Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Solve the equation $x^6-2x^5+3x^4-3x^2+2x-1=0$ Solve the equation $$x^6-2x^5+3x^4-3x^2+2x-1=0$$ Let's divide both sides of the equation by $x^3\ne0$ (as $x=0$ is obviously not a solution, we can consider $x\ne0$). Then we have $$x^3-2x^2+3x-3\cdot\dfrac{1}{x}+2\cdot\dfrac{1}{x^2}-\dfrac{1}{x^3}=0\\\left(x^3-\dfrac{1}{x... | The equation is:-
$$x^6-2x^5+3x^4-3x^2+2x-1=0$$
Continuing the given method we get:-
$$x^3-2x^2+3x-3\cdot\dfrac{1}{x}+2\cdot\dfrac{1}{x^2}-\dfrac{1}{x^3}=0\\\left(x^3-\dfrac{1}{x^3}\right)-2\left(x^2-\dfrac{1}{x^2}\right)+3\left(x-\dfrac{1}{x}\right)=0$$
$$\left(x-\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}+1\right)-2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Reciprocal sum of triangular numbers is 1. n∈N
Triangular numbers set is $$\{1,3,6,10,15,21,28,36,45,55,\dots,\frac{n(n+1)}2,\dots\}$$
n≥3
I solved the equation $$\frac 1{x_1} +\frac1{x_2} +\cdots+\frac1{x_n} =1$$ for triangular numbers:
$$\begin{align}n&=3,(3,3,3)\\
n&=4,(3,3,6,6)\\
n&=5,(3,6,6,6,6)\\
n&=6,(6,6... | Not an answer, just expanding on my ideas in comments.
Let $x_i=\frac{m_i(m_i+1)}{2}.$ Then $\frac{1}{x_i}=\frac{2}{m_i(m_i+1)}=\frac2{m_i}-\frac2{m_i+1}.$ Then your equation can be written as: $$\sum_{i=1}^{n} \frac{1}{m_i}=\frac 12+\sum_i \frac1{m_i+1}$$
If the largest $m_i$ has $m_i+1=p^k>2$ for some prime $p,$ then... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4523095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If $g(x)=ax^2+bx+c$ and $f(x)= \begin{cases} {g(x)} & {x\ge k} \\ {g'(x)} & {x< k} \end{cases} $ , $\max (k)=?$ if $f$ is a differentiable function
If $g(x)=ax^2+bx+c$ and $f(x)= \begin{cases} {g(x)} & {x\ge k} \\
{g'(x)} & {x< k} \end{cases} $. If $f(x)$ is a differentiable
function, what is the maximum value of $k... | Equating $g(x)$ and $g'(x)$ and using $c=a-b$ gives
$$ax^2+x(b-2a)=2b-a.$$
Completing the square:
$$(2ax+b-2a)^2=4a(2b-a)+(b-2a)^2=b^2+4ab.$$
We must have only one solution, that is $b=0$ or $b=-4a$. From $(2ax+b-2a)^2=0$ we get $x = 1$ or $x = 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4523509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Can we show algebraically that this is true? This problem is part of a larger proof I've been working on. Long story short, I'm trying to prove a particular explicit formula for the recursive function
$$
t(p, 0) = 0 \\
t(p, k+1) = \left\lfloor \dfrac{t(p, k)}{3} \right\rfloor + p
$$
where $p=2n+1, n \in \mathbb{N}$. Th... | Let $q = \dfrac{2n-1}{2\cdot 3^k}$, and so $\dfrac{2n-1}{2\cdot 3^{k-1}} = 3q$. Both $q$ and $3q$ are is always positive and never an integer for $k,n = 1,2,3, \ldots$.
Converting the RHS into an inequality,
$$\begin{array}{rcl}
\left(q+\dfrac12\right) -1 <& \left\lfloor q+\dfrac12\right\rfloor &\le q+\dfrac12\\
-q+\df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Inequality $xy+yz+zx-xyz \leq \frac{9}{4}.$ Currently I try to tackle some olympiad questions:
Let $x, y, z \geq 0$ with $x+y+z=3$. Show that
$$
x y+y z+z x-x y z \leq \frac{9}{4}.
$$
and also find out when the equality holds.
I started by plugging in $z=3-x-y$ on the LHS and got
$$
3y-y^2+3x-x^2-4xy+x^2y+xy^2 = 3y-... | Assume without loss of generality that $x \le 1$. Then,
$$xy+yz+zx-xyz = x(y+z) + yz(1-x) \le x(y+z) + \left(\frac{y+z}{2}\right)^2(1-x).$$
This is equal to
$$(3-x)\left( x + \frac{(1-x)(3-x)}{2} \right). $$
It is not too difficult to check that this is maximal when $x = 0$ (and $y=z=3/2)$, where it attains a value of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4527271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Integer solution of $a^2+b^2=c^2+d^2$ that relates $a,b,c$, and $d$ explicitly. I have found that the integer solution of $a^2+b^2=c^2+d^2$ is $(a,b,c,d)=(pr+qs,ps-qr,pr-qs,ps+qr)$ for integer $p,r,q,s$.
I wonder if there is an explicit relation between $a,b,c,$ and $d$? Or could you give me a hint on what topics shoul... | From https://sites.google.com/site/tpiezas/003 (10. Form $mx^2+ny^2=mz^2+nt^2$-S. Realis(complete):
$$ u^2+nv^2=x^2+ny^2 \\ \Leftrightarrow \small \{ u, v, x, y \}=\{ a^2-n(a-b)^2+n(a-c)^2, b^2-(a-b)^2+n(b-c)^2, a^2+nb^2-nc^2, c^2-(a-c)^2-n(b-c)^2 \}. $$
Substituting $n=1$;
$$ u^2+v^2=x^2+y^2 \\ \Leftrightarrow \small ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4527903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the all possible values of $a$, such that $4x^2-2ax+a^2-5a+4>0$ holds $\forall x\in (0,2)$
Problem: Find the all possible values of $a$, such that
$$4x^2-2ax+a^2-5a+4>0$$
holds $\forall x\in (0,2)$.
My work:
First, I rewrote the given inequality as follows:
$$
\begin{aligned}f(x)&=\left(2x-\frac a2\right)^2+\fr... | Your attempt is mostly correct, apart from a few fairly minor issues regarding an open set being used, and there's some work you didn't need to do. First, as you pointed out, the inequality can be rewritten as
$$f(x) = \left(2x-\frac{a}{2}\right)^2 + \frac{3a^2}{4} - 5a + 4 \gt 0 \tag{1}\label{eq1A}$$
For any fixed val... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4528746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 11,
"answer_id": 3
} |
Showing $\int_1^\infty\left(\sqrt{\sqrt{x}-\sqrt{x-1}}-\sqrt{\sqrt{x+1}-\sqrt{x}}\right)dx=\frac4{15}\left(\sqrt{26\sqrt2-14}-2\right)$ A Putnam problem asked to show that some improper integral is convergent, but I was curious to see if it can be computed in closed form and Mathematica came up with this:
$$\int_1^{\i... | To compute \begin{align}I=\int_{0}^{1}f(x)\;dx\end{align} with $f(x)=\sqrt{\sqrt{x+1}-\sqrt{x}}$ you can use change of variables: $x=\cot^2(u)$ then $I=\int_{\pi/8}^{\pi/4} g(u) du$ where \begin{align}g(u)=2\cot(2u)\left(2+2 \cot^2 ( 2u) \right) \sqrt {{\frac {1-\cos(2u)}{\sin \left( 2\,u \right) }}}=4{\frac {\cos \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4529613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 4
} |
Rationalize the denominator of $\frac{1}{1+\sqrt[3]{3}-\sqrt[3]{9}}$ Rationalize the denominator of $$\dfrac{1}{1+\sqrt[3]{3}-\sqrt[3]{9}}$$ Usually we are supposed to use one of the formulas $$x^3\pm y^3=(x\pm y)(x^2\mp xy+y^2)$$ I don't think they will work here. We can say $\sqrt[3]{3}=t\Rightarrow t^3=3$ and the gi... | Notice that $(1 + t - t^2)(2 + t + t^2) = 2 + 3t - t^4$. Since $t^4 = 3t$, this implies
$$(1 + t - t^2)(2 + t + t^2) = 2.$$
Hence
$$\frac{1}{1+\sqrt[3]{3}-\sqrt[3]{9}} = \frac{1}{2}(2 + \sqrt[3]{3} + \sqrt[3]{9}).$$
EDIT: As to Mark's comment: You know it must be something of the form $a + bt + t^2$. To get rid of the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4530664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
The locus of the intersection point of two perpendicular tangents to the curve $xy^2=1$. Find the locus of the intersection point of two perpendicular tangents to the curve $xy^2=1$.
I find out that the tangents to this curve with slope $m$ has this general form:
$y = mx+(-2m)^{\frac{1}{3}}-m \left( -\frac{1}{2m} \righ... | The parametric equations for the curve $xy^2=1$ could be $\forall \ t \neq 0$:
$\begin{cases}
x =& \displaystyle t^2 \\
y =& \displaystyle \frac{1}{t} \\
\end{cases}$
The tangent at the point $\left( t^2, \frac{1}{t} \right)$ is:
$m = \displaystyle \frac{\mathrm{d} y}{\mathrm{d}x} = \frac{ \frac{\mathrm{d}y}{\mathrm{d}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4532890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $f(7)$ given $f(x)f(y)=f(x+y)+f(x-y)$ and $f(1)=3$
Let $f:\Bbb R\to\Bbb R$ such that $f(1)=3$ and $f$ satisfies the functional equation
$$f(x)f(y) = f(x+y) + f(x-y)$$
Find the value of $f(7)$.
Attempt:
If $x=1$ and $y=0$, we find
$$f(1) f(0) = 2f(1) \implies f(0) = 2$$
If we fix $y=1$, we get the recurrence rela... | Evaluate $\pmb{f(0)}$
$f(1)=3$ and
$$
f(x)f(y)=f(x+y)+f(x-y)\tag1
$$
Setting $x=1$ and $y=0$, $(1)$ gives $f(0)=2$.
Determine and Solve a Recursion
Setting $y=1$, $(1)$ gives
$$
3f(x)=f(x+1)+f(x-1)\tag2
$$
Solving the second order linear recurrence in $(2)$ and using $f(0)=2$ and $f(1)=3$ gives, for $x\in\mathbb{Z}$,
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4536280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integration of $\int\sin^{3}2x\, dx.$ using IBP I am attempting to solve this integral:
$$\int\sin^{3}2x\, dx.$$
Using an identity, I have manipulated the integral into this:
$$\int\frac{1}{2}\left(1-\cos4x\right)\sin2x dx.$$
From here, using IBP, I let $u$ = $1-cos4x$ and $v'$ = sin2x
However, I obtained an answer of ... | Integration by parts is interesting as below:
$$
\begin{aligned}
&I=\int \sin ^3 2 x d x=-\frac{1}{2} \int \sin ^2 2 x d(\cos 2 x)\\
&=-\frac{1}{2} \sin ^2 2 x \cos 2 x+\frac{1}{2} \int 4 \sin 2 x \cos ^2 2 xdx\\
&=-\frac{1}{2} \sin ^2 2 x \cos 2 x+2 \int \sin 2 x\left(1-\sin ^2 2 x\right) d x\\
&=-\frac{1}{2} \sin ^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4537770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Closed form for $\int_{0}^{\pi/2}x\cos^n x \ dx$, $n\in\mathbb{N}$ Closed form for the definite integral $$I(n)=\int_{0}^{\pi/2}x\cos^n x \ dx$$ where $n\in\mathbb{N}$.
I tried using $\int_{0}^a f(x)\ dx= \int_{0}^a f(a-x)\ dx$ to get $$I(n)= \int_{0}^{\pi/2}\left(\frac{\pi}{2}-x\right)\sin^n x \ dx$$
So we have $$I(n)... | Utilize the expansions
\begin{align}
&\cos^{2m}x= \frac1{2^{2m}}\binom {2m}{m}+\frac1{2^{2m-1}}\sum_{k=1}^{m} \binom {2m}{m-k} \cos2kx\\
&\cos^{2m+1}x= \frac1{2^{2m}}\sum_{k=0}^{m} \binom {2m+1}{m-k} \cos(2k+1)x\\
\end{align}
and the results
\begin{align}
&\int_0^{\pi/2}x\cos2mx\ dx= \frac{(-1)^m-1}{(2m)^2}\\
&\int_0^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4538742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Is a trig substitution the only way to solve $ \int_a^b \frac{1}{\left(1 + cx^2\right)^{3/2}} \mathrm{d}x $? I have an integral that looks like the following:
$$
\int_a^b \frac{1}{\left(1 + cx^2\right)^{3/2}} \mathrm{d}x
$$
I have seen a method of solving it being to substitute $x = \frac{\mathrm{tan}(u)}{\sqrt{c}}$; h... | $$
\begin{aligned}
&\text { Let } y=\frac{1}{x^2} \text {, then } d x=-\frac{1}{2y^{\frac{3}{2}}} d y\\
I&=\int_{\frac{1}{a^2}}^{\frac{1}{b^2}} \frac{1}{\left(1+\frac{c}{y}\right)^{\frac{3}{2}}}\left(-\frac{d y}{2y \frac{3}{2}}\right)\\
&=\frac{1}{2} \int_{\frac{1}{b^2}}^{\frac{1}{a^2}} \frac{d y}{(y+c)^{\frac{3}{2}}}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4541392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
$\lim\limits_{x\to0}\frac{2(\tan x-\sin x)-x^3}{x^5}=$? Original question: $$\lim_{x\to0}\frac{2(\tan x-\sin x)-x^3}{x^5}$$
What is wrong in my solution as answer is $\frac14$ not $\frac12$:
$$\lim_{x\to0}\frac{2({\frac{\sin x}{\cos x}-\sin x)}-x^3}{x^5}$$
$$\lim_{x\to0}\frac{2(\sin x-\sin x\cos x)-x^3\cos x}{x^5\cos x... | As Stephen pointed out in the comment, the problem is taking the limit in the third line of your computation. Using $S(x)=\sin(x)/x$, by adding $-1+1$ in the nominator, the correct expression is
$$\lim_{x\rightarrow 0}\frac{S(x)S(x/2)-\cos(x)}{x^2\cos(x)}
=\lim_{x\rightarrow 0}\frac{S(x)S(x/2)+\frac{1}{2}S(x/2)x^2-1}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4542936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why am I getting $\beta=90^{\circ}$ Consider the geometry below, where the small circle is touching both semi circles of radius $5$ and the side of the square. Find the radius of the small circle.
My try: $M$ and $N$ are centers of semicircles and $G$ is center of small circle indicated below.Let $r$ be the radius of ... | This might be easier if you notice that that when two circles are tangent, then their tangent point is co-linear with their centers.
So if the lower left corner is the point $(0,0),$ then the center of your circle is $(r,y)$ where $r$ is the radius you seek, since the circle is tangent to the $y$ axis, and:
$$r^2+(y-5)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4545101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Possible "clever" ways to solve $x^4+x^3-2x+1=0$, with methodological justification
Solve the quartic polynomial :
$$x^4+x^3-2x+1=0$$
where $x\in\Bbb C$.
Algebraic, trigonometric and all possible methods are allowed.
I am aware that, there exist a general quartic formula. (Ferrari's formula). But, the author says, t... | Since this quartic has no real roots, it has two pairs of complex conjugate roots, so it must factor into two conjugate quadratics:
$$(x^2 + ax + b)(x^2 + \overline ax + \overline b) = x^4 + (a + \overline a)x^3 + (a\overline a + b + \overline b)x^2 + (a\overline b + \overline ab)x + b\overline b.$$
The $x^3$ coefficie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4545364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 0
} |
Deriving exact value of $\sin \pi/12$ using double angle identity
Deriving the exact value of $\sin \pi/12$ using double angle identity
Double angle identity - $\sin 2A = 2\sin A \cos A$
So, $\sin (2 \frac{\pi}{12}) = 2 \sin \frac{\pi}{12} \cos \frac{\pi}{12} $
$\sin^2 A + \cos^2 A = 1 $ so, $\cos \frac{\pi}{12} = \... | A double identity which gets you there more directly (perhaps not the one you are using) is $$\cos2x=1-2\sin^2x.$$ When $x=\pi/12$, we have \begin{align}\cos\frac\pi6&=1-2\sin^2\frac\pi{12}\\
2\sin^2\frac\pi{12}&=1-\frac{\sqrt3}2\\
\sin^2\frac\pi{12}&=\frac12-\frac{\sqrt3}4=\frac{2-\sqrt3}4\\
\sin\frac\pi{12}&=\pm\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4548727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
show that $f(x)=x$ where $\,\forall x,y\in\mathbb{R}^+, f(x+f(y+xy)) = (y+1)f(x+1)-1$
Let $f:\mathbb{R}^+\to\mathbb{R}^+$ be a function such that $\,\forall x,y\in\mathbb{R}^+, f(x+f(y+xy)) = (y+1)f(x+1)-1$. Prove that $f(x)=x$ for all real numbers $x>0$.
I think it might be useful to prove that f is injective and t... | I found that the answer above had some apparent issues w.r.t OP, so I'll write an answer that should not be doubtable, hopefully. It will borrow statements from the above answer, but will try to be more precise and detailed.
Let $P(x,y)$ be the assertion that $$
f(x + f(y+xy)) = (y+1)f(x+1)-1
$$
You have listed all th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4549625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Finding the value of $\int_{-\pi/4}^{\pi/4}\frac{(\pi-4\theta)\tan\theta}{1-\tan\theta}\,d\theta$. It is given that $$
I=\int_{-\pi/4}^{\pi/4}\frac{(\pi-4\theta)\tan\theta}{1-\tan\theta}\,d\theta=\pi\ln k-\frac{\pi^2}{w}
$$ and was asked to find the value $kw$. Here is my try on it:
Substituting $\theta = \dfrac{\pi}{4... | letting $\theta=\frac{\pi}{4}-x$ changes the integral into
$$
I=2 \int_0^{\frac{\pi}{2}} \frac{x(1-\tan x)}{\tan x} d x
$$
$$
\begin{aligned}
\int_0^{\frac{\pi}{2}} \frac{x}{\tan x} d x &=\int_0^{\frac{\pi}{2}} x d(\ln (\sin x)) \\
&= \left[x \ln (\sin x) \right] _0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} \ln (\sin x) d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4552842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Non-negative $f(n)$ satisfies $f(ab)=f(a)+f(b)$, $f(n)=0$ for $n$ a prime greater than $10$, and $f(1)
For each positive integer n, a non-negative integer $f (n)$ is associated so that the following three rules are met:
i) $f (ab) = f (a)+ f (b).$
ii) $f (n) = 0 $ if $n$ is a greater prime than $10$.
iii) $f (1) < f (2... | You now know that $f(2)+4<11$ and $f(2)>5$; it then follows from $f$'s integrality that $f(2)=6$. Then $f(96)=5f(2)+f(3)=5\cdot6+1=31$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4554913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
The solution of differential equation $2xy+6x+(x^2-4)y'=0$ When I solved the above DE, I reached to the step and confused how to continue
$2x(y+3)+(x^2-4)dy/dx$
$(x^2-4) dy/dx=-2x(y+3)$
$dy/(y+3)=-2x/(x^2-4) dx$
$∫dy/(y+3)=∫-2x/(x^2-4) dx$
$\ln|y+3|=-ln|x^2-4|+C$
$\ln|y+3|+ln|x^2-4|=C$
$\ln|(y+3)(x^2-4)|=C$
$|(y+... | Note from
$$ \ln|(y+3)(x^2-4)|=C$$
you have
$$ y=-3+\frac{C}{x^2-4}=\frac{-3x^2+12+C}{x^2-4}=\frac{-3x^2}{x^2-4}+\frac{12+C}{x^2-4} $$
Now define $C_1=12+C$ and you have
$$ y=\frac{-3x^2}{x^2-4}+\frac{C_1}{x^2-4} $$
which is the same as the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4555213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Why taking combinations works but conventional approach does not in this case? The question is
Let two items be chosen from a lot containing 12 items of which 4 are defective. Find the probability that atleast one item is defective.
Approach 1:
$$
\frac{4}{12}\times\frac{8}{11}+\frac{4}{12}\times\frac{3}{11} = \frac{1... | In the first approach you have to count the first term twice, representing the fact that you can either pull a working item and then a defective one or the other way around. This gives
$$2×\frac4{12}×\frac8{11}+\frac4{12}×\frac3{11}=\frac{19}{33}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4560242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How do I write a power series from a recurrence relation? I have ODE
$$y''- 2xy'+ 2ky = 0$$
From this I have found a recurrence relation
$$C_{n+2} = \dfrac {(2n+2k)}{(n+1)(n+2)}C_n$$
How would I write this out in series summation form? I am trying to relate all terms bacl to $C_0$, but am struggling to write out in sum... | Starting with $$C_{n+2} = \dfrac {(2n+2k)}{(n+1)(n+2)} \, C_n,$$ apply $n = 0,1,2, \cdots$ to determine the pattern. As seen by:
\begin{align}
C_{n+2} &= \dfrac {(2n+2k)}{(n+1)(n+2)} \, C_n \\
C_{2} &= \dfrac {(2k)}{(1)(2)} \, C_0 \\
C_{3} &= \dfrac {2k+2}{(2)(3)} \, C_1 \\
C_{4} &= \dfrac {2k+4}{(3)(4)} \, C_2 = \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4563833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation $\frac{\sqrt{4+x}}{2+\sqrt{4+x}}=\frac{\sqrt{4-x}}{2-\sqrt{4-x}}$ Solve the equation $$\dfrac{\sqrt{4+x}}{2+\sqrt{4+x}}=\dfrac{\sqrt{4-x}}{2-\sqrt{4-x}}$$ The domain is $4+x\ge0,4-x\ge0,2-\sqrt{4-x}\ne0$.
Note that the LHS is always positive, so the roots must also satisfy: $A:2-\sqrt{4-x}>0$.
Firstl... | My first instinct is to note that for $x \ge -4$ and $x \ne 0$, $$\frac{\sqrt{4+x}}{2+\sqrt{4+x}} = \frac{\sqrt{4+x}(\sqrt{4+x}-2)}{(4+x)-4} = \frac{4+x - 2\sqrt{4+x}}{x}, \tag{1}$$ and similarly, for $x \le 4$ and $x \ne 0$, $$\frac{\sqrt{4-x}}{2-\sqrt{4-x}} = \frac{\sqrt{4-x}(2+\sqrt{4-x})}{4-(4-x)} = \frac{2\sqrt{4-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4566671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Given the perimeter, find the length $\overline{\rm PQ}$ in the following setup.
Let $\overline{\rm AB}$ be a diameter of a circle $\omega$ and let $C$ be a point on $\omega$, different from $A$ and $B$. The perpendicular from $C$ intersects $\overline{\rm AB}$ at $D$ and omega at $E (\ne C)$. The circle with the cent... |
Let the radius of the circle centered at $C$ be $r$.
$$\implies \overline{\rm CP} = \overline{\rm CQ} = r$$
Therefore, in $\omega$, $\angle{CEQ} = \angle{CEP} = \theta$
And subtended by $\overline{\rm EP}, \angle{PQE} = \angle{PCE} = \delta.$ $\tag*{}$
Now in $\triangle{PEQ}$, we have:
$$\frac{\sin{\delta}}{\rm PE} = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4567197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Why is $ab+\frac{1}{2} - \frac{a+b}{2} >0$ for $0I am trying to show that the function $$ab+\frac{1}{2}-\frac{a+b}{2}$$ is positive when $0<a,b<1.$
Here is what I have done.
*
*The arithmetic-geometric mean inequality doesn't seem like the way to go because it implies $$ab < \sqrt{ab} \le \frac{a+b}{2}.$$
This implie... | You have shown that
$$
f(a, b) = ab+\frac{1}{2}-\frac{a+b}{2} < \frac{1}{2}
$$
for $0 < a, b < 1$, because
$$
ab < \sqrt{ab} \le \frac{a+b}{2}.
$$
But $ f(a, b) + f(1-a, b) = \frac 12$,
and therefore your upper bound also gives the desired lower bound:
$$
f(a, b) = \frac 12 - \underbrace{f(1-a, b)}_{<1/2} > 0 \, .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4567690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
The least value of $f(x)$=$\sqrt{x^2-2x+2}+\sqrt{x^2-4x+29}$ If $m$ is the least value of $f(x)$=$\sqrt{x^2-2x+2}+\sqrt{x^2-4x+29}$, occur at $x = α$ , then $[m]+[α]$ is equal to (where [.] denotes greatest integer function)
(A)6
(B)7
(C)5
(D)4
My approach is as follow
$f(x)=\sqrt{(x-1)^2+1}+\sqrt{(x-2)^2+5^2}$
$\alpha... | By using the Cauchy Schwarz inequality, you can easily obtain:
$$\sqrt {a^2+b^2}+\sqrt{c^2+d^2}\ge \sqrt {(a+c)^2+(b+d)^2};$$
and equality holds if $\frac{a}{b}=\frac {c}{d}$. Now, take $a=x-1$, $c=2-x$, $b=1$, $d=5$. We get:
$$\sqrt {(x-1)^2+1^2}+\sqrt{(2-x)^2+5^2}\ge \sqrt {1^2+6^2}=\sqrt {37}\implies m=\sqrt {37}.$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4570233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that: $e-\ln(10)>\sqrt 2-1.$ The author's original inequality is as follows.
Prove that:
$$e-\ln(10)>\sqrt 2-1$$
Is there a good approximation for $$e-\ln 10?$$
Actually, I am also wondering that,
Where does $\sqrt 2-1$ come from? Maybe, there exist relevant inequality?
My attempt:
$$-\ln 10>\sqrt 2-1-e\\
\ln ... | I hope the following will help.
$$\ln10=\ln2+\ln5=2\left(\frac{1}{3}+\frac{\left(\frac{1}{3}\right)^3}{3}+\frac{\left(\frac{1}{3}\right)^5}{5}+\frac{\left(\frac{1}{3}\right)^7}{7}+...\right)+$$
$$+2\left(\frac{2}{3}+\frac{\left(\frac{2}{3}\right)^3}{3}+\frac{\left(\frac{2}{3}\right)^5}{5}+\frac{\left(\frac{2}{3}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4572084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 3
} |
Find the flux of $\mathbf{F} = (2x-xy, -y, yz)$ out of the region W Question statement: Let $W$ be the 3D region under the graph of $f(x,y) = \exp(x^2+y^2)$ over the region in the plane defined by $1 \leq x^2 + y^2 \leq 2$. Find the flux of $\mathbf{F} = (2x-xy,-y,yz)$ out of the region W and check your answer using th... | There is a fourth surface: an internal cylinder with radius $1$: $$\mathbf{r}(\theta, z) = (\cos \theta,\sin \theta, z), 0 \leq \theta \leq 2 \pi, 0 \leq z \leq e.$$
Then
$\mathbf{N} = \mathbf{r}_{\theta} \times \mathbf{r}_z = (\cos \theta, \sin \theta, 0)$ and $\mathbf{F} \cdot \mathbf{N} = 2 \cos^2 \theta - \cos \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4572377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find a, real number for the following limit Find a, real number, such that $\displaystyle{\lim_{n\to\infty}{(n^3+an^2)^\frac{1}{3}-(n^2-an)^\frac{1}{2}}}=1$
I noted x=$(n^3+an^2)^\frac{1}{3}$
and y=$(n^2-an)^\frac{1}{2}$
I applied with the conjugate ($x^2-xy+y^2$) but I do not know how to continue. Any ideas?
Thank you... | Note that
$$\begin{align*}
x-y &= (x-n) + (n-y) \\
&= \frac{x^3-n^3}{x^2+xn+n^2} + \frac{n^2-y^2}{n+y} \\
&= \frac{an^2}{x^2+xn+n^2} + \frac{an}{n+y} \\
&= \frac{a}{(x/n)^2 + (x/n) + 1} + \frac{a}{1+(y/n)}\end{align*}$$
and therefore, as $\lim_{n\to\infty} \frac{x}{n} = \lim_{n\to\infty} \frac{y}{n} = 1$, $$\lim_{n\to\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4575284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How can we guess one solution of the equation $y''+\frac{2}{x}y'+y=0$? I want to solve the equation $$\frac{1}{x^2} \, \frac{d}{dx}\left(x^2\frac{dy}{dx}\right)=-y.$$ I converted it to this equation $y''+\frac{2}{x}y'+y=0$. How can we guess one solution of the equation $y''+\frac{2}{x}y'+y=0$? Is there any way to solve... | Assume a solution with power series and derivatives
$$y(x) = \sum_{n\ge0} a_n x^n \implies y'(x) = \sum_{n\ge0} (n+1) a_{n+1} x^n \implies y''(x) = \sum_{n\ge0} (n+1) (n+2) a_{n+2} x^n$$
Putting these into the ODE yields
$$\begin{align*}
0 &= y'' + \frac2x y' + y \\[1ex]
&= xy'' + 2y' + xy \\[1ex]
&= \sum_{n\ge0} (n+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4576195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How far can we go with the integral $I_n=\int_0^1 \frac{\ln \left(1-x^n\right)}{1+x^2} d x$ Inspired by my post, I decided to investigate the integral in general
$$
I_n=\int_0^1 \frac{\ln \left(1-x^n\right)}{1+x^2} d x$$
by the powerful substitution $x=\frac{1-t}{1+t} .$
where $n$ is a natural number greater $1$.
Let’s... | The odd cases are much harder to evaluate. Below are some of them:
\begin{align}
\int_0^1 \frac{\ln(1-x^5)}{1+x^2}dx= & \ \frac{9\pi}8\ln2-\frac{13}5G+\frac\pi2\ln\left(\cos\frac\pi{20}\cos\frac{3\pi}{20} \right)\\
&\>\>\>+ \frac{3\pi}{20}\ln \tan\frac{3\pi}{20}
- \frac{\pi}{20}\ln \tan\frac{\pi}{20} \\
\\
\int_0^1 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4579985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 0
} |
Is this limit solvable using Stolz’s theorem ? $\lim_\limits{n\to+\infty}\left(\frac12+\frac3{2^2}+\dots+\frac{2n−1}{2^n}\right)$ I would like to ask if this example is solvable using Stolz’s theorems, because I have a $2^n$ expression at the bottom and a quadratic expression at the top, so it doesn't work for me, but ... | Denote $S_{n}=\frac{1}{2}+\frac{3}{2^{2}}+\ldots+\frac{2n-1}{2^{n}}$.
Firstly, observe that
\begin{eqnarray*}
S_{n} & = & \left\{ \frac{2}{2}+\frac{4}{2^{2}}+\frac{2n}{2^{n}}\right\} -\left\{ \frac{1}{2}+\frac{1}{2^{2}}+\ldots+\frac{1}{2^{n}}\right\} \\
& = & 2\sum_{k=1}^{n}\frac{k}{2^{k}}-\frac{\frac{1}{2}\left[1-(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4580463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
$S_p = \dfrac{1}{(y+z) x^p} + \dfrac{1}{(z+x) y^p} + \dfrac{1}{(x+y) z^p} \geq \dfrac{3}{2}$ Main Problem: Let $x,y,z$ be positive real numbers and $xyz = 1$. For any $p$ real number, let's define
$$ S_p = \dfrac{1}{(y+z) x^p} + \dfrac{1}{(z+x) y^p} + \dfrac{1}{(x+y) z^p} $$
Prove (or disprove) that for all $p\geq 2$ v... | Oh this one is not that bad. First, substitute $x\to\frac 1x$, etc for simplicity to obtain an equivalent inequality:
$$\sum\dfrac{x^{p-1}}{y+z}\geq \frac 32.$$ Then, do the natural Cauchy-Schwarz to obtain:
$$\sum\dfrac{x^{p-1}}{y+z}\sum (y+z)\geq \left(\sum x^{\tfrac{p-1}{2}}\right)^2.$$ But if $p\geq 2$, then $\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4582784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find all the real solutions of $(1+x^2)(1+x^4)=4x^3$. As title suggests, the question is to find all the real roots to the polynomial:
$$(1+x^2)(1+x^4)=4x^3$$
This problem was asked in the Kettering University Math Olympiad a few years back, it's an interesting problem with many different ways to approach it. I'm going... | After expanding, we're trying to find the roots of the polynomial $p(x)=x^6+x^4-4x^3+x^2+1$. Factoring out obvious roots one at time gives
\begin{align*}
p(x)&=x^6+x^4-4x^3+x^2+1\\
&=(x-1)(x^5+x^4+2x^3-2x^2-x-1)\\
&=(x-1)^2(x^4+2x^3+4x^2+2x+1)\\
&=(x-1)^2\left[(x^2+x)^2+2x^2+(x+1)^2\right]
\end{align*}
at which point $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4583516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Hard triple Integral $\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{1}{2-zx^{2}-zy^{2}}dxdydz=\ln(2^{G})$ How do prove this triple integral?
$$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{1}{2-zx^{2}-zy^{2}}dxdydz=\ln(2^{G})$$
where G is Catalan's constant.
As my try I only reach to this hard single integral:
$$\int _0^1\frac... | Integrate by parts
$$I=\int _0^1\frac{\operatorname{Li}_2\left(\frac{1+x^2}{2}\right)}{1+x^2}dx
= \frac{\pi^3}{24}+\int_0^1\frac{2x\tan^{-1}x\ln\frac{1-x^2}2}{1+x^2}dx$$
Then, utilize $\int_0^1 \frac{x \tan^{-1} x \ln \left( 1-x^2\right)}{1+x^2}dx= -\frac{\pi^3}{48}-\frac{\pi}{8}\ln^2 2 +G\ln 2$ and $\int_0^1 \frac{x\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4585070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Calculating tangent at two points $(4a,8a)$ Find the equation of the tangent to the curve $ay^2=x^3$ at the points $(4a,8a)$
I have re-arranged the equation to get
$$y = \left(\frac{x^3}{a}\right)^{\frac{1}{2}}$$
Then taking its derivative I get
$$\frac{dy}{dx}=\frac{3\left(\sqrt{\frac{x^3}{a}}\right)}{2x}$$,
when $x=... | You have probably miscalculated the derivative when the chain rule was applied.
Note that $\dfrac{du^\frac{1}{2}}{du}=\dfrac{1}{2}u^{-\frac{1}{2}}$.
Hence,
$y'=\dfrac{d}{d\dfrac{x^3}{a}}(\dfrac{x^3}{a})^\frac{1}{2}\cdot\dfrac{d}{dx}(\dfrac{x^3}{a})$
$=\dfrac{1}{2}\cdot(\dfrac{x^3}{a})^{-\frac{1}{2}}\cdot\dfrac{3x^2}{a}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4591359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the remainder of 3^3^3...^3 divided by 46 Find the remainder of $3^{{{{{3^3}^3}^3}^3}^\cdots} \text{(2020 copies of 3)}$ by 46
Note that $a^{b^c}$ means $a^{(b^c)}$ not $(a^b)^c$
If there was 2 copies of 3:
$3^3\equiv27 (mod 46)$.
If there was 3 copies of 3:
$3^{3^3}\equiv3^{27}\equiv(3^9)^3\equiv(19683)^3\equiv(4... | Since $3$ is odd, for all odd $k$ we have $3^k$ is odd, and $3^k=3\pmod 4$.
Now consider
$$
x_0 = 3^{3^{3^{3^{k}}}}, x_1 = 3^{3^{3^{k}}}, x_2 = 3^{3^{k}}, x_3 = 3^{k}
$$
where $k = 3^{3^{⋰^{3}}}$ is the remaining part of the exponent.
We have
*
*$x_0$ is computed modulo $46$, and $x_0 = 3^{x_1} \pmod {46}$
*$x_1$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4592762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Basic question about floor function and limit ( $\lim \limits_{x \to 0} \lfloor{x-2}\rfloor \cdot \lfloor{x+3}\rfloor$)
$\lim \limits_{x \to 0} \lfloor{x-2}\rfloor \cdot \lfloor{x+3}\rfloor$
calculate the limit if it exists if not then prove it does not exist
I tried approaching by squeeze theorem and floor function... | $x+2 < \lfloor x+3\rfloor\leq x+3$ and $ x-3<\lfloor x-2 \rfloor\leq x-2$.
this implies $(x+2).(x-3)<\lfloor x+3 \rfloor.\lfloor x-2 \rfloor\leq (x+3).(x-2)$
so $lim_{x \to 0}(x+2).(x-3)<lim_{x \to 0}\lfloor x+3 \rfloor.\lfloor x-2 \rfloor\leq lim_{x \to 0} (x+3).(x-2)$
$\Rightarrow $$lim_{x \to 0}\lfloor x+3 \rfloor.\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4595678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
IVP variable change. I'm working in this IVP:
$$(4t^2+16t+17)y'' - 8y = 0, \quad y(-2) = 1, \quad y'(-2) = 0$$
and it's asked to do this variable change $x=t+2$.
I did as asked and found the equation:
$$ (4x^2 + 1)y'' - 8y = 0. $$
Changing the $t \to x$: then $ x = t + 2 $.
Changing $y \to s$: then $ s = y - 1 $.
With ... | From $$ (4 \, t^2 + 16 \, t + 17) \, y^{''}(t) - 8 \, y^{'}(t) = 0, \quad y(-2) = 1 \quad y^{'}(-2) = 0 $$ and knowing that the transformation $t \to x-2$ needs to be performed then:
$$ dt = dx \to \frac{dx}{dt} = 1$$ and
\begin{align}
\frac{dy}{dt} &= \frac{dx}{dt} \, \frac{dy}{dx} = \frac{dy}{dx} \\
\frac{d^2 y}{d t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4596588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to minimize $(x-1)^2+(y-1)^2+(z-1)^2$ under the constraint $xyz=s$? Let $s \in (0,1)$ be a parameter.
Can we find an exact closed form expression for
$$
F(s)=\min_{xyz=s,x,y,z>0}(x-1)^2+(y-1)^2+(z-1)^2, \tag{1}
$$
and exact formulas the minimizers $x(s) \le y(s) \le z(s)$?
I tried using Mathematica but failed (Howe... | You need to solve $$\nabla ((x-1)^2+(y-1)^2+(z-1)^2)=\lambda \nabla (xyz-s)$$ $$[2(x-1),2(y-1),2(z-1)]=[\lambda yz,\lambda xz, \lambda xy]$$ Equating $\lambda /2$ for the first and second terms, we have $$\frac {x-1}{yz}=\frac {y-z}{xz}$$ Note that the constraint $xyz=s $ guarantees that $x,y,z$ are all non-zero. A bit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4599605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Algebraic calculation with polynomial and complex root.
Let $f=X^{3}-7 X+7$ be in $\mathbb{Q}[X]$.
Let $\alpha \in \mathbb{C}$ be a root of $f$ and hence $1, \alpha, \alpha^{2}$ be a basis of the $\mathbb{Q}$ vector space $\mathbb{Q}(\alpha)$. Let $\beta=3 \alpha^{2}+4 \alpha-14$. Write $\beta^{2}$ and $\beta^{3}$ as ... | You should review your computations, there must be some mistake in them, I tried myself to compute $\beta^{2}$ and $\beta^{3}$ and found
$$
\beta^{2} = -5\alpha^{2}-7\alpha+28,
$$
$$
\beta^{3} = 21\alpha^{2} +28\alpha -105
$$
Hence
$$
f\left(\beta\right) = \beta^{3} - 7\beta +7 = 21\alpha^{2} +28\alpha -105 -7\left(3\a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4600336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Is this quadratic polynomial monotone at solutions of this cubic? Let $0<s < \frac{4}{27}$. The equation $x(1-x)^{2}=s$ admits exactly two solutions in $(0,1)$: Denote by $a,b$ be these solutions, and suppose that $a<b$.
Does
$$
(1-a)^2+2a^2<(1-b)^2+2b^2
$$
hold?
The limitation on the range of $s$ is due to $$\max_{a \... | Using my answer to your previous question, we have (after trigonometric simplifications)
$$a=\frac{4}{3} \cos ^2\left(\frac{1}{6} \left(\cos
^{-1}\left(\frac{27 s}{2}-1\right)+2 \pi \right)\right)$$
$$b=\frac{4}{3} \sin ^2\left(\frac{1}{6} \left(\cos
^{-1}\left(\frac{27 s}{2}-1\right)+\pi \right)\right)$$
Then
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4600892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find all three complex solutions of the equation $z^3=-10+5i$ Let $z\in \mathbb{C}$. I want to calculate the three solutions of the equation $z^3=-10+5i$. Give the result in cartesian and in exponential representation.
Let $z=x+yi $.
Then we have $$z^2=(x+yi)^2 =x^2+2xyi-y^2=(x^2-y^2)+2xyi$$
And then $$z^3=z^2\cdot z=[... | Hint: Write $-10+5i$ in its polar form, then recognize the exponential has period $2\pi i$. Use this fact to take the cube root of both sides and obtain all solutions. Once you have the solutions, you can find individual real and imaginary parts.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4601201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Intersection of an elliptic paraboloid and a sphere Problem: Prove that sphere $x^2+y^2+z^2=50z$ and elliptic paraboloid $\frac{x^2}{25}+\frac{y^2}{16}=2z$ intersects in two circles and find the radius and center of these circles.
I figured out that intersection of them lies in two planes $3y\pm4z$. Since every plane s... | $x^2 + y^2 + z^2 = 50 z $ and $ \dfrac{x^2}{25} + \dfrac{y^2}{16} = 2 z $
Multiplying the second equation by $25$ and subtracting, we get
$ y^2 \bigg(1 - \dfrac{25}{16} \bigg) + z^2 = 0 $
Multiply through by $16$:
$ - 9 y^2 + 16 z^2 = 0 $
which is, as you said in the question statment, the equation of the two planes
$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4602745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $20 x^2+13 x-15$ can be written as $(a x+b)(c x+d)$, where $a, b, c$ and $d$ are integers, what is $a+b+c+d$?
If $20 x^2+13 x-15$ can be written as $(a x+b)(c x+d)$, where $a, b, c$ and $d$ are integers, what is $a+b+c+d$ ?
The quadratic formula tells me that $x=\dfrac{-13 \pm 20\sqrt{3}}{40}.$
How to proceed?
| You used the quadratic formula incorrectly.
$$\begin{align*}20x^2 + 13x - 15 = 0
\implies
x &= \frac{-(13) \pm \sqrt{(13)^2 - 4(20)(-15)}}{2(20)}\\
&= \frac{-13 \pm \sqrt{169 + 1200}}{40} \\
&= \frac{-13 \pm \sqrt{1369}}{40} \\
&= \frac{-13 \pm 37}{40} \\
&= \frac{24}{40} \text{ or } - \frac{50}{40} \\
&= \frac 3 5 \t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4603501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Simplify fraction within a fraction (Precalculus) Simplify $$\frac{x-2}{x-2-\frac{x}{x-\frac{x-1}{x-2}}}$$
My attempt:
$$=\frac{x-2}{x-2-\frac{x}{\frac{x(x-2)-(x-1)}{x-2}}} \ \ = \ \ \frac{x-2}{x-2-\frac{x}{\frac{x^2-2x-x+1}{x-2}}} \ \ = \ \ \frac{x-2}{x-2-\frac{x^2-2x}{x^2-3x+1}}$$
$$ =\frac{x-2}{\frac{(x-2)(x^2-3x+1... | In the comments I explain how your answer is fine, just not simplified yet.
The way I would have simplifed this though avoids that issue. And you may like to study this technique.
$$\begin{align}
\frac{x-2}{x-2-\frac{x}{x-\frac{x-1}{x-2}}}
&=\frac{x-2}{x-2-\frac{x}{x-\frac{x-1}{x-2}}\cdot\frac{x-2}{x-2}}\\
&=\frac{x-2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4604736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Show that $(x+1)^{p}x^{1-p}-(x+1)^{1-p}x^{p}$ is strictly increasing Let $1/2<p< 1$. I am asked to show that
$$f(x)=(x+1)^{p}x^{1-p}-(x+1)^{1-p}x^{p}$$
is strictly increasing for $x\geq 0$ and to compute $\lim_{x\to\infty} f(x)$.
I first computed the derivative, but I don't see why it must be positive:
$$\frac{d f(x)}{... | As zwim just a comment :
Starting with your derivative and using the lemma 7.1 (see the reference) we have :
$$\frac{px}{x+1}\left(\left(1-p\right)^{2}+\frac{x+1}{x}p(2-p)-\frac{x+1}{x}p(1-p)\ln\frac{x+1}{x}\right)\leq p(x+1)^{p-1}x^{1-p}$$
Or :
$$\frac{p(1-p)^{2}x}{x+1}+p^{2}(2-p)-p^{2}(1-p)\ln\frac{x+1}{x}\le p(x+1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4604904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Is there another simpler method to evaluate the integral $\int_0^{2 \pi} \frac{1}{1+\cos \theta \cos x} d x , \textrm{ where } \theta \in (0, \pi)?$ Using ‘rationalization’, we can split the integral into two manageable integrals as:
$\displaystyle \begin{aligned}\int_0^{2 \pi} \frac{1}{1+\cos \theta \cos x} d x = & \i... | As a heuristic calclulation which can be proved later:
Let $I(a) = \int^{2\pi}_{0}\frac{dx}{1+a\cos(x)}$.
Using the geometric series:
\begin{align*}
\frac{1}{1+a\cos(x)} = \sum^{\infty}_{n=1} (-1)^na^n\cos^n(x)
\end{align*}
We have:
\begin{align*}
I(a) = \sum^{\infty}_{n=1} (-1)^na^n \int^{2\pi}_{0} \cos^n(x)dx
\end{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4605188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Is it possible to adjust the sizes of square and circle inside a unit circle so the combined area is half of that unit circle. A square and a circle fits together inside an unit circle symmetrically. Is it possible to adjust the sizes of circle and the square so that their combined area is half of the area of that unit... | Trying to Proving the Existence ( Answering the Question in the title ) :
Where-ever we take the Centre $X$ of the Inner Circle , we can get the Combined total Area by Drawing the Square beside it.
We can move the Point $X$ from right Extreme to left Extreme.
In the Image , we can Draw the largest Square ( Blue ) insid... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4607126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Indefinite integral $\int \frac{x^2}{\sqrt{x^2 + x^4}} \, dx $ involving $\mathrm{sgn}()$ function How to solve integrals like
$$\int \frac{x^2}{\sqrt{x^2 + x^4}} \, dx $$
If we factor out $x^2$ from denominator it become
$=\int \frac{x}{\left| x \right|} \frac{x}{\sqrt{1 + x^2}} \, dx $
$=\int \mathrm{sgn}(x) \frac... | To avoid the sign function, integrate as follows
\begin{align}
\int \frac{x^2}{\sqrt{x^2 + x^4}}dx
=\int \frac{\sqrt{x^2 + x^4}}{x^2}\overset{ibp}{dx}
- \int \frac{1}{\sqrt{x^2 + x^4}} dx
=\frac{\sqrt{x^2 + x^4}}{x}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4608994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Where is error in this solution to $ \sqrt{x} + \sqrt{-x} = 2 $? Given the equation:
$$ \sqrt{x} + \sqrt{-x} = 2 $$
The solutions are $x = \pm 2i$. This can be seen via Wolfram Alpha
$$
\left( \sqrt{x} + \sqrt{-x} \right)^2 = 2^2
$$
$$
\sqrt{x}^2 + 2\sqrt{x}\sqrt{-x} + \sqrt{-x}^2 = 4
$$
$$
x + 2\sqrt{x}\sqrt{-x} - x =... | Squaring twice,
$$ x +(-x) + 2 \sqrt{x}\sqrt{-x} =4 \to -x^2=4$$
and solutions of quadratic equation
$$ x^2+4 = 0$$
should be two in number
$$ x= \pm \sqrt {2} i~. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4609145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Solving $\tan\beta\sin\gamma-\tan\alpha\sec\beta\cos\gamma=b/a$, $\tan\alpha\tan\beta\sin\gamma+\sec\beta\cos\gamma=c/a$ for $\beta$ and $\gamma$ I am trying to solve the following tricky system of two trigonometric equations. $\alpha, a, b, c$ are all given (and $a$ is nonzero) so I am trying to solve for $\beta$ and ... | In order to see more symmetry, multiply both equations
$$
\left\{
\begin{aligned}
\tan\beta \sin\gamma - \tan\alpha \sec\beta \cos\gamma
&= \frac{b}{a} \\[3pt]
\tan\alpha \tan\beta \sin\gamma + \sec\beta \cos\gamma
&= \frac{c}{a}
\end{aligned}
\right.
$$
by $\cos \alpha$ to get
$$
\left\{
\begin{aligned}
\cos\al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4611075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$ Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$.
My effort: I tried using the fact $9^{\sqrt{2}}<9^{1.5}=27.$
Also We have $512 <729 \Rightarrow 2^9<27^2 \Rightarrow 2^{\frac{9}{2}}<27 \Rightarrow \sqrt{2}^9=2^{4.5}<27$. But both are below $27$.
| The given inequality can be transformed to various equivalent forms:
$$\begin{align}
9^{\sqrt{2}} \stackrel{?}{<} \sqrt{2}^9
&\iff (3^2)^{\sqrt{2}} \stackrel{?}{<} \sqrt{2}^{3\times 3}
\iff 3^{2\sqrt{2}} \stackrel{?}{<} {(2\sqrt{2})}^3\\
&\iff 3^{\frac13} \stackrel{?}{<} {(2\sqrt{2})}^{\frac{1}{2\sqrt{2}}}\end{align}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4611390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 5
} |
I solved this question with a Arithmetic mean ≥ Harmonic mean inequality. If there is another solution, please show me. $x,y,z>0$. Prove that:$$\frac{2xy}{x+y} + \frac{2yz}{y+z} + \frac{2xz}{x+z} ≤ x+y+z $$ My solution:$$\frac{x+y}{2}≥\frac{2xy}{x+y}$$ $$\frac{y+z}{2}≥\frac{2yz}{y+z}$$ $$\frac{x+z}{2}≥\frac{2xz}{x+z}$$... | This can even be shown without the help of any theorems.
Note that
$4 x y = (x+y)^2 - (x-y)^2$, likewise for the other terms. Substituting this into the question gives
$$
\frac{(x+y)^2 - (x-y)^2}{2 (x+y)} + \frac{(y+z)^2 - (y-z)^2}{2 (y+z)} + \frac{(x+z)^2 - (x-z)^2}{2 (x+z)} \\
≤ \frac12 ((x+y) + (y+z) + (x+z))
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4612207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Is there any other simpler method to evaluate $\int_0^{\infty} \frac{x^n-2 x+1}{x^{2 n}-1} d x,$ where $n\geq 2?$ Though $n\geq 2$ is a real number, which is not necessarily an integer, we can still resolve the integrand into two fractions,
$$\displaystyle I=\int_0^{\infty} \frac{x^n-2 x+1}{\left(1+x^n\right)\left(1-x^... | Here is maybe the simplest proof\begin{align}
&\int_0^{\infty} \frac{x^a-2 x+1}{x^{2 a}-1} d x\\
=& \int_0^{\infty} \frac{x^a-x}{x^{2 a}-1} d x
- \int_0^{\infty} \frac{x-1}{x^{2 a}-1}\overset{x\to 1/x}{ d x}\\
=& \int_0^{\infty} \frac{x^{2a-3}-x}{x^{2 a}-1} d x
- \int_0^{\infty} \frac{x^{2a-2}-x^a}{x^{2 a}-1}\overset... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4614562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 1
} |
Maximize $x^2 + y^2$ subject to $x^2 + y^2 = 1 - xy$ The equation $x^2 + y^2 = 1 - xy$ represents an ellipse. I am trying to show that its major axis is along $y=-x$ and find the vertex. To find the vertex I tried to find the vector in the ellipse with the greatest norm, which is equivalent to maximizing $x^2 + y^2$ su... | Substitutions work much faster .
Let $\thinspace x=a+b,\thinspace y=a-b$, then you have :
$$\begin{align}&x^2+y^2=1-xy\\
\iff
&3a^2+b^2=1\end{align}$$
This leads to the following :
$\underline{\rm {Global\thinspace\thinspace maximum}}$
$$\begin{align}x^2+y^2&=2(a^2+b^2)\\
&=2(a^2+1-3a^2)\\
&=2-4a^2\leq 2\end{align}$$
E... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4616478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 4
} |
Do 1.$\sum_{n=2}^{\infty} \frac{\cos{n}}{n^\frac{2}{3} + (-1)^n}$ 2. $\sum_{n=1}^\infty(e^{\cos n/n}−\cos\frac{1}n)$ converge? Does
$$ 1. \hspace{8mm} \sum_{n=2}^{\infty} \frac{\cos{n}}{n^\frac{2}{3} + (-1)^n} $$
and
$$ 2. \hspace{8mm} \sum_{n=1}^{\infty} \left(e^\frac{\cos{n}}{n} - \cos\frac{1}{n}\right) $$
converge? ... | For the first series, even though $(-1)^n$ is neglible compared with $n^{2/3}$ as $n$ gets large, the monotone hypothesis of the Dirichlet test is not met and it cannot be applied directly.
However, we have
$$\sum_{n=2}^m \frac{\cos{n}}{n^\frac{2}{3} + (-1)^n}= \sum_{n=2}^m \frac{\cos{n}}{n^\frac{2}{3} + (-1)^n}\frac{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4625043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$\int_0^1\frac{1}{7^{[1/x]}}dx$ $$\int_0^1\frac{1}{7^{[1/x]}}dx$$
Where $[x]$ is the floor function
now as the exponent is always natural, i converted it to an infinite sum
$$\sum\limits_{k=1}^{\infty} \frac{1}{7^{[1/k]}}$$
Which is an infinite GP, whose sum is 6.
However, this is wrong, why is this wrong?
| Note that when $x\in\left(\frac{1}{n+1},\frac{1}{n}\right],$ for some $n\in\mathbb{N},$ we have that $\lfloor\frac{1}{x}\rfloor=n.$ This observation motivates me to split up the given integral as such:
$$\begin{align*}
\int_0^1\frac{dx}{7^{\lfloor\frac{1}{x}\rfloor}}&=\int_\frac{1}{2}^1\frac{dx}{7}+\int_\frac{1}{3}^\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4628117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving $n(n-2)(n-1)^2$ is divisible by 12 I want to show that $n^4 - 4n^3 + 5n^2 -2n$ is divisible by $12$ whenever $n>0$.
I reduced this to $n(n-2)(n-1)^2$. My approach has been to check divisibility by $3$ and $4$.
In both cases, squares are always congruent to $0$ or $1$. So
\begin{align*}n^2 &\equiv_3 0, 1 \\ \imp... | If $n$ is even, $(n-2)$ is also even, and $n(n-2)$ is divisible by 4
If $n$ is odd, $(n-1)$ is even, and $(n-1)^2$ is divisible by 4.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4629961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Transformation of Variables (Finding The Distribution) Let $V \sim f(v) = Tv^2,$ such that $-2<v<1$ and $T$ is a constant. I want to know the distribution of $W=V^2$ and the value $T$. My attempt is that $V = \pm \sqrt{ W }.$ We know that
$$\int_{-2}^1 Kx^2 = 1 \quad \Rightarrow
3K =1, \quad K = 1/3$$
Then $V\sim f(... | The value of the constant is right. I continue later with your approach. First of all we need the cdf of $V$, since we work with inequalities. For this purpose we integrate the pdf from $-2$ (lower bound) to $v$:
$$\int_{-2}^v \frac13t^2 \ dt=\left[\frac19t^3\right]_{-2}^v=\frac{v^3}{9}+\frac{8}{9}$$
Thus the cdf is
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4631927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proof of n-times matrix multiplication I want to proof the following: $$
\left[ {\begin{array}{cc}
1 & 0 \\
1 & 2 \\
\end{array} } \right]^n= \left[ {\begin{array}{cc}
1 & 0 \\
2^n-1 & 2^n \\
\end{array} } \right]
$$
I betitle the matrix on the left with $A$ and the matrix on the right with $B$.
... | The matrix can be represented as $$I+P,\quad P=\begin{pmatrix}0&0\\ 1&1\end{pmatrix}$$ We have $P^2=P.$ Thus $$(I+P)^n=I+\sum_{k=1}^n{n\choose k}P^k=I+(2^n-1)P$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4634252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solution of complex differential equation
Finding solution of differential equation $\displaystyle \frac{xdy-ydx}{ydy-xdx}=\sqrt{\frac{x^2-y^2+1}{x^2-y^2}},$
Where $y=f(x)$ and $f(0)=0$ and $f(1)=1$
I have write $\displaystyle \int \frac{xdy-ydx}{x^2}=-\int \frac{1}{2x^2}(xdx-ydy)\sqrt{\frac{x^2-y^2+1}{x^2-y^2}}$
$... | $$\displaystyle \frac{xdy-ydx}{ydy-xdx}=\sqrt{\frac{x^2-y^2+1}{x^2-y^2}},$$
$$2x^2d\dfrac yx=-\sqrt{\frac{x^2-y^2+1}{x^2-y^2}}d(x^2-y^2)$$
Substitute $x^2-y^2=v$ and $u=\dfrac yx$:
$$2x^2du=-\sqrt{\frac{v+1}{v}}dv$$
For the $x^2$ factor we have:
$$v=x^2-y^2=x^2(1-u^2)$$
$$ \implies {x^2}=\dfrac v{1-u^2}$$
The DE become... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4637107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$2\sqrt3 h\geq$ perimeter Problem:
Let $ABC$ be an acute triangle and $h$ be the longest height. Show that $2\sqrt3h\geq$ the perimeter of $ABC$.
I have a solution using geometric arguments, where I reduce the problem into isoceles triangle and bash out the length.
Here is my problem: When I tried to use algebraic ap... | Not entirely complete but even I've tried and I am also stuck at an inequality, I took $a$ as the smallest side
$h = \frac{2\triangle}{a} \ \ \ and \ \
4R\triangle = abc\\
h = \frac{bc}{2R}\\
h = \frac{bc\ sina}{a}\\
$
According to the question
$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\sqrt3h > 2s \\\\
\ \ \ \ \ \frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4639497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Taylor expansion of $f(x)=\int_0^1 \frac{1}{\sqrt{t^3+x^3}}dt$ Let $f(x)=\int_0^1 \frac{1}{\sqrt{t^3+x^3}}dt$, for $x>0.$
Show that for all $n\in \mathbb N$, the function $g(x)=f(x)+\ln(x)$ has an expansion of order $n$. Comptue for $n=8$.
I have tried to develop the term $\frac{1}{\sqrt{t^3+x^3}}$ in the neihborhood ... | Assume for a moment that $0 < x < 1$, and write
\begin{align*}
f(x)
&= \int_{0}^{\infty} \frac{\mathrm{d}t}{\sqrt{t^3 + x^3}}
- \int_{1}^{\infty} \frac{\mathrm{d}t}{\sqrt{t^3 + x^3}}.
\end{align*}
The first integral can be handled by substituting $t = x u^{1/3}$:
\begin{align*}
\int_{0}^{\infty} \frac{\mathrm{d}t}{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4640718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that if $x,y>0$, $\left(\frac{x^3+y^3}{2}\right)^2≥\left(\frac{x^2+y^2}{2}\right)^3$ Through some rearrangement of the inequality and expansion, I have been able to show that the inequality is equivalent to
$$x^6-3x^4y^2+4x^3y^3-3x^2y^4+y^6≥0$$
However, I am not sure how to prove the above or if expansion and rear... | Since $f(x)=x^\frac{3}{2}$ is convex in $\mathbb{R}^+$ (its second derivative is positive), we have
$$\left(\frac{x^2+y^2}{2}\right)^{3/2}=f\left(\frac{x^2+y^2}{2}\right)\leq\frac{f(x^2)+f(y^2)}{2}=\frac{x^3+y^3}{2}.$$
by the definition of convexity. Taking the square of both sides gives the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4641593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Show that the Maclaurin expansion of exact ODE general solution has the same form as the power series solution So, given the ODE
$$
y''+2y'+y=0
$$
I have found a power series solution, coefficient recurrence relation and the general solution in terms of elementary functions:
$$
\begin{aligned}
a_{n+2} &= -\frac{2a_{n+... | $$y = a_0(1 - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{8}x^4 +...)+ a_1(x - x^2 + \frac{1}{2}x^3 - \frac{1}{6}x^4 + ...)$$
$$y(x)=a_0y_1(x)+a_1y_2(x)$$
There is no mistake in your answer. The recurrence relation is correct and your final answer $y(x)$ is also correct. Your calculations are correct. Now let's check th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4641688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Range of $f(x)=x \sqrt{1-x^2}$ I have to find the range of $f(x)=x\sqrt{1-x^2}$ on the interval $[-1,1]$. I have done so by setting $x=\sin\theta$ and thus finding it to be $[-0.5,0.5]$.
Let $x=\sinθ$. Then, for $x\in[-1,1]$ we get that $θ \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. Thus, $f(x)$ becomes:
$f(\theta)=\sin\thet... | You could reason like this: $f(x)$ will have its maximum value at the same point as $g(x) = (f(x))^2 = x^2 - x^4.$ This is quadratic in $x^2$ and an even function. Since $z-z^2$ takes its maximum value when $z=1/2$ (by finding the vertex of the parabola) then we know $g(x)$, and hence $f(x)$ takes its maximum value... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4643055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to solve this series $f(n) = f(n/2) + n$? Can I solve this as:
$f(n) = f(n/2) + n$
let find, $$f(n/2) = f(n/2/2) + n/2\\
f(n/2) = f(n/4) + n/2$$
Now,
$$\begin{split}
f(n) &= f(n/4) + n/2 + n\\
f(n) &= f(n/8) + n/4 + n/2 + n
\end{split}$$
hence so on.
$$\vdots$$
Now, $n = 2^i$.
$$\begin{split}
f(2^i) &= f(2^i/2^i)... | You’ve just given a recursion. A recursion needs have a base to be well defined. The base here could be the set of all $x \in (0.5, 1]$ (then any $x>0$ can be uniquely written as $2^kx_0$ for some $k\in\mathbb Z$ and $x_0\in (0.5,1]$).
So choose for one such $x_0$ a $y_0$ and define $f(x_0) := y_0$. Then suppose $n = 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4648186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Why is the volume of a sphere $\frac{4}{3}\pi r^3$? I learned that the volume of a sphere is $\frac{4}{3}\pi r^3$, but why? The $\pi$ kind of makes sense because its round like a circle, and the $r^3$ because it's 3-D, but $\frac{4}{3}$ is so random! How could somebody guess something like this for the formula?
| A complete answer using the disk method would be the following.
If you revolve $y=\sqrt{r^2-x^2}$ about the $x$-axis (and form a solid) you get the volume of a sphere.
Form a disk with height $f(x)$, and find its area.
The area of the red disk above is $\pi r^2$, or $\pi f(x)^2$, or we could say $\pi \sqrt{r^2-x^2}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "108",
"answer_count": 19,
"answer_id": 8
} |
Why does the series $\sum_{n=1}^\infty\frac1n$ not converge? Can someone give a simple explanation as to why the harmonic series
$$\sum_{n=1}^\infty\frac1n=\frac 1 1 + \frac 12 + \frac 13 + \cdots $$
doesn't converge, on the other hand it grows very slowly?
I'd prefer an easily comprehensible explanation rather tha... | The answer given by AgCl is a classic one. And possibly pedagogically best; I don't know.
I also like the following argument. I'm not sure what students who are new to the topic will think about it.
Suppose 1 + 1/2 + 1/3 + 1/4 + ... adds up to some finite total S. Now group terms in the following way:
$$1 + \frac{1}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "159",
"answer_count": 25,
"answer_id": 7
} |
Perfect numbers, the pattern continues The well known formula for perfect numbers is
$$
P_n=2^{n-1}(2^{n}-1).
$$
This formula is obtained by observing some patterns on the sum of the perfect number's divisors. Take for example $496$:
$$
496=1+2+4+8+16+31+62+124+248
$$
one can see that the first pattern is a sequence o... | What you have discovered is the special case $\rm\ a = 2,\ m = n-1\ $ of the following simple identity
$\rm\displaystyle\ a^{m}\:\frac{a^n-1}{a-1}\: =\: \frac{a^{m+n}-1}{a-1} - \frac{a^m-1}{a-1}\: =\: 1+a+\cdots+a^{m+n-1} - (1+a+\cdots+a^{m-1})$
For example, it yields $\ 11111000 = 11111111 - 111\ $ for $\rm\ a = 10,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/5432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
The Basel problem As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.
I believe many of you know some nice proofs of this, can you please share it w... | The most recent issue of The American Mathematical Monthly (August-September 2011, pp. 641-643) has a new proof by Luigi Pace based on elementary probability. Here's the argument.
Let $X_1$ and $X_2$ be independent, identically distributed standard half-Cauchy random variables. Thus their common pdf is $p(x) = \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/8337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "814",
"answer_count": 48,
"answer_id": 5
} |
What type of triangle satisfies: $8R^2 = a^2 + b^2 + c^2 $? In a $\displaystyle\bigtriangleup$ ABC,R is circumradius and $\displaystyle 8R^2 = a^2 + b^2 + c^2 $ , then $\displaystyle\bigtriangleup$ ABC is of which type ?
| This essentially means $\sin^2(A)+\sin^2(B)+\sin^2(C)=2$. (This follows from $\sin$ rule.)
Replace $C = \pi - (A+B)$ to get $\sin^2(A+B) = \cos^2(A) + \cos^2(B)$.
Expand $\sin(A+B)$ and do the manipulations to get
$2\cos^2(A)\cos^2(B) = 2\sin(A)\sin(B)\cos(A)\cos(B)$
which means $\cos(A) = 0$ or $\cos(B) = 0$ or $\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/10663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
How to compute the formula $\sum \limits_{r=1}^d r \cdot 2^r$? Given $$1\cdot 2^1 + 2\cdot 2^2 + 3\cdot 2^3 + 4\cdot 2^4 + \cdots + d \cdot 2^d = \sum_{r=1}^d r \cdot 2^r,$$
how can we infer to the following solution? $$2 (d-1) \cdot 2^d + 2. $$
Thank you
| Summation by parts gives that with the choice $a_k=2^k, b_k=k$ we have
$$ A_k = 2^1+\ldots+2^k = 2^{k+1}-2 $$ and
$$ \sum_{k=1}^{n}a_k b_k = A_n b_n - \sum_{k=1}^{n-1} A_k $$
so:
$$ \sum_{k=1}^{n} k 2^k = (2^{n+1}-2)n-\sum_{k=1}^{n-1}(2^{k+1}-2) = (2^{n+1}-2)n+2(n-1)-2(2^n-2)$$
and the RHS simplifies to $(n-1) 2^{n+1}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/11464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 9,
"answer_id": 8
} |
Proof by induction $\frac1{1 \cdot 2} + \frac1{2 \cdot 3} + \frac1{3 \cdot 4} + \cdots + \frac1{n \cdot (n+1)} = \frac{n}{n+1}$ Need some help on following induction problem:
$$\dfrac1{1 \cdot 2} + \dfrac1{2 \cdot 3} + \dfrac1{3 \cdot 4} + \cdots + \dfrac1{n \cdot (n+1)} = \dfrac{n}{n+1}$$
| Hint:
$$\frac{1}{k(k+1)}= \frac{1}{k} - \frac{1}{k+1}.$$
Hint2:
$$\frac{n}{n+1} = 1 - \frac{1}{n+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/11831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Computing the sum $\sum \frac{1}{n (2n-1)}$ I was asked to sum the given series:
*
*$\displaystyle \sum\limits_{n=1}^{\infty} \frac{1}{n\cdot (2n-1)}=\frac{1}{1} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 5} + \cdots \infty$
Here i workout the details.
\begin{align*}
\sum\limits_{n=1}^{\infty} \frac{1}{n \cdot (2n-1... | We have,
$$\sum_{n=1}^{\infty} \frac{1}{n(2n-1)} = 2 \sum_{n=1}^{\infty} \left( \frac{1}{2n-1} - \frac{1}{2n} \right)$$
The RHS has the alternating harmonic series, and its value is $\ln 2$.
So your final answer would be $2 \ln 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/18277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to find the sum of the following series How can I find the sum of the following series?
$$
\sum_{n=0}^{+\infty}\frac{n^2}{2^n}
$$
I know that it converges, and Wolfram Alpha tells me that its sum is 6 .
Which technique should I use to prove that the sum is 6?
| You can also apply the formula for the sum of a geometric series three times in a row :
$$\begin{array}{rcl}
\sum_{n=1}^\infty \frac{n^2}{2^n}
&=& \sum_{n=1}^\infty (2n-1) \sum_{k=n}^\infty \frac{1}{2^k}
= \sum_{n=1}^\infty \frac{2n-1}{2^{n-1}} \\
&=& \sum_{k=0}^\infty \frac{1}{2^k} + \sum_{n=1}^\infty 2 \sum_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/20418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Minimum and maximum There are five real numbers $a,b,c,d,e$ such that
$$a + b + c + d + e = 7$$$$a^2+b^2+c^2+d^2+e^2 = 10$$
How can we find the maximum and minimum possible values of any one of the numbers ?
Source
| I just wanted to point out that if you don't want to appeal to symmetry there is a Lagrange multiplier argument which isn't too bad: Plugging in the first equation into the second, the last 4 variables satisfy the equation
$$(7 - b - c - d - e)^2 + b^2 + c^2 + d^2 + e^2 = 10$$
Suppose you want to maximize $b$ for examp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/21029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How can I prove that $n^7 - n$ is divisible by $42$ for any integer $n$? I can see that this works for any integer $n$, but I can't figure out why this works, or why the number $42$ has this property.
| Just to give a different approach, we have
$$\begin{align}
n^7-n&=n(n^6-1)\\
&=n(n^3-1)(n^3+1)\\
&=n(n-1)(n^2+n+1)(n+1)(n^2-n+1)\\
&=n(n-1)(n+1)(n^2+n-6+7)(n^2-n-6+7)\\
&=n(n-1)(n+1)\Big((n-2)(n+3)(n+2)(n-3)+7\cdot\text{quadratic in }n\Big)
\end{align}$$
where the quadratic in $n$, to be explicit, is $2(n^2-6)+7$. Wha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/22121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 8,
"answer_id": 7
} |
Proof by contradiction: $r - \frac{1}{r} =5\Longrightarrow r$ is irrational? Prove that any positive real number $r$ satisfying:
$r - \frac{1}{r} = 5$ must be irrational.
Using the contradiction that the equation must be rational, we set $r= a/b$, where a,b are positive integers and substitute:
$\begin{align*}
&\fra... | Since you have already seen answers which complete your proof, here is another proof using continued fractions.
If $r$ was rational, it will have a finite continued fraction, say $[a_1, a_2, \dots, a_n]$, but that can be extended to $[5, a_1, a_2, \dots , a_n]$ and $[5,5,a_1, a_2, \dots a_n]$ etc, because
$$ r = 5 + \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/22423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 0
} |
Prove that if $a^{k} \equiv b^{k} \pmod m $ and $a^{k+1} \equiv b^{k+1} \pmod m $ and $\gcd( a, m ) = 1$ then $a \equiv b \pmod m $ My attempt:
Since $a^{k} \equiv b^{k}( \text{mod}\ \ m ) \implies m|( a^{k} - b^{k} )$ and $a^{k+1} \equiv b^{k+1}( \text{mod}\ \ m ) \implies m|( a^{k+1} - b^{k+1} ) $
Using binomial id... | We have $$m|(a^{k}+a^{k-1}b+a^{k-2}b^2 + \cdots + ab^{k-1}+b^k)$$ $$-t(a^{k-1}+a^{k-2}b+a^{k-3}b^2+ \cdots + ab^{k-2} + b^{k-1}) = a^k$$
which is impossible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/23461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
How does summation formula work with floor function?
Prove that if a and b are relatively prime, then
$$\sum_{n=1}^{a-1} \left\lfloor \frac{nb}{a}\right\rfloor = \frac{(a - 1)(b - 1)}{2}$$
My attempt was:
We have:
$$\sum_{i=1}^{n-1} i = \frac{n(n - 1)}{2}$$
Then,
$$\sum_{n=1}^{a-1} \left\lfloor \frac{nb}{a}\right\... | Notice for each $n$ $$\left\lfloor\frac{nb}{a}\right\rfloor=\frac{nb}{a}-\frac{r_n}{a}$$ where $r_n\equiv nb \pmod a$ and $0\leq r_n< a$.
Then since $a$ and $b$ are relatively prime, $nb\equiv mb \pmod a$ implies $n\equiv m \pmod a$ so that we conclude each remainder $r_n$ is unique. But, we know $r_n\in \{1,2,\dots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/24184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Proving the bound on a recurrence relation I am trying to prove the recurrence $2T(n-1) + 1$ has the bound $\theta(2^{n})$. $T(1) = \theta (1)$
My attempted solution:
\begin{align*}
T(n) &= 2T(n-1) + 1 \\
&= 2 \{ 2T(n-2) + 1 \} + 1 \\
&= 2 \{ 2 \{ 2T( n - 3 ) + 1 \} + 1 \} + 1 \\
& \ldots \\
&= 2^{i}T(n-i)+\sum_{i=0}^... | Maybe here?
$$\begin{align*}
T(n) &= 2T(n-1) + 1 \\
&= 2 \{ 2T(n-2) + 1 \} + 1 \\
&= 2 \{ 2 \{ 2T( n - 3 ) + 1 \} + 1 \} + 1 \\
& \ldots \\
&= 2^{i}T(n-i)+\sum_{k=1}^{i-1} 2^{k} + 1.
\end{align*}$$
Also,
$$
\begin{align*}
& 2^{n-1}T(n-(n-1)) + \sum_{i=1}^{n-2} 2^{i} + 1 \\
&= 2^{n-1}T(n-n+1) + \sum_{i=1}^{n-2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/24723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
For a fixed positive integer n, show that the determinant below is divisible by n For a fixed positive integer n, if
$D = \left|\begin{array}{ccc}
n! & (n + 1)! & (n + 2)! \\
(n + 1)! & (n + 2)! & (n + 3)! \\
(n + 2)! & (n + 3)! & (n + 4)!
\end{array} \right|$
show that $\left(\dfrac{D}{(n!)^{3}} - 4 \righ... | $$\begin{eqnarray*}
D &=&%
\begin{vmatrix}
n! & (n+1)! & (n+2)! \\
(n+1)! & (n+2)! & (n+3)! \\
(n+2)! & (n+3)! & (n+4)!%
\end{vmatrix}
\\
&=&%
\begin{vmatrix}
n! & (n+1)n! & (n+2)(n+1)n! \\
(n+1)n! & (n+2)(n+1)n! & (n+3)(n+2)(n+1)n! \\
(n+2)(n+1)n! & (n+3)(n+2)(n+1)n! & (n+4)(n+3)(n+2)(n+1)n!%
\end{vma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/24976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Equation of the complex locus: $|z-1|=2|z +1|$ This question requires finding the Cartesian equation for the locus:
$|z-1| = 2|z+1|$
that is, where the modulus of $z -1$ is twice the modulus of $z+1$
I've solved this problem algebraically (by letting $z=x+iy$) as follows:
$\sqrt{(x-1)^2 + y^2} = 2\sqrt{(x+1)^2 + y^2... | View the point $z$ as a vector in $\mathbb R^2$. Without doing any calculations, we know that $\lVert z - (1,0) \rVert^2$ and $\lVert z - (-1,0) \rVert^2$ are of the form $z^Tz + b^T z + c$ for some $b$ and $c$. So the equation of the locus is $$(z^Tz + b_1^Tz + c_1) = 4(z^Tz + b_2^Tz + c_2),$$ or $$3z^Tz + b_3^Tz + c_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/27199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 0
} |
Simple 4-cycle permutation I call a 4-cycle permutation simple if I can write it as $(i,i+1,i+2,i+3)$ so $(2,3,4,5)$ is a simple 4-cycle but $(1,3,4,5)$ is not. I want to write $(1,2,3,5)$ as a product of simple 4-cycles. So this is what I did:
$$
(1,2,3,5)=(1,2)(1,3)(1,5)
$$
but
$$\begin{align}
(1,3)&=(2,3)(1,2)(2,... | One way would be to write $(1,2,3,5) = (4,5)(1,2,3,4)(4,5)$ and try to write $(4,5)$ as a simple $4$-cycle instead of trying to do so for all the 2-cycles you came up with.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/28418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Why is $\arctan(x)=x-x^3/3+x^5/5-x^7/7+\dots$? Why is $\arctan(x)=x-x^3/3+x^5/5-x^7/7+\dots$?
Can someone point me to a proof, or explain if it's a simple answer?
What I'm looking for is the point where it becomes understood that trigonometric functions and pi can be expressed as series. A lot of the information I fin... | The derivative of the arc tangent is
$$\frac{d}{dx}\arctan(x) = \frac{1}{1+x^2}.$$
From the formula for geometric series (see for example this answer for a proof) shows that
$$1+y+y^2+y^3+\cdots = \frac{1}{1-y}\qquad\text{if }|y|\lt 1.$$
Plugging in $-x^2$ for $y$, we get that
$$\begin{align*}
\frac{1}{1+x^2} &= \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/29649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "41",
"answer_count": 10,
"answer_id": 2
} |
How to calculate Jacobi Symbol $\left(\dfrac{27}{101}\right)$? How to calculate Jacobi Symbol $\left(\dfrac{27}{101}\right)$?
The book solution
$$\left(\dfrac{27}{101}\right) = \left(\dfrac{3}{101}\right)^3 = \left(\dfrac{101}{3}\right)^3 = (-1)^3 = -1$$
My solution
$$\left(\dfrac{27}{101}\right) = \left(\dfrac{101}{27... | $\big(\frac{4}{27}\big) = +1,$ not $-1$.
You can use the formula $\big( \frac {2^b}{m} \big) = (-1)^{(m^2-1)/8}$ only when $b$ is odd. When $b$ is even, $2^b$ is a square so the value is $+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/31200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
inverse of a matrix What is the inverse of the following matrix ?
$$
\begin{bmatrix}
\binom{N}{0} &\binom{N+1}{0} &... &\binom{2N-1}{0} \\
\binom{N}{1} &\binom{N+1}{1} &... &\binom{2N-1}{1}\\
...& ... &... &... \\
\binom{N}{N-1} &\binom{N+1}{N-1} &... &\binom{2N-1}{N-1}
\end{bmatrix}
$$
| Your matrix, call it $A_N$, can be factorized as $B_N C_N$, where
$$B_N = \left[ \binom{N}{i-j} \right]_{i,j=0,\dots,N-1}$$
and
$$C_N = \left[ \binom{j}{j-i} \right]_{i,j=0,\dots,N-1}.$$
For example, for $N=5$ this will be
$$
\left(
\begin{array}{lllll}
1 & 1 & 1 & 1 & 1 \\
5 & 6 & 7 & 8 & 9 \\
10 & 15 & 21 & 28 & 36 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/33086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
How can I sum the infinite series $\frac{1}{5} - \frac{1\cdot4}{5\cdot10} + \frac{1\cdot4\cdot7}{5\cdot10\cdot15} - \cdots\qquad$ How can I find the sum of the infinite series
$$\frac{1}{5} - \frac{1\cdot4}{5\cdot10} + \frac{1\cdot4\cdot7}{5\cdot10\cdot15} - \cdots\qquad ?$$
My attempt at a solution - I saw that I co... | Here is a way that uses the Gamma function, and that was strangely the first thing to come to mind. (Warning!: It is perhaps more complicated than necessary.)
Let $I$ denote your infinite series. We can write this as $$I=\sum_{n=0}^{\infty}(-1)^{n}\prod_{m=0}^{n}\frac{3m+1}{5m+5}$$ which becomes
$$I=\sum_{n=0}^{\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/34671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 1
} |
Writing a percent as a decimal and a fraction I am having a problem understanding some manipulations with recurring decimals. The exercise is
Write each of the following as a
decimal and a fraction:
(iii) $66\frac{2}{3}$%
(iv) $16\frac{2}{3}$%
For (iii), I write a decimal $66\frac{2}{3}\% = 66.\bar{6}\% = 0.66\bar{... | The recurring decimal $0.\overline{a_1\ldots a_n}$ is equal to
$$\frac{a_1\cdots a_n}{10^n-1}.$$
E.g., $x=0.\overline{285} = 0.285285285\cdots$, then
$$x = \frac{285}{10^3-1} = \frac{285}{999}.$$
That is, you get the periodic portion divided by a number that consists of as many $9$s as the length of the periodic portio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/35631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Monotonicity of $f(n)= \frac{1}{n} \sum_{i=1}^{n-1} \frac{1}{i}$ Define $f: \mathbb{N} \rightarrow \mathbb{R}$ as $f(n)= \frac{1}{n} \sum_{i=1}^{n-1} \frac{1}{i}$.
I was wondering how to tell if $f$ is a increasing or decreasing function?
Thanks and regards!
| A formal proof would be
$$
\begin{align}
f(n+1) - f(n) &= \frac{1}{n+1} \sum_{i=1}^{n} \frac{1}{i} - \frac{1}{n} \sum_{i=1}^{n-1} \frac{1}{i}
\\
&= \frac{n}{n+1}\frac{1}{n} \left(\sum_{i=1}^{n-1} \frac{1}{i} + \frac{1}{n}\right) - \frac{1}{n} \sum_{i=1}^{n-1} \frac{1}{i}
\\
&= (\frac{n}{n+1} - 1)\frac{1}{n} \sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/36483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How do I find roots of a single-variate polynomials whose integers coefficients are symmetric wrt their respective powers Given a polynomial such as $X^4 + 4X^3 + 6X^2 + 4X + 1,$ where the coefficients are symmetrical, I know there's a trick to quickly find the zeros. Could someone please refresh my memory?
| Such a polynomial with symmetric coefficients is called a palindromic polynomial. If $x_0$ is a zero of a palindromic polynomial than $\frac{1}{x_0}$ is a zero , too.
If we divide the polynomial
$$Ax^4+Bx^3+Cx^2+Bx+A=0$$
by $A$ we get
$$x^4+ax^3+bx^2+ax+1=0 \tag{1}$$
$$(x^4+1)+a(x^3+x)+bx^2=0$$
$$x^2(x^2+\frac{1}{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/40864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 2
} |
Primes of the form $x^2 +ny^2$ where swapping $x$ and $y$ still gives a prime I am studying primes of the form $x^2+ny^2$, and i was wondering if there are any known results about the primes of this form such that when you swap $x$ and $y$ you also get a prime. ie for $y^2+nx^2$ you get another prime, I have found quit... | This could make a nice trick question:
Prove or disprove if $p=x^2+2 y^2$ is
a prime which is $3 \mod 8$, then
$y^2+2x^2$ is prime.
The answer is "disprove". The first counter-example I can find is $p=131$, which is $9^2+2 \times 5^2$, where $5^2+2 \times 9^2 = 187=11 \times 17$. As I will explain below, there i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/42829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 1,
"answer_id": 0
} |
How can I compute the integral $\int_{0}^{\infty} \frac{dt}{1+t^4}$? I have to compute this integral $$\int_{0}^{\infty} \frac{dt}{1+t^4}$$ to solve a problem in a homework. I have tried in many ways, but I'm stuck. A search in the web reveals me that it can be do it by methods of complex analysis. But I have not taken... | You can also do it by partial fraction decomposition. We have
$$ \frac{2\sqrt{2}}{1+t^4} = \frac{t + \sqrt{2}}{t^2 + \sqrt{2}t + 1} - \frac{t - \sqrt{2}}{t^2 - \sqrt{2}t + 1}. $$
We have
$$\int_0^\infty \frac{dt}{t^2 \pm \sqrt{2}t + 1} =
\int_0^\infty \frac{dt}{(t \pm \sqrt{2}/2)^2 + 1/2} =
\int_0^\infty \frac{2dt}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/43457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 7,
"answer_id": 3
} |
Effect of adding a constant to both Numerator and Denominator I was reading a text book and came across the following:
If a ratio $a/b$ is given such that $a \gt b$, and given $x$ is a positive integer, then
$$\frac{a+x}{b+x} \lt\frac{a}{b}\quad\text{and}\quad \frac{a-x}{b-x}\gt \frac{a}{b}.$$
If a ratio $a/b$ is gi... | For the first question. If $x>0$, we have:
$$a>b\Rightarrow ax>bx\Rightarrow ab+ax>ab+bx\Rightarrow a(b+x)>b(a+x)$$
$$\Rightarrow \frac{a}{b}>\frac{a+x}{b+x}\Leftrightarrow \frac{a+x}{b+x}<%
\frac{a}{b}.$$
Then $-x<0$. We thus have
$$a>b\Rightarrow -ax<-bx\Rightarrow ab-ax<ab-bx\Rightarrow a(b-x)<b(a-x)$$
$$\Rightarr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/46156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 8,
"answer_id": 4
} |
Deriving the rest of trigonometric identities from the formulas for $\sin(A+B)$, $\sin(A-B)$, $\cos(A+B)$, and $\cos (A-B)$ I am trying to study for a test and the teacher suggest we memorize $\sin(A+B)$, $\sin(A-B)$, $\cos(A+B)$, $\cos (A-B)$, and then be able to derive the rest out of those. I have no idea how to ge... | Three examples of algebraic derivations of trigonometric identities as an application of the addition and subtraction formulas.
1. Example on how to deduce the logarithmic transformation formulas (sum to product formulas) from
$$\sin (a+b)=\sin a\cos b+\sin b\cos a,\qquad (1)$$
$$\sin (a-b)=\sin a\cos b-\sin b\cos a.\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/48938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 0
} |
Trigonometric equality: $\frac{1 + \sin A - \cos A}{1 + \sin A + \cos A} = \tan \frac{A}{2}$ Can you guys give me a hint on how to proceed with proving this trigonometric equality? I have a feeling I need to use the half angle identity for $\tan \frac{\theta}{2}$. The stuff I have tried so far(multiplying numerator and... | Using the identities$$1+\cos A=2 \cos ^{2} \frac{A}{2}, 1-\sin A=2 \sin ^{2} \frac{1}{2} , \sin A=2 \sin\frac{\theta}{2} \cos\frac{\theta}{2},
$$
We have $$
\begin{aligned}
\frac{1+\sin A-\cos A}{1+\sin A+\cos A} &=\frac{2 \sin ^{2} \frac{A}{2}+2 \sin \frac{A}{2} \cos \frac{A}{2}}{2 \cos ^{2} \frac{A}{2}+2 \sin \frac{A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/50093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 8,
"answer_id": 7
} |
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.