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"Ordering" of Complex Plane I have heard that the Complex Numbers do not form an ordered field, so you can't say that one number is larger or smaller than another. I've always been fine with this, but I recently wondered what happens if one describes size as the distance along a space-filling curve? If the real numbers can be ordered, and you can map a curve through all points in 2D space, then what prevents you from ordering the points purely by how far along the curve the points are? Each point should be a unique distance, and the distance continuously increases (as far as my understanding goes). Is there perhaps some difference between being "ordered" and being an "ordered field" that I am missing? I know the basics about why they differ, but I lack formal math training above Algebra 2... Everything else (up to most of Real Analysis) I have studied on my own, so my knowledge is spotty in areas like this.
|
You can definitely put an order $\prec$ on the complex numbers. In fact, the order can be a well-order, by the well ordering theorem. The question is whether such an order "is useful."
Here are some things we probably want in the order $\prec$:
*
*if $0 \prec \alpha, \beta$, then $0 \prec \alpha \beta$.
*if $\alpha, \beta, z, w$ are complex with $\alpha \prec \beta$ and $z \prec w$, then $\alpha + x \prec \beta + y$.
In fact, an order on a field (such as the complex numbers) that satisfies these properties gives us an ordered field. See Ross Millikan's comment for why you cannot put such an order on $\mathbb C$.
|
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|
Projection and positive element in C$^*$-algebras Let $A$ be a $C^*$-algebra, $p\in A$ a projection.
Assume that $a$ is a element in $ \text{Ball}(A_+)$ such that $a\leq p$.
Q: May I say $ap=pa$? Why?
|
You have
$$
0\leq(1-p)a(1-p)\leq(1-p)p(1-p)=0.
$$
Thus $$0=(1-p)a(1-p)=(a^{1/2}(1-p))^*(a^{1/2}(1-p)),$$
and $a^{1/2}(1-p)=0$, from where $a(1-p)=0$; so $a=ap$. Taking adjoints, $a=pa$. If follows that $a=pap$.
|
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|
What is an example of a non-zero "ring pseudo-homomorphism"? By "pseudo ring homomorphism", I mean a map $f: R \to S$ satisfying all ring homomorphism axioms except for $f(1_R)=f(1_S)$.
Even if we let this last condition drop, there are only two ring pseudo-homomorphisms from $\Bbb Z$ to $\Bbb Z[\omega]$ where $\omega$ is any root of unity, for example. They are the identity homomorphism and the "$0$ pseudo-homomorphism".
I couldn't find a non-zero example. It is easy to conclude that $0=f(a)(1_S-f(1_R))$, so I know that the counterexample will have to be some $S$ with zero-divisors.
|
Every commutative example takes the following form. Suppose $f : R \to S$ is a non-unital ring homomorphism between two commutative rings. Then $f(1_R)$ is some idempotent $m \in S$, as Jendrik Stelzner remarks. $mS$ is a "non-unital" subring of $S$ (it's a subring except that its unit is $m$, not $1_S$), and $f$ is a ring homomorphism in the ordinary sense to this subring. Moreover, $S$ decomposes as a product of rings
$$S \cong mS \times (1 - m)S.$$
So darij's comment essentially exhausts all examples.
Geometrically such a morphism corresponds to a "partially defined" morphism $\text{Spec } S \to \text{Spec } R$ of affine schemes, where "partially defined" means defined on some union of connected components. There are analogous statements one can make about non-unital C*-algebra homomorphisms between commutative C*-algebras.
|
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Prove that $\sum_{n=1}^\infty \frac{1}{n(n-x)}$ converges absolutely and uniformly on each [a,b]\N Let $a<b$, $a,b\in R$ Prove that the series $\sum_{n=1}^\infty \frac{1}{n(n-x)}$ converges absolutely and uniformly on [a,b]\N
What is [a,b]\N? I am really confused. For absolute convergence I know we take the absolute of the series and for uniform we need Weistress or Abel's uniform convergence test.
Any help much appreciated!
|
Fact: Suppose $f_n$ is a sequence of real-valued functions defined on a set $E.$ Let $N\in \mathbb N.$ If $\sum_{n=N}^{\infty} f_n$ converges uniformly on $E,$ then $\sum_{n=1}^{\infty} f_n$ converges uniformly on $E.$ I'll use this below.
In our problem, it's enough to show that the series converges uniformly on $[-N,N]\setminus \mathbb N$ for each $N \in \mathbb N.$ Fix such an $N.$ Then for $n>N,$ we have
$$\left |\frac{1}{n(n-x)} \right| \le \frac{1}{n(n-N)}$$
for all $x \in [-N,N].$ Because $\sum_{n=N+1}^{\infty} 1/n(n-N) < \infty,$ we see $ \sum_{n=N+1}^{\infty} 1/n(n-x)$ converges uniformly on $[-N,N]$ by the Weierstrass M-test. This implies $ \sum_{n=1}^{\infty} 1/n(n-x)$ converges uniformly on $[-N,N]\setminus \mathbb N$ as desired.
|
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|
Intersection of all open intervals of the form $(-\frac{1}{n}, \frac{1}{n})$ is $\{0\}$? In Mathematical Analysis by Apostol he mentions that the "Intersection of all open intervals of the form $(-\frac{1}{n}, \frac{1}{n})$ is $\{0\}$"
Obviously this is a super basic question but I thought that an open interval does not include the endpoints, so from the limit as $n\to\infty$ on both sides we get (0,0), i.e. if $x$ belongs to the intersection then $0<x$ and $x<0$ which no real number satisfies.
|
$\cap_{n=1}(-\frac{1}{n}, \frac{1}{n}) = {0}$ is equivalent to the statement that for all $n \in \mathbb{N}, 0 \in (-\frac{1}{n}, \frac{1}{n})$, which is obviously true.
As $n \rightarrow\infty$, $(-\frac{1}{n}, \frac{1}{n})$ approaches the zero, but never reaches the zero
|
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Find function f(x) Find function f(x), where:
$$f(3)=3$$
$$f'(3)=3$$
$$f'(4)=4$$
$$f''(3) = \nexists$$
How to find function like this in general? What steps should I do?
|
Take a function $|x|$ which has no derivative at 0. Integrate it to get a function $x\cdot|x|$ (yes, $1\over2$ is missing, but who cares) which has no second derivative at 0. Shift it to get $(x-3)|x-3|$ which has no second derivative at 3. Add some $const\cdot x^2$ to get $f'(4)-f'(3)=1$. Then add some $const\cdot x$ to get whatever derivative you like at 3. Then add a constant to get the desired value at 3.
|
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Are $\frac{\pi}{e}$ or $\frac{e}{\pi}$ irrational? Is it clear whether $\displaystyle \frac{\pi}{e}$ or $\displaystyle \frac{e}{\pi}$ are irrational or not?
If not, then there would exist $q,p\in \mathbb{Z}$ such that $$p\cdot \pi = q\cdot e$$
|
A QUITE SIMPLE REMARK $$\begin{cases}\frac{e}{\pi}=t\\e+\pi=s\end{cases}\qquad (*)$$ would imply $$\pi=\frac{s}{t+1}\\e=\frac{st}{t+1}$$ Consequently and least one of $t$ and $s$ in (*) must be trascendental.
|
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How to evaluate $\lim _{x\to 0}\left(\frac{x\left(\sqrt{3e^x+e^{3x^2}}-2\right)}{4-\left(\cos x+1\right)^2}\right)$? I have a problem with this limit, I don't know what method to use. I have no idea how to compute it. Can you explain the method and the steps used?
$$\lim _{x\to 0}\left(\frac{x\left(\sqrt{3e^x+e^{3x^2}}-2\right)}{4-\left(\cos x+1\right)^2}\right)$$
|
You get in the denominator
$$
2^2-(1+\cos x)^2=(3+\cos x)·(1-\cos x)=(3+\cos x)·2 \sin^2(x/2)
$$
so that you can reduce the limit to
$$
\lim_{x\to 0}\frac{x^2}{(3+\cos x)·(1-\cos x)}·\lim_{x\to 0}\frac{\sqrt{3e^x+e^{3x^2}}−2}{x}
\\
=\frac12·\lim_{x\to 0}\frac1{\sqrt{3e^x+e^{3x^2}}+2}·\lim_{x\to 0}\frac{3(e^x-1)+(e^{3x^2}-1)}{x}
$$
which now has simple limits
|
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Sudoku grid guaranteed to be solvable? I want to generate random sudoku grids.
My approach is to fill the three 3x3 groups on a diagonal of the grid, each with the numbers 1-9 randomly shuffled. That looks e.g. like the grid below:
+-------+-------+-------+
| 5 6 4 | · · · | · · · |
| 1 7 2 | · · · | · · · |
| 9 8 3 | · · · | · · · |
+-------+-------+-------+
| · · · | 3 2 5 | · · · |
| · · · | 7 9 6 | · · · |
| · · · | 8 1 4 | · · · |
+-------+-------+-------+
| · · · | · · · | 1 5 9 |
| · · · | · · · | 3 2 7 |
| · · · | · · · | 6 4 8 |
+-------+-------+-------+
Then I let my sudoku solver process it until it finds a solution to fill all remaining gaps. The example above resulted in the filled grid below:
+-------+-------+-------+
| 5 6 4 | 9 3 7 | 2 8 1 |
| 1 7 2 | 5 4 8 | 9 6 3 |
| 9 8 3 | 1 6 2 | 4 7 5 |
+-------+-------+-------+
| 7 4 9 | 3 2 5 | 8 1 6 |
| 2 1 8 | 7 9 6 | 5 3 4 |
| 6 3 5 | 8 1 4 | 7 9 2 |
+-------+-------+-------+
| 8 2 6 | 4 7 3 | 1 5 9 |
| 4 5 1 | 6 8 9 | 3 2 7 |
| 3 9 7 | 2 5 1 | 6 4 8 |
+-------+-------+-------+
My question is if this approach is mathematically safe, i.e. can you prove me that when I fill my grid using the described approach (randomly assigning the first 27 numbers to fields in the groups on a diagonal line), there will be always at least one possible way to complete the grid from there on?
Or are there chances that the randomly placed numbers can interfere with each other and result in an impossible grid?
|
WLOG the top block is in the canonical order, because we can just relabel, as in 5 -> 1, 6 -> 2, … in your example. I'm not going to make that replacement through the rest of your grid, but will just pretend you started off like that.
+-------+-------+-------+
| 1 2 3 | · · · | · · · |
| 4 5 6 | · · · | · · · |
| 7 8 9 | · · · | · · · |
+-------+-------+-------+
| · · · | 3 2 5 | · · · |
| · · · | 7 9 6 | · · · |
| · · · | 8 1 4 | · · · |
+-------+-------+-------+
| · · · | · · · | 1 5 9 |
| · · · | · · · | 3 2 7 |
| · · · | · · · | 6 4 8 |
+-------+-------+-------+
Making minor-row swaps within one of the three major rows, or making minor-column swaps within one of the three major columns, doesn't change solveability of the sudoku. Therefore we may WLOG that the top-left entry of each of the three grids is 1, by rotating:
+-------+-------+-------+
| 1 2 3 | · · · | · · · |
| 4 5 6 | · · · | · · · |
| 7 8 9 | · · · | · · · |
+-------+-------+-------+
| · · · | 1 4 8 | · · · |
| · · · | 9 6 7 | · · · |
| · · · | 2 5 3 | · · · |
+-------+-------+-------+
| · · · | · · · | 1 5 9 |
| · · · | · · · | 3 2 7 |
| · · · | · · · | 6 4 8 |
+-------+-------+-------+
Similarly we may WLOG that the 5 of the middle block appears in one of two places:
+-------+
| 1 * |
| ! * |
| |
+-------+
because if it's in any of the others, we can row/column swap it into one of those. The exception is if it's in the ! position, which is actually equivalent to the top *. This is because we may transpose the entire grid, and relabel the top-left block again, without affecting the middle block's 1 or 5 except in moving them to the correct positions.
Likewise (but without the option of reflecting this time, because that could mess up the middle block) we may WLOG that the bottom-right block's 5 is in one of three positions:
+-------+
| 1 * |
| * * |
| |
+-------+
There are now $2 \times 7! \times 3 \times 7! = 152409600$ remaining cases, which are left as an exercise to the reader.
|
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Does $f_n(x)=\frac{x}{n}\log(\frac{x}{n})$ converge uniformly in $(0,1)$? $f_n(x)=\frac{x}{n}\log\frac{x}{n}$
I tried to expand the bounds to $[0,1]$ and prove that the hypotheses of Dini's uniform convergence criterion are satisfied, but I'm not even sure I can expand the bounds since $f_n(0)$ isn't defined.
|
Hint:
$$
f_n(x)=\frac{x\log x}{n}-\frac{\log n}{n}\,x
$$
and $\lim_{x\to0^+}x\log x=0$.
|
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How to calculate $\int_a^bx^2 dx$ using summation? So for this case, we divide it to $n$ partitions and so the width of each partition is $\frac{b-a}{n}$ and the height is $f(x)$.
\begin{align}
x_0&=a\\
x_1&=a+\frac{b-a}{n}\\
&\ldots\\
x_{i-1}&=a+(i-1)\frac{b-a}{n}\\
x_i&=a+i\frac{b-a}{n}
\end{align}
So I pick left point, which is $x_{i-1}$
I start with
\begin{align}
\sum\limits_{i=1}^{n} \frac{b-a}{n}f\left(a+\frac{(i-1)(b-a)}{n}\right)
&=\frac{b-a}{n}\sum\limits_{i=1}^{n} \left(a+\frac{(i-1)(b-a)}{n}\right)^2\\
&=\frac{b-a}{n}\left(na^2+ \frac{2a(b-a)}{n}\sum\limits_{i=1}^{n}(i-1)+\frac{(b-a)^2}{n^2} \sum\limits_{i=1}^{n}(i-1)^2\right)
\end{align}
Here I am stuck because I don't know what$ \sum\limits_{i=1}^{n}(i-1)^2$is (feel like it diverges). Could someone help?
|
Hint: $$\sum_{i=1}^{n}(i-1)^2 = \sum_{i=1}^{n}(i^2-2i+1) = \sum_{i=1}^{n}i^2 - 2\sum_{i=1}^{n}i + \sum_{i=1}^{n}1$$
You should be able to take it from here.
|
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|
Integrate: $\int \frac{\sin(x)}{9+16\sin(2x)}\,\text{d}x$. Integrate: $$\int \frac{\sin(x)}{9+16\sin(2x)}\,\text{d}x.$$
I tried the substitution method ($\sin(x) = t$) and ended up getting $\int \frac{t}{9+32t-32t^3}\,\text{d}t$. Don't know how to proceed further.
Also tried adding and substracting $\cos(x)$ in the numerator which led me to get $$\sin(2x) = t^2-1$$ by taking $\sin(x)+\cos(x) = t$.
Can't figure out any other method now. Any suggestions or tips?
|
To attack this integral, we will need to make use of the following facts:
$$(\sin{x} + \cos{x})^2 = 1+\sin{2x}$$
$$(\sin{x} - \cos{x})^2 = 1-\sin{2x}$$
$$\text{d}(\sin{x}+\cos{x}) = (\cos{x}-\sin{x})\text{d}x$$
$$\text{d}(\sin{x}-\cos{x}) = (\cos{x}+\sin{x})\text{d}x$$
Now, consider the denominator.
It can be rewritten in two different ways as hinted by the above information.
$$9+16\sin{2x} = 25 - 16(1-\sin{2x}) = 16(1+\sin{2x})-7$$
$$9+16\sin{2x} = 25 - 16(\sin{x}-\cos{x})^2 = 16(\sin{x}+\cos{x})^2-7$$
Also note that
$$\text{d}(\sin{x}-\cos{x})-\text{d}(\sin{x}+\cos{x}) = 2\sin{x}\text{d}x$$
By making the substitutions
$$u = \sin{x}+\cos{x}, v = \sin{x}-\cos{x}$$
The integral is transformed into two separate integrals which can be evaluated independently.
$$2I = \int \frac{\text{d}v - \text{d}u}{9+16\sin{2x}} = \int \frac{\text{d}v}{25-16v^2} + \int \frac{\text{d}u}{7-16u^2}$$
The remainder of this evaluation is left as an exercise to the reader.
|
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The sum of infinitely many $c$s is $c$ implies $c = 0$. This seems like an obvious claim, but I would like to be able to prove this rigorously.
Suppose I have $c \in \mathbb{R}$ satisfying
$$\lim_{n \to \infty}\sum_{i=1}^{n}c = c\text{.}$$
How does it follow that $c = 0$?
What I think I shouldn't do: $$c\lim_{n \to \infty}n = c$$
because this gives $0 \cdot \infty = 0$ (is this true?).
To understand the context of this question, I am working with a probability measure $\mathbb{P}$ and am trying to show that $\mathbb{P}\left(\emptyset\right) = 0$ using the definition (and in this case, $\mathbb{P}\left(\emptyset\right) = c$). Maybe infinite sums are defined differently in measure theory? I don't know.
|
We'll assume that $c \in \mathbb{R}$.
$$\lim_{n\to\infty}\sum_{i=1}^{n}c=c$$
Subtract $2c$
$$\lim_{n\to\infty}\sum_{i=1}^{n-2}c=-c$$
$$-\lim_{n\to\infty}\sum_{i=1}^{n-2}c=\lim_{n\to\infty}\sum_{i=1}^{n}c$$
But we know that $$\lim_{n\to\infty}\sum_{i=1}^{n-2}c = \lim_{n\to\infty}\sum_{i=1}^{n}c$$
Thus
$$
-\lim_{n\to\infty}\sum_{i=1}^{n}c = \lim_{n\to\infty}\sum_{i=1}^{n}c
$$
and so we conclude that
$$
\lim_{n\to\infty}\sum_{i=1}^{n}c = 0
$$
and since we have that
$$
\lim_{n\to\infty}\sum_{i=1}^{n}c=c
$$
then we finally conclude that
$$
c = 0
$$
|
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|
Prove that G is a cyclic group
Suppose that $|G| = pq$ where $p$ and $q$ are primes such that $p < q$ and $p$ does not divide $q − 1$. Prove that $G$ is a cyclic group.
A cyclic group is a group that has a unique generator element, so is the way to go with this to find that element? Or am I missing something?
|
A precedent question shows Prove that G has a normal Sylow p-subgroup that $G$ has a Sylow normal group $N$ of order $p$, Consider $m:G\rightarrow G/N$ it defines an extension $1\rightarrow N\rightarrow G\rightarrow G/N\rightarrow 1$ this extension splits since the theorem of Sylow implies that $G$ has a $q$-subgroup, but it is trivial since splitting extensions of $Z/p$ by $Z/q$ are classified by morphisms $Z/q\rightarrow Aut(Z/p)$ https://en.wikipedia.org/wiki/Group_extension#Classifying_split_extensions but any action of $Z/p$ on $Z/p$ is trivial since $p$ and $q$ are different prime numbers and $p$ does not divide $q-1$. This implies that $G=Z/p\times Z/q$.
|
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Continuity between topological spaces Let $X,Y$ be topological spaces, and $f:X\to Y$. I am trying to show that, if "$A$ closed in $Y\Rightarrow$ $f^{-1}(A)$ closed in $X$", then we have sequential continuity:
$$x_n\to x\;\text{in }X\Rightarrow f(x_n)\to f(x)\;\text{in }Y$$
I am stuck because this is "in the wrong direction": if I talk about things in $Y$ to then get to things in $X$ (from the closed-set definition) then I can't see how to prove the sequential statement.
I'd like to prove this using only the closed-sets statement, without mentioning continuity explicitly. The reason for this is that I want to show several definitions $p,q,r$ are equivalent to the usual open-set definiton of continuity. So proving "continuity" $\Rightarrow p\Rightarrow q\Rightarrow r\Rightarrow$ "continuity" seems the most efficient way to do it.
|
It is maybe easier to think about proving the contrapositive rather than proving it directly. Suppose $x_n\to x$ but $f(x_n)\not\to f(x)$ for some sequence $(x_n)$ in $X$. You want to somehow get from this a closed set $A\subseteq Y$ such that $f^{-1}(A)$ is not closed. To do this, look at what it means for $f(x_n)$ to not converge to $f(x)$. It means that there is some open set $U$ containing $f(x)$ such that there are infinitely many $n$ such that $f(x_n)\not\in U$. You can now take $A=Y\setminus U$ and show that it is exactly the $A$ you are looking for; I'll leave the details to you.
|
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How can you find $m$ in $mx^2+(m-3)x+1=0 $ so that there is only one solution How can you find $m$ in $$mx^2+(m-3)x+1=0 $$ so that there is only one solution.
I tried to solve it by quadratic equation but I end up with two solutions.
So I want it know that is there a way so that I'll only get one solution the end.
Thanks.
|
Let $x=\alpha$ be one solution of the given quadratic equation: $mx^2+(m-3)x+1=0$ then $\alpha$ & $\alpha$ be the roots of the given equation $$\text{sum of roots}=\alpha+\alpha=-\frac{m-3}{m}$$$$\alpha=\frac{3-m}{2m}\tag 1$$
$$\text{product of roots}=\alpha\cdot \alpha=\frac{1}{m}$$$$\alpha^2=\frac{1}{m}\tag 2$$
substituting the value of $\alpha$ from (2) into (1), one should get
$$\left(\frac{3-m}{2m}\right)^2=\frac 1m$$
$$m^2-10m+9=0$$
using quadratic formula,
$$m=\frac{-(-10)\pm\sqrt{(-10)^2-4(1)(9)}}{2(1)}=5\pm4$$
Thus, the given quadratic equation: $mx^2+(m-3)x+1=0$ will have only one solution for $\color{red}{m=9}\ \text{or} $ $ \ \ \color{red}{m=1}$.
|
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|
Show that the Lebesgue intergral $\int_{1}^{\infty} x^{-b} e^{\sin {x}} \sin {(2x)}\, dx$ exists iff $b>1$ Assume $b>0$. Show that the Lebesgue intergral $\int_{1}^{\infty} x^{-b} e^{\sin {x}} \sin {(2x)}\, dx$ exists iff $b>1$.
We know if $b>1$ the integrand can be bounded and it's just an integral $\int_{1}^{\infty} x^{-b}\,dx$. I don't know how to show the other direction?
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Hint: For $0 < b \leqslant 1$
$$\int_1^c |x^{-b}e^{\sin x} \sin 2x| \, dx > e^{-1}\int_1^c \frac{|\sin 2x|}{x} \, dx = \frac{1}{e}\int_2^{2c} \frac{|\sin x|}{x} \, dx $$
Show the integral on the RHS is not finite as $c \to \infty$ using:
$$\int_{n \pi}^{(n+1) \pi} \frac{| \sin x|}{x}\, dx \geqslant \frac{2}{(n+1) \pi}$$
|
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Limit of sum of periodic function Let $f_1,f_2,...,f_n$ are periodic functions,if $\lim\limits_{x\rightarrow\infty}\sum_{i=1}^n f_i(x)$ is existent and bounded.
How to show $\sum_{i=1}^n f_i(x)\equiv C$ ?
$C$ is a constant.
|
(this assumes continuity of the $f_i$, still thinking of how to remove it)
Suppose $F(x):=\sum_i f_i(x)\to C$, where $f_i(x+p_i)=f_i(x)$ for each $i$.
Now, find an sequence of positive reals $h_N$ increasing to $\infty$ which is simultaneously close to $p_i\mathbb{Z}$ for all $i$, i.e.
$$h_N=a_i^{(N)}p_i+\varepsilon_N^{(i)}, \text{where }z_N^{(i)}\in\mathbb{Z}, \varepsilon_N^{(i)}\to 0$$
as by noting Dirichlet's simultaneous approximation theorem applied to $\{\frac{1}{p_1},\frac{1}{p_2},...,\frac{1}{p_n}\}$, we can find integers $a_i^{(N)}$ and an integer $h_N\le N$ such that for each $i$:
$$\left\vert \frac{1}{p_i}-\frac{a_i^{(N)}}{h_N}\right\vert \le \frac{1}{h_N N^{1/n}}$$
Upon rearrangement, this becomes $\vert h_N - a_i^{(N)}p_i \vert \le \frac{p_i}{N^{1/n}}\to0$
Then, for any $x$:
$$\lim_{n\to\infty}F(x+h_n)=\lim_{x\to\infty}F(x)=C$$
$$\lim_{n\to\infty}F(x+h_n)=\lim_{n\to\infty}\sum_i f_i(x+h_n)=\lim_{n\to\infty}\sum_i f_i(x+\varepsilon_n^{(i)})=\sum_i f_i(x)=F(x)$$
So, $F(x)=C$ for all $x$.
Alternative Proof
Let $P(n)$ be the statement "If a sum of $n$ periodic functions has a limit $C$, then this sum is equal to $C$ for all $x$".
*
*If $f$ is $p$-periodic and tends to $C$, then for any $\varepsilon > 0$, there exists $N$ such that $x>N\implies \vert f(x) - C \vert < \varepsilon$. But periodicity gives that this is actually true for all $x$. As this is true for any $\varepsilon > 0$, we recover that $f=C$ for all $x$. So, $P(1)$ is true.
*Suppose $P(1), P(n-1)$ are true, and consider a sum of $n$ periodic functions, $F(x)=\sum_1^n f_i(x)$ with limit $C$, where in particular, $f_n$ has period $p_n$. Then $F(x+p_n)-F(x)=\sum_1^{n-1} [f_i(x+p_n)-f_i(x)]$ is a sum of $(n-1)$ periodic functions, and converges to $0$, hence is equal to $0$ by $P(n-1)$.
So, $F$ is $p_n$-periodic, and converges to $C$, and $P(1)$ tells us that it is identical to $C$ as a result, i.e. $P(n)$ is true.
Thus, by induction, $P(n)$ is true for all $n$, and any finite sum of periodic functions with a limit at $\infty$ is constant.
|
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Non-associative commutative binary operation Is there an example of a non-associative, commutative binary operation?
What about a non-associative, commutative binary operation with identity and inverses?
The only example of a non-associative binary operation I have in mind is the commutator/Lie bracket. But it is not commutative!
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The simplest example of a nonassociative commutative binary operation (but lacking an identity element) is the two-element structure $\{a,b\}$ with $aa=b$ and $ab=ba=bb=a;$ note that $a=bb=(aa)b\ne a(ab)=aa=b.$ This is the NAND (or NOR) operation of propositional logic, where $a=\text{true}$ and $b=\text{false}$ (or vice versa). See this answer or this one.
|
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Supremum proof simple I got stuck on this problem and can't figure it out, I hope somebody can help me, I also wrote my attempt. Thanks in advance!!
Question: Let $(a_n)$ be a convergent sequence in $\mathbb{R}$. $a_n
\to a^*$. Let $A=\{a_n | n \in \mathbb{N}\}$. I have to show that:
$\sup A \geq a^*$
My attempt:
Suppose $a_n \in A$.
$a_n$ is bounded because it is convergent. Because $A=\{a_n | n \in \mathbb{N}\}$ we can say that $\forall a_n \in A$ : $a_n \leq \sup A$.
We also know that $a_n \leq a^*$, because it's the limit.
|
Let $b$ be an upper bound for $A$.
Then $a_n \le b$ for all $n$, and so $\displaystyle a^*=\lim_{n\to\infty} a_n\le b$.
Since $\sup A$ is an upper bound for $A$, we can take $b=\sup A$ and conclude that $a^* \le \sup A$.
|
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Proving that if a function is a metric then it is symmetric and non negative I am trying to prove that given a metric d using only the properties that it $d(a,b)=0 iff a=b$ and $d(a,c)\le d(a,b)+d(b,c)$ that $d(a,b)=d(b,a)$ and $d(a,b) \gt 0$ I understand that it is part of the definition in most texts but it is left as an exercise in mine and I can not figure it out.
I edited my original question. I stated wrong we do not have that $d(a,a)=0$ I have that $d(a,b)=0 iff a=b$
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Note that $d:\mathbb{R}\times\mathbb{R}\to \mathbb{R}$ defined as $d(a,b)=0$ is such that :
*
*$\forall a\in\mathbb{R}, d(a,a)=0,$
*$\forall a,b,c\in\mathbb{R},d(a,c)=0\leq d(a,b)+d(a,c)=0$
but you have not that $a\neq b\implies d(a,b)>0,$ so this point should be an axiom.
For the symmetry part, consider $d':\{0,1\}\to\mathbb{R}$ defined as :
$$d'(0,0)=0,d'(1,1)=0,d'(1,0)=1 \text{ and } d'(0,1)=2.$$
Now note that :
*
*$\forall a\in\{0,1\},d'(a,a)=0,$
*$\forall a,b,c\in\{0,1\},d'(a,c)\leq d'(a,b)+d'(b,c)$
but $d'(0,1)=2\neq 1=d'(1,0),$ so that point should be an axiom too.
|
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Power series for $(a+x)^{-1}$ Is it possible to write the following expression in terms of power series?
$$ (a+x)^{-1}=\sum\limits_{k = - \infty }^\infty {{b_k}{x^k}} $$
where $0 < a < 1$ and $0 < x < 1$.
|
It can be turned into geometric series:
$$\frac{1}{a+x} = \frac{1}{a} \cdot \frac{1}{1-\left(-\frac{x}{a}\right)} = \frac{1}{a}\cdot\sum\limits_{k=0}^{\infty}\left(-\frac{x}{a}\right)^k$$
which converges whenever $|x| < |a|$.
|
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If $G$ is a finite group s.t. $|G|=4$, is it abelian ? If $G$ is a finite group s.t. $|G|=4$, is it abelian ? To me it's isomorphic to $\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z$ or $\mathbb Z/4\mathbb Z$, but a friend of me said that they are not the only group of order 4, and there exist some non-abelian. Do you have such example ?
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If $G$ is cyclic, $G$ is abelian. And if $G$ is not cyclic, the order of all elements are equal to $2$. And a group whose all elements are of order $2$ is abelian.
In any cases, a group of order $4$ is abelian.
|
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An automorphism that is not inner. Consider the group $G=SL_3(\mathbb{C})$. I want to show that the automorphism $\phi$ of $G$ given by $\phi(x)=(x^{-1})^T$ is not inner. Probably I should do this by contradiction, i can show that if $\phi(x)=RxR^{-1}$, then $R^T R$ lies in the centre of $SL_3$. How can I proceed to obtain a contradiction?
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The operator $\operatorname{tr}$ is invariant under conjugation, but $\operatorname{tr} \phi(\lambda) \not = \operatorname{tr} \lambda$ for constant $\lambda\not = 0, \pm 1$.
|
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Is $n−m$ always the largest possible number of linearly independent vectors in this vector space? Fix linear independent vectors $a_1, . . . , a_m ∈ \mathbb{R}^n$, and let $S$ be the vector space of such that
$S:=$ {$x∈ \mathbb{R}^n:a_i⋅x=0∀1≤i≤m$} .
The vector space $S$ always has at least $n − m$ linearly independent vectors (solutions to $a_i⋅x=0$).
Is $n−m$ always the largest possible number of linearly independent solutions or is it possible for me to find a counterexample?
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Consider the map from $R^n$ to $R^m$ defined by
$$
v \mapsto (v \cdot a_1, v \cdot a_2, \ldots, v \cdot a_m).
$$
What does the rank-plus-nullity theorem tell you about this linear transformation?
|
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What is a contraction on a space $(X,d)$? I have been reading some proofs on the elementary theorems of differential equations. One such proof uses the concept of a "contraction". See the definition below.
Definition 4 Let $(X,d)$ be a space equipped with a distance function $d$. A function $\Phi:X\to X$ from $X$ to itself is a contraction if there is a number $k\in (0,1)$ such that for any pair of points $x,y\in X$ we have
$$d(\Phi(x),\Phi(y))\leq kd(x,y).$$
It is important that the constant $k$ is strictly less than one.
Source. (Taken from this pdf.)
To put it into my own words, it seems that a contraction is simply a modified distance function that always yields a value less than the normal distance function for the same set of points. Am I understanding it correctly?
Would someone be able to offer an example of a contraction?
Thank you!
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No, a contraction is not a metric(aka distance function). What a contraction does is bring every pair of points closer together(in the implicit metric). For instance, $f(x)=x/2$ is a contraction in the euclidean metric of $\mathbb{R}$.
|
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For any integer $n$ greater than $1$,how many prime numbers are there greater than $n!+1$ and less than $n!+n$? For any integer $n$ greater than $1$,how many prime numbers are there greater than $n!+1$ and less than $n!+n$ ?
By trying different values of $n$ for $n=2,3,4,5,6$ I get a feeling that the number of primes in the interval is $0$,of course this might be wrong as $n$ can be any number.
I've not been able to do much progress on this problem...one idea I had was to prove the above conjecture by proving it by picking up first smaller intervals and prove it works for those intervals.
My feeling is also that Wilson's Theorem might be applicable here.
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Hint: Let $a = n! + 7$. Does $7 \mid n!$? Does $7|7$? Does $7 \mid n! + 7$? Is $n! + 7$ prime?
If it isn't what number greater than $n! + 1$ and less than or equal to $n! + n$ might be?
Can you factor anything out of the number $n! + k$?
|
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How was Zeno's paradox solved using the limits of infinite series? This is a not necessarily the exact paradox Zeno is thought to have come up with, but it's similar enough:
A man (In this photo, a dot 1) is to walk a distance of one unit from where he's standing to the wall. He, however, is to walk half the distance first, then half the remaining distance, then half of that, then half of that and so on. This means the man will never get to the wall, as there's always a half remaining.
This defies common sense. We know a man(or woman, of course) can just walk up to a wall and get to it.
My calculus book says this was solved when the idea of a limit of an infinite series was developed. The idea says that if the distances the man is passing are getting closer and closer to the total area from where he started to the wall, this means that the total distance is the limit of that.
What I don't understand is this: mathematics tells us that the sum of the infinite little distances is finite, but, in real life, doesn't walking an infinite number of distances require an infinite amount of time, which means we didn't really solve anything, since that's what troubled Zeno?
Thanks.
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If the subject walks at a constant speed, you don't need calculus to calculate exactly when he will arrive at his destination. Just use the simple speed-equals-distance-over-time formula. This formula was unknown to the ancient Greek philosophers. They had no notion of actually measuring speed. It wouldn't be for another thousand years after Zeno's time that Galileo would formulate:
$$s=\frac d{t}$$
Today, we know that in going from point $A$ to another point $B$, we will pass through infinitely many other points along the way. If we associate an event with our arrival at each of these points, then infinitely many events will have occurred in the interval. The modern mind has no trouble with this notion.
|
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Financial Mathematics, interest problem Determine the amount of interest earned from time $t=2$ to $t=4$ if $240$ is invested at $t=1$ and an additional $300$ is invested at $t=3$.
Given, $a(1)=1.2$, $a(2)=1.5$, $a(3)=2.0$, $a(4)=3.0$
I tried finding $A(0)$ using the equation $A(1)=A(0).a(1)$. Hence $A(0)=200$
Hence I found $A(2)=300$, $A(3)=400$, $A(4)=600$
$I_{[2,3]}=A(0)[\text{new } A(3)-A(2)]=80000$
$I_{[3,4]}=A(0)[A(4)-\text{new }A(3)]$
From here I am stuck.I am messed up with the different times.
|
Answer:
$$a_{(2-4)} = \frac{a(4)}{a(2)}$$
$$A(2) = A(1).\frac{a(2)}{a(1)}$$
$$A(2) =240\times\frac{1.5}{1.2} = 300$$
$$A(4) = 300\times\frac{3}{1.5} = 600$$
$$A'(4) = A'(3)\frac{a(4)}{a(3)}$$
$$A'(4) = 300\times\frac{3}{2}$$
$$A'(4) = 450$$
$$I_1 = A(4)-A(2) = 600-300 = 300$$
$$I_2 = A'(4) - A'(3) = 450-300 = 150$$
$$I = I_1 +I_2$$
The required Interest:
$$I = 300+150 = 450$$
Regards
Satish
|
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Sin(x): surjective and non-surjective with different codomain? Statement that $\operatorname{sin}(x)$ not surjective with codomain $\mathbb R$ and surjective with codomain $[-1,1]$ found here:
*
*Non-surjective: $\mathbb{R}\rightarrow\mathbb{R}: x\mapsto\operatorname{sin}(x)$
*Surjective: $\mathbb{R}\rightarrow[-1,1]: x\mapsto\operatorname{sin}(x)$
where the image $Im(f)=[-1,1]$ in both cases and the codomain is $\mathbb R$ and $[-1,1]$ for the case 1 and case 2, respectively. In the second case, $\forall x\in\mathbb R : \operatorname{sin}(x)\in [-1,1]$ where the codomain equals the image of $f$. Surjection means that the image of the function equals to the codomain.
Why is sin not surjection with one codomain and surjective with other codomain?
|
$f(x)=\sin(x)$ is not a surjection from $\Bbb R\to \Bbb R$
$f(x)=\sin(x)$ is a surjection from $\Bbb R\to [-1,1]$
The wiki page tells you that in the case of $\Bbb R\to [-1,1]$ that $\sin(x)$ is a surjection but is not an injection. It is a surjection because every possible output has a preimage. Suppose $y\in [-1,1]$. Then for $x=\arcsin (y)\in\Bbb R$ you have $f(x)=\sin(\arcsin(y))=y$.
The wiki page also tells you that in the case of $\Bbb R\to \Bbb R$ that $\sin(x)$ is not a surjection and is not an injection. It is not a surjection because there exists no preimage for, say, the point $2$ in the codomain. There is no real value of $x$ such that $\sin(x)=2$.
Neither case is an injection because you can have multiple $x$ values which map to the same output, for example $\sin(\pi)=\sin(3\pi)$
|
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Solving an integral using euler method Hy. Can someone please explain me how can I resolve $\int_{0}^{\infty} e^{-(sI-A)t}\,dt$ ? I must have the final result $(sI-A)^{-1}$. I think it has the euler form, for beta integral but I don't know how can I get to that result.
|
If you wrote the integral correctly, just use the fact that
$$\int e^{kt}=k^{-1}e^{kt}+C$$
where, of course, $k$ is a constant and $C$ is an arbitrary constant. There is no need for a fancy integral such as the beta integral. Finding the definite integral from that is easy, since the limit is easy.
All this, of course, depends on $s$, $I$, and $A$ being real or complex numbers.
|
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Showing there is a prime in a ring extension using Nakayama's lemma Here's the problem that I'm working on:
if $A \subset B$ is a finite ring extension and $P$ is a prime ideal of $A$ show there is a prime ideal $Q$ of $B$ with $Q \cap A = P$. (M. Reid, Undergraduate Commutative Algebra, Exercise 4.12(i))
There was a hint to use Nakayama's lemma to show $PB \not= B$, for that part what I've done is said $B$ is a finitely generated $A$ module, $P$ is an ideal of $A$ therefore I can apply NAK here and it tells me that: $PB = B$ implies that there exists some $r \in A$ satisfying $r \equiv 1 \pmod P$. I would need to get a contradiction from this but I wasn't able to see how.
Any guidance on whether I am approaching this correctly and how to do the part I am stuck on would be appreciated! Thank you
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For the first part, the contradiction comes from the fact that there exists $r\in A$ such $r=1+p, p\in P$ and $rB=0$ this implies $(1+p)A=0$ so for every $a\in A, (1+p)a=a+pa=0$ which implies $a\in P$ contradiction since $P$ is different of $A$.
|
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Prove that $3^x + 3^{x-2}$ ends with $0$ for any integer $x > 1$ I think that $3^x+3^{x-2}$ ends in a $0$ (i.e. is divisible by $10$) $\forall x \in \Bbb Z, x > 1$.
Examples:
$3^2+3^{2-2}=9+1=10 \\
3^3+3^{3-2}=27+3=30 \\
3^4+3^{4-2}=81+9=90 .$
In fact, I wrote a quick Python program and left it on overnight, it reported every number in the domain working.
I don't know a proof for this, though, and I also don't know if it's already a theorem or something with a fancy name that I just happened to stumble across.
Also, I don't know any really good tags for this. If you know one, please comment or edit the post.
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$3^{x} + 3^{x - 2} = 3^{x}(1 + 3^{-2}) = 3^{x}({10}/{9}) = (3^{x}/9)(10)$ The value is always a multiple of 10 since for x ≥ 2 we always have an integer value. The proof is complete.
|
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Prove $x^TAx = 0$ $\implies$ A is skew-symmetric We know for a skew-symmetric matrix A, $x^TAx = 0$. But is the converse statement true, i.e. does $x^TAx = 0$ imply A is skew-symmetric? If yes, then how to prove it?
|
$x^T A x = 0\ \forall x$ can be written as:
$\sum_i \sum_j x_j a_{ij} x_i = 0\ \forall x$
That is:
$x_1^2 a_{1,1} + x_1 a_{1,2} x_2 + ... = 0$
The only way to have $x_i x_j$ cancel out $x_j x_i$ is by: $a_{ij} = -a_{ji}$, and the remaining terms $x_i^2 a_{ii}$ must be zero (for all $x$), so we must have $a_{ii}=0$. Therefore $A$ is antisymmetric.
|
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Expanding an expression in a certain field If $\mathbb F_2$ is a field of characteristic $2$, then we have $x+x=y+y=z+z=0$ for all $x,y,z \in \mathbb F_2$. When I expand $(x+y)(y+z)(z+x)$, I get
\begin{align}
(x+y)(y+z)(z+x) &= xz^2+y^2z+yz^2+x^2y+x^2z+xy^2+2xyz \\
&= xz^2+y^2z+yz^2+x^2y+x^2z+xy^2+(x+x)yz \\
&= xz^2+y^2z+yz^2+x^2y+x^2z+xy^2.
\end{align}
I got $2xyz=0$ in $\mathbb F_2$, but how can I simplify my above expression even more?
|
In $\mathbb{F}_2$, we also have $x^2=x$ and so on. So your expression will result in $0$. Another way to look at it is that at least two of $x,y$ or $z$ will take the same value, in which case at least one of $x+y, y+z, x+z$ will be $0$.
|
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|
Is this f(z) function analytic? Is $f(z) = z^2\sin z$ an analytic function for $z \in \mathbb{C}$ ?
$z = x + iy $
I really don't know how to split this at this format $f(x,y) = P(x,y) + iQ(x,y)$, so I can prove that if this function is analytic with neccesary Cauchy–Riemann equations. Which are $ P_x = Q_y$ and $P_y = -Q_x$.
I can't get out of this $f(x+iy) = (x+iy)^2\sin(x+iy)$. How can I get rid of $i$ in $\sin(x+iy)$?
|
Using Cauchy-Riemann here is unnecessary work. It's easier (even if you haven't done so before) to show that $\sin z$ is holomorphic (analytic), and that the product of two holomorphic functions is again holomorphic.
|
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|
Reduced row echelon form and linear independence Let's say I have the set of vectors $S = \{v_1, v_2, ..., v_n\}$ where $v_j \in R^m$, $v_j = (a_{1j}, a_{2j}, ..., a_{mj})$.
If the matrix formed by each of the vectors $A=[v_1, v_2, ..., v_n]$ looks like this (I believe), which is not a square matrix:
$$A = \begin{pmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{m1} & a_{m2} & \cdots & a_{mn}
\end{pmatrix}$$
Then does A's reduced row echelon form help me determine whether the vectors of $S$ are linear dependent or independent? If so, how?
I hope I got all the indices, notation and terminology right, since I am a beginner in linear algebra, and English is not my native language.
|
Yes, if you can convert the matrix into reduced row echelon form(or even just row echelon form) without a row of $0$s,then the vectors are linearly independent.
|
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|
Factorial grow faster than Exponential - permutation case It is said that factorial grows faster than exponential, but in the case of permutation:
permutation with repetition = n^n
permutation without repetition = n!
And of course permutation with repetition has a bigger "space" than permutation without repetition which means it should grow faster.
I was looking for some convincing proof that shows this in an intuitive way.
Like for example displaying n! in a n^0.n^(m-1).n^(m-2)...n^1 and eventually shows that the sum of the exponential is < n, but did not succeed.
I am not sure if anyone can help me with the proof or maybe just help to explain the case against the "factorial grows faster than exponential" fact
Sorry for my wording, I have very poor mathematical backgrounds
|
The $n^n$ function is not an exponential as the basis varies.
To prove $n!=o(n^n)$, let's evaluate the ratio:
$$\frac{n!}{n^n}=\frac1n\cdot\frac2n\cdots\frac{n-1}n\cdot \frac nn <\frac 1n$$
since each factor $\dfrac kn$ is less than $1$ if $k<n$. This inequality implies $\;\displaystyle\lim_{n\to\infty}\frac{n!}{n^n}=0$.
|
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|
Need clarifying on basic derivatives of natural log/e So here's the question:
Find the derivative: $ y= e^{\cos(x)}$
Hint: This is a combination of the chain rule and the natural log. The derivative is $(\ln a)(a^{f(x)}) * f'(x)$
So normally without using the hint I would use the chain rule and solve it like this:
$e^{\cos(x)} * -\sin(x)$
simplified....
$-\sin x(e^{\cos(x)})$
Apparently though, according to the hint provided, there should be an $ln (e)$ which cancels out, so it should look like this:
$e^{\cos(x)} * -\sin(x)* \ln e$,
where the $\ln e$ simplifies to $1$ and has no affect on the final answer.
So when I did it, I only used the chain rule. The answer ended up the same, but I was wondering if this was a coincidence or if it is always like that. If I continue doing it my way will I be fine?
Sorry if this is a dumb question. It's been a while since I took Calculus 1 and I'm starting Calculus 2 now so I'm a bit rusty.
|
It is not a coincidence as you predicted but a special case for $e^{f(x)}$ since $ln(e) = 1$. For example: if the question were $y = a^{cos(x)}$, your answer would be $\frac{dy}{dx} = (ln(a))(a^{cos(x)})(-sin(x))$.
|
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|
If $A=\sin^{20}\theta +\cos^{48}\theta $ then identify the correct option.
If $A=\sin^{20}\theta +\cos^{48}\theta $, then for all values $\theta$
a) $A\geq 1$
b) $ 0< A\leq 1$
c) $1<A< 3$
d) None of these
$0 \leq \sin^{20}\theta \leq 1$
$0 \leq \cos^{48}\theta \leq 1 $
So I think it is $d.)$ , but I am confused.
I look for a short and simple way.
I have studied maths upto $12$th grade.
|
Hint: $0 \leq \sin^2\theta \leq 1$, $0 \leq \cos^2\theta \leq 1$, and $\sin^2\theta + \cos^2\theta = 1$.
|
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|
Proving the surprising limit: $\lim\limits_{n \to 0} \frac{x^{n}-y^{n}}{n}$ $=$ log$\frac{x}{y}$ A few months ago, while at school, my classmate asked me this curious question: What does $\frac{x^{n}-y^{n}}{n}$ tend to as $n$ tends to $0$? I thought for a few minutes, became impatient, and asked "What?" His reply, log$\frac{x}{y}$, was surprising, but his purported 'proof' was more surprising:
Consider $\lim\limits_{n \to 0}\,\int_y^x t^{n-1}\, dt$. "Pushing the limit into the definite integral", we have $$\int_y^x \lim\limits_{n \to 0}\,t^{n-1}\, dt \implies \int_y^x \frac{1}{t}\, dt \implies \mathsf{log} \frac{x}{y}$$
Leaving the fact that he had the inspiration to pull this integral out of thin air aside, is the limit allowed to pass into the definite integral? We hadn't learned Real Analysis (we were just taking a basic high school, hand-wavy single-variable calculus course), and I remember feeling very uneasy about the sorcery. I still am, hence, this question.
I've since thought about approaching it using $\mathsf{L'Hospital}$, but I still feel uneasy, since it involves differentiating with respect to different variables, which is a little bit confusing. I'd also appreciate your help in this regard.
If you have a better proof, I'll truly appreciate it.
|
Hint:
$$
\begin{align}
\lim_{n\to0}\frac{x^n-1}{n}
&=\log(x)\lim_{n\to0}\frac{e^{n\log(x)}-1}{n\log(x)}\\
&=\log(x)\lim_{u\to0}\frac{e^u-1}u\\[6pt]
&=\log(x)
\end{align}
$$
Subtract
Swapping Integral and Limit
Let $\lambda_{\rm{min}}=\min\left\{1,y^{-5/4}\right\}$ and $\lambda_{\rm{max}}=\max\left\{1,x^{-5/4}\right\}$, then for $n\in\left[-\frac14,\frac14\right]$, the Mean Value Theorem says that for some $\tau$ between $1$ and $t$, we have
$$
\frac{t^n-1}{t-1}=n\tau^{n-1}=n\,\left[\lambda_{\rm{min}},\lambda_{\rm{max}}\right]_\#
$$
where $[a,b]_\#$ represents a number in $[a,b]$.
$$
\begin{align}
\left|\int_x^y\left(t^{n-1}-t^{-1}\right)\mathrm{d}t\right|
&=\left|\int_x^y\left(t^n-1\right)t^{-1}\,\mathrm{d}t\right|\\
&=\left|n\int_x^y\left[\lambda_{\rm{min}},\lambda_{\rm{max}}\right]_\#\,(t-1)\,t^{-1}\,\mathrm{d}t\right|\\
&\le|n|\,\underbrace{\lambda_{\rm{max}}\int_x^y\,|t-1|\,t^{-1}\,\mathrm{d}t}_\text{a constant}
\end{align}
$$
Thus, in this case, it is valid to swap the integral and the limit
$$
\lim_{n\to0}\int_x^yt^{n-1}\,\mathrm{d}t=\int_x^yt^{-1}\,\mathrm{d}t
$$
|
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|
context free grammar accepted and generated language problems I'm having problems completing the following questions, I am able to attempt them but don't know if they are correct. Any help would be much appreciated.
Answer the following questions for a context free grammar $G=(N,\Sigma,P,S)$, where $N,\Sigma, P, S$ represents the set of non-terminal symbols, the set of production rules, and start symbol, respectively. Let $N=(S, A, B), \Sigma = (a,b), P= (S\rightarrow aAa|bAb, A\rightarrow aAa|bAb|B, B\rightarrow aB|bB|\epsilon)$.
$(1)$ Give a derivation of string $baabab$ in $G$
$(2)$ Give the language $L(G)$ that is generated by $G$
$(3)$ Give a context free grammar which generates the language $L(G)\cap(w \epsilon \Sigma^*| |w|_a = 2n + 1, n\geq1, where |w|_a= 2n+1, n\geq 1)$ represents the number of occurrences of $a$ in $w$.
$(1)$ $S\rightarrow bAb, A\rightarrow baAab, A\rightarrow baBab, B\rightarrow baaBab, B\rightarrow baaBbab, B\rightarrow baa\epsilon bab, \rightarrow baabab $
$(2)$ The language generated by the CFG seems to be a and b recurring to various powers eg. $aabbaaabab$. im not sure exactly what the syntax is for writing this is, but have attempted with $((a^nb^m)^*|n,m \geq 0)$
$(3)$ For this question also im not sure about the syntax for writing the language accepted by both $L(G)$ and $|w|_a=2n+1$. The way i understand it it should accept recurring a and b's like from L(G) but the a's have to be amod2=1. The following CFG seems to generate the union of these two languages.
$(S\rightarrow a^3Aa^3|bAb)$
$(A\rightarrow a^2Aa^2|bAb|B)$
$(B\rightarrow aB|bB|\epsilon)$
|
$$S\rightarrow aAa|bAb\\
A\rightarrow aAa|bAb|B\\
B\rightarrow aB|bB|\varepsilon$$
$(1)$ You want the word $baabab$ so $S\to bAb\to baAab\to baaBab\to baaBbab\to baabab$
|
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|
Is the axiom $g1 = g$ essential for a group action A group $G$ acts on a set $\Omega$ if
(1) $\omega\cdot 1_G = \omega$
(2) $(\omega \cdot g)\cdot h = \omega \cdot (gh)$
for all $\omega \in \Omega$ and $g,h \in G$. But is (1) really essential, what if we drop (1) and just require (2), then we must have
$$
\omega\cdot 1_G = \omega \cdot (gg^{-1}) = (\omega\cdot g)\cdot g^{-1}
$$
for all $g \in G$.
So do you know any example where (2) holds, but (1) fails?
|
Let $\Omega = \{1,2\}$, take $G$ to be any group and set $x \cdot g = 1$ for all $x \in \Omega$ and $g \in G$. Then of course $(x \cdot g) \cdot h = 1 \cdot h = 1 = x \cdot (gh)$, but $2 \cdot 1_G = 1 \neq 2$ .
|
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|
Limits using Epsilon Delta definition $$\lim_{x \to 3} x^{3}=27 $$
$$ 0<|x-3|<\delta$$
$$|x-3||x^{2}+3x+9|< \varepsilon$$
My teacher said that delta must be in the interval [0,3] but I don't think it's correct.
What is the correct approach to solve this kind of questions?
Thanks in advance.
|
take$ \delta = min(1,\frac{\varepsilon}{37})$
then $|x-3||x^{2}+3x+9|< 37|x-3| < \epsilon $
|
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|
An integral arising from Kepler's problem $\frac{1}{2\pi}\int_0^{2\pi}\ \frac{\sin^2\theta\ \text{d}\theta}{(1 + \epsilon\cos\theta)^2}$ I'm dealing with this integral in my spare time, since days and days, and it's really interesting. I'll provide to write what I tried until now, and I would really appreciate some help in understanding how to continue.
$$\Phi(\epsilon) = \frac{1}{2\pi}\int_0^{2\pi}\ \frac{\sin^2\theta\ \text{d}\theta}{(1 + \epsilon\cos\theta)^2}$$
Note: The integral arises in studying the Kepler's problem, which is why $\epsilon\in [0;\ 1]$.
I started with the passages to the complex plane:
$$\sin\theta = \frac{1}{2i}(z - z^{-1}) ~~~~~~~ \cos\theta = \frac{1}{2}(z + z^{-1}) ~~~~~~~ \text{d}\theta = \frac{\text{d}z}{iz}$$
thence
$$\begin{align*}
\Phi(\epsilon) & = \frac{1}{2\pi}\oint_{|z| = 1}\ \frac{-\frac{1}{4}(z - z^{-1})^2\ \text{d}z}{iz\left[1 + \epsilon\frac{(z + z^{-1}}{2}\right]^2}
\\\\
& = -\frac{1}{8\pi i}\oint_{|z| = 1}\ \frac{(z - z^{-1})^2\ \text{d}z}{z\left[\frac{2 + \epsilon(z + z^{-1})}{2}\right]^2}
\\\\
& = -\frac{1}{4\pi i}\oint_{|z| = 1}\ \frac{(z - z^{-1})^2\ \text{d}z}{z(2 + \epsilon(z + z^{-1})^2}
\\\\
& = -\frac{1}{4\pi i}\oint_{|z| = 1}\ \frac{\left(\frac{z^2 - 1}{z}\right)^2\ \text{d}z}{z\left(2 + \epsilon\frac{z^2+1}{z}\right)^2}
\\\\
& = -\frac{1}{4\pi i}\oint_{|z| = 1}\ \frac{(z^2-1)^2\ \text{d}z}{z(2z + \epsilon(z^2+1))^2}
\end{align*}$$
Poles et cetera
Poles are at $z_0 = 0$ and when $2z + \epsilon(z^2+1) = 0$, namely
$$z_1 = \frac{1+\sqrt{1 - \epsilon^2}}{\epsilon} ~~~~~~~~~~~ z_2 = \frac{1-\sqrt{1 - \epsilon^2}}{\epsilon}$$
Checking that $z_1\cdot z_2 = 1$ is ok, the question is now:
Question 1.: which one, aside $z_0$ lies into the unitary circle?
Question 2.: How to proceed?
|
I am assuming $0 < \epsilon < 1$ in the following. The cases
$\epsilon = 0$ and $\epsilon = 1$ have to be handled separately,
see below.
There is a small error in your calculation, the root of
$2z + \epsilon(z^2+1) = 0$ are
$$
z_1 = \frac{-1-\sqrt{1 - \epsilon^2}}{\epsilon} \, , \quad z_2 = \frac{-1+\sqrt{1 - \epsilon^2}}{\epsilon}
$$
Question 1.: which one, aside $z_0$ lies into the unitary circle?
From
$$
1+\sqrt{1 - \epsilon^2} > 1 > \epsilon
$$
it follows that $z_1 < -1$ and consequently, $-1 < z_2 < 0$, i.e. $z_2$
is inside the unit disk and $z_1$ outside.
Question 2.: How to proceed?
You already have that
$$
\Phi(\epsilon) = -\frac{1}{4\pi i}\oint_{|z| = 1} \frac{(z^2-1)^2\ }{z(2z + \epsilon(z^2+1))^2} \, dz =
\frac{1}{2\pi i} \oint_{|z| = 1} f(z) \, dz
$$
with
$$
f(z) := -\frac{1}{2 \epsilon^2} \frac{(z^2-1)^2\ }{z(z-z_1)^2(z-z_2)^2}
$$
From the residue theorem it follows that
$$
\Phi(\epsilon) = \text{Res}(f, 0) + \text{Res}(f, z_2) \, .
$$
$f$ has a simple pole at $z = 0$, therefore
$$
\text{Res}(f, 0) = \lim_{z \to 0} z f(z) = -\frac{1}{2 \epsilon^2} \, .
$$
$f$ has a double pole at $z = z_2$.
One possible method to compute the residue is the
limit formula for higher order poles:
$$
\text{Res}(f, z_2) = \lim_{z \to z_2} \frac{d}{dz} \bigl((z-z_2)^2 f(z) \bigr) \, .
$$
Special cases: For $\epsilon = 0$ the integral becomes
$$
\Phi(0) = -\frac{1}{4\pi i}\oint_{|z| = 1} \frac{(z^2-1)^2}{2z^3} \, dz
= -\frac{1}{8\pi i}\oint_{|z| = 1} \bigl( z - \frac 2z + \frac{1}{z^3}\bigr) \, dz
$$
which can easily be computed using the residue theorem.
Alternatively,
$$
\Phi(0) = \frac{1}{2\pi}\int_0^{2\pi} \sin^2\theta \, d\theta
$$
can be computed directly.
For $\epsilon = 1$, the integral becomes
$$
\Phi(1) = -\frac{1}{4\pi i}\oint_{|z| = 1} \frac{(z-1)^2}{z(z+1)^2} \, dz
$$
which is infinite due to the singularity at $z=-1$. Alternatively,
$$
\Phi(1) = \frac{1}{2\pi}\int_0^{2\pi} \frac{\sin^2\theta}{(1 + \cos\theta)^2} \, d\theta
= \frac{1}{2\pi}\int_0^{2\pi} \frac{(2 \sin \frac{\theta}{2} \cos \frac{\theta}{2})^2 }{(2 \cos^2 \frac{\theta}{2})^2} \, d\theta
= \frac{1}{2\pi}\int_0^{2\pi} \tan^2 \frac{\theta}{2}\, d\theta
$$
is infinite.
|
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|
Degeneracies of simplex $y$ which appears as any face of some simplex $x$ Let $K$ be simplicial set and $d_i:K_n\rightarrow K_{n-1}$, $s_i: K_n\rightarrow K_{n+1}$ ($i = 0,...,n$) face and degeneracy maps respectively.
Suppose we have some $x\in K_n$ with $d_0x = ... = d_n x = y$. How to prove that $s_0y=...=s_{n-1}y$? I think that $y$ should be an iterated degeneracy of some vertex $v\in K_0$ (probably $v = d_i^{n-1}y$), but I cannot prove it.
|
Could you expand on your motivation for this question? I think I have found a counterexample, essentially because if this identity was true, it would have to be an easy consequence of the simplicial identities.
The functor $F$ that sends a simplicial set $K_*$ to the $n$-simplices satisfying this condition is corepresented by some simplicial set $X^n$, i.e. there is a natural isomorphism $F(K) \cong \operatorname{sSet}(X^n,K)$, and by the Yoneda lemma it suffices to check this identity for the universal example $X^n$. Explicitly, it is given by the pushout
$$
\begin{array}{ccc}
\bigsqcup_{0}^n \Delta^{n-1} &\xrightarrow{\bigsqcup i\text{-th face}} & \Delta^n\\
\downarrow & & \downarrow\\
\Delta^{n-1} & \xrightarrow{\qquad} & X^n
\end{array}
$$
i.e. we identify all the faces of the "free" $n$-simplex.
By the simplicial identities, one can show that all iterated faces of the non-degenerate $n$-simplex of $X_n$ are equal. This shows that in the above pushout description, we can glue the faces of the $\Delta^{n-1}$'s together to obtain a pushout diagram
$$
\begin{array}{ccc}
\partial \Delta^{n} &\hookrightarrow & \Delta^n\\
\downarrow & & \downarrow\\
X^{n-1} & \to & X^n
\end{array}
$$
where the left-hand downward map is constructed inductively from the right-hand one. Since injections of simplicial sets are stable under pushouts, $X^{n-1}\to X^n$ is injective, and we see that $X^n$ has exactly one non-degenerate $k$-simplex for $0\le k\le n$, all of whose faces are equal to the non-degenerate $k-1$-simplex. This shows already that your simplex $y$ need not be degenerate.
There is an obvious map $X_n\to S^n = \Delta^n/\partial \Delta^n$ by collapsing $X^{n-1}$ inside $X^n$ which sends the non-degenerate $n$-simplex of $X^n$ to that of $S^n$ - on corepresented functors, this just says that an $n$-simplex all of whose faces are degeneracies of a $0$-vertex must have all faces equal. For $S^n$, it is not that difficult to see that all degeneracies of the nondegenerate $n$-simplex are pairwise different (each one has exactly two non-degenerate faces which are at different positions), so this must hold for $X^n$ as well.
Now $X^{n-1}\to X^n$ is injective, so the degeneracies of the nondegenerate $(n-1)$-simplex of $X^n$ are pairwise different. But this is "the" face of the nondegenerate $n$-simplex. So this is a counterexample once $n\ge 2$.
Here is an explicit description of the simplicial set $X^n$: its $k$-simplices are given by partitions of $[k]$ into at most $n+1$ nonempty intervals. The preimage of a nonempty interval under a monotonous function is a (possibly empty) interval; throwing away the empty preimages, this defines a functor $X^n:\Delta^{op}\to \mathrm{Set}$. Now consider the partition of $[n]$ into singletons. Each of its faces is the partition of $[n-1]$ into singletons, so these are all equal. Its degeneracies are the various partitions of $[n]$ into $n-1$ singletons and one pair, and these are all different for $n\ge 2$.
In the comments, the question was raised whether an $n$-simplex is a degeneracy of (one of) its $0$-faces if all of its degeneracies $s_k x$ are equal. More formally, this asks whether
$$
\begin{array}{ccc}
\bigsqcup_{0}^n \Delta^{n+1} &\xrightarrow{\bigsqcup i\text{-th degeneracy}} & \Delta^n\\
\downarrow & & \downarrow\\
\Delta^{n+1} & \xrightarrow{\qquad} & \Delta^0
\end{array}
$$
is a pushout diagram. This is indeed true: The simplicial identities imply that $d_0s_0 = \operatorname{id}$ and $d_0^ks_k = s_0d_0^k$, thus $d_0^k x = d_0^{k+1} s_0 x = d_0^{k+1} s_{k+1} x = s_0 d_0^{k+1} x$, and an easy induction shows that $x = d_0^0 x = s_0^n d_0^n x$, so $x$ is an $n$-fold degenerate simplex.
|
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|
Calculate this integral $\int \frac{x^2}{4x^4+25}dx$. I have to calculate this integral $\int \frac{x^2}{4x^4+25}dx$.
I dont have any idea about that.
I thought about parts integration.
Thanks.
|
Hint:
$$\int \frac { x^{ 2 } }{ 4x^{ 4 }+25 } dx=\int { \frac { x^{ 2 } }{ 4x^{ 4 }-20{ x }^{ 2 }+25-20{ x }^{ 2 } } dx } =\int { \frac { { x }^{ 2 } }{ { \left( 2{ x }^{ 2 }-5 \right) }^{ 2 }-20{ x }^{ 2 } } dx } =\int { \frac { x^{ 2 }dx }{ \left( 2x^{ 2 }-2\sqrt { 5 } x-5 \right) \left( 2x^{ 2 }+2\sqrt { 5 } x-5 \right) } } $$
|
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|
How do you solve B and C for $\frac{s-1}{s+1} \frac{s}{s^2+1} = \frac{A}{s+1} + \frac{Bs+C}{s^2+1}$? How do you solve B and C for $\frac{s-1}{s+1} \frac{s}{s^2+1} = \frac{A}{s+1} + \frac{Bs+C}{s^2+1}$ ?
$A = \left.\frac{s^2-s}{s^2+1} \right\vert_{s=-1} = \frac{1-(-1)}{1+1}=1$
|
Notice, $$\frac{s-1}{s+1}\frac{s}{s^2+1}=\frac{A}{s+1}+\frac{Bs+C}{s^2+1}$$
$$s^2-s=(A+B)s^2+(B+C)s+(A+C)$$
comparing the corresponding coefficients on both the sides, one should get $$A+B=1\tag 1$$
$$B+C=-1\tag 2$$
$$A+C=0\tag 3$$
on solving (1), (2) & (3), one can easily get $\color{red}{A=1}, \color{blue}{B=0}, \color{red}{C=-1}$
|
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How to determine (and explain) the sum of angles without measuring? Below is a photo of the angles/triangles in which I am working on determining the sum of the angles without measuring. The angles are a,b,c,d,e,f.
I understand that angles are formed my intersecting lines and I see many intersecting lines in this image. The angles in a triangle add to 180 degrees and I see 3 labeled triangles. Also, I know that if parallel lines are cut by transversal lines, then the corresponding angles are equal.
I'm confused on how to determine the sum of the angles a+b+c+d+e+f without measuring. I also don't know how I would explain my reasoning. I know the theorems but combining all of the knowledge into a reason is difficult. Where do I even begin?
|
Call the angles of the three outer triangles $a, b, g$; $c, d , h$; and $e, f, i$, respectively. Then
$$a + b + c + d + e + f = (a + b + g) + (c + d + h) + (e + f + i) - (g + h + i) = 3 \times 180^{\circ} - 180^{\circ} = 360^{\circ},$$
where we've used the fact that $g, h, i$ are also the angles of the inner triangle.
|
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Continuous function $f$ on $\mathbb R $ such that $f \notin L^1 (\mathbb R)$ but $f \in L^1([a,b]), a< b $ Give an example of a continuous function $f$ on $\mathbb R $ such that $f \notin L^1 (\mathbb R)$ but $f \in L^1([a,b]), a< b $
If $f \in L^1([a,b]), a< b$ that would mean that $$\int_{a}^{b}|f(x)|dx < \infty \text{ but $f \notin L^1 (\mathbb R)$ means} \int_{\mathbb R}|f(x)|dx\to \infty \text { or undefined.}$$
I feel like there are many examples for this, like $f(x)=x$, but this was given on a exam, so I think I am not understanding this correctly, what are your guys' thoughts on this?
Also the second part of the question was to give an example of a continuous function $f$ on $\mathbb R $ such that $f$ is Reiman integrable on $\mathbb R$ but $f \notin L^1 (\mathbb R)$ . Now this seems like it is impossible to me. We have done the other way around in class, but this I just don't know at all how I would go about, maybe constructing a set that satisfies this. Help is needed, very thankful for insight.
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How about $f(x)\equiv1$ for the first one, since then $\int_a^bf(x)dx=b-a<\infty$, but clearly $f\not\in L^1(\Bbb R)$. For the second, pick a function that oscillates enough that cancellation allows integrability but not absolute integrability. The answer below has a nice example.
EDIT: If one of $a$ or $b$ is infinite (at the appropriate end of the real line), then pick $f(x)=e^{\alpha x}$ and choose $\alpha$ so that $f\in L^1(a,b)$ but not $L^1(\Bbb R)$.
|
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If A is the range of $f(x) = ^{7-x}C_x$ then the no. of reflexive relation from A to A is... Problem :
If A is the range of $f(x) = ^{7-x}C_x$ then the no. of reflexive relation from A to A is
(a) $2^6$ (b) $2^{12}$ (c) $2^{16}$ (d)$2^{20}$
My approach :
$f(x) = ^{7-x}C_x = \frac{(7-x)!}{x!(7-2x)!} $
But having no idea how to find the range of this.
Let R be a binary relation on a set A.
R is reflexive, iff for all x $\in$ A, (x,x) $\in$ R, i.e. xRx is true.
Examples:
Equality is a reflexive relation
for any object x, x = x is true.
Please help in finding the range and no. of reflexive relation of this will be of great help thanks.
|
Assuming that $x \in \mathbb{Z}$. The binomial coefficient $\binom{7-x}{x}$ is defined for $0 \leq x \leq 7$ and $x \leq 7-x$. Thus $x \in \{0,1,2,3\}$. Now you can find possible values of this coefficient. And this gives the size of set $A$.
Any reflexive relation $R$ should have pairs of the form $(x,x)$ for all $x \in A$. So if $|A|=n$, then $R$ should have these $n$ pairs. For "other" $n^2-n$ pairs from $A \times A$, either they are in $R$ or they are not. So each such pair's fate can be decided in $2$ ways. This will help you with the number of reflexive relations.
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Roots of $f(x)+g(x)$ Question : Let $p,q,r,s \in \mathbb R$ such that $pr=2(q+s)$. Show that either $f(x)=x^2+px+q=0$ or $g(x)=x^2+rx+s=0$ has real roots .
My method :
To the contrary suppose that both $f(x)$ and $g(x)$ has no real roots.
Now observe that $f(x)+g(x)=2x^2+(p+r)x+(q+s)=0$
Since $pr=2(q+s)$ , $f(x)+g(x)=2x^2+(p+r)x+\frac{pr}{2}=0$
$\Rightarrow f(x)+g(x)=x^2+(\frac{p}{2}+\frac{r}{2})x+\frac{p}{2}\cdot \frac{r}{2}=0$
$$\Rightarrow f(x)+g(x)=\left(x-\frac{p}{2}\right)\left(x-\frac{r}{2}\right)=0$$
So $f(x)+g(x)$ has two distinct real roots.
Thus $f(x)+g(x)<0$ for $x \in (\frac{p}{2},\frac{r}{2})$ if $r>p$.
But this is a contradiction as $f(x)>0$ and $g(x)>0$ for all $x \in \mathbb R $.
Thus either $f(x)=x^2+px+q=0$ or $g(x)=x^2+rx+s=0$ has real roots .
Am I correct ?
Can this be solved directly without using contradiction ?
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(1). You did not state that $p\ne r$ and it is not needed. (2).Supposing that neither $f$ nor g has a real root, I do NOT observe that $f(x)+g(x)=0.$ Instead I observe that if neither $f$ nor $g$ has a real root then $\forall x\;(f(x)+g(x)>0).\;$ (3). I agree that $$\forall x\;[f(x)+g(x)=(x-p/2)(x-r/2)]$$ which implies that $$f(p/2)+g(p/2)=0$$ which implies that $$f(p/2)\leq 0 \lor g(p/2)\leq 0$$ which implies that $f(x)$ and $g(x)$ cannot both be always positive, which implies that at least one of $f,g$ has a root.(Because $\exists x\;(f(x)\leq 0\lor g(x)\leq 0).)$......... Another method is that if neither $f$ nor $g$ has a real root then $0>p^2-4 q$ and $0>r^2-4 s.$ Adding these, we get $0>p^2+r^2-4(q+s)=p^2+r^2-2 p r=(p-r)^2,$ which is impossible.
|
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Mensuration and similarity Cone P has a volume of 108cm^3
Calculate the volume of 2nd come , Q , whose radius is double that of cone P and its height is one-third that of cone p
Here's my working ....
$$V_Q=\frac13 \pi (2r)^2 \cdot \frac{h}{3}\\
= \frac{4}{9} \pi r^2 h\\
= \frac{1}{3}\pi r^2 h \cdot \frac{4}{3}$$
I don't understand why must I do this.
|
Hint:
In your equation, you already have $$V_Q = \left(\frac{1}{3}\pi r^2h\right) \cdot 43$$
You also know what the volume of cone $P$ is, and you know that it is equal to $\frac{1}{3}\pi r^2 h$
|
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Limit $\lim_{x \to 0} \dfrac{\sin(1-\cos x)}{ x} $ $$\lim_{x \to 0} \dfrac{\sin(1-\cos x)}{ x} $$
What is wrong with this argument: as $x$ approaches zero, both $x$ and $(1-\cos x)$ approaches $0$.
So the limit is $1$ .
*
*How can we prove that they approaches zero at same rate?
*This is not about solving the limit because I already solved it but about the rate of both functions going to zero .
m referring to $$\lim_{x \to 0} \dfrac{\sin x}{ x} =1 $$
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By your argument, $$\lim_{x\to 0}\frac{x}{x}$$ is also $0$ because as $x$ approaches zero, both $x$ and $x$ approach $0$, so the limit is $0$.
|
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Awodey's Category Theory Exercise 9.9.2 I am having problems with this question in Awodey's Category Theory book p.248:
*Show that every monoid M admits a surjection from a free monoid $F(X) → M$, by considering the counit of the free $\dashv$ forgetful adjunction.
Let $f : F(X)→M$ be a monoid homomorphism and $m \in M$. We want to show that there is $x \in F(X)$ such that $f(F(x))=m$. But now I'm kind of stuck.
I know I should use the the UMP of the counit $\epsilon : F \circ U \to 1_\mathbf{D}$ somehow, which says that, for $C \in \mathbf{C}$ and $D \in \mathbf{D}$, each $f : F(C) \to D$ determines a function $g : U(D) \to C$ uniquely up to isomorphism such that $f = F(g) \circ \epsilon_D$.
Any help would be highly appreciated!
Thanks!
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Let $f : F(X)→M$ be a monoid homomorphism and $m \in M$. We want to show that there is $x \in F(X)$ such that $f(F(x))=m$. But now I'm kind of stuck.
This looks like you're trying to show that every monoid homomorphism from a free monoid is surjective, but you only have to show that there exists a surjection from a free monoid.
You're told to use the counit of the adjunction, which is a natural transformation $FU\rightarrow id$. So its component at $M$ is a homomorphism $F(U(M))\rightarrow M$. Try to show that this is surjection.
(Hint: Find a right inverse to the underlying function $U(F(U(M)))\rightarrow U(M)$.)
|
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Why does the sum of two linearly independent solutions of a second order homogeneous ODE give a general solution? The following is a short extract from the book I am reading:
If given a Homogeneous ODE:
$$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}+5\frac{\mathrm{d} y}{\mathrm{d}x}+4y=0\tag{1}$$
Letting
$$D=\frac{\mathrm{d}}{\mathrm{d}x}$$ then $(1)$ becomes
$$D^2 y + 5Dy + 4y=(D^2+5D+4)y$$
$$\implies\color{blue}{(D+1)(D+4)y=0}\tag{2}$$
$$\implies (D+1)y=0 \space\space\text{or}\space\space (D+4)y=0$$ which has solutions $$y=Ae^{-x}\space\space\text{or}\space\space y=Be^{-4x}\tag{3}$$ respectively, where $A$ and $B$ are both constants.
Now if $(D+4)y=0$, then $$(D+1)(D+4)y=(D+1)\cdot 0=0$$
so any solution of $(D + 4)y = 0$ is a solution of the differential equation $(1)$ or $(2)$. Similarly, any solution of $(D + 1)y = 0$ is a solution of $(1)$ or $(2)$. $\color{red}{\text{Since the two solutions (3) are linearly independent, a linear combination}}$ $\color{red}{\text{of them contains two arbitrary constants and so is the general solution.}}$ Thus $$y=Ae^{-x}+Be^{-4x}$$ is the general solution of $(1)$ or $(2)$.
The part I don't understand in this extract is marked in $\color{red}{\mathrm{red}}$.
*
*Firstly; How do we know that the two solutions: $y=Ae^{-x}\space\text{and}\space y=Be^{-4x}$ are linearly independent?
*Secondly; Why does a linear combination of linearly independent solutions give the general solution. Or, put in another way, I know that $y=Ae^{-x}\space\text{or}\space y=Be^{-4x}$ are both solutions. But why is their sum a solution: $y=Ae^{-x}+Be^{-4x}$?
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If you recall from linear algebra, abstract functional spaces can be considered as vector spaces. We define the zero function to serve as the zero vector, and pointwise addition/multiplication as the vector space operations.
For a collection of normal vectors, to show linear independence, we want to show none of the chosen vectors can be written as a linear combination of any of the others; each vector describes a 'different' part of the space. For a vector space of functions, e.g. the space of differentiable functions, to show linear independence, we must show there are no non-zero scalars $a,b$ such that
\begin{align*}
af(t)+bg(t)=0
\end{align*}
for $\textit{all}$ values of $t$ in the domain. It isn't enough that we can find one or two values of $t$ where the sum equals zero, but instead for all values of the domain.
Now, to show the two functions in your problem are linearly independent. Suppose we could find two non zero numbers $a$ and $b$ such that
\begin{align*}
ae^{-t}+be^{-4t}=0
\end{align*}
For all $t$. Differentiate this expression.
\begin{align*}
-ae^{-t}-4be^{-4t}=0
\end{align*}
Adding the two equations together gives
\begin{align*}
-3be^{-4t}&=0
\end{align*}
The exponential function is strictly positive, so we must have $b=0$. This implies $ae^{-t}=0$, and similarly $a=0$.
Now, why must a linear combination also be a solution? Well the differential equation described is $\textbf{linear}$, so any element of the span of linearly independent solutions will always yield another solution. They all get mapped to zero.
|
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Interpreting functions for poset I'm stuck at a question with the following expression of poset $(\mathcal{N}, \sqsubseteq)$:
$$\forall x, y \in \mathcal{N}: x \sqsubseteq y \mbox{ iff } (x = y \vee x = \bot_{\mathcal{N}} \vee y = \top_{\mathcal{N}}).$$
$\mathcal{N} = \mathcal{N} \bigcup\{ \bot_{\mathcal{N}} , \top_{\mathcal{N}} \} $ consists of all natural numbers with two elements $ \bot_{\mathcal{N}} $ and $ \top_{\mathcal{N}} $, and I'll need to find the poset height.
I don't know how to interpret this expression though.
|
This poset is the set of all natural numbers with equality relation, so any two distinct elements $x, y \in \mathcal{N}$ are incomparable, together with the least $\bot_\mathcal{N}$ and the greatest $\top_\mathcal{N}$ elements. To find its height you should find the largest cardinality chain, that is a subset in which any two elements are comparable. Can you take it from there?
Hint: Since any two distinct $x, y \in \mathcal{N}$ are incomparable, every chain in this poset can contain at most one element from $\mathcal{N}$. You can start by trying to find the chains of cardinality $1, 2$ and so on. It can be useful to draw a Hasse diagram for this poset.
Updated. Here is some definitions (which you are assumed to be familiar with given that problem). Let $(P, \sqsubseteq)$ be a poset. Two elements $x, y \in P$ are comparable if either $x \sqsubseteq y$ or $y \sqsubseteq x$.
A chain $C$ is a subset of $P$ such that any two elements $x, y \in C$ are comparable. The height of $P$ is the largest cardinality (or size) of a chain in $P$.
So roughly, to find the height of your poset you should check all chains in it and find the chains of the largest size. Note the following:
*
*If $C$ is a chain, then $C \cup \{\top_\mathcal{N}\}$ and $C \cup \{\bot_\mathcal{N}\}$ are also chains. This is because the least and the greatest elements are comparable with any element of a poset.
*If $x, y \in \mathcal{N}$ and they are both not equal to $\top_\mathcal{N}$ and $\bot_\mathcal{N}$, then by definition of $\sqsubseteq$ we have $x \sqsubseteq y$ if and only if $x = y$.
From 1 it follows that if you have some chain, you can always add the least and the greatest elements to it obtaining new (and in general larger) chain. From 2 it follows that any chain in your poset can't contain two distinct natural numbers from $\mathcal{N}$ since they are not comparable. It means that the chain of maximal cardinality has at most $3$ element: $\top_\mathcal{N}$, $\bot_\mathcal{N}$ and some natural number $x \in \mathcal{N}$. And obviously there are such chains in your poset. So its height is $3$.
|
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Is there a rational surjection $\Bbb N\to\Bbb Q$? The question is in the title. Is there a one-dimensional rational function $f\in\Bbb R(X)$ which restricts to $\Bbb N\to\Bbb Q$, which is a surjection onto $\Bbb Q$? My guess is no.
Expanding the scope a little (at the expense of precision), are there any "nice" functions that enumerate $\Bbb Q$? Here "nice" is meant to exclude the floor function or absolute value function and related trickery. At first I thought it might work to use analytic functions here, but there is an analytic function taking any chosen values on $\Bbb N$ subject to a mild growth hypothesis (I think $f(n)\in o(n)$), by defining $f(z)=\sum_{n\in\Bbb N}a_n{\rm sinc}(\frac{z-n}\pi)$, where ${\rm sinc}(z)=\frac{\sin(z)}z$ continuously extended over zero. I will let answerers supply their own definitions of "nice" if they want to tackle the broader question.
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This is an alternative argument based on Arnaud's suggestion. Without loss of generality, you can assume the rational function $f$ is $\frac{p}{q}$ where $n = \deg(p) \ge \deg(q) = m$ (otherwise trade $f$ in for $\frac{1}{f}$). But then (using $(\frac{p}{q})' = \frac{p'q - pq'}{q^2}$) you have:
$$
f'(x) = \frac{(n-m)p_nx^{m+n-1} + \mbox{terms of lower degree}}{x^{2m} + \mbox{terms of lower degree}}
$$
where, without loss of generality, I have assumed $q$ is monic and where $p_n \neq 0$ and $n > 0$ (if $n = 0$, $f$ is constant).
So as $x$ tends to $\pm\infty$, $f'(x)$ tends to $\pm\infty$, if $n > m+ 1$, and to $p_n \neq 0$ if $n = m + 1$. So $f(x)$ is monotonic whenever $|x|$ is sufficiently large if $n > m$. If $m = n$, then as $x$ tends to $\pm\infty$, $f(x)$ tends to $\pm p_n$, so $f(x)$ is bounded. But if $f(x)$ is either monotonic for large enough $|x|$ or bounded, then $f$ cannot map $\Bbb{N}$ onto $\Bbb{Q}$.
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Generator matrix of $E_5$
Let $E_5$ denote the binary even weight code of length 5. Write down a generator matrix of $E_5$.
So I know the length $n = 5$ is the number of rows in the generator matrix and the number of columns will be the dimension of $E_5$.
So I firstly need to know how to work out the dimension and I can then attempt to construct the generator matrix. Please can someone give me a hint!
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Think of how many degrees of freedom you have for this code. How many bits can you fill in however you want, and still be able to use the remaining bits to make the word "even"?
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Suppose $a \in \mathbb{Z}.$ Prove that $5 \mid 2^na$ implies $5 \mid a$ for any $n \in \mathbb{N}$ This question is supposed to be solved by induction, however I'm unsure of where to get my base case from exactly, because the question is asking about both $a$ and $n$. I started with my base case being $n = 1$, but then I get $5\mid 2a$, and I'm unsure what to do from there. Am I just barking up the wrong tree here? Or how should I go about getting a start to this problem?
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$1)$ assume $5\nmid a$
$2)$ by the Division Algorithm $a=5q+r$, $\space\space r\lt 5$,$r<q\space\space\space\space\space\space\space 1)$
$3)$ $2^na=2^n(5q+r)=2^n(5q)+2^nr=5(2^nq)+2^nr$, $\space\space r<5\space\space\space\space\space\space\space 2)$
$4)$ $5\nmid 2^na\space\space\space\space\space\space\space1),2),3)$
$5)$ $F_0$(contradiction)$\space\space\space\space\space\space\space 4)$
$6)$ $5\nmid a\rightarrow F_0\space\space\space\space\space\space\space 5)$
$7)$ $5|a\space\space\space\space\space\space\space 6)$
|
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$3^x-2^y=17$ where $x$ and $y$ are both positive integers First, I was wondering what would the solution(s) be to the equation $3^x-2^y=17$ such that $x$ and $y$ are both positive integers. I couldn't find any small possibilities. Is there a proof that such a solution exists? How about the general case $3^x-2^y=z$? Is there exactly one x and y positive integer solution for every positive integer z (not divisible by 2 or 3 that is)? Thanks in advance!
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It is not true that there is a solution for every odd $z$ not divisible by $3$. For example, $3^x-2^y=13$ is impossible. For we can check by hand that there is no solution if $y=1$ or $y=2$. And if $y\ge 3$, then $13+2^y\equiv 5\pmod{8}$. But powers of $3$ are congruent to $1$ or $3$ modulo $8$. There are infinitely many examples of this type.
We can give a partial answer to your question about the uniqueness of the solution, if there is one. It is a theorem of Stroeker and Tijdeman that the equation $3^x-2^y=z$ has at most one solution in positive integers if $|z|\gt 13$.
For more information, and references, please see OEIS.
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$|\int\limits_a^b f(t)dt| \leq \int\limits_a^b |f(t)|dt$, where $f(t) \in \mathbb C$ Let $f: [a,b] \to \mathbb C, t \to f(t) = \text {Re } f(t) + i\text{ Im } f(t)$. Suppose $f$ is continuous.
Let $\int\limits_a^b f(t)dt = \rho e^{i\theta}$, where $\rho \geq 0$ is the module. Then:
$$|\int\limits_a^b f(t)dt| = \rho = e^{-i\theta}\int\limits_a^b f(t)dt = \text{Re }\int\limits_a^b e^{-i\theta}f(t)dt = \int\limits_a^b \text{Re }(e^{-i\theta}f)(t)dt \leq \int\limits_a^b |f(t)|dt$$
I don't understand 2 steps:
*
*$e^{-i\theta}\int\limits_a^b f(t)dt = \text{Re }\int\limits_a^b e^{-i\theta}f(t)dt$
*$\int\limits_a^b \text{Re }(e^{-i\theta}f)(t)dt \leq \int\limits_a^b |f(t)|dt$
|
For the first step, up to the sign, note that $\rho$ is real. Since it is equal to $e^{-i\theta}\int\limits_a^b f(t)dt$, the last expression is a real number and so $$\rho=e^{-i\theta}\int\limits_a^b f(t)dt= \text{Re }e^{-i\theta}\int\limits_a^b f(t)dt.$$
For the second step, write $z=x-iy$ and note that $$\text{Re }z=x\le |x|\le \sqrt{x^2}\le\sqrt{x^2+y^2}=|z|.$$
|
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Does a finite first-order theory which has a model always have a finite model? I'm curious whether this is true or not:
Let T be a finite, first-order theory. If T has a model, then T has a finite model.
I would assume the answer is 'yes', but I wanted to make sure I haven't missed something obvious. The reason I believe this to be the case is that according to my understanding of the Lowenheim-Skolem theorem, the only way a countable first-order theory can 'force' a size onto its potential models is by including a countable number of constants which the model must have representations for. Therefore, if the theory is finite, then the number of constants it can 'force' to exist must also be finite, thereby ensuring the existence of a finite model (assuming any model can exist at all). Is anything wrong with my reasoning (apart from its relative informality)?
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Let the language contain a single binary predicate symbol, to be interpreted as "less than."
Consider the theory $T$ with the following axioms:
1) There are at least two objects.
2) The relation $L$ is a strict total order.
3) For any $x$ and $y$ such that $L(x,y)$, there exists a $t$ such that $L(x,t)$ and $L(t,y)$.
Then all models of $T$ are infinite. This is the theory of dense linear order. It has nice model-theoretic properties. Cantor proved that any countable densely ordered set with no first or last element is order-isomorphic to the rationals under the usual order. Similarly, a countable densely ordered set with first element and no last is order isomorphic to the rationals in $[0,1)$, with similar results for the other two possibilities.
|
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Show that $\sum_{n=0}^\infty |a_n|<\infty.$
Let $\{a_n\}$ be a real sequence. Suppose for every convergent real sequence $\{s_n\}$, the series $\sum_{n=0}^\infty a_n s_n$ converges. Show that $\sum_{n=0}^\infty |a_n|<\infty.$
Mt attempt:
Consider the sequence $\{s_n\}$ where $s_n=1$ for all $n$. Then we have that $\sum_{n=0}^\infty a_n$ converges. But how do I show the absolute convergence? Do I have to do something using Abel summation here? Because since $\sum_{n=0}^\infty a_n$ converges, let $\sum_{n=0}^\infty a_n=s$. Then the Abel sum $\lim_ {x\rightarrow 1} \sum_{n=0}^\infty a_nx^n=s$. I am stuck here. Can somebody please help me?
|
Just elaborating Patrick's comment. Fix $N$. Let
$$
s_n =\begin{cases}
|a_n|/a_n &\text{if } n\leq N \text{ and } a_n\neq 0 \\
1 &\text{if } n\leq N \text{ and } a_n=0 \\
1 &\text{if } n> N
\end{cases}
$$
It is clear that $s_n$ is a convergent sequence. So by the hypothesis,
$$
\sum_{n=1}^{\infty} s_n a_n = \sum_{n=1}^{N} |a_n| + \sum_{n=N+1}^{\infty} a_n
$$
converges. Now let
$$
t_N = \sum_{n=1}^{N} |a_n| + \sum_{n=N+1}^{\infty} a_n
= \sum_{n=1}^{\infty} |a_n| - \sum_{n=N+1}^{\infty} |a_n| + \sum_{n=N+1}^{\infty} a_n
$$
We have shown that $t_N$ is well-defined for each $N$. Now note that
$$
\lim_{N\to\infty} \sum_{n=N+1}^{\infty} \left(a_n-|a_n|\right) \leq \lim_{N\to\infty} \sum_{n=N+1}^{\infty} \left(2 a_n \right) = 0
$$
because $\sum_{n=1}^{\infty} a_n$ converges, as you pointed out in the post. We have
$$
t_N = \sum_{n=1}^{\infty} |a_n| + p_N
$$
where $p_{N} = \sum_{n=N+1}^{\infty} (a_n-|a_n|)$. Since $p_{N}\to 0$, we can find a sufficiently large $N$ such that $|p_{N}|<1$. But then
$$
\sum_{n=1}^{\infty} |a_n| = t_{N} - p_{N} < t_{N} + 1
$$
so in particular $\sum_{n=1}^{\infty} |a_n|$ converges because $t_{N}+1$ is a finite number.
|
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|
Proving $\frac{1}{\sqrt{1-x}} \le e^x$ on $[0,1/2]$. Is there a simple way to prove $$\frac{1}{\sqrt{1-x}} \le e^x$$ on $x \in [0,1/2]$?
Some of my observations from plots, etc.:
*
*Equality is attained at $x=0$ and near $x=0.8$.
*The derivative is positive at $x=0$, and zero just after $x=0.5$. [I don't know how to find this zero analytically.]
*I tried to work with Taylor series. I verified with plots that the following is true on $[0,1/2]$:
$$\frac{1}{\sqrt{1-x}} = 1 + \frac{x}{2} + \frac{3x^2}{8} + \frac{3/4}{(1-\xi)^{5/2}} x^3 \le 1 + \frac{x}{2} + \frac{3}{8} x^2 + \frac{5 \sqrt{2} x^3}{6} \le 1 + x + \frac{x^2}{2} + \frac{x^3}{6} \le e^x,$$
but proving the last inequality is a bit messy.
|
If $f(x)=(1-x)e^{2x}$, then $f'(x)=(1-2x)e^{2x}=0$ when $x=\frac{1}{2}$. Drawing a graph/checking the second derivative shows it to be a maximum, whence $1=f(0)\le f(x)\le f(1/2)=\frac{e}{2}$ on $[0,\frac{1}{2}]$. We thus have:
$$1\le(1-x)e^{2x}\le\frac{e}{2}$$
$$\implies \frac{1}{1-x}\le e^{2x}\le\frac{e}{2(1-x)}$$
$$\implies \frac{1}{\sqrt{1-x}}\le e^x \le \sqrt{\frac{e}{2}}\frac{1}{\sqrt{1-x}}$$
on the given interval.
The reason I chose this approach is that much in the same way as young children make most of their arithmetic mistakes when dealing with fractions and negative numbers, I find myself far more at ease when fractions, square roots, inverse functions and such are all cleared out (I'm still very averse to the quotient rule for differentiation). So dealing with $(1-x)e^{2x}$ is greatly preferable for me, and is set up in such a way that the required bounds should pop out quite naturally.
|
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|
Prove with use of derivative How to prove this inequality using derivative ?
For each $x>4$ ,
$$\displaystyle \sqrt[3]{x} - \sqrt[3]{4} < \sqrt[3]{x-4} $$
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If $f(x)=x^\frac{1}{3}$ we must show that $$ f(x) < f(x-4) + f(4)
$$
Note that for any $a,\ b>0$ if we have $f(a+b)< f(a) + f(b)$, then
we are done. Note that this is equivalent to $$ f(a+b)-f(a) < f(b) -
f(0) $$
In further this is
equivalent to $$ \int_a^{a+b} f'(x)dx < \int_0^b f'(x) dx $$
So $f'(x) = \frac{1}{3} \frac{1}{x^\frac{2}{3}} $ is decreasing so
that in the above right integral is large
|
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|
When $n$ is odd, $\langle (123),(12...n)\rangle$ generates $A_{n}$ Working with the knowledge that the set of 3-cycles generates $A_{n}$, the basic idea is to express any 3-cycle as a word in $(123)$ and $(12...n)$ when $n$ is odd.
Not knowing how to progress, I decided to work with the concrete group $A_{5}$. I've listed all 3-cycles of $A_{5}$ below with each element followed by its inverse: $$(123),(132),(124),(142),(125),(152),(134),(143),(135),(153),(145),(154),(234),(243),(235),(253),(245),(254),(345),(354)$$ Now, I am able to generate the following elements $(234),(345),(451),(512)$ (and, correspondingly, their inverses) using the following rule:$$(12345)^{i}(123)(12345)^{-i}$$ for $i=1,2,3,4$ (for $i=5$ we get back $(123)$).
Promising as this looks, I am not able to extend the idea to cover the other 3-cycles in $A_{5}$.
Suggestions and hints are appreciated. Would prefer to answer myself.
|
Note that
\begin{align*}
(1\ 2\ \dots\ n)(1\ 2\ 3)(1\ 2\ \dots\ n)^{-1} &= (2\ 3\ 4)\\
(1\ 2\ \dots\ n)(2\ 3\ 4)(1\ 2\ \dots\ n)^{-1} &= (3\ 4\ 5)\\
&\ \ \vdots
\end{align*}
More generally, given a permutation $\sigma$ and a cycle $(a_1\ a_2\ \dots\ a_k)$, we have
$$\sigma(a_1\ a_2\ \dots\ a_k)\sigma^{-1} = (\sigma(a_1)\ \sigma(a_2)\ \dots\ \sigma(a_k)).$$
Finally, use the fact that $(a\ b\ c)^{-1} = (a\ c\ b)$.
|
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|
Prove that $f:\mathbb{R}^n\to\mathbb{R}$ is continuous
Let $f:\mathbb{R}^n\to\mathbb{R}$ such that for every continuous curve, $\gamma:[0,1]\to\mathbb{R}^n$: $f\circ\gamma$ is continuous. Prove that $f$ is continuous.
So I know we shall prove it by a contradiction. Let's assume that $f$ isn't continuous at $x_0$. Then, there's a sequence, $\{x_k\}$ converging to $x_0$ such that $\lim_{k\to\infty} f(x_k) \ne f(x_0)$.
Now, I need to have some curve in order to get a contradiction.
I'd be glad to get help with that.
Thanks.
|
Construct a curve "joining the points" $x_1$, $x_2$,... with $\gamma(0) = x_1$, $\gamma(1/2) = x_2$,...
|
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|
The order of pole of $\frac{1}{(2\cos z -2 +z^2)^2} $at $z=0$ What is the order of the pole of :
$$\frac{1}{(2\cos z -2 +z^2)^2}$$
at $z=0$
This is what I did :
$$\cos z = \frac{e^{iz}-e^{-iz}}{2} $$
then :
$$ \implies \frac{1}{(e^{iz}-e^{-iz}+z^2-2)^2} $$
How do I continue?
|
Hint : Let , $f(z)=2\cos z-2+z^2$. Find the order of zero of $f$.
$f'(z)=-2\sin z+2z$.
$f''(z)=-2\cos z+2$
$f'''(z)=2\sin z$
$f^{iv}(z)=2\cos z$.
Clearly , $f(0)=f'(0)=f''(0)=f'''(0)=0$ , but $f^{iv}(z)\not= 0$. So $f$ has a zero of order $4$.
So , $(2\cos z-2+z^2)^2$ has a zero of order $8$ at $z=0$.
|
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|
Is this statement about this continuous function correct? Is it true that if $f^2$ is continuous, then $\lvert f \rvert$ is continuous?
Can I say that since $f^2 \geq 0$ then $\lvert f \rvert = \sqrt{f^2}$ and that indicates that $\lvert f \rvert$ is also continuous beacuse of the arithmetic property of continuous functions?
|
Yes, the square root function is continuous in its domain, and composition of continuous functions is continuous.
|
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|
Infinite exponentials We can read a lot of about convergence of series or Infinite products.
E.g. for series.
Following series
$$\sum_{i=1}^\infty a_i$$
is convergent when
$$\lim_{n\rightarrow\infty}a_n=0$$
and
*
*D'Alembert's criterion
$$\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|<1$$
2. Cauchy's criterion
$$\limsup_{n\rightarrow\infty}\sqrt[n]{\left|a_n\right|}<1$$
3. Raabe–Duhamel's criterion
$$\lim_{n\rightarrow\infty}n\left(\frac{a_{n+1}}{a_n}-1\right)>1$$
For products we can use above criterias for series.
Following product converges
$$\prod_{i=1}^\infty a_i$$
if series
$$\sum_{i=1}^{\infty} \ln{a_i}$$
converges.
What about Infinite exponentials? (I don't know how to write it)
$$a_1^{{a_2}^{{a_3}^{{a_4}^\cdots}}} $$
I know that tetration is particular case of that for
$$
a_1=a_2=a_3=\cdots=x
$$
Then this infinite exponentials convergence for
$$
x\in\left<\frac{1}{e^e};\sqrt[e]{e}\right>
$$
Is there general theorem of that? Where can I read about results?
I'm sure that $\lim_{n\rightarrow\infty}a_n=0$ or $\lim_{n\rightarrow\infty}a_n=1$ is not required. Because for $a_i=\sqrt{2}$ we have
$${\sqrt{2}}^{{\sqrt{2}}^{{\sqrt{2}}^{{\sqrt{2}}^\cdots}}}=2 $$
|
I think you can find in WP's "tetration" the criterion for the powertower over the reals, which was established by L. Euler (but independently by others)
Over the complex numbers the range of convergence of the infinite tetration was determined by W. Thron(1957) "Convergence of Infinite Exponentials with Complex Elements" and based on his work improved by D. Shell(1961) "On the convergence of infinite exponentials" .
The article of D.Shell is online here ; the basic article of W.Thron is also online here at ams.
See also for an overview in the wiki of the tetration-forum but the Shell-article should be linked to in the WP-article too.
|
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|
Number of strings over a set $A$ How can I calculate the number of strings of length $10$ over the set $A=\{a,b,c,d,e\}$ that begin with either $a$ or $c$ and have at least one $b$ ?
Is it accomplished through some sort of combinatorial logic coupled with discrete mathematics?
|
Yes, you are correct. Use constructive counting. Begin by selecting the first letter in the string, which you said could be either $A$ or $C.$ There are $2$ ways to do this.
Now our problem becomes: construct a string of length $9$ with at least one $B.$ We count this with complementary counting - how many strings can we make without a single $B?$ We have $4$ choices for each letter, and we must select $9$ letters. There are a total of $5^{9}$ strings (without any restrictions). Therefore, the number of valid strings is $5^{9} - 4^{9}.$
Our final answer is $\boxed{2 \cdot \left(5^{9} - 4^{9}\right)}.$
|
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Huber loss vs l1 loss From a robust statistics perspective are there any advantages of the Huber loss vs. L1 loss (apart from differentiability at the origin) ? Specifically, if I don't care about gradients (for e.g. when using tree based methods), does Huber loss offer any other advantages vis-a-vis robustness ?
Moreover, are there any guidelines for choosing the value of the change point between the linear and quadratic pieces of the Huber loss ?
Thanks.
|
The Huber function is less sensitive to small errors than the $\ell_1$ norm, but becomes linear in the error for large errors. To visualize this, notice that function $| \cdot |$ accentuates (i.e. becomes sensitive to) points near to the origin as compared to Huber (which would in fact be quadratic in this region). Therefore the Huber loss is preferred to the $\ell_1$ in certain cases for which there are both large outliers as well as small (ideally Gaussian) perturbations.
The point of interpolation between the linear and quadratic pieces will be a function of how often outliers or large shocks occur in your data (eg. "outliers constitute 1% of the data"). It's common in practice to use a robust measure of standard deviation to decide on this cutoff.
Huber's monograph, Robust Statistics, discusses the theoretical properties of his estimator. For more practical matters (implementation and rules of thumb), check out Faraway's very accessible text, Linear Models with R.
|
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Finding dual basis to some given basis Can someone briefly explain me how to find dual basis to some given basis?
Let's say we've got $n$-dimensional linear space with given basis $(e_1,e_2,...,e_n)$. How to find $(e_1^*, e_2^*, ...., e_n^*)$?
|
Let $V$ be a vector space over a field $k$ with basis $e_1, ... e_n$. The dual $V^{\ast}$ is defined to be the vector space consisting of all linear transformations from $V$ to $k$. The definitions of addition and scalar multiplication in $V^{\ast}$ are obvious, as is the fact that $V^{\ast}$ is actually a vector space.
Every element in $V$ can be uniquely written as a sum $c_1e_1 + \cdots + c_ne_n$, where $c_i$ are in $k$. For each $j$, define a function $e_j^{\ast}: V \rightarrow k$ by the formula $$e_j^{\ast}(c_1e_1 + \cdots + e_nv_n) = c_j$$ You can check that for any $x, y \in V$, and any $c \in k$, that $e_j^{\ast}(x+y) = e_j^{\ast}(x) + e_j^{\ast}(y)$, and $e_j^{\ast}(cx) = c e_j^{\ast}(x)$. In other words, $e_j^{\ast}$ is a linear transformation from $V$ to $k$, so $e_j^{\ast} \in V^{\ast}$.
The basic result is that the elements $e_i^{\ast}$ form a basis for the vector space $V^{\ast}$. Is this what you're having trouble proving?
|
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How to use xor properly I need to know how to use XOR properly on more than two variables.
I have following example.
a xor b xor c
Now, the way I understand it is that:
$a$ xor $b$ = $a$ * not $b$ + not $a$ * $b$
That part is fairly simple for me to understand but how does the next part work? I imagine it goes like this.
($a$ * not $b$ + not $a$ * $b$) xor $c$
So, how can I use XOR on more than one variable?
|
NOTE: Just so you know, a XOR b is the same as $a \oplus b$. Also, here's a shortcut to XOR: If $a$ and $b$ are different, then $a \oplus b=1$. Otherwise, if they're the same, $a \oplus b=0$. Your formula works as a definition for XOR, but when you're calculating XOR in your head, I think this rule of thumb is easier.
If you know how to calculate $a \oplus b$, then you can simply calculate $a \oplus b$, set that to $d$ and then calculate $d \oplus c$. Basically, you work with two elements at a time: First, you look at the first two elements and then you take that result and use it with the third element.
Here's an example: Calculate $1 \oplus 0 \oplus 1$. Since $1$ and $0$ are different, we know that $1 \oplus 0=1$. This means that we now need to figure out $1 \oplus 1$. The first $1$ comes from the result of our first $\oplus$ calculation and the second $1$ comes from the fact that $1$ was the third element in our original expression. Since $1$ and $1$ are the same, we know that $1 \oplus 1=0$. Thus, $1 \oplus 0 \oplus 1=0$.
|
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How to find the antiderivative of this function $\frac{1}{1+x^4}$? How to find the antiderivative of the function $f(x)=\frac{1}{1+x^4}$?
My math professor suggested to use the method of partial fractions, but it doesn't seem to work, because the denominator cannot be factored at all.
Attempting to integrate with other familiar methods (integration by parts, trigonometric substitution) didn't work out as well.
Does anyone have an idea what is the best approach to this problem?
|
It's easy to see factorization in the first line.
\begin{align*}
\int \frac{1}{1+x^{4}}\mathrm{d}x &=\int \left ( \frac{\dfrac{\sqrt{2}}{4}x+\dfrac{1}{2}}{x^{2}+\sqrt{2}x+1}+\frac{-\dfrac{\sqrt{2}}{4}x+\dfrac{1}{2}}{x^{2}-\sqrt{2}x+1} \right )\mathrm{d}x\\
&=\int \left [ \frac{\dfrac{\sqrt{2}}{4}\left ( x+\dfrac{\sqrt{2}}{2} \right )+\dfrac{1}{4}}{\left ( x+\dfrac{\sqrt{2}}{2} \right )^{2}+\dfrac{1}{2}}+\frac{-\dfrac{\sqrt{2}}{2}\left ( x-\dfrac{\sqrt{2}}{2} \right )+\dfrac{1}{4}}{\left ( x-\dfrac{\sqrt{2}}{2} \right )^{2}+\dfrac{1}{2}} \right ]\mathrm{d}x \\
&=\frac{\sqrt{2}}{8}\ln\left | \frac{x^{2}+\sqrt{2}x+1}{x^{2}-\sqrt{2}x+1} \right |+\frac{\sqrt{2}}{4}\arctan\left ( \sqrt{2}x+1 \right )+\frac{\sqrt{2}}{4}\arctan\left ( \sqrt{2}x-1 \right )+C.
\end{align*}
|
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prove there is a unique set $X$ such that every set $Y$, $Y∪X = Y$ For this proof. It seems obvious that $X=∅$ such that for every set $Y$, $Y∪X = Y$ since the $Y∪X$ is just Y. How should I go about this?
Let there be sets $X,Z$
Since $Y∪X = Y$ then,
{$x| x ∈ Y ∨ x ∈ X$} = {$x| x ∈ Y$}
Since $Y∪Z = Y$ then,
{$x| x ∈ Y ∨ x ∈ Z$} = {$x| x ∈ Y$}
Then equate the two
{$x| x ∈ Y ∨ x ∈ Z$} =
{$x| x ∈ Y ∨ x ∈ X$}
Therefore conlude Z = X and prove that X is a unique set.
|
You need one more step after your last equation: If $\{x :x\in Y\lor x\in Z\}=Y=\{x:x\in Y\lor x\in X\}$ holds for every $Y$, then it holds when $Y=X$: $$X=\{x:x\in X\lor x\in X\}=\{x: x\in X\lor x\in Z\}.$$ And it holds when $Y=Z$: $$Z=\{x:x\in Z\lor x\in Z\}=\{x:x\in Z\lor x\in X\}.$$ Comparing the RHS of these, we have $X=Z.$........ More succintly we can obtain $$X=X\cup Z$$ by applying $\forall Y\;(Y=Y\cup Z)$ to the case $Y=X,$ and we can obtain $$Z=X\cup Z$$ by applying $\forall Y\;(Y=X\cup Y)$ to the case $Y=Z.$ .... Thus $X=X\cup Z=Z.$
|
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Calculate the limit $\left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}\right)$ I have to calculate the following limit:
$$\lim_{n\to\infty} \left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}\right)$$
I believe the limit equals $1$, and I think I can prove it with the squeeze theorem, but I don't really know how.
Any help is appreciated, I'd like to receive some hints if possible.
Thanks!
|
Note that we can write the term of interest as
$$\sum_{k=1}^n\frac{1}{\sqrt{n^2+k}}=\frac1n\sum_{k=1}^n\frac{1}{\sqrt{1+k/n^2}}$$
To evaluate the limit we can expand the summand using the binomial theorem as
$$\frac{1}{\sqrt{1+k/n^2}}=1-\frac{k}{2n^2} +O(k^2/n^4)$$
Therefore, we have
$$\begin{align}
\lim_{n\to \infty}\sum_{k=1}^n\frac{1}{\sqrt{n^2+k}}&=\lim_{n\to \infty}\frac1n \sum_{k=1}^n \left(1+O(k/n^2)\right)\\\\
&=\lim_{n\to \infty}\frac1n\left(n+O(1)\right)\\\\
&=1
\end{align}$$
|
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Does a right circular cone only consists of pair of straight lines, hyperbolas, parabola, circles and ellipses? I was reading about the conic sections and that a conic section includes pair of straight lines, ellipses, hyperbola, circles and parabola.
Are all these 5 components enough to form a right circular cone or are these just the parts of right circular cone? What I mean is : if I put together infinite number of circles, parabolas, hyperbolas,, ellipses and circles, will I get a right circular cone?
|
A cone is a surface and as such cannot consist of a finite number of curves.
A conic section is the intersection of a cone with a plane, so it is a planar curve. Depending on the relative positions of the cone and the plane, the section can be a single or a double line, an ellipse, a parabola or an hyperbola (the circle is just a particular case of the ellipse).
A way to reconstruct a cone is to sweep a plane over the whole space (for instance by translation or rotation), a put together all the sections so obtained.
Putting together random lines, ellipses... in any number will just result in a mess.
|
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|
Please explain, "Asymmetric is stronger than simply not symmetric". In some textbook I found a statement like, "Asymmetric is stronger than simply not symmetric".
But as I try to perceive this statement, both appear to be same to me.
For example, parentof is an asymmetric relation. If $A$ is a parentof $B$, $B$ can not be parentof $A$. We arrive at the same conclusion if we call this relation to be not symmetric.
(Liyang Yu. A Developer’s Guide to the Semantic Web.
2nd. Springer, 2014. p 226, last sentence of second paragraph from bottom.)
|
It is just the standard negation of quantifiers: Asking that a relation is never true is stronger than the negation of the relation being always true.
A relation being symmetric means that all pairs can be inverted.
The negation of this is that some pairs cannot be inverted.
Asymmetric means that all pairs cannot be inverted.
|
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|
Solving a system of ODE that arose in solving Burgers' equation Consider the Burgers' equation $$\partial_t u = \alpha u\partial_xu$$ Intend to solve this using Fourier Galerkin method. So When I convert this into $N$th Fourier partial sum, I get a system of ODE's as follows $$\partial_t u_m = \sum_{k=-N}^{k=N}iku_ku_{m-k} $$
How do I solve this system of ODE numerically or analytically?, given we know the first $N$ Fourier coefficients of the initial condition $u(x,0)$.
|
Analytic solving of the PDE :
$$\partial_t u - \alpha u\partial_xu=0$$
Thanks to the method of characteristics :
$$\frac{dt}{1}=\frac{dx}{-\alpha u}=\frac{du}{0}$$
On the characteristic curves :
\begin{cases}
u=c_1\\
\alpha c_1 dt +dx=0\:\:\rightarrow\:\: \alpha u t +x=c_2\\
\end{cases}
The general solution can be expressed on the form :
$$\Phi\left(u\:,\: \alpha t u +x\right)=0$$
where $\Phi$ is any derivable function of two variables. This is equivalent of the implicit relationship :
$$u=F\left(\alpha \;t\; u +x\right)$$
where $F$ is any derivable function.
Nothing can be done further until the boundary conditions be defined, in order to find the convenient function $F$ and then $u(x,t)$ when it is possible to explicit $u$ from the implicit equation.
|
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Logic with numbers Nick wrote each of the numbers from 1 to 9 in the cells of the 3x3 table below. Only 4 of the numbers can be seen in the figure. Nick noticed that for the number 5, the sum of the numbers in the neighboring cells is equal to 13 (neighboring cells are cells that share a side). He noticed that the same is also true for the number 6. Which number did Nick write in the shaded cell?
|
The number in the middle, shaded square abuts four numbers whose sum is at least $5+6+7+8=26$, so cannot be either the $5$ or the $6$. Hence the $5$ and $6$ each go in a side square, abutting whatever's in the middle and two corner squares. Since the total abutting sums both equal $13$, the corner abutting sums for the $5$ and $6$ must be equal. This only happens for the corner sums $1+4$ and $2+3$ (implying the $5$ and $6$ occupy the sides squares on the left and right, in some order), which means the middle, shaded square contains the $8$.
|
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|
Chenge weight function in shifted orthogonality We know that in Chebyshev orthogonal polynomial the weight function is $$\frac{1}{\sqrt{1-x^2}}$$ in interval $[-1,1]$. Do in shifted chebyshev orthogonal as example for interval $[0,1]$ the weight function changed?
|
What you need is a scaling function $K(x)$ so that
$$
\left< T_n, T_m \right>
= \int_0^1 T_n(x) T_m(x) K(x) dx
= 0 \quad \text{for } n \ne m,
$$
which is the orthogonality criterion.
In the standard case, picking $$K(x) = \frac{1}{\sqrt{1-x^2}},$$ guarantees
$$
\left< T_n, T_m \right>
= \int_{-1}^1 T_n(x) T_m(x) K(x) dx
= 0 \quad \text{for } n \ne m
$$
|
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Trig Integration, how did they get this answer?
Here is the full question for context:
And here is the full answer given:
I don't understand how they go from the first part in the red box to the next, I don't have a clue.
|
It's a $u$ substitution where they take $u = \sin(x)$. If $u = \sin(x)$, then $\mathrm{d}u = \cos(x) \;\mathrm{d}x$. So we have
$$\begin{align}
\int {\sin^{2n}(x)\cos(x)} \;\mathrm{d}x \;\;&=\;\;
\int {u^{2n}} \;\mathrm{d}u \\\;\;&=\;\;
\frac{u^{2n+1}}{2n+1} \;\;=\;\;
\frac{\sin^{2n+1}(x)}{2n+1}
\end{align}$$
|
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Show that $\psi:\mathbb{R}^3\rightarrow\mathbb{R}$ given by $\psi(x,y,z) = z$ is linear Show that $\psi:\mathbb{R}^3\rightarrow\mathbb{R}$ given by $\psi(x,y,z) = z$ is linear
I know that to be linear it must satisfy:
(i) $\psi(x+y+z)=\psi(x)+\psi(y)+\psi(z)$
(ii) $\psi(ax)=a\psi(x)$
So, this is how far I've gotten:
(i) I let $x=(x_1,x_2,x_3), y=(y_1,y_2,y_3),$ and $z=(z_1,z_2,z_3)$ to satisfy $\mathbb{R}^3$
and so $\psi(x+y+z) = ((x_1,x_2,x_3)+(y_1,y_2,y_3)+(z_1,z_2,z_3))$
which can be written as $\psi((x_1,y_1,z_1)+(x_2,y_2,z_2)+(x_3,y_3,z_3))$
and applying $\psi$ I got $(z_1+z_2+z_3)$
But this is where I got stuck, I'm not sure where to go from here. Did I approach this problem incorrectly?
(ii) I let $t=(x,y,z)$ so
$\psi(at)=\psi(a(x,y,z))=\psi(ax,ay,az)=a\psi(x,y,z)=a\psi(t)$
so $\psi(at)=a\psi(t)$
Is that a sufficient way to prove that?
|
You are the right way. But note that you can simplify what you have done. To show that it is linear you need to check
$$\psi((x_1,y_1,z_1)+(x_2,y_2,z_2))=\psi(x_1,y_1,z_1)+\psi(x_2,y_2,z_2)$$ and
$$\psi(a(x,y,z))=a\psi(x,y,z).$$ Now, it is
$$\psi((x_1,y_1,z_1)+(x_2,y_2,z_2))=\psi(x_1+x_2,y_1+y_2,z_1+z_2)=z_1+z_2=\psi(x_1,y_1,z_1)+\psi(x_2,y_2,z_2),$$ and
$$\psi(a(x,y,z))=\psi(ax,ay,az)=az=a\psi(x,y,z).$$
|
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|
# of partitions of $n$ into at most $r$ positive integers $=$ # of partitions of $n + r$ into exactly $r$ positive integers? How do I see that the number of partitions of the integer $n$ into at most $r$ positive integers is equal to the number of partitions of $n + r$ into exactly $r$ positive integers?
|
Consider a Ferrer's diagram of $n$ into at most $r$ parts.
$ooooooo\\
oooooo\\
ooo\\
o$
For example this is a partition of $17$ into $4$ parts - $17=7+6+3+1$.
So in this example we know $r\ge4$. This means we can insert a first column of size exactly $r$ to get a partition of $n+r$ with exactly $r$ parts.
We also need to establish a bijection, and we can note that, say $r=4$, then $17-4=13=(7-1)+(6-1)+(3-1)+(1-1)=6+5+2$, which can be said for all the partitions here.
Algebrically, the number of partitions of $n$ into at most $r$ part is given by the coefficient of $x^n$ in:
$$\prod_{k=1}^r\dfrac{1}{1-x^k}$$
Into exactly $r$ parts can be expressed as the coefficient of $x^{n+r}$ in:
$$x^r\prod_{k=1}^r\dfrac{1}{1-x^k}$$
|
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|
Primitive Root Mod P I have been able to answer all of the parts to the question apart from part (v).
Any tips on how?
I assume I have to show that $ m=2^n $, but I am unsure as to how.
I can't imagine it is very difficult/long.
Thanks :)
|
Theorem: If $a\in\mathbb Z$, $p$ is prime, $\gcd(a,p)=1$, and for all prime divisors $q$ of $p-1$ we have $a^{(p-1)/q}\not\equiv 1\pmod{p}$, then $a$ is a primitive root mod $p$.
First, if $p=2^{2^n}+1$ with $n\ge 2$, then $\gcd(5,p)=1$, because by Fermat's Little Theorem $$2^{2^n}\equiv \left(2^{2^{n-2}}\right)^4\equiv 1\not\equiv -1\pmod{5}$$
The only prime divisor of $p-1=2^{2^n}$ is $2$, therefore we're left with proving $5^{\frac{p-1}{2}}\not\equiv 1\pmod{p}$, which is true, because by Euler's Criterion and Quadratic Reciprocity: $$5^{\frac{p-1}{2}}\equiv \left(\frac{5}{p}\right)\equiv \left(\frac{p}{5}\right)\equiv \left(\frac{2}{5}\right)\equiv -1\pmod{p}$$
|
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If $E\backslash C$ is relatively open in $E$ then $C$ is relatively closed in $E$ Given that $E\subset \mathbb{R}^n$ and $C\subset E$, prove that if $E\backslash C$ is relatively open in $E$ then $C$ is relatively closed in $E$
The best I've so far been able to arrive at is the following:
We have that, by definition, there exists an open set $A\subset \mathbb{R}^n$, such that $E\backslash C = A\cap E$. Consider $(E\cap C^c)^c = E^c\cup C=E^c\cup A^c$. Now, $E\cap (E^c \cup C)=E\cap(E^c\cup A^c) \iff E\cap (E^c \cup C)=(E\cap E^c) \cup (E^c\cap A^c)\iff E\cap C = E\cap A^c \implies C\subset A^c \cap E.$
I'm not sure if I'm on the right track. Some hints would be appreciated.
|
Note that $E\cap C=C$.
So, following your deduction, if $E\setminus C=A\cap E$ for some open set $A$, then $C=A^c\cap E$ showing that $C$ is relatively closed in $E$.
|
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Find the value of : $\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x}$ I saw some resolutions here like $\sqrt{x+\sqrt{x+\sqrt{x}}}- \sqrt{x}$, but I couldn't get the point to find
$\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x}$.
I tried $\frac{1}{x}.(\sqrt{x+\sqrt{x+\sqrt{x}}})=\frac{\sqrt{x}}{x}\left(\sqrt{1+\frac{\sqrt{x+\sqrt{x}}}{x}} \right)=\frac{1}{\sqrt{x}}\left(\sqrt{1+\frac{\sqrt{x+\sqrt{x}}}{x}} \right)$ but now I got stuck. Could anyone help?
|
For large $x$ you have $\sqrt x<x$ and so
$$
\sqrt{x+\sqrt{x+\sqrt{x}}}<\sqrt{x+\sqrt{2x}}<\sqrt{3x}.
$$
Since $\sqrt{x+\sqrt{x+\sqrt{x}}}/x<\sqrt{3x}/x\to0$ when $x\to\infty$, your limit is zero.
|
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|
True of false: The sum of this infinite series. I'm fairly sure it is false, but I'm not quite sure about which test I should use to prove it.
$$\sum_{n=2}^\infty \ln\left(\frac{n-1}{n}\right) = -1 $$
I think using the integral test should work, but it may get kind of messy, so I'm looking for som advice. All I have to do is prove that it diverges, right?
Thanks in advance.
|
It is actually very easy. See that
$$f(N)=\sum_{n=2}^N\ln \left(\frac{n-1}{n}\right)=
\sum_{n=2}^N\ln \left(1-\frac{1}{n}\right)$$
Is decreasing.
Now $f(3)\simeq-1.09861>f(+\infty)$. So $f(+\infty)\neq -1$.
|
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Is $S^{\infty}$ contractible? Recently I was reading this post:
Unit sphere in $\mathbb{R}^\infty$ is contractible?
Then a doubt came across to me: why I can't consider the linear homotopy $H:I\times S^{\infty}\to S^{\infty}$ given by the restriction of
$$H_t= \dfrac{F_t}{|F_t|}$$
to the sphere $S^{\infty},$ where $F:I\times \mathbb{R}^{\infty}\to \mathbb{R}^{\infty}$ given by $F_t(x)=(1-t)(x_1, x_2, \ldots)+t(1,0,0,\ldots)$??
What is the need of the two steps? That is first get a homotopy between $Id$
and the shift $\sigma$ and then the homotopy between $\sigma$ and the constant map equals $(1,0,0,0, \ldots)$ ?.
|
The current answer already explains why the proposed homotopy cannot work. Let me give a geometric interpretation of the two-step homotopy on the linked answer.
Trying to contract $S^\infty$ to $(1, 0, 0,\cdots)$ directly using straightline homotopy cannot possibly work: The situation is the same as that of trying to contract $S^2$ in $\Bbb R^2$. Straightline homotopy at some point of time will run through the origin, in which case normalizing gives you undefined things.
So the point of the shift map $\sigma : S^\infty \to S^\infty$, $(x_0, x_1, x_2, \cdots) \mapsto (0, x_1, x_2, \cdots)$ is to pull $S^\infty$ up one dimension. Now you can contract the image of $\sigma$ to $(1, 0, 0, \cdots)$, because it lives in codimension one and $(1, 0, 0, \cdots)$ is just some other point outside it. The situation is the same as that of contracting $S^1 \subset \Bbb R^2$ inside $\Bbb R^3$ to a point outside the hyperplane it lives. This can easily be done using straightline homotopy.
Irrelevant to the question, but here's a different way to do it. $S^\infty$ is the same a the colimit $\bigcup_n S^n$ with $S^i \subset S^{i-1}$ being inclusion as an equator. Note that each $S^n$ bounds a disk (i.e., hemisphere) on each side in $S^\infty$. Consider the homotopy which contracts $S^n$ through those. To make this work, one needs a $[1/2^{n+1}, 1/2^n]$ trick so that the composition is continuous.
|
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|
Prove that if a set is nowhere dense iff the complement of the closure of the set is dense. I am able to prove the iff in the forward direction. But, I am having trouble proving the statement in other direction. I am trying to use the definition of dense, but I am not getting anywhere with it.
|
Assume that complement of (clA) is dense.
Then cl[complement of (clA)] = X.
So complement of (int clA) = X.
Thus int clA is empty.
Hence A is nowhere dense.
|
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Parallel Translation is Path Independent iff Manifold is Flat Problem. Let $M$ be a smooth Riemannian manifold and $\nabla$ be the Levi-Civita connection. Then the following are equivalent
*
*$R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z\equiv 0$
*For all $p,q\in M$, parallel translation along a curve segment from $p$ to $q$ is independent of the curve.
My attempt. I've tried for several hours now, but I'm out of ideas. Initially, I tried analyzing the differential equations which characterize a vector field along a curve being parallel, but I couldn't get anything out of it.
For the implication (2)$\implies$(1), it would be enough to show $R(\partial_i,\partial_j)\partial_k\equiv0$. Considering curves through some $p\in M$ which have the form $t\mapsto te_i$ in some chart around $p$ seemed promising but didn't get me far.
In particular I'm also not sure how the fact that $\nabla$ is the Levi-Civita conncetion comes into play. I know that compatibility of $\nabla$ with the metric is equivalent to the parallel translation being an isometry. Can we use that?
I would really appreciate some help.
|
For the direction: $2) \Rightarrow 1)$ use this formula, which connects the curvature tensor to parallel transport along small closed loops.
$1) \Rightarrow 2):$ (which as pointed by Anthony Carapetis hols only locally):
Any flat mnaifold is locally isometric to Euclidean space, so you can transfer the parallel transport to $(\mathbb{R}^n,e)$, do it there, and then project back to your manifold. Since parallelt transport in $(\mathbb{R}^n,e)$ is trivially path independent, we finished.
|
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Is this a mistake on my part or theirs? I'm not sure if I'm the one making the mistake, or my math book. It looks like the negative sign completely disappeared.
$$\frac{3x^2}{-\sqrt{18}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{3x^2\sqrt{2}}{-\sqrt{36}} = \frac{3x^2\sqrt{2}}{6} = \frac{x^2\sqrt{2}}{2}$$
Here is the original image.
Also, I am new to Stackexchange, so tell me if I am doing something wrong.
|
Be cautious about the negative sign,
$$\frac{3x^{2}}{-\sqrt{18}} \times \frac{\sqrt{2}}{\sqrt{2}}=- \frac{3x^{2}}{\sqrt{18}} \times \frac{\sqrt{2}}{\sqrt{2}} =- \frac{3x^{2}}{3\sqrt{2}} = -\frac{x^{2}}{\sqrt{2}}$$
Maybe, there is a typo in the textbook.
|
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If $x^2+3x+5=0$ and $ax^2+bx+c=0$ have a common root and $a,b,c\in \mathbb{N}$, find the minimum value of $a+b+c$ If $x^2+3x+5=0$ and $ax^2+bx+c=0$ have a common root and $a,b,c\in \mathbb{N}$, find the minimum value of $a+b+c$
Using the condition for common root, $$(3c-5b)(b-3a)=(c-5a)^2$$
$$3bc-9ac-5b^2+15ab=c^2+25a^2-10ac$$
$$25a^2+5b^2+c^2=15ab+3bc+ac$$
There is pattern in the equation. The coefficients of the middle terms and first terms of both sides are $(5\times 5,5)$ and $(5\times 3,3)$.
I tried to use Lagrange multipliers. But isnt there any simpler way to minimize $a+b+c$?
|
The polynomial $x^2+3x+5$ is irreducible over $\mathbb{Q}$. Hence, if $x^2+3x+5$ and $ax^2+bx+c$, with $a,b,c\in\mathbb{Q}$, have a common root, they must be proportional. That is, $$ax^2+bx+c=a\left(x^2+3x+5\right)\,.$$
The problem would be more challenging if $x^2+3x+5$ is replaced by $x^2+3x+2$.
|
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Show that the set of Hermitian matrices forms a real vector space How can I "show that the Hermitian Matrices form a real Vector Space"? I'm assuming this means the set of all Hermitian matrices.
I understand how a hermitian matrix containing complex numbers can be closed under scalar multiplication by multiplying it by $i$, but how can it be closed under addition? Wouldn't adding two complex matrices just produce another complex matrix in most instances?
Also, how can I find a basis for this space?
|
Hint:
i give you the intuition for $2\times 2$ matrices that you can extend to the general case.
An Hermitian $2\times 2$ matrix has the form:
$$
\begin{bmatrix}
a&b\\
\bar b&c
\end{bmatrix}
$$
with: $a,c \in \mathbb{R}$ and $b\in \mathbb{C}$ ($\bar b $ is the complex conjugate of $b$).
Now you can see that, for two matrices of this form, we have:
$$
\begin{bmatrix}
a&b\\
\bar b&c
\end{bmatrix}
+
\begin{bmatrix}
x&y\\
\bar y&z
\end{bmatrix}=
\begin{bmatrix}
a+x&b+y\\
\bar{b}+\bar{y}&c+z
\end{bmatrix}
$$
and, since $\bar{b}+\bar{y}=\overline{b+y}$ the result is a matrix of the same form, i.e. an Hermitian matrix.
For the product we have:
$$k
\begin{bmatrix}
a&b\\
\bar b&c
\end{bmatrix}=
\begin{bmatrix}
ka&kb\\
k\bar b&kc
\end{bmatrix}
$$
so the result matrix is Hermitian only if $k$ is a real number. Finally we see that the null matrix is hermitian and the opposite of an Hermitian matrix is Hermitian, so the set of hermitian matrix is real vector space.
For the basis:
Note that an hermitian matrix can be expressed as a linear combination with real coefficients in the form:
$$
\begin{bmatrix}
a&b\\
\bar b&c
\end{bmatrix}=
a\begin{bmatrix}
1&0\\
0&0
\end{bmatrix}+
\mbox{Re}(b)
\begin{bmatrix}
0&1\\
1&0
\end{bmatrix}
+\mbox{Im}(b)
\begin{bmatrix}
0&i\\
-i&0
\end{bmatrix}
+c
\begin{bmatrix}
0&0\\
0&1
\end{bmatrix}
$$
and ,since the four matrices at the right are linearly independent, these form a basis.
|
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"timestamp": "2023-03-29T00:00:00",
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|
A triple of pythagorean triples with an extra property I'm trying to prove the non-existance of three positive integers $x,y,z$ with $x\geq z$ such that\begin{align}
(x-z)^2+y^2 &\text{ is a perfect square,}\\
x^2+y^2 &\text{ is a perfect square,}\\
(x+z)^2+y^2 &\text{ is a perfect square.}
\end{align}
I failed tying to find such triple numerically. I tried to look at the differences, but that didn't give me much. Also, I tried the general solution of $a^2+b^2=c^2$, that is, $a=2pq$, $b=q^2-p^2$, and $c=q^2+p^2$ for some $p<q$, but this got ugly quite fast. I am not sure whether or not such $x,y,z$ exist. If it exists, one counterexample suffices to disprove my conjecture. Thanks in advance!
|
It is impossible that non-existance because there are infinitely many counterexamples. In fact, we must have by the Pythagorean triples
$$x-z=t^2-s^2;\space y=2ts\qquad (*)$$ $$x=t_1^2-s_1^2; \space y=2t_1s_1$$ $$x+z=t_2^2-s_2^2;\space \space y=2t_2s_2\qquad (**)$$ so we have from $(*)$ and $(**)$ $$x=\frac{t^2-s^2+t_2^2-s_2^2}{2}=t_1^2-s_1^2\iff t^2+t_2^2-2t_1^2=s^2+s_2^2-2s_1^2$$ On the other hand the identity
$$(m^2-n^2)^2+(2mn)^2=2(m^2+n^2)^2$$ shows infinitely many solutions to the equation $X^2+Y^2=2Z^2$.
Thus we can get an infinite set of possible $t_1,t_2,t_3$
|
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|
Calculate (or estimate) $S(x)=\sum_{k=1}^\infty \frac{\zeta(kx)}{k!}$. Let $x\in\mathbb R$, $x>1$ and
$$S(x)=\sum_{k=1}^\infty \frac{\zeta(kx)}{k!}$$
where $\zeta(x)$ is the Riemann zeta function. Calculate (or estimate) $S(x)$.
|
What one could do fairly easy (pure heuristics) is to obtain an expansion for large $x$.
The Riemann Zeta function can be approximated in this limit by $\zeta(z)\sim 1+\frac{1}{2^z}$
Therefore our sum reads
$$
S(x)=\sum_{k=1}^{\infty}\frac{\zeta(kx)}{k!}\sim_{x\rightarrow\infty}\sum_{k=1}^{\infty}\frac{1}{k!}+\frac{1}{k! 2^{kx}}=e-2+e^{\frac{1}{2^x}}\sim_{x\rightarrow\infty}e-1+\frac{1}{2x}
$$
which fits extremly well even for values as small as $x\approx 5$.
Because the Riemann Zeta function is monotonically decreasing for $z>1$ we might bound the series in question by simply estimating $\zeta(xk)<\zeta(x)$ to get
$$
(e-1)<S(x)<\zeta(x)(e-1)
$$
which by the sandwich lemma yields the same big $x$ limit as above
I would be really surprised if an closed form for this series exists, but who knows...:)
Appendix:
By a very similar reasoning it is possible to get an approximation for $x\rightarrow 1_+$. It is well known, that in the vicinity of $z=1$ the Riemann Zeta function posseses an Laurent expansion of the form
$$
\zeta(z)\sim_{z\rightarrow 1_+}=\frac{1}{z-1}+\gamma
$$
where $\gamma$ is the Euler-Marschoni constant. Therefore $S(x)$ is cleary dominated by the first term of the sum
$$
S(x)\sim_{x\rightarrow 1_+}\frac{1}{x-1}+\gamma+R(x)
$$
We might observe that the terms of the remainder $R(x)$ are given asymptotically by $ \zeta(k)/(k!)$ which is like $\gamma$ of $\mathcal{O}(1)$ . Therefore
$$
S(x)\sim_{x\rightarrow 1_+}\frac{1}{x-1}+\gamma+C
$$
turns out to be a very good approximation in this limit ($C=\sum_2^{\infty}\zeta(k)/k!)$. Even if $x=1.5$ we are only of by something like $20\text{%}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Homogeneous coordinate rings of product of two projective varieties In Ex 3.15, chapter I of Hartshorne's "Algebraic Geometry", we have shown that $ A(X \times Y) \cong A(X) \otimes A(Y)$, when X and Y are affine varieties. Is the same statement true for projective varieties, i.e. if X and Y are projective varieties then is $ S(X \times Y) \cong S(X) \otimes S(Y)$. Thank you.
|
Let $X \subset \mathbb{P}^n, Y \subset \mathbb{P}^m$ be projective varieties with homogeneous coordinate rings $S(X)$ and $S(Y)$ respectively. If we take the equations that define $X$, we see that they are homogeneous polynomials in $n+1$ variables. Naturally, they define an affine variety of $\mathbb{A}^{n+1}$, which is called the affine cone of $X$ and is denoted by $C(X)$. Now, the product $C(X) \times C(Y)$ is an affine variety of $\mathbb{A}^{n+1} \times \mathbb{A}^{m+1} \cong \mathbb{A}^{n+m+2}$ and is given by homogeneous equations. The affine coordinate ring of $C(X) \times C(Y)$ is precisely $S(X) \otimes S(Y)$. Since the equations that define $C(X) \times C(Y)$ are homogeneous, $C(X) \times C(Y)$ has the structure of a projective variety and so we can view it as a projective variety of $\mathbb{P}^{n+m+1}$ with homogeneous coordinate ring $S(X) \otimes S(Y)$. This shows that $S(X) \otimes S(Y)$ is the homogeneous coordinate ring of the projective variety that corresponds to the product of the affine cones of $X$ and $Y$.
To get a sense of what $S(X \times Y)$ is, one needs first to realize that $X \times Y$ is a projective variety of $\mathbb{P}^{(n+1)(m+1)-1}$ by means of the Segre embedding. Let us do an example using the excellent hint by Mariano. Let us take $X=Y= \mathbb{P}^1$. Then $\mathbb{P}^1 \times \mathbb{P}^1$ is a quadratic hypersurface of $\mathbb{P}^3$ given by the equation $z_{00}z_{11}-z_{10}z_{01}=0$ and so its homogeneous coordinate ring is $S(\mathbb{P}^1 \times \mathbb{P}^1) = \frac{k[z_{00},z_{01},z_{10},z_{11}]}{(z_{00}z_{11}-z_{10}z_{01})}$. On the other hand, $S(X)=k[x_0,x_1], S(Y)=k[y_0,y_1]$ and so
$S(X) \otimes S(Y) = k[x_0,x_1,y_0,y_1]$. This agrees with the fact that $C(X)\cong C(Y)=\mathbb{A}^2$.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
What is the nature of this improper integral? Consider this improper integral of first kind:
$$\int_0^{+\infty}{\frac{t\ln t}{{(t^2+1)}^{\alpha}}}\,{dt}, \quad \alpha\in\mathbb R$$
Its required to find the nature of this improper integral.
We know that the problem of this integral is only at $+\infty$ so we must study it in that neighborhood.
Ok so $t^2+1 \equiv t^2$ thus $(f(t)=\frac{t\ln t}{{(t^2+1)}^{\alpha}}) \equiv t^{1-2\alpha}\ln t$ and from here i tried to make the quotient test but didn't have a result, i had the problem even more complicated.(note:when using the quotient test we take the absolute value of $f$ because the functions must be positive)
How should I proceed from here?
|
If $a \le 1$, then your estimates give that the integrand is greater than
$$t^{1 - 2\alpha} \ge t^{-1}$$
and the integral is divergent.
If $a > 1$, then estimate the logarithm away with a very small power of $t$ - for example, adjusting the constants carefully yields for any $\beta > 0$ the estimate
$$\ln t < C t^{\beta}$$
where $C$ depends on $\beta$. Then
$$t^{1 - 2\alpha} \ln t < C t^{1 - 2\alpha + \beta}$$
Now if $\beta$ is sufficiently small, the overall exponent is less than $-1$, and we have convergence.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1631028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
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