Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Sides $\frac{|b - c|}{\sqrt{(b^2 + 1)(c^2 + 1)}}, \frac{|c - a|}{\sqrt{(c^2 + 1)(a^2 + 1)}}, \frac{|a - b|}{\sqrt{(a^2 + 1)(b^2 + 1)}}$ of a triangle.
Prove that for all pairwise distinct $a, b, c \in \mathbb R$, $$\frac{|b - c|}{\sqrt{b^2 + 1}\sqrt{c^2 + 1}}, \frac{|c - a|}{\sqrt{c^2 + 1}\sqrt{a^2 + 1}}, \frac{|a - b... | Let $a>b>c\geq0$.
Thus, easy to see that $$\frac{a-c}{\sqrt{(a^2+1)(c^2+1)}}>\frac{a-b}{\sqrt{(a^2+1)(b^2+1)}}$$ and
$$\frac{a-c}{\sqrt{(a^2+1)(c^2+1)}}>\frac{b-c}{\sqrt{(b^2+1)(c^2+1)}}$$ because $$a-c>\frac{a}{b}(b-c)$$ and it's enough to prove that:
$$\frac{a-b}{\sqrt{(a^2+1)(b^2+1)}}+\frac{b-c}{\sqrt{(b^2+1)(c^2+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Proving $\sin\frac{a\pi}{2}\,\sin\frac{b\pi}{2}=\sin\frac{ab\pi}{2}$ for $a,b\in\mathbb{N}$? I was playing with the graphs of $f:\,\mathbb{R}\to\mathbb{R},\, f(x)=\sin axt\pi$ and $g:\,\mathbb{R}\to\mathbb{R},\, g(x)=\sin xt\pi\sin at\pi$.
What I conjectured is the following:
If $t=\frac{m}{2}$ where $m$ is $1$ above ... | Suppose $a$ or $b$ is even. Without loss of generality, $a=2k$. Then
$$\sin\left(\frac{a\pi}{2}\right)\sin\left(\frac{b\pi}{2}\right)=\sin\left(\frac{2k\pi}{2}\right)\sin\left(\frac{b\pi}{2}\right)=\sin\left(k\pi\right)\sin\left(\frac{b\pi}{2}\right)=0$$
and
$$\sin\left(\frac{ab\pi}{2}\right)= \sin\left(\frac{2kb\pi}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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finding bounds for $\int_0^X\lfloor x^2\rfloor \, dx$ I am trying to find bounds for:
$$I=\int_0^X\lfloor x^2\rfloor \, dx$$ and through integrating two different ways I ended up with the sums:
$$I=(X^2+1)^{3/2}-\sum_{i=1}^{X^2+1}\sqrt{i}$$
$$I=X^3-\sum_{i=1}^{X^2}\sqrt{i}$$
I can't verify if either of these are correc... | $\newcommand{\d}{\text{d}}$Assuming $X$ is an integer, we have :
$\begin{align}
I(X)&=\int_0^X\lfloor x^2\rfloor\d x\\
&=\sum_{n=0}^{X^2-1}\int_{\sqrt{n}}^{\sqrt{n+1}}\lfloor x^2\rfloor\d x\\
&=\sum_{n=0}^{X^2-1}(\sqrt{n+1}-\sqrt{n})n\\
&=\sum_{n=0}^{X^2-1}n\sqrt{n+1}-\sum_{n=0}^{X^2-1}n\sqrt{n}\\
&=\sum_{n=1}^{X^2}(n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3808470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Advanced Inequality with Floor Functions For positive real numbers $a,$ $b,$ $c,$ and $d,$ find the minimum value of
$$\left\lfloor \frac{b + c + d}{a} \right\rfloor + \left\lfloor \frac{a + c + d}{b} \right\rfloor + \left\lfloor \frac{a + b + d}{c} \right\rfloor + \left\lfloor \frac{a + b + c}{d} \right\rfloor.$$
I'm ... | By C-S $$\sum_{cyc}\left[\frac{a+b+c}{d}\right]>\sum_{cyc}\frac{a+b+c}{d}-4=\sum_{cyc}a\sum_{cyc}\frac{1}{a}-8\geq16-8=8.$$
Now, make an example for $$\sum_{cyc}\left[\frac{a+b+c}{d}\right]=9.$$
I got that $(a,b,c,d)=(5,5,5,4)$ is valid.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $e^x > 1 + (1 + x)\log(1 + x), x > 0$ using power series expansion. Prove that $e^x > 1 + (1 + x)\log(1 + x), x > 0$ using power series expansion.
I am a bit puzzled by this statement because the power series for $\log(1+x)$ only converges iff $|x|<1$. Is the problem sound? Should it be $1>x>0$?
| For $0<x<1$, does the following proof work?
$e^x>1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}$ and $\log(1+x)<x-\dfrac{x^2}{2}+\dfrac{x^3}{3}$ and so,
$$1+(1+x)\log(1+x)<1+x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{3}$$
We observe that $$1+x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{3}<1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}$$
iff
$x^3>x^4$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3809899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the exponential form of $e^{17\pi i/60} + e^{27\pi i/60} + e^{37\pi i /60} + e^{47 \pi i /60} + e^{57 \pi i/60}$
Find the exponential form of the complex number
$$ e^{17\pi i/60} + e^{27\pi i/60} + e^{37\pi i /60} + e^{47 \pi i /60} + e^{57 \pi i/60}$$ with proof.
I know that I have to combine some of the exp... | My sol.
We first look at $e^{17 \pi i/60} + e^{57 \pi i/60}$ and represent it in polar form.
We note that $$e^{17 \pi i/60} = \cos \frac{17\pi}{60} + i\sin\frac{17\pi}{60}$$ and $$e^{57 \pi i/60} = \cos \frac{57\pi}{60} + i\sin\frac{57\pi}{60}.$$
Therefore, by adding the cosines and sines we get $$e^{17 \pi i/60} + e^{... | {
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"timestamp": "2023-03-29T00:00:00",
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Special case ($3\times 3$ and $4\times 4$) of USAMO 1998 problem #$4$ This is a special case ($3\times 3$ and $4\times 4$) of USAMO 1998 problem #4.
A $98\times 98$ chess board has the squares colored alternately black and white in the usual way. A move consists of selecting a rectangular subset of the squares (with ... | For the $3\times3$ case you could change two middle rows and get all $1$'s (check that). For the $4\times4$ case the possible sequence of moves is "$3$rd row", "$3$rd column", "$1$st row", "$1$st column".
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3814283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating double sum $\sum_{k = 1}^\infty \left( \frac{(-1)^{k - 1}}{k} \sum_{n = 0}^\infty \frac{1}{k \cdot 2^n + 5}\right)$
Find $$\sum_{k = 1}^\infty \left( \frac{(-1)^{k - 1}}{k} \sum_{n = 0}^\infty \frac{1}{k \cdot 2^n + 5}\right)$$
So far, I've gotten that the sum of the left is equal to $\log(2),$ meaning we ... | The inner sum is bounded by $\frac{2}{k},$ so the full sum is absolutely convergent, which means we can rearrange and combine terms.
If $m=2^i p$ for $p$ odd, the coefficient of $\frac1{m+5}$ is $$\frac{1}{p}-\left(\frac{1}{2p}+\frac{1}{4p}+\cdots +\frac{1}{2^ip} \right)=\frac{1}{m}.$$
So your sum is: $$\sum_{m=1}^{\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3816455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Easier way for probability problem?
Question: Bob takes a math test with $10$ questions. He has a $70\%$ chance of getting each question right. What is the probability that they get exactly $60\%$ on the test? Express your answer as a common fraction.
My bashy way: We know that the successful outcomes (numerator of p... | We know that the successful outcomes (numerator of probability)
look here: Binomial distribution.
The probability of getting exactly k successes in n independent Bernoulli trials is given by the probability mass function: $$\Pr(X=k)={\binom {n}{k}}p^{k}(1-p)^{n-k}$$
Therefore your score is the exact probability of 6 su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3819548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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solve : $xy''+2y'+xy=1$ given that $y=\frac{\sin x}{x}$ is a solution for $xy''+2y'+xy=0$, solve $xy''+2y'+xy=1$.
My try:
$\begin{aligned}xy''+2y'+xy&=0\\y''+\frac2xy'+y&=0\\\begin{pmatrix}
\frac{\sin x}x& u\\
\frac{x\cos(x)-\sin(x)}{x^2} & u'
\end{pmatrix}&=\frac2x\\\frac{\sin x }{x}u'-\frac{x\cos(x)-\sin(x)}{x^2}u&=\... | $$y''-\frac{2}{x}y'+y=0~~~(1)$$, if $y_1=\frac{\sin x}{x}$ is a solution of (1), then the other solution $y_2(x)$ is given by
$$y_2(x)=y_1(x) \int \frac{e^{\int p(x) dx}}{y^2_1(x)}dx,~ p(x)=-2/x$$
$$\implies y_2(x)= \frac{\sin x}{x} \int \frac{1}{x^2}\frac{x^2}{\sin ^2 x} dx$$
One may ignor the multiplicative $-$ sign ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3819694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find $\sum_{r=1}^{n}\frac{r}{(2r-1)(2r+1)(2r+3)}$? I tried to solve it, but the answer I got was different from the answer given.
Answer given:
$$\sum_{r=1}^{n}\frac{r}{(2r-1)(2r+1)(2r+3)} = \frac{n(2n+1)}{4(2n-1)(2n+3)}$$
My working:
$$\sum_{r=1}^{n}\frac{r}{(2r-1)(2r+1)(2r+3)}$$
$$= \sum_{r=1}^{n} \Biggl... | Hint:
Another way:
Let $$\dfrac r{(2r-1)(2r+1)(2r+3)}=f(r)-f(r+1)$$ where $f(m)=\dfrac{am+b}{(2m-1)(2m+1)}$
so that $$\sum_{r=1}^n\dfrac r{(2r-1)(2r+1)(2r+3)}=\cdots=f(n+1)-f(1)$$
Now comparing the numerator of $f(r)-f(r+1)$ with $r,$
$$2a=1, a+4b=0\iff b=-\dfrac a4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3820237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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What is the average distance between the first occurence of an element and the first occurence of another after that one? We have a set of Na elements 'a' and Nb elements 'b'.
For example for Na=2 and Nb=2 it would be (a, a, b, b).
We can generate all possible permutations with them. In our example we will end up havin... | Let $\mathbf X$ be the distance we want.
Then there are $\binom{N_a+N_b}{N_b}$ ways to choose a permutation of the as and bs to consider. Of them, $\binom{N_a+N_b}{N_b}-1$ have an a followed by a b at some point.
Among them, there are $\binom{N_a + N_b-k+1}{N_b}-1$ ways to choose a permutation in which $\mathbf X \ge k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3821307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Floor function bounding From CMC:
What is the sum of the square of the real numbers $x$ for which $x^2 - 20\lfloor x\rfloor + 19 = 0$?
We use $\lfloor x\rfloor\le x<\lfloor x\rfloor+1$ and eventually get the bounds $1\le x\le19$ and $x\ge 18,x\le 2.$ Of course, it's possible for $x$ not to be an integer, so how do we... | $$x^2 - 20 \lfloor x \rfloor + 19 = 0$$
The intuition is that the solutions does not get too far away from the solutions of $x^2-20x+19=0$, namely $x=1, 19$. So go ahead and express that intuition!
I am not used to fiddling with $x-1 < \lfloor x \rfloor \le x$. So let's go for a more granular method.
Let $n = \lfloor ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3825253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Necklace problem with Burnside Lemma How many necklaces can be made with two red beads, two green beads, and four violet beads?(8 total)
Using Burnside lemma is complicated for me due to my lack of understanding of the lemma. I want to know the method step-by-step.
| We may as well deploy PET here since we need the cycle index $Z(D_8)$ of
the dihedral group $D_8$ of order $16$ to apply Burnside. We compute
and average the number of assignments of colors to the eight slots fixed
by permutations from each conjugacy class in $D_8$, taking into account
the order of the class. This me... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3827211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How does this division work? $\frac{\;\frac{6^6}{1}\;}{2^{-3}}\cdot2^{-10}$ I came across this division and can't wrap my head around how it is solved:
$$\frac{\;\;\frac{6^6}{1}\;\;}{2^{-3}}\cdot2^{-10}$$
They subtract the exponent of $2^{-10}$ from the denominator's exponent $2^{-3}$:
$$2^{-3-(-10)}$$
Which gives us:
... | \begin{align}
\frac{\;\;\dfrac{6^6}{1}\;\;}{2^{-3}}\cdot2^{-10}
&=\frac{6^6}{2^{-3}}\cdot 2^{-10}\qquad&\text{(simplify the denominator)}\\
&=6^6\cdot 2^3\cdot 2^{-10}\qquad &(\frac{1}{2^{-3}}=2^3)\\
&=6^6\cdot 2^{3+(-10)}\qquad &(2^a2^b=2^{a+b})\\
&=6^6\cdot 2^{-7}\qquad &\text{(simplify)}\\
&=\frac{6^6}{2^7}\qquad &(... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solving for $x$ and $y$ when $(3x + y)(x + 3y)\sqrt{xy} = 14$ Solve for $x$ and $y$ when $$(3x + y)(x + 3y)\sqrt{xy} = 14$$ $$(x+y)(x^2 + 14xy + y^2) = 36.$$
I was thinking of squaring the first equation and moving on from there, but I think it'll be a bit too messy. Is there a better way to start this problem?
| Setting $p=x+y$ and $q=\sqrt{xy}$ (almost as suggested by Alexey in the comments, but $\sqrt{xy}$ looks like it will be more symmetric) and expanding gives
\begin{align}
3p^2q + 4q^3 &= 14 \\
p^3 + 12pq^2 &= 36
\end{align}
From here, a lucky coincidence is that
$$
(p + 2q)^3 = p^3 + 6p^2q + 12pq^2 + 8q^3 = 36 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solving for $x$ when $\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + x}} = x$ Suppose $x$ is a real number such that $\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + x}} = x.$ Find $x.$
I first squared to get rid of the first square root, which gave me $\sqrt{3} - \sqrt{\sqrt{3} + x} = x^2.$ However, I'm not sure how to move on from there. Can ... | Let $a = \sqrt{3}$, so $x = \sqrt{a-\sqrt{a+x}}$. Let's call this equation $(*)$.
Let $y = \sqrt{a+x}$ so $x = \sqrt{a-y}$. Therefore
*
*$x^2 = a-y$
*$y^2 = a+x$
Subtracting 2. from 1. we have $x^2 - y^2 = -(x+y)$. Let's consider the following cases:
*
*If $x+y = 0$, then $x = -\sqrt{a+x}\le 0$, but from $(*)$ we ... | {
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"timestamp": "2023-03-29T00:00:00",
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Does there exist a natural number pair $(a,b)$ such that $a^2b+a+5$ divide $ab^2+a+b$? I tried that if I can show $\gcd(a^2b+a+5,ab^2+a+b) = 1$, then there is no solution. Am I true?
Any hints /idea?
I can't factorize them out such that each is multiple of each other.
| Since $a^2b+a+5$ divides $ab^2+a+b$, we can conclude that it also divides:
$$a(ab^2+a+b)-b(a^2b+a+5)=a^2-5b$$
This means that we must have $a^2-5b=0$ or we must have $|a^2-5b| \geqslant a^2b+a+5$. In the first case, substituting $a=5k$ and $b=5k^2$ gives:
$$(125k^4+5k+5) \mid (125k^5+5k^2+5k)$$
which is obviously true.... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to integrate hyperbolic function $\frac{1-\sin(x)}{1+\sin(x)}$ I'm trying to find a primitive of $ \frac{1-\sin(x)}{1+\sin(x)} $. By changing the variable to $t=\tan(\frac{x}{2})$ and letting $\sin(x) = \frac{2t}{1+t^2}$ I get the following integral:
\begin{align}
& \int \frac{1-\sin(x)}{1+\sin(x)} \, dx \\[8pt] = ... | $\displaystyle \frac{1-sinx}{1+sinx}=\frac{2}{1+sinx}-1=\frac{2}{(sin\frac{x}{2}+cos\frac{x}{2})^{2}}-1$
$\displaystyle =\frac{1}{cos^{2}(\frac{x}{2}-\frac{\pi}{4})}-1=tan^{2}(\frac{x}{2}-\frac{\pi}{4})$
$\displaystyle \theta=\frac{x}{2}-\frac{\pi}{4}\Rightarrow 2d\theta=dx$
$\displaystyle z=tan\theta\Rightarrow dz=(1+... | {
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"timestamp": "2023-03-29T00:00:00",
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Is $(x^2+3)^3-4(x-1)^3$ an irreducible polynomial in $Q[x]$? I was trying to solve the problem below
Let $K \subset \mathbb{C}$ be a splitting field of $f(x) = x^3 − 2$ over $\mathbb{Q}$. Find a complex number $z$, such that $K = \mathbb{Q}(z)$.
All the roots of the polynomial $f(x)$ are $2^{1/3}$, $2^{1/3}\omega$ an... | Another way is to notice $h(x)$ factors as
$$(x^4 + 4x^3 + 3x^2 + 3)(x^2 + x + 2)\mbox{ in }\mathbb{Z_5}[x]\tag{1}$$
and
$$(x^3 + 4x^2 + 6x + 5)(x^3 + 3x^2 + 5x + 2)\mbox{ in }\mathbb{Z_7}[x]\tag{2}.$$
(the $2$-nd and $3$-th degrees factors are irreducible by testing finitely many roots in $\mathbb{Z_5}[x]$ and $\mathb... | {
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Find the supremum of $|z|^2+\text{Re}(\overline{z}w)$ subject to $|z|^2+|w|^2=1$. I am trying to find the operator norm of $$A=\begin{pmatrix}3 & 1 \\
1 & 1\end{pmatrix}\in M_n(\mathbb{C}).$$This is what I have so far: If $|z|^2+|w|^2=1$, then $$||A\begin{pmatrix}z\\w\end{pmatrix}||_2=\sqrt2\sqrt{1+4(|z|^2+\text{Re}(\o... | Let $z=x+iy$ and $w=u+iv$ so that $|z|^2+\Re(\overline zw)=x^2+y^2+xu+yv$ is subject to $x^2+y^2+u^2+v^2=1$. Then we have $\mathcal L=x^2+y^2+xu+yv-\lambda(x^2+y^2+u^2+v^2-1)$ so that \begin{align}\mathcal L_u&=x-2\lambda u=0\implies x=2\lambda u\\\mathcal L_v&=y-2\lambda v=0\implies y=2\lambda v\\\mathcal L_x&=2x+u-2\... | {
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"timestamp": "2023-03-29T00:00:00",
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Find all positive integer values $(x, y, n)$ such that $x^n+1=y^{n+1}$ and $gcd(x, n+1)=1$ My approach to this problem is as follows:
First, I attempt to prove that $x^n+1=y^{n+1}$ has solutions on the integers only for $n=1$. Since $y>x$, it follows that $y>1$. Thus If $n\ge2$, we have $$y^{n+1}-x^n>y^n-x^n=(y-x)(y^{... | One guy trying to prove Fermat’s Last Theorem send a telegram: “We should move $y^{n}$ to the right-hand side. The details will be sent a letter”. So let us follow this advice.
We have $x^n=y^{n+1}-1=(y-1)z$, where $z=y^n+y^{n-1}+\dots+1$. We have $\gcd(y-1,z)=(y-1,n+1)$. Since $(y-1)z=x^n$ is coprime with $n+1$, we ha... | {
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"source": "stackexchange",
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Find a mistake in a computation of a determinant. I recently have to solve one exercise in which I need to compute the determinant of the matrix
$$ A= \begin{pmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
1 & 1 & 1 & a \\
\end{pmatrix} ,$$
where $a\in\mathbb{R}$. This is how I proceed using famous prop... | We consider a square matrix $A$ and recall two rules regarding calculation with $\det A$:
*
*Multiplication of a row of $A$ with a scalar $c$ results in $c\cdot\det A$
*Addition of a scalar multiple of a row with another row leaves the value of the determinant unchanged.
Looking at OPs example we obtain
\begin{ali... | {
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"answer_id": 0
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Maximizing $\frac{y+1}{x+2}$ when $(x-3)^2 + (y-3)^2 = 6$ Suppose $x$ and $y$ are real numbers such that $(x-3)^2 + (y-3)^2 = 6.$ Than, maximize $\frac{y+1}{x+2}.$
I do in fact realize that this is a double post, but it's a 5 year old question and I don't feel as if it is appropriate to bump it. I did as the original ... | A bit of geometry.
Introduce new coordinates X,Y:
$Y := y+1;$ $X := x+2;$
1)$Y=CX;$
2)$(X-5)^2 + (Y-4)^2=6;$
Need to find the slope $C$ of the line $Y=CX$ s.t. it is tangent to the circle 2).
3)Distance of tangent line $Y=CX$ to the centre of circle $(5,4):$
$\dfrac{|-5C+4|}{\sqrt{1+C^2}}=√6;$
$(-5C+4)^2=6(1+C^2);$
A ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3837653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to prove $5^{2n+1}-3^{2n+1}-2$ is divisible by $30$ for even $n$? How to prove $5^{2n+1}-3^{2n+1}-2$ is divisible by $30$ for all positive even integer $n$?
Here are my efforts:
Let $f(n)=5^{2n+1}-3^{2n+1}-2$.
For $n=2$,
$$f(1)=5^{2\cdot 2+1}-3^{2\cdot2+1}-2=2880=30\times 96$$
So, the statement is true for $n=2$.
... | This is not an induction proof, but I'll nevertheless leave it here.
To show that $5^{2n+1}-3^{2n+1}-2$ is divisble by $30=2\cdot 3\cdot 5$, you can show that $2,3,5$ divides it.
*
*$2\mid 5^{2n+1}-3^{2n+1}-2\iff 2\mid 5^{2n+1}-3^{2n+1}$, but this is clear since the sum of two odd numbers is even.
*$3\mid 5^{2n+1}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3838423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Where did I go wrong in solving this equation?
The value of the expression $ax^2 + bx + 1$ are $1$ and $4$ when $x$ takes the values of $2$ and $3$ respectively. Find the value of the expression when $x$ takes the value of $4.$
Here is my first attempt at solving this question:
\begin{align}
2a^2 + 2b + 1 &= 1 \lefta... | We have
*
*$p(2)=1 \implies 4a + 2b + 1=1$
*$p(3)=4 \implies 9a + 3b + 1=4$
then solve for $a$ and $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3838776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Determining whether a subset is a subspace The problems are as follows:
Determine whether the following subsets of $\mathbb{R}^3$ are subspaces of $\mathbb{R}^3$.
*
*$A = \{(u^2, v^2, w^2) \,|\, u, v, w \in \mathbb{R} \}$,
*$B = \left\{(a, b, c) \,|\,
\begin{pmatrix}
a & b & c\\
1 & 2 & 0\\
0 & 1 & 2
\end{pmatrix} ... | Most of it is great.
In 1 the vectors in $A$ can not be multiplied by negative scalar, so it is not a subspace.
In 2 and 3 your reasoning is perfect, but maybe you can state it in a more concise manner, depending on what your learned already. Once you get that the subset is defined by linear equations you are done - it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3839873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
how to show that $\csc x - \csc\left(\frac{\pi}{3} + x \right) + \csc\left(\frac{\pi}{3} - x\right) = 3 \csc 3x$?
How to show that $\csc x - \csc\left(\frac{\pi}{3} + x \right) + \csc\left(\frac{\pi}{3} - x\right) = 3 \csc 3x$?
My attempt:
\begin{align}
LHS &= \csc x - \csc\left(\frac{\pi}{3} + x\right) + \csc\left(\... | We have $$\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} - x\right) \cos\left(x + \frac{\pi}{6}\right) \\$$
$$=\sin(x)[\sin(\frac{\pi}{3})\cos(x)+\cos(\frac{\pi}{3})\sin(x)]$$
$$+(\sin(\frac{\pi}{3})\cos(x)-\cos(\frac{\pi}{3})\sin(x))(\cos(\frac{\pi}{6})\cos(x)-\sin(\frac{\pi}{6})\sin(x))$$
$$=\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
ABC is a triangle and the line YCX is parallel to AB such that AX and BY are the angular bisectors of angle A and I'm interested in the following problem:
$ABC$ is a triangle and the line $YCX$ is parallel to $AB$ such that $AX$ and $BY$ are the angular bisectors of angle $A$ and angle $B$ respectively. If $AX$ meets ... | Since $XY||AB$, in the standard notation we obtain:
$\measuredangle CXD=\frac{\alpha}{2},$ $\measuredangle CYE=\frac{\beta}{2},$ $\measuredangle DCX=\beta$, $\measuredangle ECY=\alpha$, $CX=b$ and $CY=a.$
Thus, by the law of sines for $\Delta DCX$ we obtain:
$$\frac{XD}{\sin\beta}=\frac{b}{\sin\left(\beta+\frac{\alpha}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Limit of multiple absolute values Let $f(x) = \frac{x^2+2x-3|1-x^2|+1}{2|x+1|-|x^2+x|}$. Find $\lim_{x\rightarrow -1}f(x)$.
I broke up each absolute value into parts:
$|1-x^2| = \begin{cases}
1-x^2 & -1 \leq x\leq 1 \\
-(1-x^2) & x>1,x<-1
\end{cases}
$ ,
$|x+1| = \begin{cases}
x+1 & x\geq -1 \\... | Note that\begin{align}\require{cancel}f(x)&=\frac{x^2+2x-3|1-x^2|+1}{2|x+1|-|x^2+x|}\\&=\frac{|x+1|^{\cancel2}-3\cancel{|x+1|}|x-1|}{\cancel{|x+1|}(2-|x|)}\\&=\frac{|x+1|-3|x-1|}{2-|x|}\\&\to_{x\to-1}\frac{-6}1\\&=-6.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Asymptotic expansion, solving roots to an equation with dominant balance, what went wrong in my approach? So I wanted to compute the asymptotic expansion of the roots to, as $\epsilon \to 0$,
$$\epsilon x^3-x^2+2x-1=0$$
Now when I tried to find $x\sim x_0+\epsilon x_1+\epsilon^2x_2+...$ I ran into trouble as at $O(\eps... | Another way to obtain the expansion is by using series inversion. We have
\begin{align*}
\varepsilon = \frac{{(x - 1)^2 }}{{x^3 }} & = \frac{{(x - 1)^2 }}{{1 + 3(x - 1) + 3(x - 1)^2 + (x - 1)^3 }} \\ & = (x - 1)^2 (1 - 3(x - 1) + 6(x - 1)^2 - \cdots ),
\end{align*}
and thus
\begin{align*}
\pm \sqrt \varepsilon & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
The limit and Asymptotic Behavior Wallis-like Integral It is known that $$I_n=\int^{\frac{\pi}{2}}_{0} x \cos^n xdx \quad (n=0,1,2,3,\dots),$$ then
$$\lim_{n \to \infty} nI_{n}=1.$$
My question is: does there exist $k$ such that
$$\lim_{n \to \infty} n^k(nI_{n}-1)$$
converges to a non-zero real number?
| You can use Laplace's method to establish the asymptotics
$$
\int_0^{\pi /2} {x\cos ^n xdx} = \int_0^{\pi /2} {xe^{n\log \cos x} dx} = \frac{1}{n} - \frac{2}{3} \frac{1}{n^2} + \mathcal{O}\!\left( {\frac{1}{{n^{5/2} }}} \right).
$$
Thus
$$
n\cdot (nI_n - 1) \to - \frac{2}{3} ,
$$
i.e., $k=1$.
Addendum 1. A more str... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3843470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How do you calculate phase of $\text{DTFT}$ when exponential is in denominator? $$Xe^{jw}=\frac{1}{1-ae^{jw}}$$
How is the phase of this derived as $\tan ^{-1}\left(a*\sqrt{\left(1-a^2\right)}\right)$?
| $$\begin{align*} X\left(e^{j\omega}\right) &= \dfrac{1}{1-ae^{-j\omega}}\\
\\
&= \dfrac{1}{1-ae^{-j\omega}}\cdot\dfrac{1-ae^{j\omega}}{1-ae^{j\omega}} \\
\\
&= \dfrac{1-ae^{j\omega}}{1 -ae^{-j\omega} -ae^{j\omega} + a^2}\\
\\
&= \dfrac{1-ae^{j\omega}}{1 -2a\cos\omega + a^2}\\
\\
&= \dfrac{1-a\cos\omega}{1 -2a\cos\ome... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The result of $\int{\sin^3x}\,\mathrm{d}x$ $$\int{\sin^3x}\,\mathrm{d}x$$
I find that this integration is ambiguous since I could get the answer with different approaches. Are these answers are valid and true? Could someone tell me why and how? And also, is there any proof stating that these two method I use results th... | Yes, they are both valid and true. Actually,$$(\forall x\in\Bbb R):\frac13\cos^3(x)-\cos(x)=\frac1{12}\cos(3x)-\frac34\cos(x)$$since$$(\forall x\in\Bbb R):\cos(3x)=4\cos^3(x)-3\cos(x).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3845996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find a and c in the equation Find a and c in the equation
$$y=ca^x$$
when it is given that $x = \frac{1}{2}$ gives $y = 6$ and $x = 2$ gives $y = 162$
I already tried to solve when $6=ca^{\frac{1}{2}}$ and managed to get $a=\frac{36}{c^2}$ and when I insert $a$ in the equation $6=c\cdot \left(\frac{36}{c^2}\right)^{\fr... | Suppose we have $y=ca^{x}$. Then substituting the points $(x,y)=(\frac{1}{2},6)$ and $(x,y)=(2,162)$ gives $$6=ca^{\frac{1}{2}}$$
$$162=ca^{2}$$
Dividing the two we obtain $a^{\frac{3}{2}}=\frac{162}{6}=27$ so $a=9$. Now substituting into the first we obtain $$6=c(9)^{\frac{1}{2}}=3c$$
so $c=2$. Thus we have $y=2\cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3847472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to show that $\frac{n^3}{\sqrt{n^6 + 1}}$ converges to 1. I want to show that $\frac{n^3}{\sqrt{n^6 + 1}}$ converges to 1. I've tried using $\epsilon-N$ to show that it converges but I can't isolate $n$. Is there a way to bound this so I can apply the squeeze theorem? Or am I not on the right track?
| If you want to use the sqeeze theorem you can use
$\frac {n^3}{(n+1)^3} = \frac {n^3}{\sqrt{(n+1)^6}}=\frac {n^3}{n^6+ 6n^5 + .... + 1} < \frac {n^3}{\sqrt{n^6 + 1}} < \frac {n^3}{\sqrt{n^6}}=\frac {n^3}{n^3} =1$ (but you will have to prove $\frac {n^3}{(n+1)^3} \to 1$).
But to do an $\epsilon-N$ proof:
$|\frac {n^3}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3864831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $x,y,z$ are the lengths of sides of a triangle such that $x+y+z=2$. Find the range of $xy+yz+xz-xyz$ .
Let $x,y,z$ are the lengths of sides of a triangle such that $x+y+z=2$. Find the range of $xy+yz+xz-xyz$ .
What I Tried: I have tried by doing $(x+y+z)(x+y+z)=2*2 = 4.$
Also I got $2(xy+yz+xz)+(x^2+y^2+z^2)=4$.... | $xy+yz+xz-xyz=(1-x)(1-y)(1-z)+x+y+z-1=(1-x)(1-y)(1-z)+1$.
And $x+y+z=2$ gives $(1-x)+(1-y)+(1-z)=1$.
$1-x,1-y,$ and $1-z$ are all strictly positive (otherwise we couldn't make a triangle out of $x,y,$ and $z$). So let $a=1-x,b=1-y,$ and $c=1-z$. We have
$$a,b,c\in(0,1)$$
$$a+b+c=1$$
From $AM\ge GM$, their product satis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3869454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
$\sqrt{1-x^2}$ is not differentiable at $x = 1$. Please give me an easier proof if exists.
Prove that $\sqrt{1-x^2}$ is not differentiable at $x = 1$.
My proof is here:
Let $0 < h \leq 2$.
$\frac{\sqrt{1-(1-h)^2}-\sqrt{1-1^2}}{-h} = \frac{\sqrt{2h-h^2}}{-h} = \frac{2h-h^2}{-h\sqrt{2h-h^2}} = \frac{h-2}{\sqrt{2h-h^2}... | Let $g(x) = 1 - x^2$. Then
\begin{align}
g(f(x))
&= 1 - \left(\sqrt{1-x^2} \right)^2\\
&= 1 - \left(1-x^2 \right)\\
&= x^2\\
\end{align}
so
$$
(g \circ f)'(x) = 2x \tag{1}
$$ for every $x$.
$f$ is evidently differentiable for $-1 < x < 1$.
Suppose that $f$ were differentiable at $x = 1$ as well.
Then (by a version of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3870737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
How to solve $\cos x-\sin 3x=\cos 2x$? My attempt:
$$\begin{align} (\cos x- \cos 2x) - \sin 3x &= 0 \\
2\sin \frac{3x}{2}\sin \frac{x}{2} - 2 \sin \frac{3x}{2}\cos \frac{3x}{2}&=0\\
\sin\frac{3x} {2} \left(\sin\frac{x} {2} - \cos\frac{3x} {2}\right)&=0 \end{align}$$
Now either $\sin\frac{3x} {2} = 0$ or $\left(\sin\fra... | I think you can use the fact that $\sin(x) = \cos(\frac{\pi}{2}-x)$ to translate the equation into one involving only sines (or cosines) on both sides, and then compare directly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3877705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Hard limit $\lim_{x\to\infty}\Bigg((x(x+1)(x+2))^{\frac{1}{3}}-\Big(2(x+1)-\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}\Big)\Bigg)=0$ Prove that :
$$\lim_{x\to\infty}\Bigg((x(x+1)(x+2))^{\frac{1}{3}}-\Big(2(x+1)-\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}\Big)\Bigg)=0$$
I can prove that :
$$\lim_{x\to\... | We have that
$$\Bigg((x(x+1)(x+2))^{\frac{1}{3}}-\Big(2(x+1)-\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}\Big)\Bigg)=$$
$$=x\left(\left(1+\frac1x\right)^{\frac{1}{3}}\left(1+\frac2x\right)^{\frac{1}{3}}-\left(2\left(1+\frac1x\right)-\left(\left(1+\frac1x\right)^{x+1}\left(1+\frac2x\right)^{x+2}\right)^{\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3881843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Volume with spherical polar coordinates Determine the volume between the surface $z=\sqrt{4-x^2-y^2}$ and the area of the xy plane determined by $x^2+y^2\le 1,\ x+y>0,\ y\ge 0$.
I convert to spherical polar coordinates.
$$K=0\le r\le 1,\ 0\le \phi \le \frac{3\pi}{4},\ 0\le \theta \le 2\pi$$
$$\iiint_{K} (\sqrt {4-r^2\s... | This is much easier to solve in cylindrical coordinates. $$x=r\cos\theta\\y=r\sin\theta\\z=h$$
Then the limits for $r$ are $0$ and $1$, the limits for $\theta$ are from $-\frac\pi4$ to $\frac{3\pi}4$, and the limits for $h$ are $0$ and $4-r^2$.
With these, $$V=\int_{-\frac\pi4}^{\frac{3\pi}4}d\theta\int_0^1dr\cdot r\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3884663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
prove $\sum_\text{cyc}\frac{a+2}{b+2}\le \sum_\text{cyc}\frac{a}{b}$
Prove if $a,b,c$ are positive $$\sum_\text{cyc}\frac{a+2}{b+2}\le \sum_\text{cyc}\frac{a}{b}$$
My proof:After rearranging we have to prove $$\sum_\text{cyc} \frac{b}{b^2+2b} \le \sum_\text{cyc} \frac{a}{b^2+2b}$$
As inequality is cyclic:
let $a\ge b... | Your proof is nice and right.
Another way.
Let $c=\min\{a,b,c\}$.
Thus, we need to prove that:
$$\frac{a}{b}+\frac{b}{a}-2+\frac{c}{a}-\frac{b}{a}+\frac{b}{c}-1\geq\frac{a+2}{b+2}+\frac{b+2}{a+2}-2+\frac{c+2}{a+2}-\frac{b+2}{a+2}+\frac{b+2}{c+2}-1$$ or
$$\frac{(a-b)^2}{ab}+\frac{(c-a)(c-b)}{ac}\geq\frac{(a-b)^2}{(a+2)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3889118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Generalize the formula :$1=1, 3+5=8, 7+9+11=27, 13+15+17+19=64$ So the solution of this is $n^3$, as $1=1^3, 3+5=2^3, 7+9+11=3^3$
So I find that the $n+1 = m^2+3m+2 + (n_m)(m+1)$ where $n_m$ is the largest number in the previous equation and $m$ is the number of terms in last equation. How can I prove by induction that... | Approaching by pattern,
*
*We know that sum of first $k$ odd numbers is $k^2$.
*We have $1,2,3,4\ldots$ terms which may be expressed as difference of triangular number of terms
$$ 1 = \underset{1}{\underbrace{1}} - \underset{0}{\underbrace{0}} $$ $$ 3+5 = \underset{\text{first 3 odd}}{\underbrace{(1+3+5)}} - \unders... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3897138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Which positive integers $a$ and $b$ make $(ab)^2-4(a+b) $ a square of an integer? Which positive integers
$a$ and $b$ make
$(ab)^2-4(a+b)
$
a square of an integer?
I saw this in quora,
and found that
the only solutions with
$a \ge b > 0$
are
$(a, b, (ab)^2–4(a+b)) = (5, 1, 1)$
and $(3, 2, 16)$.
Another “solution” is
$a... | We can find an upper bound for their "product" in the following way:
\begin{align}
{\left( {ab} \right)^2} - 4\left( {a + b} \right) &< {\left( {ab} \right)^2} \ \ (\because a,b>0) \\
{\left( {ab} \right)^2} - 4\left( {a + b} \right) &\le {\left( {ab - 1} \right)^2} \\
(2ab-1)-4(a+b) &\le 0\\
ab - 2(a+b) - \frac 12 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3897406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prime elements in $\mathbb{Z}[\sqrt{5}]$ I need to determine which of the elements $3+2\sqrt{5}$, $9+4\sqrt{5}$ and $4-\sqrt{5}$ are prime elements in $\mathbb{Z}[\sqrt{5}]$, respectively which are associated.
My ansatz is as follows:
So let $x=3+2\sqrt{5}$ divide $ab$ for $a,b \in R$. Thus, there are $u,v \in \mathbb{... | $3+2\sqrt{5}$ has norm $-11$ and $4-\sqrt{5}$ has norm $11$. Perhaps they are associates. Indeed
$$
\frac{3+2\sqrt{5}}{4-\sqrt{5}} = 2+\sqrt{5}
$$
and $2+\sqrt{5}$ is a unit because it has norm $-1$. In fact, $(2+\sqrt{5})(-2+\sqrt{5})=1$.
$9+4\sqrt{5}$ is a unit because it has norm $1$ and so $(9+4\sqrt{5})(9-4\sqrt{5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3901397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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A unique question in combinatorics and non-commutative variables. Let $x,y$ be two variables that satisfy:
$xy=yx+1$ (they are not commutative).
Find $(xy)^2 ,(xy)^3, (yx)^2, (yx)^3$ as a linear combination in terms of $y^jx^j$.
Then find a formula for $(xy)^n$ and $(yx)^n$.
After some calculations we get:
$(xy)^2=y^2... | I found different formulas, namely
\begin{align*}
(xy)^1 & = yx+1 \\
(xy)^2 & = y^2x^2+3yx+1\\
(yx)^3 & = y^3x^3+6y^2x^2+7yx+1\\
(yx)^4 & = y^4x^4+10y^3x^3+25y^2x^2+15yx+1\\
(yx)^5 & = y^5x^5+15y^4x^4+65y^3x^3+90y^2x^2+31yx+1
\end{align*}
if correct, can you guess a general formula?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3901549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that $\sum\limits_{n=1}^{\infty}\frac{\left(1+\frac{1}{n}\right)^n}{n}(-1)^n$ converges How would you show that $\sum\limits_{n=1}^{\infty}\frac{\left(1+\frac{1}{n}\right)^n}{n}(-1)^n$ converges?
My approach
At first glance I would go for the alternate series criterion and show that $\frac{\left(1+\frac{1}{n}\ri... | HINT
We have that
$$\left(1+\frac{1}{n}\right)^n= e^{n\log\left(1+\frac{1}{n}\right)}=e^{1-\frac1{2n}+O\left(\frac1{n^2}\right)}=e-\frac e{2n}+O\left(\frac1{n^2}\right)$$
therefore
$$\frac{\left(1+\frac{1}{n}\right)^n}{n}=\frac e n-\frac e{2n^2}+O\left(\frac1{n^3}\right)$$
| {
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"source": "stackexchange",
"question_score": "2",
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Solve fourth degree inequality $x^4 - (1 + (m + 1)^2)x^2 + (m + 1)^2 \geq 0$ where $m \in \mathbb{R}$ Solve fourth degree inequality $x^4 - (1 + (m + 1)^2)x^2 + (m + 1)^2 \geq 0$ where $m \in \mathbb{R}$.
I tried using substitution as follows:
$$ a = x^2 $$
$$ b = (m + 1)^2$$
Using the substitution:
$$ a^2 - (1 + b)a +... | Here's how it's done in high school, using Vieta's relations and results on the sign of a quadratic polynomial:
First set $y=x^2$ and solve for $y$: you have the quadratic inequation
$$y^2-\bigl(1+(m+1)^2\bigr)y+(m+1)^2 \ge0,\qquad y\ge 0.$$
The discriminant is
$$\Delta=\bigl(1+(m+1)^2\bigr)^2-4(m+1)^2=\bigl(1-(m+1)^2\... | {
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"source": "stackexchange",
"question_score": "1",
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Proving the duplication formula I'm trying to prove the famous Legendre duplication formula:
$$ \Gamma(2z) = \frac{2^{2z-1} \Gamma(z) \Gamma(z+\frac{1}{2})}{\sqrt{\pi}} $$
I have to prove this using these two identities/theorems:
$$ \Gamma(z)= \lim_{n \rightarrow \infty} \frac{n! n^{z}}{z(z+1)...(z+n)}$$
and
$$ (2n)... | We have (with all limits for $n\to\infty$):
$$
\begin{aligned}
\Gamma(z)
&=
\lim \frac{n!\; n^{z}}{z(z+1)(z+2)\dots(z+n)}\ ,
\\
\Gamma\left(z+\frac 12\right)
&=
\lim
\frac{n!\; n^{z+\frac 12}}
{\left(z+\frac 12\right)\left(z+\frac 32\right)\left(z+\frac 52\right)\dots\left(z+\frac {2n+1}2\right)}\ ,
\\[3mm]
\Gamma(z... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Implicit differentiation $(\sin x +y)/(\cos y +x)=1$ I am currently stuck on this problem
Find $\frac{dy}{dx}$ when
$$\frac{\sin x+y}{\cos y+x}=1$$
I've been using the quotient rule to try and differentiate this, but my online homework rejected my answer of:
$$\frac{-x\cos x - \cos x\cos y +y+\sin x}{\cos y +x-\sin y... | The use of the Quotient Rule doesn't need to be complicated, as long as one resists differentiating any expressions that don't really need that (or writing out components until we finally need them). If we call the ratio $ \ \large{\frac{\sin x \ + \ y}{\cos y \ + \ x}} \ \normalsize{= \ \frac{f(x)}{g(x)}} \ = \ 1 \ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3910796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
System of equations involving 4 variables
If $$a + b = 6$$
$$ax + by = 10$$
$$ax^2 + by^2 = 24$$
$$ax^3 + by^3 = 62$$
then $$ax^4 + by^4 = ?$$
I got $$a(x-1) + b(y-1) = 4$$ $$ax(x-1) + by(y-1) = 14$$ $$ax^2(x-1) + by^2(y-1) = 38$$ by subtracting the given equations and also got $$a(x-1)^2 + b(y-1)^2 = 10$$ $$ax^2(x-1... | Use succssive eliminations.
From $(1)$, $b=6-a$. PLug in $(2)$ to get $y=\frac{a x-10}{a-6}$. Plug in $(3)$ to get $a=\frac{22}{3 x^2-10 x+12}$. Plug in $(4)$ and obtain
$$22\frac{x^2-3x+1}{3 x-5}=0 \implies x= ???$$
Now, go backward.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Maxima and minima of $\frac{x^2-3x+4}{x^2+3x+4}$ without calculus How to find the minimum and max values of $y=\frac{x^2-3x+4}{x^2+3x+4}$ for all real values of $x$ without using calculus?
Perhaps it could be done graphically by noting the fact that the numerator and denominator are a pair of parabolas symmetric about ... | Let the maximum of $f(x) = \frac{x^2-3x+4}{x^2+3x+4}$ be $m$. Then:
$$x^2-3x+4 = mx^2 + 3mx + 4m$$
$$(m-1)x^2 + (3m+3)x + (4m - 4) = 0$$
We want this equation to have only one real root (a double root), so:
$$\Delta = 0 \Rightarrow (3m+3)^2-4(m-1)(4m-4) = 0.$$
A similar process for the minimum ($n$) yields the same equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3915318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integration of $\frac{1}{x(x+1)(x+2)...(x+m)}$ I came across this question,
$$\int \frac{1}{x(x+1)(x+2)...(x+m)} dx$$
I tried to split the rational function into partial fractions
$$ \frac{A}{x}+\frac{B}{x+1}+\frac{C}{x+2}+...+\frac{Z}{x+m}$$
Not really sure how to proceed from here.
Can someone enlighten me with step ... | Note
$$\frac{1}{x(x+1)(x+2)...(x+m)}=\sum_{k =0}^{m} \frac{a_k}{x+k}
$$
where $a_k$ are obtained as follows
\begin{align}
a_k &=\lim_{x\to -k} \frac{x+k}{x(x+1)(x+2)...(x+k)...(x+m)}\\
&= \frac{1}{[(-k)(1-k)(2-k)(-2)(-1)]\cdot[(1)(2)...(m-k-1)(m-k)]}\\
&=\frac1{(-1)^k k!(m-k)!}
\end{align}
Thus
$$\int \frac{dx}{x(x+1)(... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove elegantly this inequality? Let $a$, $b$ and $c$ three reals greater than or equal to 1. Put $a+b+c=3u$, $ab+bc+ca=3v^2$ and $abc=w^3$. Show that
$$w-\frac{1}{w}\ge\frac{w^3}{v^2}-\frac{1}{u}$$
I am able to prove it by assuming without loss of generality that $c\ge b\ge a$ and posing $b=a+u$ and $c=a+u+v$ f... | Write the question as
$$
\sqrt[3]{abc} - \frac{1}{\sqrt[3]{abc}} \ge\!\!? \; \sqrt[3]{abc} \frac{3\sqrt[3]{a^2b^2c^2}}{ab+bc+ca} - \frac{1}{\sqrt[3]{abc}} \frac{3 \sqrt[3]{abc}}{a+b+c}
$$
By AM-GM, both following coefficients are at most $1$:
$$
A = \frac{3\sqrt[3]{a^2b^2c^2}}{ab+bc+ca} \le 1\qquad
B = \frac{3 \sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3916266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Equation of Circle touching a straight line and passing through the centroid of a triangle and a particular point If the equation $3{x^2}y + 2x{y^2} - 18xy = 0$ represent the sides of a triangle whose centroid is G, then the equation of the circle which touches the line $x-y+1=0$ and passing through G and $(1,1)$ is | Hint:
$k=\dfrac{11-2h}4$
For any circle, the distance of a tangent from the center $(h,\dfrac{11-2h}4)=$radius
$$r^2=\dfrac{(h-k+1)^2}{1^2+1^2}=\dfrac{\left(h+1-\dfrac{11-2h}4\right)^2}2$$
Again $r^2=(h-1)^2+(k-1)^2=(h-1)^2+\left(\dfrac{11-2h}4-1\right)^2$
Equate the two values of $r^2$ to find $h$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3917805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Solve in real numbers the system of equations $x^2+2xy=5$, $y^2-3xy=-2$ Solve in real numbers the system of equations $x^2+2xy=5$, $y^2-3xy=-2$
I couldn't do this question and hence I looked at the solution which goes as follows:
if $x=0$ then $x^2+2xy=0\ne5$ hence $x\ne0$
I state that $y=lx$. Hence the system becomes:... | $x^2+2xy=5$, $y^2-3xy=-2$
Multiply the first by $2$, the second by $5$ and add
$$2 x^2-11 x y+5 y^2=0$$
Divide all terms by $y^2$ and set $\frac{x}{y}=z$
$$2z^2-11z+5=0\to z=\frac{1}{2};\;z=5$$
So we have $\frac{x}{y}=\frac{1}{2}\to y=2x$ and $\frac{x}{y}=5\to x = 5y$
Plug $y=2x$ into the first equation
$$x^2+2x(2x)=5\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3922418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is $(4+\sqrt{5})$ a prime ideal of $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$? Consider the integral domain $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$. Is $(4+\sqrt{5})$ a prime ideal of $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$?
I know the following elementary facts. We have
\begin{equation}
\math... | Call $A = \mathbb Z \left[ \frac{1 + \sqrt 5}2\right]$. We can show that $A / (4+\sqrt 5) \cong \mathbb Z/11 \mathbb Z$, so that the ideal $(4 + \sqrt 5)$ is maximal.
*
*As $N(4 + \sqrt 5) = 11$, it is clear that the elements $0, 1, \ldots, 10$ are pairwise incongruent modulo $4 + \sqrt 5$.
*Every element of $A$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3923792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove $\lim_{n \to \infty} \frac{n}{1+\sqrt{n}}cos(\frac{\pi n}{n+1})= -\infty$? I tried to prove by using definition for divergence:
We say that a sequence $a_n$ diverges to $-\infty$ if for every $M>0$, there is an integer $n_0$ such that $a_n < -M$ whenever $n>n_0$
So I start:
$$\frac{n}{1+\sqrt{n}}cos(\fra... | Hi I have like no idea about this being proof enough for you but we usually do this:
$
\lim_{n \to \infty} \frac{n}{1+\sqrt{n}}\cos\left(\frac{\pi n}{n+1}\right)= -\infty
$
$
1. \frac{\pi n}{n+1} = \frac{\pi }{1+\frac{1}{n} }
$ // divide by n
for $n\to
\infty , \frac{1}{n} = 0
$
$
\frac{\pi }{1+\frac{1}{n} } = \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3924938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving Differential Equation $y'=\cos(xy)$ My question is about solving differential equation $y'=\cos(xy)$.
I tried to changing variable $u=xy$
\begin{align*}
u &= xy \\
\frac{du}{dx} &= y+ x\frac{dy}{dx}\\
\frac{dy}{dx} &= \frac{1}{x}\frac{du}{dx}-\frac{y}{x}\\
\end{align*}
then since $u=xy\to \frac{y}{x}=\frac{u}{x... | Hint:
Let $u=xy$ ,
Then $y=\dfrac{u}{x}$
$\dfrac{dy}{dx}=\dfrac{1}{x}\dfrac{du}{dx}-\dfrac{u}{x^2}$
$\therefore\dfrac{1}{x}\dfrac{du}{dx}-\dfrac{u}{x^2}=\cos u$
$\dfrac{1}{x}\dfrac{du}{dx}=\dfrac{x^2\cos u+u}{x^2}$
$(x^2\cos u+u)\dfrac{dx}{du}=x$
Let $v=x^2$ ,
$\dfrac{dv}{du}=2x\dfrac{dx}{du}$
$\therefore\dfrac{(x^2\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3927241",
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"source": "stackexchange",
"question_score": "1",
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How can I rewrite $\frac{1}{(1-z)^4}$ as a power series? I'm supposed to rewrite $\frac{1}{(1-z)^4}$ as a power series, using the Cauchy-Product. It is given that $|z| < 1$.
So far I have
\begin{aligned}
\frac{1}{(1-z)^4} &= \left( \frac{1}{(1-z)^2} \right)^2\\
&= \left( \sum_{n=0}^{\infty} (n+1)z^n \right)^2\\
&= \su... | The third line should be
\begin{eqnarray*}
&= \sum_{n=0}^{\infty} \sum_{k=0}^{n} (n-k+1)z^{n-k} \color{red}{(k+1)}z^k \\
\end{eqnarray*}
So we need to perform the sum
\begin{eqnarray*}
\sum_{k=0}^{n} (n-k+1)(k+1) = -\sum_{k=0}^{n} k^2+ n \sum_{k=0}^{n} k +(n+1) \sum_{k=0}^{n} 1.
\end{eqnarray*}
Recall that
\begin{eq... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
critical points of $f(x,y) = \frac{\sin(x)}{1+y^2}$ so I determined: $f'(x,y) = \left(\begin{array}{c} \frac{\cos\left(x\right)}{y^2+1}\\ -\frac{2\,y\,\sin\left(x\right)}{{\left(y^2+1\right)}^2} \end{array}\right)$
and: hess-f = $\left(\begin{array}{cc} -\frac{\sin\left(x\right)}{y^2+1} & -\frac{2\,y\,\cos\left(x\right... | Trying plugging in those values of $x$, you'll notice that the Hessian of $f$ doesn't change for any value of $k$.
You could consider studying the critical points of $f$ only in a $\{(x,y)\mid -\frac{\pi}2<x<\frac{\pi}2\}$ strip first, and deduce later that all the other critical points are just translated copies of th... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Integrating $\int{\frac{x\,dx}{x^2 + 3x -4}}$ I saw this "Beat the Integral" problem and wanted to be sure I was approaching it correctly.
The integral is $$\int{\frac{x}{x^2 + 3x -4}dx}$$
So I decide I want to decompose this, because it's a degree-2 polynomial in the denominator and a degree-1 in the numerator, so tha... | $$\int{\frac{xdx}{x^2 + 3x -4}}=\frac{1}{2}(\int{\frac{(2x+3)dx}{x^2 + 3x -4}}-3\int{\frac{dx}{x^2 + 3x -4}})=\frac{1}{2}(\int{\frac{(2x+3)dx}{x^2 + 3x -4}}+\frac{3}{5}(\int{\frac{dx}{x +4}}-\int{\frac{dx}{x -1}}))=\frac{1}{2}\ln{|x^2+3x-4|}+\frac{3}{5}\ln{|x+4|} -\frac{3}{5}\ln{|x-1|}+C=\frac{1}{2}\ln{|x+4|}+\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3940216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 5
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Question on limit of Riemann Sum I'm trying to follow the following calculation of the definite integral using a limit of Riemann Sums. I have a question on the reasoning for a step in derivation of the limit which I will detail.
The original post is here.
$$\int_0^3 x^3 dx = \lim \limits_{n \to \infty} \sum_{i=1}^n f... | $= \lim \limits_{n \to \infty} (\frac 3 n)^4 (\frac{n(n+1)}{2})^2 \sum_{i=1}^n i^3 = (\frac{n(n+1)}{2})^2$ should be
$= \lim \limits_{n \to \infty} (\frac 3 n)^4 (\frac{n(n+1)}{2})^2$ makes use of $\sum_{i=1}^n i^3 = (\frac{n(n+1)}{2})^2$.
(Also, I saw Riemann spelled incorrectly there.)
Addendum to answer additional q... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Combinatorial proof of $ \sum\limits_{k=0}^n(-1)^k\frac{2^{n-k}\binom{n}{k}}{(m+k+1)\binom{m+k}{m}}=\sum\limits_{k=0}^n\frac{\binom{n}{k}}{m+k+1}$ While answering the following question : Solution to a definite integral as $\int_0^{1}(x^{m}\left(1+x)^{n}dx\right)$, I came accross a strange identity involving binomials... | Here is an algebraic proof by way of enrichment. We seek to verify that
$$\sum_{k=0}^n (-1)^k
\frac{2^{n-k} {n\choose k}}{(m+k+1) {m+k\choose k}}
= \sum_{k=0}^n \frac{{n\choose k}}{m+k+1}.$$
We can re-write this as
$$\frac{m! n!}{(n+m+1)!} 2^n
\sum_{k=0}^n (-1)^k 2^{-k} {n+m+1\choose n-k}
= \sum_{k=0}^n \frac{{n\choose... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3942039",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Solve for $a, b$ if $|x^2 + ax + b| \le \frac 12, \forall 0 \le x \le 2$. This is from a recently closed question. I do not think the accepted answer is rigorous because it did not address the cases when the discriminant of the two quadratics is negative, even though the final answer was correct.
My solution relies hea... | The polynomial $x^2 - 2 x + \frac{1}{2}$ takes values $\frac{1}{2}$, $-\frac{1}{2}$, $\frac{1}{2}$ at $0$, $1$, $2$. We conclude that the linear function $(x^2-2x+ \frac{1}{2}) - (x^2+ a x + b)$ takes values $\ge 0$, $\le 0$, $\ge 0$ at $0$, $1$, $2$, so it must be the $0$ function. We get $a=-2$, $b=\frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3942299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Domain of $\sqrt{x}$ function when solving $x^2 - 2=0$ I've been going through this solution and got stuck on explaining myself why $x^2 = 2 \Rightarrow \sqrt{x^2} = \sqrt{2}\Rightarrow x = \pm \sqrt2$. Domain of $f(x)=\sqrt{x}$ is $[0,\infty)$. How taking square root both sides give $\pm \sqrt{2}$ ?
| There are two distinct concepts at play here. $\sqrt{x}$ is a function denoting the nonnegative square root of $x$. However, the general solution to $a^2 = b^2$ is $a = \pm b$. So $\sqrt{25}=5$, but if $x^2=25$ then $x=\pm5$. As Gae. S. has already mentioned, the easiest way to see that $a^2=b^2 \iff a= \pm b$ is throu... | {
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"url": "https://math.stackexchange.com/questions/3944310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determinant and inverse matrix calculation of a special matrix Is there any smart way of calculating the determinant of this kind matrix?
\begin{pmatrix}
1 & 2 & 3 & \cdots & n \\
2 & 1 & 2 & \cdots & n-1 \\
3 & 2 & 1 & \cdots & n-2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
n & n-1 & n-2 & \cdots & 1 \end{pmat... | $$\begin{vmatrix}
1 & 2 & 3 & \cdots & n \\
2 & 1 & 2 & \cdots & n-1 \\
3 & 2 & 1 & \cdots & n-2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
n & n-1 & n-2 & \cdots & 1 \end{vmatrix}$$
Notice the sum of an element from the first column and an element from the last column (with the same row) will always be equal t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3947667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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How to evaluate $\int_0^{\frac{\pi}{2}} \frac{{\rm d}\alpha}{1+\cos\alpha\cos\beta}$ without using antiderivative? Someone gives a solution here:
\begin{align*} \int_0^{\frac{\pi}{2}} \frac{{\rm d}\alpha}{1+\cos\alpha\cos\beta}&=\int_0^{\frac{\pi}{2}} \sum_{n=0}^{\infty} (-\cos\alpha\cos\beta)^n{\rm d}\alpha\\ &=\sum_{... | Using tangent half-angle substitution $t=\tan \frac{x}{2} $, we transform the integral into
$$
I=2 \int_0^1 \frac{d t}{(1-\cos \beta) t^2+1 +\cos \beta},\quad \textrm{ where }\beta \textrm{ is not a multiple of }\pi.
$$
By the half-angle formula of sine and cosine, we have
$$
\begin{aligned}
I&=\int_0^1 \frac{d t}{t^2 ... | {
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"timestamp": "2023-03-29T00:00:00",
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What is inverse of a matrix whose diagonal elements are all zero. I am a science researcher and I got a problem to find the generic inverse of the following matrix:
$$
A_n = \left(\begin{array}{ccc}
0 & a_2 & a_3 & ... & a_{n-1} & a_n \\
a_1 & 0 & a_3 & ... & a_{n-1} & a_n \\
a_1 & a_2 & 0 & ..... | Your matrix can be written as $D_n + u_n \, v_n^T$:
$$u_n = \left[ \begin{matrix} 1 & ... & 1 \end{matrix} \right]^T$$
$$v_n = \left[ \begin{matrix} a_1 & ... & a_n \end{matrix} \right]^T$$
$$D_n = - \text{diag}(v_n) = - \left[ \begin{matrix}
a_1 & ... & 0 \\
\vdots & ... & \vdots \\
0 & ... & a_n \\
\end{matrix} \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3949107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Show factorization $ x^{2n}-1=(x^2-1) \prod_{k=1}^{n-1}(x^2-2x \cos \frac{\pi k}{n} + 1) $ I'm interested in how to show that
$$
x^{2n}-1=(x^2-1) \prod_{k=1}^{n-1}(x^2-2x \cos \frac{\pi k}{n} + 1)
$$
I've seen this equality too often, but have no idea how to derive it. I've tried the following:
$$
x^{2n}-1=(x^n-1)(x^n+... | Thanks to @CalvinLin for this post. The idea was to apply direct substitution
$$
x^{2n}-1=\prod_{k=0}^{2n-1} (x-\xi^k)
$$
where $\xi$ is a base root of the equation $x^{2n}-1=0$, e.g. $\xi=\exp({i \cdot \frac{\pi}{n}})$.
It gives us the following:
$$
\prod_{k=0}^{2n-1} (x-\xi^k)=(x-\xi^0)(x-\xi^n) \prod_{k=1}^{n-1} (x-... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to integrate $\int_0^1\frac{dx}{1+x+x^2+\cdots+x^n}$ I am interested in finding a solution to the integral
$$I_n=\int_0^1\frac{dx}{\sum_{k=0}^nx^k}$$
Since the denominator is a geometric series with $a=1$ and $r=x$ and it is within the radius of convergence, we should be able to say
$$\sum_{k=0}^nx^k=\frac{1-x^{n+1... | Decompose the integrand as
$$\frac{1}{1+x+x^2+...+x^n}=\sum_{k=1}^{n}\frac{a_k}{x-x_k},\>\>\>\>\> a_k=\frac{x_k(x_k-1)}{n+1}
$$
where $x_k= e^{i \frac{2\pi k}{n+1}},\> k=1,2...n$. Then
\begin{align}
I_n&=\int_0^1\frac{dx}{1+x+x^2+...+x^n}\\
&=\int_0^1 \sum_{k=1}^{n}\frac{a_k}{x-x_k}dx
=\frac1{n+1}\sum_{k=1}^{n} x_k(x_k... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Cannot find limit using epsilon delta definition Prove
$$\lim\limits_{x\to 2} x^3 = 8$$
using epsilon delta definition.
I try as below.
Let $\varepsilon>0$. We choose $\delta>0$.
Consider that
\begin{align}
\vert x^3-8\vert &= \vert (x-2) (x^2+2x+4)\vert\\
&=\vert (x-2) \vert\vert(x^2+2x+4) \vert \\
&=\vert (x-2) \vert... | More generally,
if you want to show that
$\lim_{x \to c} x^n=c^n$,
note that
$x^n-c^n
=(x-c)\sum_{k=0}^{n-1} x^kc^{n-1-k}
$.
If
$c-r < x < c+r$
where
$0 < r< \min(1, x)$
then
$\sum_{k=0}^{n-1} x^kc^{n-1-k}
\lt \sum_{k=0}^{n-1} (c+r)^kc^{n-1-k}
\lt \sum_{k=0}^{n-1} (c+r)^k(c+r)^{n-1-k}
=n(c+r)^n
$
so
$\begin{array}\\
|x... | {
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"answer_count": 4,
"answer_id": 2
} |
Solve the equation $\sqrt{45x^2-30x+1}=7+6x-9x^2$ Solve the equation $$\sqrt{45x^2-30x+1}=7+6x-9x^2.$$
So we have $\sqrt{45x^2-30x+1}=7+6x-9x^2\iff \begin{cases}7+6x-9x^2\ge0\\45x^2-30x+1=(7+6x-9x^2)^2\end{cases}.$ The inequality gives $x\in\left[\dfrac{1-2\sqrt{2}}{3};\dfrac{1+2\sqrt{2}}{3}\right].$ I am not sure how ... | From $45x^2-30x+1=(7+6x-9x^2)^2$, you have that $$81x^{4}-108x^{3}-135x^{2}+114x+48 = 0$$
This factors as $$3(x-1)(3x+1)(9x^2-6x-16) = 0$$
So the roots of that polynomial are $x = 1, -\frac{1}{3}, \frac{1\pm\sqrt{17}}{3}$. Out of these, the only ones in the range $\left[\frac{1-2\sqrt{2}}{3} ,\frac{1+2\sqrt{2}}{3} \rig... | {
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"source": "stackexchange",
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"answer_count": 4,
"answer_id": 3
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Prove that $a ( n ) = b( n + 2)$
let $a(n)$ denotes the number of ways of expressing the positive integer $n$ as an ordered sum of 1's and 2's. Let $b(n)$ denote the number of ways of expressing n as an ordered sum of integers greater than 1. prove that $a(n) = b(n+2)$. for $n=1,2,3...$
My approach:
$a(1) = 1$ (only... | It is also possible to exhibit a bijection between compositions of $n$ using only $1$ and $2$ and compositions of $n+2$ that do not use $1$.
Suppose that $n=c_1+c_2+\ldots+c_k$, where $c_i\in\{1,2\}$ for $i=1,\ldots,k$. Let $c_0=2$; then $n+2=c_0+c_1+\ldots+c_k$. Let $i_1<\ldots<i_\ell$ be the set of indices $i$ such t... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find $\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}}$ As stated in the title.
My attempt. Dividing through $(x-2)^{\frac{2}{3}}$.
$$L=\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}}=\lim_{x \to \in... | Hint:
$$\lim_{x\to\infty}\dfrac{(x+1)^{2/3}-(x-1)^{2/3}}{(x+2)^{2/3}-(x-2)^{2/3}}=\dfrac12\cdot\lim_{t\to0}\dfrac{\dfrac{(1+t)^{2/3}-(1-t)^{2/3}}t}{\dfrac{(1+2t)^{2/3}-(1-2t)^{2/3}}{2t}}=\dfrac12\cdot\dfrac{\lim_{t\to0}f(t)}{\lim_{t\to0}f(2t)}$$
Now for $\lim_{t\to0}f(t)=\lim_{t\to0}\dfrac{(1+t)^{2/3}-(1-t)^{2/3}}t,$ ... | {
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"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Are there infinitely many $p$ such that $\frac{p_k+p_{k+1}}{2} = m^2 - n^2$ has no integer solutions? $\frac{p_k+p_{k+1}}{2} = m^2 - n^2$ with $p_k$ as the $k$-th prime has no integer solutions for $9$ values of $k \leq 25$, $17$ values of $k \leq 50$, and $33$ values of $k \leq 100$. This means that roughly $33\%$ of ... | Write $m^2-n^2=(m+n)(m-n)$ The two factors on the right have the same parity, so any number that is equivalent to $2 \bmod 4$ cannot be expressed this way. Because you divide the sum by $2$, this is any pair of primes that sum to a number that is equivalent to $4 \bmod 8$. For example, the sequential primes $5,7$ gi... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Large N limit for $P(\mathbf{x} \cdot\mathbf{y}\cdot \sum\mathbf{x}\cdot \sum \mathbf{y}>0)$ for two binary vector $\mathbf{x}$ and $\mathbf{y}$ I have two binary vectors $\mathbf{x}$, and $\mathbf{y}$, each with $N$ elements; Each element of $\mathbf{x}$ and $\mathbf{y}$ belongs to {-1, 1}, and is drawn from uniform r... | Edited
Let $a^T=[1, 1, ..., 1]$, that is an all-one vector of length $N$. Then you can rewrite the random variable as:
$Z=(Y^T X)(X^T a)(a^T Y)$,
in which $T$ represent transpose, so $Y^TX$ is the inner product of $X$ and $Y$, $X^T a$ is the sum of the elements of $X$, and $a^T Y$ is the sum of the elements of $Y$. The... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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What is the remainder when $1^{2016} + 2^{2016} + ⋯ + 2016^{2016}$ is divided by $2016$? What is the remainder when $1^{2016} + 2^{2016} + ⋯ + 2016^{2016}$ is divided by $2016$?
My Approach:
started breaking 2016 into product of pairwise coprime positive integers to get $2016 = 32 * 7 * 9$
This suggests that we want to... | Define $$S = \sum_{i=1}^{2016} i^{2016}$$
Notice that $2016 = 32\cdot 9\cdot7$
and $\phi(32) | 2016,\; \phi(9) |2016,\; \phi(7) | 2016$, so we have
$$i^{2016} \equiv \begin{cases}
1 \pmod n \text{ if } \gcd(i,n)=1\\ 0 \pmod n \text{ if } \gcd(i,n)>1 \end{cases}$$
for $n=32,9,7$.
Then each group of $n$ has $\phi(n)$ ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Problem proving $\int_{-\infty}^{\infty}x^2e^{-x^2}H_n(x)H_m(x)dx$ I am trying to show that the value of the following integral is:
$$\int_{-\infty}^{\infty}x^2e^{-x^2}H_n(x)H_m(x)dx=2^{n-1}\pi^{\frac{1}{2}}(2n+1)n!\delta_{mn}+2^n\pi^{\frac{1}{2}}(n+2)!\delta_{(n+2)m}+2^{n-2}\pi^{\frac{1}{2}}n!\delta_{(n-2)m},$$
where ... | I am going to work with the statistician's Hermite polynomials, which can be related to the physicists' (which appear in this question) via
$$
He_{\alpha}(x) = 2^{-\frac{\alpha}{2}}H_\alpha\left(\frac{x}{\sqrt{2}}\right)
$$
and the desired integral becomes
$$
I_{nm} = \int_{-\infty}^\infty x^2e^{-x^2}H_n(x)H_m(x)dx \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3972541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluate integral $\int \frac{(2-x^2)e^x}{(1-x)(\sqrt{1-x^2})}dx $ I tried to solve it by the following method:
\begin{align*}
\int \frac{(2-x^2)e^x}{(1-x)(\sqrt{1-x^2})}dx&= \int \frac{(1-x^2)e^x+\frac{e^x}{2}+\frac{e^x}{2}}{(1-x)(\sqrt{1-x^2}) }dx \\ &= \int \frac{(1-x^2)e^x+\frac{(1-x)e^x}{2}+\frac{(1+x)e^x}{2}}... | Recognize
$$\frac{2-x^2}{(1-x)(\sqrt{1-x^2})}= \frac{(1-x^2)+1}{(1-x)(\sqrt{1-x^2})}\\=\sqrt{\frac{1+x}{1-x} }+ \frac{1}{(1-x)^{3/2}\sqrt{1+x})}
= \sqrt{\frac{1+x}{1-x} }+ \left( \sqrt{\frac{1+x}{1-x} }\right)’
$$
and then apply to $f(x)= \sqrt{\frac{1+x}{1-x} }$ the general integral result
$$\int e^x(f(x)+f’(x)) dx= e... | {
"language": "en",
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"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
Find all natural numbers $n$ where $n<1000$ such that cube of sum of the digits of $n$ be equal to $n^2$
Find all natural numbers $n$ where $n<1000$ such that cube of sum of
the digits of $n$ be equal to $n^2$
For one digit numbers $n=1$ is the only answer. For two digits numbers $\overline{ab}$:
$$(10a+b)^2=(a+b)^3$... | Use @lulu's notation for $S(n)$ to denote the sum of digits of $n$. Using @NeatMath's observation that $(S(n))^3 = n^2$, we see that $n$ is a perfect cube, so it suffices to check the set $\{1^3,2^3,\dots,9^3\}$, due to the constraint $n < 1000 = 10^3$. You might include $0$ if you have an affirmative answer to Is $0... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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For what $a$ and $b$ are there explicit expressions for $I(a, b) =\int_0^1 \int_0^1 \dfrac{dx\,dy}{1-x^ay^b} $? For what $a$ and $b$
are there explicit expressions for
$I(a, b)
=\int_0^1 \int_0^1 \dfrac{dx\,dy}{1-x^ay^b}
$?
This is inspired by
the answer to
https://www.quora.com/What-is-displaystyle-int_-0-1-int_-0-1-f... | May be, we could start with
$$I(a, b)=\int_0^1 \int_0^1 \dfrac{dx\,dy}{1-x^ay^b}=\int_0^1 \, _2F_1\left(1,\frac{1}{b};\frac{1}{b}+1;x^a\right)\,dx$$ and consider
$$J_a=\int_0^1 \, _2F_1\left(1,\frac{1}{b};\frac{1}{b}+1;x^a\right)\,dx$$ and discarding the ones which contain logarithms of complex arguments, we can obtain... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Find all positive integer solutions for the following equation $5^a + 4^b = 3^c$:
Find all positive integer solutions for the following equation:
$$5^a + 4^b = 3^c$$
My first guess would be to study the equation in mod, but I tried modulo 3, 4, 5, and 9 and I can't find anything.
| Hint: $$1^a + 0 \equiv _4 (-1)^c \implies c \equiv_2 0$$
so $c=2d$, $d\in \mathbb{N}$.
So $$ 5^a = (3^d-2^b)(3^d+2^b)$$
so $3^d-2^b = 5^x$ and $3^d+2^b = 5^y$. Now add/substract both equations...
...we see that $x=0$. Now if $b>1$ then we have $$(-1)^d -0 \equiv _41 $$ so $d$ is even. So we have $$(3^m-1)(3^m+1)= 2^b... | {
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"answer_count": 2,
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Evaluate $\int\frac{x^4}{(x^2-1)^3}dx$ I tried:$$I=\int\frac{x^4}{(x^2-1)^3}dx=\int\frac{(x^2-1)(x^2+1)+1}{(x^2-1)^3}dx=\int\frac{x^2+1}{(x^2-1)^2}+\frac{1}{(x^2-1)^3}dx$$
For first fraction we can write it as $\frac{1}{x^2-1}+\frac{2}{(x^2-1)^2}$. therefor we have:
$$I=\frac12\ln\left|\frac{x-1}{x+1}\right|+\int\frac{... | Another approach: with $x=-\csc t$ the original integral becomes $\int\sec^5tdt$, then use this together with $\sec t\tan t=\frac{x}{1-x^2},\,(\sec t+\tan t)^2=\frac{2x-1-x^2}{1-x^2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Given $\frac{dS}{dt}=-\frac{SI}{S + I + R}$ and $R = -(SI + S + I)/ \frac{dS}{dt}$ Determine $\frac{d^2S}{dt^2}$ I am given a system of equations:
\begin{align}
\frac{dS}{dt} &= -\frac{SI}{S + I + R}\\
\frac{dI}{dt} &= \frac{SI}{S + I + R} -I \\
\frac{dR}{dt} &= I - R\\
\end{align}
I am to subsitute $R = -\left(SI + S ... | I acknowledge this answer is very similar to that of our colleague Ak, with perhaps some differences in formatting and added words of explanation.
We start with the given equations
$\dot S = -\dfrac{SI}{S + I + R}, \tag 1$
$\dot I = \dfrac{SI}{S + I + R} - I, \tag 2$
$\dot R = I - R; \tag 3$
we differentiate (1) (here ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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What is a more rigorous method of solving modulus function inequalities? I have the inequality
$$\vert x-3\vert < 3x-4$$
To solve this I first break the modulus containing function into the two cases using the critical value
$$\vert x-3 \vert \left\{\begin{array}{cc}
x-3 & x > 3 \\
3-x & x<3
\end{array}\right. $$
For... | Note that the inequality is true whenever $3x-4\le 0.$ Therefore assume $3x-4>0$ and square both sides to obtain $$(x-3)^2>(3x-4)^2,$$ or $$(x+3)^2-(3x-4)^2>0,$$ which gives $$(x+3-3x+4)(x+3+3x-4)>0,$$ simplifying to $(-2x+7)(4x-1)>0,$ or $$(x-3.5)(x-0.25)<0.$$ This implies $0.25<x<3.5$ for $x>4/3,$ or in short $$4/3<x... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Finding the posterior PMF using Bayes Theorem The problem is as follows
Nefeli. a student in a probability class, takes a multiple-choice test with 10 questions and 3 choices per question. For each question. there are two equally likely possibilities, independent of other questions: either she knows the answer, in whi... | Correcting a small error in your last expression, you should have $$p_{X\mid Y}(6\mid y) = \binom{10-y}{6-y}\left(\frac{1}{3}\right)^{6-y}\left(1-\frac{1}{3}\right)^{10-6}$$
You then get $$p_{Y\mid X}(y\mid 6) = \frac{\binom{10-y}{6-y}\left(\frac{1}{3}\right)^{6-y}\left(1-\frac{1}{3}\right)^{10-6} \binom{10}{y}\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3986061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Factoring of a particular polynomial It is given that $n$ is an odd integer greater than $3$ but not a multiple of $3$. Prove that $x^3 +x^2 +x$ is a factor of $(x+1)^n -x^n -1$.
I tried to write $x^3+x^2+x=x(x^2+x+1) =x(x-\omega^2)(x-\omega)$ where $\omega$ and $\omega^2$ are cube roots of unity $≠1$. Then I tried to ... | Modulo $x$, $(x+1)^n-x^n-1\equiv 1^n-0-1=0$, so $(x+1)^n-x^n-1$ is divisible by $x$.
It remains to be shown that $(x+1)^n-x^n-1$ is divisible by $x^2+x+1$. Well,
$(\omega+1)^n-\omega^n-1=(-\omega^2)^n-\omega^n-1=-(\omega^{2n}+\omega^n+1)$ for $n$ odd.
Furthermore, for $n$ not a multiple of $3$, $(\omega^{2n}+\omega^n+... | {
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"question_score": "1",
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"answer_id": 0
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Let $(G,*)$ be a group and $a,b \in G$ with $a*b = b^5*a^3$. Show that $|ba^{-1}| = |b^3a^3|$. Let $(G,*)$ be a group and $a,b \in G$ with $a*b = b^5*a^3$. Show that
$|ba^{-1}| = |b^3a^3|$.
attempt:
Write $a*b=ab$. Let $a,b \in G$.
Then, $ab=b^5a^3$. Note that
$ba^{-1} = a^{-1}b^5a^2$,
$ab^{-1} = b^2(b^3a^3)b^{-2}$, an... | You don't need to go through divisibility arguments. Since $ab^{-1}=(b^2)(b^3a^3)(b^{-2})$, $ab^{-1}$ and $b^3a^3$ are conjugate and have the same order. Using the fact about orders of inverses, the original claim follows.
| {
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"url": "https://math.stackexchange.com/questions/3986585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Matrices equation with one unknown matrix Let us consider following equation:
$$
2X^2+2X=
\left( \begin{array}{ccc}
-1 & 5 & 3 \\
-2 & 1 & 2 \\
0 & -4 & -3
\end{array} \right).
$$
I have to show that there is no real matrix which satisfy my equation.
I consider three diffrent way to do this.
*
*Direct method. Let $... | If there were a real matrix $X$ satisfying ...
Scalar-multiply the given equation with $2$ and add the identity matrix to get
$$(2X+\mathbb 1)^2 \;=\; \begin{pmatrix} -1& 10& 6\\
-4& 3& 4\\
0& -8& -5 \end{pmatrix}$$
Apply the determinant which produces $\,-25\,$ on the RHS, hence a contradiction because the resulting s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3987603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
find angle x in isosceles triangle Considering the attached image, is that possible to compute for $\angle x$ in the isosceles triangle below, with no further known values?
$\\$I have found $\angle B$ and $\angle C$ are $50^o$. So $\angle B$ is divided by $\overline{BP}$ to $10^o$ and $40^o$. But I can't find how the $... | Construct equilateral triangle $ABX$ as in the picture.
Note that the concave angle $BXA$ equals $300^\circ = 2\cdot 150^\circ = 2\angle BPA$, hence $P$ lies on the circle with center $X$ and radius $XA=XB$. It follows that $\angle PXB = 2\angle PAB = 40^\circ$.
On the other hand, $AB=AX=AC$, so $A$ is the circumcenter... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3990618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the derivative of $f(x)=x^{x^{\dots}{^{x}}}$.
Find the derivative of $f(x)$:
$$f(x)=x^{x^{\dots}{^{x}}}$$
Let $n$ be the number of overall $x's$ in $f(x)$. So for $n=1$, $f(x)=x$. I then tried to determine a pattern by solving for the derivative from $n=1$ to $n=5$. Here's what I got:
\begin{align}
n = 2 \Longri... | Let $f_1(x)=x$ and $f_n(x)=x^{f_{n-1}(x)}=x^{x^{\ldots^{x}}}$ ($n$ $x's$ on the exponent).
Then $\ln(f_n(x))=f_{n-1}(x)\cdot\ln(x)$ so: $$\frac{f'_n(x)}{f_{n}(x)}=f'_{n-1}(x)\cdot\ln(x)+\frac{f_{n-1}(x)}{x}\implies f'_{n}(x)=f_{n}(x)\cdot\left(f'_{n-1}(x)\cdot\ln(x)+\frac{f_{n-1}(x)}{x}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3992667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Divisibility Rule for 21 similar to rule for 12 I already understand the divisibility rule for 12. $$10\equiv -2 \pmod{12} \implies 10^n\equiv (-2)^n \pmod{12}$$
Then for some number $n = abcd = 1000a + 100b + 10c + d, abcd\equiv1000a + 100b + 10c + d \pmod{12}$
Then $abcd\equiv (-2)^3a + (-2)^2b + (-2)c + d$
Exampl... | For $21$, the reason this works is that $10\times(-2)=1\pmod{21}$, so $10\equiv(-2)^{-1}\pmod{21}$. Hence
$$1000a+100b+10c+d\equiv(-2)^{-3}a+(-2)^{-2}b+(-2)^{-1}c+(-2)^{0}d\pmod{21},$$
so multiplying the RHS by $(-2)^3$ implies the result.
You will observe a similar result for all teen numbers $\overline{1m}=10+m$, bec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3993281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Show $ ( \sin(\theta) - \cos(2\theta))^2 = 1+ \sin(\theta) - \sin(3 \theta) - \frac{1}{2}\cos(2 \theta) + \frac{1}{2}\cos(4 \theta) $
Show $ ( \sin(\theta) - \cos(2\theta))^2 = 1+ \sin(\theta) - \sin(3 \theta) - \frac{1}{2}\cos(2 \theta) + \frac{1}{2}\cos(4 \theta) $
If I would start from the right expression I woul... | As we know the RHS, let's try to express $\cos^2(2\theta)$ in terms of $\cos(4\theta)$
$\begin{align}(\sin\theta-\cos2\theta)^2 & = \sin^2\theta - 2\sin\theta\cos2\theta + \cos^22\theta\\& = \sin^2\theta-2\sin\theta\cos2\theta + \frac{1 + \cos4\theta}2 \end{align}$
Now $\cos2\theta = 1 - 2\sin^2\theta$ we use this to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3995554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Finding the limit of $a_{n} = \frac{1}{2}(a_{n-1}+a_{n-2})$ for arbitrary $a_{0}$ and $a_{1}$. Here is the problem statement:
Let $a$, $b$ $\in$ $\Bbb{R}$. A sequence $(a_{n})_{n \in \Bbb{N}}$ is defined recursively by
$$a_{0}:=a, \qquad a_{1}:=b, \qquad a_{n} = \frac{1}{2}(a_{n-1}+a_{n-2}) \quad \text{for} \quad n \g... | The characteristic equation is
$$\lambda^2=\frac{1}{2}(\lambda+1)$$
Its solutions are
$$\lambda=1;\;\lambda=-1/2$$
The general solution of the recurrence is
$$a_n=m+n\left(-\frac{1}{2}\right)^n$$
for $n=0$ we get $a_0=m+n=a$
for $n=1$ we have $a_1=m-\frac12n=b$
so we get
$$m= \frac{1}{3} (a+2 b),n= \frac{2 (a-b)}{3}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3995819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Show that there is only one real root of the equation $1 + a^2 + ax - x^3$ I have to show that for every $a>0$ the equation $1 + a^2 + ax - x^3 = 0$, has exactly one solution. I have made the graph of the function $f(x) = 1 + a^2 + ax - x^3$ on desmos and I can clearly see that. But, how can I prove it explicity?
Thank... | Write $f(x)=x^3-ax-a^2-1.$ Then we have that $f'(x)=3x^2-a.$ Hence the function has its local extrema at $$x=\pm\sqrt {\frac a3}.$$ It follows that there is exactly one real root if the relative extrema have equal sign. Hence we compute $$f\left(\sqrt {\frac a3}\right)f\left(-\sqrt {\frac a3}\right)$$ and confirm that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3996005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Prove that $\frac{a^{2}+1}{b}+\frac{b^{2}+1}{a} \geq 4$ $a$ and $b$ are both positive Real Numbers .
I saw this in a math olympiad, and within two days i couldn't solve it.
There is a simple case where:
$$\frac{a^{2}+1}{a}+\frac{b^{2}+1}{b} \geq 4 $$
We can solve it like this:
$$(a-1)^{2}\geq 0$$
$$\Leftrightarrow a^{2... | You can use your method, making $(a-1)^2$ and $(b-1)^2$ appearing was a nice move.
$$\frac{a^2+1}{b}+\frac{b^2+1}{a}-4=\frac 1{\underbrace{ab}_{>0}}(\underbrace{a^3+a+b^3+b-4ab}_E)$$
Now use $2ab\le a^2+b^2$
$E\ge a^3+a+b^3+b-2a^2-2b^2=a(a-1)^2+b(b-1)^2\ge 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3997184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Is it possible to solve $\frac{x(x-1)}{(x+y)(x+y-1)} =\frac12$, with $x>y$ and $100 \leq x+y\leq500$, without writing some program? I am reading a puzzle that ends up based on the given info to require the solution of the following equation:
$$\frac{x(x-1)}{(x + y)(x + y -1)} =\frac12$$
also knowing that:
$$x \gt y \qq... | This extends @Empy2's answer.
Starting from $2x(x-1) = (x+y)(x+y - 1) := z(z-1)$, we complete the squares:
$$2(x^2-x+\frac14)-\frac12 = z^2-z + \frac14 - \frac14$$
$$2(x-\frac12)^2 = (z-\frac12)^2 + \frac14$$
$$2(2x-1)^2 = (2z-1)^2+1$$
Equations of the form $x^2 - Dy^2 = \pm1$, where $D$ is not a square, are the simple... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3997845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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I have to show that the sum of this double series is $\frac{1}{2}$ i have to solve this double series. i tried it, but i am not sure, that it is enough.
$$\sum_{i=1}^{\infty} \sum_{k=1}^{\infty} \left(\left(\frac{1}{k+1} \cdot \left(\frac{k}{k+1}\right)^{i}\right) - \left(\frac{1}{k+2} \cdot \left(\frac{k+1}{k+2}\right... | Given that
$$\color{red}{\sum_{i=1}^{\infty}} \sum_{k=1}^{\infty} \bigg(\bigg(\frac{1}{k+1} \cdot \bigg(\frac{k}{k+1}\bigg)^{i}\bigg) - \bigg(\frac{1}{k+2} \cdot \bigg(\frac{k+1}{k+2}\bigg)^{i}\bigg)\bigg)$$
let's deal with the black sum first.
$$\begin{align}
&\sum_{k=1}^{\infty} \bigg(\underbrace{\bigg(\frac{1}{k+1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4000278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Ideas on how to integrate $\int \frac{dx}{\sqrt{(a-x)(x-b)(x^2+q+px)}}$ where $a>b$ I am trying to integrate $$\int \frac{dx}{\sqrt{(a-x)(x-b)(x^2+q+px)}},$$ where $a>b$, and $p$ and $q$ are real positive constants.
I would be very grateful if I can get some hints on how to proceed. Wolfram Mathematica tells me that th... | Yes, it has the structure of an Elliptic integral. You cannot solve it in terms of elementary functions, so you have to accept W. Mathematica's result, which is the only "true" result you can get.
By the way, for the curious people, the result is:
$$\frac{2 (a-x) \left(a \left(x-\sqrt{-\text{px}-q}\right)+x \sqrt{-\tex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4006685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.