Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
Let $x=\frac{1}{3}$ or $x=-15$ satisfies the equation,$\log_8(kx^2+wx+f)=2.$ Let $x=\frac{1}{3}$ or $x=-15$ satisfies the equation,$\log_8(kx^2+wx+f)=2.$If $k,w,f$ are relatively prime positive integers,then find the value of $k+w+f.$
The given equation is $\log_8(kx^2+wx+f)=2$ i.e. $kx^2+wx+f=64$
Since $x=\frac{1}{3}$ or $x=-15$ satisfies the equation,so
$\frac{k}{9}+\frac{w}{3}+f=64.............(1)$
$225k-15w+f=64............................(2)$
I need to find $k,w,f$ from these equations,but i need three equations to find three variables.I am stuck here.
|
Eliminating $f$ gives $44k=3w$. Take $w=44,k=3$. Then we get $f=49$. Note that the general solution is $k=3h,w=44h,f=64-15h$, but the requirement that the numbers are relatively prime positive integers forces $h=1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1720325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Simple asymptotic analysis problem I came across a problem that I tried to formalize as follows:
Let say i have two functions $x(t)$ and $y(t)$ such that for $t \rightarrow t_0$
$$
\left\{
\begin{array}
\;y(t) \rightarrow -\infty \\
x(t) \rightarrow -\infty \\
y(t) = o(x(t))
\end{array}
\right.
$$
Shouldn't be in such case $ 2^{y(t)} = o(2^{x(t)})$ ? If yes I struggle to prove it is true... otherwise what can we say instead? What about if we don't assume $y(t),x(t) \rightarrow - \infty$?
PS. Just to be clear.. by $y(t) = o(x(t))$ I mean that $\frac{y(t)}{x(t)} = 0$, for $t \rightarrow t_0$.
|
No, $ 2^{y(t)} = O(2^{x(t)})$ is false.
Consider the example
$$ \left\{ \begin{array} y(t) = -\frac1{|t-t_0|} \\ x(t) = -\frac2{|t-t_0|} \\ y(t) = \frac12 x(t) = O(x(t)) \end{array} \right. $$
But if we let $u = \frac1{|t-t_0|}$, then
$2^{-u}$ is not $O(2^{-2u})$, although it is $O(\sqrt{2^{-2u}})$: as $u\to\infty$, the ratio of $2^{-u}$ to $2^{-2u}$ exceeds any constant $C$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1720430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Does every topology have a subbasis? I know that every topology is generated by a basis.
Is it true that every topology has a subbasis which generates it? If not, what makes it not possible to "synthesize" a subbasis out of a basis?
Having studied linear algebra, I am intuitively comfortable with the idea of basis, but it is not clear to me why we are interested in subbasis in the first place.
|
Thm:
Any class $\mathcal{A}$ of subsets of a non-empty set $X$ is the subbase for a unique topology on $X$. That is, finite intersections of members of $\mathcal{A}$ form a base for a topology $\mathcal{T}$ on $X$.
Proof:
We show that the class $\mathcal{B}$ of finite intersections of members of $\mathcal{A}$ satisfies the two conditions for it to be a base for a topology on $X$:
$(i)$ $X = \bigcup\{B: B \in \mathcal{B} \} $
$(ii)$ For any $G, H \in \mathcal{B}$, $\:G \cap H$ is the union of members of
$\mathcal{B}$.
Note $X \in \mathcal{B}$, since $X$ by definition is the empty intersection of members of $\mathcal{A}$; so
$X = \bigcup\{B: B \in \mathcal{B} \} $
Furthermore, if $G, H \in \mathcal{B}$, then $G$ and $H$ are finite intersections of members $\mathcal{A}$. Hence $G \cap H$ is also a finite intersection of members of $\mathcal{A}$ and therefore belongs to $\mathcal{B}$.
Accordingly, $\mathcal{B}$ is a base for a topology $\mathcal{T}$ on $X$ for which $\mathcal{A}$ is a subbase.
It is a straight forward exercise to show the uniqueness.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1720524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
}
|
Limit of a spiraling sequence in $\mathbb{R}^2$ I did some serious mistakes while typing a question few minutes ago.
Let $z_0=(0,0)$ and $z_1=(1,0)$ and define the sequence $z_n$, for any $ n \ge 2 $, as the endpoint of a line drawn perpendicular from $z_{n-1}$ and which is half the distance of the line joining $z_{n-2}$ and $z_{n-1}$
And drawn in a rectangular spiral fashion.
E.g., $z_2=(1,1/2), z_3=(3/4,1/2)$ and so on
Clearly convergent.
But What is the limit of $z_{n}$?
|
HINT
You can consider each coordinate separately. Look at the vectors of changes every 2 steps, so you have for $x$, for example:
$1, -1/4, 1/16, -1/64, \ldots$
which you must sum. It is a geometric series with common factor $-1/4$ and initial value of $1$.
Similar picture will be in the $y$ direction.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1720607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Restriction of an isomorphism to an invariant subspace may fail to be surjective
I'm wondering whether the restriction of a vector space automorphism $f : V \to V$ to an invariant subspace $W \subset V$ can fail to be surjective, i.e. $f\vert_W : W \to W$ is not an automorphism.
Clearly, this can only happen if $W$ (and also $V$) is infinite dimensional, since $f\vert_W$ is injective.
I tried to look at $V=l^1(\Bbb N)$ and
$f : (x_1,x_2,x_3, x_4, \dots) \mapsto (x_2,x_1,x_4, x_3, \dots) \in \text{GL}(V)$
or at some $f$ so that $f\vert_W : (x_1,x_2,x_3, x_4, \dots) \mapsto (0,x_1,x_2,x_3, x_4, \dots)$. But I wasn't able to conclude.
Thank you for your help!
|
Try a shift operator on $\ell^1(\mathbb Z)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1720711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Prove that $a-b=b-a\Rightarrow a=b$ without using properties of multiplication. Yesterday my Honors Calculus professor introduced four basic postulates regarding (real) numbers and the operation $+$:
(P1) $(a+b)+c=a+(b+c), \forall a,b,c.$
(P2) $\exists 0:a+0=0+a=a, \forall a.$
(P3) $\forall a,\exists (-a): a+(-a)=(-a)+a=0.$
(P4) $a+b=b+a, \forall a,b.$
And of course, we can write $a + (-b) = a-b$. Then he proposed a challenge, which was to prove that $$a-b=b-a\iff a=b$$ using only these four basic properties. The $(\Leftarrow )$ is extremely easy and we can prove using only (P3), but I'm struggling to prove $(\Rightarrow )$ and I'm starting to think that it is not possible at all.
My question is how to prove $(\Rightarrow )$, or how to prove that proving $(\Rightarrow )$ isn't possible, using only (P1), (P2), (P3), (P4)?
|
You are right, you can't only from those axioms, and here is why.
Consider $A=\mathbb{Z}/2\mathbb{Z}$. Then $1-0=1=0-1$. But $0 \neq 1$.
But you can show this is true if $2 \neq 0$ and your ring is a integral domain, where $2:=1+1$. In fact,
$a-b=b-a \implies a=b-a+b \implies 0=b-a+b-a \implies 0=2b-2a \implies 0=2(b-a) \implies b-a=0 \implies b=a. $
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1720834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
}
|
Determine the quadratic character of $293 \bmod 379$. Determine the quadratic character of 293 mod 379.
Did several other problems like this with 3, 5, 60, -1 and 307 all mod 379 but still having a tough time with this problem. I can post up work from these examples if helpful. Any help is appreciated.
So far I have...
(293/379)=(379/293)=(86/293)...
I'm not really sure how to finish this out, any help is appreciated.
|
$\newcommand{kron}[2]{\left( \frac{#1}{#2} \right)}$
You are trying to determine the Legendre symbol $\kron{293}{379}$. There is a pretty general recipe when trying to compute Legendre symbol $\kron{a}{p}$ when $p$ doesn't divide $a$ and $p$ is an odd prime. The tools available are
*
*Explicitly find all the squares mod $p$, or represent $a$ as a square. (This is only worthwhile for small $p$)
*Use Euler's Criterion to compute $a^{\frac{p-1}{2}} \bmod p$. This is $1$ is $a$ is a square mod $p$ and $-1$ if $a$ is not a square mod $p$.
*Alter $a$ by adding a multiple of $p$.
*Factor $a$ into a product of primes $p_i^{k_i}$. Then $\displaystyle \kron{a}{p} = \prod_i \kron{p_i}{p}^{k_i}$, and one can consider each Legendre symbol separately.
*Use quadratic reciprocity and its supplemental laws. This means that:
*
*When $a$ is an odd prime $q$, you can use $\displaystyle \kron{q}{p} = (-1)^{\frac{p-1}{2}\frac{q-1}{2}}\kron{p}{q},$
*$\kron{2}{p} = 1$ if $p \equiv \pm 1 \pmod 8$ and $-1$ otherwise,
*and $\kron{-1}{p} = 1$ if $p \equiv 1 \pmod 4$ and $-1$ otherwise.
One can always proceed algorithmically by factoring the numerator, using quadratic reciprocity to flip each Legendre symbol, reducing the new numerators mod the denominators, and repeating until the numbers are small enough to analyze by inspection.
Here, $293$ and $379$ are prime. By quadratic reciprocity,
$$ \kron{293}{379} = \kron{379}{293}.$$
Reducing mod $293$, this becomes
$$ \kron{379}{293} = \kron{86}{293} = \kron{2}{293} \kron{43}{293}.$$
As $293 \equiv 5 \pmod 8$, this becomes
$$\kron{2}{293} \kron{43}{293} = - \kron{43}{293}.$$
Applying quadratic reciprocity and reducing mod $43$, this becomes
$$- \kron{43}{293} = - \kron{293}{43} = -\kron{35}{43}.$$
One could factor and repeat. But one might also write this as
$$-\kron{35}{43} = - \kron{-8}{43} = -\kron{-1}{43} \kron{2}{43}^3.$$
As $43$ is $3 \bmod 4$, we know $\kron{-1}{43} = -1$. And as $43$ is $3 \bmod 8$, we know that $\kron{2}{43} = -1$. So this becomes
$$-\kron{-1}{43} \kron{2}{43}^3 = -(-1)(-1)^3 = -1.$$
And so we conclude that $293$ is not a square mod $379$. $\diamondsuit$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1720939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Solving inequality in complex plane I have to graphically represent the following subset in the complex plane being z a complex number:
$A={1<|z|<2}$
However after trying to do it on WolframAlpha it says that "inequalities are not well difined in the complex plane".
What I did previously was solve it like it was a regular inequality system in the real numbers, resulting in $b<-a+4$ and $b>-a+1$
How can I solve this?
|
The geometric interpretation of this would be an open annulus centred at $0$ with inner radius $1$ and outer radius $2$.
To see this, observe that $|z|$ represents the Euclidean distance from $z$ to $0$. The condition $1 < |z|$ restricts $z$ to be outside of the unit disk, and similarly for $|z| < 2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1721039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Uniformly continuous independent of metrics? Let $(X,d)$ and $(Y,e)$ be metric spaces. A map $f:X\to Y$ is uniformly continuous if for each $\epsilon>0$ there exists $\delta >0$ such that whenever $d(x,y)<\delta$ we have $e(f(x),f(y))<\epsilon$.
Suppose $f:X\to Y$ is uniformly continuous w.r.t. $d$ and $e$. My question is, if $d'$ and $e'$ are metrics on $X$ and $Y$, resp., which induce the same topologies as $d$ and $e$, then is $f$ still uniformly continuous w.r.t. $d'$ and $e'$?
|
No. Let $d$ be the usual metric on $\mathbb R$, and let $d'$ be the metric $d'(x,y)=|\arctan(x)-\arctan(y)|$. It's not hard to check that $d'$ induces the same topology on $\mathbb R$ (use the continuity of $\arctan$). Then the identity function on $\mathbb R$ is uniormly continuous when considered as a function $(\mathbb R,d)\to(\mathbb R,d)$, but not when considered as a function $(\mathbb R,d')\to(\mathbb R,d)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1721119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Show that a set of homotopy classes has a single element This is from Munkres section 51 problem 2b
Given spaces $X$ and $Y$ let $[X,Y]$ denote the set of homotopy classes of maps of $X$ into $Y$. Show that if $Y$ is path connected, the set $[I,Y]$ has a single element. Here $I=[0,1]$.
My approach to the problem is as follows. Let $f,g \in [I,Y]$ and $f,g:I \to Y$ be continuous functions. First I define a function $F(x,t)=f(x(1-t))$. Hence $F(x,0)=f(x)$ and $F(x,1)=f(0)$. This show that $F$ is a homotopy from $f$ to $e_{f(0)}$. Likewise, define $G(x,t)=g(xt).$ Here $G(x,1)=g(x)$ and $G(x,0)=g(0)$. This shows that $G$ is a homotopy from $g$ to the constant function $e_{g(0)}$. Since $Y$ is path connected we can find a path that forces a homotopy between $e_{g(0)}$ and $e_{f(0)}$. Hence $f\cong e_{f(0)}\cong e_{g(0)} \cong g$.
Is this the right strategy?
|
Your proof is correct!
Just one comment, though: perhaps you can mention why the maps $F$ and $G$ are continuous, for the sake of more clarity.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1721229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
A net in $\mathbb{R}$ Let $\{x_j\}_{j\in J}\subset \mathbb{R}$ be a net, $J$ is a directed set.
If $\{x_j\}_{j\in J}$ does not converge to 0, then there is a subnet$\{x_b\}_{b\in B}$, $B$ is a directed set, that $x_b\rightarrow x,$ where $x$ is either $\infty,-\infty,$ or a nonzero real number.
I see this in a proof of the theorem below:
Any finite dimensional topological vector space has the usual Euclidean topology.
I can only prove that there is a convergent subnet, but not surely has nonzero limit, by using the statement below:
A net $\{x_j\}_{j\in J}$ has a cluster point y if and only if it has a subnet $\{y_\beta\}$ that converges to y.
How to prove there is a convergent subnet with nonzero limit?
Any help would be appreciated.
|
Classical fact: if every subnet (in any space $X$) of a net has itself a subnet that converges to some fixed $p$, then the original net converges to $p$. (This can be proved using the cluster point fact, if you like.)
Suppose we have net $(x_j)_{j \in J}$ that does not converge to $0$. As $[-\infty, +\infty]$ (the two point compactification of the reals) is compact, there is some subnet that converges to some $p \in [-\infty, +\infty]$ (and hence so do all of its subnets). By the first classical fact, we can in fact choose such a subnet for some $p \neq 0$, or else the net would have converged to $0$ in the first place. QED.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1721338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
finding recursive formula and show it converges to a limit Suppose we are playing cards and we start with $1000$ dollars. Every hour we lose $\frac{1}{2}$ of our money and then we buy another $100$ dollars. I am trying to find $x_n$ for the amount of money the player has after $n$ hours.
I think we can just take $x_n = \frac{x_{n-1}}{2} + 100 $
An so, let $L = \lim x_n$. Then $L = \frac{L}{2} + 100 $ and so $L = 200$
Is this correct?
|
What you did is not wrong, but it's not complete.
What you did prove:
If the sequence $x_n$ has a limit, then the limit is equal to $200$.
What you did not prove:
The sequence $x_n$ has a limit.
Also, that's not what the question is asking you. The question says you need to find a formula for $x_n$, not the limit of $x_n$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1721449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Probability for "drawing balls from urn" I'm afraid I need a little help with the following:
In an urn there are $N$ balls, of which $N-2$ are red and the remaining are blue. Person $A$ draws $k$ balls, so that the first $k-1$ are red and the $k$ ball is blue. Now Person B draws m balls:
What's the probability for Person $B$ to draw the last/second blue ball after drawing $(m-1)$ red ones? Meant is, that the drawing stops as soon as the blue ball has been drawn. So the blue one must be the last one.
My ideas:
*
*there are only $(N-k)$ balls in the urn left, of which $(N-2)-(k-1) = N-k-1$ are red and $1$ is blue.
*for $m$ we have: $m \in \{1, \dots, N-k\}$.
Shouldn't that be solved with Hypergeometric Distribution? With this I like to calculate the probability for $(m-1)=l$ successes of red, so that the $m$-th ball is blue:
$$P(X=l) = \dfrac{\dbinom{N-k-1}{m-1} \cdot \dbinom{1}{1}}{\dbinom{N-k}{m}}$$
|
The hypergeometric distribution accounts for the change in probability of success, but counts number of success in finite predetermined sample, which is not what you want. The geometric distribution, which counts number of trials until the first success works in a different setting: constant probability of success and infinitely many draws. So, follow a direct approach instead:
*
*$P(X=1)=\dfrac{1}{N-k}$. This is immediate.
*$P(X=2)$. For this you need to draw a red and then the blue, hence $$P(X=2)=\frac{N-k-1}{N-k}\cdot\frac{1}{N-k-1}=\frac{1}{N-k}$$
(where by now the pattern starts to reveal!)
*$P(X=3)$. For this you need to draw two reds and then the blue, hence $$P(X=3)=\frac{N-k-1}{N-k}\cdot\frac{N-k-2}{N-k-1}\cdot\frac{1}{N-k-2}=\frac{1}{N-k}$$
So, the answer is discrete uniform distribution on $\{1,2,\dots,N-k\}$ (you can continue up to $m=N-k$ to verify this), i.e. $$P(X=m)=\frac{1}{N-k}$$ for any $1\le m\le N-k$.
Another (equivalent) way to think of this process and reach this result is to think as follows: You will order the $N-k$ balls, and you want to know the probability that the blue ball will be in the $m-$ th position Indeed, the probability that it lands in any position from $1$ to $N-k$ is the same, hence the uniform distribution.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1721545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
}
|
Limit evaluation for oscillating function I want to evaluate the following oscillating limit x tends to infinity $$\sin(\sqrt{x+1})-\sin(\sqrt{x})$$
I tried evaluating this limit using trigonometric transformations but didn't arrive at the answer
|
$$\lim_{x\to\infty}\left(\sqrt{x+1}-\sqrt{x}\right)=\lim_{x\to\infty}\left(\sqrt{x+1}-\sqrt{x}\right)\left(\frac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}\right)=$$
$$\lim_{x\to\infty}\frac{\left(\sqrt{x+1}-\sqrt{x}\right)\left(\sqrt{x+1}+\sqrt{x}\right)}{\sqrt{x+1}+\sqrt{x}}=$$
$$\lim_{x\to\infty}\frac{1}{\sqrt{x+1}+\sqrt{x}}=\lim_{t\to\infty}\frac{1}{\sqrt{t}+\sqrt{t}}=$$
$$\lim_{t\to\infty}\frac{1}{2\sqrt{t}}=\frac{1}{2}\lim_{t\to\infty}\frac{1}{\sqrt{t}}=\frac{1}{2}\cdot0=0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1721628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
If $x_{n+1}=f(x_n)$ and $x_{n+1}-x_n\to 0$, then $\{x_n\}$ converges Let $f:[0,1] \rightarrow [0,1]$be a continuous function. Choose any point $x_0 \in [0,1]$ and define a sequence recursively by $x_{n+1}=f(x_n)$. Suppose $\lim_{n \rightarrow \infty}x_{n+1}-x_n =0$, does this sequence converge?
|
Yes. Let $I = \lim_{k \rightarrow \infty} I_k$ where $I_k$ is the closure of $\{x_k,x_{k+1},...\}$. Note that because $x_{k+1}-x_k \rightarrow 0$, $I$ is connected, thus either a singleton or an interval. If $I$ is a singleton, we are done. If $I$ is an interval $(x_-,x_+)$, then $f(x)=x$ on this interval. It then follows, unless $x_k$ is constant for sufficiently large $k$, that $x_k$ is strictly monotone in $k$; and thus that the sequence must converge to some point within $I$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1721731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
}
|
Definite integral of sign function I need to calculte the integral of $F(x)=\text{sign}(x)$ (A partial function)
between $x=-1$ and $x=2$.
Of course we need to seperate the integral between $x>0$ and $x<0$
but is it a case of improper integral ? or just seperate and calculate?
|
Notice:
*
*When $x<0$:
$$\text{sign}(x)=-1$$
*When $x>0$:
$$\text{sign}(x)=1$$
*When $x=0$:
$$\lim_{x\to0^+}\text{sign}(x)=1$$
$$\lim_{x\to0^-}\text{sign}(x)=-1$$
\begin{align}
\int_{-1}^{2}\text{F}(x)\space\text{d}x &= \int_{-1}^{2}\text{sign}(x)\space\text{d}x \\
&= \int_{-1}^{0}\text{sign}(x)\space\text{d}x+\int_{0}^{2}\text{sign}(x)\space\text{d}x \\
&= \int_{-1}^{0}-1\space\text{d}x+\int_{0}^{2}1\space\text{d}x \\
&=-\int_{-1}^{0}1\space\text{d}x+\int_{0}^{2}1\space\text{d}x \\
&= -\left[x\right]_{-1}^{0}+\left[x\right]_{0}^{2}\\ &= -\left(0-\left(-1\right)\right)+\left(2-0\right)\\ &=-1+2 \\
&=1
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1721842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
How do I factorize this numerator? $$\lim_{x\to -3} \frac 1{x+3} + \frac 4{x^2+2x-3}$$
I have the solution I just need to know how i turn that into:
$$\frac {(x-1)+4}{(x+3)(x-1)}$$
I know this might be really simple but I'm not sure how to factorise the numerator.
Thanks in advance!
|
You may use common denominator technique:
$$
\frac 1{x+3} + \frac 4{x^2+2x-3} = \frac {1}{x+3} + \frac {4}{(x+3)(x-1)} = \frac {1 \times (x-1) + 4 \times 1}{(x+3)(x-1)} = \frac{x-1+4}{(x+3)(x-1)}
$$
Take a denominator which divisible to both denominators as the common denominator and divide that by each denominator separately, then multiply its quotient by its numerator.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1721951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
}
|
Solving the following trigonometric equation: $\sin x + \cos x = \frac{1}{3} $ I have to solve the following equation:
$$\sin x + \cos x = \dfrac{1}{3} $$
I use the following substitution:
$$\sin^2 x + \cos^2 x = 1 \longrightarrow \sin x = \sqrt{1-\cos^2 x}$$
And by operating, I obtain:
$$ \sqrt{(1-\cos^2 x)} = \dfrac{1}{3}-\cos x$$
$$ 1 - \cos^2 x = \dfrac{1}{9} + \cos^2 x - \dfrac{2}{3}\cos x$$
$$ -2\cos^2 x + 2/3\cos x +\dfrac{8}{9}=0$$
$$ \boxed{\cos^2 x -\dfrac{1}{3}\cos x -\dfrac{4}{9} = 0}$$
Can I just substitute $\cos x$ by $z$ and solve as if it was a simple second degree equation and then obtain $x$ by taking the inverse cosine? I have tried to do this but I cannot get the right result. If I do this, I obtain the following results:
$$ z_1 = -0.520517 \longrightarrow x_1 = 121.4º\\
z_2= 0.8538509 \longrightarrow x_2 = 31.37º$$
I obtain $x$ from $z$ by taking the inverse cosine.
The correct result should be around 329º which corresponds to 4.165 rad. My question is if what I am doing is wrong because I have tried multiple times and I obtain the same result (or in the worst case, I have done the same mistake multiple times).
|
An even simpler way is to substitute $x \rightarrow y - \frac{\pi}{4}$
You equation is now $$\sqrt{2} \sin(y) = \frac{1}{3}$$
The solutions are
$$\begin{aligned} x & = \arcsin \left( \frac{\sqrt{2}}{6} \right) - \frac{\pi}{4} + 2\pi n \\ x & = \arccos \left( \frac{\sqrt{2}}{6} \right) - \frac{\pi}{4} + 2 \pi n
\end{aligned}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1722068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 7
}
|
$\exists\text{ set }X:X=X^X$? Given sets A and B, define the set $B^A$ to be the set of all functions A $\to$ B.
My question is: Is there a set X such that X = $X^X$?
Has this something to do with the axiom of regularity?
|
Hint:
*
*By a cardinality argument it follows $|X|=1$.
*Now, can such a $X$ satisfy $X = X^X$?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1722182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
}
|
Number of ways to color paired shoes so that each pair will have different color There are 4 pairs of shoes of different sizes. Each of the 8 shoes can be colored with one of the four colors: Black, Brown, White & Red. In how many ways can one color shoes so that in at least three pairs, the left and the right shoes do not have the same color.
My solution:
Any three pair colored in odd fashion (i.e both shoes of same pair are colored with different colors)
$C(4,3) \cdot (4 \cdot 3)^3= 4 \cdot 12^3$
All four pairs having different color:
$(4 \cdot 3)^4 = 12^4$
Adding both we get $16 \cdot 12^3$.
Wrong answer it is.
Help to find the mistake.
|
The number of ways to color the shoes so that three pairs are odd-colored is
$$
N = \text{(number of ways to pick three pairs)} * \text{(number of ways to color each odd pair)}^3 * \text{(number of ways to color the one remaining pair)}
$$
which is
$$
N = C(4,3) * (12)^3 * 4 = 4 * 12^3 * 4.
$$
It looks like you forgot the fact that the remaining pair (whose shoes are the same color) can have any one of four colors.
Combined with the case where all four pairs are odd-colored, the answer then works out to be $28 * 12^3 = 48384$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1722283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Finding ALL solutions of the modular arithmetic equation $25x \equiv 10 \pmod{40}$ I am unsure how to solve the following problem. I was able to find similar questions, but had trouble understanding them since they did not show full solutions.
The question:
Find ALL solutions (between $1$ & $40$) to the equation $25x \equiv 10 \pmod{40}$.
|
Let's use the definition of congruence. $a \equiv b \pmod{n} \iff a = b + kn$ for some integer $k$. Hence, $25x \equiv 10 \pmod{40}$ means $$25x = 10 + 40k$$ for some integer $k$. Dividing each side of the equation $25x = 10 + 40k$ by $5$ yields $$5x = 2 + 8k$$
for some integer $k$. Thus,
$$5x \equiv 2 \pmod{8}$$
Since $\gcd(5, 8) = 1$, $5$ has a multiplicative inverse modulo $8$. To isolate $x$, we must multiply both sides of the congruence $5x \equiv 2 \pmod{8}$ by the multiplicative inverse of $5$ modulo $8$. To find the multiplicative inverse, we use the extended Eucldean algorithm.
\begin{align*}
8 & = 5 + 3\\
5 & = 3 + 2\\
3 & = 2 + 1\\
2 & = 2 \cdot 1
\end{align*}
Working backwards through this partial sequence of Fibonacci numbers to solve for $1$ as a linear combination of $5$ and $8$ yields
\begin{align*}
1 & = 3 - 2\\
& = 3 - (5 - 3)\\
& = 2 \cdot 3 - 5\\
& = 2(8 - 5) - 5\\
& = 2 \cdot 8 - 3 \cdot 5
\end{align*}
Therefore, $1 \equiv -3 \cdot 5 \pmod{8}$. Hence, $-3 \equiv 5^{-1} \pmod{8}$. Since $-3 \equiv 5 \pmod{8}$, we have $5 \equiv 5^{-1} \pmod{8}$. Thus, $5 \cdot 5x \equiv x \pmod{8}$. Hence,
\begin{align*}
5x & \equiv 2 \pmod{8}\\
5 \cdot 5x & \equiv 5 \cdot 2 \pmod{8}\\
x & \equiv 10 \pmod{8}\\
x & \equiv 2 \pmod{8}
\end{align*}
What remains is for you to find the solutions of the congruence $x \equiv 2 \pmod{8}$ such that $0 \leq x < 40$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1722443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Confidence interval for the Proportion Question on my Homework:
The Pew Research Center has conducted extensive research on the young adult population (Pew Research website, November 6, 2012). One finding was that 93% of adults aged 18 to 29 use the Internet. Another finding was that 21% of those aged 18 to 28 are married. Assume the sample size associated with both findings is 500.
Round your answers to four decimal places.
A. Develop a 95% confidence interval for the proportion of adults aged 18 to 29 that use the Internet.
B. Develop a 99% confidence interval for the proportion of adults aged 18 to 28 that are married.
Hate to say it, missed this day of class and i am having a hard time working this one out. Anyone got any clue?
|
The aim of a confidence interval is to estimate an interval for a population parameter using sample information. In this case we are interested in estimating a confidence interval for the population proportion $p$ of adults aged $18$ to $29$ that use the internet. To that end, we draw a sample of size $n=500$ adults with ages in $[18,29]$ and check how many of them are using the internet. This is the sample proportion $\hat{p}$ and it appears that $\hat{p} = 0.93$. A $95\%$ confidence interval for $p$ is then given by (recheck the calculation)
$$\left[\hat{p} - z_{1-\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}},\hat{p} + z_{1-\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right] = [0.9297, 0.93026],$$
where $\alpha = 0.05$ and hence $z_{1-\alpha/2} = 1.96$. The conclusion is that you are $95\%$ confident that the population proportion $p$ lies in the above interval. Note that is not the same as saying 'the probability that $p$ lies in $[0.9297,0.93026]$ is $0.95$'. The latter statement is invalid since $p$ is a number, so it either lies in the interval or it doesn't.
Exercise B. is completely similar with a different $\alpha = 0.01$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1722557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Does the sequence $\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}$ converge? I'm trying to determine if this sequence converges as part of answering whether it's monotonic:
$$
\left\{\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}\right\}
$$
First, I tried expanding it a bit to see if I could remove common factors in the numerator and denominator:
$$
\left\{\frac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...\cdot n}{1\cdot 3\cdot 5\cdot 7\cdot 9 \cdot ...\cdot (2n-1)}\right\}
$$
Second, I tried looking at elements of the sequence with common factors removed:
$$
1, \frac{2}{3}, \frac{2}{5}, \frac{2\cdot 4}{5\cdot 7}, \frac{2\cdot 4}{7\cdot 9}, ...
$$
Third, I tried looking at the elements again as fractions without simplifications:
$$
\frac{1}{1}, \frac{2}{3}, \frac{6}{15}, \frac{24}{105}, \frac{120}{945}, ...
$$
Last, I tried searching for similar questions on Stack Exchange and I found one for $\frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)}$ but I didn't understand how that might apply to my question. So, any hints would be much appreciated.
|
The reciprocal of the term of interest is
$$\begin{align}
\frac{(2n-1)!!}{n!}&=\left(\frac{(2n-1)}{n}\right)\left(\frac{(2(n-1)-1)}{(n-1)}\right)\left(\frac{(2(n-2)-1)}{(n-2)}\right) \cdots \left(\frac{5}{3}\right)\left(\frac{3}{2}\right)\\\\
&=\left(2-\frac{1}{n}\right)\left(2-\frac{1}{n-1}\right)\left(2-\frac{1}{n-2}\right) \cdots \left(\frac{5}{3}\right)\left(\frac{3}{2}\right)\\\\
&\ge \left(\frac32\right)^{n-1}
\end{align}$$
Therefore, we see immediately that the limit of interest is $0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1722673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 0
}
|
A convex subset of normed vector space is path-connected Let $(N, \|\;\|)$ be a normed vector space and $(X,\tau)$ a convex subset of $(N,\|\;\|)$ with its induced topology. Show that $(X,\tau)$ is path-connected, and hence also connected.
What I have done so far is Let $a, b \in N$, then construct $X = \{x: x = at+ (1-t)b, 0\le t \le 1\}$. Let $f$ be a function from $[0,1]$ to $X$ with $f(0) = a$ and $f(1) = b$.
I guess if I define $f(t) = at+(1-t)b$, but can I do this without knowing the form of $\| \; \|$?
|
Let $x\in X$ and $r>0$, and consider the open ball $B(x;r)=\{y\in X:\|x-y\|<r\}$. For any $y,z\in B(x;r)$ and $t\in(0,1)$, we have
\begin{align}
\|ty + (1-t)z - x\| &= \|t(y-x) + (1-t)(z-x)\|\\
&\leqslant t\|y-x\| (1-t)\|z-x\|\\
&<r,
\end{align}
so $B(x;r)$ is convex. Write $f(t) = t(a-b) + t$, then it is clear that $f$ is an affine function. So if $B(x;r)$ is an open ball in $X$ and $u,v\in f^{-1}(B(x;r))$, then $f(tu + (1-t)v)\in B(x;r)$ for any $t\in(0,1)$, from which it follows that $f^{-1}(B(x;r))$ is convex. In particular, for any distinct $x,y\in X$, the preimage of the line segment connecting $x$ and $y$ (excluding the endpoints) $f^{-1}(\{ tx+(1-t)y:0<t<1\}=(0,1)$ is open in $[0,1]$. If $y\in\delta B(x;r)$ then there is a unique point $z\in\delta B(x;r)$ with $\|y-z\|=1$. It follows that $$B(x;r) = \bigcup_{\{y:\|x-y\|=r\}}\{ty+(1-t)z:0<t<1\} $$ is the union of open line segments, and hence $$f^{-1}(B(x;r))=(0,1).$$ It follows that $f$ is continuous and therefore $X$ is path-connected.
By the way, convex sets are actually simply connected, as if $f$ and $g$ are paths in a convex set $X$ then the map $H:[0,1]^2\to\mathbb R$ defined by $$H(s,t)=(1-t)f(s) + tg(s)$$ is a path homotopy.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1722778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
If $dF_p$ is nonsingular, then $F(p)\in$ Int$N$ Here is the problem 4-2 in John Lee's introduction to smooth manifolds:
Suppose $M$ is a smooth manifold (without boundary), $N$ is a smooth manifold
with boundary, and $F: M \to N$ is smooth. Show that if $p \in M$ is a
point such that $dF_p$ is non-singular, then $F(p)\in$ Int$N$.
To approach this problem, let's assume conversely that $F(p)\in$ $\partial N$, then there are two charts $(U,\varphi)$ and $(V, \psi)$(a boundary chart) centered at $p$ and $F(p)$ respectively. I can't see how the non-singularity of $dF_p$ contradicts to the fact that $F(p)$ is in the boundary chart. Maybe the proof by contradiction is not the right way? Thanks in advance!
|
$dF_p$ is nonsingular then there exists open set $U\subset M$ s.t. $F:U\rightarrow F(U)$ is a diffeomorphism Hence $F(p)$ is inerior point in $F(U)$ where $F(U)$ is open in $N$.
(Reference : differential forms and applications - do Carmo 60p.
If $H =\{ x|x_n\leq 0\}$ assume that $V$ is open in $H$. And $f :
V\subset H\rightarrow {\bf R}$ is differentiable if there exists
open $U \supset V$ and differentiable function $\overline{f}$ on $U$
s.t. $\overline{f}|_V=f$ Here $df_p$ is defined as $d\overline{f}_p$
That is, if $p=(0,\cdots,0),\ c(t)=(0,\cdots,0, t),\ -\epsilon \leq
t\leq 0 $, then
$$ df_p e_n:= \frac{d}{dt} f\circ c $$
And assume $F(p)$ is in the boundary That is if $\phi_M$ is chart for $M$, then we have a chart for $N$
$$f:=F\circ
\phi_M : U:=B_\epsilon (0) \subset {\bf R}^n \rightarrow N $$ where $\phi_M(0)=p$ And there exists a chart $$
f_2:=\phi_N : U_2:=B_\delta (0)\cap H \rightarrow N,\ \phi_N(0)=p $$
Let $W:=F\circ \phi_M (U)\cap \phi_N( U_2) $ Then $$ f_2^{-1}\circ f : f^{-1}( W) \rightarrow
H $$ is differentiable map Here $d (f\circ f_2^{-1} )_0 \neq 0$ so that inverse function theorem,
there exists a neighborhood $V$ at $0$ in $ f^{-1}( W)$ s.t. $ f_2^{-1}\circ f$ is diffeomorphic on $V$
That is there exists a curve $c$ in $V$ s.t. $$\frac{d}{dt} f_2^{-1}\circ f \circ c =
e_n \in T_0 H $$ That is $n$-th coordinate of
$f_2^{-1}\circ f \circ c (t)$ is positive for small $0< t$ That is it is outside of
$H$.)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1722883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Where does the basis of row space matrix comes from? So what I know: According to Strang's book the row space basis comes from where the pivots are in the upper triangular. We take those rows and therefore we have our basis. My problem lies here: "The row space of A has the same basis with the U because the two row spaces are the same".
Moreover: "It is true that A and U have different rows, but the
combinations of the rows are identical: same space!"
My $$A= \left(\begin{matrix}2 & 4&5 \\ 8& 0& 3\end{matrix} \right)$$ and $$U= \left(\begin{matrix}1 & 0&0.375 \\ 0& 1& 1.0625\end{matrix} \right)$$
When solving a matrix I found a case where my row space of U was different from A but I wanted to be sure.
So regardless from which matrix I take rows its the same?
e.x: If I say Basis: $$RS= \left(\begin{matrix}1 & 0&0.375 \\ 0& 1& 1.0625\end{matrix} \right)$$ $$=\left(\begin{matrix}2 & 4&5 \\ 8& 0& 3\end{matrix} \right)?$$
|
"It is true that $A$ and $U$ have different rows, but the combinations of the rows are identical: same space!" with
$$A= \left(\begin{matrix}2 & 4&5 \\ 8& 0& 3\end{matrix} \right)$$ and $$U= \left(\begin{matrix}1 & 0&0.375 \\ 0& 1& 1.0625\end{matrix} \right)$$
Answer: Yes: $(8,0,3) = 8 \cdot (1,0,0.375)$ and
$(2,4,5)=2\cdot(1,0,0.375)+4\cdot(0,1,1.0625)$; thus, any linear combination of $(8,0,3)$ and $(2,4,5)$ can be written as a linear combination of $(1,0,0.375)$ and $(0,1,1.6025)$.
Similarly, $\displaystyle(1,0,0.375) = {1\over 8}\cdot(8,0,3)$ and
$$(0,1,1.0625)={1\over4}\cdot (2,4,5) - {1\over16}\cdot (8,0,3). $$
Thus, any linear combination of $(1,0,0.375)$ and $(0,1,1.6025)$ can be written as a linear combination of $(8,0,3)$ and $(2,4,5)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1722979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Calculate sum of infinite series by solving a differential equation Calculate the sum of the infinite series
$$\sum_{n=0}^{\infty}\frac{1}{(3n)!}$$ by solving an aptly chosen differential equation.
I know that one can solve a differential equation by assuming that we can write the solution as a power series in the form
$$y(x)=\sum_{n=0}^\infty a_n(x-x_0)^n$$ and then find all the different values for $a_n$'s. I'm trying to figure out how to to it the other way around? How am I supposed to find the differential equation when I have the infinite sum already?
Update
I've started off supposing there exists some solution, to the differential equation of the form
$$p(x)y''(x)+q(x)y'(x)+r(x)y(x)=0,$$ that can be written as
$$y(x)=\sum_{n=0}^\infty a_n(x-x_0)^n. $$
Since I already know this solution should be of the form $$y(x)=\sum_{n=0}^\infty \frac{1}{(3n)!}$$ I know that $a_n=\frac{1}{(3n)!}, x=1, x_0=0$. Furthermore, writing $$y'(x)=\sum_{n=1}^\infty n\ a_n(x-x_0)^{n-1} \ \ and \ \ y''(x)=\sum_{n=2}^\infty n(n-1)\ a_n(x-x_0)^{n-2}. $$ and filling this in the differential equation I finally end up with the equation
$$\sum_{n=0}^\infty \left(p(x)\frac{(n+2)(n+1)}{(3(n+2))!} + q(x)\frac{(n+1)}{(3(n+1))!}+r(x)\frac{1}{(3n)!} \right)=0$$
I think that all the functions $p(x), q(x) \ $and $ r(x)$ should be evaluated at $x=1$. However, I don't know how to proceed.
|
From a polynomial or power series $f(x)$ you can "punch out" the odd degree terms by taking the even part $\frac12(f(x)+f(-x))$. In a similar fashion you can produce "holes" with period $3$ in the coefficient sequence by combining $f(e^{ik\frac{2\pi}3}x)$, $k=-1,0,1$.
As the remaining coefficients in the given series can be interpreted to be every third coefficient of the exponential series,
it can be written the evaluation at $x=1$ of $$
\frac13(e^x+e^{wx}+e^{\bar wx})
$$
with $w^3=1$, i.e., $w=e^{i\frac{2\pi}3}=\frac12(-1+i\sqrt3)$.
This can now easily be transformed into a linear ODE of order 3. (Which, arguably, is backwards from what is intended for the task. So you got to find an argument directly starting from the periodicity of $n!a_n$.)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1723062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
probability of divisibility by $5$ Let $m,n$ be $2$ numbers between $1-100$ . what is the probability that if we select any two random numbers then $5|(7^m+7^n)$ . My attempt last digit should be $5$ or $0$ so $7$ powers follow the pattern $7,9,3,1,7...$ so $m,n$ should be such that if one gives $7$ as last digit other should give $3$ as last digit. Same for $9,1$ . but this will take time and it was asked in a competitive exam with little space for Rough work so i believe there exists a simple logical way. Hope you guys help.
|
As you would have noticed by now, $7^k\equiv7^{k\bmod4}$.
And as you noted, $5|(7^m+7^n)\iff|m-n|\equiv2\pmod4$.
The probability for that in the case of $m,n\in\mathbb{N}$ is obviously $\frac14$.
Since $4|100$, the probability in the case of $m,n\in[1,100]$ is also $\frac14$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1723177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Determining the limits of integration in multiple integrals over delta-functions In using Feynman parametrisation, I have noticed different expressions given in the literature that seem to imply
$$
\int_0^1dx\int_0^1dy\int_0^1dz\delta(1-x-y-z)f(x,y,z)=\int_0^1dx\int_0^{1-x}dyf(x,y,z)|_{z=1-x-y}.
$$
However I have been unable to prove this. Is this statement true and if so why?
|
I have figured out the answer to this. The result is not totally general but may be found in each case by writing the integrals as integrals over an infinite range and putting Heaviside step functions in the integrand to restore the finite range of integration.
The Dirac-deltas change the arguments of the step functions and hence the limits of integration.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1723270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Find $a^{100}+b^{100}+ab$ $a$ and $b$ are the roots of the equation $x^2+x+1=0$.
Then what is the value of $a^{100}+b^{100}+ab$?
Here's what I found out:
$$a+b=-1$$
$$ab=1$$
but how to use this to find that I don't know! Someone please answer my query.
|
The roots of your equation are $$x = -\frac{1}{2} \pm i \frac{\sqrt{3}}{2}$$
We have $e^{\frac{2\pi i}{3}}$ and $e^{\frac{4\pi i}{3}}$, both which remain unchanged in magnitude and direction upon exponentiating 100 times as $e^{\frac{2\pi i}{3}} = e^{\frac{200 \pi i}{3}}$ and $e^{\frac{4\pi i}{3}} = e^{\frac{400 \pi i}{3}}$, telling us that $a^{100} = a$ and $b^{100} = b$ and thus $a + b + ab = 0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1723413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
|
How many 20 digit numbers have 10 even and 10 odd digits? How can I perform operations so as to get this value? Number should not have leading zeros.
|
Case 1: First digit is odd
*
*How many choices do we have for this first digit?
There are $5$ ways of choosing a digit for this place from $1,3,5,7,9$
*Now in how many ways can we build the rest of the number?
For the remaining 19 digits, we need to choose 9 positions to be odd and the rest to be even.
Number of ways to choose 9 positions out of 19 is $\binom{19}{9}$. By choosing where we want to place the odd numbers we have also fixed the positions where the even numbers should be places since there are only 10 positions remaining and we want exactly 10 even digits.
Now that we know whether a given position is even or odd, each of the $19$ positions have $5$ choices to choose from. Thus there are $5^{19}$ ways to build the number once we have fixed the positions for the odd digits.
Multiplying everything together, we get $5\binom{19}{9}5^{19} = \binom{19}{9}5^{20}$ for this case.
Case 2: First digit is even
We can do the same for the case when the first digit is even taking care of the fact that this digit cannot be $0$. Thus everything remains the same as in Case 1 except for the fact that we now have $4$ ways to choose the first digit instead of $5$.
Following the steps correctly should give you $4\binom{19}{9}5^{19}$ ways.
Final Answer
Combining our answers in both the cases we get
$$9\times\binom{19}{9}\times 5^{19} = 15857734680175781250$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1723525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Determine the point of intersection between $f(x) = x^2$ and its normal in the point $(a, a^2)$ Determine the point of intersection between $f(x) = x^2$ and its normal in the point $(a, a^2)$
Answer:
This should be easy enough...
$f'(x) = 2x$
The tangent line in the point $(a, a^2)$ is $y - a^2 = 2 (x - a) \rightarrow y = 2x + a^2 - 2a$
The equation for the normal line is: $y - a^2 = -\frac{1}{2}(x - a) \rightarrow y = -\frac{1}{2}x + a^2 + \frac{1}{2}a$
Now to determine the point of intersection we just see when $f(x)$ and the normal line is equal, i.e.
$x^2 = -\frac{1}{2}x + a^2 + \frac{1}{2}a$
But this seems like a nonsense equation...
|
The slope of the normal line is going to be: $-\frac{1}{2a}$.
set $g(x)=-\frac{1}{2a}x+(a^2+\frac{1}{2})$
You want to solve $g(x)=f(x)$.
$-\frac{1}{2a}x+(a^2+\frac{1}{2})=x^2$
$x^2+\frac{1}{2a}x+\frac{1}{16a^2}=(a^2+\frac{1}{2})+\frac{1}{16a^2}$
$x=\pm\sqrt{(a^2+\frac{1}{2})+\frac{1}{16a^2}}-\frac{1}{4a}$
Personally, I prefer vectors, but it amounts to the same thing:
$(a,a^2)+t[-2a,1]=(x,x^2)$
$a^2+t=(a(1-2t))^2$
And you can solve in this way to get a $t$, which I think is more intuitive, and less "symbol pushy"
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1723628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Calculating Edge points of a rectangle in 2D I'm building a computer game and I got stuck during a math calculation:
The game is a 2D game and is based on a Cartesian coordinate system.
I know the coordinates of E and F. From there I know the angle of EF (Also the angle of AB and CD). I also know the length of AB and the length of CD.
I'm having a hard time finding the solution of calculationg A, B, C and D.
Any help would be appreciated
Thanks!
|
Let E be $(0, 0)$:
F = $(EF\cos\theta, EF\sin\theta)$
A = $(-\frac{AB}{2}\cos\theta, \frac{AB}{2}\sin\theta)$
B = $(\frac{AB}{2}\cos\theta, -\frac{AB}{2}\sin\theta)$
C = $(EF\cos\theta + \frac{CD}{2}\cos\theta, EF\sin\theta - \frac{CD}{2}\cos\theta)$
D = $(EF\cos\theta - \frac{CD}{2}\cos\theta, EF\sin\theta + \frac{CD}{2}\cos\theta)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1723755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
Can the system of equations be extracted from its solution? While I was solving the secondary school exam of 2014 I came across a question that states:
After solving those equations: $a_{1}x + b_{1}y = c_{1}$ and $a_{2}x + b_{2}y = c_{2}$, we found that x = $\frac{-7}{\begin{vmatrix}
3 & 1 \\
1 & -2
\end{vmatrix}}$ and y = $\frac{-21}{\begin{vmatrix}
3 & 1 \\
1 & -2
\end{vmatrix}}$, then $c_{1} = .....$ and $c_{2} = .....$
The answer is $c_{1} = 6$ and $c_{2} = -5$ in the ministry's model answers on its website with no steps available ,but it was probably solved by Cramer's rule.
Is this answer right and does $c_{1}$ and $c_{2}$ have only one value?
|
In view of the main determinant:
$$\begin{vmatrix}3 & 1\\1 & -2\end{vmatrix}$$
one may infer that the system is
$$\begin{cases}3x_1+x_2&=&c_1\\1x_1-2x_2&=&c_2\end{cases}$$
Using Cramer's rule (https://en.wikipedia.org/wiki/Cramer's_rule), the two numerators can be written under the form:
$$\begin{vmatrix}c_1 & 1\\c_2 & -2\end{vmatrix}=-7 \ \ \text{and} \ \ \begin{vmatrix}3 & c_1\\1 & c_2\end{vmatrix}=-21$$
This can be transformed into a linear system:
$$\begin{cases}-2c_1-c_2&=&-7\\-c_1+3c_2&=&-21\end{cases}$$
Solving this system gives $c_1=6$ and $c_2=-5$, as indicated.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1723888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Dumb question: what is $\sum\limits_{n = 1}^k 1$ Let $\{I_n\}$ be a collection of intervals on $\Bbb {R}$, whose length denoted by $I_n = |I_n|$
Then what is $\sum\limits_{n = 1}^k (I_n + \alpha)$, where $\alpha$ is some real number?
The notation confuses me:
$$\sum_{n = 1}^k (I_n + \alpha) = \sum_{n = 1}^k I_n + \sum_{n = 1}^k \alpha = \sum_{n = 1}^k I_n + \alpha \sum_{n = 1}^k 1$$
What is $\sum\limits_{n = 1}^k 1$?
Confusion 1: $n$ was an index, there is no index in the $\sum\limits_{n = 1}^k 1$ term...
Confusion 2:$\sum\limits_{n = 1}^k 1 = 1 + 1 +\cdots + 1 = k$? or does that sum term does nothing since it is not summing over something indexed by $n$?$
|
If you translate an interval by $\alpha$, it's length stays the same. Suppose $I=[3,4]$ and $\alpha=5$, $I+\alpha=[8,9]\implies |I+\alpha|=9-8=1=|I|$. Note that $|I+\alpha|\neq |I|+\alpha$.
So
$$
\sum_{n=1}^k |I_n+\alpha|=\sum_{n=1}^k|I_n|=\sum_{n=1}^k|I_1|=|I_1|\sum_{n=1}^k1=k|I
_1|
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1724074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
How to prove the number of poles minus the number of zeros is $2-2g$? I want to show that, for all differentials on the same Riemann surface S the
number of poles minus the number of zeros, counting multiplicities, always equals $2-2g$. It says this can be deduced from the following result:
By Riemann-Roch theorem we know that the dimension of holomorphic differentials on Riemann surface $S$ equals $g(S)$, the genus of $S$, and that the $g$ elements of any basis cannot have common zeros.
I have no idea.
|
The number you want to compute is $$\deg K_S,$$ the degree of the canonical divisor. This appears in the Riemann-Roch formula: $$h^0(K_S)=\deg K_S+1-g+h^0(K_S-K_S).$$ Now you need to use:
*
*your knowledge of $h^0(K_S)=g$, and
*$h^0(K_S-K_S)=h^0(\mathcal O_S)=1$.
Then you get $\deg K_S=2g-2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1724171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Find last two digits of $33^{100} $ Find last two digits of $33^{100}$.
My try:
So I have to compute $33^{100}\mod 100$
Now by Euler's Function $a^{\phi(n)}\equiv 1\pmod{n}$
So we have $33^{40}\equiv 1 \pmod{100}$
Again by Carmichael Function : $33^{20}\equiv 1 \pmod{100}$
Since $100=2\cdot40+20$ so we have $33^{100}=1\pmod{100}$
So last two digits are $01$
Is it right?
|
33^4 = 21 mod 100 ;33^20 = 01 mod 100 ;33^100 = 01 mod 100 ;Yes the
last 02 digits are 01
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1724246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
}
|
What is $\arctan(x) + \arctan(y)$ I know $$g(x) = \arctan(x)+\arctan(y) = \arctan\left(\frac{x+y}{1-xy}\right)$$
which follows from the formula for $\tan(x+y)$. But my question is that my book defines it to be domain specific, by which I mean, has different definitions for different domains:
$$g(x) = \begin{cases}\arctan\left(\dfrac{x+y}{1-xy}\right), &xy < 1 \\[1.5ex]
\pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &x>0,\; y>0,\; xy>1 \\[1.5ex]
-\pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &x<0,\; y<0,\; xy > 1\end{cases}$$
Furthermore, When I plot the function $2\arctan(x)$, it turns out that the book definition is correct. I don't understand how such peculier definition emerges. Thank you.
|
Here is a straightforward (though long) derivation of the piece wise function description of $\arctan(x)+\arctan(y)$.
We will show that:
$\arctan(x)+\arctan(y)=\begin{cases}-\frac{\pi}{2} \quad &\text{if $xy=1$, $x\lt 0$, and $y \lt 0$} \\\frac{\pi}{2} \quad &\text{if $xy=1$, $x\gt 0$, and $y \gt 0$} \\ \arctan\left(\frac{x+y}{1-xy}\right) \quad &\text{if $xy \lt 1$} \\\arctan\left(\frac{x+y}{1-xy}\right)+\pi \quad &\text{if $xy \gt 1$ and $x \gt 0$, $y \gt 0$}\\ \arctan\left(\frac{x+y}{1-xy}\right)-\pi \quad &\text{if $xy \gt 1$ and $x \lt 0$, $y \lt 0$} \end{cases}$
working from these below statements:
*
*$\arctan$ is strictly increasing and $\arctan(0)=0$
*if $x \gt 0$ then $\arctan(x)+\arctan(\frac{1}{x})= \frac{\pi}{2}$ and if $x \lt 0$ then $\arctan(x)+\arctan(\frac{1}{x})=-\frac{\pi}{2}$
*$\text{dom}(\arctan)=\mathbb R$ and $\text{image}(\arctan)=(-\frac{\pi}{2},\frac{\pi}{2})$
Case 1: $xy=1$
Under these circumstances, you can derive that $x\gt 0$ and $y \gt 0 \implies \arctan(x)+\arctan(y)=\frac{\pi}{2}$. Alternatively, $x \lt 0$ and $y \lt 0 \implies \arctan(x)+\arctan(y) = -\frac{\pi}{2}$
Case 2: $xy \lt 1$
Under these circumstances, you can derive that $\arctan(x)+\arctan(y) \in (-\frac{\pi}{2},\frac{\pi}{2})$
Case 3: $xy \gt 1$
There are two subcases to consider. Either $x \gt 0$ and $y \gt 0$ OR $x \lt 0$ and $y \lt 0$.
If $x \gt 0$ and $y \gt 0$, the we argue as follows.
Because $x \gt 0$ and $xy \gt 1$, we necessarily have that $y \gt \frac{1}{x}$. Next, because $x \gt 0$, we must have that $\arctan(x)+\arctan(\frac{1}{x}) = \frac{\pi}{2}$. Given that $\arctan$ is a strictly increasing function, we then have that $\frac{\pi}{2}=\arctan(x)+\arctan(\frac{1}{x}) \lt \arctan(x)+\arctan(y)$. Finally, we know that for any $z \gt 0:\arctan(z) \in (0,\frac{\pi}{2})$.
With the above argument, we conclude that $\arctan(x)+\arctan(y) \in (\frac{\pi}{2}, \pi)$.
A similar argument would lead one to have that if $x \lt 0$ and $y \lt 0$, then $\arctan(x)+\arctan(y) \in (-\pi,-\frac{\pi}{2})$.
With these three cases established, we can now look at the identity:
$$\tan(\alpha+\beta)=\frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}$$
Letting $\alpha=\arctan(x)$ and $\beta=\arctan(y)$, we can rewrite the above identity as:
$$\tan(\arctan(x)+\arctan(y))=\frac{\tan(\arctan(x))+\tan(\arctan(y))}{1-\tan(\arctan(x))\tan(\arctan(y))}=\frac{x+y}{1-xy} \quad (\dagger_1)$$
From our previous work, we know that across all values of $x,y \in \mathbb R$, the sum $\arctan(x)+\arctan(y)$ can take on the values:
*
*$-\frac{\pi}{2}$ or $\frac{\pi}{2}$
*$(-\frac{\pi}{2},\frac{\pi}{2})$
*$(-\pi,-\frac{\pi}{2})$ or $(\frac{\pi}{2},\pi)$
Importantly, we know that $\tan$ is a periodic function (with period of $\pi$) defined on $\cdots (-\frac{3\pi}{2},-\frac{\pi}{2}), (-\frac{\pi}{2},\frac{\pi}{2}),(\frac{\pi}{2},\frac{3\pi}{2})\cdots \quad (\dagger_2)$
Further, by definition, $\arctan$ is the inverse function of $\tan$ when $\tan$ is restricted to $(-\frac{\pi}{2},\frac{\pi}{2})$.
With the above work in mind, how do we now precisely determine what $\arctan(x)+\arctan(y)$ equals?
Firstly, we know from Case 1 that if $x \lt 0, y\lt 0,$ and $xy=1$, then $\arctan(x)+\arctan(y)=-\frac{\pi}{2}$. Similarly, if $x \gt 0, y \gt 0,$ and $xy=1$, then $\arctan(x)+\arctan(y)=\frac{\pi}{2}$
Next, we can use the result from Case 2 and $(\dagger_1)$ to conclude that if $xy \lt 1$, then $\arctan(x)+\arctan(y)=\arctan\left(\frac{x+y}{1-xy}\right)$. This is because the argument of $\tan$, in the instance of Case 2, spans from $(-\pi/2,\pi/2)$, which implies that we are dealing with a $\tan$ function whose domain is restricted to $(-\pi/2,\pi/2)$. Therefore, when we take the $\arctan$ of $\tan(z)$, it simply computes to $z$.
So far, then, we have the piecewise function:
$\arctan(x)+\arctan(y)=\begin{cases}-\frac{\pi}{2} \quad &\text{if $xy=1$, $x\lt 0$, and $y \lt 0$} \\\frac{\pi}{2} \quad &\text{if $xy=1$, $x\gt 0$, and $y \gt 0$} \\ \arctan\left(\frac{x+y}{1-xy}\right) \quad &\text{if $xy \lt 1$}\end{cases}$
For the final part (i.e. consider when $xy \gt 1$), suppose $z=\arctan(x)+\arctan(y) \in (-\pi,-\frac{\pi}{2}) \cup (\frac{\pi}{2},\pi)$. Although it is tempting to take the $\arctan$ of both sides in $(\dagger_1)$, we must remember that $\arctan$ is only the inverse function of $\tan$ restricted to $(-\frac{\pi}{2},\frac{\pi}{2})$. It is therefore invalid to claim that $\arctan\circ \tan(z)=z$. However, what IS valid is to employ the fact that $\tan$ is periodic as identified in $(\dagger_2)$. Therefore, suppose $z \in (-\pi,-\frac{\pi}{2})$, which corresponds to $x \lt 0$ and $y \lt 0$. We know that $\tan(z)=\tan(z+\pi)$, where $z+\pi \in (0,\frac{\pi}{2})$. Returning to $(\dagger_1)$, let us then write:
$$\tan(z+\pi)=\tan(z)=\frac{x+y}{1-xy}$$
We can then apply $\arctan$ to $\tan(z+\pi)$ which yields:
$$z+\pi=\arctan\left(\frac{x+y}{1-xy}\right) \implies \arctan(x)+\arctan(y)=\arctan\left(\frac{x+y}{1-xy}\right)-\pi$$
If $z \in (\frac{\pi}{2},\pi)$, which corresponds to $x \gt 0$ and $y \gt 0$ , a similar result applied to $z-\pi$ will yield:
$$z-\pi=\arctan\left(\frac{x+y}{1-xy}\right) \implies \arctan(x)+\arctan(y)=\arctan\left(\frac{x+y}{1-xy}\right)+\pi$$
With this we can completely fill in our piecewise function as follows:
$\arctan(x)+\arctan(y)=\begin{cases}-\frac{\pi}{2} \quad &\text{if $xy=1$, $x\lt 0$, and $y \lt 0$} \\\frac{\pi}{2} \quad &\text{if $xy=1$, $x\gt 0$, and $y \gt 0$} \\ \arctan\left(\frac{x+y}{1-xy}\right) \quad &\text{if $xy \lt 1$} \\\arctan\left(\frac{x+y}{1-xy}\right)+\pi \quad &\text{if $xy \gt 1$ and $x \gt 0$, $y \gt 0$}\\ \arctan\left(\frac{x+y}{1-xy}\right)-\pi \quad &\text{if $xy \gt 1$ and $x \lt 0$, $y \lt 0$} \end{cases}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1724348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 3
}
|
Find the sum $\sum _{n=1}^{\infty }\left(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}\right)$ $$\sum _{n=1}^{\infty }\left(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}\right)$$
On their own, all three are divergent, so I thought the best way would be to rewrite it as:
$$\frac{\sqrt{n+2}-2\sqrt{n+1}-\sqrt{n}}{\sqrt{n+2}-2\sqrt{n+1}-\sqrt{n}}\cdot \left(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n}\right)=\frac{\left(\left(\sqrt{n+2}-2\sqrt{n+1}\right)^2-n\right)}{\left(\sqrt{n+2}-2\sqrt{n+1}-\sqrt{n}\right)}$$
But that doesn't really make anything simpler.
|
Render the summand as $(\sqrt{n+2}-\sqrt{n+1})-(\sqrt{n+1}-\sqrt{n})$ and use telescoping. The difference between two consecutive square roots goes to zero as the arguments go to infinity. You should be able to see the answer immediately.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1724446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Probability - determine the probability for an event I have this probability question from homework
A system consists of $N$ chips in a parallel way, such that if at least one of the chips are working the system fully operates.
The probability that throughout a work day, a chip will get broken is $\frac{1}{3}$
Note that each of the events, where a chip is broken is an independent event.
At the end of a work day, the system still operates.
What is the probability that chip number 1 is operating (not broken)?
I struggle with understanding how I should approach this, because it seems at first sight that if the events are independent and we have the probability that each chip will be broken, then we could calculate the complement and get the probability that a chip still operates at the end of the day $(1-\frac{1}{3}) = \frac{2}{3}$
But I know it's wrong because I didn't consider all the other events in my calculation which I find hard to do.
Can I get some help how should I approach this?
Thanks.
|
You are looking to calculate $\mathsf P(X_1\mid X\neq 0)$ when $X$ is the count of chips that work at the end of the day, given a rate of failure, $q=1/3$ (i.i.d. for each chip) and $X_1$ is the event that chip one works. It is the probability that a specific chip works when given that at least one chip does.
Your approach needs to determine what kind of probability distribution $X$ has (and hence its probability mass function), then use the definition of conditional probability.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1724548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
Using Fourier Transform to solve an ODE Consider the differential equation
$$f^{iv}+3f^{''}-f=g$$
I have read that taking the Fourier Transform of both sides gives
$$\left(i\lambda\right)^{4}F\left(\lambda\right)+3\left(i\lambda\right)^2F\left(\lambda\right)=G\left(\lambda\right)-F\left(\lambda\right)=G\left(\lambda\right)$$
Im not sure how they have done this. I think it involves using $\mathcal{F}\left[f^{'}\right]=i\lambda\mathcal{F}\left[f\right]$ but I don't know how.
|
You can combine
*
*Linearity of the Fourier transform:
$$\mathcal{F}\left[\sum_{\forall k} g_k\right] = \sum_{\forall k} \mathcal{F}[g_k]$$
*Differentiation becomes multiplication with the frequency:
$$\mathcal{F}[g'] = i\lambda\mathcal{F}[g]$$
First use 1) to separate each term.
Then use 2) as many times you need to get down to $F[f]$ for each individual term.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1724657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How to verify that one of equations in a polynomial system is redundant? I know that system of polynomial equations
$$
p_1(x_1,\dots,x_n)=0,..., p_N(x_1,\dots,x_n)=0
$$
has infinitely many solutions.
I computed some of them numerically and notices that they always satisfy one more polynomial equation
$$
q(x_1,\dots,x_n)=0.
$$
I would like to prove that this is always the case.
Question: Does it exist a method to prove that
that $q(x_1,\dots,x_n)=0$ follows from $p_1(x_1,\dots,x_n)=0,..., p_N(x_1,\dots,x_n)=0$?
Extra info:
in my application $p_1, p_N$ contains terms of degrees $3$ and $0$, and $q$ contains terms of degree $6$,$3$, and $0$.
|
Hint: If you compute the row echelon form of the matrix given by the polynomial system, then the all-zero rows will be the redundant ones.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1724747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
}
|
Find the sum of the series $\sum_{n=1}^{\infty}(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n})$ I need to test the convergence and find the sum of the following series:
$\sum_{n=1}^{\infty}(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n})$
But i am not really sure what kind of series is this?
Since
$2\sqrt{n+1}>\sqrt{n}+\sqrt{n+2} \forall n \in \mathbb{N}$
It means that this is the series with all negative terms, what can i do with that kind of series? Can i use the same tests as for series with all positive terms (non-negative to be precise)? If that is true, then how can i use them?
|
Just write
$$\sqrt{n+2}-2\sqrt {n+1}+\sqrt n=(\sqrt{n+2}-\sqrt {n+1})+(\sqrt n-\sqrt{n+1})$$
and then telescope.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1724923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Integrate $t^2y'=-5ty$ I'm reading the notes and it says:
$t^2y'=-5ty$ where $y=y(t)$
Using direct integration we can conclude that $y=kt^{-5}$
I don't understand how this can be integrated directly. I can see that the solution is correct, but how did they achieve it?
|
Cancel $t$ at first sight. Conveniently separate the $x,y$ variables on either side of the equation.
$$\frac{dy}{dt}=-\frac{5 y }{t} $$
$$\frac{dy}{y} + \frac{5 \,dt }{t} = 0 $$
Both are logs on integration.
$$ y t ^5 = const. $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1725006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
}
|
Even and odd functions in Laurent decomposition I have the following problem and have no idea how to approach it, could anyone give me any hint about it? Thanks!
Suppose that $f(z) = f_0(z) + f_1(z)$ is the Laurent decomposition of an analytic function $f(z)$ on the annulus $\{A < |z| < B \}$. Show that if $f(z) $ is an even function, then $f_0(z)$ and $f_1(z)$ are even functions.
|
Continuing with Martin R's answer, we have that $\lim_{z \to \infty} f_1(z) = 0$, so $f_1$ is bounded.
Also, since $f_0$ is analytic, and hence continuous, on the compact set $|z| < B$, $f_0$ is bounded. Therefore, $h(z)$ is bounded and analytic on the entire plane. Liouville's theorem implies that $h(z)$ is identically zero.
Therefore, both $f_0$ and $f_1$ are even.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1725138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
How to solve $\int \dfrac{x^5\ln\left(\frac{x+1}{1-x}\right)}{\sqrt{1-x^2}} dx$ Consider the integral
$$\int \dfrac{x^5\ln\left(\frac{x+1}{1-x}\right)}{\sqrt{1-x^2}}dx$$
How to start integrating?
Any hint would be appreciated.
|
Using the substitution $ x = \sin(t) $
The integral simplifies to
$$\int \sin^5t \cdot \ln \left[\frac{1+ \sin(t)}{1-\sin(t)}\right] dt$$
Further simplifications yields:
$$ 2\int \sin^5t \cdot \ln[\sec(t)+\tan(t)] dt $$
we recognize $\ln[\sec(t)+\tan(t)]$ as the antiderivative of $\sec(t)$ which suggests integrating by parts with $u = \ln[\sec(t)+\tan(t)]$ and $dv = \sin^5t \cdot dt$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1725465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Give an example that the following condition does not imply WARP I know how to prove that Weak Axiom of Revealed Preference (WARP) implies the following condition: if $a\in B_1, B_1 \subseteq B_2, a\in C(B_2)$, then $a\in C(B_1)$. $C$ here is a notation for choice correspondence. Could someone provide an example that the above condition does not imply WARP? Thanks.
|
Let $B_1 = \{w,x,y\}$ and $B_2 =\{x,y,z\}$. Let $C(B_1)=\{x\}$ and $C(B_2)=\{y\}$. This example satisfies your property (it holds vacuously), but fails WARP.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1725565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Is $\cot(x)=\frac{1}{\tan(x)}$? I just found in some place say that $\cot(x)=\frac{1}{\tan(x)}$.
If you think about it as a function they are definitely not same. but if you think about as a relation between angle in a right angle triangle and its side they will be equal.
so the question is its right to say they are equal or not( at least pedagogically)?
Update:-
Even wolfram alpha and Desmos say they are not equal ( have different domain).
|
The tangent is defined as $$\tan\alpha =\frac{\text{opposite}}{\text{adjacent}}=\frac{a}{b}=\frac{\sin~\alpha}{\cos~\alpha}$$
and the cotangent as
$$\cot\alpha =\frac{\text{adjacent}}{\text{opposite}}=\frac{b}{a}=\frac{\cos\alpha}{\sin\alpha} $$
what leads us to
$$\frac{\cos\alpha}{\sin\alpha} = \tan\left(\frac{\pi}{2} - \alpha\right) = \frac{1}{\tan \alpha} = \cot\alpha$$
so yes, they are equivalent.
The confusing thing about it is, that you would have to consider the roots of the cosine function, if you were evaluating the tangent function in the denominator before evaluating the fraction.
This is actually a big dissent between different (functional) programming languages, as there is actually no "right order" of evaluation.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1725619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
}
|
What is the new probability density function by generating a random number by taking the reciprocal of a uniformly random number between 0 and 1? I have a random number generator which can generate a random number between $0$ and $1$.
I attempt to generate a random number between 1 and infinity, by using that random number generator, but taking the reciprocal of that result.
Is the new generator uniform? Certainly not. Then what is the probability density function of the new generator?
|
Let the old probability density function be $f_1(x)$, and the new one be $f_2(x)$.
We have:$$
\int_1^af_2(x)\mathrm dx=\int_\frac1a^1f_1(x)\mathrm dx
$$where $a>1$.
We also know that $f_1(x)$ is uniform, and spans from $0$ to $1$. Therefore, $f_1(x)=1$ in that interval.
Therefore:$$
\int_1^af_2(x)\mathrm dx=\int_\frac1a^1\mathrm dx=1-\frac1a
$$
Differentiating both sides with respect to $a$:$$
\frac{\mathrm d}{\mathrm da}\int_1^af_2(x)\mathrm dx=\frac{\mathrm d}{\mathrm da}\left(1-\frac1a\right)
$$
Simplifying both sides:$$
f_2(a)=\frac1{a^2}
$$
$\blacksquare$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1725754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
}
|
Simplify $AC'+A'C+BCD'=AC'+A'C+ABD'$ How to prove that $$AC'+A'C+BCD'=AC'+A'C+ABD'$$
approch: a way to demonstrate is expressed in its canonical form.
Any hint would be appreciated.
|
In
$$AC'+A'C+BCD'=AC'+A'C+ABD'$$
If any term out of $AC'$ and $A'C$ becomes $1$, then, both the LHS and RHS will become $1$ and the equality holds.
So, the case left out is when both $AC'$ and $A'C$ are $0$. Then, suppose $A=1$. Then, $C'=0\Rightarrow C=1$. Similarly, if $A=0$, we have $A'=1$ which implies $C=0$.
Thus, $A$ and $C$ must have the same truth value in this case. Thus,
$BCD'$ and $ABD'$ will also have the same truth value. We can thus conclude that the inequality holds in all the cases.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1725814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Estimate the bound of the sum of the roots of $1/x+\ln x=a$ where $a>1$ If $a>1$ then $\frac{1}{x}+\ln x=a$ has two distinct roots($x_1$ and $x_2$, Assume $x_1<x_2$). Show that $$x_1+x_2+1<3\exp(a-1)$$
First I tried to estimate the place of the roots separately. I have got that $x_1\leq \frac{1}{a}$ and $\exp(a-1)<x_2<\exp(a)$. Then I tried to think about $x_1 + x_2$ as a whole. Because $1/x_1+\ln x_1=a$ and $1/x_2+\ln x_2=a$. I tried to express $x_1+x_2$ as a function of $a$, But I failed. I have no idea to solve this problem. Please help me :)
|
I can show the inequality when $a$ is close enough to $1$ (namely, $a\leq 1+ln(5/4)\approx 1.22$) or when $a$ is big enough (namely, $a \geq 1+\ln(5) \approx 2.6$). In the sequel $f(x)$ denotes $x+\ln(\frac{1}{x})$.
When $a$ is close to $1$. Let us put $w=\sqrt{e^{a-1}-1}$. The inequality then becomes $x_1+x_2\leq 2+3w^2$. Now I claim that when $w\in[0,\frac{1}{2}]$, one has
$$
\begin{array}{lcl}
x_1 & \leq & 1-\sqrt{2}w+\frac{4}{3}w^2 \\
x_2 & \leq & 1+\sqrt{2}w+\frac{5}{3}w^2 \\
\end{array}
$$
To check the first claim, let $g(w)=f(1-\sqrt{2}w+\frac{4}{3}w^2)-1-\ln(1+w^2)$. Then the derivative of $g$ is $g'(w)=\frac{w^3(\frac{4}{3}\sqrt{2}w-\frac{34}{9})}{(1+w^2)(1-\sqrt{2}w+\frac{4}{3}w^2)^2}<0$. So $g$ is decreasing on $(0,\frac{1}{2})$, whence $g(w)\leq g(0)=0$, proving the first claim.
To check the second claim, let $h(w)=f(1+\sqrt{2}w+\frac{5}{3}w^2)-1-\ln(1+w^2)$. Then the derivative of $h$ is $h'(w)=\frac{w^2(\frac{-5}{3}\sqrt{2}w^2-\frac{28}{9}w+sqrt(2)))}{(1+w^2)(1+\sqrt{2}w+\frac{5}{3}w^2)^2}$. So $h'$ has a unique zero in $(0,\frac{1}{2})$ ; since $h(0)=0$ and $h(\frac{1}{2})\approx 0.00091 >0$, we have $h(w)\geq 0$ for any $w\in(0,\frac{1}{2})$, proving the second claim.
When $a$ is big enough. Let us put $u=e^{a-1}$. The inequality then becomes $x_1+x_2+1\leq 3u$. Now I claim that when $u\geq 5$, one has
$$
\begin{array}{lcl}
x_1 & \leq & (3-e)u-1 \\
x_2 & \leq & eu \\
\end{array}
$$
To check the first claim, let $k(u)=f((3-e)u-1)-1-ln(u)$. Then the derivative of $k$ is $k'(w)=\frac{-1} {u((3-e)u-1)^2}<0$. So $k$ is decreasing on $(0,\infty)$, and since $k(5)\approx -1.05 <0$, this proves the first claim.
The second claim follows from the identity $f(eu)-1-ln(u)=\frac{1}{eu}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1725944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
}
|
Prove the solution of $f''(x)-4f(x)=0$ is $f(x)=\sum_{p=0}^{\infty} \frac{4^{p+1}}{(2p)!}x^{2p}$ I'm wondering about this question :
We have the differential equation $f''(x)-4f(x)=0$ and we want to find $f$ as a power serie with $f(0)=4$ and $f'(0)=0$. I would like to prove the only solution is $f(x)=\sum_{p=0}^{\infty} \frac{4^{p+1}}{(2p)!}x^{2p}$ as well as express it using usual functions.
With the usual method to solve differential equations with power series I arrive to a recursive relation $(n+2)(n+1)a_{n+2}-4a_n=0$. How should we proceed to come to the final result ?
Thank you
|
From
$$
(n+2)(n+1)a_{n+2}-4a_n=0 \tag1
$$ you deduce, for $n=0,1,2,\ldots$,
$$
a_{n+2}=\frac4{(n+2)(n+1)}a_n \tag2
$$ giving, with $n:=2p$,
$$
\begin{align}
a_{2(p+1)}&=\frac4{(2p+2)(2p+1)}\:a_{2p}
\\\\&=\frac4{(2p+2)(2p+1)}\cdot \frac4{2p(2p-1)}\:a_{2(p-1)}
\\\\&= \cdots
\\\\&=\frac4{(2p+2)(2p+1)}\cdot \frac4{2p(2p-1)}\cdots \frac4{4\times 3}\:\frac4{2\times 1}\:a_0
\\\\&=\frac{4^{p+2}}{(2(p+1))!}
\end{align}
$$ that is
$$
a_{2p}=\frac{4^{p+1}}{(2p)!}
$$
as announced.
Similarly, with $n=2p+1$, one gets
$$a_{2p+1}=0$$
due to $a_1=f'(0)=0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1726041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Given a $3m\times m$ table, how many ways can it be filled by $x$'s & $o$'s such that each column has at least $2$ $x$'s? Looking at this treat I thought it's a pretty easy one. Apparently not :]. Well,
Given a $ 3m \times m $ table, how many ways can we fill it with $x$'s and $o$'s, such that each column has at least $2$ $x$'s?
Well, my idea was pretty simple:
Let's have a look at all the possible ways to fill this table. We have $ 3m^2 $ cells, and 2 options for each. Then we have $ 2^{3m^2} $ ways to fill the table. Than, lets subtract all cases where $x$ appears once. This case looks this way:
$$ \prod_{i=1}^{m} \binom{3m}{1} = \binom{3m}{1} \cdot \binom{3m}{1} \cdot \binom{3m}{1}\dots\binom{3m}{1}$$
Then, we have $1$ single option to fill the table without x's at all. In conclusion, we have $$2^{3m^2}-(3m)^m - 1$$
BUT, looking at the final answers in the homework solutions, it says:
$$ (2^{3m} - 3m - 1)^m $$
Seriously, I don't understand how they got that. Does my way makes sense, or do I miss something here?
|
Given the solution (and working backwards) we understand that the solution considers each column separately:
*
*$2^{3m}$ ways to fill it (without restrictions).
*$3m$ ways with one $x$,
*$1$ way with no $x$
So, there are $(2^{3m}-3m-1)$ ways to fill one column with the given restriction. Since there are $m$ columns which you can fill independently, there are: $$(2^{3m}-3m-1)^m$$ ways to complete the task.
The mistake in your reasoning is that you consider all columns at once. You count the ways that all columns simultaneously have $1$ $x$ or $0$ $x$'s but this ignores cases where for example the first column has no $x$'s and the rest $1$ $x$. You considered the whole table at once, but since the restriction concerns columns and columns can be filled independently, you should have followed the same approach by column-wise (as in the solution).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1726128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Taking inverse Fourier transform of $\frac{\sin^2(\pi s)}{(\pi s)^2}$ How do I show that
$$\int_{-\infty}^\infty \frac{\sin^2(\pi s)}{(\pi s)^2} e^{2\pi isx} \, ds = \begin{cases} 1+x & \text{if }-1 \le x \le 0 \\ 1-x & \text{if }0 \le x \le 1 \\ 0 & \text{otherwise} \end{cases}$$
I know that $\sin^2(\pi s)=\frac{1-\cos(2\pi s)}{2}=\frac{1-(e^{2\pi i s}-e^{-2\pi i s}))/2}{2}$, so
$$\int_{-\infty}^\infty \frac{\sin^2(\pi s)}{(\pi s)^2} e^{2\pi isx} \, ds=2\int_0^\infty \frac{2e^{2\pi isx}-(e^{2\pi is(1+x)}+e^{2\pi i s(-1+x)})}{4\pi^2s^2} \, ds$$
I am also allowed to use the known identity
$$\int_{-\infty}^\infty \frac{1-\cos(a \pi x)}{(\pi x)^2} \, dx = |a|$$
for some real number $a$.
|
Note that we can write
$$\begin{align}
\int_{-\infty}^\infty\frac{\sin^2(\pi s)}{(\pi s)^2}e^{i2\pi sx}\,ds&=\frac12\int_{-\infty}^\infty\frac{1-\cos(2\pi s)}{(\pi s)^2}e^{i2\pi sx}\,ds\\\\
&=\int_0^\infty \frac{1-\cos(2\pi s)}{(\pi s)^2}\,\cos(2\pi sx)\,ds\\\\
&=\int_0^\infty \frac{\cos(2\pi sx)-\frac12\left(\cos(2\pi s(x+1))+\cos(2\pi s(x-1))\right)}{(\pi s)^2}\,ds\\\\
&=\int_0^\infty \frac{\cos(2\pi sx)-1}{(\pi s)^2}\,ds\\\\
&+\frac12\int_0^\infty \frac{1-\cos(2\pi s(x+1))}{(\pi s)^2}\,ds\\\\
&+\frac12\int_0^\infty \frac{1-\cos(2\pi s(x-1))}{(\pi s)^2}\,ds\\\\
&=-|x|+\frac12|x+1|+\frac12|x-1|
\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1726241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
}
|
How many fixed points are there for $f:[0,4]\to [1,3]$ Let , $f:[0,4]\to [1,3]$ be a differentiable function such that $f'(x)\not=1$ for all $x\in [0,4]$. Then which is correct ?
(A) $f$ has at most one fixed point.
(B) $f$ has unique fixed point.
(C) $f$ has more than one fixed point.
Here, $f:[0,4]\to [1,3]\subset [0,4]$ is continuous and $[0,4]$ is compact convex. So by Brouwer's fixed point theorem $f$ has a fixed point. But I am unable to use the condition $f'(x)\not=1$ and how I can conclude that how many fixed points are there for $f$ ?
|
Note that a derivative has the intermediate value property. Therefore if $f'(x)$ is never $1$, then $f'(x)>1$ for all $x$ or $f'(x)<1$ for all $x$. In the first case, the function $g(x)=f(x)-x$ is increasing, so it can have at most one root; in the second case it is decreasing and the same applies.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1726329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
|
How to rewrite $i^3$ I learned that: $\sqrt[3]i=(e^{\frac{\pi}{2}i+2k\pi i})^\frac{1}{3}=e^{\frac{\pi i}{6}+\frac{2}{3}k\pi i}$ for k={0, 1, 2}
Now how about this case:
$i^3=(e^{\frac{\pi}{2}i+2k\pi i})^3=e^{\frac{3\pi i}{2}+6k\pi i}$ for k={???}
Why would it be $+6k\pi i$? Why not $+2k\pi i$? It seems that for example $e^{\frac{3\pi i}{2}+2\pi i}$ should also be a solution?
And, for what $k$ is it valid?
|
When we say $z = re^{ti + 2k\pi i}$ the $2k\pi i$ is just the period and is always there much as "plus a constant"
So really $\sqrt[3]i=(e^{\frac{\pi}{2}i+2k\pi i})^\frac{1}{3}=e^{\frac{\pi i}{6}+\frac{2}{3}k\pi i + 2k'\pi i}$ But $\frac{2}{3}k\pi i$ is now significant and means more that simply "give or take a few periods". But $2k'\pi i$ is redundant. $e^{\frac{\pi i}{6}+\frac{2}{3}k\pi i + 2k'\pi i}= e^{\frac{\pi i}{6}+\frac{2k+6k'}{3}\pi i }$ so we simply don't need to write the period term. So we don't.
To go the other way should really be $i^3=(e^{\frac{\pi}{2}i+2k\pi i})^3=e^{\frac{3\pi i}{2}+6k\pi i+ 2j\pi i}$ to get the "give or take some periods" back in. But that's now excessive. If we set $j = -2k$ we just get back to $6k\pi i+ 2j\pi i=2k\pi i$.
tl;dr
If the tacked on $\gamma \pi i$ is such that $\gamma$ is not an integer multiple of $2$, then $\gamma \pi i$ is significant. If $\gamma$ is an integer multiple of $2$, then $\gamma \pi i$ is not significant.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1726436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
}
|
Reindexing Exponential Generating Function I have an exponential generating function, and I need to double check what the teacher said, because I'm having trouble coming to the same result. Also, I need to verify what I am coming up with, and the reasoning behind it.
My generating function is: $$xe^{2x} = x\sum_{n=0}^{\infty}{2^n \frac{x^n}{n!}}$$
Next, I multiply through with the x:
$$
\begin{align}
&= \sum{2^n \frac{x^{n+1}}{n!}}\\
&=\frac{1}{2}\sum{(n+1)2^{n+1}\frac{x^{n+1}}{(n+1)!}}
\end{align}
$$
And this is where I am stuck. First, how do I index the last sum? Does it go from $n=-1,\text{ or }0$?
And, for the sequence $a(n)$ such that the final sum is $\sum{a(n)\frac{x^n}{n!}}$, what do I write?
Right now, my guess is:
$$
\begin{align}
xe^{2x} &= \sum_{n=1}^{\infty}{n2^{n-1}\frac{x^n}{n!}}\\
\end{align}
$$
Where you can only get away with writing:
$$\sum_{n=0}^{\infty}{...}$$
because the modifier, $n2^{n-1}$ will be zero when $n=0$ anyways...then you can write:
$$a(n) = n2^{n-1}$$
but I think logically, the sum should start from $n=1$ once you multiply the $x$ in there, so does there need to be the step where you handle the a(0) case?
Anyhow, my teacher, in an example, wrote: $a(n) = n2^n$ on the board, skipping steps, and I can't get there without adding an extra $2$...and my sum doesn't start from zero unless I do the "it'll be zero anyway" part.
update
I went here first; then I remembered Wolfram Alpha. Anyhow, I confirmed that the sequence $a(n)$ is $n2^{n-1}$ with the following two searches:
taylor series for xe^(2x)
n(2^(n-1))/(n!) for n = {0,1,2,3,4,5}
So I guess what I am looking for is a rationale for how to index the sum at zero rather than 1 after multiplying the $x$ through...because I think that the mechanics of doing that are important for being able to define $a(n)$ for all values $n \geq 0$, and that this answer only works out nicely because $x^n$ is multiplied by $n$ (the zero thing I mentioned earlier)...
|
Here is a slightly different answer, which might also be helpful for this and similar tasks.
Situation: The generating function for the exponential series is already known.
\begin{align*}
e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}
\end{align*}
We want to find the exponential series of $xe^{2x}$, i.e. the coefficients $a_n$ of
\begin{align*}
xe^{2x}=\sum_{n=0}^{\infty}a_n\frac{x^n}{n!}\tag{1}
\end{align*}
We obtain
\begin{align*}
xe^{2x}&=x\sum_{n=0}^{\infty}\frac{(2x)^n}{n!}\tag{2}\\
&=x\sum_{n=0}^{\infty}2^n\frac{x^n}{n!}\tag{3}\\
&=\sum_{n=0}^{\infty}2^n\frac{x^{n+1}}{n!}\tag{4}\\
&=\sum_{n=1}^{\infty}2^{n-1}\frac{x^{n}}{(n-1)!}\tag{5}\\
&=\sum_{n=1}^{\infty}n2^{n-1}\frac{x^{n}}{n!}\tag{6}\\
\end{align*}
and have finally reached in (6) a represenation of the form
\begin{align*}
\sum_{n=0}^{\infty}a_n\frac{x^{n}}{n!}\qquad\text{with}\qquad a_n=n2^{n-1}
\end{align*}
Note, that the index in (6) could also start from $0$, since $a_0=0\cdot2^{-1}=0$ .
Coment:
*
*In (2) we use the generating function of $\exp$ from (1)
*In (3) we do some rearrangement to isolate $x^n$
*In (4) we multiply $x^n$ with the factor $x$
*In (5) we shift the index by one to obtain $x^n$ again
*In (6) we multiply with $\frac{n}{n}$ since we want $\frac{x^n}{n!}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1726570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Is there a sequence with an uncountable number of accumulation points? Let $(x_{n})_{n \geq 1}$ sequence in $\mathbb{R}$. Is there a sequence with an uncountable number of accumulation points?
Thank you!
|
Since the rational numbers are countable, we know there is a bijection $f:\mathbb N\to\mathbb Q$. Let $x_n=f(n)$. Then this is a sequence which contains every rational number, and its set of accumulation points is $\mathbb R$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1726710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
}
|
How to find the value of a variable in a probability distribution function? I encountered a mathematics problem that I don't know how to solve.
$P(X=x) = a(\frac{5}{6})^x$ is a probability distribution function for the probability distribution of the discrete random variable $X$ for $x = 0,1,2,3\dots$
I know that the summation of all probabilities will add up to one, but whenever I set up the summation notation, moving the a to the other side creating the fraction $\frac{1}{a}$, I end up with the summation being equal to $0$. I tried to calculate the summation using the geometric series sum formula, but using a first term of $0$ makes this equal to zero.
The answer given by the textbook is $\frac{1}{6}$, but I have no idea how to get there.
|
As you said in your question, we must have
$$ 1=\sum_{x=0}^{\infty}\mathbb{P}(X=x)=a\sum_{x=0}^{\infty}\Big(\frac{5}{6}\Big)^x$$
Using the formula for the sum of a geometric series, we obtain
$$ 1=\frac{a}{1-\frac{5}{6}}=\frac{a}{\frac{1}{6}}$$
so $a=\frac{1}{6}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1726817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Show that if G is a connected simple k-regular graph with k ≥ 2 and χ'(G) = k, then G is Hamiltonian Hi I am really lost on this problem. The notation χ'(G) = k means that the graph has a proper edge coloring of size k. I am only to the point where I know our graph G is comprised of cycles, and has an even number of vertices. If it is K color-able then every vertex has k edges of distinct colors. The edges then must be adjacent to other edges of a different colors incident to a different vertex. I feel intuitively if we follow this pattern we will get a cycle that hits all the vertices. Any insight or help would be very appreciated! Thank you!
|
After a bit of searching, I've found this graph:
which is a cubic, connected, non-Hamiltonian graph with chromatic index $3$.
Another counterexample could be also the Barnette-Bosák-Lederberg Graph which is also planar.
No wonder you can't prove it, the claim is false.
I hope this helps $\ddot\smile$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1726933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
How do I show that $\sum_{k = 0}^n \binom nk^2 = \binom {2n}n$? $$\sum_{k = 0}^n \binom nk^2 = \binom {2n}n$$
I know how to "prove" it by interpretation (using the definition of binomial coefficients), but how do I actually prove it?
|
To prove binomial identities it is often convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$. So, we can write e.g.
\begin{align*}
\binom{n}{k} = [x^k](1+x)^n\tag{1}
\end{align*}
Note, this is essentially the same approach as that of @Batominovski but with a somewhat more algebraic look and feel.
We obtain
\begin{align*}
\sum_{k=0}^{n}\binom{n}{k}^2&=\sum_{k=0}^{n}[x^k](1+x)^n[t^k](1+t)^n\tag{2}\\
&=[x^0](1+x)^n\sum_{k=0}^{n}x^{-k}[t^k](1+t)^n\tag{3}\\
&=[x^0](1+x)^n\left(1+\frac{1}{x}\right)^n\tag{4}\\
&=[x^n](1+x)^{2n}\tag{5}\\
&=\binom{2n}{n}
\end{align*}
Comment:
*
*In (2) we use the coefficient of operator twice according to (1)
*In (3) we use the linearity of the coefficient of operator and the rule $$[x^n]x^{-k}P(x)=[x^{n+k}]P(x)$$
*In (4) we use the substitution rule and apply $x\rightarrow \frac{1}{x}$
\begin{align*}
P(x)&=\sum_{k=0}^na_kx^k=\sum_{k=0}^nx^k[t^k]P(t)\\
P\left(\frac{1}{x}\right)&=\sum_{k=0}^na_kx^{-k}=\sum_{k=0}^nx^{-k}[t^k]P(t)
\end{align*}
*In (5) we use $\left(1+\frac{1}{x}\right)^n=x^{-n}(1+x)^n$ and $[x^0]x^{-n}=[x^n]$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1727088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
}
|
What is $2^{\frac{1}{4}}\cdot4^{\frac{1}{8}}\cdot8^{\frac{1}{16}}\cdot16^{\frac{1}{32}}\cdots\infty$ equal to? I came across this question while doing my homework:
$$\Large 2^{\frac{1}{4}}\cdot4^{\frac{1}{8}}\cdot8^{\frac{1}{16}}\cdot16^{\frac{1}{32}}\cdots\infty=?$$
$$\small\text{OR}$$
$$\large\prod\limits_{x=1}^{\infty} (2x)^{\frac{1}{4x}} = ?$$
My Attempt:
$\large 2^{\frac{1}{4}}\cdot4^{\frac{1}{8}}\cdot8^{\frac{1}{16}}\cdot16^{\frac{1}{32}}\cdots\infty$
$\large \Rightarrow 2^{\frac{1}{4}}\cdot2^{\frac{2}{8}}\cdot2^{\frac{3}{16}}\cdot2^{\frac{4}{32}}\cdots\infty$
$\large \Rightarrow 2^{(\frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \cdots \infty)}$
OR, $\large 2^{\space (\sum\limits_{x=1}^{\infty} \frac{x}{2^{x+1}})}$
$\cdots$ That's it ... I am stuck here ... It does not resemble any series I know...
How do you do it? Thanks!
|
For every $x\in \mathbb R$ which $|x|\lt 1$, we have: $$\sum_{n=0}^\infty x^n=\frac{1}{1-x}$$ From here: $$\sum_{n=1}^\infty nx^{n-1}=\frac{1}{(1-x)^2}$$ Now, multiplying $x^2$ in both side we get that: $$\sum_{n=1}^\infty nx^{n+1}=\frac{x^2}{(1-x)^2}$$ And so: $$\sum_{n=1}^\infty n\left(\frac{1}{2}\right)^{n+1}=1$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1727174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Parametrization of $a^2+b^2+c^2=d^2+e^2+f^2$ Is there an existing parametrization of the equation above that is similar to Brahmagupta's identity for $a^2+b^2=c^2+d^2$? I need either a reference to look it up or a hint to solve it. Thanks.
|
Above equation (a^2+b^2+c^2)=(d^2+e^2+f^2) has parametric solution.
Refer to Tito piezas on line book "collection of algebraic Identities".
Section sum of squares. The answer is given below;
(a,b,c)=[(p+q),(r+s),(t+u)] and
(d,e,f)=[(p-q),(r-s),(t-u)]
Condition is (pq+rs+tu)=0
After parametrization of (pq+rs+tu)=0, the parametric solution in
one variable is given below:
(a,b,c)=[(4k^2+12k+1),(3k-8),(7k+28)] and
(d,e,f)=[(4k^2+12k-13),(k+2),(13k+26)]
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1727303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
is there a ring homomorphism $f:H^*(T^n;R)\to H^*(T^n;R)$ of graded rings which is not induced by a continuous map $T^n\to T^n$? Let $R$ be a commutative ring with unit $1_R$ and let $\xi\in H^1(S^1;R)$ be a generator ($H^1(S^1;R)$ is the first singular cohomology group of $S^1$). Let $p_i:(S^1)^n\to S^1$ be the projection onto the i-th coordinate and $H^1(p_i):H^1(S^1;R)\to H^1((S^1)^n;R)$ it's induced map on singular cohomology. We set $H^1(p_i)(\xi)=:\xi_i$, notice that $(S^1)^n=T^n$ is the n-torus. The cohomology ring of $T^n$, $H^*(T^n;R)$, is isomorphic to the exterior algebra over $R$, $\Lambda _R[\xi_1,..,\xi_n]$ (as graded rings).
My question is: is there a ring homomorphism $f:H^*(T^n;R)\to H^*(T^n;R)$ of graded rings which is not induced by a continuous map $T^n\to T^n$, i.e. does not come from a continuous map $T^n\to T^n$?
Maybe it is sufficient to find out, if every multiplicative map $H^1(T^n;R)\to H^1(T^n;R)$ comes from a continuous map $T^n\to T^n$, because a ring homomorphism of graded rings preserves the grading. But Then I have no idea how to continue. Do you have an idea?
|
When $R=\Bbb Z$ the answer to your question is no:
Since $T^n$ is a $K(\pi_1(T^n),1)$, every group endomorphism of $\pi_1(T^n)$ is induced by a map $T^n\to T^n$. (See for example Prop 1B.9 in Hatcher). Therefore, any endomorphism of $\pi_1(T^n)^{ab}=H_1(T^n)=H^1(T^n)$ is also induced by a map of spaces.
Since $H^*(T^n)\cong \bigwedge_{\Bbb Z}\left[ \alpha_1,\cdots,\alpha_n \right]$ with $\deg \alpha_i=1$, any morphism $H^*(T^n)\to H^*(T^n)$ is determined by what it does in degree $1$, hence it is induced by a map of spaces.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1727381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
General sum of $n$th roots of unity raised to power $m$ comprime with $n$ I am trying to find a reference for the following proposition:
Let $m$ and $n$ be coprime. Then,
$$ \sum_{k=0}^{r-1} \exp\left( i \frac{2\pi}{n} k m \right) = 0 $$
if and only if $r$ is an integer multiple of $n$.
Can anyone point a basic textbook or online material in which this basic fact is proven?
|
It can be proven easily using geometric summation. Observe that
$$ \sum_{k=0}^{r-1} \exp\left(i \frac{2\pi k m}{n} \right) = \sum_{k=0}^{r-1} \exp(2\pi i m/n)^k = \frac{1-\exp(2\pi i m/n)^r}{1-\exp(2\pi i m/n)}
$$
The sum is $0$ if and only if $\exp(2\pi i m/n)^r = 1$, which corresponds to $n|rm$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1727503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Covariance of X and Y on a Quadrilateral How should I go about determining the covariance? Also, how can I use intuition to determine if it should be positive or negative?
|
If $Y$ on average gets bigger as $X$ gets bigger, then the covariance is positive; if $Y$ on average gets smaller as $X$ gets bigger, then the covariance is negative. You're just picking a random point in that quadrilateral and asking that question about the horizontal and vertical coordinates of that point.
The covariance is $\operatorname{E}((X-\mu_X)(Y-\mu_Y))$ where $\mu_X=\operatorname{E}(X)$ and $\mu_Y=\operatorname{E}(Y)$. To find $\mu_X$ and $\mu_Y$, just integrate:
\begin{align}
\mu_X & = \iint_D x \cdot \text{constant} \, dy \, dx, \\[10pt]
\mu_Y & = \iint_D y \cdot \text{constant} \, dy \, dx.
\end{align}
The density is constant because that's what it means to say the distribution is uniform. To find the "constant", you need this:
$$
\iint_D \text{constant} \, dy \, dx = 1.
$$
So the covariance is
$$
\iint_D (x-\mu_X)(y-\mu_Y)\cdot\text{constant} \, dy\,dx.
$$
However, there is a slightly quicker way, if you know this identity:
$$
\operatorname{cov}(X,Y) = \operatorname{E}(XY) - \operatorname{E}(X)\operatorname{E}(Y) = \operatorname{E}(XY) - \mu_X \mu_Y.
$$
Then you have
$$
\operatorname{cov}(X,Y) = \iint_D xy\cdot\text{constant} \, dy\,dx - \mu_X\mu_Y.
$$
Finding the "constant" is easy because the integral of a constant over a region is just that constant times the area of the region, and in this case the area is $3/2$, so you have $(3/2)\cdot\text{constant} = 1$.
Notice I am being careful about the difference between capital $X$ and $Y$ on the one hand, and on the other hand lower-case $x$ and $y$.
Since you say so little in your question, I'm omitting anything about how to find the integrals.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1727608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Poisson Distribution: What's the probability of getting a first week without any events when you are told that 5 events occurred within a month? I'm told that the probability of getting $n$ murders per month in London can be modelled as a Poisson distribution with rate $\lambda$. I'd like to calculate the probability that, in a month with 5 murders, there were no murders within the first week. I thought this would simply equal the probability that there are no murders in a week multiplied by the probability that there are 5 murders in the (remaining) three weeks. Is this correct? If not, what should I be doing instead?
|
First, you have to convert the monthly event rate to a weekly event rate (assuming that there are 4 weeks in a month). So if the monthly event rate is $\lambda$, the weekly event rate is $\lambda/4$. Then you want a conditional probability: if the random number of events per week is $$X \sim \operatorname{Poisson}(\lambda/4),$$ then you want $$\Pr[X_1 = 0 \mid X_1 + X_2 + X_3 + X_4 = 5],$$ where $X_1, X_2, X_3, X_4$ are independent and identically distributed Poisson variables with rate $\lambda/4$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1727722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Bell number as summation of Stirling numbers of the second kind
A different summation formula represents each Bell number as a sum of
Stirling numbers of the second kind
$ B_n=\sum_{k=0}^n \left\{{n\atop k}\right\} $
The Stirling number $\left\{{n\atop k}\right\} $ is the number of ways
to partition a set of cardinality n into exactly k nonempty subsets.
Thus, in the equation relating the Bell numbers to the Stirling
numbers, each partition counted on the left hand side of the equation
is counted in exactly one of the terms of the sum on the right hand
side, the one for which k is the number of sets in the partition.
I was reading wiki and found the summation starts from $0$ instead of $1$. Is not that logically incorrect? Set partition never allowed to be empty. Can anybody clarify?
Thanks :)
|
The Stirling number $n\brace k$ is actually defined for all pairs of non-negative integers. Of course ${n\brace k}=0$ when $k>n$. Since a non-empty set cannot have a partition into $0$ parts, it’s also clear that we want ${n\brace 0}=0$ for $n>0$. It turns out, however, to be convenient to set ${0\brace 0}=1$, as it makes recurrence
$${{n+1}\brace k}=k{n\brace k}+{n\brace{k-1}}$$
behave nicely.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1727835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Minimizing a sum of exponential functions I want to minimize this function:
$$ g_0(\psi)= \sum_{m=0}^{M-1}e^{j(am^2+bm)}e^{jm\psi} $$
where $a$ and $b$ are constants for which I want to minimize the function.
Can anyone help me regarding this. Will some optimization technique work?
|
In your expression:
$$g_0(\psi)= \sum_{m=0}^{M-1}e^{j(am^2+bm)}e^{jm\psi}$$
why dont you express the exponent (or more precisely the coefficient of $j$) in this way:
$$am^2+m(b+\psi)=a\left(m+\dfrac{b+\psi}{2a}\right)^2-\dfrac{(b+\psi)^2}{4a}$$
In this way, you could factor out $e^{-j\frac{(b+\psi)^2}{4a}}$ and concentrate on a simplified sum (that has something to do with a certain $\Theta$ function if $M$ is large https://en.wikipedia.org/wiki/Theta_function).
Regarding optimization, to you want to minimize $|g_0(\psi)|$ (note the $||$ I have placed). Could you give a little hint about the context of appearance of this function? It might help.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1727982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Sum of a series $\frac {1}{n^2 - m^2}$ m and n odd, $m \ne n$ I was working on a physics problem, where I encountered the following summation problem:
$$ \sum_{m = 1}^\infty \frac{1}{n^2 - m^2}$$ where m doesn't equal n, and both are odd. n is a fixed constant
We could alternatively write:
$$ \sum_{j = 1}^\infty \frac{1}{(2i - 1)^2 - (2j - 1)^2}$$ where i doesn't equal j. I used a numerical solver to solve for a few values of n, and the solution seems to be $-1/(4n^2)$ which fits with the problem, but I'd like to find a rigorous solution.
My attempts so far have involved simplifying the term in the sum in various was to see if I could find a way solve the problem, but I've had no luck.
Ex. The second equation can be simplified to
$$\sum_{j=1}^\infty \frac {1} {4(i + j - 1)(i - j)} $$
but I'm not sure where to go from here.
I've also tried splitting the problem into two sections, one where m < n, and one where m > n, but again, I'm not sure how to continue.
|
Assume $N>i\geq1$. Starting by a partial fraction decomposition, one may write
$$
\begin{align}
&\sum_{j=1,\,j\neq i}^N\frac1{(2i-1)^2-(2j-1)^2}
\\\\&=\frac1{4(2i-1)}\sum_{j=1,\,j\neq i}^N\left(\frac1{j+i-1}-\frac1{j-i}\right)
\\\\&=\frac1{4(2i-1)}\left(\sum_{j=1}^N\frac1{j+i-1}-\frac1{2i-1}-\sum_{j=1,\,j\neq i}^N\frac1{j-i}\right)
\\\\&=\frac1{4(2i-1)}\left(\sum_{\ell=i}^{N+i-1}\frac1{\ell}-\frac1{2i-1}-\sum_{j=1}^{i-1}\frac1{j-i}-\sum_{j=i+1}^N\frac1{j-i}\right)
\\\\&=\frac1{4(2i-1)}\left(H_{N+i-1}-H_{i-1}-\frac1{2i-1}+H_{i-1}-H_{N-i}\right)
\\\\&=-\frac1{4(2i-1)^2}+\frac1{4(2i-1)}\left(H_{N+i-1}-H_{N-i}\right)
\end{align}
$$ then, as $N \to \infty$,
$$
\sum_{j=1,\,j\neq i}^\infty\frac1{(2i-1)^2-(2j-1)^2}=-\frac1{4(2i-1)^2}
$$
where $H_N$ is the harmonic number and where we have used
$$
H_{N+i-1}-H_{N-i}=O\left(\frac1{N}\right)
$$ as $N \to \infty$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1728090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Conjecturing concerning $\mathrm{cl}(\mathrm{int}S) = S$ Let $S$ be a subset of $\mathbb{R}$, $\mathrm{int}S$ denote the set of interior points of $S$, $\mathrm{bd}S$ denote the set of boundary points of $S$, $S'$ denote the set of accumulation points of $S$, and $\mathrm{cl}S$ denote the closure of $S$.
The problem states that
Provide a counterexample to the following claim: If $S$ is a closed set, then $\mathrm{cl}(\mathrm{int S}) = S$.
An easy counterexample is to consider a nonempty, finite subset $S$. The $S$ is closed because $S' = \varnothing \subseteq S$. But because $\mathrm{int}S = \varnothing$, we have $\mathrm{cl}(\mathrm{int}S) = \varnothing \neq S$.
My Question
So I have a conjecture and an attempt to proving it: If $S$ is a closed, infinite set, then $\mathrm{cl}(\mathrm{int}S) = S$. I was wondering if the following proof is valid.
Suppose that $S$ is a closed, infinite set. Without loss of generality, let $a, b \in \mathbb{R}$ such that $a < b$ and $S = [a,b]$. Then
$$\begin{align} \mathrm{cl}(\mathrm{int} S)
&= \mathrm{int}S \cup \mathrm{bd}(\mathrm{int}(S)) \\
&= (a,b) \cup \{a,b\} \\
&= S
\end{align}$$
as desired.
|
You cannot assume that $S$ is an interval. There are infinite closed sets with empty interior; one interesting example is the Cantor set.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1728181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Limiting question:$\displaystyle \lim_{x\to\ 0} \frac{a^{\tan\ x} - a^{\sin\ x}}{\tan\ x -\sin\ x}$ How do I find the value of $$\lim_{x\to\ 0} \frac{a^{\tan\ x} - a^{\sin\ x}}{\tan\ x - \sin\ x}$$
in easy way.
|
As Henry W; commented, Taylor series make thigs quite simple.
$$A=a \tan(x) \implies \log(A)=\tan(x)\log(a)=\Big(x+\frac{x^3}{3}+\frac{2 x^5}{15}+O\left(x^6\right)\Big)\log(a)$$ $$A=e^{\log(a)}\implies A=1+x \log (a)+\frac{1}{2} x^2 \log ^2(a)+\frac{1}{6} x^3 \left(\log ^3(a)+2 \log
(a)\right)+O\left(x^4\right)$$ Soing the same with the second term of numerator $$B=a \sin(x) \implies B=1+x \log (a)+\frac{1}{2} x^2 \log ^2(a)+\frac{1}{6} x^3 \left(\log ^3(a)-\log
(a)\right)+O\left(x^4\right)$$ So the numerator is $$\frac{1}{2} x^3 \log (a)+O\left(x^4\right)$$ Using again the series for $\sin(x)$ and $\tan(x)$, the denominator is $$\frac{x^3}{2}+O\left(x^4\right)$$ which makes the ratio $$\frac{a^{\tan\ x} - a^{\sin\ x}}{\tan\ x -\sin\ x}=\log (a)+O\left(x^1\right)$$ Using more terms would lead to $$\frac{a^{\tan\ x} - a^{\sin\ x}}{\tan\ x -\sin\ x}=\log (a)+x \log ^2(a)+O\left(x^2\right)$$ showing the limit and how it is approached.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1728390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Density of smooth positive functions Let $\Omega$ be an open bounded set of $R^n$. For $f\in L^2(\Omega)$ such that $f>0$, a.e. in $\Omega, $ there is $(f_k)\subset W^{2,\infty}(\Omega)$ such that $f_k\to f$ in $L^2(\Omega)$. My question is:
Is it possible to chose $f_k>0,\; a.e. \; \Omega, \forall k?$
|
Yes. This can be obtained by the typical approach via mollification:
*
*Extend $f$ to $\mathbb{R}^n \setminus \Omega$ by $1$.
*Let $f_k$ be the convolution of $f$ with a smooth convolution kernel.
*This directly yields $f_k > 0$ a.e. and $f_k \to f$ in $L^2(\Omega)$ if the convolution kernels are appropriately chosen.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1728561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Intuitive explanation for why $\left(1-\frac1n\right)^n \to \frac1e$ I am aware that $e$, the base of natural logarithms, can be defined as:
$$e = \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$$
Recently, I found out that
$$\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^n = e^{-1}$$
How does that work? Surely the minus sign makes no difference, as when $n$ is large, $\frac{1}{n}$ is very small?
I'm not asking for just any rigorous method of proving this. I've been told one: as $n$ goes to infinity, $\left(1+\frac{1}{n}\right)^n\left(1-\frac{1}{n}\right)^n = 1$, so the latter limit must be the reciprocal of $e$. However, I still don't understand why changing such a tiny component of the limit changes the output so drastically. Does anyone have a remotely intuitive explanation of this concept?
|
If you know that $$\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^{n} = e\tag{1}$$ (and some books / authors prefer to define symbol $e$ via above equation) then it is a matter of simple algebra of limits to show that $$\lim_{n \to \infty}\left(1 - \frac{1}{n}\right)^{n} = \frac{1}{e}\tag{2}$$ Clearly we have
\begin{align}
L &= \lim_{n \to \infty}\left(1 - \frac{1}{n}\right)^{n}\notag\\
&= \lim_{n \to \infty}\left(\frac{n - 1}{n}\right)^{n}\notag\\
&= \lim_{n \to \infty}\dfrac{1}{\left(\dfrac{n}{n - 1}\right)^{n}}\notag\\
&= \lim_{n \to \infty}\dfrac{1}{\left(1 + \dfrac{1}{n - 1}\right)^{n}}\notag\\
&= \lim_{n \to \infty}\dfrac{1}{\left(1 + \dfrac{1}{n - 1}\right)^{n - 1}\cdot\dfrac{n}{n - 1}}\notag\\
&= \frac{1}{e\cdot 1}\notag\\
&= \frac{1}{e}
\end{align}
Using similar algebraic simplification it is possible to prove that $$\lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n} = e^{x}\tag{3}$$ where $x$ is a rational number. For irrational/complex values of $x$ the relation $(3)$ holds, but it is not possible to establish it just using algebra of limits and equation $(1)$.
Regarding the intuition about "changing a tiny component in limit expression changes the output" I think it is better to visualize this simple example. We have $$\lim_{n \to \infty}n^{2}\cdot\frac{1}{n^{2}} = 1$$ and if we change the second factor $1/n^{2}$ with $(1/n^{2} + 1/n)$ then we have $$\lim_{n \to \infty}n^{2}\left(\frac{1}{n^{2}} + \frac{1}{n}\right) = \lim_{n \to \infty} 1 + n = \infty$$ The reason is very simple. The change of $1/n$ which you see here is small but due to the multiplication with other factor $n^{2}$ its impact its magnified significantly resulting in an infinite limit. You always calculate the limit of the full expression (and only when you are lucky you can evaluate the limit of a complicated expression in terms of limit of the sub-expressions via algebra of limits) and any change in a sub-expression may or may not impact the whole expression in a significant way depending upon the other parts of the expression.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1728752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "36",
"answer_count": 17,
"answer_id": 8
}
|
Solve: $\int\frac{\sin 2x}{\sqrt{3-(\cos x)^4}}$ (and a question about $t=\tan \frac{x}{2}$) I tried substituting $$t=\tan \frac{x}{2}$$ but the nominator is $\sin {2x}$, so is there a way to get from $$\sin x=\frac{2x}{1+x^2}$$ to an expression with $\sin 2x$?
|
$$\int\frac{\sin(2x)}{\sqrt{3-\cos^4(x)}}\space\text{d}x=$$
Use $\sin(2x)=2\sin(x)\cos(x)$:
$$2\int\frac{\sin(x)\cos(x)}{\sqrt{3-\cos^4(x)}}\space\text{d}x=$$
Substitute $u=\cos(x)$ and $\text{d}u=-\sin(x)\space\text{d}x$:
$$-2\int\frac{u}{\sqrt{3-u^4}}\space\text{d}u=$$
Substitute $s=u^2$ and $\text{d}s=2u\space\text{d}u$:
$$-\int\frac{1}{\sqrt{3-s^2}}\space\text{d}s=-\int\frac{1}{\sqrt{3}\sqrt{1-\frac{s^2}{3}}}\space\text{d}s=-\frac{1}{\sqrt{3}}\int\frac{1}{\sqrt{1-\frac{s^2}{3}}}\space\text{d}s=$$
Substitute $p=\frac{s}{\sqrt{3}}$ and $\text{d}p=\frac{1}{\sqrt{3}}\space\text{d}s$:
$$-\int\frac{1}{\sqrt{1-p^2}}\space\text{d}p=-\arcsin\left(p\right)+\text{C}=-\arcsin\left(\frac{s}{\sqrt{3}}\right)+\text{C}=$$
$$-\arcsin\left(\frac{u^2}{\sqrt{3}}\right)+\text{C}=-\arcsin\left(\frac{\cos^2(x)}{\sqrt{3}}\right)+\text{C}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1728941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Mapping The Unit Disc To The Hemisphere?
Question: Can a disc drawn in the Euclidean plane be mapped to the surface of a hemisphere in Euclidean space ?
If $U$ is the unit disc drawn in the Euclidean plane is there a map, $\pi$, which sends the points of $U$ to the surface of a hemisphere, $H,$ in Euclidean space ?
Background and Motivation:
If $U$ is the unit disc centered at the origin consider $n$ chords drawn through the interior of $U$ such that no two chords are parallel and no three chords intersect at the same point. The arrangement graph $G$ induced by the discs and the chords has a vertex for each intersection point in the interior of $U$ and $2$ vertices for each chord incident to the boundary of $U.$ Naturally $G$ has an edge for each arc directly connecting two intersection points. $G$ is planar and $3-$connected. I know that $G$ has $n(n+3)/2, n(n+2)$ and $(n^2+n+4)/2$ vertices, edges and faces respectively. That $G$ is Hamiltonian-connected follows from R. Thomas and his work on Plummer's conjecture. I have conjectured that $G$ is $3-$colourable and $3-$choosable.
Independently Felsner, Hurtado, Noy and Streinu :Hamiltonicity and Colorings of Arrangement Graphs, ask if the arrangement graph of great circles on the sphere is $3-$colourable. In addition they conjecture such an arrangement is $3-$choosable.
Now I began to think the following
*
*Show my graph $G$ is $3-$colourable
*$\pi:G \to H$
*Glue $H$ to a copy of itself at the equator
then I could solve the conjecture by Felsner and his colleagues. Moreover if the map $\pi$ is bijective then any solution to Felsner's conjecture will also solve mine. The map $\pi$ does not need to preserve angles or surface areas. $\pi$ necessarily will have to map chords to great circles. See @JohnHughes excellent answer concerning the map $\pi$ and why vertical projection will not work.
|
If by a "circle" you mean the set of all points inside a circle (e.g., points whose distance from some center $C$ is less than or equal to 1), then the answer is "yes" and one solution is called "stereographic projection;" another is "vertical projection".
If you have a point $(x, y)$ in the unit disk (the "filled in circle"), the corresponding point, using vertical projection, is $(x, y, \sqrt{1 - x^2 - y^2})$.
For stereographic projection, you send the point $P = (x, y)$ to a new point $Q$:
\begin{align}
h &= x^2 + y^2 \\
Q &= (\frac{x}{h+1}, \frac{y}{h+1}, \frac{h-1}{h+1})
\end{align}
The latter has the charm that it takes chords in your disk to great-circle arcs in the hemisphere, and preserves angles of intersection, although not lengths.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1729012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
}
|
Some elements of the function field of the Fermat curve For $n>0$, consider the Fermat curve:
$$C(n): \{X^n+Y^n=Z^n\}\subset\mathbb P^2(\mathbb C)$$
the function field of $\mathbb C(n)$ can be explicitly described in the following way. It is the set of all fractions $\frac{f}{g}$ satisfying the following conditions:
*
*$f,g\in\mathbb C[X,Y,Z]$ are homogeneous polynomials of the same degree.
*The polynomial $X^n+Y^n-Z^n$ doesn't divide $g$.
*Two elements $\frac{f_1}{g_1}$ and $\frac{f_2}{g_2}$ are equivalent if $\frac{f_1g_2-f_2g_1}{g_1g_2}$ is the zero function on the open set $D(g_1g_2)$.
Let's denote with $K$ the function field of $C(n)$, then $\phi\in K$ defines a morphism $\Phi\colon C(n)\to\mathbb P^1$ in the following way:
where $\phi$ is well defined and away from the $n$ points at infinity of $C(n)$, we have:
$$\Phi(\alpha:\beta:1):=(\phi(\alpha:\beta:1):1)$$
the other points are sent to $\infty$.
During a lecture my professor said, without any explanation, that:
If $\phi=\frac{f}{g}\in\mathbb Q(x)\cap K$, then:
$$ \phi(\alpha:\beta:1)=\sum_{j=0}^{n-1}\frac{f_j(\alpha)}{g_j(\alpha)}\beta^j$$
for $f_j,g_j\in \mathbb Z[x]$ and with $g_j\neq 0$. Note that the $n$ in the summation is the same $n$ of $C(n)$.
Could you please explain how to get the above description for the elements of the function field with rational coefficients?
edit: See the comments for a better interpretation of the map $\Phi$.
Many thanks in advance.
|
Let's say you want to write
$$ (*) \qquad \frac{f(\alpha,\beta,1)}{g(\alpha,\beta,1)} = \sum_{j=0}^{n-1}\frac{p_j(\alpha)}{q_j(\alpha)}\beta^j,$$
for all points $(\alpha:\beta:1)\in C(n)$. Clearing denominators, we can suppose that $f$ and $g$ have integer coefficients. Since $\beta^n = 1-\alpha^n$, we can write
$$ f(\alpha,\beta,1) = \sum_{i= 0}^{n-1} f_i(\alpha)\beta^i$$
and
$$\beta^j g(\alpha,\beta,1) = \sum_{i= 0}^{n-1} g_{ij}(\alpha)\beta^i,\quad \forall j = 1,\ldots,n-1$$ with $f_i,g_{ij}\in \Bbb Z[x]$ for $i,j=1,\ldots,n-1$. We can rewrite equation $(*)$ as
$$\sum_{i= 0}^{n-1} f_i(\alpha)\beta^i = \sum_{i = 0}^{n-1} \bigg( \sum_{j=0}^{n-1}g_{ij}(\alpha) \frac{p_j(\alpha)}{q_j(\alpha)} \bigg) \beta^i.$$
Regarding both sides as polynomials in $\beta$, the equation is satisfied iff the coefficients are equal. Thus, we need to find $p_j$, $q_j$ such that
$$f_i(\alpha) = \sum_{j=0}^{n-1}g_{ij}(\alpha) \frac{p_j(\alpha)}{q_j(\alpha)} ,\quad \forall i = 1,\ldots,n-1.$$
You can see this as a linear system $Ax = b$ with matrix $A = \big(g_{ij}(\alpha)\big)$. The $j$th coordinate of the vector $x$ is $p_j(\alpha)/q_j(\alpha)$ and the $i$th coordinate of $b$ is $f_i(\alpha)$. If you now express $A^{-1}$ in terms of its adjugate matrix (see Wikipedia), you get a unique expression for $p_j$ and $q_j$ (with integer coefficients).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1729145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Counterexample to switching limit and integral on compact domain If $f:[a,b] \times [c,d] \to \mathbb{R}$ is continuous on the compact rectangle (and, hence, uniformly continuous) then it holds for $y_o \in [c,d]$ that
$$\lim_{y \to y_0}\int_a^b f(x,y) \,dx = \int_a^b \lim_{y \to y_0}f(x,y) \,dx =\int_a^b f(x,y_0) \,dx. $$
I would like to find some counterexamples where the switch is not permitted when, if possible, $f(\cdot,y)$ is integrable over $[a,b]$ and $f(x, \cdot)$ is continuous on $[c,d]$.
I would imagine such a function is not uniformly bounded. Otherwise, I believe the bounded convergence theorem would justify the switch.
Thank you.
|
Try $$ f(x,y) = \cases{ \dfrac{x}{y^2} \exp(-x/y) & if $y \ne 0$\cr
0 & if $y = 0$ }$$
on $[0,1] \times [0,1]$, with $y_0 = 0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1729271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Distribute 5 distinguished balls to 3 undistinguished cell On a first glimpse it looks a very easy one, but I find that a bit odd.
Assuming we have 5 distinguished balls, and 3 undistinguished cells, in how many ways can we distribute the balls in the cells, when we have at least 1 ball in each?
My first guess was to distribute the first 3 this way:
$$ \binom{5}{1,1,1} = \frac{5!}{1!} \cdot \frac{1}{3!} $$
I divided by $ 3! $ since the cells are undistinguished.
Then, For each of the remaining balls we have 3 options each, then it should be $ 3^2 $.
To sum up, it's 29.
But the answer is actually 25 :(
What am I doing wrong?
edit: I know that since this case is very simple I have 2 options to divide them: 3,1,1 and 2,2,1
But I'm actually interested in the general idea (that also works for big or abstract numbers).
Thanks
|
The numbers are small, so one can use cases: (i) $1,1,3$, and (ii) $1,2,2$.
Case (i) The "team of $3$" can be chosen in $\binom{5}{3}$ ways, and now we have no further choices.
Case (ii) The lonely one can be chosen in $\binom{5}{1}$ ways. For every such choice, the person from the remaining $4$ with the lowest student number can choose her partner in $\binom{3}{1}$ ways, for a total of $\binom{5}{1}\binom{3}{1}$.
Edit: The question was modified, and asks for information about larger numbers. The number of ways to partition a set of $n$ distinct objects into $k$ non-empty subsets is the Stirling Number of the Second Kind $S(n,k)$ There is no nice closed form for these, but there are useful recurrences. (The Stirling numbers of the second kind occur so often in combinatorics that some consider $S(n,k)$ to be a closed form.)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1729396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Alternative injection from $(0,1)$ to $p(\mathbb{N} )$ I know the standard injection is to consider the binary expansion of a number in the interval, but I was wondering if it is possible to create an injection using a decimal expansion.
To that effect first denote the decimal expansion of a $r\in (0,1)$ as $r=0.d_1d_2d_3d_4\dots$. Choose the decimal expansion of all $x\in (0,1)$ as above except that $\forall n\in \mathbb{N} \exists m>n$ such that $d_m\in \{0,1,2,3,4,5,6,7,8\}$. (This does not exclude any element in the interval).
Now consider the function $f:(0,1)\to p(\mathbb{N})$ where $f(x)=\{2^{d_1},3^{d_2},5^{d_3},\dots,p_i^{d_i},\dots\}$. I believe that Euclid's theorem and the fundamental theorem of arithmetic ensure that $f$ is an injection, but I would like some confirmation.
|
This almost works, but if you have more than one $0$ digit, you’ll get multiple copies of $1$ in the description of $f(x)$. To avoid this problem, define
$$f(x)=\left\{p_i^{d_i+1}:i\in\Bbb Z^+\right\}\;.$$
Added: As Henning Makholm notes below in the comments, the original idea of setting
$$f(x)=\left\{p_i^{d_i}:i\in\Bbb Z^+\right\}$$
actually does work; it just isn’t quite as elegant. If no multiple of some prime $p_i$ appears in $f(x)$, then $d_i=0$, so the $i$-th digit is still reconstructible. In that case the $1\in f(x)$ doesn’t actually tell us anything that we can’t already learn from $f(x)\setminus\{1\}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1729525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Is there a ring homomorphism between $\mathbb{Z}$ into $\mathbb{Z} \times \mathbb{Z}$? Is there a ring homomorphism between $\mathbb{Z}$ into $\mathbb{Z} \times \mathbb{Z}$?
Also for any rings $R_{1}$ and $R_{2}$ does there exist a ring homomorphism $\phi$ : $R_{1} \rightarrow$ $R_{1} \times R_{2}$?
Note: I am allowing 1 in both rings. So any homomorphism must map the identity on one ring to the identity on the other ring.
|
Suppose $R$, $S$ and $T$ are rings. Giving a ring homomorphism $f\colon R\to S\times T$ is the same as giving homomorphisms $g\colon R\to S$ and $h\colon R\to T$.
Let's see why. First, the projection maps $p\colon S\times T\to S$ and $q\colon S\times T\to T$ are ring homomorphisms, so if we are given $f\colon R\to S\times T$, we get $p\circ f\colon R\to S$ and $q\circ f\colon R\to T$.
Suppose instead we are given $g\colon R\to S$ and $h\colon R\to T$. Define
$$
f(x)=(g(x),h(x))
$$
It's easy to see that $f$ is a ring homomorphism and that $g=p\circ f$, $h=q\circ f$.
In general terms this is the statement that $S\times T$ is the product in the category of rings.
This answers your second question: in order to have a ring homomorphism $R\to R\times T$ you need a ring homomorphism $R\to T$ (and use, for instance, the identity as the homomorphism $R\to R$). So, for a counterexample find rings $R$ and $T$ such that there is no ring homomorphism $R\to T$; take $R$ to have characteristic $2$ and $T$ to have characteristic $3$, for instance.
In the case of $\mathbb{Z}$, it's a standard result that, for any ring $R$, there is a unique ring homomorphism $\mathbb{Z}\to R$ (when unital rings are concerned and homomorphisms are required to preserve the identity element).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1729637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
How can I prove that a linear recurrence $x_{n+1} = αx_n - β$ will contain a composite number in the sequence? I'm working on a homework problem about finite automata and I got stuck trying to prove a fact about prime numbers that I think should be true.
Given a prime $p$ and integers $α$ and $β$, can I show that the sequence $x_0 = p; x_{n+1} = αx_n - β$ will contain at least one composite number? Assume that $α$ and $β$ are such that $x_n$ is always increasing.
I think it would be very surprising for this sequence to only contain prime numbers but I am very rusty on my number theory and would appreciate some hints on how to proceed.
The motivation for this question is that I am trying to use the pumping lemma to prove that the language of strings representing prime numbers in base 10 is not a regular language. The proof revolves around showing that for any sufficiently large prime $p$ and any decomposition of its decimal representation into three substrings $p = abc$ there will be at least one number in the sequence $abc$, $abbc$, $abbbc$, $...$ that is not prime. These subsequences can be represented as a linear recurrences $x_{n+1} = αx_n - β$ with an appropriate choice of $α$ and $β$.
For example, if I take the prime 1277 and the decomposition 1277 = 1 27 7 then we will have a sequence. 1277, 127277, 12727277, and so on. 1277 and 127277 are prime but 12727277 is not (its equal to 2143 x 5939)
|
The solution of the recurrence is
$$
x_n = \alpha^n x_0 -\beta(1+\alpha+\cdots+\alpha^{n-1})
$$
Since $x_0=p$, we get
$$
x_n \equiv -\beta(1+\alpha+\cdots+\alpha^{n-1}) \bmod p
$$
If $\alpha \equiv 1 \bmod p$, then
$$
x_n \equiv -\beta n \bmod p
$$
and so $x_n \equiv 0 \bmod p$ for all $n$ that are multiples of $p$.
If $\alpha \not\equiv 1 \bmod p$, then
$$
x_n \equiv -\beta\frac{\alpha^{n}-1}{\alpha-1} \bmod p
$$
where the fraction is computed mod $p$.
If moreover $\alpha \not\equiv 0 \bmod p$, then
$x_n \equiv 0 \bmod p$ for all $n$ that are multiples of $p-1$, by Fermat.
If $\alpha \equiv 0 \bmod p$, then $x_n \equiv -\beta \bmod p$, and the argument fails. But you've mentioned in a comment that this case cannot happen.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1729755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Suppose that a sequence is Cesaro summable. Prove.... Suppose that a sequence $a_{n}$ is Cesaro summable. Prove that
$$\lim_{n \to \infty }\frac{a_{n}}{n}=0$$
|
Suppose $(a_1 + \cdots +a_n)/n \to L \in \mathbb R.$ Then
$$\frac{a_n}{n} = \frac{a_1 + \cdots + a_n}{n} - \frac{n-1}{n}\frac{a_1 + \cdots + a_{n-1}}{n-1} \to L-L = 0.$$
Let's apply this to the Cesaro sums: Suppose $(S_1 + \cdots +S_n)/n \to L\in \mathbb R,$ where $S_n = a_1 + \cdots +a_n.$ By the above, $S_n/n \to 0.$ Applying the above again shows $a_n/n \to 0.$
(NOTE: This was edited from the original, where I only included the first result. Thanks to @Bungo for pointing this out.)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1729855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Prove $f$ is not differentiable at $(0,0)$ For
$$f(x,y)=\begin{cases}
\frac{x|y|}{\sqrt{x^2+y^2}} & \text{ for }(x,y)\neq (0,0)\\
0 & \text{ for } (x,y)=(0,0)
\end{cases}$$
I'm trying to prove $f$ is not differentiable at $(0,0)$. I showed if $f$ is differentiable at $(0,0),$ then $A=Df_{(0,0)}=0.$ But I don't know how this lead to a contradiction. Anyone has ideas?
|
It's easy. You only have to see that if $f$ is differentiable in $(0,0)$, then $f'(0,0)$ is a linear transformation. So:
$$f'(0,0)\cdot (1,1)=f'(0,0)\cdot((1,0)+(0,1))=f'(0,0)\cdot(1,0)+f'(0,0)\cdot(0,1).$$
And that is a contradiction, because $f'(0,0)\cdot (1,1)\not = f'(0,0)\cdot(1,0)+f'(0,0)\cdot(0,1).$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1729978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
}
|
Vector field on manifold I've only seen a vector field $V$ on a manifold $M$ as a mapping $V:M\to TM$. Is it true that they can also be seen as a mapping $V:C^{\infty}\left(M\right)\to C^{\infty}\left(M\right)$? How would $V$ work in the second case?
|
There are a couple of things to point out here. Let $M$ be a smooth manifold.
*
*Every tangent vector at $p$ may be thought of as a derivation at $p$: indeed, if $v_p$ is a tangent vector at $p$ and $f: U \to \mathbf{R}$ is a smooth function in a neighbourhood $U \subseteq M$ of $p$, then, $v_p(f)$ can be thought of as the derivative of $f$ at $p$ in the direction of $v$. More precisely, for a curve $\gamma$ defined around $0$ (so that $\gamma(0) = p$) which represents the tangent vector $v_p$, one sets:
$$v_p(f) = \left.\frac{\mathrm{d}(f \circ \gamma)}{\mathrm{d}t}\right|_{t=0}.$$
Firstly, this is well-defined: that is, the definition of $v_p$ is independent of the curve $\gamma$ that "represents" $v_p$. (This is "per definitionem", like everything is.)
The name comes from the fact that $v_p: C^\infty(U) \to \mathbf{R}$ satisfies the identity $$v_p(fg) = f(p) v_p(g) + g(p) v_p(f).$$
*
*Equipped with the above interpretation as derivation at a point $p$, we may view a smooth vector field $X$ as a derivation $\mathcal{L}_X: C^\infty(M) \to C^\infty(M)$ on the algebra $C^\infty(M)$ by setting $\mathcal{L}_X(f)(p) = X_p(f)$. (Check that this makes sense!)
See Lee's Introduction to Smooth Manifolds, a wonderful book on differential topology.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1730074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Is there a regular hexagon with integral corners? I'm looking for a regular hexagon in $\mathbb{R}^2$, whose corners are integral, i.e. the coordinates are integers.
The hexagon cannot lie "flat" (with upper and lower line segments horizontal), since then $h = \frac{\sqrt{3}}{2} w$ with $w$ width and $h$ height of the hexagon, making the ratio $\frac{h}{w}$ irrational.
|
Yes we can have a regular hexagon with integral vettex coordinates ... in three dimensions. Think of a face-centered cubic lattice. If we set the length unit to half the edge of a cubic unit cell then all lattice points have integer coordinates -- including those forming a hexagonal close-packed plane perpendicular to any body diagonal of the cube.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1730204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
If $n = a^2 + b^2 + c^2$ for positive integers $a$, $b$,$c$, show that there exist positive integers $x$, $y$, $z$ such that $n^2 = x^2 + y^2 + z^2$.
If $n = a^2 + b^2 + c^2$ for positive integers $a$, $b$,$c$, show that there exist positive integers $x$, $y$, $z$ such that $n^2 = x^2 + y^2 + z^2$.
I feel that the problem basically uses algebraic manipulation even though it's in a Number Theory textbook. I don't realize how to show $(a^2+b^2+c^2)^2$ as the sum of three squares. I have tried algebraic manipulation but this is the stage I have reached.
$$(b^2 + c^2)^2 + a^2(a^2 + b^2 + c^2 + b^2 + c^2)$$
Could you give me some hints on how to proceed with this question?
Thanks.
|
This can be generalized a bit. it happens that every positive integer $n$ is the sum of four squares, $n = A^2 + B^2 + C^2 + D^2.$ The we get, from manipulating quaternions with integer coefficients,
$$ n^2 = (A^2 + B^2 - C^2 - D^2)^2 + (-2AC +2BD)^2 + (-2AD-2BC)^2 $$
https://en.wikipedia.org/wiki/Lagrange%27s_four-square_theorem
https://en.wikipedia.org/wiki/Euler%27s_four-square_identity
I rewrote the version I had to agree with the Euler four square formula in the beginning of the Wikipedia article, with substitutions
$$ a_1 = A, b_1 = A, a_2 = B, b_2 = B, a_3 = C, b_3 = -C, a_4 = D, b_4 = -D $$
so that the second of the four terms cancels out to zero. I think this formula is associated with Lebesgue, let me see if i can find that. It plays a big part in Jones and Pall (1939).
Yes, https://en.wikipedia.org/wiki/Pythagorean_quadruple
Well, I requested the 1951 number theory book by Trygve Nagell, evidently it is one source of the idea that Lebesgue was involved.
Hmmm; Euler's four square identity 1748, then Lagrange four square theorem 1770, then Hamilton invents quaternions 1843. The (Henri) Lebesgue we know about lived 1875-1941.
NO, on page 266 of Volume II of Dickson's History of the Theory of Numbers, we find the statement that one V. A. Lebesgue, in 1874, proved the the square of a sum of three squares is again the sum of three squares in a nontrivial manner, as
$$ (\alpha^2 + \beta^2 + \gamma^2)^2 = (\alpha^2 + \beta^2 - \gamma^2)^2 + (2\alpha \gamma)^2 + (2 \beta \gamma)^2. $$
This is formula (4) on that page.
Who is V. A. Lebesgue?
Victor-Amédée Lebesgue, sometimes written Le Besgue, (2 October 1791,
Grandvilliers (Oise) – 10 June 1875, Bordeaux (Gironde)) was a
mathematician working on number theory.
Evidently the precise reference is V. A. Lebesgue, Sur un identite qui conduit a toutes les solutions
de l'equation $t^2 = x^2 + y^2 + z^2,$ Compte Rendus de l`Academie des Sciences de Paris 66 (1868), 396-398.
A 1962 article by Spira gives a short elementary proof that all primitive solutions can be found using this formula. He says that the first correct proof of such completeness was Dickson in 1920, with another by Skolem in 1941. Spira adapts the proof of Skolem to give an algorithm for finding at least one set $A,B,C,D$ from a given primitive Pythagorean quadruple.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1730327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Can a null graph be considered k-regular, for any k ? By null graph I mean a graph without any vertices. I did not find any mention of this anywhere.
|
Looking at the definition of a k-regular graph it's obvious taht you can consider the empty graph k-regular for any k. Why? Because since it has no vertices the statement: "every vertex has degree k" is true for any k.
But I am not sure how that helps and in what scenarios.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1730405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Probability of triangle to be obtuse
Two points $A,B$ fixed on a plane(distance = 2).
C - random choosen point inside circle with radius $R$ with center at the center of $AB$
Find probability of triangle $ABC$ to be obtuse
My thoughts:
*
*If $C$ lies in the circle - $ABC$ will be rectangular
*Opposite the larger angle is large side, so using cosine theorem:
$$
b^2+c^2<a^2=4R^2,
$$
where $a$ - the diagonal, $b=AC, c=BC$
And then I don't know...
|
Any such triangle $ABC$ with $AB$ diameter and $C$ any point within the circle will be obtuse.
Proof:-
(Note:-$AO=OB=r$)
Let,C be any point in the circle.$AC$ is extended to meet the circumference at $X$.So,$\angle AXB=\alpha=90^0$ and $\angle \beta>0^0$.So,$\angle ACB=\gamma=\alpha+\beta=90$+something $>0.$So,$\angle ACB$ will always be obtuse and thus $\triangle ACB$ is always obtuse.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1730551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
}
|
Understanding One of Fermat's little theorem's proof Fermat's Little theorem :
Let $p$ be a prime which does not divide the integer $a$, then $a^{p-1} \equiv 1 (\mod p)$.
Leibniz's proof
*
*suppose that $ra$ and $sa$ are the same modulo $p$, then $r \equiv s (\mod p)$. {So the first p-1 multiples of a are distinct and non zero}.
*Therefore $a, 2a, 3a, ... (p-2)a, (p-1)a$ must be congruent to $1, 2, 3, ... p-2, p-3$ in some order
*Multiply all these congruences we get $a^{p-1}(p-1)! \equiv (p-1)! (\mod p)$
*Divide by $(p-1)!$ on both sides we get $a^{p-1} \equiv 1 (\mod p)$
My issues with a full understanding of the proof
The text I've put in between brackets {} are the parts of the proof I haven't understood
*
*first brackets: {So the first p-1 multiples of a are distinct and non zero}. Isn't that implied in the statement of the proof? Why is that part of the proof necessary?
What is the proof that the first $p-1$ multiples of $a$ will necessarily be congruent to 1, 2 ... p-1 in some order?
|
Because if $ak \equiv 0 \pmod{p}$, for some $1 \leq k \leq p-1$, then $p \mid ak$. But, by Euclid's Theorem, as $\gcd(p,k) = 1, p \mid a$, a contradiction.
Therefore, $ak \not\equiv 0 \pmod{p}$, for all $1 \leq k \leq p-1$. Then we've got $p-1$ distinct possible values of $k$. Since that, for every $k$ there must be a unique $r \in \{1,\ldots,p-1\}$ so that $ak \equiv r \pmod{p}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1730637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
suppose a sample is taken from a symmetric distribution whose tails decrease more slowly than those of a normal distribution I was wondering how to go about this question about Probability QQ Plots, the question is,
suppose a sample is taken from a symmetric distribution whose tails decrease more slowly than those of a normal distribution. what would be the qualitative shape of a normal probability plot of this sample?
|
Maybe it helps to have an example. The Laplace distribution has 'fatter' tails than a normal. We can easily generate some data
from a Laplace distribution. The difference of two exponential
distributions with the same rate is Laplace. (See Wikipedia on
'Laplace distribution', third bullet under Related Distributions. if it's not in your text.) Specifically, if $Y_1 \sim Exp(1)$ and, independently, $Y_2 \sim Exp(1),$ then $Y_1 - Y_2$ has a
Laplace distribution with mean $\mu = 0$ and SD $\sigma = \sqrt{2}.$
Below, we simulate a sample of size $n = 1000$ from such a
Laplace distribution and a separate random sample of the same
size from $Norm(\mu = 0,\, \sigma=\sqrt{2})$.
Then we compare the normal Q-Q plots of the Laplace and Normal samples.
y1 = rexp(1000); y2 = rexp(1000); x = y1 - y2 # Laplace sample
z = rnorm(1000, 0, sqrt(2))
par(mfrow=c(1,2)) # enables side-by-side plots
qqnorm(x, datax=T, main="Normal QQ-Plot of Laplace Data")
qqnorm(z, datax=T, main="Normal QQ-plot of Normal Data")
par(mfrow=c(1,1)) # return to single-panel plots
As anticipated the points on the normal probability plot (normal
QQ-plot) of normal data on the right fall pretty much in a straight
line. However, the points from the Laplace sample follow an S-shaped
curve with straggling points far the the left near the bottom and far to the right near the to top.
In case it helps you to visualize the relatively heavy tails of the Laplace distribution, here are histograms of these two samples.
(The minimum and maximum of the Laplace sample are approximately
-7.6 and 7.4, respectively. For the Normal sample the extremes are
approximately -4.8 and 4.1.)
Addendum: In response to a comment, I have added graphs of
PDFs of normal and Laplace distributions with 0 means and
equal standard deviations. If you look closely you will see
that the normal PDF (black) is below the Laplace PDF (blue) in the far tails.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1730792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Why weren't continuous functions defined as Darboux functions? When we were in primary school, teachers showed us graphs of 'continuous' functions and said something like
"Continuous functions are those you can draw without lifting your pen"
With this in mind I remember thinking (something along the lines of)
"Oh, that must mean that if the function takes two values $f(y)<f(z)$ then
for every $c$ between $f(y), f(z)$ there must be some $x\, (y<x<z)$ such that $f(x)=c$"
And that's what I thought a continuous function was. But then the $\epsilon$-$\delta$ definition appeared, which put a more restrictive condition on what a continuous function was.
So, my question is, given the fact that Darboux functions "seem continuous" (in some subjective sense, I guess), why wasn't this used as the definition of continuity? More generally, how did today's (analytical) definition of continuity appear?
|
The Darboux definition does not correspond very well with our intuition about continuity. For example, the Conway function takes on every value in every interval, and is therefore Darboux. However it is not continuous, and I don't think we want it to be continuous, because it certainly doesn't agree with your teacher's definition of drawing without lifting your pencil.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1730911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "38",
"answer_count": 5,
"answer_id": 0
}
|
Why does addition not make sense on infinite vectors? I was reading http://www.math.lsa.umich.edu/~kesmith/infinite.pdf to learn more about infinite dimensional vector spaces, and the author argues that the standard basis ($e_i$ is the sequence of all zeroes except in the i-th position, where there appears a 1), does not form a basis for $\mathbb{R}^\infty$ because the span is only defined over the sum over finitely many basis vectors. So, she argues, a vector like $(1,1,1,\dots)$ is not in the span.
If we allow the span to be defined over an infinite sum, then, the author argues, something like $(1,1,1,\dots)+(2,2,2,\dots)+(3,3,3,\dots)$ does not make sense, and thus we have to restrict the span to a finite sum.
I do not understand why this is not simply $(6,6,6,\dots).$ More fundamentally, why can't we generalize the span to include an infinite sum of the basis vectors?
|
The real problem is that infinite sums should be understood as some sort of limit of a converging sequence. But while that operation may make sense, it isn't a vector space operation.
If you add a topology, then you can introduce infinite sums and things will make perfect sense (as long as the sum converges). In this case you could start with open balls consisting of open interval in the leading dimensions combined with any possible value in the rest. Now all the infinite sums you want work perfectly - as long as there is pointwise convergence in each dimension the infinite sum is defined.
But there are a lot of topologies you can use. For example maybe your basic open balls are open intervals of length $l$ in all dimensions. Now infinite sums are defined so long as they converge uniformly in all dimensions. Now the sums that you think should work, don't.
Changing the topology will leave the vector space properties alone, but infinite sums will work out differently.
So long and short is this. Lots of mathematical structures are infinite dimensional vector spaces with infinite sums. (An important example you'll learn is Hilbert spaces.) But infinite sum is never a vector space operation.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1731030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
}
|
How do I prove that if A is a $n × n$ matrix with integer entries, then $det(A)$ is an integer? My proof:
if A is an n x n matrix with integer entries, then this means that every entry of A contains an integer. Therefore, any integer operated on by another integer is also an integer. A determinant is calculated by a series of integer operations on a matrix. Therefore, det(a) is also an integer.
Is that a sufficient proof?
|
Your approach is correct. You could extend it by giving a definition for the determinant and pointing out the closedness of the involved operations.
One such definition defines the determinant of a matrix as alternating multi-linear form of the column vectors of that matrix. This definition involves only additions and multiplications, so you stay within $\mathbb{Z}$.
$$
\DeclareMathOperator{sgn}{sgn}
\DeclareMathOperator{det}{det}
\det(A) = \det(a_1,\dotsc,a_n)
= \sum_{\pi \in S_n} \sgn(\pi) \, a_{1\pi(1)} a_{2\pi(2)} \dotsb a_{n\pi(n)}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1731132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Find he local maximum and minimum value and saddle points of the function? Find he local maximum and minimum value and saddle points of the function: $$f(x,y)=x^2-xy+y^2-9x+6y+10$$
The answer is a min of $(-4,1), f(-4,1)=73$
I got a min of $(12/5,-21/5)$
my
$$
f_x=2x-y-9\\
f_y=-x+2y+6
$$
set $f_x = 0 = f_y$ and we get $x=(9+y)/2$ so $y=-21/5$ and $x=12/5$
I found out that this was a minimum by the 2nd derivative test. There are no max or saddle points.
|
You're both wrong (or maybe you made a mistake in copying $f$): the minimum is at $(4,-1)$ where the value is $-11$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1731227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
behavior of the Linear system of an ODE model I am working on a predator-prey model and the linearization about and equilibrium point $(0,e_2)$ has Jacobian matrix as follows
$$\mathcal{J} = \begin{pmatrix}
0 & 0\\
b& -b
\end{pmatrix},$$
where the parameters $e_2$ and $b$ are positive. I never have dealt with this kind of system and I tried to find something online, still cant find anything. Can someone show me how the system behaves? or what method of analysis is implemented?
Thanks in advance
|
Such a Jacobian matrix does not suffice to determine the stability of the fixed point. Compare the behaviour of the differential systems $$x'=ax^3\qquad y'=x-y+c$$ around their fixed point $(0,c)$ for some positive $a$ (unstable) and for some negative $a$ (stable).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1731358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How to compute this double integral I'm trying to show that $\int_0^1\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}dxdy \neq \int_0^1\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}dydx$ by computing these integrals directly.
I tried using polar coordinates with no success as the bounds of integration caused problems.
I also tried the substitution $x=ytan\theta$ but ended up getting something of the form $\infty-\infty$.
Can anyone offer a hint as to how I can compute these directly please???
Thanks in advance!
|
The integral is not absolutely convergent, so we cannot change the order of integration, or the coordinates used, and expect to get the same answer.
Polar Coordinates
If we try to convert to polar coordinates,
$$
\begin{align}
\int_0^1\int_0^1\frac{x^2-y^2}{\left(x^2+y^2\right)^2}\,\mathrm{d}x\,\mathrm{d}y
&=\int_0^1\int_0^{\pi/2}\left(\cos^2(\theta)-\sin^2(\theta)\right)\frac1r\,\mathrm{d}\theta\,\mathrm{d}r\\
&+\int_1^{\sqrt2}\int_{\cos^{-1}\left(\frac1r\right)}^{\sin^{-1}\left(\frac1r\right)}\left(\cos^2(\theta)-\sin^2(\theta)\right)\frac1r\,\mathrm{d}\theta\,\mathrm{d}r\\
&=\int_0^1\int_0^{\pi/2}\left(\cos^2(\theta)-\sin^2(\theta)\right)\frac1r\,\mathrm{d}\theta\,\mathrm{d}r\\
&=\int_0^10\cdot\frac1r\,\mathrm{d}r\\[6pt]
&=0
\end{align}
$$
The integral for $1\le r\le\sqrt2$ converges absolutely and is equal to its negative under the substitution $\theta\mapsto\frac\pi2-\theta$. Therefore, the integral for $1\le r\le\sqrt2$ is $0$.
The integral for $0\le r\le1$ is equal to $0$ for each $r$, so the integral is $0$ if integrated in $\theta$ first.
Of course, the integral does not converge when integrated in $r$ first since $\int_0^1\frac1r\,\mathrm{d}r$ diverges.
Rectangular Coordinates
$$
\begin{align}
\int_0^1\int_0^1\frac{x^2-y^2}{\left(x^2+y^2\right)^2}\,\mathrm{d}x\,\mathrm{d}y
&=\int_0^1\frac1y\int_0^{1/y}\frac{x^2-1}{\left(x^2+1\right)^2}\,\mathrm{d}x\,\mathrm{d}y\\
&=\int_0^1\frac1y\int_0^{\tan^{-1}(1/y)}\left(\sin^2(u)-\cos^2(u)\right)\,\mathrm{d}u\,\mathrm{d}y\\
&=-\int_0^1\frac1y\left[\sin(u)\cos(u)\vphantom{\frac1y}\right]_0^{\tan^{-1}(1/y)}\,\mathrm{d}y\\
&=-\int_0^1\frac1{y^2+1}\,\mathrm{d}y\\[6pt]
&=-\frac\pi4
\end{align}
$$
Similarly, we get
$$
\begin{align}
\int_0^1\int_0^1\frac{x^2-y^2}{\left(x^2+y^2\right)^2}\,\mathrm{d}y\,\mathrm{d}x
&=\int_0^1\int_0^1\frac{y^2-x^2}{\left(x^2+y^2\right)^2}\,\mathrm{d}x\,\mathrm{d}y\\[6pt]
&=\frac\pi4
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1731441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
$3$ children riddle, compute the ages based on information given A man has $3$ children such that their ages add up to some number $x$, and whose ages multiply to some number $y$, such that $xy = 756$. What are the ages of the $3$ children?
Letting the ages be $a$, $b$, and $c$ of the three children, what we know is the following.
$$a+b+c = x$$
$$abc = y$$
$$xy = 756.$$
How can I go about solving this? I tried just plugging in some numbers and can get semi close such as ages $3,3,7$ which gives an $xy$ value of $819$.
Also I tried working backwards from $756$ to divide thru by factors and I got $378,189,63,21,7$, which is why I thought one of the ages might be $7$.
|
There is also another "reasonable" solution which is the ages are $3.5$, $4$, and $4.5$. Some children's ages are expressed in "halves" although it is not as common as "wholes". ($3.5 + 4 + 4.5$) * ($3.5 * 4 * 4.5$) does indeed equal $756$. It is interesting to note that this is $12 * 63$ and if you were to guess the ages were $4,4,4$, that would give you $12 * 64$ which is very close to the correct answer of $12 * 63$ so that would imply that the ages have to all be very close to $4$ meaning they are likely not all integers. Ages $3,4,5$ is also a near solution at $12*60=720$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1731520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
}
|
Solution of a riccati equation. We're given a self-adjoint equation as follows : $$ \dfrac{d}{dt}[t \frac{dx}{dt}]+(1-t)x = 0$$ We first convert this into a riccati equation , and hence we get : $$ \dfrac{du}{dt} + (\dfrac{1}{t})u^{2} +(1-t)=0$$ Now we want the solution of this riccati equation of the form $ct^{n}$. How to obtain this ?
All I know about riccati equation is that the substitution $u = f + \frac{1}{v}$ would convert this system into a known system . But for this we should have a solution $f$ to the system above, and no such solution is given in the question. Can anyone help ?
|
As you said you must have a solution to get the other one. by a simple guess you can make the guess that solution is $u=-t$. and then find the other root. the first root must be given in problem or else you must guess it.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1731685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
An inequality to find the range of an unknown coefficient Find the range of $a$ such that
$$a(x_1^2+x_2^2+x_3^2)+2x_1x_2+2x_2x_3+2x_1x_3 \geq 0, x_i\in \mathbb{R}$$
I tried to use Cauchy Inequality but it seems not...
|
Answer: $a \ge 1$
Let $a=1:$ $$x_1^2+x_2^2+x_3^2+2x_1x_2+2x_2x_3+2x_1x_3 =(x_1+x_2+x_3)^2\ge 0$$
If $a>1$ then $$(a-1)(x_1^2+x_2^2+x_3^2)+x_1^2+x_2^2+x_3^2+2x_1x_2+2x_2x_3+2x_1x_3 \ge 0$$
If $a<1, a=1- E$ then
$(x_1+x_2+x_3)^2-E(x_1^2+x_2^2+x_3^2)<0$ at $x_1=-x_2-x_3$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1731807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.