Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Quadratic Diophantine Equation$(x^2+x)(y^2-1)=240$ For whole numbers $x$ and $y$, $$x,y | (x^2+x)(y^2-1)= 240$$
Find the biggest and smallest value for $x-y$.
How do you proceed with such a question?
Are their formulas or something for that type of equation?
I'd appreciate any help.
| $(x^2+x)(y^2-1)=240$ implies that
$y = \sqrt{1 + \dfrac{240}{x(x+1)}}$
Clearly both $x$ and $x+1$ need to be divisors of $240$.
These are the divisors of $240$
\begin{align}
1 &\quad 240 \\
2 &\quad 120 \\
3 &\quad 80 \\
4 &\quad 60 \\
5 &\quad 48 \\
6 &\quad 40 \\
8 &\quad 30 \\
10 &\quad 24 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2411174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Calculate the rest of the division of $(4^{103} + 2(5^{104}))^{102}$ by $13$ I already know some techniques to solve big exponents such as, Euler/Fermat Theorem,Euler/Carmichael,successive squaring.
But these problems seem to be more dificult as they involve operations insted of a single number.
How could I solve them ... | Variants :
First congruence :
Observe first that $2$ has order $12 \bmod13$, so $4$ has order $6$, and that as $5^2\equiv -1\mod 13$, $5$ has order $4$. Thus
$$4^{103}+2\cdot 5^{104}\equiv 4^{103\bmod 6}2\cdot 5^{104\bmod 4}=4^1+2\cdot 5^0=6.$$
Now $6=2\cdot 3$ and we know $2$ has order $12 \bmod 13$, whereas $3$ has ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2411403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
To find all natural triples $(a, b, c)$ such that $2^c-1 \mid 2^a+2^b+1$ Find all positive integer triples $(a, b, c)$ such that $2^c-1 \mid 2^a+2^b+1$.
I have no important result on this one!
| Appetizers from the comments:
*
*$c=1$ is left as an exercise
*$c=2$ works when both $a$ and $b$ are even
*$c=3$ works when $a,b$ and $0$ are pairwise non-congruent modulo $3$.
Then the main course:
Claim. There are no solutions with $c>3$.
Proof. We have trivially that if $a\equiv a'\pmod c$, then
$$
2^a\equiv2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2411602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I find an unknown power in this formula? $$ \frac{1}{2^a} = 3.0988$$
How do I solve for $a$?
The original equation was:
$$\frac{1}{i^a}-\left(B-\frac{NE}{P_1-E}\right) = 0$$
I know that:
*
*$B=10000$
*$N = 50$
*$E = 15$
*$P_1 = 3000$
$$\frac1{i^a}-\left(10000-\frac{50(15)}{3000-15}\right) = 0$$
$$\fr... | Using $\frac{1}{2^a} = 3.0988$,
$$\begin{align}
\ln\left(\frac {1}{2^a}\right) &= \ln\left(3.0988\right) \\
\ln\left(1\right) - \ln\left(2^a\right) &= \ln\left(3.0988\right) \\
0 - \ln\left(2^a\right) &= \ln\left(3.0988\right) \\
\ln\left(2^a\right) &= - \ln\left(3.0988\right) \\
a\ln\left(2\right) &= -\ln\left(3.0988... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2412638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Max and min of $f(x)=\sin^2{x}+\cos{x}+2$. My first try was to look for when $\cos{x}$ and $\sin{x}$ attain their min and max respectively, which is easy using the unit circle. So the minimum of $\sin{x}+\cos{x}$ has maximum at $\sqrt{2}$ and minimum at $-\sqrt{2}.$ But if the sine-term is squared, how do I find the mi... | HINT: write $f(x)$ as
$$f(x)=1-\cos(x)^2+\cos(x)+2 \\=-\cos(x)^2+\cos(x)+3\\=-\left(\cos(x)^2-\cos(x)+\frac{1}{4}\right)+\frac{13}{4}\\=-\left(\cos(x)-\frac{1}{2}\right)^2+\frac{13}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Possible values of $a+b^2+c^3$? If $a, b$ and $c$ are rational numbers satisfying the equation $x^3+ax^2+bx+c=0$ then find the possible values of $a+b^2+c^3$.
I have found one of the possible values as 0 using some vieta's rules and substituting $c$ for $x$. But can't find other possible values. The question specifica... | Let $f(x)=x^3+ax^2+bx+c$. We are given that $f(a)=f(b)=f(c)=0$. Therefore
\begin{align*}
a+b+c & = -a\\
ab+bc+ca & =b\\
abc & = -c.
\end{align*}
The last equation suggests either $c=0$ or $ab=-1$.
If $c=0$, then from the first two equations we get $2a+b=0$ and $b(a-1)=0$. This gives us two possibilities $a=1, b=-2$ OR... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2415038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the set of points in $\mathbb{C}$ s.t. $|z| = Rez + 1$ I was given the above problem as a homework question and I've arrived at a solution but I am not sure if my approach is the best one or even a valid one. Any help would be great!
Consider $z = a + bi : a, b \in \mathbb{R}$ with $a = Rez, b = Imz$. Then we woul... | Note that $\sqrt{a^2 + b^2} = \sqrt{a^2 + 2a + 1} \implies \color{red}{b^2 = 2a + 1} \implies b=\color{red}{\pm} \sqrt{2a+1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2417148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Evaluate $ \lim_{x\to \pi/2} \frac{\sqrt{1+\cos(2x)}}{\sqrt{\pi}-\sqrt{2x}}$
Evaluate
$$ \lim_{x\to \pi/2} \frac{\sqrt{1+\cos(2x)}}{\sqrt{\pi}-\sqrt{2x}}$$
I tried to solve this by L'Hospital's rule..but that doesn't give a solution..appreciate if you can give a clue.
| Dividing numerator and denominator by $\sqrt{2}$
$$\lim_{x\to \frac{π}{2}} \frac {\sqrt{\frac{1+\cos 2x}{2}}}{\sqrt {\frac{π}{2}} -\sqrt { x}}$$
$$$$
Multiplying and dividing by $(\sqrt {\frac{π}{2} } + \sqrt {x})$
$$=\lim_{x\to \frac {π}{2}}(\sqrt {\frac{π}{2}} + \sqrt x)(\frac {\sqrt {\frac {1+ \cos 2x}{2}}}{\frac {π... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2421421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Solve $2\tan{2x}\leq3\tan{x}.$ Problem:
$$\text{Solve} \quad 2\tan{2x}\leq3\tan{x}.$$
A problem of this character will yield 5 points on an exam. However, having the correct answer does not suffice to get all the 5 points. Full stringency and mathematical accuracy, on top of a correct final answer, warrants full house.... | Duplication formula for tangent
$\dfrac{4 \tan x}{1-\tan ^2 x}\leq 3 \tan x$
Bring all in the LHS
$\dfrac{4 \tan x}{1-\tan ^2 x}-3 \tan x\leq 0$
add together
$-\dfrac{\tan x \left(3 \tan ^2 x+1\right)}{\tan ^2 x-1}\leq 0$
which is better as
$\dfrac{\tan x \left(3 \tan ^2 x+1\right)}{\tan ^2 x-1}\geq 0$
The parent... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2421541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Show that for $n\to \infty, u_n - \sqrt{n} - \frac{1}{2} \sim \frac{1}{8\sqrt{n}}$ $\forall n\in\mathbb{N}^*, u_n = \sqrt{n+u_{n-1}}$ with $u_1 = 1$.
I've already shown that $ u_n \sim \sqrt{n}$ and that $\underset{n\rightarrow \infty}{\lim} u_n - \sqrt{n} = \dfrac{1}{2}$
How to show that $n\rightarrow \infty, u_n - \s... | Letting $u_n = \sqrt{n}+\frac{1}{2}+\varepsilon_n$ with $\lim_{n\to\infty} \varepsilon_n=0$, you have
$$
\sqrt{n}+\frac{1}{2}+\varepsilon_n = \sqrt{n+\sqrt{n}+\frac{1}{2}+\varepsilon_{n-1}}
$$
for all $n$. Doing a Taylor expansion of the RHS yields, using the fact that $\varepsilon_{n-1}=o(1)$:
$$\begin{align}
\sqrt{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2421877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\lim_{x\to 1} \frac{x+2}{x^2+1}=\frac{3}{2}$ The title is quite clear. I am required to prove $\lim_{x\to 1} \frac{x+2}{x^2+1}=\frac{3}{2}$ using the epsilon-delta definition of the limit.
Given any $\varepsilon \gt 0$, there exists a $\delta =$
Such that $0 \lt \lvert x-1 \rvert \lt \delta \Rightarrow \lvert\f... | In this answer we continue where the OP left off, making one last 'normalization' change:
$\tag 1 \frac{-(x-1)(3x+1)}{2(x-1)^2 + 4(x-1) +4} = \frac{-(x-1)(3(x-1)+4)}{2(x-1)^2 + 4(x-1) +4}$
Denominator:
If $|x - 1| \lt 1/2$ then
$\quad |2(x-1)^2 + 4(x-1) +4| \gt |4(x-1) +4| \gt 2 $
Note that $4(x-1) \gt -2$.
$\text{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2424030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 3
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Is it possible to have a 3rd number system based on division by zero? Is it possible in mathematics to use a third number line based on division by zero; in addition to the real and imaginary number lines?
This is because some solutions blow up when there is a division by zero. Would it be possible to solve them with t... | $\textbf {Rules of the p number system using 1/0 and $\infty$ }$
If $p = \frac{1}{0}$ and $p \cdot 0 = 1$. Then I suggest the following rules.
$$p - p = \frac{1}{0} - \frac{1}{0} = \frac{0 \cdot 1 - 0 \cdot 1}{0 \cdot 0} = \frac{0 \cdot (1 - 1)}{0 \cdot 0} = \frac{0}{0} = 1$$
problem lies here (if it is a problem?) $$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Show $k\mid 12$ with $2^k=1\bmod 13$ Let $k$ be the smallest positive integer such that $2^k=1\bmod 13$. Show that $k\mid 12$.
I'm not very good at proofs and I'm confused as how to prove this. I started by saying $2^k-1=13n$. But I don't know where to go from there.
| $$
\begin{array}{c|c}
\hline
i & \text{ $2^i $ mod 13} \\
\hline
1&2 \\
2&4 \\
3&8 \\
4&3 \\
5&6 \\
6&12 \\
7&11 \\
8&9 \\
9&5 \\
10&10 \\
11&7 \\
12&1 \\
\hline
\end{array}
$$
First, for $0 \le k \le 12$, only $2^{12} \equiv 2^0 \equiv 1 \quad mod(13)$
Assume $k=12i+j$, where $0 \le j < 12$, th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Solve the following recurrence $f(n) = 2^n f(n - 1) + 2^n$ given $f(0)=1$. Solve the recurrence $f(n) = 2^n f(n - 1) + 2^n$ given $f(0)=1$.
The pattern I can come up with is
$2^{2n-1} f(n - 2) + 2^{3n-1}$
$2^{3n-3} f(n - 3) + 2^{6n-4}$
$2^{4n-6} f(n - 4) + 2^{10n-10}$
How do I solve the recurrence relation from this p... | We have
$f(n) = 2^n f(n - 1) + 2^n
$.
Since
$\sum_{k=1}^n k
= n(n+1)/2
$,
divide the recurrence by
$2^{ n(n+1)/2}$.
It becomes
$\dfrac{f(n)}{2^{ n(n+1)/2}}
= \dfrac{f(n - 1)}{2^{ n(n+1)/2-n}} + \dfrac1{{2^{ n(n+1)/2-n}}}
= \dfrac{f(n - 1)}{2^{ n(n-1)/2}} + \dfrac1{{2^{ n(n-1)/2}}}
$.
Now,
let
$g(n)
=\dfrac{f(n)}{2^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Solution of $\sqrt{5-2\sin x}\geq 6\sin x-1$
Solve the following inequality. $$\sqrt{5-2\sin x}\geq 6\sin x-1.$$
My tries:
As $5-2\sin x>0$ hence we do not need to worry about the domain.
Case-1: $6\sin x-1\leq0\implies \sin x\leq\dfrac{1}{6}\implies -1\leq\sin x\leq\dfrac{1}{6}\tag*{}$
Case-2:$6\sin x-1>0\implies \d... | Your solution is right.
You need only to write the answer, for which
just take on the $y$-axis the point $\frac{1}{2}$ and you need $\sin{x}\leq\frac{1}{2}$, which gives $y\leq\frac{1}{2}$ and the arc
$$\left[-\frac{7\pi}{6},\frac{\pi}{6}\right]$$
on the trigonometric circle and add $2\pi n$ in the both sides.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the coordinates of the points on $y=-(x+1)^2+4$ that have a distance of $\sqrt {14}$ to $(-1,2)$ Create a function that gives the distance between the point $(-1,2)$ the graph of $$y=-(x+1)^2+4.$$ Find the coordinates of the points on the curve that have a distance of $\sqrt {14}$ units from the point $(-1,2)$. ... | The locus of points whose distance from $(-1,2)$ is $\sqrt{14}$ is the circle
$(x+1)^2+(y-2)^2=14$
that is
$\color{red}{x^2+2 x}+y^2-4 y-9=0$
The parabola equation can be written as $\color{red}{x^2+2 x}=\color{blue}{3-y}\quad(*)$
Therefore the intersection points, the points of the parabola which are $\sqrt{14}$ from... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Compute: $\arctan{\frac{1}{7}}+\arctan{\frac{3}{4}}.$
Compute: $\arctan{\frac{1}{7}}+\arctan{\frac{3}{4}}.$
I want to compute this sum by computing one term at a time. It's clear that
$$\arctan{\frac{1}{7}}=A \Longleftrightarrow\tan{A}=\frac{1}{7}\Longrightarrow A\in\left(0,\frac{\pi}{2}\right).$$
Drawing a right tr... | Hint
$$\arctan x +\arctan y
=\arctan\left(\frac{x+y}{1-xy}\right) $$
See [Additivity of $\arctan(\frac{x+y}{1-xy})$]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How to evaluate the integral $\int_0^1\frac{\arctan x}{x}\frac{1-x^3}{1+x^3}dx$ I'm looking for the value of this integral:
$$\int_0^1\frac{\arctan x}{x}\frac{1-x^3}{1+x^3}dx$$
I try to integrate it:
\begin{align}
I&=\int_0^1\arctan x\left(\frac{1}{x}-\frac{2x^2}{1+x^3}\right)dx\\
&=\int_0^1\frac{\arctan x}{x}dx-2\int_... | $$\color{blue}{I = \frac{2}{9}\pi \ln (2 + \sqrt 3 ) - \frac{{\pi \ln 2}}{{12}} - \frac{G}{9}}$$
Let $$I_1 = \int_0^1 \frac{\ln(1+x^3)}{1+x^2}dx \quad \quad I_2 = \int_1^\infty \frac{\ln(1+x^3)}{1+x^2}dx$$
Then subsituting $x\to 1/x$ in $I_1$ gives $$\tag{1} I_1 - I_2 = -3\int_1^\infty \frac{\ln x}{1+x^2} dx = -3G$$
N... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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If $ (x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1$, show that $x+y=0$
For $\{x,y\}\subset \Bbb R$, $(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1.$
Prove that $x+y=0.$
Problem presented in a book, as being from Norway Math Olympiad 1985. No answer was presented. My developments are not leading to a productive direction. Sorry if this is... | Let $x = \tan\alpha$ and $y = \tan\beta$ for some $\alpha,\beta\in(-\pi/2,\pi/2)$. Then $\sqrt{x^2+1} = \sec\alpha$ and $\sqrt{y^2+1} = \sec\beta$, so the condition can be rewritten as
$$(\tan\alpha+\sec\alpha)(\tan\beta+\sec\beta) = 1. $$
Multiplying by $\cos\alpha\cos\beta$ yields
$$(\sin\alpha+1)(\sin\beta+1) = \cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Find the limit of $ \frac{\cos (3x) - 1}{ \sin (2x) \tan (3x) } $ without L'Hospital technique I would like to find the limit as $x$ goes to zero for the following function, but without L'Hospital technique.
$$ f(x) = \frac{ \cos (3x) - 1 }{ \sin (2x) \tan (3x)} $$
This limit will go to zero (I had tried using calcul... | \begin{align}
\lim_{x\to0} \frac{ \cos (3x) - 1 }{ \sin (2x) \tan (3x)}
&= \lim_{x\to0} \frac{ -2\sin^2 (\dfrac{3x}{2}) \cos (3x) }{ \sin (2x) \sin (3x)} \\
&= -2\lim_{x\to0} \dfrac{\sin^2 (\dfrac{3x}{2}) }{(\dfrac{3x}{2})^2} \dfrac{(2x)}{\sin (2x)} \frac{ (3x)}{ ( \sin (3x)}.\dfrac{(\dfrac{3x}{2})^2}{(2x)(3x)}\cos (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2435551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
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Find Taylor series of: $x^x$ Find the Taylor seriespolynomial form of:
$$x^x$$
My attempt:
I started calculating derivatives of $x^x$ for the series,
$$f(x)=x^x$$
$$f'(x)=x^x(\ln x+1)$$
$$f''(x)=x^{x-1}+x^x(\ln x+1)$$
$$f'''(x)=\cdots$$
$$\vdots$$
But i could get only till certain terms and got frustrated,
Attempt n... | Hint: We can find an elaboration of the $n$-th derivative of $x^x$ in an example (p. 139) of Advanced Combinatorics by L. Comtet. The idea is based upon a clever Taylor series expansion. Using the differential operator $D_x^j:=\frac{d^j}{dx^j}$ the following holds:
The $n$-th derivative of $x^x$ is
\begin{align*}
\co... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Multivariable $\epsilon-\delta$ proof verification Prove that $\lim\limits_{(x,y) \to (1,1)} xy=1$
Of course, I am aware that this is "obvious", but I want to add some rigor to it. When I searched around for multivariable limits using $\epsilon-\delta$, most of the examples had $(x,y) \rightarrow (0,0)$, but in this ca... | A shorthand you might find useful:
Since
$$(x-y)^2 \gt 0 \Rightarrow xy \lt \frac{x^2+y^2}{2} $$
Also, note that
$$\lVert (x-1, y-1) \rVert \lt \delta \Rightarrow |x-1| \lt \delta \;\land\; |y-1| \lt \delta \Rightarrow x \lt \delta+1 \;\land\; y \lt \delta+1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2438504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
$ \left( 1 - \frac{x}{n}\right)^n \leq \left( 1 - \frac{x}{n+1}\right)^{n+1} $, $\forall$ $x$ $\in$ $[0,n]$
Prove that:
$$ \left( 1 - \frac{x}{n}\right)^n \leq \left( 1 - \frac{x}{n+1}\right)^{n+1} $$ $\forall$ $x$ $\in$ $[0,n].$
I think it's easy, but I'm very stuck in this question, I need to prove that
$$\left(1... | $$\left(1-\frac{x}{n+1}\right)^{n+1}=\left(1-\frac{x}{n}+\frac{x}{n}-\frac{x}{n+1}\right)^{n+1}=$$
$$=\left(1-\frac{x}{n}+\frac{x}{n(n+1)}\right)^{n+1}=\left(1-\frac{x}{n}+\frac{x}{n(n+1)}\right)^n\left(1-\frac{x}{n}+\frac{x}{n(n+1)}\right)\geq$$
$$\geq\left(\left(1-\frac{x}{n}\right)^n+n\left(1-\frac{x}{n}\right)^{n-1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding sum of the series $\sum_{r=1}^{n}\frac{1}{(r)(r+d)(r+2d)(r+3d)}$ Find the sum: $$\sum_{r=1}^{n}\frac{1}{(r)(r+d)(r+2d)(r+3d)}$$
My method:
I tried to split it into partial fractions like: $\dfrac{A}{r}, \dfrac{B}{r+d}$ etc. Using this method, we have 4 equations in $A,B,C,D$! And solving them takes much time. I... |
And solving them takes much time.
There is a trick so solving partial fractions quickly.
$$\frac{1}{(r)(r + d)(r + 2d)(r + 3d)}
=
\frac{A}{r} + \frac{B}{r + d} + \frac{C}{r + 2d} + \frac{D}{r + 3d}$$
$$1
=
A(r + d)(r + 2d)(r + 3d) +
B(r)(r + 2d)(r + 3d) +
C(r)(r + d)(r + 3d) +
D(r)(r + d)(r + 2d)$$
The above is tr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2441296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Solving for equation of a circle when tangents to it are given. Given: $\mathrm L(x) = x - 2$, $\mathrm M(y) = y - 5$ and $\mathrm N(x, y)= 3x - 4y - 10$.
To find: All the circles that are tangent to these three lines.
Outline of the method :
If we parameterised any line $\mathrm Z$ in terms of $x(t) = pt + q$ and $y(... | converting the equation in the Hessian Normalform we have
$$\frac{|3x-4y-10|}{5}=R$$ where $$(x,y)$$ denotes the middle Point of our searched circle; and it must be $$x=2+R,y=5-R$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2442063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solve $\arcsin{(2x^2-1)}+2\arcsin{x} = -\frac{\pi}{2}$ First of, we have to restrict the domain of the equation by looking at the argument of the first term. The domain for $\arcsin$ is $[-1,1]$, so the inequality $ -1\leq 2x^2-1 \leq1$ has to hold. So $x\geq0$ and $-1\leq x\leq 1$. It follow that the domain is $[-1,1]... | Hint :
Let $\arcsin x=u\implies-\dfrac\pi2\le u\le\dfrac\pi2\iff-\pi\le2u\le\pi$ and $x=\sin u$
$\arcsin(2x^2-1)=\arcsin(-\cos2u)=-\arcsin(\cos2u)=\arccos(\cos2u)-\dfrac\pi2$
Now $\arccos(\cos2u)=\begin{cases}2u &\mbox{if } 0\le2u\le\pi \\-2u& \mbox{if } 0\le-2u\le\pi\iff-\pi\le 2u\le0&\\2\pi-2u& \mbox{if } 0\le2\pi-2u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2443040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find $y(1)$ if $y(x)=x^3+\int_0^x \sin(x-t)y(t)dt$ Consider the integral equation $y(x)=x^3+\int_0^x \sin(x-t)y(t)dt, x\in[0,\pi].$ Then the value of y(1) is
$1. \;\;19/20 \\
2. \;\;1\\
3. \;\;17/20\\
4. \;\;21/20 $
My Attempt: $y(x)=x^3+\int_0^x \sin(x-t) \ y(t)dt \\ \Rightarrow y(x)=x^3+\int_0^x(\sin x \cos t-\cos x... | In the next-to-last line, you should have $y(x) - x^3$, so that $y(x) = x^3 + \frac{x^5}{20} + c_1x + c_2$. Then since line 5 implies $y'(0) = 0$, both $c_1$ and $c_2$ are zero, so that $y(x) = x^3 + \frac{x^5}{20}$, and $y(1) = \frac{21}{20}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2444547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
$(xy+yz+zx)(1+6xyz) \geq 11xyz$ I've came across this inequality:
Let $x>$, $y>0$ and $z > 0$ with $x+y+z=1$.
Prove that
$$(xy+yz+zx)(1+6xyz) \geq 11xyz.$$
I don't know where to take it from, I've tried means inequality, but it didn't help me.
Some hints would be great!
| It's $$6xyz(xy+xz+yz)+(xy+xz+yz)(x+y+z)^3\geq11xyz(x+y+z)^2$$ or
$$\sum_{sym}(x^4y+3x^3y^2-2x^3yz-2x^2y^2z)\geq0,$$
which is true by Muirhead.
Also, we can prove the last inequality by AM-GM.
Indeed, $$\sum_{cyc}x^4y=\frac{1}{13}\sum_{cyc}(9x^4y+y^4z+3z^4x)\geq$$
$$\geq\frac{1}{13}\cdot13\sum_{cyc}\sqrt[13]{\left(x^4y\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2445128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find generating function for the sequence 0, 0, 0, 0, 3, 4, 5, 6, ...
1.Derive the generating function for the sequence $$0, 0, 0, 0, 3, 4, 5, 6, . . .$$
2.Derive the generating function for the sequence $$0, 0, −12, 36, −108, 324, .. .$$
So the first function looks like $3x^4 + 4x^5+5x^6...$ = $x^4(3+4x+5x^2...)$ ... | For $|x|<1$ we obtain:
$$3x^4+4x^5+...=x^2(3x^2+4x^3+...)=x^2(x^3+x^4+...)'=$$
$$=x^2\cdot\left(\frac{x^3}{1-x}\right)'=\frac{x^4(3-2x)}{(1-x)^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2445520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is there a $2 \times 2$ real matrix $A$ such that $A^2=-4I$? Does there exist a $2 \times 2$ matrix $A$ with real entries such that $A^2=-4I$ where $I$ is the identity matrix?
Some initial thoughts related to this question:
*
*The problem would be easy for complex matrices, we could simply take identity matrix multi... | A 2-by-2 matrix is not too bad to work out the algebraic equation for its coefficients:
Let
$
A=\begin{bmatrix} a & b \\c & d\end{bmatrix}
$. We end up with the matrix equation:
$$
A^2 = \begin{bmatrix} a^2+bc & b(a+d) \\c(a+d) & d^2 + bc\end{bmatrix}
= -4 \, I = \begin{bmatrix} -4 & 0 \\0 & -4\end{bmatrix}.
$$
S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2448604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Multivariate random vector with normal distribution Let $X=(X_1, X_2)'\in N(\mu, \Lambda),$ where
$$\mu=\begin{bmatrix}1 \\ 1 \end{bmatrix}, \Lambda=\begin{bmatrix}3 & 1 \\ 1 & 2 \end{bmatrix}.$$
Compute $P(X_1\geq 2 \mid X_2+3X_1=3).$
Don't know even where to start.
| Currently your question says
$$\mu=\begin{bmatrix}1 \\ 1 \end{bmatrix}, \Lambda=\begin{bmatrix}3 & 1 \\ 1 & 2 \end{bmatrix}.$$
Compute $P(X_1\geq 2 \mid X_2+3X_3=3).$
I will assume you just forgot $X_3.$ You have
$$
\begin{bmatrix} X_1 \\ X_2 \\ X_3 \end{bmatrix} \sim N\left( \begin{bmatrix} 1 \\ 1 \\ \mu_3 \end{bmat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2450937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the inequality for $x$: $\log_4 (x^2 − 2x + 1) < \log_2 3$ Solve the inequality for $x$: $$\log_4 (x^2 − 2x + 1) < \log_2 3$$
I got two answers and I'm not sure if I did it correctly.
1st ans: $(-2,1)\cup (1,4)$
2nd ans: $x \neq -2,4$
| By the change of base formula, $\log_4{x^2-2x+1}$ is equal to $\frac{\log_2{x^2-2x+1}}{\log_2{4}}$, or $\frac{1}{2} \log_2{x^2-2x+1}$. Since both sides are in the same base, we can directly solve the inequality:
$$\frac{1}{2} \log_2{x^2-2x+1} < \log_2{3}$$
$$\log_2{x^2-2x+1} < 2 \log_2{3}$$
$$x^2-2x+1 < (2^{\log_2{3}})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2451064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Other idea to find range of $y=x+\frac{1}{x-4}$ $$y=x+\frac{1}{x-4}$$ I tried to find range of this function as below
$$y(x-4)=x(x-4)+1\\yx-4y=x^2-4x+1\\x^2-x(4+y)+(4y+1)=0\\\Delta \geq 0 \\(4+y)^2-4(4y+1)\geq 0 \\(y-4)^2-4\geq 0\\|y-4|\geq 2\\y\geq 6 \cup y\leq2$$ this usuall way. but I am interested in finall answer... | use this fact : $|a+\frac 1a|\geq 2$
$$\quad{y=x+\frac{1}{x-4}\\y-4=x+\frac{1}{x-4}-4\\y-4=(x-4)+\frac{1}{x-4}=a+\frac 1a\\|y-4|\geq 2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2451281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Factorize polynomials function The task was:
Find the set of zeros (roots) of the following function $$f(x)=x^4-6x^2-8x+24$$
What I did:
I found the possibles roots $$\pm1,\pm2,\pm3,\pm4,\pm6,\pm8,\pm12,\pm24 $$
and I found that X=2 but I cannot get the other 3 roots
| You found that $x = 2$ is a root of $f(x) = x^4 - 6x^2 - 8x + 24$. This means $x - 2$ is a factor. Dividing $f(x)$ by $x - 2$ yields
\begin{align*}
f(x) & = x^4 - 6x^2 - 8x + 24\\
& = x^3(x - 2) + 2x^3 - 6x^2 - 8x + 24\\
& = x^3(x - 2) + 2x^2(x - 2) + 4x^2 - 6x^2 - 8x + 24\\
& = x^3(x - 2) + 2x^2(x - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2452231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
If $\frac {9\sin x-3\cos x}{2\sin x + \cos x}=2$, define $\tan x$ and angle $x$. If: $$\frac {9\sin x-3\cos x}{2\sin x + \cos x}=2$$ $(0^\circ < x < 90^\circ$), define $\tan x$ and angle $x$.
Because of this in the parenthesis I know that I have to use some relation between complementary angles ex. $3\cos x = 3\sin(90^... | \begin{align}
& \frac {9\sin x-3\cos x}{2\sin x + \cos x}=2 \\[10pt]
\text{Therefore } & \frac{9\tan x - 3}{2\tan x + 1} = 2 \\[10pt]
& \text{and so on.}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2454622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How did x become -sqrt(x^2)?
In computing the limit as $x \to -\infty$, we must remember that for $x<0$, we have $\sqrt{x^2} = |x| = -x$. So when we divide the numerator by $x<0$, we get
$$
\frac{\sqrt{2x^2+1}}{x}
= \frac{\sqrt{2x^2+1}}{-\sqrt{x^2}}
= - \sqrt{\frac{2x^2+1}{x^2}}
= - \sqrt{2 + \frac{1}{x^2}}
$$
H... | For $x<0$ $$\sqrt{x^2}=|x|=-x,$$
which says $x=-\sqrt{x^2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2454812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How can I show $z^2 = 6x^2 + 2y^2$ has no non-trivial integer solutions? EDIT: I missed a part of the question as there was a typo in my notes (this is part of my working on a proof using the Bruck–Ryser–Chowla theorem which was presented incorrectly in my version of the lecture notes). I have updated the question and ... | Assume $(x,y,z)$ is a solution to $z^2 = 6x^2 + 2y^2 $ with smallest positive $|z|$, then $\gcd(x,y) = 1$. Putting $z=2z_1$ gives
$$2z_1^ 2 = 3x^2 + y^2 $$ Since $x,y$ are both odd, RHS is divisible by $4$, so $z_1$ is even, that is $8\mid (2z_1^2)$. Hence $3x^2 + y^2 \equiv 0 \pmod{8}$. But the square of an odd number... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2455459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove that $\sqrt[n]{n} > \sqrt[n+1]{n+1}$ without calculus? I'm stuck with this sample RMO question I came across:
Determine the largest number in the infinite sequence $\sqrt[1]{1}$, $\sqrt[2]{2}$, $\sqrt[3]{3}$, ..., $\sqrt[n]{n}$, ...
In the solution to this problem, I found the solver making the assumption,
$\s... | Assume $n\geq 3.$ Start with this calculation:
$$\frac{(n+1)^n}{n^n} = \left(1+\frac{1}{n}\right)^n = 1+\binom{n}{1}\frac{1}{n} +\binom{n}{2}\frac{1}{n^2} + \binom{n}{3}\frac{1}{n^3} +\cdots + \frac{1}{n^n}.$$
In the $k$th term, the numerator of the binomial coefficient is $n(n-1)(n-2)\cdots (n-k+1)$ which is less than... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2457170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
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Find $\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2}$ assuming it exists.
Find $\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2}$ assuming it
exists.
Since limit exists, we can approach from any curve to get the limit...
if we approach (0,0) from y=x
$\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2} \Rightarrow \lim_{x... | Setting $$y=tx$$ then we get
$$\frac{x^3+t^3x^3}{x^2+t^2x^2}=x\frac{1+t^3}{1+t^2}$$ this tends to Zero if $x$ tends to zero
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2460071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Need hint for solving the following problem
If $f(x^{500}-1)=5x^{1015}+3x^{244}+7x+10$. Find the sum of coefficients of $f(x^5+1)$.
Let $x^{500}-1=y$, then $f(y)=5(y+1)^\frac{1015}{500}+3(y+1)^\frac{244}{500}+7(y+1)^\frac{1}{500}+10$.
But, I don't think that it is a right approach as each term of $f(y)$ gets transfor... | I think a good hint is to take a look at factor theorem!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2460447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding minimal polynomial of $x=a+b\sqrt[3]{2}+c\sqrt[3]{4}$ I want to find the minimal polynomial of $x=a+b\sqrt[3]{2}+c\sqrt[3]{4}$
For simple case like $x=a+b\sqrt[3]{2}$, I can find
\begin{align}
(x-a)^3 = 2b^3 \qquad \Rightarrow \qquad (x-a)^3-2b^3 =0
\end{align}
I tried to do similar tings such as
\begin{alig... | I don't get it, where is a problem, a solution is direct. Lift the starting equation
$$x-a=b\sqrt[3]{2}+c\sqrt[3]{4}$$
on $3$, so have:
$$(x-a)^3=(b\sqrt[3]{2}+c\sqrt[3]{4})^3$$
so
$$ (x-a)^3=2b^3+4c^3+6bc(\underbrace{b\sqrt[3]{2}+c\sqrt[3]{4}}_{x-a})$$
and we are done:
$$ p(x)=(x-a)^3-6bc(x-a)-2b^3-4c^3$$
So we don't... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2461150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Third order convergence of an iteration scheme Consider the iteration scheme
$x_{n+1}=\alpha x_n(3-\frac{x_n^2}{a})+\beta x_n(1+\frac{a}{x_n^2})$
For third order convergence to $\sqrt 2$, the values of $\alpha$ and $\beta$ are ......
I tried it by plugging $x_{n+1}=x_n=\sqrt a$ as $n\rightarrow\infty$ and got $\alpha ... | We start by considering two sequences -
$$ A \rightarrow x_{n+1} = \frac{x_{n}}{2}(1 + \frac{a}{x_{n}^2})$$
and
$$ B \rightarrow x_{n+1} = \frac{x_{n}}{2}(3 - \frac{x_{n}^2}{a}) $$
We can observe that by letting $\lim_{n \rightarrow \infty} x_{n} = \xi$ and $\lim_{n \rightarrow \infty} x_{n+1} = \xi$ , $\xi^2 = a$ w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2462473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove $\frac{x+1}{y+1}+\frac{y+1}{z+1}+\frac{z+1}{x+1}\leq\frac{x-1}{y-1}+\frac{y-1}{z-1}+\frac{z-1}{x-1}$ Let $x,y,z$ be real numbers all greater than $1$, then prove that
$$\frac{x+1}{y+1}+\frac{y+1}{z+1}+\frac{z+1}{x+1}\leq\frac{x-1}{y-1}+\frac{y-1}{z-1}+\frac{z-1}{x-1}$$
My Attempt:
I am trying to use $A.M>GM$ but... | Let $z=\min\{x,y,z\}$, $x-1=a$, $y-1=b$ and $z-1=c$.
Thus, $c=\min\{a,b,c\}>0$ and we need to prove that
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq\frac{a+2}{b+2}+\frac{b+2}{c+2}+\frac{c+2}{a+2}$$ or
$$\frac{a}{b}+\frac{b}{a}-2+\frac{b}{c}+\frac{c}{a}-\frac{b}{a}-1\geq\frac{a+2}{b+2}+\frac{b+2}{a+2}-2+\frac{b+2}{c+2}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2462971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Solving the equation $\tan{x} + \tan \frac{x}{4}=2$ I am trying to solve the equation $\tan{x} + \tan \frac{x}{4}=2$.
1st attempt: I use the following identity $\tan2x=\frac{2\tan x}{1-\tan^{2}x}$. Then get the following equation
$$
u^5 -2u^4-8u^3+12u^2+3u+2=0,
$$
where $u=\tan \frac{x}{4}$. But i can't find any root f... | you can use $$\tan(x)=\frac{2\tan(x/2)}{1-\tan^2(x/2)}$$ and
$$\tan(x/2)=\frac{2\tan(x/4)}{1-\tan^2(x/4)}$$
putting all things together and factorizing we get
$$\left( {u}^{2}-4\,u+1 \right) \left( {u}^{3}+2\,{u}^{2}-3\,u-2
\right)
=0$$ where $$u=\tan(x/4)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2464446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
$\sin(\frac{\pi}{3})+\sin(\frac{2\pi}{3})+...+\sin(\frac{n\pi}{3})=2\sin(\frac{n\pi}{6})\sin(\frac{(n+1)\pi}{6})$ I need to prove that
$\forall n\in\mathbb N:$
$$\sum_{i=1}^n \sin(\frac{i\pi}{3}) = 2\sin(\frac{n\pi}{6})\sin(\frac{(n+1)\pi}{6})$$
Using induction only leads me to proving quite complicated trig identitie... | Induction way:
Assume that we have $$\sum_{k=0}^n \sin \frac{k\pi}{3} = 2\sin\frac{n\pi}{6}\sin\frac{(n+1)\pi}{6}.$$
We prove that
$$\sum_{k=0}^{n+1} \sin \frac{k\pi}{3} = 2\sin\frac{(n+1)\pi}{6}\sin\frac{(n+2)\pi}{6}.$$
Indeed, one has $$LHS = \sum_{k=0}^n \sin \frac{k\pi}{3} + \sin\frac{(n+1)\pi}{3} = 2\sin\frac{n\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2464549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$|\sin z|^2+|\cos z|^2=1$ I have a task here:
Show that $|\cos z|^{2}+|\sin z|^{2} = 1$ if and only if $z\in \mathbb R$.
I've tried to different ways.
1:
Since $z\in \mathbb R$ $z^2\geq0$
$-1\leq \cos z\leq1\leftrightarrow |\cos z|\leq1$
$(\cos z)^2\leq(1)^2\leftrightarrow |\cos z|^2\leq1^2$
$(\cos z)^2\leq1\leftrighta... | Imagine a right angle triangle with sides a, b and c, with c as hypotenuse, whereby $a^2 + b^2 = c^2$ and
$sin z = \frac {a}{c}$
$cos z = \frac {b}{c}$
Thus, $(sinz)^2 + (cosz)^2)$
$=(\frac {a}{c})^2 + {\frac {b}{c}})^2$
$= \frac {a^2+b^2}{c^2}$
$=\frac {c^2}{c^2}$
$=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2464954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluate the sum ${1 - \frac{1}{2} {n \choose 1} + \frac{1}{3} {n \choose 2} + \ldots + (-1)^n \frac{1}{n+1} {n \choose n}}$ Evaluate the sum $${1 - \frac{1}{2} {n \choose 1} + \frac{1}{3} {n \choose 2} + \ldots + (-1)^n \frac{1}{n+1} {n \choose n}}.$$
I have tried comparing this to the similar problem here.
I believe ... | This is $$\sum_{k=0}^{n}\frac{(-1)^{k}}{k+1}\binom{n}{k}.$$ Then multiplying each term by $\frac{n+1}{n+1},$ we get $$\sum_{k=0}^{n}\frac{(-1)^{k}}{n+1}\binom{n+1}{k+1}=\frac{1}{n+1}\sum_{k=1}^{n+1}(-1)^{k-1}\binom{n+1}{k}.$$ Adding and subtracting $1$ and applying the Binomial Theorem gives $$\frac{1}{n+1}-\frac{1}{n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2465407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Solving $n$th order determinant I have a determinant of nth order that I am not able to convert into a triangular shape. I believe that this determinant is quite easy, but I can't find a way to fully convert one of the corners into zeros. My other idea was to use the Laplace principle, but that didn't work as well.
$$
... | Alternatively: doing row operations to get a diagonal matrix:
$$ R1\cdot (-\frac14)+R2\to R2; \ R2\cdot (-\frac13)+R3\to R3; \ etc$$
$$\begin{vmatrix}
4 & 4 & 0 & 0 & \cdots & 0 & 0 \\
1 & 4 & 4 & 0 & \cdots & 0 & 0 \\
0 & 1 & 4 & 4 & \cdots & 0 & 0 \\
\vdots & \vdots & \vdots & \vdots &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2467629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Proving that: $ (a^3+b^3)^2\le (a^2+b^2)(a^4+b^4)$ This problem is from Challenge and Thrill of Pre-College Mathematics:
Prove that $$ (a^3+b^3)^2\le (a^2+b^2)(a^4+b^4)$$
It would be really great if somebody could come up with a solution to this problem.
| $$(a^3+b^3)^2 = a^6 + 2a^3b^3 + b^6$$
But we know that $$(X-Y)^2\ge 0\Longleftrightarrow X^2+Y^2 \ge 2XY$$
taking $X= a^3$ and $Y=b^3$ we get
$$2a^3b^3 \le a^2b^4+b^2a^4 $$
so
$$(a^3+b^3)^2 = a^6 + 2a^3b^3 + b^6 \le a^6 + \color{red}{a^2b^4+b^2a^4 } + b^6 = (a^2+b^2)(a^4+b^4) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2468155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Dimension of image and kernel? If W is a subspace of $\mathbb R^4$ given by $W=\{(x,y,z,w):x+z+w=0,y+z+w=0\}$ . Then what will be the dimension of image and kernel of W.
I just learned that dimension of $W=\text{number of independent variables}-\text{number of constraints}=4-2=2$. I amcurious to know abou the dimensio... | You need a linear function for a kernel and an image. A set doesn't have one. If you consider $A=\begin{pmatrix}1 & 0 & 1& 1\\0&1&1&1\end{pmatrix}$ then it induces a linear map $L:\mathbb R^4\to\mathbb R^2$ defined by $$
L\begin{pmatrix}x\\y\\z\\w\end{pmatrix}=A\begin{pmatrix}x\\y\\z\\w\end{pmatrix}=\begin{pmatrix}x+z+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2468721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving a number is irrational From the fact that $\frac{1}{5}(3+4i)$ has infinite order in $(\mathbb{C},\cdot)$, I'm supposed to infer that $\frac{1}{\pi}\arctan{\frac{4}{3}}$ is irrational. Irrationality of $\arctan\frac{4}{3}$ follows immediately but I can't see why the irrationality of the product does. Any hints w... | Let $\alpha$ be your complex number, then $$\alpha^n = \frac{a_n}{5^{n+1}} + \frac{b_n}{5^{n+1}}i$$
and since we have $$\alpha^n = \frac{1}{5}(3+4i)\left(\frac{a_{n-1}}{5^n} + i\frac{b_{n-1}}{5^n}\right) = \frac{1}{5^{n+1}}\left(3a_{n-1} - 4b_{n-1} + i(a_{n-1} + 3b_{n-1})\right)$$ it is clear that $a_n =3a_{n-1} - 4b_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2469202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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How to find out the following integral? I want to solve the integral $$\int_{0}^{\infty} \frac{dx}{(x+b^2)^{3/2}(x+a^2)^{1/2}}$$
I tried it via partial fraction but calculations goes out of my reach. So is there any easier way to do it?
| For $x\geq 0$, if $a^2\not=b^2$, the integrand
\begin{align*}\frac{1}{(x+b^2)^{3/2}(x+a^2)^{1/2}}
&=\frac{1}{2}\left(\frac{x+a^2}{x+b^2}\right)^{-1/2}\cdot\frac{2}{(x+b^2)^2}\\
&=\frac{1}{2}\left(\frac{x+a^2}{x+b^2}\right)^{-1/2}\cdot\frac{(x+b^2)-(x+a^2)}{(x+b^2)^2}\cdot\frac{2}{b^2-a^2}
\end{align*}
is the derivati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2470637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Epsilon-delta limit does not exist I am having difficult proving that the limit $$\lim_{x\to 1} \frac{x-2}{x^4-1}$$ does not exist using the epsilon-delta definition.
Clearly this must be true since the function is unbounded near 1, but I'm having difficult formalizing this.
Any help to point me in the right direction ... | $\lim_{x\to 1} \frac{x-2}{x^4-1}
$
Let $x = 1+y$.
then
$\begin{array}\\
\lim_{x\to 1} \frac{x-2}{x^4-1}
&=\lim_{y\to 0} \frac{(1+y)-2}{(1+y)^4-1}\\
&=\lim_{y\to 0} \frac{y-1}{1+4y+6y^2+4y^3+y^4-1}\\
&=\lim_{y\to 0} \frac{y-1}{4y+6y^2+4y^3+y^4}\\
&=\lim_{y\to 0} \frac{y-1}{y(4+6y+4y^2+y^3)}\\
&=\lim_{y\to 0} \frac{-1}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2474053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Foot of perpendicular proof. I know there is few answer on Foot of perpendicular, but my doubt is different, so please don't mark this as the duplicate.
My book says:
Foot of the perpendicular from a point $(x_1,y_1)$ on the line $ax+by+c=0$ is $$\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=-\dfrac{ax_1+by_1+c}{a^2+b^2}$$
My pr... | There are many way to find the foot of the perpendicular. The bookish formula is the same as yours! Just a little bit of reordering:
I pick up from where you stopped.
$$x=x_1-\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\cdot\frac{a}{\sqrt{a^2+b^2}}$$
$$ y=y_1-\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\cdot\frac{b}{\sqrt{a^2+b^2}}$$
I do... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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If $z = \tan(x/2)$, what is $\sin(x)$ and $\cos(x)$? While reading mathematical gazette, I noticed an interesting "theorem". If $z = \tan(x/2)$, then $\sin(x) = \frac{2z}{1+z^2}$ and $\cos(x) = \frac{1-z^2}{1+z^2}$.
How can I derive these so I don't have to remember them?
| $$\frac{2\tan(x/2)}{1+ \tan^2(x/2)} = \frac{2\tan(x/2)}{\sec^2(x/2)} = 2\tan(x/2) \cdot \cos^2(x/2) = 2\sin(x/2)\cos(x/2) = \sin(2\cdot x/2) = \sin(x)$$
$$\frac{1- \tan^2(x/2)}{1+ \tan^2(x/2)} = \frac{1- \tan^2(x/2)}{\sec^2(x/2)} = 1- \tan^2(x/2) \cdot \cos^2(x/2) = \cos^2(x/2) - \sin^2(x/2) = \cos(2 \cdot x/2) = \cos(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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Last digit of $3^ {29}+11^{12}+15$ I found this exercise in Beachy and Blair: Abstract Algebra:
Find the units digit of
$$
3^{29}+11^{12}+15
$$
by choosing an appropriate modulus and reducing the sum.
I find $12$ being an appropriate one since
$$
3^{29}=3^2\cdot ((3^3)^3)^3
$$
and
$$
3^3\equiv3\ \ (mod \ 12)
$$
so
$$... | $(10+r)^n=\underbrace{10^n+C_n^1\,10^{n-1}r+C_n^2\,10^{n-2}r^2+...}_\text{all this is divisible by 10}+r^n$ by binomial formula.
So we have the identity : $(10+r)^n\equiv r^n\pmod{10}$
*
*$15\equiv 5\pmod{10}$
*$11^{12}=(10+1)^{12}\equiv 1^{12}\equiv 1\pmod{10}$
*$3^{28}=9^{14}=(10-1)^{14}\equiv(-1)^{14}\equiv 1\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2480322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Find $3a+b+3c+4d$ if $\left(\begin{smallmatrix}-4&-15\\2&7\end{smallmatrix}\right)^{100}=\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$. Let
$\begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}^{100} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$.
Find $3a + b + 3c + 4d$.
I've gotten this answer for the matrix... | Let $A=\begin{pmatrix}-4&-15\\2&7\end{pmatrix}$. By Cayley-Hamilton theorem, we get $A^2-3A+2I=O$. Thus $A^n$ can be represented as the sum of $A$ and $I$; that is, there are $a_n$ and $b_n$ such that $A^n = a_n A + b_n I$. Then
\begin{align}
A^{n+1}&=a_n A^2+b_n A\\
&=a_n(3A-2I)+b_n A\\
&=(3a_n +b_n)A-2a_n I.
\end{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2483541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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The maximum possible value of $x^2+y^2-4x-6y$ subject to the condition $|x+y|+|x-y|$ The maximum possible value of
$x^2+y^2-4x-6y$
subject to the condition $|x+y|+|x-y|$=4
My workout...
Now if we add 13 to the equation we get
$x^2+y^2-4x-6y+13-13$
or,$x^2+y^2-4x-6y+4+9-13$
or,$(x-2)^2+(y-3)^2-13$
Are there any methods... | $$x^2-4x+y^2-6x= (x-2)^2 + (y-3)^2-13$$
now this means you have to maximize distance from $(2,3)$
now lets analyze $|x-y| + |x+y|$
in quadrant 1: $x>y$
$$x-y+x+y=4 \rightarrow x=2$$
similiarly you have for $y>x$
$$y=2$$
now if you take negative of both $x$ and $y$ you will get same result, there will be symmetry around... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2484326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Parametric equation of the intersection between $x^2+y^2+z^2=6$ and $x+y+z=0$ I'm trying to find the parametric equation for the curve of intersection between $x^2+y^2+z^2=6$ and $x+y+z=0$. By substitution of $z=-x-y$, I see that $x^2+y^2+z^2=6$ becomes $\frac{(x+y)^2}{3}=1$, but where should I go from here?
| With spherical coordinate system, $r=\sqrt{6}$ and
$$z=-x-y=-\sqrt{6}\,\sin \theta \,\cos \varphi -\sqrt{6}\,\sin \theta \,\sin \varphi$$
so your parametric equation will be
$$
\begin{aligned}x&=\sqrt{6}\,\sin \theta \,\cos \varphi \\y&=\sqrt{6}\,\sin \theta \,\sin \varphi \\z&=\sqrt{6}\,\sin \theta \,\cos \varphi -\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2484914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Simplest way to get the lower bound $\pi > 3.14$ Inspired from this answer and my comment to it, I seek alternative ways to establish $\pi>3.14$. The goal is to achieve simpler/easy to understand approaches as well as to minimize the calculations involved. The method in my comment is based on Ramanujan's series $$\frac... | Equation (2) in the question is the result of plugging $x=1$ in the expansion for the arctangent, using
$$\frac{\pi}{4}=\tan^{-1}(1)$$
Instead, from
$$\frac{\pi}{6}=\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)$$
a series with faster convergence is obtained. Taking six terms,
$$\pi>2\sqrt{3}\left(1-\frac{1}{3·3}+\frac{1}{5·... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2485558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "55",
"answer_count": 11,
"answer_id": 6
} |
Factored $z^4 + 4$ into $(z^2 - 2i)(z^2 + 2i)$. How to factor complex polynomials such as $z^2 - 2i$ and $z^2 + 2i$? I was wondering how to factor complex polynomials such as $z^2 - 2i$ and $z^2 + 2i$?
I originally had $z^4 + 4$, which I factored to $(z^2 - 2i)(z^2 + 2i)$ by substituting $X = z^2$ and using the quadra... | Using difference of squares ...
$$z^2-2i= z^2 - 2e^{i \frac \pi 2} = z^2 - (\sqrt 2 e^{i \frac \pi 4})^2
\\ = (z+\sqrt 2 e^{i \frac \pi 4})(z-\sqrt 2 e^{i \frac \pi 4})$$
$$z^2+2i= z^2 - 2e^{i \frac {3\pi} 2} = z^2 - (\sqrt 2 e^{i \frac {3\pi} 4})^2
\\ = (z+\sqrt 2 e^{i \frac {3\pi} 4})(z-\sqrt 2 e^{i \frac {3\pi}4})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2488626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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Largest rectangle that can fit isnide of equilateral triangle I have an equilateral triangle with a fixed side-length $x$.
What is the largest area of a rectangle that can be put inside the triangle?
| Consider the situation as in the image..
Then the side $b$ depends on $a$ by the formula $b = \tan(60°) \frac{x-a}{2} = \sqrt{3}\frac{x-a}{2}$, so the area of the rectangle is $S = ab = a\sqrt{3}\frac{x-a}{2}$.
Now look for the maximum of $S$ as a function of $a$ (calculate the first derivative and find the zero poi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2493858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is the solution for $x^{\frac {10x}{3} }=\frac {243} {32}$? I understand that $243$ is $3^5$ and $32$ is $2^5$, but I don't know where to go from there after converting it to logarithm.
| As you said, $\frac{243}{32}=(\frac{3}{2})^5$. So why not try $x=\frac{3}{2}$? Indeed, we have $\frac{10\cdot \frac{3}{2}}{3}=5$ which makes $x=\frac{3}{2}$ a solution. We will now Show that there are no other solutions.
Obviously, we cannot have $0<x<1$ because that would imply $x^{\frac{10x}{3}}<1$. Furthermore, if w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2496411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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$x+y+z=3$, prove the inequality For $x,y,z>0$ and $x+y+z=3$ prove that $\frac{x^3}{(y+2z)^2}+\frac{y^3}{(z+2x)^2}+\frac{z^3}{(x+2y)^2}\ge \frac{1}{3}$.
QM, AM, GM, HM suggested ;)
| By C-S and C-S we obtain:
$$\sum_{cyc}\frac{x^3}{(y+2z)^2}=\frac{1}{3}\sum_{cyc}x\sum_{cyc}\frac{x^3}{(y+2z)^2}\geq\frac{1}{3}\left(\sum_{cyc}\frac{x^2}{y+2z}\right)^2\geq$$
$$\geq\frac{1}{3}\left(\frac{(x+y+z)^2}{\sum\limits_{cyc}(y+2z)}\right)^2=\frac{1}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2497192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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How to prove that $\int_{0}^{1}{x^2-1\over x^2+1}{x^{2n}\over \ln(x)}{\mathrm dx\over x}=F(n)$
$$\int_{0}^{1}{x^2-1\over x^2+1}{x^{2n}\over \ln(x)}{\mathrm dx\over x}=F(n)\tag1$$
$n\ge 1.$
How do we show that $$F(n)=(-1)^{n-1}\ln\left({\pi\over 2}\prod_{k=1}^{n-1}\left({k+1\over k}\right)^{(-1)^k}\right)?$$
$x=\tan... | For any $s > 0$, we have
\begin{align*}
F(s)
&\stackrel{u=x^2}{=} \int_{0}^{1} \frac{u-1}{u+1} \frac{u^{s-1}}{\log u} \, du
= \int_{0}^{1} \frac{u^{s-1}}{u+1} \left( \int_{0}^{1} u^x \, dx \right) du \\
&= \int_{0}^{1} \int_{0}^{1} \frac{u^{x+s-1}}{u+1} \, dudx
= \int_{0}^{1} \left( \sum_{k=0}^{\infty} (-1)^k \int_{0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2497786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Simplify: $\frac{\sqrt{\sqrt[4]{27}+\sqrt{\sqrt{3}-1}}-\sqrt{\sqrt[4]{27}-\sqrt{\sqrt{3}-1}}}{\sqrt{\sqrt[4]{27}-\sqrt{2\sqrt{3}+1}}}$ I am doing a pretty hard problem:
$$\frac{\sqrt{\sqrt[4]{27}+\sqrt{\sqrt{3}-1}}-\sqrt{\sqrt[4]{27}-\sqrt{\sqrt{3}-1}}}{\sqrt{\sqrt[4]{27}-\sqrt{2\sqrt{3}+1}}}$$
So it is a pretty long a... | \begin{align}
&=\frac{\bigg(\sqrt{\sqrt[4]{27}+\sqrt{\sqrt{3}-1}}-\sqrt{\sqrt[4]{27}-\sqrt{\sqrt{3}-1}}\bigg)\bigg(\sqrt{\sqrt[4]{27}+\sqrt{\sqrt{3}-1}}+\sqrt{\sqrt[4]{27}-\sqrt{\sqrt{3}-1}}\big)}{\sqrt{\sqrt[4]{27}-\sqrt{2\sqrt{3}+1}}\bigg(\sqrt{\sqrt[4]{27}+\sqrt{\sqrt{3}-1}}+\sqrt{\sqrt[4]{27}-\sqrt{\sqrt{3}-1}}\big... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2498512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Bernoulli Differential Equation with an $xy^3$ Term Below is my solution to a differential equation. The answer that I got was different than the book's answer so I am assuming mine is wrong. I am hoping somebody can tell me where I went wrong.
Problem:
Solve the following differential equation.
\begin{eqnarray*}
\frac... | Akiva Weinberger answered the question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2499033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solve the equation $x^{x+y}=y^{y-x}$ in positive integers.
Let $x,y$ are positive integers. Solve the equation
$$x^{x+y}=y^{y-x}$$
smartplot(x^(x+y)=y^(y-x));
I used http://www.wolframalpha.com
| Let $x,y$ be integers, then $x^{x+y}$ is integer, and so must be $y^{y-x}$. Therefore, $y \geq x$.
Now from fundamental theorem of arithemtic it's evident that $x = n^a$ and $y = n^b$ for some positive integers $n, a, b$, where $b > a$. If $n = 1$ we get solution $x=y=1$. Otherwise, rewrite original equation in terms o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2499727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How to find $x$ given $\log_{9}\left(\frac{1}{\sqrt3}\right) =x$ without a calculator? I was asked to find $x$ when: $$\log_{9}\left(\frac{1}{\sqrt3}\right) =x$$
Step two may resemble:
$${3}^{2x}=\frac{1}{\sqrt3}$$
I was not allowed a calculator and was told that it was possible. I put it into my calculator and found... | There are already some great answers to this question, but just in case you need the basics of logs...
One nice way to think about logs is that it asks for the how many of a particular factor is in a number.
$$log_5(125)=3$$
Because there are three factors of 5 in the number 125.
$$125=5 \cdot 5 \cdot 5=5^3$$
Similarl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 10,
"answer_id": 5
} |
How can I prove $\begin{equation*} \lim_{n \rightarrow \infty} \frac{n}{\sqrt{n^2 + n+1}}=1 \end{equation*}$ How can I prove(using sequence convergence definition):
$$\begin{equation*} \lim_{n \rightarrow \infty} \frac{n}{\sqrt{n^2 + n+1}}=1 \end{equation*}$$
I need to cancel the n in the numerator, any hint will be a... | For a given $\epsilon > 0$, we have: $|a_n-1|=\left|\dfrac{n}{\sqrt{n^2+n+1}}-1\right|= \left|\dfrac{n-\sqrt{n^2+n+1}}{\sqrt{n^2+n+1}}\right|= \dfrac{n+1}{\sqrt{n^2+n+1}\left(\sqrt{n^2+n+1}+n\right)}< \dfrac{n+1}{n(n+n)}< \dfrac{2n}{2n^2} = \dfrac{1}{n}< \epsilon$, when $n > 1/\epsilon$. Thus you simply choose $N_0 = ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Find $\lim_{n\rightarrow \infty}\frac{(2n-1)!!}{(2n)!!}.$ Find $$\lim_{n\rightarrow \infty}\frac{(2n-1)!!}{(2n)!!}.$$
I have tried the following:
$$(2n-1)!!=\frac{(2n)!}{2^{2n}n!}$$
$$(2n)!!=2^nn!$$
$$\lim_{n\rightarrow \infty}\frac{(2n-1)!!}{(2n)!!}=\lim_{n\rightarrow \infty}\frac{(2n)!}{2^{2n}(n!)^2}$$
Now using Stir... | We have
$$ \frac{(2n-1)!!}{(2n)!!} = \frac{1}{4^n}\binom{2n}{n} = \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)=\frac{1}{2}\prod_{k=2}^{n}\left(1-\frac{1}{2k}\right) \tag{A}$$
and by squaring both sides
$$ \left[\frac{(2n-1)!!}{(2n)!!}\right]^2 = \frac{1}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}\right)\prod_{k=2}^{n}\left(1+\... | {
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"source": "stackexchange",
"question_score": "2",
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Calculating expection of number of draws necessary to capture $m$ marked animals I'm working on the following problem, but I'm stuck and I hope anyone can provide some help:
A population of $N$ animals has had a certain number $a$ of its members captured, marked and then released. Show that the probability $p_n$ that ... | Let $X$ be a random variable representing the number of animals it is necessary to capture in order to obtain $m$ marked animals, from a total population of $N$ animals, $a$ of which are marked. Then,
$$p_n = \mathbb{P}(X=n) = \frac{a}{N}\binom{a-1}{m-1}\binom{N-a}{n-m}\left/\binom{N-1}{n-1}\right.$$
for $n = m,m+1,\ld... | {
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"source": "stackexchange",
"question_score": "2",
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$ \lim_{n \to \infty} \left(\frac 1{n^2+1}+\frac 2{n^2+2}+\frac 3{n^2+3}+\cdots +\frac n{n^2+n}\right)$ Evaluate:
$$ L=\lim_{n \to \infty} \left(\frac 1{n^2+1}+\frac 2{n^2+2}+\frac 3{n^2+3}+\cdots +\frac n{n^2+n}\right)$$
My approach:
Each term can be written as
$$ \frac k{n^2+k}=\frac {n^2+k-n^2}{n^2+k}=1-\frac {n^2}{... | HINT:
Using $\sum_{k=1}^n k=\frac{n(n+1)}{2}$ along with the estimates $n^2+1\le n^2+k\le n^2+n$ reveals
$$\frac{n(n+1)}{2(n^2+n)}\le\sum_{k=1}^n\frac{k}{n^2+k}\le \frac{n(n+1)}{2(n^2+1)}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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System of simultaneous equations $x^n+y^n+z^n=3$ ($n$=1,2,5) Does someone know the solution of the following system of simultaneous equations ($x$, $y$, $z$ are real numbers):
$$
x+y+z=3
$$
$$
x^2+y^2+z^2=3
$$
$$
x^5+y^5+z^5=3
$$
Presented at one of math competitions, don't remember when and where.
| Here is the algebraic solution. Consider z as parameter, so we have
$$
x+y=3-z
$$
$$
x^2+y^2=3-z^2
$$
Subtracting second equation from first squared gives:
$$
2xy=6-6z+2z^2
$$
or
$$
xy=z^2-3z+3
$$
Roots of the system
$$
x+y=3-z
$$
$$
xy=z^2-3z+3
$$
are the roots of polynomial
$$
t^2-(3-z)t+(z^2-3z+3)
$$
which has real ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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$x\in \mathbb{R}\setminus\{0\}$ and $x + \frac{1}{x}$ is an integer. Prove $\forall n\in \mathbb{N}$ $x^n + \frac{1}{x^n}$ is also an integer.
Let $x$ be a non-zero real number such that $x + \frac{1}{x}$ is an integer. Prove that $\forall n\in \mathbb{N}$ the number $x + \frac{1}{x}$ is also an integer.
Attempt at s... | Both, $x$ and $\dfrac 1x$ are roots of the same equation $X^2-aX+1=0$
where $a$ is an integer. It follows that any equation deduced from it is also an equation of both $x$ and $\dfrac 1x$.
We have
$$X^2=aX-1$$ $$X^3=aX^2-X=a(aX-1)-X=(a^2-1)X-a$$
$$X^4=(a^2-1)X^2-aX=(a^2-1)(aX-1)-aX=(a^3-2a)X-(a^2-1)$$
For $X^n$ on... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2506773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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Generators of hyperbola
A variable generator meets two generators of the system through extremities $B$ & $B'$ of the minor axis of the principal elliptic section of the hyperboloid $$\frac{x^2} {a^2} +\frac{y^2}{b^2} -z^2c^2=1$$ in $P$ & $P'$. Prove that $BP$. $B'P'=a^2+c^2$
My attempt :
The point of intersection of... | We have the hyperboloid $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2} {c^2} =1$
Generating lines of hyperboloid can be written in the standard form:
$$\frac{x - acos\theta}{asin\theta} = \frac{y - bsin\theta}{-bcos\theta} = \frac{z}{c} \longrightarrow (1)$$ and
$$\frac{x - acos\theta}{asin\theta} = \frac{y - bsin\theta}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2510874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Injectivity and Surjectivity of a Function We have the set $S$ that is the set of odd positive integers. A function $F:\mathbb{N} \rightarrow S$ is defined by $F(n)=k$ for each $n\in \mathbb{N}$, where $k$ is an odd positive integer for which $3n+1=2^mk$ for some non-negative integer $m$. Prove or disprove the followin... | For any integer $n$, calculate $3n+1$. Once you have $3n+1$ write that as a product by taking out as many factors of $2$ as possible. The remaining odd factor is the value of $F(n)$.
If $n=1$, then $3n+1 = 4$, and $4=2^2 \times 1$, so $m=2$ and $k=1$. Hence $F(1)=1$
If $n=2$, then $3n+1 = 7$, and $7 = 2^0 \times 7$, so... | {
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"source": "stackexchange",
"question_score": "1",
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How to expand $(x^{n-1}+\cdots+x+1)^2$ (nicely) sorry if this is a basic question but I am trying to show the following expansion holds over $\mathbb{Z}$:
$(x^{n-1}+\cdots+x+1)^2=x^{2n-2}+2x^{2n-3}+\cdots+(n-1)x^n+nx^{n-1}+(n-1)x^{n-2}+\cdots+2x+1$.
Now I can show this in by sheer brute force, but it wasn't nice and ce... | A kind of graphical proof: Consider the case of $(1+x+x^2+x^3+x^4+x^5)^2$ expanded in a square array in this way:
$$\begin{array}{|l||l|l|l|l|l|}
\hline
&\color{red}{1}&\color{red}{x}&\color{red}{x^2}&\color{red}{x^3}&\color{red}{x^4}&\color{red}{x^5}\\
\hline
\color{red}{x^5}&x^5&x^6&x^7&x^8&x^9&x^{10}\\
\hline
\colo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2511494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
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Use recurrence relation to find the probability
Ram and Shyam have agreed to bet one dollar on each flip of a fair coin and to continue
playing until one of the them wins all of other’s money. Use a recurrence relation to find
the probability that Ram will win all of Shyam’s money if Ram starts with $a$ dollars an... | Hint: the following equation is the solution:
$$p(a,b)=0.5\times\left(p(a-1,b+1)+p(a+1,b-1)\right)$$
$$p(1,1)=0.5$$
$$p(x,0) = 1, x\geq 1$$
$$p(0,y) = 0, y\geq 1$$
You can expand the equation:
$$p(a,b) = 0.5\times(p(a,b) + 0.5(p(a-2,b+2) + p(a+2,b-2)) = 0.5\times p(a,b) + \frac{1}{2^2}(p(a-2,b+2) + p(a+2,b-2)) \Right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2512161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Series and integrals for Apery's constant Two integrals for Apery's constant $\zeta(3)$ are
$$\zeta(3)=\frac{16}{3} \int_0^1 \frac{x\log^2\left(x\right)}{1+x^2}dx$$
and
$$\zeta(3)=\frac{32}{7} \int_0^1 \frac{x\log^2\left(x\right)}{1-x^4}dx$$
How can series expressions be obtained from them?
Related questions:
A series ... | A variant of Marco's answer.
$\displaystyle J=\int_0^1 \frac{x\ln^2 x}{1+x^2}\,dx$
Perform the change of variable $y=x^2$,
$\begin{align} J=\frac{1}{8}\int_0^1 \frac{\ln^2 y}{1+y}\,dy\end{align}$
Let,
$\displaystyle K=\int_0^1 \frac{\ln^2 x}{1+x}\,dx$
$\displaystyle L=\int_0^1 \frac{\ln^2 x}{1-x}\,dx$
$\begin{align}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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How do I solve Cauchy problem for $y''-2y'-3y = e^{4x}$? I have one:
$$y''-2y'-3y = e^{4x} \quad y(0) = 1 \quad y'(0) = 0$$
I've found the solution as a sum of general solution and particular one:
$$y(x) = C_1e^{-x}+C_2e^{3x}+\frac{1}{5}e^{4x}$$
Applying the first condition I got:
$$C_1+C_2+ \frac{1}{5} = 1$$
but I do ... | Using
$$y(x) = C_1e^{-x}+C_2e^{3x}+\frac{1}{5}e^{4x}$$
then differentiate to obtain
$$y'(x) = - C_1 e^{-x} + 3 C_2 e^{3x} + \frac{4}{5} e^{4x}.$$
Now set $x=0$ in these equations to obtain
\begin{align}
C_{1} + C_{2} + \frac{1}{5} &= 1 \\
- C_{1} + 3 C_{2} + \frac{4}{5} &= 0.
\end{align}
Solving this set for $C_{1}$ a... | {
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"source": "stackexchange",
"question_score": "1",
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A box contains a penny, two nickels, and a dime. If two coins are selected randomly from the box, without replacement, and if X is the sum... A box contains a penny (1¢), two nickels (5¢), and a dime (10¢). If two coins are selected randomly from the box, without replacement, and if $X$ is the sum of the values of the ... | Your answer to the first question is correct.
For the second question, the cumulative distribution function (CDF) of a random variable $X$ at $x$ is found by finding the probability that $X \leq x$. Since there are only four possible values for $X$,
\begin{align*}
P(X \leq 6) & = P(X = 6)\\
P(X \leq 10) & = P(X = 6) +... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Solve the equation $x^{2/3}-6x^{1/3} + 8 = 0$ I'm going around in circles on this one...
Solve the equation $$x^{2/3}-6x^{1/3} + 8 = 0$$
According to the book, the answer is 8, 64, but that makes no sense to me either...
Thanks
Gary
| This is just a quadratic in disguise. Don't you see that $x^{2/3}$ can be actually rewritten like this: $\left(x^{\frac{1}{3}}\right)^2$? There are typically two strategies that are used to approach solving this type of equation. You either assign your original expression under the square to a new variable and solve th... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Probability of deck of cards such that each person receives one ace
Suppose that a deck of 52 cards containing four aces is shuffled thoroughly and the cards are then distributed among four players so that each player receives 13 cards. Determine the probability that each player will receive one ace.
The answer to th... | There are
$$\binom{52}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13}$$
ways to distribute $13$ cards to each of four people.
There are $4!$ ways to distribute the aces so that each person receives one and
$$\binom{48}{12}\binom{36}{12}\binom{24}{12}\binom{12}{12}$$
ways to distribute the remaining cards so that each ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2521017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Simplify to terms of generalized power mean I have the following expression${}^1$ I'd like to simplify
$$r \equiv \left[ \frac1{x\sqrt{x}} \left( 1 - \frac{k}{x} \right) - \frac1{y\sqrt{y}} \left( 1 - \frac{k}{y} \right) \right]\left[ \frac1{\sqrt{x}} \left( 1 - \frac{k}{x} \right) - \frac1{\sqrt{y}} \left( 1 - \frac{k... | We can write
$$r=\frac 1H+\frac 1G+\color{red}{\frac{1}{M-G+\frac{GM}{k}}}$$
where
$$M=(x^{-1/2}+y^{-1/2})^{-2}$$
If we write
$$r=\frac{\frac{1}{x\sqrt x}-\frac{1}{y\sqrt y}-k\left(\frac{1}{x^2\sqrt x}-\frac{1}{y^2\sqrt y}\right)}{\frac{1}{\sqrt x}-\frac{1}{\sqrt y}-k\left(\frac{1}{x\sqrt x}-\frac{1}{y\sqrt y}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2524146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How many arrangements of 1,1,1,1,2,3,3 are there with 2 not beside either 3? How many arrangements of 1,1,1,1,2,3,3 are there with 2 not beside either 3?
If I find this by method of complementation then
Total number of arrangements without any restrictions =$\frac{7!}{4!×2!}$.
Then I should subtract total no of arran... | As you observed, there are
$$\binom{7}{4}\binom{3}{2}\binom{1}{1} = \frac{7!}{4!2!1!} = \frac{7!}{4!2!}$$
distinguishable arrangements of four $1$s, one $2$, and two $3$s. From these, we must exclude those arrangements in which a $3$ is adjacent to the $2$.
A $3$ is adjacent to the $2$: We have six objects to arrang... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2525925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Does this question on conditional probability have an incorrect solution? Going through some exercises on probability, and I chanced upon this problem.
An urn contains 6 red balls and 3 blue balls. One ball is selected at
random and is replaced by a ball of the other color. A second ball is
then chosen. What is th... | With problems like this I like to turn $P(R_1 | R_2)$ into $P(R_2 | R_1)$ if it is an easier calculation. We have,
$$\begin{align*}
P(R_1 | R_2)
&= \frac{P(R_1 \cap R_2)}{P(R_2)}\\\\
&= \frac{P(R_2 | R_1)P(R_1)}{P(R_2)}\\\\
&= \frac{\frac{5}{9}\cdot\frac{6}{9}}{\frac{6}{9}\cdot\frac{5}{9}+\frac{3}{9}\cdot\frac{7}{9}}\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2527098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$\frac{1}{x+1} + \frac{2}{x^2+1} + \frac{4}{x^4+1} + \cdots $ up to $n$ terms in terms of $x$ and $n$. There's a series which I can't seem to find a way to sum. Any help would be highly appreciated. It goes as follows $$\frac{1}{x+1} + \frac{2}{x^2+1} + \frac{4}{x^4+1} + \cdots $$ up to $n+1$ terms. The sum is t... | A supplement to the nice answer from @labbhattacharjee. We see an example of telescoping based upon
\begin{align*}
\color{blue}{\frac{2^k}{1-x^{2^k}}+\frac{2^k}{1+x^{2^k}}}
=\frac{2^k\left(1+x^{2^k}\right)+2^k\left(1-x^{2^k}\right)}{\left(1-x^{2^k}\right)\left(1+x^{2^k}\right)}
&\color{blue}{=\frac{2^{k+1}}{1-x^{2^{k+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2528731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
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What's the value of $x$ if $\frac{1}{x}+\frac{1}{y+z}=\frac{1}{2}$ and... Given $x, y, z\in \mathbb{R}$, such that $\frac{1}{x}+\frac{1}{y+z}=\frac{1}{2}$ and $ \frac{1}{y}+\frac{1}{z+x}=\frac{1}{3}$ and $\frac{1}{z}+\frac{1}{x+y}=\frac{1}{4}$. Find the value of $x$.
| I suggest to go for $X=\frac 1x$, $Y=\frac 1y$ and $Z=\frac 1z$ because it has the advantage to simplify the denominators.
The equations become
$\dfrac{XY+YZ+ZX}{Y+Z}=\dfrac 12\quad;\quad\dfrac{XY+YZ+ZX}{X+Z}=\dfrac 13\quad;\quad\dfrac{XY+YZ+ZX}{X+Y}=\dfrac 14$
And it gives a simple system to solve
$\begin{cases}
Y+Z=... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Transformation to $\log\left(\frac a2\cos x\right)$
Question: How does$$T=-\int\limits_{\tfrac {\pi}2}^{-\tfrac {\pi}2}dx\,\frac {\log\left(\tfrac {a^2}4\cos^2x\right)}{\sqrt{\tfrac {a^2}4\cos^2x}}\left(\frac a2\cos x\right)=4\int\limits_0^{\tfrac {\pi}2}dx\,\log\left(\frac a2\cos x\right)\tag1$$
I was evaluating an ... | First, $\cos(x)\ge 0$ and is even for $x\in[-\pi/2,\pi/2]$. Hence, $\sqrt{\cos^2(x)}=\cos(x)$. If $a>0$, then $\sqrt{a^2}=a$. Therefore,
$$\frac{\log\left(\frac{a^2}{4}\cos^2(x)\right)\left(\frac a2\cos(x)\right)}{\sqrt{\frac{a^2}{4}\cos^2(x)}}=\frac{\log\left(\left(\frac{a\cos(x)}{2}\right)^2\right)\left(\frac a2\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2530204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simplify the expression: $(2x+1)×\left(x^2+(x+1)^2 \right)×\left( (x^4+(x+1)^4) \right)×...×\left(x^{64}+(x+1)^{64}\right)$ Simplify the expression:
$$(2x+1)×\left(x^2+(x+1)^2 \right)×\left( (x^4+(x+1)^4) \right)×...×\left(x^{64}+(x+1)^{64}\right)$$
I used the general method:
$(2x+1)(2x^2+2x+1)(2x^4+4x^3+6x^2+4x+1)×...... | We can rewrite the product as a telescoping one. The end result is:
$$\prod_{k=0}^6 \left((1+x)^{2^k} + x^{2^k}\right) =
\prod_{k=0}^6 \frac{(1+x)^{2^{k+1}} - x^{2^{k+1}}}{(1+x)^{2^k} - x^{2^k}}
= \frac{(1+x)^{2^7} - x^{2^7}}{(1+x)^{2^0} - x^{2^0}} = (x+1)^{128} - x^{128}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2531082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Computing limit of $\sqrt{n^2+n}-\sqrt[4]{n^4+1}$ I have tried to solve this using conjugate multiplication, but I got stuck after factoring out $n^2$.
$\begin{align}
\lim_{n\rightarrow\infty}\dfrac{n^2+n-\sqrt{n^4+1}}{\sqrt{n^2+n}+\sqrt[4]{n^4+1}}
&=\lim_{n\rightarrow\infty}\dfrac{n(1+\dfrac{1}{n}-\sqrt{1+\dfrac{1}{n^... | Let $t = \frac1n$
Then use the derivative formula at $t = 0$ ,
$$ \begin{align}\lim_{n\to\infty}\sqrt{n^2+n}-\sqrt[4]{n^4+1} &= \lim_{n\to\infty} n\left(\sqrt{\frac{1}{n}+1}-\sqrt[4]{\frac{1}{n^4}+1}\right) \\&= \lim_{t\to0}\frac{1}t(\sqrt{t+1}-\sqrt[4]{t^4+1})\\&= \lim_{t\to 0}\frac{\sqrt{t+1}-1}{t} - \frac{\sqrt[4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2531365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Proving $\sum\limits_{r=1}^n \cot \frac{r\pi}{n+1}=0$ using complex numbers Let $x_1,x_2,...,x_n$ be the roots of the equation $x^n+x^{n-1}+...+x+1=0$.
The question is to compute the expression $$\frac{1}{x_1-1} + \frac{1}{x_2-1}+...+\frac{1}{x_n-1}$$
Hence to prove that $$\sum_{r=1}^n \cot \frac{r\pi}{n+1}=0$$
I ... | For the first part :
Write $$x^n+x^{n-1}+...+x+1 =(x-x_1)(x-x_2)\ldots (x-x_{n})$$
Take log both the sides
$$\log (x^n+x^{n-1}+...+x+1)= \log(x-x_1) +\log(x-x_2)+\ldots +\log(x-x_n)$$
Differentiating w.r.t. $x$, we get
$$\frac{ nx^{n-1}+(n-1)x^{n-2}+\ldots+ 1}{x^n+x^{n-1}+...+x+1}= \frac{1}{x-x_1}+\frac{1}{x-x_2}+\ld... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2532206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
How do I rationalize the following fraction: $\frac{1}{9\sqrt[3]{9}-3\sqrt[3]{3}-27}$? As the title says I need to rationalize the fraction: $\frac{1}{9\sqrt[3]{9}-3\sqrt[3]{3}-27}$. I wrote the denominator as: $\sqrt[3]{9^4}-\sqrt[3]{9^2}-3^3$ but I do not know what to do after. Can you help me?
| Compute the minimal polynomial of $\alpha=-27-3\cdot3^{1/3}+9\cdot3^{2/3}$:
$$
\begin{align}
\alpha&=-27-3\cdot3^{1/3}+9\cdot3^{2/3}\tag1\\
\alpha^2&=567+405\cdot3^{1/3}-477\cdot3^{2/3}\tag2\\
\alpha^3&=-81-25515\cdot3^{1/3}+16767\cdot3^{2/3}\tag3
\end{align}
$$
Since $(-3,405,-25515)\times(9,-477,16767)=-2214(1,81,243... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2535817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Doubts on obtaining orthonormal basis In finding the matrix $P$ that orthogonally diagonalizes $A$ and to determine $P^TAP$, where $$A =
\begin{pmatrix}
1 & -1 & 1 & -1\\
-1 & 1 & -1 & 1\\
1 & -1 & 3 & 1\\
-1 & 1 & 1 & 3\\
\end{pmatrix}
$$
I've found the eigenvalues to be $0$ and $4$.
In order... | If you have two vectors $u_1$ and $u_2$, the Gram-Schmidt algorithm produces two vectors $v_1$ and $v_2$ defined by
$$
v_1 = \frac{u_1}{\|u_1\|}
\quad
v_2 = \frac{u_2 - (u_2|v_1)v_1}{\|u_2 - (u_2|v_1) v_1\|}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2536196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding function $f(x)$ which satisfy given functional equation
Find all function $f:\mathbb{R}-\{0,1\}$ in $$f(x)+2f\left(\frac{1}{x}\right)+3f\left(\frac{x}{x-1}\right)=x$$
Attempt: put $\displaystyle x = \frac{1}{x}$, then $$f\left(\frac{1}{x}\right)+2f(x)+3f\left(\frac{1}{1-x}\right) = \frac{1}{x}$$
could some he... | I do not understand the phrase "A detailed canonical answer is required to address all the concerns" since I am unaware of any concerns about the answer given some time ago. However, here are the actual equations and how to solve them.
Let $V = \pmatrix{f(x)\\f(\frac{1}{x})\\f(\frac{1}{1-x})\\f(\frac{x}{x-1})\\f(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2536550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Convergence of $ \sum_{k=1}^{\infty} \frac {3^k}{5^k + 1}$
Show the convergence of the following series:
$$\sum_{k=1}^{\infty}\frac {3^k}{5^k + 1}$$
*
*a) Show the monotony of the partial sums
*b) estimate upwards
*c) remember the geometric series (I do not know how to use that here.)
The following... | you have $\frac{3^k}{5^k+1} < \frac{3^k}{5^k}= \left(\frac{3}{5}\right)^k$ than it follows $$ \sum_{k=1}^{\infty}a_k = \sum_{k=1}^{\infty} \frac{3^k}{5^k+1} \leq \sum_{k=1}^{\infty} \frac{3^k}{5^k} = \sum_{k=1}^{\infty} \left(\frac{3}{5}\right)^k = \sum_{k=0}^{\infty} \left(\frac{3}{5}\right)^{k+1}=\frac{3}{5} \left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2538022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\lim_{x \to 0} \frac{1-(x^2/2) -\cos (x/(1-x^2))}{x^4}$
find the limits with Using :$\lim_{x \to 0} \frac{1-\cos x}{x^2}=\frac12$
$$\lim_{x \to 0} \frac{1-\dfrac{x^2}2-\cos (\dfrac{x}{1-x^2})}{x^4}$$
My Try :
$$\lim_{x \to 0} \frac{1-\dfrac{x^2}2-\cos (\dfrac{x}{1-x^2})}{x^4}=\lim_{x \to 0} \frac{(1-\cos (\... | Use the famous formula $1-\cos 2t=2\sin^{2}t$ to transform the expression into $$2\cdot\frac{\sin^{2}t-(x/2)^{2}}{x^{4}}$$ where $t=x/(2(1-x^{2}))$. Next the numerator needs to be split by adding and subtracting $t^2$. It is easy to check that $(t^2-(x/2)^{2})/x^4\to 1/2$ and hence the desired limit is equal to $$1+2\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2541078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How can I calculate the limit $\lim\limits_{x \to 7} \frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}$ without L'Hospital's rule? I have a problem with calculation of the limit:
$$\lim\limits_{x \to 7} \frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}$$
Is there a way to calculate it? How can I do it?
| Using $x^n - y^n = (x - y)\sum_{k=0}^n x^ky^{n-k}$, we have
$$
\sqrt{x+2}-\sqrt[3]{x+20} = \frac{(x+2)^3-(x+20)^2}{\sum_{k=0}^6(x+2)^{k/2}(x+20)^{2-k/3}} \\
\sqrt[4]{x+9}-2 = \frac{x-7}{\sum_{k=0}^4(x+9)^{k/4}2^{4-k}}
$$
Notice that $(x+2)^3-(x+20)^2 = (x - 7)(x^2+12x+56)$, so
$$
\begin{aligned}
\frac{\sqrt{x+2}-\sqrt[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2543125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 5
} |
Quadrilateral, two circumcircles, relation to prove In a convex quadrilateral $ABCD$ diagonals $AC$ and $BD$ intersect at point $S$. Denote with $P$ the center of circumcircle of triangle $ABS$. Denote with $Q$ the center of circumcircle of triangle $CDS$. Prove that:
$$4 PQ \geq AB + CD.$$
I was able to calculate th... | A graph is drawn below, together with the explanation of the notations. To simplify the explanation, we only restrict to the most non-trivial case: $\angle B \geq \angle A$, $\angle C \geq \angle D$.
It is not hard to see that $\angle PSA = \frac{\theta + \angle A - \angle B}{2}$ and $\angle QSD = \frac{\theta + \angl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2548146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.