Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
Is there a minimum number of vertices that a graph must have to be of a given genus? It seems intuitive that the higher the genus of a graph the more vertices and edges it would have to have. Is it known if the minimum number of vertices must be larger for graphs of higher genus?
|
There is at least an upper bound on the genus in terms of the number of vertices and edges in the graph. Let $\gamma(G)$ be the genus of a graph $G=(V,E)$ and $\beta(G)=|E|-|V|+1$ be its Betti number. Then the following holds:
$$\gamma(G)\leq\left\lceil\frac{\beta(G)}{2}\right\rceil$$
For a reference on how to prove this see for example this paper
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2485477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Simplest way to get the lower bound $\pi > 3.14$ Inspired from this answer and my comment to it, I seek alternative ways to establish $\pi>3.14$. The goal is to achieve simpler/easy to understand approaches as well as to minimize the calculations involved. The method in my comment is based on Ramanujan's series $$\frac{4}{\pi}=\frac{1123}{882}-\frac{22583}{882^{3}}\cdot\frac{1}{2}\cdot\frac{1\cdot 3}{4^{2}}+\frac{44043}{882^{5}}\cdot\frac{1\cdot 3}{2\cdot 4}\cdot\frac{1\cdot 3\cdot 5\cdot 7}{4^{2}\cdot 8^{2}}-\dots\tag{1}$$ This is quite hard to understand (at least in my opinion, see the blog posts to convince yourself) but achieves the goal of minimal calculations with evaluation of just the first term being necessary.
On the other end of spectrum is the reasonably easy to understand series
$$\frac\pi4=1-\frac13+\frac15-\cdots\tag2$$
But this requires a large number of terms to get any reasonable accuracy. I would like a happy compromise between the two and approaches based on other ideas apart from series are also welcome.
A previous question of mine gives an approach to estimate the error in truncating the Leibniz series $(2)$ and it gives bounds for $\pi$ with very little amount of calculation. However it requires the use of continued fractions and proving the desired continued fraction does require some effort.
Another set of approximations to $\pi$ from below are obtained using Ramanujan's class invariant $g_n$ namely $$\pi\approx\frac{24}{\sqrt{n}}\log(2^{1/4}g_n)\tag{3}$$ and $n=10$ gives the approximation $\pi\approx 3.14122$ but this approach has a story similar to that of equation $(1)$.
|
Do you want a solution that is a compromise between calculation efficiency and ease of understanding? This story might help.
Many, many years ago in school, I was introduced to Fortran and had limited access to a computer. It was physically large but very low power by today's standards. I knew the famous $\frac{\pi}{4} = \tan^{-1}(1)$ formula but I also realised how slowly it converges. I would not be able to get far with the run time available to me.
The internet did not exist yet and the school's and the local public library did not help. I knew how the $\frac{\pi}{4}$ was derived so I played with other trig formulae. I managed to calculate the Taylor series of $\sin^{-1}$ by a mixture of messy differentiation and induction. I figured that if I evaluated $\frac{\pi}{6} = \sin^{-1}(\frac{1}{2})$, the convergence would be linear: twice the terms would give me twice the number of decimal places. In practice, the algorithm was quadratic since if I aimed at $n$ times as many decimal places, I would need $n$ times as many terms and each calculation would take $n$ as long.
I forget the run time that I had but the best I achieved was 500 decimal places.
Many years later but still long ago, I rewrote the program in C and ran it on an idle Unix system at work. In a month, it calculated a million decimal places. For comparison, I ran it on my laptop a few years ago, a million places took 3.5 hours. My Raspberry Pi required 44 hours.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2485558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "55",
"answer_count": 11,
"answer_id": 9
}
|
Center of mass of triangle created by parabola and two lines The parabola $y^2 = 3x$ is given.
Two perpendicular straight lines are drawn from the origin point, intersecting the parabola in points P and Q.
Find equation (in cartesian form) of the set of centers of mass of all triangles OPQ (where O is the origin point).
Please help, i managed to notice, that one line is $ax$, the second is $-\frac{1}{a}x$. Also, the centers of mass move on the shape of another parabola as we rotate the lines.
|
Step 1: calculate points $P$ and $Q$. You already have the equation for those lines, make them intersect with the parabola. Assume $a>0$. Note that $x>0$
$$ax=\sqrt{3x}$$ yields $$x=\frac{3}{a^2}$$ and $$y=\frac{3}{a}$$
The second equation is $y=-x/a$. You square it and get $$x=3a^2$$ and $$y=-3a$$
Step 2: The center of mass is at the average position of the vertices $(0,0)$, $(3/a^2,3/a)$, $(3a^2,-3a)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2485976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
}
|
Proof explanation for Order of Subgroups, by Fraleigh Theorem: $G$ a cyclic group, $H < G$, $H = \langle a^s \rangle$ then $|H|=\frac{G}{\operatorname{gcd}(s,n)}.$
Proof:
Let $|H|=| \langle a^s \rangle |=m$ for $m$ smallest such that $(a^s)^m=e$
So $(a^s)^m=e$ if and only if $n|ms$.
How do we find $m$ smallest satisfying $n|ms$? Why does Fraleigh introduce Bezout's Identity, and how does the GCD help?
|
The proof in Fraleigh uses Bezout's identity to show that $n∣ms$ iff $\frac{n}{d}\mid m\frac{s}{d}$ where $d=\gcd(s,n)$, so that $m=\frac{n}{d}$ is the smallest such $m$.
To see this another way, and perhaps see why the $\gcd$ is relevant, clearly we want the smallest $m$ such that $n\mid ms$. Now, $s$ contains some of the factors of $m$ (the product of those is $\gcd(n,s)$) and does not contain others. So to make $ms$ divisible by $n$, $m$ must be the product of those other factors. But this product is precisely $\frac{n}{\gcd(n,s)}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2486045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Shifting of Line Segment
$L_1$ and $L_2$ are skew lines in $3$-dimensional space.
$A_1 C \mathbin{\bot} L_1,L_2$. If line segment $AB$ is shifted parallelly, such that $A$ moves along $L_1$ and reaches $A_1$ while $B$ reaches $B_1$, then prove that $B_1 C \mathbin\bot A_1 C$.
It seems really intuitive but I wasn't able to get it. Can anyone prove it using only simple geometry (lines, congruency or similarity) without vectors?
|
Here's a vector solution, but coordinate-free, and really simple . . .
The advantages? It works in $n$ dimensions, for all $n \ge 2$, and requires minimal visualization.
From $A_1 C \mathbin{\bot} L_1,L_2$, we get
$$(\mathbf{A_1}-\mathbf{C})\cdot(\mathbf{A}-\mathbf{A_1})=0\tag{eq1}$$
$$(\mathbf{A_1}-\mathbf{C})\cdot(\mathbf{B}-\mathbf{C})=0\tag{eq2}$$
Since line segment $AB$ moves in parallel, ending up as line segment $A_1B_1$, it follows that
$$\mathbf{B}-\mathbf{A}=\mathbf{B_1}-\mathbf{A_1}\tag{eq3}$$
The goal is to prove $B_1 C \mathbin\bot A_1 C$, or equivalently
$$(\mathbf{B_1}-\mathbf{C})\cdot(\mathbf{A_1}-\mathbf{C})=0\tag{g}$$
Now here's the intuition . . .
The $3$ equations $(\text{eq}1),\,(\text{eq}2),\,(\text{eq}3)$, represent all the information in the problem, effectively replacing the diagram. Hence, either those $3$ equations identically imply equation $(\text{g})$, or else the implication doesn't hold, in which case, the claim of the problem is false.
With that intuition, elementary algebra (of vectors) should suffice . . .
Subtracting $(\text{eq}1)$ from $(\text{eq}2)$, we get
\begin{align*}
&
(\mathbf{A_1}-\mathbf{C})
\cdot
\bigl((\mathbf{B}-\mathbf{C})
-
(\mathbf{A}-\mathbf{A_1})
\bigr)
=0
\\[4pt]
\implies\;&
(\mathbf{A_1}-\mathbf{C})
\cdot
\bigl((\mathbf{B}-\mathbf{A})
-
(\mathbf{C}-\mathbf{A_1})
\bigr)
=0
\\[4pt]
\implies\;&
(\mathbf{A_1}-\mathbf{C})
\cdot
\bigl((\mathbf{B_1}-\mathbf{A_1})
-
(\mathbf{C}-\mathbf{A_1})
\bigr)
=0
&&\text{[by $(\text{eq}3)$]}
\\[4pt]
\implies\;&
(\mathbf{A_1}-\mathbf{C})
\cdot
(\mathbf{B_1}-\mathbf{C})
=0
\\[4pt]
\end{align*}
which proves equation $(\text{g})$, as required.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2486159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Proving limits without limit theorems I have to prove the following limit without using any limit theorems. I can only do so by using the Archimedean Proprety and the definition of a limit.
I have to prove:
$$\lim_{n\to\infty} \frac{n^2 - 2}{3n^2 + 1} = \frac13$$
|
I'll help get you started. You want to show that N>n => |(n^2 - 2)/(3n^2 + 1) -1/3| < epsilon. (Definition of convergence of a sequence).
First, get a common denominator.
|-7/(9n^2+3)|< epsilon
Then you're gonna want to solve for n in terms of epsilon and set N equal to the value you get for n.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2486254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
}
|
Is a certain inverse image under an equivariant map a submanifold? Suppose $G$ is a Lie group and $K$ a maximal compact subgroup of $G$, and suppose $G$ acts smoothly and properly on a manifold $X$.
Question 1: Suppose we are given a $G$-equivariant smooth map $f:X\rightarrow G/K$. Is $S:=f^{-1}(eK)$ necessarily a submanifold of $X$?
Question 2: If $X$ is now a manifold with boundary, then is $S:=f^{-1}(eK)$ a submanifold with boundary?
Thanks.
|
Suppose $f:X\to G/K$ is smooth and $G$-equivariant, with $G$ a Lie group and $K$ a closed subgroup. Then, in particular, $f$ must be a submersion: if $p\in X$ and $g\in G$, then $f(gp)=gf(p)$, so $f(\exp(t\xi)p)=\exp(t\xi)f(p)$.
Thus, $f^{-1}(gK)$ is a submanifold for every $gK\in G/K$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2486365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Show that MNPQ and ABCD have the same centroid. If ABCD is a quadrilateral and M,N,P,Q are exterior points such that ABM, BNC, CPD, DQA are equilateral triangles. Show that MNPQ and ABCD have the same centroid.
I tried to solve it by vectors but I don't know how can I use that the triangles are equilateral.
|
Let $\varepsilon = e^{-i{\pi\over 3}} = \cos {\pi\over 3} -
i\cdot \sin{\pi\over 3}$ and let $G$, $G'$ be gravity centers of $ABCD$
and $MNPQ$, so
$$ G ={1\over 4}(A+B+C+D)$$ and $$G'= {1\over 4}(M+N+P+Q)$$
Now since we get the vector $\vec{AM}$ with rotation of $\vec{AB}$ around $A$ for angle $-{\pi\over 3}$ we have (and similary for others):
\begin{eqnarray}
% \nonumber to remove numbering (before each equation)
M-A &=& \varepsilon(B-A) \\
N-B &=& \varepsilon(C-B) \\
P-C &=& \varepsilon(D-C) \\
Q-D &=& \varepsilon(A-D)
\end{eqnarray}
If we sun these equation and divide it with 4 we get:
$$ G'-G = 0$$
and we are done.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2486492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Natural density and Poincare's recurrence Let $(X,\mathcal{B},\mu,T)$ be a dynamical system and let $A \in \mathcal{B}$ such that $\mu(A)>0$ and $\forall x \in X$ we define $$L_x=\{n \in \Bbb{N}|T^nx \in A\}$$
Prove that $\mu(\{x:\bar{d}(L_x)>0\})>0$ where $$\bar{d}(L_x)=\limsup_n \frac{|L_x \cap \{1,2...n\}|}{n}$$.
My first thought is that i have to use somewhere Poincare's recurrence theorem.
But i cannot understand the behavior of the density function.
From Poincare we just know that for almost every $x \in A$ we have that $L_x$ is infinite.
Can some give me a hint to solve this?
I clearly do not want a full solution.
I want hint to solve this only with Poincare's Theorem and the properties of the natural density function.(Not Ergodic Theorems)
Any help is appreciated.
Thank you in advance
|
I am also assuming $\mu$ is a probability measure. For a measurable set $A$ with positive $\mu$-measure, define $A_0$ to be the set of points $x\in A$ such that $\overline{d}(L_x)=0$. Let $\chi_{A_0}(.)$ be the characteristic function over the set $A_0$ and for any $x\in X$, consider
$$
F^+(x) = \limsup_{n\to +\infty} \frac{1}{n} \sum_{j=0}^{n-1} \chi_{A_0}\circ T^j(x).
$$
From the definition of $A_0$, it is easy to prove that $F^+(x) = 0$ for every $x\in X$. Also observe that
$$
F^-(x)= \liminf_{n\to + \infty} \frac{1}{n} \sum_{j=0}^{n-1} \chi_{A_0} \circ T^j(x) \geq 0.
$$
Thus, for every $x\in X$, it is well defined
$$
F(x) = \lim_{n\to + \infty} \frac{1}{n} \sum_{j=0}^{n-1} \chi_{A_0} \circ T^j(x) = 0.
$$
Of course for every $n\in\mathbb{N}$ and every $x\in X$, the function $ \frac{1}{n} \sum_{j=0}^{n-1} \chi_{A_0}\circ T^j(x)$ takes value between $0$ and $1$. By the dominated convergence theorem, one concludes that
$$ \mu(A_0) = \lim_{n\to+\infty} \int \frac{1}{n} \sum_{j=0}^{n-1} \chi_{A_0}\circ T^jd\mu= \int \lim_{n\to+\infty} \frac{1}{n} \sum_{j=0}^{n-1} \chi_{A_0}\circ T^jd\mu= \int F d\mu=0.$$
One concludes that $\mu$-almost every point in $A$ has a positive density.
Remark: The same result is true if you change $\limsup$ for $\liminf$ in the definition of density, but I don't know how to prove that without using some ergodic theorem.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2486590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
How do I show that formal logarithm is the inverse of the formal exponential? Let $A$ be a unital commutative and associative $\mathbb{Q}$-algebra.
Define $exp(f):=\sum_{n=0}^\infty \frac{f^n}{n!}$ for each $f\in XA[[X]]$.
Define $log(f):=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} (f-1)^n$ for each $f\in 1+XA[[X]]$.
Hence, we have two maps $exp:XA[[X]]\rightarrow 1+XA[[X]]$ and $log:1+XA[[X]]\rightarrow XA[[X]]$.
I am trying to prove that $log$ map is the inverse of the $exp$ map.
The first thing I tried is to directly show $log\circ exp=id$ and $exp\circ log = id$ by checking if the identities hold for every element $f$, but this does not work well since this way involves too many calculations. For example, $$[X^n]exp(log(f))=[X^n]\sum_{k=0}^n log(f)^k/k! = \sum_{k=0}^n \frac{1}{k!} [X^n]log(f)^k= \sum_{k=0}^n \frac{1}{k!} [X^n](\sum_{l=1}^k \frac{(-1)^{l+1}}{l} (f-1)^l)^k$$.
Thus, we have to show that $$\sum_{k=0}^n \frac{1}{k!} [X^n](\sum_{l=1}^k \frac{(-1)^{l+1}}{l} (f-1)^l)^k=[X^n]f$$.
But this calculation is really a nightmare.. Is there a clever way to show this? If not, how do I wisely calculate to show the above identity?
Thank you in advance.
|
Here is one possible argument. One way to define the exponential $\exp(x)$ as a formal power series is that it is the unique formal power series $f(x)$ (over any commutative $\mathbb{Q}$-algebra) satisfying $f(0) = 1$ and
$$f'(x) = f(x).$$
Repeatedly differentiating this identity easily gives $[x^n] \exp(x) = \frac{1}{n!}$ as expected. Similarly, one way to define the logarithm $\log (1 + x)$ as a formal power series is that it is the unique formal power series $g(x)$ satisfying $g(0) = 0$ and
$$g'(x) = \frac{1}{1 + x} = \sum_{n=0}^{\infty} (-1)^n x^n.$$
So what can we say about the composite $\exp \log (1 + x)$? Well, by the formal chain rule, we have
$$\frac{d}{dx} f(g(x)) = f'(g(x)) g'(x) = f(g(x)) \frac{1}{1 + x}.$$
So $f(g(x))$ is a solution $h(x)$ to the differential equation $h'(x) = h(x) \frac{1}{1 + x}$ with initial condition $h(0) = f(g(0)) = 1$. We clearly have $h(x) = 1 + x$ is a solution, and we can appeal to a formal version of the Picard-Lindelof theorem to assert that formal solutions to ODEs exist and are unique, so we conclude that
$$\exp \log (1 + x) = 1 + x.$$
Similarly, what can we say about $\log \exp x = \log ((\exp x - 1) + 1)$? Well, again by the formal chain rule, we have
$$\frac{d}{dx} g(f(x) - 1) = g'(f(x) - 1) f'(x) = \frac{f'(x)}{f(x)} = 1.$$
So $g(f(x) - 1)$ is a solution to the ODE $\frac{d}{dx} h(x) = 1$ with initial condition $h(0) = g(f(0) - 1) = 0$. Here it's a bit simpler to see that we must have $h(x) = x$, so
$$\log \exp(x) = x.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2486706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
}
|
Prove $1+2+3+...+n<=n^2$ for all n∈N
Prove $1+2+3+...+n<=n^2$ for all n∈N
This is what I have so far. Not sure what to do now.
|
One potential stumbling block is that your "in other words" isn't quite complete. What "$P(k)$ implies $P(k+1)$" really says is that if
$$
1+2+\cdots+k \leq k^2
$$
then
$$
1+2+\cdots+k+(k+1) \leq (k+1)^2
$$
In other words (!), you use the premise that $1+2+\cdots+k \leq k^2$ as the starting point. If you add $k+1$ to both sides, you get
$$
1+2+\cdots+k+(k+1) \leq k^2+(k+1)
$$
Can you finish the line of reasoning?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2486841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Annoying primes first of all, I'm no mathematician at all. I was just playing with prime numbers and ended up with this list:
2 = 2¹
3 = 3¹
5 = 5¹
7 = 3¹ + 2²
11 = 2¹ + 3²
13 = 13¹
17 = 5¹ + 2² + 2³
19 = 2¹ + 3² + 2³
23 = 11¹ + 2² + 2³
29 = 17¹ + 2² + 2³
31 = 19¹ + 2² + 2³
37 = 37¹
41 = 5¹ + 3² + 3³
43 = 7¹ + 3² + 3³
47 = 11¹ + 3² + 3³
53 = 17¹ + 3² + 3³
59 = 23¹ + 3² + 3³
61 = 61¹
67 = 7¹ + 2² + 2³ + 24 + 25
71 = 67¹ + 2²
73 = 61¹ + 2² + 2³
79 = 3¹ + 7² + 3³
83 = 79¹ + 2²
89 = 13¹ + 7² + 3³
97 = 13¹ + 3² + 3³ + 24 + 25
101 = 97¹ + 22
103 = 11¹ + 72 + 33 + 24
107 = 103¹ + 22
109 = 17¹ + 72 + 33 + 24
Edit2: As pointed out by @Charles, 29, 67 and 97 aren't annoying.
I was just playing by trying to write the primes following this form
a1 + b2 + c3 ...
(just one rule, there must be all the exponents, or else the number must be written by itself. So, I can't have a number written in the form a1 + b3 + c4)
Edit: There are actually two rules, I forgot to say that a, b, c... must be primes.
Some of the primes, the bold ones, can only be written with exponent 1. I called then annoying primes, in lack of a better name.
Is it some property of these primes? Is there a formula to express this?
|
I can only find 6 annoying primes: 2, 3, 5, 13, 37, and 61.
You gave these and three other examples, but they don’t hold:
$$29 = 17 + 2^2 + 2^3$$
$$67 = 7 + 2^2 + 2^3 + 2^4 + 2^5$$
$$97 = 13 + 3^2 + 3^3 + 2^4 + 2^5$$
By general density arguments one would expect:
Conjecture: There are only finitely many annoying primes.
I pose this problem which would strengthen my conjecture:
Open problem: Are there finitely many numbers which cannot be expressed with greatest exponent 2, 3, 4, or 5?
Sadly I do not have a computer here to check, perhaps someone else will do so and report back.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2486943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 1,
"answer_id": 0
}
|
Let $p>3$ be a prime number. Prove that for every $a$, $1$ $\lt$ $a$ $\lt$ $p-1$ Let $p>3$ be a prime number. Prove that for every $a$, $1$ $\lt$ $a$ $\lt$ $p-1$, there is a unique b $\neq$ a , $1$ $\lt$ b $\lt$ $p-1$ such that $ab$ $\equiv$ (1 mod p)
I started off with a proof of contradiction where suppose b is not unique and b can equal a. Then by substitution we have $a*a$ $\equiv$ $1 (mod$ $p)$. Not sure how to proceed further. Just to clarify, isn't $1 (mod$ $p)$ always 1? So $a*b$ has to be $pk+1$, k=1,2,3,...?
|
Let $P= \{0, 1, \dots, p-1\}$ and consider $\mu: P \to P$ given by $x \mapsto ax \bmod p$. Then, $\mu$ is injective (*) and so is surjective, because $P$ is finite. Therefore, $\mu$ is bijection and there is a unique $b \in P$ such that $1=\mu(b)=ab \bmod p$.
This defines a map $\iota: P \to P$ given by $a \mapsto b$ such that $ab \equiv 1 \bmod p$. This map $\iota$ is its own inverse and so is a bijection. Since $\iota(1)=1$ and $\iota(p-1)=p-1$, we get $1 < a < p-1 \implies 1 < b=\iota(a) < p-1$. Finally, $\iota(a)=a$ iff $a^2\equiv 1 \bmod p$, that is, $p$ divides $a^2-1=(a-1)(a+1)$ and so $a\equiv \pm 1 \bmod p$, that is $a=1$ or $a=p-1$.
(*) If $ax \bmod p = ay \bmod p$, then $p$ divides $a(x-y)$. Since $p$ does not divide $a$, it must divide $x-y$. Since $|x-y|<p$, we must have $x-y=0$ and so $x=y$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2487049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Mean duration of the renewal intervals $\textbf{Q}$: Let $\{X_n; n = 0, 1,...\}$ be a two-state Markov chain with the transition probability matrix
$$\begin{bmatrix}
1-a & a\\
b & 1-b
\end{bmatrix}$$
where state $0$ represents an operating state of some system, while state 1 represents a repair state. We assume that the process begins in state $X_0 = 0$, and then the successive returns to state 0 from the repair state form a renewal process. Determine the mean duration of one of these renewal intervals.
$\textbf{Answer provided in the textbook}$: $\frac{1}{a}+\frac{1}{b}$
I am still new to this topic on renewal processes, and I am clueless as to how the above answer is derived, as stated in the book. I thought of deriving the above via the stationary distribution, but I am also clueless on that part. Am I on the right track? Some help will be great!
|
Let $S_0 = \inf\{n>0:X_n=0,X_{n-1}=1\}$ and $S_{k+1} = \inf\{n>S_k:X_n=0,X_{n-1}=1\}$ for $k\geqslant 1$. Then $\{S_n:n\geqslant0\}$ is a renewal process. Let $X\sim\mathrm{Geo}(a)$ and $Y\sim\mathrm{Geo}(b)$ distribution, then $S_{k+1}-S_k\stackrel d=X+Y$, so
\begin{align}
\mathbb P(S_{k-1}-S_k=m) &= \mathbb P(X+Y=m)\\
&=\sum_{j=1}^{m-1} \mathbb P(X+Y=m, X=j)\\
&=\sum_{j=1}^{m-1} \mathbb P(X=j)\mathbb P(Y=m-j)\\
&=\sum_{j=1}^{m-1} a(1-a)^{j-1}b(1-b)^{m-1-j}\\
&=\frac{ab}{1-a}(1-b)^{m-1}\sum_{j=1}^{m-1}\left(\frac{1-a}{1-b}\right)^j\\
&=\frac{ab}{a-b}\left( (1-b)^{m-1}-(1-a)^{m-1}\right).
\end{align}
In particular,
$$
\mathbb E[S_1-S_0] = \mathbb E[X] + \mathbb E[Y] = \frac1a + \frac1b.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2487203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
}
|
How to apply Kuratowski's theorem here
I'm unsure which subgraph to find and which of $K_{3,3}$ or $K_5$ I should compare it to. First this looks like $K_6$. If it was $K_6$, I could simply remove one of the vertices to get $K_5$ and I would be done.
However, there is an extra vertex $I$ in the centre. I thought of removing a vertex say $B$ and then removing edge $IE$ to get $K_5$ with an extra vertex in the middle. Then I'm unsure here because it doesn't seem that this is a subdivision of $K_5$... (the middle point seems like a subdividing vertex for more than one edge)
|
$A$ is joined to $C,E,F$, also $D$ is joined to $C,E,F$, and $B$ is joined to $C$ and $F$, and to $E$ via $I$, so there's also a $K_{3,3}$ there.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2487278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
How to prove $1+{x\over 2}-{x^2 \over 8}<\sqrt{1+x}$ for all $x>0$? How to prove $1+(1/2)x-(1/8)x^2<\sqrt{1+x}$ for $x>0$ by Taylor Expansions up to $n$ order or Mean Value Theorem? I tried to apply MVT on $\sqrt{1+x}$ and get $\displaystyle \sqrt{1+x}=1+\frac{1}{2\sqrt{1+\xi}}x$ for $\xi\in (0,x)$. How to do next?
|
Let $f(x)=\sqrt {1+x}\;.$ For $x>0$ we have $$f(x)=f(0)+xf'(0)+x^2f''(y)/2$$ where $0<y<x.$ We have $f(0)=1$ and $f'(0)=1/2.$ We have $$f''(y)=-\frac {1}{4(1+y)^{3/2}}>-\frac {1}{4} .$$ So $\sqrt {1+x}=f(x)=1+x/2-x^2f''(y)/2>1+x/2-x^2/8.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2487402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
}
|
Elegant proof of combinatorial statement: at least $l-k+1$ children who were in larger teams There are $n$ children, who play a game where in round $1$ they spilt up in $k$ teams (every team non-empty, pairwise disjoint). After a while it gets boring, so they form new teams to play round $2$; this time, they split up in $l$ teams (every team non-empty, pairwise disjoint), where $l>k$, and this beeing the only condition.
Prove that there are at least $l-k+1$ children, for which their team in the first round was of strictly larger size than in the second round.
This was a homework problem in a course last year which I had a very hard time proving, and finally, a day before the deadline, I found a proof which was three pages long and used a pretty messy induction with a whole lot of definitions and notations I had to introduce (it took me the whole day to write).
Is there an elegant proof of this statement?
If you want to see my proof, comment below and I will try to summarize it. But I guarantee you, its very messy! :)
|
Say a child in a team of size $m$ takes up $1/m$ of a team. Say child $i$ takes up $x_i$ of a team in round $1$ and $y_i$ of a team in round $2$. Now we need to show there are at least $l-k+1$ values of $i$ for which $y_i>x_i$.
But $\sum_ix_i=k$ and $\sum_iy_i=l$, so $\sum_i(y_i-x_i)=l-k$. Each term in this sum is strictly less than $1$, so there must be strictly more than $l-k$ positive terms.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2487492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
}
|
Solution to second order non constant differential equation without first derivative Cheers!
So I am studying the book ‘Introduction to quantum mechanics’ by David J. Griffiths for my introductory course of quantum mechanics. On page 51 at the bottom it introduces the differential expression:
\begin{equation}
\Phi’’ = \xi^2 \Phi
\end{equation}
Without saying much further, it presents the general solution as:
\begin{equation}
\Phi = A e^{-\xi^2/2} + B e^{\xi^2/2}
\end{equation}
Indeed, when I check this solution with the differential equation, I can see that it can match it. However I am still completely missing on how one would arrive from the differential equation to the general solution.
I have already been trying several different methods, all unsuccessful, so I would indeed appreciate any help I could get.
|
For solving $y''-(\xi^2-K)y=0$ we have solution
$$y=Ae^{\sqrt{\xi^2-K}}+Be^{-\sqrt{\xi^2-K}}~~~~~,~~~~~\xi^2-K>0$$
with $\xi>>K$
$$\sqrt{\xi^2-K}=\sqrt{-K}\sqrt{1-\dfrac{\xi^2}{K}}=\sqrt{-K}\left(1-\dfrac{\xi^2}{K}\right)^\frac12\approx\sqrt{-K}\left(1-\frac12\dfrac{\xi^2}{K}\right)\approx\sqrt{-K}\left(-\frac12\dfrac{\xi^2}{K}\right)\approx\dfrac{\xi^2}{2i\sqrt{K}}\approx\dfrac{\xi^2}{2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2487622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Solving, a Square with broken diagonal: $AB=x$? I am reading a geometry lecture note and it was given this figure below.
Where ABCD is square of size $AB=x$. $BE =12$, $EF=3$ and $FD=9$.
Please help me to find the value of $x$.
I do not know how to start. So far I know that $BD =x\sqrt 2.$
|
The hint:
Use the following.
$$\measuredangle DFB=90^{\circ}+\arctan4,$$
$$DF=\sqrt{153},$$
$$FB=9$$ and
$$DB=x\sqrt2.$$
I got $x=15$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2487760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
}
|
Choosing $x$, $y$, $z$ parts in a pumping lemma $w$ string I want to proof that $L = \left\{u0v \mid u, v \in \{0, 1\}^* \land \#_1(u) = \#_0(v) \right\} $ is not regular. But my understanding of the pumping lemma is somehow not bulletproof, so I'm not sure if I'm right in what I'm doing.
I chose $w$ to be $1^k00^k = xyz$, so $|w| \ge k$, which is correct I hope. Now I'm not really sure how to correctly choose what is $x,y$ in this string. Can $y$ be any part of $w$ (having in mind the condition $|xy| \le k$)? For example let's $x$ be equal to $1^m$, $y$ then will be $1^n$ and $z$ will be $00^{m+n}$ (where $m+n = k$). Then assuming that $xy^iz \in L$ for $i\ge0$, let $i$ be $0$. Then there is only a string $1^m00^{m+n}$ left, which is not from $L$. Is this proof correct? It makes sense in some way, but I can't say if all the steps I took were alright.
|
Inside any substring of length $>k$ you can find a pumpable substring. So taking that substring inside the $0$ part is OK. You should try to understand the pumping lemma, rather than just applying it, the idea is simple: if you go $k+1$ times to $k$ places, you have to go to the same place at some point (like bar hopping, the automaton is "place hopping").
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2487861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Is the average of two endpoint's instantaneous rates of change always the same as the average rate of change of the interval on a parabola? Is the average of two endpoint's instantaneous rates of change always the same as the average rate of change of the interval on a parabola? If not, which additional circumstances would cause it to always be the same?
EDIT: How would you verify this algebraically? Answered
EDIT 2: They both result in the same answer, but do they get their answer from the same line that touches/intercepts the graph, or do they get their answer from different lines (secant, tangent, etc.) that touch/intercept the graph, and because the two lines are always parallel to each other, they result in the same answer?
|
I'll give you a hint, but you need to know some calculus to understand the problem itself.
Recall that average value of a function $f$ on the interval $[a,b]$ is: $$\frac{f(b)-f(a)}{b-a}$$
And the average of the two endpoints instantaneous rate of change is given by:
$$\frac{f'(b) + f'(a)}{2}$$
Where $f'(x)$ is the derivative at some point $x$.
Now think about a polynomial of the form $f(x) = x^2$. How would you go about showing the two expressions above are the same? And then you could do the same kind of algebraic manipulation for any parabola of the form $f(x) = a(x-h)^2 + k$ with $a,h,k$ some constants. It's just more algebra.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2488152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
If 9 points are chosen from within the rectangle, explain why two of the points must be at most 18cm away from each other. I've been revising for an exam and the following question has stumped me. I thought it might involve pythagorean theorem but after trying it out it doesn't seem to...
"A rectangle has width 6 cm and height 12 cm. If 9 points are chosen
from within the rectangle, explain why two of the points must be at
most 18cm away from each other. [Hint: divide the rectangle into
squares of equal area.]"
Any help would be appreciated.
|
Divide the rectangle into squares $3cm\times 3 cm$, there are $2 \times 4 = 8$ of them. Two of the points will land in the same square, so the distance between them is at most the diagonal of the square $= 3\sqrt{2} cm = 4.24\ldots cm$.
Obs: Maybe the problem stated $\sqrt{18} cm$ ?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2488266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Show that if $f$ is increasing on $[a, b]$ and satisfies the intermediate value property, then $f$ is continuous on $[a, b]$ I know this question has been asked before but I feel like my approach to solving the problem is "different" (EDIT: turned out to be different because it's wrong!)
Since $[a,b]$ is closed and bounded we may conclude that, by the Heini-Borel theorem, $[a, b]$ must be compact. By definition of compactness, every sequence in $[a,b]$ must contain a subsequence that converges to a limit that is also in $[a,b]$. Consider any arbitrary increasing subsequence $\{x_n\}$ inside $[a,b]$. By the definition of compactness, we know that there exist some $c \in [a,b]$ such that $\{x_n\} \rightarrow c$.
Since $f$ satisfies the intermediate value property, we know that $f(c)$ exists and is within the range of $f$. One of the "characterizations of continuity" states that:
"For all $\{x_n\} \rightarrow c$, it follows that $f(x_n) → f(c).$".
How do I show that $f(x_n)$ converges to $f(c)$?
Thing is, I know that $f(x_n)$ has to converge to something since the range of $f$ is a compact set ( $[f(a), f(b)]$)… I'm just trying to show that $f(x_n) \rightarrow f(c)$! Any idea of how I can go about this?
|
Often it is better just to go back to the $\epsilon$-$\delta$ definitions. I will show that $f$ is right continuous first. Left continuity should be easy enough for you to show yourself. Let $\epsilon > 0$. For a fixed $x \in [a,b)$, let $\epsilon' = \min\{\epsilon, f(b)-f(x)\}$, then
$$f(x) < f(x)+\epsilon' \leq f(b)$$
By the intermediate value theorem there exists a $y'\in (x, b]$ so that
$f(y')=f(x)+\epsilon'$, hence because $f$ is increasing, $f(x)<f(y)<f(y')=f(x)+\epsilon'$ for all $y\in (x,y')$.
In other words, take $\delta=y'-x$, then for all $y$ such that $y-x < \delta$,
$$f(y)-f(x) < f(y')-f(x) =\epsilon' \leq \epsilon$$
and we are done.
Alternative Proof
Another useful (and more general) characterization of continuity is that a function is continuous if and only if the preimage of any open interval is open. We may use this as follows:
Take any open interval $(y_1,y_2) \subsetneq [f(a),f(b)]$. Given that $f$ is monotone we know that it is one-to-one and therefore permits a unique inverse $f^{-1}$ on its range. Moreover the preimage must be a subset of the interval $\left(f^{-1}(y_1), f^{-1}(y_2)\right)$. By the intermediate value theorem every value in that interval is attained, so the preimage is the interval we just defined and is therefore open. Hence $f$ is continuous.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2488398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
}
|
Why does $1+\sqrt{6-2\sqrt{5}}=\sqrt{5}$? Actually, I've already found a clumsy proof of my statement, but I want to know if there's something deeper going on here. Specifically, Mathematica's "Simplify" function doesn't seem to know how to reduce this, which caused me hours of woe before I realized that Mathematica was giving me the answer I was looking for in a shrouded form. Is there a reasonable way one could "spot" this identity? Does it fit within a broader family of square root identities? I hope my curiosity and bewilderment about this is understandable—I've just never come across this before and I'm surprised Mathematica doesn't know what to do with it.
Here's my proof: $$1+\sqrt{6-2\sqrt{5}}=x$$ $$6-2\sqrt{5}=(x-1)^2$$ $$5-2\sqrt{5}=x(x-2)$$ $$\sqrt{5}(\sqrt{5}-2)=x(x-2)$$ $$x=\sqrt{5}$$
As a more practical question, does anyone know how make Mathematica "catch" things like this? To help understand where I'm coming from, I was expecting Mathematica to return the matrix $$\begin{pmatrix}
1 & 1 \\
0 & 1
\end{pmatrix}$$ when instead it returned $$\left(
\begin{array}{cc}
\frac{\sqrt{5 \left(5+\sqrt{5}\right)}+\sqrt{50-20 \sqrt{5}}}{5 \sqrt{5-\sqrt{5}}} & \frac{1}{10} \left(\sqrt{10 \left(3+\sqrt{5}\right)}-\sqrt{5}+5\right) \\
0 & \frac{\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}}{\sqrt{10}} \\
\end{array}
\right)$$
I arrived at the identity above by playing around with the upper-left entry, and I haven't even begun with the others, but Mathematica doesn't seem to realize they are equal to 1 even though a numerical calculation suggests they are. Insight appreciated!
|
$$\sqrt{6-2\sqrt{5}}=\sqrt{5}-1$$
$$6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2$$
$$6-2\sqrt{5}=6-2\sqrt{5}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2488546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
Factored $z^4 + 4$ into $(z^2 - 2i)(z^2 + 2i)$. How to factor complex polynomials such as $z^2 - 2i$ and $z^2 + 2i$? I was wondering how to factor complex polynomials such as $z^2 - 2i$ and $z^2 + 2i$?
I originally had $z^4 + 4$, which I factored to $(z^2 - 2i)(z^2 + 2i)$ by substituting $X = z^2$ and using the quadratic formula. However, I'm not sure if there exists any method to factor $z^2 - 2i$ and $z^2 + 2i$?
I would greatly appreciate it if people could please explain how one would go about this.
|
You can also go brute force, it sometimes has advantages when the polar form is not sympathetic.
$z^2=(x+iy)^2=2i\iff x^2+2ixy-y^2=2i\iff\begin{cases}x^2-y^2=0\\xy=1\end{cases}\iff z=\pm(1+i)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2488626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
}
|
About Riesz's lemma Read about Riesz's lemma and its proof as given in the Functional Analysis text (2nd ed) by Taylor.
In the proof it's said that $X$ is the normed linear space. $X_0$ is its closed and proper subspace.
$x_1\in X-X_0$.
$x_0,x\in X_0$.
$h=||x_1-x_0||^{-1}$.
Then it's said that
$$h^{-1}x+x_0\in X_0$$
How?
|
Since $h^{-1}$ is just some scalar, $h^{-1}x+x_0$ is just a linear combination of $x$ and $x_0$. Since $x,x_0\in X_0$ and $X_0$ is a linear subspace, this means $h^{-1}x+x_0\in X_0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2488744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
The sum of the cubes of three consecutive integers is divisible by $9$ The question is in the title and my try comes here:
Let $n\in\mathbb{Z}$, then we have to look at
$$
(n-1)^3+n^3+(n+1)^3=3n(n^2+2)
$$
so if we can show that $n(n^2+2)$ is divisible by $3$ we are done.
So I divide it up into three cases
*
*$n\equiv 0 \ (\mathrm{mod} \ 3)$ means we are done.
*$n\equiv 1 \ (\mathrm{mod} \ 3)$, then $n^2\equiv 1 \ (\mathrm{mod} \ 3)$ and then $3\mid(n^2+2)$ and we are done.
*$n\equiv 2 \ (\mathrm{mod} \ 3)$, then $n^2\equiv 1 \ (\mathrm{mod} \ 3)$ and then again $3\mid(n^2+2)$ and in this case too we get the desired result.
First of all, is this correct? And secondly if it is correct, is there a more elegant way of showing this result?
Thank you!
|
Yes, your way is correct. An alternative method (which I leave you to determine whether it is more elegant or not):
$$(n+1)^3 +n^3 +(n-1)^3 = 3n(n^2+2) = 3n(n^2-1+3)\\=3n((n-1)(n+1)+3)=3(n-1)n(n+1)+9n$$
Of the consecutive numbers $(n-1), n, (n+1)$ one is divisible by $3$, which leaves that the term is divisible by 9.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2488881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
}
|
Miscellenous binomial expansion Prove That the sum of the series: $\sum_{r=0}^{10} (-1)^{r}$ $\binom{10}{r}$ [$\frac{1}{2^{r}}+
\frac{3^{r}}{2^{2r}}+ \frac{7^{r}}{2^{3r}} +\frac{15^{r}}{2^{4r}}$ ....upto m terms =$\frac{2^{10m}-1}{2^{10m}(2^{10}-1)}$
I tried to expand this term but getting more and more complicated
|
Hint:
$$\sum_{r=0}^u\binom ur(-1)^r\left(\dfrac{2^n-1}{2^n}\right)^r=\sum_{r=0}^u\binom ur\left(\dfrac{1-2^n}{2^n}\right)^r=\left(1+\dfrac{1-2^n}{2^n}\right)^u=?$$
Here $n=1,2,\cdots,m$ and $u=10$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2489011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
What's the difference between "function" and "relation" in logic? What's the difference between "function" and "relation" in logic?
Why do we cosider them apart?
I think every relation expresses a function...
|
In FOL (first-order logic) a function symbol has terms (i.e. "names") as input and output.
For every pair $n,m$ of natural numbers, $+(n,m)$ (usually writteh: $n+m$) denotes a number.
Predicate (or relation) symbols have terms as input and produce sentences.
For every pair $n,m$ of natural numbers, $<(n,m)$ (usually writteh: $n < m$) is true or false according to the fact that the number $n$ is less or not than the number $m$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2489204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
}
|
Proof that the number $e$ exists based on the following definition My book defines the number $e$ as follows (Simmons' Calculus With Analytic Geometry, 2nd edition, pg. 265):
$e$ is the number for which $\lim_{h\to0}\frac{e^h-1}{h}=1$
However, it does not provide a proof that such a number exists. How can we prove that? Thanks very much in advance.
|
As you have stated that your book defines $e$ as the number for which
$$\lim_\limits{h\to 0}\frac{e^h-1}{h}=1$$
I must say that, the definition is not mathematically sound and circular in nature and also seems confusing as to whether it uses the exponential function to define $e$.
However, what I think it intended to say is that
If, for some number $x$, we have $$\lim_\limits{h\to
0}\frac{x^h-1}{h}=1$$ then we can say that $x=e$.
This might look similar to your book's definition, but this one is comparatively clear and mathematically sound; still it is circular and I hereby prove it.
This definition can be rewritten as $$\lim_\limits{h\to 0}\frac{x^h-1}{h}=1$$
$$\Rightarrow \lim_\limits{h\to 0}\frac{x^{0+h}-x^0}{h}=1$$
$$\Rightarrow \lim_\limits{h\to 0}\frac{f(0+h)-f(0)}{h}=1 \,\,\,\,\,\, \text{where} \,\,\,f(p)=x^p$$
$$\Rightarrow \frac{d}{dp}(x^p)=1 \,\,\,\,\,\, \text{at} \,\,\,\, p=0$$
$$\Rightarrow x^p \ln x=1 \,\,\,\,\,\, \text{at} \,\,\,\, p=0$$
$$\Rightarrow \ln x=1 $$
$$\Rightarrow x=e $$
So this definition implies that you have already defined natural logarithm without defining $e$, which is quite baseless and illogical.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2489317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
}
|
Determine if $\int_{0}^{1}\sqrt{x}\ln{x} \ dx$ is convergent I'm so utterly lost when it comes to these improper integrals. Everytime i feel like I got the hang of it, I always get the wrong answer.
I can clearly see that the function is not defined for $x=0$. So I need to figure out what happens there by examining the limit
$$\lim_{x\rightarrow0_+}\sqrt{x}\ln{x}.$$
Turns out there does exists a finite value here, which is 0. But how do I show this? Is there any other way to, without even computing the integral, "quickly see" that the function is convergent or divergent?
|
I thought it would be instructive to present a way forward that relies on elementary, pre-calculus tools only. To that end we proceed.
To show that $\sqrt x\log(x) \to 0$ as $x\to 0$, we simply exploit the inequality
$$\frac{x-1}{x}\le\log(x)\le x-1 \tag1$$
which I showed using only the limit definition of the exponential function and Bernoulli's inequality in THIS ANSWER. Note for $0<x\le 1$, we have from $(1)$
$$\frac{x-1}{x}\le \log(x)\le 0 \tag2$$
Replacing $x$ with $x^\alpha$ in $(2)$, multiplying by $\sqrt x$, and using $\log(x^\alpha)=\alpha \log(x)$, we find that for $0<x\le 1$ and $\alpha >0$
$$\frac{\sqrt x-x^{1/2-\alpha}}\alpha\le \sqrt x\log(x)\le 0$$
Restricting $0<\alpha<1/2$ (e.g., Take $\alpha=1/4$.), applying the squeeze theorem yields coveted limit
$$\lim{x\to 0}\sqrt x\log(x)=0$$
as was to be shown!
Note that by enforcing the substitution $x\to x^2$ and using integration by parts, we find that
$$\begin{align}
\lim_{\epsilon\to 0}\int_{\epsilon}^1 \sqrt{x}\log(x)\,dx&=\lim_{\epsilon\to 0}\left.\left(\frac29 x^{3/2}(3\log(x)-2)\right)\right|_{\epsilon}^1\\\\
&=\lim_{\epsilon\to 0}\left(-\frac49 -\frac29 \epsilon^{3/2}(3\log(\epsilon)-2)\right)\\\\
&=-\frac49
\end{align}$$
So, the integral converges and the value to which it converges is $-\frac49$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2489469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Proof that $ 2\sqrt{x_1x_2} \le x_1+x_2 $ We have
$$ x_1, x_2,...,x_n \ge 0 $$
and $$ P(n): x_1x_2....x_n \le \left(\frac{x_1+...+x_n}{n}\right)^{n} $$
I have to prove that P(2) is valid.
$$x_1 x_2 \le \left(\frac{x_1+x_2}{2}\right)^{2} $$
I don't know how to achieve this, this is what I tried so far:
$$ \sqrt{x_1x_2} \le \frac{x_1+x_2}{2} $$
$$ 2\sqrt{x_1x_2} \le x_1+x_2 $$
Here, I don't know how to go further. Is this a good way proving this or am I completely wrong?
|
The important parts about the proof are already given in the answers up there. I want to share something different, geometrical actually (not that it works for a proof though) Consider a circle $OAB$ (the center is $T$) with a diameter $|AB|$ now inside the circle choose a point $P$ ($P\in|AB|$)such that $|AP|=x_1$ and $|BP|=x_2$ and then choose a point $K$ on the circle such that $<KPB=90^\circ$
As we know from Euclid's theorem $|KP|=\sqrt{x_1x_2}$ and as we know that if this segment $|KP|$ tends to the center $T$ it will be closer and closer to the radius which is $\dfrac{x_1+x_2}{2}$ so we can conclude that it is smaller or equal to the radius, Thus;
$$\sqrt{x_1x_2}\leq\frac{x_1+x_2}{2}$$
As I've said this is completely partial, although fun If you draw the shape and see. This also works only for two variables.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2489592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Electrical Potential Here is my question
A proton (mass $\text {m}$, charge $+e$) and an alpha particle (mass
$4\text {m}$, charge $+2e$) approach one another with the same initial
speed $\text {v}$ from an initially large distance. How close will
these two particles get to one another before turning around?
The kinetic energy of the moving particles is completely transformed into electric potential energy at the point of closest approach.
$$U_e = K$$
I want to get your helps If that's possible.
|
Read up on elastic collision. When particles collide, total momentum and total kinetic energy are both conserved. Easy equations to work with. You will have two more equations to work with from this setup.
I think you will find your scenario in this article where they have worked on it. It is not as complex as it looks just read through it.
infinitylord's solution for the two equations is almost correct. The right-hand side of the equation from conservation of momentum is wrong. Come up with your own.
P.S. In reality, it all depends on the velocity. If they are traveling too fast(close to the speed of light, their mass will start to change as well then you will need more equations) since you have Hadron Collider experiment going on.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2489713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Algebraic and geometric multiplicities of eigenvalues of a $3 \times 3$ matrix "Let A = \begin{pmatrix}-1&-1&-2\\ \:2&2&1\\ \:6&2&6\end{pmatrix}
Find the characteristic polynomial of A.
Find the distinct eigenvalues of A and their respective algebraic and geometric multiplicities."
So I calculated the value of the characteristic polynomial and got x^3 -7x^2 + 16 x -12 or (x - 3)(x - 2)(x - 2) and got the eigenvalues of 2 and 3. I believe these are the correct answers. However, I am not sure how to give the geometric and algebraic multiplicities for the eigenvalues. I thought that for the eigenvalue 3 the multiplicity would have been 1 and for the eigenvalue 2 it would have been 2. But it seems like I need two multiplicity values for each.
Any help?
|
The numbers you are quoting are the algebraic multiplicity; i.e., $\lambda$ has algebraic multiplicity $n$ when the factor $(x - \lambda)$ appears exactly $n$ times in the factored characteristic polynomial.
The geometric multiplicity of an eigenvalue $\lambda$ is the dimension of the eigenspace corresponding to $\lambda$. That is, what is the size of the largest set of linearly independent eigenvectors you can create for $\lambda$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2489836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
}
|
Prove that $\sqrt{3} \notin \mathbb{Q}(\sqrt2)$ I'm trying to prove that $\sqrt{3} \notin \mathbb{Q}(\sqrt2)$
Suposse that $\sqrt{3}=a+b\sqrt{2}$
$\begin{align*}
\sqrt{3}&=a+b\sqrt{2}\\
3&=(a+b\sqrt{2})^2\\
3&=a^2+2\sqrt{2}ab+b^2\\
(3-a^2-12b^2)^2&=(2\sqrt{2}ab)^2\\
9-6a^2-12b^2+4a^2b^2+a^4+4b^4&=8a^2b^2
\end{align*}$
But I don't know what else I can do here.
|
Note that $1$ and $\sqrt{2}$ is a basis for $\mathbb{Q}(\sqrt{2}).$
Hence from
$$3=a^2+b^2+2\sqrt{2}ab$$
We have $a^2+b^2=3$ and $2ab=0$.
From there, you should be able to see a contradiction.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2489947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
}
|
why does $2^{-n}n^{1000} $ converge by the limit comparison test? the series $\displaystyle\sum\limits_{n=1}^{\infty} 2^{-n}n^{1000}$ converges by the comparison test.
$2^{-n}\leqslant \displaystyle\frac{c}{n^{1002}}$
$\displaystyle\sum\limits_{n=1}^{\infty} 2^{-n}n^{1000} \leqslant \displaystyle\sum\limits_{n=1}^{\infty}n^{1000}\frac{c}{n^{1002}} =c\displaystyle\sum\limits_{n=1}^{\infty}\frac{1}{n^{2}} $
$2 \geqslant 1$ converges by the p-test thus $\displaystyle\sum\limits_{n=1}^{\infty} 2^{-n}n^{1000}$ converges by the comparison test.
my question is, what is to prevent someone from choosing $\displaystyle\frac{c}{n^{1001}}$ which would make the series harmonic thus divergent instead of convergent. I don't feel like it is a valid proof for the series. maybe there is something i am missing here, can someon explain to me why this is a valid proof for the convergences of the series.
|
Because for large $n$, $2^{-n}$ decreases rapidly as compared to $\frac{1}{n^p}$ for any $p>0$. This you can prove by using Binomial theorem. So there exists $n_0$ and $c$ such that $$2^{-n}<\frac{c}{n^p}, ~~\text{for } n>n_0.$$
Now if $a_n\leq b_n$, and if $\sum b_n$ is divergent, by Comparison test nothing can be derived about convergence/divergence of $\sum a_n$. Hence for $p=1001$ nothing can be derived, but for $p=1002$ Comparison test works.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2490092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
For any additive function, $f: (\mathbb{Q}, +)\to (\mathbb{Q}, +)$, why does $f(nx) = nf(x)$? For the homomorphism, $f: (\mathbb{Q}, +)\to (\mathbb{Q}, +)$, where
$\mathbb{Q}$ is the set of rational numbers.
Show that for all $x\in\mathbb{Q}$ and for all $n\in\mathbb{Z}$, we have $f(nx) = nf(x)$
I can't figure out how to prove this without a definition of $f$.
|
We have that $f(nx)=f(\underbrace{x+x+\cdots +x}_{n\:\text{times}})=\underbrace{f(x)+f(x)+\cdots+f(x)}_{n\:\text{times}}=nf(x)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2490194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Proving difference quotient is equal to derivative at 0 everyone, thankfully I was able to figure out parts 1 and parts 3. However, I am stuck on part 2. The conditions somewhat confuse me. All I so far was (first steps are similar to part 1):
Start with:
$f(a_n)=f(0)+a_n(f'(0)+c_n $
$f(b_n)=f(0)+b_n(f'(0))+d_n $
such that $|c_n/a_n| $ goes to 0 and $|dn/bn| $ goes to 0.
From there I get plugging in for $f(b_n)$ and $f(a_n)$:
$D_n=f'(0)+(d_n-c_n)/(b_n-a_n) $
So now I somehow need to show that:
$\lim (d_n-c_n)/(b_n-a_n)=0 $ to show that $D_n=f'(0)$, using the conditions $0<a_n<b_n$ and $b_n/(b_n-a_n)$. The problem is I'm not sure how to use these conditions to show that the limit goes to 0. For part 1, it was pretty simple using Squeeze Theorem. However, I can't figure out how to do it now with these new conditions. Any help would be much appreciated. Thank you.
|
Note that $$0\leq\left|\frac{d_n-c_n} {b_n-a_n} \right|\leq\frac{|d_n|+|c_n|} {b_n-a_n} =\left(\frac{|d_n|} {b_n} +\frac{|c_n|} {b_n} \right) \frac{b_n} {b_n-a_n} \leq \left(\frac{|d_n|} {b_n} +\frac{|c_n|} {a_n} \right) M$$ and now the last expression tends to $(0+0)M=0$ so that by squeeze theorem $|d_{n} - c_n|/(b_n-a_n) \to 0$ and your job is done.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2490337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Counting techniques to find all nonisomorphic graphs with six vertices, all having degree $2.$ First of all, the graphs considered are undirected & parallel edges and loops are allowed.
My attempt : By handshaking lemma, $$\text{Sum of degrees of all vertices} = 2 (\text{Number of edges})$$
Thus we get $\frac {2+2+2+2+2+2}{2}=\text{Number of edges}=6.$
Therefore we have to find all nonisomorphic graphs with six vertices and six edges.
Then I formed six cases based on number of cycles in such nonisomorphic graphs. Each case was then divided into subcases where we count graphs with number of loops and parallel edges in it.
My question is whether there exist other counting techniques in
such scenario?
|
Each one of those isomorphism classes can be represented as an integer partition of 6, that is, a way of writing positive numbers which add up to 6, where the order doesn't matter. For example your 3(i), 3(ii) and 3(iii) correspond to $2 + 2 + 2, 4 + 1 + 1$ and $3 + 2 + 1$ respectively. There is a well-understood theory of integer partitions which would allow you to, for example, compute the number of partitions of some large integer $n$ without having to explicitly write them out, and therefore to compute the number of nonisomorphic graphs with $n$ vertices all having degree 2. (In the $n = 6$ case that theory is probably overkill.)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2490502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
How can I prove the convexity of $f(x)=x^{-\frac{1}{3}}$ by using the definition I'd like to prove that the function $f(x)=x^{-\frac{1}{3}}$, with $x> 0$ is convex. Actually, I already know it's convex because I have studied its derivatives but I'd like to give a more "formal" prove by using convex definition, that is:
Let be $f:S \to \mathbb{R}$ a function. The function f is said to be convex in $S$ if $\forall x,y \in S$ and $\forall \lambda \in (0,1)$ the following holds:
$$f(\lambda x+(1-\lambda)y)\leq\lambda f(x)+(1-\lambda)f(y)$$
I know that I should prove this:
$$(\lambda x+(1-\lambda)y)^{-\frac{1}{3}} \leq \lambda x^{-\frac{1}{3}}+(1-\lambda)y^{-\frac{1}{3}} $$
But I don't know what can I do in order to do it.
Any help would be aprecciated!
|
Hint:, As $t \mapsto t^3$ is increasing, we may cube both sides and replace $x, y$ with $a^3, b^3$ to equivalently prove:
$$ \left(\frac{\lambda}a + \frac{1-\lambda}b \right)^3 (\lambda a^3 + (1-\lambda) b^3) \geqslant 1 \tag{$\star$}$$
which follows from Hölder's inequality ...
--
P.S. Hölder's inequality for our case (positive numbers $a, b, x, y$ and $p, q > 0$ s.t. $1/p + 1/q = 1$.
$$(x_1^p + x_2^p)^{1/p} (y_1^q+y_2^q)^{1/q} \geqslant x_1y_1 + x_2y_2$$
Now if we let $x_1 = (\lambda/a)^{3/4}, x_2 = ((1-\lambda)/b)^{3/4}$, $y_1 = (\lambda a^3)^{1/4}, y_2 = ((1-\lambda)b^3)^{1/4}$ with $p = 4/3, q = 4$, we get:
\begin{align}
\left(\frac{\lambda}a + \frac{1-\lambda}b \right)^{3/4} (\lambda a^3 + (1-\lambda) b^3)^{1/4} & \geqslant \left(\frac{\lambda}{a}\right)^{3/4}\left( \lambda a^3\right)^{1/4} + \left(\frac{1-\lambda}{b}\right)^{3/4}\left((1- \lambda) b^3\right)^{1/4} \\&= \lambda+(1-\lambda) = 1
\end{align}
which is essentially $(\star)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2490635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Cubed exponent equation
$$\left(2 · 3^x\right)^3 + \left(9^x − 3\right)^3 = \left(9^x + 2 · 3^x − 3\right)^3$$ Solve the equation
I got the answer to the problem, in which I evaluated the whole expression (which was quite hard) which was $0$ and $1/2$, is there a way to solve this problem in a less tedious way?
|
Using $t:=3^x$ and following the factorization by @Arthur ($(a+b)^3-a^3-b^3=3ab(a+b)$), the equation is equivalent to
$$t(t^2-3)(t^2+2t-3)=t(t-\sqrt 3)(t+\sqrt 3)(t-1)(t+3)=0.$$
Only the positive $t$ yield a solution and we immediately have
$$x=\frac12\text{ or }x=0.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2490764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
}
|
Infinite sum of squares It is known that $1 + \frac{1}{4}+\frac{1}{9}+\frac{1}{16} + \cdots = \frac{\pi^2}{6}$
. Find the sum
$1 + \frac{1}{9}+\frac{1}{25} + \frac{1}{49} + \cdots$.
What method can we use to answer this? I tried expressing the 2nd equation into 2 fractions which contain the first summation but i couldnt find one
|
$S = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} +... $
Take 1/4 common from the terms whose denominator is even.
$S = 1 + \frac{1}{9} + \frac{1}{25} + ... + \frac{1}{4} ( 1 + \frac{1}{4} + \frac{1}{9} + ... ) $
$S = 1 + \frac{1}{9} + \frac{1}{25} + ... + \frac{1}{4} S$
$ 1 + \frac{1}{9} + \frac{1}{25} + ...= \frac{3}{4} S = (3/4)(\pi^2 /6) $
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2490853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Find Matrix A by certain operations Given as such:
$4(A^T+2I)^{-1} =$ \begin{bmatrix}1&1\\-3/2&1/2\end{bmatrix}
Now to find Matrix A, would I have to inverse the matrix on the RHS to make it equal to A on the LHS and perform the opposite operations? A little stumped on the operations necessary to get the right result.
Just need some clarification, thanks.
|
Hints:
Divide both sides by $4$ to get
$$(A^T+2I)^{-1}=\left[\begin{array}{cc}
1/4 & 1/4 \\
-3/8 & 1/8 \\
\end{array}\right]$$
Then take the inverse of both sides:
$$A^T+2I=\left[\begin{array}{cc}
1/4 & 1/4 \\
-3/8 & 1/8 \\
\end{array}\right]^{-1}.$$
I'll leave it to you to find the inverse.
Then subtract $2I$ from both sides:
$$A^T=\left[\begin{array}{cc}
1/4 & 1/4 \\
-3/8 & 1/8 \\
\end{array}\right]^{-1}-2I$$
Lastly, take the transpose of both sides, and note that $(A^T)^T=A$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2490985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
How many primes do I need to check to confirm that an integer $L$ is prime? I recently saw the 1998 horror movie "Cube", in which a character claims it is humanly impossible to determine, by hand without a computer, if large (in the movie 3-digit) integers are prime powers, i.e. they are divisible by exactly one prime number. Naturally I decided to try working this problem by hand on paper.
Firstly, In order to determine if a number is a prime power, I only needed to find one prime factor. For example, $5$ is a prime factor of $555$ (because $555 = 5*111$), but the other factor ($111$), is not divisible by five, so therefore $555$ has more prime factors than just $5$, and $555$ is not a prime power.
I was initially just checking all the prime numbers less than the integer $L$, in order, to see if they were factors of $L$. For 3-digit numbers this usually doesn't take that long, but if $L$ itself is a prime, then you'll end up checking every prime less than $L$.
So, the question becomes: Given an arbitrary integer $L$, if you perform a brute force search, checking every prime number from a list, what is the minimum number of primes you would need to check before you could stop and conclude that $L$ is prime?
I decided to try to narrow down the search space. If I have a prime number $P$, where $\frac{L}{2}<P<L$, then that prime number could not possibly be a factor of $L$, because $2P>\frac{2L}{2}$, or, $2P > L$. Similarly if $\frac{L}{3}<P<L$, then, once again, $P$ could not possibly be a factor of $L$, because $3P > L$, and I've already checked that $2$ is not a factor of $L$, so $2P ≠ L$. You could make this same argument for $\frac{L}{5}, \frac{L}{7}, \frac{L}{11}$, etc.
In other words, I don't need to check every prime number, I just need to check every prime number up to the point where $\frac{L}{P_{n}} < P_{n}$, ($P_{n}$ is the nth prime number) because, at that point, all primes before $P_{n}$ have been ruled out, and since $P_{n} > \frac{L}{P_{n}}$ then $(P_{n})^2 > L$, thus $P_{n}$ is not a factor of $L$. Also, if $P_{n} > \frac{L}{P_{n}}$ then $P_{n + 1} > \frac{L}{P_{n + 1}}$, because $P_{n + 1} > P_{n}$, this same reasoning rules out all larger primes.
So, for example, when trying to find a prime factor of $607$, rather than checking every prime number less than $607$, I only need to check the first $10$ primes up to $29$, because $\frac{607}{29} < 29$. If the first $10$ primes are not factors of $607$, then $607$ has no prime factors and must be prime.
Is my reasoning valid, and, is it possible to reduce the number of primes you'd need to check even more?
|
Is my reasoning valid, and, is it possible to reduce the number of primes you'd need to check even more?
Yes, your reasoning is valid. This method of trying to deduce whether a number $N$ is prime by testing for prime factors up to $\sqrt N$ is called trial division.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2491045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 6,
"answer_id": 2
}
|
Convergence of $\sum_{n=0}^{\infty} a_n \overline{z}^n$ based on $\sum_{n=0}^{\infty} a_n z^n$. Let $\sum_{n=0}^{\infty} a_n z^n$ be a convergent series, where $\{a_n\}_{n \in \mathbb{N}} \in \mathbb{R}$ and $z \in \mathbb{C}$. Is $\sum_{n=0}^{\infty} a_n \overline{z}^n$ a convergent series?
This question popped into my mind as I was working on some homework for my analysis class - my gut says that it should be so (moreso I feel it should hold even if $\{a_n\} \in \mathbb{C}$ but that seems much more difficult to prove), but I am struggling to find a proof for this.
I feel like there's some intuitive or clever way to show that the boundedness of $\sum_{n=0}^{\infty} a_n z^n$ implies the boundedness of $\sum_{n=0}^{\infty} a_n \overline{z}^n$, which is why I restrict the $\{a_n\}$s to reals, as any nice relationship I could find between the bounds of these series is obliterated when their coefficients are complex.
I am particularly interested in either a proof or counterexample for the case of complex $\{a_n\}$s, but this problem has haunted my sleep for long enough and any proof of either real or complex $\{a_n\}$s would be more than sufficient. Any help would be greatly appreciated! Thanks so much in advance.
|
Let $z=\displaystyle \sum_{n=0}^{\infty} a_nz^n$. By definition we have that for every $\varepsilon >0$, there exists $N \in \mathbb{N}$ such that $$\left \Vert \sum_{n=0}^m a_nz^n-z \right \Vert <\varepsilon, \: \forall m \geq N.$$ Note that $$\left \Vert \sum_{n=0}^m a_n\overline{z^n}-\overline{z} \right \Vert=\left\Vert \sum_{n=0}^m \overline{a_nz^n}-\overline{z} \right\Vert=\left \Vert \overline{\sum_{n=0}^m a_nz^n}-\overline{z} \right\Vert=\left\Vert \overline{\sum_{n=0}^m a_nz^n-z} \right\Vert=\left\Vert \sum_{n=0}^m a_nz^n-z \right\Vert.$$ Hence, for $m \geq N$ we get that $$\left\Vert \sum_{n=0}^m a_n\overline{z_n}-\overline{z}\right\Vert<\varepsilon.$$ So $\sum a_n \overline{z}^n$ converges to $\overline{z}$.
Note that we used the property $\Vert w \Vert=\Vert \overline{w} \Vert$, for every $w \in \mathbb{C}$, where $\Vert w \Vert:= \sqrt{\Re w+\Im w}$ is the usual norm in the space of complex numbers.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2491152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
}
|
Prove for all all $x\in\mathbb{R}: \exp(x-1) \geq x$ i just tried to solve this question, which is a small part of a bigger one.
Prove for all all $x \in \mathbb{R}: \exp(x-1) \geq x$
My first attempt was to simplify:
$$ e^{x-1} \geq x $$ $$ \Leftrightarrow \ln(e^{x-1}) \geq \ln(x) $$ $$ \Leftrightarrow x-1 \geq \ln(x) $$ $$ \Leftrightarrow x \geq \ln(x)+1 $$ $$ \Leftrightarrow e^x \geq e^{\ln(x)}+e^1 $$ $$ \Leftrightarrow e^x \geq x+e $$
My idea is that $e$ power $x$ is of course greater equal $x+e$. Is this correct and if yes is it enough?
|
the answer Michael gave is a good answer but if you want different way,
$$e^{x-1}=x\implies x=1$$ now because both $x$ and $e^{x-1}$ are continuous you just need to check one value of $x$ that is above $1$ and one that is less. for example:$$\text{for $x<1,\,x=0$},\,e^{0-1}={1\over e}\gt0\\\text{for $x>1,\,x=2$},\,e^{2-1}=e\gt2$$
therefore $$e^{x-1}\ge1$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2491282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Does there exist a smooth approximation of $x \bmod y$? I'm looking for a function $m(x,y)$ that smoothly approximates $x \bmod y$, and I'm assuming there would be some $n$ or $\varepsilon$ in the body of $m(x,y)$ that defines the degree of approximation such that as $n$ goes to infinity or $\varepsilon$ goes to zero,
$$
m(x,y)= x \bmod y.
$$
I've looked at, for example, $\sin(x/y)$ and have been trying to figure out ways to make $\sin(x)$ more "steep" so that I can apply such techniques to $\sin(x/y)$ in order to perhaps more closely approximate $x \bmod y$.
The analogy I'm looking at is the approximation of $\mid x\mid$ as $\sqrt{x^2 + \varepsilon)}$ where epsilon is some small quantity.
Does a similarly smooth function exist for $x \bmod y$?
|
One approach I found was to find a smooth floor function, and then simply apply it to the equation:
$$
x \bmod y=x-{\lfloor}\frac{x}{y}{\rfloor}y
$$
And a definition for the smooth floor function I derived is:
$$
{\lfloor}\frac{x}{y}{\rfloor}=-\frac{1}{2} + x + a ln(1/a) -\frac{i}{2 \pi} \left[ln\left(1-e^\left(-2 i \pi \left(x+a
ln(1/a)\right)\right)+a\right) - ln\left(1-e^\left(2 i \pi \left(x+a
ln(1/a)\right)\right)+a\right) \right]
$$
As you make a smaller and smaller, this increasingly approximates the floor function.
Update
I realized after posting that @user856 had already posted a much simpler approximation quite a while ago.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2491494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
}
|
Number of conjugates of $S_3$ in $S_4$ How to quickly deduct there are 4 conjugates of $S_3$ in $S_4$? Since conjugate subgroups are isomorphic, we can have at least 4 conjugates of $S_3$ in $S_4$. But I'm not sure why there aren't more.
*I'm aware there are posts that address similar questions but most involve "labeling" schemes or graphs. I'm yet to find a concise approach to this problem.
|
Let $G = S_4$ act on its subgroups by conjugation. The conjugates of the standard copy of $S_3$ constitute an orbit for this action. The size of the orbit is the index of the stabilizer subgroup. The stabilizer $N$ is what's called the normalizer. It is the set of $\pi \in S_4$ such that
$\pi S_3 \pi^{-1} = S_3$. One has $S_3 \subseteq N \subseteq S_4$ just by definition. Show that this together with the observation that the size of the orbit is at least 4 implies that $S_3 = N$, and the size of the orbit is exactly $4$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2491599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
How do I prove that $\phi:\mathbb{Z}\rightarrow\mathbb{Z}_n$ is a homomorphism? I was wondering how one shows that $\phi(m):\mathbb{Z}\rightarrow\mathbb{Z}_n$ is a homomorphism? I know that we define $m:=qn + r$, but I don't really know how to continue from there.
|
You have $m=qn + r,$ and there you should mention that $r\in\{0,1,2,\ldots,n-1\}.$
The homomorphism would be $\varphi(m) = r.$
Showing that that is a homomorphism means showing that $\varphi(m_1+m_2) = \varphi(m_1)+\varphi(m_2).$
That means if $m_1 = q_1 n + r_1$ and $m_2 = q_2 n + r_2$ and $m_1+m_2 = q_3 n + r_3$ then $r_1+r_2\equiv r_3\pmod n.$ That means $(r_1+r_2)-r_3$ is a multiple of $n.$ So observe that
$$
r_1+r_2-r_3 = (q_1+q_2-q_3) n.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2491697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
If a vector space has a countable basis, can we construct it? Given that we cannot (?) exhibit a basis for a vector space of all real sequences (certainly that basis would be uncountable), but there is a countable basis for polynomials, i.e. $1,x,x^2,x^2,\ldots$, I would like to know,
If a vector space $V$ has a countable basis, can it be always constructed/enumerated (or whatever the correct term is)?
|
It is known that there is a computable vector space over $\mathbb{Q}$ with no computable basis. This also shows that we cannot prove constructively that every countable vector space over $\mathbb{Q}$ has a basis.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2491820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Determine constant c so that g(x,y) is continuous at every point $$g(x,y)=\begin{cases} \frac{x^3+xy^2+2x^2+2y^2}{x^2+y^2} & \text{if} & (x,y) \neq (0,0) \\
c & \text{if} & (x,y) = (0,0) \end{cases}$$
Should I set the first function equal to c and then solve using polar coordinates?
|
Hint. By using polar coordinates one gets, for $(x,y)\ne (0,0)$,
$$
g(x,y)=\frac{x^3+xy^2+2x^2+2y^2}{x^2+y^2}=\frac{r^3\cos^3 \theta+r^3\cos\theta \sin^2\theta +2r^2}{r^2},\quad r\ne0,
$$ that is
$$
g(x,y)=r\cos^3 \theta+r\cos\theta \sin^2\theta +2,\quad r\ne0,
$$ then this tends to $2$ as $r \to 0^+$.
Can you take it from here?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2491911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Computing: $ \lim_{(x,y)→(0,0)}\frac{(x^5+y^5)\ln(x^2+y^2)}{(x^2+y^2)^2} $ $$
\lim_{(x,y)→(0,0)}\frac{(x^5+y^5)\ln(x^2+y^2)}{(x^2+y^2)^2}
$$
The answer is 0. Cannot seem to understand how the answer is 0.I know that the first part is 0 but I'm confused on how to deal with the natural log?
Why is squeeze theorem not a good approach? $0<|(x^5)<(x^2)^2|=|x|$ so $|x^5+y^5|<|(x^2+y^2)^2|$ so therefor
$$0< \dfrac{|x^5+y^5|}{|(x^2+y^2)^2|}<1$$ then multiply both sides with $\ln(x^2+y^2)$. and take the $$\lim_{(x,y)\to (0,0)} \ln(x^2+y^2).$$ and since it is not $0$ but its negative infinity the limit doesn't exist by the squeeze theorem. which is not the right answer.
|
Hint. Note that by letting $x=\rho\cos(\theta)$ and $y=\rho\sin(\theta)$, we have that as $(x,y)\to(0,0)$ then $\rho\to 0$ and
$$0\leq \left|\frac{(x^5+y^5)\ln(x^2+y^2)}{(x^2+y^2)^2}\right|\leq
\frac{(|\rho\cos\theta|^5+|\rho\sin\theta|^5)|\ln(\rho^2)|}{\rho^4}
\leq \frac{\rho^5(1+1)|\ln(\rho^2)|}{\rho^4}={4\rho|\ln(\rho)|}.$$
Can you take it from here?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2492061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
}
|
simple sum inequality I have an inequality for sums that I can't proof, although I know it is true.
Let $h_{ij} = h_{ji}$ a real $n\times n$ matrix, and $h_{ijk} = - h_{ikj}$ a real $n\times n\times n$ tensor with $$\sum_{i=1}^n h_{ii} =0, \quad \sum_{j=1}^n h_{jji} = 0, \, \forall \, 1 \leq i\leq n$$
Then there should follow $$\sum_k \Big( \sum_i h_{ii} \, h_{iik} \Big)^2 \leq \sum_j h_{jj}^2 \sum_{i,k} h_{iik}^2.$$
I haven't managed to show it and I would be very grateful for help!
|
Isn't this just an application of the Cauchy-Schwarz inequality in $l^2$?
Let $a = (h_{ii})_i$, $b = (h_{iik})_i$. We have $$\bigg(\sum_{i=1}^n a_i \, b_i \bigg)^2 = (a,b)^2_{l^2} \leq \|a\|_{l^2}^2 \, \|b\|_{l^2}^2 = \sum_{i=1}^n h_{ii}^2 \sum_{i=1}^n h_{iik}^2.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2492237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Inequality proof as a part of calculus lesson As part of a calculus lesson I was required to prove that:
(1) if $\ |x-3| < \frac{1}{2},\ $ then $\ \bigg|\displaystyle{\frac{\sin(x^2 -8x+15)}{4x-7}}\bigg| < \frac{1}{2}$
So, by using $|\sin(t)| \le |t|,$ I can prove that:
$$\bigg|\frac{\sin(x^2 -8x+15)}{4x-7}\bigg| \le \frac{5}{6}|x-3|$$
After proving this inequality I assume that the initial requirement is proved, and therefore I'm done, is that it or am I missing something?
Thanks for your assistance :)
|
Because, $x^{2}-8x+15=(x-3)(x-5)\ $, you can start saying that:
First,
$$ \vert{\sin(x^{2}-8x+15)}\vert\leq {x^{2}-8x+15} $$
and then you start like this
$$ \bigg\vert\frac{\sin(x^{2}-8x+15)}{4x-7}\bigg\vert=\frac{\vert{\sin(x^{2}-8x+15)}\vert}{\vert{4x-7}\vert}\leq\frac{\vert x^{2}-8x+15\vert}{\vert{4x-7}\vert}=\frac{\vert{(x-3)(x-5)}\vert}{\vert{4x-7}\vert}$$
Suppose that $\ \displaystyle{\vert{x-3}\vert<\frac{1}{2}}\ $, then
$$ -\frac{1}{2}<x-3<\frac{1}{2}$$
$$ 3-\frac{1}{2}<x<3+\frac{1}{2} $$
$$ \frac{5}{2}<x<\frac{7}{2} $$
and with the last condition you can bound $\displaystyle{\frac{1}{\vert{4x-7}\vert}}$ and $\vert{x-5}\vert$. In fact,
$$ \frac{5}{2}<x<\frac{7}{2} \Rightarrow \frac{5}{2}-5<x-5<\frac{7}{2}-5 $$
this means
$$ -\frac{5}{2}<x-5<-\frac{3}{2} $$
and the absolute value function is decreasing in the negatives real numbers:
$$ \frac{3}{2}<\vert{x-5}\vert<\frac{5}{2}, $$
so you have $$ \vert{x-5}\vert<\frac{5}{2}. $$
Also,
$$ \frac{5}{2}<x<\frac{7}{2} \Rightarrow 10<4x<14 \Rightarrow 3<4x-7<7$$
so $\vert{4x-7}\vert>3$ and this implies:
$$ \displaystyle{\frac{1}{\vert{4x-7}\vert}}<\frac{1}{3}. $$
Then you can multiply both inequalities below:
$$ \vert{x-5}\vert<\frac{5}{2} \qquad \displaystyle{\frac{1}{\vert{4x-7}\vert}}<\frac{1}{3}$$
to get:
$$ \frac{\vert{x-5}\vert}{\vert{4x-7}\vert}<\frac{5}{6}. $$
So,
$$ \bigg\vert\frac{\sin(x^{2}-8x+15)}{4x-7}\bigg\vert\leq\frac{\vert x^{2}-8x+15\vert}{\vert{4x-7}\vert}=\frac{\vert{(x-3)(x-5)}\vert}{\vert{4x-7}\vert}<\frac{5}{6}\vert{x-3}\vert<\frac{5}{6}\cdot\frac{1}{2}=\frac{5}{12}<\frac{1}{2}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2492363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Different version of chromatic number I'm wondering whether or not there is a symbol or value such as the chromatic number of a graph that asks
What is the minimum coloring of the graph such that not only adjacent vertices have different colors, but vertices adjacent to a mutual vertices have different colors.
I'm just curious on whether this is a thing, or is it question that is easily related to the chromatic number and thus is not really thought much about.
|
I've seen this called the distance-2 coloring, where distance-k coloring asks for a vertex coloring where every two vertices of the same color are more than k steps apart. It's used in algorithms and computational results, but I'm not familiar with many of its properties.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2492488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Prove that $\mathbb Z_8$ and $\mathbb Z_{24}/\langle 8\rangle$ are isomorphic.
Use the Fundamental Homomorphism Theorem to prove that $\mathbb Z_8$ and $\mathbb Z_{24}/\langle 8\rangle$ are isomorphic.
The following function is a homomorphism from $\mathbb Z_{24}$ to $\mathbb Z_8$:
$\bigl(\begin{smallmatrix}
0& 1 & 2 & 3 & 4 &5&6&7&8&9&10&11&12&13&14&1&16&17&18&19&20&21&22&23 \\
0 & 1 & 2 &3 & 4&5&6&7&0&1&2&3&4&5&6&7&0&1&2&3&4&5&6&7
\end{smallmatrix}\bigr)$.
Thus we can say $\operatorname{ker}(f)=\{0,8,16\}$.
Thus $\mathbb Z_{24}/\langle 8\rangle$ is a homomorphism to $\mathbb Z_8$.
By the FHT, $\mathbb Z_8 \cong \mathbb Z_{24}/\langle 8\rangle$.
Is this the correct use of FHT? I thought I needed to show that group/ker $\cong$ im?
|
The general idea is correct. To use the first isomorphism theorem here you want to:
*
*Find a homomorphism $\phi:\mathbb{Z}_{24} \to \mathbb{Z}_8$
*Show that $\phi$ is surjective (i.e. $\text{Im}(\phi) = \mathbb{Z}_8$)
*Show that $\ker(\phi) = \langle 8 \rangle$
Then you may conclude that $\mathbb{Z}_{24}/\langle 8 \rangle \cong \mathbb{Z}_8$. In your proof, you claimed to have found the required homomorphism, but you still need to prove that it is indeed a homomorphism. This is the biggest missing link in your proof. Also, though it is obvious from the way you have written the map, it doesn't hurt to explicitly mention in your proof that $\phi$ is surjective, because this is indeed important.
Added: As for your last question. You don't need to show that $\mathbb{Z}_{24}/\ker(\phi) \cong \text{Im}(\phi)$. This is precisely the statement of the first isomorphism theorem. You are simply using this fact in your proof.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2492624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Computing the radioactive probability integral of a non-uniform system How do I integrate $e^{-y\alpha- x\beta- \gamma\sqrt{xy}}\,dy\,dx$, with $x$ and $y$ from $0$ to infinity, i.e.,
$$\int_0^\infty\int_0^\infty e^{-y\alpha- x\beta- \gamma\sqrt{xy}}\,dy\,dx\tag{1}$$ &
$$N\int_0^\infty\int_0^x e^{-y\alpha- x\beta- \gamma\sqrt{xy}}\,dy\,dx\tag{2}$$
where $N~=~(\text{equation } 1)^{-1}$
|
The change of variable $(2\beta x,2\alpha y)\to(x^2,y^2)$ shows that the integral to be computed is $$\iint_{x>0,y>0}e^{-\alpha y-\beta x-\gamma\sqrt{xy}}\,dxdy=\frac1{\alpha\beta}I\left(\frac\gamma{2\sqrt{\alpha\beta}}\right)$$ where, for every $|w|<1$,
$$I(w)=\iint_{x>0,y>0}e^{-Q_w(x,y)/2}\,xydxdy$$ and $$Q_w(x,y)=x^2+y^2+2wxy$$
In particular, $I(w)=-J'(w)$, where $$J(w)=\iint_{x>0,y>0}e^{-Q_w(x,y)/2}\,dxdy$$
Since $|w|<1$, there exists $\vartheta$ in $(0,\pi)$ such that $$w=\cos\vartheta$$ The quadratic form $Q_w$ is positive, the inverse of the matrix of $Q_w$ being $$C=\begin{pmatrix}1&\cos\vartheta\\ \cos\vartheta&1\end{pmatrix}^{-1}=\frac1{\sin^2\vartheta}\begin{pmatrix}1&-\cos\vartheta\\-\cos\vartheta&1\end{pmatrix}$$ with $$|C|=\frac1{\sin^2\vartheta}$$
Thus, $$J(w))=\iint_{x>0,y>0}e^{-(x,y)^*C^{-1}(x,y)/2}dxdy=2\pi\sqrt{|C|}\,P(X>0,Y>0)$$ where $(X,Y)$ is centered normal with covariance $C$. Now, $$\sin\vartheta\, (X,Y)=(X_0,\sin\vartheta Y_0-\cos\vartheta X_0)$$ where $(X_0,Y_0)$ is standard normal hence $$[X>0,Y>0]=[X_0>0,Y_0>\cot\vartheta X_0]$$ The distribution of $(X_0,Y_0)$ is rotationally invariant hence the probability of this event is the angle of the sector $x>0$, $y>\cot\vartheta x$, divided by $2\pi$. This sector is limited by the angles $\frac\pi2-\vartheta$ and $\frac\pi2$ hence
$$P(X>0,Y>0)=\frac\vartheta{2\pi}$$ and $$J(w)=\frac\vartheta{\sin\vartheta}$$
Differentiating this with respect to $\vartheta$ and then $\vartheta$ with respect to $w$, one gets $$I(w)=\frac1{\sin^2\vartheta}\left(1-\vartheta\cot\vartheta\right)$$ Finally, for every $|w|<1$, $$I(w)=\frac1{1-w^2}\left(1-\frac{w}{\sqrt{1-w^2}}\arccos w\right)$$ In particular, $I(-1)$ is infinite, as was to be expected, and one can compute $$I(1)=\frac13$$
One could even deduce $I(w)$ for $w>1$ from this, by analytic continuation... but let us stop here.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2492792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Nth Branch of Lambert W function I have a program to calculate the primary branch of the Lambert W function, how do I calculate the other branches (based off of the first one if possible)?
Example:
$$W(\ln(2)) = 0.44443609101$$
But (using 1st branch)
$$W_1(\ln(2)) = -1.91415552885386478373 + 4.2929649070568775i$$
How can I calculate $W_1(\ln(2))$ based off $W(\ln(2)?)$
And The real question: How can I calculate $W_n(x)$ based off $Wx$), Where $W_n(x)$ is the nth branch of $W(x)$
|
See https://en.wikipedia.org/wiki/Lambertw and get the first reference: Corless et. al. "On the Lambert W function" http://www.apmaths.uwo.ca/~djeffrey/Offprints/W-adv-cm.pdf. In this basic paper the branch $W_k(z)$ is computed in formula 4.20, but I guess in practice the function is calculated with iterations.
You find C++ code for computing all branches at the link given in note [32], it is a bit suboptimal but seems OK (I successfully tested a Pascal port with Maple).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2492925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Finding the Taylor polynomial of $f(x) = \frac{1}{x}$ with induction So I am asked to find the Taylor polynomial of $f(x) = \frac{1}{x}$ about the point $a=1$ for ever n$\in{N}$, and then use induction to justify the answer.
I got the Taylor polynomial which was simple enough:
$$T_{n}(x)=\sum_{n=0}^{\infty} \frac{f^n(x)(a)}{n!}(x-a)^n$$
$$f(x) = 1-(x-1)+(x-1)^2-(x-1)^3+(x-1)^4+...$$
$$T_{n}(x)=\sum_{n=0}^{\infty} (-1)^n(x-1)^n$$
That wasn't too bad. How do I justify this with induction though? I am a little confused as to how I would start this.
I tried writing out the terms as such:
$$1-(x-1)+(x-1)^2-(x-1)^3+(x-1)^4+...+(-1)^k(x-1)^k = \frac{1}{x}$$
I am not really sure how to go about doing this though... Is my step valid? I would appreciate is someone could guide me in the right direction.
|
The known form of your Taylor polynomial is $P_n(x)=\sum^{n}_{k=0} a_k(x-1)^k$ where the coefficients satisfy $a_k = \frac{f^{(k)}(1)}{k!}$
You managed to find the coefficients $a_k = (-1)^k$. What the question is probably asking is to prove that this is correct using induction.
You would take the base case and show that $a_0=\frac{f(1)}{0!}=(-1)^0$
Then you would show that if $a_n=(-1)^n$ then it follows that $a_{n+1} = (-1)^{n+1}$
Doing these two steps would show that the formula holds for all $n$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2493045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Auxiliary Epsilon - is it mathematically rigorous if it's larger than epsilon? Assume $a_n \rightarrow L$. Thus, applying the standard definition, $\forall \epsilon > 0, \exists N\in\mathbb{N},\, \forall n\geq N, |a_n-L| < \epsilon$.
I'm in an introductory real analysis class, and my professor will sometimes use an "auxiliary epsilon" $\epsilon_0$ and show that $|a_n - L| < \epsilon_0$ where $\epsilon_o$ is defined in terms of $\epsilon$.
I was having a debate with my friend about the relationship between $\epsilon_0$ and $\epsilon$.
I believe that for this proof to be rigorous, $\epsilon_0$ must be less than or equal to $\epsilon$ for it to be sufficient to show $|a_n - L| < \epsilon_0$ instead of $|a_n - L| < \epsilon$. So for example, arriving at $|a_n-L| < \epsilon_0 = c*\epsilon$ for $c > 1$ would not be mathematically rigorous. Like if you chose epsilon to be $.1$, I don't see how it's enough to say that $|a_n-L|$ must be less than $.2$, if $c$ were $2$ here.
However, my friend argued that because it’s an arbitrary positive $\epsilon$, as long as the statement is universal, it doesn’t matter because you can take a smaller epsilon to fit your range. So he's saying that in the case where $\epsilon = .1$ and $\epsilon_0 = .2$, you can choose $\epsilon =.05$ so $\epsilon_0 = .1$ which satisfies the original $\epsilon = .1$.
Which is correct? I presume I'm looking at it like $\epsilon-\delta$ proofs of limits, where you must specify a $\delta$ for which all corresponding function values are within $\epsilon$ of the limit. If $\delta$ is too wide and there's some value $x$ for which $|x-a|<\delta$ and $|f(x)-L|>\epsilon$, then the proof is invalid. I'm thinking the same logic would apply here.
|
Your friend is correct in this case. Keep in mind that in an $\varepsilon-\delta$ proof, we are never really choosing an explicit value for $\varepsilon$; rather, we are saying that for any arbitrarily small $\varepsilon$, our sequence of $\{a_n\}$ is within $\varepsilon$ of $L$ eventually. If $\varepsilon_0=c\varepsilon$, then if $\varepsilon$ is taken to be arbitrarily small, then so is $\varepsilon_0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2493324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Necessary and sufficient condition for $u(x,y,z)$ , $v(x,y,z)$ ,$w(x,y,z)$ are functionally dependent Show that necessary and sufficient condition for that $u(x,y,z)$ , $v(x,y,z)$ ,$w(x,y,z)$ are functionally dependent through equation $F(u,v,w)= 0$ is $\nabla u\cdot (\nabla v \times \nabla w )$.
Any clue on this?
What if there are $2$ variable instead of $3$ (say only $u(x,y,z)$ and $v(x,y,z)$ are functionally dependent)?
|
Hint. If $F(u(x,y,z),v(x,y,z))=0$ and $F$ is differentiable then
$$\frac{\partial F}{\partial u}\nabla u+\frac{\partial F}{\partial v}\nabla v={\bf 0}.$$
Note that $\frac{\partial F}{\partial u}$ and $\frac{\partial F}{\partial v}$ are scalars.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2493477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Given that $\operatorname{E}[Y\mid X] = 1$, show that $\operatorname{Var}(XY)\geqslant\operatorname{Var}(X)$
Given that $E[Y\mid X] = 1$, show that $\operatorname{Var}(XY)\geqslant\operatorname{Var}(X)$.
So I tried to expand $\operatorname{Var}(XY) = \operatorname{E}(X^2 Y^2) - 1$ and was stuck here.
|
We have $$\operatorname{Var}[XY] = \operatorname{E}[X^2 Y^2] - \operatorname{E}[XY]^2.$$
Try to tackle each of these two terms by first conditioning on $X$ and then taking expectation. For a full proof, hover below.
Write $\operatorname{E}[XY] = \operatorname{E}[\operatorname{E}[XY \mid X]] = \operatorname{E}[X \operatorname{E}[Y \mid X] ] = \operatorname{E}[X]$ since $\operatorname{E}[Y \mid X] = 1$. Lastly, we have $$\operatorname{E}[X^2Y^2] = \operatorname{E}[ X^2 \operatorname{E}[Y^2 \mid X]] \geq \operatorname{E}[ X^2 \operatorname{E}[Y\mid X]^2 ] = \operatorname{E}[X^2]$$ where we use Jensen's inequality for conditional expectation. Thus $$\operatorname{Var}[XY] \geq \operatorname{E}[X^2] - \operatorname{E}[X]^2.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2493754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Showing that there are no $3$-digit primes $\overline {abc}$ such that $b^2-4ac=9$.
Problem: Show that there are no $3$-digit primes $\overline {abc}$ such that $b^2-4ac=9$.
Solution. Upon the given condition, the quadratic equation $ax^2+bx+c$ can be written as $(px+q)(rx+s)$ and we have:
$$\overline{abc}=100a+10b+c=(10p+q)(10r+s)$$
i.e. $\overline{abc}$ is not a prime which is a contradiction, so there is no such prime available.
I don't get the reasoning completely.
|
We show a more general fact:
There is no $3$-digit primes $\overline {abc}$ such that $b^2-4ac=d^2$ where $d$ is a non-negative integer.
We first note that $d\leq b\leq 9$. Since $b^2-4ac=d^2$, we have that
$$P(x):=ax^2+bx+c=a\left(x-\frac{-b+d}{2a}\right)\left(x-\frac{-b-d}{2a}\right)$$
Assume that $P(10)=\overline{abc}=p$ where $p$ is some prime of three digits. Then
$$4ap=\left(20a+b-d\right)\left(20a +b+d\right)$$
which implies that $p$ divides $\left(20a+b-d\right)$ or $\left(20a+b+d\right)$.
Now $1\leq a\leq 9$, $0\leq b\leq 9$, and
$$0<\left(20a+b-d\right)<\left(20a+b+d\right)\leq 20\cdot 9+9+9\leq 198.$$
Hence $a=1$, and
$$\left(20a+b-d\right)<\left(20a+b+d\right)\leq 20\cdot 1+9+9\leq 38.$$
Therefore $p>100$ can not divide any of them and we have a contradiction!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2493861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
}
|
Integrate $\ln(1 + x^\frac{1}{2})$ from $0$ to $1$ This integral is from an Integration Contest. The substitution $x^2 = u$ allowed me to evaluate the integral but I keep getting $\frac{1}{2} - \ln2$ as the answer as opposed to just $\frac{1}{2}$ which was the posted answer. Any help would be appreciated.
https://www.wolframalpha.com/input/?i=integrate+ln(1+%2B+rootx)+from+0+to+1
Edit: Can some provide a solution?
The substitution x^2 = u trivializes the integral into 2u(1+u), integrating by parts allows me to evaluate the indefinite integral, but I seem to keep messing up plugging the bounds.
|
An alternative way, by Feynman's trick:
$$ \mathfrak{I}=\int_{0}^{1}2x\log(1+x)\,dx=\left.\frac{d}{d\alpha}\int_{0}^{1}2x(1+x)^{\alpha}\,dx\right|_{\alpha=0^+} $$
and by writing $2x$ as $2(1+x)-2$ we get:
$$ \mathfrak{I}=2\left.\frac{d}{d\alpha}\left(\frac{2^{\alpha+2}-1}{\alpha+2}-\frac{2^{\alpha+1}-1}{\alpha+1}\right)\right|_{\alpha=0^+}$$
or
$$\mathfrak{I}=2\left[-\tfrac{3}{4}+2\log 2+1-2\log 2\right] = \frac{1}{2}.$$
Yet another way: $\log(1+x)=\sum_{n\geq 1}\frac{(-1)^{n+1}}{n}\,x^n$ leads to
$$ \mathfrak{I} = 2\sum_{n\geq 1}\frac{(-1)^{n+1}}{n(n+2)} = \sum_{n\geq 1}\left[\frac{(-1)^{n+1}}{n}-\frac{(-1)^{n+3}}{n+2}\right]$$
and this is a telescopic series.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2493979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
How to formalize this last step in my proof? Let $X_k \sim N(\xi, \sigma^2)$. Let $\xi > 0$.
Consider $\frac{1}{k}X_k$, and the corresponding partial sums. I wish to show divergence to infinity, almost surely, of the partial sums.
I manage to show that $\sum_{k=1}^n X_k/k - \sum_{k=1}^n \xi/k$ converges to some finite value almost surely.
However, this can "obviously" only be true if the first term goes to infinity almost surely. If it is finite, or it oscillates, or whatever, the entire expression must converge to negative infinity, hence a contradiction. But what is the formal way to prove this last step?
|
The event $A=\{\sum_{k=1}^n X_k/k \to +\infty\}$ contains the event B = $\{\exists a, \sum_{k=1}^n X_k/k - \sum_{k=1}^n \xi/k \to a\}$. You've shown $P(B)=1$, therefore $P(A)=1$.
To show $B\subseteq A$, let $\omega\in B$, so
$$\sum_{k=1}^n X_k(\omega)/k - \sum_{k=1}^n \xi/k \to a$$
Now just work deterministically to show that $\omega\in A$. More generally, if $x_n$ and $y_n$ are sequences of real numbers, $y_n\to+\infty$ and $x_n-y_n\to a$, then we must have $x_n\to+\infty$, by applying additivity of limits to $x_n=(x_n-y_n)+y_n$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2494164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Evaluate an trigonometric limit in $0$ Find the limit
$$\lim_{x \to 0}\frac{\sin(\sqrt{x})}{x}$$
whithout using and using L'Hospital Rule
We have
$$\frac{\sin(\sqrt{x})}{x}=\frac{\sin(\sqrt{x})}{\sqrt{x}}\times\frac{1}{\sqrt{x}}
\to 1 \times (+ \infty)=+\infty$$
Is correct is approach?
|
We know that $\sin x=x-\dfrac{1}{3}x^{3}+\cdots$ for $|x|<1$, by alternating series grouping, one can see that $\sin x\geq x-\dfrac{1}{3}x^{3}$ for all small $x>0$, so considering small $x>0$, we have
\begin{align*}
\frac{\sin(\sqrt{x})}{x}\geq\frac{\sqrt{x}-3^{-1}x^{3/2}}{x}=\frac{1}{\sqrt{x}}-\frac{1}{3}\sqrt{x},
\end{align*}
taking $x\rightarrow 0^{+}$ and use Squeeze Theorem would yield the result.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2494266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
If $f,g$ are continuous functions, then $fg$ is continuous? Let $X$ be a topological space and let $f:X \to \mathbb{R}$ ,$g:X \to \mathbb{R}$ be continuous functions. Show that $fg$ is continuous.
My work:
To show $fg$ is continuous at $x$ for each $x \in X$, let $y=fg(x)$.
To show if $N_y$ is a neighborhood of $y$, then the pre-image of $y$ is a neighborhood of $x$.
I know that there exists $B_\epsilon(y) \in N_y$ so I want to show that $（fg）^{-1}(B_\epsilon(y)) \in N(x)$
Let $N_x=（fg）^{-1}(B_\epsilon(y))$, I want to find an open set in $N_x$.
Can anyone give me a hint of how to choose such open set or idea how to prove this ?
|
Here is an alternative approach:
Claim 1. If $f: X\to\mathbb{R}$ and $g: X\to\mathbb{R}$ are continuous, then $f+g: X\to\mathbb{R}$ is continuous.
Proof. Given a point $x\in X$ and $\epsilon>0$, we want to show that there exists a neighbourhood $N_{x}$ of $x$ such that $(f+g)(N_{x})\subseteq B_{\epsilon}((f+g)(x))$. By continuity of $f$ and $g$, you can find a neighbourhoods $N'_{x}$ and $N''_{x}$ such that $f(N'_{x})\subseteq B_{\epsilon/2}(f(x))$ and $f(N''_{x})\subseteq B_{\epsilon/2}(g(x))$. Now let $N_{x} := N'_{x} \cap N''_{x}$. Then for each $z\in N_{x}$ we have $$|(f+g)(z)-(f+g)(x)| = |f(z)-f(x)+g(z)-g(x)|\leq |f(z)-f(x)|+|g(z)-g(x)| < \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$ so that $(f+g)(z)\in B_{\epsilon}((f+g)(x))$. So $f+g$ is continuous.
Claim 2. If $f:X\to\mathbb{R}$ is continuous, then $f^2: X\to\mathbb{R}$ is continuous.
Here $f^2$ is the function defined by $f^2(x):=(f(x))^2$.
Proof. Given a point $x\in X$, let $\epsilon>0$. Let's assume $\epsilon<1$ for simplicity (this is okay to assume). By continuity of $f$, there exists a neighbourhood $N_{x}$ such that $f(N_{x})\subseteq B_{\epsilon}(f(x))$. Note that for every $z\in N_{x}$, we have $$|f(z)+f(x)|=|f(z)-f(x)+2 f(x)| \leq |f(z)-f(x)|+2|f(x)| \leq \epsilon+2|f(x)| < 1+2|f(x)| = C
$$
where $C$ was defined to be that constant $1+2|f(x)|$. Note that $C$ is independent of $z$, and only depends on $x$. Next, for every $z\in N_{x}$, we have
$$|f^2(z)-f^2(x)| = |f(z)-f(x)|\cdot |f(z)+f(x)| \leq \epsilon\cdot C$$
Thus, $f^2(N_{x})\subseteq B_{\epsilon}(f^2(x))$ and $f^2$ is continuous.
Claim 3. If $f:X\to\mathbb{R}$ and $g:X\to\mathbb{R}$ are continuous, then $f\cdot g: X\to\mathbb{R}$ is continuous.
Proof. Notice that
$$
f\cdot g = \frac{1}{2}\left[(f+g)^2 -f^2-g^2\right]
$$
Now apply claim $1$ to see $f+g$ is continuous. So by Claim 2, all three functions $f^2$, $g^2$ and $(f+g)^2$ are continuous. By Claim $1$ again, $f\cdot g$ is continuous (because it is a sum/difference of these functions).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2494409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
}
|
How to find indicial equation How can I find the indicial equation of $x(x-1)y''+3y'-2y=0$? I tried the method of Frobenius but I keep getting lost in the algebra. Is there any other way to get the indicial equation?
|
$y''+\dfrac{3}{x(x-1)}y'-\dfrac{2}{x(x-1)}y=0$ then $p(x)=\dfrac{3}{x(x-1)}$ and $q(x)=-\dfrac{2}{x(x-1)}$. The equation has two regular singular points $x=0$ and $x=1$. For $x=0$ we see
$$p_0=\lim_{x\to0}xp(x)=\lim_{x\to0}\dfrac{3}{x-1}=-3$$
and
$$q_0=\lim_{x\to0}x^2q(x)=\lim_{x\to0}\dfrac{-2x}{x-1}=0$$
then the indicial equation is $r(r-1)+p_0r+q_0=0$ or $r^2-4r=0$ shows $r=0$ and $r=4$. In this case $4-0\in\mathbb{Z}$ so $r=4$ gives a solution and let $y=x^4(a_0+a_1x+a_2x^2+\cdots)$ to find the answer. Do like this with point $x=1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2494482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Evaluate $\int_{0}^{\pi\over 4}{\ln(\tan(x))\over \cos^{2n}(x)}\mathrm dx$ $$\int_{0}^{\pi\over 4}{\ln(\tan(x))\over \cos^{2n}(x)}\mathrm dx=F(n)\tag1$$
$n\ge1$
$F(1)=-1$
$F(2)=-{10\over 9}$
$F(3)=-{284\over 225}$
How do we evaluate the closed form for $(1)$?
$u=\tan(x)$ then $\cos^2{(x)}\mathrm du=\mathrm dx$
$$\int_{0}^{1}{\ln(u)\over \cos^{2n-2}(x)}\mathrm du\tag2$$
$\sec^2(x)=1+\tan^2(x)$
$\sec^{2n-2}(x)=(1+\tan^2(x))^{n-1}$
$\sec^{2n-2}(x)=(1+u^2)^{n-1}$
$$\int_{0}^{1}(1+u^2)^{n-1}\ln(u)\mathrm du\tag3$$
|
Start with integration by parts:
$u = \ln(\tan x) \to du = \frac{1}{\tan x} \sec^2 x dx$,
$dv = (\sec^2 x)^n = \sec^2 x(1 + \tan^2 x)^{n-1}$.
Before continuing, we'll need to compute $v = \int dv$, and for this we'll use the substitution
$w = \tan x \to dw = \sec^2x dx$. Then by the binomial theorem,
\begin{align*}v &= \int(1 + w^2)^{n-1}dw\\ &= \int\sum_{k=0}^{n-1}\binom{n-1}{k}w^{2k}dw\\ &= \sum_{k=1}^{n-1}\binom{n-1}{k}\frac{w^{2k+1}}{2k+1} \\ &= \sum_{k=0}^{n-1}\binom{n-1}{k}\frac{\tan^{2k+1}(x)}{2k+1}\end{align*}
Now our original integral is
\begin{align*} \int_{0}^{\pi/4} udv &= \left.uv\right]_0^{\pi/4} - \int_{0}^{\pi/4}vdu\\
&= \left.\ln(\tan x)\sum_{k=0}^{n-1}\binom{n-1}{k}\frac{\tan^{2k+1}(x)}{2k+1}\right]_0^{\pi/4} - \int_{0}^{\pi/4}\left(\sum_{k=0}^{n-1}\binom{n-1}{k}\frac{\tan^{2k}(x)}{2k+1}\right)\sec^2x dx\end{align*}.
The first term requires us to take a limit as $x \to 0^+$, but this limit is, mercifully, $0$ (try breaking-off a power of $\tan x$ from the sum, then use L'Hospital's rule on $\ln(\tan x) \cdot \tan x$ as $x \to 0^+$). The remaining integral can be evaluated by making a final substitution of $y = \tan x$ to get
\begin{align*}- \int_{0}^{\pi/4}\left(\sum_{k=0}^{n-1}\binom{n-1}{k}\frac{\tan^{2k}(x)}{2k+1}\right)\sec^2x dx &= -\int_0^1 \sum_{k=0}^{n-1}\binom{n-1}{k}\frac{y^{2k}}{2k+1}dy\\
&= -\left.\sum_{k=0}^{n-1}\binom{n-1}{k}\frac{y^{2k+1}}{(2k+1)^2}\right]_0^1\\
&= -\sum_{k=0}^{n-1}\binom{n-1}{k}\frac{1}{(2k+1)^2}.\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2494640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
}
|
Convergence and non-negative supermartingales Given $ (X_n)_{n\geq0}$ is a non-negative supermartingale and $X_{\infty}$ is an almost sure limit, I want to show that $ \forall n\geq0$ $$\mathbb{E}(X_\infty \mid \mathcal{F}_n) \leq X_n$$
My approach:
First, for $(X_n)$ a supermartingale I would say we then have $$ \mathbb{E}(X_m\mid\mathcal{F}_n)\leq X_n \quad \forall m\geq n$$
Now, my assumption is that $X_m$ does not necessarily converge to $X_\infty$ in $L^1$, so $\mathbb{E}(X_m)$ does not necessarily converge to $\mathbb{E}(X_\infty)$. Hence my idea would be to show that $X_m$ is uniformly integrable, where then we could show, for $X_m \longrightarrow X_\infty$ a.s in $L^1$ then $\mathbb{E}(X_m\mid\mathcal{F}_n) = \mathbb{E}(X_\infty\mid\mathcal{F}_n)$.
However, I think my approach is way off. Could someone please help me out, it would be greatly appreciated.
|
By conditional Fatou, $E(\liminf_k X_{n+k}\mid\mathcal F_n)\leq \liminf_k E(X_{n+k}\mid\mathcal F_n)$, hence $$E(X_\infty\mid\mathcal F_n)\leq \liminf_k E(X_{n+k}\mid\mathcal F_n)$$
Since $\forall k, E(X_{n+k}\mid\mathcal F_n)\leq X_n$, we have $\liminf_k E(X_{n+k}\mid\mathcal F_n)\leq X_n$ and the result follows.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2494764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Probability when given the percentage chance There is a 30% chance that a driver will have an accident in their first year of driving. From 18 people getting their license this year in June, what is the probability that less than a quarter will have an accident before June next year?
So i had a go at it. 30/100 *18= 5.4 people have a probability of an accident. Less than 1/4 of 18 is about 4.5 people. So if 30% of 18 is 5.4 then 4.5 will be 0.3*4.5/5.4= 0.25 chance, but this is incorrect. Any ideas?
|
Assuming that the events are independent, this is equal to a binomial distribution. Calculate the chance that less than $\frac14$ of the 18 people have an accident is the same as calculating if less than or equal to 4 people have an accident. We can use a table for the cumulative distribution function, which would give us 0.333 or we could calculate it using the probability mass function
$P(X = x) = f(x) = {n \choose x}p^x(1-p)^{n-x}$
$\begin{align*}P(X \le 4) &= P(X = 0) + P(X=1) + P(X=2) + P(X=3) + P(X = 4) \\
&= {18 \choose 0} 0.7^{18} + {18 \choose 1} 0.3\cdot0.7^{17} + {18 \choose 2}0.3^2\cdot 0.7^{16} + {18 \choose 3}0.3^3 \cdot 0.7^{15} + {18 \choose 4}0.3^4\cdot 0.7^{14} \\
&\approx 0.3327
\end{align*}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2494892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How prove this definition $a\oplus b=a+b$ Define $\oplus$: if for any real numbers $a,b,c$ there have
$$\left(a\oplus b\right)\oplus c=a+b+c$$
show that
$$a\oplus b=a+b$$
|
Let $k =0\oplus 0$.
Since $(0\oplus 0)\oplus c=c$ we see that $k\oplus c=c$ for each $c$, so $k$ is left neutral.
Next we see that $\oplus $ is commutative:
$$b\oplus c=(k\oplus b)\oplus c=k+b+c=k+c+b=(k\oplus c)\oplus b=c\oplus b$$
Now we see that $k=0$:
$$(a\oplus k)\oplus k=a+2k \Rightarrow a=a+2k \Rightarrow k=0 $$
So $$a+b=a+b+0=(a\oplus b)\oplus 0=a\oplus b$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2494988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
If 'a' is divisible by 'b^2' then 'a' is divisible by 'b'. It's something that I've never really thought about before but it makes sense nonetheless. Bearing in mind that 'a' and 'b' are both positive integers, what would be the best way to go about proving this statement? Which method of proof , for example, would be the best way to solve this problem? I'm just looking for a starting block to set me going :)
|
By the definition of divisibility you have $$b^2|a\Rightarrow a=mb^2=(mb)b=nb$$ for an $m \in \mathbb{Z}$ which means that $$b|a$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2495081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
For what natural $n$ does $3^n > n^3$ hold true? Prove by induction
For what natural $n$ does $3^n > n^3$ hold true?
I figured that it holds true for all $n$ except $n = 3$. I am not sure how to prove it by induction. I proved it by $p(k) \implies p(k+1)$ but that doesn't show that $n \neq 3$
|
For $n=4$ we have
$$3^n> n^3$$ is true.
Let $$3^n> n^3$$ for all $n>3$.
Thus, $$3^{n+1}=3\cdot3^n>3n^3$$ and it's enough to prove that
$$3n^3>(n+1)^3$$ or
$$\sqrt[3]3n> n+1,$$ which is true for $n>3.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2495332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
}
|
How to find the limit of series? (What should I know?) There is a couple of limits that I failed to find:
$$\lim_{n\to\infty}\frac 1 2 + \frac 1 4 + \frac 1 {8} + \cdots + \frac {1}{2^{n}}$$
and
$$\lim_{n\to\infty}1 - \frac 1 3 + \frac 1 9 - \frac 1 {27} + \cdots + \frac {(-1)^{n-1}}{3^{n-1}}$$
There is no problem to calculate a limit using Wolfram Alfa or something like that. But what I am interested in is the method, not just a concrete result.
So my questions are:
*
*What should I do when I need a limit of infinite sum? (are there any rules of thumb?)
*What theorems or topics from calculus should I know to solve these problems better?
I am new to math and will appreciate any help. Thank you!
|
They are sums of geometric progressions.
The first it's
$$\frac{\frac{1}{2}}{1-\frac{1}{2}}=1.$$
The second it's $$\frac{1}{1+\frac{1}{3}}=\frac{3}{4}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2495413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 9,
"answer_id": 8
}
|
Find the smallest possible value of an integral Say a have an integral, like this one
$$\int_{0}^1 (x-a)^2\, dx$$
and asked to find the smallest possible value of it, as a varies. How can I do this? Besides, is there a certain rule I can use to solve this type of questions?
I will be grateful for any help.
|
\begin{align}
\int_{0}^{1} (x-a)^2 &= \frac{(x-a)^3}{3} \Bigg \rvert_0^1 \\
&=\frac{(1-a)^3}{3} - \frac{(-a)^3}{3}\\
&=a^2-a+\frac{1}3\\
&=\left(a-\frac 12\right)^2+\frac {1}{12} \ge \frac{1}{12}
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2495527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Probability of getting 18 or more bulls eye out of 20 shots An Olympic archer can hit the bulls eye an average of 9 times out of 10. The probability of the archer scoring 18 or more bulls eye from 20 shots is what? I've tried it but my answer is wrong. the average probability would be 0.9 as 9/10. But if we divide 18/20, we also get 0.9, what is that incorrect?
|
This is what is known as a binomial distribution, because we are looking at “sampling” with two possible outcomes, the probability of which we say is constant between trials. There is a prescribed formula for binomial distributions:
$$X\sim\mathrm{B}(n,p) \implies \operatorname{P}(X=x)=\binom{n}{x}p^x(1-p)^{n-x}$$
We will let $X$ be how many bull’s eyes the archer hits, $n=20$ be the number of times she shoots, and $p=0.9$ bet the probability that she successfully hits a bull’s eye.
Hence we have $$\operatorname{P}(X=x)=\binom{20}{x}(0.9)^x(0.1)^{20-x}$$
To determine the probability that she gets eighteen or more bull’s eyes, we need to add the probabilities of all the valid cases—$X=18,X=19,X=20$—which we can calculate as
$$\sum_{x=18}^{20}\binom{20}{x}(0.9)^x(0.1)^{20-x}$$
or, if you have a Texas Instrument on hand, you can use the command
$$\mathtt{1-binomcdf(20,0.9,17)}$$
The answer comes out to $$0.676\,926\,805\,189$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2495603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Why does the reduced matrix $\left(\begin{array}{ccc|c}0&1&0&-7\\ 0&0&1&10\end{array}\right)$ have infinitely many solutions? $$\left(\begin{array}{ccc|c}0&1&0&-7\\ 0&0&1&10\end{array}\right)$$
I thought the requirement for a matrix to have a unique solution was that when every variable is leading. It seems like both 1's in the above matrix are leading and the other variables are 0. So why doesn't this matrix have only a unique solution?
|
This is equivalent to the system:
$$\begin{eqnarray*}
{0x + y + 0z}&=&{-7} \\
{0x + 0y+ z}&=&{10}
\end{eqnarray*}$$
Whatever $x$ you choose will work because $0x=0$ for all $x$. If the system were:
$$\begin{eqnarray*}
{ y + 0z}&=&{-7} \\
{ 0y+ z}&=&{10}
\end{eqnarray*}$$
Then there is only one solution. It is very common for people to forget the $0x_i$ matter, a lot of people answer a different problem because they write the first system as the second system. The first system has infinite solutions, the second system as one solution.
I suggest you to work on the equivalence between matrices and systems of linear equations and see what each result in matrices means in terms of results of systems of linear equations and vice-versa.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2495770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
}
|
Finding all possible Jordan forms of an $ 8\times 8$ matrix given the minimal polynomial
Find all possible Jordan forms of an $ 8\times 8$ matrix given that $$t^2(t-1)^3$$ is the minimal polynomial.
I don't really know where to start so all help would be appreciated
|
Let $J(\alpha,k)$ be the upper Jordan block with minimal polynomial $(t-\alpha)^k$.
$$
\begin{pmatrix}
\alpha & 1 & 0 & \ldots & \ldots\\
0 & \alpha & 1 & 0 & \ldots \\
0 & 0 & \alpha & \ddots & \ddots\\
\vdots & \vdots & \ddots & \ddots & \ddots
\end{pmatrix}
$$
So, there is - up to conjugation - one possible upper Jordan form of your matrix, in dimension 5 (see the end)
$$
\begin{pmatrix}
J(0,2) & 0\\
0 & J(1,3)
\end{pmatrix}
$$
Now, you have to extend it to dimension 8. Calling $p_1,p_2,\cdots p_k$ (resp. $q_1,q_2,\cdots q_l$) the orders of the Jordan blocks for the eigenvalue $0$ (resp. the orders of the Jordan blocks for the eigenvalue $1$), one has
$$
\sum_{i=1}^k p_i+\sum_{i=1}^l q_i=8
$$
$p_i\leq 2,\ q_i\leq 3$ and at least one of the $p_i=2$ at least one of the $q_i=3$.
Hope it helps !
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2495872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Finding the pdf of a transformation of these independent random variables
Hi, I am looking at the question above. We know that $f_X(x) = \lambda^nx^{n-1}e^{-\lambda x}/\Gamma(n)$ and $f_Y(y) = \lambda e^{-\lambda y}$.
We also know that the range for each of these pdfs are $x>0$ and $y>0$.
Now, after computing we get $f_{UV}(u,v) = \lambda^n(u-v)^{n-1}e^{-\lambda(u-v)}/\Gamma(n)\hspace{3mm} \cdot \hspace{3mm} \lambda e^{-\lambda v} $
Now I have to find the pdf of U. I know I have to integrate that equation between the limits of $v=u$ and $v=0$ to find said pdf of U. But I am stuck on the integral. Any help? Thanks!
|
\begin{align}
\int_0^u f_{UV}(u, v) \mathop{dv}
&= \lambda^{n+1} e^{-\lambda u} \frac{1}{\Gamma(n)} \int_0^u (u-v)^{n-1} \mathop{dv}
\\
&= \lambda^{n+1} e^{-\lambda u} \frac{1}{\Gamma(n)} \left[- \frac{1}{n}(u-v)^n\right]_{v=0}^u
\\
&= \lambda^{n+1} e^{-\lambda u} \frac{1}{\Gamma(n)}\cdot \frac{1}{n} u^n.
\end{align}
Can you take it from here? Note $\Gamma(n) \cdot n = \Gamma(n+1)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2496081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Concise Solution for Algebra Problem Consider the following expression
$$\sum_{i=1}^n \left(\frac{p\alpha_ie^{\alpha_i\cdot x}}{1-p+pe^{\alpha_i\cdot x}}\right) = c$$
where $\alpha_i, c \in \mathbb{R}$ and $p \in (0,1).$
How do I solve for $x$ from the equation?
|
Here's what I tried and it was too long to be a comment.
By integrating both sides:
$$\sum_{i=1}^n\ln(1-p+p\;e^{\alpha_i x})=cx+c_1$$
and since this must hold for $x=0$, then $c_1=0$.
Therefore
$$\prod_{i=1}^n\left(1+p(e^{\alpha_i x}-1)\right)=e^{cx}$$
Set $y:=e^x$ to convert the above equation to a nonlinear (and seemingly unsolvable in general) equation:
$$\prod_{i=1}^n\left(1+p(y^{\alpha_i}-1)\right)=y^{c}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2496194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Proving if a function is continuous, its inverse is also continuous
Let $(X,d)$ and $(X,d')$ be metric spaces. Prove that $a)\implies b) $ and $b) \implies c)$:
a) if $x_n \to x$ in $(X,d')$ then $x_n \to x$ in $(X,d)$
b)For any metric space $(Y,p)$ and any continuous $f:(X,d)\to (Y,p), f:(X,d')\to(Y,p)$ is also continuous.
c) for any metric space $(Y,p)$ and any continuous $f:(Y,p)\to(X,d'), f:(Y,p)\to(X,d)$ is also continuous.
$f: E\to E'$ a continuous function. Prove that if $E$ is compact and $f$ is bijective then $f^{-1}:E' \to E$ is continuous.
I tried to prove $a)\implies b)$ by if $d'(x',x)<\delta$ then there exists some sequence $\{x_n\}$ such that $x_n \to x$ in $(X,d')$. Then $x_n \to x$ in $(X,d)$ and there exsits a $N$ such that $d(x_n,x)<\delta$ whenever $n>=N$. Then $d(x',x)<\delta$ for all $x\in \{x_n|n>=N\}$. (I'm not sure if my proof is valid)
For $b) \implies c)$, I wanted to prove if $f$ is continuous then so is $f^{-1}$. But I'm not sure if it's generally true.
|
Hint:
If b) is true then the function $(X,d')\to (X,d)$ prescribed by $x\mapsto x$ is continuous.
Let's denote this function with $g$.
Now if $f:(Y,p)\to(X,d')$ is continuous then so is $g\circ f:(Y,p)\to(X,d)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2496310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Galois groups of rational function fields extensions Consider the subfields $$ K_{1}:=\Bbb{C}\big(4x(1-x),4y(1-y)\big) $$
$$ K_{2}:=\Bbb{C}\Big(\frac{4x(1-x)(1-2y)}{(1-2xy)^{2}},\frac{4y(1-y)(1-2x)}{(1-2xy)^{2}}\Big) $$ of $ K:=\Bbb{C}(x,y) $.
I want to compute the Galois groups $ G_{1}:=Gal(K/K_{1}),G_{2}:=Gal(K/K_{2}) $ as well as the intersection $ K_{3}:=K_{1} \cap K_{2} $ and $ G_{3}=Gal(K/K_{3}) $.
First, I claim that $G_{1} \simeq C_{2} \times C_{2} $. To this end, define $$ \sigma_{1}: x \mapsto 1-x, y \mapsto y $$ and $$ \sigma_{2}: x \mapsto x, y \mapsto 1-y $$ Notice that both $ \sigma_{1} $ and $ \sigma_{2} $ are their own inverses and map the generators $ 4x(1-x) $ and $ 4y(1-y) $ of $ K_{1} $ to themselves so the $ \sigma_{i}'s $ induce $ K_{1} $-automorphisms of $ K $.
Let $$ p_{1}(T):=(T-x)(T-\sigma_{1}(x))=T^{2}-(x+\sigma_{1}(x))T+x\sigma_{1}(x)=T^{2}-T+x(1-x) $$
and $$ p_{2}(T):=(T-y)(T-\sigma_{2}(y))=T^{2}-(y+\sigma_{2}(y))T+y\sigma_{2}(y)=T^{2}-T+y(1-y) $$
Observe that $ p_{1}(T),p_{2}(T) \in K_{1}[T] $ and that $ p_{1}(x)=p_{2}(y)=0 $ so $ [K_{1}(x):K_{1}],[K_{1}(y):K_{1}] \leq 2 $.
On the other hand, we can't have that $ [K_{1}(x):K]=1 $ or that $ [K_{1}(y):K]=1 $ as that would mean that $ x $ or $ y $ are in $ K $ which is false because of degree considerations. So $$ [K_{1}(x):K_{1}]=[K_{1}(y):K_{1}]=2 $$ Furthermore, we have that $$ [K_{1}(x,y):K_{1}]=[K_{1}(x,y):K_{1}(x)][K_{1}(x):K_{1}]=2[K_{1}(x,y):K_{1}(x)]=4 $$ as $ y \notin K_{1}(x) $ because of the same degree considerations. So we obtain that $ K $ is a degree $ 4 $ extension of $ K_{1} $ whose Galois group is generated by $ \sigma_{1} $ and $ \sigma_{2} $, each being non-trivial $ K_{1} $-automorphisms of $ K $ of order $ 2 $. Thus $ G_{1} \simeq C_{2} \times C_{2} $.
I don't know how to go about computing $ G_{2},G_{3} $ and finding $ K_{3} $. Any help would be much appreciated. Thank you!
|
Looking at this over finite fields $\Bbb F_q$ with large $q$, one finds experimentally that the map $\phi : (x,y) \mapsto (u(x,y),v(x,y))$ (where $u$ and $v$ are the two generators of $K_2$) is basically $8$-to-$1$. For a choice of a pair $(u,v)$, either there is no solution to $(u,v) = \phi(x,y)$ (approximately $7/8$ of the time, in which case the solutions are all in $\Bbb F_{q^2}$), either there are $8$ of them (approximately $1/8$ of the time).
This is very strong evidence (and would be a proof if I knew how to get some nice effective bounds from the Weil conjectures) that $K$ is Galois over $K_2$, that the group is isomorphic to $C_2^3$, and that it can be given by a bunch of explicit transformation formulas.
Those are (after quite a bit of work),
$id : (x,y) \mapsto (x,y) \\
f : (x,y) \mapsto (1-x,-\frac y{1-2y}) \\
g : (x,y) \mapsto (- \frac x{1-2x},1-y) \\
fg : (x,y) \mapsto (\frac{1-x}{1-2x}, \frac {1-y}{1-2y}) \\
h : (x,y) \mapsto (\frac 1{2y}, \frac 1{2x}) \\
fh : (x,y) \mapsto (-\frac{1-2y}{2y}, \frac 1{2(1-x)}) \\
gh : (x,y) \mapsto (\frac 1{2(1-y)}, -\frac{1-2x}{2x}) \\
fgh : (x,y) \mapsto (\frac{1-2y}{2(1-y)},\frac{1-2x}{2(1-x)})
$
Furthermore, it looks like the group $G_3$ generated by the two Galois groups is finite (noncommutative, of order 32, looking at the size of most orbits in finite fields, maybe it's a skewed product), so if this is the case, $K$ should also be Galois over $K_3$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2496429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
Linear dependence/independence of functions $f(x) = x^2+1$, $g(x) = 1+x^3$ and $h(x) = \ln(1 + x)$. Questions :
Prove that the functions $f(x) = x^2+1$, $g(x) = 1+x^3$ and $h(x) = \ln(1 + x)$ are linearly independent on the interval $(0, 1)$. Can you use the Wronskian to do this?
I try this:
$$c_1 x f(x)+ c_2 x g(x) + c_3 x h(x) = 0.$$
but I don't know how to solve for the values of $c_1$, $c_2$, and $c_3$.
Thanks.
|
The Wronskian is \begin{align}W(f,g,h)(x)&=\begin{vmatrix}f(x)& g(x) & h(x)\\f'(x) & g'(x) & h'(x) \\f''(x) & g''(x)& h''(x)\end{vmatrix}=\begin{vmatrix}1+x^2& 1+x^3 & \ln{(1+x)}\\2x & 3x^2 & \frac1{1+x} \\2 & 6x & -\frac{1}{(1+x)^2}\end{vmatrix}\\[0.3cm]&=\left\{\text{terms without} \ln(x+1)\right\}+6x^2\ln(x+1)\end{align} which means that the Wronskian does not vanish, hence the functions are linearly independent.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2496573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Find the limit of the complex function. $$ \lim_{x\to a} \left(2- \frac{x}{a}\right)^{\left(\tan \frac{\pi x}{2a}\right)}.$$
I have simplified this limit to this extent :
$$e^{ \lim_{x\to a} \left(\left(1- \frac{x}{a}\right){\left(\tan \frac{\pi x}{2a}\right)}\right)}$$
I don't know how to simplify the limit after that. please help.
|
\begin{align*}
\log\left(2-\frac{x}{a}\right)^{\tan(\pi x/2a)}&=\left(\tan\frac{\pi x}{2a}\right)\left(\log\left(2-\frac{x}{a}\right)\right)\\
&=\sin\left(\frac{\pi x}{2a}\right)\frac{\log\left(1+\left(1-\dfrac{x}{a}\right)\right)}{\cos\left(\dfrac{\pi}{2}\left(\dfrac{x}{a}-1\right)+\dfrac{\pi}{2}\right)}\\
&=\sin\left(\frac{\pi x}{2a}\right)\frac{\left(1-\dfrac{x}{a}\right)-\dfrac{1}{2}\left(1-\dfrac{x}{a}\right)^{2}+\cdots}{\dfrac{\pi}{2}\left(1-\dfrac{x}{a}\right)-\dfrac{1}{3!}\left(\dfrac{\pi}{2}\left(1-\dfrac{x}{a}\right)\right)^{3}+\cdots}\\
&=\sin\left(\frac{\pi x}{2a}\right)\frac{1-\dfrac{1}{2}\left(1-\dfrac{x}{a}\right)+\cdots}{\dfrac{\pi}{2}-\dfrac{1}{3!}\left(\dfrac{\pi}{2}\right)^{3}\left(1-\dfrac{x}{a}\right)^{2}+\cdots},
\end{align*}
taking limit as $x\rightarrow a$, then $\log\left(2-\dfrac{x}{a}\right)^{\tan(\pi x/2a)}\rightarrow\dfrac{2}{\pi}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2496701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 0
}
|
Minimum slope of a chord to parabola A line is drawn from $(-2,0)$ to intersect
$y^2 = 4x \,\,$ in P, Q within the first quadrant, such that
$$ \frac{1}{AP} +\frac{1}{AQ} < \frac{1}{4} $$
Find the minimum value of the line slope. A is the origin.
I had basically let the coordinates of the parabola in parametric form and then used the given condition, but couldn't find the minimal slope.
|
As it is stated in the current form, this problem - as shown below - has no real solutions. I suspect that the original formulation might have been slightly different, in particular with $A$ indicating the point $(-2,0)\,\,$ and maybe with a more general equation of the parabola (again see below). In this answer, I will assume that $A=(-2,0)\,\,$ and that $P$ and $Q$ are distinct points. Also, I will firstly consider the case of the specific parabola described in the OP, and then the case of a more general parabola equation. Lastly, I will add a brief comment on the case where $A$ is the origin.
Since the line crosses the point $(-2,0)\,\,$, it has the form
$$y=ax+2a$$
The intersection points of this line with the parabola have $x$-coordinates given by
$$ \frac{2 (1 - a^2 \pm \sqrt{1 - 2 a^2})}{a^2}$$
whereas the $y$-coordinates are
$$ \frac{2 (1 - a^2 \pm \sqrt{1 - 2 a^2})}{a}+2a$$
Calculating the length of $\overline{AP}$ using standard formulas yields
$$\overline{AP}= \left[\left( \frac{2 (1 - a^2 - \sqrt{1 - 2 a^2})}{a^2} \right)^2+ \left(\frac{2 (1 - a^2 - \sqrt{1 - 2 a^2})}{a}+2a \right)^2 \right]^{1/2}$$
which can be simplified as
$$\overline{ AP}=\frac{2 \sqrt{2 - 2 a^2 - a^4 -2 \sqrt{1 - 2 a^2}}}{a^2}$$
In a similar manner we get
$$\overline { AQ}=\frac{2 \sqrt{2 -2 a^2- a^4 + 2 \sqrt{1 - 2 a^2}}}{a^2}$$
Note that since we must have $1 - 2 a^2 >0\,\,\,$ to have two separate intersection points, we get the condition
$$-\frac{1}{\sqrt{2}} < a < \frac{1}{\sqrt{2}} $$
to obtain a different real value for $\overline{AP}$ and $\overline {AQ}$.
Substituting the expressions above in
$$\frac{1}{\overline{AP}}+\frac{1}{\overline{AQ}}<\frac{1}{4}$$
and simplifying, we get
$$\frac{1}{2 \sqrt{1 + a^2}}<\frac{1}{4}$$
whose positive solution is
$$a>\sqrt{3}$$
Because this solution has no intersection with the range of $a$ needed to have $AP$ and $AQ$ real, we conclude that there are no real values of $a$ that satisfy the conditions stated in the OP.
Notably, after some searching, I found on Google a nearly identical version of this problem with no solution but four possible answers ($<\sqrt{3}\,\,$, $>\sqrt{3}\,\,$, $\geq \sqrt{3}\,\,$, none of them), with the only difference that the initial parabola has equation $y^2=4kx \,\,$. The point $A$ is explicitly defined as the point $(-2,0)\,\,$. This more general version of the problem is interesting because, after calculations that are similar to those shown above, it leads to the same final inequality $1/(2 \sqrt{1 + a^2})<1/4\;$ with solution $a>\sqrt{3}\,\,\,$ (in other words, in the final part of the calculations and simplifications, $k$ is canceled out). In contrast, $k$ is not canceled out in the condition necessary for $AP$ and $AQ$ to be distinct and real, which becomes
$$-\sqrt{\frac{k}{2}} < a < \sqrt{\frac{k}{2}} $$
allowing to find, for $k>6\,$, an intersection between this range and that stated in the solution, and giving the answer $a>\sqrt{3}\,$ as the inferior bound of the slope. Note that such condition, for $k=1\,\,$, is the same obtained above considering the problem for the specific parabola $y^2=4x\,\,$.
Lastly, repeating the same calculations for the case where $A$ is the origin leads to similar results. The final inequality in this case yields the numerical solution $a>1.23414...\;$ (apparently with no closed form). Again, because of the condition needed for $\overline{AP}$ and $\overline{AQ}$ to be real and distinct, there are no real solutions for the case with parabola equation $y^2=4x\,\,$. On the other hand, in the general case with parabola equation $y^2=4kx\,\,$, there are real solutions only for
$$k>2\cdot 1.23414...^2=3.04620...\,\,\,\,$$
and the numerical solution $a>1.23414...\,\,\,$ is the lower bound of the slope.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2496826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Find the domain and range of the function, $f(x,y) = \sqrt{x+y}$ and sketch the domain in the xy-plane. I have found the domain to be $y \geq -x$.
I have found the range to be $z \geq 0$, or $[ 0, \infty )$.
I'm not sure how to sketch the domain in the xy-plane. I figured it would be a straight line through $(0,0)$ with $-1$ slope, and the values in the domain would be everything above that line, but I seem to be wrong.
|
This is not an answer. I just needed the attach picture feature to show the domain of the function, which is the hatched region.
Hope it helps
$$...$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2496931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Examining sequence convergence with parameter $a$? For $a\in R$, let $x_1=a$ and $x_{n+1}$=$-\frac{6x_n^2+4x_n}{x_n^2-2x_n+4}$. Examine the convergence of the sequence ${(x_n)}_{n=1}^{\infty}$ for different values of $a$. Also find $\lim_{n\to\infty}x_n$ whenever it exist.
I know that if the sequence converges to $x$ then it must satisfy $x$=$-\frac{6x^2+4x}{x^2-2x+4}$, that is $x=0$.
$x_{n+1}$$-$$x_n$=-$\frac{x_n(x_n^2+4x_n+8)}{x_n^2-2x_n+4}$ and should use it to find out whether the sequence is monotone . But i don't know how to do it and how to use $a$ in all this and find $a$ so that the sequence converges. Can someone explain it ? Thanks.
I can't find any answer. I'm stuck. If someone can help ,please do it:)
|
The only possible real limit is
$0$,
since all the other roots of
$x
=-\frac{6x^2+4x}{x^2-2x+4}
$
are complex.
Around zero,
it can't be monotone
since
$x_{n+1}
=-\frac{x_n(6x_n+4)}{x_n^2-2x_n+4}
$
has the opposite sign
of $x_n$.
What is needed,
imho,
is to write
$x_{n+2} = \text{some function}(x_n)$.
This would probably show convergence,
but I'm too lazy right now
to do it.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2497062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Intersection of Radical With Semisimple Subalgebra Suppose $\mathfrak{g}$ is a Lie algebra with radical $\text{Rad }\mathfrak{g}$ and let $\mathfrak{a}\subseteq\mathfrak{g}$ be a semisimple subalgebra. Is it necssarily the case that $\text{Rad }\mathfrak{g}\cap\mathfrak{a}=\{0\}$?
I am trying to prove that maximal semisimple subalgebras are Levi subalgebras, but I can't get past this step. If I know this to be true however, I can finish off the rest of the proof.
|
Yes. $\text{Rad }\mathfrak{g}\cap\mathfrak{a}$ is an ideal of $\mathfrak{a}$, because $\mathfrak{a}$ is a subalgebra and $\text{Rad }\mathfrak{g}$ is an ideal of $\mathfrak{g}$. Also, $\text{Rad }\mathfrak{g}\cap\mathfrak{a}$ is solvable, because it is a subalgebra of the solvable $\text{Rad }\mathfrak{g}$. But since $\mathfrak{a}$ is semisimple, its only solvable ideal is $\lbrace 0 \rbrace$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2497205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Sum of positive divisors if and only if perfect square let n be a positive odd integer, prove that the sum of the positive divisors of n is odd if and only if n is a perfect square.
I know that based on the prime factorization theory that every integer n can be written as the product of primes, if their sum is odd that means that there are equal pairs of even and odd divisors. Is this enough to conclude that n must be a perfect square?
|
I don't really understand your argument: What does "there are equal pairs of even and odd divisors" mean, especially given that there are no even divisors of $n$? You really need to be precise about the objects you're considering, and careful about how it's written.
Here's an approach that you might find useful. For each $d \in \mathbb{Z}$ such that $d | n$, you can pair $d$ with $n/d$ and notice that both are odd, so that
$$2 | (d + n/d).$$
This covers all the divisors of $n$, except when something special happens....
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2497326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
}
|
Derivative of l2 norm with chain rule If $X$ is a $n$ by $d$ matrix, $\alpha$ is a $n$ by $1$ vector, let $f(\alpha) = \left\Vert X^\top\alpha \right\Vert_2^2$, what is $\frac{df}{d\alpha}$
|
${df(\alpha)\over d\alpha}={d\|X^T\alpha\|_2^2\over d\alpha}={d\|X^T\alpha\|_2^2\over dX^T\alpha}{dX^T\alpha\over d\alpha}=2\alpha^TXX^T$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2497425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Solving a first order differential equation in terms of Lambert W-function I am having great difficulty solving the following equation
$$\ \frac{ax}{(bx^2 + c)} = \frac{dx}{dt} $$
I have re-edited the question. Any help is appreciated. Thank you.
|
This was for the first edit of the post.
Welcome to the world of Lambert function !
The solution of equation $$\ x^2 + \log(x) - c = 0$$ is given by
$$x=\pm\frac{\sqrt{W\left(2 e^{2 c}\right)}}{\sqrt{2}}$$ but only the positive root must be kept because of $\log(x).
The Wikipedia page gives series espansions fot he evaluation of $W(.)$.
If you cannot use Lambert function, think that you are looking for the zero of function $$f(x)=x^2 + \log(x) - c$$ $$f'(x)=2x+\frac 1x$$ The first derivative does not cancel in the real domain and it is always positive; so, only one root that you could find numerically using any method (secant, Newton, ...). For starting, you could start iterating using $x_0=\sqrt c$.
Suppose $c=0.01$. Newton iterations would then be
$$\left(
\begin{array}{cc}
n & x_n \\
0 & 0.1000000000 \\
1 & 0.3257436366 \\
2 & 0.6013225902 \\
3 & 0.6561217076 \\
4 & 0.6564436996 \\
5 & 0.6564437054
\end{array}
\right)$$
For the new version of the post.
Considering the differential equation
$$ \frac{ax}{(bx^2 + c)} = \frac{dx}{dt}\implies \frac{dt}{dx}=\frac{b x}{a}+\frac{c}{a x}$$ we then have $$t+K=\frac{b x^2}{2 a}+\frac{c \log (x)}{a}$$ which would then lead to
$$x^2=\frac{c }{b}W\left(\frac{b }{c}e^{\frac{2 a (K+t)}{c}}\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2497559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
if A is turing-recognizable, and A is mapping reducible to complement of A, A is decidable Here $<$ denotes mapping reducibility.
Show that if $A$ is Turing-recognizable and $A < A'$, then $A$ is decidable.
How can I prove this? I'm not sure I quite understand how $A < A'$ is possible.
|
I learned about these same ideas using different notation and terms. While the ideas are the same, my notation might be off. Here is what I got.
By $A \leq_m A'$ then for some computable (halts on every input) function $f:N \rightarrow N$ and any $n\in N$ we have.
$$ n \in A \Longleftrightarrow f(n) \in A'$$
equivalently
$$ n \notin A \Longleftrightarrow f(n) \notin A'$$
Then for any $n \in N$ if we want to see if $n \in A$ then because $A$ is turing recognizable we can do this.
For any $n \in N$ to see if $n \notin A$ then because $f$ is computable we have $f(n) \in N$. Because $A$ is turing recognizable, then we can check if $f(n) \in A$. If $f(n) \in A$ then $f(n) \notin A'$, any by the second iff statement above we have $n \notin A$.
So for any $n \in N$ we can see whether $n \in A$ or $n \notin A$, or that $A$ is decidable.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2497644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Solve summation of $\sum_{j=0}^{n-2}2^j (n-j)$ Question
While Solving a recursive equation , i am stuck at this summation and unable to move forward.Summation is
$$\sum_{j=0}^{n-2}2^j (n-j)$$
My Approach
$$\sum_{j=0}^{n-2}2^j (n-j) = \sum_{j=0}^{n-2}2^j \times n-\sum_{j=0}^{n-2}
2^{j} \times j$$
$$=n \times (2^{n-1}-1)-\sum_{j=0}^{n-2}
2^{j} \times j$$
I am unable to move forward , please help me out!
|
Hint
Consider $$\sum_{i=0}^p i x^i=x\sum_{i=0}^p i x^{i-1}=x\left(\sum_{i=0}^p x^i \right)'$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2497799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Is the extension of injective bounded linear transformation a injective linear transformation Let $X$ and $Y$ be two Banach spaces over $\mathbb{C}$
Let $V$ be a dense subspace of $X$
Let $T : V \to Y$ be a bounded linear operator
We know that $T$ can be uniquely extended to a bounded linear transformation $S : X \to Y$ such that $\left\| T \right\| = \left\| S \right\|$
I would like to know if is it true that
$$
T
\text{ is injective } \Longrightarrow
S \text{ is injective }
$$
thanks.
|
No. Using a Hamel basis of an infinite dimensional Banach space $Y$ one find a discontinuous linear functional $f$ on $Y$. Then $\|y\|' = \|y\|+|f(y)|$ is a strictly finer norm on $Y$ and the identity $(Y,\|\cdot\|') \to (Y\|\cdot\|)$ extends to the completion $X$ of $(Y,\|\cdot\|')$. This extension isn't injective because otherwise its inverse would be continuous by the closed graph theorem. But its restriction to the dense subspace $Y$ is injective.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2497956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Show the limit of a sequence of real numbers Define $(a_n)_{n=1}^\infty$ as a sequence of real numbers and let a be real number. Show that
$$\lim_{n\rightarrow\infty}(a_n)=a<=>\forall\epsilon\in ]0, 10^{-6}]∃N≥10^{-6}\forall n≥N: |a_n − a|<27\epsilon$$
Where n,N are natural numbers.
Im not sure if Ive understood it correctly. Here is what I have tried:
if $(a_n)$ tends towards a then $∃\epsilon=\frac{1}{n10^7}\forall n≥10^{-6}$
which means that
$|a_n − a|<\frac{1}{n10^7}\forall n≥10^{-6}$
which means
$|a_n − a|+a_n<a_n<a+\frac{1}{n10^7}$
Taking the limit we get
$\lim_{n\rightarrow\infty}(|a_n − a|+a_n)<\lim_{n\rightarrow\infty}(a_n)<\lim_{n\rightarrow\infty}(a+\frac{1}{n10^7})$
which equals $0+a_n<a_n<a+0$
So we have $\lim_{n\rightarrow\infty}(a_n)=a$
|
I think it's more like : $\lim\limits_{n\to\infty}(a_n) = a \iff \forall\varepsilon \in \ ]0,10^{-6}]\ \exists N \geq \fbox{$10^6$} \ \forall n\geq N : |a_n - a| < 27\varepsilon$ ?
$\lim\limits_{n\to\infty}(a_n) = a$ is :
$$\forall\varepsilon > 0\ \exists N\ \forall n\geq N : |a_n - a| < \varepsilon$$
is equivalent to
$$\forall\varepsilon > 0\ \exists N' > 10^6\ \forall n\geq N : |a_n - a| < \varepsilon$$
Because if it's true for a $N < 10^6$, it's also true $\forall N' > 10^6$
Like before, it's true $\forall\varepsilon > 0$ si it's always true for $0 < \varepsilon < 10^{-6}$, and still true for $27\varepsilon$.
So now we have :
$$\lim\limits_{n\to\infty}(a_n) = a \iff \forall\varepsilon\in\ (0,10^{-6}]\ \exists N' > 10^6\ \forall n\geq N : |a_n - a| < 27\varepsilon\ \square$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2498117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Find the $L^2$ norm of $\theta_\chi(z)^2$ in $\mathbb{H}/\Gamma_0(4)$ I'd like to learn about $L^2$ norms on Hyperbolic space. My Automorphic forms textbook says this function is a "cusp form" so it's in $L^2(\mathbb{H}/\Gamma_0(4))$:
$$ \theta_\chi(z) = \sum_{n \in \mathbb{Z}} \chi(n) \, e^{-\pi n^2 \, z} $$
This is a cusp form, since $\chi(0) = 0$ for any Dirichlet character $\chi$. Can we do the integral over the fundamental region?
$$ L^2(\theta_\chi) = \int_{\mathbb{H}/\Gamma_0(4)} |\theta_\chi(z)|^2 \, \frac{dx^2 + dy^2 }{y^2} $$
Have I written the norm correctly? It's very likely that in some cases, my question is not well-posed and that can be discussed in the comments or I'll up-vote an answer.
It seems the original version of my question is problematic. At least
$\theta_\chi(z)^2$ is a weight 1 (?) cusp form over $\Gamma_0(4)$ and the question $L^2$ makes sense.
$$ \theta_\chi(z)^2 = \sum_{(m,n) \in \mathbb{Z}^2} \chi(n) \, e^{-2\pi i (m^2+n^2 )\, z} $$
I may have to name a specific Dirichlet character. Perhaps the mod 4 character with $\chi(3)=-1$? And then we are integrating the absolute value squared over hyperbolic space:
$$| \theta_\chi(z)|^4 = \sum_{(m,n) \in \mathbb{Z}^4} \chi(n) \, e^{-2\pi i (a^2+b^2) -(c^2+d^2) )\, z} $$
We obtain a theta series associated to a quadratic form $q(z)=(a^2+b^2) -(c^2+d^2) $. The area form should be corrected as well
$$ L^2(\theta_\chi^2) = \int_{\mathbb{H}/\Gamma_0(4)} |\theta_\chi(z)|^4 \, \frac{dx \, dy}{y^2} $$
Is this question well stated?
|
The standard, but non-obvious, way to compute an $L^2$ norm of a modular form $f$ is as the residue at $s=1$ of the Rankin-Selberg $L$-function (or Dirichlet series) obtained as an integral $\int_{\Gamma\\\mathfrak H} E_s\cdot |f|^2$. This is because the residue at $s=1$ of that Eisenstein series is a constant (basically the multiplicative inverse of the natural volume of $\Gamma\backslash\mathfrak H$). For half-integral weight theta series, after the corrections indicated by @Peter Humphries are made, you can identify this Rankin-Selberg thing as an already-familiar Dirichlet series, whose reside at $1$ is known in other terms.
EDIT: "The" weight-zero Eisenstein series attached to the cusp $i\infty$ for $\Gamma_0(N)$ is $\sum_{c,d} y^2/|cz+d|^{2s}$ where $c,d$ are summed over relatively prime integers congruent to $0,1$ mod $N$, and multiplicatively modulo $\pm 1$ for $N=2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2498236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
expectancy value of count of prior values in array that are bigger Let A be an Array of the length n. A is filled randomly with distinct numbers ($\forall (i,j) < n: i \neq j \implies A[i] \neq A[j] $).
What is the total expected value (amount) of pairs (indexed i,j) with $i < j \land A[i] > A[j]$ in an array of the length n?
I started to look at a certain index $i$ but couldn't figure out the expectancy value of the count of numbers with an index smaller than $i$ and bigger than the number at index $i$.
Implementing that code I figured out, that the higher n is the nearer the expectancy value is to the count of pairs, when the array is sorted descending.
I would appreciate any hint to help me wrapping my mind around it.
|
You are counting what are called "inversions of permutations".
For each $i$ and $j$ with $i < j$, the probability that $A[i] > A[j]$ is $1/2$ (since it's equally likely for $A[i]$ or $A[j]$ to be the larger of the two).
There are $n(n-1)/2$ pairs of indices $i$ and $j$, and so the expectation you want is $n(n-1)/4$.
It can also be shown that the standard deviation of the number of inversions of a permutation of $n$ elements is on the order of $n^{3/2}$, so as $n$ gets larger these distributions get more concentrated relative to their mean.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2498389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Is this NP complete? I have to say that I am newbie in graph theory, so bear with me.
Here is my question:
Assume that you are given:
1. A directed graph G
2. A positive integer K
3. A source node S in graph G.
find the minimum number of edges to be added to graph G so that there is simple path of length K from node S to all the other nodes in the graph?
Does anyone know if this is a np complete problem?
|
Set cover reduces to this for $K = 2$, so yes it is NP-hard.
The set cover problem can be formulated as this: given a graph $G$ with disjoint vertex sets $X$ and $Y$ and arcs going from $X$ to $Y$, choose a subset $X' \subseteq X$ of minimum cardinality such that for every element $y \in Y$, there is an arc going from $X'$ to $Y$. Thus, $X'$ "covers" $Y$, and is called a covering set.
To reduce this to your problem, add your source vertex $s$, add another vertex $t$, add an arc between $s$ and $t$, and add every arc from $t$ to $X$. It is not too hard to show that if there is a covering set $X'$ of size $k$, then adding $k$ arcs is enough to make $s$ at distance at most $2$ from every vertex
(add each arc from $s$ to $X'$).
Conversely, suppose that adding a set $A$ of $k$ arcs makes $s$ at distance at most $2$ from every vertex. We use $A$ to find a covering set of size at most $k$. If every arc of $A$ goes from $s$ to $X$, then the endpoint of the arcs of $A$ must form a covering set, as desired. Suppose instead that there is some arc $(x, y)$ in $A$, for some $x \in X$ and $y \in Y$. Let $x' \in X$ such that $(x', y)$ is an arc of $G$ (we may assume $x'$ exists, otherwise the set cover instance is impossible). In $A$, we may replace $(x, y)$ by $(s, x')$, and $s$ is still at distance at most $2$ from every vertex. We may repeat this for every arc of $A$ from $X$ to $Y$. Observe that we may also do this for every arc from $t$ to $Y$ in $A$, if any. We may also handle arcs of $A$ from $Y$ to $Y$ in the same manner, then arcs of $A$ from $s$ to $Y$, if any. Note that there is no point in having arcs from $Y$ to $X \cup \{s, t\}$, and arcs from $X$ to $X$ can easily be replaced by an arc from $s$ to $X$, so this takes care of every possible type of arc. After these operations, $A$ has only arcs of the form $(s, x)$ for $x \in X$, and the endpoints of $A$ must form a covering set.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2498526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Frobenius method series solution Q: $x^2y^{''}-(x^2+2)y=0$ [1]
Solving using frobenius
$y=\sum_{n=0}^{\infty}a_nx^{x+r}$ [2]
$ y'=\sum_{n=0}^{\infty}(n+r)a_nx^{x+r-1}$ [3]
$ y''=\sum_{n=0}^{\infty}(n+r)(n+r-1)a_nx^{x+r-2}$ [4]
inserting [2,4] into [1]
$x^2 \sum_{n=0}^{\infty}(n+r)(n+r-1)a_nx^{x+r-2}-(x^2+2)\sum_{n=0}^{\infty}a_nx^{x+r}$ [5]
Multiplying through but the cofficent I get
$\sum_{n=0}^{\infty}(n+r-1)(n+r)a_n x^{n+r}-\sum_{n=0}^{\infty}a_nx^{x+r+2}-2\sum_{n=0}^{\infty}a_nx^{n+r}$
setting n=0 I solve for r to get the values of $r=2 , r=-1$
Now this is where my confusion comes in
so for r=2 I rearnage the series like so
$x^2(\sum_{n=0}^{\infty}[(n+r-1)(n+r)a_n -2a_n]x^{n}-\sum_{n=0}^{\infty}a_nx^{n+2})=0$ [6]
So now what I did I want all x's to be of the same power so I did the following:
$k=n+2$
$k-2=n$
$\sum_{k=0}^{\infty}[(k+1)(k+2)a_n -2a_k]x^k-\sum_{k=2}^{\infty}a_{k-2}x^k=0$ [9]
expanding on the series $\sum_{k=0}^{\infty}[(k+1)(k+2)a_n - 2a_n]x^k$ so I can get the staring point of the series in the same postion I did the following
$k=0, 2a_0-2a_0=0 \rightarrow a_0=0$
$k=1, 6a_{1}-2a_{1} \rightarrow a_1=0$
$a_0=a_1=0$ is because [9] is equal to $0$
so now I have the equation
$\sum_{k=2}^{\infty}[(k+1)(k+2)a_n - 2a_n-a_{k-2}]x^k=0$
So I now have a recurrence relation of:
$(k+1)(k+2)a_n - 2a_n-a_{k-2}=0$
$a_n=\frac{a_{k-2}}{k(k+3)}$
Now in my book solution it has the solution of
$a_2=\frac{a_0}{2\cdot 5}$ $a_3=0$ $a_4=\frac{a_0}{2\cdot 5\cdot 4\cdot 7}$
but how is it $a_0$? I know I have gone somewhere wrong but just cant see where.
|
Your approach is fine. There is just a small mistake which causes the problem. In the following we use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series.
Let's consider again the series equation
\begin{align*}
\sum_{k=0}^\infty[(k+1)(k+2)a_k-2a_k]x^k-\sum_{k=2}^\infty a_{k-2}x^k=0\tag{1}
\end{align*}
We consider the coefficient of $x^0$
\begin{align*}
[x^0]:\qquad 2a_0-2a_0=0\qquad \text{resp.}\qquad 0=0
\end{align*}
We observe this equation provides no information at all. So $k=0$ can't be usefully used (this was the mistake) and $a_0$ is left unspecified. This becomes plausible, when we rearrange the equation (1) somewhat. We obtain
\begin{align*}
\sum_{k=0}^\infty&[(k+1)(k+2)a_k-2a_k]x^k-\sum_{k=2}^\infty a_{k-2}x^k\\
&=\sum_{k=0}^\infty(k+3)ka_kx^k-\sum_{k=2}^\infty a_{k-2}x^k\tag{1}\\
&=\sum_{k=1}^\infty(k+3)ka_kx^k-\sum_{k=2}^\infty a_{k-2}x^k\tag{2}\\
&=0
\end{align*}
Comment:
*
*In (1) we simplify $(k+1)(k+2)a_k-2a_k$ and observe the coefficient $a_0$ vanishes in the left-hand series since $(k+3)k=0$ if $k=0$.
*In (2) we can therefore start the left series with $k=1$.
From (2) we obtain the expected relations:
\begin{align*}
[x^1]:\qquad &&4a_1&=0&\qquad &\color{blue}{a_1=0}\\
[x^2]:\qquad &&10a_2-a_0&=0&\qquad &\color{blue}{a_2=\frac{a_0}{10}}\\
[x^3]:\qquad && 18a_3-a_1&=0&\qquad &\color{blue}{a_3=0}\\
[x^4]:\qquad &&28a_4-a_2&=0&\qquad &\color{blue}{a_4=\frac{a_2}{28}=\frac{a_0}{280}}
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2498647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.