Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
A limit involving $\cos x$ and $x^2$ The question is finding the value of $L = \lim_{x \to 0}\frac{1 - \cos x \cos 2x \cos 3x}{x^2}$ if it exists . I've found the answer using taylor's formula but I'm looking for other solutions maybe using trigonometric identities .
| Observe you have
\begin{align}
\cos x \cos 2x \cos 3x= \cos x(\cos 5x+\sin 2x \sin 3x)
\end{align}
and
\begin{align}
\cos x \cos 5x =\cos 6x+ \sin x\sin 5x
\end{align}
which means
\begin{align}
\frac{1-\cos x \cos 2x \cos 3x}{x^2} =&\ \frac{1-\cos 6x}{x^2}-\frac{\sin x\sin 5x}{x^2}-\frac{\cos x \sin 2x\sin 3x}{x^2}\\
=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2768782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
} |
British Maths Olympiad (BMO) 2004 Round 1 Question 1 alternative approaches? The questions states:
Solve the simultaneous equations (which I respectively label as $
> \ref{1}, \ref{2}, \ref{3}, \ref{4}$)
$$\begin{align} ab + c + d &= 3 \tag{1} \label{1} \\ bc + d + a &= 5
\tag{2} \label{2} \\ cd + a + b &= 2 \tag{3... | My approach was to set $A=a-1,B=b-1,C=c-1,D=d-1$ so that we get
\begin{align}
AB + A + B + C + D &= 0 \\
BC + A + B + C + D &= 2 \\
CD + A + B + C + D &= -1 \\
AD + A + B + C + D &= 3.
\end{align}
Letting $S = A+B+C+D$, we have $ABCD = (-S)(-1-S) = (2-S)(3-S)$, so $S=1$, and therefore
\begin{align}
AB &= -1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2770117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
How to find the particular solution of the differential equation? $$\frac{dy}{dx} = 1 + \frac{y}{x} + \left(\frac{y}{x}\right)^2 , ~~ y(1)=1$$
Thank you
| We start by setting a variable, $z$ equal to $\frac{y}{x}$
$$\frac{{\rm d}y}{{\rm d}x} = 1 + z + z^2$$
Next, find $\frac{{\rm d}z}{{\rm d}x}$.
$$\frac{{\rm d}z}{{\rm d}x} = \frac{x \frac{{\rm d}y}{{\rm d}x} - y}{x^2} = \frac{1}{x} \frac{{\rm d}y}{{\rm d}x} - \frac{y}{x^2} = \frac{1}{x}\left(\frac{{\rm d}y}{{\rm d}x} - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2771027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solving $x^2u_x + y^2u_v + z(x+y)u_z = 0$ Exercise :
Solve the PDE
$$x^2u_x + y^2u_v + z(x+y)u_z = 0$$
Attempt :
We have to solve the problem :
$$\frac{dx}{x^2} = \frac{dy}{y^2} = \frac{dz}{z(x+y)}$$
Choosing the first two fractions, we yield :
$$\frac{dx}{x^2} = \frac{dy}{y^2} \Rightarrow \int \frac{dx}{x^2} = \in... | For the first equation
$$\frac{dx}{x^2} = \frac{dy}{y^2} \Rightarrow \int \frac{dx}{x^2} = \int \frac{dy}{y^2} \Leftrightarrow -\frac{1}{x} = -\frac{1}{y} + C$$
$$ \implies K=\frac 1x- \frac 1y $$
For the last equation use this trick @Rebellos
$$\frac{dx}{x^2} = \frac{dy}{y^2} = \frac{dz}{z(x+y)}$$
$$\frac{dx-dy}{x^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2771234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Determine the region of the phase plane in which all phase paths are periodic orbits Sketch the phase portrait of the dynamical system
\begin{align} \dot{x}&=y+2xy \\ \dot{y}&=-x+x^2-y^2\end{align}
and determine the region of the phase plane in which all phase paths are periodic orbits.
I have found that the equilibriu... | Note that the system is Hamiltonian since
$$\frac{\partial}{\partial x} (y+2xy) = 2y = - \frac{\partial}{\partial y} (-x+x^2 -y^2). $$
Indeed a Hamiltonian function is given by $H(x, y) = \frac{1}{2}x^2 -\frac{1}{3} x^3 + \frac{1}{2} y^2 + xy^2$. The flow lines are given by the level curve of $H$.
Now we consider the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2771397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
show this number is not a prime number The following problem is a special case Show this number always is composite number? , but I think this special case is relatively difficult to deal with, that is, if this is solved, it may solve the general situation.
show this number
$$A=2004^{2005}+1002^{2005}\cdot 2005^{1... | $$2004^{2005}+1002^{2005} 2005^{1002}+2005^{2004}$$
$$x=2004$$
$$x^{x+1}+\left(\frac{x}{2}\right)^{x+1} (x+1)^{x/2}+(x+1)^x=2^{-x-1} \left(x^{x+1} (x+1)^{x/2}+2^{x+1} x^{x+1}+2^{x+1} (x+1)^x\right)$$
which should never be prime when $x$ is even:
Consider the portion (ignoring the extra $2^{x+1}$ for now):
$$\left(x^{x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2773035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 2,
"answer_id": 1
} |
Find the altitude of a tetrahedron whose faces are congruent triangles Any set of four identical triangles can be arranged to form a four-surface solid. If these four triangles are not equilateral, e.g. 3,4,5, how do I find the altitude of from the one used for the bottom to the apex formed by the other three.
| Elaborating on my comment ...
Let the triangle have side-lengths $a$, $b$, $c$, and area $T$. The Cayley-Menger determinant for area ---equivalently Heron's formula--- tells us
$$16 T^2 = (-a+b+c)(a-b+c)(a+b-c)(a+b+c) \tag{1}$$
The CM determinant for volume gives
$$288V^2 = \left|\begin{matrix}
0 & 1 & 1 & 1 & 1 \\
1 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2775597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
What am I doing wrong in finding the $\gcd(x^3-2x^2-x+2, x^3-4x^2+3x)$? Could someone tell me what I am doing wrong in the process of finding the gcd with Euclid's Algorithm? The gcd is supposedly $(x-1)$ but I only get $-2(x-1)$.
$f(x) = x^3-2x^2-x+2$
$g(x) = x^3-4x^2+3x$
$f(x) = 1*g(x)+2x^2-4x+2$
$g(x) = 1/2x*r_1 -(2... | $$ \left( x^{3} - 2 x^{2} - x + 2 \right) $$
$$ \left( x^{3} - 4 x^{2} + 3 x \right) $$
$$ \left( x^{3} - 2 x^{2} - x + 2 \right) = \left( x^{3} - 4 x^{2} + 3 x \right) \cdot \color{magenta}{ \left( 1 \right) } + \left( 2 x^{2} - 4 x + 2 \right) $$
$$ \left( x^{3} - 4 x^{2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2776780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding the residue of $\frac{1}{z^3 \sin z}$ at z = 0 Given the function $f(z) = \frac{1}{z^3 \sin{(z)}}$, what is the residue of $f(z)$ at $z = 0$?
I want to find $b_1$ from the Laurent expansion. So I did the following:
\begin{align*}
\frac{1}{z^3 \sin{(z)}}
&= ( \dots \frac{b_5}{z^5} + \frac{b_4}{z^4} + \frac{b_... | It's much simpler than what you do, using asymptotic analysis:
\begin{align}
\frac1{\sin z}&=\frac 1{z-\cfrac{z^3}6+\cfrac{z^5}{120}+O(z^7)}\\
&=\frac1z\cdot\frac1{1-\biggl(\underbrace{\cfrac{z^2}6-\cfrac{z^4}{120}+O(z^6)}_{=\,u}\biggr)} \\
&=\frac1z\biggl[1+
\cfrac{z^2}6-\cfrac{z^4}{120}+\biggl( \cfrac{z^2}6-\cfrac{z^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2778613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Finding every solution for equation of complex numbers I need to find every solution for:
$\ z^{3} + 3i \overline z = 0 $
So I tried was just to compare imaginary and complex part of $\ z^{3} $ and $\ 3i\overline z$
Ill spare you the alegbra, here is the result:
$$\ a^{3} - 3ab^{2} + i(3a^{2}b-b^{3}) = -3b -3ai \\a^... | there is the trivial solution $z = 0$
or
$z^3 + \bar z 3i = 0\\
z(z^3 + \bar z 3i) = 0\\
z^4 + z\bar z 3i = 0\\
z^4 + |z|^2 3i = 0\\
\frac {z^4}{|z|^2} = -3i \\
\arg z^4 = \frac {-\pi}{2}+2n\pi\\
\arg z = \frac {-\pi}{8}, \frac {-5\pi}{8},\frac {3\pi}{8},\frac {7\pi}{8}\\
|\frac {z^4}{|z|^2}| = |z^2| = 3\\
|z| = \sqrt ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find integer triples $(x,y,z)$ such that $2018^x=y^2+z^2+1$ To find the triples, I had some tries.
First , consider $x\ge 2$, then $2018^x=0\pmod 4$. But for the right side $y^2+z^2+1$. Consider 3 situations.
*
*case 1. Both $y$ and $z$ are even. Then $y^2+z^2+1=1\pmod 4$.
*case 2. One of them are even, the other... | You've done most of the work already; you might also want to explicitly exclude negative $x$. One way to find all solutions with $x=1$ without any advanced machinery is as follows:
From any solution of $2017=y^2+z^2$ we get another solution by changing the signs of $y$ and $z$ and by interchanging $y$ and $z$. So let's... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the values of $(2\sin x-1)(\cos x+1)=0$ How many different value of x from 0° to 180° for the equation $(2\sin x-1)(\cos x+1) = 0$?
The solution shows that one of these is true:
$\sin x = \frac12$ and thus $x = 30^\circ$ or $120^\circ$
$\cos x = -1$ and thus $x = 180^\circ$
Question: Inserting the $\arcsin$ of $1... | As you noted $x=120°$ is not a solution indeed
*
*$(2\sin 120°-1)(\cos 120°+1)=\left(2\frac{\sqrt3}2-1\right)\left(-\frac12+1\right)\neq 0$
but
*
*$2\sin x-1=0 \implies x=\frac16 \pi + 2k\pi,\,x=\frac56 \pi + 2k\pi$
*$\cos x+1 = 0 \implies x=\pi + 2k\pi=(2k+1)\pi$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2783129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Testing $\sum\limits_{k=1}^∞(\frac{k+1}k)^{k^2}3^{-k}$ for convergence and absolute convergence
Test $$\sum_{k=1}^{\infty}\left(\frac{k+1}{k}\right)^{k^2}3^{-k}$$ for convergence and absolute convergence.
We apply the ratio test for $\displaystyle \sum_{k=1}^{\infty}\left|\left(\frac{k+1}{k}\right)^{k^2}3^{-k}\right|... | Too long for a comment.
Consider
$$a_k=\left(\frac{k+1}{k}\right)^{k^2}3^{-k}$$ and you want to analyze $\frac{a_{k+1}}{a_k}$. It is convenient to consider first
$$\log\left(\frac{a_{k+1}}{a_k}\right)=\log\left({a_{k+1}}\right)-\log\left({a_{k}}\right)$$ where
$$\log\left({a_{k}}\right)=k^2 \log \left(\frac{k+1}{k}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Integrate $\int x\sin^2 (x) dx$ Integrate $\int x\sin^2 (x) dx$
My attempt:
$$=\int x\sin^2 (x) dx\\
=x^2\sin^2 (x) - \int 2\sin (x)\cos (x)x^2 dx\\
=x^2\sin^2 (x) - \int \sin (2x) x^2 dx.$$
| Take $\int x\sin ^2(x)dx = \int udv$, where $u = x\sin x$ and $dv = \sin (x)dx$. Then by integrating by parts we get
$$ \int x \sin ^2(x) dx = -x\sin (x)\cos x + \int (\sin x + x\cos x)\cos (x)dx =\ldots $$
Now, with $\cos ^2x = 1-\sin ^2x$ we get
$$\ldots = -x\sin (x)\cos x + \int \sin (x)\cos (x) dx + \int x\left (1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
solving differential equation $\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}=x^2+2x+2$ How would you solve this third order differential equation:
$$\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}=x^2+2x+2$$
My first thought was to take a double integral:
$$\iint\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}dxdx=\iint{x^2+2x+2}dxdx$$
so:
$$y+\frac{dy}... | Here is another simple way
$$\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}=x^2+2x+2$$
Multiply both side by $e^x$ We get
$$(y''e^x)'= (e^x(x^2+2))'$$
we can reduce the order by direct integration
$$y''e^x= e^x(x^2+2)+K_1$$
$$y"=x^2+2+K_1e^{-x}$$
Integrate
$$y'=\frac {x^3}3+2x+K_1e^{-x}+K_2$$
Integrate again
$$y=\frac {x^4}{12}+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2791300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Grade $6$ Math Problem
$10$ players are playing in a card game in which the winner is the player having the most number of cards. There are $230$ cards in total. What is the smallest number of cards the winning player could have collected, assuming that each player collected a different number of cards?
When I attem... | Extending this to $n$ cards and $k$ players, let $m$ be the minimum possible sum with $k$ players i.e. $m=\sum_{i=1}^ki$. The solution to the problem is then $\lceil \frac {n-m}k \rceil+k$.
e.g. $n=230$, $k=10$, then $m=55$. Solutions is:
$\lceil \frac {230-55}{10} \rceil+10=\lceil \frac {175}{10} \rceil+10=\lceil 17.5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2798908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Generating function for sequence $a_i={n+im-1 \choose im}$ It is well known that $\sum_{i=0}^\infty {n+i-1 \choose i}x^i=\frac{1}{(1-x)^n},$ i.e. $\frac{1}{(1-x)^n}$ is generating function for sequence $a_i={n+i-1 \choose i.}$
But I want to find generating function for sequence $a_i={n+im-1 \choose im.}$
Using The On-... | The $m$-th roots of unity $\omega_j=\exp\left(\frac{2\pi ij}{m}\right), 0\leq j<m$ have the nice property to filter elements. For $m,N>0$ we have
\begin{align*}
\frac{1}{m}\sum_{j=0}^{m-1}\exp\left(\frac{2\pi ij N}{m}\right)=
\begin{cases}
1&\qquad m\mid N\\
0& \qquad otherwise
\end{cases}
\end{align*}
If $A(x)=\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2801088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Implicit derivative problem has two algebraically unequal answers with the exact same graph (??) In my math class we were assigned to find the implicit derivative of
$\dfrac yx + \dfrac xy= 2x$
And most of the class found the answer to be
$\text{(student answer)} \to\dfrac{dy}{dx} = \dfrac{(y^3-x^2y)}{(xy^2 -x^3 -2x^2... | I'm not sure about how the two answers were reached: either you differentiate as is, getting
$$ \frac{1}{x} \frac{dy}{dx} - \frac{y}{x^2} + \frac{1}{y} - \frac{x}{y^2} \frac{dy}{dx} - 2 = 0, $$
which can be rearranged to
$$ \frac{dy}{dx} = \frac{2+y/x^2-1/y}{1/x-x/y^2} = \frac{2x^2y+y^2-x^2}{y(y^2-x^2)} $$
or you clear... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2803072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to solve the problem on number theory Find the number of positive integer pairs $x,y$ such that
$$xy+\dfrac{(x^3+y^3)}3=2007.$$
I solved the question by using factorization and further checking possible values of $x$ and $y$. But it was very lengthy as I had to check many cases for $x$ and $y$. Is there any possib... | Denote: $$\begin{cases}x+y=a \\ x-y=b\end{cases} \Rightarrow \begin{cases} x=\frac{a+b}{2}\\ y=\frac{a-b}{2}\end{cases}$$ Then:
$$xy+\dfrac{x^3+y^3}3=2007 \Rightarrow (a+2)^2+3b^2=\frac{24080}{a-1}.$$
Note that $a>1$ and $a$ can be $2,3, 5,6,8,9,11, 15,17,21$. Note that $a\ge 28 \Rightarrow b^2<0$. So, among them only ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2806555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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What is the coefficient of $x^{11}$ in the power series expansion of $\frac{1}{1-x-x^4}$? I am really stuck on this problem. I don't really understand power series expansions. However, I think this has to do with generating functions.
| $\frac{1}{1-x-x^4}=\frac{1}{1-(x+x^4)}=1+(x+x^4)+(x+x^4)^2+(x+x^4)^3+(x+x^4)^4+(x+x^4)^5+(x+x^4)^6+(x+x^4)^7+(x+x^4)^8+(x+x^4)^9+(x+x^4)^{10}+(x+x^4)^{11}+\cdot\cdot\cdot$.
Checking how many $x^{11}$s are contributed by each term and adding up, we see the answer is $0+0+0+0+0+\binom{5}{2}+0+0+\binom{8}{1}+0+0+\binom{11... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2816497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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3D vector projection If I have an arbitrary vector in the 3D coordinate plane with given azimuth angle and polar angle with radius r=1, how could I use this information to project this vector onto the three 2D planes?(xy,yz,xz) There are no specific number available because I need to represent these "component" vector... | If your vector is $$\vec{p} = \left[\begin{matrix}x\\y\\z\end{matrix}\right] = \left[\begin{matrix}r \cos\theta \cos\varphi \\
r \cos\theta \sin\varphi \\
r \sin\theta \end{matrix}\right]$$
where $-180° \le \varphi \le +180°$ is the azimuth angle, and $-90° \le \theta \le +90°$ the polar angle, then its projection on t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2817143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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If $a,b\in R$ and $a \neq 0$ and quadratic $ax^2-bx+c=0$ If $a,b\in R$ and $a \neq 0$ and quadratic $ax^2-bx+c=0$ has imaginary roots then $a+b+1$
why a+b+1 should be positive?
| The roots of $ax^2-bx+c$ are given by
$$
\frac{b\pm\sqrt{b^2-4ac}}{2},
$$
which are both complex if and only if $b^2-4ac<0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2819356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral $\int_0^{\pi} \frac{\cos(2018x)}{5-4\cos{x}}dx$ I wish to evaluate $$I(2018)=\int_{0}^{\pi}\frac{\cos(2018x)}{5-4\cos x} dx$$ Considering $$X=I(k)+iJ(k)=\int_{-\pi}^{\pi}\frac{\cos{kx}}{5-4\cos x} dx +i\int_{-\pi}^{\pi}\frac{\sin{kx}}{5-4\cos x} dx=\int_{-\pi}^{\pi}\frac{e^{ikx}}{5-4\cos x} dx$$ let us substi... | You are correct! This is an elementary approach without complex analysis.
Since $$\cos((n+1)x)+\cos((n-1)x)=2\cos(nx)\cos(x)$$ then for $n\geq 1$, we have the linear recurrence
$$\begin{align}
I(n-1)+I(n+1)&=\int_{0}^{\pi}\frac{2\cos(nx)\cos(x)}{5-4\cos(x)} dx\\
&=
-\frac{1}{2}\int_{0}^{\pi}\frac{\cos(nx)(-5+5-4\cos(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2820433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
If for a real $x, x +\frac1x$ is an integer, prove $x^{2017}+\frac1{x^{2017}}$ is also. For some real $x$, let $y=x +\frac1x$, with $y\in \mathbb{Z}$.
$y=x +\frac1x\implies x^2 -xy +1 =0\implies x= \frac{y\pm \sqrt{y^2 -4}}2$, so $x+\frac1x = \frac{y\pm \sqrt{y^2 -4}}2+ \frac2{y\pm \sqrt{y^2 -4}}$
This implies: $ y ... | Alt. hint: $\;a=x\,$ and $\,b=\dfrac{1}{x}\,$ are roots of the quadratic $\,z^2-y z + 1\,$ where $\,y = a+b\,$.
If $\,y=x+\dfrac{1}{x}\,$ is an integer, then the quadratic is a monic polynomial with integer coefficients, and therefore $\,a^n+b^n=x^n+\dfrac{1}{x^n}\,$ is an integer for all $\,\forall n\,$ by Newton's id... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2822216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Rank of a matrix in $Z_p^{r\times n}$. Let $p$ be a prime number and $r$ even such that $0<r<n<p$. Calculate rank of matrix$\begin{pmatrix}1&1&1&...&1\\
0&1&2&...&n-1\\
0^2&1^2&2^2&...&(n-1)^2\\
\vdots&\vdots&\vdots&&\vdots\\
0^{r-1}&1^{r-1}&2^{r-1}&...&(n-1)^{r-1}
\end{pmatrix} \in \mathbb Z_p^{r\times n}$
Here is a... | As already observed in the comments, this matrix is related to a Vandermonde matrix. Specifically, the $(1,1)$ minor is the determinant of a matrix that is obtained from a Vandermonde matrix with distinct factors by multiplying each column with a non-zero factor. Thus, the Laplace expansion for the first column yields ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2823048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding $C_n=-C_{n-1}+C_{n-2}$ $$\begin{cases}
C_n=-C_{n-1}+C_{n-2}\\
C_0=1\\
C_1=0
\end{cases} $$
$
\begin{pmatrix}
C_n \\
C_{n-1}
\end{pmatrix}
\begin{pmatrix}
-1 && 1 \\
1 && 0
\end{pmatrix}^n
\begin{pmatrix}
1\\
0
\end{pmatrix}$
$\begin{vmatrix}
\lambda+1 && -1 \\
-1 && \lambda
\end{vmatrix}=\lambda^2+\lambda... | Apart from an offset (shift of n to n-1) this is https://oeis.org/A039834 in the Online Encyclopedia of Integer Sequences, which provides an equivalent formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2823828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Uniform Convergence of $\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}+...$ Show that following series is uniformly convergent in $(-1,1)$.
$$\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}+...$$
I had previously solved this problem using Weierstrass M test by taking
$$|f_n| = |\frac{2^n.x^{2^n-1}}{1+x^... | The series $\sum \frac{2^n.x^{2^n-1}}{1+x^{2^n}}$ is not uniformly convergent on $(-1,1)$. If it was $v_n = \sup\limits_{x \in (-1,1)} \left\vert \frac{2^n.x^{2^n-1}}{1+x^{2^n}} \right\vert$ would be such that $v_n \to 0$.
But
$$\left\vert \frac{2^n.x^{2^n-1}}{1+x^{2^n}} \right\vert \ge 2^{n-1} \vert x \vert^{2^n-1}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2824568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Consider the next succession and prove by induction The exercise says:
Knowing the next succession,
$$a_1=1$$
$$a_{n+1}=\frac{2a_n}{(2n+2)(2n+1)}, n>=1$$
Prove by induction that $a_n=\frac{2^n}{(2n)!}$
What I've done so far is prove for $n=1$
For $n=1$:
$$a_1=\frac{2^1}{(2\times1)!}=\frac{2}{2}=1$$
Which is correct.... | When you wrote that$$a_{n+1}=\frac{2\times a_{n+1}}{(2(n+1)+2)(2n+1)},$$I suppose that you meant to write $a_{n+2}$ as the LHS.
Anyway, this is inconclusive. If you were assuming that $a_{n+1}=\frac{2^{n+1}}{(2(n+1))!}$, your conclusion should have been that $a_{n+2}=\frac{2^{n+2}}{(2(n+2))!}$.
Note that$$\frac{a_{n+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2829410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proving $\sum_{k=0}^{n-1} \text{cos}\left(\frac{2\pi k}{n} \right) = \sum_{k=0}^{n-1} \text{sin}\left(\frac{2\pi k}{n} \right) = 0$ I would like to prove the following, $$\sum_{k=0}^{n-1} \text{cos}\left(\frac{2\pi k}{n} \right) = \sum_{k=0}^{n-1} \text{sin}\left(\frac{2\pi k}{n} \right) = 0$$ This is equivalent to sho... | I assume $n \ge 2$.
Set
$\omega = \exp \left ( \dfrac{2\pi i}{n} \right ); \tag 1$
then
$\omega^n = \left ( \exp \left ( \dfrac{2\pi i}{n} \right ) \right )^n = \exp \left ( \dfrac{2 \pi i n}{n} \right ) = \exp ( 2 \pi i ) = 1; \tag 2$
we have
$( \omega - 1) \displaystyle \sum_0^{n - 1} \omega^k = \omega^n - 1 = 0; \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2829976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving the product of four consecutive integers, plus one, is a square I need some help with a Proof:
Let $m\in\mathbb{Z}$. Prove that if $m$ is the product of four consecutive integers, then $m+1$ is a perfect square.
I tried a direct proof where I said:
Assume $m$ is the product of four consecutive integers.
If $m... | \begin{eqnarray*}
m=n(n+1)(n+2)(n+3) =n^4+6n^3+11n^2+6n.
\end{eqnarray*}
So
\begin{eqnarray*}
m+ 1 =n^4+6n^3+11n^2+6n+1=(n^2+3n+1)^2.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2832986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 13,
"answer_id": 5
} |
Proving $x^2+x+1\gt0$ I was doing a question recently, and it came down to proving that $x^2+x+1\gt0$. There are of course many different methods for proving it, and I want to ask the people here for as many ways as you can think of.
My methods:
*
*$x^2+x+1=(x+\frac12)^2+\frac34$, which is always greater than $0$.
... | If $x$ is positive, then $x^2+x+1$ is clearly positive.
If $x$ is negative then $x^2-x+1$ is certainly positive. Now $$(x^2+x+1)(x^2-x+1)=x^4+x^2+1$$ is certainly positive, so $x^2+x+1$ must also be positive in this case.
If $x$ is zero then $x^2+x+1=1\gt 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2834495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 21,
"answer_id": 2
} |
Binomial coefficient identity found in Apéry's theorem In Apéry's proof of the irrationality of $\zeta(3)$, while proving the formula for the fast-converging series $\zeta(3)=\frac{5}{2} \sum_{k=1}^{\infty}{\frac{ (-1)^{k-1}} {k^{3}\binom {2k}{k}}}$ there is the following identity:
$$
\frac{(-1)^{n-1}(n-1)!^2}{n^2(n^2-... | For each integer $k$, we have $n^2 - k^2 = \left(n-k\right) \left(n+k\right)$. Thus,
\begin{align*}
\prod\limits_{k=1}^{n-1} \left(n^2-k^2\right) &= \prod\limits_{k=1}^{n-1} \left(\left(n-k\right) \left(n+k\right)\right) = \underbrace{\left(\prod\limits_{k=1}^{n-1} \left(n-k\right) \right)}_{=\left(n-1\right)!} \underb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2835970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solving a congruence of the form $a^x = b \pmod m$ without indices or primitive roots Consider $9^x\equiv 7 \mod 19$. So $9^x\equiv 26 \equiv 45$, $9^{x-1} \equiv 5 \equiv 24 \equiv 43 \equiv 62 \equiv 81$, so $x=3$, and $19 \mid 722$.
But what I really want to solve is $12^x\equiv 17 \mod 25$. Using the same method, $... | It is probably quickiest & easiest to calculate the powers of $12$ modulo $25$ ..
\begin{eqnarray*}
12,19,3,11,7,9,8,21,2,24 \\
13,6,22,14,18,16,\color{red}{17} \cdots
\end{eqnarray*}
So
\begin{eqnarray*}
12^{17}=17 \pmod{25}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2839062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Determining an orthonormal set of basis vectors for the linear space The following is example C.4 from Appendix C (Linear Spaces Review) of Introduction to Laplace Transforms and Fourier Series, Second Edition, by Phil Dyke:
Example C.4 Determine an orthonormal set of vectors for the linear space that consists of all ... | By definition of norm induced from inner product, $||x||^2 = \langle x,x\rangle$.
That is, if $||a+bx|| = 1$ then $\langle a+bx,a+bx\rangle = 1$ ,so writing this down:
$$
\int_{0}^1 (a+bx)(a+bx) dx = 1 \implies \int_0^1 (a^2 + 2abx + b^2x^2) dx = 1 \\ \implies a^2 + ab + \frac {b^2}3 = 1
$$
therefore, the statement th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2839288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Prove or disprove that $ \sum\limits_{k = 1 }^T f(k)=0 $ where $f(m)=\sum\limits_{n = 1 }^ m (-1)^n \sin(\frac{n(n+1)(2n+1)}{6}x) $ $$f(m)=\sum\limits_{n = 1 }^ m (-1)^n \sin\left(\frac{n(n+1)(2n+1)}{6} \frac{a \pi}{b}\right) \tag 1 $$
Where $a,b,m$ positive integers.
I have tested in WolframAlpha for many $a$ and $b... | 1. Settings and main results
Let $a$ and $b$ be relatively prime integers. Let $\theta, e, F $ be defined by
\begin{align*}
\theta_n = \frac{a}{b}\left(\sum_{k=1}^{n} k^2 \right) + n, \qquad
e_n = \exp\{i\pi\theta_n\}, \qquad
F_m = \sum_{n=1}^{m} e_n.
\end{align*}
(Here, we extend $\sum$ by additivity to allow non-posi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2839426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
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Solving $a! + b! = 2^n$
$a! + b! = 2^n$, find $a,b,n$
My obseravtions thus far:
*
*If $a>4$, $b<4$, otherwise $a! + b! = 0 \mod 10$
*If $a=1$, $b=1$
If we try all combination with the numbers 2 to 4, we get following solutions:
(1,1,1), (2,2,2), (2,3,3),(3,2,3)
Are there more solutions and how to find them?
| There are no other solutions in $\mathbb N = \{ 1, 2, \dots \}$. There are a few more if you allow $0 \in \mathbb N$.
Indeed, we may assume that $a\le b$. Then $2^n = a! + b! = (a!)(1+c)$ and so $a!$ divides $2^n$. Therefore, $a!=1$ or $a!=2$, and so $a=1$ or $a=2$.
If $a=1$, then $b! = 2^n-1$ is odd and so $b!=1$, tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2840866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Find the minimum value of $\frac{a+b+c}{b-a}$ Let $f(x)=ax^2+bx+c$ where $(a<b)$ and $f(x)\geq 0$ $\forall x\in R$.
Find the minimum value of $$\frac{a+b+c}{b-a}$$
If $f(x)\geq 0$ $\forall x\in R$ then $b>a>0$ and $b^2-4ac\leq 0$ implying that $c>0$. After this not able to find way out.
| If $f(x) \ge 0\to b^2-4ac \le 0$ then forming the lagrangian
$$
L(a,b,c,\lambda,\epsilon) =\phi(a,b,c) +\lambda(b^2-4ac+\epsilon^2)
$$
with
$$
\phi(a,b,c) = \frac{a+b+c}{b-a}
$$
The stationary points are computed by solving
$$
\nabla L = \left\{
\begin{array}{rcl}
\frac{a+b+c}{(b-a)^2}-4 c \lambda +\frac{1}{b-a}=0 \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
For $x\geq 0$, what is the smallest value of $\frac{4x^2+8x+13}{6(x+1)}$? I know that I have to use the AM-GM inequality.
I tried separating the fraction:
$$\frac{4x^2+2x+7}{6(x+1)} + \frac{6(x+1)}{6(x+1)}$$
However, it doesn't seem to make either side of the inequality into a number.
I would appreciate some help, tha... | $$\frac {4x^2+8x+13}{6(x+1)}$$
$$= \frac {2(x+1)}{3} + \frac {3}{2(x+1)}$$
$$ = y+\frac {1}{y} \ge 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2842418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Is there a way to show that $e^x=\lim_{n\to \infty }\left(1+\frac{x}{n}\right)^n$? I know that $$e:=\lim_{n\to \infty }\left(1+\frac{1}{n}\right)^n,$$
by definition. Knowing that, I proved successively that $$e^{k}=\lim_{n\to \infty }\left(1+\frac{k}{n}\right)^n,$$
when $k\in \mathbb N$, $k\in \mathbb Z$ and $k\in\math... | Left side:
The exponential function may be written as a Taylor series:
$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$
Right side:
$(1+\frac{x}{n})^n$ is a binomial expansion like:
$(1+y)^n=\binom{n}{0}y^0+\binom{n}{1}y^1+\binom{n}{2}y^2+...+\binom{n}{n-1}y^{n-1}+\binom{n}{n}y^n$
Where $\binom{n}{k}$ is the Binomial coeffi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2842995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Applying the quadratic Tschirnhausen transformation As per my previous question, I attempted to take dxiv's approach, though I can't seem to make much headway. Considering the simpler problem $x^3=x+a$ and the substitution $y=x^2+mx+n$, I got the following:
\begin{align}y^2&=x^4+2mx^3+(m^2+2n)x^2+2mnx+n^2\\&=(m^2+2n+1)... | I can follow the calculations all the way down to the last cubic. But when equating the following determinant to $\,0\,$ ...
$$
\scriptsize{
\left|
\begin{matrix}
1 & m & n - y\\
m^2+2n+1 & 2mn+2m+a & n^2+2am-y^2 \\
3m^2n+3m^2+3n^2+3am+3n+1 & 3mn^2+m^3+3am^2+6mn+3m+3an+2a & am^3+n^3+6amn+3am+a^2 - y^3
\end{matrix}
\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2844224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Is $\lim_{x\to -\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}=1$?
Evaluate $$\lim_{x\to-\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}$$
My attempt:
$$\lim_{x\to -\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}=\lim_{x\to-\infty}\frac{5x}{-\sqrt{4x^2}}=\frac{-5}{2}$$ According to the answer key, it actually equals $1$.
Thanks in advance.... | Let $x<0.$ Then
$$\lim_{x\to-\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}=\lim_{x\to-\infty} \dfrac{5+\dfrac{9}{x}}{3+\dfrac{2}{x}+\sqrt{4-\dfrac{7}{x^2}}}=\frac{5+0}{3+0+\sqrt{4-0}}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2849212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Which Sign do I choose using the Half Angle Formula for sin for this? I'm evaluating $\sin\left(\frac{1}{2}\sin^{-1}\left(-\frac{7}{25}\right)\right).$
The first thing I did was rewrite it as $\sin\left(\frac{\beta }{2}\right)$
Then I said that
$\sin\left(\beta \right)=-\frac{7}{25}$
Using the Pythagorean Identity I f... | Let $\sin^{-1}\left(-\frac{7}{25}\right)=t$
As for real $x,-\dfrac\pi2\le\sin^{-1}x\le\dfrac\pi2,-\dfrac\pi2\le t<0$ as $t<0$
$\implies\sin\left(\dfrac t2\right)<0$
and $\cos t=+\sqrt{1-\sin^2t}=?$
$\sin\left(\dfrac t2\right)=-\sqrt{\dfrac{1-\cos t}2}=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2850201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove there is no $x, y \in \mathbb Z^+ \text{ satisfying } \frac{x}{y} +\frac{y+1}{x}=4$ Prove that there is no $x, y \in \mathbb Z^+$ satisfying
$$\frac{x}{y} +\frac{y+1}{x}=4$$
I solved it as follows but I seek better or quicker way:
$\text{ Assume }x, y \in \mathbb Z^+\\ 1+\frac{y+1}{y}+\frac{x}{y} +\frac{y+1}{x... | Hint
$$\frac{x}{y} +\frac{y+1}{x}=4\to \frac{x}{y}+\frac{y}{x}+\frac{1}{x}=4$$
Call $x/y=t\in \Bbb Q$.
$$t+\frac{1}{t}+\frac{1}{x}=4\to xt^2+t(1-4x)+x=0$$
By Rational Roots Theorem the candidates to be a rational root, $t$, are $\{\pm1,\pm x,\pm 1/x\}$.
Now, test every root and check the value of $x$ you get.
Can you ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2853669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
How is discriminant related to real $x$? Question in my text book
Solve for range of the function, $$y=\frac{x^2+4x-1}{3x^2+12x +20}$$
Text book says, cross multiply and express the obtained equation as a quadratic equation in $x$
So I get $ (3y-1)x^2 + (12y -4)x + (20y-1) = 0$
Now it says find discriminant $D$, so we ... | $$\frac{x^2+4x-1}{3x^2+12x +20} = \frac{x^2+4x + \frac{20}{3}}{3x^2+12x +20} - \frac{ \frac{23}{3}}{3x^2+12x +20} = \frac{1}{3}- \frac{ \frac{23}{3}}{3x^2+12x +20}$$
Meanwhile, $$ 3x^2 + 12 x + 20 = 3 (x+2)^2 + 8 $$ is always positive
with a minimum value of $8$ when $x=-2,$ so the minimum value of the whole thing is
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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Prove that $\sum_{j=0}^{\infty}(-1)^j\left(\frac{1}{5+6j}+\frac{1}{1+6j} \right) = \frac{\pi}{3}$ I was trying to solve the integral $$\int_{0}^{\infty}\frac{x^4}{1+x^6}dx$$
using series. Now I'm stuck at the series below.
How to prove that
$$\sum_{j=0}^{\infty}(-1)^j\left(\frac{1}{5+6j}+\frac{1}{1+6j} \right) = \frac... | $\begin{align}J&=\int_0^{\infty}\frac{x^4}{1+x^6}\,dx\\
&=\int_0^{\infty}\frac{2x^2-1}{3(x^4-x^2+1)}\,dx+\frac{1}{3}\int_0^{\infty}\frac{1}{1+x^2}\,dx\\
&=\int_0^{\infty}\frac{2x^2-1}{3(x^4-x^2+1)}\,dx+\frac{1}{3}\Big[\arctan x\Big]_0^{\infty}\\
&=\frac{1}{3}\int_0^{\infty}\frac{2x^2-1}{x^4-x^2+1}\,dx+\frac{1}{6}\pi
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2856251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Solve $\tan {(x-\frac{\pi}{4}})=-\tan{(x+\frac{\pi}{2})}$ The question:
Without the use of a calculator, solve for all values of $x$ if $\tan {(x-\frac{\pi}{4}})=-\tan{(x+\frac{\pi}{2})}.$
Using the compound angle formula for solving equations is normally easy, but I stumbled across this problem.
The $LHS$ is easy t... | Hint: $\tan(x)$ is an odd function so $\tan(-x)=-\tan(x)$
Solution:
By above we get $\tan{(x-\pi/4)}=\tan{(-x-\pi/2)}$ so
$$x-\pi/4 \equiv -x-\pi/2 \pmod{\pi}$$
(since the tangent function has a period of $\pi$)
$$2x \equiv -\frac{\pi}{4} \equiv \frac{3\pi}{4} \equiv \frac{7\pi}{4}\pmod{\pi}$$
Solving this yields... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2856589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
What is the next limit equal to? $$\lim_{n\to\infty}(\sqrt{n^2+n}-\sqrt[3]{n^3+1})=?$$
I tried amplifying the hole substraction to form the formula $$a^3-b^3$$ but didn't worked out. Can you help me figure it out?
| The key here is the (not so) popular standard limit $$\lim_{x\to a} \frac{x^r-a^r} {x-a} =ra^{r-1}\tag{1}$$ The given expression (the one whose limit is to be evaluated) can be written as $$\dfrac{\sqrt{1+\dfrac{1}{n}}-\sqrt[3]{1+\dfrac{1}{n^3}}}{\dfrac{1}{n}}$$ Now we add and subtract $1 $ in the numerator and see tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2856819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Using inverse Laplace transform to solve differential equation The differential equation is as follows-
$$\frac{d^2 x}{dt^2} + 5 \frac{dx}{dt} + 6x = e^t $$
I use laplace transform to make it to become - $$X(s) = \frac{1}{(s-1)(s+3)(s+2)}$$
where $X(s)$ is the Laplace transform of $X(t)$
So now I am trying to find $X(... | $ s=0$ is not the best choice
There are three values to assign to $s$ which makes our life very easy.
$$1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3)$$
$$ s=1 \implies 12A=1 \implies A=\frac {1}{12}$$
$$ s=-3 \implies 4B=1 \implies B=\frac {1}{4}$$
$$ s=-2 \implies -3C=1 \implies C=\frac {-1}{3}$$
Now you proceed with t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2859772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Determine all real $x$ that satisfy $\det A=0$ I want to find all real $x$ that satisfy
$$
\textrm{det } X=
\begin{vmatrix}
x &2 &2 &2\\
2 &x &2 &2\\
2 &2 &x &2\\
2 &2 &2 &x
\end{vmatrix}\\
$$
My teacher does this by adding the three bottom rows to the top row
$$
\textrm{det } X=
\begin{vmatrix}
x+6 ... | Another way: the determinant is a fourth-degree polynomial $p(x)$ in variable $x$.
It is easy to see that $p(2) = p(-6) = 0$, like you did.
Then take the derivative:
\begin{align}
p'(x) &= \begin{vmatrix}
1 &2 &2 &2\\
0 &x &2 &2\\
0 &2 &x &2\\
0 &2 &2 &x
\end{vmatrix} + \begin{vmatrix}
x &0 &2 &2\\
2 &1 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2861589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove, using first principles, that $\lim\limits_{(x,y)\to(0,0)}{x.y^2=0}$ My try is:
$\left|x.y^2-0\right|<\left|\sqrt{x^2+y^2}.(x^2+y^2)\right|< \delta.\delta^2=\epsilon$
where $0<\sqrt{x^2+y^2}< \delta$ what did I miss here?
| Consider $x^2+y^2 <1$,
then $|y| \le 1$, and $y^2 \le |y|$.
Hence:
$|xy^2| \le |xy| \le (1/2)(x^2+y^2)$ .
Let $\epsilon >0$ be given.
Choose $\delta = \min (1,2\epsilon)$
Then $|x^2+y^2| \lt \delta$ implies
$|xy^2| \le$
$(1/2)(x^2+y^2) \lt (1/2)\delta =\epsilon$.
Used: $a^2+b^2 \ge 2|ab|$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2864450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\alpha$-limit set in $A=\{x\in \mathbb{R}^2|1<|x|<2\}$ Consider the system
$$
\dot{x}=-y+x(r^4-3r^2+1)\\
\dot{y}=x+y(r^4-3r^2+1)
$$
where $r^2=x^2+y^2$.
Let $A=\{x\in \mathbb{R}^2|1<|x|<2\}$.
Affirmation (fait accompli): $\dot{r}<0$ on the circle $r=1$ and $\dot{r}>0$ on the circle $r=2$.
My question (doubt) is: W... | Considering
$$
\left\{
\begin{array}{rcl}
\dot{x}&=&-y+x(r^4-3r^2+1)\\
\dot{y}&=&x+y(r^4-3r^2+1)
\end{array}
\right.
\Rightarrow
\left\{
\begin{array}{rcl}
x\dot{x}&=&-x y+x^2(r^4-3r^2+1)\\
y\dot{y}&=&xy+y^2(r^4-3r^2+1)
\end{array}
\right.
\Rightarrow\frac 12\left(r^2\right)' = r^2(r^4-3r^2+1)
$$
Now analyzing
$$
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2865751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Simplify a summation with indicator function I have this double summation that I want to simplify:
$$\sum^{q} _{j=-q} \sum^{q} _{i=-q} \mathbb{1}_{ \{h+j-i=0 \}}$$
A given solution says the answer is $2q+1-h$ for $|h| \leq 2q$ and $0$ otherwise.
I don't see this. For example, when I take $q=1$ and $h = 2*q = 2$ the ... | We have
$$\sum_{a=-q}^q \sum_{b=-q}^q \mathrm{1}_{h+a-b=0}
= \sum_{a=-q}^q \sum_{b=-q}^q \mathrm{Res}_{z=0} \frac{1}{z^{h+1+a-b}}
\\ = \mathrm{Res}_{z=0} \frac{1}{z^{h+1}}
\sum_{a=-q}^q \frac{1}{z^a} \sum_{b=-q}^q z^b
= \mathrm{Res}_{z=0} \frac{1}{z^{h+1}}
\left(\sum_{b=-q}^q z^b\right)^2
\\ = \mathrm{Res}_{z=0} \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2875463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How to prove $1 + \frac{1}{\sqrt{2}} +... +\frac{1}{\sqrt{n}} <\sqrt{n}\ .\left(2n-1\right)^{1/4} $
Prove that $$1 + \frac{1}{\sqrt{2}} +... +\frac{1}{\sqrt{n}} <\sqrt{n}\ .\Bigl(2n-1\Bigr)^{\frac{1}{4}} $$
My Approach :
I tried by applying Tchebychev's Inequality for two same sets of numbers;
$$1 , \frac{1}{\sqrt... | Using Cauchy Schwartz inequality with
*
*$u=(1,1,...,1)$
*$v=\left(1 , \frac{1}{\sqrt{2}} ,... ,\frac{1}{\sqrt{n}}\right)$
we have
$$\vec u \cdot \vec v\le |\vec u||\vec v| \implies 1 + \frac{1}{\sqrt{2}} +... +\frac{1}{\sqrt{n}} \le\sqrt{n}\ \cdot\left(\sqrt{1 + \frac{1}{2} +... +\frac{1}{n}}\right)$$
and thus it ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2875714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Prove $\lim_{(x,y) \to (0,0)} \frac{\exp(xy)\cdot xy\cdot(x^2-y^2)}{x^2+y^2} =0$. I want to show that
$$\lim_{(x,y) \to (0,0)} \frac{\exp(xy)\cdot xy\cdot(x^2-y^2)}{x^2+y^2} =0.$$
Is it valid to do it like this:
$$\lim_{(x,y) \to (0,0)} \left|\frac{\exp(xy)\cdot xy\cdot(x^2-y^2)}{x^2+y^2}\right| \leq \lim_{(x,y) \to ... | We may restrict the neibourhood to $x^2+y^2<1$, then $e^{xy}<2$ and
$$\left|\dfrac{e^{xy}xy(x^2-y^2)}{x^2+y^2}\right|\leqslant\dfrac{2|xy|(|x|^2+|y|^2)}{x^2+y^2}\leqslant x^2+y^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2876409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Compare $\arcsin (1)$ and $\tan (1)$
Which one is greater: $\arcsin (1)$ or $\tan (1)$?
How to find without using graph or calculator?
I tried using $\sin(\tan1)\leq1$, but how do I eliminate the possibility of an equality without using the calculator?
| One can write:
$$
\frac{1}{\sqrt{1 + \tan^21}} = \cos1 = 1 - 2\sin^2\frac{1}{2} > \frac{17639}{32768},
$$
because
$$
\sin\frac{1}{2} < \frac{1}{2} - \frac{1}{2^3\cdot3!} + \frac{1}{2^5\cdot5!} = \frac{1920 - 80 + 1}{3840} < \frac{1845}{3840} = \frac{123}{256}.
$$
On the other hand, using Archimedes's lower bound, $\pi ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 3
} |
if $a^2+b^2+c^2+2\rho(ab+bc+ca)\ge0$ then $\rho\ge-1/2$ This may be trivial but I am not able to prove that if $a^2+b^2+c^2+2\rho(ab+bc+ca)\ge0$ for $a,b,c\in\mathbb{R}$ then $\rho\ge-1/2$. Can anybody help me please?
Thanks!
| If $a=b=1$ and $c=x> -1/2$ then we have $$-2\rho\leq {x^2+2\over 2x+1}=:f(x)$$
Since $f$ achieve minimum $1$ at $x=1$ we have $\boxed{\rho\geq-{1\over 2}}$.
And if $x<-{1\over 2}$ we get $$-2\rho\geq {x^2+2\over 2x+1}$$
Since $f$ achieve maximum $-2$ for $x=-3$ we have also $-2\rho \geq -2$ so $\boxed{\rho \leq 1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Find the number of trailing zeros in 50! My attempt:
50! = 50 * 49 *48 ....
Even * even = even number
Even * odd = even number
odd * odd = odd number
25 evens and 25 odds
Atleast 26 of the numbers will lead to an even multiple (24 evens + 1 even * 1 odd) so at most 26 trailing zeros.
50 is divisible by 5: ... | There are $\lfloor \frac{50}{5}\rfloor$ numbers between $1$ and $50$ that are divisible by $5$. Similiarily $\lfloor \frac{50}{5^2}\rfloor$ and $\lfloor \frac{50}{5^3}\rfloor$ numbers divisible by $5^2$ and $5^3$ respectively.
Thus, the highest power of $5$ dividing $50!$ is
$$\lfloor \frac{50}{5}\rfloor + \lfloor \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2878495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\lim_{x\rightarrow0}\left[x\left(\frac{1+3x}{1+2x}\right)^{\frac {1}{x}}\right]$ I want to evaluate the following limit:
$$\lim_{x\rightarrow0}\left[x\left(\frac{1+3x}{1+2x}\right)^{\frac {1}{x}}\right]$$
We have
$$\left(\frac{1+3x}{1+2x}\right)^{\frac 1 x} = e^{\frac{\log\left(\frac{1+3x}{1+2x}\right)}{x}}\... | $$\lim_{x\rightarrow0}\left[x\left(\frac{1+3x}{1+2x}\right)^{\frac {1}{x}}\right]=\lim_{x\to0}x \cdot\dfrac{\left(\lim_{x\to0}\left(1+3x\right)^{1/3x}\right)^3}{\left(\lim_{x\to0}\left(1+2x\right)^{1/2x}\right)^2} =\lim_{x\to0}x \cdot\dfrac{e^3}{e^2}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2879225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Double series convergent to $2\zeta(4)$? Using a computer I found the double sum
$$S(n)= \sum_{j=1}^n\sum_{k=1}^n \frac{j^2 + jk + k^2}{j^2(j+k)^2k^2}$$
has values
$$S(10) \quad\quad= 1.881427206538142 \\ S(1000) \quad= 2.161366028875634 \\S(100000) = 2.164613524212465\\$$
As a guess I compared with fractions $\pi^p/q... | $$
\begin{align}
\frac{m^2+m\,n+n^2}{m^2\,(m+n)^2\,n^2}\, &=\frac{m^2+m\,n+n^2\,\color{red}{+m\,n-m\,n}}{m^2\,(m+n)^2\,n^2} \\[2mm]
&=\,\frac{(m+n)^2-m\,n}{(m+n)^2\,m^2\,n^2} \\[2mm]
&=\,\frac{1}{m^2\,n^2}-\frac{1}{m\,n\,(m+n)^2} \\[2mm]
&=\,\frac{1}{m^2\,n^2}-\frac{1}{m^3}\left(\frac{1}{n}-\frac{1}{m+n}-\frac{m}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2881914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 2
} |
Positive Integers to make a square How many integers $n$ make the expression $7^n + 7^3 + 2\cdot7^2$ a perfect square?
Factoring $7^2$ we have that $7^n + 7^3 + 2\cdot7^2 = 7^2\cdot(7^{n-2} + 9)$.
How do we prove that the 2nd factor only has $1$ solution when $n = 3$?
| An "equivalent" way of doing things, is to note that $7^3 + 2 \times 7^2 = 343+98=441 = 21^2$.
Now, if $7^n + 21^2 = y^2$ for some $y$, then $7^n = (y-21)(y+21)$.
So the prime factorization of $(y-21)$ and $(y+21)$ can both contain only the prime $7$, by unique factorization theorem. Therefore, $y - 21$ and $y+21$ are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2882534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding n-th term of a matrix I have to find the n-th power of the following matrix $$A=\begin{pmatrix} 2 & 0&1\\ 0 & -3& 0 \\ 1 & 0&2 \end{pmatrix}
$$ I computed a few powers $$A^2=\begin{pmatrix} 4+1& 0&4\\ 0 & 9& 0 \\ 4 & 0&1+4\end{pmatrix}$$
$$A^3 =\begin{pmatrix} 14 & 0&13\\ 0 & -27& 0 \\ 13& 0&14\end{pmatrix}$$
$... | HINT
$A$ is a real and symmetric matrix therefore we can diagonalize that is find $P$ such that
$$A=PDP^{-1} \implies A^n=\overbrace{PDP^{-1}PDP^{-1}\ldots PDP^{-1}}^{n\,terms}=PD^nP^{-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2882911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
$\lim_{x \rightarrow \infty}\frac{5^{x+1}+7^{x+1}}{5^x +7^x}$ $$\lim_{x \rightarrow \infty}\frac{5^{x+1}+7^{x+1}}{5^x +7^x}$$
I tried using L'Hospital rule, which yielded :
$$\lim_{x \rightarrow \infty}\frac{5^{x+1} \ln 5+7^{x+1} \ln 7}{5^x \ln 5 }$$
But I'm at the dead end... If I divide numerator and denominator by... | The standard method is to factor out and cancel the highest power, which is here $7^x$:
We get:
$\lim_{x\to\infty} \frac{5^{x+1}+7^{x+1}}{5^x+7^x}=\lim_{x\to\infty} \frac{7^{x+1}\left(\left(\frac57\right)^{x+1}+1\right)}{7^x\left(\left(\frac57\right)^x+1\right)}=\lim_{x\to\infty} \frac{7\left(\left(\frac57\right)^{x+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2884481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
What is the best algorithm for finding the last digit of an enormous exponent? I found most answers here not clear enough for my case such as
$$
123155131514315^{4515131323164343214547}
$$
I wrote the $n\bmod10$ in Python and execution time ran out. So I need a faster algorithm or method. Sometimes, the result is incor... | If $a=10q+b$, then $a^n \equiv b^n \bmod 10$, and so it suffices to consider $b^n$, for $b = 0, 1, \dots, 9$.
These powers repeat as follows:
$0^n = 0, 0, \dots $
$1^n = 1, 1, \dots $
$2^n = 2, 4, 8, 6, 2, 4, 8, 6, \dots $
$3^n = 3, 9, 7, 1, 3, 9, 7, 1, \dots $
$4^n = 4, 6, 4, 6, \dots $
$5^n = 5, 5, \dots $
$6^n = 6, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2884729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Proving $\frac{\sin 30^\circ}{\sin 50^\circ}= \frac{\sin 40^\circ}{\sin 80^\circ}$
Prove that
$$\frac{\sin 30^\circ}{\sin 50^\circ}= \frac{\sin 40^\circ}{\sin 80^\circ}$$
I tried making $\sin 80^\circ=\sin(50^\circ+30^\circ)$, but it didn't go well. I also tried using, maybe, trig periodicities, but I still can't get... | Let us first prove that $\sin(30^\circ) \cdot \sin(80^\circ) = \sin(40^\circ) \cdot \sin(50^\circ)$.
\begin{align}
LHS & = \sin(30^\circ) \cdot \sin(80^\circ) \\
&= \cos(60^\circ) \cdot \cos(10^\circ) \\
& = \frac{1}{2} \cos(10^\circ) \\
\end{align}
\begin{align}
RHS & = \sin(40^\circ) \cdot\sin(50^\circ) \\
& = \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2888471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Evaluate $f(x)=x\log{\left|\frac{x+2}{3-x}\right|}$ for $x\rightarrow\infty$ I have the following function:
$$f(x)=x\log{\left|\frac{x+2}{3-x}\right|}$$
I want to find the limit for $x\rightarrow+\infty$.
This is what I do. Since $x>=0$, I can remove the absolute value:
$$f(x)=x\log\left({\frac{x+2}{3-x}}\right)\sim x\... | We have for big $x > 0$
$$
\ln \left|\frac{x+2}{3-x}\right|^x = \ln \left(\frac{x+2}{x-3}\right)^x
$$
and also
$$
\lim_{x\to \infty}\left(\frac{x+2}{x-3}\right)^x = e^5
$$
hence
$$
\lim_{x\to\infty}x\ln\left|\frac{x+2}{3-x}\right| = 5
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2888587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Prove that $1 + \frac{2}{3!} + \frac{3}{5!} + \frac{4}{7!} + \dotsb = \frac{e}{2}$
Prove that $1 + \frac{2}{3!} + \frac{3}{5!} + \frac{4}{7!} + \dotsb = \frac{e}{2}$.
This is problem 4 from page 303 of S.L.Loney's 'Plane Trigonometry'.
It seems fairly obvious that the series expansion $e^x$ will be used. However, I ... | Your series is $$\sum_{n=1}^{\infty}\frac{n}{(2n-1)!}.$$
Let $a_n=\frac{n}{(2n-1)!}$, then
$$a_n=\frac{1}{2}\frac{2n}{(2n-1)!}=\frac{1}{2}\frac{2n-1+1}{(2n-1)!}
=\frac{1}{2}\left(\frac{1}{(2n-2)!}+\frac{1}{(2n-1)!}\right).$$
So $$\sum_{n=1}^{\infty}\frac{n}{(2n-1)!}=\frac{1}{2}\sum_{n=1}^{\infty}\left(\frac{1}{(2n-2)!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2894991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 0
} |
Any idea on how to find a upper bound for a limit in R^2? I'm working on this problem of continuity in $\mathbb{R}^2$ the statement is the following:
Prove by definition that
$$ \lim_{ (x,y) \to (0,0) } \frac{y^2}{3+\sin^2(x^2+y^2)+y^4}=0 $$
Given $\epsilon >0$, we have that
$$| \frac{y^2}{3+\sin^2(x^2+y^2)+y^4} | =... | $\frac{y^2}{3+\sin^2(x^2+y^2)+y^4}\leq \frac{y^2}{3}\rightarrow 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $\frac{\log x}{y - z} = \frac{\log y}{z - x} = \frac{\log z}{x - y}$, prove that $xyz = 1$ without using the following method If
$$\frac{\log x}{y-z} = \frac{\log y}{z-x} = \frac{\log z}{x-y}$$
prove that $$xyz = 1
$$.
without using this method :-
$$
\mbox{Let }\frac{\log x}{y-z} = \frac{\log y}{z-x} = \frac{\log ... | $$\frac{\log x}{y-z} = \frac{\log y}{z-x}$$
Implies
$$\frac{\log x}{y-z} = \frac{\log y}{z-x}=\frac{\log x+\log y}{(y-z)+(z-x)}$$
Therefore
$$\frac{\log xy}{y-x)}= \frac{\log z}{x-y}= \frac{-\log z}{y-x}
\Rightarrow \\
\log xy = -\log z \Rightarrow \\
\log xyz =0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2896254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to calculate $\int\frac{x}{x^2-x+1}\, dx$? $$\int \frac{x}{x^2-x+1}\, dx = \int \frac{x}{(x-\frac 1 2)^{2} + \frac 3 4}\, dx = \int \frac{x}{(x-\frac 1 2)^2 + (\frac {\sqrt{3}} {2})^2}$$
Substitute $u= \frac{2x-1}{\sqrt{3}}, du=\frac{2}{\sqrt{3}}dx$:
$$\frac {\sqrt{3}} 2 \int \frac{\frac{\sqrt{3}} {2}u + \frac 1 ... | Hello and welcome to math.stackexchange. The solution that you developed and the one given by Wolfram Alpha only differ by a constant (of integration), since
$$
\frac 1 2\log(\frac 4 3x^2-\frac 4 3 x+ \frac 4 3) =
\frac 1 2\log(x^2- x+ 1) + \frac 1 2 \log \frac 4 3 \, .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2896788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Integrate $\frac{x^2-4x+10}{x^2\sqrt x}$
Find the indefinate integral with respect to $x$ of $$\frac{x^2-4x+10}{x^2\sqrt x}$$
For this problem I first made each individual number in the numerator separate from each other for easier integration and then simplified
$$=\int \left(\frac{x^2}{x^2\sqrt x}-\frac{4x}{x^2\sqr... | HINT
Place $t=\sqrt{x}$, after you solve the integral of a rational function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2900440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Prove that $\frac{\sin x}{x}=(\cos\frac{x}{2}) (\cos\frac{x}{4}) (\cos \frac{x}{8})...$ How do I prove this identity:
$$\frac{\sin x}{x}=\left(\cos\frac{x}{2}\right) \left(\cos\frac{x}{4}\right) \left(\cos \frac{x}{8}\right)...$$
My idea is to let
$$y=\frac{\sin x}{x}$$
and
$$xy=\sin x$$
Then use the double angle iden... | Partial answer:
The Taylor series of $\sin x$ is $x - \frac{x^3}{3!} + \frac{x^5}{5!} \cdots$, so:
$$\frac{\sin x}{x} = 1 - \frac{x^2}{3!}+\frac{x^4}{5!} \cdots.$$
Meanwhile, $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!}$, and therefore:
$$\cos(\frac{x}{2^1}) = 1 - \frac{x^2}{2^2 \cdot 2!} + \frac{x^4}{2^4 \cdot 4!} - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2901407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Expressing a linear map $\phi: \>{\mathbb R}^4\to{\mathbb R}^4$ in terms of a new base In terms of the standard coordinates $x$, $y$, $z$, $t$ a linear map $\phi: \>{\mathbb R}^4\to{\mathbb R}^4$ appears as
$$\phi{(x,y,z,t)} = (t,x,y,z)\ .$$ Now we are given a new base $$R = \{ (1,0,1,0),(0,1,0,1),(0,1,1,0),(0,1,1,1) \... | The matrix $P = \begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 \\
0 & 1 & 0 & 1 \\
\end{bmatrix}$ transforms the basis $R$ to the canonical basis $E$.
Therefore
$$M_R(\phi) = P^{-1}M_E(\phi) P = \begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 \\
0 & 1 & 0 & 1 \\
\end{bmatrix}^{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2903675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\frac{b-a+1}{ab} \leq \sum_{i=a}^b \frac{1}{i^2} \leq \frac{b-a+1}{(a-1)b}$ Given $b>a>1; \ a,b \in Z^+$. Prove $\frac{b-a+1}{ab} \leq \sum_{i=a}^b \frac{1}{i^2} \leq \frac{b-a+1}{(a-1)b}. \ $
I mostly care about the second inequality. Thank you!
| For the second inequality just observe that
$$
\frac{1}{i^2} \leq \frac{1}{i(i-1)} = \frac{1}{i-1} - \frac{1}{i},
$$
hence
$$
\sum_{i=a}^b \frac{1}{i^2} \leq
\sum_{i=a}^b\left(\frac{1}{i-1} - \frac{1}{i}\right)
= \frac{1}{a-1} - \frac{1}{b} = \frac{b-a+1}{(a-1)b}\,.
$$
The first inequality can be proved by induction o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2903831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum the series $\frac{3}{1⋅2⋅4}+\frac{4}{2⋅3⋅5}+ \frac{5}{3⋅4⋅6}+...\text{(upto n terms)}$ $\frac{3}{1⋅2⋅4}+\frac{4}{2⋅3⋅5}+ \frac{5}{3⋅4⋅6}+...\text{(upto n terms)}$
The general term seems to be
$$T_r= \frac{r+2}{r(r+1)(r+3)}.$$
I see no way to telescope this because the factors of the denominator of the general term ... | \begin{align*}
T_r
&= \frac{1}{r(r+1)} - \frac{1}{r(r+1)(r+3)} \\
&= \frac{1}{r(r+1)}- \left(\frac{1}{2}\cdot \frac{(r+3)-(r+1)}{r(r+1)(r+3)}\right) \\
&= \frac{1}{r}-\frac{1}{r+1}- \frac{1}{2}\left(\frac{1}{r}-\frac{1}{r+1}\right)+ \frac{1}{2}\cdot \frac{1}{r(r+3)} \\
&= \frac{1}{2}\left(\frac{1}{r}-\frac{1}{r+1}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2904182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Integration work: $\int\sqrt{\frac{2-x}{x-3}} \ \mathrm dx$
$$\int\sqrt{\dfrac{2-x}{x-3}}\mathrm dx$$
My approach
I=$$\int\sqrt{\dfrac{2-x}{x-3}}\mathrm dx$$
I= $$\int\frac{2-x}{\sqrt{-x^2+5x-6}}\mathrm dx$$
Next I substituted 2-x =t and processed but I am not getting the answer. Can you guys help me with this
| $$\frac{2-x}{x-3}=t^2$$
$$\frac{1}{(x-3)^2}dx=2tdt$$
$$x=\frac{3t^2+2}{t^2+1}$$
$$x-3=\frac{3t^2+2}{t^2+1}-3=-\frac{1}{t^2+1}$$
$$\int t\cdot2tdt\cdot(\frac{-1}{t^2+1})^2=\int\frac{2t^2}{(t^2+1)^2}dt$$
Now solving
$${\displaystyle\int}\dfrac{t^2}{\left(t^2+1\right)^2}\,\mathrm{d}t$$
Write it as
$${\displaystyle\int}\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2904991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Which is the partial solution of the ode 4th order I want to solve the ode $$y''''(t)+3y''(t)+y(t)=e^{-t}+t^2, \ \forall t>0$$ Forst we have to find the solution for the homogeneous problem using the characteristic polynomial, right?
We have the characteristic polynomial $$\lambda^4+3\lambda^2+1=0$$ which has the solu... | With $$\lambda=\pm i\sqrt{\frac{3-\sqrt{5}}{2}} \ \text{ and } \ \lambda=\pm i\sqrt{\frac{3+\sqrt{5}}{2}}$$
You have your $$y_c=A\cos \left (\sqrt{\frac{3-\sqrt{5}}{2}} t\right )+B\sin \left (\sqrt{\frac{3-\sqrt{5}}{2}} t\right )+A\cos \left (\sqrt{\frac{3+\sqrt{5}}{2}} t\right )+B\sin \left (\sqrt{\frac{3+\sqrt{5}}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2905952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does this matrix sequence always converge? Suppose $a_0, a_1, ... , a_{n-1}$ are real numbers from $(0; 1)$, such that $\sum_{k=0}^{n-1} a_k=1$. Suppose $A = (c_{ij})$ is a $n \times n$ matrix with entries $c_{ij} = a_{(i-j)\%n}$, where $\%$ is modulo operation. Is it always true that $\lim_{m \to \infty} A^m = \frac{1... | Your matrix $A$ is a circulant matrix:
$$A = \begin{bmatrix}
a_0 & a_1 & \cdots & a_{n-1}\\
a_{n-1} & a_0 & \cdots & a_{n-2}\\
\vdots & \vdots & \cdots & \vdots \\
a_1 & a_2 & \cdots & a_0\\
\end{bmatrix}$$
$A$ is known to have eigenvalues equal to $\lambda_j = \sum_{k=0}^{n-1} a_k\omega_j^k$ with eigenvectors $\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2907338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to prove $(a+1)(ab^2+1) \geq ((ab+1)^\frac{2}{k}+1)^k$? In the paper linear forms in the logarithms of real algebraic numbers close to 1, it is
written on page 8 that-
$$(a+1)(ab^2+1) \geq ((ab+1)^\frac{2}{k}+1)^k=(ab+1)^2+ \sum_{n=0}^{k-1} \binom{k-1}{n} (ab+1)^\frac{2n}{k} \cdots (1)$$
Clearly, $(a+1)(ab^2+1) >... | The inequality is obtained in two steps.
At first, for example by Rearrangement inequality, we have that
$$(a+1)(ab^2+1)=a^2b^2+ab\cdot b+a\cdot 1+1\ge a^2b^2+ab\cdot 1+a\cdot b+1$$
$$\ge a^2b^2+2ab+1= (ab+1)^2$$
but inequality holds $\iff$ $b=1$ and since $b>1$ we have that
$$(a+1)(ab^2+1)\color{red}> (ab+1)^2$$
then,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2907461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Finding the value of $\prod_{n=0}^\infty a_n$ with $a_0=1/2$ and $a_{n}=1+(a_{n-1}-1)^2$
Putting the value of the first term, we can see that the series goes like
$$1/2, 5/4, 17/16,...$$
I am unable to calculate the general term, and so not sum of the series. Please help me to find the general term or directly the in... | Let $b_n=a_n-1$ so that $b_0=-\frac 12$ and $b_{n}=b_{n-1}^2$.
Consequently, $b_n=b_0^{2^n}$ hence
$$a_n=1+(-2)^{-2^n}$$
Let $x=\frac 12$, then
\begin{align}
(1+x)\prod_{n=0}^Na_n
&=(1+x)\prod_{n=0}^N(1+(-x)^{2^n})\\
&=(1+x)(1-x)(1+x^2)\cdots(1+x^{2^N})\\
&=(1-x^2)(1+x^2)\cdots(1+x^{2^N})\\
&=(1-x^4)(1+x^4)\cdots(1+x^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2908445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Sum of three uniformly distributed random variables. Let $X,Y,Z$ be uniformly distributed $U(0,1)$. Then I know that the density function for the random variable $A= X+Y$ is
$$f(a) = \begin{cases}
a & a\in (0,1)\\
2-a & a\in[1,2)\\
0 & \text{ otherwise}
\end{cases}.
$$
and
$$
g(z)=\begin{cases}... | So we get
\begin{equation}
f(b-z) = \begin{cases}
b-z & b-z\in (0,1)\\
2-(b-z) & b-z\in[1,2)\\
0 & \text{ otherwise}
\end{cases}.
\end{equation}
We have that
\begin{equation}
0<z<1
\end{equation}
or
\begin{equation}
b-1<b-z<b
\end{equation}
If $b < 0$, then $b - z < 0$, so $h(b) = 0$ accordi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluate $ \lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right) $ $$\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)$$
Using the L'Hospital Rule I obtained the value $-1/4$, but the answer is given to be $-1/3$. I can't find the mistake. Here's what I did; please point out the mistake.
\beg... | $$\lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right)=\lim_{x \to 0}\frac{\sin^2 x-x^2}{x^2\sin^2 x}=\lim_{x \to 0}\frac{(\sin x-x)(\sin x+x)}{x^4}$$
$$=\lim_{x \to 0}\frac{(\sin x+x)}{x}\lim_{x \to 0}\frac{x(\sin x-x)}{x^4}=\lim_{x \to 0}\frac{2x(\sin x-x)}{x^4}=\lim_{x \to 0}\frac{2(\sin x-x)}{x^3}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 0
} |
Probability of 0 in Binary Sequence A random binary sequence is produced as follows. The first coordinate equals $0$ with probability $0.6$ and equals $1$ with probability $0.4$. For any positive integer $n$, the $(n+1)^\text{th}$ coordinate is the same as the $n^\text{th}$ coordinate of the sequence, with probability ... | Starting from an initial $0$ (prob. $=0.6$) then in the following $n-1$ you must have an even number ($2k$ ) of changes to end with a $0$.
Starting with $1$ instead (prob. $=0.4$), you must have $2j+1$ changes.
That means
$$
\eqalign{
& P_{\,0} (n) = 0.6\sum\limits_{0\, \le \,k\, \le \,\left\lfloor {\left( {n - 1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2911239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Range of $y = \frac{x^2-2x+5}{x^2+2x+5}$? How do I approach this problem? My book gives answer as $[\frac{3-\sqrt{5}}{2},\frac{3+\sqrt{5}}{2}]$. I tried forming an equation in $y$ and putting discriminant greater than or equal to zero but it didn't work. Would someone please help me?
I get $x^2 (y-1) + 2x (y+1) + (5y-5... | Since
$$y = \frac{x^2 - 2x + 5}{x^2 + 2x + 5}$$
we obtain
\begin{align*}
y(x^2 + 2x + 5) & = x^2 - 2x + 5\\
yx^2 + 2yx + 5y & = x^2 - 2x + 5\\
(y - 1)x^2 + (2y + 2)x + 5(y - 1) & = 0\\
(y - 1)x^2 + 2(y + 1)x + 5(y - 1) & = 0
\end{align*}
which is equivalent to your equation $x^2(y - 1) + 2x(y + 1) + (5y - 5) = 0$. H... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2912043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find this integral $I=\int\frac{x}{(1-3x)^{3/2}(x+1)^{3/2}}dx$ Find the integral
$$I=\int\dfrac{x}{(1-3x)^{3/2}(x+1)^{3/2}}dx$$
My try: $$(1-3x)(x+1)=-3x^2-2x+1=-3(x+1/3)^2+\dfrac{4}{3}$$
Thus
$$I=\int\dfrac{x}{(\frac{4}{3}-3(x+\frac{1}{3})^2)^{3/2}}dx$$
| Hint:
For real calculus, we need $$-1<x<\dfrac13$$
$$\iff-\dfrac23<x+\dfrac13<\dfrac23$$
WLOG $x+\dfrac13=\dfrac23\cos2t$ where $0<2t<\pi,\cos t,\sin t,\sin2t>0$
$\iff3x+1=2\cos2t,3x=2\cos2t-1=4\cos^2t-3$
$3dx=-8\sin t\cos t\ dt$
$1-3x=1-(2\cos2t-1)=2(1-\cos2t)=4\sin^2t\implies(1-3x)^{3/2}=(2\sin t)^3$
$x+1=\dfrac{3x+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2915564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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How to get the smallest positive integer solutions to $2\sqrt [3] {x} =3\sqrt [5] {y}$? How do I find the smallest positive integer solutions for $x$ and $y$ in the following equation? $$2\sqrt [3] {x} =3\sqrt [5] {y}$$
| we have $$ 2\sqrt[3]{x} = 3\sqrt[5]{ y}$$
so cubing both side we have
$$2^3 x=3^3y^{\frac 35}$$
now
$$2^{ 3 \times 5}x^5=3^{ 3\times 5} y^3$$
so now we can write this
$$\frac{2^{ 3 \times 5}}{3^{ 3\times 5}}= \frac{y^3}{x^5}$$
so $$ y^3= (2^5)^3 $$
and $$ x^5= (3^3)^5 $$
so$$ y=32$$ and $$x=27$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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What does $\sum\limits_{n=1}^{\infty}\left( {\frac{1}{n}-\frac{1}{\sqrt{n^2+x^2}}} \right)$ equal? I tried calculating it in Mathematica but it can't seem to figure out a formula for this. If you define x as a specific real number then the sum always converges and gives you a number. The output seems to be logarithmic ... | For $x$ sufficiently small we have that
$$\frac{1}{\sqrt{n^2+x^2}}=\frac 1 n\frac{1}{\sqrt{1+(x/n)^2}}\sim\frac1n\left(1-\frac12\frac{x^2}{n^2}\right)=\frac 1n-\frac12\frac{x^2}{n^3}$$
thus
$$\frac{1}{n}-\frac{1}{\sqrt{n^2+x^2}} \sim \frac12\frac{x^2}{n^3}$$
and therefore
$$S(x)=\sum_{n=1}^{\infty}\left({\frac{1}{n}-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2918280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solving homogeneous differential equation $\frac{dy}{dx}=\frac{x^2+8y^2}{3xy}$.
Solve the differential equation. Use the fact that the given equation is homogeneous
$$
\frac{dy}{dx}=\frac{x^2+8y^2}{3xy}
$$
First I multiply the right side by
$$
\frac{\frac{1}{x^2}}{\frac{1}{x^2}}
$$
Then
$$\frac{dy}{dx}=\frac{1+\fra... | Hint:
First a slight miscalculation (you missed the $8$) on your part, it should be:
$$
\frac{1+8v^2}{3v}=v+x\frac{\mathrm{d}v}{\mathrm{d}x}$$
After this just separate the variables and integrate:
$$\int \frac{\mathrm{d}x}{x}=\int \frac{3v\ \mathrm{d}v}{(1+5v^2)}$$
Can you proceed?
Update:
To help you remove your confu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2918636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Number of sequences formed from $1, 1, 1, 2,3,4,5,6$ in which all three $1$s appear before the $6$ I want to find number of sequences which contains $2,3,4,5,6$ exactly once and $1$ exactly three times. Also all three $1$s should be placed before $6$.
Example $1)1,1,1,2,3,4,5,6\\2)1,2,3,4,1,1,6,5$
I think that $6$ can ... | HINT: I will explain it for sixth place and leave the rest to you:
Since $6$ is on sixth place, there are five places left before it and three of them must be filled by $1$'s. For the rest of two, we can choose from $2,3,4,5$ with $\binom{4}{2}$ and arrange them in those five places with $\frac{5!}{3!}$ (since $1$'s ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2918754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Which pairs of positive integers (,) satisfy $^2−2^=153$? My attempt: Rearrange to $x^2=2^n + 153$ and with $2^n\geq 2\ $ it follows $x^2 \geq 155\ $.
The next square number is 169, so $x = 13$ and $n = 4$. A first solution. Since $2^n$ is even and 153 is odd, $x^2$ will be odd. So any candidate solution will have an e... | Suppose $x \geq 0$. Studying the equation modulo $2$ reveals that $x$ must be odd. Studying it modulo $3$ reveals that $n$ must be even $n=2k$.
$$x^2-2^n=(x-2^k)(x+2^k)=153$$
So $x-2^k$ and $x+2^k$ are two integers that average to $x$ but multiply to $153$. But $153=51(3)=1(153)=17(9)$, so either $x=26$ or $x=77$ or $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2919551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
solve this 1st order linear equation $$(x+3)^2\frac{dy}{dx}=6-12y-4xy=6-y(12+4x)$$
a. Write it in standard form.
$$\frac{dy}{dx}+\frac{12+4x}{(x+3)^2}y=\frac6{(x+3)^2}$$
b. What is the integrating factor?
$$\frac{12+4x}{(x+3)^2} = \frac14x+\frac34 \implies$$
$$IF=e^{\int{\frac14x+\frac34}}=e^{\frac18x^2+\frac34x}$$
c. ... | $$(x+3)^2\frac{dy}{dx}=6-4y(3+x)$$
$$(x+3)^2\frac{dy}{dx}+4y(3+x)=6$$
Multiply by $(x+3)^2$
$$(x+3)^4y'+4y(3+x)^3=6(x+3)^2$$
$$((x+3)^4y)'=6(x+3)^2$$
Integrate
$$(x+3)^4y=2(x+3)^3+C$$
$$y(x)=\frac 2 {(x+3)}+\frac C {(x+3)^4}$$
As pointed out in the comment this line is not correct
$$\frac{12+4x}{(x+3)^2} = \frac14x+\f... | {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 0
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Any idea how to find $\lim_{x\to 0} \frac{\sqrt{1-\cos(x^2)}}{1-\cos(x)}$? $$\lim_{x\to 0} \frac{\sqrt{1-\cos(x^2)}}{1-\cos(x)}$$
I am trying to solve this limit for 2 days, but still cant find the solution which is $\sqrt{2}$ (that's what is written in the solution sheet)
I tried multiplying with the conjugate, tried ... | By L'Hopital's rule, we have
$$\begin{aligned}
\lim_{x \to 0}\frac{1 - \cos(x^2)}{(1 - \cos(x))^2}
&= \lim_{x \to 0}\frac{2x\sin(x^2)}{2(1 - \cos(x))\sin(x)} \\
&= \lim_{x \to 0}\frac{x}{\sin(x)}\frac{\sin(x^2)}{1 - \cos(x)} \\
&= \lim_{x \to 0}\frac{\sin(x^2)}{1 - \cos(x)} \\
\end{aligned}$$
where we have used that
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2932360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
If $a^2+b^2 \gt a+b$ and $a,b \gt 0$ Prove that $a^3+b^3 \gt a^2+b^2$ I'm not too sure about this, I have been working on for some time and I reached a solution (not really too sure about)
Question: If $a^2+b^2 \gt a+b$ and $a,b \gt 0$ Prove that $a^3+b^3 \gt a^2+b^2$
My solution: Let $a \geq b$
From $a^2+b^2 \gt a+b$ ... | Your proof is correct. Indeed:
$$\begin{cases}a^2-a \gt b-b^2\\
a\ge b>0\end{cases} \Rightarrow \\
a(a^2-a)>b(b-b^2) \Rightarrow \\
a^3-a^2>b^2-b^3 \Rightarrow \\
a^3+b^3>a^2+b^2.$$
Alternative proof. Consider $b=ax, x\ge 1$. Then:
$$a^2+b^2 \gt a+b \Rightarrow \\
a^2+a^2x^2>a+ax \stackrel{\text{divide by} \ a}{\Righta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2934061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Finding the result of the polynomial The question is that: Given that $x^2 -5x -1991 = 0$, what is the solution of $\frac{(x-2)^4 + (x-1)^2 - 1}{(x-1)(x-2)}$
I've tried to factorize the second polynomial like this:
$\frac{(x-2)^4 + (x-1)^2 - 1}{(x-1)(x-2)}
=\frac{(x-2)^4+x(x-2)}{(x-1)(x-2)}
=\frac{(x-2)((x-2)^3+x)}{(x... | Try making the question simpler.
Let $t=x-2$ then $x-1$ will become $t+1$. So the question
$$\dfrac{(x-2)^4 + (x-1)^2 - 1}{(x-1)(x-2)}$$
changes to
$$\dfrac{t^4 + (t+1)^2 - 1}{t(t+1)}$$
$$\implies\dfrac{t^4 + t²+1+2t - 1}{t(t+1)}$$
$$\implies\dfrac{t^4 + t²+2t}{t(t+1)}$$
$$\implies\dfrac{t(t²-t+2)(t+1)}{t(t+1)}$$
$$\im... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Simplifying $\frac{^nC_{1}}{1 \times 2}-\frac{^nC_{2}}{2 \times 3}+..+(-1)^{n+1}\frac{^nC_{n}}{n \times (n+1)}$ I was thinking about the following sum -
$\frac{^nC_{1}}{1 \times 2}-\frac{^nC_{2}}{2 \times 3}+\frac{^nC_{3}}{3 \times 4}+..+(-1)^{n+1}\frac{^nC_{n}}{n \times (n+1)} = 1 + \frac{1}{2}+\frac{1}{3}+\frac{1}{4... | Firstly,
$$\frac{^nC_{1}}{1 \times 2}+\frac{^nC_{2}}{2 \times 3}+\frac{^nC_{3}}{3 \times 4}+..+\frac{^nC_{n}}{n \times (n+1)} =\left[\frac{^nC_{1}}{1}+\frac{^nC_{2}}{2}+\frac{^nC_{3}}{3}+..+\frac{^nC_{n}}{n}\right]-\left[\frac{^nC_{1}}{2}+\frac{^nC_{2}}{3}+\frac{^nC_{3}}{4}+..+\frac{^nC_{n}}{(n+1)}\right] $$
Now consid... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2939125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Using induction to prove $\sum_{i = 1}^{n} i\cdot 2^i = (n - 1) \cdot 2^{n + 1} + 2$ I have to prove the following proposition:
$$\sum_{i = 1}^{n} i\cdot 2^i = (n - 1) \cdot 2^{n + 1} + 2$$
I have the following proof thus far:
Proof by Induction
Basis Step $(n = 1)$
LHS: $\sum_{i = 1}^{n} i2^i = 1 \cdot 2^1 = 2$
RHS:... | Your base case is correct, but you need more information in the inductive hypothesis.
Base Case: $(n = 1)$
$$\sum_{i=1}^{1}i2^i = 1\cdot 2^1 = 2 = (1-1) \cdot 2^{1+1} + 2 $$
Inductive Hypothesis:
Assume that $$\sum_{i=1}^{k} i2^i = (k-1) \cdot 2^{k+1} + 2 \hspace{1in} \text{IH}$$ for some integer $k >1$. You must show... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2940084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Can someone help me finish this: evaluate $S_n = \frac{x}{1-x^2}+\frac{x^2}{1-x^4}+ ... + \frac{x^{2^{n-1}}}{1-x^{2^{n}}}$ I am asked to find the closed form solution for the below.
$$S_n = \frac{x}{1-x^2}+\frac{x^2}{1-x^4}+ ... + \frac{x^{2^{n-1}}}{1-x^{2^{n}}}$$
Just writing out the $S_1, S_2, S_3$, I have managed to... | Recognize the geometric series and replace it:
$$\frac{x}{1-x^2}=x+x^3+x^5+x^7+\cdots$$
For all the rest, it's the same, just replace $x\to x^2,x^4,\ldots x^{2^n}$.
Now notice that the new terms are just filling in the missing terms:
$$S_2=x+x^2+x^3+\Box+x^5+x^6+x^7+\Box+x^9+\cdots$$
$$S_3=x+x^2+x^3+x^4+x^5+x^6+x^7+\Bo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2944491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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How do I complete the square of $y= -4x^2-2x-4$? $y = -4x^2 - 2x - 4$
I just can't figure this out, do I divide the second number and the third number by $4$ then by $2$ and then add the product to the second one and subtract it from the third one?
| First, you need to isolate x^2 from the equation by factor the -4 out, which leave the equation like this:
$$y= -4(x^2+ \frac{1}{2} x ) -4 $$
We know that the equation of square binomials is:
$(x+b)^2= x^2+2xb+b^2$
To write the binomial $x^2 + \frac{1}{2}x$ to a square binomials, we need to find b^2
We got :
$2xb = \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2944950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
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Complex number equations ( process) Among the exercises I was solving these are the ones I don't understand and I don't know if the process or solutions are ok, so please correct whenever I went wrong. Thanks everyone.
(I'm supposed to solve by converting to trigonometric form when it's convenient and then finding root... | Hint: For $c$, let $z=re^{i\theta}$
\begin{align}
z^5(1+i) &= \bar z\\
r^5e^{5i\theta}(\sqrt{2}e^{i\pi/4}) &= re^{-i\theta} \\
r^4\sqrt{2} &= e^{-6i\theta-i\pi/4}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2945487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that the quantity is an integer I want to prove that $\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6} \in \mathbb{Z}, \forall n \geq 1$.
I have thought to use induction.
Base Case: For $n=1$, $\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6}=\frac{1}{3}-\frac{1}{2}+\frac{1}{6}=0 \in \mathbb{Z}$.
Induction hypothesis: We suppose t... | If you put everything over a common denominator you get $$f(n)=\frac {2n^3-3n^2+n}6=\frac {n(n-1)(2n-1)}6$$
Now it is easier, I think, to see what is going on for a straightforward induction, or you can write $2n-1=2n-4+3$ and split up into different fractions viz $$f(n)=\frac {n(n-1)(2n-4+3)}6=\frac {n(n-1)(n-2)}3+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2946269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 7
} |
Maximum and minimum absolute value of a complex number
Let, $z \in \mathbb C$ and $|z|=2$. What's the maximum and minimum value of $$\left| z - \frac{1}{z} \right|? $$
I only have a vague idea to attack this problem.
Here's my thinking :
Let $z=a+bi$
Exploiting the fact that, $a^2+b^2=4$
We get $z-\dfrac{1}{z}=a-\df... | A (very) faster way:
We know that $z$ is in the circle centered at $0$ and radius $2$ and $1/z$ is in the circle of center $0$ and radius $1/2$. The maximum distance between a point of the former and a point of the latter is
$$2+\frac12=\frac52$$
Now, we need to show that there exists some $z$ such that this distance ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2948812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
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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.