Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Calculating the length of a curve for $ y=\ln (1-x^2) $ In calculating the length of $ y=\ln (1-x^2) $ on the interval $ [0, {3\over4}]$, I found that for whatever reason, I end up with invalid domains in the indefinite integral using the given limits of integration.
Using the formula $ \int^b_a \sqrt{1+f'(x)^2} $, ... | ..........................................................................
The hitch is here :
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How can I prove remainder side of this inequality? Let $x$, $y$ be two positive numbers such that $x^4+y^4=x^2+y^2$. Prove that
$$1\leqslant x+y\leqslant 2.$$
With $x+y\leqslant 2$. I tried
We have $$x^4+y^4\geqslant \dfrac{(x^2+y^2)^2}{2}.$$
Therefore, $$x^2+y^2\geqslant \dfrac{(x^2+y^2)^2}{2}$$
or
$$(x^2+y^2)^2-2(x^... | If $x+y\lt 1$ then $(x+y)^2\lt 1$, and therefore $x^2+y^2\lt 1$.
But from $0\lt x^2+y^2\lt 1$, we conclude that $(x^2+y^2)^2 \lt x^2+y^2$, and therefore $x^4+y^4\lt x^2+y^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$n^{th}$ derivative of $y=x^2\cos x$ I am stuck with Leibniz formula
$$D^{n}y = \sum_{k=0}^{n} \binom{n}{k} \, x^{(2k)}\cos^{(n-k)}x$$
Could someone show how to do it?
| The Leibniz formula only has a few terms. This is seen by the following.
\begin{align}
D^{n} y &= D^{n}\{ x^{2} \, \cos(ax)\} = \sum_{k=0}^{n} \binom{n}{k} \, D^{k}\{x^{2}\} \, D^{n-k}\{\cos(ax)\}
\end{align}
Since $D(x^{2}) = 2 x$, $D^{2}(x^{2}) = 2$, and $D^{3+m}(x^{2}) = 0$ for $m \geq 0$, then
\begin{align}
D^{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
How to solve the trigonometric equation $\sin x-\cos x-2(2)^{\frac 1 2}\sin x\cos x=0$ the question is:
Find the solutions of the equation: $\sin x-\cos x-2(2)^{\frac 1 2}\sin x\cos x=0$.
Let $\sin x+\cos x=u \text{ and } \sin x \cos x=v \implies \sin^2x+\cos^2x+2\sin x\cos x=u^2 \implies v=\frac {u^2-1} 2$
similarl... | my hints: $$\sin \theta=\sin \alpha\iff \theta=2k\pi+\alpha\ \text{or}\ \theta =(2k+1)\pi-\alpha$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Solve a complex equation Solve the following equation $$(4-3i)z^2-25z+31-17i= 0 $$
Dividing by 4-3i gives me $$z^2 \frac{-100z-75zi + 124 + 93i -68i -51i^2}{25}$$
which goes to $$z^2 -4z-3zi + 7+i$$
then i collect the terms so $$z - \left(\frac{(4-3i)}{2}\right)^2 = -7 -i + \left(\frac{4-3i}{2}\right)^2$$
and after t... | Notice, $$(4-3i)z^2-25z+31-17i=0$$ Solving the above quadratic equation for $z$ as follows $$z=\frac{-(-25)\pm\sqrt{(-25)^2-4(4-3i)(31-17i)}}{2(4-3i)}$$
$$z=\frac{-(-25)\pm\sqrt{(-25)^2-4(4-3i)(31-17i)}}{2(4-3i)}$$ $$z=\frac{(4+3i)(25\pm\sqrt{333+644i})}{2(16+9)}$$
$$z=\frac{(4+3i)(25\pm\sqrt{333+644i})}{50}$$
$$z=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Solve trigonometric equation $ 3 \cos x + 2\sin x=1 $ Solve trigonometric equation: $$ 3 \cos x + 2\sin x=1 $$
I tried to substitue $\cos x = \dfrac{1-t^2}{1+t^2}, \sin x = \dfrac{2t}{1+t^2}$. Yet with no results.
| If you have a sum $S=a\cos x+b \sin x$
define $r$ by $r^2=a^2+b^2$
and $\theta$ by $r\sin \theta =a$ and $r\cos \theta =b$ so that $\tan \theta =\cfrac ab$
Then $S=r\sin \theta \cos x+r\cos \theta \sin x=r\sin (x+\theta)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Solve the binomial equation Solve the binomial equation $$z^4 = -8$$
Below is the steps i have done
1: I have taken |-8| that is 8 and then done 8^(1/4) which is 2^(1/4).
2: Since $z=r(cos\alpha+isin\alpha)$ leads me to
$r^4(cos4\alpha+isin4\alpha)=-8(cos\pi/2+isin\pi/2)$
Divide by 4 since the z term is raised by fo... | $$ z^4 = -8 \Longleftrightarrow $$
$$ z^4 = |-8|e^{\arg(-8)i} \Longleftrightarrow $$
$$ z^4 = 8e^{\pi i} \Longleftrightarrow $$
$$ z^4 = 8e^{\left( \pi + 2 \pi k \right) i} \Longleftrightarrow $$
$$ z = \left( 8e^{\left( \pi + 2 \pi k \right) i} \right)^{\frac{1}{4}} \Longleftrightarrow $$
$$ z = \sqrt[4]{8}e^{\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Solve trigonometric inequality $\cos x \geq \sin^2 x - \cos^2 x $ Solve trigonometric inequality $$\cos x \geq \sin^2 x - \cos^2 x $$
My incorrect solution:
$$\cos^2 x-\sin^2 x \geq -\cos x $$
$$\cos 2x \geq \cos (\pi - x) $$
which means:
$$ 2x \geq -(\pi + x)$$
$$ x \geq -\pi $$
Which is wrong.
And
$$ 2x \leq 2\p... | \begin{align*}
\cos^2x − \sin^2x & \geq −\cos x\\
\cos^2x − \cos^2x-1 & \geq −\cos x\\
2\cos^2x + \cos x + 1 & \geq 0
\end{align*}
solving for cosx
$$\cos x = [-\infty,1] \cup \left[\frac{1}{2},1\right]$$
rejecting 1st part
$$\cos x \geq \frac{1}{2}$$
$$x \leq \frac{\pi}{3}$$
$$x \geq -\frac{\pi}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Advanced Algebra Manipulation/Inequality Proof: $\frac{4x^3(x^2+y^2)-2x(x^4+y^4)}{(x^2+y^2)^2} \leq 6|x|$ I need to show that
$$\frac{4x^3(x^2+y^2)-2x(x^4+y^4)}{(x^2+y^2)^2} \leq 6|x|$$
by starting with the left side of the inequality and working from there.
Hints from the textbook said to work from these inequality "t... | We have
$$|4x^3(x^2+y^2)-2x(x^4+y^4)|\le4|x|x^2(x^2+y^2)+2|x|(x^2+y^2)^2\\\le6|x|(x^2+y^2)^2$$
and then the result is immediate by canceling the factor $(x^2+y^2)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
a very basic question on finding the discriminant for $x^2+2(a-3)x-3a-7=0$ Sorry for asking such a basic question.
In the following quadratic equation
$$x^2+2(a-3)x-3a-7=0$$
by my calculations,
$$D=\left(\frac{b}{2}\right)^2-ac=(a-3)^2-1(-3a-7)=a^2-6a+9+3a+7=a^2-3a+16$$
But in a PDF article that describes a solution t... | Notice, compare the given equation: $x^2+2(a-3)x-3a-7=0$ with $Ax^2+Bx+C=0$, we get $$A=1, \ B=2(a-3), \ C=-3a-7$$
Hence,
discriminant is calculated as follows $$D=B^2-4AC=\left(2(a-3)\right)^2-4(1)(-(3a+7))$$ $$=4a^2-24a+36+12a+28$$ $$=4a^2-12a+64$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the integration : $\int\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx$ $$\int{\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx}$$
$$\int\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx=\int \frac{(1-\sin x)(2-\sin x)}{\sqrt{(1-\sin x)(2-\sin x)(1+\sin x)(2+\sin x)}}dx$$
I am stu... | Let $$\displaystyle I = \int\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx$$
We can write $$\displaystyle \sqrt{\frac{1-\sin x}{1+\sin x}} = \sqrt{\frac{1-\sin x}{1+\sin x}\times \frac{1+\sin x}{1+\sin x}} = \frac{\cos x}{1+\sin x}$$
So we get $$\displaystyle I = \int\frac{\cos x}{1+\sin x}\cdot \sqrt{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Value of $\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}}$ Here is the question:
$$\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}} = \;?$... | For the Calculation of $$\displaystyle \frac{\sum_{k=1}^{n^2-1}\sqrt{n+\sqrt{k}}}{\sum_{k=1}^{n^2-1}\sqrt{n-\sqrt{k}}} = $$
Let $$\displaystyle A_{n} = \sum_{k=1}^{n^2-1}\sqrt{n+\sqrt{k}}$$ and $$\displaystyle B_{n} = \sum_{k=1}^{n^2-1}\sqrt{n-\sqrt{k}}$$ , where $n>1$
Now $$\left(\sqrt{n+\sqrt{k}}-\sqrt{n-\sqrt{k}}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1397863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 7,
"answer_id": 2
} |
Simple coupled ODE Find two linearly independent solutions to the pair of coupled ODEs
$\frac{dx}{dt} = 2x + 3y$,
$\frac{dy}{ dt} =-3x+2y$.
I figured the eigenvalues/eigenvectors of the corresponding matrix to be $$e^{(2+3i)t} \quad\text{with}\quad \left[\begin{array}{r}
i \\
1
\end{array}\right] \quad\text{and} \quad ... | Notice, we have $$\frac{dx}{dt}=2x+3y\tag 1$$
$$\frac{dy}{dt}=-3x+2y\tag 2$$
*
*Diving (1) by (2), we get $$\frac{\frac{dx}{dt}}{\frac{dy}{dt}}=\frac{2x+3y}{-3x+2y}$$ $$\frac{dx}{dy}=\frac{2\frac{x}{y}+3}{-3\frac{x}{y}+2}$$ Now, let $x=uy\implies \frac{dx}{dy}=u+y\frac{du}{dy}$, setting the values we get
$$u+y\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1399043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Convergent/divergent series Is the following series divergent/convergent?
$$S=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}-\frac{1}{9}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}-...$$
I think it is divergent since
$$
\begin{align}
S&>1-\frac{1... | S is convergent ( Using Leibnitz's Test for convergence of Alternating series)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Average waiting time to guess a random number from $1$ to $100$ with equal probability and 1 second between each number is generated A random number between $1$ and $100$ is generated every second.
What would be the average waiting time for a specific number ($1$ for instance) to be generated?
Probability distribution ... | Let $X$ denote the number of seconds until $1$ is generated:
*
*$P(X=0)=\frac{1}{100}$
*$P(X=1)=\frac{99}{100}\cdot\frac{1}{100}$
*$P(X=2)=\frac{99}{100}\cdot\frac{99}{100}\cdot\frac{1}{100}$
*$\dots$
*$P(X=n)=\left(\frac{99}{100}\right)^n\cdot\frac{1}{100}$
So the expected number of seconds until $1$ is gener... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding all solutions of $x^2+2x-15\equiv0 \pmod{105}$- Proof strategy. Find all solutions of $x^2+2x-15\equiv0 \pmod{105}$. Now, I wanted to suggest a proof relying on the algorithm presented in class, and there are some parts where I could use some help or criticism.
First of all, let us write the prime factorizatio... | To solve the equation modulo $3$:
$$
\begin{align}
x^2 + 2x - 15 &\equiv 0 \pmod 3\\
x^2 + 2x &\equiv 0 \pmod 3\\
x(x + 2) &\equiv 0 \pmod 3
\end{align}
$$
Therefore $x \equiv 0 \textrm{ or } 1 \pmod 3$. Similarly, working modulo 5, we'll again get $x(x+2) \equiv 0$, so $x \equiv 0 \textrm{ or } 3 \pmod 5$. Finally, in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Factorization of a polynomials in complex number. Factorize this expression:
$$a^2+b^2+c^2-ab-bc-ca.$$
The result is
$$(a+b\Omega+c\Omega^2)(a+b\Omega^2+c\Omega)$$
How I can get $\Omega$ here?What's the approach?
| If we are given that
$a^2 + b^2 + c^2 - ab - ac - bc$
$ = (a + b\Omega + c\Omega^2)(a + b\Omega^2 + c\Omega), \tag{1}$
and we wish to find $\Omega$, we can proceed as follows: multiplying out the right-hand side, we find
that
$a^2 + b^2 + c^2 - ab - ac - bc$
$ = a^2 + \Omega^3 b^2 + \Omega^3 c^2
$
$+ (\Omega^2 + \Omeg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Proving $\binom{n}{m}+2\binom{n-1}{m}+......+(n-m+1)\binom{m}{m} = \binom{n+2}{m+2}$
For $m,n\in\mathbb{N},\;n\geq m$, prove the following:
$$
\tag{i}\binom{n}{m}+\binom{n-1}{m}+\binom{n-2}{m}+......+\binom{m}{m} = \binom{n+1}{m+1}
$$
$$
\tag{ii}\binom{n}{m}+2\binom{n-1}{m}+3\binom{n-2}{m}+......+(n-m+1)\binom{m}{m}... | Suppose $n \geq m$ and there is a set $S$ of $n + 2$ objects, denoted as $o_1, o_2, \cdots, o_{n+2}$. Your task to sample $m + 2$ objects from $S$. Let $X_i$ denote the # of ways to choose $m + 2$ objects such that the object with second minimum id is $o_i$. Easy to see that
$$
X_i = (i-1) \cdot { n + 2 - i \choose m }... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
reflecting a function about a line Say you have a function $f(x)$ and a line $g(x)=ax+b$. How do you reflect $f$ about $g$?
I am apparently supposed to write more text, but the line above is all I am after, hence I wrote this sentence as well.
| Represent the graph of f parametrically by $x=t, y=f(t)$.
If we reflect the point $(t, f(t))$ in the line $y=ax+b$ to get the point $(x,y)$, then
$\color{red}{y-f(t)=-\frac{1}{a}(x-t)}$
since the line and the line segment between $(t, f(t))$ and $(x,y)$ are perpendicular to each other.
We also have that $\frac{y+f(t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Define g(x) as a function of $x$ Let $f(x)= \begin{cases}
-1 & ,-2\leq x\leq 0 \\
\\
|x-1| & ,0<x\leq2
\end{cases}$
and $g(x)=\int_{-2}^{x}f(t) dt$.Define g(x) as a function of $x$.
I tried to solve it.I redefined $f(x)= \begin{cases}
-1 & ,-2\leq x\leq 0 \\
1-x & ,0<x<1 \\
... | For $0<x<1$ we have
\begin{align*}
g(x)&=\int_{-2}^{x}f(t)dt\\
&=\int_{-2}^0f(t)dt+\int_0^xf(t)dt\\
&=\int_{-2}^0(-1)dt+\int_0^x(1-t)dt\\
&=(-1)(0+2)+x-\frac{1}{2}x^2-0\\
g(x)&=-\frac{1}{2}x^2+x-2,\qquad \text{ for }\;\;0<x<1
\end{align*}
On the other hand, for $1\le x\le 2$ it follows
\begin{align*}
g(x)&=\int_{-2}^{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1405561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solve $y^2 + 3xy - 10x^2 + y + 5x = 0$ for y in terms of x I'm given the following equation: $y^2 + 3xy - 10x^2 + y + 5x = 0$
and asked to solve $y$ in terms of $y$.
My attempt:
$y^2 + (3x+1)\times y - 10x^2 + 5x = 0$
$\Rightarrow (y+(3x+1)/2)^2 - ((3x+1)/2)^2 = (x - (\frac 12)\times x)^2 - (\frac 12)^2 $
But I can... | $$y^2 + 3xy - 10x^2 + y + 5x = 0\Rightarrow y^2+(3x+1)y-(10x^2-5x) = 0$$
So $$\displaystyle y = \frac{-(3x+1)\pm \sqrt{(3x+1)^2+4(10x^2-5x)}}{2}$$
$$\displaystyle = \frac{-(3x+1)\pm \sqrt{49x^2-14x+1}}{2}$$
so we get $$\displaystyle y = \frac{-(3x+1)\pm \sqrt{(7x-1)^2}}{3} = \frac{-(3x+1)\pm (7x-1)}{2}$$
So we get $\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1406960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find $ab+cd$ given that $a^2+b^2=c^2+d^2=1$ and $ac+bd=0$? It is given that $a^2+b^2=c^2+d^2=1 $
And it is also given that $ac+bd=0$
What then is the value of $ab+cd$ ?
| HINT:
$$ac+bd=0\iff\dfrac ad=\dfrac b{-c}=\pm\sqrt{\dfrac{a^2+b^2}{d^2+(-c)^2}}$$
But if $a^2+b^2=d^2+c^2,$ not necessarily $=1$
$$\dfrac ad=\dfrac b{-c}=\pm1$$
So, either $a=d,b=-c$ or $a=-d,b=c$
The result should follow immediately.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 2
} |
How to Derive this Matrix Equation - Converting Lagrangian in Proximal Operator Problem Form $$ \left< Z,X-L-S \right> \quad +\quad \frac { r }{ 2 } \left\| X-L-S \right\|_F^2 \quad =\quad \frac { r }{ 2 } { \left\| L-\left( X-S+\frac {Z}{r} \right) \right\| }_F^2 $$
I think $ \left< Z,X-L-S \right>$ should be $\ope... | Note that the left side of
$$ \left< Z,X-L-S \right> \quad +\quad \frac { r }{ 2 } \left\| X-L-S \right\|_F^2 \quad =\quad \frac { r }{ 2 } { \left\| L-\left( X-S+\frac {Z}{r} \right) \right\| }_F^2 $$
can be written as
$$
\operatorname{tr}(Z^T(X-L-S)) + \frac { r }{ 2 } \operatorname{tr}( (X-L-S)^T(X-L-S) ).$$
And... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Poisson events distributed uniformly in a given time It is given that $4$ Poisson events occur between $12:00$ to $13:00 $ (interval denoted by T).
Intuitively, Why the probability of each event to occur at time $t \in T$ is uniform across $T$?
I'm well aware of the calculation for that:
Let's calculate the probabili... | It has to do with the fact that it is given that there are exactly 4 events in the hour, whereas the Poisson distribution allows any number of events in a given time period. Both $X_1$ and $X_2$ are both greater than or equal to $1$ (there is at least one event in both half hours), so we need to divide the remaining tw... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Trying to solve $\sqrt{2\cos^2(x)-\sqrt{3}}+\sqrt2 \sin(x)=0$ The equation is
$$\sqrt{2\cos^2(x)-\sqrt{3}}+\sqrt2 \sin(x)=0$$
I solve it thus:
$$
\begin{cases}
2\cos^2(x)-\sqrt3=2\sin^2(x) \\
-\sqrt2 \sin(x)\ge 0 \iff \sin(x)\le 0
\end{cases}
$$
The first equation boils down to
$$2\cos^2(x)=2(1-\cos^2(x))+\sqrt3$$
$$4c... | As Brent points out in his comment, your only mistake was applying the condition $\sin x<0$ at the end:
1) If $x=\frac{\pi}{12}+n\pi$, $\;\;\sin x>0$ for $n$ even and $\sin x<0$ for $n$ odd, so this gives
$\hspace{.4 in} x=\frac{\pi}{12}+(2n+1)\pi=\frac{13\pi}{12}+2n\pi=-\frac{11\pi}{12}+2n\pi$
2) If $x=-\frac{\pi}{12}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1411088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Asymptotic behavior of elliptic integral (first kind) I came accross some obstacles in proving that the time $T(\delta)$ taken by a pendulum to travel from $\theta=\pi-\delta$ to a considerably distant angle $\theta=\theta_0\in(0,\pi/4)$ diverges logarithmic-ally with respect to $\delta$.
Here is my work so far.
The e... | I prefer to have the singular behaviour near $0$ rather than at $\frac{\pi}{2}$, so let's make the substitution $\varphi = \frac{\pi}{2} - \theta$. We obtain
$$K(k) = \int_0^{\frac{\pi}{2}} \frac{d\varphi}{\sqrt{1 - k^2\cos^2\varphi}}.$$
Split the integral at $\frac{\pi}{4}$. The part
$$\int_{\frac{\pi}{4}}^{\frac{\pi}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1411157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Difficult sets of Equations, counting
Let $ m$ be the number of solutions in positive integers to the equation $ 4x+3y+2z=2009$, and let $ n$ be the number of solutions in positive integers to the equation $ 4x+3y+2z=2000$. Find the remainder when $ m-n$ is divided by $ 1000$.
Please no complete solutions.
$4x + 3y +... | If $(x,y,z)=(a,b,c)$ is a solution of $4x+3y+2z=2000$, then $(x,y,z)=(a+1,b+1,c+1)$ is a solution of $4x+3y+2z=2009.$
Now find the number of the solutions $(x,y,z)$ of $4x+3y+2z=2009$ such that either $x,y$ or $z$ equals $1$. (note that this is $m-n$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1411680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that the function is continuous To show that the function $f: \mathbb{R}^2 \rightarrow\mathbb{R}$ with $f=\left\{\begin{matrix}
\frac{x^3-y^3}{x^2+y^2} & , (x,y) \neq (0,0)\\
0 & , (x,y)=(0,0)
\end{matrix}\right.$ is continuous on $(0,0)$ we have to show that $|f(x,y)-f(x_0,y_0)| \leq L ||(x,y)-(x_0,y_0)||$ so ... | Hint for a different, but elegant, approach: Try polar coordinates.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Number of polyhedron diagonals Suppose that I have a polyhedron with given number of faces, edges and vertices are given. Is there a formula that gives me the number of polyhedron diagonals, http://mathworld.wolfram.com/PolyhedronDiagonal.html ?
| In my comment, I failed to recognize that you were interested in just the "space diagonals". As you suspect, eliminating face diagonals requires knowing the shapes of the various faces.
Suppose, for $k = 3, 4, \dots$, there are $f_k$ faces with $k$ sides. Arguing as in my comment, we observe that the vertices of each ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that $(1+\frac{1}{n})^n+\frac{1}{n}$ is eventually increasing I would like to find a way to show that the sequence $a_n=\big(1+\frac{1}{n}\big)^n+\frac{1}{n}$ is eventually increasing.
$\hspace{.3 in}$(Numerical evidence suggests that $a_n<a_{n+1}$ for $n\ge6$.)
I was led to this problem by trying to prove by ind... | Let
$$ \eqalign{f(n) = \dfrac{1}{n} + \left( 1 + \dfrac{1}{n}\right)^n &= \dfrac{1}{n} + \exp\left( n \ln\left(1+\dfrac{1}{n}\right)\right) \cr &=
\dfrac{1}{n} + \exp\left(1 - \dfrac{1}{2n} + \dfrac{1}{3n^2} + O\left(\dfrac{1}{n^3}\right)\right) \cr &= e - \dfrac{e-2}{2n} + \dfrac{11e}{24 n^2} + O\left(\dfrac{1}{n^3}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1413145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
} |
$x(x^2-2)=0$, The answers are $x = 0, \sqrt{2}$, how do I get there? $$x(x^2-2)=0$$
The answers are $x=\sqrt{2}, 0$ how do I get there?
| Let's start from the beginning.
You are given
$$x(x^2 - 2 ) = 0$$
What are they asking for here?
Well, they're asking for any value of $x$ which would cause the given statement to become true.
What does this mean?
It means that if you were to substitute a number in place of $x$, then solve out the equation, the resul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1413435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 5
} |
elementary problem in combinatorics Let $X=\{1,2,3,4,5,6,7,8,9,10\}$ and $R$ be a set defined by $$\{(x,y)\in X\times X: \text{$x$ and $y$ have the same remainder when divided by $3$}\}$$ Then what's the numbers of elements in $R$? My approach is
$\bar{0} = \{3,6,9\}$ so we can get $2!\binom{3}{2}$ numbers of element... | Your answer is correct.
$R\subset X\times X$ can be recognized as an equivalence relation. As comes forward in your question its equivalence classes are $\{3,6,9\}$, $\{1,4,7,10\}$ and $\{2,5,8\}$.
Then: $$R=\left[\{3,6,9\}\times\{3,6,9\}\right]\cup\left[\{1,4,7,10\}\times\{1,4,7,10\}\right]\cup\left[\{2,5,8\}\times\{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1413882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving that $\frac{1}{2}<\frac{2}{3}<\frac{3}{4}<$...$<\frac{n-1}{n}$ In an attempt to find a pattern, I did this:
Let a,b,c,d be non-zero consecutive numbers. Then we have:
$a=a$
$b=a+1$
$c=a+2$
$d=a+3$
This implies:
$\frac{a}{b}=\frac{a}{a+1}$
$\frac{b}{c}=\frac{a+1}{a+2}$
$\frac{c}{d}=\frac{a+2}{a+3}$
I don't know ... | If $a_n = \frac{n-1}{n}$,
$$a_{n+1} - a_n = \frac{n}{n+1} - \frac{n-1}{n} = \frac{n^2 - (n-1)(n+1)}{n(n+1)} = \frac{1}{n(n+1)} > 0$$
Also,
$$a_n = \frac{n-1}{n} = 1 - \frac1 n$$
It should be clear that $1/n$ is decreasing as $n$ increases, and so this sequence increases towards $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
} |
Primes of the form $x^2+n\cdot y^2$, given $n$? In an attempt to get to grips with algebra for a course I intend to follow, I was working through a bunch of exercise sheets. A series of questions got me wondering:
Given an integer $n$, is there a general method (or collection of methods) to tell which primes $p$ are o... | well, no. if you pick a discriminant of binary quadratic forms, call it $\Delta,$ and you have an odd prime $p$ that does not divide $\Delta,$ finally $(\Delta|p)= 1,$ then there is some form of that discriminant that represents the prime. If the class number is larger than one, that may not be the principal form. To f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Proof that expression is integer, $\frac{(3n)!}{6^nn!}$ Can you help me with this exercises?
Proof that expression is integer,
$$\frac{(3n)!}{6^nn!}$$
I've tried for induction!!
$p(1):\frac{(3)!}{6}=1 $
for $p(k)=\frac{(3k)!}{6^kk!}$
for $p(k+1)=\frac{(3k+3)!}{6^{k+1}(k+1)!}$
where,
$$\frac{(3k+3)(3k+2)(3k+1)(3k)!}{6^k... | Let us begin with the basic case, $n=0$. The expression $\frac{(3n)!}{6^nn!}$ is an integer for $n=0$, so the basic case is valid.
Let us assume that the statement is true for $n$, and that the integer it results in is equal to $k$. Our goal is to prove it true for $n+1$.
We first plug in $n+1$ for the expression.
\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1417965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Matrix exponential: $\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix}$ It is asked to calculate $e^A$, where
$$A=\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix}$$
I begin evaluating some powers of A:
$A^0= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\; ; A=\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix} \; ; A^2 = \begin{pmat... | By separating odd and even terms in the series, what you have found can be rewritten as:
$$
e^A=I \sum_{k=0}^\infty{(-1)^k4^k\over(2k)!}+A \sum_{k=0}^\infty{(-1)^k 4^k\over(2k+1)!}=I \cos2+A{1\over2}\sin2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1418210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Calculation of improper integral: $\int_{0}^{\infty}\frac{x^3\ln x}{(x^4+1)^3} \,dx$ One of an exam's task was to calculate the following integral: $$I=\int_{0}^{\infty}\frac{x^3\ln x}{(x^4+1)^3} \,dx$$
I tried integration by parts:
$$I=\frac{1}{4}\int_{0}^{\infty}\ln x \cdot (x^4+1)^{-3} \,d(x^4+1)$$
but then things g... | Integrating by parts,
$$
\int\frac{x^3\log x}{(1+x^4)^3}\,dx=-\frac{\log x}{8(1+x^4)^2}+\int \frac{1}{8(1+x^4)^2x}\,dx
$$
Doing partial fraction decomposition,
$$
\frac{1}{8(1+x^4)^2x}=\frac{1}{8x}-\frac{x^3}{8(1+x^4)}-\frac{x^3}{8(1+x^4)^2}.
$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1422837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove by Induction: $8^n - 3^n$ is divisible by $5$ for all $n \geq 1$ Prove by Induction that for all $n \geq 1$ we have
$$8^n - 3^n \text{is divisible by 5} ...(*)$$
My proof so far
Step 1: For $n=1$ we have $8^1 - 3^1 = 8 - 3 = 5$ which is divisible by 5.
Step 2: Suppose (*) is true for some $n=k\geq 1$ that is... | Continue like this:
$8 \cdot 8^k - 3 \cdot 3^k = (5+3) \cdot 8^k - 3 \cdot 3^k = 5 \cdot 8^k + 3 \cdot (8^k - 3^k)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1423667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
} |
Prove that the group $G$ is abelian if $a^2 b^2 = b^2 a^2$ and $a^3 b^3 = b^3 a^3$
In a Group $G$, $a^2b^2=b^2a^2$ and $a^3b^3=b^3a^3$ holds, $\forall a,b\in G$. Prove that the group $G$ is abelian.
My approach was the following:
Let $a,b\in G$
Then, $a^2b^2=b^2a^2$ and $a^3b^3=b^3a^3$ holds.
Now, $$\begin{align}
a^... | Hint: If you can somehow show that $a^2b^3=b^3a^2$. First take $(a^2b^3)^2$ and then prove it equal to $(b^3a^2)^2$. After this show that $a^2b^3=b^3a^2$ implies the group to be abelian
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1423870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the number of possible solutions in $x+y+z=30$ under conditions.
Three variables $x,y,z$ have a sum of $30$. All three are Non-Negative integers.
If any $2$ variables don't have the same value and exactly one variable has
value less than or equal to $3$, find the number of possible solutions ?
$a.)\ 98 \\
... | X+Y+Z=30 ; given any one of the number ranges from 0-3 and all other numbers start from 4. Hence consider the following equations:
*
*X=0 ; Y+Z=30
The solution of the above equation is obtained from (n-1)C(r-1) formula.
Total solutions obtained = 29 of which (y,z): (2,27)(27,2)(2,28)(28,2)(1,29)(29,1) must not be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1425021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Solving $2\cot 2x\cos2x = 1-\sin 2x$ How would I solve the following trigonometric equation?
$$2\cot 2x\cos2x = 1-\sin 2x$$
I got to this stage: $2\cos^22x = \sin2x - \sin^22x$
How do I continue?
| $$2cot2xcos2x=1-sin2x\\ \frac { \cos ^{ 2 }{ 2x } }{ \sin { 2x } } =\frac { 1-\sin { 2x } }{ 2 } \\ 2\left( 1-\sin ^{ 2 }{ 2x } \right) +\sin ^{ 2 }{ 2x } -\sin { 2x } =0\\ \sin ^{ 2 }{ 2x } +\sin { 2x } -2=0\\ \sin { 2x } =\frac { -1\pm 3 }{ 2 } \\ \sin { 2x\neq -2 } ,\sin { 2x } =1\\ \sin { 2x } =1\Rightarrow 2x=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1425737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
If $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$ then $\frac{1}{a^5}+\frac{1}{b^5}+\frac{1}{c^5}=\frac{1}{a^5+b^5+c^5}.$ Suppose that $\displaystyle\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$ . Then , prove that $\displaystyle\frac{1}{a^5}+\frac{1}{b^5}+\frac{1}{c^5}=\frac{1}{a^5+b^5+c^5}.$
Attempt :
F... | Given $$\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{a+b+c}\Rightarrow \frac{(ab+bc+ca)}{abc} = \frac{1}{a+b+c}$$
So $$\displaystyle (ab+bc+ca)(a+b+c) = abc\Rightarrow a^2b+a^2c+ab^2+b^2c^2+c^2a+3abc=abc$$
so we get $(a+b)(b+c)(c+a) = 0\Rightarrow (a+b) = 0$ or $(b+c) =0$ or $(c+a) =0$
So $a=-b$ or $b=-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1426119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
If $g(x) = x^2 + 3x + 1$ and $f \circ g = g\circ f$, then $f$ and $g$ intersect on the line $y = x$.
Let $f:\mathbb R\to \mathbb R$, $g(x) = x^2 + 3x + 1$ and $f∘g=g∘f$.
Prove that $Cf, Cg$ and the line $y=x$ have at least one common point.
My solution:
$fog=gof$ then $f=g$, $Df = Dg = \mathbb R$ and $f(x) = g(x)$
I ... | Notice that the graphs of $g(x) = x^2 + 3x + 1$ and $y = x$ intersect at precisely one point: $(-1, -1)$. In other words, we know that:
$$\boxed{
g(x) = x \iff x = -1
} \tag{$\star$}$$
To obtain our common point, it is enough to show that $f(-1) = -1$.
To this end, let $k = f(-1)$ and suppose that $f \circ g = g\circ f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1426318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Express the given quantity as a single Logarithm $\frac{1}{5}\ln(x+2)^5+\frac{1}{2}[\ln x-\ln(x^2+3+2)^2)]$ Express the given quantity as a single Logarithm
$$\frac{1}{5}\ln(x+2)^5+\frac{1}{2}[\ln x-\ln(x^2+3x+2)^2)]$$
My answer is $\ln\dfrac{[(x+2)^5]^{\frac{1}{5}}\cdot x^2}{[(x^2+3x+2)^2]^{\frac{1}{2}}}$
What have I... | Given $$\displaystyle \frac{1}{5}\ln(x+2)^5+\frac{1}{2}\left[\ln(x)-\ln(x^2+3x+2)^2\right]$$
$$\displaystyle \ln(x+2)^{\frac{5}{5}}+\ln(x)^{\frac{1}{2}}-\ln(x^2+3x+2)^{\frac{1}{2}} = \ln(x+2)+\ln (\sqrt{x})-\ln(x^2+3x+2)$$
So we get $$\displaystyle \ln\left[\frac{(x+2)\cdot \sqrt{x}}{x^2+3x+2}\right] = \ln\left[\frac{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1427533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\int(x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}dx$ $\int(x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}dx$
I tried to solve this question but no luck.
My try:
$$\int(x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}dx=\int x^4(x^8+x^4+1)(2x^8+3x^4+6)^{1/4}dx\\
\int x^4(x^8+x^4+1)x^2(2+3x^{-4}+6x^{-8})^{1/4}dx$$
Now i got stuck,please help me reach the ... | We start by factoring out $x^4$:
$$
(x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}=x^4(x^{8}+x^4+1)(2x^8+3x^4+6)^{1/4}.
$$
Next, we write
$$
x^{8}+x^4+1=\frac{1}{6}(6+3x^4+2x^8)+\frac{1}{2}x^4+\frac{2}{3}x^8,
$$
so the integrand can be written as
$$
\Bigl(\frac{1}{2}x^8+\frac{2}{3}x^{12}\Bigr)(2x^8+3x^4+6)^{1/4} +\frac{1}{6}x^4(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1429571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $1^2 - 2^2 + 3^2 - 4^2 + \cdots + (-1)^{n-1}n^2 = \frac12(-1)^{n-1} n (n + 1)$, where $n $ is a positive integer
Prove that $1^2 - 2^2 + 3^2 - 4^2 + \cdots + (-1)^{n-1}n^2 = \frac12(-1)^{n-1} n (n + 1)$, where $n $ is a positive integer
How do I prove the above expression using mathematical induction? So f... | We are to prove
$$\sum_{k=1}^n(-1)^{k-1}k^2=\frac{(-1)^{n-1}n(n+1)}{2} \tag 1$$
First, we establish a base case. So, for $n=2$, we note that the left-hand side of $(1)$ is $1^2-2^2=-3$ while the right-hand side of $(1)$ is $(-1)\frac{2\times 3}{2}=-3$. Thus, the case $n=2$ is verified.
Next, we assume that $(1)$ hol... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1433103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Adding pieces on NxN chess board sequentially The problem is:
Given a $N\times N$ chess board, if you put $B$ pieces starting on square $1,1$ and going right, then down, then left, then up when possible (thus, forming an spiral), in which coordinate the last piece would be put?
Example: $8\times8$ board, $53$ pieces
Pu... | Assuming we have completed $x$ spirals (and no more spirals can be completed)
Then we have
(Comparing number of pieces utilized by $x$ and $x+1$ spirals)
$$ n^2-(n-2x)^2\le B\le n^2-(n-2x-2)^2$$
Now,$$n^2-(n-2x)^2\le B\implies n^2-B\le (n-2x)^2\implies x\le\frac{n-\sqrt{n^2-B}}{2}$$
Or,
$$n^2-(n-2x-2)^2\ge B\implies n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1433495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\lim _{x\to 1}\left(\frac{\sqrt[3]{x}-1}{2\sqrt{x}-2}\right)$ I'm trying to evaluate the limit
$$\lim _{x\to 1} \frac{\sqrt[3]{x}-1}{2\sqrt{x}-2} .$$
I used an online limit calculator to find the result, which gives
$$\lim _{x\to 1} \frac{x^{\frac{1}{6}}+1}{2\left(\sqrt[3]{x}+x^{\frac{1}{6}}+1\right)}.$$
... | Why not to apply L-Hospital's rule for $\frac{0}{0}$ form
$$\lim_{x\to 1}\frac{\sqrt[3]{x}-1}{2\sqrt x-2}$$
$$=\lim_{x\to 1}\frac{\frac{d}{dx}\left(\sqrt[3]{x}-1\right)}{\frac{d}{dx}(2\sqrt x-2)}$$
$$=\lim_{x\to 1}\frac{\frac{1}{3}x^{-2/3}}{2\frac{1}{2}x^{-1/2}}$$
$$=\lim_{x\to 1}\frac{1}{3}x^{-1/6}=\frac{1}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1434528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Solve for $x^{11} + \frac{1}{x^{11}}$ if $x^3 + \frac{1}{x^3} = 18.$ Solve for $x^{11} + \frac{1}{x^{11}}$ if $x^3 + \frac{1}{x^3} = 18.$
I'm familiar with variants of this problem, especially those where the exponent in the given is a factor of the exponent in what we want to solve for, but this one stumped me.
| In the equation $x^3+\frac{1}{x^3}=18$, multiply everything by $x^3$ to get $x^6-18x^3+1=0$. Then let $y=x^3$ and solve $y^2-18y+1=0$, then substitute back in to get $x$. Then you can input the value into $x^{11}+\frac{1}{x^{11}}$ directly.
It winds up being $x^{11}+\frac{1}{x^{11}}=39603$ (regardless of whether you us... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1435075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Solutions to the Laplace Equation $\Delta u =0$, where $u= \log p$ Find all real solutions to the two dimensional Laplace equation $U_{xx} + U_{yy} =0$ of the form $u=\log p(x,y)$, where $p$ is a quadratic polynomial.
Solution:
Let $p(x,y) = Ax^2 + By^2 +Cxy + D$ be a quadratic polynomial such that $A, B \not= 0$. The... | The solution that is proposed makes use of the wolfram language for symbolic computations.
In the following link you can download the notebook to explore all the details:
https://community.wolfram.com/groups/-/m/t/2108153
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1435906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving quartic equations Given the following quartic equation:
$$x^4-2x^3-7x^2+8x+12=0$$
Could anyone give some techniques required to solve any quartic equation (apart from this one) if they exist?
| As an A level student, I would first look at this with the Remainder Theorem, which states that if we sub in a value, say 1, and we get zero, then (x-1) is a factor.
Let $$p(x)=x^4-2x^3-7x^2+8x+12$$
Sub in -1:
$$p(-1)=(-1)^4-2(-1)^3-7(-1)^2+8(-1)+12 = 0$$
Therefore, (x + 1) must be a factor. We can then use algebraic d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1436053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Help factorising a sixth degree polynomial I have to factorise- $$x^6+5x^3+8$$Answer is $$(x^2−x+2)(x^4+x^3−x^2+2x+4)$$.I have also used factor theorem.Please help me.Thanks in advance.
| It can be factored with help of following identities (applied twice below, marked by a '*')
$$u^3 \pm v^3 = (u \pm v)(u^2 \pm uv + v^2)$$
Let $u = x^2 + 2$, we have
$$\begin{align}
x^6 + 5x^3 + 8
&= (x^2)^3 + 2^3 + 5x^3\tag{1}\\
&\stackrel{*}{=} (x^2+2)(x^4 - 2x^2 + 4) + 5x^3\\
&= u(u^2 - 6x^2) + 5x^3\tag{2}\\
&= (u^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1437662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Diophantine Equation or Ellipse I guess this is something to do with circle
The question is:
"Given x and y are real numbers, such that $2x^2 + 3y^2 - 4x - 12y = -14$, find xy."
What is the trick ?
| Rewrite
$$2x^2-4x+3y^2-12y=-14$$
$$2(x^2-2x)+3(y^2-4y)=-14$$
$$2(x^2-2x+1-1)+3(y^2-4y+4-4)=-14$$
$$2(x^2-2x+1)-2+3(y^2-4y+4)-12=-14$$
$$2(x-1)^2+3(y-2)^2-14=-14$$
$$2(x-1)^2+3(y-2)^2=0$$
When is a sum of two squares equal to zero? Only if both squares are equal to zero. So one can conclude $x-1=0$ and $y-2=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1438277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $a^2+ab+b^2\ge 0$ How to prove that $a^2+ab+b^2\ge 0$?
Obviously the squares are positive, but how can I be sure that $ab$ doesn't become too negative with a certain combination of $a$ and $b$?
| $a^2+ab+b^2= (a+ \frac{1}{2}b)^2 + \frac{3}{4}b^2$
One may prefer symmetry:
$2(a^2+ab+b^2)= a^2 +b^2+(a+b)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1439494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
How did this result come about? I was reading Chebyshev polynomials Wiki page and I could not understand one thing
$$ T_n(x) = x^n \sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor} \binom{n}{2k} \left (1 - x^{-2} \right )^k \\$$
$$= \tfrac{n}{2} \sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor} (-1)^k \frac{(... | Suppose we have the Chebychev polynomial
$$T_n(x) = x^n \sum_{k=0}^{\lfloor n/2 \rfloor}
{n\choose 2k} \left(1-\frac{1}{x^2}\right)^k
\\ = x^n \sum_{k=0}^{\lfloor n/2 \rfloor}
{n\choose 2k} \frac{1}{x^{2k}}
\sum_{q=0}^k {k\choose q} (-1)^{k-q} x^{2q}$$
and we seek to verify that
$$[x^{n-2p}] T_n(x)
= \frac{1}{2} n (-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove $2^{2n} = \sum_{k=0}^{n}\binom{2n+1}{k}$ I'm trying to prove the following equation above. So far I have:
\begin{align}
2^{2n} &= (1+1)^{2n}\\
&= \sum_{k=0}^{2n}\binom{2n}{k}1^k1^{n-k} = \sum_{k=0}^{2n}\binom{2n}{k} & \text{(By the Binomial Theorem)}
\end{align}
I know I have to use the following identity somehow... | $$ 2^{2n}=\frac12\cdot 2^{2n+1}=\frac12\sum_{k=0}^{2n+1}{2n+1\choose k}=\frac12\sum_{k=0}^n\left({2n+1\choose k}+{2n+1\choose 2n+1-k}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Proving that $\Gamma \left(n+ \frac{1}{2}\right) = \frac{(2n)!\sqrt{\pi}}{2^{2n}n!}$. The proof I am dealing with is worded exactly as follows:
Prove $\Gamma\left(n+ \frac{1}{2}\right) = \frac{(2n)!\sqrt{\pi}}{2^{2n}n!}$.
The proof itself can be done easily with induction, I assume. However, my issue is with the doma... | We can also exploit the fact that $\Gamma(x+1)=x\Gamma(x)$ to obtain
\begin{align*}
\Gamma\left(n + \frac{1}{2}\right)
& =\left(n-1+\frac{1}{2}\right)\Gamma\left(n-1+\frac{1}{2}\right) \\
& =\left(n-1+\frac{1}{2}\right)\left(n-2+\frac{1}{2}\right)\Gamma\left(n-2+\frac{1}{2}\right) \\
& = \ldots
= \left(n-\frac{1}{2}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1444967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Limits of two variables functions I'm computing limits of two variables functions I do not manage to resolve them all.
4)$\lim_{(x,y) \to (0,0)} \dfrac{x^3-y^3}{x^2+2y^2} $
Here I use polar coordinates with $x=r\cos{\theta}$ and $y=r\sin{\theta}$ I finally get $\lim_{(r) \to (0)} \dfrac{rcos^3\theta - sin^3 \theta}{cos... | Regarding 4) you have $x^2+2y^2 \ge x^2 + y^2$, $\vert x^3 \vert \le \vert x \vert (x^2+y^2)$ and $\vert y^3 \vert \le \vert y \vert (x^2+y^2)$. Consequently: $$\vert \frac{x^3-y^3}{x^2+2y^2} \vert \le \vert \frac{x^3-y^3}{x^2+y^2} \vert \le \frac{(\vert x \vert + \vert y \vert)(x^2+y^2)}{x^2+y^2} \le \vert x \vert + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1445157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluation of $\int\frac{1}{\sin^2 x+\sin x+1}dx$
Evaluation of $\displaystyle \int\frac{1}{\sin^2 x+\sin x+1}dx$
$\bf{My\; Try::}$ Using $$\; \bullet\; x^2+x+1 = (x-\omega)\cdot (x-\omega^2)\;,$$ where $\omega,\omega^2$ are cube root of unity
So we can write Integal $$\displaystyle I = \int\frac{1}{(\sin x-\omega)\... | Substitute $y=\sec x - \tan x$, or $\sin x=\frac{1-y^2}{1+y^2}$
\begin{align}
&\int \frac{1}{\sin^2 x+\sin x+1}dx\\
=&-\int\frac{2+{2y^2}}{3+y^4}dy
= -\int\frac{\frac2{y^2}+{2}}{\frac3{y^2}+y^2}dy \\
=&\ \frac{\sqrt3+1}{\sqrt{6\sqrt3}}\tan^{-1}\frac{y-{\sqrt3}y^{-1}}{\sqrt{2\sqrt3}}
-\frac{\sqrt3-1}{\sqrt{6\sqrt3}}\cot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1445354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\sqrt{ c} − \sqrt{c − 1} \geq \sqrt{ c + 1} −\sqrt{c}$ for all real $c \geq 1$. Prove that $\sqrt{ c} − \sqrt{c − 1} \geq \sqrt{c + 1} −\sqrt{c}$ for all real $c \geq 1$.
Can anyone provide some form of guidance? So far all I have been able to think of is writing $c$ as $x^2$ for some $x$, or eliminating th... | To show
$\sqrt{ c} − \sqrt{c − 1}
\geq \sqrt{c + 1} −\sqrt{c}
$.
This is the same as
$\sqrt{c}
\ge \frac{\sqrt{c + 1}+\sqrt{c - 1}}{2}
$.
If
$f(x) = x^{1/2}
$,
then
$f'(x) = \frac12x^{-1/2}
$
and
$f''(x) = -\frac14x^{-3/2}
< 0
$
so
$f(x)$
is concave
which means that
$f(\frac{a+b}{2})
\ge \frac{f(a)+f(b)}{2}
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1447074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
If $(a,b,M)$ is a Pythagorean triple, can $(b,b+a,N)$ be another triple? Does anyone know of a pair of Pythagorean triples of the form
$$(a, b, M) \quad\text{and}\quad(b, b+a, N)$$
Is such a pair possible?
| Using Wolfram Alpha to ensure accurate algegra, we start with Euclid's formula (modified) for the OP.
$$A=m^2-n^2\qquad B=2mn+(m^2-n^2)\qquad C=m^2+n^2$$
$$A^2=m^4 - 2 m^2 n^2 + n^4$$
$$B^2=m^4 + 4 m^3 n + 2 m^2 n^2 - 4 m n^3 + n^4$$
$$C^2=m^4 + 2 m^2 n^2 + n^4$$
$$A^2+B^2-C^2=m^4 + 4 m^3 n - 2 m^2 n^2 - 4 m n^3 + n^4=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1447191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How do I solve $\lim_{x \to 1} (\frac{1}{x-1}-\frac{2}{x^3-1})$ indeterminate limit without the L'hospital rule? I've been trying to solve this limit without L'Hospital's rule as homework. So I tried rationalizing the denominator and numerator but it didn't work.
My best was: $\lim_{x \to 1} (\frac{1}{x-1}-\frac{2}{x^3... | First note that $x^3 - 1 = (x-1)(x^2 + x + 1)$. Therefore multiply $\frac{1}{x-1}$ by 1 in the form $\frac{x^2 + x + 1}{x^2 + x + 1}$ and combine the fractions as $\frac{(x^2 + x + 1) - 2}{(x-1)(x^2 + x + 1)} = \frac{x^2 + x -1}{(x-1)(x^2 + x + 1)}$. The numerator is now going to $1 + 1 - 1 = 1$, while the denominator... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1449816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Another beautiful arctan integral $\int_{1/2}^1 \frac{\arctan\left(\frac{1-x^2}{7 x^2+10x+7}\right)}{1-x^2} \, dx$ Do you think we can express the closed form of the integral below in a very nice and short way?
As you already know, your opinions weighs much to me, so I need them!
Calculate in closed-form
$$\int_{1/2}^... | Substituting $x\to\frac{1-x}{1+x}$ is usually a first reaction of mine. Here it works surprisingly well: the integral is equivalent to
$$I=\int_{1/2}^1 \frac{\tan^{-1}\left(\frac{1-x^2}{7 x^2+10x+7}\right)}{1-x^2} \, dx=\frac12\int_0^{\frac13}\frac{\tan^{-1}\left(\frac{x}{x^2+6}\right)}{x} dx.$$
Integrating by parts, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1453345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Proving tan((x + y)/2) = (sin x + sin y)/(cos x + cos y) with the angle sum and difference identities I've recently come across this problem of proving
$$ \tan \frac{x + y}{2} = \frac{\sin x + \sin y}{\cos x + \cos y} $$
Not a difficult problem, I thought. I would have rewritten the RHS using the sum-to-product identit... | I would use the formulae that express the trigonometric functions in function of the tangent of the half-angle. So set $\;t=\tan\dfrac x2,\enspace u=\tan y2$. Then:
\begin{align*}
\frac{\sin x+\sin y}{\cos x+\cos y}&=\frac{\dfrac{2t}{1+t^2}+\dfrac{2u}{1+u^2}}{\dfrac{1-t^2}{1+t^2}+\dfrac{1-u^2}{1+u^2}}=\frac{2\bigl(t(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1453588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
inverse of $\arcsin (\frac{x}{x-1})$ determine the inverse of
a) $y=\arcsin \left(\dfrac{x}{x-1}\right)$
b) $y= \dfrac{1-2e^{-x}}{4}$
I learned you the steps for finding the inverse are
1) get it in a form of $x= \dots$
2) change $x$ and $y$
a) $y=\arcsin \left(\dfrac{x}{x-1}\right)$
is $y=\arcsin \left(\dfrac{x}{x-1}... | a)
\begin{align}
\sin y &= \dfrac{x}{x-1}\\
x\sin y -\sin y &=x\\
(\sin y - 1)x&=\sin y\\
x&=\frac{\sin y}{\sin y -1}
\end{align}
So the inverse function is $x\mapsto \dfrac{\sin x}{\sin x -1}$
And for b) is similar.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1454218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Is $\sum_{n=1}^{\infty }\frac{8n\cdot\zeta (2n)}{3\cdot 2^{2n}}=\zeta (2)$?
$$\sum_{n=1}^{\infty }\frac{8n\cdot\zeta (2n)}{3\cdot 2^{2n}}=\zeta(2)$$
By using numerical calculation, I found this relationship between the values of zeta function at even integers and $\zeta(2)$, but this needs proving, any help?
| Ok, let give me a slightly different answer the Jack,
Our sum is
$$
S=\frac{8}{3}\sum_{n=1}^{\infty}\frac{n}{2^{2n}}\sum_{k=1}^{\infty}\frac{1}{k^{2n}}=\frac{8}{3}\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{n}{(2^2 k^2)^n}\underbrace{=}_{(1)}\frac{8}{3}\sum_{k=1}^{\infty}\frac{4 k^2}{\left(4 k^2-1\right)^2}\underbrace{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Let $f:\mathbb R\rightarrow \mathbb R$ be a continuous function such that $f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^{2}$ . Then $f(3)$ =? Options: (a)$4$, (b)$4f(0)$, (c)$4-f(0)$, (d)$4+f(0)$, (e)$16+f(0)$.
CORRECT ANSWER USING REDUCTION
Deep thanks to @martini and @A.S. , soo much respect .
Since we ... | Let $x = 3\cdot 2^{-n}$, giving
$$ f(3\cdot 2^{-n}) = 9 \cdot 2^{-2n} + 2 f(3 \cdot 2^{-(n+1)}) - f(3 \cdot 2^{-(n+2)}) $$
Write brevitatis causa $a_n := f(3 \cdot 2^{-n})$, we have
$$ a_{n+2} - 2a_{n+1} + a_n = 9 \cdot 2^{-2n} $$
The general solution of the homogenous recursion
$$ a_{n+2} - 2a_{n+1} + a_n = 0 $$
is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1457887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Calculus 1: limits using squeeze or sandwich theorem, verification. I have the next problem:
Suppose that a function $g$ satisfies that $-1\leq g(x) \leq 1$ for all $x\geq 0$.
Calculate $\lim_{x\to\infty}\dfrac{x+g(2x)}{g(3x)+4x}$
So: $$\lim_{x\to\infty}\dfrac{x+g(2x)}{g(3x)+4x}=\lim_{x\to\infty}\dfrac{\frac{x+g(2x)}... | You can get the limit with a single application of the squeeze theorem.
For the numerator at $x>1$,
$$0<1-\frac{1}{x}\le 1+\frac{g(2x)}{x} \le 1+\frac{1}{x}$$
and for the denominator at $x>1$,
$$0<4-\frac{1}{x} \le \frac{g(3x)}{x}+4 \le 4+\frac{1}{x}$$
So for all $x>1$
$$\frac{1-\frac{1}{x}}{4+\frac{1}{x}} \le \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1458978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $x^4 + \frac{1}{x^4}$ given that $x^2 - 3x + 1 = 0$ Determine the value of $x^4 + \frac{1}{x^4}$ given that $x^2 - 3x + 1 = 0$. I've tried forcing in a difference of squares, looked for various difference of $n$s or sum of odd powers that I could equate this to, but have yet to find a solution.
| Hint Dividing gives $$x - 3 + \frac{1}{x} = 0,$$ and then rearranging gives
$$x + \frac{1}{x} = 3.$$
Additional hint Squaring gives $$3^2 = \left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2},$$ and rearranging again gives $$x^2 + \frac{1}{x^2} = 7.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1460480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
How to evaluate the limit of $\frac {2\cos({x-1})-2}{({x²}-2\sqrt{x}+1 )}$ when $x\to1$ without using L'Hospital's rule?
How do I evaluate this without using L'Hospital's rule:$$\lim_{x \to 1}\frac {2\cos({x-1})-2}{({x²}-2\sqrt{x}+1 )}\ ?$$
Note : I used L'Hospital's Rule I find $\frac{-4}{3}$ but in wolfram alpha is... | We start by using the double-angle identity for the cosine function to write
$$2\cos (x-1)-2=-4\sin^2\left(\frac{x-1}{2}\right) \tag 1$$
Next, we factor the denominator of the term of interest to find
$$\begin{align}
x^2-2x^{1/2}+1&=(x^{1/2}-1)(x^{3/2}+x+x^{1/2}-1)\\\\
&=\frac{(x^{3/2}+x+x^{1/2}-1)(x-1)}{x^{1/2}+1} \ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1461676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Showing $\pi/(2\sqrt3)=1-1/5+1/7-1/11+1/13-1/17+1/19-\cdots$ I am struggling to show that $$\dfrac \pi{2\sqrt3}=1-\dfrac 15+\dfrac 17-\dfrac 1{11}+\dfrac 1{13}-\dfrac 1{17}+\dfrac 1{19}-\cdots$$ by using the Fourier series $$\frac \pi2-\frac x2=\sum_1^\infty \dfrac {\sin(nx)}{n}.$$
Can somebody give me any hint?
| The approach we take is to find a discrete and finite Fourier-series of the periodic sequence $f(n)$ in the sum $\sum\frac{f(n)}{n} = \frac{\color{red}{1}}{1} + \frac{\color{red}{0}}{2} + \frac{\color{red}{0}}{3} + \frac{\color{red}{0}}{4} + \frac{\color{red}{-1}}{5} + \frac{\color{red}{0}}{6} \ldots$ we are trying to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 2
} |
Binomial Theorem Question (Expansion of Three Terms) I have the term: $(1 + 2x - x^2)^4.$
The question asks me to find the coefficient of $x^5$.
My solution:
$\sum\limits_{i=0}^4 {4 \choose r} (1)^{4-r}(2x-x^2)^r$
I then factored out x from $(2x-x^2)$, getting $x(2-x)$.
Then, since the terms with the x's are being rais... | $$\begin{align}
&[x^5](1+2x-x^2)^4
\\=&[x^5]\sum_{r=0}^4\binom 4r (2x-x^2)^r
\\=&[x^5]\sum_{r=0}^4\binom 4r (2x)^r \left(1-\frac x2\right)^r
\\=&[x^5]\sum_{r=0}^4\binom 4r (2x)^r \sum_{j=0}^r \binom rj \left(-\frac x2\right)^j
\\=&\underbrace{\underbrace{\binom 43 2^3\cdot \binom 32 \left(-\frac 12\right)^2}_{r=3, j=2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1463139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Cofactor expansion to check if matrices is invertible. I have a question regarding a co-factor expansion question. I understand that an easy way to check if a matrix is invertible is to do co-factor expansion and if $A \ne 0$ then it's invertible. I'm familiar with the basic co-factor expansion with $3 \times 3$ matric... | To check if matrices are invertible, you need to check the determinant is non-zero:
To find the determinant of this matrix we look for the row or column with the most zeros and do a Laplace development on that row or column.
The first row contains the most zeros so we Laplace develop that row:
$$\det=\begin{vmatrix}0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1463350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$x^{2000} + \frac{1}{x^{2000}}$ in terms of $x + \frac 1x$. If $x + \frac{1}{x} = 1$, then what is
$$ x^{2000} + \frac{1}{x^{2000}} = ?$$
| Assume that $x=e^{i\theta}$. Then you are trying to write $2\cos(2000\cdot\theta)$ in terms of $2\cos(\theta)$, so the answer is given by Chebyshev polynomials of the first kind:
$$ \left(x^{2000}+\frac{1}{x^{2000}}\right) = 2\cdot \widetilde{T}_{\!2000}\left(x+\frac{1}{x}\right) $$
where $\widetilde{T}_{n}(z) = T_n(z/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1465320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
How does $\tan^{-1}(x-\sqrt{1+x^2})=\frac{1}{2}\tan^{-1}x+C$ directly? I'm teaching baby calculus recitation this semester, and I meet a problem to calculate the derivative of
$$y=\tan^{-1}(x-\sqrt{1+x^2})$$
Just apply the chain rule and after some preliminary algebra, I find
$$\frac{dy}{dx}=\frac{1}{2(1+x^2)}$$
What... | Let me rewrite the proof a bit differently.
We are required to prove (RTP) that $$2\tan^{-1}(y - \sqrt{1+y^2}) - \tan^{-1} (y) = -\frac{\pi}{2}$$
Proof:
Plan: We will use the formula F first, followed by the identity I.
Let us call the following identity I:
$\tan^{-1}(x) + \tan^{-1} (1/x) =$
\begin{cases}
\hfill \pi/2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1466415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 2
} |
separating equation $$
\begin{align}
\frac {dy}{dx} &= \frac{(y-1)(x+2)}{(y+1)(x-3)}\\
\frac{y+1}{y-1}dy &= \frac{x+2}{x-3}dx
\end{align}
$$
Integrate both sides :
$$
\begin{align}
\int\frac{y+1}{y-1}dy & =\int\frac{x+2}{x-3}dx\\
\int 1+\frac{2}{y-1}dy &=\int 1+\frac{5}{x-3}dx\\
y+2\ln|y-1|&=x+5\ln|x-3|+C\\
\end{align}... | One first thing, you should put the absolute values there. The correct way to write down the solution is
$$y + 2\ln \left| {y - 1} \right| = x + 4\ln \left| {x - 3} \right| + C$$
At this situation, they say that the ordinary differential equation is solved. However, there is not always a way to explicitly find $y$ in ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1470117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Exponential conversion While solving Fourier series coefficients if found this problem. Anyone help me to tell that how they converted $\frac{1}{2}e^{j\pi/4}=\frac{\sqrt{2}}{4}(1+j)$
| Hint (assuming you have $j^2=-1$): Use $e^{jx}= \cos x + j \sin x$ and $\cos \frac{\pi}{4}=\sin \frac{\pi}{4}=\frac{\sqrt{2}}{2}$
Edit- Here the completed answer after reading your comment:
$$\frac{1}{2}e^{j\pi/4}=\frac{1}{2}\left(\cos \frac{\pi}{4} + j \sin \frac{\pi}{4} \right)=
\frac{1}{2}\left(\frac{\sqrt{2}}{2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1470352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
if $\sin{(aw)}+\sin{(bw)}+\sin{(cw)}=3$ find $w$ range let $w$ is postive integer,if there exist $a,b,c(\pi\le a<b<c\le 2\pi)$such
$$\sin{(aw)}+\sin{(bw)}+\sin{(cw)}=3$$
Find the $w$ range.
My attempt：
since
$$\sin{(wa)}\le 1,\sin{(wb)}\le 1,\sin{(wc)}\le 1$$
then $$\sin{(wa)}=\sin{(wb)}=\sin{(wc)}=1$$
$$aw=\dfrac{\pi}... | we also solve if $w$ be postive real numbers,I have done This result is $$w\in [\dfrac{17}{4},\dfrac{9}{2}]\bigcup [\dfrac{21}{4},+\infty) $$
This problem is equivl this:there $p,q,r\in N^{+}$,such
$$w\pi\le 2p\pi+\dfrac{\pi}{2}<2q\pi+\dfrac{\pi}{2}<2r\pi+\dfrac{\pi}{2}\le 2w\pi$$
case1 if $w\ge 6$ it is clear,because ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1470431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Help showing $\lim\limits _{x\to -\infty \:}\left(\sqrt{4\cdot \:x^2-5\cdot \:x}+2\cdot \:x\right) = \frac{5}{4}$ I am stuck solving the following limit. I know the answer is 5/4, I just can't get it. This is the steps I have done so far.
$\lim _ \limits{x\to -\infty \:}\left(\sqrt{4\cdot \:x^2-5\cdot \:x}+2\cdot \:x\r... | Expanding DirkGently's answer,
this is the point
where multiplying
by the conjugate
is useful:
$\begin{array}\\
\lim_{x\to-\infty}2x\left(1-\sqrt{1-\frac{5}{4x}}\right)
&=\lim_{x\to-\infty}2x\left(1-\sqrt{1-\frac{5}{4x}}\right)
\frac{1+\sqrt{1-\frac{5}{4x}}}{1+\sqrt{1-\frac{5}{4x}}}\\
&=\lim_{x\to-\infty}2x\frac{\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1475580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Find two numbers, $x,y$ whose sum is $35$ and $x^2y^5$ is maximum.
Find two numbers, $x,y$ whose sum is $35$ and $x^2y^5$ is maximum.
My answer:
$$x+y=35$$
$x^2y^5$ is maximum
$$y=35-x$$
$$\frac{d}{dx} x^2(35-x)^5$$
Which rule to apply here after? I reached:
$$(35-x)^4(-5x^2+(35-x)2x)=0$$
Either $(35-4x)^4 =0$ or $x^... | Note that if $y \leq 0 \Rightarrow x^2y^5 \leq 0, \forall x \in \mathbb{R}$, thus assume $y > 0$, and since the max won't change if $x$ is replaced by $-x$, we can also assume $x > 0$. Then we have a much simpler problem to solve.
We have: $35 = x+y = \dfrac{x}{2}+\dfrac{x}{2}+ \dfrac{y}{5}+\dfrac{y}{5}+\dfrac{y}{5}+\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1475970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
What are all functions of the form $\frac{\cosh(\alpha x)}{\cosh x+c}$ self-reciprocal under Fourier transform? There are some functions that are self reciprocal under cosine Fourier transform:
\begin{equation}
\frac{1}{\cosh x}, \frac{\cosh x}{\cosh 2x},\frac{1}{1+2\cosh x}
\end{equation}
It seems that they have been ... | $$
\newcommand{\res}{\operatorname{Res}}
\newcommand{\re}{\operatorname{Re}}
$$
I cannot give you the references you are looking for but I will calculate the Fourier transform of your fifth function.
Put
$$
I = \dfrac{1}{2}\int_{-\infty}^{\infty}\dfrac{e^{iax}\cosh\frac{\sqrt{3}x}{2}}{\cosh x -\cos(\sqrt{3}\pi)}\, dx.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1477074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Calculate the integral of the function only with the immediate integrals I need to calculate this indefinite integral:
$$
\int{\frac{x^2-1}{x^2+1}}dx
$$
I realized that the technique to solve is to "transform" the numerator equal to the denominator of the fraction so I tried this:
$$
\int{\frac{x^2-1}{x^2+1}}dx = \int{... | Try using long-division on your rational expression:
$$\frac{x^2-1}{x^2+1}=\frac{(x^2+1)-2}{x^2+1}=1-\frac{2}{x^2+1}$$
So now we can integrate the $1$ (that is, $\int 1\ dx$), and subtract the integral of $\frac{2}{x^2+1}$ (which is $2\arctan(x)+C$).
Whenever you have a rational function whose numerator's degree is no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1480878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Algebraic values of sine at sevenths of the circle At the end of a calculation it turned out that I wanted to know the value of
$$\sin(2\pi/7) + \sin(4\pi/7) - \sin(6\pi/7).$$
Since I knew the answer I was supposed to get, I was able to work out that the the above equals $\sqrt{7}/2$, and I can confirm this numerically... | Solution without using trinometry, but only properties of roots of unity:
(This proof has been refined after reviewing this excellent post here. I knew there had to be a more direct approach!:))
Let $\omega=e^{i2\pi/7}$, i.e. the primitive $7$th root of unity. By definition,
$$1+\omega^1+\omega^2+\omega^3+\omega^4+\o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1482145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
How to solve a system of non linear equations of the form $2ax-2x(ax^2+by^2) = 0$ Solve this system for $x,y$:
$$2ax-2x(ax^2+by^2) = 0\tag{1}$$
$$2by-2y(ax^2+by^2) = 0\tag{2}$$
where $a$ and $b$ are constants such that $a>b>0$.
I re-arranged (1) to $x = \sqrt{\frac{a-by^2}{a}}$ and tried substituting it into (2) but ... | If $x = 0$, the first equation is ok, and:
$$
2by-2y(ax^2+by^2) = 0 \\
2by-2y(by^2) = 0 \\
by-by^3 = 0 \\
y-y^3 = 0 \\
y(1-y^2) = 0
$$
Thus, either $y=0$ or $y=1$ or $y=-1$. Here we have three solutions.
If $x \ne 0$:
$$
2ax-2x(ax^2+by^2) = 0 \\
ax = x(ax^2+by^2) \\
a = ax^2+by^2
$$
Replacing in the second equation, we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1482591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the coordinate of a point $P$ on the line $2x-y+5=0$ such that $|PA-PB|$ is maximum where $A=(4,-2)$ and $B=(2,-4)$ What is the coordinate of a point $P$ on the line $2x-y+5=0$ such that $|PA-PB|$ is maximum where $A=(4,-2)$ and $B=(2,-4)$.
Let the coordinates of the point $P$ be $(x_1,y_1).$
$P$ lies on the l... |
Now $PA=\sqrt{(x-2)^2+(y+4)^2}$ and $PB=\sqrt{(x-4)^2+(y+2)^2}$ Now using Cosine formula
We get $$\displaystyle \cos \theta = \frac{(PA)^2+(PB)^2-(AB)^2}{2\cdot (PA)\cdot (PB)}$$
Now we know that $$\displaystyle |\cos \theta | \leq 1\Rightarrow \left|\frac{(PA)^2+(PB)^2-(AB)^2}{2\cdot (PA)\cdot (PB)}\right|\leq 1$$
S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1483390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Prove by induction: $\sum\limits_{k=1}^{n}sin(kx)=\frac{sin(\frac{n+1}{2}x)sin\frac{nx}{2}}{sin\frac{x}{2}}$ $\sum\limits_{k=1}^{n}sin(kx)=\frac{sin(\frac{n+1}{2}x)sin\frac{nx}{2}}{sin\frac{x}{2}}$
Base case: For $n=1$
$sinx=\frac{sinx\cdot sin\frac{x}{2}}{sin\frac{x}{2}}=sinx$
Induction hypothesis: For $n=m$
$\sum\lim... | It's simply a change product into sum or difference of trigonometric function.Note that we have double angle formula and apply it to $$\sin(m+1)x$$,we'll get $$sin\frac{(m+1)x}2 *\sin \frac{mx}2+2 \sin\frac{ x}2 *\sin\frac{(m+1)x}2*\cos\frac{(m+1)x}2=\sin\frac{(m+1)x}2*(\sin\frac{mx}2+2\cos\frac{(m+1)x}2*\sin\frac{x}2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How to prove that the following system of equations has only one solution? $
\begin{cases}
(x - 1)^2 + (y + 1)^2 = 25 \\
(x + 5)^2 + (y + 9)^2 = 25 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}
$
I have to solve this system of equations. After substituting $y = -\frac{3}{4}x - \frac{13}{2}$ into $(x - 1)^2 + (y + 1)^... | As the other answers said, your argument already proved there is only one solution.
You can make an independent check by solving the equations graphically.
$(x−1)^2 +(y+1)^2 =25$ is the equation of a circle, radius $5$, center $(1, -1)$. The second equation is another circle, and the circles have two intersection point... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1486203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
How to solve $x^6-x^5+x^4-x^3+x^2-x+1=0$? Can anyone tell me how to solve this?
$x^6-x^5+x^4-x^3+x^2-x+1=0$
What I got to was $x^7+1=0$.
Thanks in advance.
| This equation has the same coefficients read backwards.
There is a technique of solving such equations:
If the degree is odd, then $-1$ is a root, and dividing by $x+1$ gives you an even degree equation with the same property.
For even degree: divide by $x$ to the power half degree, and make the substitution $t=x+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1487231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
A serie about $\sum_{n = 1}^\infty {\arctan \frac{{10n}}{{\left( {3{n^2} + 2} \right)\left( {9{n^2} - 1} \right)}}}$ How to prove $$\sum\limits_{n = 1}^\infty {\arctan \frac{{10n}}{{\left( {3{n^2} + 2} \right)\left( {9{n^2} - 1} \right)}}} = \ln 3 - \frac{\pi }{4}.$$
Add: Maybe we can follow this!
| Since $\arctan(x)=\arg(1+ix)$ and we can factor
$$
1+\frac{10in}{\left(3n^2+2\right)\left( 9n^2-1\right)}
=\frac{\left(1-\frac in\right)\left(1+\frac i{3n-1}\right)\left(1+\frac i{3n+1}\right)\left(1+\frac i{3n}\right)}{1+\frac2{3n^2}}
$$
we have that
$$
\begin{align}
&\arctan\left(\frac{10n}{\left(3n^2+2\right)\left( ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1487596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
What is $\angle AEB$? Let $E$ be a point inside the square $ABCD$. and $|EC|=3,|EA|=1,|EB|=2$
What is the angle $\widehat {AEB}$?
I can only find $|ED|=\sqrt{12}$
| Here is a less algebraicly intensive solution and only requires knowledge of the sine law and the cosine law, and solving quadratics of the form $x^2 = a$(super easy) all of which is material known by the average grade 10 math student in Canada.
From the sine law we know that
\begin{align}
\frac{\sin(ABE)}{1} = \frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1489850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Of sum of cosines and the $7$th roots of unity In my solution here, it was shown that
$$\omega+\omega^2+\omega^4=-\frac 12\pm\frac{\sqrt7}2\qquad\qquad (\omega=e^{i2\pi/7})$$
from which we know that
$$\sin \frac{2\pi}7+\sin \frac{4\pi}7-\sin \frac{6\pi}7=\Im (\omega+\omega^2+\omega^4)=\frac{\sqrt7}2$$
This can also b... | In that answer there is the following equality
$$\begin{align}
\sin\frac{2\pi}7+\sin\frac{4\pi}7-\sin\frac{6\pi}7
&=\Im(\omega^1+\omega^2-\omega^3)\\
&=\Im(\omega^1+\omega^2+\omega^4)\\
&=\Im(S)\\
&=\dfrac{\sqrt{7}}{2}\\
\end{align}.$$
For the real part you have
$$\begin{align}
-\dfrac{1}{2}&=
\Re(S)\\
&=\Re(\omega^1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1491566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the indefinite integral: $\int { {\sqrt{x+1}} \over {\sqrt{x+2} - \sqrt{x-2}} }dx$ Find the indefinite integral:
$$\int { {\sqrt{x+1}} \over {\sqrt{x+2} - \sqrt{x-2}} }dx$$
I don't know how to start, multiplying by ${ {\sqrt{x+2} + \sqrt{x-2}} \over {\sqrt{x+2} + \sqrt{x-2}} }$ hasn't helped neither...
| Notice, $$\int\frac{\sqrt{x+1}}{\sqrt{x+2}-\sqrt{x-2}} \ dx$$
$$=\int\frac{\sqrt{x+1}(\sqrt{x+2}+\sqrt{x-2})}{(\sqrt{x+2}+\sqrt{x-2})(\sqrt{x+2}-\sqrt{x-2})} \ dx$$
$$=\int\frac{\sqrt{x^2+3x+2}+\sqrt{x^2-x-2}}{x+2-(x-2)} \ dx$$
$$=\int\frac{\sqrt{\left(x+\frac{3}{2}\right)^2-\frac{1}{4}}+\sqrt{\left(x-\frac{1}{2}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1492821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to Calculate $x^6+x^3y^3+y^6$ Given that $x,y$ real numbers such that :
$x^2+xy+y^2=4$
And
$x^4+x^2y^2+y^4=8$
How can one calculate :
$x^6+x^3y^3+y^6$
Can someone give me hint .
| why not.
$$ x^4 + x^2 y^2 + y^4 = (x^2 + xy + y^2)(x^2 - xy+y^2). $$
Worth memorizing. So
$$ x^2 - xy + y^2 = 2. $$ Also
$$ 2xy = 2, \; \; \; xy=1. $$
And
$$ x^2 + y^2 = 3. $$
$$ 27 = (x^2 + y^2)^3 = x^6 + 3 x^4 y^2 + 3 x^2 y^4 + y^6 = x^6 + y^6 + 3 x^2 y^2 (x^2 + y^2) = x^6 + y^6 + 3 \cdot 1 \cdot 3 = x^6 + y^6 + 9 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1492923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find the equation of the circle which cuts the circle $x^2+y^2+2x+4y-4=0\;$ and the lines $xy-2x-y+2=0\;$ orthogonally The equation of the circle which cuts the circle $x^2+y^2+2x+4y-4=0\quad$ and the lines $xy-2x-y+2=0\quad$ orthogonally, is
$a.\quad x^2+y^2-2x-4y-6=0\;$
$b.\quad x^2+y^2-2x-4y+6=0\;$
$c.\quad x^2+y^2-... | 1493485
For two figures to be orthogonal vis-à-vis each other means that at each of their points of intersection their slopes are perpendicular. In terms of analytic geometry, this would mean that the slope of one is the negative reciprocal of the slope of the other -- i.e., the product of the two slopes at the interse... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1493485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$ I had an example in the book given as follows:
Find the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$ .
Solution: $~~~~~$$(\sqrt{2}+\sqrt{3})^2=5+2 \sqrt6$
$(\sqrt{2}+\sqrt{3})^4=49+20 \sqrt6$
Then $(\sqrt{2}+\sqrt{3})^4-10(\sqrt{2}+\sqrt{3}... | Assume that $m$ and $n$ are square-free coprime integers. Then, if we prove that $Q(\sqrt m +\sqrt n) = Q(\sqrt m, \sqrt n)$, we can conclude that that $f(x) = x^4 -2(m+n)x +(m-n)^2$ must be the minimal polynomial of $\sqrt m +\sqrt n$ since $Q(\sqrt m +\sqrt n)$ is a four-dimensional $Q$-vector space. But if $\xi = (\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1494018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
What line is determined by the following complex equation? $$\left | z+1-i \right |= \frac{\Re z-\Im z}{\sqrt{2}}$$
This leads to the following equation, if we write $z=x+iy$ :
$$\sqrt{\left ( x+1 \right )^{2}+\left( y-1 \right)^2}=\frac{x-y}{\sqrt{2}}$$
which according to desmos, defines a set in $\mathbb{C}$ which is... | This is indeed impossible: Using Cauchy-Schwarz, we have
$$ x-y = (1)(x+1) + (-1)(y-1) -2 \le \sqrt{2}\sqrt{(x+1)^2+(y-1)^2} - 2 $$
and hence $\frac{x-y}{\sqrt{2}} < \sqrt{(x+1)^2+(y-1)^2}$ for all $x$ and $y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1494331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proof that sum and product of geometric series is divisible by 10 I need to prove that for $x=7+7^2+\cdots+7^{2016}$, $y=8+8^2+\cdots+8^{2016}$, $10\mid x+y$ and $10\mid xy$. I think that if $10\mid(m_1+m_2)$, where $m_1$ is the remainder of $\frac{x}{10}$ etc., than $10\mid x+y$, but I can't determine the remainder. I... | $$
7 + 7^2 + 7^3 + 7^4 \equiv 0 \pmod {10}
$$
Therefore
\begin{align}
& \overbrace{7 + 7^2 + 7^3 + 7^4} + \overbrace{7^5 + 7^6 + 7^7 + 7^8} + \overbrace{7^9 + 7^{10} + 7^{11} + 7^{12}} + \cdots \\[10pt]
= {} & \Big(7 + 7^2 + 7^3 + 7^4\Big) + 7^4\Big(7 + 7^2 + 7^3 + 7^4\Big) + 7^8 \Big(7 + 7^2 + 7^3 + 7^4\Big) + \cdots ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1497562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding positive values
For how many ordered triples of unequal positive integers $(x,y,z)$
does the expression $$ \frac{x}{(x-y)(x-z)} + \frac{y}{(y-x)(y-z)} +
\frac{z}{(z-x)(z-y)} $$ take on positive values?
I started with $x=3, y=4$ and $z=5$ and got:
$\frac{3}{2}+\frac{4}{-1}+\frac{5}{2}=0$
Then I worked wit... | $\dfrac{x}{(x-y)(x-z)} + \dfrac{y}{(y-x)(y-z)} + \dfrac{z}{(z-x)(z-y)}$
$\dfrac{x}{(x-y)(x-z)} + \dfrac{-y}{(x-y)(y-z)} + \dfrac{z}{(x-z)(y-z)}$
$\dfrac{x(y-z) -y(x-z) + z(x-y)}{(x-y)(x-z)(y-z)}= \dfrac{0}{(x-y)(x-z)(y-z)} = 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1497908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving A Linear Recurrence Relation With Complex Roots Question:
For the given linear homogeneous difference equation, find the general solution:
$$y_{n+2} + y_{n+1} + y_n = 0$$
With the initial conditions of:
$$y(0)=\sqrt3, y(1) = 0$$
Attempted Answer:
I approached the problem normally as one would except by solvin... | Use generating functions. Define $A(z) = \sum_{n \ge 0} y_n z^n$, multiṕly your recurrence by $z^n$, sum over $n \ge 0$ and recognize some sums:
$\begin{align}
\frac{A(z) - y_0 -y_1 z}{z^2}
+ \frac{A(z) - y_0}{z} + A(z)
&= 0 \\
\frac{A(z) - \sqrt{3}}{z^2}
+ \frac{A(z) - \sqrt{3}}{z} + A(z)
&= 0
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1499090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Prove using induction that $2^n < \binom{2n}{n} < 4^n$ for $n \geq 2$
Trying to prove that, for $n\geq2$, $2^n < \binom{2n}{n} < 4^n$.
Inductive hypothesis: Assume $P(k)$ is true:
\begin{align}
2^k < \binom{2k}{k} < 4^n \\\\
\end{align}
Show $P(k+1)$
\begin{align}
2^{k+1} < \binom{2k+2}{k+1} < 4^{k+1} \\\\
\end{alig... | For the induction step on the first inequality, you wish to show that if
$$2^k < \binom{2k}{k}$$
then
$$2^{k + 1} < \binom{2k + 2}{k + 1}$$
Using the definition of the binomial coefficient yields
\begin{align*}
\binom{2k + 2}{k + 1} & = \frac{(2k + 2)!}{(k + 1)!(k + 1)!}\\
& = \frac{(2k + 2)(2k ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1504755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
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.