$
I just thought that this is only Pythagorean triplet with sequential integers. I know that there are no others and there are no others to higher powers due to Fermat's Last Theorem. However there are many that can be approximated.
$5^3 + 6^3 = 341 ≈ 7^3$
$7^4 + 8^4 = 6479 ≈ 9^4$
$9^5 + 10^5 = 159049 ≈ 11^5$
As you can see, there is a pattern occurring in the form of:
$(2n-1)^n + (2n)^n ≈ (2n + 1)^n \{n ∈ ℤ+; n > 0\}$
It also seems the that if you rearrange the equation and take the limit at +∞:
$\lim_{n\to +\infty} { [(2n-1)^n + (2n)^n]^{1/n} - (2n + 1) } = 2ln(1 + e^{1/2}) - 2 ≈ -0.051846 $
(Calculated using Wolfram|Alpha)
Has this been noticed before; if so, is there a proof or is this the proof?
AI: $$\left(\dfrac{2n-1}{2n+1} \right)^n = \left(1 -\dfrac2{2n+1} \right)^n \sim \dfrac1e - \dfrac1{en^2} + \mathcal{O}(1/n^4)$$
$$\left(\dfrac{2n}{2n+1} \right)^n = \left(1 -\dfrac1{2n+1} \right)^n \sim \dfrac1{\sqrt{e}}$$
Hence,
$$\left(\dfrac{2n-1}{2n+1} \right)^n + \left(\dfrac{2n}{2n+1} \right)^n \sim \dfrac1e + \dfrac1{\sqrt{e}} \approx 0.97441$$ |
H: Real Numbers as Well Defined Sets
For every construction of the reals, we define a real number to be some kind of set of rational numbers (such as cuts or sequences).
However, the number of symbols we have to formulate the description of a set is finite (analytic functions, infinite sums, etc), and I assume sets can only be defined as finite strings of logical symbols.
Every number I can think of, including transcendentals, and even the Chaitin Constant, can be described as such.
Where is the fallacy in saying, then, that the real numbers are countable (as, for instance, well defined cuts seem to be)?
AI: The number of describable numbers is countable. But there are some real numbers you cannot write a formula for.
Also, describable numbers is an ill defined concept. You get contradictions along the lines of "the smallest number not describable in less than eleven words." So you have to specify what is an allowable formula. Maybe something like: $x$ is describable if $P(x)$ is a proposition in standard set theory for which it can be proven, using the standard axioms, that there exists a unique real number for which $P(x)$ holds. |
H: Binary relations, closures and equivalences
Let $R$ be the relation on $Z$ such that $xRy \iff x-y=c$.
Well, what I have so far is $R=\{ 0,-1,1,0,-1,1,0 \cdots\}$
Is $R^* $ and equivalence relation? Why not?
This is where problems start: I don't know what the definition of $R^* $. In fact, I cannot seem to find anything related to this in my notes. First off, is my relation $R$ correct at this point?
AI: Your relation is not correct. A relation is defined to be a set of ordered pairs where the ordered pair (x,y) is included if xRy. |
H: Why $\lim_{n\to \infty} ({3^n+2^n\over 3^n-2^ni^n}) = \lim_{n\to\infty} ({3^n+2^n\over 3^n-2^n}) \ $?
This week I was introduced to the limits of complex sequences. It is actually pretty simple because it's mostly the same compared to real sequences. However, there is one thing - Why is:
$\lim_{n\to \infty} ({3^n+2^n\over 3^n-2^ni^n}) = \lim_{n\to\infty} ({3^n+2^n\over 3^n-2^n}) \ $
Where does the $i^n$ go? We have not discussed the limit of $i^n$ so far, but I should be able to explain this.
AI: It didn't magically disappear. You should show that the limit of both values is 1.
A similar question would be to ask: Why is $\cos 0 = \tan \frac{\pi}{4}$?
There isn't any special relationship, just that both sides happen to be equal to 1. |
H: Solve, $\cos(x)=\frac 25, \frac{3\pi}{2}
A. Draw and label two triangles, one containing angles $x$ and one
containing angle $y$. (can use the $x-y$ axis version of triangles)
B. List $\sin(x), \cos(x), \sin(y)$, and $\cos(y)$
c. Find $\sin(x+y)$
d. Find $\sin(2x)$
e. Find $\cos(\frac y2)$
AI: HINT:
To obtain $\cos(y)$ and $\sin(x)$ use the basic trigonometric inequality:
$$\sin^2x + \cos^2x = 1$$
But note that $x$ is in the fourth quadrant, while $y$ is in the second. What does that say about the sign?
For the three calculation they are just basic trigonometric identities. You can find a load of them here and also you can find the identities you need.
HINT 2:
For problem A. Use the fact that cosine is an even function. Because $x$ is in the fourth quadrant we can write $x = 360^{\circ} - \alpha$, where $\alpha$ is an obtuse angle so we have:
$$\cos (x) = \cos(360^{\circ} - \alpha) = \cos (- \alpha) = \cos {\alpha}$$
Now since $\cos (x) = \frac{adjacent}{hypothenuse}$, we can easy draw a right triangle with side 2 and hypothenuse 5. From this you obtain $\alpha$ and subsequently $x$.
Note that $x > \pi$, so a triangle with angle $x$ would be impossible to construct, because the sum of the interior angles in a triangle is $\pi$.
For sine use the fact that $\sin (\pi - \alpha) = \sin {\alpha}$ and $sin x = \frac{opposite}{hypothenuse}$ |
H: Volume of $n$-dimensional parallelepiped as determinant
Let $V$ be a vector space of dimension $n$ and $B:V\times V\rightarrow\mathbb{R}$ be an inner product. Let $\sigma_B:V^n\rightarrow\mathbb{R}$ be the map $$ \sigma_B(v_1,\ldots,v_n)=(\det[b_{i,j}])^{\frac12},$$ where $b_{i,j}=B(v_i,v_j)$. Show that $\sigma_B(v_1,\ldots,v_n)$ is the volume of the parallelepiped $P(v_1,\ldots,v_n)$ having $v_1,\ldots,v_n$ as adjacent edges.
I don't really see how to relate the determinant of the matrix composed of the inner products into the volume of the parallelepiped. What would be the way?
AI: This is essentially equivalent to showing that a linear operator represented by a matrix $A$ takes shapes to the domain to shapes in the range, and multiplies their volume by the determinant of $A$.
This is the way I convinced myself of this many years ago. First, it is obviously true for diagonal matrices. Next, it is also true for shear matrices, which are of the form identity matrix plus a matrix all of whose entries are zero except one entry. Finally, it is obviously true for permutation matrices. Now you merely need to prove that every matrix is a product of these kinds of matrices, and this can be done, for example, by going through the steps of the Gauss-Jordon method for solving systems of linear equations. And of course, you need to convince yourself that if a matrix $A$ multiplies volumes by a factor $a$, and another matrix multiplies volumes by a factor $b$, then $AB$ multiplies volumes by a factor $ab$. And you can see this by approximating any reasonable shape by lots of little parallelepipeds that fill in the shape. |
H: What does $\Bbb N \to \Bbb R^{\ge 0}$ mean?
What is the interpretation of this:
Functions $$a,b,c\colon\mathbb{N} \rightarrow \mathbb{R} ^{\ge 0}?$$
AI: Standard notation for functions, a function $f$ has domain $X$ and range $Y$. We write $f:X \longrightarrow Y$. In this case we have functions from $\mathbf{N}$ to $\mathbf{R}^{\geq 0}$, the non-negative reals. Note, infinite sequences, such as $(x_n)_{n \in \mathbf{N}}$ are examples of such functions, so long as $x_n \geq 0$ for all $n\geq1$. |
H: Finding the smallest relation that is reflexive, transitive, and symmetric
Find the smallest relation containing the relation $\{ (1,2),(2,1),(2,3),(3,4),(4,1) \}$ that is:
Reflexive and transitive
Reflexive, symmetric and transitive
Well my first attempt:
Reflexive: $ S_1 = \{ (1,1),(2,2),(3,3),(4,4) \}$
Symmetric: $ S_2=\{ (3,2),(4,3),(1,4)
\}$
Transitive: $S_3= ? $Is where I'm stuck.
So that $S_1\cup S_2 \cup S_3 $ would be my equivalence relation?
Also, When you're testing for transitivity, what combinations do we test for? If we take: $(1,2) \land (2,3)\land(3,4) \rightarrow(1,3)$, must it be done for the converse? Starting with $(2,1)$ rather than $(1,2)$. It seems that there are many conbinations of $x,y$ that need to be tested. Is this correct?
In fact, is my attempt correct to begin with?
AI: It is very important in logic to understand your definition as hard as you can:
$$\forall_{x,y,z\in X}\, xRy\wedge yRz\Rightarrow xRz$$
So. That tells us that if we have two elements, let's say, following your example, $(1,2)$ and $(2,3)$, there must be direct connection between $1$ and $3$. Do the same for all possible pairs from $S$, $S_1$ and $S_2$ (where $S$ is your base relation), and you will have your answer. And yes, you may have many elements in your relation after all those changes. |
H: Determine whether series is convergent or divergent $\sum_{n=1}^{\infty}\frac{1}{n^2+4}$
I still haven't gotten the hang of how to solve these problems, but when I first saw this one I thought partial fraction or limit. So I went with taking the limit but the solution manual shows them using the integral test.
Was I wrong to just take the limit?
$$\sum_{n=1}^{\infty}\frac{1}{n^2+4}$$
Next:
$$\lim_{n\to\infty}\frac{1}{n^2+4}=0$$
So converges by the test for divergence?
AI: We only have the following statement to be true:
$$\text{If $\sum_{n=1}^{\infty} a_n$ converges, then $a_n \to 0$.}$$ The converse of the above statement is not true, i.e.,
$$\text{if $a_n \to 0$, then $\displaystyle \sum_{n=1}^{\infty} a_n$ converges is an incorrect statement.}$$
For instance, $\displaystyle \sum_{n=1}^{\infty} \dfrac1n$ diverges, even though $\dfrac1n \to 0$.
To prove your statement, note that $\dfrac1{n^2+4} < \dfrac1{n^2}$ and make use of the fact that $\displaystyle \sum_{n=1}^{\infty} \dfrac1{n^2}$ converges to conclude that $\displaystyle \sum_{n=1}^{\infty}\dfrac1{n^2+4}$ converges. |
H: Precalculus in a Nutshell, Geometry, Appendix B, Section 5, Question 14.
A sphere is circumscribed about a cube. Find the ratio of the volume of the cube to the volume of the sphere.
So I drew this diagram:
Next, I want to relate s and r.
I apply Pythagorean theory to my diagram.
$$\left(\frac12 s\right)^2+\left(\frac12s\right)^2=r^2$$
Or more simply:
$$\frac12 s^2=r^2$$
This gives me the identity
$$s=r\times \sqrt2$$
So far so good, I think.
$$V_{\text{cube}}=s^3$$
$$V_{\text{sphere}}=\frac{4\pi r^3}{3}$$
Substituting for $s$,
$$V_{\text{cube}}=r^3\times2\sqrt2$$
Ratio of $V_{\text{cube}}$ to $V_{\text{sphere}}$ is:
$$r^3\times2\sqrt2/4\pi r^3/3$$
Simply:
$$\frac{6\sqrt2r^3}{4\pi r^3}=\frac{6\sqrt2}{4\pi}=\color{red}{\boxed{\displaystyle\color{black}{\frac{3\sqrt2}{2\pi}}}}$$
That's the solution I keep getting.
The answer key stays the solution is actually $\displaystyle\frac{6\sqrt2}{3\pi}$
Can someone please tell me where I'm going wrong?
AI: Let $O$ be the centre of the cube, and let $C$ be one of the corners of the cube. Drop a perpendicular from $O$ to the centre $M$ of one of the faces that contains the vertex $C$.
Then $\triangle OMC$ is right-angled at $M$.
We have, in your notation, $OM=\frac{s}{2}$. Distance $OC$ is half a face diagonal of the cube. A face diagonal has length $\sqrt{2}{s}$. Half a face diagonal has therefore length $\frac{\sqrt{2} s}{2}$.
By the Pythagorean Theorem, we have
$$(OC)^2 =\left(\frac{s}{2}\right)^2+ \left(\frac{\sqrt{2}s}{2}\right)^2.$$
It follows that $OC$, the radius of the circle, is $\frac{\sqrt{3}}{2}s$. The rest of the calculation is along lines you are familiar with.
Remark: The above argument was used because it seemed closest in spirit to your approach. To me, the following seems simpler.
Grab a cube, or something close enough. As a distant second best, draw a picture of a cube in the conventional way. Let $P$ and $Q$ be opposite each other along the long diagonal of the cube. Let $S$ be a vertex on a face that contains $P$, with $S$ diagonally across from $P$. Then $\triangle PSQ$ is right-angled at $S$. We have $(PQ)^2=(PS)^2+(SQ)^2=2s^2+s^2=3s^2$. The long diagonal of a cuve of side $s$ has length $\sqrt{3} s$. |
H: How to understand the solution?
Two days ago,I have a problem about $\sum _{n=2}^{\infty } \frac{1}{n \log (n)}$,and the @Julien Clancy give me a solution:
To see whether $\sum_2^\infty 1/(n \log n)$ converges, we can use the integral test. This series converges if and only if this integral does:
$$
\int_2^\infty \frac{1}{x \log x} dx = \left[\log(\log x)\right]_2^\infty
$$
and in fact the integral diverges.
I know the Infinite series $\sum _{n=2}^{\infty } \frac{1}{n \log (n)}$ can be see as a area of trapezoid with curved edge that the $Delta x$=1,so the $\sum _{n=2}^{\infty } \frac{1}{n \log (n)}$ is less than the Integrate[1/(x*Ln(x),{x,2,Infinity}].
However,I cannot understand why the latter integral converge can reduct the former series converge?Can someone tell me why?Thanks sincerely!
AI: The summation would converge if the integral converges. This is the integral test. If the integral does not converge, then the infinite series does not converge.
I suggest you have a look at http://en.wikipedia.org/wiki/Integral_test_for_convergence |
H: Formula for $n$-dimensional parallelepiped
What is the formula for the volume of a parallelepiped in $n$ dimensions with $v_1,v_2,\ldots,v_n$ as edges?
In an exercise it's given as $(\det[b_{i,j}])^{\frac12}$, where $b_{i,j}=v_i\cdot v_j$. But I also remember seeing somewhere that the volume is given by $|\det A|$, where $A$ is the matrix with columns $v_1,v_2,\ldots,v_n$.
Am I misunderstanding somewhere, or are these two values actually the same?
AI: Suppose that $A = (v_1,\ldots,v_n)$, and you have the second formula for volume as $|\det A|$. Then the first is derived by having $\det(A^TA) = (\det A)^2$. |
H: Construction of a continuous function which is not bounded on given interval.
Actual Question is :
On Which of the following spaces is every continuous (real valued) function Bounded?
$X_1=(0,1)$
$X_2=[0,1]$
$X_3=[0,1)$
$X_4 =\{t\in [0,1] : t \text { is irrational}\}$.
I could see that $1,3$ are spaces in which not every continuous function is bounded.
I could see $2$ is a spacein which not evry continuous function is bounded.
For First option I have seen that $f(x)=\frac{1}{x}$ works as it is continuous but not bounded.
For Second option I see that continuous image of compact set is compact i.e., $f[0,1]$ is compact thus bounded.
For third option I could see that as $1$ is exculded in the domain, function is unbounded if it has a pole at $1$ So, I see that $f(x)=\frac{1}{x-1}$ would help.
For fourth option I had no clue (what i had does not help much i guess).
I would be thankful if someone can help me in this.
As this is not a serious problem I would request users to just give me hints and not to post as an answer.
Thank you.
AI: You've got the basic idea - you can just generalize it.
Suppose $X\subset \mathbb{R}$ is such that every continuous function on $X$ is bounded. I claim that $X$ is both closed and bounded.
Suppose $c \in \overline{X}\setminus X$, then the function
$$
f(x)= \frac{1}{x-c}
$$
will be unbounded. Hence, $X$ must be closed.
If $X$ is not bounded, just take $f(x) = x$, then $f$ is unbounded.
Hence, $X$ must be compact.
Conversely, if $X$ is compact, every continuous function is bounded, so just check which of these sets are compact. |
H: Abstract Algebra: Homomorphism, Kernel, Image
To prove:
Let G be the group of affine functions from R into R, as defined in A. (A = {f_m,b: R -> R | m is not equal to 0 and f_m,b(x) = mx + b}. Define phi: G -> R^x as follows: for any function f_m,b in G, let phi(f_m,b) = m. Prove that phi is a group homomorphism and find its kernel and image.
My work:
First we need to prove that A is a group.
Let we have f_m,b and f_n,v both are in A.
=> Composition, (f_m,b) o (f_n,v) in A => (f_m,b) o (f_n,v) = f_m,b(nx + v) = m(nx + v)+b = mnx+(mv+b) = h_mn,mv+b(x) in A. => Closure.
Let we have f_1,0 => f_1,0(x) = 1x+0 = x => Identity
Let f_m,b in A and it has inverse.
Since m is not equal to 0, (f_m,b) o (f_1/m,-b/m) = f_m,b(x/m-b/m) = m(x/m-b/m)+b = x so we have inverse.
Hence A is a group.
Let we have f_m,b and f_n,v in A.
phi[(f_m,b)(f_n,v)] = phi(f_m,b o f_n,v(x)) = h_mn,mv+b(x) = mn.
phi(f_m,b)*phi(f_n,v) = mn
Since they are equal, phi is a homomorphism.
And I'm stuck to prove its kernel and image..
Any help would be very appreciated.
AI: $$
\ker(\phi) = \{f_{m,b} : m = 1\} = \{f_b : b\in \mathbb{R}\}
$$
where $f_b$ denotes the map $x\mapsto x+b$.
Define a function $\ker(\phi) \to \mathbb{R}$ given by $f_b \mapsto b$. Can you check that this is an isomorphism?
Also, for any $m \in \mathbb{R}^{\times}$, define $f(x) = mx$, then what is $\phi(f)$? What can you conclude abou the image of $\phi$? |
H: non integer p adic expansion (special case)
I need to calculate the 5 adic expansion of $\frac{1}{45}$. Since i cannot compute it normally, i expand $\frac{1}{45}$ into $\frac{1}{5}*\frac{1}{9}$.
I calculated the 5 adic expansion of $\frac{1}{9}$, but i still cannot calculate the expansion of $\frac{1}{5}$
Please give me some advice if possible.
AI: The $5$-adic expansion of $5^{-1}$ is simply $5^{-1}$. Seems you are overthinking things.
Just like how multiplying real numbers by powers of $10$ shifts their decimal expansions to the left or right as appropriate, and just like how multiplying a Laurent series $\sum a_nx^n$ by powers of $x$ shift their coefficients to the left or right, so too will multiplying a $p$-adic number by powers of $p$ shift its $p$-adic digital representations, in the same exact manner as you would expect.
Since $9^{-1}=\overline{210234}21024_5$, we get $45^{-1}=\overline{210234}2102.4_5$, or
$$\frac{1}{45}=4\cdot5^{-1}+2\cdot5^0+0\cdot5^1+1\cdot5^2+2\cdot5^3+\cdots$$ |
H: Determinant of the Sum in an Inequality
Given that: $detA > 0$ and $detB > 0$, is it the case that $det(A+B) \ge 0$?
AI: $\det\left(\begin{bmatrix}2&1\\ 1&2\end{bmatrix}+\begin{bmatrix}-1&1\\ 0&-1\end{bmatrix}\right)=\det\begin{bmatrix}1&2\\ 1&1\end{bmatrix}=-1$. |
H: If every continuous function $f$ in $X \subset \mathbb{R}^2$ is bounded then $X$ is compact.
To prove:
"If every continuous function $f$ in $X \subset \mathbb{R}^2$ is bounded then $X$ is compact."
My attempt :
In $\mathbb{R}^n$ a set $X$ is compact iff it is closed and bounded. I can show $X$ is bounded, but can not prove $X$ is closed.
The distance function $f : X \rightarrow \mathbb{R}$ defined by $f(x , y) = \sqrt{x^2 + y^2}$ is continuous on $X$. So it is bounded. Thus distance between any two points of $X$ is finite and hence $X$ is bounded.
Let $a \in \mathbb{R}^2$ be a limit point of $X$. Consider a sequence $\{a_n\}$ in $X$ converges to $a$. So for any continuous function $f$, the sequence $\{f(a_n)\}$ converges to $f(a)$ and $f(a)$ is finite. It does not imply $a \in X$.
Thank you for your help.
AI: To prove that $X$ is closed, if $(x_0,y_0)$ is a limit point of $X$ that is not in $X$, consider the function $(x,y) \mapsto 1/f(x-x_0,y-y_0)$; this is just the reciprocal of the distance between the points $(x,y)$ and $(x_0,y_0)$. |
H: Problem of Harmonic function.
If H is a harmonic function on an unit disk; And $H=0$ on $R_1\cup R_2$, here $R_1, R_2$ are radius of $D(0,1)$. The angle between $R_1$ and $ R_2$ is $r\pi$; here $r\in (0,1]$. If $r$ is an irrational number then is $H$ identically zero on $D(0,1)$?
I think if $r$ is irrational then $H\equiv0$ on unit disk $D$; and
following lemma can help us to prove it.
Lemma:
If $a,b$ are two semilines starting at the same point $O$ there exists $u \not\equiv 0$ is a harmonic function vanishing on $a,b$ if and only if there exists $r \in \mathbb{Q}$ such that the angle between a and b is $r\pi.$
Proof of the Lemma: Using the following property of harmonic function we can prove this lemma.
A harmonic function vanishing on an open set is identically zero.
(Schwarz Reflection Principle) If a harmonic function is defined in a neighborhood of a line segment contained in one of the halfplanes determined by that segment and it is continued to zero on that segment, then it can be extended harmonically to the symmetric region by the given line segment.
AI: Assuming a semiline need only exist within the domain of the function (I am unsure of the definition), we have the following proof:
Note that $R_1$ and $R_2$ are two semilines starting from the same point such that the angle between them is not a rational multiple of $\pi$, and that $H$ is harmonic on the unit disk, which is a neighborhood of these two segments. By our lemma, no non-zero harmonic function vanishes on $R_1 \cup R_2$. We conclude that $H$ must be identically zero. |
H: $2\times2$ matrices are not big enough
Olga Tausky-Todd had once said that
"If an assertion about matrices is false, there is usually a 2x2 matrix that reveals this."
There are, however, assertions about matrices that are true for $2\times2$ matrices but not for the larger ones. I came across one nice little example yesterday. Actually, every student who has studied first-year linear algebra should know that there are even assertions that are true for $3\times3$ matrices, but false for larger ones --- the rule of Sarrus is one obvious example; a question I answered last year provides another.
So, here is my question. What is your favourite assertion that is true for small matrices but not for larger ones? Here, $1\times1$ matrices are ignored because they form special cases too easily (otherwise, Tausky-Todd would have not made the above comment). The assertions are preferrably simple enough to understand, but their disproofs for larger matrices can be advanced or difficult.
AI: Any two rotation matrices commute. |
H: Logical Equivalence Of Quantified Implications
$\forall x (P(x)\rightarrow Q(x))$ is logically equivalent to $(\exists x P(x)\rightarrow\exists y Q(y))$
The proofs I've seen use logical reasoning to prove the equivalence.
Can someone supply a proof using the laws of boolean algebra?
AI: The two sentences are not logically equivalent. The sentence $\exists x P(x) \longrightarrow \exists y Q(y)$ is true in the natural numbers, if $P(x)$ says $x$ is odd, and $Q(y)$ says that $y$ is prime. The sentence $\forall x(P(x)\longrightarrow Q(x))$ is not.
Additionally, one cannot in general expect tools from Boolean algebra to deal with non-trivial assertions that involve quantification. |
H: Calculating $\det(A+I)$ for matrix $A$ defined by products
Let $b_1,\ldots,b_n\in\mathbb{R}$. I have an $n\times n$ matrix $A$ whose entry is given by $a_{ij}=b_ib_j$, and I'd like to show that $\det(A+I)=\sum_{i=1}^nb_i^2+1$.
Define $b=(b_1,\ldots,b_n)$. I know that $Ab=\left(\sum_{i=1}^nb_i^2\right)b$, and $Ac=0$ for all $c$ such that $b\cdot c=0$. So all the eigenvalues of $A$ are $\sum_{i=1}^nb_i^2, 0, 0, \ldots, 0$. What can I do next?
AI: The matrix $A=b b^T$. It is easy to see that $A b = \|b\|^2 b$, so $A$ has an eigenvalue of $\|b\|^2$ corresponding to the eigenvector $b$.
Now suppose $b^T x = 0$. Then we see that $A x = 0$.
Hence $A$ has eigenvalues $0,....,0,\|b\|^2$.
It follows that $A+I$ has eigenvalues $1,...,1,
\|b\|^2+1$.
Since $\det C = \Pi_k \lambda_k$, where $\lambda_k$ are the eigenvalues of $C$, we see that $\det (A+I) = 1 + \|b\|^2 = 1+\sum_k b_k^2$. |
H: Find four groups of order 20 not isomorphic to each other.
Find four groups of order 20 not isomorphic to each other and prove why they aren't isomorphic.
So far I thought of $\mathbb Z_{20}$, $\mathbb Z_2 \oplus\mathbb Z_{10}$, and $D_{10}$ (dihedral group), but I can't find another one. Would $U(50)$ work? I know it has order 20 and is cyclic but I'm not exactly sure how to move from there. Can someone to point me on the right direction?
AI: Note that $20 = 2^2 \cdot 5$.
By Fundamental theorem of Finitely Generated Abelian Group, there are two distinct abelian groups of order $20$: $\mathbb{Z}_{20}$ and $\mathbb{Z}_{10} \times \mathbb{Z}_2$.
Now let $G$ be a nonabelian group of order $20$. By Sylow’s Theorem, $n_5 = 1$, so that $G$ has a unique (hence normal) Sylow $5$-subgroup $H \cong \mathbb{Z}_5$. Now let $K$ be any Sylow $2$-subgroup of $G$. By Lagrange, we have $H \cap K = 1$, so that $G = HK$. By the recognition theorem for semidirect products, $G \cong H \rtimes_\varphi K$ for some $\varphi : K \rightarrow \mathsf{Aut}(H)$. Evidently, classifying the nonabelian groups of order $20$ is equivalent to determining the nonisomorphic groups constructed in this manner. To that end, let $H = \mathbb{Z}_5 = \langle y \rangle$. Note that $\mathsf{Aut}(H) = \langle \alpha \rangle \cong \mathbb{Z}_4$; where $\alpha(y) = y^2$.
Let $K = \mathbb{Z}_4 = \langle x \rangle$. There are four distinct homomorphisms $K \rightarrow \mathsf{Aut}(H)$.
If $\varphi_1(x) = 1$, then $\varphi_1$ is trivial; this contradicts the nonabelianicity of $G$.
If $\varphi_2(x) = \alpha$, then $\mathbb{Z}_5 \rtimes_{\varphi_2} \mathbb{Z}_4 $is indeed a nonabelian group of order $20$.
If $\varphi_3(x) = \alpha^2$, then $\mathbb{Z}_5 \rtimes_{\varphi_3} \mathbb{Z}_4$ is indeed a nonabelian group of order $20$. Moreover, since $\mathsf{ker}\ \varphi_3 \cong \mathbb{Z}_2$ and $\mathsf{ker}\ \varphi_2 \cong 1$,$ H \rtimes_{\varphi_3} K \not\cong H \rtimes_{\varphi_2} K$.
If $\varphi_4(x) = \alpha^3$, then $\mathsf{im}\ \varphi_4 = \mathsf{im}\ \varphi_2$. Since $\mathbb{Z}_4$ is cyclic, by a previous theorem, $H \rtimes_{\varphi_4} K \cong H \rtimes_{\varphi_2} K$.
Thus there are two distinct groups of order 20 which have a cyclic Sylow 2-subgroup.
Suppose now that $K = \mathbb{Z}_2^2 = \langle a \rangle \times \langle b \rangle$. Again, $\psi : \mathbb{Z}_2^2 \rightarrow \mathbb{Z}_4$ is determined uniquely by $\psi(a)$ and $\psi(b)$, and is indeed a homomorphism provided $|\psi(a)|$ and $|\psi(b)|$ divide $2$. We thus have $\psi(a)$, $\psi(b) \in \{ 1, \alpha^2 \}$, for a total of four choices.
If $\psi_1(a) = \psi_1(b) = 1$, then $\psi_1 = 1$, contradicting the nonabelianicity of $G$.
If $\psi_2(a) = \alpha^2$ and $\psi_2(b) = 1$, then $\mathbb{Z}_5 \rtimes_{\psi_2} \mathbb{Z}_2^2$ is indeed a nonabelian group of order $20$.
If $\psi_3(a) = 1$ and $\psi_3(b) = \alpha^2$, then $\varphi_3 = \varphi_2 \circ \theta$, where $\theta(a) = b$ and $\theta(b) = a$. Clearly $\theta$ is an automorphism of $\mathbb{Z}_2^2$. By a lemma to a previous theorem, we have $H \rtimes_{\psi_3} K \cong H \rtimes_{\psi_2} K$.
If $\psi_4(a) = \alpha^2$ and $\psi_4(b) = \alpha^2$, then $\psi_4 = \psi_2 \circ \theta$, where $\theta(a) = a$ and $\theta(b) = ab$. Clearly $\theta$ is an automorphism of $\mathbb{Z}_2^2$. By a lemma to a previous theorem, we have $H \rtimes_{\psi_4} K \cong H \rtimes_{\psi_2} K$.
Thus there is a unique nonabelian group of order 20 which has an elementary abelian Sylow 2-subgroup.
In summary, the distinct groups of order $20$ are as follows. We let $\mathbb{Z}_5 = \langle y \rangle$, $\mathbb{Z}_4 = \langle x \rangle$, and $\mathbb{Z}_2^2 = \langle a \rangle \times \langle b \rangle$.
$Z_{20}$,
$Z_{10} \times Z_2$,
$Z_5 \rtimes_{\varphi_3} Z_4$, where $\varphi_3(x)(y) = y^{-1}$.
$Z_5 \rtimes_{\varphi_2} Z_4$, where $\varphi_2(x)(y) = y^2$
$Z_5 \rtimes_\psi Z_2^2$, where $\psi(a)(y) = y^{-1}$ and $\psi(b)(y) = y$.
(Source: Crazyproject) |
H: Where to go from here...graduate school
I am an undergrad math major (minor in applied stats) set to graduate next month. I have been considering graduate school for a long time, and I know I want to pursue at least a Master's in the near future.
For a while, I wanted to pursue a Master's in Applied Statistics and aim for a corporate career in analysis. However, this past summer, I worked as an intern at a large company and came to the realization that the corporate setting was not for me.
I now know that I want to further my education in Mathematics, and it's a goal of mine to attend the University of Michigan's Master's program for Mathematics. However, it is not feasible for me to apply for U of M (for Fall 2014) as I have yet to take my Math Subject GRE, and the next test date is April of next year- which is too late.
My goal, then, is to apply for the Fall 2015 semester, but this will leave me with 1.5 years of no school in between. I guess my main question is what should I do in the meantime to better my chances of being accepted?
I have taken the following upper level math courses:
Abstract Algebra,
Combinatorics,
Numerical Analysis,
Real Analysis 1,
Stochastic Processes,
Complex Variables,
Linear Algebra,
Mathematical Modeling
and my current cumulative GPA is 3.56.
I know that I'm not up there on the GPA scale and it doesn't help that I don't have any research experience, and I'm afraid this will greatly hinder my chances of being admitted to U of M's Master's program.
Can anyone give me advice on what to do from here? How can I better my chances of getting in?
I'm not the best math student, and I can't deny that there exists a fear at the back of my mind telling me that I am not cut out for this. Is there anyway to know if you're qualified for graduate mathematics?
I apologize for the lengthy post.
Thanks in advance..
AI: First, please be assured that if you were able to complete the undergrad courses you mention, you can succeed in a masters program. To do so, you must do whatever work it takes -- it is that simple.
The best way to succeed in mathematics is to work problems. No amount of theory can replace problem solving. In fact, the problems are where you find out your didn't really understand the theory (sorry, but it's often true). For your grad courses you want to do every problem you are assigned, and then seek out more problems on the topic. Keep pounding at each topic until the graduate level problems are coming with some ease -- or at least without trauma.
In terms of the 1.5 year gap, there are a number of alternatives. One is to formally audit some graduate courses in mathematics. Most schools are happy for you to do this, and are not picky about your qualifications. If you are willing to pay them, they are willing to let you attend their classes.
If you could manage to audit at the U of M that would be best. You would meet some of the professors, and if you take a sincere interest in their courses (and make sure they know it, by asking questions), they would probably recommend you for entrance. Apart from potential recommendations, your applications for grad school would show that you did this work, which tells people you really are interested. That helps a lot.
If U of M isn't practical for auditing, go someplace else.
I suggest that you take courses you've already had, such as real analysis, or whatever most interests you. Having taken the course once, you have a good start on the graduate level of the same material, so you would likely do well (helping you round up recommendations).
Have you considered applying for winter semester 2015 instead of waiting for the fall? It may be possible -- worth asking anyway.
Have you given thought to what you want to do with your masters in math? You can use some of your time to check out possibilities that may interest you more than what you've tried so far. |
H: $\int_{y=-1}^{y=1}\int_{x=y^{2/3}}^{x=(2-y)^2}f(x,y)\ \mathrm dx \mathrm dy$ what does the region look like?
More specifically, I have a double integral
$$\int_{-1}^{1}\int_{y^{2/3}}^{(2-y)^2}f(x,y) \ \mathrm dx \mathrm dy$$
It is mostly the $y^{2/3}$ that is confusing me.
AI: I try to draw your region by sage. That's what it looks like.
The green graph is :$x=y^{\frac{2}{3}}$
The blue graph is :$x=(2-y)^2$
Code sage:
x = var('x')
p1 = parametric_plot(((x-2)^2,x),(x,-4,4),rgbcolor=hue(0.5))
p2 = parametric_plot(((x^2)^(1/3),x),(x,-4,4),rgbcolor=hue(0.4))
p3 = parametric_plot((x,-1),(x,-5,11),rgbcolor=hue(0.6))
show(p1+p2+p3)
I hope it will help you. |
H: probability and combinations with the word REGULATIONS
If the letters of the word REGULATIONS are arranged at random,what is the probability that there will be exactly 4 letters between R and E?
The answer in my book is given as 11!/(9C4 x 4! x6!x2!) .Shouldn't the answer be upside down because 11!=total number of arrangements?
AI: There are $\binom{11}{2}$ equally likely ways to choose the two positions that will be reserved for the letters R and E. (Note this does not count where R and E as individual letters go, we are just putting reserved signs on the two positions.)
If the two positions have a gap of $4$ between them, there are $6$ places where the leftmost position can be, and then the other position is determined. It follows that our probability is $\frac{6}{\binom{11}{2}}$. |
H: Show that X $\times$ Y is compact if X and Y are compact.
Show that X $\times$ Y is compact if X and Y are compact.
I know there are different solutions in this site to this question, but I want to use this exact statement:
Suppose X and Y are two topological spaces with Y compact, and $x_0$ a point in X. If M is an open set in X $\times$ Y, such that {$x_0$} $\times$ Y $\subseteq$ M, then there exists an open set W $\subseteq$ X containing $x_0$ such that W $\times$ Y $\subseteq$ M
(Using this statement, I have to show that for all x $\in$ X, we can create a tube (covered by finite amount of open sets) around {x} $\times$ Y)
So I start with an open cover for X$\times$Y, say {U$_\alpha$}$_{\alpha\in J}$ = {A$_{\alpha}$$\times$B$_{\alpha}$} where each A$_{\alpha}$ is open in X and each B$_{\alpha}$ is open in Y. What do I do next?
I know that for any open cover for X, say {A$_\alpha$}$_{\alpha\in J}$, there exists a finite subcover of X. Similarly for Y.
AI: Sets of the form $U \times V$ where $U$ is open in $X$ and $V$ is open in $Y$ form a basis for the product topology.
Hence $M = \cup_\alpha (U_\alpha \times V_\alpha)$, for some collection of open sets $\{ U_\alpha \times V_\alpha \}_\alpha$.
Let ${\cal A} = \{\alpha | (\{x_0\} \times Y) \cap (U_\alpha \times V_\alpha) \neq \emptyset \}$. Note that $Y \subset \cup_{\alpha \in {\cal A}} V_\alpha$, and $x_0 \in U_\alpha$ for all $\alpha \in {\cal A}$.
Since $Y$ is compact, we have a finite $I \subset {\cal A}$ such that $Y \subset \cup_{\alpha \in I} V_\alpha$ (hence in fact, $Y=\cup_{\alpha \in I} V_\alpha$), and since $I$ is finite, the set $W = \cap_{\alpha \in I} U_\alpha$ is open, and $x_0 \in W$. Furthermore, $W \times V_\alpha \subset U_\alpha \times V_\alpha$ for all $ \alpha \in I$, and so $\cup_{\alpha \in I} (W \times V_\alpha) = W \times (\cup_{\alpha \in I} V_\alpha) = W \times Y \subset \cup_\alpha (U_\alpha \times V_\alpha) = M$.
To show $X \times Y$ is compact:
Let $\Pi_Y: X \times Y \to Y$ be given by $\Pi_Y((x,y)) = y$. Then if $Z$ is open in $X \times Y$, $\Pi_Y(Z)$ is open in $Y$. To see this, suppose $\Pi_Y((x,y)) = y$. Since $Z$ is open, and products of open sets form a basis for the product topology, we have some $U$ open in $X$ and $V$ open in $Y$ such that $(x,y) \in U \times V \subset Z$. Since $\Pi_Y(U \times V) = V$ and $y \in V$, we see that $\Pi_Y(Z)$ is open in $Y$.
It should be clear that the corresponding $\Pi_X((x,y)) = x$ is also an open map.
Suppose $\{ Z_\alpha \}$ is an open cover of $X \times Y$. Pick some $x_0 \in X$. Let ${\cal A_{x_0}} = \{\alpha | (x_0,y) \in Z_\alpha \text{ for some } y \in Y \} $. Since $\{ Z_\alpha \}$ covers $X \times Y$, we see that$ \{ \Pi_Y(Z_\alpha) \}_{\alpha \in {\cal A_{x_0}}}$ is an open cover of $Y$, and so there is a finite subcover $\{ \Pi_Y(Z_\alpha) \}_{\alpha \in {I_{x_0}}}$, where $I_{x_0}$ is finite. The above result shows that there is some open $W_{x_0}$ such that $W_{x_0} \times Y$ is covered by $\{ Z_\alpha \}_{\alpha \in {I_{x_0}}}$.
The sets $\{W_{x_0} \}_{x_0 \in X}$ form an open cover of $X$, hence there is a finite subcover $\{ W_{x_0} \}_{x_0 \in F}$, where $F \subset X$ is finite.
Now take the collection $\{ Z_\alpha \}_{x_0 \in F, \alpha \in {I_{x_0}} } $ and note that it covers $X \times Y$ and is finite. Hence $X \times Y$ is compact. |
H: How can I solve $y-xy'-\sin(y')=0$
How can I solve $y-xy'-\sin(y')=0$? Are there any general techniques for solving ODE of the form $y=f(y')$, where $f$ is a trigonometric function?
AI: Assuming you mean $y(x)$, what you have is what is known as a Clairaut's equation.
Separate your $y$ and $y'$ and then differentiate with respect to $x$, then factor and solve.
After a bit of manipulation, you should get $y = \sin \alpha + \alpha x$ or $y = \pm(x \arccos (-x) + \sqrt{-x^2 +1})$. |
H: What is the difference between a communicating class and a closed communicating class?
What is the difference between a communicating class and a closed communicating class?
I checked the definition on Wikipedia
http://en.wikipedia.org/wiki/Markov_chain#Properties
but I couldn't see any difference.
AI: Assume that the only nonzero transitions on the state space $\{a,b,c\}$ are $a\to b$, $b\to a$, $b\to c$, and $c\to c$. Then $\{a,b\}$ and $\{c\}$ are communicating classes, $\{a,b\}$ is not a closed communicating class, and $\{c\}$ is a closed communicating class. |
H: Confusion on a problem on limit
Is it true that $$lim_{n->\infty}(1-(0.75)^n)^{2^n} = 0$$ and
$$lim_{n->\infty}(1-(0.25)^n)^{2^n} = 1$$. Why ?
AI: Let $\displaystyle y=\lim_{n\to\infty}(1-x^n)^{2^n}$ where $|x|<1$
$$\ln y=\lim_{n\to\infty}2^n\ln(1-x^n)=-\lim_{n\to\infty}\frac{\ln(1-x^n)}{-x^n}\cdot\lim_{n\to\infty}(2x)^n$$
As $\displaystyle|x|<1,\lim_{n\to\infty}x^n=0\implies \lim_{n\to\infty}\frac{\ln(1-x^n)}{-x^n}=1$
As $|x|<1, |2x|<2$
Now, if $\displaystyle|2x|<1\iff -0.5<x<0.5, \lim_{n\to\infty}(2x)^n=0\implies \ln y=0$
Else if $\displaystyle1<2x<2$ whence $\lim_{n\to\infty}(2x)^n=\infty\implies \ln y=-\infty$
Else if $\displaystyle1<-2x<2\iff-1>2x>-2$ whence $\lim_{n\to\infty}(2x)^n=-\infty\implies \ln y=+\infty$
Else $\displaystyle|2x|=1,$ I leave this as an exercise as this is not directly required in the current question.
Can you take it from here? |
H: How to prove that if $\sum_{n=0}^{\infty}a_n$ absolutely convergent $\Rightarrow \sum_{n=0}^{\infty}(a_n)^2$ convergent
I need to prove that if $\sum_{n=0}^{\infty}a_n$ absolutely convergent $\Rightarrow \sum_{n=0}^{\infty}(a_n)^2$ convergent.
Do you have any idea haw can I prove it?
Thank you!
AI: $\sum_{n = 1}^{\infty} a_n$ is absolutely convergent i.e. $\sum_{n = 1}^{\infty} |a_n|$ is convergent.
We have the result from Cauchy's criterion of convergence that $\lim_{n \rightarrow \infty} a_n = 0$
Thus $\exists$ $m \in \mathbb{N}$ s.t. $|a_n| < 1$ $\forall$ $n > m$.
Thus $(a_n)^2 < |a_n|^2 < |a_n|$
Then apply comparison test.
Let $s_k = \sum_{n = 1}^{k} |a_n|$ and $t_k = \sum_{n = 1}^{k} (a_n)^2$.
Thus $t_k < s_k$
$\{s_k\}$ is convergent as $\sum_{n = 1}^{\infty} a_n$ is convergent.
So $\{t_k\}$ is also convergent.
Thus $\sum_{n = 1}^{\infty} (a_n)^2$ is convergent. |
H: Show that a paraboloid is asurface .
That I know about paraboloid is all in the picture.
I wrote its surface patch. (Hopefully, it is correct)
From there, what do I need to do in order show that a paraboloid is a surface.
Definition of a surface: S in $\Bbb R^3$ isa surface if every point $p \in S$ admits an open sets W, $p \in W$ and a homemorphism $\phi :U\to W\cup S$ For U is an open set in $\Bbb R^2$
Note that these surface examples are in my notebook. I am asking these in order to understand better. These are not homework or else. Just for learning better. Thanks a lot:)
AI: If we can prove that $\sigma $ is a homeomorphism we are done.
Hint: Let $\pi:\mathbb{R}^3\to\mathbb{R}^2$ be a canonical projection, i.e. $$\pi(x,y,z)=(x,y)$$
Note that $\pi(u,v,u^2+v^2)=(u,v)$ hence, if we restrict $\pi$ to $S$ then, $\sigma^{-1}=\pi: S\to\mathbb{R}^2$.
Now, it remains to show that $\pi$ restricted to $S$ is continuous. |
H: Find the horizontal tangent line
Fine the $x$-coordinate of all points on the curve $y=\sin(2x)+2\sin(x)$ at which the tangent line is horizontal. Consider the domain $x=[0,2\pi)$. There are three $x$ values.
Can you explain how to find the horizontal tangent line and vertical tangent line.
Also, $x=[0,2\pi)$ means that $0$ is not included?
AI: The gradient$(m)$ of the tangent line $=f'(x)$
The tangent line will be horizontal of $y=f(x)$ if $f'(x)=0$ and will be vertical if $\displaystyle f'(x)=\infty\implies \frac1{f'(x)}=0$
Now, here $\displaystyle y=\sin2x+2\sin x\implies \frac{dy}{dx}=2\cos2x+2\cos x$
So, horizontal tangent, we need $\cos2x+\cos x=0\iff \cos2x=-\cos x=\cos(\pi-x)$
$\displaystyle\implies 2x=2n\pi\pm(\pi-x)$ where $n$ is any integer
Taking the '+' sign, $\displaystyle\implies 2x=2n\pi+(\pi-x)\implies x=\frac{(2n+1)\pi}3$
Now $0\le x<2\pi\implies 0\le \frac{(2n+1)\pi}3<2\pi\iff 0\le 2n+1<6\iff0\le n\le2$
Similarly, for the '-' sign
Now as $\displaystyle|\cos y|\le1$ for real $\displaystyle y, \frac{dy}{dx}=2\cos2x+2\cos x$ will be finite, hence no vertical tangent line
In fact, we can find the range of $\displaystyle 2\cos2x+2\cos x$ for real $x$ |
H: Reasoning why the implication $t - \epsilon \le x \le t + \epsilon$ for $\epsilon \ge 0 \Rightarrow x = t$ holds using sequences.
In texts I've seen the following reasoning used several times:
Suppose $t - \epsilon \le x \le t + \epsilon$ holds for $\epsilon \ge 0$. Then it in particular holds for $t - \frac 1 n \le x \le t + \frac 1 n$ $n = 1, 2, ..$
Because it holds for any $n \in \mathbb N$ we can apply the theory of sequences to conlude that $\lim_{n \rightarrow \infty}t - \frac 1 n \le \lim_{n \rightarrow \infty} x \le \lim_{n \rightarrow \infty} t + \frac 1 n$ to conclude that $t \le x \le t \Rightarrow t = x$.
Why is it allowed to use the theory of sequences here ? The inequlity is not exactly a sequence. Indeed it can be transformed into several real sequences, but wouldn't it be more accurate to conclude that $x = t$, because any other value of $x$ doesn't satisfy the inequality ?
Looking forward hearing your opinion.
AI: I don't like what you described either. In my opinion it is very wrong unless properly interpreted. As you said, the entities involved aren't sequences.
This is how I reinterpret that kind of reasoning when I see it:
So given $x,t\in \Bbb R$, you know what $\forall n\in \Bbb N\left(t-\dfrac 1 n\leq x \leq t+\dfrac 1 n\right)$ or equivalently $\color{blue}{\forall n\in \Bbb N\left(t-\dfrac 1 n\leq x\right)}$ and $\color{green}{\forall n\in \Bbb N\left(x \leq t+\dfrac 1 n\right)}$.
Blue tells you that $x$ is an upper bound of $\left\{t-\dfrac 1 n\colon n\in \Bbb N\right\}$, therefore it is greater than it's supremum, that is, $t=\color{grey}{\sup\left(\left\{t-\dfrac 1 n\colon n\in \Bbb N\right\}\right)}\leq x$.
Similarly green and infima gives you $x\leq t$. |
H: Algebraic Fractions, no I'm not kidding..
I know most of the people on this forum/site are very advanced, and I'm just sitting here wondering how in the world you do this equation, or "simplify" it.
I know how to do equations like these when there are only two terms, as in (a/b + c/d), but I can't figure out why the same rule doesn't apply to (a/b + c/d + e/f), can someone please help me, I've searched everywhere, I can't wait a whole day to ask my teacher, I need to do this, every where I search they just give me 2 term equations, thanks in advanced, sorry for being so un-professional.
https://i.stack.imgur.com/OlYfx.jpg
AI: You can think of the addition of adding $3$ fractions $\frac{a}{b}+\frac{c}{d}+\frac{e}{f}$, thanks to the associative property of addition, as $\left(\frac{a}{b}+\frac{c}{d}\right)+\frac{e}{f}$, so you just need to apply the rule for adding two fractions twice:
$$ \left(\frac{a}{b}+\frac{c}{d}\right)+\frac{e}{f}=\frac{ad+bc}{bd}+\frac{e}{f}=\frac{(ad+bc)f+bde}{bdf}=\frac{adf+bcf+bde}{bdf} $$
As an example, Your question gives $$\begin{align}&\frac{1}{x^2-7x+10}-\frac{2}{x^2-2x-15}+\frac{4}{x^2+x-6}\\
=&\frac{1}{(x-5)(x-2)}-\frac{2}{(x-5)(x+3)}+\frac{4}{(x+3)(x-2)} \\
=&\frac{1}{x-5}\left(\frac{1}{x-2}-\frac{2}{x+3}\right)+\frac{4}{(x+3)(x-2)} \\
=&\frac{1}{x-5}\left(\frac{x+3-2(x-2)}{(x-2)(x+3)}\right)+\frac{4}{(x+3)(x-2)} \\
=&\frac{7-x}{(x-5)(x-2)(x+3)}+\frac{4}{(x+3)(x-2)} \\
=&\frac{1}{(x-2)(x+3)}\left(\frac{7-x}{x-5}+4\right) \\
=&\frac{1}{(x-2)(x+3)}\left(\frac{7-x+4(x-5)}{x-5}\right) \\
=&\frac{3x-13}{(x-5)(x-2)(x+3)}
\end{align}$$ |
H: Prove that $Re+Rf=Re\oplus R(f-fe)$
Let $e,f$ be idempotent elements of a ring $R$. Prove that $Re+Rf=Re\oplus R(f-fe)$.
This post solves the first part of my question. How should we prove the second part, that $e,f$ are orthogonal?
Help me. Thanks a lot.
AI: First of all, it is a good idea to say what you have tried and how far you have come in your attempts. That way it is easir to provide a better answer.
That being said, here are some hints: You need to show two things. First, that $e$ and $f-fe$ spans $Re + Rf$, and secondly that $e$ and $f-fe$ are orthogonal.
For the first, notice that we certainly have a containment $\supseteq$. The opposite inclusion is also straight-forward.
Secondly, what happens if you multiply $e \cdot (f-fe)$? |
H: Limits - Direct Substitution - with 1/3 power functions
First image is the question from my textbook in Latex form.
Second image shows the actual question from my book along with the solution.
What i am actually confused about is that how did they open the power 1/3 and got the following result on each step.
Is there anyone who can help me more to understand each step of this question? Or help me understand a new way to do this question? Or something like this.
Help will be highly appreciated.
Regards,
Ballu Miaa
AI: They did the Taylor expansion of it:
$$(1+h)^{1/3}=1+\frac{1}{3}h+o(h^2)$$
$$(1-h)^{1/3}=1-\frac{1}{3}h+o(h^2)$$
Then, they use the fact that
$$\lim_{h\rightarrow0}\frac{o(h^2)}{h}=0$$
So they just keep the first power (the $1$s cancel) |
H: $\mathbb{Z}_{p}\left[x\right]/\left\langle f\right\rangle $ is a semilocal ring.
Let $p$ be a prime, $f$ be a nonconstant polynomial that is contained in $\mathbb{Z}_{p}\left[x\right]$. Prove that $\mathbb{Z}_{p}\left[x\right]/\left\langle f\right\rangle $ is a semilocal ring.
AI: Hint: If $f$ is a polynomial of degree $n,$ then $\mathbb{Z}_p[x]/\langle f\rangle$ only has $p^n$ elements, in particular a finite number. So there are only finitely many subsets... |
H: Can we make rectangle from this parts?
I have next problem:
Can we using all parts from picture (every part exactly one time) to make rectangle?
I was thinking like: we have $20$ small square, so we have three possibility: $1 \times 20$, $2 \times 10$ and $4 \times 5$. I can see clearly that $1 \times 20$ and $2 \times 10$ are not possible. And some mine intuition says that we also can't make $4 \times 5$, but I can't prove it rigorous. Any help?
AI: For the 4 by 5, suppose 4 rows and 5 columns, and consider rows 1,3 as blue and columns 1,3,5 as red. Then there are 10 blues and 12 reds. Now except for the T and L shapes, the other three contribute even numbers to either blue or red rows/columns. The L contributes an odd number to either blue or red, and the T contributes, depending on its orientation, either odd number to blue and an even number to red, or else an even number to blue and an odd number to red. So no matter how the T is oriented, we get an odd number for either the blues or the reds, which is impossible.
Easier proof: Take the 4 by 5 (or the 2 by 10) board and color it with black and white squares as in a traditional chess board. Then all but the T piece are such that, no matter where they are placed, they cover two black and two white squares. But the T shape must cover either 3 black and 1 white, or else the reverse 1 black and 3 white. So together the tiles cover either 9 black and 11 white, or else 11 black and 9 white. However both the 4 by 5 and the 2 by 10 board have 10 each of black and white. |
H: Diffie hellman and the discrete algorithm problem
Suppose Alice and Bob are exchanging keys using Diffie-Hellman Key-Exchange Algorithm.
a - Alice secret key
g - generator
p - prime
x - the public key passed from Alice to Bob.
Eve is listening to the communication and she is exposed to the three parameters g,p,x.
She's using a brute-force method to find $a$, Alice's secret key.
Thus, Eve is looking for $a$ satisfying this equation:
$${g^a}\bmod p = x$$
Now, I know (By testing) Eve can find $a' \ne a$ satisfying the equation above, and by using $a'$ she can also compute the common secret key used by Alice and Bob.
Why is it mathematically true?
AI: Let the secret number chosen by Bob be $b$. During exchange, Bob will send $y=g^b\bmod p$ to Alice.
The common secret key obtained after the protocol, $k$, is $g^{ab}\bmod p$. If Eve has another $a'$ that satisfies $g^{a'}\bmod p = g^a\bmod p$, then she can still perform what Alice would do after listening to $y$:
$$\begin{align*}
k' =& y^{a'}\bmod p\\
=& \left(g^b\right)^{a'}\bmod p\\
=& \left(g^{a'}\right)^{b}\bmod p\\
=& \left(g^a\right)^{b}\bmod p\\
=& k
\end{align*}$$
Which is the same as what Alice and Bob would get. |
H: Given T is normal trasformation, and $T^2 = \frac12 (T+T^*)$, prove that $T^2=T$.
if T is normal, than there exists a unitary matrix Q such that:
$Q^*TQ=diag(\lambda_1, \lambda_2, ... , \lambda_k)$
where $\lambda_1, \lambda_2, ... , \lambda_k$ are the Eigenvalues of T.
$Q^*TQ Q^*TQ= Q^*T^2Q = diag(\lambda_1, \lambda_2, ... , \lambda_k)diag(\lambda_1, \lambda_2, ... , \lambda_k) = diag(\lambda_1^2, \lambda_2^2, ... , \lambda_k^2)$.
but im stuck trying to show that $\lambda_i = \lambda_i^2$ for $i=1...k$.
AI: From the condition
$$T^2 = \frac{1}{2}(T+T^*) \qquad (\ast)$$ it follows that for each $i$ you have $\lambda_i^2 = \frac{1}{2} (\lambda_i + \bar\lambda_i)$ . Because the expression $\lambda_i + \bar\lambda_i$ is definitely real, so is $\lambda_i^2$ and hence either $\lambda_i$ is real or $\lambda_i$ is (purely) imaginary. If $\lambda_i \in \mathbb{R}$, then the equation tells you that $\lambda_i^2 = \lambda_i$, which is precisely what you need. Else, if $\lambda_i \in i \mathbb{R}$ then $\lambda_i ^2 = 0$ so $\lambda_i = 0$, and it is also true that $\lambda_i^2 = \lambda_i$.
Or, if you want to avoid diagonalisation, note first that the condition $(\ast)$ implies $$T^{*2} = \frac{1}{2} (T^*+T) = T^2.$$
Then, if you multiply $(\ast)$ by $T$ you get:
$$T^3 = \frac{T^2 + TT^*}{2}$$
Similarly, if you square $(\ast)$ you get:
$$T^4 = \frac{T^2 + 2 TT^* + T^{*2}}{4} = \frac{T^2 + TT^*}{2} = T^3.$$
Thus, $T^3(T-I) = 0$. Because of normality, this implies $T(T-I) = 0$, or $T^2 = T$. This might be preferable if you're not working in finite dimension. |
H: What is the minimal polynomial of $x$ over $k(x^p - x)$?
Let $k=\Bbb F_p$, and let $k(x)$ be the rational function field in one variable over $k$. Define $\phi:k(x)\to k(x)$ by $\phi(x)=x+1$. I know that $\phi$ has order $p$ in $\operatorname{Gal}(k(x)/k)$ and $k(x^p-x)$ is the fixed field of $\phi$, but I'm confused about the minimal polynomial of $x$ over $k(x^p-x)$. (This is problem 17 of Fields and Galois Theory by Patrick Morandi, page 27.)
Please explain about that minimal polynomial or calculate it!
AI: If you have absolutely no idea to get the minimal polynomial, and you know that there is an isomorphism $\Bbb F_p \to Gal(k(x)/k(x^p-x))$ given by $t \mapsto (x \mapsto x+t)$, then you know that the conjugates of $x$ are the $x+t$, and then you can simply compute the product $\prod_{t \in \Bbb F_p}(T-(x+t))$. This product is invariant by $x \mapsto x+1$ so its coefficients will be in $k(x^p-x)$.
You will find $\prod_{t \in \Bbb F_p}(T-(x+t)) = \prod_{t \in \Bbb F_p}((T-x)-t) = (T-x)^p - (T-x) = (T^p-T)-(x^p-x)$
This is a polynomial $P(T)$ with coefficients in $k(x^p-x)$, and $P(x)=0$.
More generally, if you consider the extension $k(f(x)) \subset k(x)$ with $f(x) = P(x)/Q(x)$, then $Q(T)f(x)-P(T)$ is an obvious polynomial over $k(f(x))$ annihilating $x$ |
H: Prove/refute: Every tautology is contingent
I'm asking to prove/refute the following statement:
Every tautology is contingent.
According to definition of contingent:
A statement that is neither self-contradictory nor tautological is called a contingent statement. A contingent statement is true for some truth-value assignments to its statement letters and false for others. Source.
Can I tell that the statement is not correct by definition or should I use a more formal way to refute it?
AI: Seeing as "tautology" and "contingent statement" are meta-language definitions (they cannot be expressed in the logical system itself; we're talking about the system) there isn't really a more "formal" way to do it.
You can expand your claim by specifying why a tautology cannot be contingent "by definition", but this won't add to the formality. You can do that if you want practice writing proofs (explaining your conclusions in words) but otherwise I see no added value. |
H: Kernel of surjective ring homomorphism and induced isomorphism
Problem: Let $\phi: R\to S$ be a surjective ring homomorphism (the rings are not necessarily commutative) such that for every $a\in \ker\phi$ there is a $n\in\mathbb{N},n\ge 1$ s.t. $a^n=0$.
i) The set $N:=1+\ker\phi$ is a normal subgroup of $R^*$, the group of units of $R$.
ii) $\phi$ induces an isomorphism $R^*/N\cong S^*$.
I am totally lost here.
i) I need to show that $gNg^{-1}=N,\forall g\in R^*$. We have $gNg^{-1}=1+g\cdot \ker\phi\cdot g^{-1}=N$, since $g\cdot \ker\phi\cdot g^{-1}\in \ker\phi$ because $\ker\phi$ is normal in $R$. But this means that $N$ is a normal subgroup of $R^*$. I'm confused that I didn't need the provided condition for this.
ii) Where do I go from here, what does "induced" mean in this context?
AI: The homomorphism theorem says that from a surjective homomorphism $\phi\colon R\to S$ we can define an isomorphism
$$
\tilde{\phi}\colon R/\ker\phi\to S
$$
by mapping each coset $r+\ker\phi$ to $\phi(r)$.
If $x,y\in \ker\phi$, you have to show that
$$
(1+x)(1+y)\in 1+\ker\phi
$$
and that every element in $1+\ker\phi$ has an inverse which still belongs to $1+\ker\phi$. This ensures $1+\ker\phi$ is a subgroup of $R^*$.
For normality, you have to show that, for $x\in\ker\phi$ and $r\in R^*$, $r(1+x)r^{-1}\in 1+\ker\phi$.
Next, you have to show that $\phi^*\colon R^*\to S^*$ (the restriction of $\phi$) induces a morphism $R^*/N\to S^*$, that is, $N\subseteq \ker\phi^*$.
Just for completeness, here's a sketch.
If $x,y\in\ker\phi$, then $(1+x)(1+y)=1+(x+y+xy)\in 1+\ker\phi=N$, because $x+y+xy\in\ker\phi$.
If $x\in\ker\phi$, set $y=-x$; then we know that $y^n=0$, for some $n>0$. Then
$$(1-y)(1+y+y^2+\dots+y^{n-1})=1-y^n=1$$
and so $1-y=1+x$ is invertible (the element is obviously also a left inverse).
If $x\in\ker\phi$ and $r\in R^*$, then $r(1+x)r^{-1}=1+rxr^{-1}\in N$.
Now $\phi$ induces a group morphism $\phi^*\colon R^*\to S^*$, because invertibles in $R$ are mapped to invertible elements of $S$ (because $\phi$ is surjective). It is obvious that $N\subset\ker\phi^*$, because $\phi^*(1+x)=\phi(1+x)=1+\phi(x)=1$.
Conversely, if $r\in\ker\phi^*$, then $\phi(r)=1$, so $r-1\in\ker\phi$ and $r=1+(r-1)\in N$. |
H: Some questions about the function $xe^{-x}$
I want to prove that $xe^{-x}$ is bounded by some constant for all the $x \in [0, \infty)$. So I take derivative of this guy, and found that on $x \in [0, \infty)$, its derivative is positive from $[0, 1)$, $0$ at $x=1$, and negative from (1, in infinity). So $xe^{-x}$ achieves its maximum in $[0, \infty)$ at $x=1$, which is $1/e$. However, to prove that it's bounded by $1/e$, I need to prove that as $x$ goes to infinity, $xe^{-x}$ converges to $0$. Because otherwise, if it went to some value below $(-1/e)$, $xe^{-x}$ would not be bounded by $1/e$. How can I prove that $xe^{-x}$ converges to $0$? I mean, as $x$ goes to infinity, $e^{-x}$ converges to zero, but at the same time, x diverges to infinity. How can I formally prove that the product term converges to $0$? Do I need to use the epsilon definition of convergence to prove it?
Could you just show me the formal proof of $xe^{-x}$ converges to $0$? Or is there's any other way to approach this question? My professor suggested that I should use the power series for $e^{-x}$...?
Your help is greatly appreciated! Thank you so much!
I hope you enjoy your Saturday!
AI: l'Hospital's Rule :
$$\lim_{x\to\infty}\frac x{e^x}\stackrel{\text{l'H}}=\lim_{x\to\infty}\frac1{e^x}=0$$
Taylor Series:
$$e^x=1+x+\frac{x^2}2+\ldots\implies\frac x{e^x}=\frac1{\frac1x+1+\frac x2+\ldots}\xrightarrow[x\to\infty]{}0$$
Do try to justify each step above. |
H: The inverse of $f(x)=\sqrt[3]{1-x^3}$ is itself
As I was studying this function, $f(x)=\sqrt[3]{1-x^3}$, I checked that the function is one-to-one, and so is invertible.
Then: $$y=\sqrt[3]{1-x^3}$$
$$y^3=1-x^3$$
$$y^3-1=-x^3$$
$$1-y^3=x^3$$
$$x=\sqrt[3]{1-y^3}$$
$$f^{-1}(y)=\sqrt[3]{1-y^3}$$
What kind of functions are inverse of itselfs? Thanks.
AI: A function that is its own inverse is called an involution.
ADDED: As @Mufasa puts nicely in the comment below: A functions is an involution if and only if it is symmetric about the line $y = x$, with the simplest involution given by $y = x$.
Just to nitpick: in your representation of $f^{-1}$, you need to swap $y$ and $x$ in the last step and express $f^{-1}$ in terms of $x$): $$f(x) = f^{-1}(x) = \sqrt[\large 3]{1 - x^3}$$
so that $$f(f(x)) = f(f^{-1}(x)) = x.$$ |
H: The double cone is not a surface.
My question is that
A double cone ( also named as "circular cone") is not a surface.
I know its reason. But I cannot show this mathematically.
Suppose $\sigma : U \to S\cap W$ Is a surface patch.
Because the vertex $(0,0,0,)$ is a problem, S is not a surface.
I can see it. But I cannot express it in the mathematical way.
When I remove the vertex point $(0,0,0)$, the double cone is a surface.
Please can someone show/write these mathematically?
AI: If you remove a point from an open subset in $\mathbb{R}^2$, it remains connected. A surface patch around the vertex has to give a homeomorphism to an open subset of $\mathbb{R}^2$. However, any open neighborhood of the vertex has the property that if you remove the vertex it becomes disconnected. Thus it cannot be homeomorphic to a subset of $\mathbb{R}^2$. |
H: Find $E[(2+X)^2]$
Given $E(X)=1, Var(X)=5$. Find $E[(2+X)^2]$.
I have two approaches to solve the problem, but they're not giving the same result.
First:
$E[(2+X)^2]=E(4+4X+X^2)=E(4)+4E(X)+E(X^2)=4+4+[Var(X)+E^2(X)]=4+4+(5+1)=14$
Second:
$E[(2+X)^2]=E[(3+X-1)^2]=Var(3+X)=Var(X)=5$
Which is wrong and what mistake did I make?
Thanks in advance
AI: You second approach would only work if $Y = 3+X$ was a random variable with expectation $1$, which is false. |
H: Does a clique always have a Hamiltonian path?
This isn't homework I'm just preparing for an exam and I came up with this question while I was reviewing the lecture notes.
AI: This is trivially true: the path $1-2-3-\dots-n$ always exists in the $n$-clique, and is Hamiltonian. You can even complete it in a Hamiltonian cycle. |
H: Group theory notation
What does the notation $(G,.)$ mean in group theory? I have seen in places that $.$ implies the binary operation multiplication on group $G$. But then, why do we show an abelian group as $(G, +)$? And what is additive and multiplicative notation?
AI: $(G,\cdot)$ denotes the ordered pair with first entry the underlying set $G$ of the group and second entry the law of composition $\cdot$ of the group. The ordered pair notation is very common in other parts of mathematics, for instance for metric spaces $(M,d)$ or topological spaces $(X,\tau)$ and so on (the general pattern is $(\mathrm{set},\mathrm{structure \ on \ the \ set})$; it is useful because one does not have to define the notion of equality for groups, metric spaces and topological spaces and so on separately as it can be shown that two ordered pairs $(x,y)$ and $(x',y')$ are equal as sets iff $x=x'$ and $y=y'$. Hence two groups are equal iff their underlying set and their law of composition are equal. It is however very common to denote the group $(G,\cdot)$ simply by $G$ (similarily for metric and topological spaces) if the law of composition (metric, topology) is understood. Also a group is by definition an ordered pair $(G,\cdot)$ with $G$ a set and $\cdot$ a law of composition on $G$ subject to the group axioms (some people say that a group is a set $G$ "together with a law of composition on $G$" by which they simply mean that $(G,\cdot)$ is an ordered pair).
Additive notation refers to denoting the law of composition by $+$ (multiplicative notation $\cdot$), the unit element by $0$ (multiplicative notation $1$) and the inverse of $x$ by $-x$ (multiplicative notation $x^{-1}$). As others have pointed out, this is by convention often done when the law of composition is commutative. |
H: Given $T^2=\frac12(T+T^∗)$, prove that T is normal trasformation.
using the fact that:
$T = T_1 + T_2$
$T^* = T_1^* + T_2^*$
when:
$T_1 = \frac12 (T + T^*), T_2 = \frac12 (T - T^*)$ and $T_1^* = T_1, T_2^*=-T_2$.
then:
$T^*T = (T_1^* + T_2^*)(T_1 + T_2)= T_1^*T_1 + T_1^*T_2+T_2^*T_1+T_2^*T_2 = T_1^2 + T_1T_2-T_2T_1 + T_2^2$
$TT^* = (T_1 + T_2)(T_1^* + T_2^*)= T_1T_1^* + T_1T_2^*+T_2T_1^*+T_2T_2^* = T_1^2 - T_1T_2+ T_2T_1 + T_2^2$.
how can i use the fact that $T^2=\frac12(T+T^∗)$ to show that $T_1$ and $T_2$ commute?
AI: Hint: The relation
$$T^2 = \frac12(T+T^\ast)$$
allows you to express $T^\ast$ as a polynomial in $T$: $T^\ast = P(T)$. |
H: Cayley tables and cyclic subgroups
Given a Cayley table, how can I spot the subgroups, without missing any of them? I know which orders of subgroups that I should look for because of Lagrange's theorem, but how can I create an exhaustive list of subgroups without skipping any?
AI: Subsets of the group are created by deleting elements from the group (obviously). This amounts to deleting a row and column from the corresponding Cayley table. Now, you can tell if this new Cayley table represents a group by the following criterion: If each element in the subset appears exactly once in a row and column, then the resulting Cayley table represents a group (more specifically a subgroup of the original group). Stated differently, the rows (and columns) of a Cayley table for a group are simply permutations of the elements of the group. |
H: Zeros of a Polynomial and maximum principle
Let $P: \mathbb C \to \mathbb C$ be a non-constant polynomial and $c>0$. Let $\Omega =\{z\in\mathbb C : |P(z)|<c\}$.
I can't understand how does the maximum principle implies that every connected component of $\Omega$ contains at least one zero of $P$.
AI: Let $U$ be a connected component of $\Omega$. Then we have $\lvert P(z)\rvert = c$ on $\partial U$ - if $\lvert P(z_0)\rvert \neq c$, then there is an $r > 0$ with $\lvert P(z) - P(z_0)\rvert < \bigl\lvert \lvert P(z)\rvert-c\bigr\rvert$ for $\lvert z-z_0\rvert < r$, and the disk $D_r(z_0)$ is either completely contained in $U$ or doesn't intersect $U$ at all.
Since $P$ is non-constant, $\Omega$, and hence $U$, is bounded. By definition, we have $\lvert P(z)\rvert < c$ on $U$. So if $P$ had no zero in $U$, the function
$$f(z) = \frac{1}{P(z)}$$
would be holomorphic in the bounded open set $U$, continuous on $\overline{U}$, and satisfy
$$\lvert f(z)\rvert > \frac{1}{c} = \sup_{\zeta\in\partial U} \lvert f(\zeta)\rvert,$$
for some (all) $z\in U$, which contradicts the maximum principle. |
H: Computing the integration $\int_0^{+\infty}{\frac{\sin(x)}{\sqrt{x}}}dx$.
For any $A>0$,$|\int_0^A{\sin(x)}|\le2$ and ${\frac{1}{\sqrt{x}}}$ is decreasing to $0$ as $x$ tend to $+\infty$.By Dirichlet's test,the integration $\int_0^{+\infty}{\frac{\sin(x)}{\sqrt{x}}}dx$ makes sense.
Moreover,I know the method of computing $\int_0^{+\infty}{\frac{\sin(x)}{x}}dx$ which is that considering the integration $$I(\beta)=\int_0^{+\infty}f(x,\beta)dx=\int_0^{+\infty}e^{-\beta{x}}{\frac{\sin(x)}{x}}dx$$ and $$I^{'}(\beta)=\int_0^{+\infty}f_\beta(x,\beta)dx=-\int_0^{+\infty}e^{-\beta{x}}{\sin(x)}dx.$$
We can compute $\int_0^{+\infty}e^{-\beta{x}}{\sin(x)}dx$ easily thourgh integral by parts.If we use the same method,we will obtain$$I^{'}(\beta)=-\int_0^{+\infty}e^{-\beta{\sqrt{x}}}{\sin(x)}dx.$$
Integral by parts is out of use since it will produce many $\sqrt{x}$.Can you tell me how to integrate this integration and more general case $\int_0^{+\infty}{\frac{\sin(x)}{x^a}}dx$,where $0<a<1$.Thank you in advance!
AI: Let $t^2=x:$
$$\int_0^{\infty}\frac{\sin x}{\sqrt{x}}\,dx=\int_0^{\infty} 2\sin t^2\,dt=\sqrt{\frac{\pi}{2}}$$
Where the latter integral is the Fresnel integral. |
H: Proof of an inequality in arithmetic, geometric and harmonic means
Prove that $$\left[\frac{x^2+y^2+z^2}{x+y+z}\right]^{x+y+z}\gt x^xy^yz^z\gt \left[\frac{x+y+z}{3}\right]^{x+y+z}$$
I could prove the inequality between the first two terms using the fact that $A.M\gt G.M$ in the following way. $$\left[\frac{x(\frac{x^2}{x})+y(\frac{y^2}{y})+z(\frac{z^2}{z})}{x+y+z}\right]\gt (x^xy^yz^z)^{\frac{1}{x+y+z}}$$
But I am unable to prove the inequality between the next two terms.
AI: Just to spell out what Ivan says:
Consider the sequence $a_1,a_2,a_3 = x, y, z$ and weights $w_1,w_2,w_3 = x/s, y/s, z/s$, where $s = x+y+z$. The inequality between harmonic and geometric (weighted) mean says that:
$$ \prod_{i} a_i^{w_i} \geq \frac{1}{\sum_i w_i \frac{1}{a_i}}.$$
Thus:
$$ x^{x/s}y^{y/s}z^{z/s} \geq \frac{1}{\frac{x}{s} \frac{1}{x} + \frac{y}{s} \frac{1}{y}+ \frac{z}{s} \frac{1}{z}} = \frac{1}{3 \cdot{1}{s} }= \frac{s}{3}.$$
Rising both sides to power $ s $ gives the claim:
$$ x^x y^y z^z \geq \left( \frac{x+y+z}{3} \right)^{x+y+z}.$$ |
H: Definite Integral $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\frac{(\cos(x))^{\arcsin(x)+1}}{(\cos(x))^{\arctan(x)}+(\cos(x))^{\arcsin(x)}}dx$
How can I prove that
$${\large\int_{-\pi/2}^{\pi/2}}\frac{(\cos(x))^{\arcsin(x)+1}}{(\cos(x))^{\arctan(x)}+(\cos(x))^{\arcsin(x)}}dx=1$$
AI: HINT:
Let $\displaystyle \arcsin x=\theta$
$\displaystyle\implies(i) x=\sin\theta$
and $\displaystyle(ii) -\frac\pi2\le\theta\le \frac\pi2$ based the definition of the principal value of inverse sine function
$\displaystyle\implies -x=-\sin\theta=\sin(-\theta)$ and $\displaystyle -\frac\pi2\le-\theta\le \frac\pi2$
$\displaystyle\implies -\theta=\arcsin(-x)\implies -\arcsin x=\arcsin(-x)$
Similarly, $\displaystyle -\arctan x=\arctan(-x)$
$$\text{Now,}\int_a^bf(x)dx=\int_a^bf(a+b-x)dx\implies \int_{-b}^bf(x)dx=\int_{-b}^bf(-x)dx$$
$$\implies2I=\int_{-b}^bf(x)dx+\int_{-b}^bf(-x)dx=\int_{-b}^b\left(f(x)+f(-x)\right)dx$$ |
H: Distribution of the indicator function of a Gaussian variable.
Let $\xi$ be a random variable distributed according to a Normal distribution with given mean $\mu$ and standard deviation $\sigma$. Find the probability density function of
$$
\psi = c\,\mathbb{1}_{\left\{\xi\leq 0\right\}},
$$
where $\mathbb{1}_{\left\{\cdot\right\}}$ is the indicator function.
AI: Since the random variable $\psi$ is discrete, its distribution has no density. Rather, one can describe it saying that $P[\psi=c]=p$ and $P[\psi=0]=1-p$, where $p=P[\xi\leqslant0]$, thus, $p=\Phi(-\mu/\sigma)$. |
H: Prove, in a $\Delta ABC$ with medians $BE, CF$, $BE + CF > BC\cdot\frac32$
Consider a $\Delta ABC$ with medians $BE$ and $CF$. Prove that
$$BE + CF > BC\cdot\frac32$$
Consider the following inequalities, given by the triangle inequality:
$$BE + CE > BC \implies BE + \frac{AC}{2} > BC$$
$$\implies BE > BC - \frac{AC}{2} \tag1$$
Similarly,
$$CF > BC - \frac{AB}{2} \tag2$$
Adding the $(1)$ and $(2)$ gives us:
$$BE + CF > 2BC - \left(\frac{AB + AC}{2}\right) \tag3$$
Now,
$$AB + AC > BC$$
$$\implies -\frac{AB+AC}{2} < -\frac{BC}{2}$$
So, replacing $\frac{AB + AC}{2}$ by $\frac{BC}{2}$ in $(3)$ won't work. How do I proceed?
AI: Join $EF=\frac12BC$
Let $G$ be the intersection of $BE,CF$
Now in $\triangle BGC, BG+CG>BC$ and in $\triangle EFG, GF+GE>EF=\frac12BC$
Now add them and use $BG+GE=BE$ and $CG+GF=CF$ |
H: Finding the values of a
What are the values of $a$ for which all the roots of the equation $x^4-4x^3-8x^2+a=0$ are real?
My approach:
On differentiating the polynomial and equating it to $0$, we get $(x)(x+1)(x-4)=0$. So, the derivative of the given equation has all the three roots real. Now, what should be the binding condition so as to find the interval in which $a$ lies?
AI: Hint: If you plot the polynomial with $a=0$ (you may need to change the range to see everything) you see it comes down from $+\infty$, hits a local minimum, goes up to a local maximum, hits the global minimum, and rises to $+\infty$ Changing $a$ positions the $x$ axis. You want to make sure the $x$ axis cuts the graph in four places. |
H: Finding the probability of winning the tennis game
Mark plays best when it's sunny. Mark wins a set with probability P(s) when it's sunny, and with probability P(r) when it's raining. The chance that there will be sun for the first set is P(i). Luckily for mark, whenever he wins a set, the probability that there will be sun increases by P(u) with probability P(w). Unfortunately, when mark loses a set, the probability of sun decreases by p(d) with probability p(l). What is the chance that mark will be successful in his match?
You'll be given number of sets(N).
I have no clue about how to solve it ! What are the things i need to study to solve these types of problems.And please suggest a solution by explaining in simple terms :)
Testcase :
Number of sets=1
P(s)=0.800
P(r)=0.100
P(i)=0.500
P(u)=0.500
P(w)=0.500
P(d)=0.500
P(l)=0.500
Chance that he'll win= 0.450000
Source: This is a programming contest problem....
AI: You didn't specify if the match is first to win two or first to win three. I will assume two and start, but the same approach works (with more work) for three. He can win the set in three ways: win the first two games, win-lose-win, or lose-win-win. The chance that he wins the first game is $P(i)P(s)+(1-P(i))P(r)$. The first term is the chance that it is sunny and he wins, the second is the chance it is raining and he wins. In your sentence, "whenever he wins a set, the probability that there will be sun increases by P(u) with probability P(w)" it is not clear what happens if the $P(w)$ does not occur. I will assume that the chance of sun does not change. Now if he wins the first game, the chance of sun is $P(i)+P(w)P(u)$, so the chance of winning is $(P(i)+P(w)P(u))P(s)+(1-(P(i)+P(w)P(u))P(r)$ with the same logic as above. You need to keep chasing through the possibilities, accumulating the probabilities until you are done.
Added: You are building a tree. For each set, there are four possibilities: win with P(i) changing, win with no change, lose with P(i) changing, lose with no change. After $n$ sets you have $n+1$ possibilities for how many sets he has won and for each one you have $n$ possibilities for the number of up and down changes in P(i). You can calculate the chance of each of these following the idea above. |
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