Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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$a$ and $b$ are solutions of $ \frac{1}{x^{2} - 10x-29} + \frac{1}{x^{2} - 10x-45} - \frac{2}{x^{2} - 10x-69} = 0 $, $a+b=?$ $a$ and $b$ are solutions of
$$ \frac{1}{x^{2} - 10x-29} + \frac{1}{x^{2} - 10x-45} - \frac{2}{x^{2} - 10x-69} = 0 $$
What is $a+b=?$
$$ $$
Are there better approaches than the one below?
Soluti... | Wait so at the start, you say $y=x^2-10x$ then proceed to say $y=x^2-10x-39$
Apart from that, this is probably the best solution.
| {
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"url": "https://math.stackexchange.com/questions/3231701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Hard limit involving different order radicals $\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )$ Please help me to calculate the following limit
$$\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )$$
I factored out $n$ from both radicals but it didn't work and tried to use the identity $a^2-b^2$ and $a^3-b^... | Another way to solve the problem is the following, where no Differential Calculus is needed.
By elementary algebraic manipulation of radicals, you get:
$$ \begin{split} \sqrt[3]{n^3 + 3n^2} - \sqrt{n^2 + 2n} &= \underbrace{\sqrt[6]{(n^3 + 3n^2)^2}}_{=:\sqrt[6]{a}} - \underbrace{\sqrt[6]{(n^2 + 2n)^3}}_{=:\sqrt[6]{b}} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How much ways there are to produce sum = 21 with 4 different natural numbers? 0 isn't natural number , and the sum way is important (e.g. 3+5+6+7 is different from 5+6+7+3).
I got that I have for a+b+c+d = 21 , 2024 options.
I think I need to sub the invalid numbers , so I need to relate for the cases :
a = 0 , b = 0, ... | This is the answer to "How many ways are there to distribute 21 candies to 4 children, where each child gets at least 1 candy?"
To do this, lay out 21 candies in a row:
$$ \underbrace{\bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3234407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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algorithms - When not to assume $n$ is a power of 2, while solving recurrences? In Udi Manber - Introduction to Algorithms. A Creative Approach, exercise 3.29, p.59
Although in general it is sufficient to evaluate recurrence relations only for powers of 2, that is not always the case. Consider the following relation:
... | For the case of Kenneth Rosen's exercise, I posted an answer's here. As the result of solution 2, I choose $n$ such that $2^{m-1}<n<2^{m}$, in particular, $n=2^{m-1}\cdot2^{\frac{1}{2}}=2^{m-\frac{1}{2}}$.
$$\begin{align*}
T\left(n\right)&=\frac{n^{2^{\left\lfloor \log_{2}n\right\rfloor }-1}\cdot n^{2^{\log_{2}n-\left\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3236410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Solve $\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} $ Solve $$\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} $$
My attempt:
Let $A = \sqrt{1+x}, B = \sqrt{1-x}$ and then by squaring the problematic equation we get:
$$(... | Great substitution technique. Notice that $\sf{A=\sqrt{1+x}}$ and $\sf{B=\sqrt{1-x}}$ leads to $$\sf{(A^3+B^3)\sqrt{1+AB}}=2+AB$$ and since $\sf{A^3+B^3=(A+B)(A^2-AB+B^2)}$ and $\sf{A^2+B^2=2\implies (A+B)^2=2(1+AB)}$, we get $$\sf{(A+B)\sqrt{1+AB}=\frac{2+AB}{2-AB}}\implies (1+AB)\sqrt2=1+\frac{2AB}{2-AB}$$ which can ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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"answer_id": 1
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How to prove : $\cos^32\theta + 3\cos2\theta = 4(\cos^6 \theta -\sin^6 \theta)$ How to prove : $\cos^32\theta + 3\cos2\theta = 4(\cos^6 \theta -\sin^6 \theta)$
| $4(\cos^6 \theta -\sin^6 \theta) = 4(\cos^2 \theta -\sin^2 \theta)(\cos^4 \theta + \cos^2 \theta\sin^2 \theta+\sin^4 \theta) = 4(\cos^2 \theta -\sin^2 \theta)( (\cos^2 \theta +\sin^2 \theta)^2 - 2\cos^2 \theta\sin^2 \theta + \cos^2 \theta\sin^2 \theta) = 4(\cos^2 \theta -\sin^2 \theta)( 1 - \cos^2 \theta\sin^2 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3244039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Help with $-\int_0^1 \ln(1+x)\ln(1-x)dx$ I have been attempting to evaluate this integral and by using wolfram alpha I know that the value is$$I=-\int_0^1 \ln(1+x)\ln(1-x)dx=\frac{\pi^2}{6}+2\ln(2)-\ln^2(2)-2$$
My Attempt:
I start off by parametizing the integral as $$I(a)=\int_0^1 -\ln(1+x)\ln(1-ax)dx$$
where $I=I(1)$... | Another solution, this time using Mike Spivey's alternating Euler sum:
\begin{align*}
I &=-\int_0^1\ln(1-x)\ln(1+x)\,\mathrm dx \\
&= \int_0^1\ln(1-x)\sum_{n=1}^\infty\frac{(-1)^n}{n}x^n\,\mathrm dx,\qquad\text{Mercator series}\\
&= \sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^1\ln(1-x)x^n\,\mathrm dx\\
&= \sum_{n=1}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3246020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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"answer_id": 0
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Find all numbers x which satisfy $|x^2 + 2| = |x^2 − 11|$. This question is taken from book: Exercises and Problems in Calculus, by
John M. Erdman, available online, from chapter 1.1, question $4$.
Request help, as not clear if my approach is correct.
(4) Find all numbers x which satisfy $|x^2 + 2| = |x^2 − 11|$.
Ha... | Maybe another approach might also be interesting:
$$|x^2 +2| = |x^2-11| \Leftrightarrow (x^2+2)^2 = (x^2-11)^2\Leftrightarrow (x^2+2)^2 - (x^2-11)^2 = 0$$
$$\stackrel{a^2-b^2 = (a-b)(a+b)}{\Leftrightarrow} (x^2+2 - (x^2-11))(x^2 + 2 + x^2-11)= 13(2x^2 - 9)= 0$$
$$\Leftrightarrow x^2= \frac{9}{2}\Leftrightarrow x= \pm\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3247025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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Evaluate $\int_0^{\infty} \frac {\ln(1+x^3)}{1+x^2}dx$
Prove that $$\int_0^{\infty} \frac {\ln(1+x^3)}{1+x^2}dx=\frac {\pi \ln 2}{4}-\frac {G}{3}+\frac {2\pi}{3}\ln(2+\sqrt 3)$$ Where $G$ is the Catalan's constant.
Actually I proved this using the Feynman's trick namely by introducing the parameter $a$ such that $$\x... | Noticing that
$$
\begin{aligned}
& \int_{0}^{\infty} \frac{\ln \left(1+x^{3}\right)}{1+x^{2}} d x =\underbrace{\int_{0}^{\infty} \frac{\ln (1+x)}{1+x^{2}}}_{\frac{\pi}{4} \ln 2+G} d x+\underbrace{\int_{0}^{\infty} \frac{\ln \left(1-x+x^{2}\right)}{1+x^{2}}}_{K} d x
\end{aligned}
$$
where the first integral comes from m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3247341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Find $\lim_{x\to\infty}1+2x^2+2x\sqrt{1+x^2}$ Consider the function
$$f(x)=1+2x^2+2x\sqrt{1+x^2}$$
I want to find the limit $f(x\rightarrow-\infty)$
We can start by saying that $\sqrt{1+x^2}$ tends to $|x|$ when $x\rightarrow-\infty$, and so we have that
$$\lim_{x\rightarrow-\infty}{(1+2x^2+2x|x|)}=\lim_{x\rightarrow-\... | Multiply top and bottom by a conjugate:
$$\begin{align*}f(x) & = \dfrac{1+2x^2+2x\sqrt{1+x^2}}{1}\cdot \dfrac{1+2x^2-2x\sqrt{1+x^2}}{1+2x^2-2x\sqrt{1+x^2}} \\ & = \dfrac{1}{1+2x^2-2x\sqrt{1+x^2}}\end{align*}$$
Now, as $x \to -\infty$, $1+2x^2-2x\sqrt{1+x^2} \to \infty$, so $f(x) \to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3249693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Given $f(x)=ax^3-ax^2+bx+4$ Find the Value of $a+b$ Let $f(x)=ax^3-ax^2+bx+4$. If $f(x)$ divided by $x^2+1$ then the remainder is $0$. If $f(x)$ divided by $x-4$ then the remainder is $51$. What is the value of $a+b$?
From the problem I know that $f(4)=51$.
Using long division, I found that remainder of $\frac{ax^3-ax^... | Let $a+b=c$.
Solving in pari/gp:
? lift(Mod(a*x^3-a*x^2+b*x+4,x^2+1))
%9 = (-a + b)*x + (a + 4)
?
? lift(Mod(a*x^3-a*x^2+b*x+4,x-4))
%10 = 48*a + (4*b + 4)
?
? polresultant(%9,a+b-c,a)
%11 = (-2*b + c)*x + (b + (-c - 4))
?
? polresultant(%10-51,a+b-c,a)
%12 = 44*b + (-48*c + 47)
?
? polresultant(%11,%12,b)
%13 = (52*c ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3252433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Maclaurin series for $f(x) = \frac{1}{1+x+x^2}$ I need to get the Maclaurin series and its radius of the convergence for $f(x) = \dfrac{1}{1+x+x^2}$. I tried to do it manually, with getting the derivatives, but I gave up after some time, because I thought there had to be a better way to solve this. Could anyone help? ... | Note that, if $\lvert x\rvert<1$,\begin{align}\frac1{1+x+x^2}&=\frac{1-x}{(1+x+x^2)(1-x)}\\&=\frac{1-x}{1-x^3}\\&=(1-x)\times(1+x^3+x^6+x^9+\cdots)\\&=1-x+x^3-x^4+x^6-x^7+x^9-x^{10}+\cdots\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3252733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Given $ a_{k+1} \ge \frac{k a_{k}}{(a_{k}^{2} + k-1)}, \:\: k > 0$, prove $ S_{n} = a_{1} + .. + a_{n} \ge n, \:\: n \ge 2 $ Suppose a sequence of positive real numbers with
$$ a_{k+1} \ge \frac{k a_{k}}{(a_{k}^{2} + k-1)}, \:\: k > 0$$
prove that $$ S_{n} = a_{1} + .. + a_{n} \ge n, \:\: n \ge 2 $$
Solution: I will... | Remember $S_k=a_1+a_2+\dots+a_k$, so
$$
S_1+S_2+\dots+S_m=ma_1+(m-1)a_2+\dots+a_m.
$$
Hence you have
$$
m(a_1+a_2+\dots+a_m+a_{m+1})\geq m(m+1)
$$
which gives $S_{m+1}\geq m+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3255438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Greatest value of shortest distance between axis of x and normal to ellipsoid Show that greatest value of shortest distance between axis of $x$ and a normal to ellipsoid is $$b-c$$
or in other words show that the maximum distance of all the normals to the ellipsoid is $b-c$
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{... | Observe that the same values of $x,y,z$ which maximize
$$
d(x,y,z)=\frac{|b^2-c^2|}{\sqrt {\dfrac {b^4}{y^2}+\dfrac {c^4}{z^2}}}\tag1
$$subject to the constraint
$$
\dfrac {x^2}{a^2}+\dfrac {y^2}{b^2}+\dfrac {z^2}{c^2}=1,\tag2
$$
will minimize the value of
$$
\frac {(b^2-c^2)^2}{d^2}=\dfrac {b^4}{y^2}+\dfrac {c^4}{z^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3256365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Is $\left(1+\frac1n\right)^{n+1/2}$ decreasing? Using the Cauchy-Schwarz Inequality, we have
$$
\begin{align}
1
&=\left(\int_n^{n+1}1\,\mathrm{d}x\right)^2\\
&\le\left(\int_n^{n+1}x\,\mathrm{d}x\right)\left(\int_n^{n+1}\frac1x\,\mathrm{d}x\right)\\
&=\left(n+\frac12\right)\log\left(1+\frac1n\right)
\end{align}
$$
which... | We'll prove that this sequence indeed decreases.
We need to prove that $$\left(1+\frac{1}{n+1}\right)^{n+\frac{3}{2}}<\left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}.$$Let $$f(x)=\left(x+\frac{1}{2}\right)\ln\left(1+\frac{1}{x}\right)-\left(x+\frac{3}{2}\right)\ln\left(1+\frac{1}{1+x}\right)=$$
$$=(2x+2)\ln(1+x)-\left(x+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3259175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 1
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Show $\frac{\sin x_1\sin x_2\cdots\sin x_n}{\sin(x_1+x_2)\sin(x_2+x_3)\cdots\sin(x_n+x_1)}\le\frac{\sin^n(\pi/n)}{\sin^n(2\pi/n)}$, for $\sum x_i=\pi$ Let $x_{i}>0$, ($i=1,2,\cdots,n$) and such that
$$x_{1}+x_{2}+\cdots+x_{n}=\pi.$$
Show that
$$
\dfrac{\sin{x_{1}}\sin{x_{2}}\cdots\sin{x_{n}}}{\sin{(x_{1}+x_{2})}\sin{(x... | Update:
I have been thinking about this more.
The idea is to replace $x_i$ with some M such that all the properties holds and the ratio of the $\sin$ product holds and the ratio of the products of sines is smaller than the ratio formed with $\sin(M)$. Initially I thought that selecting the Maximum among $x_i$ is suffi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3259944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
"answer_count": 7,
"answer_id": 2
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Wish to evaluate $\int_{0}^{1}\frac{x^4-3x^2+2x}{\sin(\pi x)} dx$ We want to evaluate this integral,$\newcommand{\cosec}{\operatorname{cosec}}$
$$\int_{0}^{1}\frac{x^4-3x^2+2x}{\sin(\pi x)}\mathrm dx\tag1$$
$$\int_{0}^{1}[x^4\cosec(\pi x)-3x^2\cosec(\pi x)+2x\cosec(\pi x)]\mathrm dx$$
Apply integration by parts, first ... | Here is a reasonably nice way:
Claim:
\begin{align*}
\int_0^1 x^{n-2}\frac{x^2-x}{\sin\pi x}\,\mathrm{d}x
&=
\begin{cases}
93\frac{\zeta(5)}{\pi^5}-21\frac{\zeta(3)}{\pi^3} & n=4\\
-\frac72\frac{\zeta(3)}{\pi^3} & n=3\\
-7\frac{\zeta(3)}{\pi^3} & n=2
\end{cases}.
\end{align*}
Proof (sketch):
Define
$$
f_n(a):=\int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3260759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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How to compute $\sum_{k=1}^{\infty}{\frac{1}{k^2+2k}}$? To whom this may concern,
i am struggling with partial sum formulas. I don't really get why you would need to perform a partial fraction decomposition or how you know that you have to.
I started by getting trying to get a grasp of the series:$$\sum_{k=1}^{n}{\frac... | We have the partial fractions decomposition
$$\frac{1}{k^2+2k}=\frac{1}{2k}-\frac{1}{2k+4}.$$
If you write out the first few terms, you will notice a lot of cancellation:
$$\left(\frac{1}{2}-\frac{1}{6}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+\left(\frac{1}{6}-\frac{1}{10}\right)+\left(\frac{1}{8}-\frac{1}{12}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3263815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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The integral $\int_{-\infty}^{\infty}\frac{ a^2x^2 dx}{(x^2-b^2)^2+a^2x^2}=a\pi,~ a, b \in \Re ?$ This integral
$$\int_{-\infty}^{\infty}\frac{ a^2x^2 dx}{(x^2-b^2)^2+a^2x^2}=a\pi, ~ a, b \in \Re$$
looks suspiciously interesting as it is independent of the parameter $b$. The question is: What is the best way of provin... | One way:
Denote the integral as $$I=\int_{-\infty}^{\infty}\frac{ a^2x^2 dx}{(x^2-b^2)^2+a^2x^2}.$$
$$(x^2-b^2)^2+a^2x^2=(x^2+p^2)(x^2+q^2) \Rightarrow p=(a+c)/2,q=(a-c)/2, c=\sqrt{a^2-4b^2}.$$
Then $$I=\frac{2a^2}{p^2-q^2} \int_{0}^{\infty} \left( \frac{p^2}{x^2+p^2}-\frac{q^2}{x^2+q^2} \right)dx=
\frac{2a^2}{p^2-q^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3267110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Ratio between the width of the intersection of two identical intersecting circles and radius, when the intersection is $\frac{\pi r^2}{2}$ Or more visually, if all sections of the below diagram were equal in area and the circles are identical, what is the ratio of s and r, or what is s in terms of r.
I came up with an... | Without loss of generality, we can assume $r=1$.
For convenience, position the circles vertically, with centers on the $y$-axis at the points $(0,h)$ and $(0,-h)$.
Then the equation of the lower circle is
$$x^2+(y+h)^2=1$$
Solving for $y$, and noting that
$$h=1-\frac{s}{2}$$
we get that the upper half of the lower ci... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3267865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How to show that this multivariable limit exists or not? I was a given a problem to solve (if the limit exists):
$$ \lim_{(x,y)\to(0,0)} \cos\left(\frac{x^3-y^3}{x^2+y^2}\right) .$$
My Approach:
Approaching along the path $y=0$, yield the limit to be $1$. Similarly, going along the path $y=mx$ yields the limit to be $... | The limit exists and equals $1$. This follows from the squeeze theorem and the inequalities
$$
0\leq\frac{|x^3-y^3|}{x^2+y^2}\leq |x|+|y|.
$$
Indeed, we have that
$$
|x^3-y^3|\leq |x|^3+|y|^3\leq (|x|+|y|)(x^2+y^2),
$$
where the second inequality follows by expanding out the terms on the right and observing that the ex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3268275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Finding max and min from polynomial equation
Let $a, b \in \mathbb R$ and $b \geqslant 0$. Given that $x^4 + ax^2 + b = 0$ has $2$ real solutions, find maximum for $|a-b|$ and minimum for $a+2b$.
My solution :
$$x^4+ax^2+b=0 \implies (x^2-c)(x^2-d) = 0$$
there is only 1 case which is possible,that is, both $c$ and $d... | If $b=0$ and $a\geq 0$ then $x^4+ax^2+b=0$ has just one real root, i.e. $x=0$.
If $b=0$ and $a< 0$ then $x^4+ax^2+b=0$ has three real roots, i.e. $x=0, \pm\sqrt{-a}$
If $b>0$ then $x^4+ax^2+b=0$ has two distinct real roots iff $z^2+az+b=0$ has one positive root and a negative root, which is impossible because $b>0$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3270473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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unknown polynomial divided by $x^2(x-1)$, find the remainder. I took an exam today and there's a problem stuck in my head; I still can't figure out yet.
Here's the question (just the concept as I can't remember precisely).
An unknown polynomial divided by $(x-1)^2$ leaves the remainder of $x + 3$ (not sure about the n... | We have
$$f(x)=(x-1)^2q_1(x)+x+3$$
$$f'(x)=2(x-1)q_1(x)+(x-1)^2q_1'(x)+1$$
Where for $x=1$ we have $f(1)=4$ and $f'(1)=1$
Then from
$$f(x)=x^2q_2(x)+2x+4$$
$$f'(x)=2xq_2(x)+x^2q_2'(x)+2$$
Where for $x=0$ we have $f(0)=4$ and $f'(0)=2$
Now from $$f(x)=x^2(x-1)q_3(x)+ax^2+bx+c$$ and
$$f'(x)=x^2q_3(x)+2x(x-1)q_3(x)+(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3271609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
Radical equation solve $\sqrt{3x+7}-\sqrt{x+2}=1$. Cannot arrive at solution $x=-2$ I am to solve $\sqrt{3x+7}-\sqrt{x+2}=1$ and the solution is provided as -2.
Since this is a radical equation with 2 radicals, I followed suggested textbook steps of isolating each radical and squaring:
$\sqrt{3x+7}-\sqrt{x+2}=1$
$(3x+... | Let $x+2=y$, then:
$\sqrt {3y+1}=\sqrt y +1$
squaring both sides we get:
$3y+1=y+1+2\sqrt y$
⇒ $y=\sqrt y $ ⇒ $y^2-y=y(y-1)=0$
that gives:
$y=x+2=0$ ⇒ $x=-2$
$y-1=0$ ⇒ $y=x+2=1$ ⇒ $x=-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3273876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 7
} |
Solving a Fractional Equation Involving a Logarithm I may be being stupid right now, so I've come to Stack to see if this elementary algebra holds up.
Suppose I have the equation $$\frac{\ln x}{(1+ \ln x)^2} = \frac{1}{4}$$
My chosen way to solve this would be to cross multiply and expand brackets, solve the quadratic... | $\frac ab = \frac cd$ does not mean $a = c$ and $b = d$.
Here are two examples:
$\frac 13 = \frac 26$ but $1 \ne 2$ and $3\ne 6$. Likewise if $x =2$ we would have $\frac x{x+4} = \frac 13$. That does not mean $x=1$ and $x+4 =3$.
One (not advised) way of doing this is to notice that $\frac ab = \frac {\beta a}{\beta b}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3275438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
IN $ABC$ triangle $AB=28$, angle $C=120$, what is the value of smallest side of the triangle if $AC:BC=3:5$ IN $ABC$ triangle $AB=28$, angle $C=120$, what is the value of smallest side of the triangle if $AC:BC=3:5$
I only got that smallest side should be $3x%$, how do i solve this?
| The law of cosines gives
$$28^2=(3x)^2+(5x)^2-2(3x)(5x)\cos\frac{2\pi}3=34x^2+15x^2=49x^2$$
Thus $28=7x$, $x=4$ and the smallest side of the triangle is $3x=12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3277319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\lim_{n\rightarrow \infty} \left[ \frac{1}{(n+1)(n+2)} + \frac{2}{(n+2)(n+4)} + \cdots + \frac{n}{6n^2} \right]$
Evaluate: $$\lim_{n\rightarrow \infty} \left[ \dfrac{1}{(n+1)(n+2)} + \dfrac{2}{(n+2)(n+4)} + \cdots + \dfrac{n}{6n^2} \right]$$
$\text{My Attempt:}$ breaking down the summation series into: $$\... | After revising my answer, I got: as $n\to \infty$,
$$\begin{align}\sum_{r=1}^{n} \dfrac{r}{(n+r)(n+2r)}&=\frac{1}{n}\sum_{r=1}^{n} \dfrac{\frac{r}{n}}{(1+\frac{r}{n})(1+2\frac{r}{n})}\\
&\to \int_0^1 \frac{x}{(1+x)(1+2x)}\,dx\\
&=\int_0^1\left(\frac{1}{1+x}-\frac{1}{1+2x}\right)\,dx\\
&=\left[\ln\left(\frac{1+x}{(1+2x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Difficult Recurrence Problem Find all integers $n\geq 3$ for which there exist real numbers $a_{1}, a_{2},...,a_{n+2}$ satisfying $a_{n+1}=a_{1}$ , $a_{n+2}=a_{2}$ and:
$a_{i}a_{i+1}+1=a_{i+2}$
for $i=1,2,..,n$
| Hint.
Solving backwards we have for
$$
\begin{cases}
n=2\to \{2,-1,-1,2\}\\
n=5\to \{2,-1,-1,2,\frac 12,-1,2\}\\
n=8\to \{2,-1,-1,2,\frac 12,-1,2,\frac 12,-1,2\}\\
\vdots\\
n=2+3k\to \{2,-1,\underbrace{-1,2,\frac 12}_{k},-1,2\}
\end{cases}
$$
NOTE
The real solutions arise from
$$
\left\{
\begin{array}{rcl}
a_1&=&\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3279057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Area between $x^3$ and $\sqrt[3]{x}$ I'm having some difficult to find the intersection between $x^3$ and $\sqrt[3]{x}$ for calculate the area between them. Could someone help me?
| \begin{equation}
x^3 = x^{\frac{1}{3}} \rightarrow x^9 = x \rightarrow x\left(x^8 - 1\right) = 0
\end{equation}
And so we have $x = 0$ or $x^8 - 1 = 0$. For the later we employ the identity $$a^2 - b^2 = (a + b)(a - b)$$. Thus,
\begin{equation}
x^8 - 1 = 0 \rightarrow (x^4 + 1)(x^4 - 1) = 0
\end{equation}
Assuming you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3279437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solve the equation $|2x^2+x-1|=|x^2+4x+1|$ Find the sum of all the solutions of the equation $|2x^2+x-1|=|x^2+4x+1|$
Though I tried to solve it in desmos.com and getting the requisite answer but while solving it manually it is getting very lengthy.
I tried to construct the two parabola and mirror image the region bel... | $$
|2x^2+x-1|=|x^2+4x+1|\\
(2x^2+x-1)^2=(x^2+4x+1)^2\\
(2x^2+x-1)^2-(x^2+4x+1)^2=0\\
[(2x^2+x-1)+(x^2+4x+1)]\cdot[(2x^2+x-1)-(x^2+4x+1)]=0\\
(3x^2+5x)\cdot(x^2-3x-2)=0\\
x\cdot(3x+5)\cdot(x^2-3x-2)=0\\
Solving\space for\space all\space cases,\space we\space get:\\
x=0\\
x=-\frac{5}{3}\\
x=\frac{3\pm\sqrt{17}}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3279752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Maclaurin series of $\tan(x+x^2)$ to order 3 I know that the maclaurin series of $\tan(x)$ is $\tan(x)=x+\frac{1}{3}x^3+\frac{2}{15}x^5+...$, then shouldn't be $\tan(x+x^2)=(x+x^2)+\frac{1}{3}(x+x^2)^3+\frac{2}{15}(x+x^2)^5+...$?
Mathematica actually gives me $\tan(x+x^2)=x+x^2+\frac{x^3}{3}+o(x^4)$ to order 3.
| Alternatively:
$$\begin{align}y&=\tan{(x+x^2)} \Rightarrow \color{red}{y(0)=0}\\
\arctan y&=x+x^2\\
\frac1{1+y^2}\cdot y'&=1+2x\\
y'&=(1+\color{red}y^2)(1+2x) \Rightarrow \color{blue}{y'(0)=1}\\
y''&=2\color{red}yy'(1+2x)+2(1+\color{red}y^2)\Rightarrow \color{green}{y''(0)=2}\\
y'''&=2\color{blue}{y'}^2(1+2x)+2\color{r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show $\int_0^{\pi/3} \text{tanh}^{-1}(\sin x)\, dx=\frac{2}{3}G$ Discovered the integral below
$$I=\int_0^{\pi/3} \text{tanh}^{-1}(\sin x)\, dx= \frac23G$$
which looks clean, yet challenging. Have not seen it before. Post it here in case anyone is interested.
Edit:
Here is a solution. Let $J(a)=\int_0^{\frac\pi{3}}\tan... | Note that $2\operatorname{arctanh} x=\ln\left(\frac{1+x}{1-x}\right)$. So by letting $\frac{1-\sin x}{1+\sin x}=y\,$ in the integral we get: $$\int_0^\frac{\pi}{3} \operatorname{arctanh}(\sin x)dx=-\int_0^{(2-\sqrt 3)^2} \frac{\ln y}{\sqrt y(1+y)}dy\overset{\sqrt y=\tan x}=-\int_0^\frac{\pi}{12}\ln(\tan x)=\frac23G$$
S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the sum of the infinite series given I've this problem to solve, but no answer key, I tried to do some development, but i didn't succeed :( Could anyone help me, please?
$$\sum_{k=2}^\infty \sum_{l=1}^{k-1} \frac{1}{k^2l^2}=\frac{1}{2^21^2}+\frac{1}{3^21^2}+\frac{1}{3^22^2}+\frac{1}{4^21^2}+\frac{1}{4^22^2}+\frac{... | Denoting the given double sum by $S$, we note that
$$\left(\sum_{n=1}^\infty\frac1{n^2}\right)^2=\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{m^2n^2}=\sum_{n=1}^\infty\frac1{n^4}+2S$$
Thus
$$\left(\frac{\pi^2}6\right)^2=\frac{\pi^4}{90}+2S$$
$$S=\frac{\pi^4}{120}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3283189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve for $(5x-1)(x-3)<0$ The inequality $(5x-1)(x-3)<0$ is true when
$(5x-1)<0$ and $(x-3)>0$
or
$(5x-1)>0$ and $(x-3)<0$.
If I solve for $x $ in the first scenario, $x < \frac{1}{5}$ and $x > 3$ which is wrong. But if I solve for $x$ in the second scenario, $x > \frac{1}{5}$
and $x < 3$ which is correct.
Why is ... | $f(x)=(5x-1)(x-3)$ polynomial of degree 2 has 2 zeroes: $5x-1=0$, and $x=3$.
1) $\lim_{x \rightarrow \pm \infty}f(x)= \infty$.
Hence for
$5x-1<0$, or $x <1/5$: $f(x)>0$, and
$x>3$: $f(x) >0$.
2) For $x \in (1/5,3)$:
$f(x)= 5(x-1/5)(x-3) <0$ , since $(x-1/5)>0$ and $(x-3) <0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3284981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
$A=\frac{1}{1}+\frac{1}{10}+\frac{1}{11}+\dots+\frac{1}{19}+\frac{1}{21}+\frac{1}{31}+\dots+\frac{1}{91}+\frac{1}{100}+\frac{1}{101}+\dots$ Consider the series:
$$A=\frac{1}{1}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\dots+\frac{1}{19}+\frac{1}{21}+\frac{1}{31}+\frac{1}{41}+\dots+\frac{1}{91}+\frac{1}{100}+\frac{1}{101}... | $$A\gt\sum_{k=1}^\infty \frac1{10k+1}\gt\sum_{k=1}^\infty\frac1{11k}=\infty $$
Hence the sum determining $A$ does not converge and $A$ does not exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3285331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Find cumulative distribution function of random variable given probability generating function
Let $f_c(x) := c \cdot \ln\left(1 - \frac{x}{2}\right)$.
Find for what $c \in \mathbb{R}$ $f_c$ is a probability generating function.
I wrote
$$
f_c(x)
= \sum_{k \in \mathbb{N}_0} -c \cdot \frac{x^k}{2^k \cdot k!}.
$$
For... | Using standard probability notation you have probability generating function $G_X(z) = c \cdot \ln(1-\tfrac{z}{2})$ and the corresponding random variable is $X$. From the sums in your question, I will assume that $X$ is a non-negative integer random variable. With this condition, one of the properties of the PGF is t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3287193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
On the determinant of a Toeplitz-Hessenberg matrix I am having trouble proving that
$$\det
\begin{pmatrix}
\dfrac{1}{1!} & 1 & 0 & 0 & \cdots & 0 \\
\dfrac{1}{2!} & \dfrac{1}{1!} & 1 & 0 & \cdots & 0 \\
\dfrac{1}{3!} & \dfrac{1}{2!} & \dfrac{1}{1!} & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \ddots & \v... | HINT.-By property of determinants, we lower the order of n to (n-1) as follows
$$\Delta_n=\det\begin{pmatrix}
1 &1 &0 &0 &\cdots &0\\
a_2 &1 &1 &0 &\cdots &0\\
a_3 &a_2 &1 &1 &\cdots &0\\
\vdots &\vdots &\vdots &\ddots &\ddots &\vdots\\
a_{n-1} &a_{n-2} &a_{n-3} &\cdots &1 &1\\
a_n &a_{n-1} &a_{n-2} &a_{n-3} &\cdots &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3288246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Factorization in a proof of induction I have to prove the following:
$1+3^3+ ... + (2n+1)^3=(n+1)^2(2n^2+4n+1)$ by induction.
My try:
Base case, $n=1$:
$1+3^3=(2)^2(2\cdot1^2 + 4\cdot1 + 1)$, which is true.
By inductive hypothesis, assume $n=k$:
$1 + 3^3 + ... + (2k+1)^3=(k+1)^2 (2k^2 + 4k + 1)$
For $n=k+1$
$1 + 3^3 + ... | This may not be the most elegant solution,
but $(k+1)^2(2k^2+4k+1) + (2k+3)^3= (k+2)^2 ( 2(k+1)^2 + 4(k+1) + 1)$
is true because both sides equal $2k^4+16k^3+47k^2+60k+28$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3289534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving the diophantine equation $x^3+y^3 = z^6+3$ I've the following problem:
Show that the congruence $x^3+y^3 \equiv z^6+3\pmod{7}$ has no solutions. Hence find all integer solutions if any to $x^3+y^3-z^6-3 = 0.$
We can rearrange the first equation to $z^6 \equiv (x^3+y^3-3) \mod{7}$. But $z^6\equiv 1\mod{7}$ so... | ►If $z\equiv0\pmod7$ we have $x^3+y^3\equiv 3\pmod7$ but by Fermat's little theorem $w^6\equiv1\pmod7$ if
$w\not \equiv 0\pmod7$ then $w^3\equiv\pm1\pmod7$ so either $\pm1\pm1\not\equiv3\pmod7$ or $0\pm1\not\equiv\pmod7$.
►If $z\not\equiv0\pmod7$ we have the equation $x^3+y^3\equiv 1+3=4\pmod7$ and similar reasoning ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3292195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Is $100$ the only square number of the form $a^b+b^a$? Conjecture: $100$ is the only square number of the form $a^b+b^a$ for integers $b>a>1$.
In other words, $(a,b)=(2,6)$ is the only solution. Can we prove/disprove this?
Observations:
*
*The solution mentioned should not come as a surprise, since $2^6+6^2=8^2+6^2=1... | $\newcommand{\eps}{\varepsilon}$
$\newcommand{\rad}{\mathrm{rad}}$
At least, under the abc conjecture, there can be only finitely many pairs $(a,b)$ with $b>a>1$ coprime such that $a^b+b^a$ is a square.
As a reminder, the conjecture says that to any $\eps>0$ there corresponds some $K_\eps>0$ such that whenever $u,v$, a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3292683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "36",
"answer_count": 1,
"answer_id": 0
} |
Finding the solution to $xy'' +2y' +xy=0$ around $x_{0}=0$using the method of Frobenius. We know that the solution of this ODE is like:
$$ y=\sum_{n=0}^{\infty}C_nx^{n+r}$$
Them derivative $y$ and $y'$.
$$y'=\sum_{n=0}^{\infty}(n+r)C_nx^{n+r-1}$$
$$y''=\sum_{n=0}^{\infty}(n+r-1)(n+r)C_nx^{n+r-2}$$
Replace $y$ , $y'$ an... | $$xy'' +2y' +xy=0\qquad ......(1)$$
$~x=0~$ is a regular singular point of equation $(1)$.
So the equation admits of a Frobenius series of the form $$y=\sum_{n=0}^{\infty}C_n~x^{n+r},\qquad C_0\neq 0 \qquad ..........(2)$$
which converges for all $~x~$.
From $(2)$,
$$y'(x)=\sum_{n=0}^{\infty}(n+r)C_n~x^{n+r-1};\qquad ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3292893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Value of $\prod_{n>1} \frac{1}{1-\frac{1}{n^s}}$ or $-\sum_{n=2}^{\infty} \log(1-\frac{1}{n^s} )$ I know
\begin{align}
\prod_{p~is~ prime} \frac{1}{1-\frac{1}{p^2}} = \zeta(2) = \frac{\pi^2}{6}
\end{align}
which has a convergent number.
actually I can even generalized this to
\begin{align}
\prod_{p ~is~ prime} \frac... | Let
$$P_s=\prod_{n=2}^\infty \frac{1}{1-n^{-s}}$$
Using a CAS, there are some nice expressions such as
$$\left(
\begin{array}{cc}
s & P_s \\
2 & 2 \\
3 & 3 \pi \text{sech}\left(\frac{\sqrt{3} \pi }{2}\right) \\
4 & 4 \pi \text{csch}(\pi ) \\
6 & 6 \pi ^2 \text{sech}^2\left(\frac{\sqrt{3} \pi }{2}\right)
\end{ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3294574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$\sin(x) - \sin(y) = -\frac{1}{3}$, $\cos(x) - \cos(y) = \frac{1}{2}$, what is $\sin(x+y)$? If $\sin(x) - \sin(y) = -\frac{1}{3}$ and $\cos(x) - \cos(y) = \frac{1}{2}$, then what is $\sin(x+y)$?
Attempt:
$$ \sin(x+y) = \sin(x) \cos(y) + \cos(x) \sin(y) $$
If we multiply the two "substraction identities" we get
$$ (\si... | If you start with the Sum-to-Product Identities, you can rewrite the given equations as
$$2\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)=-\frac{1}{3} \quad \text{and} \quad -2\sin\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)=\frac{1}{2}.$$
Dividing the second equation by the first one, we'll c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3296374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find the constant $k$ from the determinant
Given: $$\begin{vmatrix}(b+c)^2 &a^2&a^2\\b^2 &(c+a)^2&b^2 \\c^2&c^2& (a+b)^2\end{vmatrix}=k(abc)(a+b+c)^3$$ Find $k$.
If I directly open the determinant it will go to long I can't apply most of the row or column operation as they keep making it more complex.
| It is also simple to compute the posted determinant, $D$ for short, thus ignoring the information on the special form of the result. We are subtracting at the first step the first column from the other two.
$$
\begin{aligned}
D &=
\begin{vmatrix}
(b+c)^2 & a^2 & a^2 \\
b^2 & (c+a)^2 & b^2 \\
c^2 & c^2 & (a+b)^2
\end{vm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3297816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding the number of solutions to $\cos^4(2x)+2\sin^2(2x)=17(1+\sin 2x)^4$ for $x\in(0,2\pi)$
Number of solution of the equation
$\cos^4(2x)+2\sin^2(2x)=17(1+\sin 2x)^4\; \forall $ $x\in(0,2\pi)$
what i try
$\cos^4(2x)+2\sin^2 2x=17(1+\sin^2(2x)+2\sin 2x)^2$
$1+\sin^4 (2x)=17(1+\sin^4 2x+2\sin^2 2x+4\sin^24x+4\sin 2... | Wrong expansion jacky
$$16\sin^42x+68\sin^32x+102\sin^22x+68\sin 2x+16=0$$
$$16(\sin 2x+\csc 2x)^2+68(\sin 2x+\csc 2x)+70=0$$
$$\sin 2x +\csc 2x=-15.0479,-1.9521$$
Rejecting $-1.9521$ (Why)
Thus, $$\sin^22x+(15.0479)\sin 2x+1=0$$
Both the roots of the quadratic are negative and since their product is $1$ only one root ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3300465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to solve $\int \frac{2x^2-4x+3}{(x-1)^2}\, dx$? My book says the answer to $\int \frac{2x^2-4x+3}{(x-1)^2} \, dx$ is $2x-\frac{1}{x-1}+C$ but symbolab says it is $2x-\frac{2}{x-1}+\frac{4}{x-1}-\frac{3}{x-1}-2+C$. Who is correct and how would I get to the answer? I tried to use u-substitution but that doesn't work ... | Note that
$$2x \color{red}{-\frac{2}{x-1}+\frac{4}{x-1}-\frac{3}{x-1}} \color{green}{-2+K} = 2x \color{red}{-\frac{1}{x-1}} \color{green}{+C}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3301198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Integration of $x^2\cdot\frac{x\sec^2x+\tan x}{(x\tan x+1)^2}$ Integrate
$$\int x^2\cdot\dfrac{x\sec^2x+\tan x}{(x\tan x+1)^2}dx$$
So what is did is integration by parts taking $x^2$ as $u$ and the other part as $v$ . Now I got to use it again which then eventually leads to (integral of $\dfrac1{x\tan x+1}dx $). Can s... | Let $u:=x\sin x+\cos x$ so $u^\prime=x\cos x$. You want to integrate$$\begin{align}\frac{x^2(x+\sin x\cos x)}{u^2}&=\frac{x^2(x\cos^2x+u\sin x)}{u^2}\\&=\frac{u^\prime x^2\cos x}{u^2}+\frac{x^2\sin x}{u}\\&=\frac{d}{dx}\left(-\frac{x^2\cos x}{u}\right)+\frac{(x^2\cos x)^\prime+x^2\sin x}{u}\\&=\frac{d}{dx}\left(-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3301557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to show that $\sqrt{2+\sqrt{3}} = \dfrac{\sqrt{6}+\sqrt{2}}{2}$
$\sqrt{2+\sqrt{3}} = \dfrac{\sqrt{6}+\sqrt{2}}{2}$
How to change $\sqrt{2+\sqrt{3}}$ into $\dfrac{\sqrt{6}+\sqrt{2}}{2}$
| Assume $$\sqrt{a+\sqrt b}=\sqrt c+\sqrt d$$
where $a,b,c,d$ are all rational.
Then
$$a+\sqrt b=c+2\sqrt{cd}+d.$$
By identification
$$\begin{cases}a=c+d,\\b=4cd.\end{cases}$$
From this,
$$4c^2+4cd-4ac=4c^2-4ac+b=0.$$
This is a quadratic equation in $c$, with roots
$$c=\frac{a\pm\sqrt{a^2-b}}2.$$
So for a rational soluti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3303949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Proof Correction: $\cos ' (x) = -2 \sin (x)$ Preliminary Information
Let $$A(y) = \cfrac{y \sqrt{1 - y^2}}{2} + \int_{y}^{1} 1\sqrt{1 - t^2} dt \ \ \ \text{on $[-1, 1]$}$$
Moreover $$ A'(y) = \dfrac{-1}{2\sqrt{1 - y^2}}$$
Define $\cos x$ as $A(\cos x) = \cfrac{x}{2}$ and $\sin x = \sqrt{1 - \cos ^{2} x}$
Problem: Find... | You forgot to apply the chain rule and multiply by $(x/2)'$ when differentiating $(A^{-1})(x/2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3304965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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The infinite sum $\sum_{n=1}^\infty (-1)^{n+1} \frac{2n-1}{n^2-n+1}$ My (rather old) version of Mathematica cannot compute
$$\sum_{n=0}^\infty (-1)^{n+1}\frac{2n-1}{n^2-n+1}$$
other than re-writing it as a hypergeometric function as follows:
$$\mbox{HypergeometricPFQ}\left[\left\{\frac{1}{2}, 1, -(-1)^{1/3}, (-1)^{2/... | Using partial fraction decomposition
$$\frac{2 n-1}{(n-a) (n-b)}=\frac{2 a-1}{(a-b) (n-a)}+\frac{1-2 b}{(a-b) (n-b)}$$ So
$$\sum_{n=1}^\infty (-1)^{n+1}\frac{2 a-1}{(a-b) (n-a)}=\frac{(2 a-1) \Phi (-1,1,1-a)}{a-b}$$
$$\sum_{n=1}^\infty (-1)^{n+1}\frac{1-2 b}{(a-b) (n-b)}=-\frac{(2 b-1) \Phi (-1,1,1-b)}{a-b}$$ where app... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Number of ways to chose 6 courses out of 15 (different) courses We got 3 English courses, 6 Chinese courses, 6 Spanish courses.
Each course is different.
In how many was can we choose 6 courses such that we must chose atleast 1 course from each topic (English, Chinese, Spanish)
If possible, I would like to see a soluti... | The generating function for the number of $k$ courses such that there at least one course from each category is the following:
$$\Bigg( \binom{3}{1}x + \binom{3}{2}x^2 + \binom{3}{3}x^3 \Bigg) \Bigg(\binom{6}{1}x + \binom{6}{2}x^2 + \binom{6}{3}x^3 + \binom{6}{4}x^4 +\binom{6}{5}x^5 + \binom{6}{6}x^6 \Bigg)^2$$
Thus th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Finding product of trigonometric values (cos/sin) using complex number/roots of unity Using the ninth root of unity of 1, show that $cos\frac{π}{9}cos\frac{2π}{9}cos\frac{4π}{9}=\frac{1}{8}$.
Here is my solution.
Let $\omega=cos\theta+isin\theta$,
As $\omega^9=1$,
$$cos9\theta+isin9\theta=1$$
so $$9\theta=2nπ ... | Root of Unity Approach
Using
$$
\omega=\cos\left(\frac\pi9\right)+i\sin\left(\frac\pi9\right)
$$
we get $\omega^9=-1$
$$
\begin{align}
\overbrace{\left(\omega-\omega^8\right)}^{2\cos\left(\frac\pi9\right)}\overbrace{\left(\omega^2-\omega^7\right)}^{2\cos\left(\frac{2\pi}9\right)}\overbrace{\left(\omega^4-\omega^5\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3308303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
what is the value of $9^{ \frac{a}{b} } + 16^{ \frac{b}{a} }$ if $3^{a} = 4^{b}$ I have tried to solve this expression but I'm stuck:
$$9^{ \frac{a}{b} } + 16^{ \frac{b}{a} } = 3^{ \frac{2a}{b} } + 4^{ \frac{2b}{a} }$$
and since $3^{a} = 4^{b}$:
$$3^{ \frac{2a}{b} } + 4^{ \frac{2b}{a} } = 4^{ \frac{2b}{b} } + 3^{ \frac... | You were close. However, you simplified $\frac{2b}b$ into $b$ and $\frac{2a}a$ into $a$ in your final step, which is not right. Correct that mistake, and you should have your solution.
On a less serious note, I would do an extra step here, for clarity:
$$
3^{2a/b} + 4^{2b/a} = (3^a)^{2/b} + (4^b)^{2/a} = 4^{2a/a} + 3^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3309188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
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Prove $ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le \frac{9}{8}n^4$
Prove that for all $n \in \mathbb{N}$ the inequality $$ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le \frac{9}{8}n^... | Here is a way, which is motivated by the excellent approach of @MichaelRozenberg's. Let us write $\displaystyle a_k = \frac { k}{k+1}, b_k = 2k-1$. We then have
\begin{align*}
S =& \sum_{i\le n} a_i b_i\sum_{j\le n}\frac{b_j}{a_j} \\
=& \sum_{i,j\le n} \frac{a_i}{a_j}b_ib_j \\
=& \frac 1 2 \sum_{i,j\le n} \left(\frac {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
The natural numbers $a, b, c$, formed by the same $n$ digits $x$, $n$ digits $y$, and $2n$ digits $z$ satisfy $a^2 + b = c$
Given that the natural numbers $a, b, c$ are formed by the same $n$
digits $x$, $n$ digits $y$, and $2n$ digits $z$ respectively. For any
$n \geq 2$, find the digits $x, y, z$ such that $a^2 ... | Let
$$f(n) = \sum_{i=0}^{n-1} 10^i \tag{1}\label{eq1}$$
Thus, as you've shown,
$$x^2 f^2(n) + yf(n) = zf(2n) \implies f(n)\left(x^2 f(n) + y\right) = zf(2n) \tag{2}\label{eq2}$$
Also, note that
$$f(2n) = 10^n f(n) + f(n) = f(n)\left(10^n + 1\right) \tag{3}\label{eq3}$$
Thus, substituting that into \eqref{eq2} and divid... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find the maximum possible perimeter of a right triangle
The ratio between the perimeter of a right triangle and its area is 2:3. The sides of the triangle are integers. Find the maximum possible perimeter of the triangle.
If the sides of the triangle are $A$, $B$ and $C$ (the hypotenuse), I have deduced that:
$$A+B+... | I worked out the opposite, i.e. area:perimeter ratio. Michael Rozenberg's answer is correct.
$$ R=\frac{area}{perimeter}=\frac{AB}{2}/P=\frac{2mk(m^2-k^2)}{2(2m^2+2mk)}=\frac{mk-k^2}{2}
\qquad\qquad\qquad\qquad\qquad $$
\begin{equation}
R=\frac{mk-k^2}{2}\quad\implies k=\frac{m\pm\sqrt{m^2-8R}}{2}\quad\text{for}\quad \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3313890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Chord $x+y=1$ Subtends $45^{\circ}$ at the centre If the Chord $x+y=1$ Subtends $45^{\circ}$ at the centre of the circle $x^2+y^2=a^2$, Find $a$
Method $1$:
We have radius=$a$
If $O$ is the centre and chord be $AB$ we have
$$\angle AOB =45$$
If $OM$ is drawn perpendicular to the chord we have:
$$\angle AOM=22.5$$
In $\... | It's not that you couldn't get the second answer of the second method $\sqrt{2+\sqrt2}$ in the first method. It's that $\sqrt{2+\sqrt2}$ is wrong – drawing the associated circle and line and measuring the subtended angle gives you well over $90^\circ$. There is only one (positive) solution $a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3316133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Maximizing $\frac{a^2+6b+1}{a^2+a}$, where $a=p+q+r=pqr$ and $ab=pq+qr+rp$ for positive reals $p$, $q$, $r$
Given $a$, $b$, $p$, $q$, $r \in\mathbb{R_{>0}}$ s.t.
$$\begin {cases}\phantom{b}a=p+q+r=pqr \\ab =pq+qr+rp\end{cases} $$
Find the maximum of $$\dfrac{a^2+6b+1}{a^2+a}$$
This question is terrifying that I ... | Recently, I find a solution which is quite simple. Check it out!
Claim $1$: $a^2\ge3ab\rightarrow a\ge3b, b\le\dfrac{a}{3}$
Prove:
$$a^2-3ab=\left(x+y+z\right)^2-3\left(xy+yz+zx\right)=x^2+y^2+z^2-xy-yz-zx=\dfrac{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{2}\ge0 \\ \therefore a^2\ge 3ab$$
Claim $2$: $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3319843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Simplify $ \frac{ \sqrt[3]{16} - 1}{ \sqrt[3]{27} + \sqrt[3]{4} + \sqrt[3]{2}} $ Simplify
$$ \frac{ \sqrt[3]{16} - 1}{ \sqrt[3]{27} + \sqrt[3]{4} + \sqrt[3]{2}} $$
Attempt:
$$ \frac{ \sqrt[3]{16} - 1}{3 + \sqrt[3]{4} + \sqrt[3]{2}} = \frac{ \sqrt[3]{16} - 1}{ (3 + \sqrt[3]{4}) + \sqrt[3]{2}} \times \frac{ (3 + \sqrt[... | With the same notation as in the other answer, i.e. $x = \sqrt[3] 2$, noting that $x^3+1=3,$ you can write your quantity as
\begin{eqnarray}
\frac{x^4-1}{x^2+x+3} &=& \frac{(x-1)(x^3+x^2+x+1)}{x^2+x+x^3+1}=\\
&=&x-1.
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3321716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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What is longer? Length of $2^n$ or number of $0$ in $n!$? What is longer? Length of $2^n$ or number of $0$ at the end of $n!$ ?
Assume that $n$ is really big, big number...
Solution
Look at terms of $2^k$:
$$2,4,8,16,32,64,128,256,512,1024,2048,4096... $$
It can be seen that
$$ \lfloor\frac{n}{3} \rfloor \ge length(2^... | Denote by $l(n)$ the number of digits of $2^n$ and by $t(n)$ the number of trailing zeros of $n!$. We aim to show that $t(n) < l(n)$, at least for large $n$.
Per the update, the number of trailing zeros in $n!$ is the number of factors of $5$ in the prime factorization of $n!$, and it follows from the definition of fac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3324023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Explicit description of quotient ring of $\mathbb{Z}[x]$ I am studying for a prelim and I stumbled on this problem:
Describe explicitly the elements in the quotient ring $\dfrac{\mathbb{Z}[x]}{(3,x^3-x+1)}$. First of all I don't see why the ideal $(3,x^3-x+1)$ is a maximal ideal in $\mathbb{Z}[x]$. If there is anyone w... | Let $m$ be a maximal ideal of $\mathbb Z[x]$ with $(3,x^3-x+1)\subseteq m$. Assume tha $(3,x^3-x+1)\not=m$ and let $f\in m\setminus(3,x^3-x+1)\subseteq m$. Since $x^3-x+1$ is monic we have $f=g(x^3-x+1)+h$, where $g,h\in \mathbb Z[x]$ and $1\leqslant deg(h)\leqslant 2$. Now since $x^3-x+1\in m$, we have $h\in m$. We co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3325549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Problem with $x^{6} - 2 = 0$ compute roots in $\mathbb{C}$
I have problem with simple equation $x^{6} - 2 = $ compute roots in
$\mathbb{C}$
I will try compute roots of $x^{6} - 2 = (x^{3}-\sqrt{2})(x^{3}+\sqrt{2})=(x-2^{1/6})(x^{2}+2^{1/6}x+2^{1/3})(x^{3}+\sqrt{2})$, but this not looks good.
Maybe is better soluti... | With $\sqrt[3]2z=x^2$, we have a cubic equation
$$z^3-1=0$$ which we can factor as (using a remarkable product)
$$(z-1)(z^2+z+1)=0.$$
Besides the obvious root $z=1$, we also have
$$z=\frac{-1\pm i\sqrt 3}{2}.$$
Finally, the six roots are
$$x=\pm\sqrt[6]2,\pm\sqrt[6]2\sqrt{\frac{-1\pm i\sqrt 3}{2}}.$$
Note that $(1\pm i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3327355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 6
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Limit of a function in which square roots are involved $$\lim_{x \to 0}\frac{x+2-\sqrt{2x+4}}{3x-1+\sqrt{x+1}}$$
Could someone please help me solve this problem.
I tried multiplying by a unity factor but I end up stuck.
| $$\lim_{x \to 0}\frac{x+2-\sqrt{2x+4}}{3x-1+\sqrt{x+1}}=\lim_{x\rightarrow0}\frac{((x+2)^2-(2x+4))(3x-1-\sqrt{x+1})}{((3x-1)^2-(x+1))(x+2+\sqrt{2x+4})}=$$
$$=\lim_{x\rightarrow0}\frac{(x^2+2x)(3x-1-\sqrt{x+1})}{(9x^2-7x)(x+2+\sqrt{2x+4})}=\lim_{x\rightarrow0}\frac{(x+2)(3x-1-\sqrt{x+1})}{(9x-7)(x+2+\sqrt{2x+4})}=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3328377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Use properties of determinant and show
Let $n$ be a positive integer and
\begin{align} M =
\begin{pmatrix}
n! & (n+1)! & (n+2)! \\
(n+1)! &(n+2)! & (n+3)! \\
(n+2)! & (n+3)! & (n+4)! \\
\end{pmatrix}
\end{align}
Use properties of determinant to show that \begin{align}\left(\frac{|M|}{(n!)^3}- 4\ri... | So, they're looking for the determinant of $M$. We can make that easier to calculate by factoring a common term of $n!$ out of each row, giving us a new matrix:
$$N=\left[\begin{array}{ccc}
1 & n+1 & (n+1)(n+2)\\
n+1 & (n+1)(n+2) & \small{(n+1)(n+2)(n+3)}\\
(n+1)(n+2) & \small{(n+1)(n+2)(n+3)} & \scriptsize{(n+1)(n+2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3328655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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For how many pairs of positive integers n and m is the statement $mn - 8m + 6n =0$ true?
Each interior angle of a regular polygon with $n$ sides is $\frac{3}{4}$
of each interior angle of a second regular polygon with $m$ sides.
How many pairs of positive integers $n$ and $m$ are there for which this
statement is ... | $$mn-8m+6n=0\implies n(m+6)=8m\implies n=\frac{8m}{m+6}$$
Testing this formula from $1\le m\le 1000$, the only integer solutions found were
$$\frac{2\times8}{2+6}=2
\quad\frac{6\times8}{6+6}=4
\quad\frac{10\times8}{10+6}=5
\quad\frac{18\times8}{18+6}=6
\quad\frac{42\times8}{42+6}=7$$
If we solve the other way around, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3329331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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How can I solve prove that $8(1-a)(1-b)(1-c)\le abc$ with the conditions below? There was a homework about inequalities (that why I ask a bunch of inequality problems). But I couldn't solve the following:
If $0<a,b,c<1$ and $a+b+c=2$, prove that $8(1-a)(1-b)(1-c)\le abc$
I tried many times, and finally I used Muirhea... | Denote: $1-a=x,1-b=y,1-c=z$. Then:
$$x+y+z=1 \stackrel{AM-GM}{\Rightarrow} \color{red}1=x+y+z\color{red}{\ge 3\sqrt[3]{xyz}}$$
Hence:
$$8(1-a)(1-b)(1-c)\le abc \Rightarrow 8xyz\le (1-x)(1-y)(1-z)\iff \\
8xyz\le 1-(x+y+z)+(xy+yz+zx)-xyz \iff \\
9xyz\le xy+yz+zx\iff \\
9xyz\le 3\sqrt[3]{(xyz)^2} \stackrel{AM-GM}{\le} xy+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3329546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Find the minimum $n$ such that $x^2+7=\sum_{k=1}^n f_k(x)^2$ where $f_k(x)\in \mathbb{Q}[x]$ Recently, I have found this problem:
Find the minimum $n \in N$ such that $x^2+7=f_1(x)^2+f_2(x)^2+\cdots+f_n(x)^2$ where $f_1(x),+f_2(x),+\cdots+f_n(x)$ are polynomials with rational coefficients.
I have tried to solve this... | You cannot go on this way, because even if you take a function negative, its square will be positive. Observe that degree of all of the polynomials is less than 2, because degree of LHS and RHS should match. It is easy to see that $n=1$ is not a solution. For $n=2$, we observe,
$$(ax+b)^2+(cx+d)^2=x^2+7$$
$$\Longrighta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3330146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Finding all real solutions of $x-8\sqrt{x}+7=0$ Finding all real solutions of $x-8\sqrt{x}+7=0$.
Man, I tried subtituting $x=y^2$ but IDK things got complicated. What is the best way to figure this out? Thanks!
| Isolate the square root, then square both sides:
\begin{align*}
x-8\sqrt x+7&=0\\
\implies x+7&=8\sqrt x\\
\implies (x+7)^2 &= 64 x\\
\implies x^2 -50x + 49&=0\\
\implies (x-49)(x-1) &=0\\
\implies x = 49 \text{ or } x&=1,
\end{align*}
which are the two solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3330413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 1
} |
If $x$ and $y$ are real numbers such that $4x^2 + y^2 = 4x - 2y + 7$, find the maximum value of $5x+6y$. If $x$ and $y$ are real numbers such that $4x^2 + y^2 = 4x - 2y + 7$, find the maximum value of $5x+6y$.
I did a little bit of manipulation and got $4(x+1)(x-2) + (y+1)^2=0$. I then got $x=2$ and $y=-1$ which means ... | $$4x^2-4x + 1+y^2 + 2y +1 =9\implies (2x-1)^2+(y+1)^2=9$$
So a pair $(x,y)$ is on elipse: $${(x-{1\over 2})^2\over {9\over 4}}+{(y+1)^2\over 9}=1$$ so you can write $x= {3\over 2}\cos t +{1\over 2}$ and $y= 3\sin t -1$
So your expresion is now $${15\over 2}\cos t +18\sin t -{7\over 2}$$
This you can write as $${39\ov... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3333007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Solve the equation: $y^3 + 3y^2 + 3y = x^3 + 5x^2 - 19x + 20$ A question asks
Solve the equation: $y^3 + 3y^2 + 3y = x^3 + 5x^2 - 19x + 20$ for positive integers $x$ and $y$.
I tried factoring the LHS by adding $1$ to both sides so we get $(y+1)^3$ in the LHS. But I couldn't get any factorisation for the RHS, neither... | You can rewrite it as $$(y+1)^3 = x^3 + 5x^2 - 19x + 21$$
Now $y+1$ and $x$ differ for some integer, say $z$, so $(y+1) - x=z$, so we have $$(x+z)^3 = x^3 + 5x^2 - 19x + 21$$ or $$ x^2(3z-5)+x(3z^2+19)+(z^3-21)=0$$ so we have a quadratic equation on $x$ with an integer parameter $z$: It will have an integer solution i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3333099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
For which integers $m$ is there a solution to $3^y m - 2^x = 1$? This self-answered question is motivated by a recent question asking whether for every $m$ there is a solution $(x, y)$ of integers to the equation $$3^y m - 2^x = 1$$ for every odd positive integer $m$. As the answers there show there is not---the easies... | Reducing modulo $3^y$ and rearranging leaves $$2^x = -1 \pmod {3^y} .$$ On the other hand, since $2$ is a primitive root modulo $3^1, 3^2$, it is a primitive root modulo $3^y$ for all $y \geq 1$. So, since $(\Bbb Z / (3^y \Bbb Z))^\times$ has order $\phi(3^y) = 2 \cdot 3^{y - 1}$, for fixed $y$ the $x$ satisfying the a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3336525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Impossible hyperbolic integral I'm quite good with integral calculation, very rarely happens that I'm not able to solve an indefinite integral. I've tried for 7 days to solve this monster but in the end, I surrendered to this beast.
I'm sure that it has a solution because it was a challenge from my calculus professor... | As suggested in comments, let $x=2 \cosh ^{-1}(t)$ to end with
$$I=-2\int\frac{
\left(\sqrt[3]{\frac{t+1}{t+2}}-\sqrt[6]{\frac{t+1}{t+2}}+1\right)}{\sqrt[6]{\frac{t+1}{t+2}}-1}\,dt$$
Now
$$\sqrt[6]{\frac{t+1}{t+2}}=y \implies t=\frac{1-2 y^6}{y^6-1}$$ makes
$$I=-12 \int\frac{ y^5 (y^2-y+1)}{(y-1) \left(y^6-1\right)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3336771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
What does it mean to take the ratio of two equations?
The line joining the origin and the point of intersection of the curves $ax^2+2hxy+by^2+2gx=0$ and $a_1x^2+2h_1xy+b_1y^2+2g_1x=0$ will be mutally perpendicular if $g(a_1+b_1)=g_1(a+b)$
This is solved in my reference as
$$
ax^2+2hxy+by^2=-2gx\\
a_1x^2+2h_1xy+b_1y^2... | The condition to be deduced involves the ratio $g/g_1.$ The equations each have the same multiple of $g$ and $g_1$ on their RHSs. Thus, taking their ratio begins to get one in the direction wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3338899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that summation is less than 1 Prove that $m! \times \sum_{n=m+1}^\infty \frac{1}{n!} < 1$
I started proving this by simplifying it to:
$= m! \times \left(\frac{1}{(m+1)!} + \frac{1}{(m+2)!} + ...\right)$
$= \left(\frac{1}{(m+1)} + \frac{1}{(m+2)(m+1)} + ...\right)$
I'm not too sure how to show that the last step ... | If $m\ge 1$
$m!(\frac{1}{(m+1)!}+\frac{1}{(m+2)!}+\frac{1}{(m+3)!}+...)=\frac{1}{m+1}+\frac{1}{(m+2)(m+1)}+\frac{1}{(m+3)(m+2)(m+1)}+...\le \frac{1}{2}+\frac{1}{2^2}+...=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3339706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Inverse of a skew-symmetric matrix For $a, x, y, z \in \mathbb R$, let $$M= \left(
\begin{array}{cccc}
\cos(a) & \sin(a) \, x & \sin(a)\, y & \sin(a) \, z \\
-\sin(a) \, x & \cos(a) & \sin(a) \,z & -\sin(a)\, y \\
-\sin(a) \, y & -\sin(a) \, z & \cos(a) & \sin(a) \, x \\
-\sin(a) \, z & \sin(a) \, y & ... | $A$ has the form $A = \begin{pmatrix}J&L\\-L&J\end{pmatrix}$, where $JL+LJ = 0$, $J^2 = -x^2I$, and $L^2 = (y^2+z^2)I$. Therefore,
$$
A^2 = \begin{pmatrix}J&L\\-L&J\end{pmatrix}\begin{pmatrix}J&L\\-L&J\end{pmatrix} = \begin{pmatrix}J^2-L^2&0\\0&J^2-L^2\end{pmatrix} = -r^2I_4,
$$
where $r := \sqrt{x^2+y^2+z^2}$. Hence, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3340067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
proving that $a + b \sqrt {2} + c \sqrt{3} + d \sqrt{6} $ is a subfield of $\mathbb{R}$ The question is given below:
My questions are:
1- How can I find the general form of the multiplicative inverse of each element?
2-How can I find the multiplicative identity?
3-Is the only difference between the field and the sub... | Let $F$ be a sub-field of the real numbers $\Bbb R$. If for some integer $k \gt 1$
$\quad \sqrt k \notin F$
then using elementary algebra it can be shown that the set
$\tag 1 \{ s_1 + s_2 \sqrt {k} \;\; | \, s_1,s_2 \in F\}$
is a field and that
$$\tag 2 ( s_1 + s_2 \sqrt {k})^{-1} = \frac{s_1 - s_2 \sqrt k}{s_1^2 - ks_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3340541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Stem and leaf diagrams I have the following data:
$2.6$ $ $ $3.3$ $ $ $2.4$ $ $ $1.1$ $ $ $0.8$ $ $ $3.5$ $ $ $3.9$ $ $ $1.6$ $ $ $2.8$ $ $ $2.6$ $ $ $3.4$ $ $ $4.1$ $ $ $2.0$ $ $ $1.7$ $ $ $2.9$ $ $ $1.9$ $ $ $2.9$ $ $ $2.5$ $ $ $4.5$ $ $ $5.0$
Built stem and leaf plot:
$Stem$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $Leaf$ $ ... | The stem-and-leaf plot is equivalent to the following frequency and relative frequency table:
$$\begin{array}{c|l|c|c|с}
Stem&Leaf&f&&X&f&P(X)\\
\hline
0&8&1&&[0,1) \text{ or $0\le X<1$}&\color{red}1&\color{red}1/\color{blue}{20} \\
1&1679&4&&[1,2) \text{ or $1\le X<2$}&\color{red}4&\color{red}4/\color{blue}{20}\\
2&04... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3340924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Determining convergence of $\sum_{n=1}^\infty \frac{3n i^n}{(n+2i)^3}$ I've tried to solve this problem about convergence:
$\sum_{n=1}^\infty \frac{3n i^n}{(n+2i)^3}$
it's supposed to be solved using ratio, root tests or by testing the limit of the sumand. Anyways, I've tried both 3 and I had no success:
I get to a poi... | I just learned something new, I'd like to post it to ratify if it's true:
$$\sum \frac{3n i^n}{(n+2i)^3} = \sum 3\frac{n i^n (n-2i)^3}{(n+2i)^3(n-2i)^3} = 3\sum \frac{ni^n (n-2i)^3}{(n^2+4)^3} $$
Expanding terms:
= $$3\sum\frac{ni^n(n^3-4n^2i-4n-2n^2i-8n+8i}{(n^2+4)^3} = 3\sum\frac{n^4 i^n-6n^3 i^{n+1} -12n^2 i^n+8n i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3341578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$m$ and $n$ are odd numbers such that $|m^2 - n^2 + 1| \mid (n^2 - 1)$. Prove that $|m^2 - n^2 + 1|$ is a square number.
$m$ and $n$ are odd numbers such that $$\large |m^2 - n^2 + 1| \mid (n^2 - 1)$$. Prove that $|m^2 - n^2 + 1|$ is a square number.
I have provided the solution below, but I am uncertain about the st... | Let $p = \dfrac{n^2 - 1}{m^2 - n^2 + 1} + 1 = \dfrac{m^2}{m^2 - n^2 + 1} \implies (m^2 - n^2 + 1)(p - 1) = (n^2 - 1)$
We need to prove that $p$ is a perfect square.
Let $S$ be a set of integers pairs $(x, y)$ in that order satisfying $$(4xy + 1)p = (x + y)^2 \iff \frac{m^2}{m^2 - n^2 + 1} = \frac{(x + y)^2}{4xy + 1}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3343131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is $x+y\alpha + z\alpha^2$ a subfield? Let $\alpha = \sqrt[3]{2}$. I want to Prove that the set of all numbers $x+y\alpha+z\alpha^2$, for $x,y,z \in \mathbb{Q}$, is a subfield of $\mathbb{C}$.
I have showed that this set is a subring, but I'm not sure how to show for any $a\neq 0$ in our set has a multiplicative inver... | Recall the algebraic identity
$$x^3 + y^3 + z^3 -3xyz = (x+y+z)(x^2 + y^2+z^2 - xy -yz - zx)\\
\implies
\frac{1}{x+y+z} = \frac{x^2+y^2+z^2 - xy-yz-zx}{x^3+y^3+z^3 - 3xyz}
$$
Substitute $(x,y,z)$ by $(x,y\alpha,z\alpha^2)$, one get
$$\frac{1}{x + y\alpha + z\alpha^2}
= \frac{(x^2-2yz) + (2z^2-xy)\alpha + (y^2-zx)\alph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3343416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Trouble proving $3^2 + 3^3 + ... 3^n = 9 \cdot \frac{3^{n-1} - 1}2$ by induction So i'm supposed to prove by mathematical induction that this formula: $3^2 + 3^3 + ... 3^n = 9 \cdot \dfrac{3^{n-1} - 1}2$ holds true for all numbers greater than 2.
I started with the base case and just plugged in 2, it worked.
Then I ass... | Why did you write it as
$9(3^{k-1} -1/ 2) + 3^{k+1} = 9((3^k - 1)/2)$
Rather than $9((3^{k-1} -1)/ 2) + 3^{k+1} = 9((3^k - 1)/2)$?
Anyway
$9\frac {3^{k-1} - 1}2 + 3^{k+1}=$
$9\frac {3^{k-1} - 1}2 + \frac {2*3^{k+1}}2=$
$9\frac {3^{k-1} - 1}2 + \frac {2*9*3^{k-1}}2 = $
$9\frac {3^{k-1} -1 + 2*3^{k-1}}2=$
$9\frac {3*3^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3343635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Hyperplane conversion between vector form and cartesian form Vector form to Cartesian form
In $\mathbb{R^2}$ a $\mathbb{R^1}$ object can be expressed as
$$\text{vector form: }s\begin{pmatrix}a\\b\end{pmatrix}+\begin{pmatrix}c\\d\end{pmatrix}$$
$$\text{cartesian form: }\begin{vmatrix}a&b\\x-c&y-d\end{vmatrix}=0$$
In $\m... | Note that the vector form and the cartesian form are equivalent only if the $n-1$ vectors $(a_1\;\ldots\;a_n)^T \;\ldots\;(c_1\;\ldots\;c_n)^T$ are linearly independent. Otherwise, the vector form would produce something of degree less than $n-1,$ while the cartesian form would produce the complete $\mathbb{R}^n.$ So w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3346894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
The $n^\text{th}$ term in the sequence $\frac{1}{1},\frac{1}{2},\frac{2}{1},\frac{1}{3},\frac{2}{2},\frac{3}{1},\frac{1}{4},\frac{2}{3},\dots$ $\begin{array} {|r|r|}\hline R\text{ \ }C & 1 & 2 & 3 & \dots \\ \hline 1 & \frac{1}{1} & \frac{1}{2} & \frac{1}{3} & \\ \hline 2 & \frac{2}{1} & \frac{2}{2} & \frac{2}{3} & \... | § 1 Frame of reference
It is useful to adopt a two index identification of the fractions.
Decompose the fractions into groups according to a scheme of numerators and denominators as shown for the first 4 groups below
$$(1)(1 2)(1 2 3)(1 2 3 4)...$$
$$(1)(2 1)(3 2 1)(4 3 2 1)...$$
The position $p_s(k)$ of the start of t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3347031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the probability of getting the second ace as the $n^{th}$ card is picked from $52$ cards.
Cards are drawn one by one at random from a well shuffled full pack of $52$ cards. Find the probability of the second ace being the $n^{th}$ card.
For this what I did was, for the $n-1^{th}$ positions, we need to make a se... | We must multiply the probability that exactly one ace is selected among the first $n - 1$ cards by the probability that the $n$th card selected is an ace.
We can select $n - 1$ of the $52$ cards in the deck in $\binom{52}{n - 1}$ ways. If these cards include exactly one ace, we must select one of the four aces and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3347638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Divisiblity proof with coprimes of 6 Can someone please tell me if this proof is okay? Define "okay" as acceptable and rigorous. I wanted to avoid being to wordy so I did not prove a few trivial aspects. I feel the proof is quite verbose but I think as concise as I could get it without skipping too many details.
Suppos... | May I suggest a shorter proof? (With $a \equiv_n \pm 1$ is meant: $a \equiv 1 \mod n$ or $a \equiv -1 \mod n$ etc.)
$$3\not | a,b \Rightarrow a\equiv_3 \pm 1\mbox{ and } b\equiv_3 \pm 1 \Rightarrow a^2\equiv_3 b^2 \equiv_3 1 \Rightarrow a^2-b^2 \equiv_3 0$$
Hence, $\boxed{3|a^2-b^2}$
$$2 \not | a,b \Rightarrow a\equiv_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3347933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is this a family of similar matrices $\left(\begin{smallmatrix} 0&x\\ 0&0 \end{smallmatrix}\right)$? Is matrix $A = \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}$ similar to matrix $B=\begin{pmatrix} 0&2\\ 0&0 \end{pmatrix}$? If so, how do I prove this?
I came here from following the answer to this question:
Do similar matri... | If they are similar then
$\begin{pmatrix} a&b\\ c&d \end{pmatrix} \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}= \begin{pmatrix} 0&2\\ 0&0 \end{pmatrix} \begin{pmatrix} a&b\\ c&d \end{pmatrix} $
for some invertible matrix $P=\begin{pmatrix} a&b\\ c&d \end{pmatrix}$.
Try to calculate such matrix $P$ from the above cond... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3350616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Continuity of $y \sin(\frac{1}{x})$ Problem: for which values of $\alpha \in \mathbb{R}$ is the function
\begin{equation*}
f: \mathbb{R}^2\to\mathbb{R}^2 \qquad \begin{pmatrix} x \\ y\end{pmatrix} \mapsto \begin{cases} y \sin(\frac{1}{x}) \quad &\text{if $x\neq 0$}\\0 \quad &\text{if $x= 0$}\end{cases}
\end{equation*}... | Your solution looks fine to me.
Notice that $|y\sin{\frac{1}{x}}|=\sqrt{y^2}|\sin{\frac{1}{x}}| \leq \sqrt{x^2+y^2}$
So indeed the function is continuous at $(0,0)$ and only at this point because $\lim_{x \to 0}\sin{\frac{1}{x}}$ does not exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3355539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
A system of equations Olympiad question Find all non-zero real numbers $x,y,z$ which satisfy the system of equations:
\begin{align}
(x^2+xy+y^2)(y^2+yz+z^2)(z^2+zx+x^2)&=xyz,\\(x^4+x^2y^2+y^4)(y^4+y^2z^2+z^4)(z^4+z^2x^2+x^4)&=x^3y^3z^3
\end{align}
It's an indian olympiad question. Can you guys help me out in solving it... | You obtained $$\left(\frac{x^3+y^3}{x+y}\right)\!\!\left(\frac{y^3+z^3}{y+z}\right)\!\!\left(\frac{z^3+x^3}{z+x}\right) = x^2y^2z^2$$
which is equivalent to
$$\prod_\mathrm{cyc}(x^2-xy+y^2)=x^2y^2z^2.$$
Dividing throughout by $xy\cdot yz\cdot zx$, we get
$$\prod_{\mathrm{cyc}}\left(\frac{x}{y}-1+\frac{y}{x}\right)=1.$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3356957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Solve $(x+1)(y+1)(z+1)=144$ in primes "Solve $(x+1)(y+1)(z+1)=144$ in primes".
So far, I have concluded that the solutions are $(x,y,z)=(2,3,11)$ or $(2,5,7)$ and their permutations. I worked like this:
*
*$x \equiv 0\mod 2\Rightarrow x+1=3, 144=2^4*3^2 \Leftrightarrow (y+1)(z+1)=48=2^4*3$
*
*$y \equiv 0\mod 2\Ri... | Denote $x' := x + 1$ and analogously. As you point out, solutions are closed under permutation, so we may suppose w.l.o.g. that $x' \leq y' \leq z'$.
Since the factors $x', y', z'$ are all at least $3$, we must have $z + 1 \leq \frac{144}{3^2} = 16$. But the factors of $144 = 2^4 \cdot 3^2$ no larger than $16$ are
$1, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3360021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find argmax $\frac{1}{2b}(\frac{x}{Na})^{\frac{N-1}{2}}e^{-\frac{1}{2b}(x+Na)}\frac{1}{\sqrt{2\pi(N-1)}}(\frac{\frac{e}{b}\sqrt{Nax}}{2(N-1)})^{N-1}$ I want to find an expression for:
$$ (1) \quad \underset{x}{argmax} \Big \{ \frac{1}{2b} \Big(\frac{x}{Na}\Big)^{\frac{N-1}{2}} e^{- \frac{1}{2b} (x+Na)} I_{N-1}\Big( \fr... | We simplify slightly the expression by writing
\begin{align}
&\frac{1}{2b} \left(\frac{x}{Na}\right)^{\frac{N-1}{2}} e^{- \frac{1}{2b} (x+Na)} I_{N-1}\left( \frac{\sqrt{Nax}}{b} \right)=\frac{1}{2b}\left( Na \right)^{\frac{1-N}{2}}e^{-\frac{Na}{2b}}f(x)\\
&f(x)=x^{\frac{N-1}{2}}e^{-\frac{x}{2b}} I_{N-1}\left( \frac{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3361586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$ab+ac+bc \equiv 1 \bmod abc$ or "easy chinese remainder theorem problems" When teaching students about the Chinese remainder theorem, it is traditional to ask them questions like: "An integer $n$ is equivalent to $r_1 \bmod m_1$, to $r_2 \bmod m_2$ and to $r_3 \bmod m_3$. Compute $n \bmod m_1 m_2 m_3$." For example,
... | $(1,1,1)$, $(1,1,m)$ for $m \geq 2$ and $(2,3,5)$ and are the only solutions with three positive integers. I'll limit myself to positive solutions in the following.
Note that $(m_1, m_2, \ldots, m_k, 1)$ obeys the conditions if and only if $(m_1, \ldots, m_k)$ do, so we may reduce to studying solutions with $m_j \geq ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3363602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Generating functions (with symbol method) of special partitions of natural numbers I want to show that the generating function of the number of partitions where every summand appears at most twice of every natural number n equals the number of paritions of n into summands which are not divisible by 3.
I know that for ... | The GF of the partitions where every summand appears at most twice is
$$f(x)=(1+x+x^2)(1+x^2+x^4)\cdot\ldots = \prod_{k\geq 1}(1+x^k+x^{2k})$$
while the GF of the partitions into summands which are not divisible by 3 is
$$ g(x) = \frac{1}{1-x}\cdot\frac{1}{1-x^2}\cdot\frac{1}{1-x^4}\cdot\frac{1}{1-x^5}\cdot\ldots = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3365597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\frac{1}{m_1} +\frac{1}{n_1} +\frac{1}{m_2} +\frac{1}{n_2}+...+\frac{1}{m_{2011}} +\frac{1}{n_{2011}}$ When $a=1,2,3,...,2010,2011$ the roots of the equation $x^2 -2x-a^2-a=0$ are $(m_1,n_1 ), (m_2,n_2 ), (m_3,n_ 3),..., (m_{2010},n_{2010} ), (m_{2011},n_{2011 }) $ respectively. Evaluate $\frac{1}{m_1} +\frac... | You don’t need to compute the roots $n_a, m_a$. Just note that: $\dfrac{1}{n_a}+\dfrac{1}{m_a}=\dfrac{n_a+m_a}{n_am_a}=\dfrac{S_a}{P_a}$ where $S_a$ and $P_a$ denote the sum and the product of those two roots. And:
$\dfrac{S_a}{P_a}=\dfrac{-b}{c}=-\dfrac{2}{a(a+1)}=2\left(\dfrac{1}{a+1}-\dfrac1a\right)$. Finally:
$$\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3366852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Help with $\int_0^{\frac{\pi}{2}} \frac{x \cos(x)}{\sin^2(x)+1} \; \mathrm{d}x = \frac{1}{2} \sinh^{-1}(1)^2$ I've been trying to do the above integral using elementary methods. So far, I've reduced the integral down to evaluating $\displaystyle \int_0^1 \frac{\tan^{-1}(x)}{\sqrt{1-x^2}} \; \mathrm{d}x$ or $\displaysty... | Substitute $t=\sin x$ and then IBPs,
$$\hspace{-1cm}
\int_0^{\frac{\pi}{2}} \frac{x \cos x}{\sin^2x+1}dx
=\int_0^1\frac{\sin^{-1}t}{1+t^2}dx=\frac{\pi^2}8-I,\>\>\>\>
I=\int_0^1\frac{\tan^{-1}t}{\sqrt{1-t^2}}dt \tag1$$
Express $\tan^{-1}t=\int_0^1\frac t{1+y^2t^2}dy$ and use
$$\int_0^1\frac {t\>dt}{\sqrt{1-t^2}(1+y^2t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3369411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Definite Integral $\int{x^2+1 \over x^4+1}$ Evaluate
$$\int_0^{\infty}{x^2+1 \over x^4+1}$$
I tried using Integration by parts ,
$$\frac{{x^3 \over 3 }+x}{x^4+1}+\int\frac{{x^3 \over 3 }+x}{(x^4+1)^2}.4x^3.dx$$
First term is zero
But it got me no where.
Any hints.
| Hint: $$\frac{x^2+1}{x^4+1}=\frac1{2\left(x^2+\sqrt2x+1\right)}+\frac1{2 \left(x^2-\sqrt2x+1\right)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3370526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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If $x$ is the remainder when a multiple of $4$ is divided by $6$, and $y$ is the remainder when a multiple of $2$ is divided by $3$, maximise $x+y$. The question is: if $x$ is the remainder when a multiple of $4$ is divided by $6$, and $y$ is the remainder when a multiple of $2$ is divided by $3$, what is the greatest ... |
The greatest value of $\color{red}{x \text{ is when}}$ 4 is divided by 6, which produces a remainder of 4.
Note that $4$ divided by $6$ produces the remainder $4$. In general, division by $6$ can produce remainders $0,1,2,3,4,5$. A multiple of $4$ divided by $6$ can not produce the remainder $1,3,5$, but only $0,2,4$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3372654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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Inequality with a rational function Let $0< a_1\leq a_2\leq a_3\leq a_4\leq a_5$. Define the function $$f(x,y,z)=\cfrac{(a_1(x+y+z)^3+a_2(x)^3+a_3(x+y)^3+a_4(y+z)^3+a_5(z)^3 )^2}{(a_1(x+y+z)^2+a_2(x)^2+a_3(x+y)^2+a_4(y+z)^2+a_5(z)^2 )^3}.$$
Show that for all $x,y,z\in \mathbb R$ we have $f(x,y,z)\leq \cfrac{1}{a_1+a_2+... | Not an answer.
First, I would like to give the equivalent problem:
Problem 1: Let $0 < a_1 \le a_2 \le a_3 \le a_4 \le a_5$. Let $x, y, z \in \mathbb{R}$ such that
$$a_1(x+y+z)^2+a_2x^2+a_3(x+y)^2+a_4(y+z)^2+a_5z^2 = a_1 + a_2 + a_3.\tag{1}$$ Prove that
$$a_1(x+y+z)^3 + a_2x^3 + a_3(x+y)^3 + a_4(y+z)^3 + a_5z^3 \le a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\iint_R\big(x^2+y^2\big)\,dA$
Evaluate the following double integral:
$$\iint_R\big(x^2+y^2\big)\,dA,$$
where $R$ is the region given by plane $x^2+y^2\leq a^2$.
My attempts:
\begin{align}
\iint_{R}\big(x^2+y^2\big)\,dA
&=\int_{-a}^{a}\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\big(x^2+y^2\big)\,dy\,dx\\... | As has been hinted, polar coordinates are the way to go. But if you really need to continue, the first thing I would note is that your integrand is an even function of $x$ over a symmetric integral. Therefore, you can simplify a tad to
$$\dfrac{4}{3}\int_{0}^{a}\sqrt{a^2-x^2}\cdot\left(2x^2+a^2\right)dx. $$
Now, whenev... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3378115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Question aboutPartial Fractions for example:
$$\frac{{{x^2} + 4}}{{x\left( {x + 2} \right)\left( {3x - 2} \right)}}\, = \frac{A}{x} + \frac{B}{{x + 2}} + \frac{C}{{3x - 2}}$$
first method is:
$${x^2} + 4 = A\left( {x + 2} \right)\left( {3x - 2} \right) + Bx\left( {3x - 2} \right) + Cx\left( {x + 2} \right)$$
but it is ... | Your "second method" is the next step of the first method. At each of the zeroes of $$x(x+2)(3x-2)$$ only one of the ABC terms is non-zero so you can solve for each ABC easily.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3378438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Determinant of Large Matrix $A=\left[\begin{array}{ccccc}{-2} & {-1} & {} & {\cdots} & {-1} \\ {-1} & {-2} & {-1} & {\cdots} & {-1} \\ {} & {} & {\ddots} & {} & {} \\ {-1} & {\cdots} & {-1} & {-2} & {-1} \\ {-1} & {\cdots} & {} & {-1} & {-2}\end{array}\right] \in \mathbb{R}^{53 \times 53}$
So we want to find determinan... | We can work with general dimension $n$. The wanted determinant is $(-1)^n$ times the determinant of
$$
B=\left[\begin{array}{ccccc}{ 2} & { 1} & {} & {\cdots} & { 1} \\ { 1} & { 2} & { 1} & {\cdots} & { 1} \\ {} & {} & {\ddots} & {} & {} \\ { 1} & {\cdots} & { 1} & { 2} & { 1} \\ { 1} & {\cdots} & {} & { 1} & { 2}\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3379033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.