Sir how to generalize Powers of Matrices in summation notation?? A^n in Sigma Notation?? This question was asked by my sir when we were studying basic maths for physics and Linear Algebra(Vectors and Matrices at JEE level) chapter came.. Edit (A is a square matrix of order m)
Sir please answer to this question for a square matrix of order m.
@TedShifrin Sir, Is sheldon Axler good book for studying basics. I am currently studying Lang's Introduction to Linear Algebra ?? Or should I prefer to Gilbert Strang??
Just to make sure my understanding is correct: $\Gamma$ function has a meromorphic continuation on $\Bbb C$ with simple poles at $-\Bbb N\cup \{0\}$. $\sin(\pi z/2)$ has simple zero at $0$ so $\Gamma(z)\sin(\pi z/2)$ has a removable singularity at $0$. Since $\Gamma(z)\sin(\pi z/2)$ is analytic on $-1<Re(z)<1$ as a consequence, this conclude the above identity $$\int_0^\infty\sin(t)t^{z-1}dt = \Gamma(z)\sin\left(\pi{z\over 2}\right)$$ is valid on $-1<Re(z)<1$?
@KabirMunjal Axler is a more advanced book for people who already know the basics. Strang is great, but not good for learning theory/proofs. Of course, I like my own books best.
I think from my argument, I can only say the integral $$\int_0^\infty\sin(t)t^{z-1}\ dt$$ has an analytic continuation on $-1<Re z<1$. Showing the *identity* holds requires more than that I think. I need to show the integral is holomorphic on that region.
The question is from Stein complex analysis 6.10(b)
(b) Show that the following identity
$$\int_0^\infty\sin(t)t^{z-1}\ dt = \Gamma(z)\sin\left(\pi{z\over 2}\right)\quad 0<\operatorname{Re}z<1$$
is valid in the larger strip $-1<\operatorname{Re}z<1$.
Since $\Gamma$ has a meromorphic continuatio...
No, but it seems the integrand contains $0$ as a singularity does not matter that much. Now I'm thinking $F_{\epsilon} = \int_{\epsilon}^{1/\epsilon} \sin(t)t^{z-1}\ dt$ and letting $\epsilon\to 0$.
My way may not work as one converges for $(0,1)$ and other for $(-1,0)$, but it's worth looking at. However, integrating on $(\epsilon,1/\epsilon)$ might work
I'm nearly done: if $F(z) = \int_0^\infty\sin(t)t^{z-1}\ dt$, \begin{align*} |F(z)-F_\epsilon(z)|& \leq\int_0^\epsilon\sin(t)t^{z-1}\ dt+\int_{1/\epsilon}^\infty\sin(t)t^{z-1}\ dt\\ &\leq\int_0^{\epsilon}t^{x-1}\ dt+\int_{1/\epsilon}^\infty\sin(t)t^{x-1}\ dt\\ \end{align*} The very last integral is integrable (at least Wolfram says yes) but I don't know how to show.
Once proved, then the $F_\epsilon\to F$ is uniformly convergent.
Yes I see why $\int_0^1\sin(t)t^{z-1}\ dt$ is holomoprhic on $Re(z)>-1$ the problem I can't understand is why $\int_1^\infty\sin(t)t^{z-1}\ dt$ is holomoprhic on $Re(z)<1$
I want to show that it holds that $P_sG_k=\tilde{G}_kP_s, \ s\geq k$, where we get $\tilde{G}_k$ from $G_k$ by exchanging two entries $\ell_{j_1,k}$ and $\ell_{j_2,k}$ with $j_1,j_2\neq k$ ($j_1\neq j_2$), that are exchanged also at the permutation $P_s$.
Could you give me a hint how to show that...
@TedShifrin I believe that it is rare! It seems to have some very strong implications on the polynomial. The condition has appeared naturally in something I've been looking at, and rather than investigate it myself my advisor told me to find it in the literature (unfortunately so far unsuccessful in finding anything written about this condition), since its sure to have been investigated
In the second part "To prove the converse..." in the line $|x_n - x_0| < \frac{1}{n}$, is the $n$ in $x_n$ and $\frac{1}{n}$ the same or is it just poor use of notation to use $n$ in both the places?
@robjohn if you have time and if it's not too much of a hazard, could you briefly explain how you reasoned when you calculated the Fourier transform of $$h(x,t)=\frac{1}{2\pi}\int_{-\infty}^{\infty} \Re\{a(k)e^{i(kx-\omega t)}\}\mathrm{d}k.$$ What defining property did you use?
I realized that Tychonoff theorem is indeed needed here. I figured out the solution which is furnished as follows: By the given hypothesis, suppose that $X_a$'s are compact for all a except $a\in \{a_1, a_2,...,a_k\}=:A_k$. Now, take any $(y_a)_a\in \Pi_a X_a$. For $a\notin A_k$, $y_a\in X_a\subset X_a$ (i.e. $y_a$ has a neighborhood contained in a compact space). For $a\in A_k$, by local compactness of $X_a$, there exists a compact subspace $C_a\subset X_a$ containing a neighborhood $U_a$ of $y_a$. It follows that $(y_a)_a\in \color{blue}{\Pi_a V_a}\subset \color{green}{\Pi_a Z_a}$, where…
in general, one-point compactifying a finite topological disjoint union of spaces works by one-point compactifying each summand and then gluing all those compactifications together at the respective point at infinity
@Koro That is the point at infinity. Very roughly, I am imagining that the interval $(1,2)$ is compactified by gluing a $1$ to the left, a $2$ to the right, and then identifying those two points.
This isn't really the "right" way to think about it, but it works for things like intervals.
So, glue 1 to 2 to make a loop; glue 3 to 4 to make a second loop; and glue 5 to 6 to make a third loop. Then glue those three points together to make a wedge of three circles.
Also, if you don't want to deform your circles, you could imagine them in three dimensions.
the point is that if you remove the point at which the circles touch from Xander's picture, what remains is homeomorphic to the disjoint union of three open intervals
(and that, together with the point being closed and the space being compact, actually forces it to be the one-point compactification)
I thought that add oo to T to get the compactification Z= TU {oo}. Now there is a continuous bijection from Z to the the circles which sends oo to the point of intersection of the circles.
This continuous bijection also happens to have a continuous inverse.
I believe one possible answer could be: Every continuous function is measurable and maps defined on a countable set are by definition continuous so $f$ is measurable.
A set system is a pair $(X, \tau)$ where $\tau \subseteq \mathcal{P}(X)$. A topological space is a set system where
\begin{align*}
&X \in \tau \\
&(\forall a \in \tau)(\forall b \in \tau)(a \cap b \in \tau) \\
&(\forall S \subseteq \tau)(\mathop{\cup} S \in \tau)
\end{align*}
Is there an analogou...
If $p(x)$ describes the production cost for $x$ production units, $p'(x)$ is the rate of change of the production cost per unit at $x$ production units, so approximately the production cost for the next production unit/$p(x+1) - p(x)$.
>It's optimal for a company if $p'(x)$ is equal to $\frac{p(x)}{x}$, so the median cost per unit.
Suppose $X$ is path connected and locally path connected and $\tilde{X}$ is path connected where $p:\tilde{X}\to X$ is a covering space then does it imply that $\tilde{X}$ locally path connected?
Show that if $p:X\rightarrow Y$ is a covering map and $Y$ is locally path-connected, then so is $X$.
How do you go about proving this? I can think of two ways of doing this, either by definition of locally path-connected and covering map or by trying to use the lift.
I've tried writing out the ...
I have a doubt in the part where we prove that for subgroup we have covering. $X$ is PC,LPC, S-LSC. We defined an equivalence relation $[\gamma]\sim[\gamma']$ if $\gamma(1)=\gamma'(1)$ and $[\gamma\ast\gamma'^{-1}]\in H$. I can't prove reflexivity.
$\gamma$ and $\gamma'$ start at base point $x_0$.
So $\gamma\ast\gamma^{-1}$ is a loop at $x_0$. The book said it's the identity element of $H$ that is the constant loop at $x_0$.
So I need to show that this loop is homotopic to the constant loop. I can't show it.
Part of me prefers "It is I", but I agree that it generally sounds too formal. The linked dupe on that question suggests that French forms like "C'est moi" influenced the modern English tendency towards "It is me".
Let $Y$ be a space, $X$ a space that is homotopy equivalent to a point and $f, g : Y \to X$ two continuous maps. Prove that there is a homotopy from $f$ to $g$.
@TedShifrin Two continuous maps $f,g:Y \to X$ are homotopic if there exists a continuous map $F:Y \times [0,1] \to X$ such that $F(y,0)=f(y)$ and $F(y,1)=g(y)$ for all $y \in Y$.
Anyone know how to calculate a projected win percentage for a team based on their number of points scored compared to that of the opponents? I’m purely looking at total points scored for each team to calculate a projected win percentage?
@TedShifrin $X$ is homotopy equivalent to a $1$-point space, say $Z=\{c\}$. In other words, there exist $f:X \to Z$ and $g:Z \to X$ such that $g \circ f \simeq \text{id}_X$ and $f \circ g \simeq \text{id}_Z$.
Correct in certain things. I’m only looking at points scored because in fantasy football, I basically have no control over the amount of points my opponent scores
I could use the current win/loss records of all the teams in the league to calculate a win percentage but some people have scored so many total fantasy points but have almost no wins and I want to take that into account
I think I’ll just use the average points for all the teams and scale a 0.5 win percentage by points/avg so someone averaging 10% more points than the average will have a 55% projected win percentage
Using Row spaces and Column spaces to prove the Lagrangian multiplier huh?......my spidey senses are tingling and tell me that in my future I'm going to encounter this proof with the use of adjoints instead soemwhere....
@userunkown picture $\Bbb R$ as being homotopic to the origin. You can squeeze the whole space continuously to the origin. What does that squeeizing do to the values of the map?
DC This should remind you of section something of chapter 3.
@TedShifrin Hello, I am working on Exercise 2.2.9 of "Multivariable Mathematics" and I was wondering if you could comment on my proof of E.2.2.9(b) and perhaps offer some insight into E.2.2.9(a).
Exercise 2.2.9 asks: (a) Is it true that all the interior points of $\bar{S}$ are points of $S$? Is it true if $S$ is open?
At first I believed this to be true but I haven't been able to make much progress; perhaps if it is not true there is a set $S$ whose frontier is "wide" enough so that an entire neigbhorhood centered at one of its points could be completely contained in it but I haven't been…
You mean do the problem for you, @user. Here’s the major hint. Prove that if $X$ is homotopic to a point, then every map $Y\to X$ is homotopic to a constant msp.
I thought that if $\mathscr{F}(F)$ does not coincide with $F$ it should contain at least one element that is not in $F$ and this element should be in $F^c=(S\setminus F)\cup (S^c\setminus F)$
If I take $f:(\Omega, F)\rightarrow (\Bbb{R},\mathcal{B}(\Bbb{R}))$ to be a measurable function such that $\int |f| d\mu <\infty$.
Is it then true that $\int f d\mu$ exists?
I would say yes, and my proof would be the following.
Remark that $f=f^+-f^-$, and we know that $f$ is measurable iff $f^+,f^-$ are measurable. So $\int f d\mu=\int f^+ d\mu-\int f^- d\mu<2\int |f| d\mu<\infty$. So I would conclude that $\int fd\mu$ exists. Does this work?
I can't comment on the wedge product because I haven't encountered it yet. But I'm surprised the author knows the wedge product but is uncomfortable with the "basic algebra" part......
also while talking about it, I like the touch of thinking about the lagrangian equality with constraint equations as similar triangles. Hadn't seen it before. I guess geometry is superior....
@D.C.theIII regarding your question on the differentiability of $A^2$ perhaps you'll find my question on the same topic ( math.stackexchange.com/questions/4392596/… ) useful
@lorenzo You have a sqrt missing in your norm defn. Later on (chap 5) we get to the norm as a linear map (as opposed to the length of a very long vector).
Speaking of algebra. While solving a polynomial of 4th degree whose roots, while real, have no closed form, I thought of using the newton-raphson algorithm. But now I'm confused, many iterations do I need to make to get a "good" answer
I'm not doing it physically since I can just code it up but still, I'd like to use a reasonable number of iterations while still getting a reasonably accurate result
@Thorgott well yes that's essentially what I mean. It does have a closed form (probably) but I don't want to look at it because I know what it might look like
I instead got a numerical result with several decimal places
@TedShifrin I think I am almost done. Please help me finish. I have doubt in the last line.
@TedShifrin Let $Y$ be a space, $X$ a space that is homotopy equivalent to a point and $f, g:Y \to X$ two continuous maps.
We argue that if $X$ homotopy equivalent to a point, then any map $h:Y \to X$ is homotopic to a constant map. Assume that $X$ is homotopy equivalent to a $1$-point space, say $Z=\{c\}$. In other words, there exist $f':X \to Z$ and $g':Z \to X$ such that $$g' \circ f' \simeq \text{id}_X \text{ and } f' \circ g' \simeq \text{id}_Z\, .$$ Define $\widehat{h}=f' \circ h:Y \to Z$. Since $Z$ is a $1$-point space, we have $f'(x)=c$ for every $x \in X$. Now, for all $y \in Y$, …
@TedShifrin Is this line correct: Then, by the argument in the previous paragraph, $f,g:Y \to X$ are homtopic to a constant map, say $c:Y \to X$
@Thorgott Can you please check my proof. Here is the question: Let $Y$ be a space, $X$ a space that is homotopy equivalent to a point and $f, g:Y \to X$ two continuous maps. Prove that there is a homotopy from $f$ to $g$.
The question is from Stein complex analysis 6.10(b)
(b) Show that the following identity
$$\int_0^\infty\sin(t)t^{z-1}\ dt = \Gamma(z)\sin\left(\pi{z\over 2}\right)\quad 0<\operatorname{Re}z<1$$
is valid in the larger strip $-1<\operatorname{Re}z<1$.
Since $\Gamma$ has a meromorphic continuatio...
Show that if $p:X\rightarrow Y$ is a covering map and $Y$ is locally path-connected, then so is $X$.
How do you go about proving this? I can think of two ways of doing this, either by definition of locally path-connected and covering map or by trying to use the lift.
I've tried writing out the ...
