Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
A $k+1$-sphere containing a $k$-sphere and a point. Earlier I asked a question on whether it is possible to find a sphere passing through a circle and a point non-coplanar to it. I wanted to know whether this was possible to do in higher dimensions.
In $\mathbb{R}^{k+2}$, given a $k$-sphere and a point outside the $k$-hyperplane containing it, can I find a $(k+1)$-sphere containing them?
While the geometric constructions described in the answers to the previous question seem to go through, I am not sure if they would hold in the higher dimensions or does the same intuition fail?
|
Use the algebric approach:
Equation of the hypersphere of dimension $k$:
$\{(x_1,...,x_{k+2}) | \sum_{i=1}^{k+1} x_i^2 = R^2,x_{k+2}=0 \}$
Point $M$ outside of the plan: $ 0 \not = a_{k+2}$
Can we find $(b_1,...,b_{n+2})$ the center and $R'$ the radius of a new sphere containing both?
By definition that sphere needs to have: $\sum_{i=1}^{k+2} (x_i-b_i)^2 = R'^2$ and $\sum_{i=1}^{k+2} (a_i-b_i)^2 = R'^2$.
The first equation is equivalent to $\sum_{i=1}^{k+2} x_i^2- 2x_ib_i + b_i^2 =R^2 - 2\sum_{i=1}^{k+2}x_ib_i + \sum_{i=1}^{k+2}b_i^2 = R^2 + b_{k+2}^2 = R'^2$ after setting all the relevant $b_i$ to 0 apart from the last one.
The second equation becomes $\sum_{i=1}^{k+2} (a_i-b_i)^2 = \sum_{i=1}^{k+2} a_i^2 - 2a_{k+2}b_{k+2} + R'^2-R^2 = R'^2 \Leftrightarrow b_{k+2} = \frac{1}{2a_{k+2}}(R^2 - \sum_{i=1}^{k+2} a_i^2)$
That gives you your solutions, but you need to be careful about the different conditions hidden here and there ($R' > R$, etc.)
You'll notice it's very similar to the case in $\mathbb{R^3}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/919086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Solve $\cos{z}+\sin{z}=2$ I am trying to solve the question:
$\cos{z}+\sin{z}=2$
Where $z \in \mathbb{C}$
I think I know how to solve $\cos{z}+\sin{z}=-1$:
$1+2\cos^2{\frac{z}{2}}-1+2\sin \frac{z}{2}\cos{\frac{z}{2}}=0\\
2\cos{\frac{z}{2}}(\cos{\frac{z}{2}}+\sin{\frac{z}{2}})=0$
etc... (that is, if the double angle identity holds true when the 'angle' is a complex number - I might be wrong about this)
My other methods involve:
*
*trying to substitute $\cos{z}=\frac{e^{iz}+e^{-iz}}{2}$ and $\sin{z}=\frac{e^{iz}-e^{-iz}}{2i}$. This seems to be the most obvious method, but I can't work out the next step after $$e^{iz}-e^{-iz}+(e^{iz}+e^{-iz})i=4i$$
*substituting $2=2(\sin^2{z}+\cos^2{z})$
*substituting $\sin{z}=\cos(\frac{\pi}{2}-z)$ (again, not really sure if this can be done)
|
Note that
$$\cos(\arctan(z)) + \sin(\arctan(z)) = \frac{1}{\sqrt{1+z^2}} + \frac{z}{\sqrt{1+z^2}}.$$
So, if you can solve the equation
$$\frac{1}{\sqrt{1+z^2}} + \frac{z}{\sqrt{1+z^2}} = 2,$$
(which is quadratic) then you can solve the original.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/919230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 3
}
|
General formula of repeated roots. Prove that $$\underbrace{\sqrt{k\sqrt{k\sqrt{k\sqrt{\cdots\sqrt{k}}}}}}_{n\text { times}}=k^{1-1/2^n}$$
How do I derive this formula?
|
Take the logarithm (as already suggested in comments):
$$ \frac{1}{2}\left(\ln k + \frac{1}{2}\left(\ln k + \frac{1}{2}\left( \ln k + \ldots \right) \right)\right)$$
$$ =\left(\frac{1}{2}+ \frac{1}{2^2} + \frac{1}{2^3} + \ldots + \frac{1}{2^n}\right)\ln k$$
$$ =\left(1 - \frac{1}{2^n}\right) \ln k.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/919313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
}
|
Proof by Induction - Algebra Problem (Steps included but not understood) I do not quite understand this proof, if anyone could explain the steps for me it would be greatly appreciated. It's probably something glaringly obvious I'm not seeing, thanks in advance.
Prove that for every integer $n \ge 0,$ the number $4^{2n+1}+3^{n+2} $ is a multiple of 13.
Proof. We use induction on n, starting with $n=0$
$P(0):4^{2(0)+1}+3^{0+2}=4+9=13=13\cdot1$
Assume $P(k):4^{2k+1}+3^{k+2}=13t$ for some integer $t$. We must prove
$P(k+1): 4^{2(k+1)+1}+3^{(k+1)+2}$ is a multiple of $13$.
We have
$4^{2(k+1)+1}+3^{(k+1)+2}=4^{(2k+1)+2}+3^{(k+2)+1}$
$=4^2(4^{2k+1})+4^2(3^{k+2}-3^{k+2})+3\cdot3^{k+2}$
$=4^2(4^{2k+1}+3^{k+2})+3^{k+2}(-4^2+3)$
$=16\cdot13t+3^{k+2}\cdot(-13)$ (by $P(k)$)
$=13(16t-3^{k+2})$, proven.
|
The glaringly obvious looks like the concept of proof by induction itself. To prove a proposition $P(n)$ is true for all $n\geq 0$, it suffices to prove that $P(k) \Rightarrow P(k+1)$ (which is done by assuming $P(k)$, then showing $P(k+1)$, after which you may discharge the assumption and write down $P(k) \Rightarrow P(k+1)$) and that $P(0)$ is true. Using $P(k) \Rightarrow P(k+1)$ we then have $P(0)$ is true therefore $P(1)$ is true. $P(1)$ is true therefore $P(2)$ is true and so on.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/919380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
}
|
Matrix inversion using decomposition Why do I need $LU$, $QR$ and other matrix decomposition tech to find the inverse of a matrix?
Please if possible provide a ref for the answer.
Thanks in advance
|
If you have the $A=LU$ decomposition, then you could get the the $A^{-1}$ inverse matrix of course with $A^{-1}=U^{-1}L^{-1}$. Computer algebra systems often use the method to calculate the inverse matrix. You can read about it at LU decomposotion wikipedia article. This paper also write about it. You can find details about it also in the following book.
Gene H. Golub – Charles F. Van Loan: Matrix Computations. Third
Edition. The Johns Hopkins University Press, Baltimore and London,
1996.
at page 121.
If you have the $A=QR$ decomposition, you get analogous $A^{-1}=R^{-1}Q^{-1}$. Because $Q$ is orthogonal, you get $A^{-1}=R^{-1}Q^{-1}=R^{-1}Q^{T}$. You can read more about it in the answer.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/919464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Limit using Stirling's formula Consider the limit
$$\lim_{n \to \infty} \left[n(1-2\log2) + \sum_{k=1}^{n} \log\left(1+\frac {k}{n}\right)\right] = \frac {1}{2} \log 2.$$
This can be shown by using Stirling's formula for $n!$. My question is if this is the only way or there is also an elementary solution.
|
This is the case $f:x\mapsto\log(1+x)$ of the more general result that, for every $C^1$ function $f$ on $[0,1]$,
$$\lim_{n \to \infty} \sum_{k=1}^{n} f\left(\frac {k}{n}\right)-n\int_0^1f(x)\mathrm dx = \frac {1}{2}(f(1)-f(0)).$$
To show this, one can use Riemann sums and a first-order Taylor expansion on $f$.
More in details, one starts from the identity $$f\left(\frac {k}{n}\right)=n\int_{(k-1)/n}^{k/n}f(x)\mathrm dx+R_{k,n},\qquad R_{k,n}=n\int_{(k-1)/n}^{k/n}\left(f\left(\frac {k}{n}\right)-f(x)\right)\mathrm dx.$$ The integrand in $R_{k,n}$ is $$f\left(\frac {k}{n}\right)-f(x)=f'\left(\frac{k}n\right)\cdot\left(\frac {k}{n}-x\right)+o\left(\frac1n\right),$$ hence $$R_{k,n}=nf'\left(\frac{k}n\right)\int_{(k-1)/n}^{k/n}\left(\frac {k}{n}-x\right)\mathrm dx+o\left(\frac1{n}\right)=f'\left(\frac{k}n\right)\frac1{2n}+o\left(\frac1{n}\right).$$ Summing these from $k=1$ to $k=n$ yields $$\sum_{k=1}^{n} f\left(\frac {k}{n}\right)=n\int_0^1f(x)\mathrm dx +\frac12\frac1n\sum_{k=1}^{n}f'\left(\frac{k}n\right)+o(1),$$ and it remains to identify the sum on the RHS as a Riemann sum of $f'$ to deduce that $$\frac1n\sum_{k=1}^{n}f'\left(\frac{k}n\right)=\int_0^1f'(x)\mathrm dx+o(1)=f(1)-f(0)+o(1),$$ hence the identity above.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/919545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Upper bound on $\sum_{n=1}^N \frac{1}{n}$ are there any analytical upper bounds on
\begin{align}
\sum_{n=1}^N \frac{1}{n}
\end{align}
Clearly, one upper bound is $N$. But there has to be a better one. There is an approach with Harmonic number but it's not "analytic".
Thank you
|
$\sum_{n=1}^N {1 \over n} \le 1+ \ln N$.
This follows from $\int_{n-1}^n {1 \over t} dt = \ln n - \ln (n-1) > {1 \over n}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/919693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Why does the definition of the functional limit involve a limit point? This may be an odd question, and I'm not sure if these type of questions are at all appreciated in the maths community. But given the definition of the functional limit:
Let $f: A \to \mathbb{R}$, and let $c$ be a limit point of the domain $A$. We say that $\lim_{x \to c} f(x)=L$ provided that, for all $\epsilon > 0$, there exists a $\delta > 0$ such that whenever $0<|x-c|< \delta$ (and $x \in A$) it follows that $|f(x)-L|< \epsilon$.
why does $c$ have to be a limit point? I understand that the above is just a definition, and we can define anything we want to. But, after having gone through an entire chapter of functional limits and continuity, I have not yet figured out what would go wrong if $c$ was not a limit point.
|
To see where things go wrong take $A=\mathbb{N}$ and some function
$f:A\to\mathbb{R}$ and say you want to prove that
$$
\lim_{x\to c}f(x)=L
$$
where $c\in A$ is a natural number.
Let $\epsilon>0$ and choose $\delta(\epsilon)=\delta=\frac{1}{2}$
Indeed - For every $x\in A$ s.t $0<|x-c|<\frac{1}{2}$ we have it
that $|f(x)-L|<\epsilon$
This is true vacuously because there are no such $x\in A$ that satisfy
$0<|x-c|<\frac{1}{2}$ so we get
$$
\lim_{x\to c}f(x)=L
$$
regardless of what $L$ is, which is unreasonable
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/919807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 0
}
|
A $(3k+3)$-node graph with degrees $k+1,\dots,k+3$ has $k+3$ degree-$(k+1)$ nodes or $k+1$ degree-$(k+2)$ nodes or $k+2$ degree-$(k+3)$ nodes?
Graph $G$ has order $n=3k+3$ for some positive integer $k$ . Every vertex of $G$ has degree $k+1$, $k+2$, or $k+3$. Prove that $G$ has at least $k+3$ vertices of degree $k+1$ or at least $k+1$ vertices of degree $k+2$ or at least $k+2$ vertices of degree $k+3$
I used the proof by cases.
In case 1: $k$ is odd I can show that $G$ has at least $k+1$ vertices of degree $k+2$. Otherwise it will contradict the theorem that say every graph have even number of odd vertices.
Same reason, in case 2: $k$ is odd I can show that at least $k+2$ vertices of degree $k+3$
I just can't figure out how to show $G$ has at least $k+3$ vertices of degree $k+1$
|
Since $G$ has only vertices of degrees $k+1$, $k+2$ and $k+3$ you can split all of the vertices into 3 sets: $X$, $Y$ and $Z$ where $X$ is set of vertices with degree $k+1$, $Y$ with $k+2$ and $Z$ with $k+3$.
Now, since you already know the answers for cases:
$1)$ $|X|<k+3$ and $|Y|<k+1 \Rightarrow |Z|\geq k+2$
$2)$ $|X|<k+3$ and $|Z|<k+2 \Rightarrow |Y|\geq k+1$
we are interested in case 3, which is
$3)$ $|Y|<k+1$ and $|Z|<k+2 \Rightarrow |X|\geq k+3$
Proof for that case:
Suppose $k$ is odd. Then $|Y|$ can't equal $k$ since it is odd number and if it was equal $k$ it would contradict to the theorem that states that there is even number of vertices of odd degree so since $|Y|\leq k-1$ and $|Z| \leq k+1$, $|X| \geq k+3$.
Now suppose $k$ is even. Then $|Z|$ can't equal $k+1$ for the same reason as above (that means contradiction to the theorem that states that there is even number of vertices of odd degree), so $|Z|\leq k$ and then, since $|Y|\leq k$, $|X|$ must be equal or greater $k+3$.
For the cases
$4)$ $|X|<k+3$
$5)$ $|Y|<k+1$
$6)$ $|Z|<k+2$
the reasoning is very similar. Look at case $4)$.
If $|X|\leq k+2$ then we have at least $2k+1$ vertices for $Y$ and $Z$.
When $k$ is odd, $|Y|$ must be even and thus $|Y|\neq k$. If $|Y|<k$ then $|Z|\geq k+2$. If $|Y|>k$ then it is done, since $|Y|\geq k+1$.
When $k$ is even, $|Z|$ must be even. If $|Z|=k+2$ it is done. If not, then $|Z|\leq k$ or $|Z|\geq k+4$. In the second case it is done. If $|Z|\leq k$ then $|Y|\geq k+1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/919894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Evaluating Summation of $5^{-n}$ from $n=4$ to infinity The answer is $\frac1{500}$ but I don't understand why that is so.
I am given the fact that the summation of $x^{n}$ from $n=0$ to infinity is $\frac1{1-x}$. So if that's the case then I have that $x=\frac15$ and plugging in the values I have $\frac1{1-(\frac15)}= \frac54$.
|
The problem you have is that you do not know why the formula works to begin with. If you did the situation would be clear. Here's the thing:
$$\sum_{n=0}^{\infty}r^n=\frac{1}{1-r} \; \;\;\;\;\; |r|<1.$$
Let $r=\frac{1}{5}$, then you really have the following situation:
$$\left(\frac{1}{5}\right)^0+\left(\frac{1}{5}\right)^1+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+\left(\frac{1}{5}\right)^4+\left(\frac{1}{5}\right)^5+....$$
Let's call this infinite sum $S$ and proceed as follows,
$$S=\left(\frac{1}{5}\right)^0+\left(\frac{1}{5}\right)^1+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+\left(\frac{1}{5}\right)^4+\left(\frac{1}{5}\right)^5+....$$
then
$$ \left(\frac{1}{5}\right)S=\left(\frac{1}{5}\right)^1+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+\left(\frac{1}{5}\right)^4+\left(\frac{1}{5}\right)^5+\left(\frac{1}{5}\right)^6....\;\;\;\;$$
Subtract the second from the first,
$$S-\left(\frac{1}{5}\right)S=1$$
$$S(1-\left(\frac{1}{5}\right))=1$$
$$S=\frac{1}{1-\left(\frac{1}{5}\right)}$$
$$S=\frac{5}{4}.$$
Now recall what $S$ was and realize,
$$S=\left(\frac{1}{5}\right)^0+\left(\frac{1}{5}\right)^1+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+\left(\frac{1}{5}\right)^4+\left(\frac{1}{5}\right)^5+....=\frac{5}{4}$$
But, you want the powers to start at $n=4$, so subtract the first 4 powers(0,1,2,3) to get,
$$S-\left(1+\frac{1}{5}+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3\right)$$
which means
$$\left(\frac{1}{5}\right)^4+\left(\frac{1}{5}\right)^5+....=\frac{5}{4}-\left(1+\frac{1}{5}+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3\right)=\frac{1}{500}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/920050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
Evaluating $\lim_{x \to 0}\frac{(1+x)^{1/x} - e}{x}$ How to evaluate the following limit? $$\lim_{x \to 0}\frac{(1+x)^{1/x} - e}{x}.$$
|
You may write
$$
\begin{align}
\frac{(1+x)^{1/x} - e}{x} &= \frac{\large e^{\large\frac{\log (1+x)}{x}} - e}{x}\\\\
&= \frac{e^{\large \frac{x-\frac{x^2}{2}+{\mathcal{O}}(x^3)}{x}} - e}{x}\\\\
&= \frac{ e^{1-\frac{x}{2}+{\mathcal{O}}(x^2)} - e}{x}\\\\
&= \frac{ e \:e^{-\frac{x}{2}+{\mathcal{O}}(x^2)} - e}{x}\\\\
&= \frac{ e \:\left(1-\frac{x}{2}+{\mathcal{O}}(x^2)\right)- e}{x}\\\\
&=-\frac{e}{2}+{\mathcal{O}}(x)
\end{align}
$$
giving
$$\lim_{x \to 0}\frac{(1+x)^{1/x} - e}{x}=-\frac e2$$
where we have used
$$
\begin{align} &\log(1+x) =x-\frac{x^2}{2}+\frac{x^3}{3}+ \cdots, \quad |x|<1, \\
&e^x =1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+ \cdots .
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/920132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Showing Surjectivitity of $f(x) = x^3$ I want to show that the function $f: \mathbb{R} \to \mathbb{R},\; f(x) = x^3$ is surjective.
First Question: If a function has an inverse, it is bijective yes?
Second Question: Is my process below, for showing surjectivity of $y = x^3$, incorrect because I am assuming that the function has an inverse? i.e., by assuming it has an inverse I am assuming it is bijective and using that to show it is surjective.
Surjective statement: For a function $f: A \to B$, the function is surjective if $\forall_{ y \in B}, \; \exists_{x \in A}\;|\;f(x) = y$
So then $\forall_{y \in \mathbb{R}}, \; \exists_{x \in \mathbb{R}}\; | \; f(x) = y $
If such an real $x$ exists, then we have $x^3 = y$ and $x = y^{\frac{1}{3}}$. $x$ is a real number (how do I show this?). It follows that $f(x) = (y^{\frac{1}{3}})^3 = y$
|
(1) Yes, by definition, the inverse of a map $f: X \to Y$ is a map $f^{-1}: Y \to X$ such that $f^{-1} \circ f = id_X$ and $f \circ f^{-1} = id_Y$, so in particular $f$ must be bijective.
Note that one commonly one will say that a function $f$ defined by a rule (but without specifying the codomain) has an inverse, but in the sense of the above definition this requires that we take the codomain of $f$ to be its image.
For example, we often say that $f(x) := e^x$ is invertible, but if we take the domain of $f$ to be $\mathbb{R}$, in order for $f$ to have an inverse, we must regard $f$ as a function $\mathbb{R} \to \mathbb{R}_+$, where the codomain here is the set of positive numbers. If we instead treat it as a map $\mathbb{R} \to \mathbb{R}$ then it is not invertible, despite that it has the same rule as an invertible function.
The fact that two functions with the same rule and same domain can differ in whether they are invertible is part of the reason why a function, strictly speaking, includes as part of its definition its codomain in addition to its rule and domain.
(2) If you already know what the function we denote $x^{1/3}$ is, then your proof is sufficient to show that $x\mapsto x^3$ is bijective. On the other hand, if you're simply using $x^{1/3}$ as a symbol for the inverse of $x\mapsto x^3$ without any more information, then your answer is circular.
Much thanks to @DanZimm, who pointed out a confusion in a previous answer.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/920231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
}
|
Series $\sum_{n=0}^{\infty}\frac{1}{2^n-1+e^x}$ properties We need to prove $$f(x)=\sum_{n=0}^{\infty}\frac{1}{2^n-1+e^x}<\infty\ \ \forall_{x\in\mathbb{R}}$$
and then find all continuity points and all points in which $f$ is differentiable + calculate $f'(0)$. Especially the latter part is of difficult to me. It can be shown
$$\sum_{n=0}^{\infty}\frac{1}{2^n-1+e^x} <C+ \sum_{n=0}^{\infty}\frac{1}{2^n-3} < \infty$$ since $$\sum_{n=0}^{\infty}\frac{1}{2^n}=2$$ and $$\lim_{n\rightarrow\infty}\frac{\frac{1}{2^n-3}}{\frac{1}{2^n}} = \lim_{n\rightarrow\infty}\frac{2^n}{2^n-3} = \lim_{n\rightarrow\infty}\frac{1}{1-\frac{3}{2^n}} = 1.$$
Thus, using Weierstrass theorem, we conclude the initial sum is uniformly convergent for all $x\in\mathbb{R}$ to $f$, continuous on $\mathbb{R}$ (as a limit of uniformly convergent sequence of continuous functions).
I hope it's more or less clear and correct. But how do I prove the second part of this exercise?
|
use ratio test,let $a_n=\frac{1}{2^n-1+e^x}$ and take,$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=0.5<1 $$so$\sum_{n=1}^{\infty}a_n$ converges
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/920288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
}
|
Find vertices pointing to common vertex In a directed graph I'm interested in finding pairs of vertices pointing towards a common vertex. More in detail, from an adjacency matrix I want to derive a matrix where a positive entry denotes that the vertices represented by the entry's row and column both point to a common vertex.
For example, a graph with the following adjacency matrix:
$$
\left[\begin{matrix}
0 & 1 & 0 & 1 \\
1 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
\end{matrix}\right],
$$
would result in the following matrix:
$$
\left[\begin{matrix}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
\end{matrix}\right].
$$
Since vertex 2 and 4 points to vertex 3 and vertex 1 and 3 points to vertex 4. (Or with the diagonal as ones if one does not need unique vertices in the pairs.)
I'm thinking that there must be a simple matrix algebra solution to this, similar to using powers of the adjacency matrix to find walks. My intuition tells me:
$$
\mathbf{A}\mathbf{A}^T
$$
would do the trick, where $\mathbf{A}$ is the adjacency matrix. (In this case with a single vertex allowed to form a pair). While every example I've tested works out well, I cannot prove it will work in general. So I'm wondering: will it?
|
Yes, you're correct (nice guess!)
I find it a bit easier to visualize the $(i,j)$-entry of $AA^T$ as the standard inner product of rows $i$ and $j$ of $A$. Since all entries of $A$ are non-negative, you will got a positive value if and only if there is a column $k$ so that both entries in coordinates $(i,k)$ and $(j,k)$ are positive, i.e., there is a $k$ so that $(i,k)$ and $(j,k)$ are directed edges in the graph.
So in your example, this occurs for $i=1$ and $j=3$ (at $k=4$), and also for $i=2$ and $j=4$ (at $k=3$), which of course is what you found.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/920365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Conditional expected value of the maximum of n normally distributed variables What is the conditional expected value of the maximum of $n$ normally distributed variables $x_i$, conditionally on exceeding some threshold $z$?
That is, for $x=\max(x_1,...,x_n)$, the probability that $x$ will exceed some threshold $z$ is $P[x>z]=1-F[z]^n$. For large $n$, the distribution of $x$ can be approximated by the Gumbel distribution, and the expected value of $x$ asymptotically equals
$$E[x]=\mu + \gamma \sigma (2 \log(n))^{1/2}$$
Does there exist any such basic approximation formula, when $x_i$ are normally distributed and the exact value of $z$ is known?
|
$$E(x\mid x\gt z)=z+\frac1{1-F(z)^n}\int_z^\infty(1-F(t)^n)\mathrm dt$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/920458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How prove $g(x)$ is odd function :$g(x)=-g(-x)$ QUestion:
let $f(x),g(x)$ is continuous on $R$,and such
$$f(x-y)=f(x)f(y)-g(x)g(y)$$
and $f(0)=1$
show that: for any $x\in R$, have $g(x)=-g(-x)$
my try: let $x=y=0$,then
$$f(0)-[f(0)]^2=g(0)g(0)\Longrightarrow g(0)=0$$
and let $x=0$, note $f(0)=1,g(0)=0$,so
$$f(-y)=f(y)$$
since
$$g(x)g(y)=f(x)f(y)-f(x-y)$$
so
$$g(-x)g(y)=f(-x)f(y)-f(-x-y)=f(x)f(y)-f(x+y)$$
since
$$f(x+y)=f(x)f(-y)-g(x)g(-y)\Longrightarrow g(x)g(-y)=f(x)f(y)-f(x+y)$$
so
$$g(-x)g(y)=g(x)g(-y)$$
But I can't have $g(x)=-g(-x)$
|
First note that by using $f(x)=f(-x)$
$$
f(0) = f(x-x) = f(x)^2- g(x)^2 \\
f(0) = f(-x-(-x))=f(x)^2-g(-x)^2 \\
g(x)^2 = g(-x)^2
$$
Thus let one define into two sets (note an element may be in both)
$$
\Gamma^+=\{x \in \mathbb{R} : g(x) = g(-x) \} \\
\Gamma^-=\{x \in \mathbb{R} : g(x) = -g(-x) \} \\
$$
Clearly $\mathbb{R}=\Gamma^+ \cup \Gamma^-$, such that $g$ is even for $x\in\Gamma^+$, and odd for $x\in\Gamma^-$.
Consider $x\in\Gamma^+$, one has
$$
f(x-y)=f(x)f(-y)-g(x)g(-y) \\
\therefore f(x-y)=f(x+y)\Rightarrow f(x-x)=f(x+x) \Rightarrow f(x)=1 \:\forall x\textrm{ as }f(0)=1 \\
\Rightarrow g(x)=0 \:\forall x \in\Gamma^+
$$
Thus $x\in\Gamma^+\Rightarrow g(x)=0 \Rightarrow g(x)=-g(-x)\Rightarrow x\in\Gamma^-$. Thus $\mathbb{R}=\Gamma^-$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/920522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
}
|
Inequality: $2(p^2+q^2+r^2)+2(pq+qr+rp)\ge pqr$ I need to determine the range of $p,q,r$ such that $2(p^2+q^2+r^2)+2(pq+qr+rp)\ge pqr$. I am not given any other information except that $p,q,r\in \mathbb{R}$. I haven't solved a problem like this before, so I'm really stuck. I'm thinking that it might be true for all real numbers $p,q,r$ but then I would need a proof but I couldn't find one.
Thanks!
EDIT: Sorry, I forgot to add:
I wrote it as $(p+q)^2+(q+r)^2+(r+p)^2\ge pqr$.
|
This doesn't hold true in general.
For example, let $p=q=r$, you will get $$12p^2\geq p^3.$$ You can see it is not true when $p>12$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/920623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Limit of this recursive sequence and convergence $$a_{n+1}=\sqrt{4a_n+3}$$ $a_1=5$
I can solve simpler but I get stuck here because I cant find an upper bound or roots of the quadratic equation $a_{n+1} -a_n= \frac{4a_n+3 - a_n^2}{\sqrt{4a_n +3}+a_n}...$ to find monotony.
I tried this generic aproach but have difficulties Convergence and limit of a recursive sequence
|
Prove it using induction. Whether the sequence is increasing or decreasing depends on the value of $a_1$. Observe that $$a_n \le a_{n+1} \implies 4a_n + 3 \le 4a_{n+1} + 3 \implies a_{n+1} \le a_{n+2}$$ and similarly $$a_n \ge a_{n+1} \implies 4a_n + 3 \ge 4a_{n+1} + 3 \implies a_{n+1} \ge a_{n+2}.$$ Thus if $a_1 \le a_2$ the full sequence is nondecreasing and if $a_1 \ge a_2$ the full sequence is nonincreasing.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/920707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Function that is absolutely continuous but not $C^1$
*
*I would like to know example of a function that is absolutely
continuous on a compact subset of real line but is not $C^1$.
*Does always $C^1$ imply absolute continuity on unbounded subsets of real line? I know that it is so on compact subsets.
|
Let $-\infty<a<b<\infty$ . A function $u:[a,b]\to\mathbb{R}$ will be absolutely continuous if and only if, $u'$ does exist almost everywherem $u'\in L^1(a,b)$ and $$u(x)=u(a)+\int_a^x u'(t)dt.$$
Therefore, take any function $v\in L^1(a,b)$ and let $$u(x)=\int_a^x v(t)dt.\tag{1}$$
You can verify that $u'=v$ a.e. and therefore, $u$ is absolutely continuous. Now you have plenty of examples for your first dot. For example, let $a<c<d<b$ and let $v=\chi_{[c,d]}$. Note that $v\in L^1(a,b)$ and therefore, if $u$ is defined as in $(1)$ then, $u$ is absolutely continuous.
For your second dot, you can say for sure that the function is absolutely continuous on compact sets, however, as pointed out in the comments, if the derivative is integrable then, you can conclude more. To understand this, try to get a better understanding of the above equivalence.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/920797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Show that lcm$(a,b)= ab$ if and only if gcd$(a,b)=1$ Not sure how to begin. If gcd$(a,b)=1$ what can I deduce from that?
|
Hint $ $ note that $n\mapsto ab/n\,$ bijects the common divisors of $\,a,b\,$ with the common multiples $\le ab.$ Being order-reversing it maps the greatest common divisor to the least common multiple, i.e. $\,\gcd(a,n)\mapsto ab/\gcd(a,b) = {\rm lcm}(a,b).\,$ Thus $\,\gcd(a,b)=1\iff {\rm lcm}(a,b) = ab$.
Remark $\ $ For more on the (involution) duality between gcd and lcm see here and here.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/920978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
}
|
Second derivative of Hypergeometric function I'm looking for the following second derivative
$$
\kappa_2 := \left . \frac{d^2}{d\lambda^2} \ln \left({_2F_1}\!\left(\tfrac{1}{2},\,- \lambda;\,1;\,\alpha\right)\right) \right \vert_{\lambda = 0} ,
$$
where $\alpha$ is a real parameter in $[0,1]$. As you may have guessed I'm trying to compute the variance of a certain probability distribution.
The first moment is simple and is given by
$$
m_1 = 2 \ln \frac{1 + \sqrt{1-\alpha}}{2}.
$$
The second moment can be shown to be equal to the following
$$
m_2 = \frac{2}{\pi} \int_{0}^{1} \frac{\left [ \ln(1-\alpha y^2) \right ]^2}{\sqrt{1-y^2}} dy
$$
Mathematica can evaluate the above integral in terms of various function and polylogarithms (in fact dilogarithms). However the resulting expression is not even manifestly real. The above integral expression for $m_2$ is so far the best I found to deal with but there are many others. In essence I am trying to find a "nice" expression for that integral ($m_2$).
I'd be happy if I'm given an expression which is manifestly real. I suspect that the combination $\kappa_2 = m_2 - (m_1)^2 $ (which is equal to the first equation) might look nicer and that some identity involving dilogarithms should be used.
Added
An alternative representation for $m_2$ (obtained using the series of the Hypergeometric) is the following
$$
m_{2}=2\sum_{n=2}^{\infty}\left(\begin{array}{c}
-1/2\\
n
\end{array}\right)\frac{H_{n-1}}{n}\left(-\alpha\right)^{n} \, ,
$$
where $H_n$ are the Harmonic numbers, i.e.
$$
H_n \, = \, \sum_{p=1}^n \frac{1}{p}
$$
So far this is the best I could get:
$$
\kappa_2 \,=\, -4 \log ^2\left(\sqrt{1-\alpha }+1\right)+4 \log
\left(4-4 \sqrt{1-\alpha }\right) \log
\left(\sqrt{1-\alpha }+1\right)+4 i \pi \log
\left(\frac{2}{\sqrt{1-\alpha }+1}\right)+4 \log
(2) \log \left(\frac{1}{\alpha }\right)+4
\text{Li}_2\left(\frac{2 \left(\sqrt{1-\alpha
}+1\right)}{\alpha }\right)-4
\text{Li}_2\left(\frac{-\alpha +2 \sqrt{1-\alpha
}+2}{\alpha }\right)
$$
I would like to avoid the explicitly imaginary (third) term which is compensated by the dilogarithm to produce a real result. Does anybody know an identity for dilogarithm that can be used here?
|
The definition of $\operatorname{Li}_2$ used in the last formula is
$$\operatorname{Li}_2(z) =
-\operatorname{Li}_2 \left( \frac 1 z \right) -
\frac {3 \ln^2(-z) + \pi^2} 6,
\quad |z| > 1.$$
Therefore
$$\kappa_2 =
4 \operatorname{Li}_2 \left( \frac \alpha {\omega - \alpha} \right) -
4 \operatorname{Li}_2 \left( \frac \alpha \omega \right) +
2\ln^2 \left( \frac \omega \alpha - 1 \right) -
2\ln^2 \left( \frac \omega \alpha \right) + \\
4 \ln \left( \frac \omega 2 \right)
\ln \left( \frac {4 (4 - \omega - \alpha)} \alpha \right) -
\ln(16) \ln(\alpha), \\
\omega = 2 + 2 \sqrt {1 - \alpha}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/921062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
}
|
Are these proofs logically equivalent? Here are two proofs, firstly:
x = 0.999...
10x = 9.999...
= 9 + 0.999...
= 9 + x
9x = 9
x = 1
And secondly:
x = 1 - 1 + 1 - 1 + 1 - 1 ...
= 1 - (1 - 1 + 1 - 1 + 1 ...
= 1 - x
2x = 1
x = 1/2
The fourth line in the first and the third line in the second use the same trick. Does the legitimacy of the first proof guarantee that of the second?
|
No. The first series is absolutely convergent, so its terms can be rearranged without changing the sum, while the second series is divergent.
The number $0.\overline{9}$ can be expressed in the form
\begin{align*}
0.\overline{9} & = \sum_{k = 1}^{\infty} \frac{9}{10^k}\\
& = \frac{9}{10} \sum_{k = 0}^{\infty} \frac{1}{10^k}
\end{align*}
which is a convergent geometric series, so
\begin{align*}
0.\overline{9} & = \frac{9}{10} \cdot \frac{1}{1 - \frac{1}{10}}\\
& = \frac{9}{10} \cdot \frac{1}{\dfrac{9}{10}}\\
& = \frac{9}{10} \cdot \frac{10}{9}\\
& = 1
\end{align*}
In fact, the series
$$0.\overline{9} = \sum_{k = 1}^{\infty} \frac{9}{10^k}$$
is absolutely convergent since the series
$$\sum_{k = 1}^{\infty} \bigg| \frac{9}{10^K} \bigg| = \sum_{k = 1}^{\infty} \frac{9}{10^k}$$
converges. Any rearrangement of an absolutely convergent series has the same sum.
The series
$$1 - 1 + 1 - 1 + 1 - 1 + \cdots = \sum_{k = 0}^{\infty} (-1)^k$$
diverges since the sequence of partial sums $\{1, 0, 1, 0, 1, 0, \ldots\}$
alternates between $1$ and $0$. Since the series never converges to a limit, it is meaningless to say that
$$x = 1 - 1 + 1 - 1 + 1 - 1 + \cdots$$
A series must be absolutely convergent to guarantee that rearranging the terms of the series does not change the sum.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/921172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
golden ratio from new formula? perhaps from theory of modular units? Please consider the following infinite product series which I found by pure happenstance:
$$\frac{1+\sqrt{5}}{2}= e^{\pi/6} \prod_{k=1}^\infty \frac{1+e^{-5(2k-1)\pi}}{1+e^{-(2k-1)\pi}}$$
My question: Would this formula fit within, or be obtainable by, the theory of modular units (by Kubert and Lang)? I am entirely unfamiliar with that level of math, but I was told that the theory produces algebraic values for special infinite series products at CM-points.
Of note: The infinite product series (which adjusts the logarithmic spiral at $30$ degrees) has as its repeating term: $(1 + e^{-k\pi})$, with $k$ representing the sequence of all odd integers excluding those which are divisible by five $(1,3,7,9,11,13,17,19,21,23,27,\ldots)$.
Further, the same repeating term with $k$ representing all odd integers adjusts the logarithmic spiral at $7.5$ degrees to the fourth root of $2$.
And further, the same repeating term with $k$ representing only the odd integers that are divisible by $5$, adjusts the logarithmic spiral at $37.5$ degrees to the value of the product of the golden ratio with the fourth root of $2$. In this case, the first term of the series is so small that it is easy to see the close relation (at $e^{5\pi/24}$) with a calculator.
P.S. There doesn't seem to be an official resource for formula from infinite series, so I do not know if this equation has been recognized before or not. I found a good source containing many series formulae at:
http://pi.physik.uni-bonn.de/~dieckman/InfProd/InfProd.html
And he was kind enough to place this formula on his list.
Thanks in advance for any comments that can explain the theory of modular units, or CM-points, in a simplified manner for me, or its relation to infinite product series such as this one.
|
It turns out that OP also posted this question on my blog page. And I provide the answer I posted there.
The formula can be expressed as $$\frac{1 + \sqrt{5}}{2} = e^{\pi/6}\prod_{k = 1}^{\infty}\frac{1 + e^{-5(2k - 1)\pi}}{1 + e^{-(2k - 1)\pi}} = \frac{2^{-1/4}e^{5\pi/24}}{2^{-1/4}e^{\pi/24}}\prod_{k = 1}^{\infty}\frac{1 + e^{-5(2k - 1)\pi}}{1 + e^{-(2k - 1)\pi}}$$ If we check the definition of Ramanujan's Class Invariant $$G_{n} = 2^{-1/4}e^{\pi\sqrt{n}/24}\prod_{k = 1}^{\infty}\left(1 + e^{-(2k - 1)\pi\sqrt{n}}\right)$$ then we find that the given formula says that $$\frac{1 + \sqrt{5}}{2} = \frac{G_{25}}{G_{1}}$$ I have already proved in my post that $G_{1} = 1, G_{25} = (1 + \sqrt{5})/2$. This establishes your formula.
I have given extensive details of the theory of Ramanujan's theta functions which is needed to understand the proofs in my blog posts. All of it is based on real-analysis / calculus and does not require the deep and complicated theory of Modular forms and Complex multiplication.
Update: Based on your comments below I think the formula $$e^{\pi/6} = 2^{3/8}\prod_{k = 1}^{\infty}(1 + e^{-2k\pi})$$ is wrong. Let us put $q = e^{-2\pi}$ then we have $$\begin{aligned}\prod_{k = 1}^{\infty}(1 + e^{-2k\pi}) &= \prod_{k = 1}^{\infty}(1 + q^{k})\\
&= \prod_{k = 1}^{\infty}\frac{1 - q^{2k}}{1 - q^{k}}\\
&= \prod_{k = 1}^{\infty}\frac{1}{1 - q^{2k - 1}}\\
&= \prod_{k = 1}^{\infty}\frac{1}{1 - e^{-2(2k - 1)\pi}}\\
&= \dfrac{1}{2^{1/4}e^{-\pi/12}g_{4}}\end{aligned}$$ We have the definition $$g_{n}= 2^{-1/4}e^{\pi\sqrt{n}/24}\prod_{k = 1}^{\infty}\left(1 - e^{-(2k - 1)\pi\sqrt{n}}\right)$$ so that $g_{4} = 1/g_{1} = (2\sqrt{2})^{1/12}$ which gives the product as $2^{-3/8}e^{\pi/12}$. Your formula should consist of $e^{\pi/12}$ and not $e^{\pi/6}$. Your second product $$e^{\pi/4} = \sqrt{2(1 + \sqrt{2})}\prod_{k = 1}^{\infty}\frac{1 + e^{-4k\pi}}{1 + e^{-2(2k - 1)\pi}}$$ (which is correct) can also be calculated in similar fashion and it is linked with Ramanujan class invariants discussed in my blog post. The idea is to put $q = e^{-\pi\sqrt{n}}$ for some suitable value of $n$ and link the products with the invariants $G_{n}$ or $g_{n}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/921293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
Does the analytic continuation of $f$ always exist? Let $f(z)$ be a holomorphic funtion on region $\Omega$. Then, does the analytic continuation of $f$ always exist? Note that $f$ is always the analytic continuation of itself, so I exclude this case.
If the answer is yes, then what is the analytic continuation of $\sum_{n=0}^\infty z^n$, defined on the unit disk?
|
The function
$$f(z)=\sum_{n=0}^\infty z^{n!}$$
has no analytic continuation beyond $|z|<1$. The circle $|z|=1$ is a natural boundary.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/921377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Bounding the size of all consecutive sums of i.i.d. random variables Let $X_1, X_2, \ldots$ by an i.i.d sequence of bounded (for simplicity), mean zero random variables. For any $a<b$, call $S_{a,b} = X_{a+1} + \cdots + X_b$.
I would like to show that for any $\epsilon >0$, there is a $\delta > 0$ such that for all $N \in \mathbb{N}$,
$$
\mathbb{P} \left\{ |S_{a,b}| < \delta \; (b-a)^{1/2+\epsilon} \text{ for all } 0 \leq a < b \leq N \right\} > 1 - \epsilon.
$$
Is this a standard fact? I haven't seen anything about asymptotics of all consecutive sums of random variables before. It might follow easily from standard facts, but I don't see the trick...
Edit: This event is the same as the event that the maximum of $|S_{a,b}|$ over all $a<b$ is bounded by $\delta (b-a)^{1/2+\epsilon}$. (Also edited because I forgot the absolute value around $S_{a,b}$.)
|
Ok, I figured this out. The Bernstein inequality states that if $|X_i| \leq M$,
$$\mathbf{P} \left (\sum_{i=1}^n X_i \geq t \right ) \leq \exp \left ( -\frac{\frac{1}{2} t^2}{\sum \mathbf{E} \left[X_j^2 \right ]+\frac{1}{3} Mt} \right) $$.
This gives exponential control of any window $|S_{a,b}|$ being greater than $\delta (b-a)^{1/2+\epsilon}$, and so now a union bound over all pairs $a<b \leq n$ can be made as small as you wish.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/921477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Quick way to find eigenvalues of anti-diagonal matrix If $A \in M_n(\mathbb{R})$ is an anti-diagonal $n \times n$ matrix, is there a quick way to find its eigenvalues in a way similar to finding the eigenvalues of a diagonal matrix? The standard way for finding the eigenvalues for any $n \times n$ is usually straightforward, but may sometimes lead to painstaking computational time. Just wondering if there was a quicker way in doing so for any anti-diagonal matrix without having to resort to the standard method of finding the determinant of $A - \lambda I$, where $I$ is the $n \times n$ identity matrix, and setting it equal to $0$ and solving for $\lambda$.
|
For ease of formatting and explanation, I'll be doing everything for the $5 \times 5$ example. However, the same trick works for any $n \times n$ antisymmetric matrix (though slightly differently for even $n$).
Suppose
$$
A =
\begin{pmatrix}0&0&0&0&a_{15}\\0&0&0&a_{24}&0\\0&0&a_{33}&0&0\\0&a_{42}&0&0&0\\a_{51}&0&0&0&0 \end{pmatrix}
$$
Here's a neat trick: we note that
$$
A^2 = \pmatrix{
a_{15}a_{51}&&&&\\
&a_{24}a_{42}&&&\\
&&(a_{33})^2&&\\
&&&a_{24}a_{42}&\\
&&&&a_{15}a_{51}\\
}
$$
So, the eigenvalues of $A^2$ are precisely $\{a_{15}a_{51}, a_{24}a_{42}, (a_{33})^2\}$.
Now, note that if $\lambda$ is an eigenvalue of $A$, then $\lambda^2$ must be an eigenvalue of $A^2$. This gives you six candidates for the eigenvalues of $A$.
In fact, with more thorough analysis, we can guarantee that the eigenvalues will be precisely $\lambda = \pm \sqrt{a_{i,(n+1-i)}a_{(n+1-i),i}}$ for $i = 1,\dots,\lfloor n/2\rfloor$ and, for odd $n$, $\lambda = a_{(n+1)/2,(n+1)/2}$.
Proof that this is the case: Let $e_1,\dots,e_n$ denote the standard basis vectors. Let $S_{ij}$ denote the span of the vectors $e_i$ and $e_j$.
Note that $A$ is invariant over $S_{i(n-i)}$ for $i = 1,\dots,\lfloor n/2\rfloor$. We may then consider the restriction $A_{i(n-i)}: S_{i(n-i)} \to S_{i(n-i)}$, which can be represented by the matrix
$$
\pmatrix{0 & a_{i(n-i)}\\a_{(n-i)i} & 0}
$$
It suffices to find the eigenvalues of this transformation.
For the case of an odd $n$, it is sufficient to note that $a_{(n+1)/2,(n+1)/2}$ lies on the diagonal with zeros in its row and column.
Another explanation: denote the matrix
$S = \pmatrix{e_1 & e_{n} & e_2 & e_{n-1} & \cdots}$
Noting that $S$ is orthogonal (i.e. $S^{-1} = S^{T}$), we find that
$$
SAS^{-1} =
\pmatrix{
0&a_{1,n}\\
a_{n,1}&0\\
&&0&a_{2,n-1}\\
&&a_{n-1,2}&0\\
&&&&\ddots
}
$$
This matrix is similar, and therefore has the same eigenvalues. However, it is also block diagonal.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/921579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 3,
"answer_id": 2
}
|
To find the determinant of a matrix Given $A_{n\times n}$=$(a_{ij}),$ n $\ge$ 3, where $a_{ij}$ = $b_{i}^{2}$-$b_{j}^2$ ,$i,j = 1,2,...,n$ for some distinct real numbers $b_{1},b_{2},...,b_{n}$. I have to find the determinant of A. I can see that A is a skew-symmetric matrix. So determinant of A is $0$ when n is odd. But how to find it when n is even? Is the condition given for A implies anything other than A is skew-symmetric?
|
Hint: The coefficients of the characteristic equation satisfied by $A$ are given by the equation: $c_{n} = - \frac{1}{n} \left(c_{n-1}\mathrm{Trace}(A) + c_{n-2}\mathrm{Trace}(A^{2})+ ... + c_{1}\mathrm{Trace}(A^{n-1}) + \mathrm{Trace}(A^{n}) \right)$ with $c_{0} = 1$. Now the conditions in the problem imply that $\mathrm{Trace}(A^{k}) = 0$. Now, use the Cayley-Hamilton's Theorem to conclude that $\det(A) = 0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/921700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Inequality with prime numbers I found exercise in my book for number theory that I can't resolve. How do you show that
$$p_n < e^{1+n}$$
where $p_n$ is $n$-th prime number?
|
Let's try induction. For the base case of $n=1$, it is clear that $p_{1}=2<e^{1+1}=e^2$. Now let's suppose for all $k$ where $n \geq k >1$ that our result holds. We know by Bertrand's Postulate that $p_{k+1}<2p_{k}$ for all $k \in \mathbb{N}$ and by induction we know $2p_{k}<2e^{k+1}$. It follows that $$p_{k+1}<2e^{k+1}<(e)e^{k+1}=e^{k+2}$$ We now know our result holds for all $k \in \mathbb{N}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/921919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Any number $n>11$ is a sum of two composite numbers I have seen the following problem:
Any number $n>11$ can be formed by the sum of two composite numbers
and it says to prove it by contradiction. I have done the following:
Assume that the sum will be formed by non-composite numbers. Because a composite number is one that has at least one divisor that is not one or the same number; I can assume that it could be the sum of two prime numbers (because these are not composite). From this point I can say that adding both numbers which are primes, a and b, will give us necessarily an odd number. So:
$a=2p+1$ and $b=2p+1$, adding both I will have:
$a+b=2\cdot 2p + 2 = 2(2p+2)$ which is an even number, leaving any odd number; and giving us a contradiction that our theorem will hold for any n>11.
A similar proof can be done if we suppose that one number is composite and the other not, in this case any even number will be left out.
is this fine?
Thanks
|
No, it is not fine. If $n$ is not the sum of two composite numbers, it does not follow that it must be the sum of two primes. It might be the sum of a prime and a composite number. All you can say is that it cannot be expressed as the sum of two composite numbers. Also 1 is not regarded as either prime or composite.
Also, you are not asked to show that all numbers from some point on are the sum of two composite numbers, but specifically to show that all numbers greater than 11 are the sum of two composite numbers.
You are asked to prove it by contradiction. But you probably also need to use induction. It is hard to think of a way without induction. So the kind of thing you want to prove is that at least one of $n-4,n-6,n-8$ must be composite.
However, you did not ask for a proof, just for a critique of what you had done. So I will leave you with that hint.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/922054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
If $\space o(\ker\phi) = n$, then $\varphi$ is an $n$-to-$1$ mapping from $G$ onto $\varphi(G)$ If $G,H$ are groups and $\varphi\colon G\to H$ a group homomorphism, how can I prove that if the order of $\ker\phi$ is $n$ then $\varphi$ is an $n$-to-$1$ mapping from G onto $\varphi(G)$
|
Hint: Given $g\in G$, which $x\in G$ satisfy $\phi(xg)=\phi(g)$?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/922133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Evaluating $\int \:\sqrt{1+e^x}dx$ , why I got different answers? I've got 2 steps to evaluating $\int \:\sqrt{1+e^x}dx$ which lead to different values
first step :
$\int \:\sqrt{1+e^x}dx$
let $u\:=\:\sqrt{1+e^x}$ , $du\:=\:\frac{e^x}{2\sqrt{1+e^x}}dx$ , but $e^x\:=\:u^2-1$ and substitute $u$ into $du$, i get $dx\:=\:\frac{2u}{u^2-1}du$
thus, $\int \:\sqrt{1+e^x}dx$ $=2\int \frac{u^2}{u^2-1}du\:$
apply the following algebraic property $\frac{a}{1+a}=-\frac{1}{1+a}+1$ , then the integrand becomes
$\frac{u^2}{u^2-1}=\frac{1}{\:u^2-1}+1$ ,thus
$=2\int \:\:\left(\frac{1}{u^2-1}+1\right)du\:=\:-2arctanh\left(u\right)\:+\:2u\:+C$
substitute back $u$, thus
$\int \:\sqrt{1+e^x}dx=2\sqrt{1+e^x}-2arctanh\left(\sqrt{1+e^x}\right)+C$
2nd step :
using the same substitution for $u$ , $du$, and $dx$
$\int \:\sqrt{1+e^x}dx$ $=\int \frac{2u^2}{u^2-1}du\:$ $=\int \:\left(2\:+\:\frac{1}{u-1}-\frac{1}{u+1}\right)du$
evaluating the integrand one by one, then i get
$=2u\:+ln\left(u-1\right)-ln\left(u+1\right)\:+C\:=\:2u\:+\:ln\left(\frac{u-1}{u+1}\right)\:+\:C$
substitute back $u$, thus the answer becomes
$\int \:\sqrt{1+e^x}dx=2\sqrt{1+e^x}+ln\left(\frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1}\right)+C$
when i do derivative to both answers, i got the same answer which is $\sqrt{1+e^x}$ , it makes me confused which step is the right one? thanks for you answer. :)
EDIT :
Thank you for the answers, now i get it that both answers are correct. :)
|
It is perfectly normal. It seems just that you forgot that we can rewrite $$\tanh ^{-1}(z)=\frac{1}{2}\log\frac{1+z}{1-z}$$ Have a look at http://en.wikipedia.org/wiki/Inverse_hyperbolic_function where you will find other very useful identities for inverse hyperbolic functions.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/922226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Trisecting a Triangle Given a (non-degenerate) triangle $PQR$ in the Euclidean plane, does there exists a point $A$ in the interior of the triangle such that, the triangles $APQ$, $AQR$, and $ARP$ have same area? If it exists, is it unique?
(I thought about this question while reading the book "Proof without Words- R. B. Nelsen".)
|
An analytic solution, based on the fact that the area of the triangle with vertices at $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$ is
$$
\frac{1}{2}\det
\begin{bmatrix}
1 & 1 & 1 \\
x_1 & x_2 & x_3 \\
y_1 & y_2 & y_3
\end{bmatrix}
$$
provided the path goes counterclockwise. Without loss of generality, we can assume the vertices are $O(0,0)$, $A(a,0)$ and $B(b,c)$, with $a\ne0$ and $c\ne0$. Let $P(x,y)$ be the point we're looking for. Then we must have
$$\begin{cases}
\det\begin{bmatrix}
1 & 1 & 1 \\
a & b & x \\
0 & c & y
\end{bmatrix}
=
\det\begin{bmatrix}
1 & 1 & 1 \\
0 & a & x \\
0 & 0 & y
\end{bmatrix}
\\[3ex]
\det\begin{bmatrix}
1 & 1 & 1 \\
0 & x & b \\
0 & y & c
\end{bmatrix}
=
\det\begin{bmatrix}
1 & 1 & 1 \\
0 & a & x \\
0 & 0 & y
\end{bmatrix}
\end{cases}
$$
which becomes
$$
\begin{cases}
by-ac-cx-ay=ay\\
cx-by=ay
\end{cases}
$$
or
$$
\begin{cases}
-cx+(b-2a)y=ac\\
cx-(a+b)y=0
\end{cases}
$$
which easily gives $3ay=ac$ or $y=c/3$ and similarly $x=(a+b)/3$. The point
$$
\left(\frac{a+b}{3},\frac{c}{3}\right)
$$
is indeed the barycenter of the given triangle and is the unique solution.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/922292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Evaluate $\lim_{x\to\infty}\left[x - \sqrt[n]{(x - a_1)(x - a_2)\ldots(x - a_n)}\right]$ Evaluate the following the limit:
$$\lim_{x\to\infty}\left[x - \sqrt[n]{(x - a_1)(x - a_2)\ldots(x - a_n)}\right]$$
I tried expressing the limit in the form $f(x)g(x)\left[\frac{1}{f(x)} - \frac{1}{g(x)}\right]$ but it did not help.
|
You know that the product $(x - a_1)(a - a_2)\cdots(x - a_n)$ is equal to $x^n + \cdots + a_1a_2\cdots a_n$, where the $\cdots$ between $x^n$ and the product of the $a_i$'s is made of powers of $x$ with an exponent that is smaller than $n$. After you've seen this, it's easy to see that the only term that matters inside that root is $x^n$, because all the other ones can be left out when $x$ goes to $\infty$.
So your limit is:
$$\lim_{x\to\infty} (x - \sqrt[n]{x^n}),$$
and now it will all depend on whether $n$ is even or odd. If it's even, then $\sqrt[n]{x^n} = \lvert x\rvert$, otherwise $\sqrt[n]{x^n} = x$. So, if $x$ goes to $-\infty$, for example:
$$\lim_{x\to-\infty} (x - \sqrt[n]{x^n}) = \begin{equation}
\begin{cases}
-\infty, & \text{if $n$ is even;} \\0, & \text{if $n$ is odd.}
\end{cases}
\end{equation}$$
Otherwise, if $x$ goes to $+\infty$, it's easy to see that the limit is $0$ in both cases.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/922380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
}
|
Limit points of $(2,3)$. From this question: Constructing a set with exactly three limit points.
To answer this question
Construct a bounded set of real numbers with exactly three limit points.
But why $(1,2)\bigcup (2,3)$ can't be answer?
From my understand to limit point, $(2,3)$ only have two limit point $2$ and $3$. Any $x$ not be $2$ or $3$, can't be $(2,3)$'s limit point. Am I right?
|
No.
If you understand why $(2,3)$ is an open set then,
Lemma: Let $X$ be an open set in $\mathbb{R}^k$ , if $p\in X$ then $p$ is a limit point of $X$. (Hint: every neighborhood of $p$ contains an element other than $p$)
Thus every point in $(2,3)$ is a limit point. Also we have two additional limit points, namely $2$ and $3$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/922484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
A curve internally tangent to a sphere of radius $R$ has curvature at least $1/R$ at the point of tangency Suppose $a$ is an arc length-parametrized space curve with the property that
$\|a(s)\| \leq \|a(s_0)\| = R$ for all $s$ sufficiently close to $s_0$. Prove that
$k(s_0) \geq 1/R$.
So, I was going to consider the function $f(s) = \|a(s)\|^2$ but then the hint asks what we know about $f''(s_0)$? I am confused on how to link the two together and show from there that the curvature of $k(s_0) \geq 1/R$.
I know that we should use the second derivative test to see the local minima or maxima thus giving us if the curve points upward or downward, but I am a little lost at where to begin. I was told that $f(s_0)$ is constant, why? How come $f(s)$ is not a constant function here?
|
Some intermediate hints:
*
*We are given that $\Vert a(s) \Vert \leq \Vert a(s_0)\Vert$ for all $s$ near $s_0$. Squaring both sides tells us that $f(s) \leq f(s_0)$ for all $s$ near $s_0$. In other words, $s = s_0$ is a local max for the function $f$. What does that imply about $f''(s_0)$?
*You should compute $f''(s_0)$ as the hint says. This means that you'll have to compute $f'(s)$, then its derivative $f''(s)$, then finally plug in $f''(s_0)$. To compute these derivatives, you should use that $f(s) = \Vert a(s) \Vert^2 = \langle a(s), a(s) \rangle$, and use the product rule for inner products:
$$\frac{d}{ds}\langle F(s), G(s) \rangle = \langle F'(s), G(s) \rangle + \langle F(s), G'(s) \rangle.$$
*Finally, once you've computed $f''(s)$ and also used Hint 1, you should get an inequality. At this point, you might want to apply the Cauchy-Schwarz Inequality
$$\left|\langle v,w \rangle\right| \leq \Vert v \Vert \Vert w \Vert.$$
You'll also want to remember that since $a(s)$ is unit speed, we have $\Vert a'(s) \Vert = 1$ for all $s$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/922606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Dirac delta distribution & integration against locally integrable function I was reading the a lecture note online about distribution theory and it said: The Dirac delta distribution $\delta \in D'$ is defined as $\delta(\varphi)= \varphi(0) $, and there's no locally integrable function $f$ such that $T_f=\delta$, namely, $\int_{R^n}=f(x)\varphi(x)ds=\varphi(0)$, for all $\varphi \in D$.
I was trying to construct a proof about it but failed... Can anyone help ?
|
Suppose there exists $f \in L^1_{\text{loc}}$ such that $T_f = \delta$. Define
$$\varphi_k(x) := \begin{cases} \exp \left(- \frac{1}{1-|kx|^2} \right) & |x| < \frac{1}{k} \\0 & \text{otherwise} \end{cases}.$$
for $k \in \mathbb{N}$. As $\varphi_k \in D$, we get
$$\int \varphi_k(x) f(x) \, dx = T_f(\varphi_k) = \delta(\varphi_k) = \varphi_k(0)= e^{-1}$$
for all $k \in \mathbb{N}$. Hence,
$$e^{-1} = \left| \int \varphi_k(x) f(x) \, dx \right| \leq \underbrace{\|\varphi_k\|_{\infty}}_{\leq e^{-1}} \int_{B(0,1/k)} |f(x)| \,dx.$$
Since $f$ is locally integrable, the dominated convergence theorem yields
$$e^{-1} \leq e^{-1} \inf_{k \in \mathbb{N}} \int_{B(0,1/k)} |f(x)| \, dx =0.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/922708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
What does this function notation mean? My text tells me that the general term of a sequence can be looked at like a function:
$
f:\mathbb{N}\rightarrow \mathbb{R}
$
What does that mean translated into common english?
|
This specific case is called a sequence. The function $f$ takes in an input from the natural numbers (denoted $\mathbb N$), and gives an output in the real numbers (denoted $\mathbb R$).
In general, this notation contains three parts: the function name, the domain, and the codomain. The function name (in your case $f$) is the same $f$ appearing in the notation $f(x)$. The domain is the set of inputs to your function (in your case the natural numbers $\mathbb N$). The codomain is the set of possible outputs that the function can give (in your case the real number $\mathbb R$).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/922772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Help evaluating triple integral over tetrahedron I have a triple integral of $\iiint xyz\,dx\,dy\,dz$ over the volume of a tetrahedron with vertices $(0,0,0)$, $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$.
Normally I would just have limits 0 to 1 but that does not seem to work. How do I solve a problem like this?
|
Using the facts that the projection of the solid in the xy-plane is the triangle with vertices (0,0), (0,1), and (1,0), and that the top of the solid is the plane $x+y+z=1$, we can set up the integral as
$\displaystyle\int_{T} f(x,y,z) \;dV =\int_{0}^{1}\int_{0}^{1-x}\int_{0}^{1-x-y} f(x,y,z) \;dz dy dx$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/922879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
}
|
Circle diameter using only 3 points I've never been very good at math and confuse very easily so I need some help in doing a calculation regarding a circle.
Let's assume we're drawing the letter L.
The top point $A$ of the long leg is at $0$ the bottom point of the long leg $B$ is at $3$ feet and the point to the right $C$ is at $6$ inches.
What would the diameter of a circle be if it was drawn so that $A$ and $C$ were connected? Can you tell me in lay terms how to make that calculation or some semblance of usable formula that I might comprehend, please?
|
In your special case, you have to use the Pythagorean theorem, as the letter $L$ has a right angle.
You have then $AC^2=AB^2+BC^2$
And $AC$ is the diameter of the circumcircle.
For a more generic solution (working with any three points), you will need to find out the coordinates of the center $O$ of the circle, and the radius will be for instance $OA$.
Finding the center of the circle relies on the median of two of your segments $AC, AB$ and $BC$, whaterrve your choice.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/923029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
How to get started with solving basic Differential Equations? I've just started learning Differential Equations and am having general difficulties with a bit of concepts and on how to actually get started. The problem I have is that the books and sources I find always launch into slope fields, and while I get what a field is and how it works. I don't understand how it relates to a differential equation, and more importantly how to solve them. We started with very basic examples and I am already at a loss of what to do.
Essentially my questions boil down to something like this:
*
*What does a slope field represent? i.e. What are the lines indicating or plotting?
*How do they relate to differential equations?
*What are the general steps to solving a very basic differential equation? Such as: $\frac{dy}{dx} = y + 5 $ ($\frac{dy}{dx}$ = $y'$ right?)
Please keep in mind that I have no knowledge of differentials, just calculus, including mulitvariable. (i.e. Calc I-III in the states).
|
(1). In a slope field,
the direction of the arrow
at each point
represents the direction of the derivitive
(or slope, hence its name),
and the length sometimes
represents the magnitude of the derivative
at the point.
This enables you
to draw an approximate solution
to the differential equation
starting at any point,
just by following the arrows.
(2). As stated above,
by knowing the slope of
the derivative at each point
(assuming you can compute
$\frac{dy}{dy}$
for any $x$ and $y$),
you can get a feel
for what the solutions look like.
(3). There are many, many, many techniques.
I like integrating factors.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/923125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Find a positive number $\delta<2$ such that $|x−2| < \delta \implies |x^2−4| < 1$ I have to find a positive number $\delta<2$ such that $|x−2| < \delta \implies |x^2−4| < 1$.
I know that $ \delta =\frac{1}{|x+2|} $ has this behaviour, but it is not guaranteed for it to be less than two. I don't know what else to do. Thank you!
|
Observe that : $|x^2 - 4| = |x-2||x+2| < \delta|x+2| < \delta\left(|x - 2| + 4\right) < \delta(\delta + 4) = \delta^2 + 4\delta$. Thus all we need is to solve for $\delta$:
$\delta^2 + 4\delta < 1 \iff (\delta + 2)^2 < 5 \iff \delta + 2 < \sqrt{5} \iff \delta < \sqrt{5} - 2$. For example we can take: $\delta = \dfrac{\sqrt{5} - 2}{2}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/923244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Dot product and its representation as sum.
If I define the dot product $u\cdot v $ where $u,v\in \mathbb{R}^n$ as $u\cdot v= |u||v|\cos \theta$ where $\theta$ is the angle between $u$ and $v$. How can I get that
$$u\cdot v= u_1v_1+u_2v_2+u_3v_3 $$
with $u=u_1e_1+u_2e_2+\cdots+u_ne_n$ and $v=v_1e_1+v_2e_2+\cdots+v_ne_n$?
$e_1... e_n$ is the canonical base.
If we think in $\mathbb{R}^2$ we can suppose that, $u$ have an angle of $\alpha$ with the X axis, $v$ have an angle $\beta$ with the $X$ axis, and $\alpha\geq\beta$. Then $\theta=\alpha-\beta$. We get
$$\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta $$
Using the relation $cos\alpha= \displaystyle\frac{u_1}{|u|}, \sin \alpha=\displaystyle\frac{u_2}{|u|}$ and analogous relation for $v$
$$\cos\theta= \displaystyle\frac{u_1}{|u|}\displaystyle\frac{v_1}{|v|}+\displaystyle\frac{u_2}{|u|}\displaystyle\frac{v_2}{|v|}$$
multiplying by |u||v| we get the result.
Can we generalize this idea to $\mathbb{R}^n$?
Thanks!
|
Letting $\vec u = (u_1, u_2, \dots, u_n)$ and
$\vec v = (v_1, v_2, \dots, v_n)$,
then $\displaystyle |u|^2 = \sum_{i=1}^n u_i^2$,
$\displaystyle |v|^2 = \sum_{i=1}^n v_i^2$, and
$\displaystyle |u-v|^2 = \sum_{i=1}^n (u_i - v_i)^2$.
By the law of cosines,
$\cos \theta = \dfrac{|u|^2 + |v|^2 - |u-v|^2}{2\,|u|\,|v|}$. So
\begin{align}
u \circ v
&= |u|\,|v|\,\cos \theta \\
&= \dfrac 12(|u|^2 + |v|^2 - |u-v|^2) \\
&= \dfrac 12\left(
\sum_{i=1}^n u_i^2 +
\sum_{i=1}^n v_i^2 -
\sum_{i=1}^n (u_i^2 + v_i^2 - 2u_iv_i)\right) \\
&= \sum_{i=1}^n u_iv_i
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/923330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
How was this approximation of transcendent equation solution found? I have an equation for $\xi$:
$$\xi\gamma=\cos\xi,$$
where $\gamma\gg1$. I've tried solving it assuming that $\xi\approx0$ and approximating $\cos$ by Taylor's second order formula:
$$\xi\gamma\approx1-\frac{\xi^2}2,\tag1$$
then I get
$$\xi\approx2\left(\sqrt{\gamma^2+1}-\gamma\right).\tag2$$
The book I read gives the approximate solution of it as
$$\xi\approx\frac1\gamma\left(1-\frac1{2\gamma^2}\right).\tag3$$
When I tried to understand how they got this, my first thought was like "they made another simplification and assumed $\xi\approx\frac1\gamma$ in $(1)$". I thus thought that this approximation would be worse. But when plotting these solutions as functions of $\gamma$, I found that $(3)$ in fact converges much faster to numerical solution than $(2)$!
So I now wonder: how did they get $(3)$?
|
$f(x)=\cos x-\gamma x$ is a concave decreasing function over $(0,\pi/2)$, hence Newton's method gives that the first positive root of $f(x)$ is less than:
$$ 0-\frac{f(0)}{f'(0)} = \frac{1}{\gamma} $$
as well as it is less than:
$$\frac{1}{\gamma}-\frac{f(1/\gamma)}{f'(1/\gamma)}=\frac{1}{\gamma}-\frac{1}{2\gamma^3}+O\left(\frac{1}{\gamma^5}\right).$$
Further iterations do not change the appearance of the last asymptotics.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/923431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Relation of ellipse semi-axes with rotation angle and projection length In the following setup, assume $w$ (length of the projection of the ellipse) and $\theta$ (the rotation angle) are known. I want to know what equation(s) do I have here that helps me to derive the semi-axes ($a$ and $b$) of the ellipse based on $w$ and $\theta$.
I know the semi-axes are not unique in this problem but only the relation is needed.
|
$$\large 4(a^2\sin^2\theta+b^2\cos^2\theta)=w^2$$
Equation of an ellipse is:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
Diffrentiate:
$$\frac{dy}{dx}=-\frac{b^2}{a^2}.\frac{x}{y}$$
Polar form of ellipse:
$$P(\phi)\equiv(a\cos\phi,b\sin\phi)$$
Slope of tangent in polar form:
$$m=-\frac ba\cot\phi$$
Equation of tangent:
$$\frac xa\cos\theta+\frac yb \sin\theta=1$$
Distance between two parallel lines in form:
$$ax+by+c=0\\ax+by+c'=0$$
is:
$$d=\frac{|c-c'|}{\sqrt{a^2+b^2}}$$
*
*Firstly those two lines are tangent at two points, and they are parallel too. So in the parametric form the slope of tangent at these two points – let them be called $P(\theta_1),Q(\theta_2)$ – must be equal.
*
*slope $\propto\cot\phi$, where $\phi$ is parametric angle.
$$\implies \cot \theta_1=\cot\theta_2\implies \theta_1=\pi+\theta_2\quad\text{since}\quad\theta_1\ne\theta_2$$
*Secondly the distance between the two tangents / two parallel lines must be $w$:
*
*Tangent Lines are $$\frac xa\cos\theta_1+\frac yb\sin\theta-1=0\\\frac xa\cos\theta+\frac yb\sin\theta+1=0$$ (after putting $\theta_2=\pi+\theta_1$). So taking distance between them:
$$\frac{|1-(-1)|}{\sqrt{\frac{\cos^2\theta_1}{a^2}+\frac{\sin^2\theta_1}{b^2}}}=w$$
*Thirdly slope of tangents at these points is $\theta$ so:
$$-\frac{b^2}{a^2}\cot\theta_1=\tan\theta$$
Try the rest yourself.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/923503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
retraction induced homomorphism is surjective I am having a hard time proving this although it looks trivial...
Let $r:X\to A$ be a retraction between a topological space $X$ and $A\subset X$ such that $r(a_0)=a_0$ for $a_0\in A$
then the induced homomorphism $r_*:\pi_1(X,a_0)\to \pi_1(A,a_0)$ is surjective.
I tried to prove it as follows:
I showed that if $g$ is a loop in $A$ based at $a_0$ then it is a loop also in $X$ based at $a_0$
So, given $[g]\in \pi_1(A,a_0)$, let us take $[g]\in \pi_1(X,a_0)$ (which will stand for a different homotopy class)
So we get: $$r_*([g])=[r\circ g]=[id_A\circ g]=[g]$$ that's since $Im(g)\subset A$ and $r$ is a retraction.
I believe that something is off in that proof in the part with the homotopy classes, so help please
|
Consider $i$ canonical injection $i\colon A \to X$.
The fact that $r$ is a retract means
$$r \circ i = \mathrm{id}_A$$
From this we get
$$r_* \circ i_* = \mathrm{id}_{\pi_1(A)}$$
and this implies $r_*$ surjective since it has a right inverse.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/923573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
How do I prove that numbers not divisible by 3 can be represented as 3x+1 or 3x-1? I saw that some proofs used the fact that numbers not divisible by $3$ can be represented as $3x+1$ or $3x-1$. But how do I prove that it is true?
|
If $k=3x+1$ then $$k\equiv 1\mod 3$$ and if $k=3x-1$ then $$k\equiv 2\mod 3.$$ Moreover $$\mathbb Z/3\mathbb Z=\{[0],[1],[2]\}$$
where $$[1]=\{3x+1\mid x\in\mathbb Z\}$$ and $$[2]=\{3x+2\mid x\in\mathbb Z\}=\{3x-1\mid x\in\mathbb Z\}$$
what conclude the proof.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/923639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Relationship between Continuity and Countability This is a consequence of one of the problems in elementary real analysis that I am attempting to solve. I have this doubt.
Suppose $f$ is a continuous map from the reals to the reals. If the set $S=${$f(x)|x\in A$} is countable, is the set $S'=${$f(x+1)|x\in A$} countable? Here $A$ is just a subset of $R$.
Well, the question may seem a bit artificial but let me also share the initial question which caused this doubt.
The original question is: Suppose $f$ is continuous such that $f(x)$ is rational iff $f(x+1)$ is irrational. Show such an $f$ cannot exist.
My method:
Suppose such an $f$ exists. A quick thinking tells us that it suffices to consider a restricted domain [$0,1$] only. Consider the set of rationals, $Q$, in this domain. Clearly $f(Q)$ is countable. Consider those irrationals for which $f$ is rational. This particular range of $f$ also is countable as it is a subset of rationals. So we are left with those irrational points in this domain, at which $f$ is irrational. If we can somehow show that the range of $f$ on this particular domain is countable, then it follows that range of $f$ is countable for all real points and hence $f$ is not continuous by IVT.
This is where I am facing a hurdle. I have to prove that the range of $f$ on the final particular domain i.e. set of all irrationals at which $f$ is irrational, is also countable. I figured out that considering any point $x$ in this domain, where $x$ is irrational, if $f(x)$ is irrational, then $f(x+1)$ is rational, so the set {$f(x+1)$} is countable. Does this imply that the range of $f$ when we are concentrating only those irrational points at which $f$ is irrational, is also countable?
|
Let $f$ be zero on the nonpositive numbers and then the identity on the nonnegative numbers.
If you take $A=[-1,0]$, $|f(A)|=|\{0\}|=1$, but $|f(A+1)|=|[0,1]|=|\Bbb R|$.
(Here I abuse notation a little by writing $A+1=\{x+1\mid x\in A\}$)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/923743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Some issues concerning joint random variables Let the joint random variable $P[x;y]$ be
$P[x;y] = c[2x^2 + y^2], x=-1;0;1, y=1;2;3;4$
$=0$ $elsewhere$
So I had to find the value of $c$ that makes $P[x;y]$ a joint discrete random variable.
I think I did that right. I just add up all the probabilities where $x = -1;0;1$ and $y = 1;2;3;4$ and made it equal to 1. Then I solved for $c$ and got $c=\frac{1}{106}$. Please check this for me if you think I've done something wrong.
Now they asked to calculate the $E[y]$ and the next question asked if $x$ and $y$ are independent.
So. I was a bit confused with the whole $P[x;y]$ and how to split it up into $P[x]$ and $P[y]$ (if that's possible) and then how to work out the $E[y]$.
|
I have worked out the problem for you and I hope it helps to check with your answer. I have done it in EXCEL what other responders have alluded to. To verify your answer.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/923872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
root of the function $f(x)=\sqrt{2}-x$ by using fixed point iteration How can I find the approximate value of $\sqrt{2}$ by using the fixed point iteration? I have tried
$x-\sqrt{2}=0$,
$x^2=2$,
$x^2-2+x=x$,
$g(x)=x^2-2+x$,
$g\prime(x)=2x+1$
And i choose $x_0=-\frac{1}{2}$.
But i cant find the approximate value correctly. Where am i wrong?
|
In general, you can produce a function $g(x)$ with $\sqrt2$ as a fixed point by letting
$$g(x)=x+(x^2-2)h(x)$$
with pretty much any function $h(x)$. However, as André Nicolas pointed out, if you want $\sqrt2$ to be an attracting fixed point for $g$, which is what you need if you want to approximate $\sqrt2$ by iterating the function $g$, then you need $|g'(\sqrt2)|\lt1$. Moreover, as André pointed out, you're really best off if $g'(\sqrt2)=0$. Since
$$g'(x)=1+2xh(x)+(x^2-2)h'(x)$$
we have
$$g'(\sqrt2)=1+2\sqrt2h(\sqrt2)$$
It's convenient at this point to take $h$ to have the form $h(x)=cxk(x^2)$, because then we have
$$g'(\sqrt2)=1+4ck(2)$$
in which case we can get $g'(\sqrt2)=0$ by letting $c=-1/(4k(2))$ and all we need is to choose a function $k(x)$ such that $k(2)\not=0$. The simplest such function is $k(x)\equiv1$, which gives $c=-1/4$ and thus
$$g(x)=x-{1\over4}x(x^2-2)={6x-x^3\over4}$$
Alternatively, $k(x)=1/x$ leads to $h(x)=-1/(2x)$ which gives
$$g(x)=x-{1\over2x}(x^2-2)={x\over2}+{1\over x}$$
as in André's answer. But as I said, you can really let $k$ be anything you like (as long as $k(2)\not=0$) and get a function $g$ that has $\sqrt2$ as an attracting fixed point.
Finally, as André also pointed out, whatever $g$ you use, you need to start close enough to $\sqrt2$ so that you're within the "basin of attraction" of the fixed point. I'll leave that for someone else to discuss in detail, but merely note that $x_0=1$ works as a starting point for either $(6x-x^3)/4$ or $x/2+1/x$ as the function to be iterated.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/923994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Proper way to make transition from $k$ to $k+1$ in proofs by induction First off my apologies for asking really simple questions, but I'm having some trouble wrapping my head around proof of induction when trying to prove a statement involving a natural number n holds for all values of n.
In the induction step lets assume the following simple example as found on this wikipedia page:
Proof the formula below for all positive integers.
In the wikipedia example inductive step. They ADD k+1 on the left side, but REPLACE k with k+1 on the right side. My book adds k+1 on both the left AND the right side. Example: Wikipedia starts the right side of the inductive step with:
So the wikipedia example replaced n with k+1.
My book starts the right side of the inductive step with: $= \frac {k(k+1)}{2} + k+1$. So my book just added it to the right side. Now for this particular example it doesn't matter when you replace or add the k+1 on the right side. The outcome of the proof will be the same.
My question is if it EVER matters in proof of induction. If I want to prove Pn+1 is true assuming Pn is true. Should I just replace the n with n+1 on both sides, should I add it on both sides of the equation or does it never matter?
EDIT: Based on multiple answers I was able to get a clear understanding of how I should properly use proof of induction! Thanks a lot everyone!
|
It does matter. However for this problem both methods amount to the same thing.
In induction, you prove a base case and then prove that a next case will be true. Depending on your problem, this can take a range of approaches, as long as the approach proves the next case.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/924079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
}
|
Calculate domain of $f(x) = \sqrt{\frac{1}{x-4} + 1 }^2 -9$ I ran this through the MathWorld domain/range widget, and got $x \leq 3$ & x > 4. I understand that x=4 is undefined, therefore the domain of x includes x > 4, but I cannot for the life of me figure out how to determine the $x \leq 3$. When I set up the inequality $\dfrac{1}{x-4} + 1 \leq 0$, I get $x \geq 3$. Anyone know what I'm doing wrong?
|
Note that $$\frac{1}{x-4} + 1 \geq 0 \\ \Rightarrow \frac{1+(x-4)}{x-4} \geq 0 \\ \Rightarrow \frac{x-3}{x-4} \geq 0$$
Can you figure out the rest? (Sign chart)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/924148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Evaluating $ \lim_{x\to0} \frac{\tan(2x)}{\sin x}$ $$\lim_{x\to0} \frac{\tan(2x)}{\sin x}$$
How would I evaluate that? I was thinking changing the tan to sin/cos, but when I tried that, it did not work.
|
Since $\displaystyle\tan(2x)=\frac{2\tan x}{1-\tan^2x}$
Set $t=\frac x2$
Using tangent half-angle formula we get;
$\tan x=\frac{2t}{1-t^2}$, $\quad$$\sin x=\frac{2t}{1+t^2}$
Hence $\displaystyle\lim\limits_{x\to 0}\frac{\tan(2x)}{\sin(x)}=\displaystyle\lim\limits_{x\to 0}\frac{2\tan x}{1-\tan^2x}\cdot\frac{1}{\sin x}$
$\displaystyle=\lim\limits_{t\to 0}\frac{2\big(\frac{2t}{1-t^2}\big)}{1-\big(\frac{2t}{1-t^2}\big)^2}\cdot\frac{1+t^2}{2t}=2\lim\limits_{t\to 0}\frac{1-t^4}{1+t^4-6t^2}=2\cdot 1=2$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/924299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 7
}
|
Prove that the limit exists and then find its limit We are given a function $h(x)$ which strictly positive. The function $h$ is defined on $\mathbb{R}$ and that satisfies the following:\
$$\lim_{x\to 0}(h(x)+\frac{1}{h(x)})=2$$
The question is to prove that the limit of $h$ exists at$ 0$ and then find its limit as $x\to 0$
If we assume that the limit exists (say it's equal to $\lambda$, then we have $\lambda+\frac{1}{\lambda}=2$ and by solving this equation, we get $\lambda=1$. However, I am completely clueless on how to prove the existence of the limit of $h(x)$ when $x\to 0$.
|
Let $g(x) = x + x^{-1}$. Then your limit is
$$ \lim_{x \to 0} g(h(x)) = 2 $$
$g$ isn't quite an invertible function: for every point $a \in (2, \infty)$ there are two solutions to $g(x) = a$. But in some sense it's a continuous two-valued 'function': if I let $f^+$ and $f_-$ be the larger and smaller of the two solutions, then they are both continuous functions:
$$ \lim_{x \to 0} f_+(g(h(x))) = f_+(2) = 1 $$
$$ \lim_{x \to 0} f_-(g(h(x))) = f_-(2) = 1 $$
So in some sense, we whould be able to say
$$ \lim_{x \to 0} g^{-1}(g(h(x)) = g^{-1}(2) = \{1\} $$
$$ \lim_{x \to 0} \{ h(x), h(x)^{-1} \} = \{1\} $$
and since we just get one limit point, the individual limits should exist (i.e. $h(x) \to 1$). All that's left is finding a way to make a precisely statement that captures this idea, and then prove the theorem saying we can find limits like this. This is a fair bit of work, but a good idea in the long run since it gives us a new tool useful for dealing with more problems!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/924391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
}
|
Solve the equation:$ \bar{z}=z^{n-1}$ Solve the following equation:
$\bar{z}=z^{n-1}$
Where $\bar{z}$ is the complex conjugate of z, and n is a natural number such that $n\neq 2$.
I have tried to write z in rectangular form and polar form. I have tried to play with De Moivre's formula.
But I still do not see where to proceed from here.
Could you please point me to the right direction?
Thanks.
|
Use polar coordinates, $z = re^{i\theta}$:
$$re^{-i\theta} = r^{n-1}e^{i(n-1)\theta},$$
$$r = r^{n-1}\implies\cdots,$$
$$\cdots$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/924466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
open sets in the order topological space I have a question. I am really confused about determining if a set is open. First, the idea of a set being closed has nothing to do with homomorphic ideas of closure: if $x,y \in F$ then $x+ y$ in $F$ . (This is not the idea of closure.)
My question is from an example from the book:
Consider $I \times I =[0,1] \times [0,1]$, where $I \times I$ has the dictionary order.
Then, consider the set $Y :=\{\frac{1}{2}\} \times (1/2, 1]$.
For $Y$ is open in the subspace topology of $I \times I$ since for $a \in [0,1]$ and $b \in [0,1]$ we have $(a,b) \bigcap \{\frac{1}{2}\} \times (1/2, 1] = \left(a \bigcap \{\frac{1}{2}\}\right) \times \left(b \bigcap (\frac{1}{2},1]\right) = \{\frac{1}{2}\} \times (\frac{1}{2},1]$. So this is in $Y$ as a subspace.
But $I \times I$ is not in the order topology. Why? I just know in the subspace we have an open interval. But in the order topology, we have not an open interval, but a subset... so to speak. Need some clarification here... Thank you so much.
|
Your question isn’t entirely clear, but if you’re starting with $\Bbb R\times\Bbb R$ with the dictionary order and then giving $I\times I$ the subspace topology that it inherits from $\Bbb R\times\Bbb R$, then it’s true that $Y$ is open in $I\times I$. To see this, let $p=\left\langle\frac12,\frac12\right\rangle$ and $q=\left\langle\frac12,2\right\rangle$. Let $\preceq$ denote the dictionary order on $\Bbb R\times\Bbb R$. Then the open interval $(p,q)$ in $\langle\Bbb R\times\Bbb R,\preceq\rangle$ is by definition
$$\begin{align*}
(p,q)&=\{\langle x,y\rangle\in\Bbb R\times\Bbb R:p\precneqq\langle x,y\rangle\precneqq q\}\\
&=\left\{\left\langle\frac12,y\right\rangle:\frac12<y<2\right\}\\
&=\left\{\frac12\right\}\times\left(\frac12,2\right)\;,
\end{align*}$$
and
$$\begin{align*}
(p,q)\cap(I\times I)&=\left(\left\{\frac12\right\}\times\left(\frac12,2\right)\right)\cap\left(\left\{\frac12\right\}\times\left(\frac12,1\right]\right)\\
&=\left\{\frac12\right\}\times\left(\left(\frac12,2\right)\cap\left(\frac12,1\right]\right)\\
&=\left\{\frac12\right\}\times\left(\frac12,1\right]\\
&=Y\;.
\end{align*}$$
Thus, $Y$ is the intersection with $I\times I$ of an open set — in fact an open interval — in $\Bbb R\times\Bbb R$, so $Y$ is relatively open in the subspace $I\times I$.
(In the definition of $q$ you can replace the second coordinate by any real number greater than $1$.)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/924579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Applying angle addition formulas for subtraction The angle addition formula says that:
$\sin(\phi + \theta) = \sin(\phi) \cdot \cos(\theta) + \cos(\phi) \cdot \sin(\theta)$
Why are the following steps valid?:
$\sin(\phi − \theta) = \sin(\phi) \cdot \cos(−\theta) + \cos(\phi) \cdot \sin(−\theta)= \sin(\phi) \cdot \cos(\theta) − \cos(\phi) \cdot \sin(\theta)$
Thanks.
|
Hint:
$\cos(\theta)$ is an even function while $\sin(\theta)$ is an odd function.
Hence:
$$\cos(-\theta) = \cos(\theta),$$
while
$$\sin(-\theta) = -\sin(\theta).$$
(See this Wikipedia link on Even and odd functions for more information.)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/924774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Solvable groups in group theory If $N \unlhd G$, and $M,K \leq G$ such that $M \unlhd K$, then does it imply that $MN \unlhd KN$? If yes, how?
|
First of all, $N$ normalizes $MN$. In fact, if $n_{1}, n_{2} \in N$, and $m \in M$, then
$$
(m n_{1})^{n_{2}} = n_{2}^{-1} m n_{1} n_{2} = m (m^{-1} n_{2}^{-1} m) n_{1} n_{2} = m (n_{2}^{-1})^{m} n_{1} n_{2}\in MN,
$$
as $N$ is normal in $G$.
And then $K$ also normalizes $MN$. If $k \in K$, then
$$
(m n_{1})^{k} = m^{k} n_{1}^{k} \in M N,
$$
as $N$ is normal in $G$, and $M$ is normal in $K$.
Therefore $\langle K, N \rangle = KN$ normalizes $MN$.
I am writing $x^{y} = y^{-1} x y$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/924898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Containment of $c_0$ or $\ell_p$ Suppose that $(x_n)$ is a sequence of unit vectors in a Banach space $X$ such that $$\mbox{dist}(x_m, S_{X_n})=1$$
for all $m > n$. Here $S_{X_n}$ stands for the unit sphere $\mbox{span}\{x_1, \ldots, x_n\}$. Can we conclude that $X$ contains an isomorphic copy of $c_0$ or $\ell_p$?
This question is motivated by the observation that the canonical bases of $c_0$ and $\ell_p$ have this property. I cannot think of any other example of a Banach space with this property.
|
No.
You can find such a sequence in any infinite dimensional normed space. See the proof of Lemma 1.4.22 in Megginson's An Introduction to Banach Space Theory. But, there are infinite dimensional Banach spaces that contain neither $c_0$ nor any $\ell_p$, such as Tsirelson's space.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/925004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Partial differential equation, mixed derivatives What can be concluded from following equation:
$$\frac{\partial f(x,y)}{\partial x}-\frac{\partial g(x,y)}{\partial y} = 0$$
where $f(x,y)$ and $g(x,y)$ are functions of two independent variables $x,y$. Does it generally imply that
$$\frac{\partial f(x,y)}{\partial x}=\frac{\partial g(x,y)}{\partial y}=a(x)b(y)$$ for some functions $a(x), b(y)$? Thanks for answer.
|
No.
It implies that there exists a function $F(x,y)$ such that $\partial_x F(x,y)=g(x,y),\partial_y F(x,y)=f(x,y)$. Since
$$\partial_x\partial_y F(x,y)=\partial_y g(x,y)=\partial_x f(x,y)=\partial_y\partial_x F(x,y)$$
In general you can not separate $F(x,y)$ into $a(x)b(y)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/925083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How "big" are the mathematical disciplines? This question may be difficult to answer, but is it possible to estimate the "size" of the individual mathematical disciplines; measures of size could be number of professionals dealing with it, the number of magazines and new theorems etc.? For instance, I usually consider analysis, algebra, and topology to be the three major branches of mathematics. But how many of the total number of mathematicians are dealing with each of these? My impression is that analysis is the largest discipline, but I could be wrong.
|
For a rough graphical representation see for example here and here.
(Related questions with interesting links can be found here and here and here.)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/925298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
Making a Piecewise Function a Single Function Is there a way to turn a piecewise function into one function. For example:
$$\ f(x)=\begin{cases} g(x) & \text{if $a≤x<b $} \\ h(x) & \text{if $b≤x≤d$} \end{cases}$$
(Can you use the Heaviside Step Function? $\theta(x))$
|
You can rewrite $f(x)$ as $$f(x)=\chi_{[a,b)}(x)g(x)+\chi_{[c,d)}(x)h(x),$$ where $\chi_{[a,b)}(x)$ is $1$ if $x\in[a,b)$ and $0$ otherwise.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/925407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Are there any mathematics "problem websites" similar to Project Euler? Are there any mathematics websites similar to Projet Euler, a website which hosts math-heavy programming questions, many of which can be solved with a pen and paper?
I've become almost addicted to Project Euler's progress tracking system, and I also absolutely love their solutions forums. I love reading about other solutions—about how much quicker they are, how much more intuitive or elegant they are.
All Project Euler problems directly involve math. Are there any mathematics websites which track progress, and have solution forums, that are similar to Project Euler?
I liked Khan Academy, but I've completed its most advanced classes in school already, so it isn't much fun anymore.
|
The AwesomeMath website--www.awesomemath.com--will lead you to the journal Mathematical Reflections, each issue of which has one or two short articles and lots of problems divided into Junior, Senior, Undergraduate, and Olympiad problems. Solutions appear in later issues.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/925500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 3,
"answer_id": 2
}
|
Write $\vec{a}=(3,2,-6)$ as the sum of two vectors, one parallel, and one perpendicular, to vector $\vec{d}=(2,-4,1)$. The solution is $$\vec{a}=(-8/21)\vec{d}+(79/21,10/21,-118/21)$$
|
Consider the vector
$$
v=a-\frac{a\cdot d}{d\cdot d}d
$$
Note that
$$
\begin{align}
v\cdot d
&=a\cdot d-\frac{a\cdot d}{d\cdot d}d\cdot d\\
&=0
\end{align}
$$
Therefore,
$$
a=v+\frac{a\cdot d}{d\cdot d}d
$$
where $v$ is perpendicular to $d$ and $\frac{a\cdot d}{d\cdot d}d$ is parallel to $d$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/925576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
I'm not sure what this is exactly asking Without using words of negation, write the meaning of : "f is not an increasing function"
I did:
$$"f\ is\ not\ an\ increasing\ function" \ \equiv\ "f\ is\ a\ decreasing\ function"$$
Is this what it's asking or am I completely missing it?
|
No this is not true, for example any function from $\mathbb{R}$ to itself which increases on one interval and decreases on another is neither increasing nor decreasing.
It should be noted that "nonincreasing" is actually a perfectly valid adjective for describing a function, but that is probably not what it being sought after.
One way to write this, would to be to use the definition of increasing functions: a function $f$ is said to be increasing if $x>y\implies f(x)\geq f(y)$, or in other words $\forall x,y$ such that $x>y$, it follows that $f(x)\geq f(y)$.
The negation of this statement is therefore, $\exists x,y$ such that $x>y$, and $f(x)<f(y)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/925673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
}
|
Why are degree maps for cellular boundary formula from $S^{n-1}\to S^{n-1}$? For a CW-complex, there's the cellular boundary formula that
$$
d_n(e^n_\alpha)=\sum_\beta d_{\alpha\beta}e^{n-1}_\beta
$$
where the coefficients $d_{\alpha\beta}$ are the degrees of the map
$$
S^{n-1}\to X_{n-1}\to S^{n-1}
$$
where the first map is the attaching map of $e^n_\alpha$ to the $n-1$-skeleton, and the second map is collapsing the $X_{n-1}$ except for $e^{n-1}_\beta$.
What if you had a CW-complex consisting of a $0$-cell, with two $1$-cells attaching to it to make a figure-eight, and then attached a $2$-cell to one of the one cells to get a figure-eight with a filled in loop.
If I wanted to compute the coefficient in $d_2(e^2)$ of $e^1_\gamma$ where $e^1_\gamma$ is the $1$-cell I filled in, If I follow the composition above, the second map would collapse the unfilled loop to a point, but then it seems like I end up with $D^2$, not $S^1$. Am I doing something wrong?
|
You're left with $S^1$. You've already mapped the boundary of the $2$-cell to itself, and the other $1$-cell is collapsed to the $0$-cell.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/925739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Proving that something is irrational I'm trying to evaluate the following claim:
$$ \sqrt{2} + \sqrt{n} $$ is irrational.
This is what I tried:
Proof by contrapositive: Suppose $$ r = \sqrt{2} + \sqrt{n} $$
and r is rational. Then $$ \frac{m}{l} = \sqrt{2} + \sqrt{n} $$
$$\frac{m^2}{l^2} = 2 + 2\sqrt{2n} + n $$
I'm not sure where to proceed or if I'm even heading in the right direction.
Could anyone give me a tip?
|
Case 1: $n$ is of form $2u^2$ , $u$ being an integer
Then our number is equal to $\sqrt 2 (u+1)$, clearly this is irrational
Otherwise $2n$ is not a square, hence you get contradiction in the last identity you wrote.(since $\sqrt {2n}$ is irrational)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/925840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Recurrence Problem involving multiple dependencies. I have 3 equations :-
*
*$r_n=r_{n-1}+5m_{n-1}$
*$m_n = r_{n-1} + 3m_{n-1}$
*$p_n = 5m_{n-1}$
The initial values of the sequences are
$$r_0=3, m_0=1, p_0=0$$
How can I get the formula to get the nth term of the series without the recurrence?
A ready-made will be useful.
|
It is given that,
$$r_n=r_{n-1}+5m_{n-1}$$
$$m_n = r_{n-1} + 3m_{n-1}.$$
By subtracting we can obtain that,
$$r_n=m_n+2m_{n-1}$$
$$r_{n-1}=m_{n-1}+2m_{n-2}.$$
Hence
$$m_n=4m_{n-1}+2m_{n-2}.$$
This second order liner recurrence with $m_0=1$ and $m_1=6.$
I think you can solve this one and try to find $r_n$ and $p_n$ via substituting $m_n.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/925966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Find the order of $2$ in $\mod 2^{n} -1 $ Find the order of $2$ in $\mod 2^n-1$
I know that the order of $2$ in $\mod 2^n-1$ is the smallest positive integer $k$ such that
$$2^k \equiv 1 \pmod {2^n-1}$$
How to proceed from here ? Any help/hints ?
|
Since you we use $\mod 2^n-1$ it is clear that $k$ must be greater than $n-1$ because $2^{n-1}\leq 2^n-1$ If you try $k=n$ you see that $2^k = 2^n \equiv 1 (\mod 2^n-1)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/926094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
Spectrum of integral operator
Given $g\in C^1([0,1]\times[0,1])$, consider the operator $$Tu(x) =
\int_0^1 g(x,t) u(t) dt$$ defined on $u\in C([0,1])$. Discuss the
spectrum of T.
My attempt:
First I can show that $T$ is a compact operator. Given a sequence $u_n$ bounded in $C([0,1])$$\|u_n \|_\infty \leq M$, and let $\|g\|_\infty = N$, we have
$$|(Tu_n)(x)| \leq M\cdot N$$
And given $x_0$ we have
$$|(Tu_n)(x_0) - (Tu_n)(x)| \leq \int_0^1 |g(x_0,t) - g(x,t)| M dt$$
by continuity of $g$, we see that $(Tu_n)(x)$ is equicontinuous.
Now we know that $0$ is in the spectrum of $T$. To look for other eigenvalues, let $\lambda \neq 0$ and $f\neq 0$.
$$\lambda f(x) = \int_0^1 g(x,t)f(t) dt$$
using the Lebesgue differential theorem, I get that
$$\lambda f'(x) = \frac{d}{dx} \int_0^1 g(x,t)f(t) dt = \int_0^1 \frac{\partial}{\partial x} g(x,t)f(t) dt.$$
How would I continue from here?
Thank you very much!
|
In order to show that your operator is compact, show that it maps a bounded sequence $\{ f_{n} \}_{n=1}^{\infty}\subset C[0,1]$ to an equicontinuous sequence of functions. So, let $\{ f_{n} \}_{n=1}^{\infty}$ satisfy $\|f_{n}\|_{C[0,1]}\le M$ for all $n$ and some fixed $M$; then, for every $\epsilon > 0$, show that there is a $\delta > 0$ such that
$$
|Tf_{n}(x)-Tf_{n}(y)| < \epsilon
$$
holds for all $n$ whenever $|x-y| < \delta$. Differentiability is a stronger condition than what you need, but it makes the proof a little easier:
$$
\begin{align}
|Tf_{n}(x)-Tf_{n}(y)| & \le \int_{0}^{1}|g(x,t)-g(y,t)||f_{n}(t)|\,dt \\
& \le M \int_{0}^{1}|g(x,t)-g(y,t)|\,dt \\
& \le M \int_{0}^{1}\left|\int_{x}^{y}\frac{\partial g}{\partial u}g(u,t)\,du\right|\,dt.
\end{align}
$$
Let $L$ be a bound for $\frac{\partial g}{\partial u}$ on $[0,1]\times[0,1]$, and you get
$$
|Tf_{n}(x)-Tf_{n}(y)| \le LM|x-y|.
$$
For the $\epsilon > 0$ which was given, choose $\delta = \frac{\epsilon}{2LM}$ and you get the desired equicontinuity of $\{ f_{n} \}_{n=1}^{\infty}$, i.e., that $|Tf_{n}(x)-Tf_{n}(y)| < \epsilon$ for all $n$ whenever $|x-y| < \delta$.
As you suggest, once you know $T$ is compact, you should be able to say more about the spectrum.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/926187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
STEPS: Proving $\displaystyle \lim_{x\to a} f(x)g(x) = -\infty$ Given $\displaystyle \lim_{x \to a} f(x) = \infty$ and $\displaystyle\lim_{x \to a} g(x) = c$ where $c<0$.
Prove that $\displaystyle \lim_{x \to a} f(x)g(x) = -\infty$ only using the precise definitions of limit and infinite limit.
I get the intuitive idea. But... how do we actually write it out?
|
These exercises are almost always solved in a standard way.
*
*Write down the assumptions, using their definition. For instance, $\lim_{x \to a} f(x)=\infty$ means: for every $M>0$ there exists $\delta >0$ with the property that $0<|x-a|<\delta$ implies $f(x)>M$. Now please write down the meaning of your second assumption about $g$.
*Write down what you need to prove. In you case, you need to prove that, for every $M>0$ there exists $\delta >0$ such that $0<|x-a|<\delta$ implies $f(x)g(x)<-M$.
*Try to use the information provided by the assumptions to conclude.
To give you even more hints, you'll almost surely have to play with two numbers $\delta_1$ and $\delta_2$ coming from the assumptions. You'll probably want to select $\delta = \min \{\delta_1,\delta_2\}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/926280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Finding marginal density from a joint density when range of random variables are dependent on one another. I have two joint density problems, where I would like to find the marginal density.
The first one: $f(x,y) = 24xy, 0 \leq x \leq 1, 0 \leq y \leq 1, 0 \leq x+y \leq 1$
So, I "integrate out y" and get $f_x(x)=\int_{-\infty}^{\infty} 24xydy =12x $
I don't feel like this is right because I no longer have an handle on the extra condition of $x+y\leq1$, and if I integrate $12x$ over $[0,1]$, I just would get 6, instead of 1.
The second one: $\frac{1}{8}(y^2-x^2)e^{-y} -y \leq x \leq y, 0 < y < \infty$
Again, integrating out the y here seems to be the wrong approach because I'm not taking into consideration the relationship between x and y.
How would I proceed then in these two cases to find the marginal densities?
Thanks!
|
$$f(x,y)=24xy$$
First integrate with respect to $y$, because $x+y \leq 1$ then $y$ goes from $0$ to $1-x$
$$\int_0^{1-x} 24xy dy=12x(1-x)^2$$
Second to check it's every okay we integrate with respect to $x$:
$$\int_0^1 12x(1-x)^2dx=1$$
As you thought when you are integrating you can take the variable out just when you have independence which if I'm not a mistake only happen when you have a squared domain.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/926362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Probability of "either/or" and "neither" for two independent events This is a problem from GRE quantitative section practice book.
The probability of rain in Greg's town on Tuesday is $0.3$. The probability that Greg's teacher will give him a pop quiz on Tuesday is $0.2$. The events occur independently of each other.
Quantity A
The probability that either or both events occur
Quantity B
The probability that neither event occurs
First, let the probability of rain $P(R)$ and the probability of pop quiz $P(Q)$.
Then A is asking for $P(R \cup Q)$ so it would be 0.3+0.2-0.3*0.2=0.44
and since B is asking for the complement of A, it would be 0.56.
This is my reasoning, however the solution I'm looking at says both are 0.56. It says that A should be 0.3+0.2+0.06 since the probability of either events occuring is 0.2+0.3 and the probability of both events is 0.06. But I think the interpretation of either or both events is the union of two events. Which interpretation is correct?
|
The interpretation given by the practice exam is incorrect as adding the probability of both events happening overall double counts that event as you've probably already knew.
This interpretation is clearly bogus if you increase both event's probability to 50%: Going by the practice book's explanation, then the probability of both events happening should be 125% which is obviously not true.
Which practice book is this coming off from btw?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/926430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Summation of series Find $\sum_1^n$ $\frac {2r+1}{r^2(r+1)^2}$
Also, find the sum to infinity of the series.
I tried decomposing it into partial fractions of the form $\frac Ar$ + $\frac{B}{r^2}$ + $\frac{C}{(r+1)}$ + $\frac{D}{(r+1)^2}$ but it was getting too complicated and tedious. Is there some trick here that i'm missing?
|
$$
\frac1{r^2} - \frac1{(r+1)^2} = \frac{(r+1)^2-r^2}{r^2(r+1)^2} = \frac{(r^2+2r+1)-r^2}{r^2(r+1)^2} = \frac{2r+1}{r^2(r+1)^2}
$$
The thing that suggested this to me is that $\displaystyle 2r+1 = 2\left(r+\frac12\right)$, and $r+\dfrac12$ is half-way between $r+0$ and $r+1$, the two expressions in the denominator.
Next, you rely on the fact that the sum telescopes, so nearly all of the terms vanish.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/926557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
If $c\mid ab$ and $\gcd(c,a)=d$, then $c\mid db$ I came across this problem in my number theory text and am having a bit of trouble with it:
Prove if $c\mid ab$ and $\gcd(c,a)=d$, then $c\mid db$.
Here's what I have so far:
If $c\mid ab$, then there exists an integer $x$ such that $cx=ab$.
Because $\gcd(c,a)=d$, $d\mid c$ and $d\mid a$. Let $y$ be such that $dy=a$.
Then,
$$cx=ab=dyb,$$
so $$cx \frac{d}{a} = db.$$
But I don't know how to show that $\frac{dx}{a}$ is an integer. Can anyone please offer ideas on what I've done so far?
|
There are no mistakes in what you've done so far, but I agree that it's not immediately apparent why $dx/a$ is an integer.
I would start by noting that the answer to this problem is well known when $d = 1$, so maybe I should try to turn it into a problem about relatively prime integers. So write $a = da'$ and $c = dc'$, and consider what the gcd of $a'$ and $c'$ is. Then see if you can carry on from there by reformulating the problem using $a'$ and $c'$ rather than $a$ and $c$.
Just a note: It is possible to do this problem without appealing to Bezout's Theorem. In other words, for those familiar with rings, the fact is valid in factorial rings, not just principal ones.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/926615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
}
|
Closed form sum for the series given below? Does the following series have a closed form sum?
$$f(n,r) = \sum_{i=0}^n \binom{r+i}{r}$$
|
Pascal's Identity states that $\dbinom{r+i}{r} + \dbinom{r+i}{r+1} = \dbinom{r+i+1}{r+1}$.
Hence, $\displaystyle\sum_{i = 1}^{n}\dbinom{r+i}{r} = \sum_{i = 1}^{n}\left[\dbinom{r+i+1}{r+1} - \dbinom{r+i}{r+1}\right]$.
This sum telescopes to $\dbinom{r+n+1}{r+1} - \dbinom{r+1}{r+1} = \dbinom{r+n+1}{r+1} - 1$.
Now, add $\dbinom{r}{r} = 1$ to get $\displaystyle\sum_{i = 0}^{n}\dbinom{r+i}{r} =\dbinom{r+n+1}{r+1}$.
Note: This is commonly referred to as the Hockey-Stick Identity.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/926698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Solve system of equations with $\sin$ and $\cos$ Solve system of equations
$\begin{cases}
3x^2 + \sin 2y - \cos y - 3 = 0 \\
x^3 - 3x - \sin y - \cos 2y + 3 = 0
\end{cases}$
I tried to use substitution $x = \cos t$ or sth, but I get literally nothing
|
The equations are $x^2=A$ and $x^3-3x=B$ where
$$A=\frac{3+\cos y - \sin 2y}{3},\quad B=\sin y +\cos 2y - 3.$$
Now note that $x^3-3x=x(x^2-3),$ so one can substitute the value from $x^2=A$ here, obtaining $x(A-3)=B,$ that is, $x=B/(A-3).$ Then putting this $x$ back into $x^2=A$ gives the relation, involving only $A,B,$ that
$$B^2=A(A-3)^2.\tag{1}$$
Here $A,B$ are trig functions of $y$ admitting period $2\pi,$ and a root finder indicates there are four solutions for $y$ namely $y=0.066078,\ 0.523599,\ 2.61799,\ 3.06375.$
In degrees, the second and third appear to be 30 and 150 respectively, while the first and last seem about 3.785 and 175.54 degrees, not that "nice" as solutions. (I haven't checked analytically whether the 30 and 150 degree solutions are exactly correct.)
Anyway, for each of the four possible $y$ values, one can compute its corresponding $x$ from $x=B/(A-3)$
Added: I checked the $y$ being 30 and 150 degree solutions and they are exact, each leading to $x=1.$ That is, in radians, two of the four solutions are $(x,y)=(1,\pi/6),\ (1,5\pi/6).$ Also the two other solutions for $y$ appear to add up to $\pi$ and each of those leads to the same value of $x$ for the other two pairs $(x,y)$ of solutions, though these others are not particularly nice. Going with the equation $(1)$ and trying to say put all in terms of cosine gives a high degree polynomial to solve, I think degree 8 once one uses $\sin^2 y + \cos^2 y=1$ to eliminate the other trig function.
Edit: Correction-- the "other two $y$ solutions" (besides $\pi/6,5\pi/6$) in fact do not add to $\pi$ but to about $3.1298.$ Furthermore their corresponding $x$ values are about $1.1352$ from the smaller $y$ and $0.8481$ from the larger. I noticed this while trying (unsuccessfully) to show that $B^2-A(A-3)^2$ was symmetric around $\pi/2$; it wasn't.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/926789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
}
|
Is it necessarily true that the following intersection has a positive measure? suppose that a set $S$ in $\mathbb{R}$ is a measurable set with measure greater than zero, and let $\mathcal{O}$ be an open cover of $S$, consisting of disjoint open intervals whose existence we know. It is then necessarily true that there exists an open interval $I$ in $\mathcal{O}$ such that the measure of the intersection $S\cap I$ is non-zero? If so, how does one go about proving it? (Any hint would be appreciated.)
Thanks
|
If I understand correctly, you have $S \subset \mathcal{O} := \bigsqcup_n I_n$ (disjoint union). Since $S$ is measurable, $\mu(S)=\sum_n \mu(S \cap I_n)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/926880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
The limit of $f(x)= \sqrt{x^2+4x+3} +x$ as $x\to\infty$ The problem is to find $\lim_{x\to\infty} \sqrt{x^2+4x+3} +x$.
Do I just divide everything by $x^2$ and get limit $= \sqrt{1}+0=1$?
|
According to what was given, with the assumption that the function $f: x \mapsto \sqrt{x^{2} + 4x + 3} + x$ is defined on $\mathbb{R}$, it is concluded that for every $\varepsilon > 0$ there is a real $X$ such that if $x \geq X$ then
$$|\sqrt{x^{2} + 4x + 3} + x - l| \geq \varepsilon,$$
viz, the function in question grows indefinitely as $x$ goes beyond every bound.
For, given any $\varepsilon > 0$ we can choose a real $X$ such that
$$|\sqrt{X^{2} + 4X + 3} + X| \geq \varepsilon,$$
so that for all real $x \geq X$ we have
$$|\sqrt{x^{2} + 4x + 3} + x| \geq \varepsilon$$
that follows from the monotonicity of the function in question.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/926995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Concluding that a function is not analytic at a point? I have the function $f$ defined on $\mathbb{R}$ by:
$f(x)= \begin{cases} 0 & \text{if $x \le 0$} \\ e^{-1/x^2} & \text{if $x > 0$} \end{cases}$
I've inductively proved that $f$ is indefinitely differentiable on $\mathbb{R}$ and proved that $f^{(n)}(0)=0$ for all $n \ge 1$.
From these facts, how do I conclude that $f$ does not have a converging power series expansion for $x$ near the origin? (i.e. $f$ is not analytic at $z = 0$?)
I know that $f$ is analytic at a point $z_0 \in \Omega$ if there exists a power series $\sum a_n(z-z_0)^n$ centered at $z_0$, with positive radius of convergence, such that $$f(z) = \sum_{n=0}^{\infty}a_n(z-z_0)^n $$ for all $z$ in a neighborhood of $z_0$.
Is the conclusion as simple as: because $f^{(n)}(0)=0$ for all $n \ge 1$, a power series expansion centered at $z = 0$ does not exist, because any derivative at $z = 0$ will be equal to 0? If this is correct, then how do I state this formally? Thanks!
|
Well, if you have that $F(x)=\displaystyle \sum_{n=0}^{\infty}\frac{1}{n!}x^n f^n(0)$, then since $f^n(0)=0\;\;\forall n \in \mathbb{N}$, you would have that $f$ is identical zero, and this is not true since your function is not the zero function. So you can not have a expansion on Taylor series centred at zero.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/927109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Attempt to prove that every real number is a limit of a sequence of rational numbers Prove that given a real number $x$, there exists a rational sequence $r_n$ such that $r_n \to x$ as $n$ grows.
Proof: Suppose $x$ is a real number. Then we know by definition, there exists a rational number such that $x < q < x + \frac1n$. Using the same argument. Then, $x < r_n < x + \frac1n$.
Can I say $x\to x$, and $x + \frac1n \to x$ as $n$ grows. Thus by the sandwich theorem, $r_n \to x$?
Or should I start with, let $\varepsilon>0$. Then we need to show $|r_n - x | < \varepsilon$?
Please any feedback/hint or anything to make it better would be really appreciated.
Thank you.
|
Well, you can simply show an example of such sequence for any $r\in\mathbb{R}$:
$r_n=\dfrac{\lfloor{r\cdot10^n}\rfloor}{10^n}$, i.e., the rational number that is given by the first $n$ digits of $r$ on base $10$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/927198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
Show that $\frac{(b+c)^2} {3bc}+\frac{(c+a)^2}{3ac}+\frac{(a+b)^2}{3ab}=1$ If $a^3+b^3+c^3=3abc$ and $a+b+c=0$ show that $\frac{(b+c)^2} {3bc}+\frac{(c+a)^2}{3ac}+\frac{(a+b)^2}{3ab}=1$
|
$a+b+c=0\iff b+c=-a\implies (b+c)^2=a^2$
$\implies\dfrac{(b+c)^2}{3bc}=\dfrac{a^3}{3abc}$
Actually, $b+c=-a\implies(b+c)^3=(-a)^3$
$\implies -a^3=b^3+c^3+3bc(b+c)=b^3+c^3+3bc(-a)\iff\sum a^3=3abc$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/927310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Recursion relation and initial conditions Write a recursion relation and initial conditions for the number of words of length n using the letters A,B,C such that the number of the letter 'A' is even.
(A occur an even number of times)
|
$o_{n}=$ number of words having $n$ letters and with odd number of
letter $A$
$e_{n}=$ number of words having $n$ letters and with even number
of letter $A$
Then $o_{0}=0$ and $e_{0}=1$ (empty word counts).
$o_{n+1}=2o_{n}+e_{n}$
$e_{n+1}=o_{n}+2e_{n}$
Then $o_{n}+e_{n}=3^{n}$ leads to $e_{n+1}=3^{n}+e_{n}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/927409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
}
|
Prove that the binary representation of a number n will use floor(lg(n)) + 1 bits. I'm taking Computer Algorithms class and one of my problems is from Skiena's Algorithm Design Manual, 2-41:
Prove that the binary representation of $n \ge 1$ has $\lfloor \lg n \rfloor +1$ bits ($\lg$ is base 2)
Some base cases:
$n = 1, \lfloor \lg 1 \rfloor + 1 = 1$
$n = 2, \lfloor \lg 2 \rfloor + 1 = 2$
$n = 5, \lfloor \lg 5 \rfloor + 1 = 3$
$n = 15, \lfloor \lg 15 \rfloor + 1 = 4$
I don't know where to go from there though. Any help appreciated.
|
Hint
Note that the binary representation of $2^n$ has $n+1$ bits.
Find then that the binary representation of a sum of $2^{k_i}$ with distinct $k_i$ has $\max_i k_i + 1$ bits.
Finally conclude that any integer in the interval $[2^n, 2^{n+1})$ has a binary representation of exactly $n+1$ bits.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/927468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
}
|
linear solution of curve fitting on multiple linear functions differing by a multiplier I recently posted this question here but I thought this could be of interest also in mathematics, given I found a partially related question here
I am facing the following problem. I know nonlinear least squares can provide a solution but I am wondering if a linear way to solve this data fitting problem may exists.
This is my input dataset:
I've got three different dataset composed of scattered points
I know a linear equation in the form
$$
y(x) = bx^2 + cx
$$
can be used to explain any of my dataset. I know how to fit this function to a given dataset (e.g the blue one) using linear least square, so I could fit the above function separately for each dataset, but I am looking for something different.
In the specific case I have an additional constrain: I also know that the three functions describing the three dataset share the $b$ and $c$ parameters, while they differ by a multiplier, something like this:
$$
\begin{cases}
y(x) = (bx^2 + cx) \; \text{explains the blue data}\\
y(x) = a'(bx^2 + cx) \; \text{for red data}\\
y(x) = a''(bx^2 + cx) \; \text{for green data}
\end{cases}
$$
I am looking for a (if it exists) linear way to solve this problem globally. Obtaining $a', a'', b$ and $c$
Maybe I am missing something so also having a different point of view on how to correctly formulate the problem could help.
Also approximate solutions are welcome...
|
I think it is not possible (but I may be wrong) to solve this directly. But I would do following:
*
*Calculate $b,c$ via least squares with the blue dataset only.
Then iterate over those two steps:
*
*Take $b,c$ from previous step and calculate $a',a''$ via least squares.
*Take $a',b'$ from previous step and calculate $b,c$ via least squares.
I think this will converge pretty quickly when the datapoints are not scattered too much.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/927588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Doob's decomposition and submartingale with bounded increments Let $(X_n)_{n \geq 0}$ be a submartingale defined on some filtered probability space $(\Omega, \mathcal{F}, ({\mathcal{F}}_n)_{n \geq 0}, \mathbb{P})$. It is a standard fact that $X_n = X_0 + M_n + A_n$, where $(M_n)_{n \geq 0}$ is a martingale null at $0$ and $(A_n)_{n \geq 0}$ is an increasing previsible process, also null at $0$.
Suppose that there exists a positive constant c such that $|X_{n+1} - X_n| \leq c$ for all $n \geq 0$. Then we need to prove the following:
\begin{equation}
\bigg\{ \sup_{n \geq 0 } X_n < + \infty \bigg\} \subseteq \bigg( \big\{ (X_n) \text{ converges in } \mathbb{R} \big\} \cap \big\{A_\infty < +\infty \big\} \bigg).
\end{equation}
|
Define a sequence of stopping times $(\tau_k)_k$ by
$$\tau_k := \inf\{n \geq 0; X_n \geq k\}.$$
From
$$M_{n \wedge \tau_k} = X_{n \wedge \tau_k}-X_0 - \underbrace{A_{n \wedge \tau_k}}_{\geq 0} \leq 2k$$
it follows that $(M_{n \wedge \tau_k})_{n \in \mathbb{N}}$ is a martingale which is bounded above. Consequently, by a standard convergence theorem, the limit
$$\lim_{n \to \infty} M_{n \wedge \tau_k}$$
exists almost surely. For $\omega \in \{\sup_n X_n < \infty\}$ we have $\omega \in \{\tau_k = \infty\}$ for $k$ sufficiently large; hence, $\lim_{n \to \infty} M_n(\omega)$ exists. From
$$A_n(\omega) = X_n(\omega)-X_0(\omega)-M_n(\omega) \leq 2 \sup_n X_n(\omega) - \inf_n M_n(\omega)<\infty$$
we see that $(A_n(\omega))_n$ is bounded above. Since $(A_n(\omega))_n$ is increasing, we find that $A_{\infty}(\omega) = \lim_{n \to \infty} A_n(\omega)$ exists. Obviously, this implies that
$$\lim_{n \to \infty} X_n(\omega) = X_0(\omega) + \lim_{n \to \infty} A_n(\omega)+ \lim_{n \to \infty} M_n(\omega).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/927688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
How to show that $a+b> \sqrt{a^2+b^2-ab}, \qquad a, b >0$ How do you show that
$$a+b> \sqrt{a^2+b^2-ab}, \qquad a, b >0$$
I could write $\sqrt{a^2+b^2-ab}=\sqrt{(a+b)^2-3ab}$, but this seems to lead nowhere.
|
$$a+b> \sqrt{a^2+b^2-ab} \iff (a+b)^2>\left (\sqrt{a^2+b^2-ab}\right )^2$$ $$ \iff\quad a^2+2ab+b^2>a^2+b^2-ab$$ $$\iff\quad 3ab>0 \quad \iff\quad ab>0$$ which is correct because of the hypothesis $a,b >0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/927765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
How many of these are surjective? Let $A=\{a,b,c,d\}$ and $B=\{e,f,g\}$.
*
*How many maps are there from A to B?
*How many of these maps are surjective?
$\textbf{Part 1:}$ There are 4 elements in A and 3 elements in B. Thus there are $3^4=81$ maps from A to B.
$\textbf{Part 2:}$ I can listed in order pair form the outputs for the function $f:A \to B$. For example, $f(a)=e$, $f(b)=e$, $f(c)=e$, and $f(d)=e$ can be written $(e,e,e,e)$. Is there a better way of finding how many maps are surjective than this way?
|
If $h:A\rightarrow B$ is a function then it induces a partition $\mathcal P_h$ on $A$. Elements of $\mathcal P_h$ are the sets of the form $\{a\in A\mid h(a)=b\}$ where $b$ belongs to the image of $h$. Notations for this sets are $h^{-1}(\{b\})$ or shortly $h^{-1}(b)$. $\mathcal P_h$ and the image of $h$ have equal cardinality, so if $h$ is surjective then $|\mathcal P|=|B|$. Searching for the number of maps $h:A\rightarrow B$ that are surjective two questions rise:
*
*How many partitions $\mathcal P$ exist on $A$ with $|\mathcal P|=|B|$?
*How many surjections $h:A\rightarrow B$ exist with $\mathcal P_h=\mathcal P$?
Here we drop some generality by assuming that $A$ and $B$ are finite sets. Then the second question is not too difficult: $|B|!$ (so in your case $3!=6$). The first question is more difficult. However, in your case we are dealing with partitions of a set $A$ that contains $4$ elements and the partition itself has $3$ elements. Under these conditions any partition has a determining unique element that has $2$ elements, so the answer is $\binom{4}{2}=6$. Then the total number of surjections is: $$3!\binom{4}{2}=36$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/927871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
2 similar question about how to find the $a_n$s and $b_n$ of a Fourier series Find the terms $b_n,\ n\geq 1$ so that $$x-\frac{\pi}{2}=\sum_{n=1}^{\infty}b_n \sin nx$$ for all $x\in (0,\pi)$.
A similar one:
Find the term $a_n, \ n \geq 0$ so that $$x-\frac{\pi}{2}=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n \cos nx$$ for all $x\in (0,\pi)$.
Could any one give me just a hint because that I think it is rather an algebra problem once you get the trick.
|
How much do you know already?
A $\sin$ series of a function defined on the interval $(0,\pi)$ is the Fourier series of an odd function defined on $(-\pi,\pi)$. The $\cos$ series being the Fourier series of an even function.
If we suppose that we have the full Fourier series of a function $$f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos nx + b_n \sin nx$$
and that certain convergence criteria hold on the series, we can multiply $f(x)$ by one of the trigonometric terms, $\cos kx$ for example, and integrate on the interval $(-\pi,\pi)$.
$$\int_{-\pi}^\pi f(x)\cos kx \, \mathbb{d}x = \frac12 \int_{-\pi}^\pi a_0\cos kx \, \mathbb{d}x + \sum_{n=1}^\infty \int_{-\pi}^\pi a_n \cos nx \cos kx \, \mathbb{d}x + \sum_{n=1}^\infty \int_{-\pi}^\pi a_n \sin nx \cos kx \, \mathbb{d}x$$
But since $$\int_{-\pi}^\pi \cos nx \cos kx \, \mathbb{d}x = 0 \qquad \forall n \ne k$$and$$\int_{-\pi}^\pi \sin nx \cos kx \, \mathbb{d}x = 0 \qquad \forall n,k$$
The RHS of the previous equation is just $$a_k\int_{-\pi}^\pi \cos^2 kx \, \mathbb{d}x = \pi a_n$$
So we get $$a_n = \frac{1}{\pi} \int_{-pi}^\pi f(x)\cos kx \, \mathbb{d}x$$
And when $f(x)$ is an even function on $(-\pi,\pi)$ (as it must be when the series is a cosine series, have all $b_n = 0$, we get $$\frac{2}{\pi} \int_{0}^\pi f(x)\cos kx \, \mathbb{d}x$$
Similar work gets the equivalent result for the $b_n$ in a sine series.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/927951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
How to solve the integral $\int \frac {(x^2 +1)}{x^4- x^2 +1} dx$ I have started this problem but I'm not completely sure I'm going down the right path with it.
So far I have completed the square in the denominator.
$x^4-x^2+1= (x^2-1/2)^2+\frac{3}{4}$
Then, let $u=x^2-\frac{1}{2}$ so $x=\sqrt(u+\frac{1}{2})$
$\int\frac{x^2+1}{x^4-x^2+1}dx = \int\frac{u+\frac{1}{2}+1}{u^2+\frac{3}{4}}du
=\int\frac{u}{u^2+\frac{3}{4}} +\frac{\frac{3}{2}}{u^2+\frac{3}{4}}du$
|
$$\frac{x^2+1}{x^4-x^2+1}=\frac{1+\dfrac1{x^2}}{x^2-1+\dfrac1{x^2}}$$
Now $\displaystyle\int\left(1+\dfrac1{x^2}\right)dx=x-\dfrac1x$
and $\displaystyle x^2-1+\dfrac1{x^2}=\left(x-\dfrac1x\right)^2+2-1$
Hope you can take it from here
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/928040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
}
|
Which of the following series will converge and which one will diverge? Can anyone help me out that which of the following series will converge and which one will diverge, with some explanation?
A) $\sum_{n=1}^\infty \sin\left(\frac{\pi}n\right)$
B) $\sum_{n=1}^\infty (-1)^n \cos\left(\frac{\pi}n\right)$
Thanks a lot.
|
Euler said that if $x$ is an infinitely small positive number than $\sin x=x$. Today we way $\lim\limits_{x\to0}\dfrac{\sin x}x=1$. At any rate we know that if $x$ is a sufficiently small positive number then
$$
\frac x 2 <\sin x < x,
$$
so
$$
\frac\pi2\sum_n \frac 1 n =\sum_n \frac\pi{2n}\le\sum_n \sin\frac \pi n \le \sum_n \frac \pi n = \pi\sum_n \frac 1 n.
$$
Can you do the rest?
As for the other series, the terms don't approach $0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/928115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Alternative definition for order-embedding I realize that the traditional definition for an order-embedding f is that
$a_1 \sqsubseteq a_2 \iff f(a_1) \sqsubseteq f(a_2)$
However, is it also fair to say that if $(A, \sqsubseteq)$ is a lattice, a definition for meet and this partial-order uniquely define one another:
$a_1 \sqsubseteq a_2 \iff a_1 \sqcap a_2 = a_1$
And so an alternative/equivalent way of saying f is an order-embedding might be:
$f(a_1 \sqcap a_2) = f(a_1) \sqcap f(a_2)$
Thank you! Any advice would be helpful.
|
Embeddings are functions from one structure to another which preserve the structure (and do no add additional structure to the image either).
For partially ordered sets this means, as you suggest that $a\sqsubseteq b\iff f(a)\sqsubseteq f(b)$. Lattices have more structure, so we can expect embeddings of lattices to preserve that structure. But note that this requires additional and explicit statements.
Consider the lattices $\mathcal P(X)$ and $\mathcal P(Y)$ ordered by inclusion, where $X=\{1,2\}$ and $Y=\{1,2,3,4\}$. Now define $f\colon\mathcal P(X)\to\mathcal P(Y)$:
$$\begin{align} &f(\varnothing)=\varnothing\\ &f(\{1\})=\{1,4\}\\ &f(\{2\})=\{2,4\}\\ &f(\{1,2\})=\{1,2,3,4\}\end{align}$$
It is easy to see why $f$ is an order embedding, but the lattice structure is all skewered. $\{1\}\cap\{2\}=\varnothing$ but $f(\{1\})\cap f(\{2\})=\{4\}\neq f(\varnothing)$.
So while we can define a lattice structure on $\operatorname{Range}(f)$, this is not the structure inherited from the structure on $\mathcal P(Y)$, so in some sense it's cheating.
To sum up remember the motto, if you want more, assume more.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/928185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
If a subgroup acts transitively on a set, then the index of the subgroup equals the index of the stabilizer? I am trying to prove the following: If a subgroup $H < G$ acts transitively on a set $X$, then $[G:H] = [G_x:H_x]$ for any $x \in X$ ($H_x$ denotes the point stabilizer of $x$ in $H$.) Any hints would be appreciated.
(EDIT: I am mostly interested in the case where G and H are infinite. What if $X$ is infinite?)
|
For the case where $G, H$ can be infinite:
Consider $H< H<G \Rightarrow [H:H_x][G:H] =[G:H_x]$ and $H_x<G_x<G \Rightarrow [G_x:H_x][G: G_x]= [G:H_x]$ .
$\Rightarrow [G_x:H_x][G: G_x] = [H:H_x][G:H]$. Since $[G: G_x]=[H: H_x] = |X|$, we're done.
Note: Can I assume $[G: G_x]=[H: H_x]$ even if $X$ is not finite?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/928304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Computing a messy convolution Consider the functions
$$
x(t) = u(t - \frac{1}{2}) - u(t - \frac{3}{2})
$$
and
$$
h(t) = tu(t)
$$
where $u(t) = 1$ if $t \geq 0$ and $u(t) = 0$ if $t < 0$.
I'm trying to compute
$$
(x*h)(t) = \int\limits_{-\infty}^{\infty} x(t)h(\tau - t) \, d\tau.
$$
However, the problem gets nasty very quickly and ends up as the sum of terrible looking expressions. Could anyone give me an idea on how to simplify this problem, if there is a way?
|
An easy (and pretty standard) way of computing convolutions is by taking the Laplace transform of the functions, multiplying them (convolution transforms to multiplication in Laplace transformed space), and taking the inverse Laplace transform. As for this question $$\eqalign{
& X(s) = {{{e^{ - {s \over 2}}} - {e^{ - {{3s} \over 2}}}} \over s} \cr
& H(s) = {1 \over {{s^2}}} \cr} $$ so that $$X(s)H(s) = {{{e^{ - {s \over 2}}} - {e^{ - {{3s} \over 2}}}} \over {{s^3}}}$$ Now it is easy to see that $$x(t)*h(t) = {1 \over 2}\left( {{{\left( {t - {1 \over 2}} \right)}^2}\mathscr{U}(t - {1 \over 2}) - {{\left( {t - {3 \over 2}} \right)}^2}\mathscr{U}(t - {3 \over 2})} \right).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/928552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Does the series $\sum (1+n^2)^{-1/4}$ converge or diverge? The integral is $\int\left(\,1 + n^{2}\,\right)^{-1/4}\,{\rm d}n$ is not quite possible, so I should make a comparison test.
What is your suggestion?
EDIT:
And what about the series
$$
\sum\left(\, 1 + n^{2}\,\right)^{-1/4} \cos\left(\, n\pi \over 6\,\right)
$$
Does it converge or diverge?
|
For the second part, notice that there is a pattern to the factor of $\cos\left(\frac{\pi n}{6}\right)$ which repeats every 12 terms. Use this pattern to think of the sequence as alternating.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/928663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
}
|
If$(ab)^n=a^nb^n$ & $(|G|, n(n-1))=1$ then $G$ is abelian Let $G$ be a group. If $(ab)^n=a^nb^n$ $\forall a,b \in G$ and $(|G|, n(n-1))=1$ then prove that $G$ is abelian.
What I have proven is that:
If $G$ is a group such that $(ab)^i = a^ib^i$ for three consecutive integers $i$ for all $a, b\in G$, then $G$ is abelian.
A proof of this can be found in the answers to this old question.
|
We can assume that $n>2$. Since $(ab)^n = a^nb^n$, for all $a,b\in G$, we can write $(ab)^{n+1}$ in two different ways:
$$(ab)^{n+1} = a(ba)^nb = ab^na^nb,$$
and
$$(ab)^{n+1} = ab(ab)^n = aba^nb^n.$$
Hence,
$$ab^na^nb = aba^nb^n.$$
Cancel $ab$ on the left and $b$ on the right to obtain
$$b^{n-1}a^n = a^nb^{n-1}.$$
Note that this is true for all $a,b\in G$. (This says that the $n$th power of any element of $G$ commutes with the $(n-1)$st power of any element of $G$.)
Now let $x,y\in G$ be arbitrary; we want to show that $x$ and $y$ commute. Since the order of $G$ is prime to $n$, the $n$th power map $t\mapsto t^n$ on $G$ is bijective, so there exists $a\in G$ such that $x = a^n$. Since the order of $G$ is prime to $n-1$, there exists $b\in G$ for which $y=b^{n-1}$. Therefore, $xy = a^nb^{n-1} = b^{n-1}a^n = yx$. Because $x$ and $y$ were arbitrary, it follows that $G$ is commutative.
ADDED:
Lemma. Let $G$ be a group of finite order $m$, and let $k$ be a positive integer such that $(k,m)=1$. Then the $k$th power map $x\mapsto x^k$ on $G$ is bijective.
Proof.
Since $G$ is finite, it suffices to show that the map $x\mapsto x^k$ is surjective. To this end, let $g\in G$; we show that $g$ is the $k$th power of some element in $G$. Since the order of $g$ divides the order of the group $G$, it follows that $(\left| g\right|, k) = 1$. Therefore, $\langle g\rangle = \langle g^k\rangle$. Hence, there is an integer $r$ for which $g = (g^k)^r = g^{kr} = (g^r)^k$. This completes the proof.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/928772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
}
|
Can I pick any N in the epsilon N - definition? If I were asked to pick a $N$ that would satisfy the definition such that for any $n > N$, the distance between $f(n)$ and the limit would be smaller than a specific epsilon....
...can I just pick $N$'s at random that would satisfy this, and then say that for any $n > N$ it would be true as well?
I think no; just because it works for a given $N$, doesn't mean it'll work for any $n > N$... but then somebody told me that it would work, because, and he said it like this, "once it's close to the limit, it stays close, and only gets closer", so if the distance between $f(N)$ and the limit is smaller than a given epsilon, then that would remain true for any $n>N$.
Is this true? Is he making some assumptions that aren't always given? Or is it complete bollocks?
|
[In the general case]
Suppose you somehow found some $N$ such that for any $n > N$, we have $|f(n)-L| < \epsilon$.
Then any $N' \ge N$ will also "work," that is, for any $n > N'$, we have $|f(n)-L| < \epsilon$.
However, as Jean-Claude pointed out it is not true that $|f(n)-L|<\epsilon$ implies that for all $n' \ge n$ we have $|f(n')-L| < \epsilon$. But, as I am writing this, you have mentioned that $f$ is monotonic, so in that case, this does hold.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/928855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Prove that the the variance estimator $\widehat{\sigma}^2=MSE/(n-2)$ is biased is the simple linear regression model This is in scope of the simple linear model. Im trying to prove that $\mathbb{E}\left(\widehat{\sigma}^2\right) = \sigma^2$ for
$$\widehat{\sigma}^2 = \frac{1}{n-2}\sum^n_{i=1} \left(y_i-\widehat{y}_i\right)^2$$
where $$Y_i\sim N(\beta_0+\beta_1x_i,\sigma^2)$$ and $\widehat{y_i},i=1,2,3,...,n$ are pedricted values and $y_1,y_2,...,y_n$ is a sample from $Y_i$
|
$\newcommand{\b}{\begin{bmatrix}}\newcommand{\eb}{\end{bmatrix}}$
The vector of fitted values
$$
\hat Y = \b \hat y_1 \\ \vdots \\ \hat y_n\eb
$$
is the orthogonal projection of
$$
Y = \b y_1 \\ \vdots \\ y_n \eb
$$
onto the column space of the design matrix
$$
X = \b 1 & x_1 \\ \vdots & \vdots \\ 1 & x_n \eb.
$$
The vector $\hat\varepsilon$ is $Y-\hat Y$. That means $\hat\varepsilon$ is the projection of $Y$ onto the $(n-2)$-dimensional orthogonal complement of that column space. Notice that this latter orthogonal projection maps the expected value
$$
\mathbb E Y = \b 1 & x_1 \\ \vdots & \vdots \\ 1 & x_n \eb \b \beta_0 \\ \beta_1 \eb
$$
to the zero vector. Let $H$ (conventionally called the "hat matrix") be the matrix of the first orthogonal projection above, so that $\hat Y= HY$. Then $\varepsilon = (I-H)Y$. Recall that the $n\times n$ matrix $H$ of rank $2$ is $X(X^TX)^{-1}X^T$ and this is symmetric. It is also idempotent, i.e. $H^2=H$, or in other words, if you project onto a space, and then project that projection onto that same space, then you just get the point you had when you first projected onto the space. Similarly $I-H$ is symmetric and idempotent. So
$$
\varepsilon \sim N_n(0, (I-H)\Big(\sigma^2 I\Big)(I-H)^T) = N_n(0, \sigma^2(I-H)).
$$
You have a normally distributed random vector in an $(n-2)$-dimensional Euclidean space. Its expected value is $0$ and the normal distribution is spherically symmetric. Hence the square of its norm is distributed as $\sigma^2\chi^2_{n-2}$, so the expected value of the square of the norm is $\sigma^2(n-2)$.
Indeed, if you project $Y$ onto that $(n-2)$-dimensional space, getting $\hat\varepsilon$, and then let $U_1,\ldots, U_{n-2}$ be the coordinates of $\hat\varepsilon$ with respect to an orthonormal basis of that space then you have
$$
U_1,\ldots,U_{n-2}\sim\mathrm{i.i.d.}\ N(0,\sigma^2)
$$
and
$$\varepsilon_1^2+\cdots+\varepsilon_n^2=U_1^2+\cdots+U_{n-2}^2.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/928946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
$\nabla \cdot f + w \cdot f = 0$ Let $w(x,y,z)$ be a fixed vector field on $\mathbb{R}^3$. What are the solutions of the equation
$$
\nabla \cdot f + w \cdot f = 0 \, ?
$$
Note that if $w = \nabla \phi $, then the above equation is equivalent to
$$
\nabla \cdot (e^\phi f) = 0,
$$
for which the solutions are of the form $f = e^{-\phi} \nabla \times g$ for some arbitrary $g$.
|
You can try to proceed along the following lines:
\begin{equation*}
(\partial _{\mathbf{x}}+\mathbf{w})\cdot \mathbf{f}=0
\end{equation*}
Special case $\mathbf{w}$ is constant. Then
\begin{equation*}
\exp [-\mathbf{w\cdot x}]\partial _{\mathbf{x}}\exp [+\mathbf{w\cdot x}
]=\partial _{\mathbf{x}}+\mathbf{w}
\end{equation*}
so
\begin{eqnarray*}
\exp [-\mathbf{w\cdot x}]\partial _{\mathbf{x}}\exp [+\mathbf{w\cdot x}%
]\cdot \mathbf{f} &=&0 \\
\partial _{\mathbf{x}}\cdot \{\exp [+\mathbf{w\cdot x}]\mathbf{f}\} &=&0 \\
\exp [+\mathbf{w\cdot x}]\mathbf{f} &\mathbf{=a}&+\partial _{\mathbf{x}%
}\times \mathbf{b(x)} \\
\mathbf{f(x)} &=&\exp [-\mathbf{w\cdot x}]\{\mathbf{a}+\partial _{\mathbf{x}%
}\times \mathbf{b(x)}\}
\end{eqnarray*}
In general you have to find $\mathbf{U}(\mathbf{x})$ ($3\times 3$ matrix)
such that
\begin{equation*}
\mathbf{U}^{-1}(\mathbf{x})\cdot \partial _{\mathbf{x}}\cdot \mathbf{U}(%
\mathbf{x})=\partial _{\mathbf{x}}+\mathbf{w(x})
\end{equation*}
Let
\begin{eqnarray*}
\mathbf{U}(\mathbf{x}) &=&\exp [\mathbf{A(x)}] \\
\partial _{\mathbf{x}}\cdot \mathbf{U}(\mathbf{x}) &=&\mathbf{U}(\mathbf{x}%
)[\partial _{\mathbf{x}}+\{\partial _{\mathbf{x}}\cdot \mathbf{A(x)\}]} \\
\partial _{\mathbf{x}}\cdot \mathbf{A(x)} &=&\mathbf{w(x})
\end{eqnarray*}
Then
\begin{eqnarray*}
(\partial _{\mathbf{x}}+\mathbf{w(x)})\cdot \mathbf{f} &=&\mathbf{U}^{-1}(%
\mathbf{x})\cdot \partial _{\mathbf{x}}\cdot \{\mathbf{U}(\mathbf{x})\cdot
\mathbf{f}\}=0 \\
\partial _{\mathbf{x}}\cdot \{\mathbf{U}(\mathbf{x})\cdot \mathbf{f}\} &=&0
\\
\mathbf{U}(\mathbf{x})\cdot \mathbf{f(x)} &=&\{\mathbf{a}+\partial _{\mathbf{
x}}\times \mathbf{b(x)}\} \\
\mathbf{f(x)} &=&\mathbf{U}^{-1}(\mathbf{x})\cdot \{\mathbf{a}+\partial _{
\mathbf{x}}\times \mathbf{b(x)}\}
\end{eqnarray*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/929032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
}
|
General solution for intersection of line and circle If the equation for a circle is $|c-x|^2 = r^2$ and the equation for the line is $n \cdot x=d $, and assuming that the circle and line intersect in two points, how can I find these points?
Also as kind of a side note, is there also a general formula for obtaining the equation of a circle of the intersection of a plane and sphere in $\mathbb R^3$, using similar equations?
|
Here I present a method, I don't think it was not used yet, but don't usually find it in mathematics textbooks :
$$\begin{cases}
ax + by + c = 0 \\
(x-\alpha)² + (y-\beta)² = R_{0}^2
\end{cases}
$$
As it shows, the first equation describes the line $\Delta$ with a normal vector $\left( \begin{matrix}
a \\
b
\end{matrix} \right) $ and the second equation describes every point lying on the circumference of the circle with radius $R_0^2$ and a center point lying at $(\alpha,\beta)$.
An easy but very helpful transformation is used here : let $X = x-\alpha $ and $Y = y-\beta$, thus reducing the system of equation to :
$$\begin{cases}
aX + bY = C \\
X² + Y² = R_0^2
\end{cases}
$$
with $ C = -c - a\alpha - b \beta$. (this is equivalent to dealing with the same equations with the center of the circle being the origin of our plane).
let $\left( \begin{matrix}
X \\ Y \end{matrix} \right) \left( \begin{matrix} a & b \\ -b & a \end{matrix} \right) = \left( \begin{matrix} C \\ \gamma \end{matrix} \right)$ with $\gamma \in \mathbb R, \gamma = -bX + aY$.
This expression is equivalent to $z' = z(a-ib)$ with $z' = C + i\gamma$ and $z = X + iY$. Using a basic yet a powerful rule of complex numbers we can safely say that :
$$|z'|² = |z|²|a-ib|² \\
\implies C² + \gamma² = (X² + Y²)(a²+b²)$$
and according to our initial system of equations : $X²+Y²=R_0^2$ and $\gamma = (-bX + aY) $.
$$ \implies (-bX+aY)² = R_0^2(a²+b²)-C²$$
hence :
$$\implies (-bX+aY) = \pm \sqrt{R_0^2(a²+b²)-C²}$$
which explains why we will find
$$\begin{cases}
0 \text{ solutions} & \text{if } R_0^2(a²+b²)-C²<0 & \implies R_0<\frac{|a\alpha+b\beta+c|}{\sqrt{a²+b²}} \\
1 \text{ solution} & \text{if } R_0^2(a²+b²)-C²=0 & \implies R_0=\frac{|a\alpha+b\beta+c|}{\sqrt{a²+b²}} \\
2 \text{ solutions} & \text{if } R_0^2(a²+b²)-C²>0 & \implies R_0>\frac{|a\alpha+b\beta+c|}{\sqrt{a²+b²}}
\end{cases}$$
Back to our linear transformation we have now :
$$\left( \begin{matrix}
X \\ Y \end{matrix} \right) \left( \begin{matrix} a & b \\ -b & a \end{matrix} \right) = \left( \begin{matrix} C \\ \pm \sqrt{R_0^2(a²+b²)-C²} \end{matrix} \right)$$
or equivalently expressed as :
$$(X+iY)(a-ib)=(C \pm i \ \sqrt{R_0^2(a²+b²)-C²})$$
Dealing with complex division is way too easy comparing to matrix inversion, and that's why we keep using them, and now we can directly deduce that
$$X+iY = \frac{(C \pm i \ \sqrt{R_0^2(a²+b²)-C²})(a+ib)}{(a²+b²)} \\
\implies \begin{cases}
X = \frac{1}{a²+b²}(aC \mp b\sqrt{R_0^2(a²+b²)-C²}) \\
Y = \frac{1}{a²+b²}(\pm a\sqrt{R_0^2(a²+b²)-C²} +bC)
\end{cases}\\
\implies \begin{cases}
x = \frac{1}{a²+b²}(aC \mp b\sqrt{R_0^2(a²+b²)-C²}) + \alpha \\
y = \frac{1}{a²+b²}(\pm a\sqrt{R_0^2(a²+b²)-C²} +bC) + \beta
\end{cases} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/929193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Automorphisms of the group of integers $\mathbb Z$ Can anyone help me showing $\operatorname{Aut}(\mathbb Z)\simeq \mathbb Z_2$?
I guess I should define an homomorfism $\phi:\mathbb Z\longrightarrow S(\mathbb Z)$ with kernel $2\mathbb Z$ and image $\operatorname{Aut}(\mathbb Z)$. Here $S(\mathbb Z)$ are the bijections of $\mathbb Z$. However I wasn't able to do that.
Any help will be welcome.
|
Let $\varphi \in \operatorname{Aut}(\mathbb Z)$. We have
$$
\varphi(n) = \varphi(\underbrace{1 + \cdots + 1}_{n \text{ times}}) = \underbrace{\varphi(1) + \cdots + \varphi(1)}_{n \text{ times}} = n \ \varphi(1).
$$
Thus, $\varphi$ is completely determined by its value at $1$. Only two values make $\varphi$ surjective, $1$ and $-1$. It follows that $\operatorname{Aut}(\mathbb Z)$ consists of two elements. In other words, $\operatorname{Aut}(\mathbb Z) \cong \mathbb Z / 2\mathbb Z$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/929258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.