Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Where have I made my mistake in my integrating $\int x \sqrt{2x - 1} \, dx$? $$\int x \sqrt{2x - 1} \,dx$$
Let $u = 2x - 1$
$$\int x \sqrt{u}\, dx$$
$$\frac{du}{dx} = 2 \implies \frac{1}{2}du = dx.$$
So the integral is written as
$$\int \frac{1}{2} u^{\frac{1}{2}} \, du$$
$$ = \frac{1}{2} \left(\frac{2}{3} u^{\frac{3}{... | $$\int x \sqrt{u} dx\tag{1}$$
$$\int \frac{1}{2} u^{\frac{1}{2}} du\tag{2?}$$
What happened to the factor of $x$ going from $(1) \to (2)$?
We have that $u = 2x-1$, so $x = \dfrac{u+1}{2},\;$ and as you know, $dx = \dfrac 12 du.\,$ This gives us:
$$\int x \sqrt{2x - 1} \,dx \quad = \quad \frac{1}{2} \int \frac{(u+1)}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/345179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Prove that $f \circ f (x) = x$ $$f(x) = \pi - \sqrt{1-a^2}\int_{0}^{x} \dfrac{dt}{1+\cos\,t}$$
Using that if $u = \tan \frac t2$, then $\cos\,t = \dfrac{1-u^2}{1+u^2}$
$u = \tan \frac t2 \Rightarrow t = 2 \arctan u \Rightarrow dt = \dfrac{2}{1+u^2} du$
substituting in the integral
$$\int_0^{\tan \frac x2} \dfrac{2\,d... | Something is wrong with your
definition of $f(x)$,
because,
if $a=1$ then $f(x) = \pi$ for all $x$,
so $f(f(x)) = \pi$ for all $x$.
In particular, it seems that the $a$ is out of place.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/346231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving that $\gcd(5^{98} + 3, \; 5^{99} + 1) = 14$
Prove that $\gcd(5^{98} + 3, \; 5^{99} + 1) = 14$.
I know that for proving the $\gcd(a,b) = c$ you need to prove
*
*$c|a$ and $c|b$
*$c$ is the greatest number that divides $a$ and $b$
Number 2 is what I'm struggling with. Does anybody have any ideas?
| First of all, $$5\cdot(5^{98}+3)-1\cdot(5^{99}+1)=14$$
$$\implies (5^{98}+3,5^{99}+1)\mid 14$$
Now both the numbers are even $\implies 2\mid (5^{98}+3,5^{99}+1) $
Using Fermat’s Little theorem $5^6\equiv1\pmod 7$
$\implies 5^{98}=(5^6)^{16}\cdot5^2\equiv 25\pmod 7\equiv-3\implies 7\mid (5^{98}+3)$
and $5^{99}\equiv(5^6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/346524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
} |
What does it mean to eliminate $\theta$ from these equations? How should I do it?
Eliminate $\theta$ from the following pairs of equations:
A) $x=\sin \theta$, $y=\sin 2\theta$
B) $x=3\cos 2\theta +1$, $y=2\sin \theta$
My problem is I really don't understand what the question is asking? The answer in the book seems l... | You've done the hard part. Of A):
$$
\begin{align}
\sin\theta & = x \\ \\
\cos\theta & = {\frac y{2x}}
\end{align}$$
This is where the Pythagorean Identity (as you call it) comes in very handy, indeed:
$$\quad\quad\quad\sin^2(\theta) + \cos^2(\theta) = 1$$
$$\iff \left(\sin\theta \right)^2 + \left(\cos\theta \right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/347737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Calculate the limit $\lim \limits_{n \to \infty} |\sin(\pi \sqrt{n^2+n+1})|$ Calculate
$$\lim \limits_{n \to \infty} |\sin(\pi \sqrt{n^2+n+1})|$$
| If you know a little bit about Taylor expansions and big O notation, you can do that easily. We will use
$$
\sqrt{1+u}=1+\frac{u}{2}+O(u^2)\qquad \mbox{when } u\longrightarrow 0.
$$
Now
$$
\pi\sqrt{n^2+n+1}=\pi n\sqrt{1+\frac{1}{n}+\frac{1}{n^2}}=\pi n\left(1+\frac{1}{2n}+O\left(\frac{1}{n^2} \right)\right)=\pi n+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/349522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 1
} |
Proving $x$ is divisible by $20$ I need to prove that $x$ divisible by $20$ if and only if $x=0\pmod4$ and $x=0 \pmod 5$
proving that if $x=0 \pmod 4$ and $x=0 \pmod 5$ than $x$ divisible by $20$ is by the Chinese theorem (am I right??)
But the other way - I dont understand why its not enough to be divisible only by on... | $20 = 4*5$ so if $x$ is divisible by $20$ then it is divisible by $4$ and $5$. And hence by definition $x \equiv 0 \pmod 4$ and $x \equiv 0 \pmod 5$
Conversely, if $x \equiv 0 \pmod 4$ and $x \equiv 0 \pmod 5$ then $x = 4n$ and $x= 5m$ for some integers $n$ and $m$.
So $4n$ = $5m$.
Now since $5$ does not divide $4$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/350670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find minimum of $\frac{a+3c}{a+2b+c}+\frac{7a+6b+3c}{a+b+2c}+\frac{c-a}{2a+b+c}$ for non-negative reals
Let $a, b, c\ge 0$, not all zero. Find the minimum of
$$N = \frac{a+3c}{a+2b+c}+\frac{7a+6b+3c}{a+b+2c}+\frac{c-a}{2a+b+c}. $$
| (This initial part is not a solution. It is too long to be a comment. This demonstrates that the suggested approach of substituting the denominator doesn't work out because equality is not achieved as the condition $ a \geq 0$ is not met.)
We use the substitution $x = a + 2b + c, y = a+b+2c, z = 2a + b + c$.
This syste... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/351976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Existence and value of $\lim_{n\to\infty} (\ln\frac{x}{n}+\sum_{k=1}^n \frac{1}{k+x})$ for $x>0$ Does the limit
$$W(x)=\lim_{n\to\infty} \left(\ln\frac{x}{n}+\sum_{k=1}^n \frac{1}{k+x} \right)$$
exist for all $x>0$? If so, what is the limit
$$\lim_{x\to\infty}W(x)?$$
| We have
$$\sum_{k=1}^n \dfrac1{k+x} = \int_{1^-}^{n^+} \dfrac{d \lfloor t \rfloor}{t+x} = \left. \dfrac{\lfloor t \rfloor}{t+x} \right\vert_{t=1^-}^{t=n^+} + \int_{1^-}^{n^+} \dfrac{\lfloor t \rfloor}{(t+x)^2} dt = \dfrac{n}{n+x} + \int_{1^-}^{n^+} \dfrac{\lfloor t \rfloor}{(t+x)^2} dt$$
Now
$$\int_{1^-}^{n^+} \dfrac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/352328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Reasoning about the Chebyshev functions: How does one check an upper bound based on the second Chebyshev function? In Ramanujan's proof of Bertrand's Postulate, Ramanujan states:
$\log([x]!) - 2\log([\frac{1}{2}x]!) \le \psi(x) - \psi(\frac{1}{2}x) + \psi(\frac{1}{3}x)$
where:
$\vartheta(x) = \sum_{p \le x} \log p$
$\p... | I found the answer to my question in M. Ram Murty, Problems in Analytic Number Theory.
If $a_0 \ge a_1 \ge a_2 \ge \ldots$ is a decreasing sequence of real numbers tending to $0$, then:
$a_0 - a_1 \le \sum_{n=0}^{\infty}(-1)^{n}a_n \le a_0 - a_1 + a_2$.
This result is straight forward to show using induction.
This can ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/353794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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If $\frac{1}{2}1-\left(a_1+\frac{a_2}{2}+\cdots+\frac{a_n}{2^{n-1}}\right)$
Let $n>1$ be a positive integer and $\frac{1}{2}<a_{j}<1$ for $j=1,2,\ldots,n$. Show that $$(1-a_{1})(1-a_{2})\cdots (1-a_{n})>1-\left(a_{1}+\frac{a_{2}}{2}+\cdots+\frac{a_{n}}{2^{n-1}}\right).$$
I have no ideas.
| For another approach, we re-write as:
$$(1-a_1)(1-a_2) \dots (1-a_n) + \left(a_1 + \frac{a_2}{2} + \dots + \frac{a_n}{2^{n-1}}\right) > 1$$
Let $P = (1-a_1) (1-a_2)\dots (1-a_n)$ and $Q = \left(a_1 + \frac{a_2}{2} + \dots + \frac{a_n}{2^{n-1}}\right) $. It may be noted that for any $a_k$ we can define $p = \dfrac{P}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/354445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Fast way to calculate determinant for a block matrix I have a block matrix
$$Q_{(n+m-1)\times(n+m-1)} = \begin{pmatrix} A & -J\\-J^t & B \end{pmatrix}$$
where
$$A_{(m-1)\times(m-1)} = n*I_{(m-1)\times(m-1)} \text{ and } B_{n\times n} = m*I_{n\times n}$$
where $I$ is identity matrix. $J$ is then $(m-1)\times n$ matrix w... | See in wikipedia the page of determinat in section 3.3 on the Block matrices.
When $A$ is Invertible matrix, we have
$$
\det\begin{pmatrix}A& B\\ C& D\end{pmatrix} = \det(A) \det(D - C A^{-1} B).
$$
When $D$ is invertible, a similar identity with $\det(D)$ factored out can be derived analogously,
$$
\det\begin{pmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/355886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Choosing teams with minimum number of boys and girls. I need to find different number of teams I can make with 6 people that needs to have at least 2 girls and 2 boys. There are 8 girls and 12 boys. So the way I think is I need to find total number of different teams that I can make first and subtract all boys team, al... | Just calculate the 3 cases instead.
(b,g)=(4,2), (3,3) and (2,4)
Anyways you r also right.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/356926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Bounds for the exponential integral In Abramowitz and Stegun: Handbook of Mathematical Functions
(on page 229, property 5.1.20) it is found that
$$
\frac{1}{2} \log \left(1 + \frac{2}{x} \right) < \exp(x) E_1(x) < \log \left(1 + \frac{1}{x} \right) \qquad (x > 0)
$$
where
$$
E_1(x) = \int_x^\infty \frac{\exp(-t)}{t} ... | Here is a proof following from the theorem that, if for all $x$, $g'(x) < f'(x) < h'(x)$, and the inequality $g(x)<f(x)<h(x)$ holds at some point, then this inequality holds true for all $x$.
First we can rearrange the inequality as
$$e^{-x}\frac{\ln(x+2)-\ln(x)}{2}< \int_x^\infty \frac{e^{-t}}{t}dt<e^{-x} \big(\ln(x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/358395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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How to find the minimum of $f(x)=(\sin(x)+\cos(x)+\tan(x)+\cot(x)+\sec(x)+\csc(x))^2$? I need to find the minimum of $f(x)$ with
$$f(x)=(\sin(x)+\cos(x)+\tan(x)+\cot(x)+\sec(x)+\csc(x))^2$$
Could you help me with some clues?
| To built upon @Queayiouer if you split the function as
$$ f(x) = \left( g(x) \right)^2 = \left( \left(\sin x+\frac{1}{\sin x}\right) + \left(\cos x+\frac{1}{\cos x}\right) + \left(\tan x+\frac{1}{\tan x}\right) \right)^2 $$
where the minimum occurs if $g(x)=0$ or $g'(x)=0$. The first is not going to happen.
So now we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/360728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find the eigenvalues and Jordan canonical form of this matrix Question:
let $a_{i,j}\in R,A=(a_{i,j})_{n\times n} $,and
$a_{i,j}=\begin{cases}
1&i+j\in\{n,n+1\}\\
0&i+j\notin\{n,n+1\}
\end{cases}$
that's meaning:
$$A=\begin{bmatrix}
0&0&0&\cdots&0&1&1\\
0&0&0&\cdots&1&1&0\\
0&0&\cdots&1&1&0&0\\
\vdots&\vdots&\v... | I show below that the eigenvalues of $A$ are exactly the numbers
$-2\cos\big(j\frac{2\pi}{2n+1}\big)$ for $1\leq j\leq n$.
What makes everything work is the
identity $2\cos(\theta)\cos(k\theta)=\cos((k-1)\theta)+\cos((k+1)\theta)$.
Unfortunately the eigenvectors are a little complicated to express directly,
the best p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/363237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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The minimum and the maximum of $y=\sin^2x/(1+\cos^2x)$ I was asked to find the minimum and maximum values of the functions:
*
*$y=\sin^2x/(1+\cos^2x)$;
*$y=\sin^2x-\cos^4x$.
What I did so far:
*
*$y' = 2\sin(2x)/(1+\cos^2x)^2$
How do I check if they are suspicious extrema points? After this function ... | If you don't really need to use derivatives,
$1.$
Clearly, $y=\frac{\sin^2x}{1+\cos^2x}\ge 0$ in fact $=0$ if $\sin x=0$
$y-1=\frac{\sin^2x}{1+\cos^2x}-1=\frac{\sin^2x-(1+\cos^2x)}{1+\cos^2x}=-2\frac{\cos^2x}{1+\cos^2x}\le 0$ in fact $=0$ if $\cos x=0$
$\implies y-1\le 0\iff y\le 1\implies 0\le y\le 1$
$2.$ $$y=\sin^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/364394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inequality: $\frac{a^2+b^2}{(a+b)^4}+\frac{b^2+c^2}{(b+c)^4}+\frac{c^2+d^2}{(c+d)^4}+\frac{d^2+a^2}{(d+a)^4} \leq \frac{1}{abcd}$ If $ab+bc+cd+da\leq 8$ and $a,b,c,d \in \mathbb{R}_{+},$how can I prove the following inequality :
$$\frac{a^2+b^2}{(a+b)^4}+\frac{b^2+c^2}{(b+c)^4}+\frac{c^2+d^2}{(c+d)^4}+\frac{d^2+a^2}{(... | Multiplying $(ab+bc+cd+da)/8$ to the RHS, it suffices to show that
$$\sum \frac{(a^2+b^2)}{(a+b)^4}\le \sum\frac{1}{8ab}.$$
Very luckily we have
$$\frac{(a^2+b^2)}{(a+b)^4} \le \frac{1}{8ab},$$
which is equivalent to $(a+b)^4 \ge 8ab(a^2+b^2)$ or $(a-b)^4 \ge 0$. Q.E.D.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/364856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Sequence and equation I am having a problem finding an equation for this sequence: 2, 2+4 , 2+4+6, 2+4+6+8 .
Can someone show me the steps in solving this?
| We have $a_1 = 2 \times 1$, $a_2 = 2 \times 1 + 2 \times 2$, $a_3 = 2 \times 1 + 2 \times 2 + 2 \times 3$.
In general, we have
$$a_n = 2 \times 1 + 2 \times 2 + 2 \times 3 + \cdots + 2 \times n = 2 \times (1+2+\cdots+n) = 2 \times\dfrac{n(n+1)}2 = n(n+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/369595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $3\sum\limits_{i=0}^k\binom{n}{3i}\leq2^n+2$ If $n\in \mathbb{Z^+}$ and $k$ is the largest integer for which $3k\leq n$, then is it true that $\sum_{i=0}^k\binom{n}{3i}\leq \frac{1}{3}(2^n+2)$?
My work: We can break this into two cases: $n=3k+1$ and $n=3k+2$.If $n=3k+1$, then we need to prove that $$\sum_{i=... | Here’s a completely elementary solution. Let
$$\begin{align*}
a_n&=\sum_k\binom{n}{3k}\;,\\
b_n&=\sum_k\binom{n}{3k+1}\;,\text{ and}\\
c_n&=\sum_k\binom{n}{3k+2}\;.
\end{align*}$$
Clearly $a_n+b_n+c_n=2^n$. Moreover, it follows from Pascal’s identity that
$$\begin{align*}
a_n&=a_{n-1}+c_{n-1}\;,\\
b_n&=b_{n-1}+a_{n-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/371551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Finding generating function for the recurrence $a_0 = 1$, $a_n = {n \choose 2} + 3a_{n - 1}$ I am trying to find generating function for the recurrence:
*
*$a_0 = 1$,
*$a_n = {n \choose 2} + 3a_{n - 1}$ for every $n \ge 1$.
It looks like this:
*
*$a_0 = 1$
*$a_1 = {1 \choose 2} + 3$
*$a_2 = {2 \choose 2} + 3... | Sneaky. Write your recurrence without subtractions in indices, i.e.:
$$
a_{n + 1} = 3 a_n + \binom{n + 1}{2}
$$
Define $A(z) = \sum_{n \ge 0} a_n z^n$, multiply by $z^n$, sum over $n \ge 0$ and recognize the resulting sums, particularly:
\begin{align}
\sum_{n \ge 0} \binom{n + 1}{2} z^n
&= z \sum_{n \ge 0} \binom{n +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/372439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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cauchy schwarz inequality problemes I have to prove that for all $x,y,z>0$,
$$\left(\frac{x+y}{x+y+z}\right)^{0.5} + \left(\frac{x+z}{x+y+z}\right)^{0.5} + \left(\frac{z+y}{x+y+z}\right)^{0.5} \leq 6^{0.5}$$
using Cauchy-Schwarz inequality? How do I do that ?
I have to define an inner product but I do not what, and wha... | Something to notice is that $\left(\frac{x+y}{x+y+z}\right)+ \left(\frac{x+z}{x+y+z}\right) + \left(\frac{z+y}{x+y+z}\right)= 2 $. This suggests writing $\left(\frac{x+y}{x+y+z}\right)^{0.5} + \left(\frac{x+z}{x+y+z}\right)^{0.5} + \left(\frac{z+y}{x+y+z}\right)^{0.5} $ as the dot product of $\left(\left(\frac{x+y}{x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/373649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
(4x^2+2kx-5)/(x+2) remainder is 3 find value of k? 2 methods - first is long division by $(x+2)$, 2nd is to use remainder theorem
let $f(x) = 4x^2+2kx-5$ and $g(x) = x+2$
to find the remainder of $\frac{f(x)}{g(x)}$ where $g(x) = (x+c)$ we need to evaluate $f(-c)$
$f(-2) = 4(-2)^2+2k(-2) -5$
Because the remainder is 3,... | We have
\begin{align}
\dfrac{4x^2 + 2kx - 5}{x+2} & = \dfrac{4(x+2)^2 - 4 \cdot(4x) - 4 \cdot 4 + 2kx-5}{x+2}\\
& = 4(x+2) + \dfrac{(2k-16)x - 21}{x+2}\\
& = 4(x+2) + \dfrac{(2k-16)x - 21}{x+2}\\
& = 4(x+2) + \dfrac{(2k-16)(x+2) - 21-2(2k-16)}{x+2}\\
& = 4(x+2) + (2k-16) + \dfrac{11-4k}{x+2}
\end{align}
This means $11-... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$ax+by+cz=d$ is the equation of a plane in space. Show that $d'$ is the distance from the plane to the origin. This is a 3 part practice question I would like to get some feedback on. I think I have solved the 1st two parts, but I need a little direction for part (c) (the title is Part (a) ) which is repeated here with... | Your answer looks good. Here is an alternate method to find the minimum point.
To minimize
$$
x^2+y^2+z^2\tag{1}
$$
while maintining
$$
ax+by+cz=d\tag{2}
$$
we want to have
$$
2x\,\delta x+2y\,\delta y+2z\,\delta z=0\tag{3}
$$
for all $(\delta x,\delta y,\delta z)$ so that
$$
a\,\delta x+b\,\delta y+c\,\delta z=0\tag{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/378903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
} |
How prove this $T_{n}<\dfrac{5}{4}|b_{1}|$ let $\{b_{n}\}$ such that $b^2_{n}=b_{n+1}(1+b^2_{n})$,
and
$$T_{n}=\sum_{k=1}^{n}\dfrac{(-1)^k(k+2)}{(k+1)^2}b_{k}$$
show that
$T_{n}<\dfrac{5}{4}|b_{1}|$
my idea
$b^2_{n}=b_{n+1}(1+b^2_{n})$ then $b_{n}>0,n\ge 2$
and
$$b^2_{n}\ge 2b_{n+1}b_{n},\Longleftrightarrow b_{n+1}\l... | Suppose $b_1\ne 0$. (If $b_1=0$, then $b_n=0$ for all $n$ and the inequality does not hold.) And I also assume $n\ge 1$ in $b_n^2=b_{n+1}(1+b_n^2)$.
Lemma. $0<b_{n+1}\leq b_n/2$ for $n\ge 2$, and $0<b_2\leq |b_1|/2$.
Proof of lemma. (From the OP's argument.) $b_{n+1}=\frac{b_n^2}{1+b_n^2}>0$ for $n\ge 1$. Using the fa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/379266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Is $u_n\le(1-a)^n\forall n\in\mathbb{N}$? Consider the sequence $\{u_n\}$ where $u_0=1,u_1=1-a$ for some $0< a < 1/4$, and $u_{n+2} = u_{n+1}-au_n$. Is $u_n\le(1-a)^n\forall n\in\mathbb{N}$?
| Let $u_n = (1-a)^n b_n$
Then we have that $b_0 = 1$ and $b_1 = 1$.
Now $$b_{n+2} = \frac{b_{n+1}}{1-a} - \frac{ab_n}{(1-a)^2}$$
And so
$$b_{n+2} - b_{n+1} = \frac{ab_{n+1}}{1-a} - \frac{ab_n}{(1-a)^2} = \frac{a}{1-a}\left(b_{n+1} - \frac{b_n}{1-a}\right)$$
Now if $b_{n+1} \le b_n \le \frac{b_n}{1-a}$, we must have that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/379476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to prove that an integral doesn't exist? $$\int_{0}^{\infty}\sin^2\left(\pi \left(x + \frac{1}{x} \right) \right) dx $$
Should I use any test for convergence?
| Since $\left|\sin^2(a)-\sin^2(b)\right|=|\sin(a)+\sin(b)|\,|\sin(a)-\sin(b)|\le2|a-b|$, we have
$$
\begin{align}
&\left|\int_a^b\sin^2\left(\pi x+\frac\pi x\right)\,\mathrm{d}x
-\int_{a+1/2}^{b+1/2}\cos^2\left(\pi x+\frac\pi x\right)\,\mathrm{d}x\right|\\
&=\left|\int_a^b\sin^2\left(\pi x+\frac\pi x\right)\,\mathrm{d}x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/387245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the infinite series $\sum_{r=1}^{\infty} \frac{2^{-r}}{r^{2}}$ My teacher posed an infinite series question to me today and I'm not quite sure how to start to go about it.
$$\sum_{r=1}^{\infty} \dfrac{2^{-r}}{r^{2}}$$
Any hints would be much appreciated.
| Claim: $$\color{blue}{\boxed{\displaystyle \sum_{k=1}^{\infty} \dfrac1{2^k k^2} = \dfrac{\pi^2}{12} - \dfrac{\log^2(2)}2}}$$
Proof
What you want to evaluate is the PolyLogarithm function, $\text{Li}_2(x)$, at $x=1/2$. The PolyLogarithm function $\text{Li}_s(x)$ is defined as
$$\text{Li}_s(x) = \sum_{k=1}^{\infty} \dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/388006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Explanation of why $\int \sin^2 x\cos^2 x\; dx = 1/3 \sin^3 x - 2/5 \sin^5 x + 1/7 \sin^7 x +c$
verify the solution
$$\int \sin^2 x\cos^5 x\; dx = 1/3 \sin^3 x - 2/5 \sin^5 x + 1/7 \sin^7 x +c$$
I have hit this in my book and can't work it out. Does anyone have any ideas or a walk-through that might help?
| It's easier to start from the right hand side. Derive to get:
$$ (\sin^2 x - 2\sin^4 x + \sin^6 x)\cos x$$
Note that this is just:
$$(\sin x - \sin^3 x)^2\cos x=\sin^2 x(1-\sin^2x)^2\cos x=\sin^2 x\cos^5x$$
So, it seems you are missing a couple of powers of $\cos x$ in the integral..
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/389110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Examine the continuity of $f(x)=x^2+\frac{x^2}{(1+x^2)}+\frac{x^2}{(1+x^2)^2}+...+ \frac{x^2}{(1+x^2)^n}+....$ at $x=0$
Examine the continuity of $$f(x)=x^2+\frac{x^2}{(1+x^2)}+\frac{x^2}{(1+x^2)^2}+...+ \frac{x^2}{(1+x^2)^n}+....$$ at $x=0$
I tried to solve the problem using $$\lim_{x \to0^+}f(x)=\lim_{x \to0^-}f(x)... | If the dots imply an infinite series, then you can first consider
$$
g(x)=\sum_{n\ge0}\frac{1}{(1+x^2)^n}
$$
that converges to
$$
g(x)=\frac{1}{1-\dfrac{1}{1+x^2}}=\frac{1+x^2}{x^2}
$$
whenever
$$
\frac{1}{1+x^2}<1
$$
that is, for all $x\ne0$.
Thus your function is
$$
f(x)=
\begin{cases}
1+x^2 &\text{if $x\ne0$}\\
0&\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/389260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Determine the $n$-th power of matrix Determine the $n$-th power of the matrix. $$\pmatrix{2 & 2 & 0 \\ 1 & 2 & 1 \\1 & 2 & 1 \\ }$$ May you help me with the answer since the book states : $${1\over 6}\pmatrix{4+2\cdot4^n & 3\cdot4^n & -4+4^n \\ -2+2\cdot4^n& 3\cdot4^n & 2+4^n\\-2+2\cdot4^n & 3\cdot4^n & 2+4^n \\ }$... | I think about $$\pmatrix{\left(2^{2n-1}-\frac{2}{3}(4^n-1)\right) & 2^{2n-1} & \frac{2}{3}(4^n-1) \\ \left(2^{2n-1}-\frac{2}{3}(4^n-1)+1\right) & 2^{2n-1} & \frac{2}{3}(4^n-1)+1 \\\left(2^{2n-1}-\frac{2}{3}(4^n-1)+1\right) & 2^{2n-1} & \frac{2}{3}(4^n-1)+1 \\ }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/389560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Pell's type equation in sum I have an another observation on concatenation in sum. For instance, take $12^2 + 33^2 = 1233$. Find as many such pairs (x, y) with $x^2 + y^2 = xy\,$(here xy is concatenation)is possible. Also, discuss how this is happening and is there any connection between this kind of problem and pell's... | This is equivalent to $x^2 + y^2 = 10^n x + y$ with size constraints on $y$ (preferably $10^{n-1} \le y < 10^n$, though we will shortly see that the right-hand inequality is superfluous).
Completing the square gives $(2x - 10^n)^2 + (2y - 1)^2 = 10^{2n} + 1$, so we are looking for a way to write $10^{2n}+1$ as the sum ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/389700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Proving that $x - \frac{x^3}{3!} < \sin x < x$ for all $x>0$
Prove that $x - \frac{x^3}{3!} < \sin(x) < x$ for all $x>0$
This should be fairly straightforward but the proof seems to be alluding me.
I want to show $x - \frac{x^3}{3!} < \sin(x) < x$ for all $x>0$. I recognize this shouldn't be too difficult but perh... | Recursive integration We know that $$0\le\cos a\le 1\implies \sin t = \int_0^t\cos s ds < t$$ for $0\lt t\lt z\lt x$. Integrating over $\color{blue}{(0,z)}$ we get
$$1-\cos z=\int_0^z\sin tdt < \int_0^ztdt= \frac{z^2}{2}$$
that is for all $0<z<x$ we have,
$$\color{blue}{1-\frac{z^2}{2}< \cos z\le 1}$$
integrating again... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/390899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Solve $\sqrt[3]{x+10}-\sqrt[3]{x-10}=2$.
Solve $\sqrt[3]{x+10}-\sqrt[3]{x-10}=2$.
I tried cubing the both sides but things then go very ugly. Are there simpler way to solve it? Thanks.
p.s. The answers are $\pm 6\sqrt 3$.
| Set $x+10=t^3$ and $x-10 = u^3$. So we want to solve $t - u = 2$.
However $t^3 - u^3 = (t-u)^3 + 3tu(t-u) = 20$.
That is $2^3 + 6tu = 8 + 6tu =20$.
Now we have $tu = 2$. This means $x^2 -100 = 8$ and thus $x = \pm6\sqrt{3}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/392296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\sum_{n=0}^\infty \frac 1 {(2n+1)^4}$ using Mittag-Leffler's expansion I am trying to evaluate the following series using Mittag-Leffler's expansion theorem. What function would be useful?
$$\sum_{n=0}^\infty \frac 1 {(2n+1)^4} = \frac{\pi^4}{96}$$
I considered differentiating the following relation twice b... | You should evaluate:
$$
\lim_{z \to 0} -\frac{(\pi/2)^4}{2 z} \frac{\mathrm{d}}{\mathrm{d}z} \frac{\tanh(z)}{2z}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/392529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving two equations involving the greatest common divisor
Show or prove that $$\gcd \left(\frac{a^{2m}-1}{a+1} ,a + 1\right )=\gcd(a + 1 , 2m),$$
and that
$$\gcd \left(\frac{a^{2m + 1}+1}{a+1} , a + 1\right)=\gcd(a + 1 , 2m + 1).$$
| I will show that $$\gcd \left(\frac{a^{2m}-1}{a+1} ,a + 1\right )=\gcd(a + 1 , 2m)$$
First note that $$a^{2m}-1=(a^2)^m-1=(a^2-1)(a^{2(m-1)}+
a^{2(m-2)}+\cdots+1)$$
so that $$\frac{a^{2m}-1}{a+1}=(a-1)(a^{2(m-1)}+\cdots+1)$$
But $a\equiv -1 \mod a+1$ so we have
$$\frac{a^{2m}-1}{a+1}\equiv -2((-1)^{2(m-1)}+\cdots+1)=-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/392649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why is this derivative incorrect? We have to find the derivative of $$f(x) = \dfrac{\tan(2x)}{\sin(x)}$$
I would like to know why my approach is incorrect:
$$f'(x) = \dfrac{\sin(x) \cdot \dfrac{2}{\cos^2(2x)} - \tan(2x) \cdot \cos(x)}{\sin^2(x)}$$
$$ = \dfrac{ 2 \sin(x) - \tan(2x) \cdot \cos(x)}{\cos^2(2x) \cdot \sin... | $$f'(x) = \dfrac{\sin(x) \cdot \dfrac{2}{\cos^2(2x)} - \tan(2x) \cdot \cos(x)}{\sin^2(x)}\tag{1}$$
So far so good...
But your following line is where you made an algebraic error: you forgot to multiply both (the entire numerator (both terms)) and (the denominator) by $\cos^2(2x)$. Doing this gives us:
$$f'(x) = \dfra... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $\int \frac{1+\cos(x)}{\sin^2(x)}\,\operatorname d\!x$ I`m trying to solve this integral and I did the following steps to solve it but don't know how to continue.
$$\int \dfrac{1+\cos(x)}{\sin^2(x)}\,\operatorname d\!x$$
$$\begin{align}\int \dfrac{\operatorname d\!x}{\sin^2(x)}+\int \frac{\cos(x)}{\sin^2(x)}\... | Hints:
$1+\cos x=2 \cos^2 \dfrac{x}{2}$
$\sin^2 x=(2\sin\dfrac{x}{2} \cos \dfrac{x}{2})^2$
You expression will be $\dfrac{1}{2} \int \dfrac{1}{\sin^2\dfrac{x}{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/393375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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What would be the value of $a$ and $b$ in following rational expression? If $(5 + 2\sqrt{3})/(7 + \sqrt{3}) = (a - \sqrt{3b})$,
How do I find the value of $a$ and $b$ where $a$ and $b$ are rational numbers?
| First rewrite $5+2\sqrt{3} = (a-\sqrt{3b})(7+\sqrt{3})$.
Now suppose $b=3n^2$, then we get $5+2\sqrt{3} = (a-3|n|)(7+\sqrt{3})$, which can not be solved for rational $a$ and $n$.
Next, suppose $b=n^2$, then we get $5+2\sqrt{3} = (a-|n|\sqrt{3})(7+\sqrt{3}) = 7a-3|n|+(a-7|n|)\sqrt{3}$ and hence $5=7a-3|n|$ and $2=a-7|n|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/396487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Matrix Equation, Solving for Variables. I'm going through my exercises, and came across a problem that wasn't covered in our lectures. Here's the question:
$
\begin{align}
\begin{bmatrix}
a-b & b+c\\
3d+c & 2a-4d
\end{bmatrix}
\end{align}
=
$
$
\begin{align}
\begin{bmatrix}
8 & 1\\
7 & 6
\end{bmatrix}
\end{align}
$
Wha... | The best way is follow the solution given by @Alex Wertheim. Let me show you a tedious substitution method to solve this. From $a_{11}$ we have $$b=a-8$$ (dividing by $a$ or $b$ is invalid as we may have zeros). Now from $a_{12}$ entry we have $$c=-b+1=-a+9,$$ and using $a_{21}$ $$d=\dfrac{-c+7}{3}=\dfrac{a-2}{3}.$$ Fi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/398802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Find the modular residue of this product.. Please help me solve this and please tell me how to do it..
$12345234 \times 23123345 \pmod {31} = $?
edit: please show me how to do it on a calculator not a computer thanks:)
| As $10^1\equiv 10\pmod{31},$
$10^2=100\equiv7,$
$10^3\equiv10\cdot7\equiv8,$
$10^4\equiv49\equiv18,$
$10^5=10^2\cdot10^3\equiv 7\cdot8\equiv25,$
$10^6=(10^3)^2\equiv8^2\equiv2,$
$10^7=10^4\cdot10^3\equiv18\cdot8\equiv20,$
$$12345234=4+3\cdot10+2\cdot10^2+5\cdot10^3+4\cdot10^4+3\cdot10^5+2\cdot10^6+1\cdot10^7$$
$$\equiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/398857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve this simultaneous equation of $3$ variables. I've stuck in this equation system. No clue how to start ?
$$\begin{eqnarray}
x+y+z &=&a+b+c\tag{1} \\
ax+by+cz &=&a^{2}+b^{2}+c^{2}\tag{2} \\
ax^{2}+by^{2}+cz^{2} &=&a^{3}+b^{3}+c^{3}\tag{3}
\end{eqnarray}$$
Find the value of $x,y,z$ is in the form of $a,b$ and... | This is a slightly interesting question. Let me try and provide some motivation for the ugly values that Americo calculated.
Let me assume that $a\neq b, b \neq c, c \neq a $. We'd deal with this issue separately.
Note that the first 2 equations give the equation on a plane. Hence, their intersection is a line, which w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/401414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Let $k \geq 3$; prove $2^k$ can be written as $(2m+1)^2+7(2n+1)^2$ Prove:
If $k \geq 3$, then $2^k$ can be written as $(2m+1)^2+7(2n+1)^2$, where $k, m, n \in \mathbb{N}$.
| You need to be careful about just one thing in the induction step. Note that the first case is $$ 1^2 + 7 \cdot 1^2 = 8 = 2^3. $$
The important thing is that
$$ 1 \equiv 1 \pmod 4. $$
So, we begin with $$ a^2 + 7 b^2 = 2^k $$
with $k \geq 3$ AND
$$ a \equiv b \pmod 4, $$ then we do the induction step.
What t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/402788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Is there a simpler approach to these system of equations? I recently came across the following system of equations:
$$x + y + z = 1 \\
x^2 + y^2 + z^2 = 2 \\
x^3 + y ^3 + z^3 = 3$$
And I have two questions:
One, is there a way to prove or disprove whether there is a solution for this particular set of equations? Fur... | We will make this into a cubic equation in terms of $z$. We have $x+y = 1-z$ and $x^2 + y^2 = 2-z^2$.
Factor: $$x^3 + y^3 = ( x + y ) ( x^2 - xy + y^2 ) = ( 1-z ) ( 2-z^2 - xy )$$
$$2xy = ( x + y )^2 - ( x^2 + y^2 ) = ( 1-z )^2 - ( 2 - z^2 ) = 2z^2 - 2z - 1$$
So our third equation becomes (multiplying by $2$):
$$( 1 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/403454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 3
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what values of b_2 result in solving this system? Solve the system of equations:
$$ \begin{pmatrix}
0 & 1 & 0 & 3 \\
0 & 2 & 0 & 6 \\ \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3\\ x_4 \end{pmatrix}
= \begin{pmatrix} b_1 \\ b_2\end{pmatrix}$$
What value of $b_2$ results in this system having a solution?
I d... | Your answer is correct and your reasoning is good (although I think most people would come up with $b_2-2b_1$ rather than $2b_1-b_2$ in the reduced row echelon form). However, in this particular case, there is a conceptually simpler solution (I emphasize "particular" because in general, your way of solving the problem ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the rational values of constant for which these constants are roots of equation Problem :
Determine all rational values for which $a,b,c$ are the roots of $x^3+ax^2+bx+c=0$
Solution :
Sum of the roots $a+b+c = -a$ ........(i) ( Since , as per question $a,b,c$ are roots of equation and we have to find values )... | Neither $-1$ nor $1$ is a solution to your final cubic, so this calls for the use of the cubic formula.
$$a = 2, b = 2, c = 0, d = -1$$
$$\Delta_0 = 2^2 - 3(2)(0) = 4$$
$$\Delta_1 = 2(2^3) - 9(2)(2)(0) + 27(2^2)(-1) = 16 - 108 = -92$$
$$C = \sqrt[3]{\frac{-92 + \sqrt{(-92)^2 - 4(4^3)}}{2}} = \sqrt[3]{\frac{-92 + \sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/406313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find non-singular matrices P and Q such that PAQ is in the normal form for the matrix A. $A= \left[ \begin{array}{ccc}
1 & 2 & 3 & -2 \\
2 & -2 & 1 & 3 \\
3 & 0 & 4 & 1 \end{array} \right]$
$A=IAI$
$\left[ \begin{array}{ccc}
1 & 2 & 3 & -2 \\
2 & -2 & 1 & 3 \\
3 & 0 & 4 & 1 \end{array} \right]$ = $\left[ \begin{array}{... | If I understand what you mean by normal form here, we would do the following series of operations ($R$ is row, $C$ is Column, and note the row operations are duplicated on the RHS $3x3$ and Column Operations are duplicated on the $4x4$, while performing the operations listed below on the LHS, that is, the matrix $A$ it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/407553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Stupid question about $1 - \frac{1}{2}-\frac{1}{4}+\frac{1}{3}- \frac{1}{6}-\frac{1}{8}+\frac{1}{5}\dots$ I have
$$1 - \frac{1}{2}-\frac{1}{4}+\frac{1}{3}- \frac{1}{6}-\frac{1}{8}+\frac{1}{5}\dots$$
Partial sum $S_{3n}$ of the above is:
$$(1 - \frac{1}{2}-\frac{1}{4})+(\frac{1}{3}- \frac{1}{6}-\frac{1}{8})+(\frac{1}{5}... | By guessing the pattern hidden in the ellipsis, it seems that you consider the series
$\sum_{k=1}^\infty a_k $
where $a_{3n-2}=\frac1{2n-1}$, $a_{3n-1}=-\frac1{4n-2}$, $a_{3n}=-\frac1{4n}$ for $n\ge 1$.
Thus $a_{3n-2}+a_{3n-1}+a_{3n}=\frac1{2n-1}-\frac1{4n-2}-\frac1{4n}=\frac1{4n(2n-1)}$ and we have the partial sums
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/407745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove inequality $(x+y+z-2xyz)^2 \le 2$
Problem: Prove inequality $(x+y+z-2xyz)^2 \le 2\ (1)$ with
$x^2+y^2+z^2 = 1 \land x,y,z \in \mathbb R$
I tried expand $LHS$ and have:
$$(1)\iff 1 - 2 (xy+yz+xz) + 4 xyz(x+y+z)-(2xyz)^2 \ge 0$$
Denote: $xy = a, yz = b,xz=c \implies (1) \iff1-2\sum a+ 4 \sum ab - 2abc \ge0$
Bu... | we let $x\le y\le z$,then $$z^2\ge\dfrac{1}{3},2xy\le x^2+y^2\le\dfrac{2}{3}$$
use Cauchy-Schwarz inequality we have
$$(x+y+z-2xyz)^2=[(x+y)+z(1-2xy)]^2\le [(x+y)^2+z^2][1+(1-2xy)^2]$$
we only prove
$$[(x+y)^2+z^2][1+(1-2xy)^2]\le 2$$
since:
$$ [(x+y)^2+z^2][1+(1-2xy)^2]=(1+2xy)[1+(1-2xy)^2]=2+4x^2y^2(2xy-1)\le 2$$
By ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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How do I find the exact value of $\cos\frac{\pi}{12}\cos\frac{5\pi}{12}\cos\frac{7\pi}{12}\cos\frac{11\pi}{12}$? I know that $\cos(6\phi)\equiv32c^6-48c^4+18c^2-1$ where $c=\cos\phi$.
I also know that when $\cos(6\phi)=0$, then $\phi=\frac{k\pi}{12}$ ($k = 1,3,5,7,9,11$).
How do I find the exact value of:
$$\cos\left(\... | Mosquito-nuking linear algebraic solution: it can be shown that the tridiagonal matrix
$$\mathbf T=\frac12\begin{pmatrix}-1&1&&&&\\1&0&1&&&\\&1&0&1&&\\&&1&0&1&\\&&&1&0&1\\&&&&1&1\end{pmatrix}$$
has the eigenvalues $\cos\dfrac{(2k-1)\pi}{12},\quad k=1,\dots,6$. Since the product of the eigenvalues is equal to the determ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Correlation Coefficient - $\rho(X,Y)$. If I have two aleatory variables
$$X=\begin{pmatrix}2&3&4&5&6&7&8&9&10&11&12\\ \frac{1}{36}&\frac{2}{36}&\frac{3}{36}&\frac{4}{36}&\frac{5}{36}&\frac{6}{36}&\frac{5}{36}&\frac{4}{36}&\frac{3}{36}&\frac{2}{36}&\frac{1}{36}\end{pmatrix}$$ and $$Y=\begin{pmatrix}2&3&4&5&6&7&8&9&10&11... | You cannot find the correlation without additional information. $X$ and $Y$ could be independent, in which case the correlation is $0$, or (in this instance) $X$ might always be equal to $Y$ so the correlation would be $1$, or $X$ might (again, in this instance) be equal to $14-Y$ (as $X$ goes from $2$ to $12$, then $... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find volume using double integrals? Question: Use double integral to find the volume of the solid enclosed by the spheres $x^2+y^2+z^2=1$ and $x^2+y^2+(z-1)^2=1$
Alright so I tried to doing this by myself and I'm not sure if this is right. Could someone check over my work?
Curve of intersection:
\begin{align*}
x^2 + y^... | By the word "enclosed", I am assuming you want to find the volume bounded by these two spheres, which means the solid formed by their intersection. You claimed:
$x^2 + y^2 + (z-1)^2 = 1$ is above $x^2 + y^2 + z^2 = 1$ when $z > 1/2$.
If we are truely like to find the volume of the intersection, then this is not accur... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How many compositions of n are there with same parity Let n be a non-negative integer. How many compositions of n are there where the i-th part has the same parity as i? For example, compositions of 7 that satisfy this condition are (7), (5,2), (3,4), (1,6), (1,2,1,2,1). Express your answer as the coefficient of a simp... | Hint: The number of solutions where there is exactly 1 part is the coefficient of $x^n$ in $x + x^3 + x^5 + \ldots = \frac{x}{1-x^2}$.
Hint: The number of solutions where there are exactly 2 parts is the coefficients of $x^n$ in $(x+x^3+x^5 + \ldots ) \times (x^2 + x^4 + x^6 + \ldots ) = \frac{ x } { 1-x^2} \times \f... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Similarity and Scale Factor If I have two triangles ABC and ADE and I know that each of the triangles is equilateral, each of the angles is 60 degrees. Is there a way to determine the scale factor of the two triangles, for example:
The ratio:
Area of ADE $:$Area of ABC
Thanks in advance.
| If side $DC$$ = $ $x$, then $AC$ must be $2x$ because both triangles are equilateral. By Pythagorean theorem, $AD$ is $3^\frac{1}{2}x$. Then $AD:AC=3^\frac{1}{2}:2=\frac{3^\frac{1}{2}{}}{2}:1$. So the scale factor $k$, is $\frac{3^\frac{1}{2}}{2}$. Using the formula:
$\frac{\text{Area of }AED}{\text{Area of }ABC}=k^2$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/412357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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simplifying $\min(\max(A,B),C) $ In a larger problem, I have to make use of the following
$$\min(\max(A,\ B),\ C)$$
Please how do I simplify?
| For $a, b ∈ \Bbb R:$
$$\max\left(a,b\right) = \frac{a+b+\left|a-b\right|}{2}$$
$$\min\left(a,b\right) = \frac{a+b-\left|a-b\right|}{2}$$
Therefore, if $a,b,c ∈ \Bbb R:$
$$\min\left(\max\left(a,b\right),c\right) =
\min\left(\frac{a+b+\left| a-b\right|}{2},c\right) = $$
$$=\frac{\frac{a+b+\left| a-b\right|}{2}+c-\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/412541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculating $\sum^{10}_{k=1}\left(\sin\frac{2k\pi}{11}+i\cos\frac{2k\pi}{11}\right)$
Find the value of $$\sum^{10}_{k=1}\left (\sin\left (\frac{2k\pi}{11} \right )+i\cos\left (\frac{2k\pi}{11}\right ) \right)$$
My approach:
Since $\cos\theta + i\sin\theta = e^{i\theta}$, we can write the given equation as:
$$\begin... | I am not sure where your steps came from. You are right that you are summing
$$i \sum_{k=1}^{10} e^{-i 2 \pi k/11}$$
This is a geometric series and has value
$$i\frac{e^{-i 2 \pi/11}-e^{-i 2 \pi}}{1-e^{-i 2 \pi/11}} = i\frac{e^{-i 2 \pi/11}-1}{1-e^{-i 2 \pi/11}}= -i$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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approximate $\int_{u=0}^b e^{-\frac{u^2 + ac}{au}}du.$ I'm trying to find an approximation (or exact solution if possible) for an integral of the form:
$$\int_{u=0}^b e^{-\frac{u^2 + ac}{au}}du.$$
I was thinking of somehow applying a Gauss Hermite Quadriture Expansion, but I'm not sure how I would do this. Does anyone ... | I do not know much about numerical integration, but the following trick may help you avoid exponents diverge ad infinitum.
We are dealing with the integral
$$ I = \int_{0}^{b} \exp\left( - \frac{u}{a} - \frac{c}{u} \right) \, du. $$
With the substitution $u = \sqrt{a c} x^{2}$, for $\beta = \left( b / \sqrt{ac} \right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/412878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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why is this Markov Chain aperiodic I have this Matrix:
$$P=\begin{pmatrix} 0 & 1 \\ 0.3 & 0.7 \end{pmatrix}$$
this markov chain is said to be aperiodic, I dont understand how it comes to it. Period $\delta$ is the gcd of the set of all diagonal elements, right? if $\delta>1$, $P$ is periodic, if $\delta=1$, then aper... | Since the Jordan Normal Form of the matrix is
$$
\begin{bmatrix}
0&1\\
0.3&0.7
\end{bmatrix}
=\begin{bmatrix}
-10&3\\3&3
\end{bmatrix}
\begin{bmatrix}
-0.3&0\\0&1
\end{bmatrix}
\begin{bmatrix}
-10&3\\3&3
\end{bmatrix}^{-1}
$$
we have
$$
\begin{align}
\begin{bmatrix}
0&1\\
0.3&0.7
\end{bmatrix}^n
&=\begin{bmatrix}
-10&3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/413610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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How to prove $(a-b)^3 + (b-c)^3 + (c-a)^3 -3(a-b)(b-c)(c-a) = 0$ without calculations I read somewhere that I can prove this identity below with abstract algebra in a simpler and faster way without any calculations, is that true or am I wrong?
$$(a-b)^3 + (b-c)^3 + (c-a)^3 -3(a-b)(b-c)(c-a) = 0$$
Thanks
| Notice that this is invarient if the same constant is added to all three variables, so one of them may be set equal to zero. Set $c=0$. Then the expression becomes $(a-b)^3 -a^3 +b^3+3ab(a-b)$, easily seen to be zero.
BTW, apparently the original expression is equivalent to saying $x^3 +y^3+z^3 =3xyz$ if $x+y+z=0$.
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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"answer_id": 2
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Prove $\frac{1}{2 \pi} \int_0^{2\pi} \frac{1 - r^2}{1 + r^2 - 2r \cos{t}} dt = 1$ using contour integration The question is to solve the integral using concepts of contour integrals:
$$\frac{1}{2 \pi} \int_0^{2\pi} \frac{1 - r^2}{1 + r^2 - 2r \cos{t}} dt = 1$$
| Let $$I(r) = \dfrac{1}{2 \pi} \int_0^{2\pi} \frac{1 - r^2}{1 + r^2 - 2r \cos{t}} dt \tag{$\spadesuit$}$$
Splitting the integral from $0$ to $\pi$ and $\pi$ to $2\pi$, changing $t \to \pi-t$, in the second integral, we get
$$I(r) = \dfrac{1}{\pi} \int_0^{\pi} \frac{1 - r^2}{1 + r^2 - 2r \cos{t}} dt = \dfrac{1}{\pi} \int... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is my proof correct? $2^n=x^2+23$ has an infinite number of (integer) solutions. This is how I tried to prove it. Is it correct? Thanks!!
$2^n = x^2+23$
$x^2$ must be odd, therefore $x^2 = 4k+1$, where $k \in \mathbb{N}$.
$2^n=4k+24$
$k=2(2^{n-3}-3)$
Since $x^2=4k+1$,$ \ \ \ \ $ $k_1 = \frac{x_1^2-1}{4}$
and $k_2=\frac... | There are only finitely many solutions. Pillai's conjecture is overkill because it allows for all possible exponents simultaneously, whereas the solutions of $2^n = x^2 + 23$ must lie on one of the three curves $y^3 = x^2+23$, $2y^3 = x^2+23$, $4y^3= x^2+23$, each of which has finitely many integer points.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to integrate $\frac{x+2}{x^2+2x+2}$ with the substitution method This is what I got so far:
$$\frac{x+2}{x^2+2x+2} = \frac{x+2}{(x+1)^2+1} = \int \frac{x}{(x+1)^2+1} + 2 \times \int \frac{1}{(x+1)^2+1}$$
I know the last integral $= \arctan(x+1) + c$, where $c$ is a constant, but I don't know how to integrate the fi... | HINT:
$$\int\frac{x+2}{x^2+2x+2}dx=\int\frac{x+2}{(x+1)^2+1^2}dx$$
Put $x+1=\tan \theta$
Alternatively, put $x+2= A\frac{d(x^2+2x+2)}{dx}+B$
i.e., $x+2=A\cdot2(x+1)+B=2Ax+2A+B$
Comparing the coefficients of $x,1=2A\implies A=\frac12$
Comparing the constants, $2A+B=2\implies B=2-2A=1$
So, $$\int\frac{x+2}{x^2+2x+2}dx$... | {
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"timestamp": "2023-03-29T00:00:00",
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The system of equations $x^2 + y^2 - x - 2y = 0$ and $x + 2y = c$ I have
$(1.) \quad x^2 + y^2 - x - 2y = 0 \\
(2.) \quad x + 2y = c$
Solving for $y$ in $(2.)$ gives
$y = (c - x) / 2$
Is there a way to simplify equation $(1.)$?
Because at the end I arrive at
$c^2 - 2x - c = 0$
and can't proceed. Need to get typical fo... | If you substitute correctly, you will get the equation $5\,{x}^{2}-2\,c\,x+{c}^{2}-4\,c=0$.
If you solve this, you will get $x = \frac{1}{5}(c \pm 2\sqrt{c(5-c)})$. The other value is given by $y=\frac{1}{2}(c-x)$, but you don't need this to answer your question.
If the two sets of solutions have the same values, then ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How find $a$ such that $x^2-\sqrt{a-x}=a$ has exactly two real solutions Consider the equation
$$ x^2-\sqrt{a-x}=a.$$ I wish to determine the values of $a$ for which the above equation has exactly two real solutions (for $x$).
My idea:
$$a-x=(x^2-a)^2=x^4-2ax^2+a^2\Longrightarrow f(x)=x^4-2ax^2+x+a^2-a=0$$ and we must ... | Another idea is:
Equation is equivalent to the system
1) $\sqrt{a-x}=y$
2) $\sqrt{a+y}=|x|$
with conditions
3) $x\leq a$ and $y\geq0$, $y \geq -a$.
Are obtained by squaring
1') $ a-x = y^{2}$
2') $ a+y = x^{2}$.
Subtracting the last equation is obtained:
$(x+y)(1-x+y)=0$ si cazurile $y= -x$ si $y =x-1$.
Replace $ y $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/417098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
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How to determine the rank and determinant of $A$? let $A$ be $$A_{a} = \begin{pmatrix} a & 1 & 1 & 1 \\ 1 & a & 1 & 1\\ 1 & 1 & a & 1\\ 1 & 1 & 1 & a \end{pmatrix}$$
How can I calculate the rank of $A$ by the Gauss' methode and $\det A$?
| $\det A_{a} =\det \begin{pmatrix} a & 1 & 1 & 1 \\ 1 & a & 1 & 1\\ 1 & 1 & a & 1\\ 1 & 1 & 1 & a \end{pmatrix}$
Adding all the columns to the 1st column we have,
$\det A_{a} = \det \begin{pmatrix} a+1+1+1 & 1 & 1 & 1 \\ a+1+1+1 & a & 1 & 1\\ a+1+1+1 & 1 & a & 1\\ a+1+1+1 & 1 & 1 & a \end{pmatrix}$
Taking $a+1+1+1$ comm... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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Prove $\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2}\ge 6$, given $x+y+z=3$ and $x,y,z\ge0$
Let $x+y+z=3,x,y,z\ge 0$,show that
$$\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2}\ge 6$$
Additional information
I have seen the following problem:
$x,y,z>0,x+y+z=3$, prove that
$$\sqrt{x^2+y+2}+\sqrt{y^2+z+2}+\sqrt{z^... | Here is a sketch of another alternative approach.
Substitute variables $x=a+1,y=b+1,z=c+1$; then $$f(a,b,c)=\sqrt{4+a^2+bc+a}+\sqrt{4+b^2+ca+b}+\sqrt{4+c^2+ab+c}$$ — I exploited the fact that $a+b+c=x+y+z-3=0$; also note, that $-1\le a,b,c \le 2$
Use Taylor formula $\sqrt{4+t}=2\sqrt{1+\frac{t}{4}}=2+\frac{t}{4}-\frac{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 4,
"answer_id": 3
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Determinant of symmetric matrix Given the following matrix, is there a way to compute the determinant other than using laplace till there're $3\times3$ determinants?
\begin{pmatrix}
2 & 1 &1 &1&1 \\
1 & 2 & 1& 1 &1\\
1& 1 & 2 & 1 &1\\
1&1 &1 &2&1\\
1&1&1&1&-2
\end{pmatrix}
| You can substract the first row from every other rows and get matrix of form:
$$\begin{pmatrix}
2 & 1 &1 &1&1 \\
-1 & 1 & 0& 0 &0\\
-1& 0 & 1 & 0 &0\\
-1&0 &0 &1&0\\
-1&0&0&0&-3
\end{pmatrix}.$$
Computing the determinant is now much easier.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Evaluating $\lim_{x\to1}\frac{\sqrt{x^2+3}-2}{x^2-1}$? I tried to calculate, but couldn't get out of this:
$$\lim_{x\to1}\frac{x^2+5}{x^2 (\sqrt{x^2 +3}+2)-\sqrt{x^2 +3}}$$
then multiply by the conjugate.
$$\lim_{x\to1}\frac{\sqrt{x^2 +3}-2}{x^2 -1}$$
Thanks!
| You were right to multiply "top" and "bottom" by the conjugate of the numerator. I suspect you simply made a few algebra mistakes that got you stuck with the limit you first posted:
So we start from the beginning:
$$\lim_{x\to1}\frac{\sqrt{x^2 +3}-2}{x^2 -1}$$
and multiply top and bottom by the conjugate, $\;\sqrt{x^2... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
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Find the generating function for this set of strings Let $a(n)$ be the number of $\{0,1\}$-strings of length $n$ which contain no $4$ consecutive $1$'s and no $4$ consecutive $0$'s (don't contain "$0000$" or "$1111$"). Find the generating function for this.
Thank you.
| Let $a(n)$ denote the number of admissible binary strings of length $n$. First, note that precisely half of the strings of any length end in $\color{red}{0}$ (or $\color{green}{1}$). Now, for a string of length $n>4$, there are $8$ possible endings with the following numbers of occurrences:
$$\begin{align*}
&\overbrace... | {
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Find all $n$ such that $\sigma(n)=12$ Let $ \sigma (n) = \sum_{k|n}^{}{k} $. I need to solve $\sigma(n)=12$.
Probably the following might be of use: if $n={p_1}^{a_1}{p_2}^{a_2}...{p_s}^{a_s}$ then $\sigma(n)=\frac{{p_1}^{a_1+1}-1}{p_1-1}\frac{{p_2}^{a_2+1}-1}{p_2-1}...\frac{{p_s}^{a_s+1}-1}{p_s-1}$.
| If RHS $=12,$ as $1+p+p^2=7$ or $\ge 1+3+3^3=13,n$ must be square-free
So, $\sigma(n)=\prod(1+p_i)$ where $p_i$s are distinct prime divisors $(\ge2)$ of $n$
Also, $\sigma(n)\ge 1+n \ \ \ \ (1)$
The equality occurs if $n$ is prime i.e, here $n+1=12\implies n=11$ which is prime.
$(1)\implies n\le 11$
So, other the values... | {
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An identity involving the Beta function I'm trying to show that
$$ \int _0^1 \frac{x^{a-1}(1-x)^{b-1}}{(x+c)^{a+b}}dx = \frac{B(a,b)}{(1+c)^ac^b}$$
Where $$B(a,b) := \int _0^1 x^{a-1}(1-x)^{b-1}dx $$ is the "Beta function". I am supposed to use a substitution but I'm pretty much stuck. I am familiar with the basic ... | Let $\displaystyle \int _0^1 \frac{x^{a-1}(1-x)^{b-1}}{(x+c)^{a+b}}dx = I$
We have ,
$(1+c)^ac^bI=\displaystyle \int _0^1 \frac{(1+c)^ac^bx^{a-1}(1-x)^{b-1}}{(x+c)^{a+b}}dx $
$=\displaystyle \int _0^1 \frac{(1+c)c((1+c)x)^{a-1}(c(1-x))^{b-1}}{(x+c)^{a+b}}dx $
$=\displaystyle \int _0^1 \left(\frac{(1+c)c}{(x+c)^2}... | {
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"answer_count": 3,
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} |
About rationalizing expressions For example, rationalizing expressions like
$$\frac{1}{\pm \sqrt{a} \pm \sqrt{b}}$$
Is straightforward. Moreover cases like
$$\frac{1}{\pm \sqrt{a} \pm \sqrt{b} \pm \sqrt{c}}$$
and
$$\frac{1}{\pm \sqrt{a} \pm \sqrt{b} \pm \sqrt{c} \pm \sqrt{d}}$$
Are still easy to rationalize. But my qu... | Yes, it's always possible to rationalize such expressions.
Without loss of generality, it's enough to consider the first square root having positive sign in front of it. Then, see what happens if you multiply together terms with all possible combinations of the remaining signs:
$$\left(\sqrt{a}+\sqrt{b}\right)\left(\sq... | {
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"answer_count": 3,
"answer_id": 2
} |
Are these rational curves? I have to find the singular points of the following curves and tell if they are rational. The curves are $C=Z(x^2+y^2+x^2y^2)$ and $C=Z(x^3+y^3-1)$, and the base field is the complex one.
I think I found the singular points but I'm stuck in finding the rational morphism. Can some one help me?... | A plane affine curve is rational iff its projectivization is rational, so let's study these projectivizations since we have more tools at our disposal in the projective plane.
2) The projectivization of the second curve is the curve $x^3+y^3-z^3=0$.
This is a smooth curve of genus $\frac {(3-1)(3-2)}{2}=1$, so it i... | {
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"answer_id": 0
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Derivation of Pythagorean Triple General Solution Starting Point: I was reading on proof wiki about the derivation of the general solution to the pythagorean triple diophantine equation:
$$
x^2 + y^2 = z^2,
$$
where $x,y,z > 0$ are integers.
I came across the following general solution to the primitive function:
\begi... | Here is the way to generate all relatively prime pythagorean triples:
Theorem: Let $m$ and $n$ be positive integers so that
$$
\begin{align}
&m\gt n\tag{1}\\
&m+n\text{ is odd}\tag{2}\\
&m\text{ and }n\text{ are relatively prime}\tag{3}
\end{align}
$$
Then,
$$
\begin{align}
a &= m^2 - n^2\\
b &= 2mn\\
c &= m^2 + n^2
\... | {
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How to determine whether this series convergent or divergent? Does
$$
\dfrac{7}{19}+\dfrac{7}{19}\sqrt{\dfrac{7}{19}}+\dfrac{7}{19}\sqrt{\dfrac{7}{19}}\sqrt[3]{\dfrac{7}{19}}+\cdots+\dfrac{7}{19}\sqrt{\dfrac{7}{19}}\sqrt[3]{\dfrac{7}{19}}\cdots\sqrt[n]{\dfrac{7}{19}}+\cdots
$$
converge or diverge?
The following is my ... | Since
$$
H_n = \sum_{k=1}^{n} \frac{1}{k} = \log n + \gamma + o(1),
$$
we have, for $x > 0$,
$$
x^{H_n} \sim x^{\log n + \gamma} = x^\gamma n^{\log x}
$$
As $n \to \infty$. We may then apply the limit comparison test to conclude that
$$
\sum_{n=1}^{\infty} \frac{1}{x^{H_n}}
$$
converges when $\log x > 1 \Leftrightarro... | {
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"answer_count": 2,
"answer_id": 0
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How can evaluate $\lim_{x\to0}\frac{\sin(x^2+\frac{1}{x})-\sin\frac{1}{x}}{x}$ I don't know if I apply for this case sin (a-b), or if it is the case of another type of resolution, someone with some idea without using derivation or L'Hôpital's rule? Thank you.
$$\lim_{x\to0}\frac{\sin(x^2+\frac{1}{x})-\sin\frac{1}{x}}{x... | Using $\sin\left(a\right)-\sin\left(b\right)=2\sin\left(\frac{a-b}{2}\right)\cos\left(\frac{a+b}{2}\right)$ and $\cos\left(a+b\right)=\cos\left(a\right)\cos\left(b\right)-\sin\left(a\right)\sin\left(b\right)$: $$L=\lim_{x\rightarrow 0}\frac{2\sin\left(x^2/2\right)\cos\left(x^{-1}+x^2/2\right)}{x}\\=\lim_{x\rightarrow 0... | {
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"answer_count": 4,
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Prove that $\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}+\frac{a^2+b^2}{a+b} \ge a+b+c$ If $a,b,c$ are positive , show that $$\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b} \ge a+b+c$$
Trial: Here I proceed in this way $$\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b} \ge \dfrac{2bc}{b+c}+\d... | It suffices to show, by symmetry, that
$$\frac{a^2+b^2}{a+b}\ge \frac{a+b}{2}.$$
This is equivalent to
$$2a^2+2b^2\ge a^2+b^2+2ab\iff a^2+b^2\ge 2ab.$$
The last inequality is just the standard AM-GM inequality.
| {
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"answer_id": 2
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Linear Independence by row space.
Let $(1,1,-1), (2,1,0)$ and $(-1,0,1)$ be vectors, show if they are independent.
I wrote each vector on the rows of the matrix $A$.
$A=\begin{pmatrix} 1 & 1 & -1 \\ 2 & 1 & 0\\ -1 & 0 & 1 \end{pmatrix}$
Then I put $A$ on an echelon form.
$A=\begin{pmatrix} 1 & 1 & -1 \\ 2 & 1 & 0\\ -... | Yes, this is correct.
You could also just compute the determinant of $A$ and conclude that the rows (and columns too) are linear independent.
| {
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"answer_id": 1
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A question on differentiation : Let $$f(x)=\sin^{-1}(2x\sqrt{1-x^2})$$
I found out $f'(x)$ in three methods and got three different answers !
1) Putting $x=\cos\theta$, we get $f(x)=2\cos^{-1}x$, on differentiating this we get
$$f'(x)=\frac{-2}{\sqrt{1-x^2}}$$
2) Putting $x=\sin\theta$, we get $f(x)=2\sin^{-1}x$, on ... | One should, before starting differentiating, check where the expression is defined. The function $f$ is defined where $-1\le 2x\sqrt{1-x^2}\le1$, that is,
$$
4x^2(1-x^2)\le 1
$$
which is satisfied for all $x$. Thus the function $f$ is defined in $[-1,1]$ because of the square root.
Apply the chain rule; the derivative ... | {
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"answer_count": 2,
"answer_id": 0
} |
Evaluation of $\lim\limits_{n\to\infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2}) $ Could you, please, check if I solved it right.
\begin{align*}
\lim_{n \rightarrow \infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2})
&= \lim_{n \rightarrow \infty} \sqrt{n^2(1 + \frac1n)}
- \sqrt[3]{n^3(1 + \frac1n)})\\
&= \lim_{n \rightarrow ... | You want
$\lim_{n \rightarrow \infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2})
$.
More generally,
consider
$f_{a, c}(n)
=\sqrt[a]{n^a+n^{a-c}}
$
where
$a> 0$
and
$a > c > 0$.
I will use the
generalized binomial theorem
and generalized binomial coefficients.
We have
$\begin{array}\\
f_{a, c}(n)
&=\sqrt[a]{n^a+n^{a-c}}\\
&... | {
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"answer_count": 8,
"answer_id": 7
} |
Finding the value of $\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$ Is it possible to find the value of
$$\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$$
Does it help if I set it equal to $x$? Or I mean what can I possibly do?
$$x=\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$$
$$x^2=1+2\sq... | Not an answer of a closed form, but we can use Ramanujan's formula to approximate:
For any $n\in \mathbb{N}$ $$f(n) = 1+ n = \sqrt{1+n\sqrt{1+(n+1)\sqrt{1+(n+2)\sqrt{1+(n+3)\sqrt{1+\dots}}}}}.$$
Let the sum be $x$.
$$
\begin{aligned}
x &> \sqrt{1+2\sqrt{\color{blue}{1}+3\sqrt{\color{blue}{1}+4\sqrt{\color{blue}{1}+... | {
"language": "en",
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"source": "stackexchange",
"question_score": "51",
"answer_count": 5,
"answer_id": 0
} |
show that the function $z = 2x^2 + y^2 +2xy -2x +2y +2$ is greater than $-3$ Show that the function
$$z = 2x^2 + y^2 +2xy -2x +2y +2$$
is greater than $-3$
I tried to factorize but couldn't get more than $(x-1)^2 + (x+y)^2 +(y-1)^2 - (y)^2$.
Is there any another way to factorize or another method??
| The equation $2x^2 + y^2 +2xy -2x +2y +2=0$ is a conic in the plane; so you can use the good old methods to put it in canonical form; set $x=X+a$ and $y=Y+b$, to get
$$
2X^2+4aX+2a^2+
Y^2+2bY+b^2+
2XY+2bX+2aY+2ab
-2X-2a+
2Y+2b+
2
=2X^2+2XY+Y^2+(4a+2b-2)X+(2b+2a+2)Y+2a^2+2ab+b^2-2a+2b+2
$$
and you get zero coefficients ... | {
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"source": "stackexchange",
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"answer_count": 5,
"answer_id": 2
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Find $\cos(2\alpha)$ given $\cos(\theta -\alpha)$ and $\sin(\theta +\alpha)$ My question is:
If $\cos(\theta -\alpha) = \frac{3}{5}$ and $\sin(\theta +\alpha) =\frac{12}{13}$, find $\cos(2\alpha)$.
Attempt I:
\begin{align*}
&\cos^2(\theta -\alpha)+\sin^2(\theta +\alpha)
= \frac{9}{25} + \frac{144}{169}\\
\Rightarrow ... | HINT:
$$\cos(2\alpha)=\cos\{\theta+\alpha-(\theta-\alpha)\}=\cos(\theta+\alpha)\cos(\theta-\alpha)+\sin(\theta+\alpha)\sin(\theta-\alpha)$$
As $\cos(\theta-\alpha)=\frac35,$
$\sin(\theta-\alpha)=\pm \sqrt{1-\left(\frac35\right)^2}=\pm\frac45$
Similarly, as $\sin(\theta+\alpha)=\frac{12}{13},$ $\cos(\theta+\alpha)=\pm... | {
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"answer_count": 1,
"answer_id": 0
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Inequality $\frac{1-3ab}{1-2ac}+\frac{1-3bc}{1-2ba}+\frac{1-3ca}{1-2cb}\geq 0$
Let $a\ne 0$, $b\ne 0$ and $c\ne 0$ such that $a^2+b^2+c^2=1$. Prove that: $$\dfrac{1-3ab}{1-2ac}+\dfrac{1-3bc}{1-2ba}+\dfrac{1-3ca}{1-2cb}\geq 0.$$
My attempt to the solution: We get that $ab +bc+ca$ lies between $-0.5$ and $1$. We can us... | I will use the notation $\sum_{cyc} f(x,y,z)$ to mean that the sum is over the so-called cyclic permutations of $(x,y,z)$ (or the even ones if you prefer),
so $\sum_{cyc} f(x,y,z):=f(x,y,z)+f(y,z,x)+f(z,x,y)$ (in place of $x,y,z$ there will also be other variables).
Lemma. If $x,y,z\ge 0$ then $\sum_{cyc} (x+y-2z)(z+x)... | {
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"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
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Factor Equation Help me with this,
Question: factor $x^3y-x^3z+y^3z-xy^3+xz^3-yz^3$.
Solution:
$$\begin{eqnarray}&=&x^3y-x^3z+y^3z-xy^3+xz^3-yz^3\\
&=&x\left(z^3-y^3\right)+y\left(x^3-z^3\right)+z\left(y^3-x^3\right)\\
&=&x\left[(z-y)\left(z^2+zy+y^2\right)\right]+y\left[(x-z)\left(x^2+xz+z^2\right)\right]+z\left[(y-x)... | $x^3y-x^3z+y^3z-xy^3+xz^3-yz^3$
$=x^3(y-z)+yz(y^2-z^2)-x(y^3-z^3)$
$=x^3(y-z)+yz(y+z)(y-z)-x(y-z)(y^2+yz+z^2)$
$=(y-z)\{x^3+yz(y+z)-x(y^2+yz+z^2)\}$
Now, $x^3+yz(y+z)-x(y^2+yz+z^2)$
$=x^3+y^2z+yz^2-xy^2-xyz-z^2x$
$=x(x^2-y^2)-yz(x-y)-z^2(x-y)$
$=(x-y)\{x(x+y)-yz-z^2\}$
Now, $x(x+y)-yz-z^2$
$=x^2+xy-yz-z^2=(x+z)(x-z)+y... | {
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General solution for trigonometry equation How should I state the general solution for the equation $\sin(4\phi)=\cos(2\phi)$.
The angles are $15$, $45$, $75$ and $135$ if I restrict myself within the range $[0,360]$
| As $\cos2\phi=\sin4\phi=\cos(90^\circ-4\phi)$
$\implies 2\phi=n360^\circ\pm(90^\circ-4\phi)$ where $n$ is any integer
Taking '+' sign, $2\phi=n360^\circ+90^\circ-4\phi$
$\implies 6\phi=n360^\circ+90^\circ \implies \phi=n60^\circ+15^\circ$
As $0\le \phi<360^\circ, 0\le n60^\circ+15^\circ<360^\circ\implies 0\le n\le 5$
T... | {
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Prove that $a^3+b^3+c^3 \geq a^2b+b^2c+c^2a$
Let $a,b,c$ be positive real numbers. Prove that $a^3+b^3+c^3\geq a^2b+b^2c+c^2a$.
My (strange) proof:
$$
\begin{align*}
a^3+b^3+c^3 &\geq a^2b+b^2c+c^2a\\
\sum\limits_{a,b,c} a^3 &\geq \sum\limits_{a,b,c} a^2b\\
\sum\limits_{a,b,c} a^2 &\geq \sum\limits_{a,b,c} ab\\
a^2+b... | WOLG, Let $c$=Max{$a,b,c$}, then there is 2 cases:
case I: $0<a \le b \le c$, we want to prove $c^2(c-a) \ge a^2(b-a)+b^2(c-b)$
we have $c^2\ge b^2, c^2\ge a^2 \to $,RHS $\le c^2(b-a)+c^2(c-b)=c^2(c-a)$
case II: $0<b \le a \le c$, we want to prove $a^2(a-b)+c^2(c-a) \ge b^2(c-b)$
we have $a^2\ge b^2,c^2 \ge b^2, \to$L... | {
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"answer_id": 3
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How do I write a sum of cosines as a product of sines? I am trying to prove that
$$\cos A+\cos B+\cos C=4\sin\frac A2\sin\frac B2\sin\frac C2$$ for ABC is a triangle.
I tried up to the stage of
$$-2\sin^2 C+2\cos\frac{180-C}2 \cos\frac{A+B}2$$
but how do I proceed from here?
| PROOF: There is something wrong in the question
TO PROVE :$\cos A+\cos B+\cos C -1=4\sin\frac A2\sin\frac B2\sin\frac C2$
PROOF: You can proceed from the right hand side too
$4\sin\frac A2\sin\frac B2\sin\frac C2 = 2\sin\frac A2\sin\frac B2 2\sin\frac C2$
NOW, $2\sin\frac A2\sin\frac B2 = cos\frac{A-B}2 - \cos\frac{A+B... | {
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Prove this inequality $a^{\frac{a}{b}}b^{\frac{b}{c}}c\geq1$ Please help me to prove this inequality.
Assume $a,b,c>0$ and $abc\geq1$ then $a^{\frac{a}{b}}b^{\frac{b}{c}}c\geq1$.
Thanks.
| I think the right inequality is:
$a^{\frac{a}{b}}b^{\frac{b}{c}}c^{\frac{c}{a}} \ge 1$
to prove it is to prove $\dfrac{a}{b}\log{a}+\dfrac{b}{c}\log{b}+\dfrac{c}{a}\log{c} \ge 0$
let $abc=1 \to a=\dfrac{1}{bc}$
LHS=$(b^3-1)\log{b}+(b^3c^3-1)\log{c}$
it is trivial when $b>1$ and $c>1$, $b<1$ and $c<1$, LHS $> 0$
when ... | {
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"answer_count": 2,
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Evaluate $\int_2^\infty{\frac{3x-2}{x^2(x-1)}}$ To be shown that $\int_2^\infty{\dfrac{3x-2}{x^2(x-1)}}=1-\ln2$
My thought: $\dfrac{3x-2}{x^2(x-1)}=\dfrac{3x}{x^2(x-1)}-\dfrac{2}{x^2(x-1)}$
• $\dfrac{3x}{x^2(x-1)}=\dfrac{3}{x(x-1)}=\ldots=-\dfrac{3}{x}+\dfrac{3}{x-1}$
• $\dfrac{2}{x^2(x-1)}=\ldots=-\dfrac{2}{x^2}+\dfra... | Hint
Write the integrand as: $$\frac{3x-2}{x^2(x-1)}=\frac{Ax+B}{x^2}+\frac{C}{x-1}$$ in which $A,B$ and $C$ is unknown constants and so find the proper values for them. I think via this way you can find them easier than the way you noted. Indeed, $$\frac{Ax+B}{x^2}+\frac{C}{x-1}=\frac{(A+C)x^2+x(B-A)-B}{x^2(x-1)}$$ an... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Maximize $(a-1)(b-1)(c-1)$ knowing that : $a+b+c=abc$. If : $a,b,c>0$, and : $a+b+c=abc$, then find the maximum of $(a-1)(b-1)(c-1)$.
I noted that : $a+b+c\geq 3\sqrt{3}$, I believe that the maximum is at : $a=b=c=\sqrt{3}$. (Can you give hints).
| Hints:
Lagrange multipliers. Define
$$H(x,y,z,\lambda):=(x-1)(y-1)(z-1)+\lambda(x+y+z-xyz)$$
$$\begin{align*}H'_x&=(y-1)(z-1)+\lambda(1-yz)=0\iff \lambda=\frac{(y-1)(z-1)}{yz-1}\\
H'_y&=(x-1)(z-1)+\lambda(1-xz)=0\iff \lambda=\frac{(x-1)(z-1)}{xz-1}\\
H'_z&=(y-1)(x-1)+\lambda(1-yx)=0\iff \lambda=\frac{(y-1)(x-1)}{yx-1}... | {
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"answer_count": 4,
"answer_id": 2
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Prove that $(n+1)(n+2)(n+3)$ is $O(n^3)$ Problem
*
*Prove that $(n+1)(n+2)(n+3)$ is $O(n^3)$
Attempt at Solution
*
*$f(n) = (n+1)(n+2)(n+3)$
*$g(n) = n^3$
*Show that there exists an $n_0$ and $C > 0$ such that $f(n) \le Cg(n)$ whenever $n > n_0$
*$f(n) = n^3+6n^2+11n+6 = n^3(1 + 6/n + 11/n^2 + 6/n^3)$
*$f(... | You are almost there.
Remember that big-O (and little-O) statements
apply for all large enough $n$,
not for all $n$.
So, in
$(3 + 9/n + 6/n^2) < C$,
give some lower bound for $n$,
say $n > 3$.
Then $9/n < 3$ and
$6/n^2 < 2/3$,
so
$3 + 9/n + 6/n^2 < 3+3+2/3 < 7$,
so $C=7 $ works for $n > 3$.
Note that, as the lower boun... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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For what natural numbers is $n^3 < 2^n$? Prove by induction Problem
For what natural numbers is $n^3 < 2^n$?
Attempt @ Solution
*
*For $n=1$, $1 < 2$
*Suppose $n^3 < 2^n$ for some $n = k \ge 1$
*It looks like the inequality is true for $n = 0$, $n = 1$ and $n\ge10$
*But, how can I prove this through induction?
| You need to gather more data first:
$$\begin{array}{rcc}
n:&1&2&3&4&5&6&7&8&9&10&11\\
n^3:&1&8&27&64&125&216&343&512&729&1000&1331\\
2^n:&2&4&8&16&32&64&128&256&512&1024&2048
\end{array}$$
Notice that $n^3$ is not less than $2^n$ for $n=2,3,\ldots,9$; the fact that $1^3<2^1$ is an isolated success. The simplest reasona... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/447923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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find correlation coefficient of $f(x,y)=2$ for $0
Find the correlation coefficient for the random variables $X$ and $Y$ having joint density $f(x,y)=2$ for $0 < x \leq y<1$.
Seem like a simple problem but I'm stuck.
Since $Corr(X,Y) = \frac{Cov(X,Y)}{\sqrt{Var(X)}\sqrt{Var(Y)}}$, I figure I start with $Var(X)$ and $Var... | Your integrals look right. Note that $E(X)E(Y)=\frac{2}{9}$, so the covariance calculation is not right.
The covariance is $\frac{1}{4}-\frac{2}{9}=\frac{1}{36}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/450064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Existence of linear mapping I am studying for an exam in linear algebra and I am having trouble solving the following:
Do linear mappings $\phi : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ with the following properties exist?
$1)$ $\phi_1 \begin{pmatrix} 2 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} $, $\p... | 1) You are given the image of $3$ vectors whereas you only need $2$ to define the mapping on $\Bbb R^2$ (because $\Bbb R^2$ is of dimension $2$). So you can assume you defined $\phi$ with the first two and then check if the $3^\text{rd}$ one is true. Now $\begin{pmatrix}1\\2\end{pmatrix}=2\begin{pmatrix}1\\1\end{pmatri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/453988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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How to search quadratic function If a graph of the quadratic function $f(x)=ax^2+bx+c$, where $a$, $b$ and $c$ are constants. If this function vertex is $(13,−169)$ and the distance between the two intersection points with the $x$-axis is $26$, what is the value of $a$, $b$, and $c$?
| $$y=ax^2+bx+c \implies \left(x+\frac b{2a}\right)^2=4\cdot\frac14\left(y-\left(c-\frac{b^2}{4a}\right)\right)$$
Comparing with $(x-h)^2=4a(y-k)$ the vertex will be $\left(-\frac b{2a},c-\frac{b^2}{4a}\right)$
So, $-\frac b{2a}=13\implies b=-26a\ \ \ \ (1)$
and $c-\frac{b^2}{4a}=-169\implies b^2-4ac=169\cdot4a\ \ \ \ (... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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closed form of $\int_{0}^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^n}$ closed form of $$\int_{0}^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^n}$$
for $a,b>0$
n=1 we get
$$\int_{0}^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^1}=\frac{2\pi}{ab}$$
n=2 we get
$$\int_{0}^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^2}=\frac{\pi(a^2+b^2)}{... | In the following we assume that $n\geq 1$.
Thanks to parity, the integral can be written as
\begin{align}
I_n=\int_0^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2 x)^n}&=2\int_0^{\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2 x)^n}=\\
&=2\int_0^{\pi}\frac{dx}{(\frac{a^2+b^2}{2}+\frac{a^2-b^2}{2}\cos 2x)^n}=\\
&=\frac{1}{i}\oint_{|z|=1}\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/455147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
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Summation and proof by induction question: $\sum_{j=1}^{n}j(j+1)(j+2)=\frac{n(n+1)(n+2)(n+3)}{4}$ I can't figure this out based on examples in textbooks, etc.
Show via induction that $\sum_{j=1}^{n}j(j+1)(j+2)=\frac{n(n+1)(n+2)(n+3)}{4}$
So far, I have:
(a) base case
$P(1)= 1(1+1)(1+2) = \frac{1(1+1)(1+2)(1+3)}{4} = 6 ... | First, to improve the logic of your proof, you don't want to put the thing that you are trying to prove, $P(k+1) = (2)(3)+(2)(3)(4)+(3)(4)(5)+\cdots+(k+1)(k+2)(k+3)= \frac{(k+1)(k+2)(k+3)(k+4)}{4}$ at the beginning of (the second part of) your proof. Same goes for the base case where you you are trying to show that the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/455197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Finding the limit $\lim \limits_{n \to \infty}\ (\cos \frac x 2 \cdot\cos \frac x 4\cdot \cos \frac x 8\cdots \cos \frac x {2^n}) $ This limit seemed quite unusual to me as there aren't any intermediate forms or series expansions which are generally used in limits. Stuck on this for a while now .Here's how it goes :
... | Put $z=e^{i x/2^n}.$ Then the product becomes
$$\prod_{k=1}^n \cos\left(\frac{x}{2^k}\right)
= \frac{1}{2^n} \prod_{k=0}^{n-1} \left(z^{2^k}+z^{-2^k}\right)
= \frac{1}{2^n} z^{-\sum_{k=0}^{n-1} 2^k}
\prod_{k=0}^{n-1} \left(z^{2^{k+1}}+1\right)\\
= \frac{1}{2^n} z^{1-2^n} \frac{1}{z+1} \prod_{k=0}^n \left(z^{2^k}+1\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/455995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
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generating function Homework Question 1 This is a HW question
I am asked to find a closed form generating function for $1,1,0,1,1,0,1,1,0....$
so then $f(x)=x^0+x^1+0x^2+x^3+x^4+0x^5+x^6+x^7+0x^8$
could use some hint or help.
| Hint: See if writing it as follows helps:
$$f(x) = 1 + x + x^{3} +x^{4} + \cdots = ( 1 + x + x^{2} + x^{3} + x^{4} + x^{5} + \cdots ) - (x^{2} + x^{5} + x^{8} + \cdots) = \frac{1}{1-x} - x^{2}(1+x^{3}+x^{6}+\cdots)$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/456453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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$2+2 = 5$? error in proof $$\begin{align} 2+2 &= 4 - \frac92 +\frac92\\
&= \sqrt{\left(4-\frac92\right)^2} +\frac92\\
&= \sqrt{16 -2\times4\times\frac92 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{16 -36 + \left(\frac92\right)^2} +\frac92\\
&= \sqrt {-20 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{25-45 +\left(\fra... | How did $\sqrt{16 -2\times4\times\frac92 +(\frac92)^2}$ turn into $\left(\sqrt{16 -36 + (\frac92)^2}\right)^2$?
Then later, you seem to assume that since $\left(4-\frac92\right)^2$ is the same as $\left(5-\frac92\right)^2$, it follows that $4-\frac92=5-\frac92$. Like saying that since $3^2=(-3)^2$, it follows that $3=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/457490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 9,
"answer_id": 4
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.