Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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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.
| I doubt a closed form exists for this number, but I'll show convergence and give some bounds on the values.
Let $f_m^n = \sqrt{m\pm\sqrt{(m+1)\mp \sqrt{\cdots\pm \sqrt n}}} $
be the the version where we go from $m$ to $n$ (don't take the $\pm$'s too seriously). We have the recursive expression
$$
f_m^n = \begin{cases}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4056039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 0
} |
How to solve $\int \csc^{4} \frac{3}{2} x \,dx$ using integration by parts How to solve $\int \csc^{4} \frac{3}{2} x \,dx$
This is what I have so far:
$$\int \csc^{4} \frac{3}{2} x \,dx=\frac{3}{2} \int \csc^{2} x \csc^{2}x\,dx.$$
Let $u=\csc^{2}$ and $dv=\csc^{2} \,dx$. Then we have $du=-2\csc^{2}x\cot x\,dx$ and $v=-... | Let $t=\frac32x$ and integrate by parts as follows
\begin{align}
\int \csc^{4} \frac{3}{2} x \,dx
&= \frac23\int \csc^{4} t \,dt=- \frac29\int \sec^{2} t \>d(\cot^3 t)\\
&= -\frac29\csc^2 t\cot t+\frac49 \int\csc^2 t dt\\
&= -\frac29\csc^2 t\cot t-\frac49 \cot t +C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4056699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
locus of points with sum of distance
Find the quadratic equation for the locus of points whose sum of its distance from $(1, 0)$ and $(−1, 0)$ is $6$.
I know to start with $\sqrt{(x-1)^2 +y^2} + \sqrt{(x+1)^2+y^2}=6$, then square both sides. I'm not sure about the algebra after squaring both sides as I got that part ... | $$\sqrt{(x-1)^2 +y^2} + \sqrt{(x+1)^2+y^2}=6$$
$$\sqrt{(x+1)^2+y^2}=6-\sqrt{(x-1)^2+y^2}$$
square both sides ($LHS>0;\;RHS>0$)
$$(x+1)^2+y^2=\left(6-\sqrt{(x-1)^2+y^2}\right)^2$$
$$x^2+2 x+y^2+1=36+x^2-2 x+1+y^2-12 \sqrt{(x-1)^2+y^2}$$
bring every term but the square root in the RHS
$$12 \sqrt{(x-1)^2+y^2}=36-4 x$$
sim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4058108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $3y^2\frac{\partial z}{\partial x}-\left(z-x\right)\frac{\partial z}{\partial y}=3y^2$, I keep getting the same wrong answer For the conditions $x\left(t\right)=t^{\frac{1}{2}},\:y\left(t\right)=t^{\frac{1}{3}},\:z\left(t\right)=0,\:t>0$
I get this:
$$\frac{dx}{3y^2}=\frac{dy}{-z+x}=\frac{dz}{3y^2}$$
$$(I)\:\:... | $$\frac{dy}{-z+x}=\frac{dz}{3y^2}$$
You should use the constant $C_1$ instead. Since $C_1=x-z$. Your integration of this differential equation is not correct.
$$3y^2{dy}=C_1{dz}$$
Integrate.
$$y^3=C_1z+C_2$$
$$y^3=(x-z)z+C_2$$
$$C_2=y^3-xz+z^2$$
This line seems not correct to me:
$$y^3=-\frac{z^2}{2}+zx+C_2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4058379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Different answers after different methods in solving a limit
Evaluate $$L=\lim_{x \to 0} \frac{e^{\sin(x)}-(1+\sin(x))}{(\arctan(\sin(x)))^2}$$
Method $1$: $$\frac{h^2\left(\frac{e^{h}-1}{h^2}-\frac1{h}\right)}{(\arctan(h))^2}=1^2\left(\frac{1*1}{h}-\frac1{h}\right)=\frac1{h}-\frac1{h}=0$$ Therefore $L=0$.
The identi... | My (personal) favored approach is composition of series one piece at the time.
$$y=\frac{e^{\sin(x)}-(1+\sin(x))}{\big[\arctan(\sin(x))\big]^2}$$ Since the denominator is $\sim x^2$, let us use expansions to $O(x^4)$. Working one piece at the time
$$\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)$$
$$e^{\sin(x)}=1+x+\frac{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4058968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What is the particular solution given $y=4$ and $x=3$ for the equation: $xy \frac{dy}{dx}=\frac{x^3-x}{1-\sqrt y}$? I am almost done solving this question, however, I am stuck on the integration. This is my work; the answer is:
$$\int\left(\frac{y^2}{2}-\frac{{2y^{5/2}}}{5}\right)dy=\int \left(x^2-x\right)dx$$
and I ha... | The ODE is $$\displaystyle \small xy\frac{dy}{dx} = \frac{x^3-x}{1- \sqrt y},$$ which can be rewritten as
$$\displaystyle \small (y-y^{3/2}) \ dy = (x^2-1) \ dx. \ $$ This is the step where you have a mistake! By integrating both sides we get:
$$\displaystyle \small \frac{y^2}{2} - \frac{2}{5} y^{5/2} = \frac{x^3}{3} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4063652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find length BC of triangle with incircle and circumcircle
Some thoughts... some chord theorem to get the angle between PQ and BC... AB and AC are tangent to the circle, there has to be another theorem about that, perhaps ADE is isosceles and that helps looking at the angled PDE and AEQ, and with all this one should be... | Here are 2 solutions 1) an analytical solution 2) a solution inspired by the solution of @Quanto.
*
*First solution; Let us take coordinate axes as displayed in Fig. 1 below.
Line $PQ$ is taken as the $x$-axis, with origin and unit such that $P,D,E,Q$ have resp. abscissas $-3,-1,1,2$.
Triangle $ADE$ is isosceles, du... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4065917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Given $x^2 +px + q$ has roots -1 & 4, find the values for p & q.
Given $x^2 +px + q$ has roots $-1$ & $4$, find the values for $p$ & $q$.
Attempt:
$$
x = \frac{-p\pm\sqrt{p^2-4q}}{2}\\(p + 2x)^2 = p^2 - 4q\\
p^2 + 4px + 4x^2 = p^2 -4q
\\
q = -x^2 - px
\\
q = -(-1)^2 -p(-1)
\\
q = -1 - p
\\
q = -(4)^2-p(4)
\\
q = -16 ... | The Vieta's formulas tells you that
$$p=-(r+s),\\q=rs$$ (where $r,s$ are the roots).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4066478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to evaluate $ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$? I was given the series:
$$ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$$
Making some observations I realized that the $ a_{n} $ term would be the following:
$$ a_... | HINT:
$\frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$
Separate odd and even terms.
$\frac{1}{3} +\frac{1}{3^{3}} + \frac{1}{3^{5}} + ...$
$ \frac{2}{3^{2} } + \frac{2}{3^{4}} +\frac{2}{3^{6}} ...$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4067389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 6
} |
Find eigenvalues of $T$ and find a basis $\beta$ for $P_2(\mathbb{R})$ such that$[T]_\beta$ is a diagonal matrix. $T(f(x)=3f(x)+xf'(x)+xf''(x)$ what I did was, $T(1)=3+0+0=3,T(x)=3x+x+0=4x,T(x^2)=3x^2+2x^2+2=5x^2+2$, then $[T]_\beta=\begin{pmatrix}3&0&0\\0&4&0\\0&2&5\end{pmatrix}$ how should I continue after this?
| Were you given a problem like this without knowing what eigenvalues and eigenvectors are? That's very strange!
The eigenvalues satisfy the equation $\left|\begin{array}{ccc}3-\lambda & 0 & 0 \\ 0 & 4-\lambda & 0 \\ 0 & 2 & 5- \lambda \end{array}\right|= (3- \lambda)(4- \lambda)(5- \lambda)= 0$ so are 3, 4, and 5.
An e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4068398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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
$$I(a)=\int_0^\infty \frac{\ln(a/y)}{a^2+y^2}dy=\int_0^\infty\frac{\ln... | Noticing that
by which $$I= \left.\frac{\partial}{\partial a} I(a)\right|_{a=0} $$
where$$I(a)= \int_0^{\infty} \frac{x^a}{\left(1+x^2\right)^2} d x =\frac{1}{2} \int_0^1 y^{\frac{1-a}{2}}(1-y)^{\frac{a-1}{2}} d y\\ \quad \qquad =\frac{1}{2} B\left(\frac{3-a}{2},\frac{a+1}{2}\right) =\frac{1}{2} \Gamma\left(\frac{3-a}{... | {
"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": 5
} |
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... | Note that $\sin x\le 1$ and $\cos \ge -1.$
$$\frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}\ge{e^{x^2}-5}.$$
So
$$\lim_{x\to\infty}\frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}\ge \lim_{x\to\infty}{e^{x^2}-5}.\to\infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4069593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find the value of $T=\mathop {\lim }\limits_{n \to \infty } {\left( {1+ \frac{{1+\frac{1}{2}+ \frac{1}{3}+ . +\frac{1}{n}}}{{{n^2}}}} \right)^n}$ I am trying to evaluate
$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}}} \right)^n}.$$
My sol... | We want to show that $$\displaystyle \lim_{n \to \infty} \dfrac{H_n}{n}=0$$.
We have $\log n = \displaystyle \int_{1}^{n} \dfrac{1}{t} dt$. Also, $H_n = \displaystyle \sum_{r=1}^{n}\dfrac{1}{r} = \dfrac{1}{n}\sum_{r=1}^{n}\dfrac{1}{\frac{r}{n}}$
Thus, by comparison with the Riemann sums of $\log n$, we see that $\log n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 6
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Finding the orthogonal trajectories of the family of cissoids Find the orthogonal trajectories of the cissoids, $y^2=\frac{x^3}{a-x}$.
My solution:
$\left( a-x\right)y^2=x^3$
$\left( a-x\right)2y dy-(2x^2+y^2)dx=0$
$\left( a-x\right)2ydx + (2x^2+y^2)dy=0$
$a=\frac{x^3}{y^2}+x$
$2y \left( \frac{x^3}{y^2}\right)dx+(2x^2+... | Taking a derivative of your equation
\begin{align}
y&=\sqrt{\frac{x^3}{a-x}},\\
\frac{\mathrm dy}{\mathrm dx}&=\frac{3a-2x}{2}\sqrt{\frac{x}{(a-x)^3}},
\end{align}
and substituting $a=x^3/y^2+x$ gives
\begin{align}
\frac{\mathrm dy}{\mathrm dx}&=\frac{3x^3/y^2+x}{2}\sqrt{\frac{x}{(x^3/y^2)^3}}.
\end{align}
The orthogon... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4072056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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
$$\sqrt{(x-1) (x+1) \left(2x^2-1\right)} + \sqrt{x^2\left(2x^2-1\... | let $a=2x^4-3x^2+1,b=2x^4-x^2$ where $ a,b\ge 0$ now $\sqrt{a+b}=2x^2-1$ so $$\sqrt{a}+\sqrt{b}-4x+3=0$$ $$\to \sqrt{a}+\sqrt{b}-\sqrt{a+b}+\sqrt{a+b}-4x+3=0$$ $$\to \sqrt{a}+\sqrt{b}-\sqrt{a+b}+2{(x-1)}^2=0$$ because $\sqrt{a}+\sqrt{b}\ge \sqrt{a+b}$ $$\to x=1 \space \text{and} \space \sqrt{a}+\sqrt{b}=\sqrt{a+b}$$ $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4079527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Showing $4-\csc^2\left( \frac{n \pi}{6(n-2)} \right)$ is on the order of $1/n$ The expression I have here is part of the upper bound given to the minimum distance of any 2 points out of $n$ points embedded in the unit sphere $\mathbb{S}^2$ in $\mathbb{R}^3$. If you have $n$ points on this unit sphere then we must have ... | We wish to show that $$f(x)=\left(4-\frac kx\right)\sin^2\frac{\pi x}{6(x-2)}\le1$$ for all positive real $x$. This is evidently the case when $4-k/x\le0$ so we consider $x>k/4$ with $$f'(x)=\frac k{x^2}\sin^2\frac{\pi x}{6(x-2)}-\left(4-\frac kx\right)\frac{24\pi}{(6x-12)^2}\sin\frac{\pi x}{6(x-2)}\cos\frac{\pi x}{6(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4080364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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How to resolve this integral $J = \int{\frac{2-\sin{x}}{2+\cos{x}}}dx$? $$J = \int{\frac{2-\sin{x}}{2+\cos{x}}}dx$$
I'm already try to do this step in below:
$$\overset{split}{=} \int{\frac{2}{2+\cos{x}}dx}+\int{\frac{d(\cos{x}+2)}{2+\cos{x}}} = \int{\frac{2}{1+2\cos^2{\frac{x}{2}}}dx} + \ln({2+\cos{x}})$$
| Note
$$\int{\frac{2}{1+2\cos^2{\frac{x}{2}}}dx}=\int{\frac{2\sec^2\frac x2}{\sec^2\frac x2+2}dx}
=4\int{\frac{d(\tan\frac x2)}{\tan^2\frac x2+3}dx}
= \frac4{\sqrt3}\tan^{-1}\frac{\tan\frac x2}{\sqrt3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4082645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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what result I should consider for GCD with polynomials? I need to calculate the GCD of
$$x^4+3x^3+2x^2+x+4 \ \text{and } x^3+3x+3 \ \text{in} \ \mathbb{Z}_5$$
Using Euclid algorithm:
$$x^4+3x^3+2x^2+x+4 = (x^3+3x+3)(x+3)-3x\\ x^3+3x+3 = (-3x)(\frac{1}{3}x^2 - \frac{2}{3})+3 \\-3x = (3)(-x)+0 $$
Now I should consider th... | Over the field $\Bbb F_5$ we have
$$
x^4+3x^3+2x^2+x+4=(x^3 + 4x^2 + x + 1)(x + 4)
$$
and $x^3+3x+3$ has no root in $\Bbb F_5$ and hence is irreducible.
So is the other polynomial of degree $3$. Here we have used that a cubic polynomial over a field is reducible if and only if it has a root.
From this it is clear that ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\sin\angle HIO$ 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... | $\displaystyle \small \angle IAO = \angle IAH = \frac{\angle B - \angle C}{2}$
Now use the identity, $\small AH = 2R \cos \angle A \implies AH = R = AO \ $ (as $\angle A = 60^0$)
That means $\small \triangle HAO$ is an isosceles triangle. As $\small AI$ is angle bisector of $\small \angle HAO, IH = OI$.
Now remember th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4088790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the area of three quadrilaterals inside an equilateral triangle $\triangle ABC$ is an equilateral triangle with side length of $1$. $D,E,F$ are $\frac{1}{3}$ away from $C,A,B$. What is the total area of the three quadrilaterals 1,2 and 3 enclosed by the orange sides? I can think about finding the coordinates of th... | Denote area of $\triangle ABC$ by $\Delta$. Let areas of the three quadrilaterals each be $X$ and the blue triangles in following diagram, each be $Y$.
We'll take ratios approach, which is fun, form two linear equations in $X,Y$ and solve for it.
First notice $\triangle$s $AED$, $BFE$, $CDF$ are all $30^\circ-60^\circ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4092390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
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 $$x_{1,2}=\dfrac{a^2+b^2\pm\sqrt{(a^2... | Using the quadratic formula is a good approach, and everything you have done so far is correct. The modulus/absolute value can be replaced by normal brackets as there is a $\pm$ preceding it, and the numbers involved here are real. Notice that:
$$\frac {a^2 +b^2 + (a^2 - b^2)}{2ab} = \frac ab, \quad \frac {a^2 +b^2 - (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4092687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Minimum of $E(z)=|z+1|+|z^{2n}+1|$
Let $n \in \mathbb{N}$. Find the minimum of the expression $E(z)=|z+1|+|z^{2n}+1|$ over $\mathbb{C}$.
I found this problem in a Romanian magazine with all sorts of math problems.
(Edit: ${\color{blue}{\textrm{This was suggested for a contest for 10th graders}}}$, According to @alexa... | shymilan@AoPS's elegant proof (I rewrote it):
Let $w_k = \mathrm{e}^{\mathrm{i}\frac{(2k - 1)}{2n}\pi}, ~ k = 1, 2, \cdots, 2n$. Then $Q(z) := z^{2n} + 1
= (z - w_1)(z - w_2)\cdots (z - w_{2n})$.
We have the partial fraction decomposition
$$\frac{1}{z^{2n} + 1} = \frac{1}{Q(z)}
= \sum_{k=1}^{2n} \frac{1}{Q'(w_k)}\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4093474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 2
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Order of a binomial coefficient
I am trying to find the Big-O order of $\binom{n-1}{\frac{n}{2}-1}$ where $n$ is even.
I know that $\binom{n}{\frac{n}{2}}=O\left(\frac{2^n}{\sqrt{n}}\right)$ using Stirling's approximation. I know that $\binom{n-1}{\frac{n}{2}-1}$ is smaller than $\binom{n}{\frac{n}{2}}$, so I'm guess... | Write this as
$$
2^n\times \frac{\sqrt{n-1}}{\sqrt{n}\cdot \sqrt{n-2}}\times \frac{(n-1)^{n-1}(n-2)}{(n(n-2))^{n/2}}\\
\\\approx 2^n\times \color{#2B2}{\frac1{\sqrt{n}}}\times { (1-\frac1n)^{n-1}(1-\frac2n)\over (1-\frac1{n/2})^{n/2}}
$$
and use the limit definition of $e$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4094339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding integer solutions to the system $xy=6(x+y+z)$, $x^2+y^2=z^2$ How do you go about solving a system of equations like below for integer solutions?
$$xy=6(x+y+z)$$
$$x^2+y^2=z^2$$
Would you first try and list out a number of Pythagorean triples, then try and see which ones when multiplied are divisible by 6 (by 2 ... | Let $t = x+y+z \implies z^2 = x^2+y^2 = (x+y)^2 - 2xy = (x+y+z - z)^2 - 2\cdot 6(x+y+z)= t^2-2tz+z^2-12t\implies t^2-2tz-12t=0\implies t(t-2z-12)=0\implies t = 0$ or $t = 2z+12$. Thus $x+y=-z$ or $x+y+z=2z+12$ or $x+y=z+12$. Thus first case yields $xy = 0 \implies x = 0 $ or $y = 0$ or $x = y = 0=z$. Thus $x = 0, y = -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4096326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Minimize $3\sqrt{5-2x}+\sqrt{13-6y}$ subject to $x^2+y^2=4$
If $x, y \in \mathbb{R}$ such that $x^2+y^2=4$, find the minimum value of $3\sqrt{5-2x}+\sqrt{13-6y}$.
I could observe that we can write
$$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... | We can parametrize the constraint condition as $$(x,y) = (2 \cos t, 2 \sin t), \quad t \in [0,2\pi),$$ hence we are interested in the extrema of $$f(t) = 3\sqrt{5-4\cos t} + \sqrt{13-12\sin t}.$$ Differentiating to locate critical points, we find $$\sin t \sqrt{13 - 12 \sin t} = \cos t \sqrt{5 - 4 \cos t},$$ or $$\tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4098030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Integrate $\int\frac{3x^2-1}{\sqrt{x^2+x-1}}dx$ Integrate $\int\frac{3x^2-1}{\sqrt{x^2+x-1}}dx$
I solved this integral by euler substitution by replacing
$\sqrt{x^2+x-1}=x+t$
but it's not allowed by the problem.
p.s Is there any other method to solve with?
Thank you in advance :)
| Note
\begin{align}
\int \sqrt{x^2+x-1} \>dx
&\overset{IBP}=\frac12 x \sqrt{x^2+x-1} +\frac14\int \frac {x-2}{\sqrt{x^2+x-1}}dx\\
\end{align}
Then
\begin{align}
&\int\frac{3x^2-1}{\sqrt{x^2+x-1}}\>dx\\
=& 3 \int \sqrt{x^2+x-1} dx - \int \frac {3x-2}{\sqrt{x^2+x-1}}dx\\
=& \frac32 x \sqrt{x^2+x-1} -\frac98 \int \frac {2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4100874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Expressing a determinant in the form $k(a^3 + b^3 + c^3 - 3abc)^n$ I have been given a determinant $\begin{vmatrix} bc - a^2 & ac - b^2 & ab - c^2 \\ ac - b^2 & ab - c^2 & bc - a^2 \\ ab - c^2 & bc - a^2 & ac - b^2\end{vmatrix}$ and I want to compute it in the form $k(a^3 + b^3 + c^3 - 3abc)^n$
Now I know that for dete... | The determinant is fully symmetric and homogeneous of degree $6$, and vanishes if $c=-a-b$ since then all entries are $-a^2-ab-b^2$, and also if $a=b=c$ since then all entries are $0$, so is divisible by both $a+b+c$ and$$\tfrac12((a-b)^2+(b-c)^2+(c-a)^2)=a^2+b^2+c^2-ab-bc-ca,$$and hence by their product, $a^3+b^3+c^3-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4102920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Calculating a limit. Is WolframAlpha wrong or am I wrong? What I'm trying to solve:
$$\lim _{x\to -\infty \:}\frac{\left(\sqrt{\left(x^2+14\right)}+x\right)}{\left(\sqrt{\left(x^2-2\right)}+x\right)}$$
What I put into WolframAlpha:
(sqrt(x^2+14)+x)/(sqrt(x^2-2)+x)
My result: $1$, which I get by simply dividing bot the ... | Your statement is correct for the limit as $ \ x \ $ approaches "positive infinity",
$$\lim _{x \ \to \ +\infty \:}\frac{\left(\sqrt{\left(x^2+14\right)}+x\right)}{\left(\sqrt{\left(x^2-2\right)}+x\right)} \ \ = \ \ 1 \ \ , $$
and the means by which you calculated it is valid in that direction.
A basic issue which ari... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4105217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
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... | Both of your answers are correct. Both are valid solutions. Remember this: for any trig-problem, there exists exactly one solution in a interval of length $\pi$. So, for an interval of length $2\pi$, there will be two solutions.
Usually, such problems arise in physics while studying waves and oscillations. There, $A$ m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4105426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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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... | \begin{align}&\frac{a-c}{b+c}+\frac{b-a}{c+a}+\frac{c-b}{a+b} = 1\\
\implies&\frac{a+b-b-c}{b+c}+\frac{b+c-c-a}{c+a}+\frac{c+a-a-b}{a+b} = 1\\
\implies&\frac{a+b}{b+c}-1+\frac{b+c}{c+a}-1+\frac{c+a}{a+b}-1 = 1\\
\implies&\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b}=4\\\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4106213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
k-th smallest number of standard normal distribution There are $n$ random variables obey the standard normal distribution $N(0 ,1)$.
What is the expectation of the k-th smallest number?
My attempt
I tried to use conditional probability to solve the problem, but the situation of $n=4$ is too complex to handle.
for n = 2... | The density of the $k$-th smallest number among $n$ i.i.d standard normal random variables is given by
$$
f_{X_{(k)}}(x) = \frac{n!}{(r-1)!(n-r)!} \phi(x) \left[\Phi(x)\right]^{k-1}
\left[1-\Phi(x)\right]^{n-k},$$
where $\phi(x)$ and $\Phi(x)$ are PDF and CDF of standard normal distribution respectively. Then the task ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4106737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Function with many values in a sequence Cool problem I was reading but couldn't solve!
Problem:
Let $f(n)$ be a function satisfying the following conditions:
a) $f(1) = 1$.
b) $f(a) \leq f(b)$ where $a$ and $b$ are positive integers with $a \leq b$.
c) $f(2a) = f(a) + 1$ for all positive integers a.
Let $M$ denote the ... | (My final answer differs from yours. I believe Ross and I are right.)
Define $ k_n$ to be the maximal $k$ such that $ f(k) = n$.
Show that
*
*$f (k_n ) = f(k_n - 1 ) = \ldots = f(k_{n-1} + 1) = n$
*$k_1 = 1$
*$k_{n+1} = 2k_n $ or $ 2k_{n}+1$.
*$k_n = \lfloor \frac{k_{n+1} }{2} \rfloor $
*A sequence $f(1), f(2), ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4107294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\int \frac{x^3+4x^2+x-1}{x^3+x^2}dx$ Evaluate $\int \frac{x^3+4x^2+x-1}{x^3+x^2}dx$.
Where do I start with this integral? I can easily see that it is possible to fator $x^{2}$ out on the denominator and use partial fractions. The numerator is also factorable but it does not have any integer roots. Can someone... | Using partial fractions:
$$\frac{x^3+4x^2 +x-1 }{x^3 +x^2} = \frac ax +\frac{b}{x^2} +\frac{c}{x+1}+d \\ \implies x^3 +4x^2 +x-1 = ax(x+1) +b(x+1) +cx^2 +dx^2(x+1)$$ Put $x=0$ to get $b=-1$ and put $x=-1$ to get $c=1$. Then, compare the coefficient of $x^3$ to get $d=1$ and then that of $x^2$ to get $ a=2$.
All in al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4110066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $\frac{1}{a^2-ac-ab+bc} + \frac{2}{b^2-ab-bc+ac} + \frac{1}{c^2-ac-bc+ab}$ for $a=1.69$, $b=1.73$, and $c=0.48$ An interesting question on algebra:
If $a=1.69$, $b=1.73$, and $c=0.48$, find the value of$\\$
$$\frac{1}{a^2-ac-ab+bc} + \frac{2}{b^2-ab-bc+ac} + \frac{1}{c^2-ac-bc+ab}$$
By substituting the val... | The expression $L$ is equal to
\begin{align}
L &= \frac{1}{(a-b)(a-c)}+\frac{2}{(b-c)(b-a)}+\frac{1}{(c-a)(c-b)} \\
&= \frac{-(b-c)-2(c-a)-(a-b)}{(a-b)(b-c)(c-a)} \\
&= \frac{a-c}{(a-b)(b-c)(c-a)} \\
&=- \frac{1}{(a-b)(b-c)} \\
\end{align}
and it's easy now to calculate $L = 20$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4113410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Sums of Squares algorithm I want to express a large number as the sum of two squares, given that it is possible and given its prime factors. Let's say the number is $273097$. It's prime factors are $11^2, 37$ and $61$. Here is is easy to see $11^2=11^2+0^2$ and $37=6^2+1^2$. Through trial and error I found that $61=... | Prime number $p$ can be expressed as sum of two (non-zero) squares if $p\equiv 1 \pmod{4}$.
We know $37 = 1^2 + 6^2, 61 = 5^2 + 6^2.$
Hence $37$ and $61$ are expressed using Gaussian integers below.
Norm$(1+6i) = (1+6i)(1-6i) = 1^2 + 6^2 = 37$
Norm$(5+6i) = (5+6i)(5-6i) = 5^2 + 6^2 = 61$
Since $273097 = 11^2\cdot37\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4116345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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Solve the equation $\frac{x-13}{x-14}-\frac{x-15}{x-16}=-\frac{1}{12}$ Solve the equation $$\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).$$ This doesn't look very nice. Can... | Hint: Notice that if we visually erase the symbols "$x - 1$" in both numerators and denominators we are left with $\frac34$ and $\frac56$ respectively, which is nice because
$$
\frac34 = \frac{9}{12}\quad\text{and}\quad\frac56 = \frac{10}{12}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4117579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
If $x=\sinh{\theta}$, is it possible to express $\cosh{n\theta}$ and $\sinh{n\theta}$ in terms of $x$? We know that hyperbolic sine is:
$$\sinh \theta={\frac {e^{\theta}-e^{-\theta}}{2}}$$
and that hyperbolic cosine is
$$\cosh \theta={\frac {e^{\theta}+e^{-\theta}}{2}}$$
Let $n\in\mathbb N$.
If $x=\sinh{\theta}$, is it... | $\begin{align}\sinh(\theta)=\frac{e^\theta-e^{-\theta}}{2}=x &\Rightarrow e^{2\theta}-2xe^{\theta}-1 =0\\
&\Rightarrow e^{\theta}= \frac{2x\pm\sqrt{4x^2+4}}{2} = x\pm\sqrt{x^2+1} \\
&\Rightarrow \boxed{e^{\theta} = x+\sqrt{x^2+1}} ~~~\text{ as } e^\theta \ge0,\theta \in \Bbb R\end{align}$
So, $$\sinh(n\theta) = \frac{e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4119659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Range of $\frac{c x^2(1-c)^2}{(x^2 +c^3)(x^2+c)}$? I have the following problem:
Let $f(x) = \frac{c x^2(1-c)^2}{(x^2 +c^3)(x^2+c)}$ with $c>0$ and $x \in \mathbb{R}$. How to prove that $0 \leq f(x) \leq 1$?
I'm not sure how to check the range, I tried plotting the function but that's not as formal as the problem asks.... | First to check if $0 \leq f(x)$ everywhere: $c>0,$ anything squared is $\geq 0$, so yes. The minimum is reached at $x=0$ only, because the numerator of a fraction must be $0$ for the fraction to be $0$.
If $c=1$ then the function is $0$ everywhere. If $c$ is large we have roughly $c^3/c^4 = 1/c$ so that will be $\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4124444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the sum of $\frac{1}{\sqrt1+\sqrt3} + \frac{1}{\sqrt3+\sqrt5} + \frac{1}{\sqrt5+\sqrt7} + ... \frac{1}{\sqrt79+\sqrt81}$ Find the sum of $\frac{1}{\sqrt1+\sqrt3} + \frac{1}{\sqrt3+\sqrt5} + \frac{1}{\sqrt5+\sqrt7} + ... \frac{1}{\sqrt{79}+\sqrt{81}}$
I've thought about multiplying every fraction by 1, but like thi... | $\frac{1}{\sqrt1+\sqrt3} + \frac{1}{\sqrt3+\sqrt5} + \frac{1}{\sqrt5+\sqrt7} + ... \frac{1}{\sqrt{79}+\sqrt{81}}=\frac{1}{\sqrt3+\sqrt1} *\frac{\sqrt3-\sqrt1}{\sqrt3-\sqrt1}+ \frac{1}{\sqrt5+\sqrt3}*\frac{\sqrt5-\sqrt3}{\sqrt5-\sqrt3} + \frac{1}{\sqrt7+\sqrt5}*\frac{\sqrt7-\sqrt5}{\sqrt7-\sqrt5} + ... \frac{1}{\sqrt{81... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4126580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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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 \... | You made a small mistake near the end: the midpoint's $x$-coordinate is not the sum of those of the intersection points, but half that sum, which is still $m/2$. Then the corresponding $y$-coordinate is $m^2/2$ and $y=2x^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4128046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
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... | Given $n$-vectors $x_1,\dots,x_n\in\mathbb R^m$, the $n$-dimensional volume of the parallelepiped spanned by $x_1,\dots,x_n$ is given by $\sqrt{\det(G)}$, where $G$ is the Gramian matrix given by $G=(\langle x_i, x_j\rangle)_{ij}$.
Hence, we get
$$
A = \frac{\sqrt{\det(X^T X)}}{2}=35,\quad \text{where}\quad X=\begin{pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4131956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Circle passing through two points and tangent to a line
"Find the equation of the circle passing through the origin $(0,0)$, the point $(1,0)$ and tangent to the line $x-2y+1=0$."
What I have done:
The equation of a circle with radius $R$ and center $(x_0,y_0)$ is $(x-x_0)^2+(y-y_0)^2=R^2$. Since the circle passes th... | I plotted it on desmos, your answer does seem right. Here is the link
The idea is that the point on which the circle touches the line can be varied. This leads to two circles satisfying the other two constraints of passing through the given points.
The mistake in OP's plot: They put $x=1/4$ instead of $x=1/2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4133815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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If $p^k m^2$ is an odd perfect number, then is there a constant $D$ such that $\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{D}$? (Note: This question is an offshoot of this closely related one.)
Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.
The topic of odd perfect numbers ... | Using your idea, one has$$\frac{\sigma(m^2)}{p^k} > \frac{m^2 - p^k}{D}\implies p^k\lt \frac{6750D}{3373}$$So, I think that for any given $D\gt 0$, $\dfrac{\sigma(m^2)}{p^k} > \dfrac{m^2 - p^k}{D}$ does not hold in general.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4136207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Derive that a sum equals $\log(1+x)-\log(1-x)$ Show that $$\sum_{n=1}^{\infty}\frac{x^{2n-1}}{n-1/2}=\log(1+x)-\log(1-x)$$
I got a series for $\log(1+x)$ so by replacing the $x$ by $-x$ I got that
$$\left (\sum_{n=1}^{\infty}(-1)^{(n+1)}\frac{x^n}{n} \right )-\left ( \sum_{n=1}^{\infty}(-1)^{(n+1)}\frac{(-x)^n}{n} \r... | First observe that:
$$\begin{aligned}
\ln(1-x)
&=
\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(-x)^n}{n}
\\ \\ &=
\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(-1)^n x^n}{n}
\\ \\ &=
\sum_{n=1}^{\infty}(-1)^{2n+1}\frac{x^n}{n}
\\ \\ &=
-\sum_{n=1}^{\infty}\frac{x^n}{n}
\end{aligned}$$
Thus we can write the sum as:
$$\begin{aligned}
\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4147151",
"timestamp": "2023-03-29T00:00:00",
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Smallest value of $b$ when $0<\left\lvert \frac{a}{b}-\frac{3}{5}\right\rvert\leq\frac{1}{150}$ Problem
For positive integers $a$ and $b$, $$0<\left\lvert \dfrac{a}{b}-\dfrac{3}{5}\right\rvert\leq\dfrac{1}{150}$$
What is the smallest possible value of $b$? (BdMO 2021 Junior P10)
My approach
If $\dfrac{a}{b}>\dfrac{3}{5... | Clearing denominators, we are effectively asked to find minimal $b$ s.t. $0 < |150a-90b| \leqslant b$, for $a, b \in \mathbb N$.
Note as the gcd $(150,90)=30$, the middle term will at least be $30$ (or a larger multiple), so $b\geqslant 30$.
It remains to to check if $150 \mid (30\pm 90b)$ or equivalently if $b \equiv ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4147300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
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limit of a quotient of sequences Consider the sequence $u_n=\prod_{k=0}^{n-1}(2^{2^n}-2^{2^k})$. I want to prove that $\lim_{n\to+\infty}\frac{u_{n+1}}{u^2_n}=+\infty$. I did not manage to prove it. For this, I write
$$\ln(u_{n+1})-2\ln(u_n)=2^{n+1}\ln(2)+\sum_{k=0}^n\ln\left(1-\frac1{2^{2^{n+1}-2^k}}\right)-2\sum_{k=0... | Using that $\;2^{2^{n+1}}-2^{2^{k+1}}=\left(2^{2^n}-2^{2^{k}}\right)\left(2^{2^n}+2^{2^{k}}\right)\,$:
$$
\require{cancel}
\begin{align}
\frac{u_{n+1}}{u_n^2} &= \frac{\prod_{k=0}^{n}(2^{2^{n+1}}-2^{2^k})}{\left(\prod_{k=0}^{n-1}(2^{2^n}-2^{2^k})\right)^2}
\\ &= \left(2^{2^{n+1}}-2^{2^0}\right) \cdot \frac{\prod_{k=1}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4147974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to use polar coordinate in ODE? I don't understand how to use polar coordinate.
\begin{cases} \frac{dx(t)}{dt}=2x-y \\ \frac{dy(t)}{dt}=5x-2y \\ \end{cases}
$$ \frac{d}{dt} \left( \begin{array}{c} x \\ y \end{array} \right) = \begin{pmatrix} 2 & -1\\ 5 & -2 \end{pmatrix} \left( \begin{array}{c} x \\ y \end{array} ... | If you define, following the rows of $P^{-1}$, $u=x$ and $v=y-2x$, then the system becomes
$$
u'=x'=-v
\\
v'=y'-2x'=x=u
$$
which is the standard circle system, or rotation with constant angular speed of $-1$. That you get this structure you also found in the matrix $J$.
With this you get the constant radius
$$
R^2=u^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4151308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve the inequality $\frac{3-x}{x^2-2x-3}\le\frac{3-x}{x^2+2x-3}$ Solve the inequality $$\dfrac{3-x}{x^2-2x-3}\le\dfrac{3-x}{x^2+2x-3}$$
We have $D: \begin{cases}x^2-2x-3\ne0\Rightarrow x\ne -1;3 \\ x^2+2x-3\ne0\Rightarrow x\ne-3;1\end{cases}$
Is the given equality equivalent (in $D$) to $$x^2-2x-3\ge x^2+2x-3\\\iff x... | $$\dfrac{3-x}{x^2-2x-3}\le\dfrac{3-x}{x^2+2x-3}$$
$$\implies(3-x)\Bigl(\dfrac1{x^2-2x-3}-\dfrac1{x^2+2x-3}\Bigr)\le0$$
$$\implies(3-x)\Bigl(\dfrac{x^2+2x-3-x^2+2x+3}{(x-3)(x+1)(x+3)(x-1)}\Bigr)\le0$$
$$\implies (3-x)\Bigl(\dfrac{4x}{(x-3)(x+1)(x+3)(x-1)}\Bigr)\le0$$
$$\implies \dfrac{4x}{(x+1)(x+3)(x-1)}\ge0$$
$$\impli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4152338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Power series of a function around a point ≠ 0 Using the power series $ \sum_{k \geq 0}z^k = \frac{1}{1-z} $ for $ |z|<1 $, where it's centered around $0$. How would the series look like if someone wanted to let, for example $ z_0 = \frac{1}{2}$?
Also, how would one compute the power series of $ \frac{1}{1+z^2} $ around... | Usually, when I want a series about a point $z_0 \ne 0$, I change variables and write $w = z-z_0, z = w+z_0$ and do the series in powers of $w$.
Example
$\frac{1}{1+z^2}$ centered at $z=1$. Write $w=z-1, z=w+1$, so
$$
\frac{1}{1+z^2} = \frac{1}{1+(w+1)^2} = \frac{1}{2+2w+w^2}
$$
For this I could use partial fractions
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4157918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Family of generators for congruence subgroup $\Gamma_0(11)$ Consider the congruence subgroup
$$\Gamma_0(11)=\left\{\begin{pmatrix}a&b\\11c&d\end{pmatrix}\in M(2,\mathbb{Z}): ad-11bc=1\right\}$$
I want to prove that the family
$$\Omega=\{-\text{Id}\}\cup\left\{\begin{pmatrix}a&b\\11c&d\end{pmatrix}\in \Gamma_0(11): c>0,... | A calculation taking a tiny fraction of a second in Sage reveals that
$$\{ -\mathrm{Id}\} \cup \left\{\left(\begin{array}{rr}
18 & -5 \\
11 & -3
\end{array}\right), \left(\begin{array}{rr}
7 & -2 \\
11 & -3
\end{array}\right), \left(\begin{array}{rr}
8 & -3 \\
11 & -4
\end{array}\right)\right\}$$
is a generating set of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4158973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Expected number of steps until game stops A game starts with some $N$ and a player chooses randomly an integer in range $\left[0,N-1\right]$:
*
*If the chosen number, let's denote it by $K$, is $0$ the game stops, otherwise we move to the next round, but shrink the range to $\left[0,K-1\right]$.
*What is the expecte... | Your $N$ matrix should be
$$N = (I-Q) =
\begin{bmatrix}
1 & -1/3 & -1/3 \\
0 & 1 & -1/2 \\
0 & 0 & 1
\end{bmatrix}.$$
Remember, $I$ is the identity matrix, not the all $1$s matrix. The inverse of this is
$$(I-Q)^{-1} =
\begin{bmatrix}
1 & 1/3 & 1/2 \\
0 & 1 & 1/2 \\
0 & 0 & 1
\end{bmatrix},$$
and the sum of the first ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4159402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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} \... | $$3^{2x+1}+4\cdot 3^x=15
\\3\cdot(3^x)^2+4\cdot 3^x=15$$
let $u=3^x$:
$$3u^2+4u-15=0$$
solve the quadratic then you have:
$$x_{1,2}=\log_3(u_{1,2})$$
| {
"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": 1
} |
Given that $\int \frac{1}{x\sqrt{x^2-1}}dx=\arccos(\frac{1}{x})+C$, what is $\int\frac{1}{x\sqrt{x^2-a^2}}dx$? The following is clear:
$x\sqrt{x^2-a^2}=x\sqrt{a^2}\sqrt{\frac{x^2}{a^2}-1}=ax\sqrt{\frac{x^2}{a^2}-1}= a^2\frac{x}{a}\sqrt{\frac{x^2}{a^2}-1}$.
So I get that $$\int\frac{1}{x\sqrt{x^2-a^2}}dx=\frac{1}{a^2}\i... | You are almost there. Consider $x/a =z.$ This gives $dx = a dz$. That means one of a's will be cancelled. Note that I am assuming that the given number $a$ is nonnegative. Anyway, your solution is correct. You just need to adjust the coefficient with $cos^{-1}(x/a)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4165124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
does this nested radicle converge? $1+\sqrt{1+\sqrt{\dots}+\sqrt{1+\sqrt{\dots}}}+\sqrt{1+\sqrt{1+\sqrt{\dots}+\sqrt{1+\sqrt{\dots}}}}$ Let $a_0=1$ and $a_n=1+\sqrt{a_{n-1}}+\sqrt{1+\sqrt{a_{n-1}}}$
I want to know if the limit of $a_n$ as n goes to infinity.
$$1+\sqrt{1+\sqrt{1+\sqrt{\dots}+\sqrt{1+\sqrt{\dots}}}+\sqrt... | Starting from @Will Jagy first hint $$w^3 - 2 w^2 - w + 1 = 0\qquad \text{with} \qquad x=w^2$$ the cubic shows three real roots $(\Delta=49)$. Using the trigonometric method, they are
$$w_k=\frac{2}{3} \left(1+\sqrt{7} \cos \left(\frac{1}{3} \left(2 \pi k-\sec ^{-1}\left(2
\sqrt{7}\right)\right)\right)\right)\qquad... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4165987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\frac{a}{a+\sqrt{2013a+bc}}+\frac{b}{b+\sqrt{2013b+ca}}+\frac{c}{c+\sqrt{2013c+ab}}\leq 1$ For positive real numbers satisfying $a+b+c=2013$. Prove that
$$\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... | Using the Cauchy-Schwarz inequality, we have
$$\sqrt{(b+a)(a+c)} \geqslant \sqrt{ab}+\sqrt{ac}.$$
Therefore
$$\frac{a}{a+\sqrt{2013a+bc}} = \frac{a}{a+\sqrt{a(a+b+c)+bc}} = \frac{a}{a+\sqrt{(b+a)(a+c)}} $$
$$\leqslant \frac{a}{a+\sqrt{ab}+\sqrt{ac}} = \frac{\sqrt {a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}.$$
So
$$\sum \frac{a}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4171770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Show that $(a^2-b^2)(a^2-c^2)(b^2-c^2)$ is divisible by $12$
Let $a,b,c\in\Bbb N$ such that $a>b>c$. Then $K:=(a^2-b^2)(a^2-c^2)(b^2-c^2)$ is divisible by $12$.
My attempt : Since each $a,b,c$ are either even or odd, WLOG we may assume $a,b$ are both even or odd. For both cases, $a+b$ and $a-b$ are divisible by $2$ s... | It is possible to answer the question by looking directly at modulo $12$.
First, notice that $\forall n,\ n^2 \equiv \{0,1,4,9\} \bmod 12$. So each of $a,b,c$ must have a square with one of those residues. If any two such squares have the same residue, their difference will be $\equiv 0 \bmod 12$, and the product of th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4172927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Trouble solving exercise Exercise. Evaluate
\begin{equation*}
\int_{-\infty}^{+\infty}\frac{x^{2k}}{1+x^{2n}}dx
\end{equation*}
, where $0\leq k < n$ and $n,k \in \mathbb{N_0}$
My attempt. I have tried to solve the integral through the residues method,i.e., completing the segment $[-R,R]$ with a curve $C_R$ such that $... | I got nerdsniped and ended up solving the problem. Hope it helps you.
Let $f(z) = \dfrac{z^{2k}}{z^{2n}+1}$. If $\omega$ is any $2n$th root of $-1$, then $f$ has simple pole at $\omega$ and
\begin{align*}
\operatorname{Res}(f,\omega)
&= \lim_{z\to\omega} \frac{z^{2k}(z-\omega)}{z^{2n} - \omega^{2n}}
\\&= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4174677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Consider the solid region defined by $S = \{{ (x, y, z) \in \mathbb{R} ^3: x^2 + y^2+z^2 \leq 1 \wedge z ^ 2 \geq 3 (x ^ 2 + y ^ 2)\wedge z \geq 0}\}$ Consider the solid region defined by $S = \{{ (x, y, z) \in \mathbb{R} ^ 3: x ^ 2 + y ^ 2 + z ^ 2 \leq 1 \wedge z ^ 2 \geq 3 (x ^ 2 + y ^ 2)\wedge z \geq 0}\}$
I have to... | Projection of intersection of sphere $x ^ 2 + y ^ 2 + z ^ 2 \leqslant 1$ with cone $z ^ 2 = 3 (x ^ 2 + y ^ 2)$ gives circle $x ^ 2 + y ^ 2 = \frac{1}{4}$ on $OXY$ plane, so volume can be found by integral
$$4\int\limits_{0}^{\frac{1}{2}}\int\limits_{0}^{\sqrt{\frac{1}{4}-x^2}}\int\limits_{\sqrt{3 (x ^ 2 + y ^ 2)}}^{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4175021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Showing$ \int_{0}^{\infty} \frac{1-3\cos 2x+2\cos 3x}{x^2} dx=0$ Showing $$\int_{0}^{\infty} \frac{1-3\cos 2x+2\cos 3x}{x^2} dx=0$$
We can show this by re-writing $I$ as
$$
\implies I=6\int_{0}^{\infty}\frac{\frac{1-\cos(2x)}{2x}-\frac{1-\cos(3x)}{3x}}{x}\,\mathrm dx,
$$
which is Frullani Integral.
$$J=\int_{0}^{\infty... | The result can be found almost immediately from the work in the question statement.
Note that
\begin{align*}
\int_{0}^{\infty} \frac{1-3\cos 2x+2\cos 3x}{x^2} dx
&= 6 \underbrace{\int_0^\infty\frac{1-\cos 2x}{2x^2}dx}_{\textrm{let }x=t/2} -
6 \underbrace{\int_0^\infty\frac{1-\cos 3x}{3x^2} dx}_{\textrm{let }x=t/3} \\ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4179582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
$M_t = \frac{1}{\sqrt{T_1}} \mathbb{1} (T_1 \leq t) - 2 \sqrt{T_1 \wedge t}, t\geq 0$ is a martingale I try to solve an old exam question, but I find it difficult. Maybe someone can suggest a hint.
Let $\{ T_i | i\in \mathbb{N} \}\subseteq \mathbb{R} _{\geq 0}$ be a homogeneous Poission point process with intensity 1, ... | Note that we can write
$$ 2\sqrt{T_1 \wedge t} = \int_{0}^{t} \frac{1}{\sqrt{u}} \mathbf{1}(T_1 > u) \, \mathrm{d}u. $$
From this, for $0 \leq s \leq t$,
\begin{align*}
\mathbb{E}[ M_t - M_s \mid \mathcal{F}_s]
&= \mathbb{E}\biggl[ \frac{1}{\sqrt{T_1}} \mathbf{1}(s < T_1 \leq t) \,\biggm|\, \mathcal{F}_s \biggr] - \int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4180311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"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... | A computation without $\epsilon$ but assuming a closed formula for the sum of the first $n$ squares.
For $n\ge 1$,
$a_n=\sqrt[n]{\frac{n(n+1)(2n+1)}{6}}=\exp\left(\frac{1}{n}\ln\left(\frac{n(n+1)(2n+1)}{6}\right)\right) \to \exp(0)=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4180599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
What is the maximum "alive density" of cells in Conway's Game of Life when played on a torus? I've read that Conway's Game of Life (CGOL) can have unbounded growth from a finite initial number of alive cells (e.g. a glider gun). However, if CGOL is played on a torus, space (the number of cells) becomes finite, and glid... | In the specific case of a pattern that is fixed under the rules of the Game of Life, the maximum density is 1/2, as shown by Noam Elkies in The still-Life density problem and its generalizations.
Elkies also presents (on page 22) a simple example of a period 6 oscillator with maximum density 3/4. Here are all its phase... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4181255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
} |
How do I calculate the sum of sum of triangular numbers? As we know, triangular numbers are a sequence defined by $\frac{n(n+1)}{2}$. And it's first few terms are $1,3,6,10,15...$. Now I want to calculate the sum of the sum of triangular numbers. Let's define
$$a_n=\frac{n(n+1)}{2}$$
$$b_n=\sum_{x=1}^na_x$$
$$c_n=\sum_... | The easiest way to prove your conjecture is by induction. You already checked the case $n=1$, so I won’t do it again. Let’s assume your result is true for some $n$. Then:
$$c_{n+1}=c_n+b_{n+1}$$
$$=\frac{n(n+1)(n+2)(n+3)}{24} + \frac{(n+1)(n+2)(n+3)}{6}$$
$$=\frac{n^4+10n^3+35n^2+50n+24}{24}$$
$$=\frac{(n+1)(n+2)(n+3)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4182890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
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Prove multiplication closure for the sequence of 1,4,7,10.... I am reading the following problem:
If S = ${1, 4, 7, 10, 13, 16, 19, ...}$ and $a \in S\space$ and $b\in
S \space$ then if $a = b\cdot c\space$ prove that $c \in S$
My approach:
The elements of $S$ are of the form $1 + 3n\space$ so $a = 1 + 3\cdot x\spac... | Straightforward multiplication affords the easiest visualization. Any one member of the set which is formed by the given sequence can be represented by $3a+1$, and any second member of the set can be represented by $3b+1$. Multiplying:
$$(3a+1)(3b+1)=9ab+3b+3a+1=3(3ab+b+a)+1$$
We can readily let $c=3ab+b+a$, from which... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4183499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Probability of Y, where $Y=X^2$ and $X \sim U[-1, 1]$ I'm trying to prove that if $X \sim U[-1, 1]$ and $Y = X^2$ then $p_Y(y)=\frac{1}{2} y^{-\frac{1}{2}}$, using the change of variables theorem.
The theorem states one can show that $p_Y(y) = \frac{dx}{dy} p_X(x)$.
Given the problem above, I find that $p_X(x) = \frac{... | \begin{align}
F_Y(y) = P( Y \le y) = P(X^2 \le y) = P( - \sqrt{y} \le X \le \sqrt{y}) = \frac{\sqrt y + 1}{2} - \frac{-\sqrt y + 1}{2} = \sqrt y,
\end{align}
thus
$$
f_Y(y) = \frac{\partial }{\partial y}F_Y(y) = \frac{1}{2}y^ {-1/2}
$$
Namely, your formula works only for one-to-one functions. In this case, using the tr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4183671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve $yy' = \sqrt{y^2+y'^2}y''-y'y''$ Solve $$yy' = \sqrt{y^2+y'^2}y''-y'y''$$
First I set $p = y'$ and $p' = \frac{dp}{dy}p$ to form:
$$yp=\sqrt{y^2+p^2}\frac{dp}{dy}p-p\frac{dp}{dy}p \rightarrow y=\sqrt{y^2+p^2}\frac{dp}{dy}-\frac{dp}{dy}p$$
I am trying to come up with a clever substitution to deal with the square r... | After OP's work
$$y=\frac{dp}{dy}(\sqrt{y^2+p^2}-p)$$
This is a homogeneous equation for $p(y)$
$$\frac{dp}{dy}=\frac{y}{\sqrt{y^2+p^2}-p}=\frac{\sqrt{y^2+p^2}+p}{y}$$
Let $p=vy$, then get
$$v+y\frac{dv}{dy}=\sqrt{1+v^2}+v \implies \int \frac{dv}{\sqrt{1+v^2}}= \int \frac{dy}{y}$$
Let $v=\tan t$, thenwe get
$$\ln(\sec ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4183964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
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 $$x^4+(a+... | So we have $a+b=0$ and $ab=-3$
$$ a+b=0 => a=-b $$
Put this value of $a$ in the second equation
$$ -b \cdot b=-3 $$
$$ b^2=3 => b=\pm\sqrt{3} $$
Can you now find $a$ with help of first equation?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4187515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solving $X^2-6Y^2=Z^3$ in positive integers I’m trying to solve the Diophantine equation
$$X^2-6Y^2=Z^3 \tag{$\star$}$$
in positive integers $x,y,z$.
Brute force calculations confirm the naïve intuition that there are many [read: surely infinite!] primitive solutions; numerical observation suggests the solutions have ... | We can always construct a linear relation between x and y such as $x=ay+b; a, b \in \mathbb z$ to convert pell equations $x^2-Dy^2=1$ to a single unknown equation. We can find numerous families of solutions by this method. For Pell like equations we use rational solutions. I try to show this bellow by an example:
Let... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4188669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 3
} |
What is my mistake in finding this pythagorean triplet? Since Project Euler copyright license requires that you attribute the problem to them, I'd like to add that this is about question 9 there.
I am trying to solve this problem on only two brain cells and can't figure out what am I doing wrong. Here is the system for... | One problem is the assumption that $\quad 5^2+200^2=795^2.\quad $ There are no "small" triples with side differences of two orders of magnitude. The $b$-value is good but the following solution provides the $a$-value to "fit" your logic and the $c$-value will be shown to match your "solution" in the last equations belo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4191659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a,b,c$ are rational numbers and if $\left(a+b\sqrt[3]{2}+c\sqrt[3]{4}\right)^3$ is also rational then prove that $ab+bc+ca=0$
If $a,b,c$ are rational numbers and if $\displaystyle \left(a+b\sqrt[3]{2}+c\sqrt[3]{4}\right)^3$ is also rational then prove that $ab+bc+ca=0$
My attempt
Binomial expansion is not a good ... | By using $$(x+y+z)^3=x^3+y^3+z^3+3xy(x+y)+3xz(x+z)+3yz(y+z)+6xyz$$ we get $a^2b+2b^2c+2ac^2=0$ and $ab^2+a^2c+2bc^2=0$. Now multiply the first equation by $b$ and the second by $a$ and then subtract them. We obtain $c=0$ (and then $a=0$ or $b=0$) or $a=b=0$.
Edit. I forgot to mention that I used the following: if $p+q\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4192378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is the polynomial $x^5 - 5x^4 + 7x^3 + x^2 + x - 1$ irreducible in $\mathbb{Z} [x]$ or in $\mathbb{Q} [x]$? Is the following polynomial irreducible in $\mathbb{Z}[x]$ or in $\mathbb{Q}[x]$?
$x^5 - 5x^4 + 7x^3 + x^2 + x - 1$
If it's reducible, there should be a linear factor with degree $1, 2$ or $3.$
I try $\mathbb{Z}_... | By the Rational Root Theorem, the only possible linear factors are $(x\pm1)$, but evaluation at $\mp1$ shows that this is not the case.
Modulo $2$, we have the factorization
$$x^5+x^4+x^3+x^2+x+1\equiv (x^2+x+1)(x^3+1)\equiv (x^2+x+1)^2(x+1)$$
Hence any quadratic factor in $\Bbb Z[x]$ must $\equiv x^2+x+1\pmod 2$, i.e,... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Converting $\frac{\sqrt{2-\sqrt{3}}}{2}$ into $\frac{\sqrt{3}-1}{2\sqrt{2}}$ A few days ago I had a quiz question on calculating the sine of $15^\circ$. Using the half-angle identity for sine, I did the following:
$$\sin15^\circ=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}=\frac{\sqrt{2-\sqrt{3}}}{2}$$
But the lecturer give t... | $$\frac{\sqrt{2-\sqrt{3}}}{2}=\frac{\sqrt{2}\sqrt{2-\sqrt{3}}}{2\sqrt{2}}=\frac{\sqrt{4-2\sqrt{3}}}{2\sqrt{2}}=\frac{\sqrt{3-2\sqrt{3}+1}}{2\sqrt{2}}=\frac{\sqrt{3}-1}{2\sqrt{2}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given $x+y+z=3$, how do I maximize $xy+yz+zx-xyz$? Given non-negative numbers $x,y,z$ such that $x+y+z=3$. How do I maximize $xy+yz+zx-xyz$. I found out that the maximum is 2 and equality holds when one of the numbers is 1 and the other two sum up to 2 but had no idea how to prove it. Can someone help please?
| Since the expression is symmetric, we can assume $z \leq 1$. Note that $xy+yz+zx-xyz=xy(1-z) + z(x+y) = xy(1-z)+z(3-z)$. It means that for given fixed $z \leq 1$, the maximum is reached when $xy$ is maximized (as $1-z\geq0$), and since $x+y=3-z$ is fixed, it happens when $x=y=\frac{3-z}{2}$. Therefore the expression no... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Expected value of area of a triangle
Two (possibly equal) integers $x$ and $y$ are chosen such that $1 \le x \le 5$ and $1 \le y \le 5$. What is the expected value of the area of the triangle with vertices $(0, 0)$, $(x, 0)$, and $(0, y)$?
I ended up getting$${1\over2}\left({{1+5}\over2}\right)\left({{1+5}\over2}\rig... | I assume, you have two independent random variables $x$ and $y$ with uniform distribution on the integers between 1 and 5. Each integer has a probability of $1/5$, so the probability of the pair $(x,y)$ is
$$p(x,y) = \frac{1}{25}$$
The Triangle size $ A(x,y) = \frac{xy}{2}$ is also a random variable. Its expectation v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4200320",
"timestamp": "2023-03-29T00:00:00",
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$\lim_{x\to 1^+}{\frac{1}{x^2-1}}$(with epsilon-delta) I want to prove that $\lim_{x\to 1^+}{\frac{1}{x^2-1}}=+\infty$
So I tried to define the $\delta$:
$\frac{1}{x^2-1}=\frac{1}{(x+1)(x-1)}>\frac{1}{\delta (x+1)}(\because 0<x-1<\delta)$
My problem is How to evaluate $\frac{1}{x+1}$ and How to proceed with the disc... | By definition of $\lim_{x\to 1^{+}}\dfrac{1}{x^{2}-1}=+\infty$ we have proof
$$
(\forall \epsilon>0)(\exists \delta>0)(\forall x\in \mathbb{R})
\left(( 0<\color{red}{x-1}<\delta)\implies \left(\dfrac{1}{x^{2}-1}>\color{red}{M}\right) \right)
$$
The secret here is to go working on the expression $\left(\dfrac{1}{x^{2}-... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that, for every $a\neq0$, the equations $ax^3 - x^2 -x -(a +1)= 0$ and $ax^2 - x -(a +1)=0$ have a common root
Show that, for every real number $a\neq0$, the equations
$ax^3 - x^2 -x -(a +1)= 0$ and $ax^2 - x -(a +1)=0$
have a common root.
I want to know whether there are any mistakes in it.
Proof: If the syste... | You have almost finished the solution. But you're missing one step.
Because, we must check that the solution satisfies at least one equation.
You can also reach the result as follows:
$$\begin{align}&\begin{cases}ax^3-x^2-x-(a+1)=0\\ ax^2-x-(a+1)=0\end{cases} \\\\ \implies &\begin{cases}ax^3-x^2-x-(a+1)=0\\ ax^3-x^2-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4204010",
"timestamp": "2023-03-29T00:00:00",
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Solve $a^5 - b^5 - c^5 = 30abc$ and $a^2 = 2(b + c)$ over positive integers. I have found this system of equations in a Romanian Mathematical magazine:
Solve over positive integers:
$a^5 - b^5 - c^5 = 30abc$ and $a^2 = 2(b+c)$. I tried multiplying the first equation with an $a$ and then substituting $a^6$ with $8(b+c)^... | By Fermat's little theorem $a^5 \equiv a \pmod {5}$. And since $a^5 - a = a(a^2 + 1)(a + 1)(a - 1)$, $a^5 - a \equiv 0 \pmod 3$, $\pmod 2$. Since $2,3,5$ are all coprime, this yields $a^5 \equiv a \pmod {30}$.
If $a-b-c = 0$, then $a^2 = 2(b+c) = 2a \implies a = 2$. $(a,b,c) = (2,1,1)$ is the only case to check given t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$O$ is the circumcenter of non-right $\triangle ABC$. $\frac{|AB \cdot CO|}{|AC \cdot BO|} = \frac{|AB \cdot BO|}{|AC \cdot CO|} = 3$. Find $\tan A$ Problem: $O$ is the circumcenter of $\triangle ABC$, which is not a right triangle. $$\frac{| AB \cdot CO |}{|AC \cdot BO|} = \frac{|AB \cdot BO|}{|AC \cdot CO|} = 3$$. F... | Let us record as in the OP the values for the scalar products, then start the computation using the formulas $a=2R\sin A$ (and the similar ones for $b,c$) to express the appearing sines, and $b^2+c^2-a^2=2bc\cos A$ (and the similar ones) to express the appearing cosines.
$$
\begin{aligned}
|AB\cdot CO|
&= AB\cdot CO\cd... | {
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"timestamp": "2023-03-29T00:00:00",
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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... | Why do this by hand?
Mathematica:
Simplify[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]
(* True *)
| {
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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... | Write $a_n=2^n + 5^n + 56\cdot1^n$ and write $(x-2)(x-5)(x-1)=0$ as $x^3=8 x^2 - 17 x + 10$. Then
$$
a_{n+3} = 8 a_{n+2} -17 a_{n+1}+ 10 a_n
$$
Then $a_n \bmod 9$ is
$$
4,0,4,0,\color{red}{4,0,4},\dots
$$
Because of the linear recurrence, the sequence repeats as soon as it repeats three consecutive terms, $4,0,4$ in th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206716",
"timestamp": "2023-03-29T00:00:00",
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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 ... |
As $D$ is intersection of external bisectors of $\angle B$ and $\angle C$, $AD$ must be internal bisector of $\angle A$.
So, $\angle ADB = 90^0 - \cfrac{\angle B}{2} - \cfrac{\angle A}{2} = \cfrac{\angle C}{2}$
Similarly, $\angle ADC = \cfrac{\angle B}{2}$
Now notice that quadrilateral $AIDH$ is cyclic.
So, $\angle AH... | {
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"timestamp": "2023-03-29T00:00:00",
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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 $(... | Here's a hint to simplify the calculation. Suppose you were to fully expand $(x^2+y+2t+3k)^{10}$ and then drop all but those terms containing $x^2 t$. What results is an expression of the form
$$ax^2 t+bk x^2 t+cy x^2 t+dk^2 x^2 t+\cdots=(a+bk+cy+dk^2+\cdots)x^2 t$$
Summing up these coefficients yields $a+b+c +d+\cdots... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Baby Rudin Theorem 7.18 Here is Theorem 7.18 from Baby Rudin:
There exists a real continuous function on the real line which is nowhere differentiable.
Here is a proof of the theorem:
Define
$$\tag{34} \varphi(x) = \lvert x \rvert \qquad \qquad (-1 \leq x \leq 1) $$
and extend the definition of $\varphi(x)$ to all r... | *
*If you look at a picture of $\varphi$, it is a saw blade with a slope of $1$ in intervals of the form $(2k,2k+1)$, and a slope of $-1$ in intervals of the form $(2k+1,2(k+1))$. The approach to the proof is essentially to take an arbitrary $x$ and show that there exists a sequence $\{\delta_m\}$ such that $\delta_m ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4215030",
"timestamp": "2023-03-29T00:00:00",
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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
$$
\frac{x^2}{\sqrt{x^2+y^2}} + \frac{y^2}{\sqrt{y^2+z^2}} + \frac{z^2}{\sqrt{z^2... | Another way.
By C-S and AM-GM we obtain:
$$\sum_{cyc}\frac{x^2}{\sqrt{x^2+y^2}}=\sqrt{\sum_{cyc}\left(\frac{x^4}{x^2+y^2}+\frac{2x^2y^2}{\sqrt{(x^2+y^2)(y^2+z^2)}}\right)}\geq$$
$$\geq\sqrt{\sum_{cyc}\frac{x^4}{x^2+y^2}+\frac{2(xy+xz+yz)^2}{\sum\limits_{cyc}\frac{x^2+y^2+y^2+z^2}{2}}}=\sqrt{\sum_{cyc}\frac{x^4}{x^2+y^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4215933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the probability that the quadratic equation $ax^2+x+1=0$ has two real roots? A number $a$ is chosen at random within the interval $(-1, 1)$. What is the probability that the quadratic equation $ax^2+x+1=0$ has two real roots?
For it to have its real roots, we must guarantee that $1-4a \geq 0$, or $a\leq \frac{1... | $$a \in(-1,1) \land a \leq \frac{1}{4} \iff a \in(-1,\frac{1}{4}]$$
because $(-1,1) \cup (-\infty,\frac{1}{4}] = (-1,\frac{1}{4}]$
We can therefore, divide the length of both intervals to get the probability:
$$P(a \in(-1,\frac{1}{4}]) = \frac{|\frac{1}{4} - \left(-1\right)|}{|1 - \left(-1\right)|} = \frac{\frac{5}{4}}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How can one integral have two answers? I am currently taking a calculus class. My teacher while teaching a specific type of indefinite integral told one mysteriously beautiful of the solving the integral. The general form of the integral was -
$\int \sin^m x \cos^n x dx $
He said when both m and n are odd, for example ... | if you substitute $cos^2\theta=1-sin^2\theta$ into your first expression you get (for tidiness I'm letting $cos\theta=C$ and $sin\theta=S$)
$$ \frac{C^8}{8}-\frac{C^6}{6}+c_0$$
$$ =\frac{C^6}{2}(\frac{C^2}{4}-\frac{1}{3})+c_0$$
$$ =\frac{(1-S^2)^3}{2}(\frac{1-S^2}{4}-\frac{1}{3})+c_0$$
$$ =\frac{(1-S^2)^3}{2}(\frac{-1-... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sum\limits_{cyc}\frac{a^2}{a^2+bc}+\frac{(a+b+c)^3+9abc}{\prod\limits_{cyc}(a+b)}\geq6.$ There is the following anhduy98's problem.
Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that:
$$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}+\frac{(a+b+c)^3+9abc}{(a+b)(a+c)... | Remarks: The pqr method works well, furthermore, it leads to a SOS solution directly.
Let $p = a + b + c, ~ q = ab + bc + ca, ~ r = abc$.
We need to prove that $F(p, q, r) \ge 0$ where
\begin{align*}
F(p, q, r) &= -4\,p{q}^{4}+ \left( {p}^{3}+13\,r \right) {q}^{3}+27\,{p}^{2}r{q}^{2} \\
&\quad
- \left( 11\,{p}^{4}r ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4220809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Derivative of $\tan^{-1}\left(\sqrt{\frac{a-b}{a+b}}\tan \frac x2\right)$.
Find the derivative of $\tan^{-1}\left(\sqrt{\frac{a-b}{a+b}}\tan \frac x2\right)$.
I'm learning differentiation and this is an exercise problem from my book. I used chain rule and got the following:
$\begin{align}
\dfrac d{dx}\left[\tan^{-1}\... | Just keep going from where you stopped: Taking things one or two steps at a time, we have
$$\begin{align}
{1\over1+{a-b\over a+b}\tan^2{x\over2}}\cdot{1\over2}\sqrt{a-b\over a+b}\sec^2{x\over2}
&={a+b\over a+b+(a-b)\tan^2{x\over2}}\cdot{1\over2}\sqrt{a-b\over a+b}\sec^2{x\over2}\\
&={1\over2}\cdot{\sqrt{(a+b)(a-b)}\ove... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find k in a curve equation when equation of a line tangent to curve is given
The equation of the curve is
$$y=x\left(\frac{k}{\sqrt{x}} - 1\right)$$
Does the problem mean the curve has a slope of zero at $y = 25/4$? The problem asks to find the value of $k$ and equation of line "l" which can be seen in the graph.
| Note that the equation of the curve
$$y = x\left(\frac{k}{\sqrt x} - 1\right) \tag 1$$
is undefined at $x=0$. We can change it to the more convenient
$$y = k\sqrt x - x \tag 2$$
which is equivalent, except that it is defined at zero.
As you mention in the comments,
$$y' = \frac{dy}{dx} = \frac{k}{2\sqrt x} - 1 \tag 3$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4233240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that: $S=\frac{a^2}{(2a+b)(2a+c)}+\frac{b^2}{(2b+c)(2b+a)}+\frac{c^2}{(2c+a)(2c+b)} \leq \frac{1}{3}$ Let $a,b,c>0$:
Prove that: $S=\frac{a^2}{(2a+b)(2a+c)}+\frac{b^2}{(2b+c)(2b+a)}+\frac{c^2}{(2c+a)(2c+b)} \leq \frac{1}{3}$
My solution:
We have: $\left[\begin{matrix}\frac{1}{x}+\frac{1}{y} \geq \frac{4}{x+y} \\\... | The inequality looks familiar. In fact, I solved the inequality a few weeks ago from an inequality handout. Here is the solution:
We have $$\begin{align}\frac{9a^2}{(2a+b)(2a+c)} &= \frac{(2a+a)^2}{2a(a+b+c)+2a^2+bc}\\ & \leq \frac{(2a)^2}{2a(a+b+c)}+\frac{a^2}{2a^2+bc} \\ &= \frac{2a}{a+b+c}+\frac{a^2}{2a^2+bc}
\end{a... | {
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How to solve for when this trigonometric function intersects the line $y=1$? How can I solve for $\alpha$ in
$$4\sin\left(\frac{\alpha}{2}\right)\cos^3\left(\frac{\alpha}{2}\right)\left(t-r\right)+\sin\left(\frac{\alpha}{2}\right)=1$$
on the domain $0\leq\alpha\leq\pi$? Clearly, one solution is when $\alpha=\pi$, but t... | Solving for $\alpha$
$$4\sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{2}\right)^{3}\left(t-r\right)+\sin\left(\frac{\alpha}{2}\right)=1$$
$$
4\sin{\left(\frac{\alpha}{2}\right)}\cos{\left(\frac{\alpha}{2}\right)}\left(1-\sin{\left(\frac{\alpha}{2}\right)^2}\right)\left(t-r\right)+\sin{\left(\frac{\alpha}{2}\... | {
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"timestamp": "2023-03-29T00:00:00",
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If $T_1=7^7,T_2=7^{7^{7}},T_3=7^{7^{7^{7}}}$ and so on, what will be the tens digit of $T_{1000}$? $7^4$ ends with $2301$, so $7^{4k+r}$ ends with $7^r$ digits
$7^2\equiv1 \mod(4) $, and $7^7\equiv 3 \mod(4) $
Can we use the modulo function in exponent form? I think we will use these two properties, how can I proceed f... | As said in comments observe that $T_n = 7^{T_n}$.
We use induction to find tens digits of $T_n$.
Observe that $T_1 = 43 \mod 100$.
Assume that $T_n \equiv 43\mod 100$. So write $T_n = 43 + 100k$ for some $k$
So $T_{n+1} \mod 100 = 7^{T_n} \mod 100 = 7^{43} 7^{100k} \mod 100$.
Now see that $7^4 = 1 \mod 100$ so we have ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4236563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Remainder Theorem Technique
Determine the remainder when $(x^4-1)(x^2-1)$ is divided by $1 + x + x^2$ (HMMT 2000, Guts Round)
A. Write the division in the form:
$$(x^4-1)(x^2-1)= (1 + x + x^2)Q(x) + R(x)$$
B. Multiply both sides by $x-1$:
$$(x-1)(x^4-1)(x^2-1)= (x^3-1)Q(x) + R(x)(x-1)$$
C. Substitute $x^3=1,x\neq1$, ... | Alternate method:
In the spirit of my answer to your linked question, we recognize that $ P (x) = x^2 + x + 1$ can be "simplified" with $ Q(x) = x-1$ to give us $P(x) Q(x) = x^3 -1 $, and hence
$$ \begin{align} & ( x^4 - 1 ) (x^2 -1 ) \\
= & (x^3 -1 ) A(x) + (x-1)(x^2 - 1) & (1)\\
= & (x^3 -1 ) B(x) + (-x^2 - x + 2 ) ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving divisibility by $9$ for $10^n + 3 \times 4^{n+2} + 5$. I am trying to prove that for all $n$, $9$ divides $10^n + 3 \times 4^{n+2} + 5$. I first tried induction on $n$, but couldn't get anywhere with the induction step. I then tried to use modular arithmetic. The furthest I could get was:
As $10 \equiv 1 \mod ... | One way that is bound to succeed is to notice that this expression depends only on $n\bmod \varphi(9)$, which is $6$. This is because if $n\equiv m \bmod \varphi(9)$ then $10^n \equiv 10^m$ and $4^{n+2} \equiv 4^{m+2}$. This is a consequence of euler's theorem, so this would allow us to only need to check $6$ values.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Coefficient of $x^{10}$ in $f(f(x))$ Let $f\left( x \right) = x + {x^2} + {x^4} + {x^8} + {x^{16}} + {x^{32}}+ ..$, then the coefficient of $x^{10}$ in $f(f(x))$ is _____.
My approach is as follow
$f\left( {f\left( x \right)} \right) = f\left( x \right) + {\left( {f\left( x \right)} \right)^2} + {\left( {f\left( x \rig... | We have
$$f(f(x))=f(x)+f(x)^2+f(x)^4+f(x)^8+f(x)^{16}+\ldots$$
To find the coefficient of $x^{10}$ in this expansion, note that the smallest power of $f(x)^k$ for some $k$ is $x^k$.
Hence, to find the coefficient of $x^{10}$, we only need to consider
$$f(x)+f(x)^2+f(x)^4+f(x)^8$$
The coefficient of $x^{10}$ in $f(x)^k$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Sum of binomial coefficient with 'skips'
Find
$$\sum_{k=0}^{n}
{(-4)}^k{n+k\choose2k}$$
I have dealt with such summations with variable $n$ in ${n \choose k}$ before, but never have seen one in which some $k$'s were skipped (like $1,3,5$...).
Can someone please give me a hint as to how to proceed?
| Snake oil:
\begin{align}
\sum_{n=0}^\infty\left(\sum_{k=0}^n (-4)^k \binom{n+k}{2k}\right)z^n
&= \sum_{k=0}^\infty (-4)^k \sum_{n=k}^\infty \binom{n+k}{2k} z^n \\
&= \sum_{k=0}^\infty (-4z)^k \sum_{n=0}^\infty \binom{n+2k}{2k} z^n \\
&= \sum_{k=0}^\infty (-4z)^k \frac{1}{(1-z)^{2k+1}} \\
&= \frac{1}{1-z}\sum_{k=0}^\inf... | {
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"url": "https://math.stackexchange.com/questions/4238423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The way to prove that $σ(a) = 3^k$ has no solution? $\sigma(n)$ = sum of divisors of n is a divisors function.
How to prove there are no such $a$ and $k \ge 2$ satisfy $\sigma(a) = 3^k$.
This proplem can be simplify to the case when $a$ is a power of prime ($a=p^\alpha$) because
if $a = p_0^{\alpha_{0}}p_1^{\alpha_{1}}... | As you stated, since $\sigma()$ is a multiplicative function, we only need to check the prime powers, i.e., for prime $p$, $e \ge 1$ and $j \ge 1$, that
$$\sigma(p^e) = \sum_{i=0}^{e}p^{i} = \frac{p^{e + 1} - 1}{p - 1} = 3^{j} \tag{1}\label{eq1A}$$
First, $p = 2$ and $e = 1$ gives $\sigma(2) = 2 + 1 = 3$, so $j = 1$ wo... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $\frac{1}{2(n+2)}<\int_0^1\frac{x^{n+1}}{x+1}dx$ $\displaystyle\int\limits_0^1\frac{x^{n+1}}{x+1}dx=\left[\frac{x^{n+2}}{(n+2)(x+1)}\right]_0^1+\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx$
$\displaystyle\int\limits_0^1\frac{x^{n+1}}{x+1}dx=\frac{1}{2(n+2)}+\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx$
If ... | Hint $:$ For maxima observe that $x \mapsto \frac {x} {x+1}$ is increasing on $[0,1].$ For minima observe that $x \mapsto \frac {1} {x+1}$ is decreasing on $[0,1].$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that the product of the lengths of the perpendiculars drawn from two points to a straight line is $b^2$ Question:
Show that the product of the lengths of the perpendiculars drawn from the points $(\pm c,0)$ to the straight line $bx\cos\theta+ay\sin\theta-ab=0$ is $b^2$ when $a^2=b^2+c^2$.
My attempt:
Let, the len... | Comment
Btw, the product of the length of the perpendiculars drawn from the focal points of an ellipse $(\pm c,0)$ to the tangent straight line $bx\cos\theta+ay\sin\theta-ab=0$ is $b^2$ when $a^2=b^2+c^2$.
Can be also derived from Newton's ellipse canonical form.
| {
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"source": "stackexchange",
"question_score": "1",
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If $ax+by = a^n + b^n$ then $\left[\frac{x}{b}\right]+\left[\frac{y}{a}\right]=\left[\frac{a^{n-1}}{b}\right]+\left[\frac{b^{n-1}}{a}\right]$
Let a,b,n be positive integers such that $(a,b) = 1$.
Prove that if $(x,y)$ is a solution of the equation $ax+by = a^n + b^n$ then
$$\left[\frac{x}{b}\right]+\left[\frac{y}{a}\r... | We will prove John's final observation:
If $a, b$ are co-prime positive integers, and $ c, d, x, y$ are integers such that $ax+by = ac + bd$, then
$$\left\lfloor \frac{x}{b} \right\rfloor + \left\lfloor \frac{y}{a} \right\rfloor = \left\lfloor \frac{c}{b} \right\rfloor + \left\lfloor \frac{d}{a} \right\rfloor. $$
As... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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.