Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
When is $m^2k^2(c^2+1)^2-4mc(c^2-c+1)$ a perfect square? Suppose, $m,k,c$ are positive integers
Conjecture : The expression $$m^2k^2(c^2+1)^2-4mc(c^2-c+1)$$ is a perfect square if and only if $m=k=1$
In the case $m=k=1$ , we get $(c-1)^4$ which is a perfect square ($0$ and $1$ are allowed)
The hard part is to show th... | If we write $d=\gcd(m,4c(c^2-c+1))$, then there necessarily exist positive integers $a_1$ and $a_2$ such that $m=da_1^2$ and
$$
mk^2(c^2+1)^2-4c(c^2-c+1) = d a_2^2.
$$
Writing $t=a_1k$, we thus have
$$
t^2 (c^2+1)^2 -a_2^2 = \frac{4c(c^2-c+1)}{d}
$$
and so, taking $a_3=tda_2$ and multiplying by $t^2d^2$,
$$
\left( t^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2862545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
The limit of $S_n=\sum_{k=1}^{n} \frac{kn}{k+n^3}$ Given $n>1$ and $S_n=\sum\limits_{k=1}^{n} \frac{kn}{k+n^3}$
Calculate $\lim\limits_{n \to \infty} S_n $
Well its obvious that $\frac{n^2}{n^3+n} \leq S_n\leq \frac{n^3}{n^3+1}$
$S_n$ converge to a limit $l$ such that $0\leq l \leq 1$
How can we determine the valu... | You can also get the result if you only estimate the denominator in order to get better bounds:
$$\frac{1}{2} \stackrel{n\to\infty}{\longleftarrow}\frac{n(n+1)}{2(1+n^2)} = \frac{n}{n+n^3} \sum \limits_{k=1}^n k \leq S_n \leq \frac{n}{1+n^3} \sum \limits_{k=1}^n k = \frac{n^2(n+1)}{2(1+n^3)} \stackrel{n\to\infty}{\long... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2863962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ If the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ can be expressed as $\frac{a\sqrt b+c}{d}$.Find $a+b+c+d$
T... | The triangle has area
$$
\frac12 \times \sqrt2 \times (1- \sqrt2/2) = \frac{\sqrt2 - 1}{2},
$$
so the total polygon has area
$$
2\sqrt{2} - \frac{\sqrt2 - 1}{2} = \frac{3}{2}\sqrt{2} + \frac12 = \frac{3\sqrt{2} + 1}{2},
$$
at least in lowest terms. The result is $a + b + c + d = 8$, contrary to the 10 you claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2864266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Reduction of Order Leads to Non-Elementary Integral
If $u_1=x+1$ is a solution of $$xu''-(x+1)u'+u=0$$ find another linearly independent solution using reduction of order.
I let $u_2=(x+1)v(x)$ be the second solution. Hence
$$v''(x^2+x)-v'(x^2+1)=0$$ Let $w=v'$, so
\begin{align}
\frac{dw}{dx}(x^2+x)-w(x^2+1)&=0 \\
... | Your work is correct up to the partial fractions
$$ \frac{v''}{v'} = \frac{x^2+1}{x^2+x} = 1 + \frac{1-x}{x(x+1)} = 1 + \frac{1}{x} - \frac{2}{x+1} $$
Integrating this gives
$$ \ln(v') = x + \ln x - 2\ln(x+1) $$
$$ \implies v' = \frac{xe^x}{(x+1)^2} = \frac{e^x}{x+1}-\frac{e^x}{(x+1)^2} $$
This has the form $e^xf(x) + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2866006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Transformation restriction to commutative matrices Let $T: \Bbb M_{3x3}(\Bbb R) \rightarrow \Bbb M_{3x3}(\Bbb R)$ be a linear transformation such that $T(B) = AB$ where:
$$ A=
\begin{pmatrix}
2 & -1 & -1 \\
-1 & 2 & -1 \\
-1 & -1 & 2 \\
\end{pmatrix}
$$
now let $W = \{B|AB=BA\}$ be the group of all ... | $W$ is a vector space and not a group. Note that $T=A\otimes I_3$ -if we stack the matrices row by row-. It is easy to see that $T(W)\subset W$. Moreover $spectrum(T)=\{0,0,0,3,3,3,3,3,3\}$; cf.
https://en.wikipedia.org/wiki/Kronecker_product
We may assume that $A=diag(0,3,3)$ -because $(P\otimes I)(A\otimes I)(P\otim... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the basis for the nullspace, the row space, and the column space of the given matrix.
Find the basis for the nullspace, the row space, and the column space of the given matrix.$$\begin{bmatrix} 1 & 2 & 3 & 1 \\ 3 & 2 & 5 & 2 \\ -1 & 2 & 1 & 3 \\ 1 & 1 & 2 & 2 \end{bmatrix}$$
My Try
$$\begin{bmatrix} 1 & 2 & 3... | We don't say that the RREF is not consistent in that case.
We need another step to obtain
$$\begin{bmatrix} 1 & 2 & 3 & 1 \\ 0 & 4 & 4 & 1 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 1 \end{bmatrix}\to \begin{bmatrix} 1 & 2 & 3 & 1 \\ 0 & 4 & 4 & 1 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$
therefore a basis for the nullsp... | {
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"url": "https://math.stackexchange.com/questions/2868517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number of solution of $x^4-5x^3+(\lambda+2)x^2-5x+1=0$
Consider the bi-quadratic equation $E:x^4-5x^3+(\lambda+2)x^2-5x+1=0$ then, the real values of $\lambda$ so that $E$ has four different solutions is?
My attempts:
As $x=0$ is not a solution for any $\lambda$ hence we divide by $x^2$,\begin{align}x^2+\dfrac{1}{x^2... | Let $f(x)=-\frac{x^4-5x^3+2x^2-5x+1}{x^2}.$
We need to find all values of $\lambda,$ for which the equation
$$\lambda=f(x)$$ has four different roots.
Now, $$f'(x)=\frac{(2-x)(x-1)(2x-1)(x+1)}{x^3},$$ which says that
$$\lambda<f(-1)$$ or $$f(1)<\lambda<\min\left\{f\left(\frac{1}{2}\right),f(2)\right\},$$
which gives th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2869059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Matrix vector form. Is this in the correct form? I have this question:
Write the linear system
$$\begin{array}{rcr}-2x_1+x_2-4x_3 & = & 1 \\ x_1-2x_2 & = & -3 \\ x_1+x_2-4x_3 & = & 0 \end{array}$$
in the matrix-vector form $A\mathbf{x}=\mathbf{b}$.
Is this what they want?
$$
x_1*
\begin{bmatrix}
-2 \\ 1 \\ 1
\... | basically!
$
A = \begin{pmatrix} -2 & 1 & -4 \\ 1 & -2 & 0 \\ 1 & 1 & -4 \end{pmatrix}
$
and $b = \begin{pmatrix} 1 \\ -3 \\ 0 \end{pmatrix}$ yielding $Ax = b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2869367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Shorter Way to Solve $\lim_{x\to 0} \frac{x^3-\ln^3(1+x)}{\sin^2x-x^2}$ The limit to find is
$$\lim_{x\to 0} \frac{x^3-\ln^3(1+x)}{\sin^2x-x^2}$$
What I've tried was using the factorisation for $a^3-b^3$ and $a^2-b^2$ like so:
$$\lim_{x\to 0}\frac{(x-\ln(1+x))(x^2+x\ln(x+1)+\ln^2(x+1))}{(\sin x-x)(\sin x+x)}$$
but I do... | Using $\sin(x)=x-\frac{x^3}{6}+\mathcal{O}(x^5)$ and $\ln(1+x)=x-\frac{x^2}{2}+\mathcal{O}(x^3)$:
$$\begin{aligned}\frac{x^3-\ln^3(1+x)}{\sin^2(x)-x^2}
&=\frac{x^3-\left(x-\frac{x^2}{2}+\mathcal{O}(x^3)\right)^3}{\left(x-\frac{x^3}{6}+\mathcal{O}(x^5)\right)^2-x^2}\\
&=\frac{x^3-\left(x^3- \frac{3 x^4}{2}+\mathcal{O}(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2870180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Ratio of areas determined by a square inscribed in the corner of a right triangle I’m having trouble working out how to algebraically get to the answer of this question. (See original image below.)
A square is drawn in the corner of a right-angled triangle with side lengths $a$, $b$, and [hypotenuse] $c$, as shown.
Wh... | Let the side of the square is $x$. Then, using similarity of triangles, $\frac{b-x}{x}=\frac{x}{a-x}$ or $(b-x)(a-x)=x^2$; $ab-ax-bx+x^2=x^2$; $ab=x(a+b)$ or $x=\frac{ab}{a+b}$. Thus, the shaded area is $\frac{(ab)^2}{(a+b)^2}$. The unshaded area is $$0.5x(b-x)+0.5x(a-x)=0.5\left(\frac{ab}{a+b}(b-\frac{ab}{a+b})+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2872591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that inequality $\frac{2ab}{a+b}+\sqrt{\frac{a^2+b^2}{2}}\ge \sqrt{ab}+\frac{a+b}{2}$ Let $a;b\ge 0$. Prove that inequality $$\frac{2ab}{a+b}+\sqrt{\frac{a^2+b^2}{2}}\ge \sqrt{ab}+\frac{a+b}{2}$$
My try: $LHS-RHS=\frac{2ab}{a+b}-\frac{a+b}{2}+\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}\ge 0$
Or $-\frac{\left(a-b\right)^2... | I think this is the simplest solution:
So you are stuck here:
$$\frac{1}{2\left(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\right)}\ge \frac{1}{2\left(a+b\right)}$$
...which is equivalent to:
$$\sqrt\frac{a^2+b^2}{2}+\sqrt{ab}\le a+b\tag{1}$$
Apply Jensen inequality:
$$\frac{f(x_1)+...+f(x_n)}{n}\le f(\frac{x_1+...+x_n}{n})$$
.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2872870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$x^4+x^3+x^2+x+1$ irreducible over $\mathbb F_7$ This question came from the answer here.
The answer claims that $x^4+x^3+x^2+x+1$ is irreducible over $\mathbb F_7$. I can check that it has no roots in the field, but why can't it be written as a product of two quadratics? I tried the method of undetermined coefficients... | HINT.-If $x^4+x^3+x^2+x+1$ were reducible then we have either a linear factor with a cubic one or two quadratic factors.
We have $$x^4+x^3+x^2+x+1=-\dfrac 14(-2x^2+(\sqrt5-1)x-2)(2x^2+(\sqrt5+1)x+2)$$
Since $\mathbb F_7^2=\{1,2,4,0\}$ we see that $5$ is not a square modulo $7$. It follows $x^4+x^3+x^2+x+1$ is not a fa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2874241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
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Number of non negative integral values of $n,n\le 10$ so that a root of the equation $n^2\sin^2 x-2\sin x-(2n+1)=0$ lies in the interval $[0,\pi/2]$ Find the number of non negative integral values of $n,n\le 10$ so that a root of the equation $n^2\sin^2 x-2\sin x-(2n+1)=0$ lies in the interval $[0,\pi/2]$
As $x\in[0,\... | $n^2 t^2 -2t - 2n -1=0$ is a quadratic in $t$. Solving it for $t$, we get:
$$
\begin{align}
t &= \frac{2 \pm \sqrt{4-4n^2(-2n-1)}}{2n^2}
\\ &= \frac{1 \pm \sqrt{1+n^2(2n+1)}}{n^2}
\end{align}
$$
It can easily be checked that $n=0$ does not lead to $t \in [0,1]$ so we may assume that $n \neq 0$.
Now for $t \in [0,1]$ w... | {
"language": "en",
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Given the equation $\sum_{k=11}^{99} \left[x + \frac{k}{100} \right] = 765$ find $[10 \, x]$.
Given the equation $\sum_{k=11}^{99} \left[x + \frac{k}{100} \right] = 765$ find $[10 \, x]$.
I have tried this problem in many ways and I think it uses the identity
$$[x]+\left[x+\frac{1}{n}\right]+\left[x+\frac{2}{n}\righ... | This is a well known identity known as Hermite's identity. A simple proof is here (Although, Hermite's original proof relied on identities involving exponents)
https://en.wikipedia.org/wiki/Hermite%27s_identity
For general setup, we can also do the following.
\begin{eqnarray*}
S &=& \sum_{k=m+1}^{n-1}{\left\lfloor x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Integer solutions to $x^3=y^3+2y+1$?
Find all integral pairs $(x,y)$ satisfying $$ x^3=y^3+2y+1.$$
My approach:
I tried to factorize $x^3-y^3$ as $$(x-y)(x^2 + xy + y^2)=2y+1,$$ but I know this is completely helpless. Please help me in solving this problem.
| we first consider only $x> y\ge0$
suppose $x=y+a$
then the equation is $(y+a)^3=y^3+2y+1$
$y^3+3ay^2+3a^2y+a=y^3+2y+1$
$3ay^2+(3a^2-2)y+(a-1)=0$
however, as $a\ge1$, $3a^2-2\ge0$,$a-1\ge0$
If $y>0$, then $3ay^2+(3a^2-2)y+(a-1)>0$, so $y=0$
then, $x^3=1$
$(x,y)=(1,0)$ are only integral solution
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2882685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
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"answer_id": 4
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Let $k$ be a natural number $\ge$ 3. Find the value of the constants $x$ and $y$ such that $A^k = xA +yI$ (Both x and y depend on k). So the question states that the matrix
$ A = \begin{bmatrix} 4 & -3 \\ 2 & -1 \end{bmatrix}$
characteristic polynomial is $ p(x) = x^2 - 3x + 2.\\$
I am supposed to find x and y when k i... | Write $A^k=x_kA+y_kI$. Then $A^{k+1}=x_kA^2+y_kA=x_k(3A-2I)+y_kA=(3x_k+y_k)A-2x_kI$. In other words:
$$\begin{bmatrix}x_{k+1} \\ y_{k+1} \end{bmatrix}=\begin{bmatrix} 3 & 1 \\ -2 & 0 \end{bmatrix} \begin{bmatrix}x_{k} \\ y_k \end{bmatrix}$$
Diagonalizing the matrix above yields
$$\begin{bmatrix} 3 & 1 \\ -2 & 0 \end{b... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find all matrices that commute with $A$
Given $$A = \begin{bmatrix}
3 & 1 &0 \\
0 &3 & 1\\
0 &0 & 3
\end{bmatrix}$$ find matrices $B$ such that $AB=BA$.
Trivially $B=A^{-1}$ and $B=kI$ are the solutions
Also we have Characteristic Polynomial as
$$A^3-9A^2+27A-27I=0$$ $\implies$
$$(A-3I)^3=0$$
Is it possible to fi... | Since$$A=3\operatorname{Id}_3+\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}$$and every matrix commutes with $3\operatorname{Id}_3$, you're after the matrices that commute with$$\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}.\tag1$$A simple computation shows that\begin{multline}\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2884923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $G=\mathbb{Z}/24\mathbb{Z}\times\mathbb{Z}/6\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}$. Consider the quotient group $H=G/\langle (10,3,2)\rangle$. Determine a direct product of cyclic groups that is isomorphic to $H$.
The Smith Normal form can be used to find the invariant factors in the structure theorem for finitely... | The Smith Form has diagonal entries 1, 3, 12 so that the quotient is isomorphic to $Z_3\times Z_{12}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Possibilty of a number $n$ to be a multiple of $x$ that has a remainder of $y$ Is there a formula or solution for this :
"$n$ is a number that is a multiple of $x$ and when divided by $b$ has remainder of $y$".
Is there a fast way to finding this number?
| The short answer is "No". Here is the long answer.
$n \equiv 0 \pmod x \implies n = \alpha x \ \text{for some integer $\alpha$}$
$n \equiv y \pmod b \implies \alpha x \equiv y \pmod b$
Let $g = \gcd(x, b)$. Then there is no solution unless $g \mid y$.
If $g \mid y$...
Let $X = \dfrac xg, \quad Y = \dfrac yg, \quad B =... | {
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation $\frac{\sqrt {3}}{2}\sin(x) -\cos x=\cos^2x$ Solve the equation $\frac{\sqrt {3}}{2}\sin x-\cos x=\cos^2x$
My approach $\cos^2x=1-\sin^2x $
$\frac{\sqrt {3}}{2}\sin x-\cos^2x =\cos x$
$\frac{\sqrt {3}}{2}\sin x+\sin^2x -1=\cos x$
$\sqrt {3}\sin x+2\sin^2x-2=2\cos x$
Though the equation comes in form ... | Hint; Use the so-called Weierstrass Substitution:
$$\sin(x)=\frac{2t}{1+t^2}$$
$$\cos(x)=\frac{1-t^2}{1+t^2}$$
$$\tan(\frac{x}{2})=t$$
and you don't need square the equation.
You will get the equation
$$-\sqrt {3} \left( \sqrt {3}t+{t}^{2}+2 \right) \left( -3\,t+\sqrt {3}
\right)
=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2887674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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At what angle from the ground should it be fired so that it travels the maximum distance in the air? A projectile is going to be fired from a cannon on level ground.
At what angle from the ground should it be fired so that it travels the maximum distance in the air?
Details and Assumptions:
*
*There is downward grav... |
If the ball is released with initial speed $V$ at an angle of $\theta$
to the horizontal, the trajectory of the ball is $ y \; = \;
> x\tan\theta - \tfrac{g}{2V^2}x^2\sec^2\theta \; = \; x\tan\theta -
> \tfrac{1}{2a}x^2\sec^2\theta $ where $a = \tfrac{V^2}{g}$. Then the
ball hits the ground again when $y=0$, which... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2890340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prime Number Congruence Modulo 8 Proof I want to rigorously prove that:
$$p_n=2\Biggl(\Bigl\lfloor \frac{p_n+1}{8}\Bigr\rfloor+\Bigl\lfloor \frac{p_n+3}{8} \Bigr\rfloor+\Bigl\lfloor \frac{p_n+5}{8} \Bigr\rfloor+\Bigl\lfloor \frac{p_n+7}{8} \Bigr\rfloor\Biggr)-1+\delta(n,1) \quad\quad\quad\quad\quad\quad(0)$$
So far wh... | Here is an alternative way. It is well known that
$$\lfloor nx\rfloor =\sum_{i=1}^n\,\left\lfloor x+\frac{i-1}{n}\right\rfloor\text{ for all }x\in\mathbb{R}\text{ and }n\in\mathbb{Z}_{>0}\,.$$
Take $n:=8$ and $x:=\dfrac{p}{8}$ if $p$ is an odd prime. Thus, we get
$$p=\Biggl\lfloor 8\left(\frac{p}{8}\right)\Biggr \rfl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2890559",
"timestamp": "2023-03-29T00:00:00",
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proof with induction $\sum_{k=0}^n (-1)^k \binom{n}{k} (n-k+1)^n = n!$ prove with induction:
$$\sum_{k=0}^n (-1)^k \binom{n}{k} (n-k+1)^n = n!$$
I'm stuck on
$$n[\sum_{k=0}^{n-1} (-1)^k \binom{n-1}{k} (n-k+1)^n]= \sum_{k=0}^n (-1)^k \binom{n}{k} (n-k+1)^n = n!$$
$$n[\sum_{k=1}^{n-1} (-1)^{k-1} \binom{n-1}{k-1} (n-k+1... | Let's assume
$$\sum_{k=0}^n (-1)^k \binom{n}{k} (n-k+1)^n = n!$$
Do the following change of variable so that it looks easier:
$$k \leftarrow n-i$$
So that the equation that needs to proved is now
$$\sum_{i=0}^n (-1)^{n-i} \binom{n}{n-i} (i+1)^n = n!$$
Notice that
$$\binom{n}{n-i}=\binom{n}{i}$$
So why not write it as... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Factoring $x^n + 1$. By the Fundamental Theorem of Algebra, every polynomial of degree $n$ can be factored into a product of $n$ linear polynomials.
As an example, since the polynomial $ x^5 +1$ has the five complex roots $$\tag{1} -1,\quad e^{\frac{\pi i}{5}}, \quad e^{\frac{-\pi i}{5}}, \quad e^{\frac{3\pi i}{5}}, ... | The case $\,n=5\,$ can be solved algebraically by first factoring $\,x^5+1$ $ = (x+1)(x^4-x^3+x^2-x+1)\,$, then the second factor is a palindromic polynomial which reduces to a quadratic in $\,x+1/x\,$:
$$
\begin{align}
x^4-x^3+x^2-x+1 &= x^2\left(\left(x^2+\frac{1}{x^2}\right)-\left(x+\frac{1}{x}\right)+1\right) \\
&... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Evaluate: $\int \frac{\mathrm{d}x}{x\sqrt{x^2+x+1}}$
Evaluate
$$
\int \frac{\mathrm{d}x}{x\sqrt{x^2+x+1}} \cdotp
$$
My attempt:
$$
I
= \int \frac{\mathrm{d}x}{x\sqrt{x^2+x+1}}
= \int \frac{\mathrm{d}x}{x\sqrt{\left(x + \frac{1}{2}\right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}}
$$
I thought completing t... | For integrals of the form $$\dfrac1{(x+a)\sqrt{(x+b)^2+c^2}},$$
Choose $x+b= c
\tan y,$ to reach at an integral of the form $$\dfrac1{A\cos y+B\sin y}$$
Now $A\cos y+B\sin y=\sqrt{A^2+B^2}\sin (y+\arctan \dfrac AB)=\sqrt{A^2+B^2}\cos(y-\arctan\dfrac AB)$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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How can I graph $f(x)=\lfloor{x^2}\rfloor$ when the domain is $ℝ^{-}$.
How can I graph $f(x)=\lfloor{x^2}\rfloor$ when the domain is $ℝ^{-}$.
I know that by definition $(\lfloor{x}\rfloor=m) ≡ (m≤x<m+1)$, so it follows that $(\lfloor{x^2}\rfloor=m ≡ (m≤x^2<m+1) ≡ (\sqrt{m}≤x^2<\sqrt{m+1})$; from this I can do
$\lflo... | Well, by the trivial inequality, we have that $x^2$ is nonnegative. Note that you are solving for the floor of the $y$ outputs, not the $x$ inputs. Also, $\text{floor}(-x)^2=\text{floor}(x^2),$ so both sides are symmetrical.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Show that $0$ has multiplicity $3$ in $M-4I$? Question from an exam:
Consider the matrix $M=$
\begin{bmatrix} 5&1&1&1&1&1\\1&5&1&1&1&1\\1&1&5&1&1&1\\1&1&1&4&1&0\\1&1&1&1&4&0\\1&1&1&0&0&3\end{bmatrix}
Show that $4$ is an eigen value of the above matrix with multiplicity $3$.
I considered $M-4I$ from which I got
\begi... | If it only had multiplicity $2$, $M-4I$ would have rank $4$, i.e. four of its rows would be linearly independent. Since the top three rows are the same, that would have to be the last four rows. But $(row\ 3) - (row\ 4) - (row\ 5) + (row\ 6) = 0$, so they are linearly dependent.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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An expression involving trig functions and binomial coefficients Prove that
\begin{eqnarray}\sum_{i=0}^k5^{k-i-1}\binom{k}{i}\left(\frac{(37+62\cos\frac{2\pi}{5}+56\cos\frac{4\pi}{5})^i}{\left(2\sin\frac{\pi}{5}\right)^{2(k-i-1)}}+\frac{(37+62\cos\frac{4\pi}{5}+56\cos\frac{8\pi}{5})^i}{\left(2\sin\frac{2\pi}{5}\right)^... | Using the same approach as @drhab in his answer, I arrived to a slightly different result
$$\color{blue}{t_k=\frac{b}{5}\left(a+\frac 5b\right)^{k}+\frac{d}{5}\left(c+\frac5d\right)^{k}}$$ Now, using the values of the trigonometric terms, we have
$$a=\frac{3}{2} \left(5+\sqrt{5}\right)\qquad b=\frac{1}{2} \left(5-\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2899230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I solve 6b) and 6c) if my solution for 6a) is a consistent system of linear equations? $$\begin{cases}
2x_1-x_2= dx_1 \\
2x_1-x_2+x_3=dx_2 \\
-2x_1+2x_2+x_3=dx_3
\end{cases}
$$
a) Is it possible for the system to be inconsistent? Explain?
b) For what values of d will the system have infinitely many solutions?
... | Using the Gaussian elimination method to solve a set of linear equations,
From the equations, you have given,
\begin{cases}
2x_1-x_2= dx_1 \\
2x_1-x_2+x_3=dx_2 \\
-2x_1+2x_2+x_3=dx_3
\end{cases}
We can arrive at this augmented matrix,
\begin{bmatrix}
\begin{array}{ccc|c}
-d+2&-1&0&0\\
2&-1-d&1&0\\
-2&2&1-d&0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2901362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Closed form solution for this integral? I'm hitting a road block in finding an expression (closed form preferably) for the following integral:
\begin{equation}
\int^{+\infty}_0 x^b \left ( 1-\frac{x}{u} \right )^c \exp(-a x^3) dx
\end{equation}
where $a,b$ are positive constants; $b>1$ is an odd multiple of $0.5$, whil... | Long Comment:
Since it was explicitly asked for, Mathematica 11.3 when provided with
$$\text{Integrate}\left[e^{-a x^3} x^b \left(1-\frac{x}{u}\right)^c,\{x,0,\infty \},\text{Assumptions}\to b>1\land a>0\land u>0\right]$$
gives up the monster
$$\text{ConditionalExpression}\left[\frac{1}{6} \left(2 a^{\frac{1}{3} (-b-c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2901651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 0
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Convergence of $ \sum _ { n = 1 } ^ { \infty } \frac { n ! n ^ { n } } { ( 2 n ) ! } $ The task is to find out if this series is convergent or divergent.
$$ \sum _ { n = 1 } ^ { \infty } \frac { n ! n ^ { n } } { ( 2 n ) ! } $$
The solution uses the ratio test and says:
$ \left.\begin{aligned} \frac { a _ { n + 1 } ... |
How can all n's on the left site become 1/n?
This is a standard method when finding the limit of such sequences.
You factor out the highest power of $n$ to cancel it out.
So if we would like to involve one more step befor the one you do not follow it would be:
$\frac{(n+1)(n+1)}{(2n+2)(2n+1)}\cdot\left(\frac{n+1}{n}\... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Prove this inequality for given conditions For all $x,y>0$, $$\frac{1}{(x+1)^2} + \frac{1}{(y+1)^2} \ge \frac{1}{xy+1}$$
I can only think of substituting $x+1$ with $a$ and $y+1$ with $b$. Then the inequality turns into
$$(a^2 + b^2) (ab-a-b +2) \ge a^2b^2$$
I can proceed no further. Please help.
| After eliminating denominators and simplifying, the inequality reduces to:
$$
x^3y+xy^3-x^2y^2-2xy+1 \ge 0 \;\;\iff\;\; xy(x^2+y^2) - x^2y^2-2xy+1 \ge 0
$$
Using that $\,xy \ge 0\,$ and $\,\color{blue}{x^2+y^2 \ge 2xy}\,$, the above follows from:
$$
xy(x^2+y^2) - x^2y^2-2xy+1 \color{blue}{\ge 2x^2y^2}-x^2y^2-2xy+1 = (x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2902350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Prove equality involving binomial coefficients I was solving a probability problem and I got a different answer than the one given in the book. Seems the authors were using a different way of counting/arguing.
For the two answers to be equal, the following equality should hold true.
$$\sum_{k=0}^{b-1} \binom{a+k-1}{a-... | Starting from
$$\sum_{k=0}^{b-1} {a+k-1\choose a-1} p^a (1-p)^k
= \sum_{k=a}^{a+b-1} {a+b-1\choose k} p^k (1-p)^{a+b-k-1}$$
we simplify to
$$\sum_{k=0}^{b-1} {a+k-1\choose a-1} p^a (1-p)^k
= \sum_{k=0}^{b-1} {a+b-1\choose a+k} p^{a+k} (1-p)^{b-k-1}$$
or
$$\sum_{k=0}^{b-1} {a+k-1\choose a-1} (1-p)^k
= \sum_{k=0}^{b-1} {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2904333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving $\frac{x^2-2}{x^2+2} \leq \frac{x}{x+4}$. Solve the inequality
$\frac{x^2-2}{x^2+2} \leq \frac{x}{x+4} \Leftrightarrow$ $\frac{x^2-2}{x^2+2} - \frac{x}{x+4} \leq 0 \Rightarrow$ $\frac{x^3+4x^2-2x-8-x^3-2x}{(x^2+2)(x+4)} \geq 0 \Leftrightarrow$ $4x^2-4x-8 \geq 0 \Rightarrow$ $x \geq \frac{1}{2} \pm \sqrt{2.25} ... | Hint : You wrote $\frac{x^3+4x^2-2x-8-x^3-2x}{(x^2+2)(x+4)} \geq 0 \Leftrightarrow$ $4x^2-4x-8 \geq 0$ but you forgot to study what is on the bottom. $(x+4)$ changes sign.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Finding the sum $\sum_{k=1}^rk^2\binom {n-k}{r-k}$ I was stuck while finding the given summation.
$$\sum_{k=1}^rk^2\binom {n-k}{r-k}$$
Since $n$ and $r$ are both constants, so I have first converted the above summation into this:
$$\sum_{k=1}^rk^2\binom {n-k}{n-r}$$
I have no idea how to proceed next. Any help will be ... | We have
$$\sum_{k=1}^r k^2 {n-k\choose r-k}
= \sum_{k=0}^r k^2 [z^{r-k}] (1+z)^{n-k}
\\ = [z^r] \sum_{k=0}^r k^2 z^k (1+z)^{n-k}
= [z^r] (1+z)^n \sum_{k=0}^r k^2 z^k (1+z)^{-k}.$$
There is no contribution to the coefficient extractor
when $k\gt r$ and we may continue with
$$[z^r] (1+z)^n \sum_{k\ge 0} k^2 z^k (1+z)^{-k... | {
"language": "en",
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"source": "stackexchange",
"question_score": "11",
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Solving Homogenous First Order ODE Using Substitution Problem
Find the general solution of the following homogeneous equation.
$$ ty' = y + \sqrt { t^2 - y^2} $$
Attempt
I'm following the algorithm provided in section 1.5 of the William Adkins & Mark G. Davidson, Ordinary Differential Equations textbook on page 63, whi... | According with the suggested
$$
y'=v+\sqrt{1-v^2}
$$
with
$$
v = \frac yt
$$
so
$$
y' = v + t v'
$$
then
$$
v + t v' = v +\sqrt{1-v^2}
$$
or
$$
\frac{dv}{\sqrt{1-v^2}} = \frac{dt}{t}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2912091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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let $a_{1} = 0, a_{2} =1$ and $a_{n+2} = \frac{(n+2)a_{n+1} - a_{n}}{n+1}$. Prove that $\lim\limits_{n \to +\infty} a_{n} = e$ let $a_{1} = 0$, $a_{2} =1$ and $a_{n+2} = \frac {(n+2)a_{n+1} - a_{n}} {(n+1)}$.
Prove that $\lim a_{n} = e$
Knowing that I have proved that $a_{n+1} -a_{n} = 1/n!$, then I shall do what next... | Since you've proved that $a_{n+1}-a_n = \frac{1}{n!}$ you have that $$a_n = \frac{1}{1!}+ \ldots + \frac{1}{(n-1)!}$$
So $$\lim\limits_{n \to +\infty} a_n = \lim\limits_{n \to \infty} \left( \frac{1}{1!} + \ldots + \frac{1}{(n-1)!} \right) = \sum\limits_{n=1}^{+\infty} \frac{1}{n!} = e-1$$
Are you sure the correct answ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2915432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Area of ellipse using double integral I am trying to find the area of a quadrant of an ellipse by double integrating polar coordinates but the answer I'm getting is incorrect.
ellipse : $ x^2/a^2 + y^2/b^2 =1 $
Any point on ellipse : $ ( a\cos(\theta), b\sin(\theta)) $
At $ \theta$, taking $ d\theta $ segment, Thus $ ... | Set $x=ar\cos \theta, y=br\sin \theta.$ The Jacobian is $abr$ and we compute the area
$$\mathcal {A}= \int_0^{\pi/2} \int_0^1 abr \;d r \;d\theta,$$ which is $\frac{ab\pi}{4}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2915515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Showing that the graph of $\frac{\cos(\pi x/2)}{1-x^2}$ decreases over the interval $(0,2)$. How would one justify the behaviour of the graph of
$$\frac{\cos(\frac{\pi}{2}x)}{1-x^2}$$
over the interval $(0,2)$?
More precisely, how can it be shown that the graph is decreasing over that interval?
| The derivative of $f(x)$ takes the following form
\begin{equation}
f'(x) = \frac{g(x)}{(1-x^2)^2}
\end{equation}
\begin{equation}
g(x) = -\frac{\pi}{2} \sin (\frac{\pi}{2}x)(1-x^2)+2x( \cos (\frac{\pi}{2}x))
\end{equation}
PS: The function $f(x)$ is not defined at $1$
Since the denominator of $f'(x) > 0$ on $[0,1[ \c... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Prove that $f(x)=\frac{ax^2+x-2}{a+x-2x^2}$ has the range $ℝ$ when $x\inℝ$ if $a \in [1,3]$ Prove that $f(x)=\frac{ax^2+x-2}{a+x-2x^2}$ has the range $ℝ$ when $x\inℝ$ if $a \in [1,3]$
My working:
Let $f(x)=\frac{ax^2+x-2}{a+x-2x^2}=y$
therefore,
$$(a+2y)x^2+(1-y)x-(2+ay)=0$$
Now, for $x$ to have real values, the discr... | The minimum of the polynomial $ax^2+bx+c$ , where $a$ is positive, is $\frac{4ac - b^2}{4a}$. To see this, note that $ax^2 + bx + c = \frac{(4a^2x^2 + 4abx + b^2)}{4a} -\frac{b^2}{4a} + c = \frac 1{4a} (2ax + b)^2 + \frac{4ac - b^2}{4a}$, so the smallest value is when the square is zero, then the value is $\frac{4ac -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solving periodic equation $\cos 7\theta=\cos 3\theta+\sin 5\theta$.
Here I have this trigonometric equation $\cos 7\theta=\cos 3\theta+\sin 5\theta$.
I approached the problem as follows :
$\cos 7\theta=\cos 3\theta+\sin 5\theta$
$\implies \cos 7\theta-\cos 3\theta=\sin 5\theta$
$\implies 2\sin 5\theta\sin(-2\theta)=\... | $$\sin 5\theta=0$$
$\sin x =0$ has solutions $x=n\pi$
Thus we get $$5\theta=n \pi$$
$$\theta= \frac{n\pi}{5}\hspace{10pt} n\in Z$$
Also $$\sin 2\theta=-\dfrac{1}{2}$$
$$2\theta=2n\pi-\frac{\pi}{6} \implies \theta=n\pi-\frac{\pi}{12}\hspace{10pt} n\in Z$$
$$2\theta=(2n+1)\pi+\frac{\pi}{6} \implies \theta=n\pi+\frac{7\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Decomposition of polynomials like $1+x^4$ What is the "trick" to get from $1+x^4$ to $$(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)?$$ Of course I can calculate it's true, but I don't understand what steps to take to get from one to the other.
Next to this specific question, I am also looking for the general rules for such decom... | By the factor theorem, the linear of $x^n + 1$ are of the form $x - a$, where $a$ is one of the $n$th roots of $-1$ in $\mathbb{C}$. These form complex conjugate pairs, and when two such paired factors multiply together, we get an irreducible quadratic over $\mathbb{R}$.
We know that $-1 = e^{i\pi}$, which makes the fo... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
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Does the fact that $x^2=(x-1)(x+1)+1$ have a name? Just curious about this pattern
$$x^2 = (x-1)(x+1) +1$$
So:
$$\begin{align}
1^2 &= \phantom{1}0\cdot\phantom{1}2+1 = 1 \\
2^2 &= \phantom{1}1\cdot\phantom{1}3+1 = 4 \\
3^2 &= \phantom{1}2\cdot\phantom{1}4+1 = 9 \\
4^2 &= \phantom{1}3\cdot\phantom{1}5+1 = 16... | If you were to subtract 1 from both sides, you would get
$$ x^2 - 1 = (x + 1)(x - 1)$$
This is a simpler case of the more general factoring of the difference of squares:
$$ x^2 - b^2 = (x + b)(x - b)$$
For more information see Wikipedia's Difference of two squares article.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many integers from 1 through 1000 are divisible by 3 and by at least one of 2,5,7, and 11? So I got 294 for this question, though I'm not a hundred percent sure but let me explain my process:
I did
to find all the multiples:
(floor of each)
1000/6 = 166
1000/15 = 66
1000/21 = 47
1000/33 = 30
166+66+47+30 = 309
to ... | If you consider the range $3$ to $999$, divide by $3$, then we find that your question is equivalent to the following simpler question:
how many integers between $1$ and $333$ are divisible by at least one of $2$, $5$, $7$, or $11$?
For this question we may use inclusion-exclusion.
Multiples of $2,5,7,11$ with muc... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Laurent series for $f(z)=\frac{1}{z-2i}-\frac{1}{z+i}$ that converges at 3
I am trying to find the Laurent series for $f(z)=\frac{1}{z-2i}-\frac{1}{z+i}$ with centre $1$ that converges at $3$.
I was thinking that if we want the series to converge at $3$, we are interested in the region $\sqrt{2}<|z-1|<\sqrt{5}$. Henc... | $$
\begin{align}
f(z)
&=\frac1{z-2i}-\frac1{z+i}\\
&=\frac1{w+1-2i}-\frac1{w+1+i}\\
&=\frac1{1-2i}\sum_{k=0}^\infty(-1)^k\left(\frac w{1-2i}\right)^k
-\frac1w\sum_{k=0}^\infty(-1)^k\left(\frac{1+i}w\right)^k\\
&=-\sum_{k=0}^\infty\frac{w^k}{(-1+2i)^{k+1}}-\sum_{k=0}^\infty\frac{(-1-i)^k}{w^{k+1}}\\
&=-\sum_{k=0}^\infty... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $A = x - y + z$ if $3x + 5y + 7z = 29$ and $x , y , z \in \mathbb{Z}^+$ I've found the answer by trial and error $(x = 4 , y=2,z=1) \Rightarrow$ $A = 3$. I tried to solve it using modular arithmetic but it didn't work .
$$3x + 5y + 7z = 29 $$
$$ 3x + 5y+7z \equiv 29 \mod 3$$
$$ 2y +z \equiv 2 \mod 3 $$
$$ y= -k ,... | Starting from your 3.rd line. For some integer $t$ we have $$2y+z-2 = 3t\implies z = 2+3t-2y$$
So $$3x+5y+7(2+3t-2y) = 29\implies 3x = 15-21t+9y$$
so $$x = 5-7t+3y$$
thus $$A = 5-7t+3y -y+2+3t-2y=-4t+7$$
Since $2y+z\geq 3\implies 3t+2\geq 3\implies t\geq 1$
Since $z\leq 4$ and $y\leq 5$ we get $3t+2\leq 14$ so $t\leq 4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2934223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Integral $\int_0^1 \ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{1-x^2}{1+x^2}\right)\frac{dx}{x}$ Greetings I saw here (among the last integrals) that: $$\int_0^1 \ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{1-x^2}{1+x^2}\right)\frac{dx}{x}=\pi C$$
Where $C$ is Catalan's constant. Did this integral appear here befor... | It's always the same story,
$$J=\int_0^1 \ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{1-x^2}{1+x^2}\right)\frac{dx}{x}$$
Perform the change of variable $y=\dfrac{1-x}{1+x}$,
$$
J =2\int_0^1 \frac{\ln\left(\frac{x^2+1}{2x}\right)\ln x}{x^2-1}\,dx
$$
For $x\in [0;1]$ define the function $R$,
\begin{align}R(x)&=\int_0^x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2934903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 0
} |
How to show that $\dfrac{n^3 + 2n}{3}$ is an integer Show that for each natural number $n, \dfrac{n^3 + 2n}{3}$ is an integer
My try:
Let P(n) be the statement that $n^3 + 2n$ is divisible by $3$.
Base step:
When $n = 0$ we have $0^3 + 0 = 0 = 3 \times 0$
So, the base case true.
Inductive hypothesis:
Assume that $P(k)$... | $n^3 +2n=n(n^2+2) $, now, check that all square numbers will leave a remainder $0$(means divisible by $3$) or, $1$ when we divide $n$ by $3$. Hence, when $3|n$ we have $3|n(n^2+2)$ and when $n=3k+1$ or $3k+2$, then $3|n^2+2\implies 3|n(n^2+2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2936874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Is $(q^k n^2 \text{ is perfect }) \iff (D(q^k)D(n^2) = 2s(q^k)s(n^2))$ only true for odd perfect numbers $q^k n^2$? (Preamble: This question is an offshoot of this earlier MSE post.)
The title says it all.
Is $\bigg(q^k n^2 \text{ is perfect }\bigg) \iff \bigg(D(q^k)D(n^2) = 2s(q^k)s(n^2)\bigg)$ only true for odd per... | MY ATTEMPT
For even perfect numbers $2^{p-1}(2^p - 1)$, I get
$$D(2^p - 1)D(2^{p-1}) = (2(2^p - 1) - (2^p))(1) = 2^{p+1} - 2 - 2^p = 2^p - 2$$
$$2s(2^p - 1)s(2^{p-1}) = 2(2^p - (2^p - 1))(2^p - 1 - 2^{p-1}) = 2(1)(2^{p-1} - 1) = 2^p - 2.$$
Thus, the equation
$$D(2^p - 1)D(2^{p-1}) = 2s(2^p - 1)s(2^{p-1}) = 2^p - 2$$
is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2939057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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The eigenvalues of $\begin{pmatrix}0&A\\A^*&0\\ \end{pmatrix}$ are the singular values of $A$ along with the negative signs.
The eigenvalues of $\begin{pmatrix}0&A\\A^*&0\\ \end{pmatrix}$ are the singular values of $A$ along with the negative signs.
Here $A$ is an $n \times n$ matrix, has $n$ singular values. Here w... | Let $A$ be a $(n \times n)$-matrix and let $B$ be the matrix $$B = \begin{pmatrix} 0 & A \\ A^* & 0
\end{pmatrix}.$$
Let $A = U_1 \Sigma U_2^*$ be a singular value decomposition of $A$. Then we can decompose $B$ into a product $B = USU^*$ where $U$ is unitary and $S$ has a simpler structure than $B$. Let us show this e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2942203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Prove $\binom{n}{2}+\binom{n}{3}\cdot7+\binom{n}{4}\cdot7^2+....+7^{n-2}\in \mathbb{Z}$ for all $n\geq 2$
If $X=\{8^n-7n-1:n\in \mathbb{N}\}$ and $Y=\{49(n-1):n\in \mathbb{N}\}$ then show that $X⊂Y$
While solving the problem
$$8^n-7n-1=(1+7)^n-7n-1\\
=49\bigg[\binom{n}{2}+\binom{n}{3}\cdot7+\binom{n}{4}\cdot7^2+...+\... | Another way to expand using
$$a^n-b^n=(a-b)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+...+a^2b^{n-3}+ab^{n-2}+b^{n-1}\right) \tag{1}$$
is
$$8^n-7n-1=8^n-1-7n=\color{red}{(8-1)}\left(8^{n-1}+8^{n-2}+...+8+1\right)-\color{red}{7}n=\\
7\left(8^{n-1}+8^{n-2}+...+8+1-\color{blue}{n}\right)=...$$
there are $\color{blue}{n}$ terms $8^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2944772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Least value of $L$ for $\sqrt{x^2+ax}-\sqrt{x^2+bx}0$
If $a$ and $b$ are positive real numbers such that $a-b=2$ , then the smallest value of the constant $L$ for which $\sqrt{x^2+ax}-\sqrt{x^2+bx}<L$ for all $x>0$.
This question is similar to this one, but I don't want to apply the concept of limit. I want to use ap... | We have $$\sqrt{x^2+ax}-\sqrt{x^2+bx}=\frac{(x^2+ax)-(x^2+bx)}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}=\frac{a-b}{\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}}$$
for all $x>0$. Because $\sqrt{1+\frac{a}{x}}>1$ and $\sqrt{1+\frac{b}{x}}>1$ for all $x>0$, we get
$$\sqrt{x^2+ax}-\sqrt{x^2+bx}< \frac{a-b}{2}.$$
This shows that $L\leq \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2945698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
what is the characteristic polynomial and minimal polynomial of A and B? What is the characteristic polynomial and minimal polynomial of $A$ and $B$ ?
$A=\begin{pmatrix} a & 1 & 0 & 0 \\ 0 & a & 0 & 0 \\ 0 & 0 & a & 0 \\ 0 & 0 & 0 & a\end{pmatrix},B=\begin{pmatrix} a & 1 & 0 & 0 \\ 0 & a & 0 & 0 \\ 0 & 0 & a & 1 \... | Set
$N_A = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}; \tag 1$
then
$N_A^2 = 0; \tag 2$
further note that
$A - aI = N_A; \tag 3$
thus
$(A - aI)^2 = 0, \tag 4$
which means that
$m_A = (x-a)^2, \tag 5$
since $A$ can satisfy no polynomial of degree 1; indeed,
$cA + dI ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2946605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Probability that exactly 2 balls are white
What I did.
I let $X$ be number of withdraws before $x$ white balls. We can call our succes in this case to be gettin a white ball and probability is of course $p = \frac{3}{6} = \frac{1}{2}$. So we see $X$ is negative binomial r.v with $n=4$ trials. So,
$$ P(X=2) = { 4 - 1 ... | $P(2W|4) = \binom{4}{2}\cdot (\frac{1}{2})^4$
$ = 6\cdot \frac{1}{16} = \frac{3}{8}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2950889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Finding values to make an equation with a $x^4$ into a perfect square For what values of p and q is the expression $$x^4+ 6x^3+ 13x^2+ px + q$$ a perfect
square?
The answer I got didn't seem correct so if someone break it down to me how about going this problem it would be appreciated
| Hint: write like this
$$x^4+ 6x^3+ 13x^2+ px + q= (x^2+ ax +b)^2$$
$$=x^4+ 2ax^3+ (2b+a^2)x^2+ 2abx + b^2$$
So $a =3$ and $2b+9 = 13$ so $b= 2$. Then $p = 2ab =12$ and $q=b^2 =4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2951852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
continued fraction development of the numbers of the following form, $\sqrt{n^2-1}$
Determine the continued fraction development of the numbers of the
following form, $\sqrt{n^2-1}$, with $n>1$ an integer.
I wasn't sure how to tackle this, so I just tried to write it out and tried to find $a_1...a_n$.
$$\alpha_0=\s... | $a_0=\lfloor \sqrt {n^2-1} \rfloor = n-1$
The repeating part is $1,2(n-1)$. We know when we get to $2a_0$ in a square root we have hit the repeat. You can write
$$\sqrt{n^2-1}=n-1+(\sqrt {n^2-1}-(n-1))\\=n-1+\frac 1{\frac 1{\sqrt {n^2-1}-(n-1)}}\\=n-1+\frac 1{\frac {\sqrt{n^2-1}+n-1}{n^2-1-(n-1)^2}}\\=n-1+\frac 1{\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2957878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to get the Taylor polynomial arctan with error? We have to find Taylor polynomial $P_2(x)$ of $\arctan(x)$ of grade 2 from point $x = 0$
My answer was
If $f(x) = \arctan(x)$ then $f(0)=0$
And $f'(x)= \frac{1}{1+x^2}$ where $f'(0) = 1$
And $f''(x) = \frac{-2x}{(1+x^2)^2}$ where $f''(0) = 0$
And $P_2(x) = x - \frac{1... | A simpler approach is as follows. We know that if $f(x) = \arctan(x)$, then $f' = (1 + x^2)^{-1}$. Thus
$$ \arctan(x) = \int_0^x \frac{1}{1 + t^2} dt + C$$
for some constant $C$. As $\arctan(0) = 0$, we even have that $C = 0$.
A Taylor expansion for $(1 + t^2)^{-1}$ is straightforward, as this is a geometric series
$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2959258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Series of Nested Radicals
I can't seem to find a way, squaring the expression would make more terms and would make it harder, I guess there must be something to do with the first and last terms as they sum to 100? or maybe difference of to squares, but i can't solve it.
|
Useful fact:
$$
\sqrt{a\pm \sqrt{b}}=
\sqrt{\frac{a+\sqrt{a^2-b}}{2}} \pm
\sqrt{\frac{a-\sqrt{a^2-b}}{2}} \tag{1}$$
Proof:
\begin{align}
RHS^2 &=
\frac{a+\sqrt{a^2-b}}{2}+\frac{a-\sqrt{a^2-b}}{2} \pm 2
\sqrt{\left( \frac{a+\sqrt{a^2-b}}{2} \right)
\left( \frac{a-\sqrt{a^2-b}}{2} \right)} \\
&= a \pm 2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2960855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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How to prove $\sqrt{2} + \sqrt[3]{2}$ is an algebraic number I need to show that this is an algebraic number by showing that it is a solution to:
$x^{6} - 9x^{4} - 4x^{3} + 27x^{2} + 36x -23 = 0.$
I'm really struggling with this one, I tried squaring and cubing $\sqrt{2} + \sqrt[3]{2}$, but I only end up with more squa... | Set
$\alpha = \sqrt 2 + \sqrt[3] 2; \tag 1$
then,
$\alpha - \sqrt 2 = \sqrt [3] 2; \tag 2$
$(\alpha - \sqrt 2)^3 = 2; \tag 3$
$\alpha^3 - 3 \sqrt 2 \alpha^2 + 6 \alpha - 2\sqrt 2 = 2; \tag 4$
$-\sqrt 2(2 + 3\alpha^2) = 2 - 6\alpha - \alpha^3; \tag 5$
$2(2 + 3\alpha^2)^2 = (2 - 6\alpha - \alpha^3)^2. \tag 6$
If all we w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$4ac-b^2\leq 3a(a+b+c)$ in quadratic Suppose that the polynomial $(b+c)x^2+(a+c)x+(a+b)$ doesn't have real roots, where $a,b,c\in\mathbb{R}$. Prove that $4ac-b^2\leq 3a(a+b+c)$.
The quadratic not having real roots means that $$(a+c)^2-4(b+c)(a+b)<0$$ which translates to $(a^2+2ac+c^2)-4(b^2+ab+ac+bc)<0$, or
$$a^2+c^2-... | in standard order $a^2, b^2, c^2, bc, ca, ab,$ you are given
$$ -a^2 + 4b^2 - c^2 + 4bc +2ca+ 4 ab > 0 $$
The following is positive semidefinite (rank one Hessian matrix)
$$ 25a^2 + 4b^2 + c^2 - 4bc -10ca+ 20 ab \geq 0 $$
It is, in fact, simply $(5a+2b-c)^2$
Add to get
$$ 24a^2 + 8b^2 + 0 c^2 + 0bc -8ca+ 24 ab > 0 $$
o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
find $\cos\theta$ if $\sin\theta=\frac{3}{4}$ and $\tan\theta=\frac{9}{2}$
If $\sin\theta=\frac{3}{4}$ and $\tan\theta=\frac{9}{2}$ then find $\cos\theta$
The solution is given in my reference as $\frac{1}{6}$.
$$
\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\frac{9}{16}}=\sqrt{\frac{7}{16}}=\frac{\sqrt{7}}{4}
$$
$$
\cos... | Because you have two different problems. Obviously (from your calculations which are correct) it can't be $\sin\theta=\frac{3}{4}$ and $\tan\theta=\frac{9}{2}$ at the same time.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Limit $\lim_\limits{x\to0}{\tan{x}-\sin{x}\over x^3}$
What is the limit of:
$$\lim_\limits{x\to0}{\tan{x}-\sin{x}\over x^3}$$
So what I tried is:
$$\lim_\limits{x\to0}{{\sin{x}\over\cos{x}}-\sin{x}\over x^3}=\lim_\limits{x\to0}{{\sin{x}}({1\over\cos x}-1)\over x\cdot x^2}$$
From here, using the rule $\lim_\limits{x... | $$\frac{\tan x - \sin x}{x^3} = \frac{\sin x - \sin x\cos x}{x^3\cos x} = \frac{\sin x}x \cdot\frac{1-\cos x}{x^2}\cdot\frac 1{\cos x}\to 1\cdot \frac 12\cdot \frac 1{\cos 0} = \frac 12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2964650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 3
} |
Prove $\sum_{1}^{\infty} \frac{1}{\sqrt{n}( n + \sqrt{n})} \lt 2 $
Prove that $$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}( n + \sqrt{n})} \lt 2$$
I have found that $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}( n + \sqrt{n})} < \pi / 2 $ with integrating from $1$ to infinity but integral isn’t allowed.
| $$
\frac{1}{n\sqrt n} = \frac{1}{\sqrt n(n+\sqrt n)}+\frac{1}{n(n+\sqrt n)}
$$
hence
$$
\frac{1}{\sqrt n(n+\sqrt n)}<\frac{1}{n^{\frac 32}}
$$
NOTE
$$
\sum_{n=3}^{\infty} \frac{1}{n^{\frac 32}} < 1.24
$$
now adding $\frac{1}{\sqrt 1(1+\sqrt 1)}+\frac{1}{\sqrt 2(2+\sqrt 2)}+\sum_{n=3}^{\infty} \frac{1}{n^{\frac 32}} < ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2965654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Help find the mistake in this problem of finding limit (using L'Hopital)
Evaluate $$\lim_{x \to 0} \left(\frac{1}{x^2}-\cot^2x\right).$$
Attempt
\begin{align*}
&\lim_{x \to 0} \left(\frac{1}{x^2}-\cot^2x\right)\\
= &\lim_{x \to 0} \left(\frac{1}{x}-\cot{x}\right)\left(\frac{1}{x}+\cot{x}\right)\\
= &\lim_{x \to 0} \l... | Similar work:
$$\begin{align}\lim_{x \to 0} \left(\frac{1}{x^2}-\cot^2x\right)&=\lim_{x \to 0} \left(\frac{1}{x^2}-\frac{1-\sin^2x}{\sin^2x}\right)=\\
&=\lim_{x \to 0}\left(\frac{1}{x^2}-\frac{1}{\sin^2x}+1\right)=\\
&=\lim_{x \to 0}\left(\frac{(\sin x-x)(\sin x+x)}{x^2\sin^2x}\right)+1=\\
&=\lim_{x \to 0}\left(\frac{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2968842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
What is the coefficient of the term $x^3y^5$, as a result of the binomial expansion of the following term? We have the term
$(1+xy+y^2)^n$
If we expand it using the binomial theorem, why is the factor of the term $x^3y^5$ the following: $4{n\choose 4}$? (The binomial coefficient n choose 4 multiplied by 4)
| Yes, for n = 5, use the binomial theorem by substituting $xy + y^2$ for "$a$" and then expand $(1 + a)^5 = 1 + 5a + 10a^2 + 10a^3 + 5a^4 + a^5$.
Then substitute back $xy + y^2$ for $a$.........
Here, the only term which will give $x^3y^5$ is $5a^4$ whereby again using the binomial theorem on $5(xy + y^2)^4$ we get........ | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2969497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Inequality proof as a part of induction As a part of an induction proof, I need to prove that if $\frac{3x+2}{x+2} < x$, then $\frac{3x+2}{x+2} > \frac{11x+10}{5x+6}$. I need to prove this for $x>0$ if that matters so basically $x$ has to be greater than 2. I have tried manipulating this inequality in a number of ways ... | Hint: For $$x>0$$ we get that $$\frac{3x+2}{x+2}<x$$ is equivalent to $$0<x^2-x-2$$ and this is $$0<(x+1)(x-2)$$ so it must be $$x>2$$
and now we will Show that
$$\frac{3x+2}{x+2}>\frac{11x+10}{5x+6}$$ is true for $$x>2$$
this is $$(3x+2)(5x+6)>(11x+10)(x+2)$$
Expanding and collecting like terms we get
$$4x^2-4x-8>0$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2970681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
CAN and BAN estimators of $\sigma^2$ Let $X_1,X_2,..X_n \sim^{\text{i.i.d}} N(0,\sigma^2)$
Show that $T_n=\frac{1}{n}\sum_{i=1}^{n} X_i^2$ is a Consistent and Asymptotically Normal estimator(CAN) as well as the Best Asymptotic Normal Estimator(BAN) of $\sigma^2$.
Indeed, $E(T_n)=\sigma^2$ and $V(T_n)=\frac{2 \sigma^4}{... | CRB
Let's write down the PDF
\begin{equation}
f_{X_i}(x_i \vert \sigma^2) \sim \frac{1}{\sqrt{2\pi \sigma^2}}\exp(-\frac{x_i^2}{2\sigma^2} )
\end{equation}
Assuming independence, we have that
\begin{equation}
L(\sigma^2) = f(x_1 \ldots x_n \vert \sigma^2)
=
f(x_1 \vert \sigma^2) \ldots f(x_n \vert \sigma^2)
=
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2972400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Given that $f(x) = 1 - 3/(x+2) + 3/(x+2)$ while $x ≠- 2$, show that $f(x) = (x^2 + x + 1)/(x+2)^2$ What type of problem is this and what are the first step(s) needed to tackle it?
Given that:
$$f(x) = 1-\frac{3}{x + 2}+\frac{3}{(x +2)^2}, \quad \text{with }x ≠ -2$$
Show that:
$$f(x) = \frac{x^2 + x +1}{(x + 2)... | $$f(x) = 1-\frac{3}{x + 2}+\frac{3}{(x +2)^2}; x ≠ 2$$
$$\implies f(x) = \frac{(x+2)^2}{(x+2)^2}-\frac{3(x+2)}{(x + 2)^2}+\frac{3}{(x +2)^2}$$
$$\implies f(x) = \frac{(x+2)^2-3(x+2)+3}{(x+2)^2}$$
Just expand and then add up the values in the numerator...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2976261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to show that $x_{n+1} = \frac{x_n^4 + 1}{5x_n}$ bounded below and above by $1\over 5$ and $2$
Given a sequence
$$
\begin{cases}
x_{n+1} = \frac{x_n^4 + 1}{5x_n} \\
x_1 = 2 \\
n \in \mathbb N
\end{cases}
$$
Prove it has lower bound at $1\over 5$ and upper bound at $2$
I've tried to find a closed form for the ... | hint
write $x_{n+1}$ as
$$x_{n+1}=\frac{x_n^3}{5}+\frac{1}{5x_n}$$
and prove it by induction.
there are two fixed points satisfying
$$L^4-5L^2+1=0$$
$$L_1^2=\frac{5-\sqrt{21}}{2}\approx 0.7$$
$$L_2^2=\frac{5+\sqrt{21}}{2}\approx 4.7$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2978064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Systems of Equations degree 2 Given the system
$$\begin{aligned}
x^2+y^2+\frac{\sqrt 3}{2}xy&=32\\
x^2+z^2+\frac{1}{2}xz&=16\\
y^2+z^2&=16,\end{aligned}$$
find the value of $(xy+\sqrt{3}xz+2yz).$
The answer is $32,$ so I think there will be a really nice solution to this.
I tried completing the square for each pair $... | I guess the two factors (1/2) shouldn't be there, which mean we have instead
\begin{align}
x^2+y^2+\sqrt{3}xy&=32\tag{A}\\
x^2+z^2+xz&=16\tag{B}\\
y^2+z^2&=16\tag{C}.
\end{align}
The rationale being that it gives a much nicer set of coefficients when completing the square
\begin{align}
(x+\frac{\sqrt3}2y)^2+\frac14y^2&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2978918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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If $x,y,z>0$ and $x+y+z=1$ then $\frac{xyz}{(1-x)(1-y)(1-z)}\le\frac{1}{8}$ If $x,y,z>0$ and $x+y+z=1$,
then: $$\frac{xyz}{(1-x)(1-y)(1-z)}\le\frac{1}{8}$$
$$\frac{xyz}{(1-x)(1-y)(1-z)}=\frac{x}{(1-x)}\frac{y}{(1-y)}\frac{z}{(1-z)}$$
Let $\frac{x}{(1-x)}=a$, $\frac{y}{(1-y)}=b$, $\frac{z}{(1-z)}=c$
I am stuck here.
| You have: $$x+y+z=1$$
So, write: $$1-x=y+z$$
Using inequality $AP\geq GP$, $$y+z\geq 2\sqrt{yz}$$
So, $$(1-x)\geq2 \sqrt{yz}$$ Similarly, $$(1-y)\geq2 \sqrt{xz} \hspace{0.5cm} {and} \hspace{0.5cm} (1-z)\geq2 \sqrt{xy}$$
Multiplying all: $$(1-x)\cdot (1-y) \cdot (1-z) \geq 8 \cdot xyz$$
Divide:$$\frac{(1-x)\cdot (1-y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2980320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Probability mass function of max and min function
You repeatedly draw a number at random from the numbers $1,2,...,10$.
Let $X$ be the number of drawings until the number $1$ appears and $Y$
be the number of drawings until the number $10$ appears. What is the
joint probability mass function of $X$ and $Y$? What are th... | $X$ and $Y$ are not independent since if $P(X=1,Y=1) =0$ but $P(X=1)>0$ and $P(Y>1)>0$.
if $1\le x < y$,
$$Pr(X=x, Y=y) = (0.8)^{x-1}(0.1)(0.9)^{y-x-1}(0.1)=0.8^{x-1}(0.9)^{y-x-1}(0.1)^2$$
That is the first $x-1$ times can be anything besides the two numbers, followed by $1$, then the following $y-x-1$ times we can ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find all $2\times 2$ matrices that commute with $AX = XA$? Find all $2\times 2$ matrices that commute with $$ A = \left( \begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} \right)$$ where $AX = XA$.
Solution and ask:
$$ X = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$$
$$AX = \left( \begin{array}{cc} 1\cdot ... | Here is a synthetic solution. Let $V$ denote the vector space of $2\times2$ matrices. Let $T:V\to V$ be the linear operator $TX=AX-XA$. The kernel of $T$ consists of eigenvectors of $T$ with eigenvalue $0$. Observe that $A$ is diagonalizable with eigenvalues $0$ and $1$ (well, this is obvious since $A$ is triangula... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2984447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Simplifying fraction containgin square root Say we have an expression like
$$\frac{\sqrt{\frac{k}{g}}}{k}=\frac{1}{\sqrt{gk}}$$
How do we get from the left hand side, to the right hand side? If we simplify the square root
$$\sqrt{\frac{k}{g}}=\frac{\sqrt{k}}{\sqrt{g}}$$
Then
$$\frac{\sqrt{k}}{\sqrt{g}}\times \frac{1}{k... | $$\frac{\sqrt{\frac{k}{g}}}{k}=\frac{\sqrt \frac kg}{\sqrt\frac{k^2}{1}}=\sqrt\frac{k \over g}{k^2 \over 1}=\sqrt\frac{k}{gk^2}=\sqrt\frac{1}{gk}={\sqrt{1} \over
\sqrt{gk}}={1 \over \sqrt{gk}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2985880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Struggle with finding equality condition in an AM-GM inquality problem Given $0<x,y,z<1$. Prove that
$$\dfrac{1-x}{1+y+z}+\dfrac{1-y}{1+z+x}+\dfrac{1-z}{1+x+y}\geq 3(1-x)(1-y)(1-z).$$
I've thought that the equality holds when $x=y=z$ and I'm thinking ways to use AM-GM inequality. However, it turns out that
$$3\dfrac{1-... | AM-GM seems like an unlikely choice in this problem. I am presenting a different approach.
Note that
\begin{align}\frac{1-x}{1+y+z}-(1-x)(1-y)(1-z)&=\frac{1-x}{1+y+z}\big(1-(1-y-z+yz)(1+y+z)\big)
\\&=\frac{1-x}{1+y+z}\big(y^2(1-z)+z^2(1-y)+yz\big)\geq 0\end{align}
for all $x,y,z\in[0,1]$. Similarly,
\begin{align}\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2987384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
if 1, $\alpha_1$, $\alpha_2$, $\alpha_3$, $\ldots$, $\alpha_{n-1}$ are nth roots of unity then...
if 1, $\alpha_1$, $\alpha_2$, $\alpha_3$, $\ldots$, $\alpha_{n-1}$ are nth roots of unity then
$$\frac{1}{1-\alpha_1} + \frac{1}{1-\alpha_2} + \frac{1}{1-\alpha_3}+\ldots+\frac{1}{1-\alpha_n} = ?$$
Now this has the sol... | The sum of the roots of $a_nx^{n}+a_{n-1}x^{n-1}+\cdots +a_0$ is $-\frac {a_{n-1}} {a_n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2988339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Finding a quadratic equation using roots
If $x_1$ and $x_2$ are the roots of
$$ax^2+bx+c=0$$
then $x_1^3$ and $x_2^3$ are the roots of which equation?
So I tried by solving this for $x_{1/2}$ so I could change it in $(x-x_1^3)(x-x_2^3)$
$x_{1/2}=\large{-b\pm{\sqrt{4ac}}\over2a}$
and from here:
$$\begin{align}x_1... | HINT
We have
$$(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+x_1x_2$$
$$(x-x_1^3)(x-x_2^3)=x^2-(x_1^3+x_2^3)x+x_1^3x_2^3$$
and
$$x_1^3x_2^3=(x_1x_2)^3$$
$$x_1^3+x_2^3=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2990095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Dividing numbers with exponents and powers Can someone please help me simplify the below problem. I am trying to help my niece
$${24C^4 \over 4C^{12}}$$
I believe that the simplified answer is $6/C^8$ because $4$ goes into $24$ $6$ times and $6$ remains as the numerator. Also, you need to subtract $4$ from $C^4$ which ... | One approach I've used for someone beginning to work with this type of situation is the following, although I'd probably replace it with something similar and simpler, such as $\frac{24C^2}{4C^5},$ to actually illustrate the idea.
$$ \frac{24C^4}{4C^{12}} \; = \; \frac{24 \cdot C^4}{4 \cdot C^{12}} \; = \; \frac{24}{4}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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$a\mid b^2, b\mid c^2, c\mid a^2\Longrightarrow abc\mid (a+b+c)^k$ Is it true that if $a,b,c$ are positive integers such that $a\mid b^2, b\mid c^2, c\mid a^2$, then $abc$ always divides $(a+b+c)^6$?
If not (I couldn’t find a counterexample), then $abc\mid (a+b+c)^7$ is always true? Or in general, find the minimum valu... | If you expand out $(a + b + c)^6$, then all terms are multiples of some $a^ib^jc^k$, with $i+j+k=n$. Now, at least one of $i$, $j$, and $k$ is at least $\frac{n}{3}$.
Now, we have $abc|a^3b|a^3c^2|a^7$, and symmetrically $abc|b^7$, $abc|c^7$. Thus, we have an upper bound on the solution: $abc|(a+b+c)^{21}$.
To bring t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2997428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
$2^{2n+1} +1$ divisible by $3$ $2^{2n+1} +1$ divisible by $3$
Inducción over n
Let $n=1$, then: $$2^{2(1)+1}+1=2^{3}+1=9$$
It works!
Hip. $$2^{2n+1} +1=3k$$
Then we have to show: $$2^{2n+3} +1=3k$$
Any guess?
| HINT
$$2^{2n+3} +1=2^2 \cdot 2^{2n+1} +1\stackrel{Hyp.}=2^2(3k-1)+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2998933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find a parabola touches the line $y=x$ and $y=-x$ at $A(3,3)$ and $B(1,-1)$
If a parabola touches the line $y=x$ and $y=-x$ at $A(3,3)$ and $B(1,-1)$, then find the focus, axis of the parabola and its directrix.
What I thought: Since the 2 tangents are perpendicular,the origin must lie on the directrix and the line j... | The most general form of a parabola is $$Ax^2+Bxy + Cy^2 + Dx +Ey +F=0$$
where $$B^2 = 4AC \tag{1}$$.
Letting $(x,y)=(3,3)$, we get $$9A + 9B + 9C + 3D + 3E + F = 0 \tag{2}$$
Letting $(x,y)=(1, -1)$, we get $$A - B + C + D - E + F = 0 \tag{3}$$
\begin{align}
Ax^2+Bxy + Cy^2 + Dx +Ey +F=0
&\implies 2Ax + By + Bxy'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2999165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Number of unique ways to select $3$ balls from $4$ red balls, $2$ green balls and $1$ blue ball Suppose there is a jar with $7$ balls: $4$ red, $2$ green and $1$ blue.
In how many unique ways can we choose $3$ balls?
Just by considering all the possibilities, I found out the answer is $6$. Namely:
*
*red, red, red... | Consider the polynomial
$$\eqalign{p(x)&:=(1+x+x^2+x^3+x^4)(1+x+x^2)(1+x)\cr &=(1-x^5)(1-x^3)(1-x^2)(1-x)^{-3}\cr &=(1-x^5)(1-x^3)(1-x^2)\sum_{k\geq0}{2+k\choose k}x^k\ .\cr}\tag{1}$$
For each admissible choice of $r\in[0 .. 4]$ red balls, $g\in[0 .. 2]$ green balls, and $b\in[0 .. 1]$ blue balls the expansion of thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3001464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Help with a system of inequalities with absolute values I'm trying to solve this system on inequalities
$$
\left\{
\begin{array}{c}
|x-3|<2x \\
|2x+5|>3
\end{array}
\right.
$$
The steps I'm taking are:
Finding the absolute values sings, so for
$x-3 \geq 0$ we have $x \geq 3$ therefore
$$|x-3| =
\left\{
\begin{a... | Since $0\le\left|x-3\right|$ and $\left|x-3\right|\le2x$ we conclude $2x\ge0$ and then $x\ge0$.
From $\left|x-3\right|<2x$ and since $x\ge0$, we get $-2x< x-3<2x$ and therefore we have $\left\{\begin{array}{c}x-3<2x \\x-3>-2x\end{array}\right.$ and therefore $\left\{\begin{array}{c}x>-3 \\3x>3\end{array}\right.$ and th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3005222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the limit of $\lim_{n\to \infty} (1 - \frac{1}{4})(1 - \frac{1}{9})(1 - \frac{1}{16}) \cdots (1 - \frac{1}{(n+1)^2})$? What is the evaluation of the following infinite series?
$$\lim_{n\to \infty} \left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right)\left(1 - \frac{1}{16}\right) \cdots \left(1- \frac{1}{(n+1... | $\frac{1\times 3}{2\times 2}\cdot\frac{2\times 4}{3\times 3}\cdot\frac{3\times 5}{4\times 4}\cdot\frac{4\times 6}{5\times 5}\cdots$
Do you see it?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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solve $\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35}$ I need to find
$$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35}$$
Looking at the graph, I know the answer should be $\frac{20}{17}$, but when I tried solving it, I reached $0$.
Here are the two ways I approached this:
WAY I:
$$\lim_{x\rightarrow -5} \frac{2... | Hint: Try factorization!
$$
\frac{2x^2-50}{2x^2+3x-35}=\frac{2(x^2-25)}{(1/2)(4x^2+6x-70)}=\frac{4(x-5)(x+5)}{(2x+10)(2x-7)}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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Proving $(n+1)^2+(n+2)^2+...+(2n)^2= \frac{n(2n+1)(7n+1)}{6}$ by induction My question: $(n+1)^2+(n+2)^2+(n+3)^2+...+(2n)^2= \frac{n(2n+1)(7n+1)}{6}$
My workings
*
*LHS=$2^2$ =$4$ RHS= $\frac{24}{6} =4 $
*$(k+1)^2+(k+2)^2+(k+3)^2+...+(2k)^2= \frac{k(2k+1)(7k+1)}{6}$
*LHS (subsituting $n= k+1$)----> $(k+2)^2+(k+3)^... | This is 2nd tough-degree exercise in induction. Observe that both the first and last elements change, so if $\;k=n\;$ we get
$$(n+1)^2+(n+2)^2+\ldots+(2n)^2=\frac{n(2n+1)(7n+1)}6\;\;(**)$$
then for $\;k=n+1\;$ we get
$$(n+1+1)^2+(n+3)^2+\ldots+(2n)^2+(2n+1)^2+\overbrace{(2n+2)^2}^{=(2(n+1))^2}=\frac{(n+1)(2n+3)(7n+8)}6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3011450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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$\lim_{x\to -\infty} x+\sqrt{x^2-3x}$ Hey so I'm having a bit of a hard time understanding this one.
$\lim_{x\to -\infty} x+\sqrt{x^2-3x}$
1) $x+\sqrt{x^2-3x}$ * $(\frac{x-\sqrt{x^2-3x}}{x-\sqrt{x^2-3x}})$
2) $\frac{x^2-(x^2-3x)}{x-\sqrt{x^2-3x}}$
3) $\frac{3x}{x-\sqrt{x^2(1-\frac{3}{x}})}$
4) $\frac{3x}{x-\sqrt{x^2}*... | $a^2-b^2=(a-b)(a+b).$
$y:=-x$ , and consider $\lim y \rightarrow + \infty.$
$\sqrt{y^2+3y}-y= \dfrac{(y^2+3y)-y^2}{\sqrt{y^2+3y}+y}=$
$\dfrac{3y}{\sqrt{y^2+3y}+y}= \dfrac{3y}{y(\sqrt{1+3/y}+1)}$
$=\dfrac{3}{\sqrt{1+3/y}+1}.$
Take the limit.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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General solution of $\sin 2x+\cos x=0$
Solve the trigonometric equation $\sin 2x+\cos x=0$
My Attempt
$$
2\sin x\cos x+\cos x=0\implies\cos x=0 \text{ or }\sin x=\frac{-1}{2}=\sin\frac{-\pi}{6}\\
x=(2n+1)\frac{\pi}{2} \text{ or }x=n\pi+(-1)^n\frac{-\pi}{6}\\
x=n\pi+\frac{\pi}{2} \text{ or }x=2n\pi-\frac{\pi}{6}\text{... | $$\sin(2x)+\cos(x)=0$$
$$2\sin(x)\cos(x)+\cos(x)=0$$
$$\cos(x)\left(2\sin(x)+1\right)=0$$
so you have two sets of solutions:
$$\cos(x)=0,\,\sin(x)=-\frac{1}{2}$$
EDIT:
firstly they have:
$$x=2n\pi-\pi/2=\pi(2n-1/2)$$
and you have:
$$x=(2n+1)\pi/2$$
let:
$$\pi/2(4n-1)=\pi/2(2m+1)$$
so:
$$4n-1=2m+1$$
$$m=(4n-2)/2=2n-1$$
... | {
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"url": "https://math.stackexchange.com/questions/3013564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove for all $a, b, c \in \mathbb {R^+}$ ${a\over\sqrt{a+b}} + {b\over\sqrt{b+c}} + {c\over\sqrt{c+a}} \gt \sqrt{a+b+c}$ is true
Prove for all $a, b, c \in \mathbb {R^+}$ $${a\over\sqrt{a+b}} + {b\over\sqrt{b+c}} + {c\over\sqrt{c+a}} \gt \sqrt{a+b+c}$$ is true
I have been studying about inequalities and how to prove... | You may proceed as follows using convexity of $\frac{1}{\sqrt{x}}$:
*
*$\sum_{k=1}^3 \lambda_k f(x_k) \geq f\left( \sum_{k=1}^3 \lambda_k x_k\right)$ with $\sum_{k=1}^3 \lambda_k = 1$, $\lambda_1,\lambda_2,\lambda_3 \geq 0$
$$
\begin{eqnarray*} {a\over\sqrt{a+b}} + {b\over\sqrt{b+c}} + {c\over\sqrt{c+a}}
& = & (a+b+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3013932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Values of $a$ for which $p(x)$ has a complex roots? Let $p(x) = x^4 + ax^3 + bx^2 + ax + 1$. Suppose $x = 1$ is a root of $p(x)$, then find the range of values of $a$ such that $p(x)$ has complex (non-real) roots?
My approach: Using the fact that $1$ is a root, I am able to deduce $b = -2(a+1)$. Then I wrote $$p(x) = (... | When a polynomial is "palindromic", reading the same way forwards and backwards, it must also have palindromic factors. Thus
$x^4+ax^3+bx^2+ax+1=(x^2+cx+1)(x^2+dx+1)$
If $x=1$ is to be a root then it must give $c=-2$ or $d=-2$ in one of the factors, as no other choice has the root $x=1$. Taking $c$ as $-2$ we then ge... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Little inequality After deconditioning and calculating some stuff I obtained this inequality, which looks simple but I can't get to group the termens conveniently.
$x^3 + y^3 + z^3 + 6(x^2z + y^2x + z^2y) ≥ 12xyz + 3(x^2y + y^2z+z^2x)$ for $x,y,z≥0$.
I tried breaking it in more inequalities , but the last one is not al... | Buffalo Way helps here very well.
Indeed, let $x=\min\{x,y,z\}$, $y=x+u$ and $z=x+v$.
Thus,
$$\sum_{cyc}(x^3+6x^2z-3x^2y-4xyz)=6(u^2-uv+v^2)x+u^3-3u^2v+6uv^2+v^3\geq0.$$
It's interesting that your first step gives a right inequality because after using AM-GM
$$\sum_{cyc}4x^2z\geq12xyz$$ it's enough to prove that
$$\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3020003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality for $a,c,b$ positive real numbers with $a + b + c = 1$ Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that $$\frac{1}{2a^2+bc}+\frac{1}{2b^2+ac}+\frac{1}{2c^2+ab}\geq \frac{1}{\sqrt{a^2-ab+b^2}}+\frac{1}{\sqrt{b^2-bc+c^2}}+\frac{1}{\sqrt{c^2-ca+a^2}}$$
I have a solution
$$
\left(2 a^{2}+b c... | Another way, but in one line :)
$$\sum_{cyc}\frac{1}{2a^2+bc}-\sum_{cyc}\frac{1}{\sqrt{a^2-ab+b^2}}=$$
$$=\sum_{cyc}\left(\frac{a+b+c}{2a^2+bc}-\frac{2}{b+c}\right)+\sum_{cyc}\left(\frac{2}{a+b}-\frac{1}{\sqrt{a^2-ab+b^2}}\right)=$$
$$=\sum_{cyc}\frac{b^2+c^2+ab+ac-4a^2}{(2a^2+bc)(b+c)}+\sum_{cyc}\frac{2\sqrt{a^2-ab+b^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Series convergence: $\sin (n \frac{\pi}{2})$
Determine whether the following series :
$$\sum_{n=1}^\infty \sin \left(\frac{n\pi}{2}\right) \frac{n^2+2}{n^3 +n}$$
converges absolutely, conditionally or diverges.
I know that for even natural numbers the expression will equal zero and that for odd values of $n$ the ... | HINT
*
*Note that
$$
\frac{n^2+2}{n^3+n}
= \frac{1}{n} \times \frac{n^2+2}{n^2+1}
= \frac{1}{n} \left[ 1 + \frac{1}{n^2+1} \right]
= \Theta(1/n)
$$
Does this series converge?
*Adding $\sin(n\pi/2)$ in the front effectively kills all the even-$n$ terms and makes an alternating series out of the odd ones -- does ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3024368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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$\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\tan^{-1}\frac{x+y+z-xyz}{1-xy-yz-zx}$ true for all $x$?
$\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\tan^{-1}\dfrac{x+y+z-xyz}{1-xy-yz-zx}$ true for all $x$ ?
This expression is found without mentioning the domain of $x,y,z$, but I don't think its true for all $x,y,z$ as the case with the exp... | We have that by addition formula
$$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}$$
then
$$(\tan^{-1}x+\tan^{-1}y)+\tan^{-1}z=\tan^{-1}\frac{\frac{x+y}{1-xy}+z}{1-\frac{(x+y)z}{1-xy}}$$
the simplify to the given identity, which could be not well defined for (to check)
*
*$1-xy=0$
and is certainly not well defined f... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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For all $n>1$ there are positive $a+b=n$ such that $a+ab+b\in\mathbb P$ For all integers $n>1$ there are positive integers $a,b$ such that $a+b=n$ and such that $a+ab+b\in\mathbb P$.
Tested for all $n\leq 1,000,000$. Hopefully, someone can explore and explain the heuristics about this conjecture.
This question is rela... | This is only a partial answer. Reformulating according to the comment of Crostul, we want to find $AB-1 \in \mathbb P$. Other than the even prime $2$, all primes are odd, so at least one of $A,B$ must be even. For $A+B=N,\ N\ge 6$, any prime generated will have the form $6m\pm 1$. So it is necessary (but not sufficient... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
sum of $\sum^{n}_{k=0}(-1)^k\cdot \frac{\binom{n}{k}}{\binom{k+3}{k}}$
Finding sum of $\displaystyle \sum^{n}_{k=0}(-1)^k\cdot \frac{\binom{n}{k}}{\binom{k+3}{k}}$
Try: Using
$$\int^{1}_{0}x^m\cdot (1-x)^ndx = \frac{m!\cdot n!}{(m+n+1)!}=\frac{1}{(m+n+1)\binom{m+n}{n}}.$$
So $$ \frac{1}{(k+4)\binom{k+3}{k}}=\int^{1}... | I think it is simpler to utilize binomial identities:
$$\dfrac{\binom nk}{\binom{k+3}k}=\dfrac{n!3!}{(n-k)!(k+3)!}=\dfrac6{(n+1)(n+2)(n+3)}\binom{n+3}{k+3}$$
$$\sum_{k=0}^n\dfrac{\binom nk}{\binom{k+3}k}=\dfrac6{(-1)^3(n+1)(n+2)(n+3)}\sum_{k=0}^n\binom{n+3}{k+3}(-1)^{k+3}$$
$$\sum_{k=0}^n\binom{n+3}{k+3}(-1)^{k+3}$$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3029516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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$\text{SL}_2(\mathbb Z)$ acts on upper plane $\mathbb H$. What kind of points have non-trivial stabilizer? And how many orbits are there? $\text{SL}_2(\mathbb Z)$ acts on upper plane $\mathbb H= \{z \in \mathbb{C} | \Im(z) > 0 \}$ via Mobius transformation. $$
\text{ For } \gamma =\begin{bmatrix}
a &b \\c&d \... | At some point you have $b=-c(x^2+y^2)$ and $a=2cx+d$. If $c=0$, then $b=0$, in which case the element $\begin{pmatrix} a&b\\c&d\end{pmatrix}$ of the projective special linear group $\operatorname{PSL}_2(\Bbb Z)$ of this form is the identity element (recalling that $\operatorname{PSL}_2(\Bbb Z)=\operatorname{SL}_2(\Bb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3032808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
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obtain matrix $ A$ if $X$ and $b$ given
for the right and detailed answer refer to user9077 answer
| Since $\begin{pmatrix}1\\0\end{pmatrix}+c\begin{pmatrix}0\\1\end{pmatrix}$ are all solutions to $Ax=\begin{pmatrix}1\\3\end{pmatrix}$, then we have
$$
\begin{pmatrix}1\\3\end{pmatrix}=Ax=A\begin{pmatrix}1\\0\end{pmatrix}+cA\begin{pmatrix}0\\1\end{pmatrix}.
$$
The above is true for any $c$. This can only happen when $A\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The distributive law $4\left (x+y \right)=4x+4y $ because $4\left (x+y \right) =\left (x+y \right) +\left (x+y \right) +\left (x+y \right) +\left (x+y \right)$ , but why is $\left (x+y \right) \left (x+y \right) =xx+xy+yx+yy$?
|
$4\left (x+y \right)=4x+4y $ because $4\left (x+y \right) =\left (x+y \right) +\left (x+y \right) +\left (x+y \right) +\left (x+y \right)$
This works because $4$ is a (positive) integer; but for non-integer factors you can't use the "repeated addition"-argument.
but why is $\left (x+y \right) \left (x+y \right) =xx+... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Proving $\lim_{n\to\infty}\sqrt{n^2+n}-n=\frac{1}{2}$
Prove using the definition of a limit that
$$\displaystyle{\lim_{n\to\infty}\underbrace{\sqrt{n^2+n}-n}_{s_n}}=\frac{1}{2}.$$
Proof:
Let $\epsilon>0$ and $n\in\mathbb{N}$, then
$$\left|\sqrt{n^2+n}-n-\frac12\right|\stackrel{(*)}{=}\left|\frac{n-\sqrt{n^2+n}}{2(... | We know that $$\sqrt{n^2+n}-n{=(\sqrt{n^2+n}-n){\sqrt{n^2+n}+n\over \sqrt{n^2+n}+n}\\={n\over \sqrt{n^2+n}+n}\\={1\over 1+\sqrt{1+{1\over n}}}}$$therefore$$\left|{1\over 1+\sqrt{1+{1\over n}}}-{1\over 2}\right|{={1\over 2}-{1\over 1+\sqrt{1+{1\over n}}}<\epsilon}$$which means that $${ 1+\sqrt{1+{1\over n}}}<{1\over {1\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3040207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Subsets and Splits
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.