Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
$a+n^2$ is the sum of two perfect squares $a+n^2$ is the sum of two perfect squares for a given positive integer $a$ and all positive integers $n$.
Show that $a$ is a perfect square.
At first I thought of putting a bound on the difference between perfect squares up to a point, like maybe choosing $n=c$ so that one of t... | The general solution of the equation
$$x^2+y^2=z^2+w^2$$ is given by the identity with four arbitrary parameters
$$(tX+sY)^2+(tY-sX)^2=(tX-sY)^2+(tY+sX)^2$$
Let $n$ be any integer so we have $$a+n^2=z^2+w^2$$Making
$$n=tY-sX\\z=tX-sY\\w=tY+sX$$ we have three equations with four unknowns which in general have infinitel... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4013311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Angle between tangents to circle given by $x^2 + y^2 -12x -16y+75=0$?
Given the circle: $C(x,y)=x^2 + y^2 -12x -16y+75=0$, find the two tangents from origin
First, I get the line which passes through point of contact of tangents from origin using result here which is :
$$ -12x-16y-2 \cdot 75 = 0$$
Or,
$$ 6x + 8y -75 ... | Here is a simple solution: Find the center of the given circle, which is (6, 8). The line from the center of the circle to the point of contact of one of the tangents is perpendicular to the tangent. Use this fact to get the equation of a new circle, which is $x^2-6x+y^2-8x=0$. Find where both circles meet. Find the an... | {
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"url": "https://math.stackexchange.com/questions/4013658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If$x^2+y^2+z^2+t^2=x(y+z+t)$prove $x=y=z=t=0$
If
$$x^2+y^2+z^2+t^2=x(y+z+t)$$ Prove $x=y=z=t=0$
I added $x^2$ to both side of the equation:
$$x^2+x^2+y^2+z^2+t^2=x(x+y+z+t)$$
Then rewrite it as:
$$x^2+(x+y+z+t)^2-2(xy+xz+xt+yz+yt+zt)=x(x+y+z+t)$$
$$x^2+(x+y+z+t)(y+z+t)=2(xy+xz+xt+yz+yt+zt)$$
But it doesn't seem usefu... | Note the beautiful identity $$y^2+z^2+t^2=\frac{1}{3}\left({(y+z+t)}^2+{(y-z)}^2+{(z-t)}^2+{(t-y)}^2\right)$$ thus we have $$x^2+\frac{1}{3}{(y+z+t)}^2+\frac{{(y-z)}^2+{(z-t)}^2+{(t-y)}^2}{3}=x(y+z+t)$$ $${\left (x-\frac{(y+z+t)}{2}\right)}^2+ \frac{{(y+z+t)}^2}{12}+\frac{{(y-z)}^2+{(z-t)}^2+{(t-y)}^2}{3}=0$$ which m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4014634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Expand $\frac{\Gamma\left(\frac{x}{2}\right)}{\Gamma\left(\frac{x-1}{2}\right)}$ at $x=\infty$ I used Wolfram Alpha for this problem and it gives me a "Puiseux expansion": $$\sqrt{x}-\frac{3 \sqrt{\frac{1}{x}}}{8}-\frac{7}{128}\left(\frac{1}{x}\right)^{3 / 2}-\frac{9\left(\frac{1}{x}\right)^{5 / 2}}{1024}+O\left(\left(... | Note that the expansion you gave is for $\Gamma(x)/\Gamma(x-1/2)$. Using the beta function and an appropriate change of integration variables, we obtain
\begin{align*}
&\frac{{\Gamma \left( {\frac{x}{2}} \right)}}{{\Gamma \left( {\frac{{x - 1}}{2}} \right)}} = \frac{{x - 1}}{{2\sqrt \pi }}\frac{{\Gamma \left( {\frac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4016560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Divisibility by 7 Proof by Induction Prove by Induction that
$$7|4^{2^{n}}+2^{2^{n}}+1,\; \forall n\in \mathbb{N}$$
Base case:
$$
\begin{aligned}
7&|4^{2^{1}}+2^{2^{1}}+1,\\
7&|7\cdot 3
\end{aligned}$$ Which is true.
Now, having $n=k$, we assume that:
$$7|4^{2^{k}}+2^{2^{k}}+1,\;\; \forall k\in \mathbb{N}$$
We have to ... | $ b - a = 4^{2^{k+1}} + 2^{2^{k+1}} - 4^{2^k}-2^{2^k}$
An intermediate step: notice $4=2^2$, so we have that $b-a = 2^{2^{k+2}} + 2^{2^{k+1}} - 2^{2^{k+1}}-2^{2^k} = 2^{2^{k+2}} -2^{2^k} $
Notice that each term is divisible by $2^{2^k}$, and $7$ does not divide $2^{2^k}$.
Now $\frac{b-a}{2^{2^k}} = 2^{2^{k+2} - 2^k} -2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Solving $\tan(x) = \cos(x)$ I've been trying to solve the following trigonometric equation unsuccessfully. My intuition was to reduce the degree of the equation from $4$ to $2$ so I could solve it as a quadratic equation, but my attempt so far have proved unsuccessful. Any help would be appreciated.
$$\begin{align}
\ta... |
We could also construct the right triangle suggested by the equation $ \ \tan x \ = \ \cos x \ \ , $ where $ \ x \ $ is an angle (in the first quadrant). We may set the length of the side opposite $ \ x \ $ equal to $ \ 1 \ \ , $ and write the ratios $ \ \tan x \ = \ \frac{1}{y} \ \ $ and $ \ \cos x \ = \ \frac{y}{y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4020649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Problem solving complex equation with high power I have to solve :$$(i+z)^{16}-(7+7i)(i+z)^8+25i=0$$
I consider this equation like a quadratic equation and find its delta and solutions. But i can't solve two equations :$$(i+z)^8=4+3i$$
$$(i+z)^8=3+4i$$
I really need some helps to solve completely this equation. Thanks.... | If you don't want to use polar coordinates, the eighth roots can be found by repeated square root.
We get the square root
$$ (x+iy)^2 = a+ib $$
when
$$ x = \frac{1}{2} \left( \sqrt{ \sqrt{a^2+b^2} + b} + \sqrt{\sqrt{a^2+b^2} - b} \right) $$
$$ y = \frac{1}{2} \left( \sqrt{ \sqrt{a^2+b^2} + b} - \sqrt{\sqrt{a^2+b^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4021240",
"timestamp": "2023-03-29T00:00:00",
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Is the triangle equilateral? (resolve without using trigonometry) I have been asked the following question. Take a circle, and three arcs of magnitude $\dfrac{\pi}{3}$ (or you could say the corresponding centered angles are of $60$ degrees) each, that do not intersect with each other, say arc $AB$, arc $CD$ and arc ... | It is not difficile but something tedious solve this problem without trigonometry. Let a circle of equation $x^2+y^2=r^2$ and put $A=(x_A,y_A),\cdots,F=(x_F,y_F)$. The triangles $\triangle{OAB},\triangle{OCD}$ and $\triangle{OEF}$ are clearly equilateral so we have $$\overline{AB}^2=r^2=(x_A-x_B)^2+(y_A-y_B)^2\Rightar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4025690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many ways can $10$ different kids be distributed to $4$ distinct classes? I want to find the number ways to distribute 10 different kids to 4 different classes, such that each class has at least 2 kids.
My attempt: First distribute 8 kids to 4 classes such that each class has exactly 2 kids. Then there are 16 optio... | Note that there are 2 different size types for classes, $3,3,2,2$ or $4,2,2,2$. For the first one, choose firstly which classes will have $3$ kids, $\binom{4}{2}$, then distribute the kids, $\binom{10}{3} \cdot \binom{7}{3} \cdot \binom{4}{2} \cdot \binom{2}{2}$. Do the same for the other size type and you will arrive ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integration with trig substitution Trying to evaluate this using trig substitution:
$$\int \frac {1}{49x^2 + 25}\mathrm{d}x $$
Here's how I'm going about it, using $x = 5/7(\tan\theta)$
$$\int \frac {1}{49\left(x^2 + \frac{25}{49}\right)}\mathrm{d}x $$
$$=\int \frac {1}{49\left(\frac{25}{49}\tan^2\theta + \frac{25}{49}... | We can calculate a more general integral of the form
$$I =\int\frac{1}{a^2x^2+b^2}\,\mathrm{d}x.$$
In your example $a=7$ and $b=5$. First of all do the non-trigonometrical substitution $u=ax/b$. That will give you
$$\int\frac{b}{a(b^2u^2+b^2)}\,\mathrm{d}x=\frac{1}{ab}\int\frac{1}{u^2+1}\,\mathrm{d}u.$$
You should be f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4033228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $s_{n+1}=s_n^2-2$ diverges. Let $s_1=5$ and for $n\geq 1$, $s_{n+1}=s_n^2-2$. Do the following steps to show that $(s_n)$ diverges.
1.If $\lim s_n = a$ then $a^2 -2 = a$.
$a=\lim s_{n+1}=\lim {s_n^2-2}=\lim {s_n^2}-\lim (2)=(\lim {s_n})^2-\lim (2)=a^2-2$
2. If $\lim s_n = a$ then a has to be either $−1$ or $... | More convenient to begin with an $s_0.$
Suppose we find the real number $A>1$ for which
$$ s_0 = A + \frac{1}{A} $$
From $s_{n+1} = s_n^2 - 2$ we find
$$ s_1 = A^2 + \frac{1}{A^2} \; , \; \; $$
$$ s_2 = A^4 + \frac{1}{A^4} \; , \; \; $$
$$ s_3 = A^8 + \frac{1}{A^8} \; , \; \; $$
generally
$$ s_n = A^{2^n} + \fr... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac1{\sin A}+\frac1{\sin B}+\frac1{\sin C}-\frac12(\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}) \ge \sqrt{3}$ Let $ABC$ be a triangle. Prove that: $$\frac{1}{\sin A}+\frac{1}{\sin B}+\frac{1}{\sin C}-\frac{1}{2}\left(\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}\right) \ge \sqrt{3} $$
My attempt:
$$P=\fr... | Let $p$ is the semi-perimeter, $R$ is circumradius and $r$ is radius.
We have
$$\frac{1}{\sin A}+\frac{1}{\sin B}+\frac{1}{\sin C} = \frac{2R}{a} + \frac{2R}{b} + \frac{2R}{c} = \frac{4R+r}{2p}+\frac{p}{2r},$$
and
$$\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2} = \sum \frac{r}{b+c-a} = \frac{4R+r}{p}.$$
The inequalit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4035416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Bound on Truncated Alternating Harmonic Series I'm trying to prove the following inequality
$$\sum _{n=1}^{2N}\frac{(-1)^n}{n}+\log (2)<\frac{1}{4N+1}.$$
Most of the traditional inequalities I've seen for harmonic numbers aren't tight enough to prove this, unless I'm doing something wrong.
| As the taylor series of $\ln(1+x) = \sum_{n=1}^{+\infty} \frac{(-1)^{n-1}x^n}{n}$, we have
$$\ln(2)=\sum_{n=1}^{+\infty} \frac{(-1)^{n-1}}{n}$$
then
\begin{align}
\sum_{n=1}^{2N} \frac{(-1)^{n-1}}{n} + \ln(2) &= \sum_{n=2N+1}^{+\infty} \frac{(-1)^{n-1}}{n} \\
&= \sum_{k=N}^{+\infty} \left( \frac{(-1)^{2k}}{2k+1}+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4037331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A recursive sequence $c_n =\sqrt{2+\sqrt{2+...+\sqrt{2}}}$ We have a recursive sequence $$c_n =\sqrt{2+\sqrt{2+...+\sqrt{2}}}$$ with $n$ square roots.
We can obtain a recursive formula: $$c_n =\sqrt{2+c_{n-1}}\\
c_{n+1}=\sqrt{2+c_n}$$
Now we show that the sequence is increasing:
$$c_{n+1} \geq c_n \rightarrow\text{ Th... | We you define the recursive formula you need to define $c_0 = \sqrt 2$
you don't need to define both $c_n$ and $c_{n+1}$. Defining $c_n = \sqrt{2 + c_{n-1}}$ when $n>1$ is enough. Or defining $c_{n+1}=\sqrt{2 + c_n}$ is enough.
We can streamline our prove that $c_n$ is increasing and bounded by $2$ in one fell induct... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The angle between the pair of tangents drawn from a point P to the circle and its locus The angle between the pair of tangents drawn from a point P to the circle ${x^2} + {y^2} + 4x - 6y + 9{\sin ^2}\alpha + 13{\cos ^2}\alpha = 0$ is $2\alpha$ . Then the equation of the locus of the point P is
A- $x^2+y^2+4x−6y+4=0$
... |
The choices should alert you that a simpler method is possible.
On drawing a diagram, we see that $P$ is at a constant distance from center $O=(-2,3)$. You can now finish.
| {
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Find the minimum of $\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$ Find the minimum of $$f(x,y)=\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$$ It seems that $f_x=f_y=0$ is very hard to compute. Is there any easier idea?
| Recall the well-known fact: If $g(u)$ is a convex function on $\mathbb{R}^n$, then
$$g(u) \ge g(v) + \nabla g(v)^\mathsf{T}(u - v), \ \forall u, v \in \mathbb{R}^n.$$
Clearly, $f(x, y)$ is a convex function on $\mathbb{R}^2$. Thus, we have
$$f(x, y) \ge f(\tfrac{5}{12}, \tfrac{4}{3}) + \tfrac{\partial f}{\partial x}(\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4043163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Find general solution of differential equation $x\frac{dy}{dx} + y = a \sin^2x.$ and $a \in \mathbb{R}$ The real number $a$ and the differential equation are given $$x\frac{dy}{dx} + y = a \sin^2x.$$
a) Find a general solution of the equation.
b) Find all $a \in \mathbb{R}$ that solutions of $y (x)$ of the differential... | $$\begin{align}x\frac{dy}{dx} + y &= a\sin^2(x) \\ x\frac{dy}{dx} + \frac{dx}{dx}y &= a\sin^2(x) \\ \frac{d}{dx}(xy) &= a\sin^2(x)\\ \therefore xy = \int a\sin^2x\ dx &= \frac{ax}{2} - \frac{a\sin(2x)}{4} + c\\ \implies y &= \frac{a}{2} - \frac{a\sin(2x)}{2x}+ \frac{c}{x}\end{align} $$
Therefore, as $ \lim_{x \to \inf... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\forall x,y, \, \, x^2+y^2+1 \geq xy+x+y$ Prove that $\forall x,y\in \mathbb{R}$ the inequality $x^2+y^2+1 \geq xy+x+y$ holds.
Attempt
First attempt: I was trying see the geometric meaning, but I´m fall.
Second attempt: Consider the equivalent inequality given by $x^2+y^2\geq (x+1)(y+1)$
and then compare ... | If you look at the inequality as a quadratic inequality with respect to the variable $x$, then we have
$$x^2+y^2+1 \geq xy+x+y$$
$$\implies x^2+y^2+1 -xy-x-y≥0$$
$$\implies x^2-x(y+1)+(y^2-y+1)≥0$$
$$\implies \left( x -\frac{y+1}{2}\right)^2-\left(\frac{y+1}{2}\right)^2+y^2-y+1≥0$$
$$\implies \left( x -\frac{y+1}{2}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4054130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 2
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Evaluating $\sqrt{1+\sqrt{2 - \sqrt{3 + \sqrt{4 - \cdots}}}}$
$$x =\sqrt{1+\sqrt{2 - \sqrt{3 + \sqrt{4 - \sqrt{5+ \sqrt{6 - \cdots}}}}}}$$
Find $x$.
I am not sure how to proceed.
Is this a sort of Arithmetico-Geometric Progression? Will this converge at a point?
Any help would be sincerely appreciated.
| This is just an empirical answer. Consider
$$f(n) = \sqrt{t+\sqrt{t+1 - \sqrt{t+2 + \sqrt{t+3 - \cdots}}}}$$
We need to find $f(1)$. But
\begin{align}
f(1) &= \sqrt{1+\sqrt{2 - f(3)}}\\
f(3) &= \sqrt{3+\sqrt{4 - f(5)}}\tag{Similarly moving on }\\
f(n) &= \sqrt{n+\sqrt{n+1 - f(n+2)}}\\
\end{align}
After squaring and rea... | {
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"url": "https://math.stackexchange.com/questions/4056039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 1
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Find the sum of the following geometric series $\sum_{k=2}^{\infty} \frac{5}{2^k} = \frac{5}{2}$
Find the sum of the following geometric series
$$\sum_{k=2}^{\infty} \frac{5}{2^k} = \frac{5}{2}$$
Attempt:
First I test with the root criterion if its divergent or convergent... $\frac {1}{2} < 1$ so it's convergent...
N... | $$\sum_{k=2}^{\infty} \frac{1}{2^k} = \sum_{k=0}^{\infty} \frac{1}{2^k} - (1 +\frac{1}{2}) = \frac{1}{1 - \frac{1}{2}} -\frac{3}{2} = \frac{1}{2}$$So the answer is $5\times\frac{1}{2} = \frac{5}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4057980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solution to $y' + \frac xy = 1$ $y' + \frac xy = 1$
I've identified it as a homogeneous differential equation and transformed it to $xv' + v = 1-\frac 1v$ (where $v = \frac yx$) but am stuck here.
| $$
\begin{aligned}
y' + \frac{x}{y} = 1 &\Leftrightarrow
\left|
\begin{aligned}
y(x) &= z(x)\cdot x \\
y'(x) &= z'(x)x + z(x)
\end{aligned}
\right| \Leftrightarrow \\
&\Leftrightarrow z'x + z + \frac{x}{zx} = 1 \Leftrightarrow \\
&\Leftrightarrow z'x + z + \frac{1}{z} = 1 \Leftrightarrow \\
&\Leftrightarrow z' = -\fra... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the particular solution given $y=4$ and $x=3$ for the equation: $xy \frac{dy}{dx}=\frac{x^3-x}{1-\sqrt y}$? I am almost done solving this question, however, I am stuck on the integration. This is my work; the answer is:
$$\int\left(\frac{y^2}{2}-\frac{{2y^{5/2}}}{5}\right)dy=\int \left(x^2-x\right)dx$$
and I ha... | Okay, so you have to solve the equation:
$$\left(xy\right) \frac{dy}{dx}=\frac{x^3-x}{1-\sqrt y}$$
The first step is to move things around, which you did properly.
$$y\frac{dy}{dx}=\frac{1}{x}\cdot\frac{x^3-x}{1-\sqrt{y}}$$
Now we can move the $dx$ to the other side and write the equation as,
$$y(1-\sqrt y)dy=\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4063652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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factor $1/(x\pm y)$ as $f(x)g(y)$ Is it possible to rewrite
$$\frac{1}{x\pm y}$$
as multiplication of an expression containing only $x$ and another containing only $y$ ($x$ and $y$ are real independent variables), i.e.
$$\frac{1}{x\pm y}\stackrel{?}{=}f(x)g(y)$$
If it is possible, what are $f(x)$ and $g(y)$?
| Sure it's possible. Take $A=\frac{a}{a+b}$ and take $B=0$. Then it should be trivial to show that $$\frac Aa + \frac Bb = \frac{\frac{a}{a+b}}{a} + \frac0b = \frac{1}{a+b}.$$
In fact, taking any value of $B$, you can set $A=a\cdot\left(\frac{1}{a+b} - \frac Bb\right)$ and get the equality
$$\frac{A}{a} + \frac Bb = \fr... | {
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Computing $\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$ I'm trying to compute
$$I=\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$$
The following is my effort,
$$I(a)=\int_0^\infty\frac{\ln x}{x^2+a^2}dx$$
Let $x=a^2/y$ so that $dx=-(a^2/y^2)dy$ which leads to
$$I(a)=\int_0^\infty \frac{\ln(a/y)}{a^2+y^2}dy=\int_0^\infty\frac{\ln... | $$\begin{align*}
I &= \int_0^\infty \frac{\ln x}{(x^2+1)^2} \, dx \\[1ex]
&= \int_0^1 \frac{\ln x}{(x^2+1)^2} \, dx + \int_1^\infty \frac{\ln x}{(x^2+1)^2} \, dx \\[1ex]
&= \int_0^1 \frac{1-x^2}{(x^2+1)^2} \ln(x) \, dx \tag{1} \\[1ex]
&= 2 \int_0^1 \frac{\ln(x)}{(x^2+1)^2} \, dx - \int_0^1 \frac{\ln(x)}{x^2+1} \, dx \t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4069094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 6
} |
Evaluate the following limit using Taylor Evaluate the following limit:
\begin{equation*}
\lim_{x\to 0} \frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}.
\end{equation*}
I know the Taylor series of $e^x$ at $a=0$ is $\sum_{k=0}^{\infty} \frac{x^k}{k!}$. And if we substitute $x$ with $x^2$ we get $e^{x^2}=\sum_{k=0}^{\infty} \fr... | Use $\sin x \approx x$ in the denominator, it becomes $x^4$, next expand numerator complete up to $x^4$.Then use $e^{z}=1+z+z^2/2+O(z^3)$, $\cos z =1-z^2/2+z^4/24+O(z^6)$ when $|z|$ is very small
$$L=\lim_{x\to\infty} \frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}=\lim_{x\to 0} \frac{1+x^2+x^4/2+2(1-x^2/2+x^4/(24)-3}{x^4}$$.
... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Solve the following inequality $\frac{x-a}{x^2} + x \ge 2\left(1 - \frac{a}{x}\right)$ Solve the following inequality with a real parameter a:
$\frac{x-a}{x^2} + x \ge 2\left(1 - \frac{a}{x}\right)$
I am not sure if my answer is correct. Please feel free to share your thoughts.
First, we need to multiply both sides of... | Your argument is fine up to the point
$$
f(x) = x^3-2x^2+(2a+1)x - a \ge 0
$$
Then, your roots are wrong. So one needs a curve analysis. We have
$$
f´(x) = 3x^2-4x+(2a+1)
$$
so extreme points are $x = \frac13(2 \pm \sqrt{1 - 6 a})$. So for $a > \frac16$, there are no extremal points at all, which means $f(x)$ is risin... | {
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"answer_count": 1,
"answer_id": 0
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How can I find the limit of this fraction? ($0/0$ type) Suppose I have two complex numbers $g_0$ and $g_1$, when $g_0\rightarrow1$ and $g_1\rightarrow0$, how can I evaluate
$$
\sqrt\frac{g_1^2}{1-g_0^2}
$$
Intuitively I think that should be infinity, but I'm not pretty sure how can I justify that? Thanks!!
| Let
$g_1 = a$
and
$g_0 = 1-b$
so
$\sqrt\frac{g_1^2}{1-g_0^2}
=\sqrt\frac{a^2}{1-(1-b)^2}
=\sqrt\frac{a^2}{b(2-b)}
$
where
$a, b \to 0$.
If
$a=b$ this is
$\sqrt\frac{b^2}{b(2-b)}
=\sqrt\frac{b}{(2-b)}
\to 0
$.
If $a = b^{1/4}$ this is
$\sqrt\frac{b^{1/2}}{b(2-b)}
=\sqrt\frac{1}{b^{1/2}(2-b)}
\to \infty
$.
If $a = cb^{1/... | {
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"source": "stackexchange",
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"answer_count": 2,
"answer_id": 0
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Which is the smallest parameter of the polynomial of the smallest degree such that it has $\frac{\sqrt{3}}{2},\frac{\sqrt{2}}{3}$ as roots? Which is the smallest parameter of the polynomial of the smallest degree such that it has $\frac{\sqrt{3}}{2},\frac{\sqrt{2}}{3}$ as roots and has only integral parameters?
Since i... | The polynomial with integer coefficients with least degree that has $\frac{\sqrt{3}}{2}$ as a root is $4 x^2 - 3$.
Every polynomial with rational coefficients that has $\frac{\sqrt{3}}{2}$ as a root is a rational polynomial multiple of $4 x^2 - 3$.
Likewise, for $\frac{\sqrt{2}}{3}$, the polynomial is $9 x^2 - 2$.
Sinc... | {
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"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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What is $\arcsin(\cos{x})$ (Spivak Calculus exercise 15-18)? In the first part of the exercise I proved that $\sin(x+\frac{\pi}2)=\cos{x}$.
The second part is asking what $\arcsin(\cos{x})$ is. Presumably, they want a rule for unrestricted $x$.
We have $\arcsin(\cos{x})=\arcsin(\sin(x+\frac{\pi}2))$. It was said that t... | here's my solution based on @YvesDaoust's answer:
$\cos{x}=\sin\left({x}+\frac{\pi}2\right), x \in R$, hence $\arcsin(\cos(x)) = \arcsin\left(\sin\left({x}+\frac{\pi}2\right)\right)$
The co-domain of $\arcsin$ is $[-\frac{\pi}2,\frac{\pi}2]$, and we must restrict the domain of $\sin$ to the same interval, there are 2 c... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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solving $\sqrt{2 x^4-3 x^2+1}+\sqrt{2x^4-x^2}=4 x-3.$ I am trying to solve this equation
$$\sqrt{2 x^4-3 x^2+1}+\sqrt{2x^4-x^2}=4 x-3.$$
By using Mathematica, I know that, the equation has unique solution $x=1$.
I tried to write the equation in the form
$$\sqrt{(x-1) (x+1) \left(2x^2-1\right)} + \sqrt{x^2\left(2x^2-1\... | Note that the domain of the equation is $x\ge1$ and
$$\sqrt{2x^4-x^2} - (2x^2-1)= \sqrt{2x^2-1} \left( x-\sqrt{2x^2-1}\right)
= -\frac{\sqrt{2x^2-1} (x^2-1)}{x+\sqrt{2x^2-1} }
$$
Then, factorize the equation as follows
\begin{align}
&\sqrt{2 x^4-3 x^2+1}+\sqrt{2x^4-x^2}-(4 x-3)\\
=& \sqrt{(x^2-1) (2x^2-1)} +\left( \sqr... | {
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What is the maximal area between hyperbola and chord $AB$?
Suppose $A$ and $B$ are variable points on upper branch of hyperbola $\mathcal{H}:\;x^2-y^2=-4$ such that $AB = 1$. What is the maximal area between $AB$ and $\mathcal{H}$?
Clearly $y= \sqrt{x^2+4}$. If we set $A\big(t,\sqrt{t^2+4}\big)$ then $B\big(t+h, \sq... | Parametrize the hyperbola as $x=2\sinh t $, $y= 2\cosh t$ and let the endpoints of the chord be $A(2\sinh a, 2\cosh a)$, $B(2\sinh b, 2\cosh b)$. Then, the equation of chord $AB$ is
$$y= x\tanh\frac{a+b}2 +2\frac{\cosh \frac{a-b}2}{\cosh\frac{a+b}2 }$$
and the area between the chord and hyperbola is
\begin{align}
K =&... | {
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"answer_count": 3,
"answer_id": 0
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Prove that $\sqrt{(x-2)^2+y^2}+\sqrt{(x-4)^2+(y-2)^2}+\sqrt{(x-2)^2+(y-5)^2}+\sqrt{(x-1)^2+(y-2)^2}\geq8$ Prove that $\sqrt{(x-2)^2+y^2}+\sqrt{(x-4)^2+(y-2)^2}+\sqrt{(x-2)^2+(y-5)^2}+\sqrt{(x-1)^2+(y-2)^2}\geq8$ for $x,y\in\mathbb{R}$. I used this Proof of an inequality about sequences to solve it and it worked, but I ... | Let's $A(2,0), B(4,2),C(2,5), D(1,2)$ and $G(x,y)$ on the plane $Oxy$. The left hand side of the inequality is the sum $$\mathbf{S} =GA+GB+GC+GD$$
The inequality becomes: find $G$ such that the sum $\mathbf{S}$ is minimized.
We have
$$GA+GC \ge AC$$ $$GB+GD \ge BD$$
then $$S = GA+GB+GC+GD \ge AC+BD = \sqrt{2^2+5^2}+3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4083902",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving that $3^n \gt n^4 \ \forall n \gt 8$ Regarding this question, I found no answer that combines mine, so I want to know if my proof is valid as well:
Problem: Prove that:$\ 3^n \gt n^4 \ \forall n \ge 8$
Base step (n=8): is true because$$P(8): 6561 \gt 4096 \\$$
Inductive step ($P(n) \implies P(n+1)$): $\ 3^{n+... | Rewrite the expression as $e^{n \log 3 - 4 \log n}$. This is your induction step. Now you need to show it for $n+1$.
$$
e^{(n+1)\log 3 - 4 \log (n+1)} = e^{n \log 3 - 4 \log n} \cdot e^{\log 3 - \log (1+\frac{1}{n})}
$$
The first term is greater that $1$ by the inductive argument. All is left to show that the second te... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Aproximation of the Normal Distribution by the Normal Density Function In Feller's introduction to probability the next lemma is stated:
"As $x\rightarrow \infty$
$\tag1 \frac{1-R(x)}{x^{-1}n(x)} \rightarrow 1$
Where $R(x)$ is the normal distribution and $n(x)$ is the normal density function. And more precisely:
$\tag1... | About the statement
$$ \frac{1-R(x)}{\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}(\frac{1}{x}-\frac{1}{x^3}+...+(-1)^k\frac{1*3*7*...*(2k-1)}{x^{(2k+1)}})}\xrightarrow{x \to +\infty} 1$$
In fact, it holds true, even for the general form like this one
$$ \frac{1-R(x)}{\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}(\frac{1}{x}+\su... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Prove $x^n -1 \ge (x-1)^n \ \forall n \in \mathbb{N}\ \land \ \forall x\ \ge 1 $ Does my proof make sense?
Problem: prove $x^n -1 \ge (x-1)^n \ \forall n \in \mathbb{N}\ \land \ \forall x\ \ge 1 $
Base step (P(1)):$\ \ \ \ \ \ \ \ x-1 \ge x-1$
Hypotesis (P(n)): $ \ \ \ \ \ x^n -1 \ge (x-1)^n$
Thesis (P(n+1)):$ \ \ \... | $(x-1)^n=\binom{n}{0}(-1)^n+\binom{n}{1}(-1)^{n-1}x+\cdots+\binom{n}{n}x^n$ by the Binomial Theorem.
Evaluating the above at $x=1$ we see
$\sum\limits_{i=0}^n(-1)^{n-i}\binom{n}{i}=0, \sum\limits_{i=0}^{n-1}(-1)^{n-i}\binom{n}{i}+\binom{n}{n}=0 \therefore (\color{red}{*})\sum\limits_{i=0}^{n-1}(-1)^{n-i}\binom{n}{i}=-1... | {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
$\sin\angle HIO$ given $\triangle ABC$ with $(\overline{AB},\overline{BC},\overline{CA})=(8,13,15)$? In triangle $ABC, AB = 8, BC = 13$ and $CA = 15$.
Let $H, I, O$ be the orthocenter, incenter and circumcenter of triangle $ABC$ respectively. Find $\sin$ of angle $HIO$ .
My attempt: By using law of cosine I can see ang... |
Given $a=13$, $b=15$, $c=8$, we
can find semiperimeter, area, inradius and circumradius
ot the triangle:
\begin{align}
\rho&=\tfrac12(a+b+c)=18
,\\
S_{ABC}&=
\tfrac14\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}
=
30\sqrt3
,\\
r&=\frac{S_{ABC}}\rho=\tfrac53\sqrt3
,\\
R&=\frac{abc}{4S_{ABC}}=
\tfrac{1}3\sqrt3
\end{align}
And use kn... | {
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"url": "https://math.stackexchange.com/questions/4088790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Does $2x^5 - x^4 - 22x^3 - 23x^2 + 22x +24 = 0$ have exact solutions in radicals?
Does $2x^5 - x^4 - 22x^3 - 23x^2 + 22x +24 = 0$ have exact solutions in radicals?
A mysterious commenter said on Youtube this was the "easiest quintic equation of my life," and I'm suspecting a troll is afoot. In fact, I believe the opp... | Your polynomial $2x^5-x^4-22x^3-23x^2+22x+24$ has no solution in radical it's Galois group is the symmetric group if $5$ objects, on the other hard, $x^5-2x^4-22x^3-23x^2+22x+24$ factors into $(x-1)(x+2)(x^3-3x^2-17x-12)$
Since the polynomials are related in shape
$$2x^5-x^4-22x^3-23x^2+22x+24 = x^5-2x^4-22x^3-23x^2+22... | {
"language": "en",
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"source": "stackexchange",
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Prove that $\arctan(x) = \frac{\pi}{2} - \frac{1}{x} + O\left(\frac{1}{x^3}\right)$ for $x \to +\infty$ Want to prove that $$\arctan(x) = \frac{\pi}{2} - \frac{1}{x} + O\left(\frac{1}{x^3}\right) \ \text{as}~~~ x \to +\infty$$
Wolfram says that it's true but I'm trying to find some formal prove of this equality.
Here'... | To show $$\arctan(x) = \frac{\pi}{2} - \frac{1}{x} + O\left(\frac{1}{x^3}\right) \ \text{as}~~~ x \to +\infty,$$
you have to prove (by definition)
$$\limsup_{ x\to +\infty}\left|\frac{\arctan(x)-\frac{\pi}{2} + \frac{1}{x}}{\frac{1}{x^3}}\right|<\infty.
$$
This can be easily done by calculating $\lim_{ x\to +\infty}\f... | {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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If $\tan A, \tan B, \tan C$ are roots of $x^3-ax^2+b=0$, find $(1+\tan^2 A)(1+\tan^2B)(1+\tan^2C)$ Here are all the results I got
$$\tan(A+B+C)=a-b$$
And
$$(1+\tan^2A)(1+\tan^2B)(1+\tan^2C)=(\frac{1}{\cos A\cos B\cos C})^2$$
And
$$\cot A+\cot B + \cot C=0$$
How should I use these results?
| Write $x=\tan A, y=\tan B, z=\tan C$.
By Vieta’s formula, we have
$$x+y+z=a,\ \ xy+yz+zx=0,\ \ xyz=-b.$$
Now you want to find the value of $$(1+x^2)(1+y^2)(1+z^2)=1+x^2+y^2+z^2+x^2y^2+y^2z^2+z^2x^2+x^2y^2z^2.$$
Note that
$$a^2=(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx),$$
so $x^2+y^2+z^2=a^2$. Similarly,
$$0=(xy+yz+zx)^2=x^2y^2... | {
"language": "en",
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"question_score": "5",
"answer_count": 5,
"answer_id": 1
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Factorize $abx^2-(a^2+b^2)x+ab$ Factorize the quadratic trinomial $$abx^2-(a^2+b^2)x+ab.$$
The discriminant of the trinomial is $$D=(a^2+b^2)^2-(2ab)^2=\\=(a^2+b^2-2ab)(a^2+b^2+2ab)=(a-b)^2(a+b)^2=\\=\left[(a-b)(a+b)\right]^2=(a^2-b^2)^2\ge0 \text{ } \forall a,b.$$
So the roots are $$x_{1,2}=\dfrac{a^2+b^2\pm\sqrt{(a^2... | For the title question. The factorization is given by
$$
(ax-b)(bx-a)=abx^2-(a^2+b^2)x+ab.
$$
We can find it by multiplying out $(\rho_1x+\rho_2)(\rho_3x+\rho_4)$ and comparing coefficients. This is easier than computing a discriminant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4092687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Harmonic numbers as ratio of two Determinants Provide a proof to this interesting identity:
$$\frac{\begin{vmatrix} 1^0 & 1^2 & 1^3 & \cdots & 1^n \\ 2^0 & 2^2 & 2^3 & \cdots & 2^n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ n^0 & n^2 & n^3 & \cdots & n^n \end{vmatrix}}{\begin{vmatrix} 1^1 & 1^2 & 1^3 & \cdots & 1... | As @achillehui suggested, we can consider the Vandermonde determinant
$$
P(x) = \begin{vmatrix}
x^0 & x^1 & x^2 & \cdots & x^n \\
1^0 & 1^1 & 1^2 & \cdots & 1^n \\
2^0 & 2^1 & 2^2 & \cdots & 2^n \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
n^0 & n^1 & n^2 & \cdots & n^n
\end{vmatrix}
= \prod_{k=1}^n (x-k)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4092976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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Coefficient of $x^{12}$ in $(1+x^2+x^4+x^6)^n$ I need to find the coefficient of $x^{12}$ in the polynomial $(1+x^2+x^4+x^6)^n$.
I have reduced the polynomial to $\left(\frac{1-x^8}{1-x^2}\right)^ n$ and tried binomial expansion and Taylor series, yet it seems too complicated to be worked out by hand.
What should I do?... | $(1+x^2+x^4+x^6)^n = (1+x^4)^n(1+x^2)^n$
the $r^{th}$ term in first series: ${n \choose r} x^{4r}$
the $l^{th}$ term in second series: ${n \choose l} x^{2l}$
so we have $4r+2l = 12 \implies r=0, l=6; r=1, l=4; r=2, l=2; r=3, l=0$
So we have coefficient: ${n \choose 0}{n \choose 6} +{n \choose 1}{n \choose 4} + {n \cho... | {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
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Finding integer solutions to the system $xy=6(x+y+z)$, $x^2+y^2=z^2$ How do you go about solving a system of equations like below for integer solutions?
$$xy=6(x+y+z)$$
$$x^2+y^2=z^2$$
Would you first try and list out a number of Pythagorean triples, then try and see which ones when multiplied are divisible by 6 (by 2 ... | This is the first way that came to my mind. Maybe there is a shorter way.
$$(x+y)^2=z^2+12z+12(x+y)$$
$$t^2-12t-(z^2+12z)=0, ~t=x+y$$
$$\Delta_{\text{half}}=36+z^2+12z=(z+6)^2$$
$$t_{1,2}=6±|z+6|=6±z+6=12±z$$
$$x+y=12±z$$
$$xy=6(x+y+z)=6(z+12±z)$$
If $x+y=12-z$, then $xy=72$.
Check all integers $x,y$ such that $xy=72$... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Minimize $3\sqrt{5-2x}+\sqrt{13-6y}$ subject to $x^2+y^2=4$
If $x, y \in \mathbb{R}$ such that $x^2+y^2=4$, find the minimum value of $3\sqrt{5-2x}+\sqrt{13-6y}$.
I could observe that we can write
$$3\sqrt{5-2x}+\sqrt{13-6y}=3\sqrt{x^2+y^2+1-2x}+\sqrt{x^2+y^2+9-6y}$$
$\implies$
$$3\sqrt{5-2x}+\sqrt{13-6y}=3\sqrt{(x-1... | Nice problem! Here is a solution using geometry.
For the moment, lets forget $P$ is lying on the circle $x^2+y^2=4$. Call the origin $O=(0,0)$. $A=(1,0), B=(0,3)$. So $OA=1, OB=3$ and $\angle BOA = 90^\circ$. Draw the quadrilateral $OAPB$.
Note that we have to minimize
$$3\cdot PA+1\cdot PB=OB\cdot PA + OA\cdot PB$$
W... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the integer solution of the equation $x^3+y^3=x^2+y^2+42xy$ Find the integer solution of the equation $x^3+y^3=x^2+y^2+42xy$
I try $x=0$ We have: $y^3-y^2=0 \Longrightarrow \left\{\begin{array}{l}
y=0 \\
y=1
\end{array}\right.$
I think, this equation only $(x,y)\in ${ $(0,0),(0,1),(1,0)\}$ but I can't prove that.
| With the equation
$$x^3 + y^3 = x^2 + y^2 + 42xy \tag{1}\label{eq1A}$$
you've already handled the cases of $x = 0$ or $y = 0$. For the non-zero cases, have
$$\gcd(x,y) = d, \; x = de, \; y = df, \; \gcd(e, f) = 1 \tag{2}\label{eq2A}$$
Then \eqref{eq1A} becomes
$$\begin{equation}\begin{aligned}
(de)^3 + (df)^3 & = (de)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4100375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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When we express $\sin x - \cos x$ as $A \sin (x+c)$, how many solutions are there for $c \in [0, 2\pi)$? This is a problem from problem set 1 of MIT OCW 18.01SC:
express $\sin x - \cos x$ in the form $A \sin (x+c)$.
Their solution is $\sqrt{2} \sin (x - \frac{\pi}{4})$
I found two solutions (for $c \in [0, 2\pi]$):
S... | We can use sum to product formula to confirm your solution:
\begin{align}\sin x - \cos x = \sin x - \sin (\frac{\pi}{2}- x)=2\cos \frac{\pi}{4}\sin(x-\frac{\pi}{4})\end{align}\begin{align}=\sqrt 2\sin(x-\frac{\pi}{4})=-\sqrt 2\sin(\frac{\pi}{4}-x)\\=-\sqrt 2\sin(\pi-(\frac{\pi}{4}-x))=-\sqrt 2\sin(x+\frac{3\pi}{4})\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4105426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $\lim \sqrt{1-\frac{1}{n}} = 1$ by definition Is my attempt correct?
Proof. Let $\varepsilon > 0$. Take $N > \frac{1}{\varepsilon}$ and let $n \geq N$. Then
$\begin{align}\displaystyle\left\lvert \sqrt{1- \frac{1}{n}} -1 \right\rvert
&= \left\lvert \frac{\sqrt{n-1}}{\sqrt{n}} -1 \right\rvert\\\\
&= \left\lve... | One way to shorten this proof would be to use this lemma which says that for $x,y\in[0,\infty)$, we have $$\left|\sqrt{x}-\sqrt{y}\right|\leq\sqrt{\big||x|-|y|\big|}\leq\sqrt{|x-y|}$$
Then your proof would go $$\left|\sqrt{1-\frac{1}{n}}-\sqrt{1}\right|\leq\sqrt{\left|1-\frac{1}{n}-1\right|}=\frac{1}{\sqrt{n}}\leq\frac... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find $\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b}$? For numbers $a,b,c$ we know that $\frac{a-c}{b+c}+\frac{b-a}{c+a}+\frac{c-b}{a+b}=1$ What is the value of $\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b}$ ?
$1)3\qquad\qquad2)4\qquad\qquad3)6\qquad\qquad4)2$
Because it is a multiple choice question at firs... | $\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b} = \frac{a+b+c}{b+c} - \frac{c}{b+c} +\frac{b+c+a}{c+a} -\frac{a}{c+a}+\frac{c+a+b}{a+b} - \frac{b}{a+b}$
$= \frac{a}{b+c} + 1 - \frac{c}{b+c}+ \frac{b}{c+a} + 1 -\frac{a}{c+a} + \frac{c}{a+b} + 1 - \frac{b}{a+b}$
$= \frac{a-c}{b+c}+1 + \frac{b-a}{c+a}+1 + \frac{c-b}{a+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4106213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving multivariable 4 grade equation Find the integer solutions of the following equation:
$$8x^4+y^4=6x^2y^2$$
I tried to reconfigure the equation using the notable identity $(a-b)^2=a^2+b^2-2ab$, but due to the numeric coeficients I was not able
| You can factor
$8x^4-6x^2y^2+y^4$ as
$9x^4-6x^2y^2+y^4-x^4=(3x^2-y^2)^2-(x^2)^2=(4x^2-y^2)(2x^2-y^2)$
So the equation becomes
$(4x^2-y^2)(2x^2-y^2)=0$
At least one of these two factors has to be equal to 0
But $2x^2=y^2$ is impossible, because doubling a square changes the exponent of $2$ present in the prime factoriza... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is $5^2-3^2=16$ the only example for $(P_n)^2 - (P_{n-1})^2 \bmod 6 ≠ 0$ Is $5^2-3^2=16$ the only example for $(P_n)^2 - (P_{n-1})^2 \bmod 6 ≠ 0$ where $P$ is an element of the consecutive list of prime numbers (without $2$): ${3,5,7,11,13,17...P_n}$
As a web developer and a self math learner, out of curiosity I have c... | Yes. First, we claim that every prime $>3$ is congruent to $\pm 1 \bmod{6}$. This is easy to see since a prime bigger than $3$ cannot be congruent to $0,2,3,4 \bmod{6}$ since it would then be divisible by either $2$ or $3$, which is impossible.
Next, assume that $p_{n-1}>3$. Then,
$$p_{n-1} \equiv \pm 1 \pmod{6} \impli... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Are the Sylow $p$-subgroups of $S_4$ also Sylow $p$-subgroups of $S_5$? Here is my thinking:
$|S_4| = 4! = 1 \times 2 \times 3 \times 4 = 2^3 \times 3$.
$|S_5| = 5! = 1 \times 2 \times 3 \times 4 \times 5 = 2^3 \times 3 \times 5$.
Since $2^3$ is the maximal power of $2$ which divides the order of $S_4$ and $S_5$ and si... | Hint: Embed $S_4\hookrightarrow S_5$ and then compose with the inclusion $P\hookrightarrow S_4$. Once you've a subgroup of the right order, it's a Sylow subgroup!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\sum_{n=1}^{\infty}(\sqrt{n^2+3}-\sqrt{n^2-1})=\infty$ I want to prove $$\sum_{n=1}^{\infty}\left(\sqrt{n^2+3}-\sqrt{n^2-1}\right)=\infty$$
If $\displaystyle\lim_{n\to\infty}\left(\sqrt{n^2+3}-\sqrt{n^2-1}\right)\neq0$, I can prove it.
In fact, however, $$\displaystyle\lim_{n\to\infty}\left(\sqrt{n^2+3}-\sqrt{n^... | $\sqrt{n^2+3}-\sqrt{n^2-1}=$
$=\dfrac{\left(\sqrt{n^2+3}-\sqrt{n^2-1}\right)\left(\sqrt{n^2+3}+\sqrt{n^2-1}\right)}{\sqrt{n^2+3}+\sqrt{n^2-1}}=$
$=\dfrac{n^2+3-(n^2-1)}{\sqrt{n^2+3}+\sqrt{n^2-1}}=\dfrac4{\sqrt{n^2+3}+\sqrt{n^2-1}}>$
$>\dfrac4{\sqrt{n^2+3n^2}+\sqrt{n^2}}=\dfrac4{\sqrt{4n^2}+\sqrt{n^2}}=$
$=\dfrac4{3n}>\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4115871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Converting equation into short Weierstrass form I have the curve $y^2 + 4y = x^3 + 3x^2 −x + 1$. I need to find a transformation of the form $X=x+a$ and $Y=y+b$
that turns this curve into the standard form of the elliptic curve: $$Y^2=X^3+AX+B.$$
I completed the square for the left side and now have $(y+2)^2= x^3 + 3x^... | Hint: Observe that
\begin{align}
&& y^2+4y&=(y+2)^2-4 \\
&\text{and}& \\
&&\qquad x^3+3x^2&=(x+1)^3-3x-1
\end{align}
Can you continue?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Determine the largest and smallest of the rational numbers $x$, $y$, $z$ if $x^2+4y^2+z^2-4xy-4yz+2xz-2x+2z+2=0$
Determine what number out of three rational numbers $x$, $y$, $z$ is the biggest and smallest one if following equality is true:
$$x^2+4y^2+z^2-4xy-4yz+2xz-2x+2z+2=0$$
I think it may be somewhat connected ... | Let's start by rewriting the equality as
$$
(x+z-2y)^2 + 2 = 2(x-z)
$$
Now complete the square on the LHS first by adding $2(x+z-2y)$ to both sides:
$$
(x+z-2y+1)^2 + 1 = 2(x-z) + 2(x+z-2y) = 4(x-y)
$$
And again complete the square by subtracting $2(x+z-2y)$:
$$
(x+z-2y-1)^2 + 1 = 2(x-z) - 2(x+z-2y) = 4(y-z)
$$
Since L... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the equation $\frac{x-13}{x-14}-\frac{x-15}{x-16}=-\frac{1}{12}$ Solve the equation $$\dfrac{x-13}{x-14}-\dfrac{x-15}{x-16}=-\dfrac{1}{12}.$$
For $x\ne14$ and $x\ne 16$ by multiplying the whole equation by $$12(x-14)(x-16)$$ we get: $$12(x-16)(x-13)-12(x-14)(x-15)=-(x-14)(x-16).$$ This doesn't look very nice. Can... | It's easier if you subtract $1$ from each term on the LHS, i.e. $$-\frac{1}{12}=\frac{1}{x-14}-\frac{1}{x-16}=\frac{-2}{(x-14)(x-16)},$$ which is equivalent to $$(x-14)(x-16)=24.$$ This can be rearranged to $(x-10)(x-20)=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Evaluating $\int_{0}^{\infty} \left( \text{coth} (x) - x \text{csch}^2 (x) \right) \left( \ln \left( \frac{4 \pi^2}{x^2} + 1 \right) \right) \, dx$ How can the following improper integral be evaluated?
$$\int_{0}^{\infty} \left( \text{coth} (x) - x \text{csch}^2 (x) \right) \left( \ln \left( \frac{4 \pi^2}{x^2} + 1 \ri... | Partial solution
We can present the second integral as a series (imaginary part of the integral forms a series)
$$I= \int_{0}^{\infty} \frac{x \text{coth} (x) - 1}{x^2 (2 \pi + i x)} \, dx=\int_{0}^{\infty} \frac{(x \text{coth} (x) - 1)(2\pi-ix)}{x^2 ((2 \pi)^2 + x^2)} \, dx$$
$$x\coth x=1+2\sum_{k=1}^{\infty}\frac{x^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the equation of the locus of the mid point of AB as m varies I am working through a pure maths book as a hobby. This question puzzles me.
The line y=mx intersects the curve $y=x^2-1$ at the points A and B. Find the equation of the locus of the mid point of AB as m varies.
I have said at intersection:
$mx = x^2-1 \... | If $x_1$ and $x_2$ are x-coordinates of intersection points, x-coordinate of midpoint,
$x_m = \frac{x_1 + x_2}{2} = \frac{m}{2}$
$y = mx \implies y_m = 2 x_m^2$ so locus of midpoint is $y = 2 x^2$.
Your approach is correct but note that you can also get to it quickly using Vieta's formula for quadratic equation.
If $ax... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to show a sequence of functions is increasing How does one show that for all $x\geq 0$ that the following sequence of functions is increasing where $f_n$ is defined by
$$f_n(x)= x\left(1+\frac{x^2}{n}\right)^n$$
using the fact that for all $y\geq 0$
$$\frac{y}{y+1}\leq \ln(1+y) \leq y$$
I have been trying different... | Proof only using AM$-$GM: if $x\geq 0$ then proving $f_{n+1}(x)\geq f_{n}(x)$ is equivalent to prove $$\left(1+\frac{x^2}{n}\right)^{n}$$
is an incresing sequence. Using AM$-$GM inequality we have that
\begin{align*}
\left(1+\frac{x^2}{n}\right)^{\frac{n}{n+1}}&=\sqrt[n+1]{1\cdot\underbrace{\left(1+\frac{x^2}{n}\right)... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Area of a Triangle in $\mathbb{R}^4$ I find this question quite tricky, and I don't know if this kind of treatment is right.
I was asked about the area of a triangle formed by the points: $A:(1,2,-3,3)$; $B:(3,-6,-4,2)$; and $C:(-3,-16,-4,0)$. The only way I could make a reason out of this would be getting all the $2\t... | As @Unit commented, Herons formula might be an easier bet, we have:
$$\overrightarrow{AB}=\begin{pmatrix}3-1\\-6-2\\-4+3\\2-3\end{pmatrix}=\begin{pmatrix}2\\-8\\-1\\-1\end{pmatrix}\Rightarrow\left|\overrightarrow{AB}\right|=\sqrt{2^2+8^2+1^2+1^2}=\sqrt{70}$$
$$\overrightarrow{BC}=\begin{pmatrix}-6\\-10\\0\\-2\end{pmatr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solve over the positive integers: $7^x+18=19^y.$ Solve over the positive integers:
$$7^x+18=19^y.$$
Progress:-
I first took $\mod 7,$ so we get $5^y\equiv 4 \mod 7$ since $5$ is a primitive root of $7$ and $5^2\equiv 4\mod 7.$ So we get $y\equiv 2\mod 6.$
And then took $\mod 9$
So we get $7^x\equiv 1\mod 9.$ Since resi... | Excluding the obvious solution $(x, y) = (0, 1)$, an argument mod $7$ shows that $y$ is even (which you have done). Similarly, an argument mod $19$ shows that $x$ is a multiple of $3$.
Thus we may look at the elliptic curve $Y^2 = X^3 + 18$. A computer algebra system such as Sage can be used to find all integral points... | {
"language": "en",
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Circle passing through two points and tangent to a line
"Find the equation of the circle passing through the origin $(0,0)$, the point $(1,0)$ and tangent to the line $x-2y+1=0$."
What I have done:
The equation of a circle with radius $R$ and center $(x_0,y_0)$ is $(x-x_0)^2+(y-y_0)^2=R^2$. Since the circle passes th... | There is a very neat way of doing it if you are familiar with the idea of family of circles.
Equation of family of circles passing through $O(0, 0)$ and $A(1, 0)$ is given by
$$(x-0)(x-1)+(y-0)(y-0)+ky=0$$
where k is the parameter.
On simplification we get
$x^2+y^2-x+ky=0$
So the centre of above circle is $\left(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4133815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why $\lim\limits_{n \to +\infty} \bigg(\dfrac{n+1}{n+2}\bigg)^n = \frac{1}{e}$? I tried to solve this limit: $\lim\limits_{n \to +\infty} \bigg(\dfrac{n+1}{n+2}\bigg)^n$.
My approach was to re-write it as $\lim\limits_{n \to +\infty} \bigg(\dfrac{n}{n+2} + \dfrac{1}{n+2}\bigg)^n$, and since $\dfrac{n}{n+2}$ tends to 1 ... | You must know that $\left(1+\frac{1}{n}\right)^n \xrightarrow{n\rightarrow \infty} e$. So, we are going to play wiht it.
$\left(\frac{n+1}{n+2}\right)^n=\left[\left(\frac{n+2}{n+1}\right)^n\right]^{-1}= \left[\left(1+\frac{1}{n+1}\right)^n\right]^{-1}=\left[\left(1+\frac{1}{n+1}\right)^{(n+1)}\cdot\left(1+\frac{1}{n+1}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding sum of the roots of $4(x-\sqrt x)^2-7x+7\sqrt x=2$
Find sum of the roots of $$4(x-\sqrt x)^2-7x+7\sqrt x=2$$
By substituting $t=x-\sqrt x$ we have $4t^2-7t-2=0$
$$4t^2-8t+t-2=0$$
$$(4t+1)(t-2)=0$$
So we get $x-\sqrt x=2$ Hence $x=4$.
Or $x-\sqrt x=-\frac14$ then $x-\sqrt x+\frac14=0$ and $(\sqrt x-\frac12)^2=... | We are asked to find the sum of roots of this function:
$$4(x-\sqrt{x})^2-7x+7\sqrt{x}=2 \Leftrightarrow \\ 4(x-\sqrt{x})^2-7x+7\sqrt{x}-2=0$$
$\text{As you are saying by substituting } t=x-\sqrt{x} \text{ (1)}\text{ ,we have:}$
$$4t^2-7t-2=0 \Leftrightarrow (4t+1)\cdot(t-2)=0$$
$\text{And the solution we get from t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4135580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $p^k m^2$ is an odd perfect number, then is there a constant $D$ such that $\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{D}$? (Note: This question is an offshoot of this closely related one.)
Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.
The topic of odd perfect numbers ... | This is a partial answer.
If $D=3$, then $p=5$ and $k=1$ both hold.
Suppose that $D=3$.
$$\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{3}$$
$$2 > I(m^2)=\frac{\sigma(m^2)}{m^2} > \frac{p^k \bigg(1 - \dfrac{p^k}{m^2}\bigg)}{3}$$
$$\implies 6 > p^k \bigg(1 - \frac{p^k}{m^2}\bigg) \geq p^k \bigg(1 - \frac{2}{3375}\bigg) = \... | {
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Prove that all elements of sequence $a_{n}=\frac{ \left(1+\sqrt{n^4-n^2+1}\right)^{n} + \left(1-\sqrt{n^4-n^2+1}\right)^{n}}{2^{n}}$ are integers. $\textbf{PROBLEM}$:
Prove or disprove that all elements of sequence $a_{n}=\frac{ \left(1+\sqrt{n^4-n^2+1}\right)^{n} + \left(1-\sqrt{n^4-n^2+1}\right)^{n}}{2^{n}}$ are inte... | Let $x_1$ and $x_2$ denote $\left(1 + \sqrt{n^4 - n^2 + 1} \right)$ and $\left(1 - \sqrt{n^4 - n^2 + 1} \right)$. We want to prove that $\frac{x_1^n+x_2^n}{2^n}$ is integer.
Proof can be obtained by induction on n.
Firstly, let's fix $x_1$ and $x_2$. By that I mean: choose any fixed $n = n_0$ and let $x_1 = \left(1 + ... | {
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Smallest value of $b$ when $0<\left\lvert \frac{a}{b}-\frac{3}{5}\right\rvert\leq\frac{1}{150}$ Problem
For positive integers $a$ and $b$, $$0<\left\lvert \dfrac{a}{b}-\dfrac{3}{5}\right\rvert\leq\dfrac{1}{150}$$
What is the smallest possible value of $b$? (BdMO 2021 Junior P10)
My approach
If $\dfrac{a}{b}>\dfrac{3}{5... | Once you get that $\frac{a}{b}$ is between $\frac{89}{150}$ and $\frac{91}{150}$, you may simply consider the continued fractions of these rational numbers:
$$ \frac{89}{150}=[0;1,1,2,5,1,1,2] $$
$$ \frac{91}{150}=[0;1,1,1,1,5,2,2] $$
In particular we are looking for a fraction of the form $[0;1,1,2,\alpha]$ with $\alp... | {
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"timestamp": "2023-03-29T00:00:00",
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Locus of point M such that its projections onto two fixed lines are always at the same distance. A point moves so that the distance between the feet of the perpendiculars drawn from it to the lines $ax^2+2hxy+by^2=0$ is a constant $c$. Prove that the equation of its locus is
$$4(x^2+y^2)(h^2-ab)=c^2(4h^2+(a-b)^2).$$
Wh... | Please note that $OAPB$ is cyclic quadrilateral and $OP$ is the diameter of the circle. Also both lines pass through origin.
So if $(x, y)$ is the coordinates of $P$, radius of the circle is $ \displaystyle r = \frac{\sqrt{x^2 + y^2}}{2}$.
If angle between two lines is $\theta$, the angle subtended by chord $AB$ on the... | {
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Sufficient criteria for specific constrained extrema I'm trying to solve the next problem using lagrange multipliers:
Find the distance from the origin to the curve $z^2=x^2+y^2$, $x-2z = 3$.
We define the Lagrangian
$$
\mathcal{L}(x,y,z,\lambda,\mu) = x^2+y^2+z^2 + \lambda(x^2+y^2-z^2)+\mu(x-2z-3)
$$
and found two so... | The surface $z^2 = x^2 + y^2$ is a cone that forms both above $z = 0$ and below it, and z-axis is the axis of cone. The plane $x-2z = 3$ is parallel to y-axis. It is easy to visualize where the maximum and minimum distance to origin occur. Here is a simple approach -
The curve is given by intersection of $z^2=x^2+y^2, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4155629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the equations of circles passing through $(1, -1)$ touching the lines $4x+3y+5=0$ and $3x-4y-10=0$
Find the equations of circles passing through $(1,-1)$ touching the lines $4x+3y+5=0$ and $3x-4y-10=0$
The point of intersection of the lines is $(\frac25,-\frac{11}5)$
If we want this point of intersection to be $... | Geometric approach.
The centers of all inscribed circles are located on a bisector line,
which equation is $y=7x-5$.
Construct auxiliary inscribed circle, say centered at $O=(2,9)$
with the radius $r=8$.
The line $XP$ intersect this circle at
points $D(2,1)$ and $E(\tfrac{42}5,\tfrac{49}5)$
and we have two scaling coe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4157083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Power series of a function around a point ≠ 0 Using the power series $ \sum_{k \geq 0}z^k = \frac{1}{1-z} $ for $ |z|<1 $, where it's centered around $0$. How would the series look like if someone wanted to let, for example $ z_0 = \frac{1}{2}$?
Also, how would one compute the power series of $ \frac{1}{1+z^2} $ around... | Just manipulate the fraction
$$
\dfrac{1}{1-z} = \dfrac{1}{1-z+1/2-1/2} = \dfrac{1}{1/2-z+1/2} = \dfrac{1}{1/2-(z-1/2)} = \dfrac{1}{\dfrac{1}{2} \Big(1 - 2(z-1/2) \Big)}.
$$
And
$$
\dfrac{1}{\dfrac{1}{2} \Big(1 - 2(z-1/2) \Big)} = 2 \sum_{k=0}^{\infty} \Big(2 \, (z - 1/2) \Big)^k = \sum_{k=0}^{\infty} 2^{k+1} \Big(z - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4157918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve the equation: $ 3^{2x+1}+4 \cdot 3^x = 15 $ Solve the equation: $ 3^{2x+1}+4 \cdot 3^x = 15 $ where $x$ is a real number.
Background: Doing Olympiad question and got one from the book.
Attempt:
Let $3^x$ be $u$.
\begin{align*}
3^{2x+1} + 4 \cdot 3x - 15 &= 0 \\
3^{2x} \cdot 3^1 + 4 \cdot 3^x - 15 &= 0 \\
3^{2x} \... | It depends. It seems as if you already have a short solution and you may consider taking base-$3$ logarithm to make a shorter one, but most of the time, logarithms are of base-$e$, which is exactly what you did.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4164980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Mean independent A random variable $X$ is said to be mean independent of another random variable $Y$
if its conditional expectation given $Y$ is equal to its unconditional expectation, that is, $E[X \mid Y]=E(X)$.
My question is, for $\phi$ Borel function we can say that $E[\phi(X) \mid Y]=E[\phi(X)]$ ?
Thanks.
| Let $R$ be the result of rolling a fair die.
Define the random variable $X$ by
$$
X=(R\;\text{mod}\;3)
\qquad\qquad\qquad\qquad\;\;\,
$$
and define the random variable $Y$ by
\begin{cases}
Y=0&\text{if}\;R\in\{2,3,5,6\}
\qquad\qquad\;\;\;\,
\\[4pt]
Y=1&\text{if}\;R\in\{1,4\}\\
\end{cases}
Then we get
$$
\left\lbrace
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4165745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How many five-digit numbers can be made from the digits $0, 1, 2, 3, 4$, provided that each digit can be repeated twice? The 10,000th place can be filled in $4$ different ways, the 1000th place can be filled in $5$ different ways, the 100th place can be filled in $4$ different ways, seeing the double repetition of nume... | $\color{red}{HINT=}$There are three situation such that zero is used once , zero is used two times and zero is not used.
Zero is not used : Our generaitng function is $(1+ x+ \frac{x^2}{2})^4$ , then find the coefficient of $x^5$ and multiply it by $5!$
Zero is used once= Our generating function is $(1+ x+ \frac{x^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4167300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Algebra question to find all ordered pairs to a set of equations Find all ordered pairs of real numbers ($x,y$) for which
$$(1+x)(1+x^{2})(1+x^{4})=1+y^{7}$$
$$(1+y)(1+y^{2})(1+y^{4})=1+x^{7}$$
I am not sure on how to solve this. There is a clear symmetry in the problem. Also on expanding the LHS, we will get the GP
$$... | Observe that
*
*If $ x = y$, then we have the solutions $ (0, 0), (-1, -1)$.
*We will show that there are no solutions with $ x \neq y$.
*If $ x = 0$, then from the first equation $ y = 0$.
*If $ x > 0, y < 0 $ then $ 1 + x + x^2 + \ldots + x^7 > 1 > 1 + y^7 $, so no solutions. Likewise for $ x < 0, y > 0$.
*If $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4169885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Prove that $\frac{a}{a+\sqrt{2013a+bc}}+\frac{b}{b+\sqrt{2013b+ca}}+\frac{c}{c+\sqrt{2013c+ab}}\leq 1$ For positive real numbers satisfying $a+b+c=2013$. Prove that
$$\frac{a}{a+\sqrt{2013a+bc}}+\frac{b}{b+\sqrt{2013b+ca}}+\frac{c}{c+\sqrt{2013c+ab}}\leq 1$$
This is my attempt.
We have
$$\frac{a}{a+\sqrt{2013a+bc}}+\fr... | We have for $y>1$ and $a\geq b$ and $a\geq c$: $$p=\left(\frac{\left(abc\right)^{y}}{a^{y}+b^{y}+c^{y}}\right)^{\frac{1}{y}}\leq bc$$
So we have with $ap=q$:
$$ f(a)+f(b)+f(c)=\frac{a}{a+\sqrt{2013a+\frac{q}{a}}}+\frac{b}{b+\sqrt{2013b+\frac{q}{b}}}+\frac{c}{c+\sqrt{2013c+\frac{q}{c}}}$$
The function :
$$g(x)=\frac{x}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4171770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Computing $\lim_{x\to-1}\frac{2x+\sqrt{3-x}}{x^2+x}$ I have this limit as my question to solve:
$$\lim_{x\to-1}\frac{2x+\sqrt{3-x}}{x^2+x}$$
My procedure:
$$\lim_{x\to-1}\frac{(2x+\sqrt{3-x})(2x-\sqrt{3-x})}{(x^2+x)(2x-\sqrt{3-x})}$$
$$\lim_{x\to-1}\frac{4x^2+x-3}{(x^2+x)(2x-\sqrt{3-x})}$$
$$\lim_{x\to-1}\frac{(x+1)(x-... | It's because $\displaystyle4x^2+x-3=(x+1)(4x-3)$ rather than $\displaystyle(x+1)\left(x-\frac34\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4172689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Evaluating $\int_0^1\frac{\ln(x)\ln(1-x)\ln(1+\sqrt{1-x^2})}{x}\mathrm{d}x$ While I was working on computing $\sum_{n=1}^\infty\frac{{2n\choose n}H_{2n}^{(2)}}{4^n n^2}$, I came across the integral:
$$I=\int_0^1\frac{\ln(x)\ln(1-x)\ln(1+\sqrt{1-x^2})}{x}\mathrm{d}x.$$
I tried $x=\sin(u)$ then $\tan(u/2)=t$ and got
$$I=... | Bonus: Differentiate both sides of $\int_0^1 x^{2n-1}\ln(1-x)dx=-\frac{H_{2n}}{2n}$, we get
$$\int_0^1 x^{2n-1}\ln(x)\ln(1-x)dx=\frac{H_{2n}}{4n^2}+\frac{H_{2n}^{(2)}}{2n}-\frac{\zeta(2)}{2n}.$$
Multiply both sides by $\frac{2n\choose n}{4^nn}$ then sum up from $n=1$ to $\infty$, we obtain
$$\frac14\sum_{n=1}^\infty\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4176803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 0
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$\int_0^{\infty} \frac{\sin x}{x}\frac{\sin \frac{x}{3}}{\frac{x}{3}} dx = \frac{\pi}{2}$ and more general results? Problem: evaluate that $\int_0^{\infty} \frac{\sin x}{x}\frac{\sin \frac{x}{3}}{\frac{x}{3}} dx = \frac{\pi}{2}$ and prove if a more general case, $\int_0^{\infty} \frac{\sin x}{x}\frac{\sin \frac{x}{3}}{... | These are Borwein Integrals and the behavior of exactly equaling $\pi/2$ stops at $2n+1 = 15$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4178486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Number of integer solutions of $a^2+b^2=10c^2$ Find the number of integer solutions of the equation $a^2+b^2=10c^2$.
I can only get by inspection that $a=3m, b=m,c=m$ satisfies for any $m \in Z$.
Is there a formal logic to find all possible solutions? Any hint?
Also i tried taking $a=p^2-q^2$, $b=2pq$ and $10c^2=(p^2+q... | $$\color{magenta}{a=3p^2-2pq-3q^2, \quad b=-p^2-6pq+q^2, \quad c=p^2+q^2}$$
or, same thing
$$\color{green}{a=3x^2-2xy-3y^2, \quad b=-x^2-6xy+y^2, \quad c=x^2+y^2}$$
When $p,q$ are coprime, the primes that can still divide $\gcd(a,b,c)$ are $2$ and $5.$ The proof of all (primitive, integral) solutions is just showig t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4180314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
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How to find local maxima of $\frac{x^2+2x-3}{x^2+1}$ quickly?
What is the value of relative maximum of the function
$f(x)=\dfrac{x^2+2x-3}{x^2+1}$ ?
$1)-1+\sqrt5\qquad\qquad2)1+\sqrt5\qquad\qquad3)-1+\sqrt3\qquad\qquad4)1+\sqrt3$
It is a problem from a timed exam so I'm looking for the fastest approaches. Here is my ... | $$\frac{2x-4}{x^2+1}=u$$
$$\begin{align} &\implies ux^2-2x+(u+4)=0\\
&\implies \Delta=1-u(u+4)≥0 \\
&\implies u^2+4u-1≤0 \\
&\implies (u+2)^2-5≤0 \\
&\implies |u+2|≤\sqrt 5 \\
&\implies \max \left\{u\right\}=\sqrt 5-2 \\
&\implies \max \left\{f(x)\right\}=\sqrt 5-1.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4180440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
A proof in $\varepsilon$-language for $\lim \sqrt[n]{1^2+2^2+...+n^2} = 1$ I found a proof that $\lim \sqrt[n]{1^2+2^2+...+n^2}=1$ by $\varepsilon$-language, but I think it's quite complicated and not sure that it's correct.
My question is:
1- Is my proof correct?
2- Is there another simpler proof in the sense of $\var... | Note that $\sum_{k=1}^{n} k^a \sim \frac{n^{a+1}}{a+1}$
because $$\lim_{n\to \infty} \frac{1}{n}\sum_{k=1}^{n} (k/n)^a =\int_{0}^{1} x^a dx=\frac{1}{a+1}$$
So $$L=\lim_{n \to \infty} \left [\sum_{k=1}^{n} k^2 \right]^{1/n}=\lim_{n \to \infty} (n^3/3)^{1/n}= \lim_{n\to \infty} \exp[\frac{1}{n} \ln (n^3/3)]=\lim_{n \to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4180599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove multiplication closure for the sequence of 1,4,7,10.... I am reading the following problem:
If S = ${1, 4, 7, 10, 13, 16, 19, ...}$ and $a \in S\space$ and $b\in
S \space$ then if $a = b\cdot c\space$ prove that $c \in S$
My approach:
The elements of $S$ are of the form $1 + 3n\space$ so $a = 1 + 3\cdot x\spac... | Since $a,b\in S$ we have $a=3x+1$ and $b=3y+1$ for some integers $x,y$. So $$3x+1 = (3y+1)c$$ and thus $$c = 3\underbrace{(x-yc)}_k +1 = 3k+1 \implies c\in S$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4183499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find $a$ & $b$ , s.t. $a+b=0$ and $ab=-3$? I've a problem in this question :
In the polynomial identity $x^6+1=(x^2+1)(x^2+ax+1)(x^2+bx+1)$ , find $ab$ $? $
MY APPROACH
We have : $$x^6+1=(x^2)^3+1=(x^2+1)(x^4-x^2+1)$$
Now according to the Problem : $$x^6+1=(x^2+1)(x^2+ax+1)(x^2+bx+1)=(x^2+1)(x^4-x^2+1)$$ Or $$x^4+(a+... | Since, $a+b=0$
$\implies a=-b$
And, $ab=-3$
$\implies -b^2=-3$
$\implies b=\pm\sqrt{3}$
So, $a+b=0$
$\implies a\pm\sqrt{3}=0$
$\implies a=\mp\sqrt{3}$
$\therefore (a,b)=(\pm\sqrt{3},\mp\sqrt{3})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4187515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How can i solve this integral by differentiating under integral sign or by any other method $$\int_{1}^{\infty}\frac{dx}{(x+a)^n\sqrt{x-1}}$$
where $\Im(a)\neq0$ and $\Re\ge-1$
Here's how i tried to solve
$$I(a)=\int_{1}^{\infty}\frac{dx}{(x+a)\sqrt{x-1}}=\frac{\pi}{\sqrt{a+1}}$$
Differentiating wrt $a$
$$I'(a)=\int_{1... | Upon differentiating, you should notice that
$$(n-1)! \int_1^{\infty}\frac{dx}{(x+a)^n\sqrt{x-1}} = \frac{(2n-3)\cdot(2n-5) \cdots1}{2\cdot2\cdots2}\frac{\pi}{(a+1)^{n-\frac{1}{2}}}$$
where there are $(n-1) \; 2's$ in product in the denominator.
Now the RHS can be manipulated a bit by multiplying and dividing by $(2n-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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What is my mistake in finding this pythagorean triplet? Since Project Euler copyright license requires that you attribute the problem to them, I'd like to add that this is about question 9 there.
I am trying to solve this problem on only two brain cells and can't figure out what am I doing wrong. Here is the system for... | $a = p^2 - q^2\\
b = 2pq\\
c = p^2 + q^2$
For any $(p,q:p>q), (a,b,c)$ will be a pythagorean tripple
$p^2 - q^2 + 2pq + p^2 + q^2 = 1000\\
2p^2 + 2pq = 1000\\
p(q+p) = 500$
$p$ is a factor of $500$
If $p$ is too small, $q$ is too big, and $p^2 - q^2 < 0$
And $p$ is too big, $q< 0$
$\sqrt{250}<p<\sqrt{500}$
$p = 20,$ wo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4191659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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How do I have to use this assumption?
Let be $V$ a vector space over $\mathbb{R}$ with dimension 4. Let be
$T$ a linear operator in $V$. In case that a base $\mathcal{B}$ of $V$
exists, such that the associated matrix of $T$ relative to the basis
$\mathcal{B}$ is: $$ \left [ T \right ]_{\mathcal{B}}=\begin{pmatrix}
1... | Note that $a^2 + b^2 + c^2 = 0$, for $a, b, c \in \mathbb{R}$, only holds if $a = b = c = 0$.
Your error comes in this step:
\begin{align*}
a \cdot x_2&=0& & &x_2=0 & &\\
b \cdot x_3&=0& &\Longrightarrow &x_3=0 & &\wedge& &x_1=r\\
c \cdot x_4&=0& & &x_3=0 & &
\end{align*}
Because in this step, you've assumed that $a, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4193017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Can we obtain the relation between $x,y$ in terms of $a,b$ for which $ (a+x-y)(x+y)>b $? I have this condition
$$ (a+x-y)(x+y)>b ,$$
where $x,y$ are two positive parameters and $a,b$ are fixed positive numbers.
Are we able to obtain other simpler alternative relations between $x,y$ in terms of $a,b$ under which the gi... | Rearranging the expression:
\begin{align*}
(a+x-y)(x+y) &> b\\
a(x+y)+(x-y)(x+y) &> b\\
ax+ay+x^2-y^2 &> b\\
(x^2+ax)-(y^2-ay) &> b\\
(x^2+ax+a^2/4-a^2/4)-(y^2-ay+a^2/4-a^2/4) &> b\\
(x^2+ax+a^2/4)-a^2/4-(y^2-ay+a^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4200469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving $2((a+b)^4+(a+c)^4+(b+c)^4)+4(a^4+b^4+c^4+(a+b+c)^4)=3(a^2+b^2+c^2+(a+b+c)^2)^2$ in another way? How do I prove the following identity without expanding both sides directly.
$$2((a+b)^4+(a+c)^4+(b+c)^4)+4(a^4+b^4+c^4+(a+b+c)^4)\\=3(a^2+b^2+c^2+(a+b+c)^2)^2$$
I expanded both sides directly and it is true. Howeve... | You are trying to prove that $F=G$ where $F$ and $G$ are symmetric homogeneous 4th degree polynomials in three variables. Fully expanded, such a polynomial is in the form $$A\left(a^4+b^4+c^4\right)+B\left(a^3(b+c)+b^3(a+c)+c^3(a+b)\right)\\+C\left(a^2bc+b^2ac+c^2ab\right)+D\left(a^2b^2+a^2c^2+b^2c^2\right)$$
Imagine t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove that $2^n + 5^n + 56$ is divisible by $9$, where $n$ is an odd integer
Prove that $9 \mid2^n + 5^n + 56$ where n is odd
I have proved this by using division into cases based on the value of $n\bmod3$ but it seems a little bit clumsy to me and I wonder if there are other ways to prove it, probably by using modul... | I wanted to offer a different approach, using a bit of algebra.
$$\begin{align}2^n+5^n+56 &\equiv \\ &\equiv2^n+5^n+54+2\\
&\equiv 2^n+5^n+2\\
&\equiv 10^n-1+2^{2n}+2\times 2^n+1\\
&\equiv2^{2n}+2\times 2^n+1\\
&\equiv \left(2^n+1\right)^2\\
&\equiv 0\thinspace \thinspace \thinspace(\text{mod 9}) \end{align}$$
Becaus... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 7
} |
Find the area of the right triangle with a right triangle I came up with a problem myself:
What is the area/sides of the largest right triangle?
Question: Point $D$ is the midpoint of $AB$ and point $E$. It is also known that $\triangle AEF$ is also a right triangle. The goal is to find the area and sides of the large... |
Let
\begin{align}
|BC|&=a=6
,\quad
|AC|=b=8
,\quad
|AB|=|BE|=|BF|=c=10
,\\
|AD|&=|BD|=|CD|=\tfrac c2=5
,\\
\angle BCA&=
\angle FEA=
\angle DGC=
\angle DHE=
90^\circ
.
\end{align}
Then $\triangle EDH\sim\triangle CDG$
and
\begin{align}
\frac{|EH|}{|CG|}&=
\frac{|DE|}{|DC|}
=\frac{|DC|+|CE|}{|DC|}
\tag{1}\label{1}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4208108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find the size of the segment joining the foot of the perpendiculars of a scalene triangle The sides of a scalene triangle measure 13, 14, and 15 units. Two outer bisectors of different angles are drawn and the third vertex is drawn perpendicular to these bisectors. Calculate the size of the segment joining the foot of ... | The way you had started does lead to a solution, although it is a little calculative. Note that:
$$AH=13 \cos \left(\frac {\beta}{2}\right)$$
$$AI=15 \cos \left(\frac {\gamma}{2} \right)$$
Also, $\angle AHI=\alpha +\frac {\beta+\gamma}{2}=90°+\frac {\alpha}{2}$.
Now, let $HI=x$. From the cosine law on $\Delta HAI$, we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4209269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all positive integers $ a $ and $ b $ such that $7^a-3(2^b)=1$ I am looking for all positive integers satisfying the equation
$$7^a-3(2^b)=1$$
I used the fact that $ (1,1) $ is a solution and got
$$(7^a-7)=3(2^b-2)$$
I thaught about using Gauss Theorem in vain.
Thanks in advance for any idea.
| An argument mod $7$ shows that $b+2$ is divisible by $3$, then let $b=3n-2.$
We can take the two cases $a=2m, a=2m+1.$
The problem can be reduced to finding the integer points on elliptic curves as follows.
$\bullet\ a=2m$
Let $X=3\cdot2^{n}, Y=6\cdot7^{m}$, then we get
$Y^2 =X^3 + 36.$
According to LMFDB, this ellipti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4212397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$(1+A)^n \ge (1+a_1)\cdot(1+a_2)\cdots(1+a_n) \ge (1+G)^n$ $A$ and $G$ are arithmetic mean and the geometric mean respectively of $n$ positive real numbers $a_1$,$a_2$,$\ldots$,$a_n$ . Prove that
*
*$(1+A)^n$$\ge$$(1+a_1)\cdot(1+a_2)\cdots(1+a_n)$$\ge$$(1+G)^n$
*if $k$$\gt$$0$, $(k+A)^n$$\ge$$(k+a_1)\cdot(k+a_2)\cdo... | I think there is a much faster way.
*
*$\log(k+x)$ is concave, hence by Jensen's inequality $\log(k+A)\geq\frac{1}{n}\sum_{m=1}^{n}\log(k+a_m)$. Exponentiation leads to $(k+A)^n \geq \prod_{m=1}^{n}(k+a_m)$.
*Let $b_m=\log(a_m)$ or, equivalently, $a_m=e^{b_m}$. We have that $\log(k+e^x)$ is convex, since $\frac{d^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4213478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
What is the sum of the coefficients of the terms containing $x^2t^3$ in the given expansion.
It is given that $(x^2 +y +2t +3k)^{10}$. What is the sum of the coefficients of the terms containing $x^2t^3$ in the given expansion. For example the terms are $x^2y^4t^3k^2$ etc.
I said that let $(x^2 +2t)$ be $"a"$ and $(... | As an alternative, we can use the multinomial expansion formula:
$$(x_1+x_2+\cdots +x_m)^n=\sum_{i_1+i_2+\cdots +i_m=n} {n\choose i_1,i_2,...,i_m} \prod_{t=1}^m x_t^{i_t}, \\
\quad \text{where} \quad {n\choose i_1,i_2,...,i_m}=\frac{n!}{i_1!i_2!\cdots i_m!}\\
$$
$$ (x^2 +y +2t +3k)^{10}=\sum_{i_1+i_2+i_3+i_4=10} {10\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4213641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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For which prime $p$, is the group units $R^*$ cyclic? I'm studying for my qualifying exam and found the following question in the question bank of previous years paper:
Let $p$ be a prime and $\mathbb{F}_p$ the field with $p$ elements. Set $R=\mathbb{F}_p[x]/(x^3)$. For which prime $p$ is the group of units $R^*$ cycl... | First, note that if $ax^2 + bx + c$ has $c \neq 0,$ then $\gcd(x^3, ax^2 + bx + c) = 1$ in $\mathbb{F}_p[x]$ (since the only prime factor of $x^3$ is $x,$ which won't divide $ax^2+bx+c$ for $c\neq 0$) and so $ax^2+bx+c$ is invertible mod $x^3.$ It's easy to see that nothing else can be invertible mod $x^3$ from a simil... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4214034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding the value of $n\in\mathbb{Z}$ in the equation $16^n-25\times 12^{n-1}+9^n=0$ If $n$ is satisfy the equation $16^n-25\times 12^{n-1}+9^n=0$. Find the value of $n$.
Here is my work:
$$16^n-2\times12\times 12^{n-1}+9^n=12^{n-1}$$
$$(4^n-3^n)^2=12^{n-1}$$
For $n\in\mathbb{Z}^+\cup\{0\}$ we can see the LHS is an odd... | The original expression $16^n-25\times 12^{n-1}+9^n=0$ when taken $\dfrac{1}{12}$ out common reduces to $\dfrac{1}{12}(4\cdot 3^n-3 \cdot 4^n)(3^{n+1}-4^{n+1})=0$
This results in 2 cases:
Case 1:
$(3^{n+1}-4^{n+1})=0 \implies 3^{n+1}=4^{n+1}$
This is one possible when both have power $0$
So $n+1=0 \implies n=-1$
Case 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4214879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Show that $ \frac{x^2}{\sqrt{x^2+y^2}} + \frac{y^2}{\sqrt{y^2+z^2}} + \frac{z^2}{\sqrt{z^2+x^2}} \geq \frac{1}{\sqrt{2}}(x+y+z) $ Show that for positive reals $x,y,z$ the following inequality holds and that the constant cannot be improved
$$
\frac{x^2}{\sqrt{x^2+y^2}} + \frac{y^2}{\sqrt{y^2+z^2}} + \frac{z^2}{\sqrt{z^2... | When $x=y=z$, lhs=$\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}+\frac{z}{\sqrt{2}}=\frac{x+y+z}{\sqrt{2}}$=rhs, which means that $\sqrt{2}$ is best constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4215933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Solve the system of differential equations and plot the curves given the initial conditions. We are given the following system of ordinary differential equations:
$$\dot{x} = x - 4y \quad \dot{y} = x - 2y -4.$$
Thus, the slope of the trajectories is given by
$$\frac{dy}{dx} = \frac{\dot{y}}{\dot{x}} = \frac{x - 2y -4}{... | Here is an approach, from the second equation, we have
$$x = y' + 2 y + 4, ~~\mbox{so}~~ x' = y'' + 2 y'$$
Substituting these two into the first equation
$$ y'' + y'+2 y = 4$$
For initial conditions, we are given $y(0) = 5$ and use the second equation to find $y'(0) = x(0) - 2 y(0) - 4 = 1$.
This DEQ and ICs lets you f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4217864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Color ball drawer probability question There is a ball drawer.
Seven color balls will be drawn with the same probability ($1/7$).
(black, blue, green, yellow, white, pink, orange)
If Anson attempts $9$ times,
what is the probability that he gets all $7$ different color balls?
My work:
I separate the answer to $3$ ways.... | Since there are seven choices for each of the nine balls Anson selects, there are $7^9$ possible sequences of colors.
Method 1: If each color appears among the nine balls, there are two possibilities:
*
*One color is selected three times and each of the other colors is selected once.
*Two colors are each selected t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4220311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find the second real root for cubic $x^3+1-x/b=0$ A cubic of the form $$x^3+1-x/b=0$$ has has three real roots
Using the Lagrange inversion theorem one of the roots is given by
$$x = \sum_{k=0}^\infty \binom{3k}{k} \frac{b^{3k+1} }{(2)k+1} $$
How do you find the second one? I cannot find any info online. I am looking... | Too long for a comment.
Looking for truncated series expansions, you can directly solve for the roots solving the cubic using the trigonometric method.
The solutions are given by
$$x_k=\frac{2 \cos \left(\frac{2 \pi }{3}k-\frac{1}{3} \cos
^{-1}\left(-\frac{3\sqrt{3}}{2} b^{3/2}\right)\right)}{\sqrt{b}\sqrt{3}}\qqu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4224731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
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.