Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Finding the radius of the smallest circle that can circumscribe an equilateral triangle Q:A puzzle board is in the form of an equilateral triangle that has an area of $7\sqrt{3}$ if the board is placed on a circular table, what should be the min area of the table so that the whole board fits inside the table. | Consider this image:
The distance from the center of the triangle (a.k.a. apothem) to the midpoint of one of its side is $h/3$, where $h$ is the height of the triangle. So the radius of the circle r is the distance from the center of the triangle to a vertex, and this distance is $2h/3$.
Finding the side of the triang... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1461168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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If $t = \tan (x/2)$, find expressions for $\sin x, \cos x$ in terms of $t$. Hence, solve the equation $3\sin x - 4\cos x = 2$.
If $$t = \tan \frac{x}{2},$$ find expressions for $\sin x, \cos x$ in terms of $t$. Hence, solve the equation $$3\sin x - 4\cos x = 2.$$
Attempt:
I have been solving a lot of trig questions l... | Indicated Solution
We can derive the Weierstrass Substitution:
Using the tangent double angle formula:
$$
\tan(x)=\frac{2t}{1-t^2}\tag{1}
$$
Then writing $\sec^2(x)$ in terms of $\tan^2(x)$
$$
\begin{align}
\sec^2(x)
&=1+\tan^2(x)\\
&=1+\frac{4t^2}{1-2t^2+t^4}\\
&=\frac{1+2t^2+t^4}{1-2t^2+t^4}\\
&=\left(\frac{1+t^2}{1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1468410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Multivariable continuity in ball How do you show continuity for ball-based functions such as
$$f:B[(0,0),1)]\rightarrow\mathbb{R}, \space f(x,y) = \sqrt{(1-(x^2+y^2)}$$
| If you want an $\varepsilon$-argument:
Let $\varepsilon > 0$; let $(x',y') \in \mathbb{R}^{2}$ such that $x'^{2} + y'^{2} < 1$; then
$$
|\sqrt{1-x^{2}-y^{2}} - \sqrt{1- x'^{2}-y'^{2}}| \leq \frac{|x-x'||x+x'| + |y-y'||y+y'|}{\sqrt{1-x^{2}-y^{2}} + \sqrt{1-x'^{2}-y'^{2}}} < \frac{|x-x'||x+x'| + |y-y'||y+y'|}{\sqrt{1-x'^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1468524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to prove that $(1+x^2)(1+y^2)(1+z^2)$ is a square of an integer? $x,y,z \in \Bbb Z$, if $xy+yz+xz=1$ then prove that $(1+x^2)(1+y^2)(1+z^2)$ is a square of an integer
$(1+x^2)(1+y^2)(1+z^2)=1+x^2+y^2+z^2+(xy)^2+(yz)^2+(xz)^2+(xyz)^2$
$(xy)^2+(yz)^2+(xz)^2+2xyz(x+y+z)=1$
| Use Brahmagupta–Fibonacci identity, $$(1+x^2)(1+y^2)=(1\pm xy)^2+(x\mp y)^2$$
Again, $$\{(1-xy)^2+(x+y)^2\}(1+z^2)=\{(1-xy)-z(x+y)\}^2+\{(1-xy)z+x+y\}^2$$
Do you see the destination?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1471345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $\lim_{x\to \infty}\left(x^2\ln\left(\cos \frac{4}{x}\right)\right)$ How do you evaluate $$\lim_{x\to\infty}\left(x^2\ln\left(\cos\frac{4}{x}\right)\right)$$
I know that you have to rewrite the expression as a quotient and use L'Hospital's Rule but I cant seem to figure it out.
| In this approach, we use some standard inequalities along with the squeeze theorem. First, recall that the log function satisfies the inequalities
$$\frac{x-1}{x}\le\log x\le x-1 \tag 1$$
for $x>0$. Then using $(1)$, it is straightforward to see that
$$x^2\frac{\left(\cos\left(\frac 4x\right)-1 \right)}{\cos\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1475550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding a permutation given a composite of permutations Given the permutations $x = (1, 2)(3, 4)$ and $y = (5, 6)(1, 3)$ find a permutation $a$ such that $a^{−1}xa = y$.
I have an answer to this but I would like to know the process involved in getting the answer.
My thoughts thus far are that $a^{-1}(12)(34)a =(56)(13)... | Let $\sigma=[i_1,\ldots, i_n]$ be a cycle. Use the fact that $\tau\sigma\tau^{-1}=[\tau(i_1),\ldots,\tau(i_n)]$. So, what permutation takes $1$ to $5$, $2$ to $6$, $3$ to $1$ and $4$ to $3$. The inverse of this permutation is the answer.
To find this, we can expess it as a matrix:
$$a^{-1}=\begin{pmatrix}
1 & 2 & 3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1478866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Trigonometric equation with parameter
Find $p$ for which $\cos^2(x) - \cos(x) + p + 1 = 0$ has EXACTLY two solutions for $0 \le x \le 2\pi$
I tried to substitute $t = \cos(x)$ and then I got two solutions, but I don't know what to do next.
$t_1 = \frac{1 + \sqrt{-3 - 4p}}{2}$
$t_2 = \frac{1 - \sqrt{-3 - 4p}}{2}$
| Since $\cos x=T$ has at most two real solutions in $0\le x\le 2\pi$ for a given real constant $T$, we have three cases.
(1) $\cos x=t_1$ has no solution and $\cos x=t_2$ has two solutions.
(2) $\cos x=t_1$ has exactly one solution and $\cos x=t_2$ has exactly one solution.
(3) $\cos x=t_1$ has two solutions and $\cos x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1479658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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$2=1$ Paradoxes repository I really like to use paradoxes in my math classes, in order to awaken the interest of my students. Concretely, these last weeks I am proposing paradoxes that achieve the conclusion that 2=1. After one week, I explain the solution in the blackboard and I propose a new one. For example, I poste... | Another enjoyable "paradox": We first denote
$$S:=\sum_{n\in\mathbb N}\dfrac{(-1)^{n+1}}{n}$$
The fact that $0\neq S\in\mathbb R$ can be established using elementary tools.
We then write:
$S=\frac{1}{1}-\frac {1}{2}+\frac{1}{3}-...+...-...$
$2S = 2(\frac{1}{1}-\frac {1}{2}+\frac{1}{3}-...+...-...)=\frac{2}{1}-\frac {2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1480488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 15,
"answer_id": 1
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Regarding Leibniz formula for $\pi/4$ proof and its convergence \begin{align}
\frac{\pi}{4} & = \arctan(1)\;=\;\int_0^1 \frac 1{1+x^2} \, dx \\[8pt]
& = \int_0^1\left(\sum_{k=0}^n (-1)^k x^{2k}+\frac{(-1)^{n+1}\,x^{2n+2} }{1+x^2}\right) \, dx \\[8pt]
& = \sum_{k=0}^n \frac{(-1)^k}{2k+1}
+(-1)^{n+1}\int_0^1\frac{x^{2n+2... | Using long division, the integrand $\frac{1}{1+x^2}$ can be written
$$\begin{align}
\frac{1}{1+x^2}&=1-x^2+x^4-x^6+\cdots +(-1)^nx^{2n}+\frac{(-1)^{n+1}x^{2n+2}}{1+x^2}\\\\
&=\sum_{k=0}^n (-1)^kx^{2k}+\frac{(-1)^{n+1}x^{2n+2}}{1+x^2}
\end{align}$$
Thus, the integral of interest is
$$\begin{align}
\int_0^1\frac{1}{1+x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1482529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Olympiad inequality problem with $a+b+c+abc=4$ If $a,b,c \in \mathbb R_{> 0}$ and $a+b+c+abc=4$, prove that
$$({a\over {\sqrt {b+c}}}+{b\over {\sqrt {c+a}}}+{c\over {\sqrt {a+b}}})^2(ab+bc+ca) \ge {\frac 12}(4-abc)^3$$
This can be solved by AM-GM-HM or the Cauchy-Schwarz inequality. I'd tried for some hours but couldn'... | Here is a possible approach without using Hölder's inequality. Assume without loss of generality that $a\ge b \ge c$. Then
$$\left({a\over {\sqrt {b+c}}}+{b\over {\sqrt {c+a}}}+{c\over {\sqrt {a+b}}}\right)^2\ge \frac{1}{2a}\left(a+b+c\right)^2$$
Now
$$ab+bc+ca=A=\frac{A+A}{2}=\frac{1}{2}(a(b+c)+b(c+a)+c(a+b))\ge c(a+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1483425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Prove by induction: $\sum\limits_{k=1}^{n}(-1)^{k+1}{n\choose k}\frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$ $\sum\limits_{k=1}^{n}(-1)^{k+1}{n\choose k}\frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$
For $n=1$ equality is true.
For $n=m$
$m-{m\choose 2}\frac{1}{2}+...+(-1)^{m+1}\frac{1}{m}=1+\frac{1}{... | We have:
$$ \sum_{k=1}^{n}\binom{n}{k}(-1)^{k+1} x^k = 1-(1-x)^n\tag{1}$$
hence:
$$ \sum_{k=1}^{n}\binom{n}{k}(-1)^{k+1}\frac{1}{k} = \int_{0}^{1}\frac{1-(1-x)^n}{x}\,dx= \int_{0}^{1}\frac{1-x^n}{1-x}\,dx\tag{2}$$
and:
$$ \sum_{k=1}^{n}\binom{n}{k}(-1)^{k+1}\frac{1}{k} = \int_{0}^{1}\left(1+x+\ldots+x^{n-1}\right)\,dx ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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For f(x) = ((3x+7)^8)((4x-5)^3) , find f'(x) and use this answer to find the value(s) of x at which the graph of f(x) has a horizontal tangent line I know that $f'(x) = (24(3+7)^7)*(4x-5)^3+((3x+7)^8)*12(4x-5)^2$, but is there any easier way to find the horizontal tangent line without expanding the terms using pascal's... | $$f(x)=(3x+7)^8(4x-5)^3$$
$$f'(x)=8(3x+7)^7.3(4x-5)^3+(3x+7)^83.(4x-5)^2.4$$
$$f'(x)=(3x+7)^7(4x-5)^2[24(4x-5)+12(3x+7)]$$
$$f'(x)=(3x+7)^7(4x-5)^2[132x-36]$$
$$f'(x)=12(3x+7)^7(4x-5)^2[11x-3]$$
to find the tangent point
$$(3x+7)^7=0$$
$$(4x-5)^2=0$$
$$(11x-3)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1487919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving a recurrence relation with no real roots? $a_n - 2a_{n-1} -12a_{n-2} - 14a_{n-3} -5a_{n-4} = 0 $
I've tried a few method:
*
*setting the denominator to $1 -2x -12x^2 -14x^3 -5x^4$ and then finding the numerator by multiplying $(1 -2x -12x^2 -14x^3 -5x^4)*\sum a_nx^{n}$
*finding the roots of C(y) = $y^4 -4y... | A start: The given recurrence has characteristic equation $x^4-2x^3-12x^2-14x-5=0$. This has $x=-1$ as a solution.
Dividing our polynomial by $x+1$, we get $x^3-3x^2-9x-5$. Note that $-1$ is a zero of this polynomial. Continue.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1491007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sequence of positive integers such that their reciprocals are in arithmetic progression Let $m_1 < m_2 < \ldots < m_k$ be $k$ distinct positive integers such that their reciprocals $\dfrac{1}{m_i}$ are in arithmetic progression.
*
*Show that $k < m_1 + 2$.
*Give an example of such a sequence of length $k$ for any ... | Hint for the bound: Use the fact that the common difference $d$ of the AP is $\ge \frac{1}{m_1}-\frac{1}{m_1+1}$. Now if we take $(n-1)$ steps of length down from $a=\frac{1}{m_1}$, the result must be $\gt 0$.
Hint for the construction: The first interesting length is $k=3$. There we have the familiar AP $\frac{1}{2},... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1495367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why does $\lim\limits_{n \to \infty } ((1 + x)(1 + {x^2})(1 + {x^4})\ldots(1 + {x^{{2^n}}})) = \frac{1}{{1 - x}}$? Let $\left| x \right| < 1$.
Why does $$\lim\limits_{n \to \infty } ((1 + x)\cdot(1 + {x^2})\cdot(1 + {x^4})\cdot\ldots\cdot(1 + {x^{{2^n}}})) = \frac{1}{{1 - x}}$$
| $$(1+x)(1+x^2)(1+x^4)(1+x^8).....=1+x+x^2+x^3+x^4+x^5......$$
then you can use the fact of geometric series
$$\frac{1}{1-x}=1+x+x^2+x^3+x^4+x^5......$$
$$|x|<1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1495750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Write the expression in Euler's formula
$a=2+2i$ , $b=5e^{i\frac{\pi}{3}}$
such that $$\frac{b^5}{a^3}=Re^{\theta i}$$
find: $R$ and $\theta$
$$R=\sqrt{2^2+2^2}=\sqrt{8}$$,
$$tan^{-1}=\frac{2}{2}\rightarrow \theta=\frac{\pi}{4}$$
$$a=\sqrt{8}e^{\frac{\pi}{4}i}, a^3=({\sqrt{8}e^{\frac{\pi}{4}i}})^3=(\sqrt{8})^3e^{\fr... | Write $a = 2\sqrt 2 e^{\pi/4 i}$ Then
$$\frac {b^5}{a^3} = \frac{5^5e^{5i\pi/3}}{(\sqrt 8)^3 e^{3/4\pi i}} = \frac{5^5}{(\sqrt 8)^3} e^{ (5/3 - 3/4) i \pi} = \frac{5^5}{(\sqrt 8)^3} e^{ \frac {11}{12} i \pi}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1497408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Write on my own my first mathematical induction proof I am trying to understand how to write mathematical induction proofs. This is my first attempt.
Prove that the sum of cubic positive integers is equal to the formula
$$\frac{n^2 (n+1)^2}{4}.$$ I think this means that the sum of cubic positive integers is equal to a... | You want to prove that $$\sum_{i=1}^{n}i^3 = \frac{n^2 (n+1)^2}{4}$$ using induction.
For $n=1$, $$\sum_{i=1}^{1}i^3 = 1^3=1=\frac{1^2(1+1)^2}4=1$$
So the formula does work for the base case $n=1$.
Now, assume the formula works for $n=k$ and show that this implies that the formula is correct for $n=k+1$ which will acco... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1499897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Hypergeometric series for $\sin(az)$ I am having trouble trying to show that: $$F\left(\frac{1}{2}+\frac{1}{2}a,\frac{1}{2}-\frac{1}{2}a;\frac{3}{2};\sin^2z\right) = \sin{az}.$$
I have found that the coefficients in the series are given by: $$\frac{\left(\frac{1}{2}+\frac{1}{2}a\right)_n \left(\frac{1}{2}-\frac{1}{2}a\... | By comparing the Chebyshev differential equation with the hypergeometric differential equation you may check that:
$$ \phantom{}_{2}F_1\left(\frac{1+a}{2},\frac{1-a}{2};\frac{3}{2};z^2\right)=\frac{\sin(a\arcsin z)}{a z} \tag{1}$$
from which it follows that:
$$ \phantom{}_{2}F_1\left(\frac{1+a}{2},\frac{1-a}{2};\frac{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1500794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Reading cycle notation I need a little help understanding how we read cycle notation. I know it is a function so if we have something like
$\begin{pmatrix}1 & 2 & 3 & 4 & 5 \\
2 & 5 & 4 & 3 & 1
\end{pmatrix} = (1 \quad 2 \quad 5)(3 \quad 4)=(3 \quad 4)(1 \quad 2 \quad 5)=(3 \quad 4)(5\quad 1 \q... | I think my original comment may have led to some confusion in regards to my intentions behind it--your function
\begin{pmatrix}1 & 2 & 3 & 4 & 5 \\\tag{1}
2 & 5 & 4 & 3 & 1
\end{pmatrix}
may, perhaps, be more effectively communicated as
\begin{pmatrix}1 & 2 & 3 & 4 & 5 \\
\downarrow & \downarrow & \downarrow ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1500878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Understanding telescoping series? The initial notation is:
$$\sum_{n=5}^\infty \frac{8}{n^2 -1}$$
I get to about here then I get confused.
$$\left(1-\frac{3}{2}\right)+\left(\frac{4}{5}-\frac{4}{7}\right)+...+\left(\frac{4}{n-3}-\frac{4}{n-1}\right)+...$$
How do you figure out how to get the $\frac{1}{n-3}-\frac{1}{n-1... | $$S = \sum_{n=5}^\infty \frac{8}{n^2 -1}=
\lim_{N\to\infty} \sum_{n=5}^N \frac{8}{n^2 -1}
= \lim_{N\to\infty} \sum_{n=5}^N \left\{ {4\over n-1} - {4\over n+1}\right\} = \lim_{N\to\infty} \sum_{n=5}^N {4\over n-1} - \sum_{n=5}^N {4\over n+1}$$
Now reindex to get
$$S = \lim_{N\to\infty} \sum_{n=4}^{N-1} {4\over n} - \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1502309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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How to Completely Simplify the Derivative of $\sqrt{16-x^2}-4\cos^{-1}(x/4)$ I am trying to completely simplify the derivative of the following function:
So far, I have gotten the answer:
Apparently this is not simplified enough.
Does anyone know how to simplify this further, if it is at all possible?
All help is ap... | \begin{align}
& \frac{d}{dx} \left(\sqrt{16-x^2}-4\cos^{-1}(\frac{x}{4})\right) \\[10pt]
= {} &\frac{d}{dx} \left(\sqrt{16-x^2}\right)-\frac{d}{dx} \left(4\cos^{-1} \left(\frac{x}{4}\right)\right) \\[10pt]
= {} & \frac{1}{2\sqrt{16-x^2}}\cdot(-2x)+\frac{4}{\sqrt{1-\frac{x^2}{16}}} \cdot \left(\frac{-1}{4}\right) \\[10p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1507277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find the last two digits of the number $7^{100}-8^{100}$. So I tried and found that $$7^{100} \equiv 1 \pmod{100}$$ but I got stuck with $8^{100}$. Help me out please.
| First, note that $7^2\equiv8^2\pmod{5}$. So $$7^{100}-8^{100}=(7^2-8^2)\left((7^2)^{49}+(7^2)^{48}(8^2)+\cdots+(8^2)^{49}\right)\equiv 0\pmod{25},$$
since the second term consists of $50$ summands of the same modulo $5$.
Second $7^{100}-8^{100}\equiv 7^{100}\pmod{4}\equiv 1\pmod{4}$.
Chinese remainder theorem then imp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1511477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solve the differential equation $\frac{dy}{dx}=\frac{2x+y}{y}$ Solve
$$\frac{dy}{dx}=\frac{2x+y}{y}$$
Let
$$\frac{dx}{dt}=y \\ \frac{dy}{dt}=2x+y$$
Plugin $\frac{dx}{dt}=y$ into $\frac{dy}{dt}=2x+y$ I get $$2x+\frac{dx}{dt}=\frac{dy}{dt}$$
Using the fact that $\frac{d^2x}{dt^2}=\frac{dy}{dt}$ I get the 2nd order system... | I'm a bit rusty with differential equations, so I'll just assume that we have the required number of constants of integration...
$$\begin{align}
x & = Ae^{2t} + Be^{−t}\\
y & = \frac{dx}{dt}\\
y & = 2Ae^{2t} - Be^{−t}\\
x+y & = 3Ae^{2t}\\
2x-y & = 3Be^{−t}\\
e^{2t} = \frac{x+y}{3A} & = \left(\frac{3B}{2x-y}\right)^2\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1511567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solving $\sqrt\frac{x^2+2x-2}{x^2+3x+2} + \sqrt\frac{x^2+3x+2}{x^2+2x-2}=\frac52 $ Got these two problems on an exam recently and was unsure if I managed to answer them correctly:
$$\sin^4x + \cos^4x \ge \frac58, \ x \in \left\{-\frac\pi4,\frac\pi4\right\}$$
Second:
$$\sqrt\frac{x^2+2x-2}{x^2+3x+2} + \sqrt\frac{x^2+3x+... | I'll solve both in one shot here using the simplest methods I can think of:
For the first, we use the proof here
$$\sin^4x + \cos^4x \ge \frac58, \ x \in \left\{-\frac\pi4,\frac\pi4\right\}$$
$$\Rightarrow \displaystyle \frac {\cos (4 x)- 4 \cos (2 x)+3 } 8 + \frac {\cos(4x)+4\cos(2x) + 3} 8 \geq \frac{5}{8}$$
$$\Rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1513586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve $ \left|\frac{x}{x+2}\right|\leq 2 $ I am experiencing a little confusion in answering a problem on Absolute Value inequalities which I just started learning. This is the problem: Solve:
$$
\left|\frac{x}{x+2}\right|\leq 2
$$
The answer is given to be $x\leq-4$ or $x \geq-1$
This is my attempt to solve the probl... | To solve this kind of problem systematically. If you want to solve $|f(x)|\le c$ or $|f(x)|\geq c$ (for some constant $c\geq 0$), you have to study the sign of $f(x)$. It can be longer for "simple" problem like yours, but it has the advantage to always give you the solution for more complex ones.
For your particular ex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1515986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Prove that $ \int_0^1 x^2 \psi(x) \, dx = \ln\left(\frac{A^2}{\sqrt{2\pi}} \right) $ Basically what the title says:
Prove that $ \displaystyle \int_0^1 x^2 \psi(x) \, dx = \ln\left(\dfrac{A^2}{\sqrt{2\pi}} \right). $
where $A\approx 1.2824$ denotes the Glaisher–Kinkelin constant and $\psi(x) $ denote the digamma fun... | Let $(A_n)$ and $(B_n)$ by
$$ A_n = \frac{1^1 2^2 \cdots n^n}{n^{n^2/2+n/2+1/12} e^{-n^2/4}}
\quad \text{and} \quad
B_n = \frac{n!}{e^{-n}n^{n+1/2}}. $$
In view of the Wikipedia page, we know that $A_n \to A$ and $B_n \to \sqrt{2\pi}$. We can also equivalently state these definitions as follows:
\begin{align*}
\sum_{k=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1517085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Evaluation of $\int \sqrt{1+\cot x}dx$?
What is $$\int \sqrt{1+\cot x}dx$$
My friends and I tried using all possible trigonometric formula. We couldn't find a way to solve it
Please help me solve it.
| Put $(1+cotx)=t^2$ in the given integral and then do the reqd. Substitutions to get :
$$\int -\frac {2x^2dx}{x^4-2x^2+2}$$
I think we can avoid complex numbers.
Write it as:
$$\int -2(\frac {(x^2+ \sqrt2+x^2-\sqrt2)dx}{x^4-2x^2+2})$$
Split it into two integrals,
$$\int\frac {(x^2+\sqrt2)dx}{x^4-2x^2+2}$$
And
$$\int \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1517274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Sign table in $2^k$ factorial experiment I have a factorial experiment with four factors $\{A,B,C,D\}$ ($k=4$) , just to ilustrate the case with $k=3$, the signal table is
\begin{matrix}
& I & A & B & C & AB & AC & BC & ABC\\
(1) & +1 & -1 & -1 & -1 & +1 & +1 & +1 & -1\\
a & +1 & +1 & -1 & -1 & -1 & -1 & +1 & +1\\
b &... | Let's start by isolating the basis:
$$\begin{array} {r|r|rrr|rrrr}
& I & A & B & C & AB & AC & BC & ABC\\
(1) & +1 & -1 & -1 & -1 & +1 & +1 & +1 & -1\\
a & +1 & +1 & -1 & -1 & -1 & -1 & +1 & +1\\
b & +1 & -1 & +1 & -1 & -1 & +1 & -1 & +1\\
ab & +1 & +1 & +1 & -1 & +1 & -1 & -1 & -1\\
c & +1 & -1 & -1 & +1 & +1 & -1 & -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1517460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Determine the smallest disc in which all the eigen values of a given matrix lie Let , $$A=\left[\begin{matrix}1&-2&3&-2\\1&1&0&3\\-1&1&1&-1\\0&-3&1&1\end{matrix}\right]$$Which of the following is the smallest disc in $\mathbb C$ which contains all eigen values of $A$.
*
*$|z-1|\le 7$
*$|z-1|\le 6$
*$|z-1|\le 4$.
... | (Too long for a comment.)
This looks like a trick question and I don't know the trick, but you may try the following approach. Let $B=A-I$. The characteristic polynomial of $B$ is $x^4 + 15x^2 - 14x + 16$. As $15x^2-14x+16>0$ over $\mathbb R$, $B$ has no real eigenvalues. Since $B$ has zero trace and nonreal eigenvalue... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1517537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find the limit? How to calculate this limit ?
$$\lim _{x\to 1}\frac{\sqrt{x^2-1}}{\sqrt[3]{x-1}}$$
An image for clarification.
| We can rewrite the function:
$$\frac{\sqrt{x^{2}-1}}{\sqrt[3]{x-1}}=(x+1)^{\frac{1}{2}}(x-1)^{\frac{1}{2}}(x-1)^{-\frac{1}{3}}=(x+1)^{\frac{1}{2}}(x-1)^{\frac{1}{2}-\frac{1}{3}}$$
So that
$$\lim_{x\to 1}\frac{\sqrt{x^{2}-1}}{\sqrt[3]{x-1}}=\lim_{x\to 1}(x+1)^{\frac{1}{2}}(x-1)^{\frac{1}{6}}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1519346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Modified Leibnitz integral: $\lim\limits_{a \to\infty}\frac1a\int _0^\infty\frac{(x^2+ax+1)\arctan(\frac{1}{x})}{1+x^4}dx=?$ $\lim\limits_{a \to \infty} \frac{1}{a} \int _0^\infty\frac{(x^2+ax+1)\arctan(\frac{1}{x})}{1+x^4}dx $ ,where $a$ is a parameter.
ATTEMPT:- Let $I(a)=\frac{1}{a} \int _0^\infty \frac{(x^2+ax+1)\... | It is not necessary to calculate $I(a)$. Noting that $\lim_{a\to\infty}\frac{x^2+ax+1}{a(x^4+1)}=\frac{x}{x^4+1}$ and
$$ \frac{x^2+ax+1}{a(x^4+1)}-\frac{x}{x^4+1}=\frac{x^2+1}{a(x^4+1)} $$
and $\int_0^\infty\frac{(x^2+1)\arctan\frac1x}{x^4+1}dx$ converges, we have, for big $M>0$,
\begin{eqnarray}
&&\bigg|\int_0^\infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1520120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Express $-a^3 b + a^3 c+ab^3-a c^3-b^3c+bc^3$ as a product of linear factors.
Express
$$-a^3 b + a^3 c+ab^3-a c^3-b^3c+bc^3$$
as a product of linear factors.
I have tried rewriting the expression as:
$$ab^3-a^3b + a^3c-ac^3 +bc^3-b^3c$$
$$= ab(b^2-a^2)+ac(a^2-c^2)+bc(c^2-b^2)$$
$$= ab(b-a)(b+a) + ac(a-c)(a+c)+bc(... | The key is to look for symmetry, so write it as symmetrically as possible:$$a^3(c-b)+b^3(a-c) +c^3(b-a).$$This has cyclic symmetry; so, if it has linear factors, then they must be invariant under cyclic permutation of $a$, $b$, and $c$. The factors have to include negative signs, so try the simplest first: say $b-c$. P... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1521015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Prove an inequality I want to show that following:
$$\left(\frac{n^2-1}{n^2}\right)^n\sqrt{\frac{n+1}{n-1}}\leq 1; ~~n\geq 2$$ and $n$ is an integer.
After some simplifications, I got left hand-side as
$$LHS:\left(1-\frac{1}{n}\right)^{n-\frac{1}{2}} \left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}$$
It is clear that the 1s... | Consider on $(-1,1)$ the function (with motivation $x=\frac1n$)
$$
f(x)=\ln\bigl(1-x^2\bigr)+\frac x2\bigl(\ln(1+x)-\ln(1-x)\bigr)-\frac16\ln\bigl(1-x^4\bigr)
$$
Then
$$
f'(x)=-\frac{x}{1-x^2}+\frac12\bigl(\ln(1+x)-\ln(1-x)\bigr)+\frac16\frac{4x^3}{1-x^4}
$$
where we see that $f(0)=0=f'(0)$. The next derivative is
\beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1522216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Give bases for col(A) and null(A) I have
A= $\begin{bmatrix}1&1&-3\\0&2&1\\1&-1&-4\end{bmatrix}$
I row reduce it to
$\begin{bmatrix}1 &0& -3.5\\0&1&.5\\0&0&0\end{bmatrix}$
How do I find col(A) from the above info? Is it the pivot points correspond to the columns?
So col(A) would be $\begin{bmatrix}1\\0\\1\end{bmatrix}... | To clear up confusion, work out the steps.
*
*Form the augmented matrix
$$
% A I
\left[
\begin{array}{c|c}
\mathbf{A} & \mathbf{I}_{3} \\
\end{array}
\right]
%
=
%
\left[
\begin{array}{rcr|ccc}
1 & 0 & 1 & 1 & 0 & 0 \\
1 & 2 & -1 & 0 & 1 & 0 \\
-3 & 1 & -4 & 0 & 0 & 1 \\
\end{array}
\right]
$$
*Clear column 1.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1522796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Using hypergeometric functions to solve this integral After looking at calculations, I realized that the exponent needs to be 1/4 instead of -1/4
I have this equation and I am trying to solve the integral of it.
$$((R^2) - (y^2))^{1/4} dy$$
I tried to put it into wolfram alpha, and I got an answer, but I wanted to kno... | I'm assuming the integral in question is
$$\int_0^R (R^2-y^2)^{-\frac{1}{4}}\,dy$$
as these types are quite common. If not, a very modest change can be made (just let the upper bound be $x$ instead of $R$ in the following analysis and leave the hypergeometric function in terms of $x$ and ignore the stuff concerning Gau... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1525408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Uniform convergence of $\sum_{n \ge 0} (-1)^n(1-x)x^n$ Let $f_n: [0,1] \to \mathbb{R}$, $f_n(x)=(-1)^n (1-x)x^n, \forall x\in [0,1], n \ge 0$. Prove that $\sum\limits_{n \ge 0} f_n(x)$ convergenes uniformly.
Let $S_n(x) = \sum\limits_{k = 0}^n f_k(x), \forall x \in [0,1], n \ge 0$.
Then,
$S_n(x)=1-x-x+x^2+x^2-x^3-x^3+... | Hints : with your notation, $g_n'(x) = \frac{ x^{n-1}(-nx^2+n-2x)}{(x+1)^2}$. The sign of $g_n'(x)$ on $[0,1]$ is the same as the sign of $-nx^2+n-2x$ and consequently is non-negative on $\left[0, \frac{\sqrt{n^2+1}-1}{n}\right]$ and non-positive on $\left[\frac{\sqrt{n^2+1}-1}{n},1\right]$. Consequently, the maximum o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1526972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How do I complete the proof of proving the $\lim _{ x\rightarrow \infty }{ \frac { x^{ 2 }+1 }{ (x+1)^{ 2 } } =1 } $ $$\lim _{ x\rightarrow \infty }{ \frac { x^{ 2 }+1 }{ (x+1)^{ 2 } } =1 } $$
Proof: Let $\epsilon > 0$
Then, $$ \left| \frac { x^{ 2 }+1 }{ (x+1)^{ 2 } } -1 \right| <\epsilon $$
$$\Longleftrightarrow \le... | Given $\epsilon > 0$ we want to find an $M$ such that $x > M \Rightarrow | \cdots | < \epsilon$.
Remember you just need to find an $M$; make life as easy for yourself as you can and don't care about finding a low $M$.
Hence, I would simplify and write for all $x > 0$,
$$ \left| \frac{x^2+1}{(x+1)^2} -1 \right| = \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How can I manipulate $\frac { \sqrt { x+1 } }{ \sqrt { x } +1 } $ to find $M>0$ to prove a limit? Given the following limit, find such an $M>0$ that for every $x>M$, the expression is $\frac { 1 }{ 3 }$ close to the limit. In other words find $M>0$ that for every $x>M:\left| f(x)-L \right| <\frac { 1 }{ 3 }$ for the fo... | The usual rationalization:
$\begin{array}\\
\sqrt{x+1}-\sqrt{x}
&=(\sqrt{x+1}-\sqrt{x})\frac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}\\
&=\frac{1}{\sqrt{x+1}+\sqrt{x}}\\
&< \frac1{2\sqrt{x}}
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Homework: Sum of the cubed roots of polynomial Given $7X^4-14X^3-7X+2 = f\in R[X]$, find the sum of the cubed roots.
Let $x_1, x_2, x_3, x_4\in R$ be the roots. Then the polynomial $X^4-2X^3-X+ 2/7$ would have the same roots. If we write the polynomial as $X^4 + a_1X^3 + a_2X^2 +a_3X + a_4$ then per Viete's theorem:
$a... | Use Newton's relations between sums of powers $p_k=\sim_ix_i^k$ and the elementary symmetric functions:
\begin{align*}
p_1&=\sigma_1,\\
p_2&=\sigma_1p_1-2\sigma_2,\\
p_3&=\sigma_1p_2-\sigma_2p_1+3\sigma_3.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How should I integrate $\int _{ 0 }^{ \frac{1}{\sqrt{3}} } \frac { dx }{ (1+x^2)\sqrt{1-x^2}\ } $? How should I integrate $\int _{ 0 }^{ \frac{1}{\sqrt{3}} } \frac { dx }{ (1+x^2)\sqrt{1-x^2}\ } $ ?
I tried substituting $x=1/t$,but it is not working.Could someone help?Thanks.
| Let $x=\sin \theta$ and then
\begin{eqnarray}
\int _{ 0 }^{ \frac{1}{\sqrt{3}} } \frac { dx }{ (1+x^2)\sqrt{1-x^2}\ }&=&\int _{ 0 }^{ \arcsin\frac{1}{\sqrt{3}} } \frac {d\theta }{1+\sin^2\theta}\\
&=&\frac{1}{\sqrt 2}\arctan(\sqrt{2}\tan\theta)\bigg|_{ 0 }^{\arcsin\frac{1}{\sqrt{3}}}\\
&=&\frac{1}{\sqrt 2}\arctan(\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Computing common volume of sphere and cylinder using triple integrals Find the volume of material cut from the solid sphere $x^2 + y^2 + z^2 \leq 9$ by the cylinder $r = 3\sin(\theta)$ using cylindrical coordinates.
| In cylindrical coordinates the equation of the sphere is: $r^2+z^2=9$ so, since $r\le 3 \sin \theta$ the volume is limited by the inequalities:
$$
0\le \theta \le 2\pi \qquad 0\le r \le 3 \sin \theta \qquad-\sqrt{9-r^2}\le z \le \sqrt{9-r^2}
$$
so, given the symmetry, the volume is:
$$
V=4 \int_0^{\frac{\pi}{2}}\int_0^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1536964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to prove this infinite product identity? How can I prove the following identity?
$$\large\prod_{k=1}^\infty\frac1{1-2^{1-2k}}=\sum_{m=0}^\infty\left(2^{-\frac{m^2+m}{2}}\prod_{n=1}^\infty\frac{1-2^{-m-n}}{1-2^{-n}}\right)$$
Numerically both sides evaluate to
$$2.38423102903137172414989928867839723877...$$
| The identity is in fact true for any $0 < x < 1$: $$\prod_{k=1}^{\infty}\frac{1}{1-x^{2k-1}}=\sum_{m=0}^{\infty}\left(x^{\frac{m^2+m}{2}}\prod_{n=1}^{\infty}\frac{1-x^{m+n}}{1-x^{n}}\right) = \sum_{m=0}^\infty\left(x^{\frac{m^2+m}{2}}\prod_{n=1}^{m}\frac{1}{1-x^{n}}\right)$$
We note the telescopic product: $$\displayst... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1539082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Distance between points - equation of a line I have worked on this particular example:
The distance between the point $M_1(3,2)$ and $A$ is $2\sqrt5$ and the distance between the point $M_2(-2, 2)$ and $B$ is $\sqrt5$. Come up with a equation for the line going through $A$ and $B$.
I have tried playing around with th... | Point on the circle with radius $2\sqrt{5}$ and center $(3,2)$ is at:
$$ A = (2\sqrt{5} \cos(a) + 3, 2\sqrt{5} \sin(a) + 2)$$
The other circle's point is at:
$$ B = (\sqrt{5} \cos(b) - 2, \sqrt{5} \sin(b) + 2) $$
Then slope of the line is:
$ m = \frac{B_y - A_y}{B_x-A_x} = \frac { \sqrt{5} \sin(b) + 2 - (2\sqrt{5} \sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1544804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Are these trigonometric expressions for the ceiling and floor functions correct? I believe that I have found a trigonometric expression for both the ceiling and floor function, and I seek confirmation that it is, indeed, correct.
Update.
$$\begin{align}
\lfloor x \rfloor &= x - \frac12+f(x) \\[4pt]
\lceil x \rceil &... | If you're using the single-valued $\arctan$, i.e. $-\frac{\pi}{2} \leq \arctan(x) \leq \frac{\pi}{2}$, which is what all computer languages I've seen use, this works perfectly fine.
Proof:
Let $x = s(a + b)$, where $a \in \mathbb{N}$, $0 \leq b < 1$ and $s \in \{-1,1\}$. We can call $a$ the absolute integer part, $b$ t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1545006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Polynomial Equations Root Finding If one of the roots of the Equation :
$$2000x^6 + 100x^5 + 10x^3 + x - 2 = 0$$ ;
Is of the form $$(m + √n)/r$$. Where m is a non zero integer and n and r are relatively coprime.
We have to find $$(m+n+r)/100$$ .
I am not able to factorise this as I am not able to guess any roots.
I... | Given $$2000x^6+100x^5+10x^3+x-2=0\Rightarrow (2000x^6-2)+(100x^5+10x^3+x)=0$$
Now Divide it by $x^3\;,$ Where $x\neq 0\;,$ We get
$$2\left[(10x)^3-\left(\frac{1}{x}\right)^3\right]+\left[(10x)^2+1+\frac{1}{x^2}\right]=0$$
So $$2\left[(10x)^3-\left(\frac{1}{x}\right)^3\right]+\frac{(10x)^3-\frac{1}{x^3}}{\left[10x-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1547392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Proving that $\binom{n}{0}+\binom{n-1}{1}+\binom{n-2}{2}+\cdots =F_{n+1}$ where $F_{n+1}$ is the $n+1$ th Fibonacci number I have to proove this this identity which connects Fibonacci sequence and Pascal's triangle:
$$\begin{pmatrix}n\\0\end{pmatrix}+\begin{pmatrix}n-1\\1\end{pmatrix}+\dotsm+\begin{pmatrix}n-\lfloor\fr... | Another way is using generating functions. The g.f. of $\displaystyle a_n = \sum_{k=0}^{\lfloor n/2 \rfloor} {n-k \choose k} $ is
$$ \eqalign{G(x) & =\sum_{n=0}^\infty \sum_{k=0}^{\lfloor n/2 \rfloor} {n-k \choose k} x^n\cr &= \sum_{k=0}^\infty \sum_{n=2k}^\infty {n-k \choose k} x^n\cr
&= \sum_{k=0}^\infty x^{2k} \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Inverse Trigonometry Question (Stuck in Algebraic Simplification) The original question:
If x and y are of same sign, then find the value of $$\cfrac{x^3}{2} \csc^2 \left(\cfrac{1}{2}\tan^{-1} \cfrac{x}{y}\right) + \cfrac{y^3}{2} \sec^2 \left(\cfrac{1}{2}\tan^{-1} \cfrac{y}{x} \right) $$
This is my attempt to the qu... | i will take $x,y$ to have the same sign. using double angle formula, we have $$\tan^{-1}(y/x) = 2t, \tan^{-1}(x/y) = \pi/2 - 2t, r = \sqrt{x^2 + y^2}.$$ then
$$\begin{align} \cfrac{x^3}{2} \csc^2 \left(\cfrac{1}{2}\tan^{-1} \cfrac{x}{y}\right) + \cfrac{y^3}{2} \sec^2 \left(\cfrac{1}{2}\tan^{-1} \cfrac{y}{x} \right) &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549449",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\tan\frac{\pi}{16}+\tan\frac{5\pi}{16}+\tan\frac{9\pi}{16}+\tan\frac{13\pi}{16}$ Find the value of the expression $\tan\frac{\pi}{16}+\tan\frac{5\pi}{16}+\tan\frac{9\pi}{16}+\tan\frac{13\pi}{16}$
I identified that $\frac{\pi}{16}+\frac{13\pi}{16}=\frac{5\pi}{16}+\frac{9\pi}{16}=\frac{14\pi}{16}$
$\tan(\frac{\pi}{16}+... | As $\tan4\left(\dfrac\pi4+x\right)=\tan(\pi+4x)=\tan4x,$
If $\tan4x=\tan4A,4x=n\pi+4A\implies x=\dfrac{n\pi}4+A$ where $n=0,1,2,3$
as $$\tan4x=\dfrac{4\tan x-\binom41\tan^3x}{1-\binom42\tan^2x+\tan^4x}$$
$$\dfrac{4\tan x-\binom41\tan^3x}{1-\binom42\tan^2x+\tan^4x}=\tan4A$$
$$\iff\tan4A\tan^4x+4\tan^3x-6\tan4A\tan^2x-4... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int \frac{x+1}{(x^2-x+8)^3}\, dx$ Could you give me a hint on how to find $$\int \frac{x+1}{(x^2-x+8)^3}\, dx$$
It doesn't seem like partial fractions are the way to go with here and using the integration by parts method seems to be tedious.
I have also tried substituting $(x^2-x+8)$ but it gets even more c... | There is nothing simple about this. My take would be to write
$$
\frac{x+1}{x^2-x+8}=\frac{x+1}{((x-1/2)^2+31/4)^3}
=\frac12\,\frac{2(x-1/2)+3}{((x-1/2)^2+31/4)^3}\\
=\frac12\,\frac{2(x-1/2)}{((x-1/2)^2+31/4)^3}
+
\frac12\,\frac{3}{((x-1/2)^2+31/4)^3}
$$
Now the first term is a straightforward substitution. The second ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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Infinite series equality $\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\cdots$ Prove the following equality ($|x|<1$).
$$\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\cdots\\
=\frac{1}{1-x}+\frac{3x^2}{1-x^3}+\frac{5x^4}{1-x^5}+\frac{7x^6}{1-x^7}+\cdots\\$$
| Check that for $|x| < r < 1$,
$$\frac{nx^{n-1}}{1+x^n}, \frac{nx^{n-1}}{1-x^n}, \frac{2nx^{2n-1}}{1-x^n}$$ are all $O(nr^n)$, which is summable.
Then, you can switch the order of summation in the following infinite sum :
$$0 = \sum_{n \ge 1} (\frac{ nx^{n-1}}{1+x^n} - \frac {nx^{n-1}}{1-x^n} + \frac {2nx^{2n-1}}{1-x^... | {
"language": "en",
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"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
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What is the integral of $\frac{x-1}{(x+3)(x^2+1)}$? I've worked with partial fractions to get the integral in the form $$\int\frac{A}{x+3} + \frac{Bx + C}{x^2+1}\,dx$$ Is there a quicker way?
| Here is a slightly different way.
We start with
$$
\frac{x - 1}{(x+3)(x^2+1)} =
\frac{A}{x+3} + \frac{B}{x+i} + \frac{B^*}{x-i},
\qquad (1)
$$
where $B^*$ is the complex conjugate of $B$. The coefficient for $(x-i)^{-1}$ must be $B^*$: since the left-hand side is real, the right-hand side is real too, and hence is inva... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Simplify $\sqrt[3]{162x^6y^7}$ The answer is $3x^2 y^2 \sqrt[3]{6y}$
How does $\sqrt[3]{162x^6y^7}$ equal $3x^2 y^2\sqrt[3]{6y}$?
| \begin{align}
\sqrt[3]{162x^6y^7}&=\sqrt[3]{27x^6y^6*6y}\\
&=\sqrt[3]{27x^6y^6}\times\sqrt[3]{6y}\\
&=3x^2y^2\sqrt[3]{6y}
\end{align}
| {
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"answer_id": 0
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How to prove $\sum_{s=0}^{m}{2s\choose s}{s\choose m-s}\frac{(-1)^s}{s+1}=(-1)^m$? Question: How to prove the following identity?
$$
\sum_{s=0}^{m}{2s\choose s}{s\choose m-s}\frac{(-1)^s}{s+1}=(-1)^m.
$$
I'm also looking for the generalization of this identity like
$$
\sum_{s=k}^{m}{2s\choose s}{s\choose m-s}\frac{(-1... | Suppose we seek to verify that
$$\sum_{s=0}^m {2s\choose s} {s\choose m-s}
\frac{(-1)^s}{s+1} = (-1)^m$$
without using the generating function of the Catalan numbers.
Re-write the sum as follows:
$$\sum_{s=0}^m {2s\choose m} {m\choose s}
\frac{(-1)^s}{s+1}$$
which is
$$\frac{1}{m+1}
\sum_{s=0}^m {2s\choose m} {m+1\ch... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1559612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 1
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Order of $5$ in $\Bbb{Z}_{2^k}$ Is it true that the order of $5$ in $\Bbb Z_{2^k}$ is $2^{k-2}$? I was unable solve the congruence $5^n\equiv 1\pmod {2^{k}}$ nor see why $5^{2^{k-2}}\equiv 1\pmod {2^{k}}$. I'm not sure if this is the correct approach to this problem.
| Lemma 1: If $a$ is odd, $k\ge 3$, then $a^{2^{k-2}}\equiv 1\pmod{2^k}$.
Proof: Use induction. Base case: $a^2\equiv 1\pmod{8}$ for all odd $a$,
because if $a=2k+1$ for some $k\in\mathbb Z$, then $a^2-1=4k(k+1)$,
where $k(k+1)$ is a product of two consecutive integers, so $2\mid k(k+1)$.
If $a^{2^{m-2}}=2^mt+1$ for some... | {
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Determinants: divisibility by 6 without remainder and $n\times n$ matrix
*
*Compute the determinant of \begin{pmatrix}
1 & 2 & 3 & ...& n\\
-1 & 0 & 3 & ...& n\\
-1 & -2 & 0 & ...& n\\
...& ...& ...& ...& \\
-1 & -2 & -3 & ...& n
\end{pmatrix}
after some elementary raw operation, one can reach:
\begin{pmatrix}... | Regarding (2), you can reduce the matrix $A$ modulo $2$ to obtain
$$ A' = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{pmatrix}. $$
Clearly, $\det(A') = 0$ which shows that $\det(A) \mod 2 = \det(A') = 0$. Similarly, reducing $A$ modulo 3, we obtain
$$ A'' = \begin{pmatrix} 1 & 2 & 1 \\ 1 & 1 & 1 \\ 1 & 2 &... | {
"language": "en",
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"source": "stackexchange",
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Why is my $\epsilon$-$\delta$ proof incorrect? Prove that $f(x)= \sqrt{4+x^2}$ is continuous at $x_0$ using the $\epsilon -\delta$ definition of continuity.
Textbook proof:
$|\sqrt{4+x^2}-\sqrt{4+x^2}|=\frac{|4+x^2-(4+{x_0}^2)|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}=\frac{|x^2-{x_0}^2|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}=\underbr... | As others have pointed out the things that matter, I provide an argument for your reference; it is more succinct:
Let $x_{0} \in \Bbb{R}$. If $x \in \Bbb{R}$, then
$$
|\sqrt{4+x^{2}} - \sqrt{4+x_{0}^{2}}| = \frac{|x-x_{0}||x+x_{0}|}{\sqrt{4+x^{2}} + \sqrt{4 + x_{0}^{2}}} < |x-x_{0}||x+x_{0}|;
$$
if in addition $|x-x_{0... | {
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"source": "stackexchange",
"question_score": "1",
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Solving $\int \left ( \frac{5x^2+3x-2}{x^3+2x^2} \right )$ via partial fractions I'm having issues with the Partial Fractions method:
$$\int \left ( \frac{5x^2+3x-2}{x^3+2x^2} \right )dx$$
I am doing this in a way I was taught, which somehow I feel is a bit different from other methods online.
So, first of all, we must... | You got $A$ correct.
You have incorrectly evaluated $C$. When you sub in $x=-2$ you should get:
$$5(-2)^2+3(-2)-2=(-2)^2C$$
$$12=4C$$
$$C=3$$
Then to evaluate $B$ you can pick an arbitrary value for $x$ then solve for $B$. E.g. you selected $x=1$ this gives:
$$5\cdot1^2+3\cdot1-2=A\cdot(1+2)+B\cdot(1+2)\cdot1+C\cdot1^2... | {
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"source": "stackexchange",
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If $a_{n+1}=\frac {a_n^2+5} {a_{n-1}}$ then $a_{n+1}=Sa_n+Ta_{n-1}$ for some $S,T\in \Bbb Z$. Question
Let $$a_{n+1}:=\frac {a_n^2+5} {a_{n-1}},\, a_0=2,a_1=3$$
Prove that there exists integers $S,T$ such that $a_{n+1}=Sa_n+Ta_{n-1}$.
Attempt
I calculated the first few values of $a_n$: $a_2=7,a_3=18, a_4=47$ so I'd ... | From the recurrence relation we have
$$
a_{n+1} a_{n-1} - a_n^2 = 5
=
a_{n+2} a_{n} - a_{n+1}^2.
$$
So
$$
a_{n+1} \, (a_{n-1} + a_{n+1}) = a_n \, (a_n + a_{n+2}).
$$
or
$$
\frac{a_{n-1} + a_{n+1}}{a_n} = \frac{a_n + a_{n+2}}{a_{n+1}},
$$
This means
$$
c_n \equiv \frac{a_{n-1} + a_{n+1}}{a_n}.
$$
satisfies $c_n = c_{n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1566162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
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Sequence corresponding to the generating function
Find the sequence corresponding to the generating function $$G(x) = \frac{2x^4}{2x^3-x^2-2x+1}$$
First of all, I wrote this equation like that;
$\sum\limits_{n=0}^\infty(a_n)x^n = \frac{2x^4}{2x^3-x^2-2x+1}$
Then, I think right hand side should be;
$\frac{2x^4}{2x^3-... | Assuming the partial fraction decomposition is correct (I haven't checked it, but I suppose you have), remember that $\frac{1}{1-x}=\sum_{n=0}^\infty x^n$. Hence $$\frac{1}{2x-1}=-\frac{1}{1-2x}=-\sum_{n=0}^\infty (2x)^n=-\sum_{n=0}^\infty 2^nx^n$$
and $$\frac{1}{x-1}=-\frac{1}{1-x}=-\sum_{n=0}^\infty x^n$$
and $$\frac... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove that $\sqrt{2} + \sqrt[3]{3}$ is irrational $\sqrt{2} + \sqrt[3]{3}$ is irrational ?
These are my steps -
$\sqrt{2} + \sqrt[3]{3} = a$
$3 = (a-\sqrt{2})^{3}$
$3 = a^{3} -3a^{2}\sqrt{2} + 6a -2\sqrt{2}$
$3a^{2}\sqrt{2}+2\sqrt{2} = a^{3}+6a-3$
$\sqrt{2}(3a^{2}+2) = a^{3}+6a-3$
Then, $\sqrt{2}$ in the left side is i... | Taking powers of $\alpha=\sqrt2+\sqrt[\large3]{3}$ and putting them into matrix form, we get
$$
\begin{bmatrix}
\alpha^0\\\alpha^1\\\alpha^2\\\alpha^3\\\alpha^4\\\alpha^5\\\alpha^6
\end{bmatrix}
=
\begin{bmatrix}
1&0&0&0&0&0\\
0&1&1&0&0&0\\
2&0&0&2&1&0\\
3&2&6&0&0&3\\
4&12&3&8&12&0\\
60&4&20&15&3&20\\
17&120&90&24&60&1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1571961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Circle of radius of Intersection of Plane and Sphere The plane $x+2y-z=4$ cuts the sphere $x^2+y^2+z^2-x+z-2=0$ in a circle of radius?
I tried putting value of y from plane in sphere but then I get a $zx$ term. How to proceed?
| $x+2y-4=z$ and $x^2+y^2-x=-(z^2+z-2)$
$$x^2+y^2-x=(z+2)(1-z)$$
Substituting, $$x^2+y^2-x=(x+2y-2)(1-(x+2y-4))$$
$$x^2+y^2-x=-(x+2y-2)(x+2y-5)$$
And this way you get an equation in $x(s)$ and $y$(s).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1572097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
The roots of the equation $z^n=(z+3)^n$
Prove that the roots of the equation $z^n=(z+3)^n$ are collinear
Taking modulus on both sides, $$\vert z-0\vert =\vert z-3\vert$$
This represents the perpendicular bisector of line joining $x=0$ and $x=3$
That was easy. But I tried to solve it using algebra:
$$\frac{z+3}{z}=1^{... | $\left(\dfrac{z+3}{z}\right)=(1)^n= e^{\dfrac{2\pi i k}{n}}$ where k=0,1,2.....(n-1)
$\left(1+\dfrac{3}{z}\right)=(1)^n= \alpha^k$ where $\alpha=e^{\dfrac{2\pi i}{n}}$
$\implies z=\dfrac{3}{\alpha^k-1} ;k\neq0$
roots of your equation will be $z=\dfrac{3}{\alpha-1},\dfrac{3}{\alpha^2-1},.......\dfrac{3}{\alpha^{n-1}-1}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1573341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Solving a simultaneous equation How can I solve the following simultaneous equations:
$$3x^4+3x^2y^2-6xy = 0\tag 1$$
$$-2x^3y+3x^2-y^2=0\tag 2$$
I have tried rearranging for $y$ in eq(1) and plugging it into eq(2), but the result did not give me the right answer.
| Apart the obvious solution $(x=0,y=0)$ :
$$x^3+xy^2-2y=0$$
$$3x^2-y^2-2x^3y=0$$
From eq.(2) : $y^2=3x^2-2x^3y$ that is plugg into eq.(1)
$x^3+x(3x^2-2x^3y)-2y=0$
$$y=\frac{2x^3}{x^4+1}$$
Plugging it into eq.(1) leads to :
$$x^8+2x^4-3=0$$
$$x^4=-1\pm 2$$
$$x=(-1\pm 2)^{1/4}$$
$$y=\frac{2x^3}{x^4+1}=(-1\pm 2)^{3/4}$$
S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1577952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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How do you evaluate this sum of multiplied binomial coefficients: $\sum_{r=2}^9 \binom{r}{2} \binom{12-r}{3} $? We have to find the value of x+y in:
$$\sum_{r=2}^9 \binom{r}{2} \binom{12-r}{3} = \binom{x}{y} $$
My approach:
I figured that the required summation is nothing but the coefficient of $x^3$ is the following e... | $$\begin{align}\sum_{r=2}^9 \binom{r}{2} \binom{12-r}{3}
&=\sum_{r=2}^9\binom r{r-2}\binom {12-r}{9-r}\\
&=\sum_{r=2}^9(-1)^{r-2}\binom{-3}{r-2}(-1)^{9-r}\binom {-4}{9-r}&&\text{(upper negation)}\\\
&=-\sum_{r=2}^9\binom {-3}{r-2}\binom{-4}{9-r}\\
&=-\binom {-7}7&&\text{(Vandermonde)}\\
&=-(-1)^7\binom {13}7&&\text{(u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1579691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving $(ax+by-1)^2 \ge (x^2+y^2-1)(a^2+b^2-1)$
Let $x$, $y$, $a$, $b$ be real numbers such that $a^2+b^2 \leq 1$ and $x^2+y^2 \leq 1$. Show that $$(ax+by-1)^2 \ge (x^2+y^2-1)(a^2+b^2-1)$$
I am unable to find a solution to this problem. My initial thoughts were to have a trigonometric substitution of variables, but ... | This is a quadratic inequality in $x,y$, so all you have to do
is complete the squares.
Formally, if $\Delta=(ax+by-1)^2 -(x^2+y^2-1)(a^2+b^2-1)$ then you
have
$$
(1-a^2)\Delta=(1-a)^2(1+ay-bx)^2+(1-(a^2+b^2))(x-a)^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1584284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
Finding the value of an expression given three equations. $$x+y+2z=22$$
$$3x-2y+z=6$$
$$7x+3y-5z=1$$
above are three equations and we have to find the value of $$x^2+y^2+z^2=?$$
| $$
\begin{cases}
x+y+2z=22\\
3x-2y+z=6\\
7x+3y-5z=1
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
y=22+(-x-2z)\\
3x-2y+z=6\\
7x+3y-5z=1
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
y=22+(-x-2z)\\
3x-2(22+(-x-2z))+z=6\\
7x+3(22+(-x-2z))-5z=1
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
y=22+(-x-2z)\\
-4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1584601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Generating functions - deriving a formula for the sum $1^2 + 2^2 +\cdots+n^2$ I would like some help with deriving a formula for the sum $1^2 + 2^2 +\cdots+n^2$ using generating functions.
I have managed to do this for $1^2 + 2^2 + 3^2 +\cdots$ by putting
$$f_0(x) = \frac{1}{1-x} = 1 + x + x^2 + x^3 +\cdots$$
$$f_1(x)... | Brute force:
$$f(x) \triangleq \sum_{r=0}^{n+1}x^r=\frac{x^{n+2}-1}{x-1} $$
$$f'(x) = \sum_{r=0}^{n+1}rx^{r-1} = \sum_{r=0}^{n}(r+1)x^r$$
$$f''(x) = \sum_{r=0}^{n}r(r+1)x^{r-1} \therefore xf''(x)=\sum_{r=0}^{n}r(r+1)x^{r} = \sum_{r=0}^n\left(r^2+r\right)x^r$$
$$f'(x)-(f(x)-x^{n+1}) = \sum_{r=0}^{n}rx^r$$
So $$\sum_{r=... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 0
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find $\lim_{x\to 2}\frac{f(x^2-2)-f(f(x)-3)}{(x-2)^2}$ If the equation of the tangent to the graph of the function $y=f(x)$ at $x=2$ is $4x-y-3=0$ and at this point tangent cuts the graph also,then find $\lim_{x\to 2}\frac{f(x^2-2)-f(f(x)-3)}{(x-2)^2}$
$$\lim_{x\to 2}\frac{f(x^2-2)-f(f(x)-3)}{(x-2)^2}$$
As the denomin... | The numerator is approaching 0 as you can see
$$\lim_{x\to 2}\frac{f(x^2-2)-f(f(x)-3)}{(x-2)^2} = \lim_{x\to 2}\frac{f(2^2-2)-f(f(2)-3)}{(x-2)^2} = \lim_{x\to 2}\frac{f(2)-f(5-3)}{(x-2)^2} = \lim_{x\to 2}\frac{f(2)-f(2)}{(x-2)^2}$$
By applying L'Hôpital's rule you have $$\lim_{x\to 2}\frac{f(x^2-2)-f(f(x)-3)}{(x-2)^2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1585390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Approximation of $\pi$ using Brahmagupta's Identity Brahmagupta, an ancient Indian Mathematician, gave an pretty efficient algorithm for finding integer solutions to the famous Pell's Equation, far before Fermat propounded this before the European mathematicians' community.
Brahmagupta's Identity:
If $(x_1,y_1)$ is a... | The procedure for calculating the square root of the number can be used to calculate the number $\pi$ to arbitrary precision. You can use the ratio $$\tan\dfrac x2 = \dfrac{\tan x} {\sqrt{\tan^2 x+1} + 1}$$ with initial data $\tan{\dfrac{\pi}4} = 1$.
After $n$ iterations you will have $\tan\dfrac{\pi}{2^{n+2}}$ and the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1585534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why are the factors of some solutions to a Pell equation also a solution? I came across this observation while trying to answer this post using the Pell equation $x^2-2y^2=1$. Define,
$$P(m) = \frac{ (3+2\sqrt{2})^m+(3-2\sqrt{2})^m}{2}$$
$$Q(m) = \frac{ (3+2\sqrt{2})^m-(3-2\sqrt{2})^m}{2\sqrt{2}}$$
such that $P(m)^2-2Q... | Note that $x^2-y^2=(x-y)(x+y)$ and by induction $$x^{2^{n+1}}-y^{2^{n+1}}=(x^{2^n}-y^{2^n})(x^{2^n}+y^{2^n})=(x-y)\prod_{k=0}^n(x^{2^k}+y^{2^k}).$$ For roots $x_1, x_2$ of a quadratic equation and $x_1\leq x_2$ we have $x_2-x_1 = \sqrt{D}/a$. This proves your identity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1587852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Sum of factors of a huge number. I recently appeared in a math olympiad and it had this one question which had me stumped. This was a few weeks back and I have been looking for a way to find its answer ever since, but with no success. Searched the internet for the solution, but couldn't find any on it too! Anyway, here... | Using $a^2-b^2 = (a+b)(a-b)$
\begin{align*}
2^{96}-3^{16}
={} &
(2^{48}+3^8)(2^{48}-3^8)
\\
={} &
(2^{48}+3^8)(2^{24}+3^4)(2^{24}-3^4)
\\
={} &
(2^{48}+3^8)(2^{24}+3^4)(2^{12}+3^2)(2^{12}-3^2)
\\
={} &
(2^{48}+3^8)(2^{24}+3^4)(2^{12}+3^2)(2^6+3)(2^6-3)
\end{align*}
and $2^6+3=67$, $2^6-3=61$, ...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1587900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
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Existence of Solutions to a $2-$Equation System of Congruences Do there exist $a, b> 1$, such that
$$ a^4 \equiv 1 \pmod{b^2}$$ and $$ b^4 \equiv 1 \pmod{a^2}.$$
| If the given $2$ relations hold true, then
$$b^2\big|a^4-1 \Rightarrow b^2\big|(a^2+1)(a+1)(a-1)$$
and
$$a^2\big|b^4-1 \Rightarrow a^2\big|(b^2+1)(b+1)(b-1)$$
Also we have the relations below for suitable positive integers $k,l$ that
$$a^4=kb^2+1$$
$$b^4=la^2+1$$
Hence $$a^4-b^4=kb^2-la^2$$ or $$a^2(a^2+l)=b^2(b^2+k)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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How to prove $\sum\left(\frac{a}{b+c}\right)^2\ge \frac34\left(\frac{a^2+b^2+c^2}{ab+bc+ca}\right)$ The question is to prove:
$$\left(\frac{a}{b+c}\right)^2+\left(\frac{b}{c+a}\right)^2+\left(\frac{c}{a+b}\right)^2\ge \frac34\left(\frac{a^2+b^2+c^2}{ab+bc+ca}\right)$$
$$a,b,c>0$$
I tried Cauchy, AM-GM, Jensen, etc. but... | You can actually do it with Cauchy:
$$
\left(\sum_{cyc}\left(\frac{a}{b+c}\right)^2\right)\left(\sum_{cyc}\left(a(b+c)\right)^2\right)≥\left(a^2+b^2+c^2\right)^2
$$
Thus it is enough to prove:
$$
\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)≥\frac{3}{4}\sum_{cyc}\left(a(b+c)\right)^2=\frac{3}{2}\sum_{cyc}a^2b^2+\frac{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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If $x-y = 5y^2 - 4x^2$, prove that $x-y$ is perfect square Firstly, merry christmas!
I've got stuck at a problem.
If x, y are nonzero natural
numbers with $x>y$ such that
$$x-y = 5y^2 - 4x^2,$$
prove that $x - y$ is perfect square.
What I've thought so far:
$$x - y = 4y^2 - 4x^2 + y^2$$
$$x - y = 4(y-x)(y+x) ... | Solve,
$$x - y=5y^2 - 4x^2\tag1$$
hence,
$$x = \frac{-1\pm\sqrt{1+16y+80y^2}}{8}$$
Let,
$$1+16y+80y^2 = \big(\tfrac{2p}{q}y-1\big)^2$$
Expand and factor to get,
$$y = \frac{pq+4q^2}{p^2-20q^2}$$
with the relevant $x$ as,
$$x = \frac{pq+5q^2}{p^2-20q^2}$$
and $x,y$ will be integers if $p,q$ satisfy the Pell equation,
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
How to solve the equation $\sin^3x+\sin^3(\frac{2\pi}{3}+x)+\sin^3(\frac{4\pi}{3}+x)+\frac{3}{4}\cos2x=0$. Solve the equation $$\sin^3x+\sin^3\left(\frac{2\pi}{3}+x\right)+\sin^3\left(\frac{4\pi}{3}+x\right)+\frac{3}{4}\cos2x=0$$
May I have a hint on how to solve this equation?
| Using the identity $\sin^3 \theta = \dfrac{3}{4}\sin\theta-\dfrac{1}{4}\sin 3\theta$, we get:
$\sin^3 x + \sin^3(x+\tfrac{2\pi}{3}) + \sin^3(x+\tfrac{4\pi}{3})$
$= \dfrac{3}{4}\left[\sin x + \sin(x+\tfrac{2\pi}{3})+\sin(x+\tfrac{4\pi}{3})\right] - \dfrac{1}{4}\left[\sin 3x + \sin(3x+2\pi) + \sin(3x+4\pi)\right]$
$= \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Evaluate $\int{\sin^{13}x}dx$ Evaluate $\int \sin^{13}x\,dx$.
My solution so far:
$$\int\sin^{13}x\,dx=\int\sin x\sin^{12}x\,dx=\int\sin x(\sin^{2}x)^6\,dx$$
Let $t=\cos x$, then $dt=-\sin x\,dx$.
Now we have: $$\int\sin^{13}x\,dx=-\int(1-t^2)^6,\ dt$$.
How do I proceed from here? I need detailed answer.
| \begin{align}
\int(1-t^2)^6 &= \int \sum_{n=0}^6 {6 \choose n} (-t^2)^n \\
&= \sum_{n=0}^6 {6 \choose n} (-1)^n \int t^{2n} \\
&= \sum_{n=0}^6 {6 \choose n} (-1)^n \left( \frac{t^{2n+1}}{2n+1}\right) + C \\
&= \frac{t^{13}}{13} - \frac{6 t^{11}}{11} + \frac{5t^9}{3} - \frac{20t^7}{7} + 3 t^5 - 2t^3 + t + C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
$3^{n+1}$ divides $2^{3^n}+1$ Describe all positive integers,n such that $3^{n+1}$divides $2^{3^n}+1$.
I am little confused about what the question asks-if it asks me to find all such positive integers, or if it asks me to prove that for every positive integer n,$3^{n+1}$ divides $2^{3^n}+1$. Kindly clarify this doubt ... | We can easily verify that $n=1$ is a solution.
Assume for some $k$ that,
$3^{k+1}|2^{3^{k}}+1$, hence there exists an integer $m$, such that,
$m(3^{k+1})=2^{3^{k}}+1$.
Then we have,
$2^{3^{k+1}}+1=(2^{3^{k}})^{3}+1=(2^{3^{k}}+1)[(2^{3^{k}})^{2}-2^{3^{k}}+1]=m(3^{k+1})[(2^{3^{k}})^{2}-2^{3^{k}}+1]$. (*)
Now, $3^{k}$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 3
} |
Solve this equation $xy-\frac{(x+y)^2}{n}=n-4$ Let $n>4$ be a given positive integer. Find all pairs of positive integers $(x,y)$ such that
$$xy-\dfrac{(x+y)^2}{n}=n-4$$
What I tried is to use
$$nxy-(x+y)^2=n^2-4n\Longrightarrow (n-2)^2+(x+y)^2=nxy+4$$
| If you frame the problem as a quadratic equation in $y$, the condition that $y$ is an integer implies that
\begin{equation*}
n(n-4)(x^2-4)=\Box
\end{equation*}
By parameterising this quadric and searching for integer pairs, we get what looks like an infinite number of pairs $(x,y)$ for each $n$.
Investigating these we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
What will be the remainder when $2^{31}$ is divided by $5$? The question is given in the title:
Find the remainder when $2^{31}$ is divided by $5$.
My friend explained me this way:
$2^2$ gives $-1$ remainder.
So, any power of $2^2$ will give $-1$ remainder.
So, $2^{30}$ gives $-1$ remainder.
So, $2^{30}\times 2$ or ... | From Fermat's Little Theorem-
$a^{p-1}\equiv1\pmod p$
Putting $a=2$ and $p=5$ we have,
$2^4\equiv 1\pmod 5$
$\implies 2^{28}\equiv 1\pmod 5$
Now,$2^{31}=2^{28}\times 2^{3}$.So,it is equivalent to finding the remainder when $2^3$ is divided by $5$..which is $3$.So,the remainder when $2^{31}$ is divided by $5$ is $3$..
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
Find all $(x,y)$ satisfying $(\sin^2x+\frac{1}{\sin^2 x})^2+(\cos^2x+\frac{1}{\cos^2 x})^2=12+\frac{1}{2}\sin y$ Find all pairs $(x,y)$ of real numbers that satisfy the equation $(\sin^2x+\frac{1}{\sin^2 x})^2+(\cos^2x+\frac{1}{\cos^2 x})^2=12+\frac{1}{2}\sin y$
I supposed $a=\sin^2x$ and $b=\cos^2x$
So the equation b... | As JMoravitz notes in the comments, it's probably the case that the minimum value of the LHS is 12.5. Let's prove this.
Using the inequality $a^2+b^2\ge\frac{(a+b)^2}{2}$, note that
\begin{align*}\left(\sin^2x + \frac{1}{\sin^2x}\right)^2 + \left(\cos^2x + \frac{1}{\cos^2x}\right)^2&\ge \frac{1}{2}\left(\sin^2x + \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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$C$ be curve of intersection of hemisphere $x^2+y^2+z^2=2ax$ and cylinder $x^2+y^2=2bx$ ; to evaluate $\int_C(y^2+z^2)dx+(x^2+z^2)dy+(x^2+y^2)dz$ $C$ be the curve of intersection of the hemisphere $x^2+y^2+z^2=2ax$ and the cylinder $x^2+y^2=2bx$ , where $0<b<a$ ; how to evaluate $\int_C(y^2+z^2)dx+(x^2+z^2)dy+(x^2+y^2)... | There are at least to ways you can parametrize curve $C$.
Method 1. Since the curve is at the intersection of the sphere and the cylinder, it belongs to the cylinder, which has radius $b$ and is shifted $b$ units along the $x$ axis, so you can write:
$$
x=b\cos\theta+b, \quad y =b\sin\theta, \quad 0\le \theta \le 2\pi,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that if $a,b,$ and $c$ are positive real numbers, then $\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a} \geq ab + bc + ca$.
Prove that if $a,b,$ and $c$ are positive real numbers, then $\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a} \geq ab + bc + ca$.
I tried AM-GM and it doesn't look like AM-GM or Cauchy-Schwarz work... | your inequalitiy is equivalent to
$$a^4c+b^4a+c^4b-a^2b^2c-a^2bc^2-ab^2c^2\geq 0$$
this is equivalent to
$$(a^2-b^2)(a^2c-bc^2)+(b^2-c^2)(ab^2-bc^2)\geq 0$$ we assume that $$a\geq b\geq c$$ thus $$a^2\geq b^2$$ and $$a^2\geq bc$$ and $$b^2\geq c^2$$ and $$ab\geq c^2$$
in the case $$a\geq c\geq b$$ we have
$$(a^2-c^2)(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 4
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How to solve the functional equation $ f(f(x))=ax^2+bx+c $ Find all real numbers $a,b,c\in\mathbb{R}$ for which there exists a function $f:\mathbb{R}\to\mathbb{R}$ such that:
$$
f(f(x))=ax^2+bx+c
$$
for all $x\in\mathbb{R}$.
The only thing I could deduce is:
$$
f(ax^2+bx+c)=af(x)^2+bf(x)+c
$$
Which doesn't help much. H... | Here are some reflections for continuous and differentiable $f$. It turns out that there is a unique solution if some conditions are satisfied (you can look at the end for the final answer).
*
*$$
f(f(x))=ax^2+bx+c\implies f(ax^2+bx+c)=af(x)^2+bf(x)+c
$$
*If $f(x)=f(y)$, $x\neq y$ then:
$$
a(x+y)=b.
$$
If $a(x+y)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 1
} |
Trigonometric Ratios Of Multiple Angles If $2\tan A=3\tan B$ then prove that $$\tan(A-B)=\frac{\sin2B}{5-\cos 2B}.$$
I found that $\tan A=\frac{3}{2} \tan B$ and after that used the formula of $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$ but could not reach to the required answer.
| Notice, here is another method: $$RHS=\frac{\sin 2B}{5-\cos 2B}$$
$$=\frac{\frac{2\tan B}{1+\tan^2B}}{5-\frac{1-\tan^2 B}{1+\tan^2A}}$$
$$=\frac{2\tan B}{4+6\tan^2B}$$
$$=\frac{\tan B}{2+3\tan^2B}$$
$$=\frac{(3\tan B)-2\tan B}{2+(3\tan B)\tan B}$$
setting $3\tan B=2\tan A$,
$$=\frac{2\tan A-2\tan B}{2+2\tan A\tan B}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
solution of nested radical $\sqrt{7+2\sqrt{7-2\sqrt{7-2x} } } =x$ This question is from my friend. he think that there is trigonometry involved to this equation.
$\sqrt{7+2\sqrt{7-2\sqrt{7-2x} } } =x$
that is from $\ \ \ x = \sqrt{7+2\sqrt{7-2\sqrt{7-2\sqrt{7+2\sqrt{7-2\sqrt{7-2\sqrt...} } }} } }$
I try to compare... | I hope and wish that you will get a simpler solution. For the time being, here is mine.
I started writing $$\sqrt{7+2\sqrt{7-2\sqrt{7-2y} } } =x$$ which I solved for $y$; this gives $$y=\frac{1}{128} \left(-x^8+28 x^6-238 x^4+588 x^2+7\right)$$ Replacing $y$ by $x$ and expanding leads to $$x^8-28 x^6+238 x^4-588 x^2+12... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Find the summation $\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+ \cdots$ What is the value of the following sum?
$$\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+ \cdots$$
The possible answers are:
A. $e$
B. $\frac{e}{2}$
C. $\frac{3e}{2}$
D. $1 + \frac{e}{2}$
I tried to expand the options using the series represen... | You want calculate $\sum_{k=1}^\infty\frac{k(k+1)}{2(k!)}=\frac{1}{2}\sum_{k=1}^\infty\frac{k+1}{(k-1)!}=\frac{1}{2}\sum_{k=0}^\infty\frac{k+2}{k!}$...(*)
But, as $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$, then $x^2e^x=x^2+x^3+\frac{x^4}{2!}+\frac{x^5}{3!}+\cdots$. Deriving both sides give us $x^2e^x+2xe^x=2x+3x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 4
} |
Prove that for positive real numbers $a,b,c$ we have $\frac{a}{b+c}+ \frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}.$
Prove that for positive real numbers $a,b,c$ we have $$\dfrac{a}{b+c}+ \dfrac{b}{a+c}+\dfrac{c}{a+b} \geq \dfrac{3}{2}.$$
Attempt
I tried using AM-GM and got $ \dfrac{a}{2\sqrt{bc}}+\dfrac{b}{2\sqrt{ac}... | $$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = \frac{a^2}{ab+ac} + \frac{b^2}{ab+bc} + \frac{c^2}{ac+bc} \geq \frac{(a+b+c)^2}{2(ab+bc+ac)}$$
Recall Buniakowsky inequality:
$$ (a^2+b^2+c^2)(x^2+y^2+z^2) \geq (ax+by+cz)^2$$
by expanding and regrouping the terms of:
$$(ay-bx)^2 + (az-cx)^2 + (bz - cy)^2 \geq 0$$
So ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How can one show that : $|u_{n+1}-\sqrt{2}|\le\frac{1}{4}|u_n-\sqrt{2}|$ $U_n$ numerical sequence such that :
( For all natural numbers $n$ ) $U_{n+1}=1+\dfrac{1}{1+U_n}$ and $U_0=1$
How can one show that : $|U_{n+1}-\sqrt{2}|\le\frac{1}{4}|U_n-\sqrt{2}|$
I arrived to show that
$$|U_{n+1}-U_n|\le\frac{1}{4}|U_n-U_{n-1... | Let $x=\sqrt{2}$ .
Note that this isn't an ordinary number , it's the root of $t=1+\frac{1}{1+t}$ which looks exactly as the recurrence :
$$U_{n+1}-x=1+\frac{1}{1+U_n}-\left (1+\frac{1}{1+x} \right )=\frac{x-U_n}{(1+x)(1+U_n)}$$
Now we need only to prove that :
$$(1+x)(1+U_n) \geq 4$$
But $U_n \geq 1$ for every $n$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1598681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Find Complex Roots
Question:
Find the complex roots of $$ {(z^{12} -1)\over (z^4-1)(z^3-1)} = 0 $$
What I have attempted:
$$ {(z^{12} -1)\over (z^4-1)(z^3-1)} = 0 $$
$$ {(z^{6} -1)(z^{6} +1)\over (z^2-1)(z^2+1)(z^3-1)} = 0 $$
$$ {(z^{3} -1)(z^3 +1)(z^6+1)\over (z+1)(z-1)(z^2+1)(z^3-1)} = 0 $$
$$ {(z^3 +1)(z^6+... | Only $$z^{12}-1$$ must equal 0.
Also, find the roots of $$z^{4}-1$$ and $$z^{3}-1$$ and remove them from your final solution, since the denominator ≠ 0
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can I integrate this? (for calculate value of L-function ) I want to calculate the definite integral:
$$
\int_{0}^{1} \frac{x+x^{3}+x^{7}+x^{9}-x^{11}-x^{13}-x^{17}-x^{19}}{x(1-x^{20})}dx.
$$
Indeed, I already know that $\int_{0}^{1} \frac{x+x^{3}+x^{7}+x^{9}-x^{11}-x^{13}-x^{17}-x^{19}}{x(1-x^{20})}dx=L(\chi,1)=\f... | Noting that
$$ t^4+3t^2+1=(t^2+\frac{3}{2})^2-(\frac{\sqrt5}{2})^2,$$
and
$$ \frac{t^2+1}{t^4+3t^2+1}=\frac{5-\sqrt{5}}{10 \left(t^2-\frac{\sqrt{5}}{2}+\frac{3}{2}\right)}-\frac{-5-\sqrt{5}}{10\left(t^2+\frac{\sqrt{5}}{2}+\frac{3}{2}\right)}$$
we have
\begin{eqnarray}
&&\int_{-\infty}^0\frac{t^2+1}{t^4+3t^2+1}dt\\
&=&\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
If $a^2 + b^2 = 1$, show there is $t$ such that $a = \frac{1 - t^2}{1 + t^2}$ and $b = \frac{2t}{1 + t^2}$ My question is how we can prove the following:
If $a^2+b^2=1$, then there is $t$ such that $$a=\frac{1-t^2}{1+t^2} \quad \text{and} \quad b=\frac{2t}{1+t^2}$$
| As its an equation and open end it will have infinite solutions so Assume $x=\sin(2\theta)$, $y=\cos(2\theta)$ as we all know that sum of their squares is $1$ now by double angle $\sin(2\theta)=\frac{2t}{1+t^2}$, $\cos(2\theta)=\frac{1-t^2}{1+t^2}$ and now you can prove it trigonometrically.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
$\lim\frac{\cos{2x^2}-1}{x^2\sin{x^2}}$ as $x$ goes to $0$
Calculate $\displaystyle \lim_{x \to 0}\frac{\cos{2x^2}-1}{x^2\sin{x^2}}$
I tried L'hopital, but the denominator gets more and more complicated.
How does one calculate this limit?
| Notice, you can find this limit by using 'Hôpital's rule (do it twice):
$$\lim_{x\to0}\frac{\cos(2x^2)-1}{x^2\sin(x^2)}=\lim_{x\to0}\frac{\frac{\text{d}}{\text{d}x}\left(\cos(2x^2)-1\right)}{\frac{\text{d}}{\text{d}x}\left(x^2\sin(x^2)\right)}=$$
$$\lim_{x\to0}\frac{-4x\sin(2x^2)}{2x^3\cos(x^2)+2x\sin(x^2)}=-2\left[\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
De Moivre Theorem Express the following function in the form of $a+ib$, $z$ being written for $\cos \theta+i \sin \theta$
$$\frac{1}{1+z^2}$$
My attempt,
$$\frac{1}{1+z^2}=\frac{1}{1+(\cos \theta +i \sin \theta)^2}
=\frac{1}{1+\cos 2\theta+i \sin 2\theta}$$
How to proceed? The given answer is $\frac{1}{2}(1-i \tan \... | Continuing from where you stopped:
$$\begin{align}
\frac 1 {1+\cos 2\theta + {\rm i} \sin 2\theta} = \\
\frac {1+\cos 2\theta - {\rm i} \sin 2\theta} {(1+\cos 2\theta + {\rm i} \sin 2\theta) (1+\cos 2\theta - {\rm i} \sin 2\theta)} = \\
\frac {1+\cos 2\theta - {\rm i} \sin 2\theta} {(1+\cos 2\theta)^2 + \sin^2 2\theta}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is the difference between the largest and smallest possible values of any of these numbers?
If $x,y,z$ are real numbers such that $x+y+z = 5$ and $xy+yz+zx = 3$, what is the difference between the largest and smallest possible values of any of these numbers?
What is wrong with this approach?
We have that $(x+y+z... | $(x,y,z)$ is not $\bf{three}$ points. It is a single point in three dimensional space, where $x,y,z$ are the coordinates.
Now think of a sphere of radius $\sqrt{19}$. If the $x$ coordinate is $\sqrt{19}$, then is it possible for any other coordinate to be $-\sqrt{19}$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
long division with no numbers Im stumped by the following brain teaser:
Solve the following long division problem. Each letter represents a unique digit ($0$ - $9$)
$$
\require{enclose}
\begin{array}{r}
B \\[-3pt]
XZD \enclose{longdiv}{BUMG} \\[-3pt]
\underline{APZK} \\[-3pt]
... | You can separate it into cases on $B$ (it is easy to see that $B\not=0,1,5,9$), and find the values in the following order : $B\rightarrow A\rightarrow X\rightarrow Z,D,K\rightarrow P\rightarrow U,M,G$.
For $B=2$, we have $A=0,1$. But if $A=0$, then
$$(BU)=(2UMG)-(PZK)\ge 2000-999=1001$$
So, $A=1$. Also, if $X\le 8$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Real numbers $x,y$ satisfies $x^2+y^2=1.$If the minimum and maximum value of the expression $z=\frac{4-y}{7-x}$ are $m$ and $M$ Real numbers $x,y$ satisfies $x^2+y^2=1.$If the minimum and maximum value of the expression $z=\frac{4-y}{7-x}$ are $m$ and $M$ respectively,then find $2M+6m.$
Let $x=\cos\theta$ and $y=\sin\... | Let $\displaystyle k=\frac{4-y}{7-x}\Rightarrow 7k-kx=4-y\Rightarrow kx-y = 7k-4$ and given $x^2+y^2=1$
Now using the Cauchy-Schwarz inequality, we get $$[k^2+(-1)^2](x^2+y^2)\geq (kx-y)^2$$
So $$k^2+1\geq (7k-4)^2\Rightarrow 49k^2+16-56k\leq k^2+1$$
So $$48k^2-56k+15\leq 0\Rightarrow (4k-3)(12k-5)\leq 0$$
So we get $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Find $\int_{1}^{2}\frac{x-1}{x^2\sqrt{x^2+(x-1)^2}}$ Find $\int_{1}^{2}\frac{x-1}{x^2\sqrt{x^2+(x-1)^2}}$
I tried to solve it by using the property $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$
Let $I=\int_{1}^{2}\frac{(x-1)dx}{x^2\sqrt{x^2+(x-1)^2}}$
$I=\int_{1}^{2}\frac{(2-x)dx}{(3-x)^2\sqrt{(3-x)^2+(2-x)^2}}$
But thi... | First write $$x - 1 = x(2x - 1) - (2x^2 - 2x + 1).$$ Then our integrand becomes: $$\frac{x-1}{x^2 \sqrt{2x^2 - 2x + 1}} = \frac{2x-1}{x\sqrt{2x^2 - 2x + 1}} - \frac{\sqrt{2x^2 - 2x + 1}}{x^2}.$$ Now with the choice $$u = \sqrt{2x^2 - 2x + 1}, \quad v = \frac{1}{x},$$ we have $$u' = \frac{2x-1}{\sqrt{2x^2-2x+1}}, \qu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
What is the maximum value of $\sqrt6 xy + 4yz$ given $x^2 + y^2 + z^2 = 1?$ Problem:
Let $x$,$y$ and $z$ be real numbers such that $x^2 + y^2 + z^2 = 1.$ Find the maximum possible value of $\sqrt6 xy + 4yz$
I don't know how to proceed with the question. Applying AM-GM inequality doesn't work because the second equatio... | We can use AM-GM inequality alone to settle the question. To this end, we find the constants $A,B,C,D$ such that: $\sqrt{6}xy \leq A^2x^2+B^2y^2$, and $4yz \leq C^2y^2+D^2z^2$ and $A^2 = B^2+C^2 = D^2, AB = \dfrac{\sqrt{6}}{2}, CD = 2 = AC$. Thus $A^4 = A^2\cdot A^2 = A^2(B^2+C^2) = (AB)^2 + (AC)^2 = \left(\dfrac{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Rational solutions to $e^x - \frac{1}{2} = \sqrt{ x^3 + 1/4}$ Show that the only rational solution to the title curve is $x=0$.
My attempt: Squaring both sides we have $e^{2x} - e^x + \frac{1}{4} = x^3 + \frac{1}{4}$, which yields $e^{2x} - e^x = x^3$. I suspect that the left hand side is irrational for any rational $... | We can write equation as $$e^x(e^x-1)=x^3$$ therefore $e^x-1=\frac{x^3}{e^x}$ now we know that $e$ is irrational(2.781...) so any value of $x$ else than $0$ makes the the equation irrational so $0$ is the only rational solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to calculate this definite integral? How can I calculate this
$$\int_1^{+\infty} \frac {dx}{x^3\sqrt{x^2+x}}$$
I have no idea what to do with it, integration by parts or by substitution doesn't work for me.
| First of all, I've checked my answer with Mathematice and I'm right.
$$\int_{1}^{\infty}\frac{1}{x^3\sqrt{x^2+x}}\space\text{d}x=$$
$$\lim_{n\to\infty}\int_{1}^{n}\frac{1}{x^3\sqrt{x^2+x}}\space\text{d}x=$$
$$\lim_{n\to\infty}\int_{1}^{n}\frac{1}{x^3\sqrt{\left(x+\frac{1}{2}\right)^2-\frac{1}{4}}}\space\text{d}x=$$
Su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
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