Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Inequality $\left(\frac{2x-1}{2x+1}\right)^2>0$ The inequality
$$\left(\frac{2x-1}{2x+1}\right)^2>0$$
has the same solution set as
a) $(2x-1)^2 > 0$
b) $2x-1 \neq 0$
c) $2x-1 > 0$
d) None of the above
Is this equivalent to$$\frac{2x-1}{2x+1}>0 \ ? $$
| The answer for your question is shortly "No", because $$\left( \frac { 2x-1 }{ 2x+1 } \right) ^{ 2 }>0\quad \Rightarrow x\in R-\left\{ \pm \frac { 1 }{ 2 } \right\} \\ \\ \frac { 2x-1 }{ 2x+1 } >0\quad \Rightarrow x\in \left( -\infty ;-\frac { 1 }{ 2 } \right) \cup \left( \frac { 1 }{ 2 } ;+\infty \right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2296429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show that $\int_{-\alpha}^{\alpha}\arccos\left({x\over \alpha}\right)\ln(\alpha+x)\mathrm dx=\alpha \pi \ln\left({\alpha \over 2}\right)?$ Given that:
Where $\alpha >0$
$$\int_{-\alpha}^{\alpha}\arccos\left({x\over \alpha}\right)\ln(\alpha+x)\mathrm dx=\alpha \pi \ln\left({\alpha \over 2}\right)\tag1$$
Setting... | $\begin{align} J= \int_{-1}^{1} \arccos x\ln(1+x)\,dx&=\Big[(x+1)(\ln(1+x)-1)\arccos x\Big]_{-1}^1+\int_{-1}^{1}\dfrac{(x+1)(\ln(1+x)-1)}{\sqrt{1-x^2}}dx\\
=&\int_{-1}^{1}\sqrt{\dfrac{1+x}{1-x}}(\ln(1+x)-1)dx\\
\end{align}$
Perform the change of variable $y=\sqrt{\dfrac{1+x}{1-x}}$,
$\begin{align}
J&=8\int_0^{+\infty} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2296856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$S_n=\sum_{k=1}^n\frac{1}{k}$. then $S_{2^n}$=? Let $S_n$=$\sum_{k=1}^n\frac{1}{k}$. which of the following is true?
*
*$S_{2^n}\ge\frac{n}{2}$ for every n$\ge1$.
*$S_n$ is a bounded sequence.
*$|S_{2^n}-S_{2^{n-1}}|\to0$ as n$\to\infty$.
*$\frac{S_n}{n}\to1$ as n$\to\infty$.
I have a confusion that whether $S_... | The second interpretation is correct: $S_{2^n}=1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{2^n}$.
Hint for your questions: note that
$$S_{2^{n+1}}=S_{2^{n}}+\sum_{k=1}^{2^n}\frac{1}{2^n+k}\geq S_{2^{n}}+\sum_{k=1}^{2^n}\frac{1}{2^n+2^n}=S_{2^{n}}+\frac{1}{2}.$$
Hence $S_{2^{n+1}}-S_{2^{n}}\geq 1/2$ and 3) is false.
Show ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2297523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to prove an inequality I didn't know where can I start because it doesn't fit to any theories or formulas.
If $a>0$, $b>0$, $c>0$ and $a+b+c=1$, prove that:
$$\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}\geq \frac{1}{2}$$
| Note that using AM-GM,
$$\frac{a^3}{a^2+b^2} = a - \frac{ab^2}{a^2+b^2} \geqslant a - \frac{ab^2}{2ab} = a - \frac{b}2 $$
Hence $$\sum_{cyc} \frac{a^3}{a^2+b^2} \geqslant \frac{a+b+c}2 = \frac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2297838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding solutions to $2^x+17=y^2$
Find all positive integer solutions $(x,y)$ of the following equation:
$$2^x+17=y^2.$$
If $x = 2k$, then we can rewrite the equation as $(y - 2^k)(y + 2^k) = 17$, so the factors must be $1$ and $17$, and we must have $x = 6, y = 9$.
However, this approach doesn't work when $x$ is ... | I think that there are not solutions for $x$ odd.
Indeed,
$y^2=2^x+17=2^{2k+1}+17=2 \cdot 4^k +17 \equiv 2+17 \equiv3 \mod4$, which cannot happen.
(If $x$ is of the form $x=2k$, then the equation can be easily written $(y-2^k)(y+2^k)=17$, and so we go on).
EDIT: Wrong solution!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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"answer_id": 3
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Show that $\int_{0}^{\pi/2}{\ln[\cos^2(x)+\sin^2(x)\tan^4(x)]\over \sin^2(x)}\mathrm dx=\pi$ Given that:
$$\int_{0}^{\pi/2}{\ln[\cos^2(x)+\sin^2(x)\tan^4(x)]\over \sin^2(x)}\mathrm dx=\pi\tag1$$
$$1-\sin^2(x)+\sin^2(x)\tan^4(x)$$
$$=1+\sin^2(x)[\tan^4(x)-1]$$
$$=1+\sin^2(x)[\tan^2(x)-1][\tan^2(x)+1]$$
$$=1+\tan^2(x)[... | $J=\displaystyle \int_{0}^{\infty}{\ln(1-t^2+t^4)\over t^2}\mathrm dt$
$\begin{align}
J&=\Big[-\dfrac{\ln(1-t^2+t^4)}{t}\Big]_{0}^{+\infty}+\int_{0}^{+\infty}\dfrac{-2+4t^2}{1-t^2+t^4}dt\\
&=\int_{0}^{+\infty}\dfrac{-2+4t^2}{1-t^2+t^4}dt
\end{align}$
Observe that,
$\displaystyle\int_{0}^{+\infty}\dfrac{t^2}{1-t^2+t^4}d... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Find the value of $√(4+6\sqrt{5}i) + √(4-6\sqrt{5}i)$
Find the value of $\sqrt{(4+6\sqrt{5}i)} + \sqrt{(4-6\sqrt{5}i)}$
$\sqrt{(4+6\sqrt{5}i)} = ± (3+\sqrt{5}i)$
$\sqrt{(4-6\sqrt{5}i)} = ± (3-\sqrt{5}i)$
There are two solutions to each which implies on adding them together, we will get four combinations and four dif... | I think the book means the following.
$$\left(\sqrt{4+6\sqrt5i}+\sqrt{4-6\sqrt5i}\right)^2=8+2\cdot14=36$$
and the book gets $\sqrt{4+6\sqrt5i}+\sqrt{4-6\sqrt5i}=6$.
This reasoning is wrong, of course.
| {
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Condition for three roots The equation $$2\sin ^3 x +(2\lambda -3)\sin ^2 x -(3\lambda +2)\sin x -2\lambda=0$$ has excatly three roots in $(0,2\pi)$ then what can be the value of $\lambda$ .
I thought for three roots the differentiation of the equation should have two roots . but from that I am not getting anya graph .... | The equation can be factored as follows:
$$
(\sin x - 2)(2\sin x + 1)(\sin x + \lambda) = 0
$$
For the lefthand side to be $0$, one of the factors needs to be $0$.
$\sin x - 2$ cannot be $0$.
$2\sin x + 1 = 0$ has two solutions ($\frac{7\pi}{6}$ and $\frac{5\pi}{3}$). The number of solutions of $\sin x + \lambda = 0$ d... | {
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"url": "https://math.stackexchange.com/questions/2304117",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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How to integrate $\int_a^b t \cdot \sqrt{\frac{t - a}{b - t}} \,dt$? How to integrate $\displaystyle \int_a^b t \cdot \sqrt{\frac{t - a}{b - t}} \,dt$?
I literally have no idea how to integrate this integral.
I've tried all basic methods, but it seems like a quite hard problem for a beginner.
| Do the following substitution
$$t=a\cos^2 \theta+b \sin^2 \theta \implies \mathrm dt=(b-a) \sin 2 \theta \, \mathrm d \theta$$
Your function $\displaystyle \sqrt{\frac{t - a}{b - t}}$ will be tremendously simplified to $\tan \theta$. Hence, your integral will be simplified :
\begin{align}
\int_a^b t \cdot \sqrt{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2304904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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How to find the area of the surface enclosed between a sphere $x^2+y^2+z^2=4a^2$ and a cylinder $x^2+(y-a)^2=a^2$?
Find the area of the surface enclosed between a sphere $x^2+y^2+z^2=4a^2$ and a cylinder $x^2+(y-a)^2=a^2$. The correct answer should be $(8\pi-16)a^2$.
This is an illustration:
First, there's the follo... | $$\frac{r^2}{4a^2-r^2}+1=\frac{r^2+4a^2-r^2}{4a^2-r^2}\implies r\sqrt{\frac{r^2}{4a^2-r^2}+1}=\frac r{\sqrt{4a^2-r^2}}\implies$$
$$\left.\int_0^{2a\sin\theta}\frac r{\sqrt{4a^2-r^2}}dr=-\sqrt{4a^2-r^2}\right|_0^{2a\sin\theta}=-\sqrt{4a^2-4a^2\sin^2\theta}+2a=-2a\cos\theta+2a$$
and etc. Yet I think the final answer is d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2305680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Predicting if a $3 \times 3$ matrix has a real square root Here are a few $3 \times 3$ matrices
A)$$
\begin{pmatrix} 1& 0& 0\\
0 & 1 & 0\\
0& 0 & 1\end{pmatrix}$$B)$$
\begin{pmatrix} 1& 0& 0\\
0 & 1 & 0\\
0& 0 & -1\end{pmatrix}$$C)$$
\begin{pmatrix} 1& 0& 0\\
... | A necessary condition for a real matrix to have a real square root is to have a non-negative determinant.
Indeed, if $A=B^2$, then $\det(A)=\det(B)^2\geqslant 0$.
Method. If $A$ is diagonalizable and has positive eigenvalue, a square root can be found taking the square root of its eigenvalues, namely if $A=P\Lambda P^{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How to calculate $\int{\frac {x^2}{ ( x\cos x -\sin x ) ^2}}\,{\rm d}x$? When I want to calculate
$$\int{\frac {x^2}{ ( x\cos x -\sin x ) ^2}}\,{\rm d}x$$
I have tested with software and get
$${\frac {x\sin \left( x \right) +\cos \left( x \right) }{x\cos \left( x
\right) -\sin \left( x \right) }}$$
But I can not come ... | We can tackle the integral by integration by parts as below:
$$
\begin{aligned}
& \int \frac{x^{2}}{(x \cos x-\sin x)^{2}} d x \\
=& \int \frac{x}{\sin x} d\left(\frac{1}{x \cos x-\sin x}\right) \\
=& \frac{x}{\sin x(x \cos x-\sin x)} -\int \frac{1}{x \cos x-\sin x} \cdot \frac{\sin x-x \cos x}{\sin ^2x}dx\\=& \frac{x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2307673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
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Calculus Spivak Chapter 2 problem 16(c) The question asks to prove that if $\frac mn \lt \sqrt{2}$, then there is another rational number $\frac {m'}{n'}$ with $\frac mn \lt \frac {m'}{n'} \lt \sqrt{2}$.
Intuitively, it's clear that such a number exists, but I don't understand the solution to this problem. It states: l... | Actually we have the following estimates:
\begin{gather*}
\frac{m}{n}<\frac{2(m+n)}{m+2n}<\sqrt{2},
\end{gather*}
provided $m,n$ are positive integers with $\frac{m}{n}<\sqrt{2}.$
Verification: Since $0<\frac{m}{n}<\sqrt{2},$ we have $\frac{n}{m}>\frac{1}{\sqrt{2}},$ and so
\begin{gather*}
\frac{2(m+n)}{m+2n}=\frac{m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2308909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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$1 + \frac{1}{3}\frac{1}{4} + \frac{1}{5}\frac{1}{4^2} + \frac{1}{7}\frac{1}{4^3} + ..........$ $1 + \frac{1}{3}\frac{1}{4} + \frac{1}{5}\frac{1}{4^2} + \frac{1}{7}\frac{1}{4^3} + ......$
Can anyone help me out how to solve this.
My try : I was thinking about the expansion of $tan^{-1}x$. But in that series positive ... | $$\sum_{r=0}^\infty\dfrac1{(2r+1)}\left(\dfrac12\right)^{2r}=2\sum_{r=0}^\infty\dfrac{x^{2r+1}}{(2r+1)}$$ where $2x=1$
Now $-\ln(1-x)=\sum_{n=0}^\infty\dfrac{x^n}n$ for $-1\le x<1$
$\ln(1+x)-\ln(1-x)=?$
Alternatively, $$\sum_{r=0}^\infty\dfrac{x^{2r+1}}{(2r+1)}=\int\sum_{r=0}^\infty(x^2)^rdx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2313339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Calculating sample size using Central Limit Theorem How many times do we have to throw a dice to achieve the situation, that the proportion of sixes to all numbers is between $\frac{9}{60}$ and $\frac{11}{60}$ with the probability $\leq\frac{1}{100}$?
This is my solution, that gives a wrong answer.
I know, that number ... | for a single die
$\mu = \frac 16\\
\sigma^2 = \frac {5}{36}$
the expected number of sixes in $n$ throws
$\mu = \frac 16n\\
\sigma^2 = \frac 5{36}n\\
\sigma = \sqrt{\frac{5n}{36}}$
The average number of sixes in $n$ throws
$\mu = \frac 16\\
\sigma = \frac 1n \sqrt{\frac{5n}{36}}\\
\sigma = \sqrt{\frac{5}{36n}}$
The expe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2313603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Taylor series expansion of $\ln(5 + x)$. If I start with the function $1/(5+x)$ I can find its Taylor series about 0 by using the familiar geometric series $1/(1 + x)$.
\begin{align}
\frac{1}{5 + x} &= \frac{1}{5 (1 + x/5)} \\
&= \frac15 \cdot \frac{1}{1 + (1/5) x} \\
&= \frac15 - \frac{x}{25} + \frac{x^2}{125} - \frac... | First the correct answer:
If you do the transformation without integration you get
$$ \ln(5 + x) = \ln \left(5 \left(1 + \frac15 x\right) \right) = \ln 5 + \ln \left(1 + \frac15 x\right). $$
Then using the series for $\ln(1 + x)$ you get
$$ \ln(5 + x) = \ln 5 + \sum_{n = 1}^\infty (-1)^{n + 1} \frac{x^n}{n \cdot 5^n}.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2317442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A question on the Euler prime of odd perfect numbers A number $N \in \mathbb{N}$ is said to be perfect if $\sigma(N)=2N$, where $\sigma=\sigma_{1}$ is the classical sum-of-divisors function. For example, $\sigma(6)=1+2+3+6=2\cdot{6}$, so that $6$ is perfect. (Note that $6$ is even.)
It is currently unknown whether th... | It appears that the numerators of the lower bounds form OEIS sequence A016789. Hence, the lower bounds have the closed form
$$\frac{3i + 2}{i + 1}.$$
Consequently,
$$\frac{3i + 2}{i + 1}=3-\frac{1}{i + 1},$$
so that the lower bounds never reach $3$.
| {
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"url": "https://math.stackexchange.com/questions/2318266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $A$ and $B$, find matrix $C$ such that $CA=B$ Hoffman and Kunze, Linear Algebra, ch. 1
$A=\left[\begin{matrix} 1 & -1 \\ 2 & 2 \\ 1 & 0 \end{matrix}\right], B=\left[\begin{matrix} 3 & 1 \\ -4 & 4 \end{matrix}\right]$
Is there a matrix $C$ such that $CA=B$?
I assumed $C=\left[\begin{matrix} a & b & c\\ d & e & f ... | We have
$$
B = C A
$$
where $C$ consists of two unknown row vectors. I am more used to having the unknows on the right hand side ($Ax=b$) so I would, like Siong Thye Goh did, look at the transposed equations:
$$
B^T = (b_1^T, b_2^T)= A^T C^T = A^T (c_1^T, c_2^T)
$$
which can be interpreted as two linear systems
$$
A^T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2321930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integrating rational functions. $\int\frac{x^2-3x+1}{x^4-x^2+1}\mathrm{d}x$ Is there really another way of solving this problem $$\int\frac{x^2-3x+1}{x^4-x^2+1}\mathrm{d}x$$
avoiding partial fractions decomposition? I have tried the partial fractions decomposition, but it is not serving me right.
| We have that $$\int\frac{x^2-3x+1}{(x^2-\sqrt{3}x+1)(x^2+\sqrt{3}x+1)}dx=\\=\int\frac{x^2-\sqrt{3}x+1}{(x^2-\sqrt3{x}+1)(x^2+\sqrt3{x}+1)}-\int\frac{(3-\sqrt{3})x}{x^4-x^2+1}=\\\int\frac{dx}{x^2+\sqrt{3}x+1}-\frac{3-\sqrt{3}}{2}\int\frac{2x}{x^4-x^2+1}dx$$
Now putting $u=x^2$ for the second integral
$$\int\frac{dx}{(x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2322455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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To find the last digit of the number $4^{7^5}$? To find the last digit of the number $4^{7^5}$, I use the following method.
first, we know $4\equiv_{10}4,4^2\equiv_{10}6,4^3\equiv_{10}4,...,4^{2n}\equiv_{10}6,4^{2n+1}\equiv_{10}4$
Then$7^1\equiv_21,7^2\equiv_21,7^3\equiv_21,...,7^5\equiv_21$
Therefore, $4^{7^5}\equiv4... | $$4^{7^5}=4\cdot\left(4^2\right)^{\frac{7^5-1}{2}}\equiv4\cdot6(\mod10)\equiv4(\mod10)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Integer solutions for $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}+\frac{1}{xyz}=1$ Is there a beautiful solution for this equation over the integers?
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}+\frac{1}{xyz}=1$$
| $$\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right)=1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xy}+\frac{1}{xz}+\frac{1}{yz}+\frac{1}{xyz}\color{red}{=1+1=2}$$
Rewriting $1+\frac{1}{x}=\frac{x+1}{x}$ and mutliplying both sides by $xyz$ (assumed nonzero), we get $$(x+1)(y+1)(z+1)=2xyz$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Find $P$ such that $P^TAP = D$ where $D$ is a diagonal matrix.
Let $$A = \left(\begin{array}{cc} 2&3 \\ 3&4 \end{array}\right) \in
M_n(\mathbb{C})$$
Find $P$ such that $P^TAP = D$ where $D$ is a diagonal matrix.
How can I find $P$? I am doing Gauss but it does not work?$$A = \left(\begin{array}{cc|cc} 2&3&1&0\\ 3&4&... | You need to perform simulatenous row and column operations on the left hand side while performing only column operations on the right hand side. Then, when the left side becomes diagonal, the right side will be your $P$. In your case,
$$ \left(\begin{array}{cc|cc} 2&3&1&0\\ 3&4&0&1 \end{array}\right) \xrightarrow[C_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is Matrix Multiplication Commutative? I am doing some self-study in math. Below is a problem that I did but I am not sure if I have the right answer. I am hoping that somebody could check it for me.
Thanks
Bob
Problem:
If $A$ and $B$ are two by two matrices, does it imply that $A * B = B * A$?
Answer:
\begin{eqnarray*}... | Your approach is correct, but there is an easier way:
You can prove that something is not true by giving a counterexample (the claim is that matrix multiplication is not commutative, so it is sufficient to find at least one example where commutativity fails). So, a much easier approach is:
$$\begin{pmatrix} 1 &0\\0&0\e... | {
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"url": "https://math.stackexchange.com/questions/2326629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate the sum of the series $\sum_{1\leq aCalculate the sum of the series:
$$S = \sum_{1\leq a<b<c \\a,b,c\in \mathbb{N}} \frac{1}{2^a 3^b 5^c}$$
My attempt:
$$S = \sum_{1\leq a<b<c \\a,b,c\in \mathbb{N}} \frac{1}{2^a 3^b 5^c} = \sum _{c=3}^{\infty } \sum _{b=2}^{c-1} \sum _{a=1}^{b-1} \frac{1}{2^a 3^b 5^c}$$
Is it... | \begin{eqnarray*}
\sum_{1 \leq a < b < c} \frac{1}{2^a 3^b 5^c} = \sum_{a=1}^{\infty} \frac{1}{2^a} \sum_{ b=a+1}^{\infty} \frac{1}{3^b} \sum_{c=b+1}^{\infty} \frac{1}{5^c}
\end{eqnarray*}
\begin{eqnarray*}
= \sum_{a=1}^{\infty} \frac{1}{2^a} \sum_{ b=a+1}^{\infty} \frac{1}{3^b} \frac{1}{5^b \times 4}
\end{eqnarray*}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2327827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
What would the solution be if I cannot simplify my solution to find more basic variables(pivots)? 4x3 matrix $$x - 5y + 4z = -3$$
$$2x - 7y + 3z = -2$$
$$-2x + y + 7z = -1$$
The furthest I was able to reach before coming to the conclusion that I could not move any further was:
Row 1 => $x - 5y + 4z = -3$
Row 2 => $9y ... | Here is the reduction of the augmented matrix.
Column 1
$$
\left[
\begin{array}{rcc}
1 & 0 & 0 \\
-2 & 1 & 0 \\
2 & 0 & 1 \\
\end{array}
\right]
%
\left[
\begin{array}{rrc|ccc}
1 & -5 & 4 & 1 & 0 & 0 \\
2 & -7 & 3 & 0 & 1 & 0 \\
-2 & 1 & 7 & 0 & 0 & 1 \\
\end{array}
\right]
%
=
%
\left[
\begin{array}{crr|rcc}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2327923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Let $f(x)$ be a quadratic expression which is positive for all real values of $x.$ If $g(x)=f(x)+f'(x)+f''(x)$
Let $f(x)$ be a quadratic expression which is positive for all real values of $x$. If $g(x)=f(x)+f'(x)+f''(x)$, then for any real $x$,
$$(A)\, g(x)<0,\quad (B)\,g(x)>0,\quad (C)\, g(x)=0,\quad (D)\;g(x)\ge ... | $f(x)=ax^2+bx+c$. $f>0 \Rightarrow a>0$.
$f(x)+f^\prime(x)+f^{\prime\prime}(x)=ax^2+(2a+b)x+c+b+2a=g(x)$
Reformulating $g$ in terms of $x+1$ gives
$$a(x+1)^2+ (2a+b)(x+1)+c+b+2a-(2ax+a)-(2a+b) = $$
$$a(x+1)^2+ (2a+b)(x+1)+c-2a(x+1)+a = $$
$$a(x+1)^2+b(x+1)+c+a=g(x)$$.
So $g(x) = f(x+1)+a$, so $g(x)$ is $f(x)$ translat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $x^3 - 5x^2+ x=0$ then find the value of $\sqrt {x} + \dfrac {1}{\sqrt {x}}$ If $x^3 - 5x^2+ x=0$ then find the value of $\sqrt {x} + \dfrac {1}{\sqrt {x}}$
My Attempt:
$$x^3 - 5x^2 + x=0$$
$$x(x^2 - 5x + 1)=0$$
Either,
$x=0$
And,
$$x^2-5x+1=0$$
??
| For a more brut'ish force alternative, note that the problem assumes $x \ne 0$ for $\sqrt {x} + 1 / \sqrt {x}$ to be defined, so that $x$ must be a root of $x^2-5x+1=0 \iff x = \frac{1}{2}\left(5 \pm \sqrt{21}\right)\,$.
By known radical denesting techniques, $\sqrt{5 \pm \sqrt{21}}=\frac{1}{2}\left(\sqrt{14} \pm \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Polar Coordinates (Area enclosed)
Find the common area enclosed by the curves
$$r= 3 - 2 \cos \theta$$
$$r= 2$$
My attempt,
Area$$=4\pi-\int_{\frac{\pi}{3}}^{\frac{5\pi}{3}}\frac{1}{2}(4-(3-2\cos \theta)^2)d\theta$$
$$=36.753$$
Am I correct?
|
Area:
$$\begin{align}
2\left(\int_0^{\pi/3} \frac 12\big(\overbrace{3-2\cos\theta}^{\color{red}{r_1}}\big)^2 d\theta +\int_{\pi/3}^{\pi}\frac 12\cdot (\overbrace{\;\;2\;\;}^{\color{blue}{r_2}})^2 d\theta\right)
&=2\left(\frac {11}2\bigg(\frac {\pi}3-\frac{\sqrt{3}}2\bigg)+ \frac {4\pi}3\right)\\
&=11\bigg(\frac {\pi}3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\frac{1}{2\pi i}\int_\mathcal{C} |1+z+z^2+\cdots+z^{2n}|^2~dz =2n$ where $\mathcal{C}$ is the unit circle On the generalization of a recent question, I have shown, by analytic and numerical means, that
$$\frac{1}{2\pi i}\int_\mathcal{C} |1+z+z^2+\cdots+z^{2n}|^2~dz =2n$$
where $\mathcal{C}$ is the unit circ... | You can simplify things as follows. We have:
$$f(z) = \sum_{k=0}^{2n}z^k = \frac{z^{2n+1}-1}{z-1}$$
Then on the unit circle, we have:
$$\left|f(z)\right|^2 = f(z)f^*(z) = f(z)f(z^*) = f(z)f\left(z^{-1}\right) = \frac{\left(z^{2n+1}-1\right)^2}{z^{2n}(z-1)^2}$$
The integral over the unit circle is then given by the co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
$\sin(2\arcsin(\frac{1}{3}))=? $ A broad one I have written it as $$\sin(2\arcsin(\frac{1}{3}))=2\sin(\arcsin(\frac{1}{3}))\cos(\arcsin(\frac{1}{3})).$$Then, solved for $$\cos(\arcsin(\frac{1}{3}))=\frac{\sqrt{10}}{3}$$ $$\arcsin(\frac{1}{3})=\theta$$$$\sin\theta=\frac{1}{3}\Rightarrow\cos\theta=\frac{\sqrt{10}}{3}.$$B... | Your mistake is that $\cos\theta=\dfrac{2\sqrt{2}}{3}$, not $\dfrac{\sqrt{10}}{3}$.
\begin{eqnarray}
\cos\left(2\arcsin\left(\frac{1}{3}\right)\right)&=&2\sin\left(\arcsin\left(\frac{1}{3}\right)\right)\cos\left(\arcsin\left(\frac{1}{3}\right)\right)\\
&=&2\left(\frac{1}{3}\right)\cdot\frac{2\sqrt{2}}{3}\\
&=&\frac{4\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2334116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Changing the order of double integration The function is $\int_0^8 \int_{y^{1/3}}^2 \sqrt{x^4 + 1}dxdy$ and the prompt is to change the order of integral and evaluate it.
Plotting the graphs for the limits $$y^{\frac{1}{3}} < x < 2$$
$$0 < y < 8$$
Using the graph we can identify, changing the order of the limits will ... | Yes, you changed your order of integration correctly. Good Job! However, your final result is wrong because you integrated incorrectly in the last steps: $$\int_{x=0}^{x=2}x^3\sqrt{x^4+1}dx=\frac{1}{4}\int_{{\color{red}{t=1}}}^{t=17}\sqrt{t}dt=\frac{1}{6}\left(17\sqrt{17}-1\right)$$
where you mistakenly treated the low... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2334216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Significance of the Triangle Inequality After working through a few problems regarding the Cauchy Integral Formula, I'm still a little confused on the significance of the triangle inequality. Why do we use it and what information does it tell us?
See the following example below.
Determine $\int_{0}^{\infty} \frac{x^... | The reason you're confused is that you should be using the reverse triangle inequality here. What we should have is the following; $$|z^2+1|^2 \ge ||z|^2-1|^2 = (|z|^2-1)^2$$
Where the last equality is because we will consider the semicircular contour letting $R=|z| \to \infty$ (so certainly bigger than $1$).
Then w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
(Dice), population mean and variance of the odd numbers.
A fair die is rolled 100 times. Find the population mean and variance of the sum of the odd numbers that appear.
I am aware that the expected value for discrete multinomial distributions is $\Sigma \Sigma ... g(x_1 , x_2,...)f(x_1 ,x_2...)$. In this case, to f... | For the mean, we find:
$$E[100X] = 100 E[X] = 100 \bigg(\frac{1}{6} \cdot 1 + \frac{1}{6} \cdot 3 + \frac{1}{6} \cdot 5 + \frac{3}{6} \cdot 0\bigg) = 100 \cdot \frac{3}{2} = 150$$
For the variance, we find:
$$Var[100X] = 100 Var[X] = 100 \cdot \frac{\big(1-\frac{3}{2}\big)^2 + \big(3-\frac{3}{2}\big)^2 + \big(5-\frac{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2337319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the series of $f(x) = x^3\arctan(x^3)$ around $x=0$ Am I correct?
$$x^3\arctan(x^3)$$
$$\frac{d}{dx}\arctan(x^3) = \frac{3x^2}{x^6+1} = \sum_{n=0}^{\infty}3x^2(x^6)^n$$
$$x^3\sum_{n=0}^{\infty}3x^2(x^6)^n$$
$$f(x) = x^3\arctan(x^3) = 3x^5\sum_{n=0}^{\infty}(x^6)^n$$
| For $X $ such that $X^2 <1$,
$$\frac {1}{1+X^2}=\sum_{n=0}^{+\infty}(-1)^nX^{2n} $$
and by integration
$$\arctan (X)=\arctan (0)+\sum_{n=0}^{+\infty}\frac {(-1)^nX^{2n+1}}{2n+1} $$
replace $X $ by $x^3$ and you can finish and get
$$f (x)=\sum_{n=0}^{+\infty}\frac {(-1)^nx^{6n+6}}{2n+1} $$
for $x $ such that $|x|<1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2337568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving trigonometric identity $\cos^6A+\sin^6A=1-3 \sin^2 A\cos^2A$ Show that
$$\cos^6A+\sin^6A=1-3 \sin^2 A\cos^2A$$
Starting from the left hand side (LHS)
\begin{align}
\text{LHS} &=(\cos^2A)^3+(\sin^2A)^3 \\
&=(\cos^2A+\sin^2A)(\cos^4A-\cos^2A\sin^2A+\sin^4A)\\
&=\cos^4A-\cos^2A\sin^2A+\sin^4A
\end{align}
Can a... | HINT:
$$a^3+b^3=(a+b)(a^2-ab+b^2)$$ can also be written as
$$a^3+b^3=(a+b)^3-3ab(a+b)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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If the point of minima of the function ,$f(x)=1+a^2x-x^3$ satisfy the inequality $\frac{x^2+x+1}{x^2+5x+6}<0$, If the point of minima of the function ,$f(x)=1+a^2x-x^3$ satisfy the inequality $\frac{x^2+x+1}{x^2+5x+6}<0$,then $a$ must lie in the interval:
$(A)(-3\sqrt3,3\sqrt3)$
$(B)(-2\sqrt3,-3\sqrt3)$
$(C)(2\sqrt3,3\... | Finding the $x$-value of the minimum point, you have assumed without warrant that $a>0$. We know that the minimum occurs at $x=\pm\frac{a}{\sqrt{3}}$, but we still don't know if $a$ is positive or negative.
Then, you find out that $f''(x)=-6x$, which must be positive at the minimum value. That means we have two cases f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve three equations with three unknowns Solve the system:
$$\begin{cases}a+b+c=6\\ab+ac+bc=11\\abc=6\end{cases}$$
The solution is:
$a=1,b=2,c=3$
How can I solve it?
| As an alternative to the other derivations, here's my approach.
In the second equation, move everything to the left side, factor $ac+bc$ into $(a+b)c$, multiply everything by $c$, replace $a+b$ with $6-c $ from the first equation, and $abc$ with $6$ from the third. Then you have $6-11c+(6-c)c^2 = 0$, which simplifies t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
Proving inequality $\ 1+\frac14+\frac19+\cdots+\frac1{n^2}\le 2-\frac1n$ using induction Question:
Prove $$\ 1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{n^2}\le 2-\frac{1}{n},
\text{ for all natural } n$$
My attempt:
Base Case: $n=1$ is true:
I.H: Suppose $1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{k^2}\le 2-\frac{1}{k}... | You may use creative telescoping for proving a much tighter inequality.
If we set $H_n^{(2)}=\sum_{k=1}^{n}\frac{1}{k^2}$, for any $n\geq 3$ we have:
$$\begin{eqnarray*} H_n^{(2)} &=& \sum_{k=1}^{n}\frac{1}{k(k+1)}+\sum_{k=1}^{n}\frac{1}{k^2(k+1)}\\&=&\left(1-\frac{1}{n+1}\right)+\frac{1}{2}+\sum_{k=1}^{n-1}\frac{1}{(k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2340164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
The sequence $a_1 = \frac{1}{2}, a_2 = 1, a_{n+1} = \frac{na_n+1}{a_{n-1}+n}$ is decreasing
Consider the sequence $\{a_n\}$ defined by $$a_1 = \frac{1}{2}, a_2 = 1, a_{n+1} = \frac{na_n+1}{a_{n-1}+n}, \forall n\ge 2.$$
Prove that $\{a_n\}_{n\ge 3}$ is decreasing.
I get the first $200$ values of $\{a_n\}$ and recogn... | From comments: it would appear that part III is not relevant for the original question. Plus, if we drop part III, the remaining induction hypotheses take effect with smaller $n.$ On the other hand, finding III is what allowed me to solve the problem; together, parts III and IV give a quite precise rate of convergence.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2340962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
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How to find this simplification integral involving products of roots? Consider the following integral
$$
f = \int_0^1 \frac{1}{\sqrt{-\frac{1}{2} \, t^{2} + 1} \sqrt{-t^{2} + 1}} \mathrm \,dt.
$$
If we change variable by letting $x^2=t^2/(2-t^2)$, then we have
$$
f = \int_0^1 \sqrt{2} \cdot\sqrt{\frac{1}{1-x^{4}}} \mat... | $\int\frac{dt}{\sqrt{1-t^2}}=\arcsin(t)$, $\int\frac{dt}{\sqrt{1-\frac{t^2}{2}}}=\sqrt{2}\arcsin\left(\frac{t}{\sqrt{2}}\right)$, hence two candidate substitutions for simplifying things are $t=\sin\theta$ and $t=\sqrt{2}\sin\frac{\theta}{\sqrt{2}}$. Let us try the first one:
$$ I = \int_{0}^{\pi/2}\frac{d\theta}{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Proving right angle triangle A triangle $A$$B$$C$ is integer sided and has inradius $1$.prove that it is right angled. Please help.give some hints please.
| Let the segments of tangents from the vertices to the incircle be $x,y,z$ (what is popularly known as the Ravi-substitution.
We have $(x+y), (y+z), (z+x)$ are all integers and so is $2(x+y+z)$ so $x+y+z = \frac{n}{2}, n \in \mathbb{N}$. Now its easy to see that $x,y,z$ are in the form $\frac{a}{2}, \frac{b}{2}, \frac{c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2343094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Compute $\mathbb{E}[\text{max}(x,0) \text{max}(y,0)]$ where $(x,y)$ is jointly gaussian with given covariance and nonzero mean
Is there a closed form expression for the expectation of $g(x,y) = \text{max}(x,0) \text{max}(y,0)$ where $(x,y)$ is jointly gaussian with the bivariate gaussian distribution, when the mean is... | The calculation in question is pretty straightforward and can be done using elementary methods. Assume for simplicity that the variables have variance one and mean zero. Then the joint pdf reads:
\begin{eqnarray}
\rho(x,y) = \frac{1}{2\pi \sqrt{1-\rho^2}} \exp\left[ -\frac{1}{2} \frac{1}{(1-\rho^2)} (x^2+y^2-2 \rho x y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2344676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
A limit involving $\sum_{k=1}^{n}\frac{2^k}{k}$ I would like to show that
$\lim_{n \to \infty} \frac{n}{2^n} \sum_{k=1}^n \frac{2^k}{k} = 2$
I have seen proofs that $\sum_{k=1}^n \frac{2^k}{k} \sim \frac{2^{n+1}}{n}$ using the Euler-Maclaurin method but instead I tried with the squeeze theorem.
For one side I showed th... | Note that
$$\frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k} = \frac{n}{2^n}\sum_{k=0}^{n-1} \frac{2^{n-k}}{n-k} = \sum_{k=0}^{n-1} \frac{1}{2^k} + \sum_{k=1}^{n-1} \frac{k}{2^k(n-k)}.$$
The limit of the first sum on the RHS is $\sum_{k=0}^\infty 2^{-k} = 2$ and we can show that the limit of the second sum is $0$ giving the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2345610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
convergence and divergence of $\iiint\limits_{x^2+y^2+z^2\geq 1}(x^2+y^2+z^2)^\alpha \ln(x^2+y^2+z^2)\,dx\,dy\,dz$ I want to study the divergence or convergence of this integral :
$$\iiint\limits_{x^2+y^2+z^2\geq 1}(x^2+y^2+z^2)^\alpha \ln(x^2+y^2+z^2)\,dx\,dy\,dz$$
I think that if $\alpha\geq-3$, then in the domain: ... | The surface area of $x^2+y^2+z^2=\rho^2$ is given by $4\pi\rho^2$, hence the given integral equals
$$ \int_{1}^{+\infty} 4\pi\rho^2 \cdot \rho^{2\alpha}\cdot 2\log\rho\,d\rho $$
i.e. $\frac{8\pi}{(2\alpha+3)^2}$ as soon as $\alpha<-\frac{3}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
MCQ The nth derivative of $f(x)=\frac{1+x}{1-x}$ Let $f(x)=\dfrac{1+x}{1-x}$ The nth derivative of f is equal to:
*
*$\dfrac{2n}{(1-x)^{n+1}} $
*$\dfrac{2(n!)}{(1-x)^{2n}} $
*$\dfrac{2(n!)}{(1-x)^{n+1}} $
by Leibniz formula
$$ {\displaystyle \left( \dfrac{1+x}{1-x}\right)^{(n)}=\sum _{k=0}^{n}{\binom {n}{k}}\ (... | $$y=\dfrac{1+x}{1-x}=\dfrac{2-(1-x)}{1-x}=\dfrac2{1-x}-1$$
$$\implies\dfrac{dy}{dx}=\dfrac{2(-1)}{(1-x)^2}$$
$$\dfrac{d^2y}{dx^2}=\dfrac{2(-1)(-2)}{(1-x)^3}=\dfrac{2(-1)^22!}{(1-x)^{2+1}}$$
Can you follow the pattern?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2353190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Common complex roots
If the equations $ax^2+bx+c=0$ and $x^3+3x^2+3x+2=0$ have two common roots then show that $a=b=c$.
My attempts:
Observing $-2$ is a root of $x^3+3x^2+3x+2=0\implies x^3+3x^2+3x+2=(x+2)(x^2+x+1)=0$
Hence $ax^2+bx+c=0$ can have complex roots in common, comming from $(x^2+x+1)=0$
Both the roots of $... | Yes, I just check your calculus and are all correct. Your polynomial of grade $3$ defined as $x^3+3x^2+3x+2=0$ has one real root $x_{1}=-2$ and two complex roots $x_{2,3}=-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$ so it can be written as $P(x)=(x+2)(x-(\frac{1}{2}-\frac{\sqrt{3}}{2}i))(x-(\frac{1}{2}+\frac{\sqrt{3}}{2}i))$. T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2353367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
If $a + \frac{1}{a} = -1$, then the value of $(1-a+a^2)(1+a-a^2)$ is?
If $a + \frac{1}{a} = -1$ then the value of $(1-a+a^2)(1+a-a^2)$ is?
Ans. 4
What I have tried:
\begin{align}
a + \frac{1}{a} &= -1 \\
\implies a^2 + 1 &= -a \tag 1 \\
\end{align}
which means
\begin{align}
(1-a+a^2)(1+a-a^2) &=(-2a)(-2a^2) \\
&=4a^... | Just solve the quadratic $(1)$ to get $a$, then substitute $a$ back in the original expression or $4a^3$.
Also, since $a^3-1=(a-1)(a^2+a+1),$ $a^3=1$.
In more detail, from $(1)$,
$$\begin{align}a&=\frac{-1\pm\sqrt{1-4}}{2} \\
&=\frac{-1\pm i\sqrt{3}}{2}.\end{align}$$
Now
$$\begin{align}
a^2&=\frac{-1\pm i\sqrt{3}}{2}\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2353620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If a is a non real root of $x^7 = 1$, find the equation whose roots are $ a + a^6 , a^2 + a^5, a^3 + a^4$ If a is a non real root of $ x^7 = 1$, find the equation whose roots are $a + a^6 , a^2 + a^5, a^3 + a^4$. This is one of the questions I have encountered while preparing for pre rmo. I feel the question requires t... | WLOG let $a=e^{\frac{2i\pi}{7}}$. Then expand
$$(x-(a+a^6))(x-(a^2+a^5))(x-(a^3+a^4))$$
using the facts that $a^7=1$ and $a^6+a^5+a^4+a^3+a^2+a+1=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2353710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $ a=\frac{\sqrt{x+2} + \sqrt {x-2}}{\sqrt{x+2} -\sqrt{x-2}}$ then what is the value of $a^2-ax$ is equal to
If $$ a=\frac{\sqrt{x+2} + \sqrt {x-2}}{\sqrt{x+2} -\sqrt{x-2}}$$ then
the value of $a^2-ax$ is equal to:
a)2 b)1 c)0 d)-1
Ans. (d)
My attempt:
Rationalizing $a$ we get,
$ x+ \sqrt {x^2-4}$
$a^2=(x+\sqrt{... | Rationalizing $a$, you should get $$\frac{x+\sqrt{x^2-4}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2356126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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$(xyz)_b$ is divisible by $n$ if and only if $z+3y-4x$ is divisible by $n$
Determine all natural numbers $n > 1$ with the following property: there exists a base $b \geq 5$ such that any three digit-number $(xyz)_b$ is divisible by $n$ if and only if $z+3y-4x$ is divisible by $n$.
Suppose that $b$ is such a base for ... | Immediately, we only have
$$n\mid xb^2+yb+z\iff n\mid z+3y-4x\qquad \text{for }0\le x,y,z\le b-1$$
and not necessarily a congruence in general.
Nevertheless, $1+3\cdot 1-4\cdot1=0$ implies that $n\mid 111_b=b^2+b+1$.
By a similar argument, $n\mid 104_b=b^2+4$ (note that $4$ is a valid digit), hence also $n\mid b-3$.
T... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Can anybody solve the following using binomial theorem? If $x,y \in\mathbb{R}$ such that $x^2 + y^2 - 6x + 8y + 24 = 0$ then what is the greatest value of $$\frac{16\cos^2(\sqrt{x^2+y^2})}{5} - \frac{24\sin(\sqrt{x^2+y^2})}{5}?$$
| Hint. Note that $x^2 + y^2 - 6x + 8y + 24 = 0$ is equivalent to
$$(x-3)^2 + (y-4)^2 = 1$$
that is the circle $C$ centered in $(3,4)$ of radius $1$. Hence for any point $(x,y)$ on this circle its distance from the origin, that is $\sqrt{x^2+y^2}$, attains by continuity all the values in the interval $[\sqrt{3^2+4^2}-1,\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to integrate: $\int_0^{\infty} \frac{1}{x^3+x^2+x+1}dx$ I want to evaluate $\int_0^{\infty} \frac{1}{x^3+x^2+x+1}$. The lecture only provided me with a formula for $\int_{-\infty}^{\infty}dx \frac{A}{B}$ where $A,B$ are polynomials and $B$ does not have real zeros. Unfortunately, in the given case $B$ has a zero at... | Yes there is , integrate the following function
$$f(z) = \frac{\log(z)}{z^3+z^2+z+1}$$
Around a key-hole contour. Due to the evaluation of the lines on the positive x-axis the $\log$ will cancel. What is remaining is the evaluation of the residues of the three poles.
More explanation
we have three poles $\pm i , -1$. I... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Middle binomial coefficient size How can we prove that $$\lim_{n\rightarrow\infty}\frac{\dbinom{2n}{n}^2}{\dbinom{4n}{2n}}=0?$$
From plotting this for some values of $n$ it is clear, but using an approximation like $\binom{n}{k}\approx \left(\frac{ne}{k}\right)^k$ gives a value of $1$, since both numerator and denomina... | $$ \binom{2n}{n}= \frac{2^n (2n-1)!!}{n!} = 4^n \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)\tag{1}$$
hence:
$$\begin{eqnarray*} \binom{2n}{n}^2\binom{4n}{2n}^{-1}&=&\prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)^2 \prod_{k=1}^{2n}\left(1-\frac{1}{2k}\right)^{-1}\\&=&\frac{1}{2n}\prod_{k=2}^{n}\left(1-\frac{1}{(2k-1)^2}\righ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Confused by Proof by induction Im just confused by this equation and how it was used.
Assume
$$P(n) = 1 + 2 + 3 + 4 + \cdots + n = \frac{n(n+1)}2$$
Inductive step
$$P(n+1) = 1 + 2 + 3 + 4 + \cdots + n + (n+1) = \frac{(n+1)(n+2)}2$$
Why is the inductive step not written like this?
$$P(n+1) = 1 + 2 + 3 + 4 + \cdots + (n+... | You assume that $P(n)$ is true and prove that $P(n+1)$ is also true from that.
So assume $1+2+3+\cdots+n=\frac {n(n+1)}2$.
Then from this we get $1+2+3+\cdots+n+(n+1)=\frac {n(n+1)}2 + (n+1)=(n+1)(\frac n2+1)=\frac {(n+1)(n+2)}2.$(Which the answer has avoided to mention here)
This is exactly what is done in that step y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$
Evaluate
$$\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$$
I assumed $x=\frac{1}{y}$ we get
$$L=\lim_{y \to 0}\frac{\left( \tan^{-1}\left(\frac{1+y}{1+4y}\right)-\frac{\pi}{4... | Using the formula for the tangent of a sum,
$$
\tan\left(x-\frac\pi4\right)=\frac{\tan(x)-1}{\tan(x)+1}
$$
we get
$$
\tan^{-1}(x)-\frac\pi4=\tan^{-1}\left(\frac{x-1}{x+1}\right)
$$
Thus,
$$
\begin{align}
\lim_{x\to\infty}\left(\tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac\pi4\right)x
&=\lim_{x\to\infty}\tan^{-1}\left(\fr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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To find number of ways of solving $x + y + z = 12$, with $0 \leq x, y, z \leq 6$, using generating functions I am learning generating functions so I tried to solve the below question using generating functions.
Number of ways in which value of three variables add up to 12.
$x + y + z = 12$ and $0 \leq x,y,z \leq 6$.
T... | You can write $\displaystyle\left(\frac{1-x^7}{1-x}\right)^3=(1-x^7)^3(1-x)^{-3}=(1-3x^7+3x^{14}-x^{21})\sum_{n=0}^{\infty}\binom{n+2}{2}x^n,$
so the coefficient of $x^{12}$ is given by $\displaystyle\binom{14}{2}-3\binom{7}{2}=28.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2366390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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integral $\int_{}^{}\frac{dx}{1+x^4+x^8} $ looking for help for the following integral -
$$
\int_{}^{}\frac{dx}{1+x^4+x^8}
$$
what I tried to do:
$$\int_{}^{}\frac{dx}{1+x^4+x^8} = \int_{}^{}\frac{dx}{\frac14+x^4+x^8 + \frac{3}{4}}= \int_{}^{}\frac{dx}{\left(x^4+\frac{1}{2}\right)^2 + \frac{3}{4}} $$
and now I am stu... | I want to present another method such that we calculate 2 partial fractions incited of 4 partial fractions. By substitution $\frac{1}{x}=t$ we have $dx=-\frac{1}{t^2}dt$ and we obtain the form $J$ that is equal the form $I$ as follows:
$$
\begin{array}{l}
I=\int\,\frac{dx}{1+x^4+x^8} \\
\\
J=\int\, \frac{-\frac{1}{t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2368798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the number of five digit numbers of the form $d_1d_2d_3d_4d_5$ and satisfying $d_1Find the number of five digit numbers of the form $d_1d_2d_3d_4d_5$ and satisfying $d_1<d_2\le d_3<d_4\le d_5.$
I considered $d_1,d_2,d_3,d_4,d_5$ as 5 baskets and these have to be filled with 0 to 9 apples.I found total ways as $\b... | Method 1: The leading digit must be at least $1$, so $d_1, d_2, d_3, d_4, d_5 \in \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.
There are three possibilities:
*
*All the digits are different: There are $\binom{9}{5}$ such cases since selecting five of the nine digits completely determines the number since $d_1 < d_2 < d_3 < d_4... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Horizontal line(s) that intersect $f(x)=x-2+\frac{5}{x}$ in two points. Compute exactly the value(s) of $q$ for which the horizontal line $y = q$
intersects the graph of $f(x)$ in two points that are located on a distance $4$ from each other.
While I found the two lines as $y = 4$ and $y = -8$
, I do not know how to fi... | We want to values of $x$ (of distance $4$ apart) such that $x-2 + \frac{5}{x} = q$, but this is equivalent to $x^2 - 2x + 5 = qx$ by multiplying throughout by $x$, more simply: $x^2 - (2+q)x + 5 = 0$.
Now say that the smaller point is $\alpha$, then the other one is necessarily $\alpha + 4$. Hence $2\alpha + 4= 2 +q$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find $k$ such that $(\alpha-1)^3+(\beta-2)^3+(\gamma-3)^3=0$ holds for the roots of $x^3-6x^2+kx+k=0$
If $\alpha$, $\beta$ and $\gamma$ are the three roots of the equation $x^3-6x^2+kx+k=0$, find the values of $k$ such that $(\alpha-1)^3+(\beta-2)^3+(\gamma-3)^3=0$. For each of the possible values of $k$ solve the equ... | Since $\alpha-1+\beta-2+\gamma-3=0$ and $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc),$$
we see that $$(\alpha-1)(\beta-2)(\gamma-3)=0$$ and we have three following cases:
*
*$\alpha=1$, which gives $k=\frac{5}{2}$ and our equation is
$$2x^3-12x^2+5x+5=0$$ or
$$2x^3-2x^2-10x^2+10x-5x+5=0$$ or
$$(x-1)(2x^2-10x-5)=... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I rotate 3 groups of 4 people into teams of 3 so that each person in one group works with each person in the other groups? I have three teams of four people. I would like to create rotating groups of three, where each group has one person from each team, and those people rotate on a staggered schedule so that ea... | \begin{eqnarray*}
\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
a&b&c&d& &e&f&g&h&& i&j&k&l\\ \hline
\color{red}{1}&0&0&0& &\color{red}{1}&0&0&0& &\color{red}{1}&0&0&0 \\ \hline
0&\color{red}{1}&0&0& &0&\color{red}{1}&0&0& &0&\color{red}{1}&0&0 \\ \hline
0&0&\color{red}{1}&0& &0&0&\color{red}{1}&0& &0&0&\color{red}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2371562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How can I solve $y''=x^{-1/3}y^{3/2}$? I am trying to solve the differential equation.
$$y''=x^{-1/3}y^{3/2}$$
I tried the solution: $y=Ax^s$
$$\implies y'=Asx^{s-1} \\ \iff y''=As(s-1)x^{s-2}$$
$$\implies As(s-1)x^{s-2}= x^{-1/3}(Ax^s)^{3/2} \\ \iff As(s-1)x^{s-2}=x^{-1/3}A^{3/2}x^{\frac{3s}{2}} \\ \iff A^{5/2}s(s-1)... | This is in the form of the Emden-Fowler equation, $y'' = Ax^ny^m$. For $m\neq1$, the solution is
\begin{equation*}
y = Cx^{\frac{n+2}{1-m}}
\end{equation*}
for a constant $C$ that depends on $A$, $n$, and $m$. In your case, the solution is $y = Cx^{\frac{5/3}{-1/2}} = Cx^{-10/3}$. You would have gotten this constant al... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Common root between a cubic and a quadratic The equation $x^2-ax+b=0$ & $x^3-px^2+qx=0$, have one common root & the second equation has two equal roots. Prove that $2(q+b) = ap$
($b\neq0 q\neq0$)
I am not able to think of any way to solve this any hint would be appreciated.
| Counterexample
Let $a = 1, b = 0, p = 4, q = 4$. Then, $x^2 - ax + b = x(x-1)$ and $x^3 - 4x^2 + 4x = x(x-2)^2$ have one common root, $0$, and the second equation has two equal roots, $2$. However,
$$
2(q+b) = 2(4+0) = 8\neq 4 = 1\cdot 4 = ap.
$$
Perhaps you're missing some restrictions?
Edit
Given that $b,q\neq 0$, ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Pythagorean Triple: $\text{Area} = 2 \cdot \text{perimeter}$ Find the unique primitive Pythagorean triple whose area is equal to twice the perimeter.
So far I set the sides of the triangle to be $a, b,~\text{and}~c$ where $a$ and $b$ are the legs of the triangle and c is the hypotenuse.
I came up with 2 equations whic... | With Euclid's formula $F(m,k)$, we can solve $R=$area/perimeter for $k$ and test a defined range of $m$-values to see which yield integers.
$$ \quad A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2\quad$$
$$R=\frac{area}{perimeter}=\frac{AB}{2P} =\frac{2mk(m^2-k^2)}{2(2m^2+2mk)}=\frac{mk-k^2}{2}$$
\begin{equation}
R=\frac{mk-k^2}... | {
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"url": "https://math.stackexchange.com/questions/2381494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\sum_{k=0}^n \frac{A_k}{x+k}=\frac{n!}{x(x+1)(x+2)(x+3)\cdots(x+n)}$ Then find $A_7$ If $$\sum_{k=0}^n \frac{A_k}{x+k}=\frac{n!}{x(x+1)(x+2)(x+3)\cdots(x+n)}$$ Then find $A_7$
My Try:
I have considered a function
$$f(x)=x^{A_0}(x+1)^{A_1}(x+2)^{A_2}\cdots(x+n)^{A_n}$$
taking natural log on both sides and then diff... | Note that:
$$\frac1x\frac1{\binom{x+n}n}=\frac{n!}{x(x+1)(x+2)\dots(x+n)}$$
And thus,
$$\frac1x\left[\frac1{\binom{x+n}n}-\frac1{\binom{x+{n-1}}{n-1}}\right]=\frac{A_n}{x+n}$$
Which gives the general solution immediately.
| {
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"url": "https://math.stackexchange.com/questions/2381927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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How to bound above $| \frac{n}{\sqrt{n^2+n} +n}-\frac{1}{2} |$
Find and Justify $\lim\limits_{n \rightarrow \infty} S_n$
$$S_n = \sqrt{n^2+n} - n = $$
$$\lim\limits_{n \rightarrow \infty} \sqrt{n^2+n} - n
= \lim\limits_{n \rightarrow \infty} \frac{n}{\sqrt{n^2+n} +n}
= \lim\limits_{n \rightarrow \infty} \frac{1}{... | Your method of proof for me it is correct.
Now the bound of $N$ you need is $$N=[\frac{1}{16 \epsilon^2}]+1$$
This $N$ is the least natural number with the property:
$$ \frac{\sqrt{1/n}}{4} < \epsilon, \forall n\geqslant N $$
As for answering the comment below,you proved that $N> \frac{1}{16\epsilon^2}$.
We use the ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A nice but somewhat challenging binomial identity
When working on a problem I was faced with the following binomial identity valid for integers $m,n\geq 0$:
\begin{align*}
\color{blue}{\sum_{l=0}^m(-4)^l\binom{m}{l}\binom{2l}{l}^{-1}
\sum_{k=0}^n\frac{(-4)^k}{2k+1}\binom{n}{k}\binom{2k}{k}^{-1}\binom{k+l}{l}
=\frac{... | Let us complete the OP's work, started with
$$ \frac{1}{2k+1}\stackrel{\text{Binomial transform}}{\longleftrightarrow} \frac{4^k}{(2k+1)\binom{2k}{k}}\tag{$d=0$}$$
by computing first the binomial transform of $\frac{1}{2k+3}$. We have:
$$\begin{eqnarray*}\sum_{k=0}^{n}\frac{(-1)^k}{2k+3}\binom{n}{k}=\int_{0}^{1}x^2(1-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Center of mass using vectors I encountered a physics problem in this book where I was given trees located at points $A,B,C,D,E$.
The problem tells me to move from $A$ to $B$, but only covering half the distance. From my current location, I would then move to $C$, but only covering one-third the distance. This pattern c... | Remember that the center of mass (assuming all trees have the same mass) is $\frac{1}{5}(A + B + C + D + E)$ (treating the locations as vectors).
After the first step, we end up with the point $\frac{1}{2}(A + B)$, found by taking their midpoint. Now from here, we travel $\frac{1}{3}$ of the distance to $C$. We take th... | {
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Find the real and imaginary part of z let $z=$ $$ \left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} \right)^n$$
Rationalizing the denominator:
$$\frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta}\cdot\left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta + i\cos\theta}\right) =... | The denominator is $2+2\sin\theta$, the numerator is, setting $2\alpha=\theta$,
$$
1+e^{2i\alpha}=2e^{i\alpha}\cos\alpha
$$
Then your number is
$$
\left(\frac{\cos\alpha}{1+\sin2\alpha}\right)^{n}e^{ni\alpha}
$$
| {
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"url": "https://math.stackexchange.com/questions/2387172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 3
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Solving $\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \frac{1}{(x-4)}$. Why is my solution wrong? I'm following all hitherto me known rules for solving equations, but the result is wrong. Please explain why my approach is not correct.
We want to solve:
$$\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \f... | A hint to make your life much simpler.
Always simplify your variables to start with. Here you can easily make the substitution $x-4 = y$ to get an easier equation in $y$.
Then noting that $\frac 1y - \frac 1{y+1} = \frac 1{y(y+1)}$, your can very easily see that the numerators on both sides when combining the rationa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2388701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 1
} |
Find all positive integers such that: $n^2-3|3n^3-5n^2+7$
Find all positive integers such that: $$n^2-3|3n^3-5n^2+7.$$
I did the following:
$$n^2-3|3n^3-5n^2+7,n^2-3\Rightarrow n^2-3|3n^3-5n^2+7,3n^3-9n$$
$$\Rightarrow n^2-3|5n^2-9n-7,5n^2-15\Rightarrow n^2-3|9n-8$$
Now we must have: $9n-8\geq n^2-3$.We should deter... | Yes, $n^2-9n+5\leq0$, which gives $1\leq n\leq8$ and check it.
$n=2$ is valid.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2389546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Zeroes of function with real exponents Let real $x \geq 0 $ and real $p > 2$.
Let $ f(x) = (x - 1)(x + 1)^{p - 1} - x^p + 1$.
Show that, for the given range of $x$ and $p$, $f(x)=0$ only for $x=0$ and $x=1$.
Since $f(x)$ is not convex, I find it difficult to show that $f(x) < 0$ for $0<x<1$ and $f(x) > 0$ for $x>1$.... | We will show that $f(x) > 0$ for $x>1$ and $f(x) < 0$ for $0<x<1$.
Let's start with the first one, $f(x) > 0$ for $x>1$.
For real $p > 2$, we can use Bernoulli's inequality (which holds for real exponents!):
$(1 + w)^r \ge 1 + r w $, for real $w > -1$ and real $r > 1$.
This gives for the first exponent in $f(x)$:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2389644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Finding the Equation of Circle, Given is its Differential Equation I was studying for some quizzes when I stumbled upon this question.
It goes like this:
Find the equation of the circle whose differential equation is
$y'' = (1 + (y')^2)^{\frac{3}{2}}$ and which passes through the points
(0, 0) and (1,1).
My wo... | $$\frac{y'}{\sqrt{(y')^2 + 1}} = x + c$$
$$\frac{y'^2}{(y')^2 + 1} = (x + c)^2 \quad\to\quad
y'^2 = \frac{(x + c)^2}{(1-(x + c)^2)}$$
$$y' =\pm \frac{x + c}{\sqrt{1-(x + c)^2}}$$
$$y =\pm \int\frac{x + c}{\sqrt{1-(x + c)^2}}dx=\mp\sqrt{1-(x + c)^2}+C$$
$$(y-C)^2+(x+c)^2=1$$
I suppose that you can take it from here to f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Determine the Polynomial $P$ for which $16P(x^2) = P(2x)^2$ Determined the Polynomial $P$ for which $16P(x^2) = P(2x)^2$
Okay, this problem was given as an example on an Olympiad training material I'm Self-studying, however I do not understand most of the arguments in the solution. Here's the solution.
Plugging $x = 0... | I assume you understood how the relation $4xQ(x^2)=xQ(2x)^2+16Q(2x)$ was obtained. Now you just have to substitute $Q(x)=x^nR(x)$.
In particular, $4xQ(x^2)=4x^{2n+1}R(x^2)$, $xQ(2x)^2=2^{2n}x^{2n+1}R(2x)^2$ and $16Q(2x)=2^{n+4}x^nR(2x)$.
After simplifying $x^n$ on both sides, you find $4x^{n+1}R(x^2)=2^{2n}x^{n+1}R(2x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2392017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Where the object that moves along the intersection of $x^2+y^2=1$ and $y+z=1$ needs to be if we want the sum $x+2y+z$ to be max/min?
An object is moving along the curve which is derived from the intersection of the cylinder $x^2+y^2=1$ and the plane $y+z=1$. Where does the object need to be located if we want to maxim... | Express $z$ and we are interested in maximum/minimum of $E = x+y+1$. By the Cauchy inequality we have: $$|x+y|\leq \sqrt{2(x^2+y^2)} = \sqrt{2}$$ So $E_{\max} = 1+\sqrt{2}$ which is reached at $x=y=1/\sqrt{2}$ and $E_{\min} = 1-\sqrt{2}$ which is reached at $x=y=-1/\sqrt{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2393232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove a matrix has an inverse Prove the matrix
\begin{bmatrix}
A & B \\
B^T & C
\end{bmatrix}
has the inverse
\begin{bmatrix}
D & -DBC^{-1} \\
-C^{-1}B^TD & C^{-1} + C^{-1}B^TDBC^{-1}
\end{bmatrix}
whenever C and D = $(A - BC^{-1}B^T)^{-1}$ are nonsingular.
I tried to see if their multiplication results in $I_{2}$,... | Note that
\begin{align*}
&\begin{bmatrix}
A & B \\ B^T & C
\end{bmatrix}
\begin{bmatrix}
D & -DBC^{-1} \\ -C^{-1}B^TD & C^{-1}+C^{-1}B^TDBC^{-1}
\end{bmatrix}
\\ &=
\begin{bmatrix}
AD - BC^{-1}B^TD & -ADBC^{-1} + BC^{-1} + BC^{-1}B^TDBC^{-1} \\
B^TD + -CC^{-1}B^TD & -B^TDBC^{-1}+CC^{-1}+CC^{-1}B^TDBC^{-1}
\end{bmatrix}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2394467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Is it possible to solve for $\theta$ in this multivariable equation? Question: Given the equation $$c^2=a^2+(\frac{pq}{\sqrt{q^2\sin^2{\theta}+p^2\cos^2{\theta}}})^2-2a\frac{pq}{\sqrt{q^2\sin^2{\theta}+p^2\cos^2{\theta}}}\cos{\theta},$$
is it possible to solve for $\theta$ in terms of $a,c,p,$ and $q$?
Original Problem... | EDIT: After the OP has been edited, there is no hope to find a closed form solution, since you are likely to obtain at least fourth degree powers in $\cos\theta$ when squaring :(
First try to isolate the $\theta$:
$$\frac{c^2-a^2}{pq}=\frac{1-2a\cos \theta}{\sqrt{q^2 \sin^2\theta+p^2\cos^2\theta}}$$
now take squares in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2394546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Expression for the element $(a+u)^{-1}$ in the extension field $G = F(u)$, where $u$ is a root of $x^{2}+px +q$ over a field $F$ Let $f = x^{2}+px+q$ be an irreducible polynomial over a field $F$ and let $G = F(u)$ be the extension of $F$ by a root $u$ of $f$. Every element of $G$ can be written in the normal form $a+b... | Looks like a sign error in the very last line.
The top multiplies out to be $a^2-pa-pu-u^2$, but then $-pu-u^2=q$, so that it is equal to the denominator.
In the other example you gave,
$$ \frac{a^{2}-ab-b^{2}u-b^{2}u^{2}}{a^{2}-ab+b^{2}} \\ = \frac{a^{2}-ab-b^{2}(u+u^{2})}{a^{2}-ab+b^{2}}=\frac{a^2-ab-b^2(-1)}{a^2-ab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2394866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find $x\in \Bbb Z$ such that $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$
Find $x\in \Bbb Z$ such that $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$
Tried (without sucess) two different approaches: (a) finding $x^3$ by raising the right expression to power 3, but was not able to find something useful in the result t... | You can solve these types of problems by denesting each radical and then simplifying the terms that cancel out. In general, we have$$\sqrt[n]{A+B\sqrt[m]C}=a+b\sqrt[m]{C}$$From which you raise both sides to the nth power and expand using Newton's binomial theorem.
In this case, we have$$\sqrt[3]{2\pm\sqrt5}=a\pm b\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2395060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 3
} |
Integral $\int\cos (2nx)\,\log \sin(x)\; dx $ Please help me out wrt this question -
$$\int_0^\frac{\pi}2 \cos (2nx)\; \log \sin(x)\; dx = -\frac {\pi}{4n}$$
Here $n>1$.
I tried doing integration by parts but then how to calculate
$$\int_0^\frac{\pi}2 \cot (x) \,\sin (2nx)\, dx $$
The question is given under Impro... | A different approach.
By integrating by parts, one has
$$
\begin{align}
\int_0^{\Large \frac{\pi}2} \cos (2nx) \log \sin (x)\; dx&=\left[ \frac{\sin (2nx)}{2n}\cdot \log \sin (x)\right]_0^{\Large \frac{\pi}2} -\frac{1}{2n}\int_0^{\Large \frac{\pi}2} \sin (2nx)\: \frac{\cos (x)}{\sin (x)}\; dx
\\&=\color{red}{0}-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Partial fractions - integration $$\int \frac{4}{(x)(x^2+4)} $$
By comparing coefficients,
$ 4A = 4 $,
$A = 1$
$1 + B = 0 $,
$B= -1 $
$xC= 0 $,
$C= 0 $
where $\int \frac{4}{x(x^2+4)}dx =\int \left(\frac{A}{x} + \frac{Bx+C}{x^2 + 4}\right)dx$.
So we obtain $\int \frac{1}{x} - \frac{x}{x^2+4} dx$.
And my final an... | Your partial fraction expansion was correct.$$\frac 4{x(x^2+4)}=\frac 1x-\frac x{x^2+4}$$However, the error lies in integrating the second term of the right-hand side. To simplify$$\int\frac x{x^2+4}\, dx$$Make a substituting $z=x^2+4$. The derivative is $dz=2x\, dx$ so $x\, dx=dz/2$. Therefore, the integral transforms... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Obtain the value of $\int_0^1f(x) \ dx$, where $f(x) = \begin{cases} \frac {1}{n}, & \frac{1}{n+1}
Is the function Riemann integrable? If yes, obtain the value of $\int_0^1f(x) \ dx$
$f(x) =
\begin{cases}
\frac {1}{n}, & \frac{1}{n+1}<x\le\frac{1}{n}\\
0, & x=0
\end{cases}$
My attempt
$f$ is bounded and monotonically ... | Since
$$
(0,1]=\bigcup_{n=1}^\infty\left(\dfrac{1}{n+1},\dfrac{1}{n}\right]=\bigcup_{n=1}^\infty I_n,
$$
and
$$
I_n\cap I_m=\emptyset\quad \forall m,n\in \mathbb{N},
$$
we have
\begin{eqnarray}
\int_0^1f(x)\,dx&=&\int_{[0,0]}f(x)\,dx+\int_{(0,1]}f(x)\,dx\\
&=&\sum_{n=1}^\infty \int_{I_n}f(x)\,dx\\
&=&\sum_{n=1}^\infty\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Prove or disprove : If $a\equiv b$ mod $m$, when $a,b,m\in \mathbb{Z}$ , then $a^3\equiv b^3$ mod $m^2$. Prove or disprove : If $a\equiv b$ mod $m$, when $a,b,m\in \mathbb{Z}$ , then
$a^3\equiv b^3$ mod $m^2$
My attempt:
since $a\equiv b$ mod $m$ then $a=b+mk$ for some $k$
Now $a^3=m^3k^3+b^3+3m^2k^2b+3mkb^2$
$a^3-b^... | For every $\color{Red}{m \notin \{ \pm 1, \pm 3 \}} $ ,
we will construct a
$\color{Red}{\text{counter-example}}$!
Remark(I):
Let $\color{Red}{m \neq \pm 3}$,
and $\color{Red}{m \neq \pm 1}$;
in this case note that $m^2 \color{Red}{\nmid} 3m$.
Let $a:=m+1$ and $b:=1$;
then one can see easily that:
$$a^3-b^3=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the area of the triangle defined in the figure provided Find the area of the triangle defined in the figure below
This question appeared in a math olympiad contest and was considered invalid later on without any specific reason. Is it solvable? If yes, provide the answer!
|
In $\triangle ABC$,
$|BC|=a$, $|CA|=b$, $|AB|=c$,
points $D,E,F$ are midpoints,
$|D_1D_2|=|E_1E_2|=|F_1F_2|=2\,R$.
Using the power of a points $D,E$ and $F$ with respect to
the circumscribed circle with radius $R$, we have
\begin{align}
\tfrac{a}2\cdot\tfrac{a}2&=2\cdot(2\,R-2)
,\\
\tfrac{b}2\cdot\tfrac{b}2&=1\cdot(2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2399632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How prove infinitely many postive integers triples $(x,y,z)$ such $(x+y+z)^2+2(x+y+z)=5(xy+yz+zx)$
show that there exsit infinitely many postive integers triples $(x,y,z)$
such
$$(x+y+z)^2+2(x+y+z)=5(xy+yz+zx)$$
May try it is clear $(x,y,z)=(1,1,1)$ is one solution，and
$$(x+y+z+1)^2=5(xy+yz+xz)+1$$
| Lemma($\color{Green}{\text{Vieta's formula}}$):
Let $\alpha_1$
be the root of the quadratic polynomial equation
$aY^2+bY+c=0$;
then we have: $\alpha_2= \color{Blue}{\dfrac{-b}{a}}-\alpha_1$.
Proof:
Only notice that
$\alpha_1 + \alpha_2 = \dfrac{-b}{a}$.
$
\color{Purple}
{
\text{Let's to look at one of the}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2399830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Integration with logarithmic expression. I'm trying to solve an integration problem from the book which is the $$\int\frac{\sqrt{9-4x^2}}{x}dx$$ using trigonometric substitution. The answer from the book is $$3\ln\left|\frac{3-\sqrt{9-4x^2}}{x}\right|+\sqrt{9-4x^2}+C.$$
I have almost the same solution where there's a ... | I tried to follow your steps. Here we are assuming that $0<x\leq 3/2$. Let $x=3/2\sin(t)$ then $$3\cos(t)=3\sqrt{1-(2x/3)^2}=\sqrt{9-4x^2}$$ and
\begin{align*}\int\frac{\sqrt{9-4x^2}}{x}dx&=\int\frac{3\cos(t)}{3/2\sin(t)}d(3/2\sin(t))\\&
=3\int\frac{\cos^2(t)}{\sin(t)}dt=3\int\frac{1-\sin^2(t)}{\sin(t)}dt\\&=3\int\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2400006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding the quadratic equation from its given roots.
If $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + c
=0$ , then form an equation whose roots are:
$\alpha+\dfrac{1}{\beta},\beta+\dfrac{1}{\alpha}$
Now, using Vieta's formula,
For new equation,
Product of roots ($P$) = $\dfrac{a^2+c^2+2ac}{ac}$
Su... | For an alternative approach, note that $\alpha \beta=c/a\,$, so the two new roots are $\alpha+1/\beta= \alpha(1+a/c)$ and $\beta+1/\alpha= \beta(1+a/c)$.
If $P(x)=ax^2+bx+c\,$ has roots $\alpha,\beta\,$, then the polynomial with roots $\lambda \alpha, \lambda\beta\,$ is:
$$
P\left(\frac{x}{\lambda}\right) = \frac{a}{\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Proving $\sqrt{3+\sqrt{13+4\sqrt{3}}} = 1+\sqrt{3}$. How would I show $\sqrt{3+\sqrt{13+4\sqrt{3}}} = 1+\sqrt{3}$?
I tried starting from the LHS, and rationalising and what-not but I can't get the result...
Also curious to how they got the LHS expression from considering the right.
| $$\sqrt{3+\sqrt{13+4\sqrt{3}}}= \sqrt{3+\sqrt{{(2\sqrt{3}+1)}^2}} = \sqrt{4+ 2\sqrt{3}} = \sqrt{{(\sqrt{3}+1)}^2} = \sqrt{3}+1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Infinite power summation where no common ratio can be seen Problem in book states to evaluate the sum:
$$\sum_{i=0}^\infty \frac{i^2}{4^i}$$
I can come up with the series S = $\frac{1^2}{4^1} + \frac{2^2}{4^2} + \frac{3^2}{4^3} + \frac{4^2}{4^4} + \frac{5^2}{4^5} + ...$
And 4S = $ 1 + \frac{2^2}{4^1} + \frac{3^2}{4^2} ... | $S=\sum_{i=1}^{\infty}\frac{i^2}{4^i} \rightarrow 4S=\sum_{i=1}^{\infty}\frac{i^2}{4^{i-1}}=1+\sum_{i=1}^{\infty}\frac{(i+1)^2}{4^i}$
Thus we have $3S=1+\sum_{i=1}^{\infty}\frac{2i+1}{4^i}=1+\frac{1}{3}+2\sum_{i=1}^{\infty}\frac{i}{4^i}$
$T=\sum_{i=1}^{\infty}\frac{i}{4^i} \rightarrow 4T=\sum_{i=1}^{\infty}\frac{i}{4^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
$\frac{1}{2^n}\binom{n}{n}+\frac{1}{2^{n+1}}\binom{n+1}{n}+...+\frac{1}{2^{2n}}\binom{2n}{n}=1$: short proof? The identity $\frac{1}{2^n}\binom{n}{n}+\frac{1}{2^{n+1}}\binom{n+1}{n}+...+\frac{1}{2^{2n}}\binom{2n}{n}=1$ arises from a question on probability in my textbook. A proof by induction on $n$, which exploits the... | Here is an answer based upon generating functions. It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. This way we can write e.g.
\begin{align*}
[z^k](1+z)^n=\binom{n}{k}
\end{align*}
We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^n\binom{n+k}{k}\frac{1}{2^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Express $(\sqrt{3}+i)^{6-i}$ in the form $x+iy$ I have used that $2e^{i\pi /6}=\sqrt{3}+i $ and I have reached that $(\sqrt{3}+i)^{6-i}= 2^{6-i}(-e^{\pi /6}) $ but I do not know how to pass $2^{6-i}$ to the form $x + iy$, could anyone help me? Thank you.
| $$\left( \sqrt { 3 } +i \right) ^{ 6-i }=\left( 2{ e }^{ \frac { \pi }{ 6 } i } \right) ^{ 6-i }={ 2 }^{ 6-i }{ e }^{ \pi i }{ e }^{ \frac { \pi }{ 6 } }=64{ e }^{ \frac { \pi }{ 6 } }\left( { e }^{ i\left( \pi -\ln { 2 } \right) } \right) =\\ =64{ e }^{ \frac { \pi }{ 6 } }\left( \cos { \left( \pi -\ln { 2 } ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solve in positive integers, $ x^6+x^3y^3-y^6+3xy(x^2-y^2)^2=1$ Solve in positive integers, $$ x^6+x^3y^3-y^6+3xy(x^2-y^2)^2=1$$
My attempt :
$ x^3y^3+x^6-y^6+3xy(x^2-y^2)^2$
$=x^3y^3+3x^2y^2(x^2-y^2)+3xy(x^2-y^2)^2+(x^2-y^2)^3$
$=(xy+(x^2-y^2))^3$
$=(x^2+xy-y^2)^3 = 1$
so $x^2+xy-y^2 = 1$
Please suggest how to proceed.... | It's obvious that $(1,1)$ is one of solutions.
Let $y-1=m(x-1)$ be an equation of the line which has other solutions.
Hence, $m\in\mathbb Q$ and after substitution in $x^2+xy-y^2=1$ we obtain:
$$x=\frac{m^2-2m+2}{m^2-m-1}$$ and
$$y=\frac{2m-1}{m^2-m-1}.$$
I hope it will help.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2412268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Constrained Optimisation Problem: Confusion with Algebra/Contradiction I completed the following constrained optimisation problem:
Maximum and minimum values of $f(x,y) = x^2 + 2xy + y^2$ on the ellipse $g(x,y) = 2x^2 + y^2 - xy - 4x = 0$.
$\nabla f = \lambda \nabla g$
$\therefore (2x + 2y, 2x + 2y) = \lambda(4x - y ... | $(x+y)^2\geq0$ for all real variables.
The equality occurs for $x=y=0$ for example.
Thus, $0$ is a minimal value.
For $x=y=2$ we'll get a value $16$.
We'll prove that it's a maximal value.
Indeed, we need to prove that
$$(x+y)^2\leq16$$ or
$$16-(x+y)^2\geq0$$ or
$$16-(x+y)^2+4(2x^2+y^2-xy-4x)\geq0$$ or
$$3(x-y)^2+4(x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Sequence : $f(f(n))+f(n+1)=2n$ Does there exist a function $f : \mathbb{Z}^+ \to \mathbb{Z}^+$ such that $f(f(n))+f(n+1)=2n$, $\forall n \in \mathbb{Z}^+$ ?
My attempt :
Substitute $n=1$, $f(f(1))+f(2)=2$ so $f(f(1))=f(2)=1$
Substitute $n=2$, $f(f(2))+f(3)=4$ so $f(1)+f(3)=4$
Substitute $n=3$, $f(f(3))+f(4)=6$
Assu... | For a positive integer $n$, let $P(n)$ denote the condition that $f\big(f(n)\big)+f(n+1)=2n$. The condition $P(1)$ implies that $f(2)=1$. Therefore, $P(2)$ leads to $f(1)\in\{1,2,3\}$. Hence, there are only three functions satisfying the condition, and they are listed below.
If $f(1)=1$, then $P(2)$ gives $f(3)=3$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Proving/disproving $∀n ∈ \text{positive integers}$, $\left\lceil{\frac{4n^2+1}{n^2}}\right \rceil = 5$? Is my proof getting anywhere for proving/disproving $∀n ∈ \text{postive integers}$, $$\left\lceil{\frac{4n^2+1}{n^2}}\right \rceil = 5\text{?}$$
$$\left\lceil{\frac{4n^2}{n^2}}+\frac{1}{n^2}\right \rceil = 5$$
$$\lef... | For any $n\in\mathbb{N}$ we have that
$$4=\dfrac{4n^2}{n^2}<\color{red}{\dfrac{4n^2+1}{n^2}}\le\dfrac{4n^2+n^2}{n^2}=\dfrac{5n^2}{n^2}=5.$$ So, we have that
$$\left\lceil \frac{4n^2+1}{n^2} \right\rceil=5$$ and
$$\left\lfloor \frac{4n^2+1}{n^2} \right\rfloor=4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Find all of the points of the form $(x, −1)$ which are $4$ units from the point $(3, 2)$
Find all points of the form $(x, −1)$, which are $4$ units from the point
$(3, 2)$.
I understand the distance formula, I think. Also, I don't know how to format for math on here yet so I apologize for that.
\begin{align}
&\t... | Your computations are correct up to the factoring
$$ (x+\text{something})(x-\text{somethign}).$$
In general, there is no reason to expect that your polynomial will factor like a difference of squares. In fact, the presence of a linear term (the $-6x$) indicates that this won't work. Instead, you need to either factor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2416760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Alternative method to find this infinite sum?
Find where m is an integer:
$$S = 1+\sum_{n=1}^{\infty}\frac{n!}{(mn)!}$$
When we rewrite this sum as one plus a function of $x$ and multiply
the $n^{th}$ term of the series by $x^{mn-1}$, the resulting function
solves for the differential equation $$\frac{d^{m-1}f(x... | Considering $$S_m = \sum_{n=1}^{\infty}\frac{n!}{(mn)!}$$ beside the case $m=2$ for which $$S_2=\frac{1}{2} \sqrt[4]{e} \sqrt{\pi } \text{erf}\left(\frac{1}{2}\right)$$ as Patrick Stevens commented, we are left with hypergeometric functions which apparently do not simplify at all.
They show quite nice patterns
$$S_3=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2417415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve this trigonometric equation ${\tan ^2}x + {\cot ^2}x - 3(\tan x - \cot x) = 0$
Solve the following equation. $${\tan ^2}x + {\cot ^2}x - 3(\tan x - \cot x) = 0.$$
I can't figure out the way to solve this equation. This was my attempt
\begin{array}{l}
{\tan ^2}x + {\cot ^2}x - 3(\tan x - \cot x) = 0\\
\Leftrigh... | Let $\tan{x}-\cot{x}=t$.
Thus, $$t^2+2-3t=0.$$
For $t=1$ we obtain
$$\tan^2x-\tan{x}-1=0,$$
which gives $x=\arctan\frac{1\pm\sqrt5}{2}+\pi k$, where $k\in\mathbb Z$.
For $t=2$ we obtain
$$\tan^2x-2\tan{x}-1=0,$$ which gives
$x=\arctan(1\pm\sqrt2)+\pi k$, where $k\in\mathbb Z$.
Actually, $\arctan(1+\sqrt2)=\frac{3\pi}{8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2417589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Probability distribution: finding 'a' Consider the random variable X with probability distribution,
$$
P(x) =
\begin{cases}
x^2/a & -2,-1,0,1,2 \\
0 & \text{otherwise}
\end{cases}
$$
a) Find $a$ and $E(X)$.
b) What is the probability distribution of random variable $Z = (X - E[X])^2$?
c) Using part b), compute the var... |
a) a = 10, E(X) = 0
Yes. $
{~\\{P_X(x) =
\begin{cases} 0/a^2 & : & x=0 \\ 1/a^2 &:& x\in\{-1,1\} \\ 4/a^2 & : & x\in\{-2,2\}\\ 0 & : & \text{otherwise}
\end{cases} }\\~\\ {\therefore ~a=10, \mathsf E(X)=0}}
$
b) ${P(Z = z) = \begin{cases}
\frac{1}{5} & z = 0 \\
\frac{2}{5} & z = 1,4
\end{cases}}$
N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2418287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find all triples $(a, b, c) \in \mathbb{R} : ab + c = bc + a = ca + b = 4$ Any help on this question would be great. Ive found that (1,1,3), (3,1,1), (1,3,1) but am confused on how to find more anymore (if there are any).Any help would be appreciated.
| $ab + c = bc + a = ca + b=4$
By symmetry any solution for any any one of $a,b,c$ can be a solution for any other.
$ab +c = bc +a$ means $ab - bc = b(a-c) = a-c$ So either $b=1$ or $a=c$.
So by symmetry:
if $a\ne 1$ then $b=c$ and if $b\ne 1$ then $a = c = b$. So it is not possible that only one of $a,b,c$ is equal t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2420692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.