Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Does the sequence $x_{n+1}=x_n+\frac{1}{x_n}$ converge or diverge? Set $x_1=a$, where $a > 0$ and let $x_{n+1}= x_n + \frac{1}{x_n}$. Determine if the sequence $ \lbrace X_n \rbrace $ converges or diverges.
I think this sequence diverges since Let $x_1= a > 0$ and $x_{n+1}= x_n + (1/x_n)$ be given. Let $x_1=2$ since $a... | It is easy to see (by induction) that $x_{n+1}>x_n \ge a >0$ for all $n$.
We assume that $(x_n)$ is convergent and that $l$ is the limmit. Then $l \ge a>0$.
From $x_{n+1}= x_n + (1/x_n)$ we get $l= l+ 1/l$, which is impossible. Hence the sequence is divergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2190592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $f'(1)$ given $f(x)$ and $f(1)$ Question:
Let $$f(x)=\int\frac{x-1}{x+1}\frac{dx}{\sqrt{x^3+x^2+x}}$$ such that $f(1)=\frac{2\pi}{3}$
Then $f'(1)$ is equal to
A) $0$
B) $\cfrac \pi3$
C) $\cfrac\pi4$
D) $2$
Attempt:
\begin{align}f(x)&=\int\frac{x-1}{x+1}\frac{1}{\sqrt{x^3+x^2+x}}dx\\
f(1)&=\int0dx\\
f(1)&=0+c\\
... | Does $f'(x)$ mean $\frac{df}{dx}$. Isnt $f'(x)$ then simply the integrated term, with the integration removed? ie, $\frac{x-1}{x+1}\times \frac{1}{\sqrt{x + x^2 + x^3}}$ ? Am I missing something?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2190884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}=\frac{\sqrt{7}}{8}$.
Prove that
$\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}=\frac{\sqrt{7}}{8}$.
What I've tried doing :
If $\theta=\frac{\pi}{7}:$
$$
3\theta+4\theta=\pi
$$
This allowed me to prove that :
$$
\tan^2\frac{\pi}{7}+\tan^2\fra... | You can show that $\sin^2(\pi/7)$, $\sin^2(2\pi/7)$, and $\sin^2(3\pi/7)$ are the three roots of $64x^3-112x^2+56x-7 = 0$, as follows:
Take $\zeta_7$ a $7$'th root of unity (a root of $(x^7-1)/(x-1) = x^6+x^5+x^4+x^3+x^2+x+1 = 0$), then $2 \cos(k 2\pi/7) = \zeta_7^k + \zeta_7^{-k}$, and with $\sin^2+\cos^2=1$ you can n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2191058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find alternating series that converges to $ \int_0^{1/2}x\log(1+x^3)dx $ I need to find the alternating series that converges to $ \int_0^{1/2}x\log(1+x^3)\,dx $
Here's what I did:
$$
\frac{d}{dx}[\log(1+x^3)]=\frac{1}{1+x^3}=\frac{1}{1-(-x)^2}=\sum_{n=1}^\infty(-x^3)^{n-1}=1-x^3+x^6-x^9+-...
$$
$$\begin{align}
f(x)&=x... | The logarithm is $\sum_{k=1}^\infty\frac{\left( -1\right)^{k-1}}{k}x^{3k}$. Making sure to multiply by $x$ before we integrate, the final result is $\sum_{k=1}^\infty\frac{\left( -1\right)^{k-1}}{2^{3k+2}k\left( 3k+2\right)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2191358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Evaluate the integral $\int_{0}^{\pi}\frac{\log(1+a\cos x)}{\cos x}\,dx$ Please help me to evaluate the integral $\int_{0}^{\pi}\frac{\log(1 + a\cos x)}{\cos x}\,dx$. In the question they said to use the formula $\frac{d}{dy}\int_{a}^{b}f(x,y)dx=\int_{a}^{b}\frac{\partial }{\partial y}f(x,y)dx$.
| Well, we have:
$$\frac{\text{d}}{\text{d}\text{a}}\left\{\int_0^\pi\frac{\ln\left(1+\text{a}\cos\left(x\right)\right)}{\cos\left(x\right)}\space\text{d}x\right\}=\int_0^\pi\frac{\frac{1}{\text{a}+\sec\left(x\right)}}{\cos\left(x\right)}\space\text{d}x=$$
$$\int_0^\pi\frac{1}{1+\text{a}\cos\left(x\right)}\space\text{d}x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2192443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
The remainder when $33333\ldots$ ($33$ times) is divided by $19$ $A= 33333\ldots$ ($33$ times). What is the remainder when $A$ is divided by $19$?
I don't know the divisibility rule of $19.$
What I did was
$32\times(33333\times100000)/19$ and my remainder is not zero and this is completely divisible by $19.$
This is a... | From the definition $A=(10^{33}-1)/3$
(if that is what you meant - a $33$-digit number, all threes)
$\bmod 19$, we have
$10^2\equiv 5$,
$10^3\equiv 12$,
$10^4\equiv 6$,
$10^5\equiv 3$,
$10^6 \equiv 11$,
$10^9\equiv 12\cdot 11 \equiv 18\equiv -1$,
$10^{18}\equiv (10^9)^2\equiv 1$,
$10^{33}\equiv 10^{15}\equiv -1\cdot 11... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
} |
Function notation with term before f(x) I saw an exam question using this notation:
$$x^2\,f(x) = x^2 + 7x + 3\,.$$
I didn't understand it, so I checked the answer and I saw that was equal to:
$$f(x) = \frac{x^2 + 7x + 3}{x^2}\,.$$
Is the former a common function notation? Where or how can I find more information about... | Assuming you have copied it verbatim, it's a simple redistribution of terms. Suppose for example that you wanted to subtract 3 from both sides. You would then have $$x^2 f(x) - 3 = x^2 + 7x.$$
That was just for the sake of example. Let's go back to $$x^2 f(x) = x^2 + 7x + 3.$$ Now we want to divide both sides by $x^2$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2196743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find the orthogonal trajectories of the family of curves
Find the orthogonal trajectories of the family of curves $y^5=kx^2$.
I start by knowing that I need a differential equation satisfying all members of family, which means $k$ needs to be eliminated.
Differentiating both sides with respect to $x$:
\begin{align*}
... | I made a mistake when taking $\int ydy$. It should be $\frac{y^2}{2}$ not $y$.
This means that,
$$
\frac{y^2}{2}=-\frac{5}{4}x^2 + C\\
\frac{y^2}{2}+\frac{5}{4}x^2=C\\
y^2+\frac{5}{2}x^2=2C=C_2
$$
Which is the correct answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2200074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int_0^{\frac{\pi}{2}} \frac{\sin^3(x)}{\cos^3(x)+\sin^3(x)} dx$
Evaluate $\int_0^{\frac{\pi}{2}} \frac{\sin^3(x)}{\cos^3(x)+\sin^3(x)} dx$
$$\int_0^{\frac{\pi}{2}} \frac{\sin^3(x)}{\cos^3(x)+\sin^3(x)} dx=\int_0^{\frac{\pi}{2}} 1-\frac{\cos^3(x)}{\cos^3(x)+\sin^3(x)} dx=\int_0^{\frac{\pi}{2}} 1-\frac{1}{1+... | Let $x=\frac{\pi}2-u$. Then $dx=-du$ and the integral becomes:
$$I=-\int_{\pi/2}^0 \frac{\sin^3(\pi/2-u)}{\sin^3(\pi/2-u)+\cos^3(\pi/2-u)}du\\=\int_0^{\pi/2}\frac{\cos^3(u)}{\cos^3 (u)+\sin^3(u)}du$$
So your integral satisfies:
$$2I=\int_0^{\pi/2} 1dx=\frac{\pi}2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2200716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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If $\dfrac {1}{a+b} +\dfrac {1}{b+c}=\dfrac {3}{a+b+c}$, prove that $\angle B=60^\circ $ If $\dfrac {1}{a+b}+\dfrac {1}{b+c}=\dfrac {3}{a+b+c}$, prove that $\angle B=60^\circ$.
My Attempt
$$\dfrac {1}{a+b}+\dfrac {1}{b+c}=\dfrac {3}{a+b+c}$$
$$\dfrac {a+2b+c}{(a+b)(b+c)}=\dfrac {3}{a+b+c}$$
$$a^2-ac-b^2+c^2=0$$.
How t... | By the cosine rule, $\cos B = \frac{1}{2}$ iff $b^2=a^2+c^2-ac$. That's the result you need to try to obtain by rearrangement.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2201496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Prove that $(2n+1)(x+y+z)+6$ divides $x^3+y^3+z^3$
Let $n,x,y,z$ be positive integers such that $$(x-y)^2+(y-z)^2+(z-x)^2 = (2n+1)xyz.$$ Prove that $(2n+1)(x+y+z)+6$ divides $x^3+y^3+z^3$.
The original equation doesn't have $x^3,y^3,z^3$, so how can we show that $(2n+1)(x+y+z)+6$ divides $x^3+y^3+z^3$? Expanding the ... | This is linked to a well-known identity (a standard exercise in high-school):
\begin{align}x^3+y^3+z^3-3xyz&=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\\&=\tfrac12(x+y+z)\bigl((x-y)^2+(y-z)^2+(z-x)^2\bigr)\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2203156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
System of Equation. (Complicated) Given that $x,y \in \mathbb{R}$. Solve the system of equation.
$$5x(1+\frac{1}{x^2+y^2})=12$$
$$5y(1-\frac{1}{x^2+y^2})=4$$
My attempt,
I made them become $x+\frac{x}{x^2+y^2}=\frac{12}{5}$ and $y-\frac{y}{x^2+y^2}=\frac{4}{5}$.
I don't know how to continue from here. Hope anyone woul... | Just to provide another way.
Use polar coordinates $$x=\rho \cos(\theta) \qquad y=\rho \sin(\theta)$$ to make the equations to be $$\frac{5 \left(\rho ^2+1\right) \cos (\theta )}{\rho }=12$$ $$\frac{5 \left(\rho ^2-1\right) \sin (\theta )}{\rho }=4$$ So,$$\cos(\theta)=\frac{12 \rho }{5 \left(\rho ^2+1\right)}\qquad \si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2203287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Is the sequence defined by $a_{n+1} = \frac{1}{a_n} + 1$ convergent? The sequence defined by $a_1=1$ and
$$a_{n+1} = \frac{1}{a_n} + 1$$
is increasing. And I can get the limit which is equal to $$\frac{1+\sqrt{5}}{2}$$
But $a_2 = 2$ which is larger than the limit. So is the sequence convergent or divergent?
| $\begin{array}\\
a_{n+1}
&=1+ \frac{1}{a_n}\\
&=1+ \frac{1}{1+ \frac{1}{a_{n-1}}}\\
&=1+ \frac{1}{1+ \frac{1}{1+ \frac{1}{a_{n-2}}}}\\
&...\\
&=1+ \frac1{1+ \frac1{1+ \frac1{1+...}}}
\qquad n \text{ deep}\\
\end{array}
$
But that continued fraction
is
$\dfrac{F_{n+1}}{F_n}
\to \phi
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2205883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Prove this $1+2abcxyz \geq a^2x^2 + b^2y^2+c^2z^2$ Let $a,b,c,x,y,z\in\mathbb{R}$ such that \begin{align}\{a,b,c,x,y,z\}&\subset[-1,1]\\
1 + 2abc &\geq a^2+b^2+c^2\\
1+2xyz&\geq x^2+y^2+z^2\end{align}
Prove that:
$$1+2abcxyz \geq a^2x^2 + b^2y^2+c^2z^2$$
| Let $A=\left(\begin{matrix}1 & a & b \\ a & 1 & c \\ b & c & 1 \end{matrix}\right)$. Due to assumptions on $a,b,c$, all principal minors of $A$ are nonnegative. Therefore $A$ is positive-semidefinite.
Analogously, $B = \left(\begin{matrix}1 & x & y \\ x & 1 & z \\ y & z & 1 \end{matrix}\right)$ is positive-semidefinit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2208334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Is $\displaystyle\lim_{x\rightarrow y}\dfrac{\cos x-\cos y}{x^2-y^2}$ equal to $-\dfrac 12$ or just $-\dfrac{\sin y}{2y}$ Question:
$$\lim_{x\rightarrow y}\dfrac{\cos x-\cos y}{x^2-y^2}=?$$
Here is my try:
\begin{align}\lim_{x\rightarrow y}\dfrac{\cos x-\cos y}{x^2-y^2}&=\lim_{x\rightarrow y}\dfrac{-2 \sin (\frac 12(x+... | we know by definition that
$$\lim_{x\rightarrow y}\left(\frac{f(x) - f(y)}{x - y}\right) = \frac{df}{dy}$$
so the given can be expressed as
$$\lim_{x\rightarrow y}\left(\frac{\cos(x) - \cos(y)}{(x - y)(x + y)}\right)$$
that is equal to
$$\left(\frac{df}{dy}\right)\left(\lim_{x\rightarrow y}\frac{1}{x + y}\right)$$
at ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2208465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
If the roots of $9x^2-2x+7=0$ are $2$ more than the roots of $ax^2+bx+c=0$, then value of $4a-2b+c=$?
If the roots of $9x^2-2x+7=0$ are $2$ more than the roots of $ax^2+bx+c=0$, then value of $4a-2b+c=$?
My approach is: Let the roots of $ax^2+bx+c=0$ are $\alpha$ and $\beta$. So, $$\alpha+\beta=-\frac{b}{a}\\ \alpha\... | My way:
If $p$ is a root of $9x^2-2x+7=0, p-2$ will be a root of $ax^2+bx+c=0$
Now writing $p-2=q\iff p=q+2$ in $9x^2-2x+7=0,$
we get $0=9(q+2)^2-2(q+2)+7=9q^2+34q+39$
So, we need $$\dfrac a9=\dfrac b{34}=\dfrac c{39}\ \ \ \ (1)$$
So, with the given conditions, $a,b,c$ can assume any set of non-zero finite values th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2209958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Question on calculating $\int_{\partial B_{3/2}(1)} \frac{z^7 + 1}{z^2(z^4 + 1)}dz$ My task is to calculate $$\int_{\partial B_{3/2}(1)} \frac{z^7 + 1}{z^2(z^4 + 1)}dz$$ using the Cauchy integral formula. My question is: Is there a simple trick or do I have to perform a partial fraction decomposition?
| Yes there is a trick you don't have to use partial fraction decomposition
Note that the degree of nomenator is higher than the degree of the denomenator.
We use partial fraction usually when the degree of the nomenator is lower than the degree of the
denomenator.
I prefer solving this using another method than using ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2213090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to do partial fraction decomposition. I have a to solve.$$\int\dfrac{1}{(a-2x)^2(b-x)}dx$$
But I don't even know how to start it. Please help.
| Starting with
$$
\dfrac{1}{(a-2x)(b-x)}=\dfrac{1}{a-2b}\left(\dfrac{-2}{a-2x}+\dfrac{1}{b-x}\right),
$$
we get
\begin{eqnarray}
\dfrac{1}{(a-2x)^2(b-x)}&=&\dfrac{1}{a-2b}\left[\dfrac{-2}{(a-2x)^2}+\dfrac{1}{(a-2x)(b-x)}\right]\\
&=&\dfrac{1}{a-2b}\left[\dfrac{-2}{(a-2x)^2}+\dfrac{1}{a-2b}\left(\dfrac{-2}{a-2x}+\dfrac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2215556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$c_n = 1 + \frac{1}{1!} + \frac{1}{2!} +...+\frac{1}{n!}$ so prove $e - c_n \le \frac{1}{n! * n}$ $c_n = 1 + \frac{1}{1!} + \frac{1}{2!} +...+\frac{1}{n!}$, so $e - c_n \le \frac{1}{n! * n}$
I absolutely have no idea how to solve it, could anyone tell me the approach?
| The Taylor series for $e^x$ is given by
$$e^x=\sum_{k=0}^\infty\frac{x^k}{k!}\tag 1$$
Using $(1)$ with $x=1$, we have
$$\begin{align}
e-c_n&=\sum_{k=n+1}^\infty \frac{1}{k!}\\\\
&\le\sum_{k=n+1}^\infty \frac{1}{k!}\left(1+\color{blue}{\underbrace{\frac{1}{k-1}-\frac1k}_{\text{Positive Terms}}}\right)\\\\
&=\sum_{k=n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2221084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Calculate $\lim_{n\rightarrow \infty}\sum_{k=1}^{n}\cos\left(\frac{2\pi k}{2n+1}\right)$. Calculate
$$\lim_{n\rightarrow \infty}\sum_{k=1}^{n}\cos\left(\frac{2\pi k}{2n+1}\right)$$
Question: I want to verify that my next attempt is correct, I do it too exhausted and in that state I do not trust my abilities.
My attempt... | There is an error is the calculation. We have
$$\sum_{k=1}^n\cos\left(\frac{2k\pi}{2n+1}\right)=-\frac12 \tag 1$$
and hence the limit is $-1/2$. To show that $(1)$ is correct, we follow the approach taken in the OP.
Then, we have
$$\begin{align}
\sum_{k=1}^n e^{i\left(\frac{2k\pi}{2n+1}\right)}&=\frac{e^{i\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2221618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Integral $\int {t+ 1\over t^2 + t - 1}dt$
Find : $$\int {t+ 1\over t^2 + t - 1}dt$$
Let $-w, -w_2$ be the roots of $t^2 + t - 1$.
$${A \over t + w} + {B \over t+ w_2} = {t+ 1\over t^2 + t - 1}$$
I got $$A = {w - 1\over w - w_2} \qquad B = {1- w_2\over w - w_2}$$
$$\int {t+ 1\over t^2 + t - 1}dt = A\int {1\over t+ w} ... | Note that we have
$$\begin{align}
& {\sqrt{5} + 1\over 2\sqrt{5}}\log|t + 1/2 - \sqrt{5}/2| + {\sqrt{5} - 1 \over 2\sqrt{5}}\log|t + 1/2 + \sqrt{5}/2| + C\\[10pt]
&= \frac12\log\left(|t + 1/2 - \sqrt{5}/2|\,|t + 1/2 + \sqrt{5}/2|\right)\\[10pt]
&{}+\frac{1}{2\sqrt{5}}\left(\log\left(|t + 1/2 - \sqrt{5}/2|\right)-\log\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2222679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Using the Lagrange interpolation formula to find a polynomial f with $f(-1)=-6, f(0)=2, ...$ The exercise goes as follows:
Use the Lagrange interpolation formula to find a polynomial $f$ with real coefficients such that $f$ has degree $\leq3$ and $f(-1)=-6, f(0)=2, f(1)=-2, f(2)=6$.
So to start, I wrote down the formu... | I would like to propose an alternative way to solve the problem. Coefficients of Lagrange interpolation polynomial can be found if one uses a determinant form of Lagrange interpolation presented in "Beginner's guide to mapping simplexes affinely", section "Lagrange interpolation" (you may check for concrete example in ... | {
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"url": "https://math.stackexchange.com/questions/2226123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Showing that $f(x)=\frac x{x+1}$ is the unique function satisfying $f(x)+f\left(\frac1x\right)=1$ and $f(2x)=2f\big(f(x)\big)$ We are given a function $ f : \mathbb Q ^ + \to \mathbb Q ^ + $ such that $$ f ( x ) + f \left( \frac 1 x \right) = 1 $$
and
$$ f ( 2 x ) = 2 f \big( f ( x ) \big) \text . $$
Find, with proof, ... | As j___d does, I will attempt to prove that $f(\frac pq) = \frac{p}{p+q}$ by strong induction on $p+q$, starting with the base case $p+q=2$ where $f(\frac11) = \frac12$.
Now assume that $f(\frac{p}{q}) = \frac{p}{p+q}$ holds when $p+q<k$, and consider fractions $\frac pq$ with $p+q=k$.
Whenever $p<q$, we have $f(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2227678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\left[ -(2/K) \tanh^{-1} ( \frac{A-(B-1)\tan(\theta/2)}{K}) \right]_0^{2\pi}$ where $K = \sqrt{A^2 + B^2 -1}$ and $A^2 + B^2 << 1$. I am having difficulty in evaluating the following integrand:
$$\left[ -(2/K) \tanh^{-1} ( \frac{A-(B-1)\tan(\theta/2)}{K})
\right]_0^{2\pi}$$
where $K = \sqrt{A^2 + B^2 -1}$ a... | Following comments by user mickep let us look at correcting the partitions used...
So, to avoid operating across the singularity at $\frac{\pi}{2}$, let us break the integration into the two continuous ranges $0...\pi$ and $\pi...2\pi$ thus:-
$$I_{A=B=0}=$$
$$ -\left(\frac{2}{i}\right) \tanh^{-1} \left( \frac{\tan\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2230692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to prove that $\int_{0}^{\infty}{\ln[x(1+x)]\ln\left(x\over 1+x\right)\over (x+2)^2}\mathrm dx={1\over 2}\ln^2(2)?$ Log integral
$$\int_{0}^{\infty}{\ln[x(1+x)]\ln\left(x\over 1+x\right)\over (x+2)^2}\mathrm dx={1\over 2}\ln^2(2)\tag1$$
Making an attempt:
Following from my previous post making $u={x\over 1+x}$ it d... | On the path of Jack D'Aurizio,
$I_1=\displaystyle \int_0^{+\infty} \dfrac{(\ln(x))^2}{(x+2)^2}dx$
In $I_1$ perform the change of variable $y=\dfrac{x}{2}$,
$\begin{align} I_1&=2\int_0^{+\infty} \dfrac{(\ln(2x))^2}{(2x+2)^2}dx\\
&=\dfrac{1}{2}\int_0^{+\infty} \dfrac{(\ln 2+\ln x)^2}{(1+x)^2}dx\\
&=\dfrac{1}{2}(\ln 2)^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2230910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find tan(C/2) in The triangle ABC . $ABC$ is a triangle.
$\tan\frac{A}{2} = 0.5$
$\tan\frac{B}{2} =\frac{1}{3}$.
Find $\tan\frac{C}{2}$.
I tried to find it :
when $A+B+C = 180^{\circ}$
So $\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = 90^{\circ}$.
How can I complete on these information?
| The following identity is very useful.
$$\tan\frac{A}{2}\tan\frac{B}{2}+\tan\frac{A}{2}\tan\frac{C}{2}+\tan\frac{B}{2}\tan\frac{C}{2}=1$$
From here $C=90^{\circ}$
We can get this result by geometric way.
Let $P(0,0)$,$Q(2,1)$ and $R(3,-1)$. Draw it!
Hence $PQ=QR$, $\measuredangle Q=90^{\circ}$ and from here $$\arctan{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2231705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Shortest universal bit string: One string to contain all others Let $s$ be a string of bits. Treat it as a cycle, with the first bit following
the last.
Say that $s$ is universal for $n$ if all the $2^n$ strings
of $n$ bits can be found in $s$ as consecutive, left-to-right bits
(with wrap-around).
Define $u(n)$ as the ... | What you're looking for is related to De Bruijn sequences. The formula seems to be very simple: $u(k,n) = k^n$ (and the special case $u(n) = 2^n$).
(source: Wikipedia, by Eviatar Bach)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Floor function of powers of $2$ Is there any way to know the exact value of
$$\left(\left\lfloor 2^{\frac{n}{2}} \right\rfloor\right)^2$$
for $n$ an integer $n>0$?
When $n$ is even, the solution is trivial, since we do not have to face any fractional part. On the other hand, for $n$ an odd number, it seems difficult t... | Using identity:
$$\lfloor x\rfloor =x-\frac{1}{2}+\frac{\sum _{k=1}^{\infty } \frac{\sin (2 k \pi x)}{k}}{\pi }$$
Solve sum ,substitute x=2^(n/2) and raise to the power 2 we have:
$$\left\lfloor 2^{n/2}\right\rfloor ^2=\frac{\left(\left(-1+2^{1+\frac{n}{2}}\right) \pi -i \log \left(1-e^{-i
2^{1+\frac{n}{2}} \pi }\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
A proof that $\frac{\log^k(1)+\log^k(2)+\dotsb +\log^k(n)}{1^k+2^k+\dotsb +n^k} \to 0$ Let $\left(a_n\right)$ be the following sequence:
$$a_n =\frac{\log^k\left(1\right)+\log^k\left(2\right)+\dotsb +\log^k\left(n\right)}{1^k+2^k+\dotsb +n^k},$$ for a fixed $k \in \mathbb{N}$. Prove that $a_n \to 0$. There are many pr... | Fix $k\in \mathbb N.$ Let $a_n$ be the numerator, $b_n$ the denominator. Then $b_n \to \infty.$ Time to think about Stolz-Cesaro, which suggests we consider
$$\tag 1 \frac{a_{n+1}- a_n}{b_{n+1}- b_n} = \frac{\ln^k (n+1)}{(n+1)^k} = \left (\frac{\ln (n+1)}{n+1} \right )^k.$$
Since $[\ln (n+1)]/(n+1) \to 0,$ the right s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2236082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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$\sqrt{x^2+1}$ uniformly continuous on (0, 1)? $\sqrt{x^2+1}$ uniformly continuous on (0, 1)?
How to deal with such problems? Please help.
I know the definition of U C. but unable to handle the problem.
| Hint: Observe
\begin{align}
\left|\sqrt{x^2+1}-\sqrt{y^2+1}\right| =& \frac{|x^2-y^2|}{\sqrt{x^2+1}+\sqrt{y^2+1}} \\
=&\ \frac{|x+y|}{\sqrt{x^2+1}+\sqrt{y^2+1}}|x-y|\\
\leq&\ \frac{|x|+|y|}{2}|x-y| \leq |x-y|
\end{align}
when $x, y \in (0, 1)$.
Edit: This is more or less the solution. However, you should try to use t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2236204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given a $2\times 2$ matrix $A$, compute $A^7$. Let $A =
\begin{pmatrix}
e^{2x} & -1 \\
0 & e^{2x}-1 \\
\end{pmatrix}
$. Compute $A^7$.
I've tried the obvious way of multiplying $A$ with $A$, then $A^2$ with $A^2$, but I arrived at a messy result in the top right member of the matrix. ... | You have $A=B+C$, where
$$
B=\begin{bmatrix} e^{2x}&0\\0&e^{2x}-1\end{bmatrix},\ \ \ C=\begin{bmatrix}0&-1\\0&0\end{bmatrix}.
$$
The key is that $C^2=0$, and that $CB^kC=0$. Also,
\begin{align}
B^kCB^m&=\begin{bmatrix} e^{2kx}&0\\0&(e^{2x}-1)^k\end{bmatrix}\begin{bmatrix}0&-1\\0&0\end{bmatrix}
\begin{bmatrix} e^{2mx... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2237097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Prove that the sequence $\left\{\frac{4n^2}{2n^3-5}\right\}$ converges to $0$. Prove that the sequence $$\left\{\frac{4n^2}{2n^3-5}\right\}$$ converges to $0$.
The first thing I would like to do is further bound the sequence so I can get rid of the $-5$ in the bottom of the fraction. How can I do this? For negatives it... | Two-stage bounding seems unavoidable. To simplify the works, I would do something like the following. If $n \geq 3$, then
$$
\frac{4n^{2}}{2n^{3}-5} < \frac{4n^{2}}{2n^{3}-n^{2}} = \frac{4}{2n-1}.
$$
Given any $\varepsilon > 0$, we have $\frac{4}{2n-1} < \varepsilon$ if in addition $n > (\frac{4}{\varepsilon+1})/2$. So... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2238649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Simple inequality $\left|\frac{3x+1}{x-2}\right|<1$
$$\left|\frac{3x+1}{x-2}\right|<1$$
$$-1<\frac{3x+1}{x-2}<1$$
$$-1-\frac{1}{x-2}<\frac{3x}{x-2}<1-\frac{1}{x-2}$$
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2}$$
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2} \text{ , }x \neq 2$$
$${-x+1}<{3x}<{x-3} \text{ , ... | When you multiply by $x-2$, you have to worry about whether it's negative, in which case the direction of the inequalities would be reversed. So break things into two cases: $x>2$ and $x<2$.
If $x>2$, you can do what you did or be a little more efficient:
$$-(x-2) < 3x+1 < x-2$$
$$-x +1 < 3x < x-3$$
$$1<4x \mbox{ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 3
} |
Prove expression is not prime $$
(n + 4)^4 + 4
$$
If n is natural number, how to prove that above expression is not prime?
I am stuck here
$$
(n+4)^2 \cdot (n+4)^2 + 2 . 2
$$
$$
\left(\left(n^2+4^2\right) \cdot 2\right)\left(\left(n^2+4^2\right) \cdot 2\right)
$$
| Try to write
a more general form
as the difference of two squares
which we know how to factor.
$\begin{array}\\
x^4 + a
&=(x^2+u)^2-b^2x^2\\
&=x^4+2ux^2+u^2-b^2x^2\\
&=x^4+(2u-b^2)x^2+u^2\\
\text{and}\\
x^4 + a
&=(x^2+u+bx)(x^2+u-bx)\\
\end{array}
$
so
$u^2=a$
and
$b^2 = 2u$.
Therefore
$b^4 = 4u^2
= 4a
$.
Writing $c^2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the limits for the average triple integral I am trying to do the following problem:
Calculate the average of $f(x,y,z) = x^2 + y^2 + z^2$ over the set of points satisfying $|x| + |y| + |z| ≤ 1$.
So,
$\frac{1}{volume of D}\iiint Fdv$ =
$\frac{1}{volume of D}\iiint (x^2 + y^2 + z^2 )dz dy dx$
My problem is how to de... | To simplify computations, note that the integrand does not vary when the variables change sign. Because both the integrand and the region are symmetric in the three variables, the integral will be the same when restricted to each of the eight octants. Thus we may assume that $x,y,z\geq0$, and then multiply our integral... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2243030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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On Chinese remainder theorem Suppose we know $x\equiv 3 \bmod 11$ and $x\equiv 7\bmod 13$ and $0<x<143$ holds then CRT gives that $x=3\times 13[13^{-1}\bmod 11] + 7\times 11[11^{-1}\bmod 13]=39\times 6 + 77\times 6=696$ gives a solution but it is not within $0$ and $143$. So we take $696\bmod 143$ and choose $124$ as ... | The direct way to $124$ would be to start with $7$ and see how many $13$s to add to make it into a number that is $3\bmod 11$.
Since $13\equiv \color{red}2 \bmod 11$, we can see that we need $(3-7)/\color{red}2 = -2\equiv 9\bmod 11$ copies of $13$. $7+9\cdot 13 = 7+117 = 124$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2244885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
I spent a lot of time trying to solve this and, having consulted some ... | If you rewrite your proof as:
$3 ( x^2+y^2+z^2 ) \ge ( x^2+y^2+z^2 ) + ( 2xy+2yz+2zx ) = (x+y+z)^2 = 1$.
You would find that it is not so unintuitive after all.
Since you know that $x+y+z = 1$, a natural thing to do to the original equation is to homogenize it; namely make all terms have the same degree. This gives u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 15,
"answer_id": 1
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In a triangle $ABC$, if its angles are such that $A=2B=3C$ then find $\angle C$? In a triangle $ABC$, if its angles are such that $A=2B=3C$ then find $\angle C$?
$1$. $18°$
$2$. $54°$
$3$. $60°$
$4$. $30°$
My Attempt:
$$A=2B=3C$$
Let $\angle A=x$ then $\angle B=\dfrac {x}{2}$ and $\angle C=\dfrac {x}{3}$
Now,
$$x+\dfr... | It looks like neither of the options is correct. $$\angle C = \frac{x}{3} = \frac{180 \times 6}{11 \times 3} = \frac{180 \times 2}{11} = \frac{360}{11} = 32 \frac{9}{11} $$
You can also calculate $A+B+C = 3C + \frac{3}{2}C + C = \frac{11}{2} C$ for all possibile answers to show that none of them gives $180^\circ$ for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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There exist $3$ positive numbers $x,y,z$ such that $x\cdot y\cdot z=100$ but $x^2+y^2+z^2<65$. I have to show this using a form of calculus.
I know the answer is the cube root of $100$. I did this using intuition and upper and lower bounds but I have no idea how to show it from a calculus standpoint. Maybe showing the ... | Let $x= 2\sqrt{5}$ , $y= 2\sqrt{5}$ $z=5 $.
Then of course $xyz=100$
And $x^2+y^2+z^2=20+20+25=65$.
Now set $x=y$ so $z= \frac{100}{x^2}$ therefore $x^2+y^2+z^2=2x^2+\frac{100}{x^4}$. Now define $$f(x)=2x^2+\frac{100}{x^4}$$
By differentiating we get $$f'(x)=\frac{4x^6-4\cdot100^2}{x^5}$$
so $x=100^{1/3}$ is the minimu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Express the value of $s\left(m\right)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}$ in terms of $m$. The previous question was: Find the range such that the equation $|x^2 -3x +2|=mx $ has 4 distinct real solutions: $a,b,c,d$, and that turned out to be $0<m<3-2\sqrt{2}$. The book says that the solution is $\... | The quadratic equation $x^2+px+q=0$ has two distinct roots, namely $$r_1=\frac{-p+\sqrt{p^2-4q}}{2}\qquad\text{and}\qquad r_2=\frac{-p-\sqrt{p^2-4q}}{2}\qquad\text{whenever }\,p^2-4q>0$$
Then $$r_1r_2=q\qquad\text{and}\qquad r_1^2+r_2^2=p^2-2q$$
It follows that $$\frac1{r_1^2}+\frac1{r_2^2}=\frac{r_1^2+r_2^2}{(r_1r_2)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2249090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
locus of $z=w - \frac{1}{w}$ If $|w|=2$ , then set of points $z=w -\frac{1}{w}$ is equal to ?
One of my friend helped me like this:
$$|z| = \left| w - \frac{1}{w}\right| \leq |w| + \frac{1}{|w|} = 2 + 0.5 = 2.5 \\ \implies |z| \le 2.5$$
After that I am unable to proceed. Can anybody help me?
| Given $w \overline{w} = |w|^2=4\,$, it follows that $z = w - \cfrac{1}{w} = w - \cfrac{\overline{w}}{4}\,$.
Let $z=x+iy$ and $w=u+iv\,$ with $x,y,u,v \in \mathbb{R}\,$, then the above can be written as:
$$
x+iy = u+iv - \frac{1}{4}(u - iv) = \frac{3}{4}u + i\,\frac{5}{4}v \;\;\implies\;\; x = \frac{3}{4}u, \;\;y = \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2249297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the Maximum value of $\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}$ if $x$, $y$ and $z$ are positive real numbers such that $x+y+z=4$ Find the maximum value of $$S=\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}$$
I tried as follows.
The given expression can be rewritten as
$$S=\s... | I think it means that $x$, $y$ and $z$ are non-negatives such that $xy+xz+yz\neq0$.
If $x=3$, $y=1$ and $z=0$ then $S=\frac{5}{2}$.
We'll prove that it's a maximal value.
Indeed, we need to prove that:
$$\sum_{cyc}\frac {x}{\sqrt {x+y}}\leq\frac{5}{4}\sqrt{x+y+z}.$$
By Cauchy-Schwarz
$$\left(\sum_{cyc}\frac {x}{\sqrt ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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What am I doing wrong in computing this Taylor Series of $f(x) = \frac{1}{4+x^{2}}$ about a = 0? $$f(x) = \frac{1}{4+x^{2}}$$
What I did was notice that this is similar to the geometric series:
$$f(x) = \frac{1}{1-x} = 1 + x^{2} + x^{3} + x^{4} + ....$$
So I altered my original function from:
$$f(x) = \frac{1}{4 + x^{2... | Everything is perfect, just one thing:
You have
$$ \frac{1}{4} \cdot \frac{1}{1-\left(-\frac{1}{4} x^2\right)} $$
so you need to take the series and multiply by $\frac14$, not $4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2251680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that there is no primitive Pythagorean triple $(a,b,c)$ where one side differs from another by three I tried using all the Theorems I know about Pythagorean triples, but nothing is working out. I know I should probably go for a proof by contradiction.
| A primitive Pyth. triplet is of the form $(a,b,c)=(m^2-n^2, 2mn, m^2+n^2)$ where $m>n>0$ and $\gcd (m.n)=1 ,$ and where $m,n$ are not both odd. (BTW: If $m,n$ were both odd then $a,b,c$ would all be even.) We have $\gcd (a,b)=1.$
(1). We have $c-a=(m^2+n^2)-(m^2-n^2)=2n^2\neq 3.$
(2). We have $c-b=(m^2+n^2)-2mn=(m-n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2252640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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If $\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$ then the value of $\alpha^6+\beta^6$ is
If $\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$ then the value
of $\alpha^6+\beta^6$ is
I know that, here, $\alpha\beta=4$ and $\alpha + \beta = 2$ and use that result to find $\alpha^2 + \beta^2$ using the expansion... | Well, in general:
$$\text{a}\cdot x^2+\text{b}\cdot x+\text{c}=0\space\Longleftrightarrow\space x=\frac{-\text{b}\pm\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}}{2\cdot\text{a}}\tag1$$
So, for your example we can set:
*
*$$\alpha=x_+=\frac{-\text{b}+\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}}{2\cdot\text{a}}\tag... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2256142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Evaluate the following $(3-2)(5-3)(7-5)........(1995-1993)$
Evaluate the following
$(3-2)(5-3)(7-5)....(1995-1993)$
I have problem in counting the terms since the answer will be $2^n$ , $n$ is the number of terms.
So I count them as follows:
$1995=2n-1$
$1996=2n \to\ n= 998$ ; but since the first term equals $1$... | Since the first factor is indeed $3-2=1$, we will simply ignore it and start by counting $5-3$ as factor #1.
Factor $\#\color{red}{1}$ is $5-3$, which is $(2 \cdot \color{red}{2} + 1) - (2 \cdot \color{red}{2} - 1)$.
Factor $\#\color{red}{2}$ is $7-5$, which is $(2 \cdot \color{red}{3} + 1) - (2 \cdot \color{red}{3} - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2256232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Solve the equation in integers $a,b$: $20a^3-b^3=1$
Solve the equation in integers $a,b$: $$20a^3-b^3=1.$$
Assume that $a \neq 0$. Then simplifying and rearranging the equation gives $$20a^3 = 2^2 \cdot 5 \cdot a^3 = b^3+1 = (b+1)(b^2-b+1).$$ Note that neither factors can be divisible by $3$, since both of them must ... | From your work, after $b+1=20n^3$ and $b^2-b+1=m^3$, multiply second by $64$
$$64b^2-64b+64=(4m)^3 \ \ \ \Longrightarrow \ \ \ (8b-4)^2+48=(4m)^3$$
All solutions to $y^2+48=x^3$ in integers are known: $(x, y)= (4,\pm4), (28,\pm148)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2256544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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Solve the inequality $\sqrt { x + 2} - \sqrt { x + 3} < \sqrt { 2} - \sqrt { 3}$ Solve for $x$ real the inequality $$\sqrt { x + 2} - \sqrt { x + 3} < \sqrt { 2} - \sqrt { 3}.$$ Obviously $x\ge-2$. After that I tried to square the whole inequality, which led me to $x < - \frac { 18} { 4\sqrt { 6} - 5}$. Now, the answer... | One approach:
\begin{align}
\sqrt{ x + 2} - \sqrt{ x + 3} &< \sqrt{ 2} - \sqrt{ 3} \\
\sqrt{ x + 2} + \sqrt{ 3} &< \sqrt{ x + 3} + \sqrt{ 2} \tag{1}\\
(x + 2) + 3 + 2\sqrt{ 3}\sqrt{ x + 2} &< (x + 3) + 2 +2\sqrt{ 2}\sqrt{ x + 3} \tag{2}\\
2\sqrt{ 3}\sqrt{ x + 2} &< 2\sqrt{ 2}\sqrt{ x + 3} \tag{3}\\
3(x + 2) &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2260395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
To prove $\sum_{cyc}\frac{1}{a^3+b^3+abc} \le \frac{1}{abc}$ if $a,b,c$ are non zero positive reals prove $$\sum_{cyc}\frac{1}{a^3+b^3+abc} \le \frac{1}{abc}$$ I have used A.M G.M inequality as follows:
$$a^3+b^3+c^3 \ge 3abc$$ adding $abc$ both sides we get
$$a^3+b^3+abc \ge 4abc-c^3=c(4ab-c^2)$$ so
$$\frac{1}{a^3+b... | $$\sum_{cyc}\frac{abc}{a^3+b^3+abc}\leq\sum_{cyc}\frac{abc}{a^2b+ab^2+abc}=\sum_{cyc}\frac{c}{a+b+c}=1$$
because by Rearrangement $a^3+b^3=a^2\cdot a+b^2\cdot b\geq a^2b+b^2a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2261565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Matrix determinant as Dickson polynomial $\frac{x^{n+1}-y^{n+1}}{x-y}$
Given matrix $$
A=\begin{bmatrix}
x+y&xy&0& .&.&. &0\\
1&x+y&xy&0& .&.&0 \\
0&1&x+y&xy&.&.&. \\
.&.&.&.&.&.&. \\
.&.&.&.&.&.&0 \\
.&.&.&.&.&.&xy \\
0&.&.&.&0&1&x+y
\end{bmatrix}
$$
prove by induction that $$|A|=\frac{x^{n+1}-y^{n+1}}{x-y}$$ $x \neq... | Your base cases are good. For the induction step, assume that
$$D_n=\frac{x^{n+1}-y^{n+1}}{x-y}\quad \text{and}\quad D_{n-1}=\frac{x^{n}-y^{n}}{x-y}.$$
Then you have that
$$D_{n+1}=(x+y)D_n-xyD_{n-1},$$so you only have to show that
$$(x+y)\frac{x^{n+1}-y^{n+1}}{x-y}-xy\frac{x^{n}-y^{n}}{x-y}=\frac{x^{n+2}-y^{n+2}}{x-y}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2262488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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pythagorean theorem extensions are there for a given integer N solutions to the equations
$$ \sum_{n=1}^{N}x_{i} ^{2}=z^{2} $$
for integers $ x_i $ and $ z$
an easier equation given an integer number 'a' can be there solutions to the equation
$$ \sum_{n=1}^{N}x_{i} ^{2}=a^2 $$
for N=2 this is pythagorean theorem
| Just to avoid notation bloat, let's let $N=4$ and let the implied generalization take care of the rest of the question. The question can be restated as asking whether there are rational points on the $4$-sphere $\mathbb S_4:x_1^2 + x_2^2 +x_3^2+x_4^2=1$. We know that $(1,0,0,0)$ is a (trivial) rational point on
$\math... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2265828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Prove by induction that $2^n + 4^n \leq 5^n$ I'm trying to prove by induction that $2^n + 4^n \leq 5^n$. Through some value plugging I've established that the induction must start from $n = 2$ because $2^2 + 4^2 \leq 5^2 \equiv 20 \leq 25$; for $n = 1$ it doesn't hold since $2 + 4 \geq 5$.
Now I assume that $2^k + 4^... | Assume that $2^k + 4^k \leq 5^k$ is true.
Then
\begin{align} 2^{k+1} + 4^{k+1} & \leq 2 ( 2^{k} + 4^{k}) + 2\cdot 4^k \\
& \le 2 \cdot 5^k + 2 \cdot 4^k \\
& \le 2 \cdot 5^k + 2 \cdot 5^k \\
& \le 4 \cdot5^k \\
& \le 5^{k+1}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2266291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
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Prove $k^8 \geq (k^2 - 1)^4$ for $k \geq 1$ Can anyone please help me with following inequality:
$$k^8 \geq (k^2 - 1)^4, \text{where} \ k \geq 1.$$
I tried induction but get stuck at step $k=n+1$ and can't progress anywhere.
| Since $k^8 = (k^2)^4$, we just need to show that (for $a,b\geq0$) $a\geq b \implies a^4 \geq b^4$. Indeed, if $a\geq b$, then $a-b\geq 0$, and so $a^4-b^4 = (a-b)(a^3+a^2b+ab^2+b^3) \geq 0$. Thus, $a^4 \geq b^4$.
Clearly, $k^2>k^2-1$, and so $k^8 > (k^2-1)^4$ since $k^2, k^2-1 \geq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2268309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq 3$ I have to prove that $$\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq 3 $$
is always true for real numbers $a, b, c>0$ with $abc=1$.
Using the AM-GM inequality I got as far as $$\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq \frac{b}... | Hint: $\frac{1+ab}{1+a}=\frac{abc+ab}{1+a}=ab\big(\frac{1+c}{1+a}\big)$. So what is $\big(\frac{1+ab}{1+a}\big)\big(\frac{1+bc}{1+b}\big)\big(\frac{1+ca}{1+c}\big)$?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Express $\sin^8\theta+\sin^6\theta+\sin^4\theta+\sin^2\theta-2$ as a single term in terms of $\sin\theta$
If $\cos^2\theta+\cos\theta = 1$, express $\sin^8\theta+\sin^6\theta+\sin^4\theta+\sin^2\theta-2$ as a single term in terms of $\sin\theta$.
We have \begin{align*}\sin^8\theta+\sin^6\theta+\sin^4\theta+\sin^2\the... | $$\cos^2 (t)+\cos (t)=1$$
$$\implies \sin^2 (t)=\cos(t)$$
$$\implies \sin^4 (t)=\cos^2 (t) $$
$$\implies \sin^4 (t)+\sin^2 (t)=1 $$
$$\implies \sin^8 (t)+\sin^6 (t)=\sin^4 (t) $$
$$\implies f(t)=\sin^4 (t)+1-2=$$
$$=\cos^2 (t)-1=-\sin^2 (t ) $$ which is your result .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2269077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $\frac{\tan \alpha + \sqrt{5}\sin \alpha}{2}(\sqrt{5}\cos \alpha -1)$ is equivalent to $\frac{5}{4}\sin(2\alpha)-\frac{1}{2}\tan \alpha$
Show that $\frac{\tan \alpha + \sqrt{5}\sin \alpha}{2}(\sqrt{5}\cos
\alpha -1)$ is equivalent to $\frac{5}{4}\sin(2\alpha)-\frac{1}{2}\tan
\alpha$.
I tried:
$$\frac{\fra... | Problem:
$$
f(t) = \frac{1}{2} \left(\sqrt{5} \cos t-1\right) \left(\sqrt{5} \sin t+\tan t\right)
$$
FOIL Components:
$$
%
\begin{align}
%
F
&= \frac{1}{2} \sqrt{5} \cos t
\left(\sqrt{5} \sin t\right)
= \frac{5}{2} \sin t \cos t\\[7pt]
%
O &= \frac{1}{2} \sqrt{5} \cos t \left( \tan t\right)
= \frac{\sqrt{5}}{2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2269193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\frac{xy}{z+1}+\frac{yz}{x+1}+\frac{zx}{y+1}$ is an integer Let $x$ be a positive integer. Does there always exist integers $y,z$ such that the sum
$$\frac{xy}{z+1}+\frac{yz}{x+1}+\frac{zx}{y+1}$$
is an integer, but none of the three terms of the sum is an integer?
For instance: Take $x=1$. Then we need the sum $$\fra... | Let $y=z=2x(x+1)-1$. Then
$$
\frac{xy}{z+1}=\frac{zx}{y+1}=x-\frac{1}{2(x+1)}
$$
and
$$
\frac{yz}{x+1}=4x^2(x+1)-4x+\frac{1}{x+1}.
$$
Since $x+1\geq2$, neither of these are integers, but
$$
\frac{xy}{z+1}+\frac{yz}{x+1}+\frac{zx}{y+1}=4x^2(x+1)-2x
$$
is an integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2270843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How can we tackle this integral $\int_{0}^{1}{2x^2-2x+\ln[(1-x)(1+x)^3]\over x^3\sqrt{1-x^2}}\mathrm dx=-1?$ Something is wrong with this integral (in terms of splitting them out)
$$\int_{0}^{1}{2x^2-2x+\ln[(1-x)(1+x)^3]\over x^3\sqrt{1-x^2}}\mathrm dx=\color{blue}{-1}\tag1$$
My try:
Splitting the integral
$$\int_{0... | I will present an answer without Gamma functions and without series' expansions. Instead, a more general problem is solved, where the OP's question is a special case.
Let
$$
I(a) = \int_{0}^{1}{2a^2x^2-2ax+\ln[(1-ax)(1+ax)^3]\over x^3\sqrt{1-x^2}}\mathrm dx
$$
The answer to the OP's question is then given by $I(a = 1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2271011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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Proving that $a_n=\frac{n^2+2n+6}{n^3-3}\to 0$ as $n\to\infty$ $$a_n=\frac{n^2+2n+6}{n^3-3}$$
So I want to show that "$a_n\to a\iff\forall \epsilon>0,\quad\exists N\in\mathbb{N}:n\geq N\implies |a_n-a|<\epsilon$"
Then my rough working:
$|a_n-0| =\left|\frac{n^2+2n+6}{n^3-3}\right|<\epsilon$
Estimate $\frac{n^2+2n+6}{n^... | If i were you, i would just multiplicate numerator and denominator by $1/n^2$ both, so that: $a_n = (1 + 2/n + 6/n^2)/(n -3/n^2)$. From the defenition of infinite number, we have $2/n$, $6/n^2$ and $-3/n^2$ all equal to 0: $a_n = (1 + 0 + 0)/(n + 0 + 0) = 1/n$ which is 0 from the same defenition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2275344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
showing $ 1-\frac{3}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{3}{6}+\frac{1}{7}+\frac{1}{8}+\cdots=0 $ How to show that the following infinite series
$$
1-\frac{3}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{3}{6}+\frac{1}{7}+\frac{1}{8}+\cdots=0?
$$
The above series is of the form $\sum_{n \ge 1} \frac{f(n)}{n}$,... | I'm a little wary of doing this, since it's not absolutely convergent, but can you rewrite each period as
$$
\left(\frac{1}{4k-3}-\frac{1}{4k-2}+\frac{1}{4k-1}-\frac{1}{4k}\right) -
\left(\frac{2}{4k-2}-\frac{2}{4k}\right)
$$
which can in turn be rewritten as
$$
\left(\frac{1}{4k-3}-\frac{1}{4k-2}+\frac{1}{4k-1}-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2275508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If $\tan ^{-1}y=\tan ^{-1}x+\tan ^{-1}\left( \frac{2x}{1-x^2} \right) $ where $|x| < \frac {1}{\sqrt {3}}$, then what is $y$? If $\tan ^{-1}y=\tan ^{-1}x+\tan ^{-1}\left( \dfrac{2x}{1-x^2} \right)$ where $|x| < \dfrac {1}{\sqrt {3}}$ then find the value of $y$.
. . .
Let $\tan^{-1} y= A$
$$y=\tan A$$
$$\tan^{-1} x=B$$
... | We know that $$tan2A= \frac{2tanA}{1-tan^2A}$$
Now, substitute $tanA=x\implies tan^{-1}x=A$
Plugging this into first equation, $tan2(tan^{-1}x)= \frac{2x}{1-x^2}$
Again taking $tan^{-1}$ on both sides we get
$$2tan^{-1}x = tan^{-1}\frac{2x}{1-x^2}$$
Plugging this into the question,
$$tan^{-1}y= tan^{-1}x + 2tan^{-1}x =... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the value of $\cos \tan^{-1} \sin \cot^{-1} (x)$ . Find the value of $\cos \tan^{-1} \sin \cot^{-1} (x)$ .
...
Let $\cot^{-1} x=z$
$$x=\cot z$$
Then,
$$\sin \cot^{-1} (x)$$
$$=\sin z$$
$$=\dfrac {1}{\csc z}$$
$$=\dfrac {1}{\sqrt {1+\cot^2 z}}$$
$$=\dfrac {1}{\sqrt {1+x^2}}$$
| Then we have $$\cos\left(\left(\tan^{-1}\dfrac1{\sqrt{1+x^2}}\right)\right)$$
If $\tan^{-1}\dfrac1{\sqrt{1+x^2}}=y,\dfrac\pi4\le y\le\dfrac\pi2\implies\tan y=\dfrac1{\sqrt{1+x^2}}$ and $\cos y\ge0$
$\cos\left(\tan^{-1}\dfrac1{\sqrt{1+x^2}}\right)=\cos y=+\dfrac1{\sqrt{1+\tan^2y}}=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2275999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve for $x$ in $\sqrt[4]{57-x}+\sqrt[4]{x+40}=5$ Solve for $x$ in $$\sqrt[4]{57-x}+\sqrt[4]{x+40}=5$$
i have done in a lengthy way:
By inspection we observe that $x=41$ and $x=-24$ are the solutions
we have
$$\sqrt[4]{57-x}=5-\sqrt[4]{x+40}$$ squaring both sides we get
$$\sqrt{57-x}=25+\sqrt{x+40}-10 \sqrt[4]{x+40}$$... | Firstly, I would recommend you to make the following substitution $a = \sqrt[4]{57 - x}$ and $b = \sqrt[4]{x+40}$. Thus we have $a + b = 5$ and $a^{4} + b^{4} = 57 + 40 = 97$. Therefore:
\begin{align*}
a^{4}+b^{4} & = (a^{2}+b^{2})^{2} - 2a^{2}b^{2} = [(a+b)^{2} - 2ab]^{2} - 2a^{2}b^{2} = (5^{2} - 2ab)^{2} - 2a^{2}b^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2277360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Functions and Sequences Problem
The function $F(k)$ is defined for positive integers as $F(1) = 1$,
$F(2) = 1$, $F(3) = -1$ and $F(2k) = F(k)$, $F(2k + 1) = F(k)$ for $k \geq
2$. Then $$F(1) + F(2) + \dotsb + F(63)$$ equals
$\begin{array}{lr}
(\text{A}) & 1 \\
(\text{B}) & -1 \\
(\text{C}) & -32 \\
(\text{D})... | For $k \ge 2$ we have:
\begin{align}
F(2k) &= F(k) \\
F(2k+1) &= F(k)
\end{align}
This means
$$
F(n) = F(\lfloor n / 2 \rfloor) \quad (*)
$$
for $n \ge 4$.
If we are using base 2 numbers this means we have a word of length $m$ with at least three bits
$$
n = (1 b_{m-1} \dotsb b_2 b_1)_2 \quad (b_i \in \{0, 1\})
$$
and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2278951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Argument and modules of complex numbers Can you help me to find the argument and modules of this complex number please?
$$z=\frac{2-i}{-\sqrt {3}-2i}$$
| Argument of $z$,
\begin{align}
\arg(z)&=\arg\left(\dfrac{2-i}{-\sqrt{3}-2i}\right)\\
&=\arg(2-i)-\arg(\sqrt{3}-2i)\\
&=\left[-\tan^{-1}\left(\dfrac{1}{2}\right)\right]-\left[-\pi+\tan^{-1}\left(\dfrac{2}{\sqrt3}\right)\right]\\
&=\pi-\left[\tan^{-1}\left(\dfrac{1}{2}\right)+\tan^{-1}\left(\dfrac{2}{\sqrt3}\right)\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2279037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Algebra: Prove inequality $\sum_{n=1}^{2015} \frac1{n^3} < \frac 54$ Can someone prove inequality (n is natural):$$\sum_{n=1}^{2015} \frac{1}{n^3} < \frac 5 4$$
I have tried some predictions like $a^3 > a(a - 1)(a - 2) $ but couldn't get anything out of them.
| You can use (for $n \geq 2$)
$$\frac{1}{n^3} < \frac{1}{n^2(n-1)} < \frac12\frac{2(n-1) +1}{n^2 (n-1)^2} =\frac{1}{2} \left( \frac{1}{(n-1)^2} - \frac{1}{n^2}\right).$$
Write your sum as
$$1 + \frac{1}{2^3} + \sum_{2015\geq n\geq 3} \frac{1}{n^3} < \frac{9}{8} + \sum_{n\geq 3} \frac{1}{2} \left( \frac{1}{(n-1)^2} - \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2279549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
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Bernoulli equation for pressure (with the vortex) An incompressible inviscid fluid of constant density ρ moves subject to a gravitational acceleration −gk.
The vortex caused by water flowing down a plughole is modelled by the steady velocity field
when $x^2+y^2\le a^2$
$$u=\left(\frac{Γy}{a^2},-\frac{Γx}{a^2},0 \rig... | For inviscid, rotational flow a weaker form of the Bernoulli equation $p + \frac{1}{2} \rho | \mathbf{u}|^2 + \rho g z = C(\psi)$ is valid, where $C(\psi)$ is constant along a streamline. You can work with this or solve for the pressure field directly from the governing equations for inviscid flow, which reveals the B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2280972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Recurrence relation with variable $x$ in the denominator. A recurrence is defined such that $$f_n(x)=f_1(f_{(n-1)}(x)); x\ge2$$
$f_1(x)$ is defined as
$$f_1(x)= \frac 23 -\frac 3{3x+1}$$
How can I find the values of $x$ for which the following holds true? $$f_{1001}(x)= x-3$$
I was not able to deduce any kind of patt... | Notice that $$f_2(x) = \frac{2}{3}-\frac{3}{3\cdot\left(\frac{2}{3}-\frac{3}{3x+1}\right)+1} = \frac{2}{3}-\frac{3x+1}{3x-2}$$
$$f_3(x) = \frac{2}{3}-\frac{3}{3\cdot\left(\frac{2}{3}-\frac{3x+1}{3x-2}\right)+1} = \frac{2}{3}+\frac{3x-2}{3}$$
$$f_4(x) = \frac{2}{3}-\frac{3}{3\cdot\left(\frac{2}{3}+\frac{3x-2}{3}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2282977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is it true that for $n \ge5$, ${{3n} \choose {2n}} > \frac{6^n}{n}$ For $n \ge 5, \frac{9n^2 - 9n + 2}{4n^2-2n} > 2$
$9n^2 -9n +2 -8n^2+4n = n^2 -5n +2 \ge 5^2-25+2$
Here's the basis step:
$${{9} \choose {6}} = 84 > \frac{6^3}{3} = 72$$
Here's the inductive step:
$${{3n} \choose {2n}} = {{3(n-1)}\choose {2(n-1)}}\... | $$\begin{eqnarray*}\log\binom{3n}{2n} &=& \log\Gamma(3n+1)-\log\Gamma(n+1)-\log\Gamma(2n+1)\\ &=& -\log n-\log B(n,2n+1)\end{eqnarray*}$$
and
$$ B(n,2n+1) = \int_{0}^{1}\left[x(1-x)^2\right]^n\,\frac{dx}{x} $$
is a log-convex function by the Cauchy-Schwarz inequality.
You don't even need induction.
| {
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"timestamp": "2023-03-29T00:00:00",
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Compute the sum of the power serie $\sum_{1}^{\infty } \frac{(n^{2} + n)x^{n-1}}{2^{n-1}}$
I have found convergence interval for series: $\left | x \right | < 2$,
but I have no idea how to find the sum of the power series.
| $$\sum\limits_{n=1}^\infty{(n^2+n)x^{n-1}\over2^{n-1}} = 4{d^2\over dx^2}\sum\limits_{n=1}^\infty{x^{n+1}\over2^{n+1}} = 4{d^2\over dx^2}{{x^2\over2^2}\over1-{x\over2}} = 2{d^2\over dx^2}{x^2\over2-x} = 2{d^2\over dx^2}{x^2-4+4\over2-x} = 2{d^2\over dx^2}\left(-x-2 + {4\over2-x}\right) = {8\cdot1\cdot2\over(2-x)^3} = {... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Similar Triangles and Incircle The incircle of trianlge ABC touches the sides AB, AC and BC at points P, N, and M respectively. Denote AP=AN=x, BM=BP=y and CM=CN=z. Segment UV is tangent to the incircle at point X and parallel to the side AC.
Prove $\cfrac{UV}{AC}$= $\cfrac{y}{x+y+z}$
What I have so far
$\triangle$ ... | Let $a$, $b$, $c$ be the sidelengths of $BC$, $CA$, $AB$, let $h_1$ be the height from $B$ to $AC$, and let $h_2$ be the height from $B$ to $UV$. Note that $\dfrac{UV}{AC} = \dfrac{h_2}{h_1}$ because of the similar triangles you mentioned. Also, $h_2 = h_1 - 2r$ where $r$ is the inradius of triangle $ABC$. Hence $$\... | {
"language": "en",
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"source": "stackexchange",
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Matlab Question: How to compute all partitions of [1,2,...,n] for a fixed number of parts? For example: [1,2,3,4] and k=3 parts should yield
[1],[2],[3,4]
[1],[3],[2,4]
[1],[4],[2,3]
[2],[3],[1,4]
[2],[4],[1,3]
[3],[4],[1,2]
as an output.
For k=2 with the same set we would get
[1],[2,3,4]
[2],[1,3,4]
[3],[1,2,4]
[4],[... | I will solve problem for $n=4$ special case. I hope that you can find the solution method for bigger values of $n$.
For $k=1$, there is only $1$ partition: $[1234]$
For $k=2$, there are two groups that $[ab],[cd]$ and $[a],[bcd]$. In the case $[ab],[cd]$: $\dbinom{4}{2} \cdot \dbinom{2}{1}$. But $[ab],[cd]$ and $[cd],[... | {
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solve the set of simultaneous congruences using Chinese Remainder Theorem 2x= 1 (mod 5), 3x= 9 (mod 6), 4x = 1 (mod 7), 5x = 9 (mod 11) Solve the set of simultaneous congruences using Chinese Remainder Theorem
$\begin{cases}
2x \equiv 1 \pmod{5} \\
3x \equiv 9 \pmod{6} \\
4x \equiv 1 \pmod{7} \\
5x \equiv 9 \pmod{11} \... | $\begin{cases}
2x \equiv 1 \pmod{5} & \times 3 \quad 6\equiv 1\pmod 5 & x\equiv \color{green}3 \pmod 5\\
3x \equiv 9 \pmod{6} & /3 \text{ whole equation} & x\equiv 3\equiv \color{green}1\pmod 2\\
4x \equiv 1 \pmod{7} & \times 2 \quad 8\equiv 1\pmod 7 & x\equiv \color{green}2\pmod 7\\
5x \equiv 9 \pmod{11} & \times 9 \q... | {
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"timestamp": "2023-03-29T00:00:00",
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Factorise $x^5+x+1$
Factorise $$x^5+x+1$$
I'm being taught that one method to factorise this expression which is $=x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1$
$=x^3(x^2+x+1)-x^2(x^2+x+1)+x^2+x+1$
=$(x^3-x^2+1)(x^2+x+1)$
My question:
Is there another method to factorise this as this solution it seems impossible to invent it?
| standard trick from contests: $5 \equiv 2 \pmod 3.$ Therefore, if $\omega^3 = 1$ but $\omega \neq 1,$ we get
$$ \omega^5 = \omega^2 $$
$$ \omega^5 + \omega + 1 = \omega^2 + \omega + 1 = 0 $$
Which means, various ways of saying this, $x^5 + x + 1$ must be divisible by
$$ (x - \omega)(x - \omega^2) = x^2 + x + 1. $$
T... | {
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The eigenvalues and the eigenvectors of $y^{(4)} = \lambda y,~y(0)=0,y(1)=0,y'(0)=0,y'(1)=0$ To find the eigenvalues and the eigenvectors of $$y^{(4)} = \lambda y,~y(0)=0,y(1)=0,y'(0)=0,y'(1)=0,$$ I proceed as follows
$$\begin{align}
\bigg(\frac{d^{
4}}{dt^{4}} - \lambda \bigg) y &= \bigg( \frac{d^{2}}{dt^{2}} + \sqrt{... | This is my solution of the problem. Please let me know if you have any comments (Ahmed).
To find the eigenvalues and the eigenvectors of $$y^{(4)} = \lambda y,~y(0)=0,y(1)=0,y'(0)=0,y'(1)=0,$$ we proceed as follows
\begin{align*}
\bigg(D^4 - \lambda \bigg) y &= \bigg( D^2 + \sqrt{\lambda} \bigg) \bigg(D^2 - \sqrt{\lamb... | {
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Intersection of two parabolas Given $a>0$ and $b>0$, I want to find the points of intersection of the two parabolas
\begin{align}
y&=1-ax^2 \\x&=1-by^2
\end{align}
Clearly I can just eliminate one of the variables, and I'll get a quartic equation, whose general solutions will be an enormous mess (according to Mathe... | Perhaps this is an improvement.
$$
y = 1 + a x^{2}
\tag{1}
$$
$$
x = 1 + b y^{2}
\tag{2}
$$
Substitute $(2)$ into $(1)$ to obtain
$$
y = 1 + a \left(b y^2+1\right)^2
$$
and solve for $y$.
$$
y = \color{blue}{\pm} \frac{1}{2} \sqrt{-\frac{4 \sqrt[3]{2} (4 a+3)}{3 \sqrt[3]{\beta -3 \sqrt{3} \sqrt{\alpha }}}-\frac{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2295710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Manipulating a Summation Series
Summation: How does$$\begin{align*} & \frac 12\sum\limits_{k=1}^n\frac 1{2k-1}+\frac 12\sum\limits_{k=1}^n\frac 1{2k+1}-\frac 12\sum\limits_{k=1}^n\frac 1k+\frac n{2n+1}\tag1\\ & =\sum\limits_{k=1}^n\frac 1{2k-1}-\frac 12\sum\limits_{k=1}^n\frac 1k\tag2\\ & =\sum\limits_{k=1}^{2n}\frac ... | When you face problem like this, always try to decompose the summation into simple addition form. If you decompose your summation into simple series addition form these summations are settled down easily.
Step 1-2:
\begin{align*}
&\frac 12\sum\limits_{k=1}^n\frac 1{2k-1}+\frac 12\sum\limits_{k=1}^n\frac 1{2k+1}\\&=\fra... | {
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limit $\lim_{n\to\infty}\frac{1}{\sqrt n}\left(\frac{1}{\sqrt 2+\sqrt4}+\frac{1}{\sqrt4+\sqrt6}+\cdots+\frac{1}{\sqrt{2n}+\sqrt{2n+2}}\right)$
$$\lim_{n\to\infty}\frac{1}{\sqrt n}\left(\frac{1}{\sqrt 2+\sqrt4}+\frac{1}{\sqrt4+\sqrt6}+\cdots+\frac{1}{\sqrt{2n} +\sqrt{2n+2}}\right)$$
To find the limit I think I can use... | $$\lim _{ n\to \infty } \frac { 1 }{ \sqrt { n } } \left( \frac { 1 }{ \sqrt { 2 } +\sqrt { 4 } } +\frac { 1 }{ \sqrt { 4 } +\sqrt { 6 } } +.........+\frac { 1 }{ \sqrt { 2n } +\sqrt { 2n+2 } } \right) =\\ =\lim _{ n\to \infty } \frac { 1 }{ \sqrt { n } } \left( \frac { \sqrt { 2 } -\sqrt { 4 } }{ -2 } +\frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2296393",
"timestamp": "2023-03-29T00:00:00",
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Derivative of vector consisting of euclidean distances I have $g: \Bbb R^2 \to \Bbb R^4$ given by $g(x) = (\|c_1 - x\|, \dots, \|c_4 - x\|)$, where $c_1, \dots, c_4 \in \Bbb R^2$.
I want to find $\left( \dfrac {\partial g} {\partial x_1}, \dfrac {\partial g} {\partial x_2} \right)$. Any hints?
| You are esssentially asking about the partial derivatives of the Euclidean distance function. Given that $\|c - x\| = \sqrt {(c_1 - x_1)^2 + (c_2 - x_2)^2}$, it follows that
$$\begin{eqnarray} \frac {\partial \|c - x\|} {\partial x_1} = \frac {x_1 - c_1} {\sqrt {(c_1 - x_1)^2 + (c_2 - x_2)^2}} \\
\frac {\partial \|c - ... | {
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Find values of constants a and b such that the given improper integral converges Find values of constants a and b such that:$$\int_{3}^\infty \left(\frac{ax+2}{x^2+3x}-\frac{b}{3x-2}\right) dx=k$$ then by partial fractions we get:
$$\lim_{N \to \infty} \int_{3}^N \left(\frac{2}{3}\frac{1}{x}+\frac{a-\frac{2}{3}}{x+3}-\... | Hint. From your third line, one may write, as $x \to \infty$,
$$
\begin{align}
&\color{blue}{\frac{2}{3}}\ln(x)+\color{blue}{\left(a-\frac{2}{3}\right)}\ln (x+3)\color{blue}{-\frac{b}{3}}\ln(3x-2)
\\=&\color{blue}{\left(\frac{2}{3}+a-\frac{2}{3}-\frac{b}{3}\right)}\ln x+\left(a-\frac{2}{3}\right)\ln \left(1+\frac 3x\ri... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\prod_{n=2}^\infty \frac{1}{e^2}\left(\frac{n+1}{n-1}\right)^n=\frac{e^3}{4\pi}$ I friend of mine sent me this problem a while ago, but I still can't figure it out. (He can't figure it out either.) I figured here would be a good place to ask for help.
Prove:
$$\prod_{n=2}^\infty \frac{1}{e^2}\left(\frac{... | $$\prod_{n=2}^\infty \frac{1}{e^2}\left(\frac{n+1}{n-1}\right)^n=\frac{e^3}{4\pi}$$
$$\log\left(\prod_{n=2}^\infty \frac{1}{e^2}\left(\frac{n+1}{n-1}\right)^n\right) = \log(\frac{e^3}{4\pi})$$
$$\sum_{n=2}^\infty \log\left(\frac{1}{e^2}\left(\frac{n+1}{n-1}\right)^n\right) = \log(\frac{e^3}{4\pi})$$
$$\sum_{n=2}^\infty... | {
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"timestamp": "2023-03-29T00:00:00",
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Sum of Squares of odd numbers I just wanted to know how we get from the first step to the next. I spent too much time on this but could not get to the next step. This seems a simple sum of squares of odd numbers but was not able to get the next step.
$$ \frac{2 \mathcal{E}_p}{M} \left(1^2 + 3^2 + 5^2 + \dots + (M-1)^2\... | By Faulhaber's formulas, $$1^2+2^2+\cdots + (M-1)^2=\frac{(M-1)M(2M-1)}{6}$$
Since $M-1$ is odd, we must have $M-2$ even. We now calculate, using the same formula,
$$2^2+4^2+\cdots+(M-2)^2=2^2(1^2+2^2+\cdots+(\frac{M-2}{2})^2)=2^2\frac{\frac{M-2}{2}\frac{M}{2}(M-1)}{6}=\frac{(M-2)M(M-1)}{6}$$
Subtract the sum of the e... | {
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Inclusion exclusion distribution problem Prompt: How many ways are there to give 20 different presents to 4 different children, so that no child gets exactly 6 presents. All presents are different.
Here's what I tried doing
Since there are 20 presents to be distributed among 4 children,
$$C_1 + C_2 + C_3 + C_4 = 20$$... | Your use of combinations with repetition would be appropriate if the presents are identical. However, they are different, so it matters which child receives which present.
If there were no restrictions, there would be four possible recipients for each of the twenty presents, so there would be $4^{20}$ ways to distribu... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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every integer can be represented in the form $x^2+y^2-z^2$ Every integer can be represented in the form $x^2+y^2-z^2$ and show that $6$ actually requires all three terms.
I put
*
*$z=y+1$
*$x=n^2+3$
*$y=3n^2+4+(n^4-n)/2$
what does it mean that $6$ actually requires all three terms?
| Any integer number that is not of the form $4m+2$ can be represented as a difference of two squares, since $n=y^2-z^2=(y-z)(y+z)$ has a solution as soon as $n$ has a divisor $d$ such that $d$ and its complementary divisor $\frac{n}{d}$ have the same parity. That is the same as stating "$n$ is not twice an odd number". ... | {
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"timestamp": "2023-03-29T00:00:00",
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Egyptian fraction for $\varphi- {F(2n+2) \over F(2n+1)}$ The sum of the reciprocals of the ${2^n}$th Fibonacci numbers is known to be $\dfrac{3-\sqrt{5}}{2}$.
https://math.stackexchange.com/a/746678/134791
This may be written as the following closed form for an Egyptian fraction.
$$\varphi=2-\sum_{k=0}^\infty \frac{1}{... | From a different expansion for the expression given by @achillehui $\frac{\alpha-\beta}{\alpha^{2m}+1}$, as a geometric series, particular cases are
$$\varphi = 1+\sqrt{5}\sum_{k=1}^\infty (-1)^{k+1}\left(\frac{2}{1+\sqrt{5}}\right)^{2k}$$
$$\varphi = \frac{3}{2}+\sqrt{5}\sum_{k=1}^\infty (-1)^{k+1}\left(\frac{2}{1+\sq... | {
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"url": "https://math.stackexchange.com/questions/2308987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove that $\sqrt{-3-2i}+\sqrt{-3+2i} = \sqrt{2(\sqrt{13}-3)}$? Is there a trick to show that
$$\sqrt{-3-2i}+\sqrt{-3+2i} = \sqrt{2(\sqrt{13}-3)}$$
is true ?
| Note that the right hand side is real, so if we take complex conjugates it remains unchanged.
$$
\begin{align}
\sqrt{-3-2i}+\sqrt{-3+2i} &= \sqrt{2(\sqrt{13}-3)} \\
& = \sqrt{-3+2i}+\sqrt{-3-2i} \\
\end{align}
$$
Multiply the 2 conjugate forms to obtain
$$
\begin{align}
&(\sqrt{-3-2i}+\sqrt{-3+2i})\cdot (\sqrt{-3+2i}... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate series $\sum\limits_{n=1}^{\infty}\frac{x^{2^{n-1}}}{1-x^{2^n}}$
Determine the value of
$$\sum_{n=1}^{\infty}\frac{x^{2^{n-1}}}{1-x^{2^n}}$$
or $$\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8}+\cdots$$
for $x\in\mathbb{R}$.
The answer is $\dfrac{x}{1-x}$ for $x\in(0,1)$. To prove this, notice
$$\f... | Adding term by term:
$$S=\frac{x(1+x^2)+x^2}{1-x^4}+\frac{x^2}{1-x^8}+...=$$
$$\frac{(x+x^2+x^3)(1+x^4)+x^4}{1-x^8}+\frac{x^8}{1-x^{16}}+...=$$
$$\frac{\sum_{n=1}^{\infty} x^n}{1-x^{2\cdot \infty}}=$$
$$\frac{x}{1-x}$$
because $0<x<1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Hensel Lifting Explanation
I'm trying to understand this example for my exam tomorrow, I understand everything apart from where the recurrence relation comes from, is there a set formula for this? Any help is greatly appreciated.
| Let's do this slowly. At each stage we have a number $a_n$ with both $a_n\equiv 3\pmod 5$ and with $a_n^3\equiv2\pmod{5^n}$. We try $a_{n+1}=a_n+5^nt$. Then
$$a_{n+1}^3\equiv a_n^3+3\times 5^na_n^24
=2+(a_n^3-2)+3\times 5^na_n^2t\pmod{5^{n+1}}.$$
Now
$$3a_n^2\equiv3\times 3^2\equiv 2\pmod5$$
so
$$3\times 5^na_n^2\equiv... | {
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Prove that $n=\frac{5^{125}-1}{5^{25}-1}$ is a composite number
Prove that $n=\dfrac{5^{125}-1}{5^{25}-1}$ is a composite number
My attempt,
Let $x=5^{25}$, so that $5^{125}-1=x^5-1=(x-1)(x^4+x^3+x^2+x+1)$
$=(x^4+9x^2+1+6x^3+6x+2x^2-5x^3-10x^2-5x)(x-1)$
$=((x^2+3x+1)^2-5x(x+1)^2)(x-1)$
I'm stuck at this point and don... | The polynomial $\Phi_5(x)=x^4+x^3+x^2+x+1$ fulfills an interesting identity.
We have that $4\cdot \Phi_5(x)$ is pretty close to the square of $2x^2+x+2$, and indeed:
$$ 4 \Phi_5(x) = (2x^2+x+2)^2 - 5x^2 \tag{1}$$
as well as:
$$ \Phi_5(x) = (x^2+3x+1)^2 - 5x(x+1)^2 \tag{2} $$
so if $x=5^{2k+1}$, $\Phi_5(x)$ is the diffe... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Maximum rotation angle of a rectangle inside a given rectangle I want to calculate the maximum rotation angle of a rectangle which is rotating with the center on the center of a bigger rectangle.
Here is a figure for better understanding: fig.
I have tried it out already but my solution is not plausible.
Thanks in ... | Both rectangle have the same center due to their symmetry. There will be four contact points on the outer rectangle if the rectangles have same aspect ratio, else only two.
Let the inner rectangle have dimensions $(2a,2b)$. The y coordinate of contact point is $B_0/2$. Radius of inner rectangle is $ \sqrt{(a^2+b^2)} $ ... | {
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Finding all primes $p,q$ with $p^2+q^2=9pq-13$ The last month I was trying to solve a problem of a magazine, and I found the following equation $$p^2+q^2=9pq-13,$$
Where $p$ and $q$ are primes. We need to get solutions when $p$ and $q$ are odd, because if any of them is even, the only solution that works is $(2,17)$. A... | This answer will find all integer answers. Don't know if there is an easy way to find prime answers.
Multiplying by $4$ we get $4p^2-36pq+4q^2=-52$ or $(2p-9q)^2-77q^2=-52$.
Now, if there is a solution to $x^2-77y^2=-52$, with $x,y>0$ then we have that $x,y$ must have the same parity, and we take.
$$\begin{align}x_1+y... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Two dice-related games, do the outcomes have the same distribution? In this codegolf question, it is effectively asked to find the distribution for $X$, defined below:
*
*Throw $2$ dice; is the sum $7$ then $X=2$, else continue.
*Throw $4$ dice; is the sum $14$ then $X=4$, else continue.
*...
*Throw $2n$ dice; is... | $X$ and $Y$ have different distributions. To see this, consider the following:
$$P[X=2] = P[Y=2] = \frac{6}{36} = \frac{1}{6}$$
For $X$ to equal 4, we first must not hit 7 with the first two dice, and then hit 14 with the four new dice. To throw these last four, we can first throw two new dice and then another two dice... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2321146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is $\lim_{k\to\infty}\frac{\sum_{n=1}^{k} 2^{2\times3^{n}}}{2^{2\times3^{k}}}=1$? Look at this limit. I think, this equality is true.But I'm not sure.
$$\lim_{k\to\infty}\frac{\sum_{n=1}^{k} 2^{2\times3^{n}}}{2^{2\times3^{k}}}=1$$
For example, $k=3$, the ratio is $1.000000000014$
Is this limit mathematically correct?
... | I have taken a simpler approach to solve this.
$$\frac{\sum_{n=1}^k 2^{2X3^n}}{2^{2X3^k}}$$ can be written as:
$$\frac{\sum_{n=1}^k 4^{3^n}}{4^{3^k}}$$ which further can be written as :
$$\frac{4^{3^1}}{4^{3^k}} + \frac{4^{3^2}}{4^{3^k}} +\frac{4^{3^3}}{4^{3^k}}......... \frac{4^{3^{k-1}}}{4^{3^k}} +\frac{4^{3^k}}{4^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2322481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 6
} |
Balls in bins, probability that exactly two bins are empty If we throw randomly $n+1$ balls into $n$ bins, what is the probability that exactly two bins are empty?
I tried to do this problem but I didn't get right solution, I would be very grateful if someone would point where am I making mistake, and how should I thin... | This is a probability question, so you have a choice of considering ways to arrange indistinguishable balls or ways to arrange balls that are all different. You just have to be consistent in how you add up the probability of each arrangement.
The number of ways to arrange $n+1$ balls (all different) into the $n$ bins i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2322858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Prove or disprove that where $a, b \in Z$, the number a+b is even if and only if a-b is even.
Prove or disprove that where $a, b \in Z$, the number a+b is even if and only if a-b is even.
Can some help me improve on this or check it please?
If $a + b$ is even, then there is some number $k \in Z$ such that $a + b = 2k... | We have $a+b$ even iff $a+b=2k$ for some $k$ iff $a+b-2b = 2k-2b=2(k-b)$ even iff $a-b=2(k-b)$ even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How prove this inequality $(a+b)^rb^{2n-r}\le\dfrac{n}{4}+r,$
let $a,b>0$,and $n\ge 4$ be postive integers, such $(a+b)^{2n}=2n+\dfrac{n}{4},a^{2n}=\dfrac{n}{4}$
show that
$$(a+b)^rb^{2n-r}\le\dfrac{n}{4}+r,\forall r\in[0,2n]$$
it seem hard for $n=4$ case.show this inequality $3^{\frac{n}{4}}\cdot (3^{\frac{1}{4}}-... | Another version ...
From $a=\sqrt[2n]{\frac{n}{4}}$, $a+b=\sqrt[2n]{2n+\frac{n}{4}}$
we have $a\geq1$ and $a+b >1$, then:
$$(a+b)^{r}b^{2n-r}=\left(2n+\frac{n}{4}\right)^{\frac{r}{2n}}\left(b^{2n}\right)^{\frac{2n-r}{2n}} \tag{1}$$
However
$$b=\sqrt[2n]{2n+\frac{n}{4}}-\sqrt[2n]{\frac{n}{4}}<a \tag{2}$$
because
$$\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Discuss the monotonicity of the following function without using differentiation. Can I discuss the monotonicity of the following function without using differentiation?
$$f(x) = x + \frac{9}{x}$$
Could anyone help me?
| If $f$ is strictly increasing on an interval $I$, then $f(x)> f(y)$ for $x,y\in I$ with $x>y$.
Note that $f$ is undefined when $x=0$
\begin{align}
f(x)-f(y)&=x-y+\frac{9}{x}-\frac{9}{y}\\
&=(x-y)\left(1-\frac{9}{xy}\right)\\
&=\frac{(x-y)\left(x-\frac{9}{y}\right)}{x}\\
\end{align}
For $x>0$, $\displaystyle f(x)-f(3)=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
Find the ratio of lengths in a triangle with cevians, without using mass-points In $\triangle ABC$, $D$ is the midpoint of $BC$ and $E$ is the midpoint of $AD$. $CE$ is extended to meet $AB$ at $F$.
The question is to find the ratio of the lengths of $AF$ to $FB$.
This is trivial using mass-points geometry:
$$C=1\to B=... | Let the points $A,B,C$ corresponds to vectors $\mathbf a,\mathbf b,\mathbf c$.
The vector $\vec{AD}$ is
$$\begin{align*}
\vec{AB} + \vec{BD}
&= \vec{AB} + \frac12\vec{BC}\\
&= \mathbf b - \mathbf a+\frac12(\mathbf c-\mathbf b)\\
&= -\mathbf a+\frac12\mathbf b + \frac12\mathbf c
\end{align*}$$
The vector $\vec{AE}$ is h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2325647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to solve problem of maximum and minimum in two variables? I have function
$$f(x,y)=\frac{3(x+y)-2}{x^2+y^2}$$
and need maximum and minimum in domain
$$D=\left\{(x,y)\in\Bbb R^2 ; x^2+y^2=8\right\}\cup\left\{(x,y)\in\Bbb R^2;|x|+|y|=1\right\}$$
I thought of Lagrange multipliers but that square above is hard: it is e... | With $|x|+|y|=1$ we obtain:
$$\frac{3(x+y)-2}{x^2+y^2}\leq\frac{3(|x|+|y|)-2}{\frac{(|x|+|y|)^2}{2}}=2$$
The equality occurs for $x=y=\frac{1}{2}$.
With $x^2+y^2=8$ we get a smaller value.
In another hand, for $x^2+y^2=8$ we obtain:
$$\frac{3(x+y)-2}{x^2+y^2}\geq\frac{-3\sqrt{(x+y)^2}-2}{x^2+y^2}\geq$$
$$\geq\frac{-3\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving a relation related to quadratic equation Question:If $α$ and $β$ be the roots of $ax^2+2bx+c=0$ and $α+δ$, $β+δ$ be those of $Ax^2+2Bx+C=0$, prove that, $\frac{b^2-ac}{a^2}=\frac{B^2-AC}{A^2}$.
My Attempt: Finding the sum of roots and product of roots for both the equations we get,
$α+β=\frac{-2b}{a}$
$αβ=\... | Here is the LHS:
$$\frac{b^2-ac}{a^2} = \left(\frac{b}{a}\right)^2-\frac {c}{a}$$
$$=\left(\frac {-(\alpha+\beta)} {2}\right)^2-\alpha\beta$$
$$=\frac {{\alpha}^2+{\beta}^2+2\alpha\beta}{4}-\frac {4\alpha\beta}{4}$$
$$=\frac{{\alpha}^2+{\beta}^2-2\alpha\beta} {4}$$
$$=\left(\frac{\alpha-\beta} {2}\right)^2$$
And here i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2327883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
} |
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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.